diff --git a/README.md b/README.md
index fe17cb76..7e94c277 100644
--- a/README.md
+++ b/README.md
@@ -1,6 +1,10 @@
 # Formalization of a generalized Carleson's theorem
 A (WIP) formalized proof of a generalized Carleson's theorem in Lean
 
+* [Blueprint](http://florisvandoorn.com/carleson/blueprint/)
+* [Blueprint as pdf](http://florisvandoorn.com/carleson/blueprint.pdf)
+* [Doc pages for this repository](http://florisvandoorn.com/carleson/docs/)
+
 ## Contribute
 
 To get the repository, make sure you have [installed Lean](https://leanprover-community.github.io/get_started.html). Then get the repository using `git clone https://github.com/fpvandoorn/carleson.git` and run `lake exe cache get!` inside the repository. Run `lake build` to build all files in this repository. See the README of [my course repository](https://github.com/fpvandoorn/LeanCourse23) for more detailed instructions.
@@ -13,25 +17,5 @@ you've worked have no errors (having `sorry`'s is of course fine).
 
 ## Build the blueprint
 
-To build the web version of the blueprint, you need a working LaTeX installation.
-Furthermore, you need some packages:
-```
-sudo apt install graphviz libgraphviz-dev
-pip3 install invoke pandoc
-cd .. # go to folder where you are happy clone git repos
-git clone git@github.com:plastex/plastex
-pip3 install ./plastex
-git clone git@github.com:PatrickMassot/leanblueprint
-pip3 install ./leanblueprint
-cd carleson
-```
-
-To actually build the blueprint, run
-```
-lake exe cache get
-lake build
-inv all
-```
-
-To view the web-version of the blueprint locally, run `inv serve` and navigate to
-`http://localhost:8000/` in your favorite browser.
\ No newline at end of file
+To test the Blueprint locally, you can compile `print.tex` using XeLaTeX (i.e. `xelatex print.tex` in the folder `blueprint/src`). If you have the Python package `invoke` you can also run `inv bp` which puts the output in `blueprint/print/print.pdf`.
+If you feel adventurous and want to build the web version of the blueprint locally, you need to install some packages by following the instructions [here](https://pypi.org/project/leanblueprint/). But if the pdf builds locally, you can just make a pull request and use the online blueprint.
\ No newline at end of file
diff --git a/blueprint/src/chapter/main.tex b/blueprint/src/chapter/main.tex
index 8601c14d..bc449b1f 100644
--- a/blueprint/src/chapter/main.tex
+++ b/blueprint/src/chapter/main.tex
@@ -22,53 +22,30 @@
 
 \chapter{Introduction}
 
-In \cite{carleson}, L. Carleson
-answered a classical question on convergence of
-Fourier series of continuous functions by proving that they converge pointwise almost everywhere.
-Theorem \ref{classical Carleson} is a version of his result.
+In \cite{carleson}, L. Carleson addressed a classical question regarding the convergence of Fourier series of continuous functions by proving their pointwise convergence almost everywhere. Theorem \ref{classical Carleson} represents a version of this result.
 
-Let $f$ be a complex valued, $2\pi $-periodic bounded Borel measurable function on the real line and given an integer $n$, define the Fourier coefficient
+Let $f$ be a complex-valued, $2\pi$-periodic bounded Borel measurable function on the real line, and for an integer $n$, define the Fourier coefficient as
 \begin{equation}
     \widehat{f}_n:=\frac {1}{2\pi} \int_0^{2\pi} f(x) e^{- i nx} dx .
 \end{equation}
-Define for $N\ge 0$ the partial Fourier sum
+Define the partial Fourier sum for $N\ge 0$ as
 \begin{equation}
     S_Nf(x):=\sum_{n=-N}^N \widehat{f}_n e^{i nx}\ .
 \end{equation}
-\begin{theorem}\label{classical Carleson}
-\uses{metric space Carleson}
-Let $f$ be a $2\pi $-periodic complex valued uniformly continuous function an $\R$ that satisfies the bound
-$|f(x)|\le 1$ for all $x\in \R$. For all $0<\epsilon\le 2\pi$,
-there exists a Borel set $E\subset [0,2\pi]$ with Lebesgue measure
-$|E|\le \epsilon$ and a positive integer $N_0$ such that for all
-$x\in [0,2\pi]\setminus E$ and all integers $N>N_0$ we have
+
+\begin{theorem}[classical Carleson]
+\label{classical Carleson}
+\uses{piecewise constant approximation, convergence for piecewise constant,
+control approximation effect,metric space Carleson}
+Let $f$ be a $2\pi$-periodic complex-valued uniformly continuous function on $\mathbb{R}$ satisfying the bound $|f(x)|\le 1$ for all $x\in \mathbb{R}$. For all $0<\epsilon\le 2\pi$, there exists a Borel set $E\subset [0,2\pi]$ with Lebesgue measure $|E|\le \epsilon$ and a positive integer $N_0$ such that for all $x\in [0,2\pi]\setminus E$ and all integers $N>N_0$, we have
 \begin{equation}\label{aeconv}
 |f(x)-S_N f(x)|\le \epsilon.
 \end{equation}
 \end{theorem}
 
-Note that mere continuity implies uniform continuity
-in the setting of this theorem. Applying this theorem
-with a sequence of $\epsilon_n:= 2^{-n}\delta$ for $n\ge 1$
-and taking the union of corresponding exceptional sets $E_n$, we see that
-outside a set of measure $\delta$, the partial Fourier sums
-converge pointwise for $N\to \infty$. Applying this with a sequence
-of $\delta$ shrinking to zero and taking the intersection of the corresponding exceptional
-sets, which has measure zero, we see that the Fourier series converges outside
-a set of measure zero.
-
-The purpose of this paper is twofold.
-On the one hand, it prepares computer verification of Theorem \ref{classical Carleson} by presenting a very detailed proof as a blueprint for coding in Lean.
-We pass through a bound for a
-generalization of the so-called Carlson operator
-to doubling metric measure spaces. This generalization is new
-and proving these bounds is the second purpose of this paper.
-This generalization incorporates several results in the recent literature, most prominently bounds for the polynomial Carleson
-operator of V. Lie \cite{lie-polynomial}
-as well as its generalization \cite{zk-polynomial}.
-A computer verification of our theorem will also be a computer
-verification for the bulk of the work in these results.
+Note that mere continuity implies uniform continuity in the setting of this theorem. By applying this theorem with a sequence of $\epsilon_n:= 2^{-n}\delta$ for $n\ge 1$ and taking the union of corresponding exceptional sets $E_n$, we see that outside a set of measure $\delta$, the partial Fourier sums converge pointwise for $N\to \infty$. Applying this with a sequence of $\delta$ shrinking to zero and taking the intersection of the corresponding exceptional sets, which has measure zero, we see that the Fourier series converges outside a set of measure zero.
 
+The purpose of this paper is twofold. On the one hand, it prepares computer verification of Theorem \ref{classical Carleson} by presenting a very detailed proof as a blueprint for coding in Lean. We pass through a bound for a generalization of the so-called Carleson operator to doubling metric measure spaces. This generalization is new, and proving these bounds constitutes the second purpose of this paper. This generalization incorporates several results from the recent literature, most prominently bounds for the polynomial Carleson operator of V. Lie \cite{lie-polynomial} as well as its generalization \cite{zk-polynomial}. A computer verification of our theorem will also entail a computer verification for the bulk of the work in these results.
 
 
 We proceed to introduce the setup for our general theorem.
@@ -107,11 +84,10 @@ \chapter{Introduction}
     2d_{B_1}(\mfa,\mfb)
 \le d_{B_2}(\mfa,\mfb) .
 \end{equation}
-For every ball $B$ in $X$ and
-every $d_B$-ball of radius $2R$ in $\Mf$, there is a collection $\mathcal{B}$ of
-at most $2^a$ many  $d_B$-balls of radius $R$ covering $B$, that is,
+For every ball $B$ in $X$ and every $d_B$-ball $\tilde B$ of radius $2R$ in $\Mf$, there is a collection $\mathcal{B}$ of
+at most $2^a$ many  $d_B$-balls of radius $R$ covering $\tilde B$, that is,
 \begin{equation}\label{thirddb}
-    B\subset \bigcup \mathcal{B}.
+    \tilde B\subset \bigcup \mathcal{B}.
 \end{equation}
 
 
@@ -124,7 +100,7 @@ \chapter{Introduction}
 (1+d_B(\mfa,\mfb))^{-\frac{1}{a}},
 \end{equation}
 %where $\C^\nu$ denotes the homogeneous H\"older space.
-where $\|\cdot\|_{\Lip(B)}$ denotes the inhomogeneous Lipschitz norm on $B$,
+where $\|\cdot\|_{\Lip(B)}$ denotes the inhomogeneous Lipschitz norm on $B$:
 $$
     \|\varphi\|_{\Lip(B)} = \sup_{x \in B} |\varphi(x)| + R \sup_{x,y \in B, x \neq y} \frac{|\varphi(x) - \varphi(y)|}{\rho(x,y)}\,.
 $$
@@ -161,9 +137,9 @@ \chapter{Introduction}
 
 Our main result  is the following restricted weak type estimate for $T$ in the range $1<q\le 2$, which by interpolation techniques recovers $L^q$ estimates for the open range
 $1<q<2$.
-\begin{theorem}
+\begin{theorem}[metric space Carleson]
 \label{metric space Carleson}
-\uses{finitary Carleson, discrete Carleson, antichain operator, forest operator, Holder van der Corput, Hardy Littlewood, kernel summand, R truncation}
+\uses{ball metric,finitary Carleson, discrete Carleson, antichain operator, forest operator, Holder van der Corput, Hardy Littlewood, kernel summand, R truncation}
     For all  integers $a \ge  4$ and  real numbers $1<q\le 2$
     % there exists a positive  such that
     the following holds.
@@ -183,60 +159,19 @@ \chapter{Introduction}
         \end{equation}
 \end{theorem}
 
- In the one dimensional Euclidean setting with $K$ the Hilbert kernel
- \begin{equation*}
- K(x,y)=(x-y)^{-1}
- \end{equation*}
-and $\Mf$ the class of linear functions, the operator  \eqref{def main op} is the classical
-Carleson operator that has a role in proving
-almost everywhere convergence of Fourier series
-\cite{carleson}, \cite{fefferman}, \cite{lacey-thiele}.
-The supremum in $R_1,R_2$ is classically often omitted but
-the case including the maximal truncations can easily be
-reduced to the case without these truncations.
-
-Replacing $\Mf$ by the class of polynomials vanishing at $0$
-up to some fixed but arbitrary degree,
-we obtain the polynomial Carleson operator of Lie,
-\cite{lie-quadratic} (quadratic case) and \cite{lie-polynomial}.
- The case of the class of polynomials with
-vanishing linear coefficient is simpler and was estimated in \cite{stein-wainger}.
-The polynomial Carleson operator was
-generalized to the high dimensional Euclidean setting in
-\cite{zk-polynomial} for $K$ a Calder\'on--Zygmund kernel with some H\"older regularity.
-
-Doubling metric measure spaces are instances of
-spaces of homogeneous type. Indeed, by changing from a quasi metric to an equivalent metric, every space of homogeneous type can be viewed as a doubling metric measure space (cf. \cite{MaciasSegovia}). Spaces of homogeneous type were introduced
-by \cite{MR0499948} as a natural setting for Calder\'on-Zygmund theory. We refer to the textbook \cite{stein-book} for an account on these spaces.
-
-Our concept of a compatible collection $\mathcal{Q}$ as a natural class of phase functions on a doubling metric measure space does not
-appear in \cite{stein-book} but is implicitly anticipated in \cite{zk-polynomial} and subsequent
-work of \cite{mnatsakanyan}, who proves a Carleson type theorem
-for the Malmquist-Takenaka series, which leads to classes of phases related to Blaschke products.
-A generalization of \eqref{def main op} from the previously mentioned Euclidean setting into the anisotropic setting that
-was suggested in \cite{zk-polynomial} is included in our theory.
-The polynomial Carleson operator also plays a role in the
-study of maximally modulated singular Radon transforms
-along the parabola, see \cite{ramos} and \cite{becker2024maximal}.
-
-
-
-
-
+In the one-dimensional Euclidean setting, with $K$ representing the Hilbert kernel:
+\begin{equation*}
+K(x,y)=(x-y)^{-1}
+\end{equation*}
+and $\Mf$ denoting the class of linear functions, the operator \eqref{def main op} is the classical Carleson operator, which plays a crucial role in proving the almost everywhere convergence of Fourier series \cite{carleson}, \cite{fefferman}, \cite{lacey-thiele}. The supremum in $R_1$ and $R_2$ is often omitted in classical treatments, but considering the maximal truncations can easily be reduced to the case without these truncations.
 
+By replacing $\Mf$ with the class of polynomials vanishing at $0$ up to some fixed but arbitrary degree, we obtain the polynomial Carleson operator of Lie \cite{lie-quadratic} (quadratic case) and \cite{lie-polynomial}. The case of the class of polynomials with vanishing linear coefficient is simpler and was estimated in \cite{stein-wainger}. The polynomial Carleson operator was generalized to the high-dimensional Euclidean setting in \cite{zk-polynomial} for $K$ being a Calderón-Zygmund kernel with some Hölder regularity.
 
+Doubling metric measure spaces are instances of spaces of homogeneous type. Indeed, by changing from a quasi-metric to an equivalent metric, every space of homogeneous type can be viewed as a doubling metric measure space (cf. \cite{MaciasSegovia}). Spaces of homogeneous type were introduced by \cite{MR0499948} as a natural setting for Calderón-Zygmund theory. We refer to the textbook \cite{stein-book} for an account of these spaces.
 
+Our concept of a compatible collection $\Mf$ as a natural class of phase functions on a doubling metric measure space does not appear in \cite{stein-book} but is implicitly anticipated in \cite{zk-polynomial} and subsequent work of \cite{mnatsakanyan}, who proves a Carleson-type theorem for the Malmquist-Takenaka series, which leads to classes of phases related to Blaschke products. A generalization of \eqref{def main op} from the previously mentioned Euclidean setting into the anisotropic setting that was suggested in \cite{zk-polynomial} is included in our theory. The polynomial Carleson operator also plays a role in the study of maximally modulated singular Radon transforms along the parabola, see \cite{ramos} and \cite{becker2024maximal}.
 
-For the proof of Theorem \ref{metric space Carleson}, we largely follow \cite{zk-polynomial}, which in turn was inspired by \cite{lie-polynomial}.
-We do suitable modifications into our more general setting and have a few technical improvements of the proof.  In particular, we explicitly divide
-in Section \ref{overviewsection} the main work of the proof into mutually independent
-Sections \ref{thmfromproplinear}, \ref{christsection}
-\ref{proptopropprop},
-\ref{antichainboundary},
-\ref{treesection}, \ref{liphoel}, and \ref{sec hlm}.
-Some of the sections follow a similar pattern, they start with a Subsection dividing the
-proof into further mutually independent subsections.
-This modularization of our proof was strongly endorsed in personal communication by the author of \cite{zk-polynomial}.
+For the proof of Theorem \ref{metric space Carleson}, we largely follow \cite{zk-polynomial}, which in turn was inspired by \cite{lie-polynomial}. We make suitable modifications to adapt to our more general setting and have made a few technical improvements in the proof. In particular, in Section \ref{overviewsection}, we explicitly divide the main work of the proof into mutually independent sections \ref{thmfromproplinear}, \ref{christsection}, \ref{proptopropprop}, \ref{antichainboundary}, \ref{treesection}, \ref{liphoel}, and \ref{sec hlm}. Some of these sections follow a similar pattern, starting with a subsection dividing the proof into further mutually independent subsections. This modularization of our proof was strongly endorsed in personal communication by the author of \cite{zk-polynomial}.
 
 
 
@@ -249,23 +184,17 @@ \chapter{Introduction}
 
 
 \noindent \textit{Acknowledgement.}
-L.B., F. v. D., R.S., and C.T. were funded by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) under Germany's Excellence Strategy -- EXC-2047/1 -- 390685813.
+L.B., F.v.D., R.S., and C.T. were funded by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) under Germany's Excellence Strategy -- EXC-2047/1 -- 390685813.
 L.B. , R.S., and C.T. were also supported by SFB 1060.
-A.J. is funded by the T\"UBITAK (Scientific and Technological Research Council of T\"urkiye) under the 1001-project 123F122.
+A.J. is funded by the T\"UBITAK (Scientific and Technological Research Council of T\"urkiye) under Grant Number 123F122.
 
 \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overview}
 \label{overviewsection}
 
 
-This section organizes the proof of Theorem
-\ref{metric space Carleson} into sections \ref{thmfromproplinear}, \ref{christsection}, \ref{proptopropprop}, \ref{antichainboundary}, \ref{treesection}, \ref{liphoel} and \ref{sec hlm}. These sections are mutually independent except for referring to
-the statements formulated in the present section.
-Section \ref{thmfromproplinear} proves the main Theorem \ref{metric space Carleson}, while sections
-\ref{christsection}, \ref{proptopropprop}, \ref{antichainboundary}, \ref{treesection}, \ref{liphoel} and \ref{sec hlm} each prove one proposition that is stated in the present section.
-The present section  also introduces all definitions used across these sections.
+This section organizes the proof of Theorem \ref{metric space Carleson} into sections \ref{thmfromproplinear}, \ref{christsection}, \ref{proptopropprop}, \ref{antichainboundary}, \ref{treesection}, \ref{liphoel}, and \ref{sec hlm}. These sections are mutually independent except for referring to the statements formulated in the present section. Section \ref{thmfromproplinear} proves the main Theorem \ref{metric space Carleson}, while sections \ref{christsection}, \ref{proptopropprop}, \ref{antichainboundary}, \ref{treesection}, \ref{liphoel}, and \ref{sec hlm} each prove one proposition that is stated in the present section. The present section also introduces all definitions used across these sections.
 
-Subsection \ref{global auxiliary lemmas} proves some auxiliary lemmas that are used in more than one
-of the sections 3-9.
+Subsection \ref{global auxiliary lemmas} proves some auxiliary lemmas that are used in more than one of the sections 3-9.
 
 Let $a, q$ be given as in Theorem \ref{metric space Carleson}.
 
@@ -285,10 +214,10 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv
     Z := 2^{12a}\,.
 \end{equation}
 Let
- $\psi:\R \to \R$ be the unique compactly supported, piece-wise affine, linear, continuous function with corners precisely at $\frac 1{4D}$, $\frac 1{2D}$, $\frac 14$ and $\frac 12$ which satisfies
+ $\psi:\R \to \R$ be the unique compactly supported, piece-wise linear, continuous function with corners precisely at $\frac 1{4D}$, $\frac 1{2D}$, $\frac 14$ and $\frac 12$ which satisfies
  \begin{equation}
     \label{eq psisum}
-    \sum_{s\in \mathbb{Z}} \psi(D^{-s}x)=1 .
+    \sum_{s\in \mathbb{Z}} \psi(D^{-s}x)=1
 \end{equation}
 for all  $x>0$. This function vanishes outside $[\frac1{4D},\frac 12]$, is constant one on
 $[\frac1{2D},\frac 14]$, and is Lipschitz
@@ -302,7 +231,7 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv
 
 Let a doubling metric measure space $(X,\rho,\mu, a)$ be given.
 Let a cancellative compatible collection $\Mf$ of functions on $X$ be given.
-Let $o\in X$ be a point such that $\mfa(o)=1$
+Let $o\in X$ be a point such that $\mfa(o)=0$
 for all $\mfa\in \Mf$.
 
 
@@ -310,7 +239,8 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv
 the triangle inequality part, without specific reference.
 It allows us to call $d_B$ a metric.
 
-\begin{lemma}
+\begin{lemma}[ball metric]
+\label{ball metric}
     For any ball $B$ in $X$, the local oscillation
 $d_{B}$ is a metric on $\Mf$.
 \end{lemma}
@@ -322,12 +252,11 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv
 d_{B(x,r)}(\mfa,\mfb)=0
 \end{equation}
 implies $\mfa(y)=\mfb(y)$ for all $y\in X$. Assume $\eqref{dvanish}$.
-Let
+Let $y\in X$ and let
 \begin{equation}
     R=1+\rho(x,o)+\rho(x,y)\, .
 \end{equation}
-By an iterated application of
-the comparability of the norms \eqref{firstdb} or \eqref{seconddb}
+By an iterated application of \eqref{firstdb} or \eqref{seconddb},
 \begin{equation}
     d_{B(x,R)}(\mfa,\mfb)=0.
 \end{equation}
@@ -354,7 +283,8 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv
  that a function from a measure space to a finite set is measurable if the pre-image of each of the elements in the range is measurable.
 
 
-\begin{prop}\label{finitary Carleson}
+\begin{prop}[finitary Carleson]
+\label{finitary Carleson}
 \uses{linearized truncation, grid existence, tile structure, tile sum operator}
 Let ${\sigma_1},\sigma_2\colon X\to \mathbb{Z}$ be measurable functions with finite range and ${\sigma_1}\leq  \sigma_2$. Let $\tQ\colon X\to \Mf$ be a measurable function with finite range. Let $F,G$ be bounded Borel sets in $X$. Then there is a Borel set $G'$ in $X$ with $2\mu(G')\leq  \mu(G)$ such that
 for all Borel functions $f:X\to \C$ with $|f|\le \mathbf{1}_F$.
@@ -468,7 +398,7 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv
 \end{equation}
 
 
-\begin{prop}
+\begin{prop}[discrete Carleson]
 \label{discrete Carleson}
 \uses{tile sum operator, exceptional set, forest union, forest complement}
 Let $(\mathcal{D}, c, s)$ be a grid structure and \begin{equation*}
@@ -559,7 +489,10 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv
 
 The following proposition is proved in Section \ref{antichainboundary}.
 
-\begin{prop}\label{antichain operator}
+\begin{prop}[antichain operator]
+\label{antichain operator}
+
+\uses{dens2 antichain,dens1 antichain}
 For any antichain $\mathfrak{A} $ and  for all $f:X\to \C$ with $|f|\le \mathbf{1}_F$ and all $g:X\to\C$ with $|g| \le \mathbf{1}_G$
 \begin{equation} \label{eq antiprop}
     |\int  \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)|
@@ -587,14 +520,14 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv
 \begin{equation}\label{forest1}
 4\fp\lesssim 1\fu.
 \end{equation}
-For each $\fu, \in \fU$ and each $\fp,\fp''\in \fT(\fu)$ and $\fp'\in \fP$
+For each $\fu \in \fU$ and each $\fp,\fp''\in \fT(\fu)$ and $\fp'\in \fP$
 we have
 \begin{equation}\label{forest2}
     \fp, \fp'' \in \mathfrak{T}(\fu), \fp \leq \fp' \leq \fp'' \implies \fp' \in \mathfrak{T}(\fu).
 \end{equation}
 We have
 \begin{equation}\label{forest3}
-   \|\sum_{\fu\in \fU} \mathbf{1}_{\sc(\fu))}\|_\infty \leq 2^n\,.
+   \|\sum_{\fu\in \fU} \mathbf{1}_{\sc(\fu)}\|_\infty \leq 2^n\,.
 \end{equation}
 We have for every $\fu\in \fU$
 \begin{equation}\label{forest4}
@@ -611,7 +544,8 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv
 
 
 The following proposition is proved in Section \ref{treesection}.
-\begin{prop}\label{forest operator}
+\begin{prop}[forest operator]
+\label{forest operator}
 For any $n\ge 0$ and any $n$-forest $(\fU,\fT)$ we have for all $f: X \to \mathbb{C}$ with $|f| \le \mathbf{1}_F$ and all bounded $g$ with bounded support
 $$
     | \int \overline{g(x)} \sum_{\fu\in \fU} \sum_{\fp\in \fT(\fu)} T_{\fp} f(x) \, \mathrm{d}\mu(x)|
@@ -639,10 +573,10 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv
     \|\varphi\|_{C^\tau(B)} = \sup_{x \in B} |\varphi(x)| + R^\tau \sup_{x,y \in B, x \neq y} \frac{|\varphi(x) - \varphi(y)|}{\rho(x,y)^\tau}\,.
 \end{equation}
 
-\lars{fractions in exponents everywhere}
 
-\begin{prop}
+\begin{prop}[Holder van der Corput]
     \label{Holder van der Corput}
+    \uses{Lipschitz Holder approximation}
      Let $z\in X$ and $R>0$ and set $B=B(z,R)$.
      Let $\varphi: X \to \mathbb{C}$ by
      supported on $B$ and satisfy $\|{\varphi}\|_{C^\tau(B)}<\infty$.
@@ -667,8 +601,9 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv
 \end{equation}
 Define further $M_{\mathcal{B}}:=M_{\mathcal{B},1}$.
 
-\begin{prop}\label{Hardy Littlewood}
-\uses{first exception}
+\begin{prop}[Hardy Littlewood]
+\label{Hardy Littlewood}
+\uses{layer cake representation,covering separable space}
    Let $\mathcal{B}$ be a finite collection of balls in $X$.
 If for some $\lambda>0$ and some measurable function $u:X\to [0,\infty)$ we have
 \begin{equation}\label{eq ball assumption}
@@ -689,13 +624,12 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv
     \label{eq ball av}
     \frac{1}{\mu(B)} \int_{B}  |w(y)| \, \mathrm{d}\mu(y) \le Mw(x)
 \end{equation}
-and for all $1 < p \le \infty$
+and for all $1 < p_1 \le p_2 \le \infty$
 \begin{equation}
     \label{eq hlm 2}
-    \|Mw\|_p \le 2^{4a} \frac{p}{p-1}\|w\|_p\,.
+    \|M(w^{p_1})^{\frac{1}{p_1}}\|_{p_2} \le 2^{4a} \frac{p_2}{p_2-p_1}\|w\|_{p_2}\,.
 \end{equation}
 
-
 \end{prop}
 
 This completes the overview of the proof of Theorem \ref{metric space Carleson}.
@@ -706,25 +640,25 @@ \section{Auxiliary lemmas}
 
 First, we record an estimate for the metrical entropy numbers of balls in the space $\Mf$ equipped with any of the metrics $d_B$, following from the doubling property \eqref{thirddb}.
 
-\begin{lemma}
-    \label{lem entropy Theta}
-    Let $B \subset X$ be a ball. Let $r > 0$, $\mfa \in \Mf$ and $k \in \mathbb{N}$. Suppose that $\mathcal{Z} \subset B_B(\mfa, r2^k)$ \rs{Double notation} satisfies that for all $z, z' \in \mathcal{Z}$ with $z \ne z'$, we have $d_B(z,z') \ge r$. Then
+\begin{lemma}[ball metric entropy]
+    \label{ball metric entropy}
+    Let $B' \subset X$ be a ball. Let $r > 0$, $\mfa \in \Mf$ and $k \in \mathbb{N}$. Suppose that $\mathcal{Z} \subset B_{B'}(\mfa, r2^k)$  satisfies that for all $z, z' \in \mathcal{Z}$ with $z \ne z'$, we have $d_{B'}(z,z') \ge r$. Then
     $$
         |\mathcal{Z}| \le 2^{ka}\,.
     $$
 \end{lemma}
 
 \begin{proof}
-    By applying property \eqref{thirddb}  $k$ times, we obtain a collection $\mathcal{Z}' \subset \Mf$ with $|\mathcal{Z}'| \le 2^{ka}$ and
+    By applying property \eqref{thirddb}  $k$ times, we obtain a collection $\mathcal{Z}' \subset \Mf$ with $|\mathcal{Z}'| = 2^{ka}$ and
     $$
-        B_{B}(\mfa,r2^k) \subset \bigcup_{z' \in \mathcal{Z}'} B_B(z', \frac{r}{2})\,.
+        B_{B'}(\mfa,r2^k) \subset \bigcup_{z' \in \mathcal{Z}'} B_{B'}(z', \frac{r}{2})\,.
     $$
     Then each $z \in \mathcal{Z}$ is contained in one of the balls $B(z', \frac{r}{2})$, but by the separation assumption no such ball contains more than one element of $\mathcal{Z}$. Thus $|\mathcal{Z}| \le |\mathcal{Z}'| = 2^{ka}$.
 \end{proof}
 
 The next lemma concerns monotonicity of the metrics $d_{B(c(I), \frac 14 D^{s(I)})}$ with respect to inclusion of cubes $I$ in a grid.
 
-\begin{lemma}
+\begin{lemma}[monotone cube metrics]
     \label{monotone cube metrics}
     \uses{wiggle order 2, forest separation}
     Let $(\mathcal{D}, c, s)$ be a grid structure. Denote for cubes $I \in \mathcal{D}$
@@ -770,7 +704,8 @@ \section{Auxiliary lemmas}
 
 We also record the following basic estimates for the kernels $K_s$.
 
-\begin{lemma}\label{kernel summand}
+\begin{lemma}[kernel summand]
+\label{kernel summand}
     Let $-S\le s\le S$ and $x,y,y'\in X$.
     If $K_s(x,y)\neq 0$, then we have
     \begin{equation}\label{supp Ks}
@@ -781,7 +716,8 @@ \section{Auxiliary lemmas}
        \label{eq Ks size}
         |K_s(x,y)|\le \frac{2^{102 a^3}}{\mu(B(x, D^{s}))}\,
     \end{equation}
-    and \begin{equation}
+    and %if $2\rho(y,y')\leq \rho(x,y)$
+    \begin{equation}
         \label{eq Ks smooth}
         |K_s(x,y)-K_s(x, y')|\le \frac{2^{150a^3}}{\mu(B(x, D^{s}))}
         \left(\frac{ \rho(y,y')}{D^s}\right)^{\frac 1a}\,.
@@ -808,14 +744,14 @@ \section{Auxiliary lemmas}
     \end{equation}
     Using $a\ge 4$ proves \eqref{eq Ks size}.
 
-
+    \lars{The case where $\rho(y,y') > \rho(x,y)/2$ is missing}
     Similarly, we obtain with  \eqref{eqkernel y smooth} and the lower bound in
     \eqref{supp Ks}
     \begin{equation}
         |K_s(x,y)-K_s(x, y')|\le \frac{2^{a^3}}{\mu(B(x, \frac 14 D^{s-1}))}
         \left(\frac{ \rho(y,y')}{\frac 14 D^{s-1}}\right)^{\frac 1a}\,.
     \end{equation}
-    Using $a \ge 4$, this is estimated by
+    As above, this is estimated by
     \begin{equation}
        \le \frac{4D 2^{2a+101a^3}}{\mu(B(x,  D^{s}))}
         \left(\frac{ \rho(y,y')}{D^{s}}\right)^{\frac 1a}
@@ -836,7 +772,8 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})}
     X=\bigcup_{R>0}B(o,R),
 \end{equation} because every point of $X$
 has finite distance from $o$.
-\begin{lemma}\label{R truncation}
+\begin{lemma}[R truncation]
+\label{R truncation}
  \uses{S truncation}
 For all integers $R>0$
 $$
@@ -854,25 +791,11 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})}
 \end{equation}
 \end{lemma}
 
-We first see how the Lemma \ref{R truncation} implies
-Theorem \ref{metric space Carleson}. When $R$ tends to $\infty$, the integrand of the left-hand-side of \eqref{Rcut}
-grows monotonically to the integrand of the
-left-hand side of \eqref{resweak} for all $x$.
-By Lebesgue's monotone convergence theorem, the left-hand side of \eqref{Rcut} converges to the
-left-hand side of \eqref{resweak}. This proves Theorem \ref{metric space Carleson}.
+We first show how Lemma \ref{R truncation} implies Theorem \ref{metric space Carleson}. As $R$ tends to $\infty$, the integrand of the left-hand side of \eqref{Rcut} grows monotonically toward the integrand of the left-hand side of \eqref{resweak} for all $x$. By Lebesgue's monotone convergence theorem, the left-hand side of \eqref{Rcut} converges to the left-hand side of \eqref{resweak}. This verifies Theorem \ref{metric space Carleson}.
 
-It remains to prove Lemma \ref{R truncation}.
-Fix an integer $R>0$.  Replacing
-$G$ by $G\cap B(o,R)$ if necessary, it suffices to show
-\eqref{Rcut} under the assumption that $G$ is contained in $B(o,R)$. We make this assumption.
-For every $x\in G$, the domain of integration
-in \eqref{TRR} is contained in $B(o,2R)$.
-Replacing
-$F$ by $F\cap B(o,2R)$ if necessary, it suffices to show
-\eqref{Rcut} under the assumption that $F$ is contained in $B(o,2R)$. We make this assumption.
+It remains to prove Lemma \ref{R truncation}. Fix an integer $R>0$. By replacing $G$ with $G\cap B(o,R)$ if necessary, it suffices to show \eqref{Rcut} under the assumption that $G$ is contained in $B(o,R)$. We make this assumption. For every $x\in G$, the domain of integration in \eqref{TRR} is contained in $B(o,2R)$. By replacing $F$ with $F\cap B(o,2R)$ if necessary, it suffices to show \eqref{Rcut} under the assumption that $F$ is contained in $B(o,2R)$. We make this assumption.
 
-With the definition \eqref{defks} of $K_s$
-and the partition of unity \eqref{eq psisum}, we write \eqref{TRR} as the sum of
+Using the definition \eqref{defks} of $K_s$ and the partition of unity \eqref{eq psisum}, we express \eqref{TRR} as the sum of
 \begin{equation}\label{middles}
 {T}_{1,s_1,s_2,\mfa}f(x):=\sum_{s_1 \le s\le s_2}
 \int K_s(x,y)  f(y) e(\mfa(y)) \, \mathrm{d}\mu(y)
@@ -883,25 +806,15 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})}
 \int_{R_1 <  \rho(x,y) < R_2}  K_s(x,y) f(y) e(\mfa(y)) \,
  \mathrm{d}\mu(y),
 \end{equation}
-where $s_1$ is the smallest integer such that $D^{s_1-2}R_2>\frac 1{4D}$ and $s_2$
-is the largest integer so that $D^{s_2+2}R_1<\frac 12$. Here we restricted the summation index $s$
-by omitting the summands with $s<s_1-2$
-or $s>s_2+2$ because for these summands the function $K_s$ vanishes on the domain of integration, and we have omitted the restriction in the integral
-in the  summands in \eqref{middles} because in these summands the support of $K_s$ is contained in
-the set described by this restriction.
+where $s_1$ is the smallest integer such that $D^{s_1-2}R_2>\frac 1{4D}$ and $s_2$ is the largest integer such that $D^{s_2+2}R_1<\frac 12$. We restrict the summation index $s$ by excluding summands with $s<s_1-2$ or $s>s_2+2$ because for these summands, the function $K_s$ vanishes on the domain of integration. We also omit the restriction in the integral for the summands in \eqref{middles} because in these summands, the support of $K_s$ is contained in the set described by this restriction.
 
 
 
-We apply the triangle inequality and estimate the
-versions of \eqref{Rcut} separately with $T_{R_1,R_2,\mfa}$ replaced by
-\eqref{middles} and by each summand of \eqref{boundarys}.
-To deal with the case \eqref{middles} we use the following lemma, where we use that if
-$\frac 1R\le R_1\le R_2\le R$, then $s_1,s_2$
-as in \eqref{middles} are in an interval $[-S,S]$ for some
-sufficiently large $S$ depending on $R$.
+We apply the triangle inequality and estimate the versions of \eqref{Rcut} separately with $T_{R_1,R_2,\mfa}$ replaced by \eqref{middles} and by each summand of \eqref{boundarys}. To handle the case \eqref{middles}, we employ the following lemma. Here, we utilize the fact that if $\frac 1R\le R_1\le R_2\le R$, then $s_1$ and $s_2$ as in \eqref{middles} are in an interval $[-S,S]$ for some sufficiently large $S$ depending on $R$.
 
 
-\begin{lemma}\label{S truncation}
+\begin{lemma}[S truncation]
+\label{S truncation}
 \uses{finitary S truncation}
 For all integers $S>0$
 $$
@@ -915,7 +828,7 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})}
 where $T_{1, s_1,s_2,\mfa}$ is defined in \eqref{middles}.
 \end{lemma}
 
-To reduce Lemma \ref{R truncation} to Lemma \ref{S truncation},  we need estimates for the summands in \eqref{boundarys}. Using Lemma \ref{kernel summand}, we have for arbitrary $s$ the inequality
+To reduce Lemma \ref{R truncation} to Lemma \ref{S truncation}, we need estimates for the summands in \eqref{boundarys}. Using Lemma \ref{kernel summand}, we obtain for arbitrary $s$ the inequality
 \begin{multline}
 \left|\int_{R_1 <  \rho(x,y) < R_2}  K_s(x,y) f(y) e(\mfa(y)) \,  \mathrm{d}\mu(y)\right|\\
 \leq \frac{2^{102 a^3}}{\mu(B(x, D^{s}))}
@@ -923,17 +836,16 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})}
 \leq 2^{102 a^3} M\mathbf{1}_F(x),
 \end{multline}
 where $M\mathbf{1}_F$ is as defined in Proposition \ref{Hardy Littlewood}.
-Now the left hand side of \eqref{Rcut}, with $T_{R_1,R_2,\mfa}$ replaced by a summand of \eqref{boundarys}, can be estimated using Hölder's inequality and Proposition \ref{Hardy Littlewood} by
+Now, the left-hand side of \eqref{Rcut}, with $T_{R_1,R_2,\mfa}$ replaced by a summand of \eqref{boundarys}, can be estimated using Hölder's inequality and Proposition \ref{Hardy Littlewood} by
 $$
     2^{102 a^3}\int \mathbf{1}_{G}(x) M\mathbf{1}_F(x)\, d\mu(x)
     \leq \frac{2^{102 a^3+4a}q}{q-1}\mu(G)^{1-\frac{1}{q}} \mu(F)^{\frac{1}{q}}\,.
 $$
-Using the triangle inequality to estimate the left hand side of \eqref{Rcut} by contributions from the summands in \eqref{middles} and \eqref{boundarys}, using Lemma \ref{S truncation} to control the first term, and the above to estimate the contribution from the four summands in \eqref{boundarys}, combined with $a>4$ and $q < 2$ completes the reduction of
-Lemma \ref{R truncation} to Lemma \ref{S truncation}.
+Applying the triangle inequality to estimate the left-hand side of \eqref{Rcut} by contributions from the summands in \eqref{middles} and \eqref{boundarys}, using Lemma \ref{S truncation} to control the first term, and the above to estimate the contribution from the four summands in \eqref{boundarys}, combined with $a\geq 4$ and $q < 2$, completes the reduction of Lemma \ref{R truncation} to Lemma \ref{S truncation}.
 
 It remains to prove Lemma \ref{S truncation}. Fix $S>0$.
-
-\begin{lemma}\label{finitary S truncation}
+\begin{lemma}[finitary S truncation]
+\label{finitary S truncation}
 \uses{linearized truncation}
 For all finite sets $\tilde{\Mf}\subset \Mf$
 $$
@@ -949,10 +861,10 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})}
 We reduce Lemma \ref{S truncation} to Lemma \ref{finitary S truncation}.
 By the Lebesgue monotone convergence theorem,
 applied to an increasing sequence of finite sets $\tilde{\Mf}$, inequality \eqref{Sqcut}
-continues to hold for  countable $\tilde{\Mf}$.
+continues to hold for countable $\tilde{\Mf}$.
 
 Let $\epsilon=\frac{1}{2S+1}$. Pick some $\mfa_0 \in \Mf$.
-Let for $k \ge 0$ the set $\tilde{\Mf}_k$ be a subset of $B_{B(o,2R)}(\mfa, k)$ of maximal size, such that for all $\mfa, \mfb \in \tilde{\Mf}_k$, it holds that $d_{B(o, 2R)}(\mfa, \mfb) \ge \epsilon$. Such a set exists, since by Lemma \ref{lem entropy Theta} there exists an upper bound for the size of such subsets in $B_{B(o, 2R)}(\mfa_0, k)$. Let
+For $k \ge 0$, let the set $\tilde{\Mf}_k$ be a subset of $B_{B(o,2R)}(\mfa_0, k)$ of maximal size, such that for all $\mfa, \mfb \in \tilde{\Mf}_k$, it holds that $d_{B(o, 2R)}(\mfa, \mfb) \ge \epsilon$. Such a set exists, since by Lemma \ref{ball metric entropy} there exists an upper bound for the size of such subsets in $B_{B(o, 2R)}(\mfa_0, k)$. Define
 $$
     \tilde{\Mf} := \bigcup_{k \in \mathbb{N}} \tilde{\Mf}_k\,.
 $$
@@ -961,7 +873,7 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})}
     d_{B(o, 2R)}(\mfb, \mfa) < \epsilon\,.
 $$
 
-For every $\mfa\in \Mf$ we have
+For every $\mfa\in \Mf$, we have
 \begin{equation}
 \left| T_{1, s_1,s_2,\mfa} \mathbf{1}_{F}(x)\right|=
 \left|\sum_{s_1 \le s\le s_2}
@@ -969,54 +881,46 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})}
 \end{equation}
 Moreover, there is a $\tilde{\mfa}
 \in \tilde{\Mf}$ with $d_{B(o,2R)}(\mfa,\tilde{\mfa})\le \epsilon$. Hence,
-$$
-    \left| T_{1, s_1,s_2,\mfa} \mathbf{1}_{F}(x)\right|-\left|T_{1, s_1,s_2,\tilde{\mfa}} \mathbf{1}_{F}(x) \right|
-$$
-$$
-  \leq\sum_{s_1 \le s\le s_2} \int |K_s(x,y)|  \mathbf{1}_F(y) |e(\mfa(y)-{\mfa(x)})-e({\tilde{\mfa}(y)}-{\tilde{\mfa}(x)})| \, \mathrm{d}\mu(y)
-$$
-$$
-    \leq \sum_{s_1 \le s\le s_2} \epsilon \int |K_s(x,y)|  \mathbf{1}_F(y) \, \mathrm{d}\mu(y)
-$$
+\begin{align*}
+    &\left| T_{1, s_1,s_2,\mfa} \mathbf{1}_{F}(x)\right|-\left|T_{1, s_1,s_2,\tilde{\mfa}} \mathbf{1}_{F}(x) \right|\\
+  &\leq\sum_{s_1 \le s\le s_2} \int |K_s(x,y)|  \mathbf{1}_F(y) |e(\mfa(y)-{\mfa(x)})-e({\tilde{\mfa}(y)}-{\tilde{\mfa}(x)})| \, \mathrm{d}\mu(y)\\
+    &\leq \sum_{s_1 \le s\le s_2} \epsilon \int |K_s(x,y)|  \mathbf{1}_F(y) \, \mathrm{d}\mu(y)
+\end{align*}
 Using Lemma \ref{kernel summand}, we can estimate the above expression by
-$$
-    \sum_{s_1 \le s\le s_2} \frac{2^{102 a^3}}{\mu(B(x, D^{s}))}
-\epsilon \int_{B(x, D^s)} \mathbf{1}_F(y)  \, \mathrm{d}\mu(y)
-$$
-$$
-    \leq (2S+1)\epsilon 2^{102 a^3} M\mathbf{1}_F(x)\le 2^{102 a^3}M\mathbf{1}_F(x)
-$$
-% \begin{equation}
-%   \leq  (2S+1) \epsilon M\mathbf{1}_F(x)\lesssim  M\mathbf{1}_F(x)
-% \end{equation}
+\begin{align*}
+    &\sum_{s_1 \le s\le s_2} \frac{2^{102 a^3}}{\mu(B(x, D^{s}))}
+\epsilon \int_{B(x, D^s)} \mathbf{1}_F(y)  \, \mathrm{d}\mu(y)\\
+    &\leq (2S+1)\epsilon 2^{102 a^3} M\mathbf{1}_F(x)\le 2^{102 a^3}M\mathbf{1}_F(x)
+\end{align*}
 We estimate the left-hand-side of \eqref{Scut} by
-the sum of left hand side of \eqref{Sqcut} and
-  \begin{equation}
-    \int \mathbf{1}_{G}(x)
+the sum of left-hand-side of \eqref{Sqcut} and
+\begin{align*}
+    &\int \mathbf{1}_{G}(x)
 \max_{-S<s_1\le s_2<S}\sup_{\mfa\in {\Mf}}
 \inf_{\tilde{\mfa}\in\tilde{\Mf}}
-(| T_{1, s_1,s_2,\mfa}|-|T_{1, s_1,s_2,\tilde{\mfa}}  |)\mathbf{1}_{F}(x)\, d\mu(x)\,,
-\end{equation}
+(| T_{1, s_1,s_2,\mfa}|-|T_{1, s_1,s_2,\tilde{\mfa}}  |)\mathbf{1}_{F}(x)\, d\mu(x)\,,\\
+\end{align*}
 which, as we have just shown, is estimated by
-\begin{equation}
-     2^{102 a^3}\int \mathbf{1}_{G}(x)
+\begin{align*}
+     &2^{102 a^3}\int \mathbf{1}_{G}(x)
 M\mathbf{1}_{F}(x)\, d\mu(x).
-\end{equation}
+\end{align*}
 By Hölder's inequality and Proposition \ref{Hardy Littlewood}
- (more precisely, \eqref{eq hlm 2} with $p=q$) the above is no greater than
-\begin{equation}
-\frac{2^{102 a^3+4a}q}{q-1} \mu(G)^{1-\frac{1}{q}} \mu(F)^{\frac{1}{q}}.
-\end{equation}
+ (more precisely, \eqref{eq hlm 2} with $p=q$), the above is no greater than
+\begin{align*}
+&\frac{2^{102 a^3+4a}q}{q-1} \mu(G)^{1-\frac{1}{q}} \mu(F)^{\frac{1}{q}}.
+\end{align*}
 Combining this with Lemma \ref{finitary S truncation} and the fact that
-\begin{equation*}
+\begin{align*}
     \frac{2^{102 a^3+4a}q}{q-1}\leq \frac{2^{265a^3}}{(q-1)^5}
-\end{equation*}
+\end{align*}
 proves Lemma \ref{S truncation}.
 
