From abdfd92b480409e1a922a384f59df8786e290ea5 Mon Sep 17 00:00:00 2001
From: Floris van Doorn <fpvdoorn@gmail.com>
Date: Wed, 3 Apr 2024 17:05:09 +0200
Subject: [PATCH] full blueprint

---
 blueprint/src/chapter/main.tex | 3955 +++++++++++++++++++++++++++-----
 1 file changed, 3396 insertions(+), 559 deletions(-)

diff --git a/blueprint/src/chapter/main.tex b/blueprint/src/chapter/main.tex
index 5d71b291..4709ac2e 100644
--- a/blueprint/src/chapter/main.tex
+++ b/blueprint/src/chapter/main.tex
@@ -32,7 +32,7 @@ \chapter{Introduction}
     S_Nf(x):=\sum_{n=-N}^N \widehat{f}_n e^{i nx}\ .
 \end{equation}
 \begin{theorem}\label{classical}
-\uses{thm main 1, thm main 1}
+\uses{thm main 1}
 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 measure
@@ -60,7 +60,7 @@ \chapter{Introduction}
 to doubling metric measure spaces \rs{Should also mention polynomials/general class of functions}. This generalization is of its own interest as new result and we state it as our main novel theorem in this paper.
 
 The maximally truncated generalized Carleson operator $T$, which we  simply refer to as Carleson operator, shall be defined by
- \begin{equation}
+    \begin{equation}
         \label{def main op}
         Tf(x):=\sup_{\mfa\in\Mf} \sup_{0 < R_1 < R_2}\left| \int_{R_1 <  \rho(x,y) < R_2}  K(x,y) f(y) e(\mfa(y)) \, \mathrm{d}\mu(y) \right|\, ,
 \end{equation}
@@ -87,7 +87,7 @@ \chapter{Introduction}
 \end{equation}
 where we have denoted by $B(x,R)$ the open ball of radius $R$ centred at $x$:
 \begin{equation}\label{eq define ball}
- B(x,R):=\{y\in X: \rho(x,y)<R\}. \end{equation}
+    B(x,R):=\{y\in X: \rho(x,y)<R\}. \end{equation}
 
 
 A  collection $\Mf$ of real valued continuous functions on the doubling metric measure space $(X,\rho,\mu,a)$ is called compatible, if there is a point $o\in X$ where all the functions are equal to $0$, and the  local oscillation, defined for a Borel set $\borel\subset X$ by
@@ -136,31 +136,31 @@ \chapter{Introduction}
 
 
 A one-sided $\tau$-Calder\'on--Zygmund kernel $K$ on  the doubling metric measure space $(X, \rho, \mu, A)$ is a measurable function
- \begin{equation}\label{eqkernel0}
+    \begin{equation}\label{eqkernel0}
 K:X\times X\to \mathbb{C}
- \end{equation}
+    \end{equation}
 such that for all $x,y',y\in X$ with $x\neq y$, we have
 \begin{equation}\label{eqkernel size}
     |K(x,y)| \leq \frac{2^{a^3}}{V(x,y)}
-  \end{equation}
- and if $2\rho(y,y') \leq \rho(x,y)$, then
-  \begin{equation}
-  \label{eqkernel y smooth}
-       |K(x,y) - K(x,y')| \leq \left(\frac{\rho(y,y')}{\rho(x,y)}\right)^{\tau}\frac{2^{a^3}}{V(x,y)},
+    \end{equation}
+    and if $2\rho(y,y') \leq \rho(x,y)$, then
+    \begin{equation}
+    \label{eqkernel y smooth}
+        |K(x,y) - K(x,y')| \leq \left(\frac{\rho(y,y')}{\rho(x,y)}\right)^{\tau}\frac{2^{a^3}}{V(x,y)},
 \end{equation}
 where  \[V(x,y):=\mu(B(x,\rho(x,y))).\]
 %Observe that $V$ is essentially symmetric in the sense that $V(x,y)$ and $V(y,x)$ are comparable, with the implied constant depending only on $A$.
-Define the  truncated non-tangential singular integral $T_{R_1,R_2}$ associated with $K$ by
+Define the maximally truncated non-tangential singular integral $T_{*}$ associated with $K$ by
 \begin{equation}
     \label{def tang unm op}
-    T_{R_1,R_2}f(x):=\sup_{\rho(x,x')<R_1} \left|\int_{R_1< \rho(x,y) < R_2}  K(x,y) f(y)  \, \mathrm{d}\mu(y) \right|\,.
+    T_{*}f(x):=\sup_{R_1 < R_2} \sup_{\rho(x,x')<R_1} \left|\int_{R_1< \rho(x,y) < R_2}  K(x,y) f(y)  \, \mathrm{d}\mu(y) \right|\,.
 \end{equation}
 
 \ct{I changed this definition removing the sup over radii.
 I think to remember lacking smoothness in x we cannot remove the other sup. The treatment of the sup should be
 done carefully in section 7, there was a widening
 of the angle that needs to be done carefully. one has to juggle Cotlar there anyways, so maybe one can do the sup
-in radii there. Avoids the sup in the last section}
+in radii there. Avoids the sup in the last section} \lars{Changed it back}
 Our main result  is the following restricted weak type estimate in the range $1<q\le 2$, which by interpolation techniques recovers $L^q$ estimates for the open range
 $1<q<2$.
 \begin{theorem}
@@ -169,12 +169,11 @@ \chapter{Introduction}
     $0 < \tau\leq 1$,  and  $1<q\le 2$,
     there exists a positive $C_{a,\tau, q}$ such that the following holds.
     Let $(X,\rho,\mu,a)$ be a doubling metric measure space. Let  $\Mf$ be a
-    $\tau$-cancellative compatible  collection of functions and let $K$ be a one-sided $\tau$-Calder\'on--Zygmund kernel on $(X,\rho,\mu,a)$. Assume  that for every $0<R_1<R_2$
-    and every bounded measurable function $g$ on $X$ supported on a set of finite measure we have
+    $\tau$-cancellative compatible  collection of functions and let $K$ be a one-sided $\tau$-Calder\'on--Zygmund kernel on $(X,\rho,\mu,a)$. Assume  that for every bounded measurable function $g$ on $X$ supported on a set of finite measure we have
     \begin{equation}\label{nontanbound}
-        \|T_{R_1,R_2}g\|_{2} \leq 2^a \|g\|_2\,,
+        \|T_{*}g\|_{2} \leq 2^a \|g\|_2\,,
     \end{equation}
-    where $T_{R_1,R_2}$ is defined in
+    where $T_{*}$ is defined in
 \eqref{def tang unm op}.
     Then for all Borel sets $F$ and $G$ in $X$ and
     all Borel functions $f:X\to \C$ with
@@ -241,7 +240,7 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 \ref{thm main 1} into the subsequent five
 mutually independent sections.
 It formulates four auxiliary propositions, each proved
- in one sections. The fifth section proves Theorem \ref{thm main 1}.
+    in one sections. The fifth section proves Theorem \ref{thm main 1}.
 The present section  also introduces all definitions and statements used across boundaries of these five sections.
 
 
@@ -250,7 +249,7 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 For two complex quantities $X,Y$, usually depending on
 $A, q, \tau$ and some further parameters,
 we write $X\lesssim Y$ if there exists
- \[C(a,\tau,q)>0\]
+    \[C(a,\tau,q)>0\]
 such that for all values of the further parameters
 \[|X|\le C(a,\tau,q)|Y|.\] Note that $\lesssim$ is transitive.
 
@@ -267,8 +266,8 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 \ct{fill in after completing  grid section}.
 
 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
- \begin{equation}
+    $\psi:\R \to \R$ 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
+    \begin{equation}
     \label{eq psisum}
     \sum_{s\in \mathbb{Z}} \psi(D^{-s}x)=1 .
 \end{equation}
@@ -326,16 +325,16 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 \end{equation}
 so that for each $x, y \in X$ with $x\neq y$  we have
 $$K(x,y)=\sum_{s\in\mathbb{Z}}K_s(x,y).$$
- In Section \ref{thmfromproplinear}, we prove Theorem \ref{thm main 1}
- from the more finitary version, Proposition \ref{prop-linear} below. We call a function from a measure space to a finite set measurable if the pre-image of each of the elements in the range is measurable.
- \lars{Use consistent notation $\mathbf{1}$ or $1$ for indicator functions}
+    In Section \ref{thmfromproplinear}, we prove Theorem \ref{thm main 1}
+    from the more finitary version, Proposition \ref{prop-linear} below. We call a function from a measure space to a finite set measurable if the pre-image of each of the elements in the range is measurable.
+    \lars{Use consistent notation $\mathbf{1}$ or $1$ for indicator functions}
 
 \begin{prop}\label{prop-linear}
 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 $\mu(G')\leq \frac 12 \mu(G)$ such that
 for all Borel functions $f:X\to \C$ with $|f|\le \mathbf{1}_F$.
 \begin{multline}\label{eq-linearized}
-\int_{G \setminus G'} \left|\sum_{s={\sigma_1}(x)}^{{\sigma_2}(x)} \int K_s(x,y) f(y) \tQ(x)(y)  \, \mathrm{d}\mu(y) \right| \mathrm{d}\mu(x)  \lesssim \mu(G)^{\frac 1{q'}}
- \mu(F)^{\frac 1 q}\,.
+\int_{G \setminus G'} \left|\sum_{s={\sigma_1}(x)}^{{\sigma_2}(x)} \int K_s(x,y) f(y) e(\tQ(x)(y))  \, \mathrm{d}\mu(y) \right| \mathrm{d}\mu(x)  \lesssim \mu(G)^{\frac 1{q'}}
+    \mu(F)^{\frac 1 q}\,.
 \end{multline}
 \end{prop}
 Let measurable functions ${\sigma_1}\leq \sigma_2\colon X\to \mathbb{Z}$ with finite range be given. let a measurable function
@@ -349,7 +348,7 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 
 In Section \ref{christsection},
 we prove Proposition \ref{prop-linear}
- using  a
+    using  a
 bound for a dyadic model formulated in Proposition
 \ref{prop dyadic} below.
 
@@ -377,7 +376,7 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 x\in I.
 \end{equation}
 For any dyadic cube $I$,
- \begin{equation}
+    \begin{equation}
         \label{eq vol sp cube}
         c(I)\in B(c(I), \frac{1}{4} D^{s(I)}) \subset I \subset B(c(I), 4 D^{s(I)})\,.
     \end{equation}
@@ -406,9 +405,9 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 \pc &\colon \fP\to X\\
 \ps &\colon \fP\to \mathbb{Z}
 \end{align*}
-with $\sc$ surjective and $\mathcal{P}(\Mf)$ denoting the power set of $\Mf$ such that the five Properties \eqref{eq dis freq cover}, \eqref{eq dis freq cover},
-\eqref{eq freq comp ball}, \ref{tilecenter}, and
-\ref{tilescale} hold.
+with $\sc$ surjective and $\mathcal{P}(\Mf)$ denoting the power set of $\Mf$ such that the five Properties \eqref{eq dis freq cover}, \eqref{eq freq dyadic},
+\eqref{eq freq comp ball}, \eqref{tilecenter}, and
+\eqref{tilescale} hold.
 
 For each dyadic cube $I$, the restriction of the  map $\Omega$ to the set
 \begin{equation}\label{injective}
@@ -431,7 +430,7 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 \begin{equation}
     B_{\fp} := \{\mfb \in \Mf \, : \, d_{\fp}(\mfa, \mfb) < R\,\} ,
 \end{equation}
- and
+    and
 \begin{equation}\label{defdp}
 d_{\fp} := d_{B(\pc(\fp),\frac 14 D^\ps(\fp))}\, .
 \end{equation}
@@ -459,14 +458,14 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 \end{equation}
 and
 \begin{equation}\label{definetp}
-    T_{\fp} f(x)= 1_{E(\fp)}(x) \int   K_{\ps(\fp)}(x,y) f(y) \tQ(x)(y)\overline{\tQ(x)(x)}\, d\mu(y).
+    T_{\fp} f(x)= 1_{E(\fp)}(x) \int   K_{\ps(\fp)}(x,y) f(y) e(\tQ(x)(y)-\tQ(x)(x))\, d\mu(y).
 \end{equation}
 Then there exists a Borel set $G'\subset G$ with $\mu(G') \leq 1/2\mu(G)$ such that for all $f:X\to \C$ with $|f|\le \mathbf{1}_F$and all
 $g:X\to \C$ with $|g|\le \mathbf{1}_{G\setminus G'}$
 we have
 \begin{equation}
     \label{disclesssim}
-   \int g(x) \sum_{\fp \in \fP} T_{\fp} f (x) \, \mathrm{d}\mu(x)  \lesssim \mu(G)^{1/q'} \mu(F)^{1/q}\,.
+    \int g(x) \sum_{\fp \in \fP} T_{\fp} f (x) \, \mathrm{d}\mu(x)  \lesssim \mu(G)^{1/q'} \mu(F)^{1/q}\,.
 \end{equation}
 \end{prop}
 
@@ -489,7 +488,7 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 \begin{equation}\label{straightorder}
     \fp\le \fp'
 \end{equation}
- on $\fP\times \fP$ meaning
+    on $\fP\times \fP$ meaning
 $\sc(\fp)\subset \sc(\fp')$ and
 $\Omega(\fp')\subset \Omega(\fp)$.
 We further define for $\lambda,\kappa>0$
@@ -546,11 +545,11 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 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)|
-  \end{equation}
-  \begin{equation}
-      \le  2^{200a^3}({q}-1)^{-1} \tau^{-1}\dens_1(\mathfrak{A})^{\frac {(q-1)\tau^2}{8a^2}}\dens_2(\mathfrak{A})^{\frac 1{q}-\frac 12}  \|f\|_2\|g\|_2\, .
-  \end{equation}
+    |\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)|
+    \end{equation}
+    \begin{equation}
+        \le  2^{200a^3}({q}-1)^{-1} \tau^{-1}\dens_1(\mathfrak{A})^{\frac {(q-1)\tau^2}{8a^2}}\dens_2(\mathfrak{A})^{\frac 1{q}-\frac 12}  \|f\|_2\|g\|_2\, .
+    \end{equation}
 \end{prop}
 
 Let $n\ge 0$.
@@ -579,7 +578,7 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 \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}
@@ -596,7 +595,7 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 
 %For $\fp\in \fP$ and $Q\in\mathcal{Q}$, we define their {separation} to be
 %\[\Delta(\fp, Q):=d_{\sc(\fp)}(Q(\fp), Q)+1.\]
- %A pair of boundary parts $\mathfrak{B}_1, \mathfrak{B}_2$ is \emph{$\Delta$-separated}  if
+    %A pair of boundary parts $\mathfrak{B}_1, \mathfrak{B}_2$ is \emph{$\Delta$-separated}  if
 %    for $i \neq j$, $\fp \in \mathfrak{B}_i$ and $I(\fp) \subset I(\tp(\mathfrak{B}_j))$ implies $\Delta(\fp, Q({\tp(\mathfrak{B}_j)})) > \Delta$.
 
 
@@ -638,15 +637,15 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 $$
 \begin{prop}
     \label{lem vdc regularity}
-     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 $\|{\varphi}\|_{C^\tau(B)}<\infty$.
-     Let $\mfa, \mfb \in \mathcal{Q}$. Then
+        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 $\|{\varphi}\|_{C^\tau(B)}<\infty$.
+        Let $\mfa, \mfb \in \mathcal{Q}$. Then
     \begin{equation}
         \label{eq vdc cond tau 2}
         |\int e(\mfa(x)-{\mfb(x)})\varphi(x) dx|\le
-         2^{4a} \mu(B) \|{\varphi}\|_{C^\tau(B)}
-       (1 + d_{B}(\mfa,\mfb))^{-\tau^2/(2+a)}
+            2^{4a} \mu(B) \|{\varphi}\|_{C^\tau(B)}
+        (1 + d_{B}(\mfa,\mfb))^{-\tau^2/(2+a)}
     \,.
     \end{equation}
     \end{prop}
@@ -664,16 +663,16 @@ \chapter{Overview of the proof of Theorem \ref{thm main 1}}
 Define further $M_{\mathcal{B}}:=M_{\mathcal{B},1}$.
 
 \begin{prop}\label{prop hlm}
-   Let $\mathcal{B}$ be a finite collection of balls in $X$.
+    Let $\mathcal{B}$ be a finite collection of balls in $X$.
 If for some $\lambda>0$ and some measurable function $h:X\to [0,\infty)$ we have
 \begin{equation}\label{eq ball assumption}
 \int_{B'} h(x)\, d\mu(x)\ge \lambda \mu(B')
 \end{equation}
-   for each $B'\in \mathcal{B}$,
-   then
-   \begin{equation}\label{eq besico}
+    for each $B'\in \mathcal{B}$,
+    then
+    \begin{equation}\label{eq besico}
 \lambda \mu(\bigcup \mathcal{B}) \le 2^{2a}\int_X h(x)\, d\mu(x)\, .
-       \end{equation}
+        \end{equation}
 \rs{For the union of balls, I would write $\bigcup_{B\in\mathcal{B}} B$, here and elsewhere}
 For every measurable function $h$ \rs{Choose a different name for the function as $h$ is already used above}
 and $1\le q'<q$ we have \rs{$q'$ is used for the dual exponent to $q$. Use $\Tilde{q}$, here and in the proof}
@@ -744,7 +743,7 @@ \chapter{T. \ref{thm main 1} from finitary P.
 \begin{equation}\label{boundarys}
 \sum_{s=\sigma_1-2,\sigma_1-1, \sigma_2+1,\sigma_2+2}
 \int_{R_1 <  \rho(x,y) < R_2}  K_s(x,y) f(y) e(Q(y)) \,
- \mathrm{d}\mu(y),
+    \mathrm{d}\mu(y),
 \end{equation}
 where $\sigma_1$ is the smallest integer such that $D^{\sigma_1-2}R_2>\frac 1{4D}$ and $\sigma_2$
 is the largest integer so that $D^{\sigma_2+2}R_1<\frac 12$. Here we restricted the summation index $s$
@@ -785,7 +784,7 @@ \chapter{T. \ref{thm main 1} from finitary P.
 \rs{where $M$ is the Hardy Littlewood maximal function.}
 Now \eqref{Rcut} with $T_{R_1,R_2,\mfa}$ replaced by a summand of \eqref{boundarys}
 follows from
- \begin{equation}
+    \begin{equation}
     \int 1_{G}(x) M1_F(x)\, d\mu(x)
 \lesssim \mu(G)^{1/q'} \mu(F)^{1/q},
 \end{equation}
@@ -828,17 +827,17 @@ \chapter{T. \ref{thm main 1} from finitary P.
     \left| \Tilde{T}_{\sigma_1,\sigma_2,\mfa} 1_{F}(x)\right|-\left|\Tilde{T}_{\sigma_1,\sigma_2,\tilde{\mfa}} 1_{F}(x) \right|
 \end{equation}
 \begin{equation}
-  \lesssim
-  \sum_{\sigma_1 \le s\le \sigma_2}
+    \lesssim
+    \sum_{\sigma_1 \le s\le \sigma_2}
 \int |K_s(x,y)|  1_F(y) |\mfa(y)-{\mfa(x)}
 -{\tilde{Q}(y)}+{\tilde{Q}(x)}| \, \mathrm{d}\mu(y)
 \end{equation}
 \begin{equation}
-  \lesssim  (2S+1) \epsilon M1_F(x)\lesssim  M1_F(x)
+    \lesssim  (2S+1) \epsilon M1_F(x)\lesssim  M1_F(x)
 \end{equation}
 We estimate the left-hand-side of \eqref{Scut} by
 the sum of \eqref{Sqcut} and
-  \begin{equation}
+    \begin{equation}
     \int 1_{G}(x)
 \max_{-S<\sigma_1\le \sigma_2<S}\sup_{\mfa\in {\Mf}}
 \inf_{\mfa\in\tilde{\Mf}}
@@ -847,7 +846,7 @@ \chapter{T. \ref{thm main 1} from finitary P.
 \end{equation}
 \begin{equation}
     \lesssim
-     \int 1_{G}(x)
+        \int 1_{G}(x)
 M1_{F}(x)\, d\mu(x)
 ,
 \end{equation}
@@ -864,7 +863,7 @@ \chapter{T. \ref{thm main 1} from finitary P.
     \begin{equation}  \label{Sqlin}
     \int 1_{G}(x)
 \left| \Tilde{T}_{\underline{\sigma},\overline{\sigma},\tQ}1_F(x)
- \right|\, d\mu(x)
+    \right|\, d\mu(x)
 \lesssim \mu(G)^{1/q'} \mu(F)^{1/q},
 \end{equation}
 with
@@ -918,8 +917,8 @@ \chapter{T. \ref{thm main 1} from finitary P.
 Applying Proposition \ref{prop-linear} 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
- \begin{equation}
+    and
+    \begin{equation}
     \int 1_{G_{n}\setminus G_{n+1}}(x)
 \max_{-S<\sigma_1\le \sigma_2<S}\max_{\mfa\in\tilde{\Mf}}
 \left| \Tilde{T}_{\sigma_1,\sigma_2,\mfa} 1_{F}(x) \right|\, d\mu(x)
@@ -939,7 +938,7 @@ \chapter{T. \ref{thm main 1} from finitary P.
 \left| \Tilde{T}_{\sigma_1,\sigma_2,\mfa} 1_{F}(x) \right|,
 \end{equation}
 which by interchange of summation and integration is seen to be integrable, by Lebesgue's dominated convergence theorem we obtain
- \begin{equation} \label{Sqcut3}
+    \begin{equation} \label{Sqcut3}
     \int 1_{G}(x)
 \left| \Tilde{T}_{\underline{\sigma},\overline{\sigma},\tilde{Q}}1_F(x) \right|\, d\mu(x)
 \lesssim \mu(G)^{1/q'} \mu(F)^{1/q}.
@@ -989,7 +988,7 @@ \chapter{P. \ref{prop-linear}
 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)}
-    \int   K_{s}(x,y) f(y) tQ(x)(y)\overline{\tQ(x)(x)}\, d\mu(y).
+    \int   K_{s}(x,y) f(y) e(\tQ(x)(y)-\tQ(x)(x))\, d\mu(y).
 \end{equation}
 \end{lemma}
 \begin{proof}
@@ -999,10 +998,10 @@ \chapter{P. \ref{prop-linear}
 \begin{equation}\label{outsump}
     \sum_{\fp\in \fP: \ps(\fp)=s}T_{\fp} f(x)=0
 \end{equation}
- if $s\not\in [\sigma_1(x), \sigma_2(x)]$ and
+    if $s\not\in [\sigma_1(x), \sigma_2(x)]$ and
 \begin{equation}\label{insump}
     \sum_{\fp\in \fP: \ps(\fp)=s}T_{\fp} f(x)=
-    \int   K_{s}(x,y) f(y) tQ(x)(y)\overline{\tQ(x)(x)}\, d\mu(y).
+    \int   K_{s}(x,y) f(y) e(\tQ(x)(y) - \tQ(x)(x))\, d\mu(y).
 \end{equation}
 if $s\in [\sigma_1(x),\sigma_2(x)]$.
 If $s\not\in [\sigma_1(x), \sigma_2(x)]$, then by definition of $E(\fp)$ we have
@@ -1020,14 +1019,14 @@ \chapter{P. \ref{prop-linear}
 
 We now estimate with Lemma \ref{lemmasump} and Proposition \ref{prop dyadic}
 \begin{equation}
- \int_{G \setminus G'} \left|\sum_{s={\sigma_1}(x)}^{{\sigma_2}(x)} \int K_s(x,y) f(y) \tQ(x)(y)  \, \mathrm{d}\mu(y)\right| \mathrm{d}\mu(x)
+    \int_{G \setminus G'} \left|\sum_{s={\sigma_1}(x)}^{{\sigma_2}(x)} \int K_s(x,y) f(y) e(\tQ(x)(y))  \, \mathrm{d}\mu(y)\right| \mathrm{d}\mu(x)
 \end{equation}
 \begin{equation}
- =\int_{G \setminus G'} \left|\sum_{s={\sigma_1}(x)}^{{\sigma_2}(x)} \int K_s(x,y) f(y) \tQ(x)(y)  \, \overline{\tQ(x)(x)}\mathrm{d}\mu(y)\right| \mathrm{d}\mu(x)
+    =\int_{G \setminus G'} \left|\sum_{s={\sigma_1}(x)}^{{\sigma_2}(x)} \int K_s(x,y) f(y) e(\tQ(x)(y) - \tQ(x)(x))\mathrm{d}\mu(y)\right| \mathrm{d}\mu(x)
 \end{equation}
 \begin{equation}
- =\int_{G \setminus G'} \left|\sum_{\fp\in \fP}T_{\fp} f(x)\right| \mathrm{d}\mu(x)
- \lesssim \mu(G)^{1/q'} \mu(F)^{1/q} \,.
+    =\int_{G \setminus G'} \left|\sum_{\fp\in \fP}T_{\fp} f(x)\right| \mathrm{d}\mu(x)
+    \lesssim \mu(G)^{1/q'} \mu(F)^{1/q} \,.
 \end{equation}
 This proves \eqref{eq-linearized} for the chosen set $G'$ and arbitrary $f$ and thus completes the proof of Proposition
 \ref{prop-linear}.
@@ -1042,7 +1041,7 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 Let $-S\le k\le S$.Let $Y\subset X$ 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 for any further $y'\in Y$ with $y\neq y'$ we have
 \begin{equation} \label{eq disj yballs}
@@ -1056,19 +1055,19 @@ \section{Proof of L.\ref{dyadiclemma}, 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 A^j \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 und thus $B(y,D^k)$ has positive measure.
-
- Applying \eqref{jballs} for $j'$ the smallest integer larger than $\ln_2(8D^{2S})$, 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}
+    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 A^j \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 und thus $B(y,D^k)$ has positive measure.
+
+    Applying \eqref{jballs} for $j'$ the smallest integer larger than $\ln_2(8D^{2S})$, 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 disjointness of the balls in \eqref{eq disj yballs}, we obtain
 \begin{equation}
 |Y|\mu(B(o,4D^S))\le A^{j'}\sum_{y\in Y}\mu(B(y,D^k))
@@ -1090,9 +1089,9 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 order  $<$ on $Y_{k}$.
 
 \begin{lemma}\label{ballcover}
-  For each $-S\le k\le S$, the ball
-  $B(o, 4D^S-D^k)$ is contained
-  in the union of the balls $B(y,2D^k)$ with $y\in Y_k$.
+    For each $-S\le k\le S$, the ball
+    $B(o, 4D^S-D^k)$ is contained
+    in the union of the balls $B(y,2D^k)$ with $y\in Y_k$.
 \end{lemma}
 
 \begin{proof}
@@ -1134,35 +1133,35 @@ \section{Proof of L.\ref{dyadiclemma}, 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{firstgridlemma}
-   For each $-S\le k\le S$ and $1\le j\le 3$
-   the following holds.
+    For each $-S\le k\le S$ and $1\le j\le 3$
+    the following holds.
 