 It remains to prove Lemma \ref{finitary S truncation}.
 Fix a finite set
 $\tilde{\Mf}$.
-\begin{lemma}\label{linearized truncation}
+\begin{lemma}[linearized truncation]
+\label{linearized truncation}
 Let $\sigma_1,\sigma_2\colon X\to \mathbb{Z}$ be measurable functions with finite range
 $[-S,S]$ and $\sigma_1\leq \sigma_2$.
 Let $\tQ\colon X\to {\tilde{\Mf}}$ be a measurable function. Then we have
@@ -1032,7 +936,7 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})}
 \end{equation}
 \end{lemma}
 
-We reduce Lemma \ref{finitary S truncation}  to
+We reduce Lemma \ref{finitary S truncation} to
 Lemma \ref{linearized truncation}.
 For each $x$, let $\sigma_1(x)$ be the
 minimal element $s'\in [-S,S]$ such that
@@ -1050,7 +954,7 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})}
 % \Tilde{T}_x|
 % \]
 \[\max_{\mfa\in\tilde{\Mf}}
-| {T}_{1, {\sigma}_1(x), s_2,\mfa} \mathbf{1}_{F}(x)|
+| {T}_{1, {\sigma}_1(x), s'',\mfa} \mathbf{1}_{F}(x)|
 =
 {T}_{1,x}\,.
 \]
@@ -1069,7 +973,7 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})}
 
 It remains to prove Lemma \ref{linearized truncation}.
 Fix $\sigma_1$, $\sigma_2$,
-and $\tQ$ as in  the lemma.
+and $\tQ$ as in the lemma.
 Applying Proposition \ref{finitary Carleson} recursively, we obtain
 a sequence of sets $G_n$ with $G_0=G$ and,
 for each $n\ge 0$, $\mu(G_{n})\le 2^{-n} \mu(G)$ and
@@ -1111,17 +1015,16 @@ \chapter{Proof of finitary Carleson (Prop \ref{finitary Carleson})}
 measurable functions ${\sigma_1},\sigma_2,  \tQ$ and Borel sets $F,G$. We have also
 defined $S$ to be the smallest
 integer such that the ranges of
-$\underline{\sigma}$ and $ \overline\sigma$ are contained in $[-S,S]$ and $F$ and $G$ are contained
+$\sigma_1$ and $\sigma_2$ are contained in $[-S,S]$ and $F$ and $G$ are contained
 in the ball $B(o, D^S)$.
 
 
 The proof of the next lemma is done in Subsection \ref{subsecdyadic},
 following the construction of dyadic cubes in \cite[\S 3]{christ1990b}.
 
-\begin{lemma}\label{grid existence}
-\uses{counting balls,cover big ball,cover by cubes}
-
-    There exists a grid structure $(\mathcal{D}, c,s)$.
+\begin{lemma}[grid existence]
+\label{grid existence}
+\uses{counting balls,cover big ball,cover by cubes}     There exists a grid structure $(\mathcal{D}, c,s)$.
 \end{lemma}
 
 
@@ -1130,9 +1033,9 @@ \chapter{Proof of finitary Carleson (Prop \ref{finitary Carleson})}
 The next lemma, which we prove in Subsection \ref{subsectiles}, should be compared
 with the construction in \cite[Lemma 2.12]{zk-polynomial}.
 
-\begin{lemma}\label{tile structure}
- \uses{frequency ball cover, disjoint frequency cubes,,
-frequency cube cover}
+\begin{lemma}[tile structure]
+\label{tile structure}
+ \uses{frequency ball cover, disjoint frequency cubes,frequency cube cover}
     For a given grid structure $(\mathcal{D}, c,s)$, there exists a tile structure
     $(\fP,\sc,\fc,\fcc,\pc,\ps)$.
 \end{lemma}
@@ -1140,9 +1043,10 @@ \chapter{Proof of finitary Carleson (Prop \ref{finitary Carleson})}
 Choose a grid structure $(\mathcal{D}, c,s)$ with Lemma \ref{grid existence} and a tile structure for this
 grid structure $(\fP,\sc,\fc,\fcc,\pc,\ps)$ with Lemma \ref{tile structure}.
 Applying Proposition \ref{discrete Carleson}, we obtain a Borel set $G'$ in $X$ with $2\mu(G')\leq \mu(G)$ such that for all Borel functions $f:X\to \C$ with $|f|\le \mathbf{1}_F$
-we have \eqref{eq-linearized}.
+we have \eqref{disclesssim}.
 
-\begin{lemma} \label{tile sum operator}
+\begin{lemma}[tile sum operator]
+\label{tile sum operator}
 We have for all $x\in G\setminus G'$
 \begin{equation}\label{eq sump}
     \sum_{\fp\in \fP}T_{\fp} f(x)= \sum_{s=\sigma_1(x)}^{\sigma_2(x)}
@@ -1191,21 +1095,21 @@ \chapter{Proof of finitary Carleson (Prop \ref{finitary Carleson})}
 
 
 
-\section{Proof of L.\ref{grid existence}, dyadic structure}
+\section{Proof of Lemma \ref{grid existence}, dyadic structure}
 \label{subsecdyadic}
 
 We begin with the construction of the centers of the dyadic cubes.
-\begin{lemma}\label{counting balls}
-Let $-S\le k\le S$. Let $Y\subset X$ such that for any $y\in Y$
-we have
+\begin{lemma}[counting balls]
+\label{counting balls}
+Let $-S\le k\le S$. Consider $Y\subset X$ such that for any $y\in Y$, we have
 \begin{equation}\label{ybinb}
-  y\in B(o,4D^S-D^k)
+  y\in B(o,4D^S-D^k),
 \end{equation}
-and further, for any $y'\in Y$ with $y\neq y'$ we have
+furthermore, for any $y'\in Y$ with $y\neq y'$, we have
 \begin{equation} \label{eq disj yballs}
     B(y,D^k)\cap B(y',D^k)=\emptyset.
 \end{equation}
-Then  the cardinality of $Y$ is estimated by
+Then the cardinality of $Y$ is bounded by
 \begin{equation}\label{boundY}
     |Y|\le 2^{3a + 200Sa^3}\, .
 \end{equation}
@@ -1213,20 +1117,17 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 
 
 \begin{proof}
-   Let $k$ and $Y$ be given.
-  By applying the doubling property \eqref{doublingx} inductively, we have for each integer $j\ge 0$
-  \begin{equation}\label{jballs}
-      \mu(B(y,2^{j}D^k))\le 2^{aj} \mu(B(y,D^k))\, .
-  \end{equation}
- As $X$ is the union of the balls $B(y,2^{j}D^k)$ and $\mu$ is not zero, at least one of
- the balls $B(y,2^{j}D^k)$ has positive measure and thus $B(y,D^k)$ has positive measure.
+Let $k$ and $Y$ be given. By applying the doubling property \eqref{doublingx} inductively, we have for each integer $j\ge 0$
+\begin{equation}\label{jballs}
+    \mu(B(y,2^{j}D^k))\le 2^{aj} \mu(B(y,D^k))\, .
+\end{equation}
+Since $X$ is the union of the balls $B(y,2^{j}D^k)$ and $\mu$ is not zero, at least one of the balls $B(y,2^{j}D^k)$ has positive measure, thus $B(y,D^k)$ has positive measure.
 
- Applying \eqref{jballs} for $j' = \ln_2(8D^{2S}) = 3 + 2S \cdot 100a^2$ by \eqref{defineD}, using $-S\le k\le S$
- and $y\in B(o,4D^S)$ and the triangle inequality, we have
-   \begin{equation}
-     B(o, 4D^S) \subset B(y, 8D^S) \subset B(y,2^{j'}D^k) \, .
-  \end{equation}
-Using disjointedness of the balls in \eqref{eq disj yballs}, \eqref{ybinb} and the triangle inequality for $\rho$, we obtain
+Applying \eqref{jballs} for $j' = \ln_2(8D^{2S}) = 3 + 2S \cdot 100a^2$ by \eqref{defineD}, using $-S\le k\le S$, $y\in B(o,4D^S)$, and the triangle inequality, we have
+\begin{equation}
+    B(o, 4D^S) \subset B(y, 8D^S) \subset B(y,2^{j'}D^k) \, .
+\end{equation}
+Using the disjointedness of the balls in \eqref{eq disj yballs}, \eqref{ybinb}, and the triangle inequality for $\rho$, we obtain
 \begin{equation}
 |Y|\mu(B(o,4D^S))\le 2^{j'a}\sum_{y\in Y}\mu(B(y,D^k))
 \end{equation}
@@ -1246,7 +1147,7 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 For each $-S\le k\le S$, choose an enumeration of the points in the finite set $Y_k$ and thus a total
 order  $<$ on $Y_{k}$.
 
-\begin{lemma}
+\begin{lemma}[cover big ball]
 \label{cover big ball}
 \uses{basic grid structure}
   For each $-S\le k\le S$, the ball
@@ -1273,7 +1174,7 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 \end{equation}
 
 
-Assume the sets ${I}_j(y',k')$ already defined  for $j=1,2,3$ if $k'<k$ and if  $k=k$ and $y'<y$.
+Assume the sets ${I}_j(y',k')$ have already been defined  for $j=1,2,3$ if $k'<k$ and if  $k=k'$ and $y'<y$.
 
 
 
@@ -1292,12 +1193,12 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 \end{equation}
 with
 \begin{equation}
-      X_{k}:=\bigcup\{I_1(y', k):y'\in Y_{k} .\}
+      X_{k}:=\bigcup\{I_1(y', k):y'\in Y_{k}\}.
 \end{equation}
 
 
-\begin{lemma}\label{basic
-grid structure}
+\begin{lemma}[basic grid structure]
+\label{basic grid structure}
 \uses{dyadic property}
    For each $-S\le k\le S$ and $1\le j\le 3$
    the following holds.
@@ -1323,82 +1224,22 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 \end{lemma}
 %\rs{Write}
 \begin{proof}
-We prove these statements simultaneously by induction on the ordered set of pairs $(y,k)$.
-Let $-S\le k\le S$.
-
-We first consider \eqref{disji} for $j=1$.
-If $k=-S$, disjointedness of the sets $I_1(y,-S)$ follows by definition of $I_1$ and $Y_k$. If $k>-S$,
-assume $x$ is in $I_1(y_m,-k)$ for $m=1,2$.
-Then, for $m=1,2$, there is $z_m\in Y_{k-1}\cap B(y_m,D^k)$ with $x\in I_3(z_m,k-1)$.
-Using \eqref{disji} inductively for $j=3$, we
-conclude $z_1=z_2$. This implies that the balls
-$B(y_1, D^k)$ and $B(y_2, D^k)$ intersect. By construction of $Y_k$, this implies $y_1=y_2$.
-This proves \eqref{disji} for $j=1$,
-
-We next consider \eqref{disji} for $j=3$.
-Assume $x$ is in $I_3(y_m,k)$ for $m=1,2$ and $y_m\in Y_k$. If $x$ is in $X_k$, then by definition
-\eqref{definei3}, $x\in I_1(y_m,k)$ for $m=1,2$.
-As we have already shown \eqref{disji} for $j=1$,
-we conclude $y_1=y_2$. This completes the proof in
-case $x\in X_k$ and we may assume $x$ is not in $X_k$. By definition \eqref{definei3}, $x$ is not
-in $I_3(z,k)$  for any $z$ with $z<y_1$ or $z<y_2$.
-Hence neither $y_1<y_2$ nor $y_2<y_1$, and by totality
-of the order of $Y_k$ we have $y_1=y_2$. This completes the proof of \eqref{disji} for $j=3$.
-
-We  show \eqref{unioni}  for $j=2$.
-In case $k=-S$, this follows from Lemma \ref{cover big ball}.
-Assume  $k>-S$. Let $x$ be a point of $B(o, 4D^S-2D^k$).
-By induction, there is $y'\in Y_{k-1}$ such that
-$x\in I_3(y',k-1)$. Using the inductive statement
-\eqref{squeezedyadic}, we obtain $x\in B(y',4D^{k-1})$.
-As $D>4$, by applying the triangle inequality with
-the points, $o$, $x$, and $y'$ we obtain that  $y'\in B(o, 4D^S-D^k)$.
-By Lemma {\ref{cover big ball}}, $y'$ is in $B(y,2D^k)$
-for some $y\in Y_k$. It follows that  $x\in I_2(y,k)$.
-This proves \eqref{unioni}  for $j=2$.
-
-We  show \eqref{unioni} for $j=3$.
-Let $x\in  B(o, 4D^S-2D^k)$. In case $x\in X_k$,
- then by definition of $X_k$ we have $x\in I_1(y,k)$ for some $y\in Y_k$ and thus $x\in I_3(y,k)$. We may thus assume $x\not\in X_k$. As we have already seen
-\eqref{unioni} for $j=2$,
- there is $y\in Y_k$ such that $x\in I_2(y,k)$.
-We may assume this $y$ is minimal with respect to the order in $Y_k$
-Then $x\in I_3(y,k)$.
- This proves \eqref{unioni}  for $j=3$.
-
-Next we  show the first inclusion in \eqref{squeezedyadic}.
-Let $x\in B(y,\frac 1{2}D^k)$.
-As $I_1(y,k)\subset I_3(y,k)$,
-it suffices to show $x\in I_1(y,k)$.
-If $k=-S$, this follows immediately from
-the assumption on $x$ and definition of $I_1$.
-Assume $k>-S$. By inductive statement \eqref{unioni}
-and $D>4$, there is a
-$y'\in Y_{k-1}$ such that $x\in I_3(y',k-1)$.
-By inductive statement \eqref{squeezedyadic},
-we conclude $x\in B(y',4D^{k-1})$.
-By the triangle inequality with points $x$, $y$, $y'$ and $D>4$, we have
-$y'\in B(y,D^k)$. It follows by definition
-\eqref{defineij} that
-$I_3(y',k-1)\subset I_1(y,k)$ and thus
-{$x\in I_3(y,k)$}. This proves the first inclusion
-in \eqref{squeezedyadic}.
-
-
-We show the second inclusion in \eqref{squeezedyadic}.
-Let $x\in I_3(y,k)$. As $I_1(y,k)\subset I_2(y,k)$
-directly from the definition \eqref{defineij},
-it follows by definition \eqref{definei3} that
-$x\in I_2(y,k)$. By definition
-\eqref{defineij}, there is $y'\in Y_{k-1}\cap B(y,2D^k)$
-with $x\in I_3(y',k-1)$. By induction,
-$x\in B(y', 4D^{k-1})$. By the triangle inequality
-applied to the points $x,y',y$ and $D>4$, we conclude
-$x\in B(y,4D^k)$.
-This shows the second inclusion in \eqref{squeezedyadic} and completes the proof of the lemma.
+We prove these statements simultaneously by induction on the ordered set of pairs $(y,k)$. Let $-S\le k\le S$.
+
+We first consider \eqref{disji} for $j=1$. If $k=-S$, disjointedness of the sets $I_1(y,-S)$ follows by definition of $I_1$ and $Y_k$. If $k>-S$, assume $x$ is in $I_1(y_m,k)$ for $m=1,2$. Then, for $m=1,2$, there is $z_m\in Y_{k-1}\cap B(y_m,D^k)$ with $x\in I_3(z_m,k-1)$. Using \eqref{disji} inductively for $j=3$, we conclude $z_1=z_2$. This implies that the balls $B(y_1, D^k)$ and $B(y_2, D^k)$ intersect. By construction of $Y_k$, this implies $y_1=y_2$. This proves \eqref{disji} for $j=1$.
+
+We next consider \eqref{disji} for $j=3$. Assume $x$ is in $I_3(y_m,k)$ for $m=1,2$ and $y_m\in Y_k$. If $x$ is in $X_k$, then by definition \eqref{definei3}, $x\in I_1(y_m,k)$ for $m=1,2$. As we have already shown \eqref{disji} for $j=1$, we conclude $y_1=y_2$. This completes the proof in case $x\in X_k$, and we may assume $x$ is not in $X_k$. By definition \eqref{definei3}, $x$ is not in $I_3(z,k)$ for any $z$ with $z<y_1$ or $z<y_2$. Hence, neither $y_1<y_2$ nor $y_2<y_1$, and by totality of the order of $Y_k$, we have $y_1=y_2$. This completes the proof of \eqref{disji} for $j=3$.
+
+We show \eqref{unioni} for $j=2$. In case $k=-S$, this follows from Lemma \ref{cover big ball}. Assume $k>-S$. Let $x$ be a point of $B(o, 4D^S-2D^k)$. By induction, there is $y'\in Y_{k-1}$ such that $x\in I_3(y',k-1)$. Using the inductive statement \eqref{squeezedyadic}, we obtain $x\in B(y',4D^{k-1})$. As $D>4$, by applying the triangle inequality with the points, $o$, $x$, and $y'$, we obtain that $y'\in B(o, 4D^S-D^k)$. By Lemma \ref{cover big ball}, $y'$ is in $B(y,2D^k)$ for some $y\in Y_k$. It follows that $x\in I_2(y,k)$. This proves \eqref{unioni} for $j=2$.
+
+We show \eqref{unioni} for $j=3$. Let $x\in B(o, 4D^S-2D^k)$. In case $x\in X_k$, then by definition of $X_k$ we have $x\in I_1(y,k)$ for some $y\in Y_k$ and thus $x\in I_3(y,k)$. We may thus assume $x\not\in X_k$. As we have already seen \eqref{unioni} for $j=2$, there is $y\in Y_k$ such that $x\in I_2(y,k)$. We may assume this $y$ is minimal with respect to the order in $Y_k$. Then $x\in I_3(y,k)$. This proves \eqref{unioni} for $j=3$.
+
+Next, we show the first inclusion in \eqref{squeezedyadic}. Let $x\in B(y,\frac 12 D^k)$. As $I_1(y,k)\subset I_3(y,k)$, it suffices to show $x\in I_1(y,k)$. If $k=-S$, this follows immediately from the assumption on $x$ and the definition of $I_1$. Assume $k>-S$. By the inductive statement \eqref{unioni} and $D>4$, there is a $y'\in Y_{k-1}$ such that $x\in I_3(y',k-1)$. By the inductive statement \eqref{squeezedyadic}, we conclude $x\in B(y',4D^{k-1})$. By the triangle inequality with points $x$, $y$, $y'$, and $D>4$, we have $y'\in B(y,D^k)$. It follows by definition \eqref{defineij} that $I_3(y',k-1)\subset I_1(y,k)$, and thus $x\in I_3(y,k)$. This proves the first inclusion in \eqref{squeezedyadic}.
+
+We show the second inclusion in \eqref{squeezedyadic}. Let $x\in I_3(y,k)$. As $I_1(y,k)\subset I_2(y,k)$ directly from the definition \eqref{defineij}, it follows by definition \eqref{definei3} that $x\in I_2(y,k)$. By definition \eqref{defineij}, there is $y'\in Y_{k-1}\cap B(y,2D^k)$ with $x\in I_3(y',k-1)$. By induction, $x\in B(y', 4D^{k-1})$. By the triangle inequality applied to the points $x,y',y$ and $D>4$, we conclude $x\in B(y,4D^k)$. This shows the second inclusion in \eqref{squeezedyadic} and completes the proof of the lemma.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[cover by cubes]
 \label{cover by cubes}
 \uses{dyadic property}
 Let  $-S\le l\le k\le S$ and
@@ -1421,12 +1262,10 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 We conclude \eqref{3coverdyadic} by induction.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[dyadic property]
 \label{dyadic property}
 \uses{transitive boundary}
-Let  $-S\le l\le k\le S$ and
-$y\in Y_k$ and  $y'\in Y_l$.
-with
+Let  $-S\le l\le k\le S$ and $y\in Y_k$ and  $y'\in Y_l$ with
 $I_3(y',l)\cap I_3(y,k)\neq \emptyset$.  Then
 \begin{equation}
     \label{3dyadicproperty}
@@ -1437,43 +1276,15 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 \end{lemma}
 
 \begin{proof}
-Let $l,k,y,y'$ as in the Lemma.
-Pick $x\in I_3(y',l)\cap I_3(y,k)$.
-Assume first $l=k$. By \eqref{disji} of Lemma
-\ref{basic grid structure}, we conclude $y'=y$
-and thus \eqref{3dyadicproperty}.
-Now assume $l<k$. By induction, we may assume
- that the statement of the lemma is proven for $k-1$ in place of $k$.
-
-By Lemma \ref{cover by cubes}, there is
-a $y''\in Y_{k-1}$ such that $x\in I_3(y'',k-1)$.
-By induction, we have $I_3(y',l)\subset I_3(y'',k-1)$. If $l=k-1$, then by the disjointedness property
-of Lemma \ref{basic grid structure} we see that
-that $y'=y''$ and hence \eqref{3dyadicproperty} follows. For $l<k-1$, it remains to prove
+Let $l,k,y,y'$ be as in the lemma. Pick $x\in I_3(y',l)\cap I_3(y,k)$. Assume first $l=k$. By \eqref{disji} of Lemma \ref{basic grid structure}, we conclude $y'=y$, and thus \eqref{3dyadicproperty}. Now assume $l<k$. By induction, we may assume that the statement of the lemma is proven for $k-1$ in place of $k$.
+
+By Lemma \ref{cover by cubes}, there is a $y''\in Y_{k-1}$ such that $x\in I_3(y'',k-1)$. By induction, we have $I_3(y',l)\subset I_3(y'',k-1)$. If $l=k-1$, then by the disjointedness property of Lemma \ref{basic grid structure}, we see that $y'=y''$, and hence \eqref{3dyadicproperty} follows. For $l<k-1$, it remains to prove
 \begin{equation}\label{wyclaim}
 I_3(y'',k-1)\subset I_3(y,k).
 \end{equation}
+We make a case distinction and assume first $x\in X_k$. By Definition \eqref{definei3}, we have $x\in I_1(y,k)$. By Definition \eqref{defineij}, there is a $v\in Y_{k-1}\cap B(y,D^k)$ with $x\in I_3(v,k-1)$. By \eqref{disji} of Lemma \ref{basic grid structure}, we have $v=y''$. By Definition \eqref{defineij}, we then have $I_3(y'',k-1)\subset I_1(y,k)$. Then \eqref{wyclaim} follows by Definition \eqref{definei3} in the given case.
 
-We make a case distinction and assume first $x\in X_k$.
-By Definition \eqref{definei3}, we have
- $x\in I_1(y,k)$. By Definition \eqref{defineij}, there is a $v\in Y_{k-1}\cap B(y,D^k)$ with $x\in I_3(v,k-1)$.
-By \eqref{disji} of Lemma \ref{basic grid structure}, we have $v=y''$.
-By Definition \eqref{defineij}, we then have
-$I_3(y'',k-1)\subset I_1(y,k)$.
-Then \eqref{wyclaim} follows by Definition \eqref{definei3} in the given case.
-
-Assume now the case $x\notin X_k$.
-By \eqref{definei3}, we have
- $x\in I_2(y,k)$. Moreover, for any $u<y$ in
- $Y_k$ we have $x\not\in I_3(u,k)$.
- Let $u<y$. By transitivity of the order in $Y_k$, we conclude $x\not \in I_2(u,k)$.
-By \eqref{defineij} and the disjointness property of Lemma \ref{basic grid structure}, we have
-$I_3(y'',k-1)\cap  I_2(u,k)= \emptyset$.
-Similarly, $I_3(y'',k-1)\cap  I_1(u,k)= \emptyset$.
-Hence $I_3(y'',k-1)\cap  I_3(u,k)=\emptyset$.
-As $u<y$ was arbitrary, we conclude with
-\eqref{definei3} the claim in the given case.
-This completes the proof of \eqref{wyclaim}, and thus also \eqref{3dyadicproperty}.
+Assume now the case $x\notin X_k$. By \eqref{definei3}, we have $x\in I_2(y,k)$. Moreover, for any $u<y$ in $Y_k$, we have $x\not\in I_3(u,k)$. Let $u<y$. By transitivity of the order in $Y_k$, we conclude $x\not \in I_2(u,k)$. By \eqref{defineij} and the disjointedness property of Lemma \ref{basic grid structure}, we have $I_3(y'',k-1)\cap  I_2(u,k)= \emptyset$. Similarly, $I_3(y'',k-1)\cap  I_1(u,k)= \emptyset$. Hence $I_3(y'',k-1)\cap  I_3(u,k)=\emptyset$. As $u<y$ was arbitrary, we conclude with \eqref{definei3} the claim in the given case. This completes the proof of \eqref{wyclaim}, and thus also \eqref{3dyadicproperty}.
 \end{proof}
 
 
@@ -1486,7 +1297,8 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 \end{equation}
 
 
-\begin{lemma}\label{transitive boundary}
+\begin{lemma}[transitive boundary]
+\label{transitive boundary}
 \uses{small boundary}
 Assume $-S\le k''< k'< k\le S$ and
 $y''\in Y_{k''}$, $y'\in Y_{k'}$, $y\in Y_k$.
@@ -1499,17 +1311,16 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 \end{lemma}
 
 \begin{proof}
-As $x\in I_3(y'',k'')\cap I_3(y',k')$ and $k''< k'$.
-we have by Lemma \ref{dyadic property} that
+As $x\in I_3(y'',k'')\cap I_3(y',k')$ and $k''< k'$, we have by Lemma \ref{dyadic property} that
 $I_3(y'',k'')\subset I_3(y',k')$. Similarly,
 $I_3(y',k')\subset I_3(y,k)$.
 Pick $x'\in X\setminus I_3(y,k)$ such that
 \begin{equation}\label{yppxp}
     \rho(y'',x')< 6D^{k''}\, ,
 \end{equation}
-which exists as $(y'',k''|y,k)$. As $x'$ is also in
-$x'\in X\setminus I_3(y',k')$, we conclude
-$(y'',k''|y',k')$.
+which exists as $(y'',k''|y,k)$. As
+$x'\in X\setminus I_3(y',k')$ as well, we conclude
+$(y'',k''| y',k')$.
 By the triangle inequality, we have
 \begin{equation}
 \rho(y',x')\le  \rho(y',x)+\rho(x,y'')+\rho(y'',x')
@@ -1519,12 +1330,13 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 \begin{equation}
 <  4D^{k'}+4D^{k''}+6D^{k''}\le 6D^{k'}\, ,
 \end{equation}
-where we have used $D<4$ and $k''<k'$.
+where we have used $D>5$ and $k''<k'$.
 We conclude $(y',k'|y,k)$.
 \end{proof}
 
 
-\begin{lemma}\label{small boundary}
+\begin{lemma}[small boundary]
+\label{small boundary}
 \uses{smaller boundary}
  Let $K = 2^{4a+1}$. For each $-S+K\le k\le S$ and $y\in Y_k$ we have
     \begin{equation}
@@ -1538,8 +1350,8 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 
 Pick $k'$ so that $k-K\le k'\le k$.
 For each $y''\in Y_{k-K}$ with $(y'',k-K| y,k)$,
-By Lemma \ref{cover by cubes} and Lemma \ref{dyadic property}, there is a unique $y'\in Y_{k'}$ such that
-\begin{equation}
+by Lemma \ref{cover by cubes} and Lemma \ref{dyadic property}, there is a unique $y'\in Y_{k'}$ such that
+\begin{equation}\label{4.31}
     I_3(y'',k-K)\subset I_3(y',k')\subset I_3(y,k)\, .
 \end{equation}
 Using Lemma \ref{transitive boundary}, $(y',k'|y,k)$.
@@ -1547,14 +1359,14 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 We conclude using the disjointedness property of
 Lemma \ref{basic grid structure} that
 \begin{equation}\label{scalecompare}
-    \sum_{y''\in Y_{k-K}: (y'',k-K|y,k)}\mu(I_3(y'',k-K))
-\end{equation}
-\begin{equation}
-    \le
-\sum_{y': (y',k'|y,k)}\left(
- \sum_{y'': (y'',k-K|y',k')}\mu(I_3(z,k-K))\right)
-\end{equation}
-   \begin{equation}
+    \sum_{y'': (y'',k-K|y,k)}\mu(I_3(y'',k-K))
+%\end{equation}
+%\begin{equation}
+%    \le
+%\sum_{y': (y',k'|y,k)}\left(
+% \sum_{y'': (y'',k-K|y',k')}\mu(I_3(z,k-K))\right)
+%\end{equation}
+%   \begin{equation}
    \le
 \sum_{y': (y',k'|y,k)}
 \mu(I_3(y',k'))    \, .
@@ -1575,13 +1387,13 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 Each ball $B(y', \frac 14 D^{k'})$ occurring in
 \eqref{sumcompare1} is contained in $I_3(y',k')$
 by \eqref{squeezedyadic} and in turn contained in
-$I_3(y,k)$. Assume for the moment all these balls are pairwise disjoint. Then
+$I_3(y,k)$ by \eqref{4.31}. Assume for the moment all these balls are pairwise disjoint. Then
 by additivity of the measure,
 \begin{equation}
     K\sum_{y'': (y'',k-K|y,k)}
     \mu(I_3(y'',k-K))
     \le 2^{4a}
-\mu(I_3(y,k))\, .
+\mu(I_3(y,k))\,
 \end{equation}
 which by $K=2^{4a+1}$ implies \eqref{new small boundary}.
 
@@ -1593,21 +1405,20 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 $ B(u, \frac 14 D^l)$ and $B({u'}, \frac 14 D^{l'})$
 be the corresponding balls. If
 $l=l'$, then the balls are equal or disjoint by
-\eqref{squeezedyadic} and \eqref{disji} of Lemma \ref{basic grid structure}. Assume then without loss of generality that $l'<l$. Assume to get a contradiction there is $x''\in  B(u, \frac 14 D^l)\cap B({u'}, \frac 14 D^{l'})$.
-By the triangle inequality and $D>32$,
+\eqref{squeezedyadic} and \eqref{disji} of Lemma \ref{basic grid structure}. Assume then without loss of generality that $l'<l$. Towards a contradiction, assume that
 \begin{equation}\label{bulbul}
-    B(u', \frac 13 D^{l'})\subset B(u, \frac 38 D^l).
+    B(u, \frac 14 D^l)\cap B({u'}, \frac 14 D^{l'})\neq \emptyset
 \end{equation}
 As $(u',l'|y,k)$, there is a point  $x$ in
-$X\setminus I_3(y,k)$ with $\rho(x,u')<4D^{k'}$.
-Using $D>32$, we conclude from \eqref{bulbul} that
+$X\setminus I_3(y,k)$ with $\rho(x,u')<6D^{l'}$.
+Using $D>25$, we conclude from the triangle inequality and \eqref{bulbul} that
 $x\in B(u,\frac 12D^l)$. However, $B(u,\frac 12 D^l)\subset I_3(u,l)$,
 and $I_3(u,l)\subset I_3(y,k)$, a contradiction to
 $x\not\in I_3(y,k)$.
 This proves the lemma.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[smaller boundary]
 \label{smaller boundary}
 \uses{boundary measure}
 Let $K = 2^{4a+1}$
@@ -1621,9 +1432,9 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 \begin{proof}
     We prove this by induction on $n$. If $n=0$,
     both sides of \eqref{very new small} are equal to
-    $\mu(I_3(y,k)$. If $n=1$, this follows from Lemma \ref{small boundary}.
+    $\mu(I_3(y,k))$ by \eqref{disji}. If $n=1$, this follows from Lemma \ref{small boundary}.
 
-    Assume $n>1$ and \eqref{very new small} has been proven with  $n-1$ in place of $n$.
+    Assume $n>1$ and \eqref{very new small} has been proven for  $n-1$.
 We write  \eqref{very new small}
      \begin{equation}
              \sum_{y''\in Y_{k-nK}: (y'',k-nK|y,k)}\mu(I_3(y'',k-nK))
@@ -1636,10 +1447,11 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 =  \sum_{y'\in Y_{k-K}:(y',k-K|y,k)}   2^{1-n}\mu(I_3(y',k-K))
     \end{equation}
 Applying \eqref{new small boundary} gives
-\eqref{very new small} and proves the lemma.
+\eqref{very new small}, and proves the lemma.
 \end{proof}
 
-\begin{lemma}\label{boundary measure}
+\begin{lemma}[boundary measure]
+\label{boundary measure}
  For each $-S\le k\le S$ and $y\in Y_k$ and $0<t<1$
  with $tD^k\ge D^{-S}$ we have
     \begin{equation}
@@ -1663,7 +1475,7 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 $$
     \le 4D^{k'} + D^{k - nK} < 6D^{k'}\,.
 $$
-Thus $(y',k'|y,k)$, and we have shown that
+Together with \eqref{3dyadicproperty} thus $(y',k'|y,k)$. We have shown that
 $$
     \{x \in I_3(y,k) \ : \ \rho(x, X \setminus I_3(y,k)) \leq t D^{k}\}
 $$
@@ -1700,65 +1512,63 @@ \section{Proof of L.\ref{grid existence}, dyadic structure}
 \eqref{coverball}
 for $(\mathcal{D},c,s)$ and $x$.
 
-We first show \eqref{coverdyadic}.
-Assume to get a contradiction that \eqref{coverdyadic}
-is false. Then there is an $I$ violating the conclusion of
+We first show \eqref{coverdyadic} by contradiction. Then there is an $I$ violating the conclusion of
 \eqref{coverdyadic}. Pick such $I=I_3(y,l)$ such that $l$ is minimal.
 {By assumption, we have $-S\le k<l$; in particular $-S<l$}.
 % By definition, $I$ is contained in $\tilde{B}(x,k)$, which
 % is contained in the union of $I(y,l-1)$ with $y\in X_{l-1}$.
 By definition, $I_3(y, l)$ is contained in $I_1(y, l)\cup I_2(y, l)$, which is contained in the union of $I_3(y',l-1)$ with $y'\in Y_{l-1}$.
-By minimality if $l$, each such $I_3(y',l-1)$ is contained in the union of
+By minimality of $l$, each such $I_3(y',l-1)$ is contained in the union of
 all $I_3(z,k)$ with $z\in Y_k$. This proves \eqref{coverdyadic}.
 
 We now show \eqref{dyadicproperty}. Assume to get a contradiction that
 there are non-disjoint $I, J\in \mathcal{D}$ with $s(I)\le s(J)$
 and $I \not \subset J$. We may assume the existence of such $I$ and $J$ with minimal
 $s(J)-s(I)$. Let $k=s(I)$. Assume first $s(J)=k$. Let $I=I_3(y_1,k)$ and $J=I_3(y_2,k)$ with $y_1,y_2\in Y_k$.
-{If $y_1=y_2$, then $I=J$, a contradiction to $I\subset J$.
-If $y_1<y_2$, then $I\cap J=\emptyset$ by definition \eqref{definei3} of $J=I_3(y_2, k)$ and the disjointedness of $I_1(y_1, k)$ and $I_1(y_2, k)$ by Lemma \ref{basic grid structure}, a contradiction.
-Similarly, if $y_2<y_1$, then $I\cap J=\emptyset$, a contradiction again.
+{If $y_1=y_2$, then $I=J$, a contradiction to $I\not \subset J$.
+If $y_1\neq y_2$, then $I\cap J=\emptyset$ by \eqref{disji}, a contradiction to the non-disjointedness of $I,J$.% of $J=I_3(y_2, k)$ and the disjointedness of $I_1(y_1, k)$ and $I_1(y_2, k)$ by Lemma \ref{basic grid structure}, a contradiction.
+%Similarly, if $y_2<y_1$, then $I\cap J=\emptyset$, a contradiction again.
 Assume now $s(J)>k$. Choose $y\in I\cap J$. By property \eqref{coverdyadic},
 there is $K\in \mathcal{D}$ with $s(K)=s(J)-1$ and $y\in K$. By construction
-of $J$, and pairwise disjointness of all $I(w,s(J)-1)$ that we have already seen,
+of $J$, and pairwise disjointedness of all $I_3(w,s(J)-1)$ that we have already seen,
 we have $K\subset J$. By minimality of $s(J)$, we have $I\subset K$.
 This proves $I\subset J$ and thus \eqref{dyadicproperty}.
 
 We next establish \eqref{coverball}. Let $-S\leq k\leq S$. Using \eqref{unioni} for $j=3$, we get \begin{equation}
 x\in B(o, D^S)\subset B(o, 4D^S-2D^k)\subset \bigcup_{y\in Y_k} I_3(y,k)\, .
-\end{equation} Thus, there exists a dyadic cube $I=I_3(y', k)$ with $s(I)=k$ and $x\in I$. This proves \eqref{coverball}.
+\end{equation} Thus, there exists a dyadic cube $I=I_3(y, k)$ with $s(I)=k$ and $x\in I$. This proves \eqref{coverball}.
 
 
-\section{Proof of L.\ref{tile structure}, tile structure}
+\section{Proof of Lemma \ref{tile structure}, tile structure}
 \label{subsectiles}
 Choose a grid structure $(\mathcal{D}, c, s)$.
 Let $I \in \mathcal{D}$. Suppose that
 \begin{equation}
     \label{eq tile Z}
-    Z \subset \bigcup_{\mfa \in \tQ(X)} B_{I^\circ}(\mfa, 1)
+    \mathcal{Z} \subset \bigcup_{\mfa \in \tQ(X)} B_{I^\circ}(\mfa, 1)
 \end{equation}
-is such that for any $\mfa, \mfb \in Z$ we have
+is such that for any $\mfa, \mfb \in \mathcal{Z}$ we have
 \begin{equation}
     \label{eq tile disjoint Z}
     B_{I^\circ}(\mfa, 0.3) \cap B_{I^\circ}(\mfb, 0.3) = \emptyset\,.
 \end{equation}
-By Lemma \ref{lem entropy Theta} applied to each of the balls $B_{I^\circ}(\mfa, 1)$, $\mfa \in \tQ(X)$, we have
+By Lemma \ref{ball metric entropy} applied to each of the balls $B_{I^\circ}(\mfa, 1)$, $\mfa \in \tQ(X)$, we have
 $$
-    |Z| \le 2^{2a} |\tQ(X)|\,.
+    |\mathcal{Z}| \le 2^{2a} |\tQ(X)|\,.
 $$
-In particular, there exists a set $Z$ satisfying both \eqref{eq tile Z} and \eqref{eq tile disjoint Z} of maximal cardinality among all such sets. We pick for each $I \in \mathcal{D}$ such a set $Z(I)$.
+In particular, there exists a set $\mathcal{Z}$ satisfying both \eqref{eq tile Z} and \eqref{eq tile disjoint Z} of maximal cardinality among all such sets. We pick for each $I \in \mathcal{D}$ such a set $\mathcal{Z}(I)$.
 
-\begin{lemma}
+\begin{lemma}[frequency ball cover]
 \label{frequency ball cover}
     For each $I \in \mathcal{D}$, we have
     \begin{equation}
         \label{eq tile cover}
-        \tQ(X) \subset  \bigcup_{\mfa \in \tQ(X)} B_{I^\circ}(\mfa, 1) \subset \bigcup_{z \in Z(I)} B_{I^\circ}(z, 0.7)\,.
+        \tQ(X) \subset  \bigcup_{\mfa \in \tQ(X)} B_{I^\circ}(\mfa, 1) \subset \bigcup_{z \in \mathcal{Z}(I)} B_{I^\circ}(z, 0.7)\,.
     \end{equation}
 \end{lemma}
 
 \begin{proof}
-    To show \eqref{eq tile cover} note that the first inclusion is obvious. For the second inclusion let $\mfb \in  \bigcup_{\mfa \in \tQ(X)} B_{I^\circ}(\mfa, 1)$. By maximality of $Z(I)$, there must be a point $z \in Z(I)$ such that $B_{I^\circ}(z, 0.3) \cap B_{I^\circ}(\mfb, 0.3) \ne \emptyset$. Else, $Z(I) \cup \{\mfb\}$ would be a set of larger cardinality than $Z(I)$ satisfying \eqref{eq tile Z} and \eqref{eq tile disjoint Z}. Fix such $z$, and fix a point $z_1 \in B_{I^\circ}(z, 0.3) \cap B_{I^\circ}(\mfb, 0.3)$. By the triangle inequality, we deduce that
+    To show \eqref{eq tile cover} note that the first inclusion is obvious. For the second inclusion let $\mfb \in  \bigcup_{\mfa \in \tQ(X)} B_{I^\circ}(\mfa, 1)$. By maximality of $\mathcal{Z}(I)$, there must be a point $z \in \mathcal{Z}(I)$ such that $B_{I^\circ}(z, 0.3) \cap B_{I^\circ}(\mfb, 0.3) \ne \emptyset$. Else, $\mathcal{Z}(I) \cup \{\mfb\}$ would be a set of larger cardinality than $\mathcal{Z}(I)$ satisfying \eqref{eq tile Z} and \eqref{eq tile disjoint Z}. Fix such $z$, and fix a point $z_1 \in B_{I^\circ}(z, 0.3) \cap B_{I^\circ}(\mfb, 0.3)$. By the triangle inequality, we deduce that
     $$
         d_{I^\circ}(z,\mfb) \le d_{I^\circ}(z,z_1) + d_{I^\circ}(\mfb, z_1) < 0.3 + 0.3 = 0.6\,,
     $$
@@ -1767,25 +1577,25 @@ \section{Proof of L.\ref{tile structure}, tile structure}
 
 We define
 $$
-    \fP = \{(I, z) \ : \ I \in \mathcal{D}, z \in Z(I)\}\,,
+    \fP = \{(I, z) \ : \ I \in \mathcal{D}, z \in \mathcal{Z}(I)\}\,,
 $$
 $$\sc((I, z)) = I\qquad \text{and} \qquad \fcc((I, z)) = z.$$ We further set $$\ps(\fp) = s(\sc(\fp)),\qquad \qquad \pc(\fp) = c(\sc(\fp)).$$ Then \eqref{tilecenter}, \eqref{tilescale} hold by definition.
 