-   If $j\neq 2$ and for some $x\in X$ and $y_1,y_2\in Y_k$ we have
-   \begin{equation}\label{disji}
-       x\in I_j(y_1,k)\cap I_j(y_2,k),
-   \end{equation}
+    If $j\neq 2$ and for some $x\in X$ and $y_1,y_2\in Y_k$ we have
+    \begin{equation}\label{disji}
+        x\in I_j(y_1,k)\cap I_j(y_2,k),
+    \end{equation}
 then $y_1=y_2$.
 
-   If $j\neq 1$, then
+    If $j\neq 1$, then
 \begin{equation}\label{unioni}
 B(o, 4D^S-2D^k)\subset \bigcup_{y\in Y_k} I_j(y,k)\, .
 \end{equation}
 We have for each $y\in Y_k$,
-   \begin{equation}\label{squeezedyadic}
+    \begin{equation}\label{squeezedyadic}
     B(y,\frac 12 D^k) \subset I_3(y,k)\subset
-     B(y,4D^k).
+        B(y,4D^k).
 \end{equation}
 
 
 
 \end{lemma}
 \begin{proof}
-We prove the statement by induction on the ordered set of pairs $(y,k)$.
+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 $j=1$.
@@ -1198,12 +1197,12 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 
 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
+    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 $x\in I_2(y,k)$.
+    there is $y\in Y_k$ such that $x\in I_2(y,k)$.
 We may assume this $y$ is minimal with respect ot the order in $Y_k$
 Then $x\in I_3(y,k)$.
- This proves \eqref{unioni}  for $j=3$.
+    This proves \eqref{unioni}  for $j=3$.
 
 Next we  show the first inclusion in \eqref{squeezedyadic}.
 Let $x\in B(y,\frac 1{2}D^k)$.
@@ -1242,7 +1241,7 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 $y\in Y_k$.
 We have
 \begin{equation}\label{3coverdyadic}
-   I_3(y,k)\subset \bigcup_{y'\in Y_l} I_3(y',l)\, .
+    I_3(y,k)\subset \bigcup_{y'\in Y_l} I_3(y',l)\, .
 \end{equation}
 \end{lemma}
 \begin{proof}
@@ -1278,7 +1277,7 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 \ref{firstgridlemma}, we conclude $y'=y$
 and thus \eqref{3dyadicproperty}.
 Now assume $l<k$. By induction, we may assume
- withe statement of the lemma is proven for $k-1$ in place of $k$.
+    withe statement of the lemma is proven for $k-1$ in place of $k$.
 
 By Lemma \ref{icoveri}, there is
 a $y''\in Y_{k-1}$ such that $x\in I_3(y'',k-1)$
@@ -1290,7 +1289,7 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 
 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 $x\in I_3(v,k-1)$.
+    $x\in I_1(y,k)$. 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{firstgridlemma}, we have $v=w$.
 By Definition \eqref{defineij}, we then have
 $I_3(w,k-1)\subset I_1(y,k)$.
@@ -1298,9 +1297,9 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 
 Assume now the case $x\notin X_k$.
 By \eqref{definei3}, we have
- $x\in I_2(y,k)$. Morover, 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 $x\not \in I_2(y,k)$.
+    $x\in I_2(y,k)$. Morover, 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 $x\not \in I_2(y,k)$.
 By \eqref{defineij} and the disjointness property of Lemma \ref{firstgridlemma}, we have
 $I_3(w,k-1)\cap  I_2(u,k)= \emptyset$.
 Similarly, $I_3(w,k-1)\cap  I_1(u,k)= \emptyset$.
@@ -1365,8 +1364,8 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 \end{proof}
 
 \begin{lemma}\label{david}
- Let $K$ be the smallest integer larger than
- $2A^6$. For each $-S+K\le k\le S$ and $y\in Y_k$ we have
+    Let $K$ be the smallest integer larger than
+    $2A^6$. For each $-S+K\le k\le S$ and $y\in Y_k$ we have
     \begin{equation}
         \label{new small boundary}
         \sum_{z\in Y_{k-K}: (z,k-K|y,k)}\mu(I_3(z,k-K)) \le \frac 12 \mu(I_3(y,k))\,.
@@ -1374,34 +1373,32 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 \end{lemma}
 
 \begin{proof}
-Let $K$ be as in the lemma and let $-S+K\le k\le S$ and $y\in Y_k$.
+Let $K$ be as in the lemma. Let $-S+K\le k\le S$ and $y\in Y_k$.
 
 Pick $k'$ so that $k-K\le k'\le k$.
-For each $y''\in k-K$ with $(y'',k-K| y,k)$
-By Lemma \ref{3coverdyadic} and Lemma \ref{3dyadiclemma}, there is a unique $y'\in Y_{k'}$ such that
+For each $y''\in Y_{k-K}$ with $(y'',k-K| y,k)$,
+By Lemma \ref{icoveri} and Lemma \ref{3dyadiclemma}, there is a unique $y'\in Y_{k'}$ such that
 \begin{equation}
     I_3(y'',k-K)\subset I_3(y',k')\subset I_3(y,k)\, .
 \end{equation}
-Using Lemma \ref{bdtrans},
+Using Lemma \ref{bdtrans}, \rs{Add: $(y',k'|y,k)$}
 
-
-
-We conclude with disjointness property of
-Lemma \ref{firstgridlemma} for
+We conclude using the disjointedness property of
+Lemma \ref{firstgridlemma} 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)
+    \sum_{y'': (y'',k-K|y',k')}\mu(I_3(z,k-K))\right)
 \end{equation}
-   \begin{equation}
-   \le
-\sum_{u: (y',k'|y,k)}
+    \begin{equation}
+    \le
+\sum_{y': (y',k'|y,k)}
 \mu(I_3(y',k'))    \, .
-   \end{equation}
-Adding over $k-K<k'\le k$ and using
+    \end{equation}
+Adding over $k-K<k'\le k$, and using
 \[\mu(I_3(y',k'))\le A^4 \mu(B(y', \frac 14 D^{k'}))\]
 from the doubling property \eqref{doublingx} and
 \eqref{squeezedyadic} gives
@@ -1411,14 +1408,14 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 \end{equation}
 \begin{equation}\label{sumcompare1}
     \le A^4 \sum_{k-K<k'\le k}
-   \left[ \sum_{y': (y',k'|y,k)}
+    \left[ \sum_{y': (y',k'|y,k)}
 \mu(B(y', \frac 14 D^{k'}))\right]
 \end{equation}
-Each ball $B(y', \frac 12 D^{k'})$ occurring in
+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
-y additivity of the measure,
+by additivity of the measure,
 \begin{equation}
     K\sum_{y'': (y'',k-K|y,k)}
     \mu(I_3(y'',k-K))
@@ -1454,7 +1451,7 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 Let $K$ be the smallest integer larger than $2A^6$
 and let $n\ge 0$ be an integer. Then
 for each $-S+nK\le k\le S$ we have
-   \begin{equation}
+    \begin{equation}
         \label{very new small}
         \sum_{y'\in Y_{k-nK}: (y',k-nK|y,k)}\mu(I_3(y',k-nK)) \le 2^{-n} \mu(I_3(y,k))\,.
     \end{equation}
@@ -1466,14 +1463,14 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 
     Assume $n>1$ and \eqref{very new small} has been proven with  $n-1$ in place of $n$.
 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))
+        \begin{equation}
+                \sum_{y''\in Y_{k-nK}: (y'',k-nK|y,k)}\mu(I_3(y'',k-nK))
     \end{equation}
-     \begin{equation}
+        \begin{equation}
 =  \sum_{y'\in Y_{k-K}:(y',k-K|y,k)}   \left[   \sum_{y''\in Y_{k-nK}: (y'',k-nK|y',k-K)}\mu(I_3(y'',k-nK)) \right]
     \end{equation}
 Applying the induction hypothesis, this is bounded by
-         \begin{equation}
+            \begin{equation}
 =  \sum_{y'\in Y_{k-K}:(y',k-K|y,k)}   2^{1-n}\mu(I_3(y',k-K))
     \end{equation}
 Applying Lemma \ref{new small boundary} gives
@@ -1481,8 +1478,8 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 \end{proof}
 
 \begin{lemma}
- For each $-S\le k\le S$ and $y\in Y_k$ and $0<t<1$
- with $tD^k>D^{-S}$ we have
+    For each $-S\le k\le S$ and $y\in Y_k$ and $0<t<1$
+    with $tD^k>D^{-S}$ we have
     \begin{equation}
         \label{old small boundary}
         \mu(\{x \in I(y,k) \ : \ \rho(x, X \setminus I(y,k)) \leq t D^{k}\}) \le 4A^2 t^\kappa \mu(I)\,.
@@ -1490,64 +1487,58 @@ \section{Proof of L.\ref{dyadiclemma}, dyadic structure}
 \end{lemma}
 \begin{proof}
 Let $x\in I(y,k)$ with  $\rho(x, X \setminus I(y,k)) \leq t D^{k}$. Let
-
+\rs{This is still needed to show \eqref{eq small boundary}}
 \end{proof}
 
-
-
-
-
-
-
-
-
-
-
-
-
-Let $\mathcal{D}$ the set of all $I_3(x,k)$ with $k\in [-S,S]$ and
+Let $\mathcal{D}$ be the set of all $I_3(y,k)$ with $k\in [-S,S]$ and
 $y\in Y_k$. Define
 \begin{equation}
-s(I_3(x,k)):=k
+s(I_3(y,k)):=k
 \end{equation}
 \begin{equation}
 c(I_3(y,k)):=y
 \end{equation}
 We show that $(\mathcal{D},c,s)$ constitutes a grid structure. Property \eqref{eq vol sp cube}
 follows from \eqref{squeezedyadic}.
-To see \eqref{coverdyadic}, let $I\in \mathcal{D}$
-and $-S\le k< s(I)$. Let $x\in I$.
+\rs{\sout{To see \eqref{coverdyadic}, let $I\in \mathcal{D}$
+and $-S\le k< s(I)$. Let $x\in I$.}}
 
 
-We show properties
+\rs{Let $x\in B(o, D^S)$.} We show properties
 \eqref{coverdyadic},
 \eqref{dyadicproperty},
-\eqref{eq vol sp cube}, and \eqref{eq small boundary}
-for this $\mathcal {D}$, $s$, and $x$.
+\eqref{coverball}, and \eqref{eq small boundary}
+for this $\mathcal {D}$, \rs{\sout{$s$}}, and $x$.
 
 We first show \eqref{coverdyadic}.
 Assume to get a contradiction that \eqref{coverdyadic}
 is false. Then there is a $I$ violating the conclusion of
-\eqref{coverdyadic}. Pick such $I=I(x,l)$ such that $l$ is minimal.
-Let $-S\le k<l$, in particular $-S<l$.
+\eqref{coverdyadic}. Pick such $I=I(x,l)$ \rs{$I=I_3(y,l)$} such that $l$ is minimal.
+Let $-S\le k<l$, in particular $-S<l$. \rs{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 minmiality if $l$, each such $I(y,l-1)$ is contained in the union of
-all $I(z,k)$ with $z\in X_k$. This proves \eqref{coverdyadic}.
+\rs{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 $I(y,l-1)$ \rs{$I_3(y',l-1)$} is contained in the union of
+all $I(z,k)$ with $z\in X_k$ \rs{$I_3(z,k)$ 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 such a counterexample with minimal
-$s(J)-s(I)$ and set $k=s(I)$. Assume first $s(J)=k$. Let $I=I(x,k)$ and $J=I(y,k$
-with $x,y\in X_k$. If $x=y$, then $I=J$, a contradiction to $I\subset J$.
+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 $I=I(x,k)$ and $J=I(y,k)$ with $x,y\in X_k$ \rs{$I=I_3(y_1,k)$ and $J=I_3(y_2,k)$ with $y_1,y_2\in Y_k$}
+If $x=y$, \rs{If $y_1=y_2$} then $I=J$, a contradiction to $I\subset J$.
 If $x<y$, then $I\cap J=\emptyset$ by definition of $J$, a contradiction.
-If $y<x$, then $I\cap J=\emptyset$ by definition of $I$, a contradiction.
+\rs{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{firstgridlemma}, a contradiction.}
+If $y<x$, then $I\cap J=\emptyset$ by definition of $I$, a contradiction. \rs{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 $s(K)=s(J)-1$ with $y\in K$. By construction
 of $J$, and pairwise disjointness of all $I(w,s(J)-1)$ that we have already seen,
-we have $K\subset J$. By minimality if $s(J)$, we have $I\subset K$.
+we have $K\subset J$. By minimality if $s(J)$, we have $I\subset K$. \rs{Ask}
 This proves $I\subset J$ and thus \eqref{dyadicproperty}.
 
+\rs{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}.}
+
 
 \section{Proof of L.\ref{tilelemma}, tile structure}
 \label{subsectiles}
@@ -1569,7 +1560,7 @@ \section{Proof of L.\ref{tilelemma}, tile structure}
     I^\circ = B(c(I), \frac{1}{4} D^{s(I)})\,.
 $$
 
-\lars{The definitions of the doubling roperties changed, the cubes do not have to have the same centers. Check if this is easier to prove now.}
+\lars{The definitions of the doubling properties changed, the cubes do not have to have the same centers. Check if this is easier to prove now.}
 \begin{lemma}
     \label{lem cube monotone}
     Let $I, J \in \mathcal{D}$ with $I \subset J$.
@@ -1631,7 +1622,7 @@ \section{Proof of L.\ref{tilelemma}, tile structure}
     $$
         B_{I^\circ}(q,1) \subset \bigcup_{j \in J(q)} B_{I^\circ{}}(j, \frac{1}{4})\,.
     $$
-    Thus for each $z \in Z$, there exists a $j(z) \in \bigcup_{q \in Q(x)} J(q)$ such that $d_{I^\circ}(z,j(z)) < \frac{1}{4} <0.3$ and hence $j(z) \in B_{I^\circ}(z, 0.3)$. By the assumption \eqref{eq tile disjoint Z}, it follows that the map $z \mapsto j(z)$ is injective. This establishes the Lemma, since then
+    Thus for each $z \in Z$, there exists a $j(z) \in \bigcup_{q \in Q(x)} J(q)$ such that $d_{I^\circ}(z,j(z)) < \frac{1}{4} <0.3$ and hence $j(z) \in B_{I^\circ}(z, 0.3)$. By the assumption \eqref{eq tile disjoint Z}, it follows that the map $z \mapsto j(z)$ is injective. This establishes the lemma, since then
     $$
         |Z| \le |\bigcup_{q \in Q(x)} J(q)| \le \sum_{q \in Q(X)}|J(q)| \le \sum_{q \in Q(X)} 2^{2a} \le |Q(X)|2^{2a}\,.
     $$
@@ -1657,27 +1648,30 @@ \section{Proof of L.\ref{tilelemma}, tile structure}
 
 We define
 $$
-    \fP = \{(I, z) \ : \ I \in \mathcal{D}, z \in Z(I)\}\,.
+    \fP = \{(I, z) \ : \ I \in \mathcal{D}, z \in Z(I)\}\,,
 $$
-We define $\sc((I, z)) = I$ and $\fcc((I, z)) = z$. We further set $s(\fp) = s(\sc(\fp))$, $c(\fp) = c(\sc(\fp))$. Then \eqref{tilecenter}, \eqref{tilescale} hold.
+$$\sc((I, z)) = I\qquad \text{and} \qquad \fcc((I, z)) = z.$$ We further set $$s(\fp) = s(\sc(\fp)),\qquad \qquad c(\fp) = c(\sc(\fp)).$$ Then \eqref{tilecenter}, \eqref{tilescale} hold by definition.
 
-It remains to construct the map $\Omega$. We first construct an auxiliary map $\Omega_1$. For each $I \in \mathcal{D}$, we pick an enumeration of the finite set $Z(I)$
+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 $Z(I)$
 $$
     Z(I) = \{z_1, \dotsc, z_M\}\,.
 $$
-We define
+We define \rs{$\Omega_1:\fP \mapsto \mathcal{P}(\Mf) $ as below}. Set
 $$
     \Omega_1((I, z_1)) = B_{I^\circ}(z_1, 0.7) \setminus \bigcup_{z \in Z(I)} B_{I^\circ}(z, 0.3)\,,
 $$
+\rs{Should be $$
+    \Omega_1((I, z_1)) = B_{I^\circ}(z_1, 0.7) \setminus \bigcup_{z \in Z(I)\setminus \{z_1\}} B_{I^\circ}(z, 0.3)\,,?
+$$}
 and then we iteratively define
 \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))\,.
 \end{equation}
-
 \begin{lemma}
     \label{lem omega1 disj}
-    For each $I \in \mathcal{D}$, the sets $\Omega_1(\fp), \fp \in \fP(I)$ are pairwise disjoint.
+    For each $I \in \mathcal{D}$, the sets $\Omega_1(\fp), \fp \in \fP(I)$ are pairwise disjoint. \rs{and $\fp_1, \fp_2\in \fP(I)$, if $$\Omega_1(\fp_1)\cap \Omega_1(\fp_2)\neq \emptyset,$$ then $\fp_1=\fp_2$.}
 \end{lemma}
 
 \begin{proof}
@@ -1692,7 +1686,7 @@ \section{Proof of L.\ref{tilelemma}, tile structure}
     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 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}
@@ -1745,7 +1739,7 @@ \section{Proof of L.\ref{tilelemma}, tile structure}
 \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 $I \subsetneq J$.
 
-    Suppose that $q \in \Omega(\fp)$. By \eqref{eq it omega}, there exists $z \in Z(J) \cap \Omega_1(\fp)$ with $q \in \Omega(J,z)$. Using the triangle inequality and \eqref{eq omega1 incl}, we obtain
+    Suppose that $q \in \Omega(\fp)$. \rs{If $q\in B_{\fp}(\mathcal{Q}(\fp), 0.2)$, then since $B_{\fp}(\mathcal{Q}(\fp), 0.2)\subset B_{\fp}(\mathcal{Q}(\fp), 1)$, we conclude that $q\in B_{\fp}(\mathcal{Q}(\fp), 0.7)$. If not, } By \eqref{eq it omega}, there exists $z \in Z(J) \cap \Omega_1(\fp)$ with $q \in \Omega(J,z)$. Using the triangle inequality and \eqref{eq omega1 incl}, we obtain
     $$
         d_{I^\circ}(\fcc(\fp),q) \le d_{I^\circ}(\fcc(\fp), z) + d_{I^\circ}(z, q) \le 0.7 + d_{I^\circ}(z, q)\,.
     $$
@@ -1759,20 +1753,20 @@ \section{Proof of L.\ref{tilelemma}, tile structure}
 
     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{lem omega1 disj}. To obtain the inclusion in \eqref{eq dis freq cover} one combines the inclusions \eqref{eq tile cover} and \eqref{eq omega1 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 $x \in \fc(\fp) \cap \fc(\fp')$.
+    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 disjointness of the sets $\fc(\fp), \fp \in \fP(I)$ we pick two tiles $\fp, \fp' \in \fP(I)$ and $q \in \fc(\fp) \cap \fc(\fp')$.
     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 $x \in \Omega(J, z)$, and there exists $z' \in Z(J) \cap \Omega_1(\fp')$ such that $x \in \Omega(J, z')$. By the induction hypothesis, that \eqref{eq dis freq cover} holds for $J$, we must have $z = z'$. By Lemma \ref{lem omega1 disj}, we must then have $\fp = \fp'$.
+    1. There exist $z \in Z(J) \cap \Omega_1(\fp)$ such that $q \in \Omega(J, z)$, and there exists $z' \in Z(J) \cap \Omega_1(\fp')$ such that $q \in \Omega(J, z')$. By the induction hypothesis, that \eqref{eq dis freq cover} holds for $J$, we must have $z = z'$. By Lemma \ref{lem omega1 disj}, we must then have $\fp = \fp'$.
 
-    2. There exists $z \in Z(J) \cap \Omega_1(\fp)$ such that $x \in \Omega(J,z)$, and $x \in B_{\fp'}(\fcc(\fp'), 0.2)$.  Using the triangle inequality, Lemma \ref{lem cube monotone} and \eqref{eq freq comp ball}, we obtain
+    2. There exists $z \in Z(J) \cap \Omega_1(\fp)$ such that $q \in \Omega(J,z)$, and $q \in B_{\fp'}(\fcc(\fp'), 0.2)$.  Using the triangle inequality, Lemma \ref{lem cube monotone} and \eqref{eq freq comp ball}, we obtain
     $$
-        d_{\fp'}(\fcc(\fp'),z) \le d_{\fp'}(\fcc(\fp'), x) + d_{\fp'}(z, x) \le 0.2 + 2^{-95a} \cdot 1 < 0.3\,.
+        d_{\fp'}(\fcc(\fp'),z) \le d_{\fp'}(\fcc(\fp'), q) + d_{\fp'}(z, q) \le 0.2 + 2^{-95a} \cdot 1 < 0.3\,.
     $$
     Thus $z \in \Omega_1(\fp')$ by \eqref{eq omega1 incl}. By Lemma \ref{lem omega1 disj}, it follows that $\fp = \fp'$.
 
-    3. There exists $z' \in Z(J) \cap \Omega_1(\fp')$ such that $x \in \Omega(J,z')$, and $x \in B_{\fp}(\fcc(\fp), 0.2)$. This case is the same as case 2., after swapping $\fp$ and $\fp'$.
+    3. There exists $z' \in Z(J) \cap \Omega_1(\fp')$ such that $q \in \Omega(J,z')$, and $q \in B_{\fp}(\fcc(\fp), 0.2)$. This case is the same as case 2., after swapping $\fp$ and $\fp'$.
 
-    4. We have $x \in B_{\fp}(\fcc(\fp), 0.2) \cap B_{\fp'}(\fcc(\fp'), 0.2)$. 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{lem omega1 disj}.
+    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{lem omega1 disj}.
 
     To show the inclusion in \eqref{eq dis freq cover}, let $q \in Q(X)$. By the induction hypothesis, there exists $\fp \in \fP(J)$ such that $q \in \Omega(\fp)$. 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 $d_{J^\circ}(Q(x), z) \le 1$. By Lemma \ref{lem cube monotone}, it follows that $d_{I^\circ}(Q(x), z) \le 1$.
     Thus, by \eqref{eq tile cover}, there exists $z' \in Z(I)$ with $z \in B_{I^\circ}(z', 0.7)$. Then by \eqref{eq omega1 cover} there exists $\fp' \in \fP(I)$ with $z \in Z(J) \cap \Omega_1(\fp')$. Consequently, by \eqref{eq it omega}, $q \in \fc(\fp')$. This completes the proof of \eqref{eq dis freq cover}.
@@ -1823,11 +1817,11 @@ \section{Organisation of the tiles}\label{subsectilesorg}
 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(G)}\, ,
 \end{equation}
 but there does not exists a $J\in \mathcal{D}$ with $I\subset J$ and
 \begin{equation}\label{muhj2}
-   {\mu(G \cap J)} > 2^{-k}{\mu(J)}\,.
+    {\mu(G \cap J)} > 2^{-k}{\mu(J)}\,.
 \end{equation}
 Let
 \begin{equation}
@@ -1835,18 +1829,18 @@ \section{Organisation of the tiles}\label{subsectilesorg}
     \fP(k)=\{\fp\in \fP \ : \ \sc(\fp)\in \mathcal{C}(G,k)\}
 \end{equation}
 Define $ {\mathfrak{M}}(n,k)$ to be the set of  $\fp \in \fP(k)$ such that
- \begin{equation}\label{ebardense}
+    \begin{equation}\label{ebardense}
     \mu({E_1}(\fp))  > 2^{-n}  \mu(\sc(\fp))
- \end{equation}
+    \end{equation}
 and there does not exists $\fp'\in \fP(k)$ with
 $\fp'\neq \fp$ and  $\fp\le \fp'$  such that
- \begin{equation}\label{mnkmax}
+    \begin{equation}\label{mnkmax}
     \mu({E_1}(\fp'))  > 2^{-n}  \mu(\sc(\fp')).
- \end{equation}
+    \end{equation}
 Define for a collection $\fP'\subset \fP(k)$
 \begin{equation}
     \label{eq densdef}
-   \dens_k' (\fP'):= \sup_{\fp'\in \fP'}\sup_{\lambda \geq 2} \lambda^{-a} \sup_{\fp \in \fP(k): \lambda \fp' \lesssim \lambda \fp}
+    \dens_k' (\fP'):= \sup_{\fp'\in \fP'}\sup_{\lambda \geq 2} \lambda^{-a} \sup_{\fp \in \fP(k): \lambda \fp' \lesssim \lambda \fp}
     \frac{\mu({E}_2(\lambda, \fp))}{\mu(\sc(\fp))}\,.
 \end{equation}
 Sorting by density, we define
@@ -1859,15 +1853,15 @@ \section{Organisation of the tiles}\label{subsectilesorg}
 Following Fefferman \cite{fefferman}, we
 define for $\fp \in \fP(k)$
     \begin{equation}\label{defbfp}
-         \mathfrak{B}(\fp) := \{ \mathfrak{m} \in \mathfrak{M}(n,k) \ : \ 2 \fp \lesssim 100 \mathfrak{m}\}
+            \mathfrak{B}(\fp) := \{ \mathfrak{m} \in \mathfrak{M}(n,k) \ : \ 2 \fp \lesssim 100 \mathfrak{m}\}
     \end{equation}
 and
 \begin{equation}\label{defcnkj}
-       \fC_1(n,k,j) := \{\fp \in \fC(n,k) \ : \ 2^{j} \leq |\mathfrak{B}(\fp)| < 2^{j+1}\}\,.
+        \fC_1(n,k,j) := \{\fp \in \fC(n,k) \ : \ 2^{j} \leq |\mathfrak{B}(\fp)| < 2^{j+1}\}\,.
 \end{equation}
 and
 \begin{equation}\label{defl0nk}
-       \fL_0(n,k) := \{\fp \in \fC(n,k) \ : \ |\mathfrak{B}(\fp)| <1\}\,.
+        \fL_0(n,k) := \{\fp \in \fC(n,k) \ : \ |\mathfrak{B}(\fp)| <1\}\,.
 \end{equation}
 Together with the following removal of minimal layers, the splitting into $\fC_1(n,k,j)$ will lead to a separation of  trees.
 Define recursively for $0\le l\le Z(n+1)$
@@ -1891,7 +1885,7 @@ \section{Organisation of the tiles}\label{subsectilesorg}
 prospective tree tops, which we define now.
 Define
 \begin{equation}\label{defunkj}
-     \fU_1(n,k,j)
+        \fU_1(n,k,j)
 \end{equation}
 to be the set of all
 $\fu \in \fC_1(n,k,j)$ such that
@@ -1913,7 +1907,7 @@ \section{Organisation of the tiles}\label{subsectilesorg}
 \begin{equation}
 \label{eq C3 def}
 \fC_3(k,n,j):=\fC_2(k,n,j)
-  \setminus \fL_2(k,n,j)\, .
+    \setminus \fL_2(k,n,j)\, .
 \end{equation}
 