 It remains to construct the map $\Omega$, and verify properties \eqref{eq dis freq cover}, \eqref{eq freq dyadic} and
-\eqref{eq freq comp ball}. We first construct an auxiliary map $\Omega_1$. For each $I \in \mathcal{D}$, we pick an enumeration of the finite set $Z(I)$
+\eqref{eq freq comp ball}. We first construct an auxiliary map $\Omega_1$. For each $I \in \mathcal{D}$, we pick an enumeration of the finite set $\mathcal{Z}(I)$
 $$
-    Z(I) = \{z_1, \dotsc, z_M\}\,.
+    \mathcal{Z}(I) = \{z_1, \dotsc, z_M\}\,.
 $$
 We define {$\Omega_1:\fP \mapsto \mathcal{P}(\Mf) $ as below}. Set
 $$
-    \Omega_1((I, z_1)) = B_{I^\circ}(z_1, 0.7) \setminus \bigcup_{z \in Z(I)\setminus \{z_1\}} B_{I^\circ}(z, 0.3)
+    \Omega_1((I, z_1)) = B_{I^\circ}(z_1, 0.7) \setminus \bigcup_{z \in \mathcal{Z}(I)\setminus \{z_1\}} B_{I^\circ}(z, 0.3)
 $$
 and then define iteratively
 \begin{equation}
     \label{eq def omega1}
-    \Omega_1((I, z_k)) = B_{I^\circ}(z_k, 0.7) \setminus \bigcup_{z \in Z(I) \setminus \{z_k\}} B_{I^\circ}(z, 0.3) \setminus \bigcup_{i=1}^{k-1} \Omega_1((I, z_i))\,.
+    \Omega_1((I, z_k)) = B_{I^\circ}(z_k, 0.7) \setminus \bigcup_{z \in \mathcal{Z}(I) \setminus \{z_k\}} B_{I^\circ}(z, 0.3) \setminus \bigcup_{i=1}^{k-1} \Omega_1((I, z_i))\,.
 \end{equation}
-\begin{lemma}
+\begin{lemma}[disjoint frequency cubes]
     \label{disjoint frequency cubes}
     \uses{dyadic frequency cubes}
     For each $I \in \mathcal{D}$, and $\fp_1, \fp_2\in \fP(I)$, if $$\Omega_1(\fp_1)\cap \Omega_1(\fp_2)\neq \emptyset,$$ then $\fp_1=\fp_2$.
@@ -1794,18 +1604,18 @@ \section{Proof of L.\ref{tile structure}, tile structure}
 \begin{proof}
     By the definition of the map $\sc$, we have
     $$
-        \fP(I) = \{(I, z) \, : \, z \in Z(I)\}\,.
+        \fP(I) = \{(I, z) \, : \, z \in \mathcal{Z}(I)\}\,.
     $$
     By \eqref{eq def omega1}, the set $\Omega_1((I, z_k))$ is disjoint from each $\Omega_1((I, z_i))$ with $i < k$. Thus the sets $\Omega_1(\fp)$, $\fp \in \fP(I)$ are pairwise disjoint.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[frequency cube cover]
 \label{frequency cube cover}
 \uses{dyadic frequency cubes}
     For each $I \in \mathcal{D}$, it holds that
     \begin{equation}
     \label{eq omega1 cover}
-         \bigcup_{z \in Z(I)} B_{I^\circ}(z, 0.7)\subset \bigcup_{\fp \in \fP(I)} \Omega_1(\fp)\,.
+         \bigcup_{z \in \mathcal{Z}(I)} B_{I^\circ}(z, 0.7)\subset \bigcup_{\fp \in \fP(I)} \Omega_1(\fp)\,.
     \end{equation}
     For every $\fp \in \fP$, it holds that
     \begin{equation}
@@ -1819,16 +1629,16 @@ \section{Proof of L.\ref{tile structure}, tile structure}
     The second inclusion in \eqref{eq omega1 incl} then follows from \eqref{eq def omega1} and the equality $B_{\fp}(\fcc(\fp), 0.7) = B_{I^\circ}(z, 0.7)$, which is true by definition.
     For the first inclusion in \eqref{eq omega1 incl} let $\mfa \in B_{\fp}(\fcc(\fp),0.3)$. Let $k$ be such that $z = z_k$ in the enumeration we chose above. It follows immediately from \eqref{eq def omega1} and \eqref{eq tile disjoint Z} that
     $\mfa \notin \Omega_1((I, z_i))$ for all $i < k$. Thus, again from \eqref{eq def omega1}, we have
-    $q \in \Omega_1((I,z_k))$.
+    $\mfa \in \Omega_1((I,z_k))$.
 
-    To show \eqref{eq omega1 cover} let $\mfa \in \bigcup_{z \in Z(I)} B_{I^\circ}(z,0.7)$.
-    If there exists $z \in Z_1(I)$ with $q \in B_{I^\circ}(z,0.3)$, then
+    To show \eqref{eq omega1 cover} let $\mfa \in \bigcup_{z \in \mathcal{Z}(I)} B_{I^\circ}(z,0.7)$.
+    If there exists $z \in \mathcal{Z}(I)$ with $\mfa \in B_{I^\circ}(z,0.3)$, then
     $$
         z \in \Omega_1((I, z)) \subset \bigcup_{\fp \in \fP(I)} \Omega_1(\fp)
     $$
     by the first inclusion in \eqref{eq omega1 incl}.
 
-    Now suppose that there exists no $z \in Z(I)$ with $\mfa \in B_{I^\circ}(z, 0.3)$. Let $k$ be minimal such that $\mfa \in B_{I^\circ}(z_k, 0.7)$. Since $\Omega_1((I, z_i)) \subset B_{I^\circ}(z_i, 0.7)$ for each $i$ by \eqref{eq def omega1}, we have that $\mfa \notin \Omega_1((I, z_i))$ for all $i < k$. Hence $\mfa \in \Omega_1((I, z_k))$, again by \eqref{eq def omega1}.
+    Now suppose that there exists no $z \in \mathcal{Z}(I)$ with $\mfa \in B_{I^\circ}(z, 0.3)$. Let $k$ be minimal such that $\mfa \in B_{I^\circ}(z_k, 0.7)$. Since $\Omega_1((I, z_i)) \subset B_{I^\circ}(z_i, 0.7)$ for each $i$ by \eqref{eq def omega1}, we have that $\mfa \notin \Omega_1((I, z_i))$ for all $i < k$. Hence $\mfa \in \Omega_1((I, z_k))$, again by \eqref{eq def omega1}.
 \end{proof}
 
 Now we are ready to define the function $\Omega$. For all cubes $I \in \mathcal{D}$ such that there exists no $J \in \mathcal{D}$ with $I \subset J$ and $I \ne J$, we define for all $\fp \in \fP(I)$
@@ -1839,23 +1649,23 @@ \section{Proof of L.\ref{tile structure}, tile structure}
 For cubes $I \in \mathcal{D}$ for which there exists $J \in \mathcal{D}$ with $I \subset J$ and $I \ne J$, we define $\Omega$ by recursion. We can pick an inclusion minimal $J \in \mathcal{D}$ among the finitely many cubes such that $I \subset J$ and $I \ne J$. This $J$ is unique: Suppose that $J'$ is another inclusion minimal cube with $I \subset J'$ and $I \ne J'$. Without loss of generality, we have that $s(J) \le s(J')$. By \eqref{dyadicproperty}, it follows that $J \subset J'$. Since $J'$ is minimal with respect to inclusion, it follows that $J = J'$.  Then we define
 \begin{equation}
     \label{eq it omega}
-    \fc(\fp) = \bigcup_{z \in Z(J) \cap \Omega_1(\fp)} \Omega((J, z)) \cup B_{\fp}(\fcc(\fp),0.2)\,.
+    \fc(\fp) = \bigcup_{z \in \mathcal{Z}(J) \cap \Omega_1(\fp)} \Omega((J, z)) \cup B_{\fp}(\fcc(\fp),0.2)\,.
 \end{equation}
 
-\begin{lemma}
+\begin{lemma}[dyadic frequency cubes]
 \label{dyadic frequency cubes}
 
     With this definition, \eqref{eq dis freq cover}, \eqref{eq freq dyadic} and \eqref{eq freq comp ball} hold.
 \end{lemma}
 
 \begin{proof}
-    First, we prove \eqref{eq freq comp ball}. If $I \in \mathcal{D}$ is maximal in $\mathcal{D}$ with respect to set inclusion, then \eqref{eq freq comp ball} holds for all $\fp \in \fP(I)$ by \eqref{eq max omega} and\eqref{eq omega1 incl}. Now suppose that $I$ is not maximal in $\mathcal{D}$ with respect to set inclusion. Then we may assume by induction that for all $J \in \mathcal{D}$ with $I \subset J$ and all $\fp' \in \fP(J)$, \eqref{eq freq comp ball} holds. Let $J$ be the unique minimal cube in $\mathcal{D}$ with $I \subsetneq J$.
+    First, we prove \eqref{eq freq comp ball}. If $I \in \mathcal{D}$ is maximal in $\mathcal{D}$ with respect to set inclusion, then \eqref{eq freq comp ball} holds for all $\fp \in \fP(I)$ by \eqref{eq max omega} and \eqref{eq omega1 incl}. Now suppose that $I$ is not maximal in $\mathcal{D}$ with respect to set inclusion. Then we may assume by induction that for all $J \in \mathcal{D}$ with $I \subset J$ and all $\fp' \in \fP(J)$, \eqref{eq freq comp ball} holds. Let $J$ be the unique minimal cube in $\mathcal{D}$ with $I \subsetneq J$.
 
     Suppose that $\mfa \in \Omega(\fp)$. If $\mfa\in B_{\fp}(\mathcal{Q}(\fp), 0.2)$, then since
     \begin{equation*}
         B_{\fp}(\mathcal{Q}(\fp), 0.2)\subset B_{\fp}(\mathcal{Q}(\fp), 1)\, ,
     \end{equation*}
-    we conclude that $\mfa\in B_{\fp}(\mathcal{Q}(\fp), 0.7)$. If not, by \eqref{eq it omega}, there exists $z \in Z(J) \cap \Omega_1(\fp)$ with $\mfa \in \Omega(J,z)$. Using the triangle inequality and \eqref{eq omega1 incl}, we obtain
+    we conclude that $\mfa\in B_{\fp}(\mathcal{Q}(\fp), 0.7)$. If not, by \eqref{eq it omega}, there exists $z \in \mathcal{Z}(J) \cap \Omega_1(\fp)$ with $\mfa \in \Omega(J,z)$. Using the triangle inequality and \eqref{eq omega1 incl}, we obtain
     $$
         d_{I^\circ}(\fcc(\fp),\mfa) \le d_{I^\circ}(\fcc(\fp), z) + d_{I^\circ}(z, \mfa) \le 0.7 + d_{I^\circ}(z, \mfa)\,.
     $$
@@ -1867,34 +1677,34 @@ \section{Proof of L.\ref{tile structure}, tile structure}
 
     Next, we show \eqref{eq dis freq cover}. Let $I \in \mathcal{D}$.
 
-    If $I$ is maximal with respect to inclusion, then disjointness of the sets $\fc(\fp)$ for $\fp \in \fP(I)$ follows from the definition \eqref{eq max omega} and Lemma \ref{disjoint frequency cubes}. To obtain the inclusion in \eqref{eq dis freq cover} one combines the inclusions \eqref{eq tile cover} and \eqref{eq omega1 cover}
+    If $I$ is maximal with respect to inclusion, then disjointedness of the sets $\fc(\fp)$ for $\fp \in \fP(I)$ follows from the definition \eqref{eq max omega} and Lemma \ref{disjoint frequency cubes}. To obtain the inclusion in \eqref{eq dis freq cover} one combines the inclusions \eqref{eq tile cover} and \eqref{eq omega1 cover}
     of Lemma \ref{frequency cube cover} with  \eqref{eq max omega}.
 
-    Now we turn to the case where there exists $J \in \mathcal{D}$ with $I \subset J$ and $I\ne J$. In this case we use induction: It suffices to show \eqref{eq dis freq cover} under the assumption that it holds for all cubes $J \in \mathcal{D}$ with $I \subset J$. As shown before definition \eqref{eq it omega}, we may choose the unique inclusion minimal such $J$. To show disjointness of the sets $\fc(\fp), \fp \in \fP(I)$ we pick two tiles $\fp, \fp' \in \fP(I)$ and $\mfa \in \fc(\fp) \cap \fc(\fp')$.
+    Now we turn to the case where there exists $J \in \mathcal{D}$ with $I \subset J$ and $I\ne J$. In this case we use induction: It suffices to show \eqref{eq dis freq cover} under the assumption that it holds for all cubes $J \in \mathcal{D}$ with $I \subset J$. As shown before definition \eqref{eq it omega}, we may choose the unique inclusion minimal such $J$. To show disjointedness of the sets $\fc(\fp), \fp \in \fP(I)$ we pick two tiles $\fp, \fp' \in \fP(I)$ and $\mfa \in \fc(\fp) \cap \fc(\fp')$.
     Then we are by \eqref{eq it omega} in one of the following four cases.
 
-    1. There exist $z \in Z(J) \cap \Omega_1(\fp)$ such that $\mfa \in \Omega(J, z)$, and there exists $z' \in Z(J) \cap \Omega_1(\fp')$ such that $\mfa \in \Omega(J, z')$. By the induction hypothesis, that \eqref{eq dis freq cover} holds for $J$, we must have $z = z'$. By Lemma \ref{disjoint frequency cubes}, we must then have $\fp = \fp'$.
+    1. There exist $z \in \mathcal{Z}(J) \cap \Omega_1(\fp)$ such that $\mfa \in \Omega(J, z)$, and there exists $z' \in \mathcal{Z}(J) \cap \Omega_1(\fp')$ such that $\mfa \in \Omega(J, z')$. By the induction hypothesis, that \eqref{eq dis freq cover} holds for $J$, we must have $z = z'$. By Lemma \ref{disjoint frequency cubes}, we must then have $\fp = \fp'$.
 
-    2. There exists $z \in Z(J) \cap \Omega_1(\fp)$ such that $\mfa \in \Omega(J,z)$, and $\mfa \in B_{\fp'}(\fcc(\fp'), 0.2)$.  Using the triangle inequality, Lemma \ref{monotone
+    2. There exists $z \in \mathcal{Z}(J) \cap \Omega_1(\fp)$ such that $\mfa \in \Omega(J,z)$, and $\mfa \in B_{\fp'}(\fcc(\fp'), 0.2)$.  Using the triangle inequality, Lemma \ref{monotone
     cube metrics} and \eqref{eq freq comp ball}, we obtain
     $$
         d_{\fp'}(\fcc(\fp'),z) \le d_{\fp'}(\fcc(\fp'), \mfa) + d_{\fp'}(z, \mfa) \le 0.2 + 2^{-95a} \cdot 1 < 0.3\,.
     $$
     Thus $z \in \Omega_1(\fp')$ by \eqref{eq omega1 incl}. By Lemma \ref{disjoint frequency cubes}, it follows that $\fp = \fp'$.
 
-    3. There exists $z' \in Z(J) \cap \Omega_1(\fp')$ such that $\mfa \in \Omega(J,z')$, and $\mfa \in B_{\fp}(\fcc(\fp), 0.2)$. This case is the same as case 2., after swapping $\fp$ and $\fp'$.
+    3. There exists $z' \in \mathcal{Z}(J) \cap \Omega_1(\fp')$ such that $\mfa \in \Omega(J,z')$, and $\mfa \in B_{\fp}(\fcc(\fp), 0.2)$. This case is the same as case 2., after swapping $\fp$ and $\fp'$.
 
-    4. We have $q \in B_{\fp}(\fcc(\fp), 0.2) \cap B_{\fp'}(\fcc(\fp'), 0.2)$. In this case it follows that $\fp = \fp'$ since the sets $B_{\fp}(\fcc(\fp), 0.2)$ are pairwise disjoint by the inclusion \eqref{eq omega1 incl} and Lemma \ref{disjoint frequency cubes}.
+    4. We have $\mfa \in B_{\fp}(\fcc(\fp), 0.2) \cap B_{\fp'}(\fcc(\fp'), 0.2)$. In this case it follows that $\fp = \fp'$ since the sets $B_{\fp}(\fcc(\fp), 0.2)$ are pairwise disjoint by the inclusion \eqref{eq omega1 incl} and Lemma \ref{disjoint frequency cubes}.
 
-    To show the inclusion in \eqref{eq dis freq cover}, let $\mfa \in \tQ(X)$. By the induction hypothesis, there exists $\fp \in \fP(J)$ such that $\mfa \in \Omega(\fp)$. By definition of the set $\fP$, we have $\fp = (J, z)$ for some $z \in Z(J)$. By \eqref{eq tile Z}, there exists $x \in X$ with $d_{J^\circ}(\tQ(x), z) \le 1$. By Lemma \ref{monotone cube metrics}, it follows that $d_{I^\circ}(\tQ(x), z) \le 1$.
-    Thus, by \eqref{eq tile cover}, there exists $z' \in Z(I)$ with $z \in B_{I^\circ}(z', 0.7)$. Then by Lemma \eqref{frequency cube cover} there exists $\fp' \in \fP(I)$ with $z \in Z(J) \cap \Omega_1(\fp')$. Consequently, by \eqref{eq it omega}, $\mfa \in \fc(\fp')$. This completes the proof of \eqref{eq dis freq cover}.
+    To show the inclusion in \eqref{eq dis freq cover}, let $\mfa \in \tQ(X)$. By the induction hypothesis, there exists $\fp \in \fP(J)$ such that $\mfa \in \Omega(\fp)$. By definition of the set $\fP$, we have $\fp = (J, z)$ for some $z \in \mathcal{Z}(J)$. By \eqref{eq tile Z}, there exists $x \in X$ with $d_{J^\circ}(\tQ(x), z) \le 1$. By Lemma \ref{monotone cube metrics}, it follows that $d_{I^\circ}(\tQ(x), z) \le 1$.
+    Thus, by \eqref{eq tile cover}, there exists $z' \in \mathcal{Z}(I)$ with $z \in B_{I^\circ}(z', 0.7)$. Then by Lemma \eqref{frequency cube cover} there exists $\fp' \in \fP(I)$ with $z \in \mathcal{Z}(J) \cap \Omega_1(\fp')$. Consequently, by \eqref{eq it omega}, $\mfa \in \fc(\fp')$. This completes the proof of \eqref{eq dis freq cover}.
 
     Finally, we show \eqref{eq freq dyadic}. Let $\fp, \fq \in \fP$ with $\sc(\fp) \subset \sc(\fp)$ and $\fc(\fp) \cap \fc(\fq) \ne \emptyset$. If we have $\ps(\fp) \ge \ps(\fq)$, then it follows from \eqref{dyadicproperty} that $I = J$, thus $\fp, \fq \in \fP(I)$. By \eqref{eq dis freq cover} we have then either $\fc(\fp) \cap \fc(\fq) = \emptyset$ or $\fc(\fp) = \fc(\fq)$. By the assumption in \eqref{eq freq dyadic} we have $\fc(\fp) \cap \fc(\fq) \ne \emptyset$, so we must have $\fc(\fp) = \fc(\fq)$ and in particular $\fc(\fq) \subset \fc(\fp)$.
 
     So it remains to show \eqref{eq freq dyadic} under the additional assumption that $\ps(\fq) > \ps(\fp)$. In this case, we argue by induction on $\ps(\fq)-\ps(\fp)$. By \eqref{coverdyadic}, there exists a cube $J \in \mathcal{D}$ with $s(J) = \ps(\fq) - 1$ and $J \cap\sc(\fp) \ne \emptyset$. We pick one such $J$. By \eqref{dyadicproperty}, we have $\sc(\fp) \subset J \subset \sc(\fq)$.
 
     By \eqref{eq tile Z}, there exists $x \in X$ with $d_{\fq}(\tQ(x), \fcc(\fq)) \le 1$. By Lemma \ref{monotone cube metrics}, it follows that $d_{J^\circ}(\tQ(x), \fcc(\fq)) \le 1$.
-    Thus, by \eqref{eq tile cover}, there exists $z' \in Z(J)$ with $\fcc(\fq) \in B_{J^\circ}(z', 0.7)$. Then by Lemma \eqref{frequency cube cover} there exists $\fq' \in \fP(J)$ with $\fcc(\fq) \in\Omega_1(\fq')$.
+    Thus, by \eqref{eq tile cover}, there exists $z' \in \mathcal{Z}(J)$ with $\fcc(\fq) \in B_{J^\circ}(z', 0.7)$. Then by Lemma \eqref{frequency cube cover} there exists $\fq' \in \fP(J)$ with $\fcc(\fq) \in\Omega_1(\fq')$.
     By \eqref{eq it omega}, it follows that $\Omega(\fq) \subset \Omega(\fq')$. Note that then $\sc(\fp) \subset \sc(\fq')$ and $\fc(\fp) \cap \fc(\fq') \ne \emptyset$ and $\ps(\fq') - \ps(\fp) = \ps(\fq) - \ps(\fp) - 1$. Thus, we have by the induction hypothesis that $\Omega(\fq') \subset \Omega(\fp)$. This completes the proof.
 \end{proof}
 
@@ -1904,21 +1714,8 @@ \chapter{Proof of discrete Carleson (Prop.  \ref{discrete Carleson}) }
 %\section{Decomposition of grid cubes and first exceptional set}
 
 
-Let a grid structure $(\mathcal{D}, c, s)$  and a tile structure $(\fP,\sc,\fc,\fcc)$
-for this grid structure be given.
-In Subsection \ref{subsectilesorg}, we decompose the
-set $\fP$ of tiles into subsets.
-Each subset will be controlled by one of three methods.
-The guiding principle of the decomposition is
-to be able to apply the forest estimate
-of Proposition \ref{forest operator} to the final subsets
-defined in \eqref{defc5}. This application is done in Subsection \ref{subsecforest}.
-The miscellaneous subsets along the construction of the
-forests will be either thrown into exceptional sets,
-which are defined and controlled in Subsection
-\ref{subsetexcset}, or will be controlled by
-the antichain estimate of Proposition \ref{antichain operator},
-which is done in Subsection \ref{subsecantichain}. Subsection \ref{subsec lessim aux} contains some auxiliary lemmas needed for the proofs in Subsections \ref{subsecforest}-\ref{subsecantichain}.
+Let a grid structure $(\mathcal{D}, c, s)$ and a tile structure $(\fP, \sc, \fc, \fcc)$ for this grid structure be given. In Subsection \ref{subsectilesorg}, we decompose the set $\fP$ of tiles into subsets. Each subset will be controlled by one of three methods. The guiding principle of the decomposition is to be able to apply the forest estimate of Proposition \ref{forest operator} to the final subsets defined in \eqref{defc5}. This application is done in Subsection \ref{subsecforest}. The miscellaneous subsets along the construction of the forests will either be thrown into exceptional sets, which are defined and controlled in Subsection \ref{subsetexcset}, or will be controlled by the antichain estimate of Proposition \ref{antichain operator}, which is done in Subsection \ref{subsecantichain}. Subsection \ref{subsec lessim aux} contains some auxiliary lemmas needed for the proofs in Subsections \ref{subsecforest}-\ref{subsecantichain}.
+
 %The oganisation and summation of these various
 %estimates is done in Subsection \ref{subsecaddingup}.
 
@@ -1927,16 +1724,16 @@ \chapter{Proof of discrete Carleson (Prop.  \ref{discrete Carleson}) }
 
 \section{Organisation of the tiles}\label{subsectilesorg}
 
-In the following chain of definitions, $k, n$, and
+In the following definitions, $k, n$, and
 $j$ will be nonnegative integers.
 Define
 $\mathcal{C}(G,k)$ to be the set of $I\in \mathcal{D}$
 such that there exists a $J\in \mathcal{D}$ with $I\subset J$
 and
 \begin{equation}\label{muhj1}
-   {\mu(G \cap J)} > 2^{-k-1}{\mu(G)}\, ,
+   {\mu(G \cap J)} > 2^{-k-1}{\mu(J)}\, ,
 \end{equation}
-but there does not exists a $J\in \mathcal{D}$ with $I\subset J$ and
+but there does not exist a $J\in \mathcal{D}$ with $I\subset J$ and
 \begin{equation}\label{muhj2}
    {\mu(G \cap J)} > 2^{-k}{\mu(J)}\,.
 \end{equation}
@@ -1949,7 +1746,7 @@ \section{Organisation of the tiles}\label{subsectilesorg}
  \begin{equation}\label{ebardense}
     \mu({E_1}(\fp))  > 2^{-n}  \mu(\sc(\fp))
  \end{equation}
-and there does not exists $\fp'\in \fP(k)$ with
+and there does not exist $\fp'\in \fP(k)$ with
 $\fp'\neq \fp$ and  $\fp\le \fp'$  such that
  \begin{equation}\label{mnkmax}
     \mu({E_1}(\fp'))  > 2^{-n}  \mu(\sc(\fp')).
@@ -2017,7 +1814,7 @@ \section{Organisation of the tiles}\label{subsectilesorg}
 \fL_2(k,n,j)
 \end{equation}
 to be the set of all $\fp\in \fC_2(k,n,j)$ such that there
-does not exists
+does not exist
 $\fu\in \fU_1(k,n,j)$
 with $\sc(\fp)\neq \sc(\fu)$ and $2\fp\lesssim \fu$.
 Define
@@ -2087,7 +1884,7 @@ \section{Organisation of the tiles}\label{subsectilesorg}
 and define
 \begin{equation}\label{definegone2}
     G_2:=
-\bigcup_{0\le  k}\bigcup_{k< n}
+\bigcup_{k\ge  0}\bigcup_{k< n}
 A(2n+6,k,n)\, .
 \end{equation}
 Define
@@ -2101,7 +1898,8 @@ \section{Organisation of the tiles}\label{subsectilesorg}
 Define $G'=G_1\cup G_2 \cup G_3$
 The following bound of the measure of $G'$ will be proven in
 Subsection \ref{subsetexcset}.
-\begin{lemma}\label{exceptional set}
+\begin{lemma}[exceptional set]
+\label{exceptional set}
 \uses{first exception,dense cover,pairwise disjoint,dyadic union,tree count,boundary measure}
 We have
 \begin{equation}
@@ -2112,7 +1910,8 @@ \section{Organisation of the tiles}\label{subsectilesorg}
 In Subsection \ref{subsecforest}, we identify each set $\fC_5(k,n,j)$ outside $G'$ as forest and use Proposition
 \ref{forest operator} to prove the following lemma.
 
-\begin{lemma}\label{forest union}
+\begin{lemma}[forest union]
+\label{forest union}
 \uses{relation geometry,C6 forest, C6 convex, forest separation, forest stacking}
   Let
   \begin{equation}
@@ -2131,12 +1930,13 @@ \section{Organisation of the tiles}\label{subsectilesorg}
 \ref{forest union} and apply the antichain estimate of
 Proposition \ref{antichain operator} to prove the following lemma.
 
-\begin{lemma}\label{forest complement}
-\uses{antichain decomposition,L0 antichain,l2 antichain,L1 L3 antichain}
+\begin{lemma}[forest complement]
+\label{forest complement}
+\uses{antichain decomposition,L0 antichain,L2 antichain,L1 L3 antichain}
   Let
   \begin{equation}
-      \fP_2 =\fP\setminus \left(\bigcup_{k\ge 0}\bigcup_{n\ge k}
-      \bigcup_{0\le j\le 2n+3}\fC_5(k,n,j)\right)\,.
+      \fP_2 =\fP\setminus \fP_1 %\left(\bigcup_{k\ge 0}\bigcup_{n\ge k}
+     % \bigcup_{0\le j\le 2n+3}\fC_5(k,n,j)\right)\,.
   \end{equation}
  For all $f:X\to \C$ with $|f|\le \mathbf{1}_F$ we have
 \begin{equation}
@@ -2147,8 +1947,9 @@ \section{Organisation of the tiles}\label{subsectilesorg}
 Proposition \ref{discrete Carleson} follows by applying
 triangle inequality to \eqref{disclesssim}
 according to the splitting in Lemma \ref{forest union}
-and \ref{forest complement} and using both Lemmas as well
+and Lemma \ref{forest complement} and using both Lemmas as well
 as the bound on the set $G'$ given by Lemma \ref{exceptional set}.
+\asgar{Perhaps $\frac{2^{260 a^3+1}}{(q-1)^4}$ by triangle inequality.}
 
 
 
@@ -2162,7 +1963,9 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 
 The bound for $G_1$ is follows from the Vitali covering lemma, Proposition \ref{Hardy Littlewood}.
 
-\begin{lemma}\label{first exception}
+\begin{lemma}
+[first exception]\label{first exception}
+\uses{Hardy Littlewood}
 We have
 \begin{equation}
     \mu(G_1)\le 2^{-4}\mu(G)\, .
@@ -2183,23 +1986,24 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 $$
     \mathcal{B} = \{B(\pc(\fp),r(\fp)) \ : \ \fp \in \fP_{F,G}\}
 $$
-and the function $h = \mathbf{1}_F$, we obtain
+and the function $u = \mathbf{1}_F$, we obtain
 $$
     \mu(\bigcup \mathcal{B}) \le 2^{2a+1} K^{-1}  \mu(F)\,.
 $$
 We conclude  with \eqref{eq vol sp cube} and $r(\fp)>4D^{\ps(\fp)}$
 $$
-    \mu(G_1)\le \mu(\bigcup_{\fp\in \fP_{F,G}} \sc(\fp))
+    \mu(G_1)= \mu(\bigcup_{\fp\in \fP_{F,G}} \sc(\fp))
     \le \mu(\bigcup \mathcal{B})\le 2^{2a+1} K^{-1} \mu (F) = 2^{-4}\mu(G)\,.
 $$
 \end{proof}
 
 
-We turn to the bound of $G_2$, which relies on the dyadic covering Lemma \ref{dense cover} and the
-John-Nirenberg Lemma \ref{john nirenberg} below.
+We turn to the bound of $G_2$, which relies on the Dyadic Covering Lemma \ref{dense cover} and the
+John-Nirenberg Lemma \ref{John Nirenberg} below.
 
-\begin{lemma}\label{dense cover}
-\uses{john nirenberg}
+\begin{lemma}[dense cover]
+\label{dense cover}
+\uses{John Nirenberg}
 For  each $k\ge 0$, the union of all dyadic cubes
 in $\mathcal{C}(G,k)$ has measure at most $2^{k+1} \mu(G)$ .
 \end{lemma}
@@ -2216,7 +2020,7 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 \sum_{J\in \mathcal{M}^*(k)}\mu(J)
 \end{equation}
 Using the definition of $\mathcal{M}(k)$ and then
-the  pairwise disjointness of elements in
+the  pairwise disjointedness of elements in
 $\mathcal{M}^*(k)$,
 we estimate \eqref{cbymstar} by
 \begin{equation}
@@ -2230,8 +2034,9 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 
 
 
-\begin{lemma}\label{pairwise disjoint}
-\uses{john nirenberg}
+\begin{lemma}[pairwise disjoint]
+\label{pairwise disjoint}
+\uses{John Nirenberg}
      If $\fp, \fp' \in {\mathfrak{M}}(k,n)$ and
      \begin{equation}\label{eintersect}
      {E_1}(\fp)\cap {E_1}(\fp')\neq \emptyset,
@@ -2239,7 +2044,7 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
      then $\fp=\fp'$.
 \end{lemma}
 \begin{proof}
-Let $\fp,\fp'$ as in the lemma. As by definition of $E_1$
+Let $\fp,\fp'$ be as in the lemma. By definition of $E_1$,
 we have
 $E_1(\fp)\subset \sc(\fp)$ and analogously for $\fp'$, we conclude from \eqref{eintersect} that  $\sc(\fp)\cap \sc(\fp')\neq \emptyset$. Let without loss of generality $\sc(\fp)$ be maximal in
 $\{\sc(\fp),\sc(\fp')\}$, then $\sc(\fp')\subset \sc(\fp)$.
@@ -2250,8 +2055,9 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 \end{proof}
 
 
-\begin{lemma}\label{dyadic union}
-\uses{john nirenberg}
+\begin{lemma}[dyadic union]
+\label{dyadic union}
+\uses{John Nirenberg}
 For each  $x\in A(\lambda,k,n)$,
 there is a dyadic cube $I$
 that contains $x$ and is
@@ -2260,21 +2066,16 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 \end{lemma}
 
 \begin{proof}
-
-Fix $k,n,\lambda,x$ as in the lemma such that
-$x\in A(\lambda,k,n)$. Let
-$\mathcal{M}$ be the set of dyadic cubes
- $\sc(\fp)$
-with $\fp$ in  $\mathfrak{M}(k,n)$
-and $x\in \sc(\fp)$. By definition of
-$A(\lambda,k,n)$, the cardinality of $\mathcal{M}$
-is at least $\lambda$. Let $I$ be an cube of
-smallest scale in $\mathcal{M}$. Then
-$I$ is contained in all cubes of $\mathcal{M}$.
+Fix $k,n,\lambda,x$ as in the lemma such that $x\in A(\lambda,k,n)$.
+Let $\mathcal{M}$ be the set of dyadic cubes $\sc(\fp)$ with $\fp$ in $\mathfrak{M}(k,n)$ and $x\in \sc(\fp)$.
+By definition of $A(\lambda,k,n)$, the cardinality of $\mathcal{M}$ is at least $1+\lambda 2^{n+1}$.
+Let $I$ be a cube of smallest scale in $\mathcal{M}$.
+Then $I$ is contained in all cubes of $\mathcal{M}$.
 It follows that $I\subset A(\lambda,k,n)$.
 \end{proof}
 
-\begin{lemma}\label{john nirenberg}
+\begin{lemma}[John Nirenberg]
+\label{John Nirenberg}
 \uses{second exception, top tiles}
     For all integers $k,n,\lambda\ge 0$, we have
     \begin{equation}\label{alambdameasure}
@@ -2290,7 +2091,7 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 We prove the lemma by induction on $\lambda$.
 For $\lambda=0$, we use that $A(\lambda)$ by definition of $\mathfrak{M}(k,n)$ is contained in the union of elements in $ \mathcal{C}(G,k)$. Lemma \ref{dense cover} then completes  the base of the induction.
 
-Now assume that the statement of Lemma \ref{john nirenberg}
+Now assume that the statement of Lemma \ref{John Nirenberg}
 is proven for some integer $\lambda\ge 0$.
 The set $A(\lambda+1)$ is contained in the set $A(\lambda)$.
 Let  $\mathcal{M}$ be the set of dyadic cubes which are a subset of $A(\lambda)$. By Lemma \ref{dyadic union}, the union of $\mathcal{M}$ is $A(\lambda)$.
@@ -2298,9 +2099,9 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 
 Let $L\in \mathcal{M}^*$. For each $x\in L$, we have
 \begin{equation}\label{suminout}
- \sum_{\fp \in {\mathfrak{M}}(k,n)} \mathbf{1}_{I(\fp)}(x)=
-   \sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \subset L} \mathbf{1}_{I(\fp)}(x)+
-  \sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \not \subset L} \mathbf{1}_{I(\fp)}(x)\, .
+ \sum_{\fp \in {\mathfrak{M}}(k,n)} \mathbf{1}_{\sc(\fp)}(x)=
+   \sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \subset L} \mathbf{1}_{\sc(\fp)}(x)+
+  \sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \not \subset L} \mathbf{1}_{\sc(\fp)}(x)\, .
 \end{equation}
 If the second sum on the right-hand-side is not zero, there is
 an element of $\mathcal{D}$  strictly containing $L$.
@@ -2315,7 +2116,7 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 the left-hand-side of \eqref{suminout} is at least
 $1+(\lambda+1) 2^{n+1}$, so we have by the triangle inequality for the first sum on the right-hand side
 \begin{equation}\label{mnkonl}
-\sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \subset L} \mathbf{1}_{I(\fp)}(x)\ge 2^{n+1}\, .\end{equation}
+\sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \subset L} \mathbf{1}_{\sc(\fp)}(x)\ge 2^{n+1}\, .\end{equation}
 By Lemma \ref{pairwise disjoint}, we have
 \begin{equation}
 \sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \subset L} \mu({E_1}(\fp)) \leq \mu(L)\, .
@@ -2331,7 +2132,7 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 \end{equation}
 \begin{equation}
 \le
-    \int \sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \subset L} \mathbf{1}_{I(\fp)} d\mu
+    \int \sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \subset L} \mathbf{1}_{\sc(\fp)} d\mu
 \le 2^n \mu(L)\, .
 \end{equation}
 Hence
@@ -2344,7 +2145,8 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 \eqref{alambdameasure} for $\lambda+1$ and completes the proof of the lemma.
 \end{proof}
 
-\begin{lemma}\label{second exception}
+\begin{lemma}[second exception]
+\label{second exception}
 We have
 \begin{equation}
     \mu(G_2)\le 2^{-4} \mu(G)\, .
@@ -2352,7 +2154,7 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 \end{lemma}
 \begin{proof}
 
-We use Lemma \ref{john nirenberg} and sum twice a geometric series
+We use Lemma \ref{John Nirenberg} and sum twice a geometric series
 to obtain
 \begin{equation}
     \sum_{0\le  k}\sum_{k< n}
@@ -2367,7 +2169,8 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 
 We turn to the set $G_3$.
 
-\begin{lemma}\label{top tiles}
+\begin{lemma}[top tiles]
+\label{top tiles}
     We have
     \begin{equation}\label{eq musum}
         \sum_{\mathfrak{m} \in \mathfrak{M}(k,n)} \mu(\sc(\mathfrak{m}))\le 2^{n+1}2^{k+1}\mu(G).
@@ -2379,7 +2182,7 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
     \int \sum_{\mathfrak{m} \in \mathfrak{M}(k,n)} \mathbf{1}_{\sc(\mathfrak{m})}(x) \, d\mu(x) \le
 2^{n+1} \sum_{\lambda=1}^{|\mathfrak{M}|}\mu(A(\lambda, k,n))\,.
 \end{equation}
-Using Lemma \ref{john nirenberg}
+Using Lemma \ref{John Nirenberg}
 %\ref{alambdameasure}
 and then summing a geometric series, we estimate this by
 \begin{equation}
@@ -2393,7 +2196,8 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 \end{proof}
 
 
-\begin{lemma}\label{tree count}
+\begin{lemma}[tree count]
+\label{tree count}
 \uses{third exception}
 Let $k,n,j\ge 0$. We have for every $x\in X$
 \begin{equation}
@@ -2466,7 +2270,8 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 Inserting this into \eqref{usumbymsum} proves the lemma.
 \end{proof}
 
-\begin{lemma}\label{boundary exception}
+\begin{lemma}[boundary exception]
+\label{boundary exception}
 \uses{third exception}
 Let $\mathcal{L}(\fu)$ be as defined in \eqref{eq L def}. We have for each $\fu\in \fU_1(k,n,l)$,
 \begin{equation}
@@ -2513,7 +2318,8 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets}
 
 
 
-    \begin{lemma}\label{third exception}
+    \begin{lemma}[third exception]
+    \label{third exception}
 
        We have
 \begin{equation}
@@ -2584,7 +2390,7 @@ \section{Auxiliary lemmas}
 \label{subsec lessim aux}
 Before proving Lemma \ref{forest union} and Lemma \ref{forest complement}, we collect some useful properties of $\lesssim$.
 