 
@@ -1937,7 +1931,7 @@ \section{Organisation of the tiles}\label{subsectilesorg}
 \begin{equation}
 \label{eq C4 def}
 \fC_4(k,n,j):=\fC_3(k,n,j)
-  \setminus \bigcup_{0 \le l \le Z(n+1)} \fL_3(k,n,j,l)\,.
+    \setminus \bigcup_{0 \le l \le Z(n+1)} \fL_3(k,n,j,l)\,.
 \end{equation}
 
 Finally, we remove the boundary pairs relative to the prospective tree tops. Define
@@ -1959,7 +1953,7 @@ \section{Organisation of the tiles}\label{subsectilesorg}
 with $\sc(\fp) \subset \bigcup \mathcal{L}(\fu)$, and define
 \begin{equation}\label{defc5}
 \fC_5(k,n,j):=\fC_4(k,n,j)
-  \setminus \fL_4(k,n,j)\, .
+    \setminus \fL_4(k,n,j)\, .
 \end{equation}
 
 
@@ -1992,7 +1986,7 @@ \section{Organisation of the tiles}\label{subsectilesorg}
         \bigcup_{0\le j\le 2n+3}
         \bigcup_{\fp \in \fL_4 (n,k,j)}
         \sc(\fp)\, .
-     \end{equation}
+        \end{equation}
 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}.
@@ -2009,12 +2003,12 @@ \section{Organisation of the tiles}\label{subsectilesorg}
 \lars{Add constants in the following two lemmas, after figuring out the exact constants in tree section}
 
 \begin{lemma}\label{subsecflemma}
-  Let
-  \begin{equation}
-      \fP'=\bigcup_{k\ge 0}\bigcup_{n\ge k}
-      \bigcup_{0\le j\le 2n+3}\fC_5(k,n,j)
-  \end{equation}
-  For all $f:X\to \C$ with $|f|\le \mathbf{1}_F$ we have
+    Let
+    \begin{equation}
+        \fP'=\bigcup_{k\ge 0}\bigcup_{n\ge k}
+        \bigcup_{0\le j\le 2n+3}\fC_5(k,n,j)
+    \end{equation}
+    For all $f:X\to \C$ with $|f|\le \mathbf{1}_F$ we have
 \begin{equation}
     \label{disclesssim1}
     \int_{G \setminus  G'} \left|\sum_{\fp \in \fP'} T_{\fp} f \right|\, \mathrm{d}\mu  \lesssim \mu(G)^{1/q'} \mu(F)^{1/q}\,.
@@ -2027,12 +2021,12 @@ \section{Organisation of the tiles}\label{subsectilesorg}
 Proposition \ref{antichainprop} to prove the following lemma.
 
 \begin{lemma}\label{subsecalemma}
-  Let
-  \begin{equation}
-      \fP'=\fP\setminus \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
+    Let
+    \begin{equation}
+        \fP'=\fP\setminus \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}
     \label{disclesssim2}
     \int_{G \setminus  G'} \left|\sum_{\fp \in \fP'} T_{\fp} f\right| \, \mathrm{d}\mu  \lesssim \mu(G)^{1/q'} \mu(F)^{1/q}\,.
@@ -2066,7 +2060,7 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 For each $\fp\in \fP_{F,G}$ pick a
 $r(\fp)>4D^{\ps(\fp)}$  with
 $$
-  {\mu(F\cap B(\pc(\fp),r(\fp)))}\ge 2^{-k_0-1}{\mu(B(\pc(\fp),r(\fp)))}\, .
+    {\mu(F\cap B(\pc(\fp),r(\fp)))}\ge 2^{-k_0-1}{\mu(B(\pc(\fp),r(\fp)))}\, .
 $$
 This ball exists by definition of $\fP_{F,G}$
 and $\dens_2$. By applying Proposition \ref{prop hlm} to the collection of balls
@@ -2122,11 +2116,11 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 
 
 \begin{lemma}\label{pairwise disjoint}
-     If $\fp, \fp' \in {\mathfrak{M}}(n,k)$ and
-     \begin{equation}\label{eintersect}
-     {E_1}(\fp)\cap {E_1}(\fp')\neq \emptyset,
-     \end{equation}
-     then $\fp=\fp'$.
+        If $\fp, \fp' \in {\mathfrak{M}}(n,k)$ and
+        \begin{equation}\label{eintersect}
+        {E_1}(\fp)\cap {E_1}(\fp')\neq \emptyset,
+        \end{equation}
+        then $\fp=\fp'$.
 \end{lemma}
 \begin{proof}
 Let $\fp,\fp'$ as in the lemma. As by definition of $E_1$
@@ -2153,7 +2147,7 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 Fix $k,n,\lambda,x$ as in the lemma.
 Let  $x\in A(\lambda,n,k)$. Let
 $\mathcal{M}$ be the set of dyadic intervals
- $\sc(\fp)$
+    $\sc(\fp)$
 with $\fp$ in  $\mathfrak{M}(n,k)$
 and $x\in \sc(\fp)$. By definition of
 $A(\lambda,n,k)$, the cardinality of $\mathcal{M}$
@@ -2186,9 +2180,9 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 
 Let $L\in \mathcal{M}^*$. For each $x\in L$, we have
 \begin{equation}\label{suminout}
- \sum_{\fp \in {\mathfrak{M}}(n,k)} 1_{I(\fp)}(x)=
-   \sum_{\fp \in {\mathfrak{M}}(n,k):\sc(\fp) \subset L} 1_{I(\fp)}(x)+
-  \sum_{\fp \in {\mathfrak{M}}(n,k):\sc(\fp) \not \subset L} 1_{I(\fp)}(x)\, .
+    \sum_{\fp \in {\mathfrak{M}}(n,k)} 1_{I(\fp)}(x)=
+    \sum_{\fp \in {\mathfrak{M}}(n,k):\sc(\fp) \subset L} 1_{I(\fp)}(x)+
+    \sum_{\fp \in {\mathfrak{M}}(n,k):\sc(\fp) \not \subset L} 1_{I(\fp)}(x)\, .
 \end{equation}
 If the second sum on the right-hand-side is not zero, there is
 an element of $\mathcal{D}$  containing $L$.
@@ -2215,7 +2209,7 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 We then have with \eqref{mnkonl} and \eqref{mnkintl}
 \begin{equation}
 2^{n+1}\mu(A(\lambda+1)\cap L) =
- \int_{A(\lambda+1)\cap L} 2^{n+1} d\mu
+    \int_{A(\lambda+1)\cap L} 2^{n+1} d\mu
 \end{equation}
 \begin{equation}
 \le
@@ -2247,7 +2241,7 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 \mu(A(2n+6,k,n))\le \sum_{0\le  k}\sum_{k< n} 2^{k-5-2n}\mu(G)
 \end{equation}
 \begin{equation}
-   \le \sum_{0\le  k} 2^{-k-5}\mu(G)\le 2^{-4}\mu(G)\, .
+    \le \sum_{0\le  k} 2^{-k-5}\mu(G)\le 2^{-4}\mu(G)\, .
 \end{equation}
 This proves the lemma.
 \end{proof}
@@ -2285,7 +2279,7 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
     \sum_{\fu\in \fU_1(n,k,j)} 1_{\sc(\fu)}(x)
     \le 2^{1-j}
     2^{4a} \sum_{\mathfrak{m}\in \mathfrak{M}(n,k)}
-     1_{\sc(\mathfrak{m})}(x)
+        1_{\sc(\mathfrak{m})}(x)
 \end{equation}
 \end{lemma}
 
@@ -2298,7 +2292,7 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 with $2\fu \lesssim 100\mathfrak{m}$ and in particular
 $x\in \sc(\mathfrak{m})$. Hence
 \begin{equation}\label{ubymsum}
-     1_{\sc(\fu)}(x)
+        1_{\sc(\fu)}(x)
     \le 2^{1-j}\sum_{\mathfrak{m} \in \mathfrak{M}(n,k): 2\fu\lesssim 100 \mathfrak{m}} 1_{\sc(\mathfrak{m})}(x)\, .
 \end{equation}
 Conversely, for each $\mathfrak{m}\in \mathfrak{M}(n,k)$
@@ -2310,7 +2304,7 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 $(\fu,\mathfrak{m})$ with $2\fu \lesssim 100\mathfrak{m}$
 differently gives
 \begin{equation}\label{usumbymsum}
-     \sum_{\fu\in \fU_1(n,k,j)} 1_{\sc(\fu)}(x)
+        \sum_{\fu\in \fU_1(n,k,j)} 1_{\sc(\fu)}(x)
     \le 2^{1-j}\sum_{\mathfrak{m} \in \mathfrak{M}(n,k)}
     \sum_{\fu \in \fU(\mathfrak{m})} 1_{\sc(\mathfrak{m})}(x)\, .
 \end{equation}
@@ -2322,7 +2316,7 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 Then by definition of
 $\fU(\mathfrak{m})$
 \begin{equation}\label{dby2}
-     d_{\fu}(\fcc(\fu),\fcc(\mathfrak{m}))\le 2\, .
+        d_{\fu}(\fcc(\fu),\fcc(\mathfrak{m}))\le 2\, .
 \end{equation}
 If $\fu'$ is a further element in $\fU(\mathfrak{m})$ with $\fu\neq \fu'$, then
 \begin{equation}
@@ -2373,7 +2367,7 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 \lars{Write consistently $\ps$ or $s$.}
 
 \begin{proof}
-  Let $\fu\in \fU_1(n,k,l)$.
+    Let $\fu\in \fU_1(n,k,l)$.
 Let $I \in \mathcal{L}(\fu)$. Then we have $s(I) = s(\fu) - Z(n+1) - 1$ and $I \subset \sc(\fu)$ and $B(c(I), 8D^{s(I)}) \not \subset \sc(\fu)$.
 By \eqref{eq vol sp cube}, the set $I$
 is contained in $B(c(I), 4D^{s(I)})$.
@@ -2382,8 +2376,8 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 \begin{equation}
     X(\fu):=\{x \in \sc(\fu) \, : \, \rho(x, X \setminus \sc(\fu)) \leq 12 D^{s(\fu) - Z(n+1)-1}\}\,.
 \end{equation}
- By the small boundary property \eqref{eq small boundary}, we have
-   $$
+    By the small boundary property \eqref{eq small boundary}, we have
+    $$
         \mu(X(\fu)) \le
         2^{2a+2}(12 D^{-Z(n+1)-1})^\kappa
         \mu(\sc(\mathfrak{u})).
@@ -2406,7 +2400,7 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 
     \begin{lemma}\label{g3bound}
 
-       We have
+        We have
 \begin{equation}
     \mu(G_3)\le 2^{-4} \mu(G)\, .
 \end{equation}
@@ -2424,7 +2418,7 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 \mu(\bigcup_{I \in \mathcal{L} (\fu)}I).
 \end{equation}
 Using Lemma \ref{lulemma} and the Lemma \ref{countu}, we estimate this further
- by
+    by
 \begin{equation}
     \le \sum_{\fu\in \fU_1(n,k,j)}
     2^{2a+2} D^{-\kappa Z(n+1)}
@@ -2432,7 +2426,7 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 \end{equation}
 \begin{equation}
     \le 2^{6a+3-j} \sum_{\mathfrak{m}\in \mathfrak{M}(n,k)}
-     D^{-\kappa Z(n+1)}
+        D^{-\kappa Z(n+1)}
     \mu(\sc(\mathfrak{m}))\,.
 \end{equation}
 %\begin{equation}
@@ -2441,10 +2435,10 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
 %    \mu(\sc(\mathfrak{m}))\, ,
 %\end{equation}
 Using Lemma \ref{musumlemma}, we estimate this by
-  \begin{equation}
-     \le
+    \begin{equation}
+        \le
 2^{6a + 3-j}  D^{-\kappa Z(n+1)}
-     2^{n+1}2^{k+1}\mu(G)\, .
+        2^{n+1}2^{k+1}\mu(G)\, .
 \end{equation}
 Now we estimate $G_3$ defined in \eqref{defineg3} by
 \begin{equation}
@@ -2471,12 +2465,12 @@ \section{Proof of Lemma \ref{allgbound}, the exceptional sets}
     \le \sum_{k\ge 0} 2^{6a + 7 + 2k}  D^{-\kappa Z(k+1)}\mu(G)
 \end{equation}
 \begin{equation}
-   \le 2^{6a + 8}  D^{-\kappa Z}\mu(G)
+    \le 2^{6a + 8}  D^{-\kappa Z}\mu(G)
 \end{equation}
 Using $D = 2^{100a^2}$ and $a \ge 4$ and $\kappa Z \ge 1$ proves the lemma.
 \end{proof}
 
-\section{Auxiliary lemmas}
+\section{Auxilliary lemmas}
 
 Before proving Lemma \ref{subsecflemma} and Lemma \ref{subsecalemma}, we collect some useful properties of $\lesssim$.
 
@@ -2615,7 +2609,7 @@ \section{Auxiliary lemmas}
 \end{lemma}
 
 \begin{proof}
-     Let $\fp \le \fp' \le \fp''$ with $\fp, \fp'' \in \fC_2(n,k,j)$. By \eqref{eq C2 def}, we have $\fC_2(n,k,j) \subset \fC_1(n,k,j)$. Combined with Lemma \ref{lem convexity C1}, it follows that $\fp' \in \fC_1(n,k,j)$. Suppose that $\fp' \notin \fC_2(n,k,j)$. By \eqref{eq C2 def}, this implies that there exists $0 \le l' \le Zn$ \lars{Z} with $\fp' \in \fL_1(n,k,j,l')$. By the definition \eqref{eq L1 def} of $\fL_1(n,k,j,l')$, this implies that $\fp$ is minimal with respect to $\le$  in $\fC_1(n,k,j) \setminus \bigcup_{l < l'} \fL_1(n,k,j,l)$. Since $\fp \in \fC_2(n,k,j)$ we must have $\fp \ne \fp'$. Thus $\fp \le \fp'$ and $\fp \ne \fp'$. By minimality of $\fp'$ it follows that $\fp \notin \fC_1(n,k,j) \setminus \bigcup_{l < l'} \fL_1(n,k,j,l)$. But by \eqref{eq C2 def} this implies $\fp \notin \fC_2(n,k,j)$, a contradiction.
+        Let $\fp \le \fp' \le \fp''$ with $\fp, \fp'' \in \fC_2(n,k,j)$. By \eqref{eq C2 def}, we have $\fC_2(n,k,j) \subset \fC_1(n,k,j)$. Combined with Lemma \ref{lem convexity C1}, it follows that $\fp' \in \fC_1(n,k,j)$. Suppose that $\fp' \notin \fC_2(n,k,j)$. By \eqref{eq C2 def}, this implies that there exists $0 \le l' \le Zn$ \lars{Z} with $\fp' \in \fL_1(n,k,j,l')$. By the definition \eqref{eq L1 def} of $\fL_1(n,k,j,l')$, this implies that $\fp$ is minimal with respect to $\le$  in $\fC_1(n,k,j) \setminus \bigcup_{l < l'} \fL_1(n,k,j,l)$. Since $\fp \in \fC_2(n,k,j)$ we must have $\fp \ne \fp'$. Thus $\fp \le \fp'$ and $\fp \ne \fp'$. By minimality of $\fp'$ it follows that $\fp \notin \fC_1(n,k,j) \setminus \bigcup_{l < l'} \fL_1(n,k,j,l)$. But by \eqref{eq C2 def} this implies $\fp \notin \fC_2(n,k,j)$, a contradiction.
 \end{proof}
 
 \begin{lemma}
@@ -2667,25 +2661,25 @@ \section{Auxiliary lemmas}
     and \eqref{muhj2}.
 
 We show \eqref{muhj1}.
- As $\fp\in \fP(\fP')$, there exists
+    As $\fp\in \fP(\fP')$, there exists
 $\fp''\in \fP'$ with $\sc(\fp')\subset \sc(\fp'')$. By assumption on $\fP'$, we have  $\fp''\in \fP(k)$ and there exists
 $J\in \mathcal{D}$ with
-   $\sc(\fp'')\subset J$ and
-   \begin{equation}
-       \mu(G\cap J)>2^{-k-1} \mu(J).
-   \end{equation}
+    $\sc(\fp'')\subset J$ and
+    \begin{equation}
+        \mu(G\cap J)>2^{-k-1} \mu(J).
+    \end{equation}
 Then also $\sc(\fp')\subset J$, which proves
 \eqref{muhj1} for $\fp$.
 
 We show \eqref{muhj2}. Assume to get a contradiction that
 there exists $J\in \mathcal{D}$ with
-   $\sc(\fp)\subset J$ and
-   \begin{equation}\label{mugj}
-       \mu(G\cap J)>2^{-k} \mu(J).
-   \end{equation}
-   As $\lambda\fp'\lesssim \lambda\fp$, we have $\sc(\fp')\subset \sc(\fp)$, and therefore
+    $\sc(\fp)\subset J$ and
+    \begin{equation}\label{mugj}
+        \mu(G\cap J)>2^{-k} \mu(J).
+    \end{equation}
+    As $\lambda\fp'\lesssim \lambda\fp$, we have $\sc(\fp')\subset \sc(\fp)$, and therefore
     $\sc(\fp')\subset J$. This contradicts
-   $\fp'\in \fP'\subset \fP(k)$. This proves
+    $\fp'\in \fP'\subset \fP(k)$. This proves
 \eqref{muhj2} for $\fp$.
 \end{proof}
 
@@ -2780,7 +2774,7 @@ \section{Proof of Lemma \ref{subsecflemma}, the forests}
 Define for each $\fu\in \fU_3(n,k,j)$
 \begin{equation}\label{definesv}
 \fT_2(\fu):=
-   \bigcup_{\fu\sim \fu'}\mathfrak{T}_1(\fu')\cap \fC_6(k,n,j)\, .
+    \bigcup_{\fu\sim \fu'}\mathfrak{T}_1(\fu')\cap \fC_6(k,n,j)\, .
 \end{equation}
 
 \begin{lemma}
@@ -3094,7 +3088,7 @@ \section{The density  arguments}\label{sec TT* T*T}
 Let $x\in X$.
 Then
 \begin{equation}\label{hlmbound}
-  | \sum_{\fp \in \mathfrak{A}}T_{\fp} f(x)|\le 2^{107 a^3} M_{\mathcal{B}} f (x) \, .
+    | \sum_{\fp \in \mathfrak{A}}T_{\fp} f(x)|\le 2^{107 a^3} M_{\mathcal{B}} f (x) \, .
 \end{equation}
 \end{lemma}
 
@@ -3103,11 +3097,11 @@ \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 $\fp \in \mathfrak{A}$
 such that
- $T_{\fp} f(x)$ is not zero.
- If there is no such $\fp$, the estimate \eqref{hlmbound} follows.
+    $T_{\fp} f(x)$ is not zero.
+    If there is no such $\fp$, the estimate \eqref{hlmbound} follows.
 
- Assume there is such a $\fp$.
- By definition of $T_{\fp}$ we have $x\in E(\fp)\subset \sc(\fp)$ and  by the squeezing property \eqref{eq vol sp cube}
+    Assume there is such a $\fp$.
+    By definition of $T_{\fp}$ we have $x\in E(\fp)\subset \sc(\fp)$ and  by the squeezing property \eqref{eq vol sp cube}
 \begin{equation}\label{eqtttt0}
     \rho(x, \pc(\fp))\le 4D^{\ps(\fp)}\, .
 \end{equation}
@@ -3115,8 +3109,8 @@ \section{The density  arguments}\label{sec TT* T*T}
 Let $y\in X$ with  $K_{\ps(\fp)}(x,y)\neq 0$. By Definition \eqref{defks} of $K_{\ps(\fp)}$
 we have
 \begin{equation}\label{supp Ks1}
-   \frac{1}{4} D^{\ps(\fp)-1}
-   \leq \rho(x,y) \leq \frac{1}{2} D^{\ps(\fp)}\, .
+    \frac{1}{4} D^{\ps(\fp)-1}
+    \leq \rho(x,y) \leq \frac{1}{2} D^{\ps(\fp)}\, .
 \end{equation}
 As by the squeezing property \eqref{eq vol sp cube}, we have
 \begin{equation}
@@ -3137,10 +3131,10 @@ \section{The density  arguments}\label{sec TT* T*T}
 \begin{equation}
 \le \frac{2^{5a+101a^3}}{\mu(B(x,  8D^{\ps(\fp)}))}\, .
 \end{equation}
- Using that $|\mfa|$ is bounded by $1$
- for every $\mfa\in \Mf$, we estimate with the triangle inequality and the above information
- \begin{equation}
-  | T_{\fp} f(x)|
+    Using that $|\mfa|$ is bounded by $1$
+    for every $\mfa\in \Mf$, we estimate with the triangle inequality and the above information
+    \begin{equation}
+    | T_{\fp} f(x)|
     \le \frac{2^{5a+101 a^3}}{\mu(B(x, 8D^{\ps(\fp)}))} \int _{\mu(B(x, 8D^{\ps(\fp)}))} |f(y)|\, dy   \end{equation}
 This together with $a\ge 1$ proves the Lemma.
 \end{proof}
@@ -3154,8 +3148,8 @@ \section{The density  arguments}\label{sec TT* T*T}
 \label{lem decay t}
 
 \begin{equation}\label{eqttt9}
-  |\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)|\le
-  2^{111a^2}({q}-1)^{-1} \dens_2(\mathfrak{A})^{\frac 1{\tilde{q}}-\frac 12} \|f\|_2\|g\|_2\, .
+    |\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)|\le
+    2^{111a^2}({q}-1)^{-1} \dens_2(\mathfrak{A})^{\frac 1{\tilde{q}}-\frac 12} \|f\|_2\|g\|_2\, .
 \end{equation}
 \end{lemma}
 \begin{proof}
@@ -3176,7 +3170,7 @@ \section{The density  arguments}\label{sec TT* T*T}
 Taking the maximum over all $B'$ containing $x$, we obtain
 \begin{equation} \label{eqttt1}
     M_{\mathcal{B}}|f|\le
-     \left(M_{\mathcal{B},\frac {2{\tilde{q}}}{3{\tilde{q}}-2} } f\right)
+        \left(M_{\mathcal{B},\frac {2{\tilde{q}}}{3{\tilde{q}}-2} } f\right)
 \dens_2(\mathfrak{A})^{\frac 1{\tilde{q}}-\frac 12}\, .
 \end{equation}
 
@@ -3187,25 +3181,25 @@ \section{The density  arguments}\label{sec TT* T*T}
 
 Using $1<\tilde{q}\le 2$ estimates the last display by
 \begin{equation}\label{eqttt2}
- 2^{2a+2} (\tilde{q}-1)^{-1}  \|f\|_2\, .
+    2^{2a+2} (\tilde{q}-1)^{-1}  \|f\|_2\, .
 \end{equation}
 
 
 We obtain with Cauchy-Schwarz
 and then Lemma \ref{lem hlmbound}
- \begin{equation}
-  |\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)|
+    \begin{equation}
+    |\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)|
 \end{equation}
- \begin{equation}
-  \le \|g\|_2 \| \sum_{\fp \in \mathfrak{A}} T_{\fp} f \|_2
+    \begin{equation}
+    \le \|g\|_2 \| \sum_{\fp \in \mathfrak{A}} T_{\fp} f \|_2
 \end{equation}
- \begin{equation}
-  \le 2^{107a^2}\|g\|_2 \|  M_{\mathcal{B}}f \|_2
+    \begin{equation}
+    \le 2^{107a^2}\|g\|_2 \|  M_{\mathcal{B}}f \|_2
 \end{equation}
 With \eqref{eqttt1} and
 \eqref{eqttt2} we can estimate the last display by
 \begin{equation}
-  \le 2^{107a^2+2a+2}(\tilde{q}-1)^{-1} \|g\|_2 \|f\|_2\dens_2(\mathfrak{A})^{\frac 1{\tilde{q}}-\frac 12}
+    \le 2^{107a^2+2a+2}(\tilde{q}-1)^{-1} \|g\|_2 \|f\|_2\dens_2(\mathfrak{A})^{\frac 1{\tilde{q}}-\frac 12}
 \end{equation}
 Using $a\ge 4$ and
 $q-1=2(\tilde{q}-1)$
@@ -3216,10 +3210,10 @@ \section{The density  arguments}\label{sec TT* T*T}
 
 
 \begin{lemma}\label{lem decay n}
- We have
+    We have
     \begin{equation}\label{eqttt3}
-  |\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)|\le
-   \tau^{-1}2^{200a^3}\dens_1(\mathfrak{A})^{\frac 1{2p}} \|f\|_2\|g\|_2
+    |\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)|\le
+    \tau^{-1}2^{200a^3}\dens_1(\mathfrak{A})^{\frac 1{2p}} \|f\|_2\|g\|_2
 \end{equation}
 \end{lemma}
 
@@ -3227,45 +3221,45 @@ \section{The density  arguments}\label{sec TT* T*T}
 
 \begin{proof}
 
- We write for the left-hand side of \eqref{eqttt3}
+    We write for the left-hand side of \eqref{eqttt3}
 \begin{equation}
-  \sum_{\fp \in \mathfrak{A}}\int \overline{g(x)} 1_{E(\fp)}(x)
-  {K_{\ps(\fp)}(x,y)}{\tQ(x)(y)}
-    \overline{\tQ(x)(x)}
-   f(y)\, d\mu (y)d\mu(x)
+    \sum_{\fp \in \mathfrak{A}}\int \overline{g(x)} 1_{E(\fp)}(x)
+    {K_{\ps(\fp)}(x,y)}e(\tQ(x)(y) -
+    \tQ(x)(x))
+    f(y)\, d\mu (y)d\mu(x)
 \end{equation}
 \begin{equation}
-  =\int \sum_{\fp \in \mathfrak{A}} \overline{T_{\fp} ^*g(y)}   f(y)\, d\mu(y)
+    =\int \sum_{\fp \in \mathfrak{A}} \overline{T_{\fp} ^*g(y)}   f(y)\, d\mu(y)
 \end{equation}
 with the adjoint operator
 \begin{equation}\label{eq tstarwritten}
-    T_{\fp}^*g(y)=\int_{E(\fp)} \overline{K_{\ps(\fp)}(x,y)}\overline{\tQ(x)(y)}
-    \tQ(x)(x)g(x)\, d\mu(x)\, .
+    T_{\fp}^*g(y)=\int_{E(\fp)} \overline{K_{\ps(\fp)}(x,y)}e(-\tQ(x)(y)+
+    \tQ(x)(x))g(x)\, d\mu(x)\, .
 \end{equation}
 
 
 
 
 
- We have by expanding the square
-      \begin{equation}
- \int |\sum_{\fp\in \mathfrak{A}}T^*_{\fp}g(y)|^2\, d\mu(y)=
- \int \left(\sum_{\fp\in \mathfrak{A}}\int T^*_{\fp}g(y)\right)
- \left(\overline{\sum_{\fp'\in \mathfrak{A}}T^*_{\fp}g(y)}\right)\, d\mu(y)
- \end{equation}\label{eqtts1}
- \begin{equation}
- \le \sum_{\fp\in \mathfrak{A}} \sum_{\fp'\in \mathfrak{A}}
-  \Big|\int T^*_{\fp}g(y)\overline{T^*_{\fp'}g(y)}\, d\mu(y)\Big|
-         \end{equation}
+    We have by expanding the square
+        \begin{equation}
+    \int |\sum_{\fp\in \mathfrak{A}}T^*_{\fp}g(y)|^2\, d\mu(y)=
+    \int \left(\sum_{\fp\in \mathfrak{A}}\int T^*_{\fp}g(y)\right)
+    \left(\overline{\sum_{\fp'\in \mathfrak{A}}T^*_{\fp}g(y)}\right)\, d\mu(y)
+    \end{equation}\label{eqtts1}
+    \begin{equation}
+    \le \sum_{\fp\in \mathfrak{A}} \sum_{\fp'\in \mathfrak{A}}
+    \Big|\int T^*_{\fp}g(y)\overline{T^*_{\fp'}g(y)}\, d\mu(y)\Big|
+            \end{equation}
 We split the sum into the terms with $s(\fp')\le s(\fp)$
 and $s(\fp)< s(\fp')$. Using the symmetry of each summand,
 we may switch $\fp$ and $\fp'$ in the second sum. Using further positivity
 of each summand to replace the condition $s(\fp')< s(\fp)$
 by $s(\fp')\le  s(\fp)$ in the second sum, we estimate \eqref{eqtts1} by
 \begin{equation}\label{eqtts2}
- \le2 \sum_{\fp\in \mathfrak{A}} \sum_{\fp'\in \mathfrak{A}: \ps(\fp')\le \ps(\fp)}
-  \Big|\int T^*_{\fp}g(y)\overline{T^*_{\fp'}g(y)}\, d\mu(y)\Big|
-         \end{equation}
+    \le2 \sum_{\fp\in \mathfrak{A}} \sum_{\fp'\in \mathfrak{A}: \ps(\fp')\le \ps(\fp)}
+    \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*}
@@ -3306,18 +3300,18 @@ \section{The density  arguments}\label{sec TT* T*T}
     \mu(B(\fp))\le 2^{16a} \mu(\sc(\fp))\, .
 \end{equation}
 
- Using Lemma \ref{lem basic TT*}, and \eqref{eqttt4} we estimate \eqref{eqtts2} by
-  \begin{equation}\label{eqtts3}
- \le  2^{162a^3+16a+1} \sum_{\fp\in \mathfrak{A}}
- \int_{E(\fp)}|g|(y) h(\fp)\, dy
-         \end{equation}
+    Using Lemma \ref{lem basic TT*}, and \eqref{eqttt4} we estimate \eqref{eqtts2} by
+    \begin{equation}\label{eqtts3}
+    \le  2^{162a^3+16a+1} \sum_{\fp\in \mathfrak{A}}
+    \int_{E(\fp)}|g|(y) h(\fp)\, dy
+            \end{equation}
 with $h(\fp)$ defined as
 \begin{equation}\label{def hp}
     \frac 1{\mu(B(\fp))}\int \sum_{\fp'\in \mathfrak{A}(\fp)}
 {(1+d_{\fp'}(\fcc(\fp'), \fcc(\fp))^{-\tau^2/(2+a)}}(1_{E(\fp')}|g|)(y')\, dy'\,.
 \end{equation}
 
- The following lemma will be proved in Subsection \ref{subsec geolem}.
+    The following lemma will be proved in Subsection \ref{subsec geolem}.
 \begin{lemma}
 \label{lem antichain 1}
 Set $p:=4a^2/\tau^2$ and let $p'$ be the dual exponent of $p$, that is $1/p+1/p'=1$.
@@ -3333,18 +3327,18 @@ \section{The density  arguments}\label{sec TT* T*T}
 \end{lemma}
 
 Let $p$ be as stated in the lemma.
-   We estimate $h(\fp)$ as defined in \eqref{def hp} with H\"older using  $|g|\le 1_G$ and $E(\fp')\subset B(\fp)$ by
+    We estimate $h(\fp)$ as defined in \eqref{def hp} with H\"older using  $|g|\le 1_G$ and $E(\fp')\subset B(\fp)$ by
 
-   \begin{equation}
+    \begin{equation}
 \frac{\|g1_{B(\fp)}\|_{p'}}{\mu(B(\fp))}
- \|\sum_{\fp\in\mathfrak{A}(\fp)}(1+d_{\fp}(\fcc(\fp), \fc(\fp'))^{-\tau^2/(2+a)}\mathbf{1}_{E(\fp)}1_G\|_{p}\, .
- \end{equation}
+    \|\sum_{\fp\in\mathfrak{A}(\fp)}(1+d_{\fp}(\fcc(\fp), \fc(\fp'))^{-\tau^2/(2+a)}\mathbf{1}_{E(\fp)}1_G\|_{p}\, .
+    \end{equation}
 Then we apply Lemma \ref{lem antichain 1} to estimate this by
 \begin{equation}\label{eqttt5}
     \le  \tau^{-2}
 2^{102a}
     \frac{\|g1_{B(\fp)}\|_{p'}}{\mu(B(\fp))}
-   \dens_1(\mathfrak{A})^{\frac 1p}\mu(B(\fp))^{\frac 1p}\, .
+    \dens_1(\mathfrak{A})^{\frac 1p}\mu(B(\fp))^{\frac 1p}\, .
 \end{equation}
 Let $\mathcal{B}$ be the collection of all balls
 $B(\fp)$ with $\fp\in \mathfrak{A}$. Then
@@ -3360,39 +3354,39 @@ \section{The density  arguments}\label{sec TT* T*T}
     \le  \tau^{-2}
 2^{102a}
     (M_{\mathcal{B}, p'}g)
-   \dens_1(\mathfrak{A})^{\frac 1p}\, .
+    \dens_1(\mathfrak{A})^{\frac 1p}\, .
 \end{equation}
 With this estimate of $h(\fp)$,
 using $E(\fp)\subset B(\fp)$ by construction of $B(\fp)$, we estimate
 \eqref{eqtts3} by
- \begin{equation}\label{eqtts4}
- \le  \tau^{-2}2^{162a^3+118a + 1} \sum_{\fp\in \mathfrak{A}}
- \int_{E(\fp)}|g|(y)M_{\mathcal{B}, p'}g(y) \, dy
-         \end{equation}
+    \begin{equation}\label{eqtts4}
+    \le  \tau^{-2}2^{162a^3+118a + 1} \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},
 the last display is observed to be
 \begin{equation}\label{eqtts4}
 =  \tau^{-2}2^{162a^3+118a + 1}
- \int |g|(y)(M_{\mathcal{B}, p'}g)(y) \, dy
-         \end{equation}
+    \int |g|(y)(M_{\mathcal{B}, p'}g)(y) \, dy
+            \end{equation}
 Applying Cauchy-Schwartz and using Proposition \eqref{prop hlm} estimates the last display by
 \begin{equation}
-  \tau^{-2}2^{162a^3+118a + 1}
- \|g\|_2 \|M_{\mathcal{B}, p'} g\|_2
-         \end{equation}
+    \tau^{-2}2^{162a^3+118a + 1}
+    \|g\|_2 \|M_{\mathcal{B}, p'} g\|_2
+            \end{equation}
 \begin{equation}
- \le  \tau^{-2}2^{162a^3+118a + 3}\frac{2}{2-p'}
- \|g\|_2 ^2
-          \end{equation}
+    \le  \tau^{-2}2^{162a^3+118a + 3}\frac{2}{2-p'}
+    \|g\|_2 ^2
+            \end{equation}
 \begin{equation}\le  \tau^{-2}2^{162a^3+118a + 3}\frac{2}{2-p'}
 \|g\|_2^2
-         \end{equation}
+            \end{equation}
 Using $p>4$ and thus $1<p'<\frac 43$, we estimate the last display by
 \begin{equation}\le  \tau^{-2}2^{162a^3+118a + 5}
 \|g\|_2^2
-         \end{equation}
+            \end{equation}
 Now lemma \ref{lem decay n} follows by applying Cauchy-Schwartz on the left-hand side and using
- $a>4$.
+    $a>4$.
 
 
 
@@ -3407,12 +3401,12 @@ \section{The density  arguments}\label{sec TT* T*T}
 Multiplying  the $(2-q)$-th power of  \eqref{eqttt9} and the $(q-1)$-th power of \eqref{eqttt3}
 and estimating gives after simplification of some factors gives
 \begin{equation}\label{eqttt8}
-  |\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)|
-  \end{equation}
-  \begin{equation}
-      \le  2^{200a^3}({q}-1)^{-1} \tau^{-1}\dens_1(\mathfrak{A})^{\frac {q-1}{2p}}\dens_2(\mathfrak{A})^{\frac 1{q}-\frac 12}  \|f\|_2\|g\|_2\, .
-  \end{equation}
- With the definition of $p$, this  implies
+    |\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)|
+    \end{equation}
+    \begin{equation}
+        \le  2^{200a^3}({q}-1)^{-1} \tau^{-1}\dens_1(\mathfrak{A})^{\frac {q-1}{2p}}\dens_2(\mathfrak{A})^{\frac 1{q}-\frac 12}  \|f\|_2\|g\|_2\, .
+    \end{equation}
+    With the definition of $p$, this  implies
 Proposition \ref{antichainprop}.
 
 
@@ -3422,11 +3416,11 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 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}
-  \frac{1}{4} D^{s-1} \leq \rho(x,y) \leq \frac{1}{2} D^s\, .
+    \frac{1}{4} D^{s-1} \leq \rho(x,y) \leq \frac{1}{2} D^s\, .
 \end{equation}
 We have
 \begin{equation}
-   \label{eq Ks size}
+    \label{eq Ks size}
     |K_s(x,y)|\le \frac{2^{102 a^3}}{\mu(B(x, D^{s}))}\,
 \end{equation}
 and \begin{equation}
@@ -3463,9 +3457,9 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 \end{equation}
 Using $\tau\le 1$, this is estimated by
 \begin{equation}
-   \le \frac{4D 2^{2a+101a^3}}{\mu(B(x,  D^{s}))}
+    \le \frac{4D 2^{2a+101a^3}}{\mu(B(x,  D^{s}))}
     \left(\frac{ \rho(y,y')}{D^{s}}\right)^{\tau}
-     = \frac{2^{2+2a+100a^2+101a^3}}{\mu(B(x,  D^{s}))}
+        = \frac{2^{2+2a+100a^2+101a^3}}{\mu(B(x,  D^{s}))}
     \left(\frac{ \rho(y,y')}{D^{s}}\right)^{\tau}\,.
 \end{equation}
 Using $a\ge 4$, this proves  \eqref{eq Ks smooth}.
@@ -3475,8 +3469,8 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 \begin{lemma}\label{lem ksquare}
 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)}
- K_{s_2}(x_2, y) \, .
+    \varphi(y) :=  \overline{K_{s_1}(x_1, y)}
+    K_{s_2}(x_2, y) \, .
 \end{equation}
 If $\varphi(y)\neq 0$, then
 \begin{equation}\label{eqt10}
@@ -3484,9 +3478,9 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 \end{equation}
 Moreover,
 \begin{equation}\label{eqt11}
-  \|\varphi\|_{C^\tau(B(x_1, 5D^{s_1})}\le
+    \|\varphi\|_{C^\tau(B(x_1, 5D^{s_1})}\le
 \frac{2^{154 a^3}}{\mu(B(x_1, D^{s_1}))\mu(B(x_2, D^{s_2}))}
-      \, .
+        \, .
 \end{equation}
 
 \end{lemma}
@@ -3503,23 +3497,23 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 and for $y'\neq y$
 \begin{equation}
     |\varphi(y)-\varphi(y')|
- \end{equation}
- \begin{equation}
- \le
- |K_{s_1}(x_1,y)-K_{s_1}(x_1,y'))||
- K_{s_2}(x_2, y)|
+    \end{equation}
+    \begin{equation}
+    \le
+    |K_{s_1}(x_1,y)-K_{s_1}(x_1,y'))||
+    K_{s_2}(x_2, y)|
 \end{equation}
- \begin{equation}+| \overline{K_{s_1}(x_1, y')}|
- |K_{s_2}(x_2, y) - K_{s_2}(x_2, y'))|
+    \begin{equation}+| \overline{K_{s_1}(x_1, y')}|
+    |K_{s_2}(x_2, y) - K_{s_2}(x_2, y'))|
 \end{equation}
 \begin{equation}
-      \le \frac{2^{152 a^3}}{\mu(B(x_1, D^{s_1}))\mu(B(x_2, D^{s_2}))}
-       \left(\left(\frac{ \rho(y,y')}{D^{s_1}}\right)^{\tau}+
-       \left(\frac{ \rho(y,y')}{D^{s_2}}\right)^{\tau}\right)
+        \le \frac{2^{152 a^3}}{\mu(B(x_1, D^{s_1}))\mu(B(x_2, D^{s_2}))}
+        \left(\left(\frac{ \rho(y,y')}{D^{s_1}}\right)^{\tau}+
+        \left(\frac{ \rho(y,y')}{D^{s_2}}\right)^{\tau}\right)
 \end{equation}
 \begin{equation}\label{holderpart}
-      \le \frac{2^{153 a^3}}{\mu(B(x_1, D^{s_1}))\mu(B(x_2, D^{s_2}))}
-       \left(\frac{ \rho(y,y')}{D^{s_1}}\right)^{\tau}
+        \le \frac{2^{153 a^3}}{\mu(B(x_1, D^{s_1}))\mu(B(x_2, D^{s_2}))}
+        \left(\frac{ \rho(y,y')}{D^{s_1}}\right)^{\tau}
 \end{equation}
 Adding the estimates \eqref{suppart} and \eqref{holderpart} gives \eqref{eqt11}.
 This proves the lemma.
@@ -3530,7 +3524,7 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 $\ps({\fp_1})\leq \ps({\fp_2})$. For each $x_1\in E(\fp_1)$ and
 $x_2\in E(\fp_2)$  we have
 \begin{equation}\label{tgeo}
-  1+d_{\fp_1}(\fcc(\fp_1), \fcc(\fp_2))\le
+    1+d_{\fp_1}(\fcc(\fp_1), \fcc(\fp_2))\le
     2^{5a}(1 + d_{B(x_1, D^{\ps(\fp_1)})}(\tQ(x_1),\tQ(x_2)))\, .
 \end{equation}
 \end{lemma}
@@ -3563,7 +3557,7 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 By the triangle inequality, we obtain from \eqref{dponetwo} and
 \eqref{tgeo0.5}
 \begin{equation}\label{tgeo1}
-     1+d_{\fp_1}(\fcc(\fp_1), \fcc(\fp_2))\le  2+2^{4a} +d_{\fp_1}(\tQ(x_1), \tQ(x_2))\, .
+        1+d_{\fp_1}(\fcc(\fp_1), \fcc(\fp_2))\le  2+2^{4a} +d_{\fp_1}(\tQ(x_1), \tQ(x_2))\, .
 \end{equation}
 As $x_1\in \sc(\fp_1)$ by Definition \eqref{defineep} of $E$, we have by the squeezing property  \eqref{eq vol sp cube}
 \begin{equation}
@@ -3590,9 +3584,9 @@ \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{equation}\label{tstargnot0}
-     T_{\fp} g^*(y)\neq 0
+        T_{\fp} g^*(y)\neq 0
 \end{equation}
-   implies
+    implies
 \begin{equation}\label{ynotfar}
     y\in  B(\pc(\fp),5D^{\ps(\fp)})\, .
 \end{equation}
@@ -3601,8 +3595,8 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 Fix $\fp$ and $y$ with \eqref{tstargnot0}.
 Then there exists $x\in E(\fp)$ with
 \begin{equation}
-   \overline{K_{\ps(\fp)}(x,y)}\overline{\tQ(x)(y)}
-    \tQ(x)(x)g(x) \neq 0\, .
+    \overline{K_{\ps(\fp)}(x,y)}e(-\tQ(x)(y)
+    +\tQ(x)(x))g(x) \neq 0\, .
 \end{equation}
 As $E(\fp)\subset \sc(\fp)$ and by the squeezing property
 \eqref{eq vol sp cube}, we have
@@ -3622,20 +3616,20 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 
 We expand the left-hand side of \eqref{eq basic TT* est} as
 \begin{equation}\label{tstartstar}
-\left|\int \int_{E(\fp_1)} \overline{K_{\ps(\fp_1)}(x_1,y)}\overline{\tQ(x_1)(y)}
-    \tQ(x_1)(x_1)g(x_1)\, d\mu(x_1) \right.
+\left|\int \int_{E(\fp_1)} \overline{K_{\ps(\fp_1)}(x_1,y)}e(-\tQ(x_1)(y)+
+    \tQ(x_1)(x_1))g(x_1)\, d\mu(x_1) \right.
 \end{equation}
 \begin{equation}
- \times \left.\int_{E(\fp_2)} {K_{\ps(\fp_2)}(x_2,y)}{\tQ(x_2)(y)}
-    \overline{\tQ(x_2)(x_2)}\overline{g(x_2)}\, d\mu(x_2)\, d\mu(y)\right|\, .
+    \times \left.\int_{E(\fp_2)} {K_{\ps(\fp_2)}(x_2,y)}e(\tQ(x_2)(y)
+    -\tQ(x_2)(x_2))\overline{g(x_2)}\, d\mu(x_2)\, d\mu(y)\right|\, .
 \end{equation}
 
 By Fubini and the triangle inequality and
-the fact $|\tQ(x_i)(x_i)|=1$ for $i=1,2$, we can estimate
+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)} |\int
-\overline{\tQ(x_1)(y)}{\tQ(x_2)(y)}\varphi_{x_1,x_2}(y)
+e(-\tQ(x_1)(y)+\tQ(x_2)(y))\varphi_{x_1,x_2}(y)
 \,dy|\,|g(x_1)g(x_2)|\, dx_1dx_2\,.
 \end{equation}
 We estimate for fixed $x_1\in E(\fp_1)$ and
@@ -3649,23 +3643,23 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 \end{equation}
 \begin{equation}
     |\int
-\overline{\tQ(x_1)(y)}{\tQ(x_2)(y)}\varphi_{x_1,x_2}(y)
+e(-\tQ(x_1)(y)+\tQ(x_2)(y))\varphi_{x_1,x_2}(y)
 \,dy|
 \end{equation}
 \begin{equation}
- \le     2^{4a} \mu(B') \|{\varphi}\|_{C^\tau(B')}
-       (1 + d_{B'}(\tQ(x_1),\tQ(x_2)))^{-\tau^2/(2+a)}
+    \le     2^{4a} \mu(B') \|{\varphi}\|_{C^\tau(B')}
+        (1 + d_{B'}(\tQ(x_1),\tQ(x_2)))^{-\tau^2/(2+a)}
 \end{equation}
 \begin{equation}
- \le     \frac{2^{154a^3}}
- {\mu(B(x_2, D^{\ps(\fp_2)}))}
-       (1 + d_{B'}(\tQ(x_1),\tQ(x_2)))^{-\tau^2/(2+a)}
+    \le     \frac{2^{154a^3}}
+    {\mu(B(x_2, D^{\ps(\fp_2)}))}
+        (1 + d_{B'}(\tQ(x_1),\tQ(x_2)))^{-\tau^2/(2+a)}
 \end{equation}
 Using Lemma \ref{lem tgeo} and $a\ge 1$ and $\tau \le 1$ estimates the last display by
 \begin{equation}\label{eqa2}
- \le     \frac{2^{159a^3}}
- {\mu(B(x_2, D^{\ps(\fp_2)}))}
-       (1+d_{\fp_1}(\fcc(\fp_1), \fcc(\fp_2)))^{-\tau^2/(2+a)}
+    \le     \frac{2^{159a^3}}
+    {\mu(B(x_2, D^{\ps(\fp_2)}))}
+        (1+d_{\fp_1}(\fcc(\fp_1), \fcc(\fp_2)))^{-\tau^2/(2+a)}
 \end{equation}
 As $x_2\in \sc(\fp_2)$ by Definition \eqref{defineep} of $E$, we have by \eqref{eq vol sp cube}
 \begin{equation}
@@ -3689,15 +3683,15 @@ \section{Proof of L. \ref{lem basic TT*}, the tile correlation bound }\label{sec
 \end{equation}
 By the triangle inequality and Lemma \ref{lem tstarsupport}, 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)}\, .
+    \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)}\, .
 \end{equation}
 By the squeezing property \eqref{eq vol sp cube} and the triangle inequality,
 we conclude
 \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{lem basic TT*}.
 
 
 
@@ -3726,16 +3720,16 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
     By the squeezing property \eqref{eq freq comp ball}
     and by assumption, we have
 \begin{equation}\label{ageo0}
-   B(\pc(\fp), \frac 14 D^{\ps(\fp)})\subset \sc(\fp)\subset \sc(\fp')
+    B(\pc(\fp), \frac 14 D^{\ps(\fp)})\subset \sc(\fp)\subset \sc(\fp')
     \subset B(\pc(\fp'),  4D^{\ps(\fp')})\, .
 \end{equation}
     Applying the doubling property \eqref{firstdb} four times, and the monotonicity in
     the set in Definition \eqref{definedE} gives with
     \eqref{eqassumedismfap}
 \begin{equation}
-     d_{\fp}(\fcc(\fp'),\mfa)
-     \le 2^{4a} d_{\fp'}(\fcc(\fp'),\mfa)'
-     \le  2^{4a+N} \, .
+        d_{\fp}(\fcc(\fp'),\mfa)
+        \le 2^{4a} d_{\fp'}(\fcc(\fp'),\mfa)'
+        \le  2^{4a+N} \, .
 \end{equation}
 Together with \eqref{eqassumedismfa} and the triangle inequality, we obtain
 \begin{equation}\label{eqdistqpqp}
@@ -3750,13 +3744,13 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 \end{equation}
 We have by \eqref{ageo0}
 \begin{equation}
- \pc(\fp)\in
+    \pc(\fp)\in
 B(\pc(\fp'),4D^{\ps(\fp')})\, .
 \end{equation}
 Hence by the triangle inequality
 \begin{equation}
- B(\pc(\fp), 4D^{\ps(\fp')})
- \subset
+    B(\pc(\fp), 4D^{\ps(\fp')})
+    \subset
 B(\pc(\fp'),8D^{\ps(\fp')})\, .
 \end{equation}
 Together with \eqref{ageo1} and   monotonicity of the Definition \eqref{definedE}
@@ -3846,12 +3840,12 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 
 \begin{lemma}\label{lem antichain-.5}
 Let $\mfa\in\Mf$ and  be
-an integer. Let $\fp_\mfa$ be a tile with $\mfa\in \fc(\fp_\mfa)$.
+an integer. Let $\fp_{\mfa}$ be a tile with $\mfa\in \fc(\fp_{\mfa})$.
 Then we have
 \begin{equation}\label{eqanti-0.5}
-    \sum_{\fp\in\mathfrak{A}_{\mfa,N}: \ps(\fp_\mfa)<\ps(\fp)}\mu(E(\fp)\cap G \cap \sc(\fp_\mfa))
-    \le  \mu (E_2(2^{4a+N+3},\fp_\mfa))
- \, .
+    \sum_{\fp\in\mathfrak{A}_{\mfa,N}: \ps(\fp_{\mfa})<\ps(\fp)}\mu(E(\fp)\cap G \cap \sc(\fp_{\mfa}))
+    \le  \mu (E_2(2^{4a+N+3},\fp_{\mfa}))
+    \, .
 \end{equation}
 
 
@@ -3861,36 +3855,36 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 \begin{proof}
 
 
-Let $\fp$ be any tile in $\mathfrak{A}_{\mfa,N}$ with $\ps(\fp_\mfa)<\ps(\fp)$. By definition of
+Let $\fp$ be any tile in $\mathfrak{A}_{\mfa,N}$ with $\ps(\fp_{\mfa})<\ps(\fp)$. By definition of
 $E$, the tile contributes zero to the sum on the left-hand side of \eqref{eqanti-0.5} unless
- $\sc(\fp)\cap \sc(\fp_{\mfa}) \neq \emptyset$, which we may assume. With $\ps(\fp_\mfa)<\ps(\fp)$
+    $\sc(\fp)\cap \sc(\fp_{\mfa}) \neq \emptyset$, which we may assume. With $\ps(\fp_{\mfa})<\ps(\fp)$
 and the dyadic property
 \eqref{dyadicproperty} we conclude $\sc(\fp_{\mfa})\subset  \sc(\fp)$.
 By the squeezing property
 \eqref{eq freq comp ball},
 we conclude from
-$\mfa\in \fc(\fp_\mfa)$
+$\mfa\in \fc(\fp_{\mfa})$
 that
 \begin{equation}
-    \mfa\in B(\fcc(\fp_\mfa), 1)\, .
+    \mfa\in B(\fcc(\fp_{\mfa}), 1)\, .
 \end{equation}
 We conclude from $\fp \in \mathfrak{A}_{\mfa,N}$ that
 \begin{equation}
     \mfa \in B(\fcc(\fp), 2^{N+1})\, .
 \end{equation}
 With Lemma \ref{lem a geo}, we conclude
- \begin{equation}
-     2^{4a+N+3}\fp_\mfa  \lesssim  2^{4a+N+3}\fp \, .
- \end{equation}
+    \begin{equation}
+        2^{4a+N+3}\fp_{\mfa}  \lesssim  2^{4a+N+3}\fp \, .
+    \end{equation}
 By the squeezing property \eqref{eq freq comp ball}
 and $a\ge 1$ and $N\ge 0$, we conclude
 \begin{equation}
-    \fcc(\fp)\subset B(2^{4a+N+1}, \fcc(\fp_\mfa)\, .
+    \fcc(\fp)\subset B(2^{4a+N+1}, \fcc(\fp_{\mfa})\, .
 \end{equation}
 By Definition \eqref{definee2} of $E_2$,
-  we conclude
+    we conclude
 \begin{equation}
-E(\fp)\cap G \subset E_2(2^{4a+N+3},\fp_\mfa)\, .
+E(\fp)\cap G \subset E_2(2^{4a+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}.
 This proves the lemma.
@@ -3902,33 +3896,33 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 \begin{equation}\label{eqanti00}
     \sum_{\fp\in\mathfrak{A}_{\mfa,N}}\mu(E(\fp)\cap G)
     \le
- 2^{101a^3+Na}\dens_1(\mathfrak{A})\mu\left(\cup_{\fp\in\mathfrak{A}}I_{\fp}\right)\, .
+    2^{101a^3+Na}\dens_1(\mathfrak{A})\mu\left(\cup_{\fp\in\mathfrak{A}}I_{\fp}\right)\, .
 \end{equation}
 \end{lemma}
 
 
 
 \begin{proof}
-   Fix $\mfa$ and $N$ and let
+    Fix $\mfa$ and $N$ and let
 $\mathfrak{A}'$ for the set of $\fp\in=\mathfrak{A}_{\mfa,N}$ such that $\sc(\fp)\cap G$ is not empty.
 