-\begin{lemma}
+\begin{lemma}[wiggle order 1]
     \label{wiggle order 1}
     \uses{wiggle order 3}
     If $n\fp \lesssim m\fp'$ and
@@ -2595,7 +2401,7 @@ \section{Auxiliary lemmas}
     This follows immediately from the definition \eqref{wiggleorder} of $\lesssim$ and the two inclusions $B_{\fp}(\fcc(\fp), n) \subset B_{\fp}(\fcc(\fp), n')$ and $B_{\fp'}(\fcc(\fp'), m') \subset B_{\fp'}(\fcc(\fp'), m)$.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[wiggle order 2]
     \label{wiggle order 2}
     \uses{wiggle order 3}
     Let $n, m \ge 1$.
@@ -2623,7 +2429,7 @@ \section{Auxiliary lemmas}
     Thus $B_{\fp'}(\fcc(\fp'),n) \subset B_{\fp}(\fcc(\fp),n + 2^{-95a}m)$. Combined with $\sc(\fp) \subset \sc(\fp')$, this yields \eqref{eq wiggle2}.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[wiggle order 3]
 \label{wiggle order 3}
 \uses{C convex, relation geometry}
     The following implications hold for all $\fq, \fq' \in \fP$:
@@ -2653,7 +2459,7 @@ \section{Auxiliary lemmas}
     \fp \le \fp' \le \fp'' \ \text{and} \ \fp, \fp'' \in \mathfrak{A} \implies \fp' \in \mathfrak{A}\,.
 \end{equation}
 
-\begin{lemma}
+\begin{lemma}[P convex]
     \label{P convex}
     \uses{C convex}
     For each $k$, the collection $\fP(k)$ is convex.
@@ -2667,7 +2473,7 @@ \section{Auxiliary lemmas}
     and $\mu(G \cap J) > 2^{-k-1} \mu(G)$. Thus \eqref{muhj1} holds for $\sc(\fp')$. On the other hand, by \eqref{muhj2}, there exists no $J \in \mathcal{D}$ with $\sc(\fp) \subset J$ and $\mu(G \cap J) > 2^{-k} \mu(G)$. Since $\sc(\fp) \subset \sc(\fp')$, this implies that \eqref{muhj2} holds for $\sc(\fp')$. Hence $\sc(\fp') \in \mathcal{C}(G,k)$, and therefore by \eqref{eq defPk} $\fp' \in \fP(k)$.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[C convex]
     \label{C convex}
     \uses{C1 convex}
     For each $k,n$, the collection $\fC(k,n)$ is convex.
@@ -2685,7 +2491,7 @@ \section{Auxiliary lemmas}
     Thus $\fp' \in \fC(k,n)$.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[C1 convex]
     \label{C1 convex}
     \uses{C2 convex}
     For each $k,n,j$, the collection $\fC_1(k,n,j)$ is convex.
@@ -2699,7 +2505,7 @@ \section{Auxiliary lemmas}
     thus $\fp' \in \fC_1(k,n,j)$.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[C2 convex]
     \label{C2 convex}
     \uses{C3 convex}
     For each $k,n,j$, the collection $\fC_2(k,n,j)$ is convex.
@@ -2712,7 +2518,7 @@ \section{Auxiliary lemmas}
      \end{equation*} Combined with Lemma \ref{C1 convex}, it follows that $\fp' \in \fC_1(k,n,j)$. Suppose that $\fp' \notin \fC_2(k,n,j)$. By \eqref{eq C2 def}, this implies that there exists $0 \le l' \le Zn$ with $\fp' \in \fL_1(k,n,j,l')$. By the definition \eqref{eq L1 def} of $\fL_1(k,n,j,l')$, this implies that $\fp$ is minimal with respect to $\le$  in $\fC_1(k,n,j) \setminus \bigcup_{l < l'} \fL_1(k,n,j,l)$. Since $\fp \in \fC_2(k,n,j)$ we must have $\fp \ne \fp'$. Thus $\fp \le \fp'$ and $\fp \ne \fp'$. By minimality of $\fp'$ it follows that $\fp \notin \fC_1(k,n,j) \setminus \bigcup_{l < l'} \fL_1(k,n,j,l)$. But by \eqref{eq C2 def} this implies $\fp \notin \fC_2(k,n,j)$, a contradiction.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[C3 convex]
     \label{C3 convex}
     \uses{C4 convex}
     For each $k,n,j$, the collection $\fC_3(k,n,j)$ is convex.
@@ -2722,7 +2528,7 @@ \section{Auxiliary lemmas}
     Let $\fp \le \fp' \le \fp''$ with $\fp, \fp'' \in \fC_3(k,n,j)$. By \eqref{eq C3 def} and Lemma \ref{C2 convex} it follows that $\fp' \in \fC_2(k,n,j)$. Suppose that $\fp' \notin \fC_3(k,n,j)$. Then, by \eqref{eq C3 def} and \eqref{eq L2 def}, there exists $\fu \in \fU_1(k,n,j)$ with $2\fp' \lesssim \fu$ and $\sc(\fp') \ne \sc(\fu)$. Together this gives $\sc(\fp') \subsetneq \sc(\fu)$. From $\fp' \le \fp$, \eqref{eq sc1} and transitivity of $\lesssim$ we then have $2\fp \lesssim \fu$. Also, $\sc(\fp) \subset \sc(\fp') \subsetneq \sc(\fu)$, so $\sc(\fp) \ne \sc(\fu)$. But then $\fp \in \fL_2(k,n,j)$, contradicting by \eqref{eq C3 def} the assumption $\fp \in \fC_3(k,n,j)$.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[C4 convex]
     \label{C4 convex}
     \uses{C5 convex}
     For each $k,n,j$, the collection $\fC_4(k,n,j)$ is convex.
@@ -2732,7 +2538,7 @@ \section{Auxiliary lemmas}
     Let $\fp \le \fp' \le\fp''$ with $\fp, \fp'' \in \fC_4(k,n,j)$. As before we obtain from the inclusion $\fC_4(k,n,j) \subset \fC_3(k,n,j)$ and Lemma \ref{C3 convex} that $\fp' \in \fC_3(k,n,j)$. Thus, if $\fp' \notin \fC_4(k,n,j)$ then by \eqref{eq L3 def} there exists $l$ such that $\fp' \in \fL_3(k,n,j,l)$. Thus $\fp'$ is maximal with respect to $\le$ in $\fC_3(k,n,j) \setminus \bigcup_{0 \le l' < l} \fL_3(k,n,j,l')$.  Since $\fp'' \in \fC_4(k,n,j)$ we must have $\fp' \ne \fp''$. Thus $\fp' \le \fp''$ and $\fp'\ne \fp''$. By minimality of $\fp'$ it follows that $\fp'' \notin \fC_3(k,n,j) \setminus \bigcup_{l < l'} \fL_3(k,n,j,l)$. But by \eqref{eq C4 def} this implies $\fp'' \notin \fC_4(k,n,j)$, a contradiction.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[C5 convex]
     \label{C5 convex}
     \uses{C6 convex}
     For each $k,n,j$, the collection $\fC_5(k,n,j)$ is convex.
@@ -2743,7 +2549,7 @@ \section{Auxiliary lemmas}
     By \eqref{eq L4 def}, there exists $\fu \in \fU_1(k,n,j)$ with $\sc(\fp') \subset \bigcup \mathcal{L}(\fu)$. Then also $\sc(\fp) \subset \bigcup \mathcal{L}(\fu)$, a contradiction.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[dens compare]
     \label{dens compare}
     \uses{C dens1}
     We have for every $k\ge 0$ and $\fP'\subset \fP(k)$
@@ -2787,7 +2593,7 @@ \section{Auxiliary lemmas}
 \eqref{muhj2} for $\fp$.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[C dens1]
     \label{C dens1}
     For each set $\mathfrak{A} \subset \mathfrak{C}(k,n)$, we have
     $$
@@ -2835,7 +2641,7 @@ \section{Proof of Lemma \ref{forest union}, the forests}
 if $\fu=\fu'$ or there exists $\fp$ in $\mathfrak{T}_1(\fu)$
 with $10 \fp\lesssim \fu'$.
 
-\begin{lemma}
+\begin{lemma}[relation geometry]
     \label{relation geometry}
     \uses{forest geometry,equivalence relation,forest inner}
     If $\fu \sim \fu'$, then $\sc(u) = \sc(u')$ and
@@ -2866,7 +2672,7 @@ \section{Proof of Lemma \ref{forest union}, the forests}
 \end{proof}
 
 
-\begin{lemma}
+\begin{lemma}[equivalence relation]
 \label{equivalence relation}
 For each $k,n,j$, the relation $\sim$ on
 $\fU_2(k,n,j)$ is an equivalence relation.
@@ -2929,7 +2735,7 @@ \section{Proof of Lemma \ref{forest union}, the forests}
    \bigcup_{\fu\sim \fu'}\mathfrak{T}_1(\fu')\cap \fC_6(k,n,j)\, .
 \end{equation}
 
-\begin{lemma}
+\begin{lemma}[C6 forest]
 \label{C6 forest}
 We have
 \begin{equation}
@@ -2941,7 +2747,7 @@ \section{Proof of Lemma \ref{forest union}, the forests}
     By \eqref{eq C4 def} and \eqref{defc5}, we have $\fp \in \fC_3(k,n,j)$. By \eqref{eq L2 def} and \eqref{eq C3 def}, there exists $\fu \in \fU_1(k,n,j)$ with $2\fp \lesssim \fu$ and $\sc(\fp) \ne \sc(\fu)$, that is, with $\fp \in \mathfrak{T}_1(\fu)$. Then $\mathfrak{T}_1(\fu)$ is clearly nonempty, so $\fu \in \fU_2(k,n,j)$. By the definition of $\fU_3(k,n,j)$, there exists $\fu' \in \fU_3(k,n,j)$ with $\fu \sim \fu'$. By \eqref{definesv}, we have $\fp \in \mathfrak{T}_2(\fu')$.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[C6 convex]
     \label{C6 convex}
     \uses{forest convex}
     Let $\fu \in \fU_3(k,n,j)$. If $\fp \le \fp' \le \fp''$ and $\fp, \fp'' \in \mathfrak{T}_2(\fu)$, then $\fp' \in \mathfrak{T}_2(\fu)$.
@@ -2953,7 +2759,7 @@ \section{Proof of Lemma \ref{forest union}, the forests}
 \end{proof}
 
 
-\begin{lemma}
+\begin{lemma}[forest geometry]
     \label{forest geometry}
     For each $\fu\in \fU_3(k,n,j)$,
     the set $\mathfrak{T}_2(\fu)$
@@ -2976,7 +2782,7 @@ \section{Proof of Lemma \ref{forest union}, the forests}
     Together with $\sc(\fp) \subset \sc(\fu) = \sc(\fu')$, this gives $4\fp \lesssim 1\fu'$, which is \eqref{forest1}.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[forest convex]
     \label{forest convex}
     For each $\fu\in \fU_3(k,n,j)$,
     the set $\mathfrak{T}_2(\fu)$
@@ -2991,7 +2797,7 @@ \section{Proof of Lemma \ref{forest union}, the forests}
     It follows that $\fp' \in \mathfrak{T}_2(\fu')$, which shows \eqref{forest2}.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[forest separation]
     \label{forest separation}
     For each $\fu,\fu'\in \fU_3(k,n,j)$ with $\fu\neq \fu'$ and
     and each $\fp \in \fT_2(\fu)$
@@ -3019,7 +2825,7 @@ \section{Proof of Lemma \ref{forest union}, the forests}
     The Lemma follows by combining the two displays with the fact that $95 a \ge 1$.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[forest inner]
     \label{forest inner}
     For each $\fu\in \fU_3(k,n,j)$
     and each $\fp \in \mathfrak{T}_2(\fu)$
@@ -3042,7 +2848,7 @@ \section{Proof of Lemma \ref{forest union}, the forests}
 \end{proof}
 
 
-\begin{lemma}
+\begin{lemma}[forest stacking]
     \label{forest stacking}
     It holds that
     \begin{equation}
@@ -3098,7 +2904,7 @@ \section{Proof of Lemma \ref{forest complement}, the antichains}
 \label{subsecantichain}
 
 Define $\fP_{X \setminus G'}$ to be the set of all $\fp \in \fP$ such that $\sc(\fp) \not \subset G'$.
-\begin{lemma}
+\begin{lemma}[antichain decomposition]
 \label{antichain decomposition}
     We have that
     \begin{align}
@@ -3119,7 +2925,7 @@ \section{Proof of Lemma \ref{forest complement}, the antichains}
     Since $\fp \in \fP_{X \setminus G'}$, we have in particular $\fp \not \subset A(2n + 6, k, n)$, so there exist at most $1 + (4n + 12)2^n < 2^{2n+4}$ tiles $\mathfrak{m} \in \mathfrak{M}(k,n)$ with $\fp \le \mathfrak{m}$. It follows that $\fp \in \fL_0(k,n)$ or $\fp \in \fC_1(k,n,j)$ for some $1 \le j \le 2n + 3$. In the former case we are done, in the latter case the inclusion to be shown follows immediately from the definitions of the collections $\fC_i$ and $\fL_i$.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[L0 antichain]
 \label{L0 antichain}
     We have that
     $$
@@ -3171,7 +2977,7 @@ \section{Proof of Lemma \ref{forest complement}, the antichains}
     Thus, by \eqref{straightorder}, $2\fp_0 \leq 100\mathfrak{m}$, a contradiction to $\fp_0 \notin \fC(k,n)$.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[L2 antichain]
 \label{L2 antichain}
     Each of the sets $\fL_2(k,n,j)$ is an antichain.
 \end{lemma}
@@ -3181,7 +2987,7 @@ \section{Proof of Lemma \ref{forest complement}, the antichains}
     If we have $\fp_l \in \fU_1(k,n,j)$, then it follows from $2\fp \lesssim 200 \fp_l \lesssim \fp_l$ and \eqref{eq L2 def} that $\fp \not\in \fL_2(k,n,j)$, a contradiction. Thus, by the definition \eqref{defunkj}  of $\fU_1(k,n,j)$, there exists $\fp_{l+1} \in \fC_1(k,n,j)$ with $\sc(\fp_l) \subsetneq \sc(\fp_{l+1}) $ and $\mfa \in B_{\fp_l}(\fcc(\fp_l), 100) \cap B_{\fp_{l+1}}(\fcc(\fp_{l+1}), 100)$. Using the triangle inequality and Lemma \ref{monotone cube metrics}, one deduces that $200 \fp_l \lesssim 200\fp_{l+1}$. This contradicts maximality of $l$.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[L1 L3 antichainf]
 \label{L1 L3 antichain}
     Each of the sets $\fL_1(k,n,j,l)$ and $\fL_3(k,n,j,l)$ is an antichain.
 \end{lemma}
@@ -3236,13 +3042,13 @@ \chapter{Proof of  Proposition \ref{antichain operator}}
 as the geometric mean of two inequalities,
 each involving one of the two densities.
 One of these two inequalities will need a careful estimate formulated in
-Lemma \ref{lem basic TT*} of
+Lemma \ref{tile correlation} of
 the $TT^*$ correlation between two tile operators of two tiles.
-Lemma \ref{lem basic TT*} will be proven in
+Lemma \ref{tile correlation} will be proven in
 Subsection \ref{sec tile operator}.
 The summation of the contributions of these individual correlations will require a
-geometric Lemma \ref{lem antichain 1} counting the relevant tile pairs.
-Lemma \ref{lem antichain 1} will be proven in Subsection
+geometric Lemma \ref{antichain tile count} counting the relevant tile pairs.
+Lemma \ref{antichain tile count} will be proven in Subsection
 \ref{subsec geolem}.
 
 
@@ -3252,9 +3058,9 @@ \chapter{Proof of  Proposition \ref{antichain operator}}
 
 \section{The density  arguments}\label{sec TT* T*T}
 
-We begin with the following crucial disjointness property of the sets $E(\fp)$ with $\fp \in \mathfrak{A}$.
-\begin{lemma}
-\label{lem antichain -1}
+We begin with the following crucial disjointedness property of the sets $E(\fp)$ with $\fp \in \mathfrak{A}$.
+\begin{lemma}[tile disjointness]
+\label{tile disjointness}
 Let $\fp,\fp'\in \mathfrak{A}$.
 If there exists an $x\in X$ with $x\in  E(\fp)\cap E(\fp')$,
 then $\fp= \fp'$.
@@ -3277,7 +3083,8 @@ \section{The density  arguments}\label{sec TT* T*T}
 \end{equation}
 with $\fp\in \mathfrak{A}$ and recall the definition of
 $M_{\mathcal{B}}$ from Proposition \ref{Hardy Littlewood}.
-\begin{lemma}\label{lem hlmbound}
+\begin{lemma}[maximal bound antichain]\label{maximal bound antichain}
+\uses{tile disjointness}
 Let $x\in X$.
 Then
 \begin{equation}\label{hlmbound}
@@ -3288,7 +3095,7 @@ \section{The density  arguments}\label{sec TT* T*T}
 
 
 \begin{proof}
-Fix $x\in X$.  By Lemma \ref{lem antichain -1}, there is at most one $\fp \in \mathfrak{A}$
+Fix $x\in X$.  By Lemma \ref{tile disjointness}, there is at most one $\fp \in \mathfrak{A}$
 such that
  $T_{\fp} f(x)$ is not zero.
  If there is no such $\fp$, the estimate \eqref{hlmbound} follows.
@@ -3333,8 +3140,9 @@ \section{The density  arguments}\label{sec TT* T*T}
     \tilde{q}=\frac {2q}{1+q}\,.
 \end{equation}
 Since $1< q\le 2$, we have $1<\tilde{q}<q\le 2$.
-\begin{lemma}
-\label{lem decay t}
+\begin{lemma}[dens2 antichain]
+\label{dens2 antichain}
+\uses{,Hardy Littlewood,maximal bound antichain}
 We have that
 \begin{equation}\label{eqttt9}
   \left|\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)\right|\le
@@ -3371,7 +3179,7 @@ \section{The density  arguments}\label{sec TT* T*T}
  2^{2a+2} (\tilde{q}-1)^{-1}  \|f\|_2\, .
 \end{equation}
 We obtain with Cauchy-Schwarz
-and then Lemma \ref{lem hlmbound}
+and then Lemma \ref{maximal bound antichain}
  \begin{equation}
      |\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)|
 \end{equation}
@@ -3392,7 +3200,8 @@ \section{The density  arguments}\label{sec TT* T*T}
 \end{proof}
 
 
-\begin{lemma}\label{lem decay n}
+\begin{lemma}[dens1 antichain]\label{dens1 antichain}
+\uses{Hardy Littlewood, tile disjointness, tile correlation,antichain tile count}
 Set $p:=4a^4$. We have
     \begin{equation}\label{eqttt3}
   \left|\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)\right|\le
@@ -3444,8 +3253,9 @@ \section{The density  arguments}\label{sec TT* T*T}
     \Big|\int T^*_{\fp}g(y)\overline{T^*_{\fp'}g(y)}\, d\mu(y)\Big|\,.
 \end{equation}
 The following basic $TT^*$ estimate will be proved in Subsection \ref{sec tile operator}.
-\begin{lemma}
-    \label{lem basic TT*}
+\begin{lemma}[tile correlation]
+    \label{tile correlation}
+    \uses{Holder van der Corput,correlation kernel bound,tile uncertainty,tile range support}
     Let $\fp, \fp'\in \fP$ with
     $\ps({\fp'})\leq \ps({\fp})$.
     Then
@@ -3476,7 +3286,7 @@ \section{The density  arguments}\label{sec TT* T*T}
 \begin{equation}\label{eqttt4}
     \mu(B(\fp))\le 2^{6a} \mu(\sc(\fp))\, .
 \end{equation}
-Using Lemma \ref{lem basic TT*} and \eqref{eqttt4}, we estimate \eqref{eqtts2} by
+Using Lemma \ref{tile correlation} and \eqref{eqttt4}, we estimate \eqref{eqtts2} by
 \begin{equation}\label{eqtts3}
      \le  2^{255a^3+6a+1} \sum_{\fp\in \mathfrak{A}}
     \int_{E(\fp)}|g|(y) h(\fp)\, d\mu(y)
@@ -3487,8 +3297,9 @@ \section{The density  arguments}\label{sec TT* T*T}
     {(1+d_{\fp'}(\fcc(\fp'), \fcc(\fp))^{-1/(2a^2+a^3)}}(\mathbf{1}_{E(\fp')}|g|)(y')\, d\mu(y')\,.
 \end{equation}
 The following lemma will be proved in Subsection \ref{subsec geolem}.
-\begin{lemma}
-    \label{lem antichain 1}
+\begin{lemma}[antichain tile count]
+    \label{antichain tile count}
+    \uses{global antichain density}
     Set $p:=4a^4$ and let $p'$ be the dual exponent of $p$, that is $1/p+1/p'=1$.
     For  every $\mfa\in\Mf$ and every subset $\mathfrak{A}'$ of $\mathfrak{A}$ we have
     \begin{equation}
@@ -3508,7 +3319,7 @@ \section{The density  arguments}\label{sec TT* T*T}
     \frac{\|g\mathbf{1}_{B(\fp)}\|_{p'}}{\mu(B(\fp))}
     \Big\|\sum_{\fp\in\mathfrak{A}(\fp)}(1+d_{\fp}(\fcc(\fp), \fc(\fp'))^{-1/(2a^2+a^3)}\mathbf{1}_{E(\fp)}\mathbf{1}_G\Big\|_{p}\, .
 \end{equation}
-Then we apply Lemma \ref{lem antichain 1} to estimate this by
+Then we apply Lemma \ref{antichain tile count} to estimate this by
 \begin{equation}\label{eqttt5}
     \le  2^{104a}
     \frac{\|g\mathbf{1}_{B(\fp)}\|_{p'}}{\mu(B(\fp))}
@@ -3537,7 +3348,7 @@ \section{The density  arguments}\label{sec TT* T*T}
  \le  2^{255a^3+110a + 1} { \dens_1(\mathfrak{A})^{\frac 1p}}\sum_{\fp\in \mathfrak{A}}
  \int_{E(\fp)}|g|(y)M_{\mathcal{B}, p'}g(y) \, dy\,.
          \end{equation}
-Using Lemma \ref{lem antichain -1},
+Using Lemma \ref{tile disjointness},
 the last display is observed to be
 \begin{equation}\label{eqtts4a}
 =  2^{255a^3+110a + 1}
@@ -3557,7 +3368,7 @@ \section{The density  arguments}\label{sec TT* T*T}
     \le  2^{255a^3+110a + 5} \dens_1(\mathfrak{A})^{\frac 1p}
     \|g\|_2^2\,.
 \end{equation}
-Now Lemma \ref{lem decay n} follows by applying Cauchy-Schwarz on the left-hand side and using
+Now Lemma \ref{dens1 antichain} follows by applying Cauchy-Schwarz on the left-hand side and using
 $a\ge 4$.
 \end{proof}
 We have
@@ -3576,11 +3387,11 @@ \section{The density  arguments}\label{sec TT* T*T}
 Proposition \ref{antichain operator}.
 
 
-\section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec tile operator}
+\section{Proof of Lemma \ref{tile correlation}, the tile correlation bound }\label{sec tile operator}
 
 The next lemma prepares an application of
 Proposition \ref{Holder van der Corput}.
-\begin{lemma}\label{lem ksquare}
+\begin{lemma}[correlation kernel bound]\label{correlation kernel bound}
 Let $-S\le s_1\le s_2\le S$ and let $x_1,x_2\in X$.
 Define \begin{equation}
  \varphi(y) :=  \overline{K_{s_1}(x_1, y)}
@@ -3633,7 +3444,8 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 This proves the lemma.
 \end{proof}
 The next lemma is a geometric estimate for two tiles.
-\begin{lemma}\label{lem tgeo}
+\begin{lemma}\label{tile uncertainty}
+\uses{monotone cube metrics}
     Let $\fp_1, \fp_2\in \fP$ with
 $\ps({\fp_1})\leq \ps({\fp_2})$. For each $x_1\in E(\fp_1)$ and
 $x_2\in E(\fp_2)$  we have
@@ -3681,10 +3493,10 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 
 
 
-\begin{lemma}\label{lem tstarsupport}
-    For each $\fp\in \fP$, and each $y\in X$, we have that
+\begin{lemma}[tile range support]\label{tile range support}
+    For each  $\fp\in \fP$, and each $y\in X$, we have that
 \begin{equation}\label{tstargnot0}
-     T_{\fp} g^*(y)\neq 0
+     T_{\fp}^* g(y)\neq 0
 \end{equation}
    implies
 \begin{equation}\label{ynotfar}
@@ -3712,7 +3524,7 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 \end{proof}
 
 
-We now prove Lemma \ref{lem basic TT*}. We begin with  \eqref{eq basic TT* est}.
+We now prove Lemma \ref{tile correlation}. We begin with  \eqref{eq basic TT* est}.
 
 We expand the left-hand side of \eqref{eq basic TT* est} as
 \begin{equation}\label{tstartstar}
@@ -3727,31 +3539,34 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 the fact $|e(\tQ(x_i)(x_i))|=1$ for $i=1,2$, we can estimate
 \eqref{tstartstar} from above by
 \begin{equation}\label{eqa1}
-    \int_{E(\fp_1)} \int_{E(\fp_2)} \left|\int
+    \int_{E(\fp_1)} \int_{E(\fp_2)} {\bf I}\, d\mu(x_1)d\mu(x_2)\,.
+\end{equation}
+with
+\begin{equation}
+    {\bf I}:=
+    \left|\int
     e(-\tQ(x_1)(y)+\tQ(x_2)(y))\varphi_{x_1,x_2}(y)
-    d\mu(y) \, g(x_1)g(x_2)\right| d\mu(x_1)d\mu(x_2)\,.
+    d\mu(y) \, g(x_1)g(x_2)\right|
 \end{equation}
+
+
 We estimate for fixed $x_1\in E(\fp_1)$ and
 $x_2\in E(\fp_2)$ the inner integral of \eqref{eqa1} with
 Proposition \ref{Holder van der Corput}. The function
 $\varphi:=\varphi_{x_1,x_2}$ satisfies the assumptions of
-Proposition \ref{Holder van der Corput} with $z = x_1$ and $R = D^{s_1}$ by Lemma \ref{lem ksquare}.
+Proposition \ref{Holder van der Corput} with $z = x_1$ and $R = D^{s_1}$ by Lemma \ref{correlation kernel bound}.
 We obtain with $B':= B(x_1, D^{\ps(\fp_1)})$,
-\begin{equation}
-    \left|\int
-    e(-\tQ(x_1)(y)+\tQ(x_2)(y))\varphi_{x_1,x_2}(y)
-    \,d\mu(y)\right|
-\end{equation}
-\begin{equation}
- \le     2^{8a} \mu(B') \|{\varphi}\|_{C^\tau(B')}
+\begin{equation*}
+ {\bf I}\le     2^{8a} \mu(B') \|{\varphi}\|_{C^\tau(B')}
        (1 + d_{B'}(\tQ(x_1),\tQ(x_2)))^{-1/(2a^2+a^3)}
-\end{equation}
+\end{equation*}
 \begin{equation}
+\label{eqa1.5}
  \le     \frac{2^{254a^3+8a}}
  {\mu(B(x_2, D^{\ps(\fp_2)}))}
        (1 + d_{B'}(\tQ(x_1),\tQ(x_2)))^{-1/(2a^2+a^3)}\,.
 \end{equation}
-Using Lemma \ref{lem tgeo} and $a\ge 1$ estimates the last display by
+Using Lemma \ref{tile uncertainty} and $a\ge 1$ estimates \eqref{eqa1.5} by
 \begin{equation}\label{eqa2}
  \le     \frac{2^{254a^3 + 8a + 1}}
  {\mu(B(x_2, D^{\ps(\fp_2)}))}
@@ -3777,7 +3592,7 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 \begin{equation}
     T^*_{\fp}g(y)\overline{T^*_{\fp'}g(y)}\neq 0
 \end{equation}
-By the triangle inequality and Lemma \ref{lem tstarsupport}, we conclude
+By the triangle inequality and Lemma \ref{tile range support}, we conclude
 \begin{equation}
    \rho(\pc(\fp),\pc(\fp'))\le  \rho(\pc(\fp),y) +\rho(\pc(\fp'),y)
    \le 5D^{\ps(\fp)}+5D^{\ps(\fp')}\le 10 D^{\ps(\fp)}\, .
@@ -3787,17 +3602,18 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 \begin{equation}
     \sc(\fp') \subset B(\pc(\fp), 15D^{\ps(\fp)})\, .
 \end{equation}
-   This completes the proof of Lemma  \ref{lem basic TT*}.
+   This completes the proof of Lemma  \ref{tile correlation}.
 
 
 
 
 
-\section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
+\section{Proof of Lemma \ref{antichain tile count}, the geometric estimate}
 \label{subsec geolem}
 
 
-\begin{lemma}\label{lem a geo}
+\begin{lemma}[tile reach]\label{tile reach}
+\uses{monotone cube metrics}
 Let $\mfa\in \Mf$ and $N\ge0$ be an integer.
 Let $\fp, \fp'\in \fP$ with
 \begin{equation}\label{eqassumedismfa}
@@ -3870,7 +3686,8 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 \end{equation}
 
 
-\begin{lemma}\label{lem samel}
+\begin{lemma}[stack density]
+\label{stack density}
 Let $\mfa \in \Mf$ and $N\ge 0$ and
 $L\in \mathcal{D}$. Then
 \begin{equation}\label{eqanti-1}
@@ -3921,7 +3738,8 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 \end{proof}
 
 
-\begin{lemma}\label{lem antichain-.5}
+\begin{lemma}[local antichain density]\label{local antichain density}
+\uses{tile disjointness,tile reach}
 Let $\mfa\in\Mf$ and {$N$} be
 an integer. Let $\fp_{\mfa}$ be a tile with $\mfa\in \fc(\fp_{\mfa})$.
 Then we have
@@ -3955,7 +3773,7 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 \begin{equation}
     \mfa \in B(\fcc(\fp), 2^{N+1})\, .
 \end{equation}
-With Lemma \ref{lem a geo}, we conclude
+With Lemma \ref{tile reach}, we conclude
 \begin{equation}
     2^{N+3}\fp_{\mfa}  \lesssim  2^{N+3}\fp \, .
 \end{equation}
@@ -3963,11 +3781,12 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 \begin{equation}
     E(\fp)\cap G \subset E_2(2^{N+3},\fp_{\mfa})\, .
 \end{equation}
-Using disjointness of the various $E(\fp)$ with $\fp\in \mathfrak{A}$  by Lemma \ref{lem antichain -1}, we obtain \eqref{eqanti-0.5}.
+Using disjointedness of the various $E(\fp)$ with $\fp\in \mathfrak{A}$  by Lemma \ref{tile disjointness}, we obtain \eqref{eqanti-0.5}.
 This proves the lemma.
 \end{proof}
-\begin{lemma}
-\label{lem antichain 0}
+\begin{lemma}[global antichain density]
+\label{global antichain density}
+\uses{stack density,local antichain density}
 Let $\mfa\in\Mf$ and let $N\ge 0$  be
 an integer. Then we have
 \begin{equation}\label{eqanti00}
@@ -4016,7 +3835,7 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 \begin{equation}
     \sum_{\fp\in\mathfrak{A}':\sc(\fp)=L}\mu(E(\fp)\cap G\cap L)\,.
 \end{equation}
-Thus in this case \eqref{eqanti0} is a direct consequence of Lemma \ref{lem samel} and $a \ge 4$.
+Thus in this case \eqref{eqanti0} is a direct consequence of Lemma \ref{stack density} and $a \ge 4$.
 
 We now assume that there exists a cube $J \in \mathcal{D}$ with $L \subsetneq J$.
 By \eqref{coverdyadic}, there is an
@@ -4033,7 +3852,7 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 \end{equation}
 
 We first estimate \eqref{eqanti1}
-with Lemma \ref{lem samel} by
+with Lemma \ref{stack density} by
 \begin{equation}\label{equanti1.5}
     \le \sum_{\fp\in\mathfrak{A}':\sc(\fp)=L'}\mu(E(\fp)\cap G\cap L')\le 2^{a(N+5)}\dens_1(\mathfrak{A})\mu(L')\, .
 \end{equation}
@@ -4067,7 +3886,7 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 \begin{equation}
     \mfa \in B(\fcc(\fp''), 2^{N+1})\, .
 \end{equation}
-By Lemma \ref{lem a geo}, we conclude
+By Lemma \ref{tile reach}, we conclude
 \begin{equation}
     2^{N+3}\fp''  \lesssim  2^{N+3}\fp_{\mfa} \, .
 \end{equation}
@@ -4083,7 +3902,7 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 $s(L)<\ps(\fp)$ and thus $s(L')\le \ps(\fp)$. By the dyadic property
    \eqref{dyadicproperty} again, we have $L'\subset \sc(\fp)$.
 As $L'\neq \sc(\fp)$, we conclude $s(L)<\ps(\fp)$.
-By Lemma \ref{lem antichain-.5}, we can thus estimate \eqref{eqanti2} by
+By Lemma \ref{local antichain density}, we can thus estimate \eqref{eqanti2} by
 \begin{equation}\label{eqanti0.5}
     \sum_{\fp\in\mathfrak{A}':\sc(\fp)\neq L'}\mu(E(\fp)\cap G\cap L')
     \le  \mu (E_2(2^{N+3},\fp_{\mfa}))\, .
@@ -4111,7 +3930,7 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 
 
 
-We turn to the proof of Lemma \ref{lem antichain 1}.
+We turn to the proof of Lemma \ref{antichain tile count}.
 
 
 
@@ -4127,7 +3946,7 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 \le \sum_{N\ge 0} \left\|\sum_{\fp\in \mathfrak{A}_{\mfa,N}} 2^{-N/(2a^2+a^3)}\mathbf{1}_{E(\fp)} \mathbf{1}_G\right\|_{p}
 \end{equation}
 We consider each individual term in this sum and estimate it's $p$-th power.
-   Using that for each $x\in X$  by Lemma \ref{lem antichain 0} there is at most one $\fp\in \mathfrak{A}$ with $x\in E(\fp)$,
+   Using that for each $x\in X$  by Lemma \ref{global antichain density} there is at most one $\fp\in \mathfrak{A}$ with $x\in E(\fp)$,
  we have
  \begin{equation}
      \left\|\sum_{\fp\in \mathfrak{A}_{\mfa,N}} 2^{-N/(2a^2+a^3)}\mathbf{1}_{E(\fp)} \mathbf{1}_G\right\|_{p}^p
@@ -4142,7 +3961,7 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
   =   2^{-pN/(2a^2+a^3)} \sum_{\fp\in\mathfrak{A}_{\mfa,N}}\mu(E(\fp)\cap G)
 \end{equation}
 
-Using Lemma \ref{lem antichain 0}, we estimate the last display by
+Using Lemma \ref{global antichain density}, we estimate the last display by
 \begin{equation}\label{eqanti21}
     \le  2^{-pN/(2a^2+a^3)+101a^3+Na}\dens_1(\mathfrak{A})\mu\left(\cup_{\fp\in\mathfrak{A}}\sc(\fp)\right)
 \end{equation}
@@ -4179,7 +3998,7 @@ \chapter{Proof of Proposition \ref{forest operator}}
 \label{treesection}
 
 \section{The pointwise tree estimate}
-Fix a forest $(\fU, \fT)$. The main result of this subsection is Lemma \ref{lem pointw tree estimate}, we begin this section with some definitions necessary to state the lemma.
+Fix a forest $(\fU, \fT)$. The main result of this subsection is Lemma \ref{pointwise tree estimate}, we begin this section with some definitions necessary to state the lemma.
 
 For $\fu \in \fU$ and $x\in X$, we define
 $$
@@ -4192,8 +4011,8 @@ \section{The pointwise tree estimate}
 $$
     \underline{\sigma} (\fu, x) := \min\sigma(\fT(\fu), x)\,.
 $$
-\begin{lemma}
-\label{lem sigma convex}
+\begin{lemma}[convex scales]
+\label{convex scales}
     For each $\fu \in \fU$, we have
     $$
         \sigma(\fu, x) = \mathbb{Z} \cap [\underline{\sigma} (\fu, x), \overline{\sigma} (\fu, x)]\,.
@@ -4220,8 +4039,8 @@ \section{The pointwise tree estimate}
 $$
 to be the collection of dyadic cubes $L \in \mathcal{D}$ such that $s(L) = -S$, or there exists $\fp \in \mathfrak{S}$ with $L \subset \sc(\fp)$ and there exists no $\fp \in \mathfrak{S}$ with $\sc(\fp) \subset L$. We define $\mathcal{L}(\mathfrak{S})$ to be the collection of inclusion maximal cubes in $\mathcal{L}_0(\mathfrak{S})$.
 
-\begin{lemma}
-    \label{lem partition}
+\begin{lemma}[dyadic partitions]
+    \label{dyadic partitions}
     For each $\mathfrak{S} \subset \fP$, we have
     \begin{equation}
         \label{eq J partition}
@@ -4267,8 +4086,9 @@ \section{The pointwise tree estimate}
 
 The following pointwise estimate for operators associated to sets $\fT(\fu)$ is the main result of this subsection.
 
-\begin{lemma}
-    \label{lem pointw tree estimate}
+\begin{lemma}[pointwise tree estimate]
+    \label{pointwise tree estimate}
+    \uses{first tree pointwise,second tree pointwise,third tree pointwise}
     Let $\fu \in \fU$ and $L \in \mathcal{L}(\fT(\fu))$. Let $x, x' \in L$.
     Then for all bounded functions $f$ with bounded support
     $$
@@ -4304,11 +4124,11 @@ \section{The pointwise tree estimate}
         \label{eq term C}
         + \Bigg| \sum_{s \in \sigma(\fu, x)} \int K_s(x,y) (f(y) - P_{\mathcal{J}(\fT(\fu))} f(y)) \, \mathrm{d}\mu(y) \Bigg|\,.
     \end{equation}
-    The proof is completed using the bounds for these three terms proven in Lemma \ref{lem term A}, Lemma \ref{lem term B} and Lemma \ref{lem term C}.
+    The proof is completed using the bounds for these three terms proven in Lemma \ref{first tree pointwise}, Lemma \ref{second tree pointwise} and Lemma \ref{third tree pointwise}.
 \end{proof}
 
-\begin{lemma}
-    \label{lem term A}
+\begin{lemma}[first tree pointwise]
+    \label{first tree pointwise}
     For all $\fu \in \fU$, all $L \in \mathcal{L}(\fT(\fu))$, all $x, x' \in L$ and all bounded $f$ with bounded support, we have
     $$
         \eqref{eq term A} \le 10 \cdot 2^{105a^3} M_{\mathcal{B}, 1}P_{\mathcal{J}(\fT(\fu))}|f|(x')\,.
@@ -4365,8 +4185,8 @@ \section{The pointwise tree estimate}
     This completes the estimate for term \eqref{eq term A}.
 \end{proof}
 
-\begin{lemma}
-    \label{lem term B}
+\begin{lemma}[second tree pointwise]
+    \label{second tree pointwise}
     For all $\fu \in \fU$, all $L \in \mathcal{L}(\fT(\fu))$, all $x, x' \in L$ and all bounded $f$ with bounded support, we have
     $$
         %\eqref{eq term B}
@@ -4379,8 +4199,8 @@ \section{The pointwise tree estimate}
     The lemma now follows from the definition of $T_{\mathcal{N}}$.
 \end{proof}
 
-\begin{lemma}
-    \label{lem term C}
+\begin{lemma}[third tree pointwise]
+    \label{third tree pointwise}
     For all $\fu \in \fU$, all $L \in \mathcal{L}(\fT(\fu))$, all $x, x' \in L$ and all bounded $f$ with bounded support, we have
     \begin{equation*}
         \Bigg| \sum_{s \in \sigma(\fu, x)} \int K_s(x,y) (f(y) - P_{\mathcal{J}(\fT(\fu))} f(y)) \, \mathrm{d}\mu(y) \Bigg|
@@ -4433,8 +4253,9 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate}
 
 In this subsection we prove the following estimate on $L^2$ for operators associated to trees.
 
-\begin{lemma}
-    \label{TreeEstimate}
+\begin{lemma}[tree projection estimate]
+    \label{tree projection estimate}
+    \uses{dyadic partitions,pointwise tree estimate,nontangential operator bound,boundary operator bound}
     Let $\fu \in \fU$.
     Then we have for all $f, g$ bounded with bounded support
     $$
@@ -4446,18 +4267,20 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate}
     \end{equation}
 \end{lemma}
 
-Below, we deduce Lemma \ref{TreeEstimate} from Lemma \ref{lem pointw tree estimate} and the following estimates for the operators in Lemma \ref{lem pointw tree estimate}.
+Below, we deduce Lemma \ref{tree projection estimate} from Lemma \ref{pointwise tree estimate} and the following estimates for the operators in Lemma \ref{pointwise tree estimate}.
 
-\begin{lemma}
-    \label{lem nontangential}
+\begin{lemma}[nontangential operator bound]
+    \label{nontangential operator bound}
+  \uses{Hardy Littlewood}
     For all bounded $f$ with bounded support
     $$
         \|T_{\mathcal{N}} f\|_2 \le 2^{103a^3} \|f\|_2\,.
     $$
 \end{lemma}
 
-\begin{lemma}
-    \label{lem L2 Su estimate}
+\begin{lemma}[boundary operator bound]
+    \label{boundary operator bound}
+    \uses{Hardy Littlewood,boundary overlap}
     For all $\fu \in \fU$ and all bounded functions $f$ with bounded support
     \begin{equation}
         \label{eq S bound}
@@ -4465,7 +4288,7 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate}
     \end{equation}
 \end{lemma}
 
-\begin{proof}[Proof of Lemma \ref{TreeEstimate}]
+\begin{proof}[Proof of Lemma \ref{tree projection estimate}]
     For each $L \in \mathcal{L}(\fT(\fu))$, choose a point $x'(L) \in L$ such that for all $y \in L$
     $$
         (M_{\mathcal{B},1}+S_{1,\fu})P_{\mathcal{J}(\fT(\fu))}|f|(x')+|T_{\mathcal{N}}P_{\mathcal{J}(\fT(\fu))}f(x')|
@@ -4475,7 +4298,7 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate}
         \le 2 ((M_{\mathcal{B},1}+S_{1,\fu})P_{\mathcal{J}(\fT(\fu))}|f|(y)+|T_{\mathcal{N}}P_{\mathcal{J}(\fT(\fu))}f(y)|)\,.
     \end{equation}
     This point exists since \eqref{eq x'L almost inf} is non-negative for each $y$.
-    Then we have by Lemma \ref{lem pointw tree estimate} for each $L \in \mathcal{L}(\fT(\fu))$
+    Then we have by Lemma \ref{pointwise tree estimate} for each $L \in \mathcal{L}(\fT(\fu))$
     $$
         \int_L |g(y)| \Bigg| \sum_{\fp \in \fT(\fu)} T_{\fp} f(y) \Bigg| \, \mathrm{d}\mu(y)
     $$
@@ -4493,7 +4316,7 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate}
     $$
         \Bigg| \int \bar g(y) \sum_{\fp \in \fT(\fu)} T_{\fp} f(y)  \, \mathrm{d}\mu(y) \Bigg| = \Bigg| \int_{\bigcup_{\fp \in \fT(\fu)} \sc(\fp)} \bar g(y) \sum_{\fp \in \fT(\fu)} T_{\fp} f(y)  \, \mathrm{d}\mu(y) \Bigg|\,.
     $$
-    Since $\mathcal{L}(\fT(\fu))$ partitions $\bigcup_{\fp \in \fT(\fu)} \sc(\fp)$ by Lemma \ref{lem partition},
+    Since $\mathcal{L}(\fT(\fu))$ partitions $\bigcup_{\fp \in \fT(\fu)} \sc(\fp)$ by Lemma \ref{dyadic partitions},
     we get from the triangle inequality
     $$
         \le \sum_{L \in \mathcal{L}(\fT(\fu))} \int_L |g(y)| \Bigg| \sum_{\fp \in \fT(\fu)} T_{\fp} f(y) \Bigg| \, \mathrm{d}\mu(y)
@@ -4509,7 +4332,7 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate}
     \begin{multline*}
         2^{151a^3 + 1} \|P_{\mathcal{L}(\fT(\fu))}|g|\|_2 \times \\(\|M_{\mathcal{B},1}P_{\mathcal{J}(\fT(\fu))}|f|\|_2 + \|S_{1,\fu}P_{\mathcal{J}(\fT(\fu))}|f|\|_2 + \|T_{\mathcal{N}}P_{\mathcal{J}(\fT(\fu))}f(y)|\|_2)\,.
     \end{multline*}
-    By Proposition \ref{Hardy Littlewood}, Lemma \ref{lem nontangential} and Lemma \ref{lem L2 Su estimate}, the second factor is at most
+    By Proposition \ref{Hardy Littlewood}, Lemma \ref{nontangential operator bound} and Lemma \ref{boundary operator bound}, the second factor is at most
     $$
         (2^{2a+1} + 2^{12a})\|P_{\mathcal{J}(\fT(\fu))}|f|\|_2 + 2^{103a^3} \|P_{\mathcal{J}(\fT(\fu))}f\|_2\,.
     $$
@@ -4522,7 +4345,7 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate}
 
 Now we prove the two auxiliary lemmas. We begin with the nontangential maximal operator $T_{\mathcal{N}}$.
 
-\begin{proof}[Proof of Lemma \ref{lem nontangential}]
+\begin{proof}[Proof of Lemma \ref{nontangential operator bound}]
     Fix $s_1, s_2$. By \eqref{eq psisum} we have for all $x \in (0, \infty)$
     $$
         \sum_{s = s_1}^{s_2} \psi(D^{-s}x) = 1 - \sum_{s < s_1} \psi(D^{-s}x) - \sum_{s > s_1} \psi(D^{-s}x)\,.
@@ -4569,14 +4392,14 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate}
     and the doubling property \eqref{doublingx} $6 +100a^2$ times estimates
     the last display by
     \begin{equation}
-    \label{pf lem nontangential imeq}
+    \label{pf nontangential operator bound imeq}
         \le \frac{2^{6a+101a^3}}{\mu(B(x',  16D^{s_1}))}\, .
     \end{equation}
     By the triangle inequality and \eqref{eq vol sp cube}, we have
     $$
         B(x', 8D^{s_1}) \subset B(c(I_{s_1}(x)), 16D^{s(I_{s_1}(x))})\,.
     $$
-    Combining this with \eqref{pf lem nontangential imeq}, we conclude that \eqref{eq lower bound term} is at most
+    Combining this with \eqref{pf nontangential operator bound imeq}, we conclude that \eqref{eq lower bound term} is at most
     $$
         2^{6a + 101a^3} M_{\mathcal{B},1} f(x)\,.
     $$
@@ -4609,8 +4432,8 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate}
 
 We need the following lemma to prepare the $L^2$-estimate for the auxiliary operators $S_{1, \fu}$.
 
-\begin{lemma}
-    \label{lem aux overlap}
+\begin{lemma}[boundary overlap]
+    \label{boundary overlap}
     For every cube $I \in \mathcal{D}$, there exist at most $2^{8a}$ cubes $J \in \mathcal{D}$ with $s(J) = s(I)$ and $B(c(I), 16D^{s(I)}) \cap B(c(J), 16 D^{s(J)}) \ne \emptyset$.
 \end{lemma}
 
@@ -4624,7 +4447,7 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate}
     If $\mathcal{C}$ is any finite collection of cubes $J \in \mathcal{D}$ satisfying $s(J) = s(I)$ and
     \begin{equation*}
         B(c(I), 16 D^{s(I)}) \cap B(c(J), 16 D^{s(J)}) \ne\emptyset\ ,
-    \end{equation*} then it follows from \eqref{eq vol sp cube} and pairwise disjointness of cubes of the same scale \eqref{dyadicproperty} that the balls $B(c(J), \frac{1}{4} D^{s(J)})$ are pairwise disjoint. Hence
+    \end{equation*} then it follows from \eqref{eq vol sp cube} and pairwise disjointedness of cubes of the same scale \eqref{dyadicproperty} that the balls $B(c(J), \frac{1}{4} D^{s(J)})$ are pairwise disjoint. Hence
     \begin{align*}
         \mu(B(c(I), 32 D^{s(I)})) &\ge \sum_{J \in \mathcal{C}} \mu(B(c(J), \frac{1}{4}D^{s(J)}))\\
         &\ge |\mathcal{C}| 2^{-8a} \mu(B(c(I), 32 D^{s(I)}))\,.
@@ -4634,7 +4457,7 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate}
 
 Now we can bound the operators $S_{1, \fu}$.
 
-\begin{proof}[Proof of Lemma \ref{lem L2 Su estimate}]
+\begin{proof}[Proof of Lemma \ref{boundary operator bound}]
     Note that by definition, $S_{1,\fu}f$ is a finite sum of indicator functions of cubes $I \in \mathcal{D}$ for each locally integrable $f$, and hence is bounded, has bounded support and is integrable. Let $g$ be another function with the same three properties. Then $\bar g S_{1,\fu}f$ is integrable, and we have
     $$
         \Bigg|\int \bar g(y) S_{1,\fu}f(y) \, \mathrm{d}\mu(y)\Bigg|
@@ -4648,18 +4471,17 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate}
     \end{multline*}
     Changing the order of summation and using $J \subset B(c(I), 16 D^{s(I)})$ to bound the first average integral by $M_{\mathcal{B},1}|g|(y)$ for any $y \in J$, we obtain
     \begin{align}
-    \label{eq lem L2 Su estimate 1}
+    \label{eq boundary operator bound 1}
         \le \sum_{J\in\mathcal{J}}\int_J|f(y)|  M_{\mathcal{B},1}|g|(y) \, \mathrm{d}\mu(y) \sum_{I \in \mathcal{D} \, : \, J\subset B(c(I),16 D^{s(I)})} D^{(s(J)-s(I))/a}.
     \end{align}
-    By \eqref{eq vol sp cube} and \eqref{defineD} the condition $J \subset B(c(I), 16 D^{s(I)})$ implies $s(I) \ge s(J)$. By Lemma \ref{lem aux overlap}, there are at most $2^{8a}$ cubes $I$ at each scale with $J \subset B(c(I), D^{s(I)})$.
+    By \eqref{eq vol sp cube} and \eqref{defineD} the condition $J \subset B(c(I), 16 D^{s(I)})$ implies $s(I) \ge s(J)$. By Lemma \ref{boundary overlap}, there are at most $2^{8a}$ cubes $I$ at each scale with $J \subset B(c(I), D^{s(I)})$.
     By convexity of $t \mapsto D^t$ and since $D \ge 2$, we have for all $-1 \le t \le 0$
     $$
         D^t \le 1 + t\left(1 - \frac{1}{D}\right) \le 1 + \frac{1}{2}t\,,
     $$
     so $(1 - D^{-1/a})^{-1} \le 2a \le 2^a$.
-    \rs{Ds convex}
     Using this estimate for the sum of the geometric series,
-    we conclude that \eqref{eq lem L2 Su estimate 1} is at most
+    we conclude that \eqref{eq boundary operator bound 1} is at most
     $$
         2^{9a} \sum_{J\in\mathcal{J}}\int_J|f(y)|  M_{\mathcal{B},1}|g|(y) \, \mathrm{d}\mu(y)\,.
     $$
@@ -4679,8 +4501,9 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate}
 
 The main result of this subsection is the following quantitative bound for operators associated to trees, with decay in the the densities $\dens_1$ and $\dens_2$.
 
-\begin{lemma}
-    \label{lem tree est}
+\begin{lemma}[densities tree bound]
+    \label{densities tree bound}
+    \uses{tree projection estimate,local dens1 tree bound,local dens2 tree bound}
     Let $\fu \in \fU$. Then for all $f,g$ bounded with bounded support
     \begin{equation}
         \label{eq cor tree est}
@@ -4693,10 +4516,11 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate}
     \end{equation}
 \end{lemma}
 
-Below, we deduce this lemma from  Lemma \ref{TreeEstimate} and the following two estimates controlling the size of support of the operator and its adjoint.
+Below, we deduce this lemma from  Lemma \ref{tree projection estimate} and the following two estimates controlling the size of support of the operator and its adjoint.
 
-\begin{lemma}
-    \label{lem 1density estimate tree}
+\begin{lemma}[local dens1 tree bound]
+    \label{local dens1 tree bound}
+    \uses{monotone cube metrics}
     Let $\fu \in \fU$ and $L \in \mathcal{L}(\fT(\fu))$. Then
     \begin{equation}
     \label{eq 1density estimate tree}
@@ -4704,15 +4528,15 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate}
     \end{equation}
 \end{lemma}
 
-\begin{lemma}
-    \label{lem 2density estimate tree}
+\begin{lemma}[local dens2 tree bound]
+    \label{local dens2 tree bound}
     Let $J \in \mathcal{J}(\fT(\fu))$ be such that there exist $\fq \in \fT(\fu)$ with $J \cap \sc(\fq) \ne \emptyset$. Then
     $$
         \mu(F \cap J) \le 2^{200a^3 + 19} \dens_2(\fT(\fu))\,.
     $$
 \end{lemma}
 
-\begin{proof}[Proof of Lemma \ref{lem tree est}]
+\begin{proof}[Proof of Lemma \ref{densities tree bound}]
     Denote
     $$
         \mathcal{E}(\fu) = \bigcup_{\fp \in \fT(\fu)} E(\fp)\,.
@@ -4721,7 +4545,7 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate}
     $$
         \left| \int_X \bar g \sum_{\fp \in \fT(\fu)} T_{\fp} f \, \mathrm{d}\mu \right|  = \left| \int_X \overline{ g\mathbf{1}_{\mathcal{E}(\fu)}}  \sum_{\fp \in \fT(\fu)} T_{\fp} f \, \mathrm{d}\mu \right|\,.
     $$
-    By Lemma \ref{TreeEstimate}, this is bounded by
+    By Lemma \ref{tree projection estimate}, this is bounded by
     \begin{equation}
         \label{eq both factors tree}
         \le 2^{104a^3}\|P_{\mathcal{J}(\fT(\fu))}|f|\|_2 \|P_{\mathcal{L}(\fT(\fu))} |\mathbf{1}_{\mathcal{E}(\fu)}g|\|_2\,.
@@ -4731,11 +4555,11 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate}
     $$
         \|P_{\mathcal{L}(\fT(\fu))} |\mathbf{1}_{\mathcal{E}(\fu)}g|\|_2 = \left( \sum_{L \in \mathcal{L}(\fT(\fu))} \frac{1}{\mu(L)} \left(\int_{L \cap \mathcal{E}(\fu)} |g(y)| \, \mathrm{d}\mu(y)\right)^2 \right)^{1/2}\,.
     $$
-    By Cauchy-Schwarz and Lemma \ref{lem 1density estimate tree} this is at most
+    By Cauchy-Schwarz and Lemma \ref{local dens1 tree bound} this is at most
     $$
         \le \left( \sum_{L \in \mathcal{L}(\fT(\fu))} 2^{101a^3} \dens_1(\fT(\fu)) \int_{L \cap \mathcal{E}(\fu)} |g(y)|^2 \, \mathrm{d}\mu(y) \right)^{1/2}\,.
     $$
-    Since cubes $L \in \mathcal{L}(\fT(\fu))$ are pairwise disjoint by Lemma \ref{lem partition}, this is
+    Since cubes $L \in \mathcal{L}(\fT(\fu))$ are pairwise disjoint by Lemma \ref{dyadic partitions}, this is
     \begin{equation}
         \label{eq factor L tree}
          \le 2^{51 a^3} \dens_1(\fT(\fu))^{1/2} \|g\|_2\,.
@@ -4750,7 +4574,7 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate}
     $$
         \le \left( \sum_{J \in \mathcal{J}(\fT(\fu))} \int_J |f(y)|^2 \, \mathrm{d}\mu(y) \right)^{1/2}\,.
     $$
-    Since cubes in $\mathcal{J}(\fT(\fu))$ are pairwise disjoint by Lemma \ref{lem partition}, this at most
+    Since cubes in $\mathcal{J}(\fT(\fu))$ are pairwise disjoint by Lemma \ref{dyadic partitions}, this at most
     \begin{equation}
         \label{eq factor J tree}
         \|f\|_2\,.
@@ -4761,7 +4585,7 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate}
     $$
         \left( \sum_{J \in \mathcal{J}(\fT(\fu))} \int_J |f(y)|^2 \, \mathrm{d}\mu(y) \right)^{1/2} = \left( \sum_{J \in \mathcal{J}(\fT(\fu))} \int_{J \cap F} |f(y)|^2 \, \mathrm{d}\mu(y) \right)^{1/2}\,.
     $$
-    We estimate as before, using now Lemma \ref{lem 2density estimate tree} and Cauchy-Schwarz, that the right hand side is
+    We estimate as before, using now Lemma \ref{local dens2 tree bound} and Cauchy-Schwarz, that the right hand side is
     $$
         \le 2^{100a^3 + 10} \dens_2(\fT(\fu))^{1/2} \|f\|_2\,.
     $$
@@ -4770,7 +4594,7 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate}
 
 Now we prove the two auxiliary estimates.
 
-\begin{proof}[Proof of Lemma \ref{lem 1density estimate tree}]
+\begin{proof}[Proof of Lemma \ref{local dens1 tree bound}]
     If the set on the right hand side is empty, then \eqref{eq 1density estimate tree} holds. If not, then there exists $\fp \in \fT(\fu)$ with $L \cap \sc(\fp) \ne \emptyset$.
 
     Suppose first that there exists such $\fp$ with $\ps(\fp) \le s(L)$. Then by \eqref{dyadicproperty} $\sc(\fp) \subset L$, which gives by the definition of $\mathcal{L}(\fT(\fu))$ that $s(L) = -S$ and hence $L = \sc(\fp)$. Let $\fq \in \fT(\fu)$ with $E(\fq) \cap L \ne \emptyset$. Since $s(L) = -S \le \ps(\fq)$ it follows from \eqref{dyadicproperty} that $\sc(\fp) = L \subset \sc(\fq)$. We have then by Lemma \ref{monotone cube metrics}
@@ -4807,7 +4631,7 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate}
     This completes the proof.
 \end{proof}
 
-\begin{proof}[Proof of Lemma \ref{lem 2density estimate tree}]
+\begin{proof}[Proof of Lemma \ref{local dens2 tree bound}]
     Suppose first that there exists a tile $\fp \in \fT(\fu)$ with $\sc(\fp) \subset B(c(J), 100 D^{s(J) + 1})$. By the definition of $\mathcal{J}(\fT(\fu))$, this implies that $s(J) = -S$, and in particular $\ps(\fp) \ge s(J)$. Using the triangle inequality and \eqref{eq vol sp cube} it follows that $J \subset B(\pc(\fp), 200 D^{\ps(\fp) + 1})$. From the doubling property \eqref{doublingx}, $D=2^{100a^2}$ and \eqref{eq vol sp cube}, we obtain
     $$
         \mu(\sc(\fp)) \le 2^{100a^3 + 9} \mu(J)
@@ -4860,7 +4684,7 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate}
 
 \section{Almost orthogonality of separated trees}
 
-The main result of this subsection is the almost orthogonality estimate for operators associated to distinct trees in a forest in Lemma \ref{SeparatedTrees} below. We will deduce this Lemma from Lemmas \ref{lem big sep tree} and \ref{lem small sep tree}, which are proven in Subsections \ref{subsec big tiles} and \ref{subsec rest tiles}, respectively. Before stating the lemma, we introduce some relevant notation.
+The main result of this subsection is the almost orthogonality estimate for operators associated to distinct trees in a forest in Lemma \ref{correlation separated trees} below. We will deduce this Lemma from Lemmas \ref{correlation distant tree parts} and \ref{correlation near tree parts}, which are proven in Subsections \ref{subsec big tiles} and \ref{subsec rest tiles}, respectively. Before stating the lemma, we introduce some relevant notation.
 
 The adjoint of the operator $T_{\fp}$ defined in \eqref{definetp} is given by
 \begin{equation}
@@ -4868,8 +4692,8 @@ \section{Almost orthogonality of separated trees}
     T_{\fp}^* g(x) = \int_{E(\fp)} \overline{K_{\ps(\fp)}(y,x)} e(-\tQ(y)(x)+ \tQ(y)(y)) g(y) \, \mathrm{d}\mu(y)\,.
 \end{equation}
 
-\begin{lemma}
-    \label{lem Tp support adjoint}
+\begin{lemma}[adjoint tile support]
+    \label{adjoint tile support}
     For each $\fp \in \fP$, we have
     $$
         T_{\fp}^* g = \mathbf{1}_{B(\pc(\fp), 5D^{\ps(\fp)})} T_{\fp}^* \mathbf{1}_{\sc(\fp)} g\,.
@@ -4899,8 +4723,9 @@ \section{Almost orthogonality of separated trees}
     The second claimed equation follows now since $\sc(\fp) \subset \sc(\fu)$ and by \eqref{forest6} $B(\pc(\fp), 5D^{\ps(\fp)}) \subset \sc(\fu)$.
 \end{proof}
 
-\begin{lemma}
-    \label{lem tree adjoint bound}
+\begin{lemma}[adjoint tree estimate]
+    \label{adjoint tree estimate}
+    \uses{densities tree bound}
     For all bounded $g$ with bounded support, we have that
     $$
         \left\| \sum_{\fp \in \fT(\fu)} T_{\fp}^* g\right\|_2 \le 2^{155a^3} \dens_1(\fT(\fu))^{1/2} \|g\|_2\,.
@@ -4908,7 +4733,7 @@ \section{Almost orthogonality of separated trees}
 \end{lemma}
 
 \begin{proof}
-    By Cauchy-Schwarz and Lemma \ref{lem tree est}, we have for all bounded $f,g$ with bounded support that
+    By Cauchy-Schwarz and Lemma \ref{densities tree bound}, we have for all bounded $f,g$ with bounded support that
     $$
         \left| \int_X \overline{\sum_{\fp\in \fT(\fu)} T_{\fp}^* g} f \,\mathrm{d}\mu \right| = \left| \int_X g \sum_{\fp \in \fT(\fu)} T_{\fp} f \,\mathrm{d}\mu \right|
     $$
@@ -4923,8 +4748,9 @@ \section{Almost orthogonality of separated trees}
 $$
     S_{2,\fu}g := \left|\sum_{\fp \in \fT(\fu)} T_{\fp}^*g \right| + M_{\mathcal{B},1}g + |g|\,.
 $$
-\begin{lemma}
-    \label{lem L2 Sg estimate}
+\begin{lemma}[adjoint tree control]
+    \label{adjoint tree control}
+    \uses{Hardy Littlewood,adjoint tree estimate}
     We have for all bounded $g$ with bounded support
     $$
         \|S_{2, \fu} g\|_2 \le 2^{156a^3} \|g\|_2\,.
@@ -4932,14 +4758,15 @@ \section{Almost orthogonality of separated trees}
 \end{lemma}
 
 \begin{proof}
-    This follows immediately from Minkowski's inequality, Proposition \ref{Hardy Littlewood} and Lemma \ref{lem tree adjoint bound}, using that $a \ge 4$.
+    This follows immediately from Minkowski's inequality, Proposition \ref{Hardy Littlewood} and Lemma \ref{adjoint tree estimate}, using that $a \ge 4$.
 \end{proof}
 
 
 Now we are ready to state the main result of this subsection.
 
-\begin{lemma}
-    \label{SeparatedTrees}
+\begin{lemma}[correlation separated trees]
+    \label{correlation separated trees}
+    \uses{adjoint tile support,correlation distant tree parts,correlation near tree parts}
     For any $\fu_1 \ne \fu_2 \in \fU$ and all bounded $g_1, g_2$ with bounded support, we have
     \begin{equation}
         \label{eq lhs sep tree}
@@ -4952,17 +4779,17 @@ \section{Almost orthogonality of separated trees}
     \rs{Constant changed from $2^{550a^3}$, check}
 \end{lemma}
 
-\begin{proof}[Proof of Lemma \ref{SeparatedTrees}]
-    By Lemma \ref{lem Tp support adjoint} and \eqref{dyadicproperty}, the left hand side \eqref{eq lhs sep tree} is $0$ unless $\sc(\fu_1) \subset \sc(\fu_2)$ or $\sc(\fu_2) \subset \sc(\fu_1)$. Without loss of generality we assume that $\sc(\fu_1) \subset \sc(\fu_2)$.
+\begin{proof}[Proof of Lemma \ref{correlation separated trees}]
+    By Lemma \ref{adjoint tile support} and \eqref{dyadicproperty}, the left hand side \eqref{eq lhs sep tree} is $0$ unless $\sc(\fu_1) \subset \sc(\fu_2)$ or $\sc(\fu_2) \subset \sc(\fu_1)$. Without loss of generality we assume that $\sc(\fu_1) \subset \sc(\fu_2)$.
 
     Define
     \begin{equation}
         \label{def Tree S set}
          \mathfrak{S} := \{\fp \in \fT(\fu_1) \cup \fT(\fu_2) \ : \ d_{\fp}(\fcc(\fu_1), \fcc(\fu_2)) \ge 2^{Zn/2}\,\}.
     \end{equation}
-    Lemma \ref{SeparatedTrees} follows by combining the definition \eqref{defineZ} of $Z$ with the following two lemmas.
-    \begin{lemma}
-        \label{lem big sep tree}
+    Lemma \ref{correlation separated trees} follows by combining the definition \eqref{defineZ} of $Z$ with the following two lemmas.
+    \begin{lemma}[correlation distant tree parts]
+        \label{correlation distant tree parts}
         We have for all $\fu_1 \ne \fu_2 \in \fU$ with $\sc(\fu_1) \subset \sc(\fu_2)$ and all bounded $g_1, g_2$ with bounded support
         \begin{equation}
             \label{eq lhs big sep tree}
@@ -4974,8 +4801,8 @@ \section{Almost orthogonality of separated trees}
         \end{equation}
         \rs{Constant changed from $2^{550a^3}$, check}
     \end{lemma}
-    \begin{lemma}
-        \label{lem small sep tree}
+    \begin{lemma}[correlation near tree parts]
+        \label{correlation near tree parts}
         We have for all $\fu_1 \ne \fu_2 \in \fU$ with $\sc(\fu_1) \subset \sc(\fu_2)$ and all bounded $g_1, g_2$ with bounded support
         \begin{equation}
             \label{eq lhs small sep tree}
@@ -4990,8 +4817,8 @@ \section{Almost orthogonality of separated trees}
 
 In the proofs of both lemmas, we will need the following observation.
 
-\begin{lemma}
-    \label{lem T1 S}
+\begin{lemma}[overlap implies distance]
+    \label{overlap implies distance}
     Let $\fu_1 \ne \fu_2 \in \fU$ with $\sc(\fu_1) \subset \sc(\fu_2)$. If $\fp \in \fT(\fu_1) \cup \fT(\fu_2)$ with $\sc(\fp) \cap \sc(\fu_1) \ne \emptyset$, then $\fp \in \mathfrak{S}$. In particular, we have $\fT(\fu_1) \subset \mathfrak{S}$.
 \end{lemma}
 
@@ -5011,15 +4838,15 @@ \section{Almost orthogonality of separated trees}
     so $\fp \in \mathfrak{S}$.
 \end{proof}
 
-To simplify the notation, we will write at various places throughout the proof of Lemmas \ref{lem big sep tree} and \ref{lem small sep tree} for a subset $\fC \subset \fP$
+To simplify the notation, we will write at various places throughout the proof of Lemmas \ref{correlation distant tree parts} and \ref{correlation near tree parts} for a subset $\fC \subset \fP$
 $$
     T_{\fC} f := \sum_{\fp \in \fC} T_{\fp} f\,, \quad\quad T_{\fC}^* g := \sum_{\fp\in\fC} T_{\fp}^* g\,.
 $$
 
-\section{Proof of Lemma \ref{lem big sep tree}: Tiles with large separation}
+\section{Proof of Lemma \ref{correlation distant tree parts}: Tiles with large separation}
     \label{subsec big tiles}
 
-Lemma \ref{lem big sep tree} follows from the van der Corput estimate in Proposition \ref{Holder van der Corput}. We apply this proposition in Subsubsection \ref{subsubsec van der corput}. To prepare this application, we first, in Subsubsection \ref{subsubsec pao}, construct a suitable partition of unity, and show then, in Subsubsection \ref{subsubsec holder estimates} the Hölder estimates needed to apply Proposition \ref{Holder van der Corput}.
+Lemma \ref{correlation distant tree parts} follows from the van der Corput estimate in Proposition \ref{Holder van der Corput}. We apply this proposition in Subsubsection \ref{subsubsec van der corput}. To prepare this application, we first, in Subsubsection \ref{subsubsec pao}, construct a suitable partition of unity, and show then, in Subsubsection \ref{subsubsec holder estimates} the Hölder estimates needed to apply Proposition \ref{Holder van der Corput}.
 
 \subsubsection{A partition of unity}
 \label{subsubsec pao}
@@ -5037,7 +4864,7 @@ \subsubsection{A partition of unity}
     \end{lemma}
 
     \begin{proof}
-        By Lemma \ref{lem partition}, it remains only to show that each $J \in \mathcal{J}(\mathfrak{S})$ with $J \cap \sc(\fu_1) \ne \emptyset$ is in $\mathcal{J}'$. But if $J \notin \mathcal{J}'$, then by \eqref{dyadicproperty} $\sc(\fu_1) \subsetneq J$. Pick $\fp \in \fT(\fu_1) \subset \mathfrak{S}$.  Then $\sc(\fp) \subsetneq J$. This contradicts the definition of $\mathcal{J}(\mathfrak{S})$.
+        By Lemma \ref{dyadic partitions}, it remains only to show that each $J \in \mathcal{J}(\mathfrak{S})$ with $J \cap \sc(\fu_1) \ne \emptyset$ is in $\mathcal{J}'$. But if $J \notin \mathcal{J}'$, then by \eqref{dyadicproperty} $\sc(\fu_1) \subsetneq J$. Pick $\fp \in \fT(\fu_1) \subset \mathfrak{S}$.  Then $\sc(\fp) \subsetneq J$. This contradicts the definition of $\mathcal{J}(\mathfrak{S})$.
     \end{proof}
 
     For cubes $J \in \mathcal{D}$, denote
@@ -5248,7 +5075,7 @@ \subsubsection{Hölder estimates for adjoint tree operators}
             \eqref{T*Hölder1b}+\eqref{T*Hölder1} \le \frac{2^{151a^3}}{\mu(B(\pc(\fp), 4D^{\ps(\fp)}))} \left(\frac{\rho(y,y')}{D^{\ps(\fp)}}\right)^{1/a} \int_{E(\fp)}|g(x)| \, \mathrm{d}\mu(x)\,.
         $$
 
-        Next, if $y,y' \notin B(\pc(\fp), 5D^{\ps(\fp)})$, then $T_{\fp}^*g(y) = T_{\fp}^*g(y') = 0$, by Lemma \ref{lem Tp support adjoint}. Then \eqref{T*Hölder2} holds.
+        Next, if $y,y' \notin B(\pc(\fp), 5D^{\ps(\fp)})$, then $T_{\fp}^*g(y) = T_{\fp}^*g(y') = 0$, by Lemma \ref{adjoint tile support}. Then \eqref{T*Hölder2} holds.
 
         Finally, if $y \in B(\pc(\fp), 5D^{\ps(\fp)})$ and $y' \notin B(\pc(\fp), 5D^{\ps(\fp)})$, then
         $$
@@ -5323,7 +5150,7 @@ \subsubsection{Hölder estimates for adjoint tree operators}
     \end{lemma}
 
     \begin{proof}
-        For the first estimate, assume that $\ps(\fp) < s(J)$, then in particular $\ps(\fp) \le \ps(\fu_1)$. Since $\fp \notin \mathfrak{S}$, we have by Lemma \ref{lem T1 S} that $\sc(\fp) \cap \sc(\fu_1) = \emptyset$.
+        For the first estimate, assume that $\ps(\fp) < s(J)$, then in particular $\ps(\fp) \le \ps(\fu_1)$. Since $\fp \notin \mathfrak{S}$, we have by Lemma \ref{overlap implies distance} that $\sc(\fp) \cap \sc(\fu_1) = \emptyset$.
         %by \eqref{dyadicproperty}.
         Since $B\Big(c(J), \frac{1}{4} D^{s(J)}\Big) \subset \sc(J) \subset \sc(\fu_1)$, this implies
         $$
@@ -5405,7 +5232,7 @@ \subsubsection{Hölder estimates for adjoint tree operators}
     \end{lemma}
 
     \begin{proof}
-        By Lemma \ref{lem T1 S}, we have that in both cases, $\fC \subset \mathfrak{S}$. If $\fp \in \fC$ with $B(\sc(\fp)) \cap B(J) \neq \emptyset$ and $\ps(\fp) < s(J)$, then $\sc(\fp) \subset B(c(J), 100 D^{s(J) + 1})$. Since $\fp \in \mathfrak{S}$, it follows from the definition of $\mathcal{J}'$ that $s(J) = -S$, which contradicts $\ps(\fp) < s(J)$.
+        By Lemma \ref{overlap implies distance}, we have that in both cases, $\fC \subset \mathfrak{S}$. If $\fp \in \fC$ with $B(\sc(\fp)) \cap B(J) \neq \emptyset$ and $\ps(\fp) < s(J)$, then $\sc(\fp) \subset B(c(J), 100 D^{s(J) + 1})$. Since $\fp \in \mathfrak{S}$, it follows from the definition of $\mathcal{J}'$ that $s(J) = -S$, which contradicts $\ps(\fp) < s(J)$.
     \end{proof}
 
     \begin{lemma}
@@ -5429,7 +5256,7 @@ \subsubsection{Hölder estimates for adjoint tree operators}
         Note that \eqref{TreeUB} follows from \eqref{TreeHölder}, since for $y'\in B^\circ{}(J)$, by the triangle inequality,
         $$\left(\frac{\rho(y,y')}{D^{s(J)}}\right)^{1/a}\le \Big(8 + \frac{1}8\Big)^{1/a}\le 2^{a^3}.$$
 
-        By the triangle inequality, Lemma \ref{lem Tp support adjoint} and Lemma \ref{lem tile Hölder}, we have for all $y, y' \in B(J)$
+        By the triangle inequality, Lemma \ref{adjoint tile support} and Lemma \ref{lem tile Hölder}, we have for all $y, y' \in B(J)$
         \begin{equation}
             \label{eq C Lip}
             |e(\fcc(\fu)(y)) T_{\fC}^* g(y) - e(\fcc(\fu)(y')) T_{\fC}^* g(y')|
@@ -5447,7 +5274,6 @@ \subsubsection{Hölder estimates for adjoint tree operators}
         $$
             \le 2^{152a^3} \left(\frac{\rho(y,y')}{D^s}\right)^{1/a} \sum_{S \ge s \ge s(J)} D^{(s(J) - s)/a} \inf_J M_{\mathcal{B},1} |g|\,.
         $$
-        \rs{Ds convex}
         By convexity of $t \mapsto D^t$ and since $D \ge 2$, we have for all $-1 \le t \le 0$
         $$
             D^t \le 1 + t(1 - \frac{1}{D}) \le 1 + \frac{1}{2}t\,.
@@ -5487,7 +5313,7 @@ \subsubsection{Hölder estimates for adjoint tree operators}
     \begin{proof}[Proof of Lemma \ref{lem hHölder}]
         Let $P$ be the product on the right hand side of \eqref{hHölder}, and $h_J$ as defined in \eqref{def hj}.
 
-        By \eqref{eq pao 2} and Lemma \ref{lem Tp support adjoint}, the function $h_J$ is supported in $B(J) \cap \sc(\fu_1)$.
+        By \eqref{eq pao 2} and Lemma \ref{adjoint tile support}, the function $h_J$ is supported in $B(J) \cap \sc(\fu_1)$.
         By \eqref{eq pao 2} and Lemma \ref{lem sep tree aux 1}, we have for all $y \in B(J)$:
         $$
             |h_J(y)| \le 2^{308a^3} P\,.
@@ -5536,7 +5362,7 @@ \subsubsection{The van der Corput estimate}
     \end{lemma}
 
     \begin{proof}
-    Since $\emptyset \ne \fT(\fu_1) \subset \mathfrak{S}$ by Lemma \ref{lem T1 S}, there exists at least one tile $\fp \in \mathcal{S}$ with $\sc(\fp) \subsetneq \sc(\fu_1)$. Thus $\sc(\fu_1) \notin \mathcal{J}'$, so $J \subsetneq \sc(\fu_1)$. Thus there exists a cube $J' \in \mathcal{D}$ with $J \subset J'$ and $s(J') = s(J) + 1$, by \eqref{coverdyadic} and \eqref{dyadicproperty}. By definition of $\mathcal{J'}$ and the triangle inequality, there exists $\fp \in \mathfrak{S}$ such that
+    Since $\emptyset \ne \fT(\fu_1) \subset \mathfrak{S}$ by Lemma \ref{overlap implies distance}, there exists at least one tile $\fp \in \mathcal{S}$ with $\sc(\fp) \subsetneq \sc(\fu_1)$. Thus $\sc(\fu_1) \notin \mathcal{J}'$, so $J \subsetneq \sc(\fu_1)$. Thus there exists a cube $J' \in \mathcal{D}$ with $J \subset J'$ and $s(J') = s(J) + 1$, by \eqref{coverdyadic} and \eqref{dyadicproperty}. By definition of $\mathcal{J'}$ and the triangle inequality, there exists $\fp \in \mathfrak{S}$ such that
     $$
         \sc(\fp) \subset B(c(J'), 100 D^{s(J') + 1}) \subset B(c(J), 128 D^{s(J)+2})\,.
     $$
@@ -5551,13 +5377,13 @@ \subsubsection{The van der Corput estimate}
     which gives the lemma using $a \ge 4$.
     \end{proof}
 
-    Now we are ready to prove Lemma \ref{lem big sep tree}.
-    \begin{proof}[Proof of Lemma \ref{lem big sep tree}]
+    Now we are ready to prove Lemma \ref{correlation distant tree parts}.
+    \begin{proof}[Proof of Lemma \ref{correlation distant tree parts}]
     We have
     $$
         \eqref{eq lhs big sep tree} = \left| \int_{X} T_{\fT(\fu_1)}^* g_1 \overline{T_{\fT(\fu_2) \cap \mathfrak{S}}^* g_2 }\right|\,.
     $$
-    By Lemma \ref{lem Tp support adjoint}, the right hand side is supported in $\sc(\fu_1)$. Using \eqref{eq pao 1} and the definition \eqref{def hj} of $h_J$, we thus have
+    By Lemma \ref{adjoint tile support}, the right hand side is supported in $\sc(\fu_1)$. Using \eqref{eq pao 1} and the definition \eqref{def hj} of $h_J$, we thus have
     $$
         \le  \sum_{J \in \mathcal{J}'} \left|\int_{B(J)} e(\fcc(\fu_2)(y) - \fcc(\fu_1)(y)) h_J(y) \, \mathrm{d}\mu(y) \right|\,.
     $$
@@ -5590,16 +5416,16 @@ \subsubsection{The van der Corput estimate}
     $$
         \eqref{eq big sep 1} \le 2^{650a^3} 2^{-Zn/(4a^2 + 2a^3)} \int_X \prod_{j=1}^2 ( |T_{\fT(\fu_j)}^* g_j|(x) +  M_{\mathcal{B},1} g_j(x)) \, \mathrm{d}\mu(x)\,.
     $$
-    Applying the Cauchy-Schwarz inequality, Lemma \ref{lem big sep tree} follows.
+    Applying the Cauchy-Schwarz inequality, Lemma \ref{correlation distant tree parts} follows.
     \end{proof}
 
-\section{Proof of Lemma \ref{lem small sep tree}: The remaining tiles}
+\section{Proof of Lemma \ref{correlation near tree parts}: The remaining tiles}
     \label{subsec rest tiles}
-    Recall that
+    We define
     $$
-        \mathcal{J}' = \{J \in \mathcal{J}(\fT(\fu_1)) \, : \, J \subset \sc(\fu_1)\}\,.
+        \mathcal{J}' = \{J \in \mathcal{J}(\fT(\fu_1)) \, : \, J \subset \sc(\fu_1)\}\,,
     $$
-    \rs{This lemma has been proven before}
+    note that this is different from the $\mathcal{J}'$ defined in the previous subsection.
     \begin{lemma}
         \label{lem J' partition 2}
         We have
@@ -5609,10 +5435,10 @@ \section{Proof of Lemma \ref{lem small sep tree}: The remaining tiles}
     \end{lemma}
 
     \begin{proof}
-        By Lemma \ref{lem partition}, it remains only to show that each $J \in \mathcal{J}(\fT(\fu_1))$ with $J \cap \sc(\fu_1) \ne \emptyset$ is in $\mathcal{J}'$. But if $J \notin \mathcal{J}'$, then by \eqref{dyadicproperty} $\sc(\fu_1) \subsetneq J$. Pick $\fp \in \fT(\fu_1)$.  Then $\sc(\fp) \subsetneq J$. This contradicts the definition of $\mathcal{J}(\fT(\fu_1))$.
+        By Lemma \ref{dyadic partitions}, it remains only to show that each $J \in \mathcal{J}(\fT(\fu_1))$ with $J \cap \sc(\fu_1) \ne \emptyset$ is in $\mathcal{J}'$. But if $J \notin \mathcal{J}'$, then by \eqref{dyadicproperty} $\sc(\fu_1) \subsetneq J$. Pick $\fp \in \fT(\fu_1)$.  Then $\sc(\fp) \subsetneq J$. This contradicts the definition of $\mathcal{J}(\fT(\fu_1))$.
     \end{proof}
 
-    Lemma \ref{lem small sep tree} follows from the following key estimate.
+    Lemma \ref{correlation near tree parts} follows from the following key estimate.
 
     \begin{lemma}
         \label{lem sep tree small 1}
@@ -5623,10 +5449,10 @@ \section{Proof of Lemma \ref{lem small sep tree}: The remaining tiles}
         $$
     \end{lemma}
 
-    We prove this lemma below. First, we deduce Lemma \ref{lem small sep tree}.
+    We prove this lemma below. First, we deduce Lemma \ref{correlation near tree parts}.
 
-    \begin{proof}[Proof of Lemma \ref{lem small sep tree}]
-        By Lemma \ref{TreeEstimate} and Lemma \ref{lem Tp support adjoint}, we have
+    \begin{proof}[Proof of Lemma \ref{correlation near tree parts}]
+        By Lemma \ref{tree projection estimate} and Lemma \ref{adjoint tile support}, we have
         \begin{align*}
             \eqref{eq lhs small sep tree} \le 2^{104a^3} \|P_{\mathcal{L}(\fT(\fu_1))} |g_1\mathbf{1}_{\sc(\fu_1)}| \|_2 \|P_{\mathcal{J}(\fT(\fu_1) )}|T_{\fT(\fu_2) \setminus \mathfrak{S}}^* g_2|\|_2\,.
         \end{align*}
@@ -5733,13 +5559,13 @@ \section{Proof of Lemma \ref{lem small sep tree}: The remaining tiles}
     $$
         \le \sum_{s = -S}^S \left( \sum_{J \in \mathcal{J}'} \frac{1}{\mu(J)} \left|\int_J \sum_{\substack{\fp \in \fT(\fu_2) \setminus \mathfrak{S}\\ \ps(\fp) = s}} T_{\fp}^* g_2 \, \mathrm{d}\mu(y) \right|^2\right)^{1/2}\,.
     $$
-    By Lemma \ref{lem Tp support adjoint}, the integral in the last display is $0$ if $J \cap B(\sc(\fp)) = \emptyset$. By Lemma \ref{lem sep aux 3}, it follows with $s_1 := \frac{Zn}{202a^3} - 2$:
+    By Lemma \ref{adjoint tile support}, the integral in the last display is $0$ if $J \cap B(\sc(\fp)) = \emptyset$. By Lemma \ref{lem sep aux 3}, it follows with $s_1 := \frac{Zn}{202a^3} - 2$:
     \begin{equation}
     \label{eq sep tree small 1}
         = \sum_{s = s_1}^{s_1 + 2S} \Bigg( \sum_{J \in \mathcal{J}'} \frac{1}{\mu(J)} \Bigg|\int_J \sum_{\substack{\fp \in \fT(\fu_2) \setminus \mathfrak{S}\\ \ps(\fp) = s(J) - s\\
         J \cap B(\sc(\fp)) \ne \emptyset}} T_{\fp}^* g_2 \, \mathrm{d}\mu(y) \Bigg|^2\Bigg)^{1/2}\,.
     \end{equation}
-    We have by Lemma \ref{lem Tp support adjoint} and \eqref{eq Ks size}
+    We have by Lemma \ref{adjoint tile support} and \eqref{eq Ks size}
     $$
         \int |T_{\fp}^* g_2|(y) \, \mathrm{d}\mu(y)
     $$
@@ -5772,7 +5598,7 @@ \section{Proof of Lemma \ref{lem small sep tree}: The remaining tiles}
     $$
         \le 2^{103a^3} \int_J M_{\mathcal{B},1} |g_2|(y) \mathbf{1}_{B(I)}(y) \, \mathrm{d}\mu(y)\,.
     $$
-    By Lemma \ref{lem T1 S}, we have $\sc(\fp) \cap \sc(\fu_1) = \emptyset$ for all $\fp \in \fT(\fu_2) \setminus \mathfrak{S}$.
+    By Lemma \ref{overlap implies distance}, we have $\sc(\fp) \cap \sc(\fu_1) = \emptyset$ for all $\fp \in \fT(\fu_2) \setminus \mathfrak{S}$.
     Thus we can estimate \eqref{eq sep tree small 1} by
     $$
         2^{103a^3} \sum_{s = s_1}^{s_1 + 2S} \Bigg( \sum_{J \in \mathcal{J}'} \frac{1}{\mu(J)} \Bigg|\int_J \sum_{\substack{I \in \mathcal{D}, s(I) = s(J) - s\\ I \cap \sc(\fu_1) = \emptyset\\
@@ -5795,7 +5621,6 @@ \section{Proof of Lemma \ref{lem small sep tree}: The remaining tiles}
     $$
         \le 2^{116a^3} D^{-s_1 \kappa /2} \frac{1}{1 - D^{-\kappa/2}} \|\mathbf{1}_{\sc(\fu_1)} M_{\mathcal{B},1} |g_2|\|_2\,.
     $$
-    \rs{Ds convex}
     By convexity of $t \mapsto D^t$ and since $D \ge 2$, we have for all $-1 \le t \le 0$
     $$
         D^t \le 1 + t(1 - \frac{1}{D}) \le 1 + \frac{1}{2}t\,.
@@ -5833,7 +5658,7 @@ \section{Forests}
     $$
     with inclusion maximal $\sc(\fu)$. Properties \eqref{forest1}, -\eqref{forest6} for $(\fU_j, \fT|_{\fU_k})$ follow immediately from the corresponding properties for $(\fU, \fT)$, and the cubes $\sc(\fu), \fu \in \fU_j$ are disjoint by definition. The collections $\fU_j$ are also disjoint by definition.
 