 
 
-   Let $\mathcal{L}$ be the collection dyadic cubes $I\in\mathcal{D}$ such that $I\subset \sc(\fp)$ for some $\fp\in\mathfrak{A}'$ and if $\sc(\fp)\subset I$ for some $\fp\in\mathfrak{A}'$, then $\ps(\fp)=-S$. By \eqref{coverdyadic}, for each $\fp \in \mathfrak{A}'$
-   and each $x\in \sc(\fp)\cap G$, there is a $I\in \mathcal{D}$ with $s(I)=-S$ and $x\in I$. By \eqref{dyadicproperty},
-   we have $I\subset \sc(\fp)$. Hence
-   \begin{equation}
-       \sc(\fp)\subset \bigcup\{I\in \mathcal{D}: s(I)=-S, I\subset \sc(\fp)\}\subset \bigcup \mathcal{L}\, .
-   \end{equation}
+    Let $\mathcal{L}$ be the collection dyadic cubes $I\in\mathcal{D}$ such that $I\subset \sc(\fp)$ for some $\fp\in\mathfrak{A}'$ and if $\sc(\fp)\subset I$ for some $\fp\in\mathfrak{A}'$, then $\ps(\fp)=-S$. By \eqref{coverdyadic}, for each $\fp \in \mathfrak{A}'$
+    and each $x\in \sc(\fp)\cap G$, there is a $I\in \mathcal{D}$ with $s(I)=-S$ and $x\in I$. By \eqref{dyadicproperty},
+    we have $I\subset \sc(\fp)$. Hence
+    \begin{equation}
+        \sc(\fp)\subset \bigcup\{I\in \mathcal{D}: s(I)=-S, I\subset \sc(\fp)\}\subset \bigcup \mathcal{L}\, .
+    \end{equation}
 As each $I\in \mathcal{L}$ satisfies $I\subset \sc(\fp)$ for some $\fp$ in $\mathfrak{A'}$, we conclude
-     \begin{equation}
+        \begin{equation}
 \bigcup\mathcal{L}=\bigcup_{\fp \in \mathfrak{A}'}\sc(\fp)\, .
-   \end{equation}
+    \end{equation}
 Let $\mathcal{L}^*$ be the set of maximal elements on $\mathcal{L}$ with respect to set inclusion.
 By \eqref{dyadicproperty}, the elements in $\mathcal{L}^*$ are pairwise disjoint and we have
- \begin{equation}\label{eqdecAprime}
+    \begin{equation}\label{eqdecAprime}
 \bigcup\mathcal{L}^*=\bigcup_{\fp \in \mathfrak{A}'}\sc(\fp)\, .
-   \end{equation}
+    \end{equation}
 Using the partition \eqref{eqdecAprime} into elements of $\mathcal{L}$ in \eqref{eqanti0}, it suffices to show for each $L\in \mathcal{L}^*$
 \begin{equation}\label{eqanti0}
     \sum_{\fp\in\mathfrak{A}'}\mu(E(\fp)\cap G \cap L)
@@ -3942,6 +3936,7 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 in $\mathfrak{A}$ with minimal $\ps(\fp')$. As $\sc(\fp')\not \subset L$ by definition of $L$, we have
 with \eqref{dyadicproperty} that $s(L)< \ps(\fp')$. In particular $s(L)<S$.
 
+\lars{Is the next sentence correct? I think parents do not have to exist when $L$ is not contained in $B(o, D^S)$}
 By \eqref{coverdyadic}, there is an
 $L'\in \mathcal{D}$ with $s(L')=s(L)+1$ and $c(L)\in L'$. By \eqref{dyadicproperty}, we have
 $L\subset L'$.
@@ -3952,7 +3947,7 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 \end{equation}
 \begin{equation}\label{eqanti2}
     +
-     \sum_{\fp\in\mathfrak{A}':\sc(\fp)\neq L'}\mu(E(\fp)\cap G\cap L)\, ,
+        \sum_{\fp\in\mathfrak{A}':\sc(\fp)\neq L'}\mu(E(\fp)\cap G\cap L)\, ,
 \end{equation}
 
 We first estimate \eqref{eqanti1}
@@ -3966,7 +3961,7 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 We turn to \eqref{eqanti2}.
 Consider the element $\fp'\in \mathfrak{A}'$ as above
 with $L\subset \sc(\fp')$ and $s(L)<s(\fp')$.
-As $L\subset L'$, we conclude with the dyadic property that $L'\subset \sc(\fp')$.
+As $L\subset L'$, we conclude twith the dyadic property that $L'\subset \sc(\fp')$.
 By maximality of $L$, we have
 $L'\not\in \mathcal{L}$.
 This together with the existence of the given $\fp'\in \mathfrak{A}$
@@ -3977,12 +3972,12 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 
 
 
-By the covering property \eqref{eq dis freq cover}, there exists a unique $\fp_\mfa$ with $\sc(\fp_\mfa)=L'$
-such that $\mfa\in \fc(\fp_\mfa)$.
+By the covering property \eqref{eq dis freq cover}, there exists a unique $\fp_{\mfa}$ with $\sc(\fp_{\mfa})=L'$
+such that $\mfa\in \fc(\fp_{\mfa})$.
 Note that
 that
 \begin{equation}
-    \mfa\in B(\fcc(\fp_\mfa), 1)
+    \mfa\in B(\fcc(\fp_{\mfa}), 1)
 \end{equation}
 and as  $\fp'' \in \mathfrak{A}_{\mfa,N}$ that
 \begin{equation}
@@ -3992,30 +3987,30 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 
 By Lemma \ref{lem a geo}, we conclude
 \begin{equation}
-     2^{4a+N+3}\fp''  \lesssim  2^{4a+N+3}\fp_\mfa \, .
- \end{equation}
+        2^{4a+N+3}\fp''  \lesssim  2^{4a+N+3}\fp_{\mfa} \, .
+    \end{equation}
 As $\fp''\in \mathfrak{A}'$, we have by Definition
 \eqref{definedens1} of $\dens_1$ that
 \begin{equation}\label{pmfadens}
-   {\mu(E_2(2^{4a+N+3}, \fp_\mfa)}\le 2^{4a^2+Na+3a}\dens_1(\mathfrak{A}) {\mu(L')}\, .
+    {\mu(E_2(2^{4a+N+3}, \fp_{\mfa})}\le 2^{4a^2+Na+3a}\dens_1(\mathfrak{A}) {\mu(L')}\, .
 \end{equation}
 Now let $\fp$ be any tile in the summation set in \eqref{eqanti2}, that is, $\fp\in \mathfrak{A}'$ and $\sc(\fp)\neq L'$.
 Then $\sc(\fp)\cap L\neq \emptyset$. It follows by the dyadic property \eqref{dyadicproperty}
 and the definition of $L$ that
 $L\subset \sc(\fp)$ and $L\neq \sc(\fp)$. By the dyadic property \eqref{dyadicproperty}, we have
 $s(L)<\ps(\fp)$ and thus $s(L')\le \ps(\fp)$. By the dyadic property
-   \eqref{dyadicproperty} again, we have $L'\subset \sc(\fp)$.
+    \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 estimate \eqref{eqanti2} by
+    By Lemma \ref{lem antichain-.5}, we estimate \eqref{eqanti2} by
 
-  \begin{equation}\label{eqanti0.5}
+    \begin{equation}\label{eqanti0.5}
     \le \sum_{\fp\in\mathfrak{A}':\sc(\fp)=\neq L'}\mu(E(\fp)\cap G\cap L')
-    \le  \mu (E_2(2^{4a+N+3},\fp_\mfa))
- \, .
+    \le  \mu (E_2(2^{4a+N+3},\fp_{\mfa}))
+    \, .
 \end{equation}
 Using the decomposition
 into \eqref{eqanti1} and
@@ -4030,7 +4025,7 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 
 
 
-  Using $s(L')=s(L)+1$ and $D=2^{100a^2}$ and the
+    Using $s(L')=s(L)+1$ and $D=2^{100a^2}$ and the
 squeezing property \eqref{eq vol sp cube}
 and the doubling property \eqref{doublingx} $100a^2+4$ times , we obtain
 \begin{equation}
@@ -4040,13 +4035,6 @@ \section{Proof of Lemma \ref{lem antichain 1}, the geometric estimate}
 
 Inserting in \eqref{eqanti3.14} and using $a>4$ gives \eqref{eqanti0}.
 This completes the proof of the lemma.
-
-
-
-
-
-
-
 \end{proof}
 
 
@@ -4067,19 +4055,19 @@ \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\tau^2/(2+a)}1_{E(\fp)} 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 $x\in E(\fp)$,
- we have
- \begin{equation}
-     \left\|\sum_{\fp\in \mathfrak{A}_{\mfa,N}} 2^{-N\tau^2/(2+a)}1_{E(\fp)} 1_G\right\|_{p}^p
- \end{equation}
+    Using that for each $x\in X$  by Lemma \ref{lem antichain 0} there is at most one $\fp\in \mathfrak{A}$ with $x\in E(\fp)$,
+    we have
+    \begin{equation}
+        \left\|\sum_{\fp\in \mathfrak{A}_{\mfa,N}} 2^{-N\tau^2/(2+a)}1_{E(\fp)} 1_G\right\|_{p}^p
+    \end{equation}
 \begin{equation}
     =\int_G(\sum_{\fp\in\mathfrak{A}}2^{-N\tau^2/(2+a)}\mathbf{1}_{E(\fp)}(x))^p\, d\mu(x)
 \end{equation}
 \begin{equation}
-  =   \int _G\sum_{\fp\in\mathfrak{A}}2^{-(p-1)N\tau^2/(2+a)}\mathbf{1}_{E(\fp)}(x)\, d\mu(x)
+    =   \int _G\sum_{\fp\in\mathfrak{A}}2^{-(p-1)N\tau^2/(2+a)}\mathbf{1}_{E(\fp)}(x)\, d\mu(x)
 \end{equation}
 \begin{equation}
-  =   2^{-(p-1)N\tau^2/(2+a)} \sum_{\fp\in\mathfrak{A}_{\mfa,N}}\mu(E(\fp)\cap G)
+    =   2^{-(p-1)N\tau^2/(2+a)} \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
@@ -4118,73 +4106,62 @@ \chapter{Proof of Proposition \ref{forestprop}}
 
 \label{treesection}
 
-\section{Tree Estimate}
-%%%%%% Previous discussion about non-tangential Cotlar maximal inequality
-% Given a measurable function $\sigma$ mapping $X$ to the set of finite convex subsets of $\mathbb{Z}^d$, we define an associated truncated singular maximal operator
-% \begin{equation}
-%     \label{def Tsigma}
-%     T_\sigma f(x):=\sum_{s\in\sigma(x)}\int K_s(x,y) f(y)\,dy.
-% \end{equation}
-% From the proof of Cotlar's inequality, it can be seen that the non-tangentially maximally truncated operator
-% \begin{equation}
-%     \label{def non-tang max op}
-%     T_{\mathcal{N}}f(x):=\sup_{\rho(x,x')<CD^{\min\, \sigma(x)}}|T_\sigma f(x')|
-% \end{equation}
-% is bounded on $L^p(X)$ for $p\in (1,\infty).$
-% In practice for us, $\sigma$ will be associated to the sets of scales given by trees $\mathfrak{T}$ (as in \cite{zk-polynomial}).
-
-\lars{The following things are already proven in the previous section. To make sections independent I would have to prove them again, but I don't like that because it might make some Lean guy waste an hour proving things again that are already there.}
-\lars{Put Lemma 6.7 in "auxilliary Lemmas" section}
+\section{The pointwise tree estimate}
+
+\lars{This Lemma is copied from section 6, we should put it in a global auxiliary Lemma section}
+
+\begin{lemma}
+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 t}
+    \frac{1}{4} D^{s-1} \leq \rho(x,y) \leq \frac{1}{2} D^s\, .
+\end{equation}
 We have
 \begin{equation}
-    \label{eq Ks supp t}
-    K_s(x,y)=0
+    \label{eq Ks size t}
+    |K_s(x,y)|\le \frac{2^{102 a^3}}{\mu(B(x, D^{s}))}\,
 \end{equation}
-unless $\frac{1}{4} D^{s-1} \leq \rho(x,y) \leq \frac{1}{2} D^s$; and by \eqref{eqkernel size} and \eqref{eqkernel y smooth}
-\begin{equation}
-   \label{eq Ks size t}
-    |K_s(x,y)|\lesssim \frac{1}{\mu(B(x, D^{s}))}\,;
+and \begin{equation}
+    \label{eq Ks smooth t}
+    |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)^{\tau}\,.
 \end{equation}
+\end{lemma}
+
+\begin{proof}
+By Definition \eqref{defks}, the function $K_s$ is the product of
+$K$ with a function which is supported in the set of all
+$x,y$ satisfying \eqref{supp Ks t}. This proves \eqref{supp Ks t}.
+
+Using \eqref{eqkernel size} and the lower bound in \eqref{supp Ks t}
+we obtain
 \begin{equation}
-    \label{eq Ks smooth t}
-    |K_s(x,y)-K_s(x, y')|\lesssim \left(\frac{\rho(y,y')}{D^s}\right)^{\tau}\frac{1}{\mu(B(x, D^{s}))}\,.
+|K_s(x,y)|\le \frac{2^{a^3}}{\mu(B(x,\frac 14 D^{s-1}))}
 \end{equation}
-Given $s_{\min}, s_{\max} \in \mathbb{Z}$, we define
+Using $D=2^{100a^2}$
+and the doubling property \eqref{doublingx} $2 +100a^2$ times estimates
+the last display by
 \begin{equation}
-    \label{eq TN def}
-    T_{\mathcal{N},s_{\min}, s_{\max}} f(x) := \sup_{s_{\min} \le s_1 < s_2 \le s_{\max}} \sup_{\rho(x,x') \leq 8 D D^{s_1}} \left| \sum_{s = s_1}^{s_2}  \int K_s(x',y) f(y)  \, \mathrm{d}\mu(y) \right|\,.
+\label{eq Ks aux}
+\le \frac{2^{2a+101a^3}}{\mu(B(x,  D^{s}))}\, .
 \end{equation}
-\begin{lemma}
-    \label{lem nontangential}
-    For all $s_{\min}, s_{\max}$, it holds that
-    $$
-        \|T_{\mathcal{N},s_{\min}, s_{\max}} f\|_2 \le ...\|f\|_2\,.
-    $$
-\end{lemma}
+Using $a\ge 4$ proves \eqref{eq Ks size t}.
 
-\lars{I dont know how to prove this from our current assumptions. I think it is easier to check the stronger assumption for the Hilbert transform. (At least I know how to do that in one page)}
 
-\begin{proof}
-    Fix $f$, a point $x$ and scales $s_1 < s_2$. We split
-    $$
-        f = f\mathbf{1}_{B(x, (8D + 1)D^{s_1})} + f\mathbf{1}_{X \setminus B(x, (8D + 1)D^{s_1})} =: f_1 + f_2\,.
-    $$
-    By \eqref{eq Ks supp t} and the triangle inequality, we have for all $x' \in B(x, 8D \cdot D^{s_1})$ and all $s \le s_1$ that
-    \begin{equation}
-        \label{eq outside}
-        \int K_s(x', y) f_2(y) \, \mathrm{d}\mu(y) = 0\,.
-    \end{equation}
-    Also by \eqref{eq Ks supp t}, the triangle inequality and the fact that $D \ge 100$, we have for all $x' \in B(x, 8D \cdot D^{s_1})$ and all $s \ge s_1 + 3$
-    \begin{equation}
-        \label{eq outside}
-        \int K_s(x', y) f_1(y) \, \mathrm{d}\mu(y) = 0\,.
-    \end{equation}
-    Thus, we have for all $x' \in B(x, 8D \cdot D^{s_1})$
-    \begin{equation}
-        \sum_{s = s_1}^{s_{\max}} \int K_s(x',y) f(y) \, \mathrm{d}\mu(y) = \sum_{s = s_{\min}}^{s_{\max}} \int K_s(x',y) f_2(y) \, \mathrm{d}\mu(y) + \sum_{s = s_1}^{s_1 + 2} \int K_s(x', y) f_1(y) \, \mathrm{d}\mu(y)
-    \end{equation}
-    ... first term is bounded by assumption, up to applications of Hardy Littlewood at the boundary (maybe write those as separate Lemma) and up to the fact that the aperture is wrong.
-    ... second term is bounded by the Hardy-Littlewood maximal function.
+Similarly, we obtain with  \eqref{eqkernel y smooth} and the lower bound in
+\eqref{supp Ks t}
+\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)^{\tau}\,.
+\end{equation}
+Using $\tau\le 1$, 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)^{\tau}
+        = \frac{2^{2+2a+100a^2+101a^3}}{\mu(B(x,  D^{s}))}
+    \left(\frac{ \rho(y,y')}{D^{s}}\right)^{\tau}\,.
+\end{equation}
+Using $a\ge 4$, this proves  \eqref{eq Ks smooth t}.
 \end{proof}
 
 Fix a forest $(\fU, \fT)$.
@@ -4210,58 +4187,81 @@ \section{Tree Estimate}
     Let $s \in \mathbb{Z}$ with $\underline{\sigma} (\fT(\fu), x) \le s \le \overline{\sigma} (\fT(\fu), x)$. There exists $\fp \in \fT(\fu)$ with $s(\fp) = \underline{\sigma} (\fT(\fu), x)$ and $x \in E(\fp)$, and there exists $\fp'' \in \fT(\fu)$ with $s(\fp'') = \overline{\sigma} (\fT(\fu), x)$ and $x \in E(\fp'') \subset \sc(\fp'')$. By \eqref{coverdyadic}, there exists a cube $I \in \mathcal{D}$ of scale $s$ with $x \in I$. By \eqref{eq dis freq cover}, there exists a tile $\fp' \in \fP(I)$ with $Q(x) \in \fc(\fp')$. By \eqref{dyadicproperty}  we have $\sc(\fp) \subset \sc(\fp') \subset \sc(\fp'')$, and by \eqref{eq freq dyadic}, we have $\fc(\fp'') \subset \fc(\fp') \subset \fc(\fp)$. Thus $\fp \le \fp' \le\fp''$, which gives with \eqref{forest2} that $\fp' \in \fT(\fu)$, so $s \in \sigma(\fT(\fu), x)$.
 \end{proof}
 
+Given a point $x \in B(o, D^S)$ and a scale $-S \le s\le S$, we denote by $I_s(x) \in \mathcal{D}$ a cube of scale $s$ containing $x$. This cube exists by \eqref{coverball} and is unique by \eqref{coverdyadic}. We define
+\begin{equation}
+    \label{eq TN def}
+    T_{\mathcal{N}} f(x) := \sup_{-S \le s_1 < s_2 \le S} \sup_{x' \in I_{s_1}(x)} \left| \sum_{s = s_1}^{s_2}  \int K_s(x',y) f(y)  \, \mathrm{d}\mu(y) \right|\,.
+\end{equation}
+
 For a nonempty collection of tiles $\mathfrak{S} \subset \fP$ we define
 $$
-    \mathcal{J}(\mathfrak{S})
+    \mathcal{J}_0(\mathfrak{S})
 $$
-to be the collection of maximal dyadic cubes $J$ such that
+to be the collection of all dyadic cubes $J \in \mathcal{D}$ such that $s(J) = -S$ or
 $$
-    \sc(\fp) \not\subset B(x(J), 100D^{s(J) + 1})
+    \sc(\fp) \not\subset B(c(J), 100D^{s(J) + 1})
 $$
-for all $\fp \in \mathfrak{S}$.
+for all $\fp \in \mathfrak{S}$. We define $\mathcal{J}(\mathfrak{S})$ to be the collection of inclusion maximal cubes in $\mathcal{J}_0(\mathfrak{S})$.
 
-\lars{There is an annoying problem here: If some $\fp \in \mathfrak{S}$ has scale $-S$, then the cubes in $\mathcal{J}(\mathfrak{S})$ contained in $\sc(\fp)$ would have scale $-S - 2$, but there are no cubes of that scale in our construction. So we have to leave some more room.}
 We further define
 $$
-    \fL(\mathfrak{S})
+    \mathcal{L}_0(\mathfrak{S})
 $$
-to be the collection of maximal dyadic cubes $L$ such that $L \subsetneq \sc(\fp)$ for some $\fp \in \mathfrak{S}$ and $\sc(\fp) \not\subset L$ for all $\fp \in \mathfrak{S}$. \lars{The same applies here. How was this solved in section 6?}
+to be the collection of dyadic cubes $L \in \mathcal{D}$ such that there exists $\fp \in \mathfrak{S}$ with $L \subset \sc(\fp)$, and there exists no $\fp \in \mathfrak{S}$ with $\sc(\fp) \subset L$ or $s(L) = -S$. 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}
+    For each $\mathfrak{S} \subset \fP$, we have
+    \begin{equation}
+        \label{eq J partition}
+        \bigcup_{I \in \mathcal{D}} I = \dot{\bigcup_{J \in \mathcal{J}(\mathfrak{S})}} J
+    \end{equation}
+    and
+    \begin{equation}
+        \label{eq L partition}
+        \bigcup_{\fp\in \mathfrak{S}} \sc(\fp) = \dot{\bigcup_{L \in \mathcal{L}(\mathfrak{S})}} L\,.
+    \end{equation}
+\end{lemma}
+
+\begin{proof}
+    Since $\mathcal{J}(\mathfrak{S})$ is the set of inclusion maximal cubes in $\mathcal{J}_0(\mathfrak{S})$, cubes in $\mathcal{J}(\mathfrak{S})$ are pairwise disjoint by \eqref{dyadicproperty}. The same applies to $\mathcal{L}(\mathfrak{S})$.
 
+    If $x \in \bigcup_{I \in \mathcal{D}} I$, then there exists by \eqref{coverdyadic} a cube $I \in \mathcal{D}$ with $x \in I$ and $s(I) = -S$. Then $I \in \mathcal{J}_0(\mathfrak{S})$. There exists an inclusion maximal cube in $\mathcal{J}_0(\mathfrak{S})$ containing $I$. This cube contains $x$ and is contained in $\mathcal{J}(\mathfrak{S})$. This shows one inclusion in \eqref{eq J partition}, the other one follows from $\mathcal{J}(\mathfrak{S}) \subset \mathcal{D}$.
 
-%\begin{remark}
-%    The scales of adjacent cubes in $\mathcal{J}(\mathfrak{S})$ can differ at most by one. More precisely, if $J,J'\in \mathcal{J}$ with $\text{dist }(J,J'):=\inf_{x\in J,x'\in J'} \rho(x,x')\leq 10 \max(D^{s(J)}, D^{s(J')})$, then $|s(J)-s(J')|\leq 1$. This is because if $J,J'\in\mathcal{J}$ with $s(J)\leq s(J')-2$ and $\text{dist }(J,J')\leq 10 D^{s(J')}$, then $B(x_{\hat{J}}, 100D^{s(\hat{J}+1})\subseteq B(x_{J'}, 100D^{s(J')+1})$ and hence does not contain $I_{\fp}$ for any $\fp\in\fP$. This contradicts the maximality of $J$.
-%\end{remark}
-For a collection of pairwise disjoint cubes $\mathcal{C}$, we also define the projection operator
+    The proof of the two inclusions in \eqref{eq L partition} is similar.
+\end{proof}
+
+For a finite collection of pairwise disjoint cubes $\mathcal{C}$, we define the projection operator
 $$
     P_{\mathcal{C}}f(x) :=\sum_{J\in\mathcal{C}}\mathbf{1}_J(x) \frac{1}{\mu(J)}\int_J f(y) \, \mathrm{d}\mu(y)\,.
 $$
 Define for each $\fu \in \fU$ the auxiliary operator
 \begin{equation}
     \label{eq def S op}
-    S_{\fu}f(x):=\sum_{I\in\mathcal{D}} \mathbf{1}_{I}(x) \sum_{\substack{J\in \mathcal{J}(\fT(\fu))\\
-    J\subset B(c(I), D^{s(I)})}} \frac{D^{(s(J) - s(I))\tau}}{\mu(B(c(I), 8D^{s(I)}))}\int_J |f(y)| \, \mathrm{d}\mu(y)\,.
+    S_{1,\fu}f(x):=\sum_{I\in\mathcal{D}} \mathbf{1}_{I}(x) \sum_{\substack{J\in \mathcal{J}(\fT(\fu))\\
+    J\subset B(c(I), 16 D^{s(I)})}} \frac{D^{(s(J) - s(I))\tau}}{\mu(B(c(I), 16D^{s(I)}))}\int_J |f(y)| \, \mathrm{d}\mu(y)\,.
 \end{equation}
 Let
 $$
-    \mathcal{B} = \{B(c(I), 8D^{s(I)}) \ : \ I \in \mathcal{D}\}\,.
+    \mathcal{B} = \{B(c(I), 16D^{s(I)}) \ : \ I \in \mathcal{D}\}\,.
 $$
 
-\begin{lemma}[Tree estimate]
-    \label{TreeEstimate}
+\begin{lemma}
+    \label{lem pointw tree estimate}
     Let $\fu \in \fU$ and $L \in \mathcal{L}(\fT(\fu))$. Let $x, x' \in L$.
-    Then we have
+    Then we have \lars{conditions $f$}
     $$
-        |\sum_{\fp \in \fT(\fu)} T_{\fp}[ e(-Q(\fu))f](x)|
+        \left|\sum_{\fp \in \fT(\fu)} T_{\fp}[ e(-Q(\fu))f](x)\right|
     $$
     \begin{equation}
         \label{eq LJ ptwise}
-        \leq 2^{157a^3}(M_{\mathcal{B},1}+S)P_{\mathcal{J}(\fT(\fu))}|f|(x')+|T_{\mathcal{N}}P_{\mathcal{J}(\fT(\fu))}f(x')|,
+        \leq 2^{151a^3}(M_{\mathcal{B},1}+S_{1,\fu})P_{\mathcal{J}(\fT(\fu))}|f|(x')+|T_{\mathcal{N}}P_{\mathcal{J}(\fT(\fu))}f(x')|,
     \end{equation}
 \end{lemma}
 
 
 \begin{proof}
-    By \eqref{definetp}, if $T_{\fp} h(x) \ne 0$, then $x \in E(\fp)$. Combining this with $|e(Q(\fu)(x)-Q(x)(x))| = 1$, we obtain
+    By \eqref{definetp}, if $T_{\fp}[ e(-Q(\fu))f](x) \ne 0$, then $x \in E(\fp)$. Combining this with $|e(Q(\fu)(x)-Q(x)(x))| = 1$, we obtain
     $$
         |\sum_{\fp \in \fT(\fu)} T_{\fp}[ e(Q(\fu))f](x)|
     $$
@@ -4283,24 +4283,29 @@ \section{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}.
+\end{proof}
 