-    Now we show by induction on $j$ that each point is contained in at most \rs{at least?} $2^n - j$ cubes $\sc(\fu)$ with $\fu \in \fU \setminus \bigcup_{j' \le j} \fU_{j'}$. This implies that $\bigcup_{j = 1}^{2^n} \fU_j = \fU$, which completes the proof of the Lemma. For $j = 0$ each point is contained in at most $2^n$ cubes by \eqref{forest3}. For larger $j$, if $x$ is contained in any cube $\sc(\fu)$ with $\fu \in \fU \setminus \bigcup_{j' < j} \fU_{j'}$, then it is contained in a maximal such cube. Thus it is contained in a cube in $\sc(\fu)$ with $\fu \in \fU_j$. Thus the number $\fu \in \fU \setminus \bigcup_{j' \le j} \fU_{j'}$ with $x\in \sc(\fu)$ is zero, or is less than the number of $\fu \in \fU \setminus \bigcup_{j' \le j-1} \fU_{j'}$ with $x \in \sc(\fu)$ by at least one.
+    Now we show by induction on $j$ that each point is contained in at most $2^n - j$ cubes $\sc(\fu)$ with $\fu \in \fU \setminus \bigcup_{j' \le j} \fU_{j'}$. This implies that $\bigcup_{j = 1}^{2^n} \fU_j = \fU$, which completes the proof of the Lemma. For $j = 0$ each point is contained in at most $2^n$ cubes by \eqref{forest3}. For larger $j$, if $x$ is contained in any cube $\sc(\fu)$ with $\fu \in \fU \setminus \bigcup_{j' < j} \fU_{j'}$, then it is contained in a maximal such cube. Thus it is contained in a cube in $\sc(\fu)$ with $\fu \in \fU_j$. Thus the number $\fu \in \fU \setminus \bigcup_{j' \le j} \fU_{j'}$ with $x\in \sc(\fu)$ is zero, or is less than the number of $\fu \in \fU \setminus \bigcup_{j' \le j-1} \fU_{j'}$ with $x \in \sc(\fu)$ by at least one.
 \end{proof}
 
 We pick a decomposition of the forest $(\fU, \fT)$ into $2^n$ $n$-rows
@@ -5857,7 +5682,7 @@ \section{Forests}
 \end{lemma}
 
 \begin{proof}
-    By Corollary \ref{lem tree est} and the density assumption \eqref{forest4}, we have for each $\fu \in \fU$ and all bounded $f$ of bounded support that
+    By Corollary \ref{densities tree bound} and the density assumption \eqref{forest4}, we have for each $\fu \in \fU$ and all bounded $f$ of bounded support that
     \begin{equation}
         \label{eq explicit tree bound 1}
         \left\|\sum_{\fp \in \fT(\fu)} T_{\fp} f \right\|_{2}  \le 2^{155a^3} 2^{(4a+1-n)/2}  \|f\|_2\,
@@ -5867,7 +5692,7 @@ \section{Forests}
         \label{eq explicit tree bound 2}
         \left\|\sum_{\fp \in \fT(\fu)} T_{\fp} \mathbf{1}_F f \right\|_{2} \le 2^{256a^3} 2^{(4a + 1-n)/2} \dens_2(\fT(\fu))^{1/2} \|f\|_2\,.
     \end{equation}
-    Since for each $j$ the top cubes $\sc(\fu)$, $\fu \in \fU_j$ are disjoint, we further have for all bounded $g$ of bounded support by Lemma \ref{lem Tp support adjoint}
+    Since for each $j$ the top cubes $\sc(\fu)$, $\fu \in \fU_j$ are disjoint, we further have for all bounded $g$ of bounded support by Lemma \ref{adjoint tile support}
     $$
         \left\|\mathbf{1}_F \sum_{\fu \in \fU_j} \sum_{\fp \in \fT(\fu)} T_{\fp}^* g\right\|_2^2 = \left\|\mathbf{1}_F \sum_{\fu \in \fU_j} \sum_{\fp \in \fT(\fu)} \mathbf{1}_{\sc(\fu)} T_{\fp}^* \mathbf{1}_{\sc(\fu)} g\right\|_2^2
     $$
@@ -5879,7 +5704,7 @@ \section{Forests}
     $$
         \le 2^{256a^3} 2^{(4a+1-n)/2} \max_{\fu \in \fU_j}\dens_2(\fT(\fu))^{1/2} \sum_{\fu \in \fU_j} \left\| \mathbf{1}_{\sc(\fu)} g\right\|_2^2\,.
     $$
-    Again by disjointness of the cubes $\sc(\fu)$, this is estimated by
+    Again by disjointedness of the cubes $\sc(\fu)$, this is estimated by
     $$
         2^{256a^3} 2^{(4a+1-n)/2} \max_{\fu \in \fU_j}\dens_2(\fT(\fu))^{1/2} \|g\|_2^2\,.
     $$
@@ -5903,14 +5728,14 @@ \section{Forests}
     $$
         T_{\fC}^* = \sum_{\fp \in \fC} T_{\fp}^*\,.
     $$
-    We have by Lemma \ref{lem Tp support adjoint} and the triangle inequality that
+    We have by Lemma \ref{adjoint tile support} and the triangle inequality that
     $$
         \left| \int \sum_{\fu \in \fU_j} \sum_{\fu' \in \fU_{j'}} \sum_{\fp \in \fT_j(\fu)} \sum_{\fp' \in \fT_{j'}(\fu')} T^*_{\fp} g_1 \overline{T^*_{\fp'} g_2} \, \mathrm{d}\mu \right|
     $$
     $$
         \le   \sum_{\fu \in \fU_j} \sum_{\fu' \in \fU_{j'}} \left| \int   T^*_{\fT_j(\fu)} (\mathbf{1}_{\sc(\fu)} g_1) \overline{T^*_{\fT_{j'}(\fu')} (\mathbf{1}_{\sc(\fu')} g_2)} \, \mathrm{d}\mu \right|\,.
     $$
-    By Lemma \ref{SeparatedTrees}, this is bounded by
+    By Lemma \ref{correlation separated trees}, this is bounded by
     \begin{equation}
         \label{eq S2uu'}
          2^{650a^3-3n} \sum_{\fu \in \fU_j} \sum_{\fu' \in \fU_{j'}} \|S_{2,\fu} g_1\|_{L^2(\sc(\fu')\cap \sc(\fu)} \|S_{2, \fu'} g_2\|_{L^2(\sc(\fu')\cap\sc(\fu))}\,.
@@ -5926,7 +5751,7 @@ \section{Forests}
         \left(\sum_{\fu \in \fU_j} \sum_{\fu' \in \fU_{j'}} \|S_{2,\fu'} g_2\|_{L^2(\sc(\fu')\cap\sc(\fu))}^2 \right)^{1/2}\,.
     $$
     \rs{Constant changed form 550 to 650, check}
-    By pairwise disjointness of the sets $\sc(\fu)$ for $\fu \in \fU_j$ and of the sets $\sc(\fu')$ for $\fu' \in \fU_{j'}$, we have
+    By pairwise disjointedness of the sets $\sc(\fu)$ for $\fu \in \fU_j$ and of the sets $\sc(\fu')$ for $\fu' \in \fU_{j'}$, we have
     $$
         \sum_{\fu \in \fU_j}\sum_{\fu' \in \fU_{j'}} \|S_{2,\fu} g_1\|_{L^2(\sc(\fu')\cap \sc(\fu))}^2
         = \sum_{\fu \in \fU_j}\sum_{\fu' \in \fU_{j'}} \int_{\sc(\fu) \cap \sc(\fu')} |S_{2,\fu} g_1(y)|^2 \, \mathrm{d}\mu(y)
@@ -5939,7 +5764,7 @@ \section{Forests}
         \le 2^{650a^3-3n} \|S_{2,\fu}g_1\|_2 \|S_{2,\fu'} g_2\|_2\,.
     $$
     \rs{Constant changed from 550 to 650, check}
-    The lemma now follows from Lemma \ref{lem L2 Sg estimate}.
+    The lemma now follows from Lemma \ref{adjoint tree control}.
 \rs{Constant changed, check.}
 \end{proof}
 
@@ -5997,7 +5822,7 @@ \section{Forests}
     $$
         2^{862a^3} (2^{-n} + 2^{n}2^{-3 n}) \sum_{j = 1}^n \|\mathbf{1}_{E_j} g\|_2^2\,,
     $$
-    and by disjointness of the sets $E_j$, this is at most
+    and by disjointedness of the sets $E_j$, this is at most
     $$
         2^{862a^3 - n} \|g\|_2^2\,.
     $$
@@ -6006,7 +5831,7 @@ \section{Forests}
         \label{eq forest bound 1}
         \left\|\sum_{\fu \in \fU} \sum_{\fp \in \fT(\fu)} T_{\fp} f\right\|_2 \le 2^{431a^3-\frac{n}{2}} \|f\|_2\,.
     \end{equation}
-    On the other hand, we have by disjointness of the sets $E_j$
+    On the other hand, we have by disjointedness of the sets $E_j$
     $$
         \left\|\sum_{\fu \in \fU} \sum_{\fp \in \fT(\fu)} T_{\fp} f\right\|_2^2 =  \left\|\sum_{j=1}^{2^n} \mathbf{1}_{E_j} T_{\mathfrak{R}_{j}} f\right\|_2^2 = \sum_{j = 1}^{2^n} \|\mathbf{1}_{E_j} T_{\mathfrak{R}_{j}} f\|_2^2\,.
     $$
@@ -6021,14 +5846,14 @@ \section{Forests}
     Proposition \ref{Holder van der Corput} follows by taking the product of the $(2 - \frac{2}{q})$-th power of \eqref{eq forest bound 1} and the $(\frac{2}{q} - 1)$-st power of \eqref{eq forest bound 2}.
 \end{proof}
 
-\chapter{Proof of P. \ref{Holder van der Corput}, the H\"older cancellative condition}
+\chapter{Proof of Prop. \ref{Holder van der Corput}, the H\"older cancellative condition}
 \label{liphoel}
 
 We need the following auxiliary lemma.
 Recall that $\tau = 1/a$.
 
-\begin{lemma}
-    \label{lem regularity aux}
+\begin{lemma}[Lipschitz Holder approximation]
+    \label{Lipschitz Holder approximation}
     Let $z\in X$ and $R>0$. Let $\varphi: X \to \mathbb{C}$ be a function supported in the  ball
     $B:=B(z,R)$ with finite norm $\|\varphi\|_{C^\tau(B)}$. Let $0<t \leq 1$. There exists a function $\tilde \varphi : X \to \mathbb{C}$, supported in $B(z,2R)$, such that for every $x\in X$
     \begin{equation}\label{eq firstt}
@@ -6167,7 +5992,7 @@ \chapter{Proof of P. \ref{Holder van der Corput}, the H\"older cancellative cond
 \end{equation}
 Combining \eqref{eql52} and  \eqref{eql226} using $a\ge 4$ and $t\le 1$ and
 adding \eqref{eql42} proves \eqref{eq secondt} and completes the proof
-of Lemma \ref{lem regularity aux}.
+of Lemma \ref{Lipschitz Holder approximation}.
 \end{proof}
 
 
@@ -6178,7 +6003,7 @@ \chapter{Proof of P. \ref{Holder van der Corput}, the H\"older cancellative cond
 \begin{equation}\label{eql69}
     t:=(1+d_B(\mfa,\mfb))^{-\frac{\tau}{2+a}}
 \end{equation}
-and define $\tilde{\varphi}$ as in Lemma \ref{lem regularity aux}. Let $\mfa$ and $\mfb$ in $\Mf$.
+and define $\tilde{\varphi}$ as in Lemma \ref{Lipschitz Holder approximation}. Let $\mfa$ and $\mfb$ in $\Mf$.
 Then
    \begin{equation}\label{eql60}
        \left|\int e(\mfa(x)-{\mfb(x)}) \varphi (x)\, \mathrm{d}\mu(x)\right|
@@ -6263,12 +6088,12 @@ \chapter{Proof of P. \ref{Holder van der Corput}, the H\"older cancellative cond
 
 
 
-\chapter{Proof of P. \ref{Hardy Littlewood}, Vitali covering and Hardy Littlewood}
+\chapter{Proof of Prop. \ref{Hardy Littlewood}, Vitali covering and Hardy Littlewood}
 \label{sec hlm}
-\rs{Checked}
 
-We begin with a classical representation of the Lebesgue norm. \ct{touch base with Floris about infinite integral boundaries. open/closed.}
-\begin{lemma}\label{lem layercake}
+
+We begin with a classical representation of the Lebesgue norm.
+\begin{lemma}[layer cake representation]\label{layer cake representation}
 Let $1\le p< \infty$.  Then for any measurable function $u:X\to [0,\infty)$ on the measure space $X$
 relative to the measure $\mu$
 we have
@@ -6309,7 +6134,7 @@ \chapter{Proof of P. \ref{Hardy Littlewood}, Vitali covering and Hardy Littlewoo
 If there is no such ball, stop the selection and set
 $i'':=i$.
 
-By  disjointness of the chosen balls and since $0 \le u$, we have
+By  disjointedness of the chosen balls and since $0 \le u$, we have
 \begin{equation}
 \sum_{0\le i<i''}\int_{B_i} u(x)\, d\mu(x) \le \int_X  u(x)\, d\mu(x)\, .
 \end{equation}
@@ -6362,7 +6187,7 @@ \chapter{Proof of P. \ref{Hardy Littlewood}, Vitali covering and Hardy Littlewoo
 
 We turn to the proof of \eqref{eq hlm}. We first consider the case $p_1=1$ and recall $M_{\mathcal{B}}=M_{\mathcal{B},1}$.
 We write for the $p_2$-th power of left-hand side of \eqref{eq hlm}
-with Lemma \ref{lem layercake}
+with Lemma \ref{layer cake representation}
 and a change of variables
 \begin{equation}
     \|M_{\mathcal{B}}u(x)\|_{p_2}^{p_2}
@@ -6418,7 +6243,7 @@ \chapter{Proof of P. \ref{Hardy Littlewood}, Vitali covering and Hardy Littlewoo
    2^{2a}
     \int u_\lambda (x)\, dx\, .
 \end{equation}
-With Lemma \ref{lem layercake},
+With Lemma \ref{layer cake representation},
 \begin{equation}\label{eqbesi12}
     \lambda \mu(\{x: M_{\mathcal{B}}u(x)\ge 2\lambda\})\le
    2^{2a}
@@ -6445,7 +6270,7 @@ \chapter{Proof of P. \ref{Hardy Littlewood}, Vitali covering and Hardy Littlewoo
 We split the integral  into $\lambda\ge \lambda'$ and $\lambda<\lambda'$ and resolve the
 maximum correspondingly.
 We have for $\lambda\ge \lambda'$
-with Lemma \ref{lem layercake}
+with Lemma \ref{layer cake representation}
 \begin{equation}
     \int_0^\infty \lambda^{p_2-2}
    \int_0^\lambda
@@ -6461,7 +6286,7 @@ \chapter{Proof of P. \ref{Hardy Littlewood}, Vitali covering and Hardy Littlewoo
    =p_2^{-1} \|u\|_{p_2}^{p_2}\, .
 \end{equation}
 We have for $\lambda< \lambda'$
-with Fubini and Lemma \ref{lem layercake}
+with Fubini and Lemma \ref{layer cake representation}
 \begin{equation}
     \int_0^\infty \lambda^{p_2-2}
    \int_\lambda^\infty
@@ -6510,8 +6335,8 @@ \chapter{Proof of P. \ref{Hardy Littlewood}, Vitali covering and Hardy Littlewoo
 
 Now we construct the operator $M$ satisfying \eqref{eq ball av} and \eqref{eq hlm 2}.
 
-\begin{lemma}
-    \label{lem separable}
+\begin{lemma}[covering separable space]
+    \label{covering separable space}
     For each $r > 0$, there exists a countable collection $C(r) \subset X$ of points such that
     $$
         X \subset \bigcup_{c \in C(r)} B(c, r)\,.
@@ -6543,7 +6368,7 @@ \chapter{Proof of P. \ref{Hardy Littlewood}, Vitali covering and Hardy Littlewoo
 $$
     \mathcal{B}_\infty = \{B(c, 2^k) \ : \ c \in C(2^k), k \in \mathbb{N}\}\,.
 $$
-By Lemma \ref{lem separable}, this is a countable collection of balls. We choose an enumeration $\mathcal{B}_\infty = \{B_1, \dotsc\}$ and define
+By Lemma \ref{covering separable space}, this is a countable collection of balls. We choose an enumeration $\mathcal{B}_\infty = \{B_1, \dotsc\}$ and define
 $$
     \mathcal{B}_n = \{B_1, \dotsc, B_n\}\,.
 $$
@@ -6567,21 +6392,15 @@ \chapter{Proof of P. \ref{Hardy Littlewood}, Vitali covering and Hardy Littlewoo
 
 \chapter{Proof of Theorem \ref{classical Carleson}}
 
-\lars{TODO: Subsection 4 has to be rewritten to prove the bound for the correct maximal operator.
-
-Subsection 6 is not written at all.
 
-Subsection 7 is not complete.
-
-I have checked all other subsections, and I have checked Subsectino 7 as far as it is written.}
 
 The convergence of partial Fourier sums is proved in
 Subsection \ref{10classical} in two steps. In the first step,
-one establishes convergence on a suitable dense subclass of functions. We choose piece-wise constant functions as subclass, the convergence is stated in Lemma \ref{lem piece} and proved in Subsection \ref{10piecewise}.
+one establishes convergence on a suitable dense subclass of functions. We choose piece-wise constant functions as subclass, the convergence is stated in Lemma \ref{convergence for piecewise constant} and proved in Subsection \ref{10piecewise}.
 In the second step, one controls the relevant error of approximating a general function by a function in
 the subclass by a bound of the Carleson maximal operator. This approximation result is stated
-in Lemma \ref{lem diff} and proved in Subsection \ref{10difference}.
-This latter proof refers to the main Theorem \ref{metric space Carleson}. Two assumptions in Theorem \ref{metric space Carleson} require more work, this is prepared in Lemmas \ref{lem hilbert} and \ref{lem vdc} and proved in Subsections \ref{10hilbert}
+in Lemma \ref{control approximation effect} and proved in Subsection \ref{10difference}.
+This latter proof refers to the main Theorem \ref{metric space Carleson}. Two assumptions in Theorem \ref{metric space Carleson} require more work, this is prepared in Lemmas \ref{lem hilbert} and \ref{van der Corput} and proved in Subsections \ref{10hilbert}
 and \ref{10vandercorput}. Subsections \ref{10piecewise},
 \ref{10hilbert}, \ref{10vandercorput}, and \ref{10difference}
 are mutually independent.
@@ -6590,15 +6409,15 @@ \chapter{Proof of Theorem \ref{classical Carleson}}
 \section{The classical Carleson theorem}
 \label{10classical}
 
-Let a uniformly continuous $2\pi$-periodic function $f:\R\to \mathbb{C}$ bounded in absolute value by $1$ be given. Let $\epsilon>0$ be given.
+Let a uniformly continuous $2\pi$-periodic function $f:\R\to \mathbb{C}$ bounded in absolute value by $1$ be given. Let $\epsilon>0$ be given, without loss of generality we also assume $\epsilon \le 1$.
 
 By uniform  continuity of $f$, there is a $0<\delta<\pi$
 such that for all $x,x' \in \R$ with $|x-x'|\le \delta$
 we have
 \begin{equation}\label{uniconbound}
-|f(x)-f(x')|\le 2^{-2^{50}}\epsilon\, .
+|f(x)-f(x')|\le 2^{-2^{50}}\epsilon^2\, .
 \end{equation}
-Let $K$ be the Gaussian bracket of $2\pi/\delta$, that is the unique integer with
+Let $K$ be the Gaussian bracket of $\frac{2\pi}\delta+1$, that is the unique integer with
 \begin{equation}
     K-1\le  \frac{2\pi}{\delta} < K\, .
 \end{equation}
@@ -6611,12 +6430,15 @@ \section{The classical Carleson theorem}
 \begin{equation}\label{def fzero}
 f_0(x):=f\left(\frac{2\pi k(x)}{K}\right)\, .
 \end{equation}
-\begin{lemma}
+\begin{lemma}[piecewise constant
+approximation]
+\label{piecewise constant
+approximation}
 The function $f_0$ is measurable.
 The function $f_0$ is $2\pi $- periodic.
-The function $f_0$ satisfies  for all $x\in \R$,
+The function $f_0$ satisfies  for all $x\in \R$:
 \begin{equation}\label{eq ffzero}
-|f(x)-f_0(x)|\le  2^{-2^{50}} \epsilon\, ,
+|f(x)-f_0(x)|\le  2^{-2^{50}} \epsilon^2\, ,
 \end{equation}
 \begin{equation}\label{eq ffzero1}
 |f_0(x)|\le  1\, .
@@ -6641,9 +6463,14 @@ \section{The classical Carleson theorem}
 
 
 To see that $f_0$ is $2\pi$-periodic, we observe by
-adding $2\pi$ to \eqref{definekx} that
+applying \eqref{definekx} to $x+2\pi$ and subtracting $K$
+that
+\begin{equation}\label{definekxshifted}
+k(x+2\pi)-K\le \frac{Kx}{2\pi}<k(x+2\pi)-K+1\,
+\end{equation}
+and hence
 \begin{equation}
- k(x+2\pi)=k(x)+K
+ k(x+2\pi)-K=k(x)
 \end{equation}
 and hence by Definition \eqref{def fzero}
 \begin{equation}
@@ -6659,13 +6486,14 @@ \section{The classical Carleson theorem}
 
 
 
-To see \eqref{eq ffzero}, observe that by \eqref{definekx} and by definition of $K$ we have
+To see \eqref{eq ffzero}, observe that by \eqref{definekx} used
+twice and then by definition of $K$ we have
 \begin{equation}
-0\le x-\frac{2\pi k(x)}{K} \le \frac{2\pi k(x)}{K}\le \delta\, .
+0\le x-\frac{2\pi k(x)}{K} \le \frac{2\pi }{K}\le \delta\, .
 \end{equation}
 Hence, by choice of $\delta$, for all $x\in \R$, we obtain \eqref{eq ffzero}.
 
-Finally \eqref{eq ffzero} follows as $|f|$ by assumption is bounded by $1$.
+Finally \eqref{eq ffzero} follows because $|f|$ by assumption is bounded by $1$.
 This completes the proof of the lemma.
 \end{proof}
 
@@ -6680,8 +6508,9 @@ \section{The classical Carleson theorem}
 \end{equation}
 
 We prove in Subsection \ref{10piecewise}:
-\begin{lemma}
-\label{lem piece}
+\begin{lemma}[convergence for piecewise constant]
+\label{convergence for piecewise constant}
+\uses{constant function,flat interval partial sum}
     For all $N>\frac {2^{25} K^2}{\epsilon ^3}$ and
     \begin{equation}
         x\in [0,2\pi)\setminus E_1
@@ -6693,15 +6522,18 @@ \section{The classical Carleson theorem}
 \end{lemma}
 
 
-We prove in Subsection \ref{10difference}: \lars{The proof of this Lemma is still missing completely}
-\begin{lemma}\label{lem diff}
+We prove in Subsection \ref{10difference}:
+\begin{lemma}[control approximation effect]
+\label{control approximation effect}
+\uses{real Carleson, dirichlet kernel,lower secant bound}
     There is a set $E_2 \subset \R$ with Lebesgue measure
     $|E_2|\le \frac \epsilon 2$ such that for all
     \begin{equation}
-        x\in [0,1)\setminus E_2
+        x\in [0,2\pi)\setminus E_2
     \end{equation}
    we have
     \begin{equation}
+    \label{eq max partial sum diff}
         \sup_{N\ge 0} |S_Nf(x)-S_Nf_0(x)|
         \le \frac \epsilon 4\,.
     \end{equation}
@@ -6715,70 +6547,84 @@ \section{The classical Carleson theorem}
 \begin{equation}
     |E|\le |E_1|+|E_2|\le \frac \epsilon 2 +\frac \epsilon 2 \le \epsilon\, .
 \end{equation}
-Let $N_0$ be as in Lemma \ref{lem piece}.
+Let $N_0$ be the unique integer such that
+\begin{equation}
+    N_0- 1\le {2^{25} K^2}{\epsilon ^{-3}}<N_0\, .
+\end{equation}
 For every
 \begin{equation}
 x\in [0,1)\setminus E\, ,
 \end{equation}
 and every $N>N_0$ we have by the triangle inequality
-\begin{equation}
+\begin{equation*}
     |f(x)-S_Nf(x)|
-    \end{equation}
-    \begin{equation}
+    \end{equation*}
+    \begin{equation}\label{epsilonthird}
     \le |f(x)-f_0(x)|+ |f_0(x)-S_Nf_0(x)|+|S_Nf_0(x)-S_N f(x)|\, .
 \end{equation}
-Using \eqref{uniconbound} and Lemmas \ref{lem piece}
-and \ref{lem diff}, we estimate the last display by
+Using \eqref{uniconbound} and Lemmas \ref{convergence for piecewise constant}
+and \ref{control approximation effect}, we estimate \eqref{epsilonthird} by
 \begin{equation}
-    \le 2^{-2^{50}} \epsilon +\frac \epsilon 4 +\frac \epsilon 4\le \epsilon\, .
+    \le 2^{-2^{50}} \epsilon^2 +\frac \epsilon 4 +\frac \epsilon 4\le \epsilon\, .
 \end{equation}
 This shows  \eqref{aeconv} for the given $E$ and $N_0$
 and completes the proof of Theorem \ref{classical Carleson}.
 
-The proof of Lemma \ref{lem diff} will essentially be
+The proof of Lemma \ref{control approximation effect} will essentially be
 an application of
 the following variant of Theorem \ref{metric space Carleson}.
-Let  $k:\R\to \R$ be the function defined by
-by $k(0)=0$ and for $x\neq 0$
+Let  $\kappa:\R\to \R$ be the function defined by
+$\kappa(0)=0$ and for $0<|x|<1$
 \begin{equation}\label{eq hilker}
-k(x)=\frac {\max (1-|x|, 0)}{1-e^{ix}}\, .
+\kappa(x)=\frac { 1-|x|}{1-e^{ix}}\,
+\end{equation}
+\ct{make sure you never divide anything by zero... dangerous if $x=2\pi m$}
+and for $|x|\ge 1$
+\begin{equation}\label{eq hilker1}
+\kappa(x)=0\, .
 \end{equation}
+Note that this function is continuous at every point $x$ with $|x|>0$.
 The following lemma is proved in Subsection \ref{10carleson}.
-\begin{lemma}\label{lem rcarleson}
+\begin{lemma}\label{real Carleson}
+\uses{metric space Carleson,real line metric,real line measure,real line doubling,frequency ball doubling,frequency ball growth,frequency ball cover,real van der Corput,Hilbert kernel bound,Hilbert kernel regularity}
     Let $F,G$ be Borel subsets of $\R$ with finite measure. Let $f$ be a bounded measurable function on $\R$ with $|f|\le \mathbf{1}_F$. Then
 \begin{equation}
     |\int _G Tf(x) \, dx| \le 2^{2^{40}}|F|^{\frac 12} |G|^{\frac 12} \, ,
 \end{equation}
 where
 \begin{equation}
+    \label{define T carleson}
     T f(x)=\sup_{n\in \mathbb{Z}}
-    \sup_{r>0}\left|\int_{r<|x-y|<1} f(y)k(x-y) e^{iny}\, dy\right|\, .
+    \sup_{r>0}\left|\int_{r<|x-y|<1} f(y)\kappa(x-y) e^{iny}\, dy\right|\, .
 \end{equation}
 \end{lemma}
 
 
-The following lemma is proved in Subsection \ref{10hilbert}. \lars{This used to be the assumption of our main theorem. Now that we changed it, we should change this Lemma to prove the bound for the full maximal operator, i.e. with another sup over $r$}
+The following lemma is proved in Subsection \ref{10hilbert}.
+\ct{Lars was right. I changed the following so that it exactly fits what we need. This makes 10.3 mildly more laborous, but that's where one can get synergies....
+The $2^8$ can be a bit larger if needed}
 \begin{lemma}\label{lem hilbert}
-    For every $0<r<\pi$ and every  bounded measurable function $g$ with bounded support we have
+    For every  bounded measurable function $g$ with bounded support we have
 \begin{equation}
-    \|T_rg\|_2\le 2^8\|g\|_2,
+    \|T_*g\|_2\le 2^8\|g\|_2,
 \end{equation}
 where
-\begin{equation}
-    T_r g(x)=\sup_{|x-x'|<r}\frac 1{2\pi} \left|\int_{r}^{2\pi -r}
-g(x'-y) \frac 1{1-e^{iy}}\, dx\right|\, .
+\begin{equation}\label{concretetstar}
+    T_* g(x):=\sup_{0<r_1<r_2<1}\sup_{|x-x'|<r_1}\frac 1{2\pi} \left|\int_{r_1<|x-y|<r_2}
+g(y) \kappa(x-y)\, dy\right|\, .
 \end{equation}
 \end{lemma}
-The following lemma is proved in Subsection \ref{10vandercorput}. \lars{The assumptions $0 \le a$ and $b \le 2\pi$ are unnecessary, and we need the version without them later. If nobody disagrees I will delete}
-\begin{lemma}\label{lem vdc}
-    Let $0\le a<b\le 2\pi$. Let $g:\R\to \C$ be a  measurable function and assume
+The following lemma is proved in Subsection \ref{10vandercorput}.
+\begin{lemma}[van der Corput]
+\label{van der Corput}
+    Let $\alpha<\beta$ be real numbers. Let $g:\R\to \C$ be a  measurable function and assume
     \begin{equation}
-        \|g\|_{Lip(a,b)}:=\sup_{a\le x\le b}|g(x)|+|b-a|
-        \sup_{a\le x<y\le b} \frac {|g(y)-g(x)|}{|y-x|}<\infty\, .
+        \|g\|_{Lip(\alpha,\beta)}:=\sup_{\alpha\le x\le \beta}|g(x)|+|\beta-\alpha|
+        \sup_{\alpha\le x<y\le \beta} \frac {|g(y)-g(x)|}{|y-x|}<\infty\, .
     \end{equation}
-    Then for any $0\le a<b\le 2\pi$ we have
+    Then for any $0\le \alpha<\beta\le 2\pi$ we have
     \begin{equation}
-        \int _{a}^{b} g(x) e^{-inx}\, dx\le 2\pi |b-a|\|g\|_{Lip(a,b)}(1+|n||b-a|)^{-1}\, .
+        \int _{\alpha}^{\beta} g(x) e^{-inx}\, dx\le 2\pi |\beta-\alpha|\|g\|_{Lip(\alpha,\beta)}(1+|n||\beta-\alpha|)^{-1}\, .
     \end{equation}
 
 \end{lemma}
@@ -6788,10 +6634,11 @@ \section{The classical Carleson theorem}
 
 
 
-We close this section with three lemmas that are used
+We close this section with four lemmas that are used
 across the following subsections.
 
-\begin{lemma}\label{lem expintegral}
+\begin{lemma}[mean zero oscillation]
+\label{mean zero oscillation}
 Let $n\in \mathbb{Z}$ with   $n\neq 0$, then
 \begin{equation}
 \int_0^{2\pi} e^{inx}\, dx=0
@@ -6805,10 +6652,11 @@ \section{The classical Carleson theorem}
 
 \end{proof}
 
-\begin{lemma}\label{dirichlet}
+\begin{lemma}[Dirichlet kernel]
+\label{dirichlet kernel}
 We have for every $2\pi$-periodic bounded measurable $f$ and every $N\ge 0$
 \begin{equation}
-    S_Nf(x)=\frac 1{2\pi}\int_{0}^{2\pi}f(y) K_N(x-y)\, dy
+    S_Nf(x)=\frac 1{2\pi}\int_{-\pi}^{\pi}f(y) K_N(x-y)\, dy
 \end{equation}
 where $K_N$ is the $2\pi$-periodic continuous function of
 $\R$ given by
@@ -6831,11 +6679,11 @@ \section{The classical Carleson theorem}
         S_Nf(x)=\sum_{n=-N}^N \widehat{f}_n e^{inx}
     \end{equation*}
        \begin{equation*}
-    =\sum_{n=-N}^N \frac 1{2\pi}\int_{0}^{2\pi}
+    =\sum_{n=-N}^N \frac 1{2\pi}\int_{-\pi}^{\pi}
     f(x) e^{in(x-y)}\, dy
     \end{equation*}
  \begin{equation}\label{eq expsum}
-     \frac 1{2\pi}\int_{-\pi}^\pi
+     =\frac 1{2\pi}\int_{-\pi}^\pi
     f(y) \sum_{n=-N}^N e^{in(x-y)}\, dy\, .
  \end{equation}
 This proves the first statement of the lemma.
@@ -6853,9 +6701,30 @@ \section{The classical Carleson theorem}
 This  proves the second part of the lemma.
 \end{proof}
 
+\begin{lemma}[lower secant bound]
+\label{lower secant bound}
+Let $\eta>0$ and $-2\pi +\eta \le  x\le 2\pi-\eta$ with $|x|\ge \eta$. Then
+\begin{equation}
+    |1-e^{ix}|\ge \frac {\eta} 8
+\end{equation}
+\end{lemma}
+\begin{proof}
+
+We have
+$$
+    |1 - e^{ix}| = \sqrt{(1 - \cos(x))^2 + \sin^2(x)} \ge |\sin(x)|\,.
+$$
+If $0 \le x \le \frac{\pi}{2}$, then we have from concavity of $\sin$ on $[0, \pi]$ and $\sin(0) = 0$ and $\sin(\frac{\pi}{2}) = 1$
+$$
+    |\sin(x)| \ge \frac{2}{\pi} x \ge \frac{2}{\pi} \eta\,.
+$$
+When $x\in \frac{m\pi}{2} + [0, \frac{\pi}{2}]$ for $m \in \{-4, -3, -2, -1, 1, 2, 3\}$ one can argue similarly.
+ \end{proof}
+
 The following lemma will be proved in Subsection \ref{10projection}.
 
-\begin{lemma}\label{lem l2sn}
+\begin{lemma}[spectral projection bound]
+\label{spectral projection bound}
     Let $f$ be a bounded $2\pi$-periodic measurable function. Then, for all $N\ge 0$
    \begin{equation}\label{snbound}
    \|S_Nf\|_2\le \|f\|_2.
@@ -6880,11 +6749,12 @@ \section{Piecewise constant functions.}
 We first compute the partial Fourier sums for constant functions.
 
 
-\begin{lemma}\label{constant}
+\begin{lemma}[constant function]
+\label{constant function}
 If $h$ satisfies $h(x)=h(0)$ for all $x\in \R$, then for all $N\ge 0$ we have $S_Nh=h$.
  \end{lemma}
 \begin{proof}
-    We compute with Lemma \ref{lem expintegral} for $n\in \mathbb{Z}$ with $n\neq 0$,
+    We compute with Lemma \ref{mean zero oscillation} for $n\in \mathbb{Z}$ with $n\neq 0$,
     \begin{equation}
         \widehat{h}_n=\frac 1{2\pi}\int_0^{2\pi}h(y)e^{-iny}\, dy=\frac {h(0)}{2\pi}\int_0^{2\pi}e^{-iny}\, dy=0\, .
     \end{equation}
@@ -6895,27 +6765,10 @@ \section{Piecewise constant functions.}
 \end{equation}
 This proves the lemma.
 \end{proof}
-\begin{lemma}\label{expbound}
-Let $\eta>0$ and $-2\pi +\eta \le  x\le 2\pi-\eta$ with $|x|\ge \eta$. Then
-\begin{equation}
-    |1-e^{ix}|\ge \frac {\eta} 8
-\end{equation}
-\end{lemma}
-\begin{proof}
-\ct{check with Floris what the library has here} \lars{I wrote something that Lean can definitely do.}
-\lars{(And to not forget it, the sharp constant is really $\frac{2}{\pi}$, because $|1 - e^{ix}|$ is concave on $[0, \pi]$)}
-We have
-$$
-    |1 - e^{ix}| = \sqrt{(1 - \cos(x))^2 + \sin^2(x)} \ge |\sin(x)|\,.
-$$
-If $0 \le x \le \frac{\pi}{2}$, then we have from concavity of $\sin$ on $[0, \pi]$ and $\sin(0) = 0$ and $\sin(\frac{\pi}{2}) = 1$
-$$
-    |\sin(x)| \ge \frac{2}{\pi} x \ge \frac{2}{\pi} \eta\,.
-$$
-When $x\in \frac{m\pi}{2} + [0, \frac{\pi}{2}]$ for $m \in \{-4, -3, -2, -1, 1, 2, 3\}$ one can argue similarly.
- \end{proof}
 
-\begin{lemma}\label{lem diri}
+\begin{lemma}[Dirichlet kernel bound]
+\label{dirichlet kernel bound}
+\uses{lower secant bound}
 Let $\eta>0$. Let
 \begin{equation}
     -2\pi +\eta \le a<b \le 2\pi-\eta
@@ -6933,7 +6786,7 @@ \section{Piecewise constant functions.}
 \end{lemma}
 \begin{proof}
  Note first that the integral in \eqref{dirichletint}
-    is well defined as the denominator of the integrand is bounded away from zero by Lemma \ref{expbound}
+    is well defined as the denominator of the integrand is bounded away from zero by Lemma \ref{lower secant bound}
     and thus the integrand is continuous.
     We have with partial integration
 \begin{equation*}
@@ -6950,7 +6803,7 @@ \section{Piecewise constant functions.}
 We estimate the two summands in
 \eqref{eq part int} separately. We have
 for the first summand in \eqref{eq part int}
-with Lemma \ref{expbound},
+with Lemma \ref{lower secant bound},
 \begin{equation*}
 \left|   \frac 1 {-iN}\int_a^b
 {e^{-iNx}}\frac{ie^{ix}}
@@ -6964,7 +6817,7 @@ \section{Piecewise constant functions.}
     \le  \frac {64|b-a|}{\eta^2N}
 \end{equation}
 We have for the second summand in \eqref{eq part int}
-with Lemma \ref{expbound} again
+with Lemma \ref{lower secant bound} again
 \begin{equation*}
    \left|\left[ \frac 1{-iN}
     \frac {e^{-iNx}} {1-e^{ix}}\right]_a^b\right|
@@ -6990,7 +6843,9 @@ \section{Piecewise constant functions.}
 
 
 
-\begin{lemma}\label{lem g with zero}
+\begin{lemma}[flat interval partial sum]
+\label{flat interval partial sum}
+\uses{dirichlet kernel,dirichlet kernel bound}
 Let $g\in \mathcal{G}$. Assume
 $x\in [0,2\pi)\setminus E_1 $ and $g(x)=0$.
 Then for all $N>\frac {2^{25} K^2}{\epsilon^3}$
@@ -7005,10 +6860,10 @@ \section{Piecewise constant functions.}
 
 
 
-   With Lemma \ref{dirichlet}, breaking up the domain of integration into a
+   With Lemma \ref{dirichlet kernel}, breaking up the domain of integration into a
    partition of subintervals,
    \begin{equation*}
-      |S_N(g)|=\left|\int_{0}^{2\pi} g(y)K_N(x-y)\, dy\right|
+      |S_N(g)|=\left|\int_{-\pi}^{\pi} g(y)K_N(x-y)\, dy\right|
    \end{equation*}
    \begin{equation}\label{eqcf1}
   =  \left|\sum_{k=0}^{K-1}\int_{\frac{2\pi k}K}^{\frac{2\pi (k+1)}K} g(y)K_N(x-y)\, dy\right|\, .
@@ -7053,7 +6908,7 @@ \section{Piecewise constant functions.}
 
 
 
-With Lemma \ref{dirichlet} and the triangle inequality,
+With Lemma \ref{dirichlet kernel} and the triangle inequality,
 it follows that
 \begin{equation}
    \left|\int_{x-2\pi (k+1)/K}^{x-2\pi k/K} K_N(y)\, dy\right|
@@ -7062,7 +6917,7 @@ \section{Piecewise constant functions.}
    \left|\int_{a}^{b}
    \frac{e^{-iNy}}{1-e^{iy}}\, dy\right|
 \end{equation}
-Using that the two integrals on the right-hand side are complex conjugates of each other, and using Lemma \ref{lem diri} and $\epsilon <2\pi$, this is bounded by
+Using that the two integrals on the right-hand side are complex conjugates of each other, and using Lemma \ref{dirichlet kernel bound} and $\epsilon <2\pi$, this is bounded by
 \begin{equation}
    \le 2\left|\int_{a}^{b}
    \frac{e^{iNy}}{1-e^{-iy}}\, dy\right|
@@ -7073,18 +6928,18 @@ \section{Piecewise constant functions.}
 the lemma.
 \end{proof}
 
-We now prove  Lemma \ref{lem piece}
+We now prove  Lemma \ref{convergence for piecewise constant}
 
 Let $x\in [0,2\pi)\setminus E_1 $ and $N>\frac {2^{25} K^2}{\epsilon ^3}$.
 Let $h$ be the function which is constant equal to $f_0(x)$ and let $g=f_0-h$.
 By the bound on $f$ and the triangle inequality,
 $|g|$ is bounded by $2$. Hence $g$ is in the class $\mathcal{G}$. We also have $g(x)=0$.
-Using Lemma \ref{constant}, we obtain
+Using Lemma \ref{constant function}, we obtain
 \begin{equation}
     S_Nf_0(x)- f_0(x)=S_N(g+h)(x)- S_Nh(x)=S_Ng(x)\, .
 \end{equation}
-Lemma \ref{lem piece}
-now follows by Lemma \ref{lem g with zero}.
+Lemma \ref{convergence for piecewise constant}
+now follows by Lemma \ref{flat interval partial sum}.
 