-    We first control the term \eqref{eq term B}.
-    By the definition of $\mathcal{L}(\fT(\fu))$, we have that $s(\fp) > s(L)$ for all $\fp \in \fT(\fu)$ with $L \cap \sc(\fp) \ne \emptyset$. In particular, we have that $\underline{\sigma}(\fu, x) > s(L)$. By \eqref{eq vol sp cube}, we have $\rho(x,x') \le 8D^{s(L)} \le 8D^{\underline{\sigma}(\fu, x) - 1}$, and by Lemma \ref{lem sigma convex} the set $\sigma(\fu, x)$ is an interval. Hence, \lars{write correct name of nontangential operator}
+\begin{lemma}
+    \label{lem term A}
+    For all $\fu \in \fU$, all $L \in \mathcal{L}(\fT(\fu))$, all $x, x' \in L$ and all $f$ \lars{conditions}, we have
     $$
-        \eqref{eq term B} \le T_{\mathcal{N}, } P_{\mathcal{J}(\fT(\fu))} f(x')\,.
+        \eqref{eq term A} \le 10 \cdot 2^{105a^3} M_{\mathcal{B}, 1}P_{\mathcal{J}(\fT(\fu))}|f|(x')\,.
     $$
+\end{lemma}
 
-    Next, we control \eqref{eq term A}. If $K_s(x,y)\neq 0$, then by \eqref{eq Ks supp t} $\rho(x,y)\leq 1/2 D^s$. By $1$-Lipschitz continuity of the function $t \mapsto \exp(it) = e(t)$, it follows that
+\begin{proof}
+    By \eqref{supp Ks t}, if $K_s(x,y)\neq 0$, then $\rho(x,y)\leq 1/2 D^s$. By $1$-Lipschitz continuity of the function $t \mapsto \exp(it) = e(t)$, it follows that
     \begin{multline*}
         |e(-Q(\fu)(y)+Q(x)(y)+Q(\fu)(x)-Q(x)(x))-1|\\
         \leq d_{B(x, 1/2 D^{s})}(Q(\fu), Q(x))\,.
     \end{multline*}
     Let $\fp_s \in \fT(\fu)$ be a tile with $s(\fp_s) = s$ and $x \in E(\fp_s)$, and let $\fp'$ be a tile with $s(\fp') = \overline{\sigma}(\fu, x)$ and $x \in E(\fp')$.
-    Using \eqref{firstdb}, \eqref{eq vol sp cube} and Lemma \ref{lem cube monotone}, we obtain
+    Using \eqref{firstdb}, \eqref{eq vol sp cube} and Lemma \ref{lem cube monotone}, we can bound the previous display by
     $$
-        \le 2^a d_{\fp_s}(Q(\fu), Q(x)) \le 2^{a} 2^{s - \overline{\sigma}(\fu, x)} d_{\fp'}(Q(\fu), Q(x))\,.
+        2^a d_{\fp_s}(Q(\fu), Q(x)) \le 2^{a} 2^{s - \overline{\sigma}(\fu, x)} d_{\fp'}(Q(\fu), Q(x))\,.
     $$
-    Since $Q(\fu) \in B_{\fp'}(Q(\fp'), 4)$ by \eqref{forest1} and $Q(x) \in \Omega(\fp') \subset B_{\fp'}(Q(\fp'), 1)$ by \eqref{eq freq comp ball}, we can estimate this by
+    Since $Q(\fu) \in B_{\fp'}(Q(\fp'), 4)$ by \eqref{forest1} and $Q(x) \in \Omega(\fp') \subset B_{\fp'}(Q(\fp'), 1)$ by \eqref{eq freq comp ball}, this is estimated by
     $$
         \le 5 \cdot 2^{a} 2^{s - \overline{\sigma}(\fu, x)} \,.
     $$
@@ -4308,33 +4313,59 @@ \section{Tree Estimate}
     $$
         \eqref{eq term A} \le 5\cdot 2^{103a^3} \sum_{s\in\sigma(x)}2^{s - \overline{\sigma}(\fu, x)} \frac{1}{\mu(B(x,D^s))}\int_{B(x,0.5D^{s})}|f(y)|\,\mathrm{d}\mu(y)\,.
     $$
-    Since the collection $\mathcal{J}$ is a partition of $X$, we can estimate this by
+    By \eqref{eq J partition}, the collection $\mathcal{J}$ is a partition of $X$, so this is estimated by
     $$
-         5\cdot 2^{103a^3} \sum_{s\in\sigma(x)}2^{s - \overline{\sigma}(\fu, x)} \frac{1}{\mu(B(x,D^s))}\sum_{\substack{J \in \mathcal{J}(\fT(\fu))\\J \cap B(x, 0.5D^s) \ne \emptyset} }\int_{J}|f(y)|\,\mathrm{d}\mu(y)\,.
+            5\cdot 2^{103a^3} \sum_{s\in\sigma(x)}2^{s - \overline{\sigma}(\fu, x)} \frac{1}{\mu(B(x,D^s))}\sum_{\substack{J \in \mathcal{J}(\fT(\fu))\\J \cap B(x, 0.5D^s) \ne \emptyset} }\int_{J}|f(y)|\,\mathrm{d}\mu(y)\,.
     $$
-    This expression does not change if we replace $|f|$ by $P_{\mathcal{J}(\fT(\fu))}|f|$. Further, if $J \in \mathcal{J}(\fT(\fu))$ with $B(x, 0.5 D^s) \cap J \ne \emptyset$, then $B(c(\fp_s), 4.5D^s) \cap J \ne \emptyset$ by the triangle inequality. If $s(J) \ge s - 1$, then it follows from the triangle inequality, \eqref{eq vol sp cube} and \eqref{defineD} that $\sc(\fp) \subset B(c(J), 100 D^{s(J)+1})$, contradicting $J \in \mathcal{J}(\mathfrak{T}(\fu))$. Thus $s(J) < s - 1$. By the triangle inequality and \eqref{eq vol sp cube}, we conclude that $J \subset B(x, D^s)$. Thus we can continue our chain of estimates with
+    This expression does not change if we replace $|f|$ by $P_{\mathcal{J}(\fT(\fu))}|f|$.
+
+    If $J \in \mathcal{J}(\fT(\fu))$ with $B(x, 0.5 D^s) \cap J \ne \emptyset$, then $B(c(\fp_s), 4.5D^s) \cap J \ne \emptyset$ by the triangle inequality. If $s(J) \ge s$ and $s(J) > -S$, then it follows from the triangle inequality, \eqref{eq vol sp cube} and \eqref{defineD} that $\sc(\fp_s) \subset B(c(J), 100 D^{s(J)+1})$, contradicting $J \in \mathcal{J}(\mathfrak{T}(\fu))$. Thus $s(J) \le s - 1$ or $s(J) = -S$. If $s(J) = -S$ and $s(J) > s - 1$, then $s = -S$. Thus always $s(J) \le s$. It then follows from the triangle inequality and \eqref{eq vol sp cube} that $J \subset B(c(\fp_s), 16 D^s)$.
+
+    Thus we can continue our chain of estimates with
     $$
-        5\cdot 2^{103a^3} \sum_{s\in\sigma(x)}2^{s - \overline{\sigma}(\fu, x)} \frac{1}{\mu(B(x,D^s))}\int_{B(x,D^s)}|P_{\mathcal{J}(\fT(\fu))}f(y)|\,\mathrm{d}\mu(y)\,.
+        5\cdot 2^{103a^3} \sum_{s\in\sigma(x)}2^{s - \overline{\sigma}(\fu, x)} \frac{1}{\mu(B(x,D^s))}\int_{B(c(\fp_s),16 D^s)}P_{\mathcal{J}(\fT(\fu))}|f(y)|\,\mathrm{d}\mu(y)\,.
     $$
-    We have $B(x, D^s) \subset B(\sc(\fp_s), 8D^s)$ and $B(\sc(\fp_s), 8D^s)) \subset B(x, 16D^s)$, both by \eqref{eq vol sp cube} since $x \in \sc(\fp)$. Combining this with the doubling property \eqref{doublingx}, we obtain
+    We have $B(c(\fp_s), 16D^s)) \subset B(x, 32D^s)$, by \eqref{eq vol sp cube} and the triangle inequality, since $x \in \sc(\fp)$. Combining this with the doubling property \eqref{doublingx}, we obtain
     $$
-        \mu(B(c(\fp_s), 8D^s)) \le 2^{4a} \mu(B(x, D^s))\,.
+        \mu(B(c(\fp_s), 16D^s)) \le 2^{5a} \mu(B(x, D^s))\,.
     $$
-    Hence \eqref{eq term A} is bounded by
+    Since $a \ge 4$, it follows that \eqref{eq term A} is bounded by
     $$
-        5\cdot 2^{107a^3} \sum_{s\in\sigma(x)}2^{s - \overline{\sigma}(\fu, x)} \frac{1}{\mu(B(c(\fp_s),8D^s))}\int_{B(c(\fp_s),8D^s)}|P_{\mathcal{J}(\fT(\fu))}f(y)|\,\mathrm{d}\mu(y)\,.
+        2^{104a^3} \sum_{s\in\sigma(x)}2^{s - \overline{\sigma}(\fu, x)} \frac{1}{\mu(B(c(\fp_s),16D^s))}\int_{B(c(\fp_s),16D^s)}P_{\mathcal{J}(\fT(\fu))}|f(y)|\,\mathrm{d}\mu(y)\,.
     $$
-    Since $\rho(x,x') \le 8 D^{s - 1}$, as shown above, we have $x' \in B(x, D^s) \subset B(\sc(\fp_s), 8D^s)$. Thus
+    Since $L \in \mathcal{L}(\fT(\fu))$, we have $s(L) \le s(\fp)$ for all $\fp \in \fT(\fu)$. Since $x\in L \cap \sc(\fp_s)$, it follows by \eqref{dyadicproperty} that $L \subset \sc(\fp_s)$, in particular $x' \in \sc(\fp_s) \subset B(c(\fp_s), 16D^s)$. Thus
     $$
-        \le 5\cdot 2^{107a^3} \sum_{s\in\sigma(x)}2^{s - \overline{\sigma}(\fu, x)} M_{\mathcal{B}, 1}|P_{\mathcal{J}(\fT(\fu))}f|(x')
+        \le 2^{104a^3} \sum_{s\in\sigma(x)}2^{s - \overline{\sigma}(\fu, x)} M_{\mathcal{B}, 1}P_{\mathcal{J}(\fT(\fu))}|f|(x')
     $$
     $$
-        \le 10 \cdot 2^{107a^3} M_{\mathcal{B}, 1}|P_{\mathcal{J}(\fT(\fu))}f|(x')\,.
+        \le 2^{105a^3} M_{\mathcal{B}, 1}P_{\mathcal{J}(\fT(\fu))}|f|(x')\,.
     $$
     This completes the estimate for term \eqref{eq term A}.
+\end{proof}
+
+\begin{lemma}
+    \label{lem term B}
+    For all $\fu \in \fU$, all $L \in \mathcal{L}(\fT(\fu))$, all $x, x' \in L$ and all $f$ \lars{conditions}, we have
+    $$
+        \eqref{eq term B} \le T_{\mathcal{N}} P_{\mathcal{J}(\fT(\fu))} f(x')\,.
+    $$
+\end{lemma}
+
+\begin{proof}
+    Let $s = \underline{\sigma}(\fu, x)$. By definition, there exists a tile $\fp \in \fT(\fu)$ with $s(\fp) = s$ and $x \in E(\fp)$. Then $x \in \sc(\fp) \cap L$. By \eqref{dyadicproperty} and the definition of $\mathcal{L}(\fT(\fu))$, it follows that $L \subset \sc(\fp)$, in particular $x' \in \sc(\fp)$, so $x = I_s(x')$.
+    The lemma now follows from the definition of $T_{\mathcal{N}}$.
+\end{proof}
 
+\begin{lemma}
+    \label{lem term C}
+    For all $\fu \in \fU$, all $L \in \mathcal{L}(\fT(\fu))$, all $x, x' \in L$ and all $f$ \lars{conditions}, we have
+    $$
+        \eqref{eq term C} \le 2^{151a^3} S_{1,\fu} P_{\mathcal{J}(\fT(\fu))}|f|(x')\,.
+    $$
+\end{lemma}
 
-    Finally, we estimate \eqref{eq term C}. We have for $J \in \mathcal{J}(\fT(\fu))$:
+\begin{proof}
+    We have for $J \in \mathcal{J}(\fT(\fu))$:
     $$
         \int_J K_{s}(x,y)(1 - P_{\mathcal{J}(\fT(\fu))})f(y) \, \mathrm{d}\mu(y)
     $$
@@ -4346,93 +4377,2899 @@ \section{Tree Estimate}
     $$
         |K_s(x,y) - K_s(x,z)| \le \frac{2^{150a^3}}{\mu(B(x, D^s))} \left(\frac{8 D^{s(J)}}{D^s}\right)^\tau\,.
     $$
-    Using this, \eqref{eq canc comp}, and the fact, shown in the estimate for term \eqref{eq term A}, that $J \subset B(x, D^s)$ if it intersects the support of $K_s(x,y)$, we estimate \eqref{eq term C} by
+    Suppose that $s \in \sigma(\fu, x)$.
+    If $K_s(x,y) \ne 0$ for some $y \in J \in \mathcal{J}(\fT(\fu))$ then, by \eqref{supp Ks t}, $y \in B(x, 0.5 D^s) \cap J \ne \emptyset$. Let $\fp \in \fT(\fu)$ with $s(\fp) = s$ and $x \in E(\fp)$. Then $B(c(\fp_s), 4.5D^s) \cap J \ne \emptyset$ by the triangle inequality. If $s(J) \ge s$ and $s(J) > -S$, then it follows from the triangle inequality, \eqref{eq vol sp cube} and \eqref{defineD} that $\sc(\fp) \subset B(c(J), 100 D^{s(J)+1})$, contradicting $J \in \mathcal{J}(\mathfrak{T}(\fu))$. Thus $s(J) \le s - 1$ or $s(J) = -S$. If $s(J) = -S$ and $s(J) > s - 1$, then $s = -S$. So in both cases, $s(J) \le s$. It then follows from the triangle inequality and \eqref{eq vol sp cube} that $J \subset B(x, 16 D^s)$.
+
+    Thus, we can estimate \eqref{eq term C} by
     $$
-        2^{150a^3 + 3\tau}\sum_{\fp\in \mathfrak{T}}\frac{\mathbf{1}_{E(\fp)}}{\mu(B(x,D^{s(\fp)}))}(x)\sum_{\substack{J\in \mathcal{J}\\J\subset B(x,  D^{s(\fp)})}}  D^{\tau(s(J) - s(\fp))} \int_J |f|.
+        2^{150a^3 + 3\tau}\sum_{\fp\in \mathfrak{T}}\frac{\mathbf{1}_{E(\fp)}}{\mu(B(x,D^{s(\fp)}))}(x)\sum_{\substack{J\in \mathcal{J}(\fT(\fu))\\J\subset B(x,  16D^{s(\fp)})}}  D^{\tau(s(J) - s(\fp))} \int_J |f|\,.
     $$
     $$
-        = 2^{150a^3 + 3\tau}\sum_{I \in \mathcal{D}} \sum_{\substack{\fp\in \mathfrak{T}\\ \sc(\fp) = I}}\frac{\mathbf{1}_{E(\fp)}(x)}{\mu(B(x, D^{s(I)}))}\sum_{\substack{J\in \mathcal{J}\\J\subset B(x,  D^{s(\fp)})}}  D^{\tau(s(J) - s(\fp))} \int_J |f|.
+        = 2^{150a^3 + 3\tau}\sum_{I \in \mathcal{D}} \sum_{\substack{\fp\in \mathfrak{T}\\ \sc(\fp) = I}}\frac{\mathbf{1}_{E(\fp)}(x)}{\mu(B(x, D^{s(I)}))}\sum_{\substack{J\in \mathcal{J}(\fT(\fu))\\J\subset B(x,  16 D^{s(\fp)})}}  D^{\tau(s(J) - s(\fp))} \int_J |f|\,.
     $$
     By \eqref{eq dis freq cover} and \eqref{definedE}, the sets $E(\fp)$ for tiles $\fp$ with $\sc(\fp) = I$ are pairwise disjoint.
-    If $x \in E(\fp)$ then in particular $x \in \sc(\fp)$, so by \eqref{eq vol sp cube} $B(c(I),8D^{s(I)}) \subset B(x, 16D^{s(I)})$. By the doubling property \eqref{doublingx}
+    If $x \in E(\fp)$ then in particular $x \in \sc(\fp)$, so by \eqref{eq vol sp cube} $B(c(I),16D^{s(I)}) \subset B(x, 32D^{s(I)})$. By the doubling property \eqref{doublingx}
     $$
-        \mu(B(c(I), D^{s(I)})) \le 2^{4a} \mu(B(x, D^{s(I)}))\,.
+        \mu(B(c(I), 16D^{s(I)})) \le 2^{5a} \mu(B(x, D^{s(I)}))\,.
     $$
-    Hence we can complete our estimate with
+    Since $a \ge 4$ and $\tau \le 1$ we can continue our estimate with
     $$
-        \le 2^{157a^3}\sum_{I \in \mathcal{D}} \frac{\mathbf{1}_{I}(x)}{\mu(B(c(I),  8D^{s(I)}))}\sum_{\substack{J\in \mathcal{J}\\J\subset B(x,  D^{s(\fp)})}}  D^{\tau(s(J) - s(\fp))} \int_J |f|\,.
+        \le 2^{151a^3}\sum_{I \in \mathcal{D}} \frac{\mathbf{1}_{I}(x)}{\mu(B(c(I),  16D^{s(I)}))}\sum_{\substack{J\in \mathcal{J}(\fT(\fu))\\J\subset B(x,  16 D^{s(\fp)})}}  D^{\tau(s(J) - s(\fp))} \int_J |f|\,.
     $$
     Finally, it follows from the definition of $\mathcal{L}(\fT(\fu))$ that $x \in \sc(\fp)$ if and only if $x' \in \sc(\fp)$, thus this equals
     $$
-         2^{157a^3} S P_{\mathcal{J}(\fT(\fu))}|f|(x')\,.
+            2^{151a^3} S_{1,\fu} P_{\mathcal{J}(\fT(\fu))}|f|(x')\,.
     $$
     This completes the proof.
 \end{proof}
 
+\section{Auxiliary \texorpdfstring{$L^2$}{L2} estimates}
+
 \begin{lemma}
-    \label{lem aux overlap}
-    For every cube $I \in \mathcal{D}$, there exist at most $2^{4a}$ cubes $J \in \mathcal{D}$ with $s(J) = s(I)$ and $B(c(I), D^{s(I)}) \cap B(c(J), D^{s(J)}) \ne \emptyset$.
+    \label{lem nontangential}
+    It holds that \lars{assumptions on f}
+    $$
+        \|T_{\mathcal{N}} f\|_2 \le 2^{103a^3} \|f\|_2\,.
+    $$
 \end{lemma}
 
 \begin{proof}
-    Suppose that $B(c(I), D^{s(I)}) \cap B(c(J), D^{s(J)}) \ne \emptyset$ and $s(I) = s(J)$. Then $B(c(I), 2D^{s(I)}) \subset B(c(J), 4D^{s(J)})$. Hence by the doubling property \eqref{doublingx}
+    Fix $s_1, s_2$. By \eqref{eq psisum} we have for all $x \in (0, \infty)$
+    $$
+        \sum_{s = s_1}^{s_2} \psi(D^{-s}x) = 1 - \sum_{s < s_1} \psi(D^{-s}x) - \sum_{s > s_1} \psi(D^{-s}x)\,.
+    $$
+    Since $\psi$ is supported in $[\frac{1}{4D}, \frac{1}{2}]$, the two sums on the right hand side are zero for all $x \in [\frac{1}{2}D^{s_1-1}, \frac{1}{4} D^{s_2 - 1}]$, hence
+    $$
+        x \in  [\frac{1}{2}D^{s_1-1}, \frac{1}{4} D^{s_2}] \implies \sum_{s = s_1}^{s_2} \psi(D^{-s}x) = 1\,.
+    $$
+    Since $\psi$ is supported in $[\frac{1}{4D}, \frac{1}{2}]$, we further have
+    $$
+        x \notin [\frac{1}{4}D^{s_1 - 1}, \frac{1}{2}D^{s_2}] \implies \sum_{s = s_1}^{s_2} \psi(D^{-s}x) = 0\,.
+    $$
+    Finally, since $\psi \ge 0$ and $\sum_{s \in \mathbb{Z}} \psi(D^{-s}x) = 1$, we have for all $x$
+    $$
+        0 \le \sum_{s = s_1}^{s_2} \psi(D^{-s}x) \le 1\,.
+    $$
+    Let $x' \in I_{s_1}(x)$. By the triangle inequality and \eqref{eq vol sp cube}, it holds that $\rho(x,x') \le 8D^{s_1}$. We have
     $$
-        2^{4a}\mu(B(c(J), \frac{1}{4}D^{s(J)})) \ge \mu(B(c(I), 2D^{s(I)}))\,,
+        \Bigg|\sum_{s = s_1}^{s_2} \int K_s(x',y) f(y) \, \mathrm{d}\mu(y)\Bigg|
     $$
-    and by the triangle inequality, the ball $B(c(J), \frac{1}{4}D^{s(J)})$ is contained in $B(c(I), 2D^{s(I)})$. If $\mathcal{C}$ is any finite collection of cubes $J \in \mathcal{D}$ satisfying $s(J) = s(I)$ and $B(c(I), D^{s(I)}) = B(c(J), D^{s(J)})$, 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
     $$
-        \mu(B(c(I), 2D^{s(I)})) \ge \sum_{J \in \mathcal{C}} \mu(B(c(J), \frac{1}{4}D^{s(J)})) \ge |\mathcal{C}| 2^{-4a} \mu(B(c(I), 2D^{s(I)}))\,.
+        = \Bigg|\int \sum_{s = s_1}^{s_2} \psi(D^{-s}\rho(x',y)) K(x',y) f(y) \, \mathrm{d}\mu(y)\Bigg|
     $$
-    \lars{Maybe prove this somewhere}
-    It follows from the fact that $\mu$ is doubling and nonzero that $\mu(B(c(I), 2D^{s(I)})) > 0$. The lemma follows.
+    \begin{equation}
+        \label{eq sharp trunc term}
+        \le \Bigg| \int_{8D^{s_1} \le \rho(x',y) \le \frac{1}{4}D^{s_2}} K(x',y) f(y) \, \mathrm{d}\mu(y) \Bigg|
+    \end{equation}
+    \begin{equation}
+        \label{eq lower bound term}
+        + \int_{\frac{1}{4}D^{s_1-1} \le \rho(x',y) \le 8D^{s_1}} |K(x', y)| |f(y)| \, \mathrm{d}\mu(y)
+    \end{equation}
+    \begin{equation}
+        \label{eq upper bound term}
+        + \int_{\frac{1}{4}D^{s_2} \le \rho(x',y) \le \frac{1}{2}D^{s_2}} |K(x', y)| |f(y)| \, \mathrm{d}\mu(y)\,.
+    \end{equation}
+    The first term \eqref{eq sharp trunc term} is at most $T_* f(x)$.
+
+    The other two terms will be estimated by the Hardy-Littlewood maximal function, for this purpose let
+    $$
+        \mathcal{B} = \{B(c(I), 16D^{s(I)}) \ : \ I \in \mathcal{D}\}\,.
+    $$
+    For the second term \eqref{eq lower bound term} we use that by \eqref{eqkernel size} for all $y$ with $\rho(x', y) \ge \frac{1}{4}D^{s_1 - 1}$, we have
+    $$
+        |K(x', y)| \le \frac{2^{a^3}}{\mu(B(x', \frac{1}{4}D^{s_1 - 1}))}\,.
+    $$
+    Using $D=2^{100a^2}$
+    and the doubling property \eqref{doublingx} $6 +100a^2$ times estimates
+    the last display by
+    \begin{equation}
+        \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, \eqref{eq lower bound term} is at most
+    $$
+        2^{6a + 101a^3} M_{\mathcal{B},1} f(x)\,.
+    $$
+
+    For \eqref{eq upper bound term} we argue similarly. We have for all $y$ with $\rho(x', y) \ge \frac{1}{4}D^{s_2}$
+    $$
+        |K(x', y)| \le \frac{2^{a^3}}{\mu(B(x', \frac{1}{4}D^{s_2}))}\,.
+    $$
+    Using the doubling property \eqref{doublingx} $6$ times estimates
+    the last display by
+    \begin{equation}
+        \le \frac{2^{6a + a^3}}{\mu(B(x,  16 D^{s_2}))}\, .
+    \end{equation}
+    Note that by \eqref{dyadicproperty} we have $I_{s_1}(x) \subset I_{s_2}(x)$, in particular $x' \in I_{s_2}(x)$.
+    By the triangle inequality and \eqref{eq vol sp cube}, we have
+    $$
+        B(x', 8D^{s_2}) \subset B(c(I_{s_2}(x)), 16D^{s(I_{s_2}(x))})\,.
+    $$
+    Combining this, \eqref{eq lower bound term} is again at most
+    $$
+        2^{6a} M_{\mathcal{B},1} f(x)\,.
+    $$
+
+    Using $a \ge 4$, taking a supremum over all $x' \in I_{s_1}(x)$ and then a supremum over all $-S \le s_1 < s_2 \le S$, we obtain
+    $$
+        T_{\mathcal{N}} f(x) \le T_*f(x) + 2^{102a^3}  M_{\mathcal{B},1} f(x)\,.
+    $$
+    The Lemma now follows from assumption \eqref{nontanbound}, Proposition \ref{prop hlm} and $a \ge 4$.
+\end{proof}
+
+\begin{lemma}
+    \label{lem aux 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}
+
+\begin{proof}
+    Suppose that $B(c(I), 16 D^{s(I)}) \cap B(c(J), 16 D^{s(J)}) \ne \emptyset$ and $s(I) = s(J)$. Then $B(c(I), 32 D^{s(I)}) \subset B(c(J), 64 D^{s(J)})$. Hence by the doubling property \eqref{doublingx}
+    $$
+        2^{8a}\mu(B(c(J), \frac{1}{4}D^{s(J)})) \ge \mu(B(c(I), 32 D^{s(I)}))\,,
+    $$
+    and by the triangle inequality, the ball $B(c(J), \frac{1}{4}D^{s(J)})$ is contained in $B(c(I), 32 D^{s(I)})$.
+
+    If $\mathcal{C}$ is any finite collection of cubes $J \in \mathcal{D}$ satisfying $s(J) = s(I)$ and $B(c(I), 16 D^{s(I)}) \cap B(c(J), 16 D^{s(J)}) \ne\emptyset$, 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
+    \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)}))\,.
+    \end{align*}
+    Since $\mu$ is doubling and $\mu \ne 0$, we have $\mu(B(c(I), 32D^{s(I)})) > 0$. The lemma follows after dividing by $2^{-8a}\mu(B(c(I), 32D^{s(I)}))$.
 \end{proof}
 
 \begin{lemma}
+    \label{lem L2 Su estimate}
     For all $\fu \in \fU$ and all $f \in L^2(X)$, we have
     \begin{equation}
         \label{eq S bound}
-        \|S_{\fu}f\|_2 \le ... \|f\|_2\,.
+        \|S_{1,\fu}f\|_2 \le \frac{2^{10a+1}}{\tau} \|f\|_2\,.
     \end{equation}
 \end{lemma}
 
 \begin{proof}
-    We have for every $g \in L^2(X)$
-    \begin{align*}
-        &\quad \big|\int g Sf\big|\\
-        &= \sum_{I\in\mathcal{D}} \frac{1}{\mu(B(c(I), 8D^{s(I)}))} \int_I g(y) \, \mathrm{d}\mu(y) \sum_{J\in \mathcal{J}:J\subseteq B(c(I), D^{s(I)})} D^{(s(J)-s(I))\tau}\int_J |f(y)| \,\mathrm{d}\mu(y)\\
-        &\le \sum_{I\in\mathcal{D}} \frac{1}{\mu(B(c(I), 8D^{s(I)}))} \int_{B(c(I), 8D^{s(I)})} |g(y)| \, \mathrm{d}\mu(y) \sum_{J\in \mathcal{J}:J\subseteq B(c(I), D^{s(I)})} D^{(s(J)-s(I))\tau}\int_J |f(y)| \,\mathrm{d}\mu(y)\,.
-    \end{align*}
-    Changing the order of summation and using $J \subset B(c(I), D^{s(I)})$ to bound the first average integral by $M_{\mathcal{B},1}|g|(y)$ for any $y \in J$, we obtain
+    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
+    $$
+        \Bigg|\int \bar g(y) S_{1,\fu}f(y) \, \mathrm{d}\mu(y)\Bigg|
+    $$
+    \begin{multline*}
+        = \Bigg|\sum_{I\in\mathcal{D}} \frac{1}{\mu(B(c(I), 16 D^{s(I)}))} \int_I \bar g(y) \, \mathrm{d}\mu(y)\\
+        \times \sum_{J\in \mathcal{J}\,:\,J\subseteq B(c(I), 16 D^{s(I)})} D^{(s(J)-s(I))\tau}\int_J |f(y)| \,\mathrm{d}\mu(y)\Bigg|
+    \end{multline*}
+    \begin{multline*}
+        \le \sum_{I\in\mathcal{D}} \frac{1}{\mu(B(c(I), 16D^{s(I)}))} \int_{B(c(I), 16D^{s(I)})} | g(y)| \, \mathrm{d}\mu(y)\\ \times \sum_{J\in \mathcal{J}\,:\,J\subseteq B(c(I), 16 D^{s(I)})} D^{(s(J)-s(I))\tau}\int_J |f(y)| \,\mathrm{d}\mu(y)\,.
+    \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*}
-        \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), D^{s(I)})} D^{(s(J)-s(I))\tau}.
+        \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))\tau}.
     \end{align*}
-    By \eqref{eq vol sp cube} and \eqref{defineD} the condition $J \subset B(c(I), D^{s(I)})$ implies $s(I) \ge s(J)$. By Lemma \ref{lem aux overlap}, there are at most $2^{4a}$ cubes $I$ at each scale with $J \subset B(c(I), D^{s(I)})$. Summing the geometric series using that $D \ge e$ and hence $(1 - D^\tau)^{-1} \le \tau^{-1}$, we obtain
+    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 $J \subset B(c(I), D^{s(I)})$. Summing the geometric series using that $D \ge e$ and hence $(1 - D^\tau)^{-1} \le \tau^{-1}$, we obtain
     $$
-        \le \frac{2^{4a}}{\tau} \sum_{J\in\mathcal{J}}\int_J|f(y)|  M_{\mathcal{B},1}g(y) \, \mathrm{d}\mu(y)\,.
+        \le \frac{2^{8a}}{\tau} \sum_{J\in\mathcal{J}}\int_J|f(y)|  M_{\mathcal{B},1}|g|(y) \, \mathrm{d}\mu(y)\,.
     $$
     The collection $\mathcal{J}$ is a partition of $X$, so this equals
     $$
-        \frac{2^{4a}}{\tau} \int_X|f(y)|  M_{\mathcal{B},1}g(y) \, \mathrm{d}\mu(y)\,.
+        \frac{2^{8a}}{\tau} \int_X|f(y)|  M_{\mathcal{B},1}|g|(y) \, \mathrm{d}\mu(y)\,.
     $$
     Using Cauchy-Schwarz and Proposition \ref{prop hlm} we conclude
     $$
-        \left|\int g Sf \, \mathrm{d}\mu \right| \le  2\frac{2^{6a}}{\tau} \|g\|_2\|f\|_2\,.
+        \left|\int \bar g S_{1,\fu}f \, \mathrm{d}\mu \right| \le  2\frac{2^{8a}}{\tau} \|g\|_2\|f\|_2\,.
     $$
-    The lemma now follows from duality. \lars{Is this duality lemma in Lean? Also, this proof is not really finitary, because the integrals are not. Is it necessary in Lean to prove in each step that all integrals converge?}
+    The lemma now follows by choosing $g = S_{1,\fu}f$ and dividing on both sides by the finite $\|S_{1,\fu}f\|_2$.
 \end{proof}
 
+\lars{Write everywhere the correct assumptions on the functions. Probably bounded, bounded support.}
+
 \begin{lemma}[Tree estimate]
     \label{TreeEstimate}
     Let $\fu \in \fU$.
     Then we have for all $f, g$
+    $$
+        \Bigg|\int \sum_{\fp \in \fT(\fu)} \bar g(y) T_{\fp}f(y) \, \mathrm{d}\mu(y)  \Bigg|
+    $$
     \begin{equation}
         \label{eq tree est}
-        \int |\sum_{\fp \in \fT(\fu)} gT_{\fp}f \, \mathrm{d}\mu |\le ... \|P_{\mathcal{J}(\fT(\fu))}|f|\|_{2}\|P_{\mathcal{L}(\fT(\fu))}|f|\|_{2}.
+            \le (2^{104a^3} + \frac{2^{11a}}{\tau}) \|P_{\mathcal{J}(\fT(\fu))}|f|\|_{2}\|P_{\mathcal{L}(\fT(\fu))}|g|\|_{2}.
     \end{equation}
 \end{lemma}
 