 
 
@@ -7092,13 +6947,11 @@ \section{The truncated Hilbert transform}
 \label{10hilbert}
 
 \lars{I am skipping this subsection because it has to be rewritten anyway}
+\rs{To be discussed: 1. Using $L^2$ norms in Cotlar leads to a weak $(2,2)$ estimates for $T_*$}
+\rs{2. Obtaining $L^2$ estimate for $T_0$ using uniform $L^2$ bound for truncated kernels}
+\rs{3. Passing to $\mathbb{R}$ from periodic functions}
 
-
-
-
-
-
-
+\rs{Change norms}
 
 
 Let $M_n$ be the modulation operator
@@ -7117,16 +6970,16 @@ \section{The truncated Hilbert transform}
 \begin{lemma}
 We have for every bounded measurable $2\pi$-periodic function $g$
 \begin{equation}\label{lnbound}
-    \|L_Ng\|_2\le \|g\|_2
+    \|L_Ng\|_{L^2[-\pi, \pi]}\le \|g\|_{L^2[-\pi, \pi]}
 \end{equation}
 \end{lemma}
 \begin{proof}
     We have
     \begin{equation}\label{mnbound}
-        \|M_ng\|_2^2=\int _0^{2\pi} |e^{inx}g(x)|^2\, dx
-        =\int _0^{2\pi} |g(x)|^2\, dx=\|g\|_2^2\, .
+        \|M_ng\|_2^2=\int_{-\pi}^{\pi} |e^{inx}g(x)|^2\, dx
+        =\int_{-\pi}^{\pi} |g(x)|^2\, dx=\|g\|_{L^2[-\pi, \pi]}^2\, .
     \end{equation}
-     We have by the triangle inequality, the square root of Identity \eqref{mnbound}, and Lemma \ref{lem l2sn}
+     We have by the triangle inequality, the square root of the identity in\eqref{mnbound}, and Lemma \ref{spectral projection bound}
     \begin{equation*}
         \|L_ng\|_2=\|\frac 1N\sum_{n=0}^{N-1}
        M_{-n-N} S_{N+n}M_{N+n}g\|_2
@@ -7147,14 +7000,16 @@ \section{The truncated Hilbert transform}
 
 \begin{lemma}\label{lem shift}
 \lean{Function.Periodic.intervalIntegral_add_eq, intervalIntegral.integral_comp_sub_right}
+\leanok
 Let $f$ be a bounded $2\pi$-periodic function. We have for any
 $0 \le x\le 2\pi$ that
 \begin{equation}
  \int_0^{2\pi} f(y)\, dy= \int_{-x}^{2\pi -x}   f(y)\, dy
- =\int_{0}^{2\pi}   f(y-x)\, dy
+ =\int_{-\pi}^{\pi}   f(y-x)\, dy
 \end{equation}
 \end{lemma}
 \begin{proof}
+\leanok
     We have  by periodicity and change of variables
     \begin{equation}\label{eqhil9}
  \int_{-x}^{0} f(y)\, dy=\int_{-x}^{0} f(y+2\pi)\, dy= \int_{2\pi -x}^{2\pi}   f(y)\, dy\, .
@@ -7178,19 +7033,19 @@ \section{The truncated Hilbert transform}
 \begin{lemma}\label{young}
     Let $f$ and $g$ be two bounded non-negative measurable $2\pi$-periodic functions on $\R$. Then
     \begin{equation}\label{eqyoung}
-        \left(\int_0^{2\pi} \left(\int_0^{2\pi}
+        \left(\int_{-\pi}^{\pi} \left(\int_{-\pi}^{\pi}
         f(y)g(x-y)\, dy\right)^2\, dx\right)^{\frac 12}\le \|f\|_2 \|g\|_1\, .
     \end{equation}
     \end{lemma}
 \begin{proof}
 Using Fubini and Lemma \ref{lem shift}, we observe
 \begin{equation*}
-  \int_0^{2\pi}\int_0^{2\pi}f(y)^2g(x-y)\, dy
-    \, dx=\int_0^{2\pi}f(y)^2\int_0^{2\pi}g(x-y)\, dx
+  \int_{-\pi}^{\pi}\int_{-\pi}^{\pi}f(y)^2g(x-y)\, dy
+    \, dx=\int_{-\pi}^{\pi}f(y)^2\int_{-\pi}^{\pi}g(x-y)\, dx
     \, dy
 \end{equation*}
 \begin{equation}\label{eqhil4}
-=\int_0^{2\pi}f(y)^2\int_0^{2\pi}g(x) \, dx
+=\int_{-\pi}^{\pi}f(y)^2\int_{-\pi}^{\pi}g(x) \, dx
      dy
 =\|f\|_2^2\|g\|_1\, .
 \end{equation}
@@ -7201,19 +7056,19 @@ \section{The truncated Hilbert transform}
    \eqref{eqyoung} with Cauchy-Schwarz and then with
    \eqref{eqhil4} by
        \begin{equation*}
-         \int_0^{2\pi} (\int_0^{2\pi}f(y)h(x-y)h(x-y)\, dy)^2\, dx
+         \int_{-\pi}^{\pi} (\int_{-\pi}^{\pi}f(y)h(x-y)h(x-y)\, dy)^2\, dx
    \end{equation*}
 \begin{equation}
-    \le \int_0^{2\pi}\left(\int_0^{2\pi}f(y)^2g(x-y)\, dy\right)
-    \left(\int_0^{2\pi}g(x-y)\, dy\right)\, dx
+    \le \int_{-\pi}^{\pi}\left(\int_{-\pi}^{\pi}f(y)^2g(x-y)\, dy\right)
+    \left(\int_{-\pi}^{\pi}g(x-y)\, dy\right)\, dx
    = \|f\|_2^2\|g\|_1^2\, .
 \end{equation}
 Taking square roots, this proves the lemma.
 \end{proof}
 
-For $0<r\le \frac \pi 2$, Define the kernel $k_r$ to be the $2\pi$-periodic function
+For $0<r<1$, Define the kernel $k_r$ to be the $2\pi$-periodic function
 \begin{equation}
-    |k(x)|:=\min \left(r^{-1}, 1+\frac r{|1-e^{ix}|^2}\right)\, ,
+    |k_r(x)|:=\min \left(r^{-1}, 1+\frac r{|1-e^{ix}|^2}\right)\, ,
 \end{equation}
 where the minimum is understood to be $r^{-1}$ in case $1=e^{ix}$.
 \begin{lemma}\label{krbound}
@@ -7225,7 +7080,7 @@ \section{The truncated Hilbert transform}
 \end{equation}
 Let
 \begin{equation}
-   h(x)= \int_0^{2\pi} f(y)g(x-y)\, dy \, .
+   h(x)= \int_{-\pi}^{\pi} f(y)g(x-y)\, dy \, .
 \end{equation}
 Then
 \begin{equation}
@@ -7237,46 +7092,46 @@ \section{The truncated Hilbert transform}
 \begin{proof}
 We compute with Lemma \ref{lem shift}
 \begin{equation}
-    \|g\|_1\le  \int_0^{2\pi}k_r(x)\, dx=\int_{-\pi}^\pi k_r(x)\, dx
+    \|g\|_1\le  \int_{-\pi}^{\pi}k_r(x)\, dx=\int_{-\pi}^\pi k_r(x)\, dx
 \end{equation}
 Using the symmetry
 $k_r(x)=k_r(-x)$ and the assumption,  the last display
 is equal to
 \begin{equation*}
-    =  2 \int_0^\pi \min\left(\frac 1r, \frac r{|1-e^{ix}|^2}\right)\, dx
+    =  2 \int_0^\pi \min\left(\frac 1r, 1+\frac r{|1-e^{ix}|^2}\right)\, dx
 \end{equation*}
 \begin{equation*}
-    \le 2\int_0^{r} \frac 1r \, dy+2\int_r^{\pi}1+\frac {8r}{x^2}\, dr
+    \le 2\int_0^{r} \frac 1r \, dx+2\int_r^{\pi}1+\frac {64r}{x^2}\, dx
 \end{equation*}
 \begin{equation}
-    \le 2+2\pi + 2\left(\frac {8r}r-\frac {8r}{\pi}\right)
+    \le 2+2\pi + 2\left(\frac {64r}r-\frac {64r}{\pi}\right)
     \le 2^{5}\, .
 \end{equation}
     Together with Lemma \ref{young}, this proves the lemma.
 \end{proof}
-
+\rs{Change power of $2$}
 \begin{lemma}\label{lem dirichlet2}
-Let $0<r<\pi$. Let $N$ be the smallest
+Let $0<r<1$. Let $N$ be the smallest
 integer larger than $\frac 1r$.
 There is a   $2\pi$-periodic continuous function
- ${L'}$ on $\R$ that satisfies for all $0\le x\le 2\pi$
+ ${L'}$ on $\R$ that satisfies for all $-\pi\le x\le \pi$
 and all $2\pi$-periodic bounded measurable functions $f$ on $\R$
 \begin{equation}\label{lthroughlprime}
-    L_Nf(x)=\frac 1{2\pi}\int_{0}^{2\pi}f(y) {L'}(x-y)\, dy
+    L_Nf(x)=\frac 1{2\pi}\int_{-\pi}^{\pi}f(y) {L'}(x-y)\, dy
 \end{equation}
 and
 \begin{equation}\label{eqdifflhil}
-    |L'(x)-\frac{\mathbf{1}_{[r, 2\pi -r]}(x)}{1-e^{-ix}}|\le 2^{10}k_r(x)\, .
+    \left|L'(x)-\frac{\mathbf{1}_{\{y:\, r<|y|<1\}}(x)\max(1-|x|, 0)}{1-e^{-ix}}\right|\le 2^{10}k_r(x)\, .
 \end{equation}
 \end{lemma}
 
 
 \begin{proof}
-We have by definition and Lemma \ref{dirichlet}
+We have by definition and Lemma \ref{dirichlet kernel}
 \begin{equation}
     L_Ng(x)=
     \frac 1N\sum_{n=0}^{N-1}
-       \int_0^{2\pi} e^{-i(N+n)x} K_{N+n}(x-y) e^{i(N+n)y}g(y)
+       \int_{-\pi}^{\pi} e^{-i(N+n)x} K_{N+n}(x-y) e^{i(N+n)y}g(y)
 \, dy \, .\end{equation}
 This is of the form \eqref{lthroughlprime} with
 the continuous function
@@ -7284,18 +7139,23 @@ \section{The truncated Hilbert transform}
     {L'}(x)=  \frac 1N\sum_{n=0}^{N-1}
       K_{N+n}(x) e^{-i(N+n)x}\, .
 \end{equation}
-With \eqref{eqksumexp} of Lemma \ref{dirichlet}
+With \eqref{eqksumexp} of Lemma \ref{dirichlet kernel}
 we have $|K_N(x)|\le N$ for every $x$ and thus
 \begin{equation}\label{eqhil13}
     |{L'}(x)|\le  \frac 1N\sum_{n=0}^{N-1}
       (N+n) \le 2N\le 2^{10} r^{-1}\, .
 \end{equation}
-
-
+Therefore, for $|x|\in [0, r)\cup (1, \pi]$, we have
+\begin{equation}
+    \label{eqdiffzero}
+    \left|L'(x)-\frac{\mathbf{1}_{\{y:\, r<|y|<1\}}(x)\max{(1-|x|, 0)}}{1-e^{-ix}}\right|=|L'(x)|\leq 2^{10} r^{-1}.
+\end{equation}
+This proves \eqref{eqdifflhil} for $|x|\in [0, r)$ since $k_r(x)=r^{-1}$ in this case.
+% The case $x\in [2\pi-r, 2\pi]$ also follows by the symmetry $k_r(-x)=k_r(x)$ and $2\pi$ periodicity of $k_r$.
 For $e^{ix'}\neq 1$
 and may use the expression
 \eqref{eqksumhil} for $K_N$
-in Lemma \ref{dirichlet} to obtain
+in Lemma \ref{dirichlet kernel} to obtain
 \begin{equation*}
     {L'}(x)=  \frac 1N\sum_{n=0}^{N-1}
      \left(\frac{e^{i(N+n)x}}{1-e^{-ix}}
@@ -7314,14 +7174,24 @@ \section{The truncated Hilbert transform}
 \end{equation}
 and thus
 \begin{equation}
-  L''(x):={L'}(x)  -\frac{1}{1-e^{-ix}}=\frac 1N \frac {e^{-i2Nx}}{1-e^{ix}}
+\label{eq L'L''}
+  {L'}(x)  -\frac{\mathbf{1}_{{\{y:\, r<|y|<1\}}}(x)\max{(1-|x|, 0)}}{1-e^{-ix}}=L''(x)+ \frac{1-\mathbf{1}_{{\{y:\, r<|y|<1\}}}(x)(1-|x|)}{1-e^{-ix}},
+\end{equation}
+where
+$$L''(x):=\frac 1N \frac {e^{-i2Nx}}{1-e^{ix}}
      \sum_{n=0}^{N-1}
-    {e^{-i2nx}}
+    {e^{-i2nx}}.$$
+For $x\in [-\pi, r]\cup [r, \pi]$, we have using Lemma \ref{lower secant bound} that
+\begin{equation*}
+   \left|\frac{1-\mathbf{1}_{{\{y:\, r<|y|<1\}}}(x)(1-|x|)}{1-e^{-ix}} \right|=\left|\frac{\min(|x|, 1)}{1-e^{-ix}} \right|\leq \frac{8\min(|x|, 1)}{|x|}
+\end{equation*}
+\begin{equation}
+ \label{eq diffzero2}
+    \leq 2^3\cdot 1\leq  2^{8} k_r(x).
 \end{equation}
-We need to estimate $L''(x)$
+% Using the symmetry $k_r(-x)=k_r(x)$ and the $2\pi$ periodicity of $k_r$, we also conclude the same for $x\in [\pi, 2\pi-r)$.
 
-
-If the real part of
+Next, we need to estimate $L''(x)$. If the real part of
 $e^{ix}$ is negative, we have
 \begin{equation}
   1\le   |1-e^{ix}|\le 2\, .
@@ -7346,7 +7216,7 @@ \section{The truncated Hilbert transform}
            \frac 1N \frac {e^{-i2Nx}}{1-e^{ix}}
      \frac{1-e^{-i2Nx}}{1-e^{-2ix}}\, .
 \end{equation}
-Hence, with Lemma \ref{expbound} and nonnegativity of the real part of $e^{ix}$
+Hence, with Lemma \ref{lower secant bound} and nonnegativity of the real part of $e^{ix}$
 \begin{equation*}
     |L''(x)|
  \le \frac 2 N \frac {1}{|1-e^{ix}|}
@@ -7357,75 +7227,552 @@ \section{The truncated Hilbert transform}
      \frac{1}{|1+e^{ix}|}\le
  \frac {4r}{|1-e^{ix}|^2}\le 2^{10} \left (1+\frac {r}{|1-e^{ix}|^2}\right)
 \end{equation}
-Inequalities \eqref{eqhil13}, \eqref{eqhil12}, and \eqref{eqhil11} prove \eqref{eqdifflhil}. This completes the proof of the lemma.
+Inequalities \eqref{eqhil13}, \eqref{eqdiffzero}, \eqref{eq L'L''}, \eqref{eq diffzero2}, \eqref{eqhil12}, and \eqref{eqhil11} prove \eqref{eqdifflhil}. This completes the proof of the lemma.
+\end{proof}
+% \begin{proof}
+% We have by definition and Lemma \ref{dirichlet kernel}
+% \begin{equation}
+%     L_Ng(x)=
+%     \frac 1N\sum_{n=0}^{N-1}
+%        \int_0^{2\pi} e^{-i(N+n)x} K_{N+n}(x-y) e^{i(N+n)y}g(y)
+% \, dy \, .\end{equation}
+% This is of the form \eqref{lthroughlprime} with
+% the continuous function
+% \begin{equation}
+%     {L'}(x)=  \frac 1N\sum_{n=0}^{N-1}
+%       K_{N+n}(x) e^{-i(N+n)x}\, .
+% \end{equation}
+% With \eqref{eqksumexp} of Lemma \ref{dirichlet kernel}
+% we have $|K_N(x)|\le N$ for every $x$ and thus
+% \begin{equation}\label{eqhil13}
+%     |{L'}(x)|\le  \frac 1N\sum_{n=0}^{N-1}
+%       (N+n) \le 2N\le 2^{8} r^{-1}\, .
+% \end{equation}
+
+
+% For $e^{ix'}\neq 1$
+% and may use the expression
+% \eqref{eqksumhil} for $K_N$
+% in Lemma \ref{dirichlet kernel} to obtain
+% \begin{equation*}
+%     {L'}(x)=  \frac 1N\sum_{n=0}^{N-1}
+%      \left(\frac{e^{i(N+n)x}}{1-e^{-ix}}
+%       +\frac {e^{-i(N+n)x}}{1-e^{ix}}\right) e^{-i(N+n)x}
+% \end{equation*}
+% \begin{equation*}
+%     =  \frac 1N\sum_{n=0}^{N-1}
+%     \left(\frac{1}{1-e^{-ix}}
+%       +\frac {e^{-i2(N+n)x}}{1-e^{ix}}\right)
+% \end{equation*}
+% \begin{equation}\label{eqhil3}
+%     =  \frac{1}{1-e^{-ix}} +
+%      \frac 1N \frac {e^{-i2Nx}}{1-e^{ix}}
+%      \sum_{n=0}^{N-1}
+%     {e^{-i2nx}}
+% \end{equation}
+% and thus
+% \begin{equation}
+% \label{eq L'L''}
+%   {L'}(x)  -\frac{\mathbf{1}_{[r, 2\pi -r]}(x)\max{(1-|x|, 0)}}{1-e^{-ix}}=L''(x)+ \frac{1-\mathbf{1}_{[r, 1]}(x)(1-x)}{1-e^{-ix}},
+% \end{equation}
+% where
+% $$L''(x):=\frac 1N \frac {e^{-i2Nx}}{1-e^{ix}}
+%      \sum_{n=0}^{N-1}
+%     {e^{-i2nx}}.$$
+% We first show
+% \begin{equation}
+%     \label{est L''}
+%     |L''(x)|\leq 2^8|k_r(x)|.
+% \end{equation}
+% If the real part of
+% $e^{ix}$ is negative, we have
+% \begin{equation}
+%   1\le   |1-e^{ix}|\le 2\, .
+% \end{equation}
+% and hence
+% \begin{equation}\label{eqhil12}
+%     |L''(x)|\le
+%      \frac 1N
+%      \sum_{n=0}^{N-1}
+%     1=1\le 2^{8}\left(1+\frac r{|1-e^{ix}|^2}\right)\, .
+% \end{equation}
+% If the real part of $e^{ix}$ is positive and in particular while still $e^{ix}\neq \pm 1$, then  we have by telescoping
+% \begin{equation}
+%  (1-e^{-2ix})
+%      \sum_{n=0}^{N-1}
+%     {e^{-i2nx}}=1-e^{-i2Nx}\, .
+% \end{equation}
+% As $e^{-2ix}\neq 1$, we may divide by $1-e^{-2ix}$ and insert this into
+% \eqref{eqhil3} to obtain
+% \begin{equation}
+%  L''(x)=
+%            \frac 1N \frac {e^{-i2Nx}}{1-e^{ix}}
+%      \frac{1-e^{-i2Nx}}{1-e^{-2ix}}\, .
+% \end{equation}
+% Hence, with Lemma \ref{lower secant bound} and nonnegativity of the real part of $e^{ix}$
+% \begin{equation*}
+%     |L''(x)|
+%  \le \frac 2 N \frac {1}{|1-e^{ix}|}
+%      \frac{1}{|1-e^{-2ix}|}
+%  \end{equation*}
+% \begin{equation}\label{eqhil11}
+%     = \frac 2 N \frac {1}{|1-e^{ix}|^2}
+%      \frac{1}{|1+e^{ix}|}\le
+%  \frac {4r}{|1-e^{ix}|^2}\le 2^{8} \left (1+\frac {r}{|1-e^{ix}|^2}\right)
+% \end{equation}
+% Inequalities \eqref{eqhil13}, \eqref{eqhil12}, and \eqref{eqhil11} prove \eqref{est L''}. Next, we estimate the second term on the right in \eqref{eq L'L''}. For $x\in [r, 1]$, we have using Lemma \ref{lower secant bound} that
+% \begin{equation*}
+% \left|\frac{1-\mathbf{1}_{[r, 1]}(x)(1-x)}{1-e^{-ix}}\right|=
+% \frac{|x|}{|1-e^{-ix}|}\leq \min\left(\frac{8|x|}{|x|}, \frac{8}{r}\right)\leq 2^8 k_r(x).
+% \end{equation*}
+% \rs{Argue for other cases}
+% Combining the above with \eqref{est L''} and the triangle inequality completes the proof of the lemma.
+% \end{proof}
+
+For $r\in (0,1)$, $x\in \mathbb{R}$ and $g$ a bounded, measurable function on $\mathbb{R}$ with bounded support, we let
+\begin{equation}
+\label{def H_r}
+H_r g(x):= \int_{r<|x-y|<1}
+g(y) \kappa(x-y)\, dy.
+\end{equation}
+We also define
+\begin{equation}
+\label{def Tr}
+     T_r g(x):=\sup_{|x-x'|<r}|H_r g(x')|=\sup_{|x-x'|<r}  \left|\int_{r<|x'-y|<1}
+g(y) \kappa(x'-y)\, dy\right|\,.
+\end{equation}
+Observe that
+\begin{equation*}
+T_*g(x)=\sup_{0<r_1<r_2<1}\sup_{|x-x'|<r_1}\left|H_{r_1} g(x')-H_{r_2} g(x')\right|\,,
+\end{equation*}
+so that
+\begin{equation}
+    \label{eq 1 sup}
+    T_*g(x)\leq 2\sup_{0<r<1} T_r g(x).
+\end{equation}
+Further, let
+\begin{equation}
+    T_0f(x):=\int f(y)\kappa(x-y) \, dy.
+\end{equation}
+
+\begin{lemma}
+    \label{lem R transfer}
+    Let $0<r<1$. Let $f$ be a bounded, measurable function on $\mathbb{R}$ with bounded support. Then
+    \begin{equation}
+        \label{eq Hr L2 bound}
+        \|H_rf\|_{2}\leq 2^{16} \|f\|_2.
+    \end{equation}
+\end{lemma}
+
+\begin{proof}
+    Suppose first that $f$ is supported in the interval $[-\pi, \pi$ with $f(-\pi)=f(\pi)=0$. Let $\tilde{f}$ be the $2\pi$-periodic extension of $f$ to $\mathbb{R}$. Let $N$ be the smallest
+    integer larger than $\frac 1r$. Then, by Lemma \ref{lem dirichlet2} and the triangle inequality, for $x\in [-\pi, \pi]$ we have
+    \begin{equation*}
+        |H_r \tilde{f}(x)|\leq |L_N \tilde{f}(x)|+2^{10}\left|\int_{-\pi}^{\pi}k_r(x-y)\tilde{f}(y)\, dy\right|.
+    \end{equation*}
+    Taking $L^2$ norm over the interval $[-\pi, \pi]$ and using its sub-additivity, we get
+    \begin{equation*}
+        \|H_r \tilde{f}\|_{L^2[-\pi, \pi]}\leq  \|L_N \tilde{f}\|_2\, + \left(\int_{-\pi}^{\pi} \left|\int_{-\pi}^{\pi}k_r(x-y)\tilde{f}(y)\, dy\right|^2\, dx\right)^{\frac{1}{2}}.
+    \end{equation*}
+Using Lemma \ref{krbound} and \ref{lem dirichlet2}, and passing back to $f$, we conclude
+\begin{equation}
+\|H_r f\|_{2}\leq \|f\|_2 + 2^{15}\|f\|_2.
+\end{equation}
+\rs{Pass to $\mathbb{R}$}
 \end{proof}
 
+\begin{lemma}
+    \label{lem kappa op L2 bound}
+    Let $f$ be a bounded, measurable function on $\mathbb{R}$ with bounded support. Then
+    \begin{equation}
+        \label{eq kappa op L2 bound}
+        \|T_0 f\|_2\leq 2^{17}\|f\|_2.
+    \end{equation}
+\end{lemma}
+\begin{proof}
+\rs{Write}
+\end{proof}
 
 \begin{lemma}\label{lem dirichlet3}
-Let $0<r<\pi$.  Let $0\le x\le 2\pi$ and let $|x'-x|\le r$.
+\rs{Replaced by the next lemma}
+Let $0<r<1$.  Let $x\in \mathbb{R}$ and let $|x'-x|\le r$. Then
 \begin{equation}\label{eqdifftrans}
-    |\frac{\mathbf{1}_{[r, 2\pi -r]}(x)}{1-e^{-ix}}-\frac{\mathbf{1}_{[r, 2\pi -r]}(x')}{1-e^{-ix'}}|\le 2^{10}k_r(x)\, .
+    \left|\frac{\mathbf{1}_{[r, 2\pi -r]}(x)\max{(1-|x|, 0)}}{1-e^{-ix}}-\frac{\mathbf{1}_{[r, 2\pi -r]}(x')\max{(1-|x'|, 0)}}{1-e^{-ix'}}\right|\le 2^{10}k_r(x)\, .
 \end{equation}
 \end{lemma}
 
 \begin{proof}
-We have by Lemma \ref{expbound} for each $0<x\le \pi$
+We have by Lemma \ref{lower secant bound} for each $0<x\le \pi$
 \begin{equation}
-    |\frac{\mathbf{1}_{[r, 2\pi -r]}(x)}{1-e^{-ix}}|\le \frac 8x\le \frac 8r\, .
+    \left|\frac{\mathbf{1}_{[r, 2\pi -r]}(x)\max(1-|x|, 0)}{1-e^{-ix}}\right|\le \frac 8x\le \frac 8r\, .
 \end{equation}
-By symmetry $k_r(x)=k_r(-x)$, the analogous inequality holds for $-\pi \le x\le 0$
+By symmetry $k_r(x)=k_r(-x)$, the analogous inequality holds for $-\pi \le x\le 0$.
 By periodicity, the analogous inequality holds for $-\pi\le x\le 3\pi$.
 Hence
 \begin{equation}\label{eqhil16}
-    |\frac{\mathbf{1}_{[r, 2\pi -r]}(x)}{1-e^{-ix}}-\frac{\mathbf{1}_{[r, 2\pi -r]}(x')}{1-e^{-ix'}}|\le 2^{3}r^{-1}\, .
-\end{equation}
-It remains to show
-\begin{equation}\label{eqhil17}
-    |\frac{\mathbf{1}_{[r, 2\pi -r]}(x)}{1-e^{-ix}}-\frac{\mathbf{1}_{[r, 2\pi -r]}(x')}{1-e^{-ix'}}|\le 2^{10}\left(1+\frac{r}{|1-e^{ix}|^2}\right)\, .
+    \left|\frac{\mathbf{1}_{[r, 2\pi -r]}(x)\max(1-|x|, 0)}{1-e^{-ix}}-\frac{\mathbf{1}_{[r, 2\pi -r]}(x')\max(1-|x'|, 0)}{1-e^{-ix'}}\right|\le 2^{3}r^{-1}\, .
 \end{equation}
+It remains to show the the left hand side above is no greater than $2^{10}\left(1+\frac{r}{|1-e^{ix}|^2}\right)$,
+% \begin{equation}\label{eqhil17}
+%     \left|\frac{\mathbf{1}_{[r, 2\pi -r]}(x)\max(1-|x|, 0)}{1-e^{-ix}}-\frac{\mathbf{1}_{[r, 2\pi -r]}(x')\max(1-|x'|, 0)}{1-e^{-ix'}}\right|\le 2^{10}\left(1+\frac{r}{|1-e^{ix}|^2}\right)\, .
+% \end{equation}
 for those $x$ such that
 \begin{equation}\label{eqhil15}
  2^{10}\left(1+\frac{r}{|1-e^{ix}|^2}\right)\le 2^{3}r^{-1}\, .
 \end{equation}
- For those $x$, the left-hand side  of \eqref{eqhil17} becomes
+We consider four cases. If both $x, x'\not\in [r, 1]$, then the left hand side of \eqref{eqhil16} is zero and we are done. If $\{x, x'\}\cap [r, 1]=\{x\}$, then by the triangle inequality, either $|x|\leq 2r$ or $|x|\geq 1-r$. In the first instance, the left hand side of \eqref{eqhil16} is
+\begin{equation*}
+    \left|\frac{\max(1-|x|, 0)}{1-e^{-ix}}\right|\leq \frac{|1-e^{-ix}|}{|1-e^{-ix}|^2}.
+\end{equation*}
+This can be estimated by
+\begin{equation*}
+    \leq \frac{|x|}{|1-e^{-ix}|^2}\leq \frac{2r}{|1-e^{-ix}|^2}\leq 2^{10}\left(1+\frac{r}{|1-e^{ix}|^2}\right),
+\end{equation*}
+which is the desired inequality. Alternatively, when $|x|\geq 1-r$, we have
+\begin{equation*}
+    \left|\frac{\max(1-|x|, 0)}{1-e^{-ix}}\right|\leq \frac{r |1-e^{-ix}|}{|1-e^{-ix}|^2}.
+\end{equation*}
+This can again be estimated by
+\begin{equation*}
+    \leq \frac{2 r}{|1-e^{-ix}|^2}\leq 2^{10}\left(1+\frac{r}{|1-e^{ix}|^2}\right).
+\end{equation*}
+The case when $\{x, x'\}\cap [r, 1]=\{x'\}$ is dealt with similarly.
+
+Finally, when both $|x|, |x'|\in [r, 1]$, the left-hand side  of \eqref{eqhil16} becomes
  \begin{equation}
-    |\frac{1}{1-e^{-ix}}-\frac{1}{1-e^{-ix'}}|\, .
+    \left|\frac{1-x}{1-e^{-ix}}-\frac{1-x'}{1-e^{-ix'}}\right|\, .
 \end{equation}
-If $x\ge x'$, we can write this as
+If $x'\ge x$, using the fundamental theorem of calculus, we can write this as
 \begin{equation*}
-    |\int_{x'}^x \frac{-ie^{-iy}}{(1-e^{-iy})^2}\, dy|
+\left|\int_{x}^{x'} \frac{-iye^{-iy}+(1-e^{-iy})}{(1-e^{-iy})^2}\, dy\right|\leq 4|x-x'| \sup_{x\leq y\leq x'<1} \frac{1}{|1-e^{-iy}|^2}.
 \end{equation*}
+Using the decreasing property of $\frac{1}{|1-e^{-iy}|^2}$ on $[0,1]$, we deduce that the right hand side above can be estimated by
+$$\frac{4r}{|1-e^{-ix}|^2}\leq 2^{10}\left(1+\frac{r}{|1-e^{-ix}|^2}\right).$$
+\rs{The case when $x'<x$}
+% \begin{equation*}
+%     \left|\int_{x'}^x \frac{-ie^{-iy}}{(1-e^{-iy})^2}\, dy\right|
+% \end{equation*}
+% \begin{equation}
+%     \le \int_{x'}^x \frac{64}{r^2}\, dy=\frac{64(x-x')}{r^2}\le \frac{64}r\, .
+% \end{equation}
+\end{proof}
+
+\begin{lemma}
+    \label{lem non tang diff}
+    Let $0<r<1$ and  $x\in \mathbb{R}$. Let $g$ be a bounded measurable function with bounded support on $\mathbb{R}$. Then, there exists a measurable function $Mg: \mathbb{R}\to [0, \infty)$ satisfying
+$$\|Mg\|_2\leq 2^{17}\|g\|_2\,,$$ such that for all $x'$ with $|x-x'|<\frac{r}{2}$.
 \begin{equation}
-    \le \int_{x'}^x \frac{64}{r^2}\, dy=\frac{64(x-x')}{r^2}\le \frac{64}r\, .
+\label{eq non-tang diff}
+|H_rg(x)-H_rg(x')|\leq 2^{12} Mg(x)\, .
 \end{equation}
+\rs{Upgrade to $|x-x'|<r$}
+\end{lemma}
 
+\begin{proof}
+We have
+$$|H_rg(x)-H_rg(x')|$$ $$\leq  \int_{\{y:\, r<|x-y|<1\}\cup \{y:\, r<|x'-y|<1\}} |\kappa(x-y)-\kappa(x'-y)||g(y)|\, dy.$$
+For $y$ with $|x'-y|>r$, we have
+$$|x-y|\geq |x'-y|-|x-x'|>r-\frac{r}{2}=\frac{r}{2}.$$
+Therefore the domain of integration above is contained in the set
+$\{y:\, |x-y|<\frac{r}{2}\}.$ Decomposing this set into dyadic ranges, we see that the integral above is estimated by
+\begin{equation}
+    \label{eq rKer dyadic sum}
+    \sum_{j=0}^{\lfloor \log_2(\frac{1}{r})\rfloor+1}\int_{2^{j-1}r\leq |x-y|<2^{j}r}|\kappa(x-y)-\kappa(x'-y)||g(y)|\, dy.
+\end{equation}
+% Since $|x-x'|<\frac{r}{2}<\frac{|x-y|}{2}$, we can argue as in Lemma \ref{Hilbert kernel regularity} to get the estimate
+We claim that
+\begin{equation}
+\label{eq kappa reg Tr}
+    |\kappa(x-y)-\kappa(x'-y)|\leq 2^{10}\frac{1}{|x-y|} \frac{|x-x'|}{|x-y|}\,,
+\end{equation}
+%Detailed proof
+\rs{This proof is repeated several times so can be a lemma at the beginning of the section}
+To see the above, let $z=x-y$ and $z'=x-y'$. Since $|z-z'|<\frac{r}{2}<\frac{|z|}{2}$, we conclude that $z, z'$ have the same sign. Since $\kappa(z)=\overline{\kappa(-z)}$, we may assume without loss of generality that they are both positive. Then our assumption implies that
+    $$
+        \frac{z}{2} \le z' \,.
+    $$
+    We now consider four cases. If $z, z' \le 1$, then we have
+    $$
+        |\kappa(z)-\kappa(z')| = \left|  \frac{1 - z}{1- e^{-iz}} - \frac{1 - z'}{1- e^{-iz'}}\right|
+    $$
+    and by the fundamental theorem of calculus
+    $$
+        = \left| \int_{z'}^{z} \frac{-1 + e^{-it} + i(1-t)e^{it}}{(1 - e^{-it})^2} \,dt \right|\,.
+    $$
+    Using $z' \ge \frac{z}{2}$ and Lemma \ref{lower secant bound}, we bound this by
+    $$
+        \le |z - z'| \sup_{\frac{z}{2} \le t \le 1} \frac{3}{|1 - e^{-it}|^2}  \le 3 |z-z'| (8 \frac{2}{z})^2 \le 2^{10} \frac{|z-z'|}{|z|^2}\,.
+    $$
+    If $z \le 1$ and $z' > 1$, then $\kappa(z')= 0$ and we have from the first case
+    $$
+        |\kappa(z)-\kappa(z')| = |\kappa(z)| \le 2^{10} \frac{|z-1|}{|z|^2} \le 2^{10} \frac{|z-z'|}{|z|^2}\,.
+    $$
+    The case when $z > 1$ and $z' \le 1$ can be handled similarly.
+    Finally, if $z, z' > 1$ then
+    $$
+        |\kappa(z)-\kappa(z')| = 0 \le 2^{10} \frac{|z-z'|}{|z|^2}\,.
+    $$
+
+Plugging \eqref{eq kappa reg Tr} into \eqref{eq rKer dyadic sum}, we get
+\begin{equation}
+    \label{eq rker HL}
+    |H_rg(x)-H_rg(x')|\leq 2^{10} \sum_{j=0}^{\lfloor \log_2(\frac{1}{r})\rfloor+1}\frac{r}{2}(2^{j-1}r)^{-2}\int_{|x-y|<2^{j}r}|g(y)|\, dy.
+\end{equation}
 
+By Proposition \ref{Hardy Littlewood}, there exists a measurable function $Mg: \mathbb{R}\to [0, \infty)$ such that
+for each $j$ in the above sum,
+$$\frac{1}{2^{j}r}\int_{|x-y|<2^{j}r}|g(y)|\, dy\leq  Mg(x),$$
+and
+$$\|Mg\|_2\leq 2^{17}\|g\|_2.$$
+In combination with \eqref{eq rker HL}, this yields
+\begin{equation*}
+ |H_rg(x)-H_rg(x')|\leq 2^{10} \sum_{j=0}^{\lfloor \log_2(\frac{1}{r})\rfloor+1}2^{-j+1}Mg(x)\leq 2^{12} Mg(x).
+\end{equation*}
 \end{proof}
 
-We now prove Lemma \ref{lem hilbert}.
-We have with $r$ and $N$ as above for every $x$
+
+
+
+% \begin{lemma}\label{lem dirichlet3}
+% Let $0<r<\pi$.  Let $0\le x\le 2\pi$ and let $|x'-x|\le r$.
+% \begin{equation}\label{eqdifftrans}
+%     |\frac{\mathbf{1}_{[r, 2\pi -r]}(x)}{1-e^{-ix}}-\frac{\mathbf{1}_{[r, 2\pi -r]}(x')}{1-e^{-ix'}}|\le 2^{10}k_r(x)\, .
+% \end{equation}
+% \end{lemma}
+% \rs{Change to the correct kernel}
+% \begin{proof}
+% We have by Lemma \ref{lower secant bound} for each $0<x\le \pi$
+% \begin{equation}
+%     \left|\frac{\mathbf{1}_{[r, 2\pi -r]}(x)}{1-e^{-ix}}\right|\le \frac 8x\le \frac 8r\, .
+% \end{equation}
+% By symmetry $k_r(x)=k_r(-x)$, the analogous inequality holds for $-\pi \le x\le 0$
+% By periodicity, the analogous inequality holds for $-\pi\le x\le 3\pi$.
+% Hence
+% \begin{equation}\label{eqhil16}
+%     |\frac{\mathbf{1}_{[r, 2\pi -r]}(x)}{1-e^{-ix}}-\frac{\mathbf{1}_{[r, 2\pi -r]}(x')}{1-e^{-ix'}}|\le 2^{3}r^{-1}\, .
+% \end{equation}
+% It remains to show
+% \begin{equation}\label{eqhil17}
+%     |\frac{\mathbf{1}_{[r, 2\pi -r]}(x)}{1-e^{-ix}}-\frac{\mathbf{1}_{[r, 2\pi -r]}(x')}{1-e^{-ix'}}|\le 2^{10}\left(1+\frac{r}{|1-e^{ix}|^2}\right)\, .
+% \end{equation}
+% for those $x$ such that
+% \begin{equation}\label{eqhil15}
+%  2^{10}\left(1+\frac{r}{|1-e^{ix}|^2}\right)\le 2^{3}r^{-1}\, .
+% \end{equation}
+%  For those $x$, the left-hand side  of \eqref{eqhil17} becomes
+%  \begin{equation}
+%     |\frac{1}{1-e^{-ix}}-\frac{1}{1-e^{-ix'}}|\, .
+% \end{equation}
+% If $x\ge x'$, we can write this as
+% \begin{equation*}
+%     |\int_{x'}^x \frac{-ie^{-iy}}{(1-e^{-iy})^2}\, dy|
+% \end{equation*}
+% \begin{equation}
+%     \le \int_{x'}^x \frac{64}{r^2}\, dy=\frac{64(x-x')}{r^2}\le \frac{64}r\, .
+% \end{equation}
+
+
+% \end{proof}
+
+For $x\in \mathbb{R}$ and $r\in (0, 1)$, we shall make use of the following decomposition of $g$.
 \begin{equation}
-|T_rg(x)|\le |L_Ng(x)|+ |H_rg(x)-L_Ng(x)|+ |T_rg(x)-H_rg(x)|\, .
+    \label{eq g dec}
+    g_{x, 1}:=g\cdot \mathbf{1}_{\{y: |x-y|<r \}},\qquad g_{x, 2}:=g-g_{x,1}\,.
 \end{equation}
-With Lemma \ref{lem dirichlet2} and Lemma \ref{dirichlet}, we estimate this by
 
+\begin{lemma}
+\label{lem Tr to T}
+Let $0<r<1$ and  $x\in \mathbb{R}$. Let $g$ be a bounded measurable function on $\mathbb{R}$ with bounded support. Then, for all $x'\in \mathbb{R}$, we have
+\begin{multline}
+\label{Tr est1}
+|T_rg(x)|\leq  |T_0g_{x,2}(x)-T_0g_{x,2}(x')|+|T_0 g(x')|+|T_0 g_{x,1}(x')|+2^{12} Mg(x)\,,
+%\\+2^{10}|\int_0^{2\pi} g(y) k_r(x-y)\, dy|\, .
+\end{multline}
+\end{lemma}
+where $Mg$ is as in Lemma \ref{lem non tang diff}.
+\begin{proof}
+For every $x$ under consideration, we have
+\begin{equation*}
+|T_rg(x)|\le |H_rg(x)|+ |T_rg(x)-H_rg(x)|\, .
+\end{equation*}
+Using Lemma \ref{lem non tang diff}, we obtain the estimate
+\begin{equation*}
+\label{eq TrHr}
+|T_rg(x)|\leq |H_rg(x)|+ 2^{12}\left|Mg(x)\right|\, .
+\end{equation*}
+Observe that
+\begin{equation*}
+    \label{gg2}
+    H_r g(x)= T_0g_{x,2}(x).
+\end{equation*}
+Using the triangle inequality twice, we estimate
+\begin{equation*}
+\label{eq diff xx'}
+|T_0g_{x,2}(x)|\leq |T_0g_{x,2}(x)-T_0g_{x,2}(x')|+|T_0g_{x,1}(x')| + |T_0g(x')|.
+\end{equation*}
+Putting the above inequalities together, we obtain \eqref{Tr est1}.
+% For the last term on the right above, another application of triangle inequality yields
+% \begin{equation*}
+% |H_rg(x')|\leq  |L_N g(x')|+|H_rg(x')-L_N g(x')|.
+% \end{equation*}
+% Combining the above with Lemma \ref{lem dirichlet2}, we have
+% \begin{equation}
+% \label{eq diff x'g}
+% |H_rg(x')|\leq |L_N g(x')|+ 2^{10}\left|\int_0^{2\pi} g(y) k_r(x'-y)\, dy\right|.
+% \end{equation}
+% Similarly, we can estimate
+% \begin{equation}
+% \label{eq diff x'g2}
+% |H_rg_{x,1}(x')|\leq  |L_N g_{x,1}(x')|+ 2^{10}\left|\int_0^{2\pi} g_{x,1}(y) k_r(x'-y)\, dy\right|.
+% \end{equation}
+\end{proof}
+
+\begin{lemma}
+    \label{lem cotlar}
+    Let $x\in \mathbb{R}$. Let $g$ be a bounded, measurable function with bounded support on $\mathbb{R}$. Then, there exists $x'$ with $|x-x'|<\frac{r}{2}$ such that
+such that
 \begin{equation}
-\le |L_Ng(x)|+ 2^{11}|\int_0^{2\pi} g(y) k_r(x-y)\, dy|\, .
+    \label{eq cotlar}
+    |T_0g_{x,1}(x')|+|T_0g(x')|\leq  2^3 [(M(|T_0g|^{\frac{3}{2}})(x))^{\frac{2}{3}}+(M(|g|^2)(x))^{\frac{1}{2}}],
 \end{equation}
-
-By monotonicity and sub-additivity of the $L^2$ norm, we obtain
+where $(M(|T_0g|^{\frac{3}{2}})$ and $M(|g|^2)$ are measurable functions satisfying \rs{The second one will only satisfy a weak (2, 2) type inequality}
+\end{lemma}
+\begin{proof}
+For $\alpha>0$, we have
 \begin{equation}
-    \|T_rg\|_2\le \|L_N\|_2+2^{11}\|g*k_r\|_2\ .
+    \mu\left(\{x'\in \mathbb{R}: |x'-x|<\frac,\, |T_0g(x')|>\alpha\}\right)\leq \alpha^{-\frac{3}{2}}\int_{|x-y|<\frac{r}{2}} |T_0g(y)|^2\, dy.
 \end{equation}
+Using Proposition \ref{Hardy Littlewood} (more precisely, \eqref{eq hlm 2} with $w=|T_0g|^{\frac{3}{2}}$), we conclude that there exists a measurable function $M(|T_0g|^{\frac{3}{2}})$ such that the right hand side above can be estimated by
+\begin{equation}
+    \label{eq HL1}
+    \alpha^{-{\frac{3}{2}}} M(|T_0g|^{\frac{3}{2}})(x)\mu(\{x'\in \mathbb{R}: |x-x'|<\frac{r}{2}\}).
+    %= 2\frac{r}{2} \alpha^{-2} M(|L_Ng|^2)(x).
+\end{equation}
+Moreover, this function satisfies the estimate
+$$\|M(|T_0g|^{\frac{3}{2}})^{\frac{2}{3}}\|_2\leq 2^{\frac{2.16}{3}}\left(\frac{2}{2-\frac{3}{2}}\right)^{\frac{2}{3}}\|T_0g\|_2\leq 2^{12} \|T_0g\|_2.$$
+Similarly, we have
+\begin{equation*}
+    \mu\left(\{x'\in \mathbb{R}: |x'-x|<\frac{r}{2},\, |T_0g_{x,1}(x')|>\alpha\}\right)\leq \alpha^{-2}\int_{|x-y|<\frac{r}{2}} |T_0g_{x,1}(y)|^2\, dy.
+\end{equation*}
+This time, we use Lemma \ref{lem kappa op L2 bound} to estimate the above by
+\begin{equation*}
+    2\alpha^{-2}\int |T_0g_{x,1}(y)|^2\, dy \leq 2^2 \alpha^{-2}\int |g_{x,1}(y)|^2\, dy
+    = 2^2 \alpha^{-2}\int_{|y-x'|<\frac{r}{2}} |g(y)|^2 \,dy.
+\end{equation*}
+Using Proposition \ref{Hardy Littlewood} again (more precisely, \eqref{eq hlm 2} with $w=|g|^2$), we obtain another measurable function $M(|g|^2)$ such that the above can be estimated by
+\begin{equation}
+    \label{eq HL2}
+    2^2\alpha^{-2} M(|g|^2)(x)\mu(\{x'\in \mathbb{R}: |x-x'|<\frac{r}{2}\}).
+    %= 2^3 \frac{r}{2} \alpha^{-2} M(|g|^2)(x).
+\end{equation}
+From \eqref{eq HL1} and \eqref{eq HL2}, we deduce that for
+\begin{equation*}
+    \label{eq alph choice}
+    \alpha=\max\left(4 (M(|T_0g|^{\frac{3}{2}})(x))^{\frac{2}{3}}, 4 (M(|g|^2)(x))^{\frac{1}{2}} \right),
+\end{equation*}
+we have the estimate
+\begin{multline*}
+    \mu(\{x'\in \mathbb{R}: |x-x'|<\frac{r}{2}, |T_0g_{x,1}(x')|>\alpha \text{ or }|T_0g(x')|>\alpha\}\\\leq \frac{1}{2} \mu(\{x'\in \mathbb{R}: |x-x'|<\frac{r}{2}\}.
+\end{multline*}
+Making this choice of $\alpha$, we conclude that there exists an $x'$ with $|x-x'|<\frac{r}{2}$ such that \eqref{eq cotlar} is true.
+\end{proof}
 
-By Lemma \ref{lnbound} and Lemma\ref{krbound}, we obtain
-
+\begin{lemma}
+\label{lem xx' g2 diff}
+Let $0<r<1$ and  $x\in \mathbb{R}$. Let $g$ be a bounded measurable function with bounded support on $\mathbb{R}$. Then, there exists a measurable function $Mg: \mathbb{R}\to [0, \infty)$ satisfying
+$$\|Mg\|_2\leq 2^{17}\|g\|_2\,,$$ such that for all $x'$ with $|x-x'|<\frac{r}{2}$.
 \begin{equation}
-    \|T_rg\|_2\le 2^{30}\|g\|_2\ .
+\label{eq xx' g2 diff}
+|T_0g_{x,2}(x)-T_0g_{x,2}(x')|\leq 2^{11} Mg(x)\, .
 \end{equation}
+\end{lemma}
+\begin{proof}
+We have
+$$|T_0g_{x,2}(x)-T_0g_{x,2}(x')|\leq \int_{r<|x-y|<1} |\kappa(x-y)-\kappa(x'-y)||g(y)|\, dy.$$
+Decomposing the range of the integral into dyadic ranges, we see the above is estimated by
+\begin{equation}
+    \label{eq hilbker dyadic sum}
+    \sum_{j=0}^{\lfloor \log_2(\frac{1}{r})\rfloor}\int_{2^jr\leq |x-y|<2^{j+1}r}|\kappa(x-y)-\kappa(x'-y)||g(y)|\, dy.
+\end{equation}
+% Since $|x-x'|<\frac{r}{2}<\frac{|x-y|}{2}$, we can argue as in Lemma \ref{Hilbert kernel regularity} to get the estimate
+We claim that
+\begin{equation}
+\label{eq kappa reg}
+    |\kappa(x-y)-\kappa(x'-y)|\leq 2^{10}\frac{1}{|x-y|} \frac{|x-x'|}{|x-y|}\,,
+\end{equation}
+%Detailed proof
+To see the above, let $z=x-y$ and $z'=x-y'$. Since $|z-z'|<\frac{r}{2}<\frac{|z|}{2}$, we conclude that $z, z'$ have the same sign. Since $\kappa(z)=\overline{\kappa(-z)}$, we may assume without loss of generality that they are both positive. Then our assumption implies that
+    $$
+        \frac{z}{2} \le z' \,.
+    $$
+    We now consider four cases. If $z, z' \le 1$, then we have
+    $$
+        |\kappa(z)-\kappa(z')| = \left|  \frac{1 - z}{1- e^{-iz}} - \frac{1 - z'}{1- e^{-iz'}}\right|
+    $$
+    and by the fundamental theorem of calculus
+    $$
+        = \left| \int_{z'}^{z} \frac{-1 + e^{-it} + i(1-t)e^{it}}{(1 - e^{-it})^2} \,dt \right|\,.
+    $$
+    Using $z' \ge \frac{z}{2}$ and Lemma \ref{lower secant bound}, we bound this by
+    $$
+        \le |z - z'| \sup_{\frac{z}{2} \le t \le 1} \frac{3}{|1 - e^{-it}|^2}  \le 3 |z-z'| (8 \frac{2}{z})^2 \le 2^{10} \frac{|z-z'|}{|z|^2}\,.
+    $$
+    If $z \le 1$ and $z' > 1$, then $\kappa(z')= 0$ and we have from the first case
+    $$
+        |\kappa(z)-\kappa(z')| = |\kappa(z)| \le 2^{10} \frac{|z-1|}{|z|^2} \le 2^{10} \frac{|z-z'|}{|z|^2}\,.
+    $$
+    The case when $z > 1$ and $z' \le 1$ can be handled similarly.
+    Finally, if $z, z' > 1$ then
+    $$
+        |\kappa(z)-\kappa(z')| = 0 \le 2^{10} \frac{|z-z'|}{|z|^2}\,.
+    $$
+
+Plugging \eqref{eq kappa reg} into \eqref{eq hilbker dyadic sum}, we get
+\begin{equation*}
+    \label{eq hilbker HL}
+    |T_0g_{x,2}(x)-T_0g_{x,2}(x')|\leq 2^{10} \sum_{j=0}^{\lfloor \log_2(\frac{1}{r})\rfloor}\frac{r}{2}(2^jr)^{-2}\int_{|x-y|<2^{j+1}r}|g(y)|\, dy.
+\end{equation*}
+
+By Proposition \ref{Hardy Littlewood}, there exists a measurable function $Mg: \mathbb{R}\to [0, \infty)$ such that
+for each $j$ in the above sum,
+$$\frac{1}{2^{j+1}r}\int_{|x-y|<2^{j+1}r}|g(y)|\, dy\leq  Mg(x),$$
+and
+$$\|Mg\|_2\leq 2^{17}\|g\|_2.$$
+In combination with \eqref{eq hilbker HL}, this yields
+\begin{equation*}
+ |T_0g_{x,2}(x)-T_0g_{x,2}(x')|\leq 2^{10} \sum_{j=0}^{\lfloor \log_2(\frac{1}{r})\rfloor}2^{-j}Mg(x)\leq 2^{11} Mg(x).
+\end{equation*}
+\end{proof}
+
+
+\begin{proof}[Proof of Lemma \ref{lem hilbert}]
+We now prove Lemma \ref{lem hilbert}. \rs{Write}
+\end{proof}
+
+% We now prove Lemma \ref{lem hilbert}.
+% We have with $r$ and $N$ as above for every $x$
+% \begin{equation}
+% |T_rg(x)|\le |L_Ng(x)|+ |H_rg(x)-L_Ng(x)|+ |T_rg(x)-H_rg(x)|\, .
+% \end{equation}
+% With Lemma \ref{lem dirichlet2} and Lemma \ref{dirichlet kernel}, we estimate this by
+
+% \begin{equation}
+% \le |L_Ng(x)|+ 2^{11}|\int_0^{2\pi} g(y) k_r(x-y)\, dy|\, .
+% \end{equation}
+
+% By monotonicity and sub-additivity of the $L^2$ norm, we obtain
+% \begin{equation}
+%     \|T_rg\|_2\le \|L_N\|_2+2^{11}\|g*k_r\|_2\ .
+% \end{equation}
+
+% By Lemma \ref{lnbound} and Lemma \ref{krbound}, we obtain
+
+% \begin{equation}
+%     \|T_rg\|_2\le 2^{30}\|g\|_2\ .
+% \end{equation}
 
 
-\section{The proof of the van der Corput Lemma \ref{lem vdc}}
+\section{The proof of the van der Corput Lemma \ref{van der Corput}}
 \label{10vandercorput}
 
 Let $g$ be a Lipschitz continuous function as in the lemma. We have
@@ -7434,38 +7781,38 @@ \section{The proof of the van der Corput Lemma \ref{lem vdc}}
 $$
 Using this, we write
 $$
-    \int_a^b g(x) e^{-inx} \, \mathrm{d}x
-    = \frac{1}{2} \int_a^b g(x) e^{inx} \, \mathrm{d}x - \frac{1}{2} \int_a^b g(x) e^{in(x + \pi/n)}) \, \mathrm{d}x\,.
+    \int_\alpha^\beta g(x) e^{-inx} \, \mathrm{d}x
+    = \frac{1}{2} \int_\alpha^\beta g(x) e^{inx} \, \mathrm{d}x - \frac{1}{2} \int_\alpha^\beta g(x) e^{in(x + \pi/n)}) \, \mathrm{d}x\,.
 $$
-We split the the first integral at $a + \frac{\pi}{n}$ and the second one at $b - \frac{\pi}{n}$, and make a change of variables in the first part of the second integral to obtain
+We split the the first integral at $\alpha + \frac{\pi}{n}$ and the second one at $\beta - \frac{\pi}{n}$, and make a change of variables in the first part of the second integral to obtain
 $$
-    = \frac{1}{2} \int_{a}^{a + \frac{\pi}{n}} g(x) e^{inx} \, \mathrm{d}x - \frac{1}{2} \int_{b - \frac{\pi}{n}}^{b} g(x) e^{in(x + \pi/n)} \, \mathrm{d}x
+    = \frac{1}{2} \int_{\alpha}^{\alpha + \frac{\pi}{n}} g(x) e^{inx} \, \mathrm{d}x - \frac{1}{2} \int_{\beta - \frac{\pi}{n}}^{\beta} g(x) e^{in(x + \pi/n)} \, \mathrm{d}x
 $$
 $$
-    + \frac{1}{2} \int_{a + \frac{\pi}{n}}^{b} (g(x) - g(x - \frac{\pi}{n}) e^{inx} \, \mathrm{d}x\,.
+    + \frac{1}{2} \int_{\alpha + \frac{\pi}{n}}^{\beta} (g(x) - g(x - \frac{\pi}{n}) e^{inx} \, \mathrm{d}x\,.
 $$
 The sum of the first two terms is by the triangle inequality bounded by
 $$
-    \frac{\pi}{n} \sup_{x \in [a,b]} |g(x)|\,.
+    \frac{\pi}{n} \sup_{x \in [\alpha,\beta]} |g(x)|\,.
 $$
 The third term is by the triangle inequality at most
 $$
-    \frac{1}{2} \int_{a + \frac{\pi}{n}}^b |g(x) - g(x - \frac{\pi}{n})| \, \mathrm{d}x
+    \frac{1}{2} \int_{\alpha + \frac{\pi}{n}}^\beta |g(x) - g(x - \frac{\pi}{n})| \, \mathrm{d}x
 $$
 $$
-    \le \frac{\pi}{2n} \sup_{a \le x < y \le b} \frac{|g(x) - g(y)|}{|x-y|} |b-a|\,.
+    \le \frac{\pi}{2n} \sup_{\alpha \le x < y \le \beta} \frac{|g(x) - g(y)|}{|x-y|} |\beta-\alpha|\,.
 $$
 Adding the two terms, we obtain
 $$
-    \left|\int_a^b g(x) e^{-inx} \, \mathrm{d}x\right| \le \frac{\pi}{n} \|g\|_{\mathrm{Lip}(a,b)}\,.
+    \left|\int_\alpha^\beta g(x) e^{-inx} \, \mathrm{d}x\right| \le \frac{\pi}{n} \|g\|_{\mathrm{Lip}(\alpha,\beta)}\,.
 $$
 By the triangle inequality, we also have
 $$
-    \left|\int_a^b g(x) e^{-inx} \, \mathrm{d}x\right| \le |b -a| \sup_{x \in [a,b]} |g(x)| \le |b-a| \|g\|_{\mathrm{Lip}(a,b)}\,.
+    \left|\int_\alpha^\beta g(x) e^{-inx} \, \mathrm{d}x\right| \le |\beta -\alpha| \sup_{x \in [\alpha,\beta]} |g(x)| \le |\beta-\alpha| \|g\|_{\mathrm{Lip}(\alpha,\beta)}\,.
 $$
 This completes the proof of the lemma, using that
 $$
-    \min\{|b-a|, \frac{\pi}{n}\} \le 2 \pi |b-a|(1 + n|b-a|)^{-1}\,.
+    \min\{|\beta-\alpha|, \frac{\pi}{n}\} \le 2 \pi |\beta-\alpha|(1 + n|\beta-\alpha|)^{-1}\,.
 $$
 
 
@@ -7475,21 +7822,22 @@ \section{The proof of the van der Corput Lemma \ref{lem vdc}}
 \section{Partial sums as orthogonal projections}
 \label{10projection}
 
+This subsection proves Lemma \ref{spectral projection bound}
 
 
 
 
 
 
-
-\begin{lemma}\label{lem projection}
+\begin{lemma}[partial sum projection]
+\label{partial sum projection}
   Let $f$ be a bounded $2\pi$-periodic measurable function. Then, for all $N\ge 0$
    \begin{equation}\label{projection}
    S_N(S_N f)=S_Nf\, .
    \end{equation}
    \end{lemma}
 \begin{proof}
-Let $N>0$ be given. With $K_N$ as in Lemma \ref{dirichlet},
+Let $N>0$ be given. With $K_N$ as in Lemma \ref{dirichlet kernel},
 \begin{equation*}
 S_N (S_Nf) (x)=
 \int_0^{2\pi} S_Nf(y)K_N(x-y)\, dy
@@ -7498,7 +7846,7 @@ \section{Partial sums as orthogonal projections}
 =
 \int_0^{2\pi} \int_0^{2\pi} f(y')K_N(y-y') K_N(x-y)\, \, dy' dy\, .
 \end{equation}
-We have by Lemma \ref{dirichlet}
+We have by Lemma \ref{dirichlet kernel}
 \begin{equation*}
 \int_0^{2\pi} K_N(y-y') K_N(x-y)\,  dy
 \end{equation*}
@@ -7510,7 +7858,7 @@ \section{Partial sums as orthogonal projections}
 =\sum_{n=-N}^N\sum_{n'=-N}^N
 e^{i(n'x-ny')}\int_0^{2\pi} e^{i(n-n')y}\,  dy\, .
 \end{equation}
-By Lemma \ref{lem expintegral}, the summands for $n\neq n'$ vanish.
+By Lemma \ref{mean zero oscillation}, the summands for $n\neq n'$ vanish.
 We obtain for \eqref{eqhil6}
 \begin{equation}\label{eqhil2}
 =\sum_{n=-N}^N
@@ -7524,26 +7872,28 @@ \section{Partial sums as orthogonal projections}
 \end{equation}
 This proves the lemma.
 \end{proof}
-\begin{lemma}\label{selfadjoint}
+\begin{lemma}[partial sum selfadjoint]
+\label{partial sum selfadjoint}
+\uses{dirichlet kernel}
     We have for any $2\pi$-periodic bounded measurable $g,f$ that
     \begin{equation}
        \int_0^{2\pi} \overline{S_Nf(x)} g(x)=\int_0^{2\pi} \overline{f(x)} S_Ng(x)\, dx\, .
     \end{equation}
 \end{lemma}
 \begin{proof}
-  We have with $K_N$ as in Lemma \ref{dirichlet} for every $x$
+  We have with $K_N$ as in Lemma \ref{dirichlet kernel} for every $x$
   \begin{equation}
       \overline{K_N(x)}=\sum_{n=-N}^N\overline{ e^{in x}}=
       {\sum_{n=-N}^N e^{-in x}}=K_N(-x)\, .
   \end{equation}
- Further, with Lemma \ref{dirichlet} and Fubini
+ Further, with Lemma \ref{dirichlet kernel} and Fubini
 \begin{equation*}
 \int_0^{2\pi} \overline{S_Nf(x)} g(x)
-= \frac 1{2\pi} \int_0^{2\pi} \int_{0}^{2\pi}\overline{f(y) K_N(x-y)} g(x)\, dy dx
+= \frac 1{2\pi} \int_0^{2\pi} \int_{-\pi}^{\pi}\overline{f(y) K_N(x-y)} g(x)\, dy dx
  \end{equation*}
  \begin{equation}
 =
-\frac 1{2\pi} \int_0^{2\pi} \int_{0}^{2\pi}\overline{f(y)} K_N(y-x)
+\frac 1{2\pi} \int_0^{2\pi} \int_{-\pi}^{\pi}\overline{f(y)} K_N(y-x)
 g(x)\, dx dy
 =\int_0^{2\pi} \overline{f(x)} S_Ng(x)\, dx
 \, .
@@ -7552,8 +7902,10 @@ \section{Partial sums as orthogonal projections}
 \end{proof}
 
 
+We turn to the proof of Lemma
+\ref{spectral projection bound}.
 
-We have with Lemma \ref{selfadjoint}, then Lemma \ref{lem projection} and the Lemma\ref{selfadjoint} again
+We have with Lemma \ref{partial sum selfadjoint}, then Lemma \ref{partial sum projection} and the Lemma \ref{partial sum selfadjoint} again
 \begin{equation*}
  \int_0^{2\pi}  S_Nf(x)\overline{S_Nf(x)}\, dx
  \int_0^{2\pi}  f(x)\overline{S_N(S_Nf)(x)}\, dx
@@ -7602,75 +7954,173 @@ \section{Partial sums as orthogonal projections}
 
 \section{The error bound}
 \label{10difference}
+We prove Lemma \ref{control approximation effect}. Define
+$$
+    E_2 := \{x \in [0, 2\pi) \ : \ \sup_{N > 0} |S_N f(x) - S_N f_0(x)| > \frac{\epsilon}{4} \}\,.
+$$
+Then \eqref{eq max partial sum diff} clearly holds, and it remains to show that $|E_2| \le \frac{\epsilon}{2}$. This will follow from Lemma \ref{real Carleson}.
+
+Let $x \in E_2$. Then there exists $N > 0$ with
+$$
+    |S_N f(x) - S_N f_0(x)| > \frac{\epsilon}{4}\,,
+$$
+pick such $N$. We have with Lemma \ref{dirichlet kernel}
+$$
+    \frac{\epsilon}{4} < |S_N f(x) - S_N f_0(x)| = \frac{1}{2\pi} \left| \int_0^{2\pi} (f(y) - f_0(y)) K_N(x-y) \, dy\right|\,.
+$$
+We make a change of variables, replacing $y$ by $x -y$. Then we use $2\pi$-periodicity of $f$, $f_0$ and $K_N$ to shift the domain of integration to obtain
+$$
+    = \frac{1}{2\pi} \left|\int_{-\pi}^{\pi} (f(x-y) - f_0(x-y)) K_N(y) \, dy\right|\,.
+$$
+Using the triangle inequality, we split this as
+$$
+    \le \frac{1}{2\pi} \left|\int_{-\pi}^{\pi} (f(x-y) - f_0(x-y)) \max(1 - |y|, 0) K_N(y) \, dy\right|
+$$
+$$
+    + \frac{1}{2\pi} \left|\int_{-\pi}^{\pi} (f(x-y) - f_0(x-y)) \min(|y|, 1) K_N(y) \, dy\right|\,.
+$$
+Note that all integrals are well defined, since $K_N$ is by \eqref{eqksumexp} bounded by $2N+1$.
+Using \eqref{eqksumhil} and the definition \eqref{eq hilker} of $k$, we rewrite the two terms and obtain
+\begin{equation}
+    \label{eq diff singular}
+    \frac{\epsilon}{4} < \frac{1}{2\pi} \left| \int_{-\pi}^{\pi} (f(x-y) - f_0(x-y)) (e^{iNy} \overline{k}(y) + e^{-iNy}k(y)) \, dy\right|
+\end{equation}
+\begin{equation}
+    \label{eq diff integrable}
+    + \frac{1}{2\pi} \left|\int_{-\pi}^{\pi} (f(x-y) - f_0(x-y)) ( e^{iNy} \frac{\min\{|y|, 1\} }{1 - e^{-iy}} + e^{-iNy} \frac{\min\{|y|, 1\} }{1 - e^{iy}}) \, dy \right|\,.
+\end{equation}
+By Lemma \ref{lower secant bound} with $\eta = |y|$, we have for $-1 \le y \le 1$
+$$
+    |e^{iNy}\frac{\min\{|y|, 1\} }{1 - e^{-iy}}| = \frac{|y|}{|1 - e^{iy}|} \le 8\,.
+$$
+By Lemma \ref{lower secant bound} with $\eta = 1$, we have for $1 \le |y| \le \pi$
+$$
+    |e^{iNy}\frac{\min\{|y|, 1\} }{1 - e^{-iy}}| = \frac{1}{|1 - e^{iy}|} \le 8\,.
+$$
+Thus we obtain using the triangle inequality and  \eqref{eq ffzero}
+$$
+    \eqref{eq diff integrable} \le \frac{16}{2\pi} \int_{-\pi}^{\pi} |f(x-y) - f_0(x-y)| \, dy \le 2^{4-2^{50}} \epsilon^2\,.
+$$
+Consequently, we have that
+$$
+    \frac{\epsilon}{8} \le \frac{1}{2\pi} \left| \int_{-\pi}^{\pi} (f(x-y) - f_0(x-y)) (e^{iNy} \overline{k}(y) + e^{-iNy}k(y)) \, dy\right|\,.
+$$
+By dominated convergence and since $k(y) = 0$ for $|y| > 1$, this equals
+$$
+     = \frac{1}{2\pi} \lim_{r \to 0^+} \left| \int_{r < |y| < 1} (f(x-y) - f_0(x-y)) (e^{iNy} \overline{k}(y) + e^{-iNy}k(y)) \, dy\right|\,.
+$$
+Let $h = (f - f_0) \mathbf{1}_{[-\pi, 3\pi]}$. Since $x \in [0, 2\pi]$, the above integral does not change if we replace $(f - f_0)$ by $h$. We do that, apply the triangle inequality and bound the limits by suprema
+$$
+    \le \frac{1}{2\pi} \sup_{r > 0} \left| \int_{r < |y| < 1} h(x-y) e^{-iNy} k(y) \, dy\right|
+$$
+$$
+    + \frac{1}{2\pi} \sup_{r > 0} \left| \int_{r < |y| < 1} \overline{h}(x-y) e^{-iNy} k(y) \, dy\right|\,.
+$$
+By the definition \eqref{define T carleson} of $T$, this is
+$$
+    \le \frac{1}{2\pi} (Th(x) + T\bar{h}(x))\,.
+$$
+Thus for each $x \in E_2$, at least one of $Th(x)$ and $T\bar h(x)$ is larger than $\frac{\epsilon}{16}$.
+Thus at least one of $Th$ and $T\bar h$ is $\ge \frac{\epsilon}{16}$ on a subset $E_2'$ of $E_2$ with $2|E_2'| \ge |E_2|$. Without loss of generality this is $Th$. By assumption \eqref{eq ffzero}, we have $|2^{2^{50}}\epsilon^{-2} h| \le \mathbf{1}_{[-\pi, 3\pi]}$. Applying Lemma \ref{real Carleson} with $F = [-\pi, 3\pi]$ and $G = E_2'$, it follows that
+$$
+    \frac{\epsilon}{16} |E_2'| \le \int_{E_2'} Th(x) \, dx = 2^{-2^{50}}\epsilon^2 \int_{E_2'} T(2^{2^{50}}\epsilon^{-2} h)(x) \, dx
+$$
+$$
+    \le 2^{-2^{50}}  \epsilon^2 \cdot 2^{2^{40}}  |F|^{\frac{1}{2}} |E_2'|^{\frac{1}{2}} \le  2^{-2^{40}} \frac{\epsilon^2}{16} |E_2'|^{\frac{1}{2}}\,.
+$$
+Rearranging, we obtain
+$$
+    |E_2'| \le 2^{-2^{41}}\epsilon^2 \le \frac{\epsilon}{4}\,.
+$$
+This completes the proof using $|E_2| \le 2|E_2'|$.
+
+
+
+
 
 \section{Carleson on the real line}
 \label{10carleson}
 
-We prove Lemma \ref{lem rcarleson}.
+We prove Lemma \ref{real Carleson}.
 
 Consider the standard distance function
 \begin{equation}
     \rho(x,y)=|x-y|
 \end{equation}
 on the real line $\R$.
-\begin{lemma}
+\begin{lemma}[real line metric]
+\label{real line metric}
 The space $(\R,\rho)$ is a complete locally compact metric space.
 \end{lemma}
 \begin{proof}
     This is part of the Lean library.
 \end{proof}
-\begin{lemma}\label{lem ball int}
+\begin{lemma}
+\label{real line ball}
     \lean{Real.ball_eq_Ioo}
+    \leanok
     For $x\in R$ and $R>0$, the ball $B(x,R)$ is the interval $(x-R,x+R)$
 \end{lemma}
 \begin{proof}
+\leanok
 Let $x'\in B(x,R)$. By definition of the ball,
 $|x'-x|<R$. It follows that $x'-x<R$ and $x-x'<R$.
 It follows $x'<x+R$ and $x'>x-R$. This implies
 $x'\in (x-R,x+R)$.
 Conversely, let $x'\in (x-R,x+R)$. Then
 $x'<x+R$ and $x'>x-R$. It follows that
-$x'-x<R$ and $x-x'<R$. It follows that $|x'-x|<R$.
+$x'-x<R$ and $x-x'<R$. It follows that $|x'-x|<R$,
 hence $x'\in B(x,R)$.
-This implies the lemma.
+This proves the lemma.
 \end{proof}
 We consider the Lebesgue measure $\mu$ on $\R$.
-\begin{lemma}
+\begin{lemma}[real line measure]
+\label{real line measure}
     The measure $\mu$ is a sigma-finite non-zero
     Radon-Borel measure on $\R$.
 \end{lemma}
 \begin{proof}
     This is part of the Lean library.
 \end{proof}
-\begin{lemma}\label{lem ball meas}
+\begin{lemma}[real line ball measure]
+\label{real line ball measure}
+\uses{real line ball}
     \lean{Real.volume_ball}
+    \leanok
     We have for every $x\in \R$ and $R>0$
     \begin{equation}
         \mu(B(x,R))=2R\, .
     \end{equation}
 \end{lemma}
 \begin{proof}
-We have with lemma \ref{lem ball int}
+\leanok
+We have with Lemma \ref{real line ball}
 \begin{equation}
     \mu(B(x,R))=\mu((x-R,x+R))=2R\, .
 \end{equation}
-Where the last identity is taken from the Lean library. \lars{I dont think we have to write this, we dont do it for other trivial manipulations either.}
 \end{proof}
 
-\begin{lemma}\label{lem r doubling}
+\begin{lemma}[real line doubling]
+\label{real line doubling}
     We have for every $x\in \R$ and $R>0$
     \begin{equation}
         \mu(B(x,2R))=2\mu(B(x,R))\, .
     \end{equation}
 \end{lemma}
 \begin{proof}
-    We have with Lemma \ref{lem ball meas}
+    We have with Lemma \ref{real line ball measure}
 \begin{equation}
     \mu(B(x,2R)=4R=2\mu(B(x,R)\, .
 \end{equation}
 This proves the lemma.
 \end{proof}
 
+
+The preceeding four lemmas show that $(\R, \rho, \mu, 4)$ is a doubling metric measure space. Indeed, we even show
+that $(\R, \rho, \mu, 1)$ is a doubling metric measure space, but we may relax the estimate in Lemma \ref{real line doubling} to conclude that $(\R, \rho, \mu, 4)$
+is a doubling metric measure space.
+
+
 For each $n\in \mathbb{Z}$ define
 $\mfa_n:\R\to \R$ by
 \begin{equation}
@@ -7680,9 +8130,11 @@ \section{Carleson on the real line}
 
 Let $\Mf$ be the collection $\{\mfa_n, n\in \mathbb{Z}\}$.
 Note that for every $n\in \mathbb{Z}$ we have $\mfa_n(0)=0$.
-Define $d$ as in \eqref{definedE}. Note that
-with for
+Define $d_E$ as in \eqref{definedE}. Note that
+for
 $n,m\in \mathbb{Z}$ and $x\in \R$ and $x>0$
+we have
+
 \begin{equation}
     d_{B(x,R)}(\mfa_n,\mfa_m)=\sup_{y,y'\in B(x,R)}|ny-ny'-my+my'|
 \end{equation}
@@ -7707,7 +8159,8 @@ \section{Carleson on the real line}
 
 
 
-\begin{lemma}\label{lem fdb1}
+\begin{lemma}[frequency ball doubling]
+\label{frequency ball doubling}
   For any $x,x'\in \R$ and $R>0$ with
    $x\in B(x',2R)$  and any $n,m\in \mathbb{Z}$, we have
 \begin{equation}\label{firstdb1}
@@ -7718,19 +8171,20 @@ \section{Carleson on the real line}
 With \eqref{eqcarl4}, both sides of \eqref{firstdb1} are equal to $4R|n-m|$. This proves the lemma.
 \end{proof}
 
-\begin{lemma}\label{lem sdb1}
+\begin{lemma}[frequency ball growth]
+\label{frequency ball growth}
     For any $x,x'\in \R$ and $R>0$ with
    $B(x,R)\subset B(x',2R)$  and any $n,m\in \mathbb{Z}$, we have
 \begin{equation}\label{seconddb1}
-    2d_{B(x,R)}(\mfa_n,\mfa_m)\le 2 d_{B(x',2R)}(\mfa_n,\mfa_m) \, .
+    2d_{B(x,R)}(\mfa_n,\mfa_m)\le  d_{B(x',2R)}(\mfa_n,\mfa_m) \, .
 \end{equation}
 \end{lemma}
 \begin{proof}
 With \eqref{eqcarl4}, both sides of \eqref{firstdb1} are equal to $4R|n-m|$. This proves the lemma.
 \end{proof}
-\lars{Here, all the doubling properties are checked with $a = 1$. It is obvious that they are then also satisfied for larger $a$, but we do not prove that. In the end we will have to choose $a \ge 4$}
 
-\begin{lemma}\label{lem tdb1}
+\begin{lemma}[frequency ball cover]
+\label{frequency ball cover}
     For every $x\in \R$ and $R>0$ and every
     $n\in \mathbb{Z}$ and $R'>0$,
     there exist $m_1, m_2, m_3\in \mathbb{Z}$
@@ -7761,7 +8215,7 @@ \section{Carleson on the real line}
 Let $\mfa_{n'}\in B'$, then with \eqref{eqcarl4},
 we have
 \begin{equation}\label{eqcarl6}
-    R|n-n'|\le R'\, .
+    2R|n-n'|< 2R'\, .
 \end{equation}
 
 Assume first $n'\le n-R'/2$. By definition of $m_1$,
@@ -7771,14 +8225,14 @@ \section{Carleson on the real line}
     R|m_1-n'|=R(m_1-n')=R(m_1-n)+R(n-n')
 \end{equation*}
 \begin{equation}
-    \le -\frac{R'}2+R'=-\frac{R'}2\, .
+    < -\frac{R'}2+R'=-\frac{R'}2\, .
 \end{equation}
 We conclude $\mfa_{n'}\in B_1$.
 
 Assume next  $n-R'/2<n'<n+R'/2$. Then
 $\mfa_{n'}\in B_2$.
 
-Assume finally that $n+R'/2<n'$
+Assume finally that $n+R'/2\le n'$
 By definition of $m_3$,
 we have $m_3\le n'$.  With \eqref{eqcarl6}
 we have
@@ -7786,14 +8240,16 @@ \section{Carleson on the real line}
     R|m_3-n'|=R(n'-m_3)=R(n'-n)+R(n-m_3)
 \end{equation*}
 \begin{equation}
-    \le R' -\frac{R'}2=-\frac{R'}2\, .
+    < R' -\frac{R'}2=-\frac{R'}2\, .
 \end{equation}
 We conclude $\mfa_{n'}\in B_1$.
 This completes the proof of the lemma.
 \end{proof}
 
 
-\begin{lemma}
+\begin{lemma}[real van der Corput]
+\label{real van der Corput}
+\uses{van der Corput}
     For any $x\in \R$ and $R>0$ and any
     function $\varphi: X\to \C$ supported on $B'=B(x,R)$
     such that
@@ -7813,10 +8269,10 @@ \section{Carleson on the real line}
 Set $n'=n-m$. Then we have to prove
 \begin{equation}
     \label{eq vdc cond2}
-    \left|\int_{x-R}^{x+R} e^{in'y}\varphi(y) dy\right|\le 4\pi  R\|\varphi\|_{\Lip(B)}
+    \left|\int_{x-R}^{x+R} e^{in'y}\varphi(y) dy\right|\le 4\pi  R\|\varphi\|_{\Lip(B')}
 (1+2R|n'|)^{-1}\, .
 \end{equation}
-This follows from Lemma \ref{lem vdc} with $a = x - R$ and $b = x + 2R$.
+This follows from Lemma \ref{van der Corput} with $\alpha = x - R$ and $\beta = x + 2R$.
 %We do a case distinction whether $Rn'\le \pi$
 %or $Rn'> \pi$.
 %Assume first  $Rn'\le \pi$. We estimate the left-hand side  of \eqref{eq vdc cond2} by
@@ -7824,7 +8280,7 @@ \section{Carleson on the real line}
 %    2R\sup_{y\in B'}|\varphi(y)|\le 2^4R\|\varphi\|_{\Lip(B)}(1+2R|n'|)^{-1}\, .
 %\end{equation}
 %This proves \eqref{eq vdc cond2} in this case.
-
+%
 %Assume now $Rn'> \pi$.
 %We write
 %\begin{equation}\label{eqcarl10}
@@ -7878,52 +8334,91 @@ \section{Carleson on the real line}
 %with the triangle inequality proves \eqref{eq vdc cond2} in the given case and completes the proof of the lemma.
 \end{proof}
 
-With $k$ as in \eqref{eq hilker}, define
+The preceding three lemmas establish that $\Mf$ is a cancellative, compatible collection of functions on $(\R, \rho, \mu, 4)$. Again, some of the statements in these lemmas
+are stronger than what is needed for $a=4$, but can be relaxed to give the desired conclusion for $a=4$.
+
+
+
+
+With $\kappa$ as near \eqref{eq hilker}, define
 the function $K:\R\times \R\to \mathbb{C}$ by
 \begin{equation}
-    K(x,y):=k(x-y)\, .
+    K(x,y):=\kappa(x-y)\, .
 \end{equation}
 The function $K$ is continuous outside the diagonal
 $x=y$ and vanishes on the diagonal. Hence it is measurable.
 
-\begin{lemma}
+\begin{lemma}[Hilbert kernel bound]
+\label{Hilbert kernel bound}
     For $x,y\in \R$ with $x\neq y$ we have
     \begin{equation}\label{eqcarl30}
         |K(x,y)|\le 2^4(2|x-y|)^{-1}\, .
     \end{equation}
 \end{lemma}
 \begin{proof}
-    Fix $x\neq y$. We have
+    Fix $x\neq y$. If $K(x,y)$ is zero, then \eqref{eqcarl30} is evident. Assume $K(x,y)$ is not zero, then $0<|x-y|<1$.
+    We have
 \begin{equation}\label{eqcarl31}
-|K(x,y)|=\frac {\max (1-|x-y|, 0)}{1-e^{i(x-y)}}\, .
-\end{equation}
-We make the case distinction whether
-$|x-y|>1$ or $|x-y|\le 1$.
-Assume first $|x-y|>1$. Then \eqref{eqcarl31}
-vanishes and \eqref{eqcarl30} holds.
-Now assume $|x-y|\le 1$. The we estimate
-with Lemma \ref{expbound}
+|K(x,y)|=|\kappa(x-y)|=\left|\frac {1-|x-y|}{1-e^{i(x-y)}}\right|\, .
+\end{equation}
+We estimate
+with Lemma \ref{lower secant bound}
 \begin{equation}\label{eqcarl311}
 |K(x,y)|\le \frac {1}{|1-e^{i(x-y)}|}\le \frac 8{|x-y|}\, .
 \end{equation}
 This proves \eqref{eqcarl30} in the given case and completes the proof of the lemma.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[Hilbert kernel regularity]
+\label{Hilbert kernel regularity}
+\uses{lower secant bound}
     For $x,y,y'\in \R$ with $x\neq y,y'$ and
     \begin{equation}
+        \label{eq close hoelder}
         2|y-y'|\le |x-y|\, ,
     \end{equation}
     we have
     \begin{equation}\label{eqcarl301}
-        |K(x,y) - K(x, y')|\le 2^6\frac{|y-y'|}{2|x-y|}\, .
+        |K(x,y) - K(x, y')|\le 2^{10}\frac{1}{|x-y|} \frac{|y-y'|}{|x-y|}\, .
     \end{equation}
 \end{lemma}
 
-\lars{I checked until here, this subsection is not complete either.}
-
 \begin{proof}
-    ...
+    Since $K(x,y) = k(x-y) = K(0, y-x)$ we can assume that $x = 0$. Then the assumption \eqref{eq close hoelder} implies that $y$ and $y'$ have the same sign. Since $K(0,y) = k(y) = \bar k(-y) = \bar K(0, -y)$ we can assume that they are both positive. Then it follows from \eqref{eq close hoelder} that
+    $$
+        \frac{y}{2} \le y' \,.
+    $$
+    We distinguish four cases. If $y, y' \le 1$, then we have
+    $$
+        |K(0,y) - K(0,y')| = \left|  \frac{1 - y}{1- e^{-iy}} - \frac{1 - y'}{1- e^{-iy'}}\right|
+    $$
+    and by the fundamental theorem of calculus
+    $$
+        = \left| \int_{y'}^{y} \frac{-1 + e^{-it} + i(1-t)e^{it}}{(1 - e^{-it})^2} \,dt \right|\,.
+    $$
+    Using $y' \ge \frac{y}{2}$ and Lemma \ref{lower secant bound}, we bound this by
+    $$
+        \le |y - y'| \sup_{\frac{y}{2} \le t \le 1} \frac{3}{|1 - e^{-it}|^2}  \le 3 |y-y'| (8 \frac{2}{y})^2 \le 2^{10} \frac{|y-y'|}{|y|^2}\,.
+    $$
+    If $y \le 1$ and $y' > 1$, then $K(0, y') = 0$ and we have from the first case
+    $$
+        |K(0,y) - K(0,y')| = |K(0,y) - K(0,1)| \le 2^{10} \frac{|y-1|}{|y|^2} \le 2^{10} \frac{|y-y'|}{|y|^2}\,.
+    $$
+    Similarly, if $y > 1$ and $y' \le 1$, then $K(0, y) = 0$ and we have by the same computation as for the first case
+    $$
+        |K(0,y) - K(0,y')| = |K(0,y') - K(0,1)| \le 2^{10} \frac{|y-1|}{|y|^2} \le 2^{10} \frac{|y-y'|}{|y|^2}\,.
+    $$
+    Finally, if $y, y' > 1$ then
+    $$
+        |K(0,y) - K(0,y')| = 0 \le 2^{10} \frac{|y-y'|}{|y|^2}\,.
+    $$
 \end{proof}
 
+By the previous two lemmas, $K$ is a one-sided Calder\'on--Zygmund kernel on $(\R,\rho,\mu,4)$.
+
+
+The operator $T^*$ defined in \eqref{def tang unm op} coincides in our setting with the operator $T^*$ defined in \eqref{concretetstar}. By  Lemma \ref{lem hilbert}, this operator satisfies the bound \eqref{nontanbound}.
+
+Thus the assumptions of Theorem \ref{metric space Carleson} are all satisfied. Applying the Theorem, Lemma \ref{real Carleson} follows.
+
 \printbibliography
diff --git a/blueprint/src/preamble/common.tex b/blueprint/src/preamble/common.tex
index 0cc74e74..3beff4e1 100644
--- a/blueprint/src/preamble/common.tex
+++ b/blueprint/src/preamble/common.tex
@@ -46,25 +46,25 @@
 \DeclareMathOperator{\bd}{\operatorname{bd}}
 \DeclareMathOperator*{\esssup}{\operatorname{ess\,sup}}
 
-\def \fp {\mathfrak p}
-\def \fP {\mathfrak P}
-\def \fu {\mathfrak u}
-\def \fU {\mathfrak U}
-\def \fv {\mathfrak v}
-\def \fq {\mathfrak q}
-\def \fQ {\mathfrak Q}
-\def\fT{\mathfrak T}
-\def\fL{\mathfrak L}
-\def\fC{\mathfrak C}
-\def\pc{\mathrm{c}}
-\def\ps{\mathrm{s}}
-\def \AD{{\bf s}}
-\def \fc{\Omega}
-\def\borel{\mathcal{E}}
-\def\borelb{\mathcal{G}}
-\def\sc{\mathcal{I}}
-\def \tQ{{Q}}
-\def\mfa{\vartheta}
-\def\mfb{\theta}
-\def\Mf{\Theta}
-\def\fcc{\mathcal{Q}}
+\newcommand{\fp}{\mathfrak p}
+\newcommand{\fP}{\mathfrak P}
+\newcommand{\fu}{\mathfrak u}
+\newcommand{\fU}{\mathfrak U}
+\newcommand{\fv}{\mathfrak v}
+\newcommand{\fq}{\mathfrak q}
+\newcommand{\fQ}{\mathfrak Q}
+\newcommand{\fT}{\mathfrak T}
+\newcommand{\fL}{\mathfrak L}
+\newcommand{\fC}{\mathfrak C}
+\newcommand{\pc}{\mathrm{c}}
+\newcommand{\ps}{\mathrm{s}}
+\newcommand{\AD}{{\bf s}}
+\newcommand{\fc}{\Omega}
+\newcommand{\borel}{\mathcal{E}}
+\newcommand{\borelb}{\mathcal{G}}
+\newcommand{\sc}{\mathcal{I}}
+\newcommand{\tQ}{{Q}}
+\newcommand{\mfa}{\vartheta}
+\newcommand{\mfb}{\theta}
+\newcommand{\Mf}{\Theta}
+\newcommand{\fcc}{\mathcal{Q}}