 \begin{proof}
-    Combine the last two lemmas.
-\end{proof}
+    For each $L \in \mathcal{L}(\fT(\fu))$, choose a point $x'(L) \in L$ such that for all $y \in L$
+    $$
+        (M_{\mathcal{B},1}+S_{1,\fu})P_{\mathcal{J}(\fT(\fu))}|f|(x')+|T_{\mathcal{N}}P_{\mathcal{J}(\fT(\fu))}f(x')|
+    $$
+    \begin{equation}
+        \label{eq x'L almost inf}
+        \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 $L \in \mathcal{L}(\fT(\fu))$
+    $$
+        \int_L |g(y)| \Bigg| \sum_{\fp \in \fT(\fu)} T_{\fp} f(y) \Bigg| \, \mathrm{d}\mu(y)
+    $$
+    $$
+        \le 2^{151a^3} \int_L |g(y)| ((M_{\mathcal{B},1}+S_{1,\fu})P_{\mathcal{J}(\fT(\fu))}|f|(x')+|T_{\mathcal{N}}P_{\mathcal{J}(\fT(\fu))}f(x')| ) \, \mathrm{d}\mu(y)
+    $$
+    \begin{multline*}
+        \le 2^{151a^3} \int_L |g(y)| \, \mathrm{d}\mu(y)\times\\
+            \int_L 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)| ) \, \mathrm{d}\mu(y)
+    \end{multline*}
+    \begin{multline*}
+        =  2^{151a^3 + 1} \int_L P_{\mathcal{L}(\fT(\fu))}|g|(y)\times \\((M_{\mathcal{B},1}+S_{1,\fu})P_{\mathcal{J}(\fT(\fu))}|f|(y)+|T_{\mathcal{N}}P_{\mathcal{J}(\fT(\fu))}f(y)| ) \, \mathrm{d}\mu(y)\,.
+    \end{multline*}
+    By \eqref{definetp}, we have $T_{\fp} f = \mathbf{1}_{\sc(\fp)} T_{\fp} f$ for all $\fp \in \fP$, so
+    $$
+        \Bigg| \int \bar g(y) \sum_{\fp \in \fT(\fu)} T_{\fp} f(y)  \, \mathrm{d}\mu(y) \Bigg| = \Bigg| \int_{\bigcup_{\fp \in \fT(\fu)} \sc(\fp)} \bar g(y) \sum_{\fp \in \fT(\fu)} T_{\fp} f(y)  \, \mathrm{d}\mu(y) \Bigg|\,.
+    $$
+    Since $\mathcal{L}(\fT(\fu))$ partitions $\bigcup_{\fp \in \fT(\fu)} \sc(\fp)$ by Lemma \ref{lem partition},
+    we get from the triangle inequality
+    $$
+        \le \sum_{L \in \mathcal{L}(\fT(\fu))} \int_L |g(y)| \Bigg| \sum_{\fp \in \fT(\fu)} T_{\fp} f(y) \Bigg| \, \mathrm{d}\mu(y)
+    $$
+    which by the above computation is bounded by
+    \begin{multline*}
+        2^{151a^3 + 1} \sum_{L \in \mathcal{L}(\fT(\fu))} \int_L P_{\mathcal{L}(\fT(\fu))}|g|(y) \times \\((M_{\mathcal{B},1}+S_{1,\fu})P_{\mathcal{J}(\fT(\fu))}|f|(y)+|T_{\mathcal{N}}P_{\mathcal{J}(\fT(\fu))}f(y)| ) \, \mathrm{d}\mu(y)
+    \end{multline*}
+    \begin{multline*}
+        = 2^{151a^3 + 1} \int_X P_{\mathcal{L}(\fT(\fu))}|g|(y)\times \\((M_{\mathcal{B},1}+S_{1,\fu})P_{\mathcal{J}(\fT(\fu))}|f|(y)+|T_{\mathcal{N}}P_{\mathcal{J}(\fT(\fu))}f(y)| ) \, \mathrm{d}\mu(y)\,.
+    \end{multline*}
+    Applying Cauchy-Schwarz and Minkowski's inequality, this is bounded by
+    \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{prop hlm}, Lemma \ref{lem nontangential} and Lemma \ref{lem L2 Su estimate}, the second factor is at most
+    $$
+        (2^{2a+1} + \frac{2^{10a + 1}}{\tau})\|P_{\mathcal{J}(\fT(\fu))}|f|\|_2 + 2^{103a^3} \|P_{\mathcal{J}(\fT(\fu))}f\|_2\,.
+    $$
+    By the triangle inequality we have for all $x \in X$ that $|P_{\mathcal{J}(\fT(\fu))}f|(x) \le P_{\mathcal{J}(\fT(\fu))}|f|(x)$, thus we can further estimate the above by
+    $$
+        (2^{2a+1} + \frac{2^{10a + 1}}{\tau} + 2^{103a^3}) \|P_{\mathcal{J}(\fT(\fu))}|f|\|_2\,.
+    $$
+    This completes the proof since $a \ge 4$.
+\end{proof}
+
+\section{The \texorpdfstring{$L^2$}{L2} tree estimate}
+
+\begin{lemma}
+    \label{lem 2density estimate tree}
+    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}
+    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 $s(\fp) \ge s(J)$. Using the triangle inequality and \eqref{eq vol sp cube} it follows that $J \subset B(c(\fp), 200 D^{s(\fp) + 1})$. From the doubling property \eqref{doublingx}, $D=2^{100a^2}$ and \eqref{eq vol sp cube}, we obtain
+    $$
+        \mu(\sc(\fp)) \le 2^{100a^3 + 9} \mu(J)
+    $$
+    and hence
+    $$
+        \mu( B(c(\fp), 200 D^{s(\fp) + 1})) \le 2^{200a^3 +19} \mu(J)\,.
+    $$
+    With the definition \eqref{definedens2} of $\dens_2$ it follows that
+    $$
+        \mu(J \cap F) \le \mu( B(c(\fp), 200 D^{s(\fp) + 1}) \cap F)
+    $$
+    $$
+        \le \dens_2(\fT(\fu)) \mu( B(c(\fp), 200 D^{s(\fp) + 1})) \le 2^{200a^3 +19} \dens_2(\fT(\fu))\mu(J)\,.
+    $$
+
+    Now suppose that there does not exist a tile $\fp \in \fT(\fu)$ with $\sc(\fp) \subset B(c(J), 100 D^{s(J) + 1})$. If we had $s(\fq) \le s(J)$, then by \eqref{dyadicproperty} and \eqref{eq vol sp cube}  $\sc(\fq) \subset J \subset B(c(J), 100 D^{s(J) + 1})$, contradicting our assumption. Thus $s(\fq) > s(J)$, so there exists by \eqref{coverdyadic} and \eqref{dyadicproperty} some cube $J' \in \mathcal{D}$ with $s(J') = s(J) + 1$ and $J \subset J'$. By definition of $\mathcal{J}(\fT(\fu))$ there exists some $\fp \in \fT(\fu)$ such that $\sc(\fp) \subset B(c(J'), 100 D^{s(J') + 1})$.  From the doubling property \eqref{doublingx}, $D=2^{100a^2}$ and \eqref{eq vol sp cube}, we obtain
+    \begin{equation}
+        \label{eq measure comparison 1}
+        \mu(B(c(\fp), 4D^{s(\fp)})) \le 2^{4a} \mu(\sc(\fp)) \le 2^{200a^3 + 14} \mu(J)\,.
+    \end{equation}
+    If $J \subset B(c(\fp), 4 D^{s(\fp)})$, then we bound
+    $$
+        \mu(J \cap F) \le \mu(B(c(\fp), 4D^{s(\fp)}) \cap F)
+    $$
+    and use the definition \eqref{definedens2}
+    $$
+        \le \dens_2(\fT(\fu)) \mu(B(c(\fp), 4D^{s(\fp)})) \le 2^{200a^3 + 14} \mu(J)\,.
+    $$
+    From now on we assume $J \not \subset B(c(\fp), 4 D^{s(\fp)})$.
+    Since $c(\fp) \in \sc(\fp) \subset B(c(J'), 100 D^{s(J') + 1})$, we have by \eqref{eq vol sp cube} and the triangle inequality
+    $$
+        J \subset J' \subset B(c(J'), 4D^{s(J')}) \subset B(c(\fp), 104 D^{s(J') + 1})\,.
+    $$
+    In particular this implies $104 D^{s(J') + 1} > 4D^{s(\fp)}$. By the triangle inequality we also have
+    $$
+        B(c(\fp), 104 D^{s(J') + 1}) \subset B(c(J), 204 D^{s(J') + 1})\,,
+    $$
+    so from the doubling property \eqref{doublingx}
+    $$
+        \mu( B(c(\fp), 104 D^{s(J') + 1})) \le 2^{200a^3 + 10} \mu(J)\,.
+    $$
+    From here one completes the proof as in the other cases.
+\end{proof}
+
+
+\begin{lemma}
+    \label{lem 1density estimate tree}
+    Let $\fu \in \fU$ and $L \in \mathcal{L}(\fT(\fu))$. Then
+    \begin{equation}
+    \label{eq 1density estimate tree}
+        \mu(L \cap \bigcup_{\fp \in \fT(\fu)} E(\fp)) \le 2^{101a^3} \dens_1(\fT(\fu)) \mu(L)\,.
+    \end{equation}
+\end{lemma}
+
+\begin{proof}
+    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 $s(\fp) \le s(L)$. 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 $s(L) = -S \le s(\fq)$ it follows from \eqref{dyadicproperty} that $\sc(\fq) = L \subset \sc(\fq)$. We have then by Lemma \ref{lem cube monotone}
+    \begin{align*}
+        d_{\fp}(Q(\fp), Q(\fq)) &\le d_{\fp}(Q(\fp), Q(\fu)) + d_{\fp}(Q(\fq), Q(\fu))\\
+        &\le d_{\fp}(Q(\fp), Q(\fu)) + d_{\fq}(Q(\fq), Q(\fu))\,.
+    \end{align*}
+    Using that $\fp, \fq \in \fT(\fu)$ and \eqref{forest1}, this is at most $8$. Using again the triangle inequality and Lemma \ref{lem cube monotone}, we obtain that for each $q \in B_{\fq}(Q(\fq), 1)$
+    $$
+        d_{\fp}(Q(\fp), q) \le d_{\fp}(Q(\fp), Q(\fq)) + d_{\fq}(Q(\fq), q) \le 9\,.
+    $$
+    Thus $L \cap E(\fq) \subset E_2(9, \fp)$. We obtain
+    $$
+        \mu(L \cap \bigcup_{\fq \in \fT(\fu)} E(\fq)) \le \mu(E_2(9, \fp))\,.
+    $$
+    By the definition of $\dens_1$, this is bounded by
+    $$
+        9^a \dens_1(\fT(\fu)) \mu(\sc(\fp)) =9^a \dens_1(\fT(\fu)) \mu(L)\,.
+    $$
+    Since $a \ge 4$, \eqref{eq 1density estimate tree} follows in this case.
+
+    Now suppose that for each $\fp \in \fT(\fu)$ with $L \cap E(\fp) \ne \emptyset$, we have $s(\fp) > s(L)$. Since there exists at least one such $\fp$, there exists in particular at least one cube $L'' \in \mathcal{D}$ with $L \subset L''$ and $s(L'') > s(L)$. By \eqref{coverdyadic}, there exists $L' \in \mathcal{D}$ with $L \subset L'$ and $s(L') = s(L) + 1$. By the definition of $\mathcal{L}(\fT(\fu))$ there exists a tile $\fp'' \in \fT(\fu)$ with $\sc(\fp'') \subset L'$. Let $\fp'$ be the unique tile such that $\sc(\fp') = L'$ and such that $\Omega(\fu) \cap \Omega(\fp') \ne \emptyset$. Since by \eqref{forest1} $s(\fp') = s(L') \le s(\fp) < s(\fu)$, we have by \eqref{dyadicproperty} and \eqref{eq freq dyadic} that $\Omega(\fu) \subset \Omega(\fp')$. Let $\fq \in \fT(\fu)$ with $L \cap E(\fq) \ne \emptyset$. As shown above, this implies $s(\fq) \ge s(L')$, so by \eqref{dyadicproperty} $L' \subset \sc(\fq)$. If $q \in B_{\fq}(Q(\fq), 1)$, then by a similar calculation as above, using the triangle inequality, Lemma \ref{lem cube monotone} and \eqref{forest1}, we obtain
+    $$
+        d_{\fp'}(Q(\fp'), q) \le d_{\fp'}(Q(\fp'), Q(\fq)) + d_{\fq}(Q(\fq), q) \le 6\,.
+    $$
+    Thus $L \cap E(\fq) \subset E_2(6, \fp')$. Since $\sc(\fp'') \subset \sc(\fp') \subset \sc(\fp)$ and $\fp'', \fp \in \fT(\fu)$, we have $\fp' \in \fP'(\fT(\fu))$. We deduce using the definition \eqref{definedens1} of $\dens_1$
+    $$
+        \mu(L \cap \bigcup_{\fq \in \fT(\fu)} E(\fq)) \le \mu(E_2(6, \fq')) \le 6^a \dens_1(\fT(\fu)) \mu(L')\,.
+    $$
+    Using the doubling property \eqref{doublingx}, \eqref{eq vol sp cube}, and $a \ge 4$ this is estimated by
+    $$
+        6^a 2^{100a^3 + 5}\dens_1(\fT(\fu)) \mu(L) \le 2^{101 a^3} \dens_1(\fT(\fu))\mu(L)\,.
+    $$
+\end{proof}
+
+
+We are now ready to prove the main estimate for trees, showing that the $1$-density and $2$-density control the norm of a tree operator on $L^2$.
+
+\begin{cor}
+    \label{cor tree est}
+    Let $\fu \in \fU$. Then for all $f,g \in L^2(X)$  \lars{add correct assumption on $f$. supported in $1_F$, bounded?} \lars{No, I also need this without (quantitative) support constraints on $f$.}
+    \begin{equation}
+        \label{eq cor tree est}
+        \left|\int_X g \sum_{\fp \in \fT(\fu)} T_{\fp }f \, \mathrm{d}\mu \right| \le (2^{155a^3} + \frac{2^{52a^3}}{\tau}) \dens_1(\fT(\fu))^{1/2} \|f\|_2\,.
+    \end{equation}
+    If $f \le \mathbf{1}_F$, then we have
+    \begin{equation}
+        \label{eq cor tree est F}
+        \left| \int_X g \sum_{\fp \in \fT(\fu)} T_{\fp }f\, \mathrm{d}\mu \right| \le (2^{256a^3} + \frac{2^{152a^3}}{\tau}) \dens_1(\fT(\fu))^{1/2} \dens_2(\fT(\fu))^{1/2} \|f\|_2\,.
+    \end{equation}
+\end{cor}
+
+\begin{proof}
+    Denote
+    $$
+        \mathcal{E}(\fu) = \bigcup_{\fp \in \fT(\fu)} E(\fp)\,.
+    $$
+    Then we have
+    $$
+        \left| \int_X \bar g \sum_{\fp \in \fT(\fu)} T_{\fp} f \, \mathrm{d}\mu \right|  = \left| \int_X \overline{ g\mathbf{1}_{\mathcal{E}(\fu)}}  \sum_{\fp \in \fT(\fu)} T_{\fp} f \, \mathrm{d}\mu \right|\,.
+    $$
+    By Lemma \ref{TreeEstimate}, this is bounded by
+    \begin{equation}
+        \label{eq both factors tree}
+        \le (2^{104a^3} + \frac{2^{11a}}{\tau})\|P_{\mathcal{J}(\fT(\fu))}|f|\|_2 \|P_{\mathcal{L}(\fT(\fu))} |\mathbf{1}_{\mathcal{E}(\fu)}g|\|_2\,.
+    \end{equation}
+    We bound the two factors on the right hand side separately.
+    We have
+    $$
+        \|P_{\mathcal{L}(\fT(\fu))} |\mathbf{1}_{\mathcal{E}(\fu)}g|\|_2 = \left( \sum_{L \in \mathcal{L}(\fT(\fu))} \frac{1}{\mu(L)} \left(\int_{L \cap \mathcal{E}(\fu)} |g(y)| \, \mathrm{d}\mu(y)\right)^2 \right)^{1/2}\,.
+    $$
+    By Cauchy-Schwarz and Lemma \ref{lem 1density estimate tree} this is at most
+    $$
+        \le \left( \sum_{L \in \mathcal{L}(\fT(\fu))} 2^{101a^3} \dens_1(\fT(\fu)) \int_{L \cap \mathcal{E}(\fu)} |g(y)|^2 \, \mathrm{d}\mu(y) \right)^{1/2}\,.
+    $$
+    Since cubes $L \in \mathcal{L}(\fT(\fu))$ are pairwise disjoint by Lemma \ref{lem partition}, this is
+    \begin{equation}
+        \label{eq factor L tree}
+            \le 2^{51 a^3} \dens_1(\fT(\fu))^{1/2} \|g\|_2\,.
+    \end{equation}
+
+    Similarly, we have
+    \begin{equation}
+        \label{eq cor tree proof}
+        \|P_{\mathcal{J}(\fT(\fu))}|f|\|_2  = \left( \sum_{J \in \mathcal{J}(\fT(\fu))} \frac{1}{\mu(J)} \left(\int_J |f(y)| \, \mathrm{d}\mu(y)\right)^2 \right)^{1/2}\,,
+    \end{equation}
+    by Cauchy-Schwarz this is
+    $$
+        \le \left( \sum_{J \in \mathcal{J}(\fT(\fu))} \int_J |f(y)|^2 \, \mathrm{d}\mu(y) \right)^{1/2}\,.
+    $$
+    Since cubes in $\mathcal{J}(\fT(\fu))$ are pairwise disjoint by Lemma \ref{lem partition}, this at most
+    \begin{equation}
+        \label{eq factor J tree}
+        \|f\|_2\,.
+    \end{equation}
+    Combining \eqref{eq both factors tree}, \eqref{eq factor L tree} and \eqref{eq factor J tree} and using $a \ge 4$ gives \eqref{eq cor tree est}.
+
+    If $f \le \mathbf{1}_F$ then $f = f\mathbf{1}_F$, so
+    $$
+        \eqref{eq cor tree proof} = \left( \sum_{J \in \mathcal{J}(\fT(\fu))} \int_{J \cap F} |f(y)|^2 \, \mathrm{d}\mu(y) \right)^{1/2}\,.
+    $$
+    We estimate as before, using now Lemma \ref{lem 2density estimate tree} and Cauchy-Schwarz
+    $$
+        \le 2^{100a^3 + 10} \dens_2(\fT(\fu))^{1/2} \|f\|_2\,.
+    $$
+    Combining this with \eqref{eq both factors tree}, \eqref{eq factor L tree} and $a \ge 4$ gives \eqref{eq cor tree est F}.
+\end{proof}
+
+\begin{lemma}
+    The adjoint of the operator $T_{\fp}$ defined in \eqref{definetp} is given by
+    \begin{equation}
+        \label{definetp*}
+        T_{\fp}^* g(x) = \int \overline{K_{s(\fp)}(y,x)} e(-Q(y)(x)+ Q(y)(y)) g(y) \, \mathrm{d}\mu(y)\,.
+        \end{equation}
+\end{lemma}
+
+\begin{proof}
+    This follows immediately from the definition \eqref{definetp}.
+\end{proof}
+
+\begin{lemma}
+    \label{lem Tp support adjoint}
+    For each $\fu \in \fU$ and each $\fp \in \fT(\fu)$, we have
+    $$
+        T_{\fp}^* g = \mathbf{1}_{\sc(\fu)} T_{\fp}^* \mathbf{1}_{\sc(\fu)} g\,.
+    $$
+\end{lemma}
+
+\begin{proof}
+    By \eqref{forest1}, $E(\fp) \subset \sc(\fp) \subset \sc(\fu)$. Thus by \eqref{definetp*}
+    $$
+            T_{\fp}^* g(x) =  T_{\fp}^* (\mathbf{1}_{\sc(\fp)} g)(x)
+    $$
+    $$
+        = \int \overline{K_{s(\fp)}(y,x)} e(-Q(y)(x) +  Q(y)(y)) \mathbf{1}_{\sc(\fp)}(y) g(y) \, \mathrm{d}\mu(y)\,.
+    $$
+    If this integral is not $0$, then there exists $y \in \sc(\fp)$ such that $K_{s(\fp)}(y,x) \ne 0$. By \eqref{supp Ks t} and \eqref{eq vol sp cube}, it follows that $x \in B(c(\fp), 5 D^{s(\fp)})$. By \eqref{forest6}, $x \in \sc(\fu)$. So
+    $$
+        T_{\fp}^* g(x) = \mathbf{1}_{\sc(\fu)} T_{\fp}^* \mathbf{1}_{\sc(\fu)} g(x)\,.
+    $$
+\end{proof}
+
+\begin{lemma}
+    \label{lem tree adjoint bound}
+    For all bounded $g$ with bounded support, we have that
+    $$
+        \left\| \sum_{\fp \in \fT(\fu)} T_{\fp}^* g\right\|_2 \le (2^{155a^3} + \frac{2^{52a^3}}{\tau}) \dens_1(\fT(\fu))^{1/2} \|g\|_2\,.
+    $$
+\end{lemma}
+
+\begin{proof}
+    By Cauchy-Schwarz and Lemma \ref{cor tree est}, we have for all bounded $f,g$ with bounded support that
+    $$
+        \left| \int_X \overline{\sum_{\fp\in \fT(\fu)} T_{\fp}^* g} f \,\mathrm{d}\mu \right| = \left| \int_X g \sum_{\fp \in \fT(\fu)} T_{\fp} f \,\mathrm{d}\mu \right|
+    $$
+    \begin{equation}
+        \label{eq adjoint bound}
+        \le (2^{155a^3} + \frac{2^{52a^3}}{\tau}) \dens_1(\fT(\fu))^{1/2} \|g\|_2 \|f\|_2\,.
+    \end{equation}
+    Let $f = \sum_{\fp \in \fT(\fu)} T_{\fp}^* g$. If $g$ is bounded and has bounded support, then the same is true for $f$. In particular $\|f\|_2 < \infty$. Dividing \eqref{eq adjoint bound} by $\|f\|_2$ completes the proof.
+\end{proof}
+
+\section{Separated Trees}
+Define \lars{define $\mathcal{B}$}
+$$
+    S_{2,\fu} := \left|\sum_{\fp \in \fT(\fu)} T_{\fp}^* \right| + M_{\mathcal{B},1}\,.
+$$
+\begin{lemma}
+    \label{lem L2 Sg estimate}
+    We have for all bounded $g$ with bounded support
+    $$
+        \|S_{2, \fu} g\|_2 \le (2^{156a^3} + \frac{2^{52a^3}}{\tau}) \|g\|_2\,.
+    $$
+\end{lemma}
+
+\begin{proof}
+    This follows immediately from Minkowski's inequality, Proposition \ref{prop hlm} and Lemma \ref{lem tree adjoint bound}.
+\end{proof}
+
+We will need the following Hölder continuity estimate for adjoints of operators associated to tiles. We define
+$$
+    \tau' := \min\{\tau, \frac{1}{a}\}\,.
+$$
+\begin{lemma}
+    \label{lem tile Hölder}
+    Let $\fu \in \fU$ and $\fp \in \fT(\fu)$.  Then for all $y, y' \in X$ we have
+    $$
+        |e(Q(\fu)(y)) T_{\fp}^* g(y) - e(Q(\fu)(y')) T_{\fp}^* g(y')|
+    $$
+    \begin{equation}
+        \label{T*Hölder2}
+        \le \frac{2^{151a^3}}{\mu(B(c(\fp), 4D^{s(\fp)}))} \left(\frac{\rho(y, y')}{D^{s(\fp)}}\right)^{\tau'}  \int_{E(\fp)} |g(x)| \, \mathrm{d}\mu(x)\,.
+    \end{equation}
+\end{lemma}
+
+\begin{proof}
+    By \eqref{definetp*}, we have
+    $$
+        |e(\tQ(\fu)(y)) T_{\fp}^* g(y) - e(\tQ(\fu)(y')) T_{\fp}^* g(y')|
+    $$
+    \begin{multline*}
+        =\bigg| \int e(\tQ(x)(x) - \tQ(x)(y) + Q(\fu)(y)) \overline{K_{s(\fp)}(x, y)} (\mathbf{1}_{E(\fp)} g)(x) \\
+        -  e(\tQ(x)(x) - \tQ(x)(y') + Q(\fu)(y')) \overline{K_{s(\fp)}(x, y')} (\mathbf{1}_{E(\fp)} g)(x) \, \mathrm{d}\mu(x)\bigg|
+    \end{multline*}
+    \begin{multline*}
+        \leq\int_{E(\fp)} |g(x)| |e(\tQ(x)(y) - \tQ(x)(y') - \tQ(\fu)(y) + \tQ(\fu)(y'))\overline{K_{s(\fp)}(x, y)}\\
+        - \overline{K_{s(\fp)}(x, y')}| \, \mathrm{d}\mu(x)
+    \end{multline*}
+    \begin{multline}
+        \leq\int_{E(\fp)} |g(x)| |e(-\tQ(x)(y) + \tQ(x)(y') + Q(\fu)(y) - Q(\fu)(y')) - 1||\overline{K_{s(\fp)}(x, y)}|\, \mathrm{d}\mu(x)\\
+        + \int_{E(\fp)} |g(x)| |\overline{K_{s(\fp)}(x, y)} - \overline{K_{s(\fp)}(x, y')} |\, \mathrm{d}\mu(x)\,.\label{T*Hölder1}
+    \end{multline}
+    We have
+    $$
+        |-\tQ(x)(y) + \tQ(x)(y') + Q(\fu)(y) - Q(\fu)(y')|
+    $$
+    $$
+        \le d_{B(y, \rho(y,y'))}(\tQ(x), \tQ(\fu))\,.
+    $$
+    Suppose that $y, y' \in B(c(\fp), 5D^{s(\fp)})$, so that $\rho(y,y') \le 10D^{s(\fp)}$. Let $k \in \mathbb{Z}$ be such that $2^{ak}\rho(y,y') \le 10D^{s(\fp)}$ but $2^{a(k+1)} \rho(y,y') > 10D^{s(\fp)}$. In particular, $k \ge 0$. Then, by \eqref{firstdb}:
+    $$
+        \le 2^{-k} d_{B(c(\fp), 10 D^{s(\fp)})}(\tQ(x), \tQ(\fu)) \le 2^{6a - k} d_{\fp}(\tQ(x), \tQ(\fu))\,.
+    $$
+    Since $x \in E(\fp)$ we have $\tQ(x) \in \Omega(\fp) \subset B_{\fp}(\tQ(\fp), 1)$, and since $\fp \in \fT(\fu)$ we have $\tQ(\fu) \in B_{\fp}(\tQ(\fp), 4)$, thus this is bounded by $5 \cdot 2^{6a - k}$.
+    By definition of $k$, we have
+    $$
+        k \le \frac{1}{a} \log_2\left(\frac{10 D^{s(\fp)}}{\rho(y,y')}\right)\,,
+    $$
+    which gives
+    $$
+        |-\tQ(x)(y) + \tQ(x)(y') + Q(\fu)(y) - Q(\fu)(y')| \le 5 \cdot 2^{6a} \left(\frac{\rho(y,y')}{10 D^{s(\fp)}}\right)^{1/a}\,.
+    $$
+    For all $x \in \sc(\fp)$, we have by \eqref{doublingx} that
+    $$
+        \mu(B(x, D^{s(\fp)})) \ge 2^{-3a} \mu(B(c(\fp), 4D^{s(\fp)}))\,.
+    $$
+    Combined with \eqref{eq Ks size t} and \eqref{eq Ks smooth t},
+    we obtain
+    $$
+        \eqref{T*Hölder1} \le \frac{2^{3a}}{\mu(B(c(\fp), 4D^{s(\fp)}))} \int_{E(\fp)}|g(x)| \, \mathrm{d}\mu(x) \times
+    $$
+    $$
+        (2^{102a^3} \cdot 5 \cdot 2^{6a} \left(\frac{\rho(y,y')}{ D^{s(\fp)}}\right)^{1/a} + 2^{150a^3} \left(\frac{\rho(y,y')}{D^{s(\fp)}}\right)^\tau)
+    $$
+    Since $\rho(y,y') \le 10 D^{s(\fp)}$ and $\max\{\tau - \tau', 1/a - \tau'\} \le 1$, we conclude
+    $$
+        \eqref{T*Hölder1} \le \frac{2^{151a^3}}{\mu(B(c(\fp), 4D^{s(\fp)}))} \left(\frac{\rho(y,y')}{D^{s(\fp)}}\right)^{\tau'} \int_{E(\fp)}|g(x)| \, \mathrm{d}\mu(x)\,.
+    $$
+
+    If $y,y' \notin B(c(\fp), 5D^{s(\fp)})$, then $T_\fp^*g(y) = T_\fp^*g(y') = 0$, by \eqref{eq Ks supp t} and the triangle inequality. Then \eqref{T*Hölder2} holds.
+
+    Finally, if $y \in B(c(\fp), 5D^{s(\fp)})$ and $y' \notin B(c(\fp), 5D^{s(\fp)})$, then
+    $$
+        |e(Q(\fu)(y)) T_{\fp}^* g(y) - e(Q(\fu)(y')) T_{\fp}^* g(y')| = |T_\fp^* g(y)|
+    $$
+    $$
+        \le \int_{E(\fp)} |K_{s(\fp)}(x,y)| |g(x)| \, \mathrm{d}\mu(x)\,.
+    $$
+    By the same argument used to prove \eqref{eq Ks aux}, this is bounded by
+    $$
+        \le 2^{102a^3} \int_{E(\fp)} \frac{1}{\mu(B(x, D^s))} \psi(D^{-s} \rho(x,y)) |g(x)| \, \mathrm{d}\mu(x)\,.
+    $$
+    It follows from the definition of $\psi$ that
+    $$
+        \psi(x) \le \max\{0, (2  - 4x)^{\tau'}\}\,.
+    $$
+    Thus, by the triangle inequality and \eqref{eq vol sp cube} for all $x\in E(\fp)$
+    $$
+        \psi(D^{-s(\fp)}\rho(x,y)) \le \max\{0, (2  - 4D^{-s(\fp)}\rho(y, c(\fp)) + 4 D^{-s(\fp)}\rho(x, c(\fp)))^{\tau'}\}
+    $$
+    $$
+        \le \max\{0, (18 - 4 D^{-s(\fp)} \rho(y, c(\fp)))^{\tau'}\}
+    $$
+    $$
+        \le \max\{0, 4 D^{-s(\fp)}\rho(y,y') - 2\}^{\tau'} \le 4 (D^{-s(\fp)}\rho(y,y'))^{\tau'}\,.
+    $$
+    Further, we obtain from the doubling property \eqref{doublingx} and \eqref{eq vol sp cube} that
+    $$
+        \mu(B(x, D^s)) \ge 2^{-2a} \mu(c(\fp), 4D^s)\,.
+    $$
+    Plugging this in and using $a \ge 4$, we get
+    $$
+        |T_\fp^* g(y)| \le   \frac{2^{103a^3}}{\mu(B(c(\fp), 4D^{s(\fp)}))} \left(\frac{\rho(y,y')}{D^{s(\fp)}}\right)^{\tau'} \int_{E(\fp)} |g(x)| \, \mathrm{d}\mu(y)\,,
+    $$
+    which completes the proof of the lemma.
+\end{proof}
+
+
+\begin{lemma}
+    \label{SeparatedTrees}
+    There exists $\delta > 0$ \lars{explicit} such that, for any $\fu_1 \ne \fu_2 \in \fU$, we have
+    \begin{equation}
+        \label{eq lhs sep tree}
+        \left| \int_X \sum_{\fp_1 \in \fT(\fu_1)} \sum_{\fp_2 \in \fT(\fu_2)} T^*_{\fp_1}g_1 \overline{T^*_{\fp_2}g_2 }\,\mathrm{d}\mu \right|
+    \end{equation}
+    \begin{equation}
+        \label{eq rhs sep tree}
+        \le ... 2^{-Zn\delta} \prod_{j =1}^2 \| S_{2, \fu_j} g_j\|_{L^2(\sc(\fu_1) \cap \sc(\fu_2))}\,.
+    \end{equation}
+\end{lemma}
+
+\begin{proof}
+    By Lemma \ref{lem Tp support adjoint} \eqref{eq rhs sep tree} is
+    $$
+        \left| \int_{\sc(\fu_1) \cap \sc(\fu_2)} T^*_{\mathfrak{T}_1}g_1 \overline{T^*_{\mathfrak{T}_2}g_2 }\,\mathrm{d}\mu \right|\,.
+    $$
+    Applying Cauchy-Schwarz and using that for $j=1,2$
+    $$
+            \left|\sum_{\fp \in \fT(\fu_j)} T_{\fp}^*g_j \right| \le S_{2,\fu_j} g_j
+    $$
+    gives estimate \eqref{eq rhs sep tree} for $Zn\delta \le ...$.
+
+    Thus we may, and will, assume from now on that $Zn\delta > ...$. 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 $\sc(\fu_1) \subset \sc(\fu_2)$.
+
+    Define
+    $$
+        \eta = ...
+    $$
+    and set
+    $$
+        \mathfrak{S} = \{\fp \in \fT(\fu_1) \cup \fT(\fu_2) \ : \ d_{\fp}(Q(\fu_1), Q(\fu_2)) \ge 2^{Zn\delta(1 - \eta)}\,.
+    $$
+    Lemma \ref{SeparatedTrees} follows then by combining
+    \begin{lemma}
+        \label{lem big sep tree}
+        We have
+        \begin{equation}
+            \label{eq lhs big sep tree}
+            \left| \int_X \sum_{\fp_1 \in \fT(\fu_1)} \sum_{\fp_2 \in \fT(\fu_2) \cap \mathfrak{S}} T^*_{\fp_1}g_1 \overline{T^*_{\fp_2}g_2 }\,\mathrm{d}\mu \right|
+        \end{equation}
+        \begin{equation}
+            \label{eq rhs big sep tree}
+            \le ... 2^{-Zn\delta} \prod_{j =1}^2 \| S_{2, \fu_j} g_j\|_{L^2(\sc(\fu_1) \cap \sc(\fu_2))}\,.
+        \end{equation}
+    \end{lemma}
+    and
+    \begin{lemma}
+        \label{lem small sep tree}
+        We have
+        \begin{equation}
+            \label{eq lhs small sep tree}
+            \left| \int_X \sum_{\fp_1 \in \fT(\fu_1)} \sum_{\fp_2 \in \fT(\fu_2) \setminus \mathfrak{S}} T^*_{\fp_1}g_1 \overline{T^*_{\fp_2}g_2 }\,\mathrm{d}\mu \right|
+        \end{equation}
+        \begin{equation}
+            \label{eq rhs small sep tree}
+            \le ... 2^{-Zn\delta} \prod_{j =1}^2 \| S_{2, \fu_j} g_j\|_{L^2(\sc(\fu_1) \cap \sc(\fu_2))}\,.
+        \end{equation}
+    \end{lemma}
+\end{proof}
+
+\begin{proof}[Proof of Lemma \ref{lem big sep tree}]
+    Define
+    $$
+        \mathcal{J}' = \{J \in \mathcal{J}(\mathfrak{S}) \ : \ J \subset \sc(\fu_1)\}\,.
+    $$
+    \lars{
+    Define
+    $$
+        \mathfrak{S} := \{\fp \in \mathfrak{T}_1 \cup \mathfrak{T}_2 \, : \, d_{\sc(\fp)}(Q_{\mathfrak{T}_1}, Q_{\mathfrak{T}_2}) \geq \Delta^{1- \eta}\}\,,
+    $$
+    where $0 < \eta < 1$ will be fixed later.
+    If $\fp \in \mathfrak{T}_1 \cup \mathfrak{T}_2$ and $\sc(\fp) \subset I_0$, then by $\Delta$-separation and the triangle inequality we have
+    $$
+        d_{\sc(\fp)}(Q_{\mathfrak{T}_1}, Q_{\mathfrak{T}_2}) \geq (\Delta - 1) - 4\,.
+    $$
+    If $I_0 \subset \sc(\fp)$, then the same holds, since the norms $\|\cdot\|_I$ are increasing in $I$.
+    Thus, for $\Delta$ sufficiently large, we have that $\sc(\fp) \cap I_0 = \emptyset$ for all $\fp \in (\mathfrak{T}_1 \cup \mathfrak{T}_2) \setminus \mathfrak{S}$. In particular, we have $\mathfrak{T}_1 \subset \mathfrak{S}$.
+
+    Define $\mathcal{J} := \{J \in \mathcal{J}(\mathfrak{S}) \, : \, J \subset I_0\}$, this is a partition of $I_0$. Let $\mathbf{1}_{I_0} := \sum_{J \in \mathcal{J}} \chi_J$ be a partition of unity adapted to $\mathcal{J}$, satisfying proprties (1) and (2) in Lemma \ref{lem pou}.
+
+    We denote $\Delta_J = d_J(Q_{\mathfrak{T}_1}, Q_{\mathfrak{T}_2})$ for $J \in \mathcal{J}$. If $J \in\mathcal{J}$, then there exists $\fp \in \mathfrak{S}$ such that $\sc(\fp) \subset B(x_J, 100 D^{s(J) + 2})$. Thus, by assumption 1),
+    \begin{multline}
+        \label{DeltaJEqn}
+        \Delta_J = d_J(Q_{\mathfrak{T}_1}, Q_{\mathfrak{T}_2}) \gtrsim d_{B(x_J, 100 D^{s(J) + 2})}(Q_{\mathfrak{T}_1}, Q_{\mathfrak{T}_2})\\ \geq d_{\sc(\fp)}(Q_{\mathfrak{T}_1}, Q_{\mathfrak{T}_2}) \geq \Delta^{1-\eta}
+    \end{multline}
+    for all $J \in \mathcal{J}$.}
+    \begin{lemma}
+        \label{lem J' partition}
+        We have that
+        $$
+            \sc(\fu_1) = \dot{\bigcup_{J \in \mathcal{J}'}} J\,.
+        $$
+    \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} $\sc(\fu_1) \subsetneq J$. \lars{Are trees defined to be nonempty? they should be} Pick $\fp \in \fT(\fu) \subset \mathfrak{S}$. Then $\sc(\fp) \subsetneq J$. This contradicts the definition of $\mathcal{J}(\mathfrak{S})$.
+    \end{proof}
+
+    \begin{lemma}
+        \label{lem partition of unity}
+        There exists a family of functions $\chi_J$, $J \in \mathcal{J}'$ such that $$
+            \mathbf{1}_{I_0} = \sum_{J \in \mathcal{J}'} \chi_J\,,
+        $$
+        and for all $J \in \mathcal{J}'$ and all $y,y' \in J$
+        $$
+            0 \leq \chi_J(y) \leq \mathbf{1}_{B(c(J), 8 D^{s(J)})}(y)\,,
+        $$
+        $$
+            |\chi_J(x) - \chi_J(y)| \le ...  \frac{\rho(x,y)}{D^{s(J)}}\,.
+        $$
+    \end{lemma}
+    \begin{proof}
+        For each cube $J \in \mathcal{J}$ let
+        $$
+            \tilde\chi_J(x) = \max\{0, 8 - D^{-s(J)} \rho(x, x_J)\}\,.
+        $$
+        Then the functions
+        \[
+            \chi_J(x) = \mathbf{1}_{I_0} \frac{\tilde \chi_J(x)}{\sum_{J' \in \mathcal{J}} \tilde \chi_{J'}(x)}=: \mathbf{1}_{I_0} \frac{\tilde \chi_J(x)}{a(x)}
+        \]
+        form a partition of unity on $I_0$ and satisfy \ref{SepTreeClaim1}.
+
+        Note that, if $B(x_J, 2D^{s(J)}) \cap B(x_{\tilde J}, 2D^{s(\tilde J)}) \neq \emptyset$, then $|s(J) - s(\tilde J)| \leq 2$. Indeed, if $s(\tilde J) > s(J) + 2$ then, since $J \in \mathcal{J}$, there exists a tile $\fp \in \mathfrak{S}$ with
+        $$
+        I_{\fp} \subset 100 D \hat{J} = B(x_{\hat J}, 100 D^{s(J) + 2}) \subset B(x_{\tilde J}, 100 D^{s(\tilde J) + 1})\,.
+        $$
+        This contradicts $\tilde J \in \mathcal{J}$. Thus, there are only boundedly many cubes $\tilde J \in \mathcal{J}$ for which the supports of $\tilde \chi_{\tilde J}$ and $\tilde \chi_J$ intersect, and each satisfies $|\tilde \chi_{\tilde J}(x) - \tilde \chi_{\tilde J}(y)| \lesssim  \frac{\rho(x,y)}{D^{s(J)}}$ for all $x, y \in I_0$. Hence, for all $x, y \in B(x_J, 2 D^{s(J)}) \cap I_0$ we have
+        $$
+            |a(x) - a(y)| \lesssim \frac{\rho(x,y)}{D^{s(J)}}\,.
+        $$
+        Since the cubes $J \in \mathcal{J}$ cover $I_0$, it also holds for all $x \in I_0$ that $a(x) \geq 1$. Hence, \ref{SepTreeClaim2} holds. \qedsymbol
+    \end{proof}
+
+
+
+    \begin{lemma}
+        \label{lem sep tree aux 2}
+        We have \lars{make explicit}
+        \begin{equation}
+            \fp \in \mathfrak{T}_2 \setminus \mathfrak{S}, J \in \mathcal{J}, \sc(\fp)^* \cap J \neq \emptyset \implies s(\fp) \leq s(J) + s_0\,.
+        \end{equation}
+    \end{lemma}
+
+    \begin{proof}
+        Assume that $s(\fp) > s(J) + s_0$. Since $J \in \mathcal{J}$, there exists some $\fp' \in \mathfrak{S}$ such that $\sc(\fp') \subset B(x_{\hat J}, 100 D^{s(J) + 2})$. On the other hand, since $\sc(\fp)^* \cap J \neq \emptyset$, it holds that $B(x_J, D^{s(J) + s_0}) \subset B(x_{\sc(\fp)}, 10 D^{s(\fp)})$. Hence
+        \begin{align*}
+            \Delta^{1 - \eta} \leq d_{\sc(\fp')} ( Q_{\mathfrak{T}_1}, Q_{\mathfrak{T}_2}) \lesssim D^{s_0/\log A} d_{\sc(\fp)} ( Q_{\mathfrak{T}_1}, Q_{\mathfrak{T}_2}) \leq D^{s_0/\log A} \Delta^{1 - \eta}\,.
+        \end{align*}
+        If $s_0$ is chosen sufficiently large, this is a contradiction, showing our claim.
+    \end{proof}
+
+    \begin{lemma}
+        \label{lem sep tree aux 3}
+        If $\fp \in \fT(\fu_2) \setminus \mathfrak{S}$ then for all $J \in \mathcal{J}'$
+        $$
+            \sup_{B(c(J), 2^{-3} D^{s(J)}} |T_{\mathfrak{T}_2 \setminus\mathfrak{S}}^* g| \le ... \inf_J Mg
+        $$
+    \end{lemma}
+
+    \begin{proof}
+        Recall that if $\fp\in \mathfrak{T}_2 \setminus \mathfrak{S}$, then $\sc(\fp) \cap I_0 = \emptyset$. Thus if $\sc(\fp)^* \cap B(x_J, 2^{-3} D^{s(J)}) \neq \emptyset$ for some $J \in \mathcal{J}$, then $s(\fp) \geq s(J) - s_1$, for an absolute constant $s_1$. From this and the claim, it follows that
+        \begin{align*}
+            \sup_{B(x_J, 2^{-3}D^{s(J)})} |T_{\mathfrak{T}_2 \setminus\mathfrak{S}}^* g|
+            &\leq \sup_{B(x_J, 2^{-3}D^{s(J)})}\sum_{\fp \in \mathfrak{T}\setminus \mathfrak{S}, \sc(\fp)^* \cap J \neq \emptyset} |T_{\fp}^*g|\\
+            &\leq \sup_{B(x_J, 2^{-3} D^{s(J)})} \sum_{s = s(J)-s_1}^{s(J) + s_0} \sum_{\fp \in \fP, s(\fp) = s} |T_{\fp}^* g|\\
+            &\lesssim \sum_{s = s(J)-s_1}^{s(J) + s_0} \sup_{y \in J}  \sum_{\fp \in \fP, s(\fp) = s, y \in \sc(\fp)^*} \frac{1}{\mu(\sc(\fp))} \int |g\mathbf{1}_{E(\fp)}| \\
+            &\leq \sum_{s = s(J)-s_1}^{s(J) + s_0} \sup_{y \in J}  \sum_{I \in \mathcal{D}_s, y \in I^*} \frac{1}{\mu(I)} \int_I |g| \\
+            &\lesssim \inf_{J} Mg\,.
+        \end{align*}
+        Here the last inequality holds because there are only boundedly many intervals $I$ in the sum, and for each we have $I \subset B(y, CD^{s(J)})$ and $\mu(I) \sim \mu(B(y, CD^{s(J)}))$.
+    \end{proof}
+
+
+    \begin{lemma}
+        \label{lem sep tree aux 1}
+        Assume that a cube $J \in \mathcal{D}$ satisfies
+        \begin{equation}
+            \label{SepTreesProp}
+            \fp \in \mathfrak{T}, \sc(\fp)^* \cap B(x_J, 2D^{s(J)}) \neq \emptyset \implies s(\fp) \geq s(J)\,.
+        \end{equation}
+        Then
+        \begin{align}
+            \label{TreeUB}
+            \sup_{B(x_J, 2D^{s(J)})} |T_{\mathfrak{T}}^*g| \leq \inf_{B(x_J, 2^{-3}D^{s(J)})} |T^*_{\mathfrak{T}} g| + C \inf_{J} Mg\,.
+        \end{align}
+    \end{lemma}
+
+    \begin{proof}
+        For all $y, y' \in B(x_J, 2D^{s(J)})$, it holds that:
+        $$
+            |e(-Q_{\mathfrak{T}}(y)) T_{\mathfrak{T}}^* g(y) - e(-Q_{\mathfrak{T}}(y')) T_{\mathfrak{T}}^* g(y')|
+        $$
+        $$
+            \leq \sum_{\fp \in \mathfrak{T}, \sc(\fp)^* \cap B(x_J, 2D^{s(J)}) \neq \emptyset} |e(-Q_{\mathfrak{T}}(y)) T_{\fp}^* g(y) - e(-Q_{\mathfrak{T}}(y')) T_{\fp}^*g(y')|
+        $$
+        $$
+            \lesssim \rho(y, y')^{\tau'} \sum_{s \geq s(J)} D^{-s\tau' }\sum_{\fp \in \mathfrak{T}, s(\fp) = s, \sc(\fp)^* \cap B(x_J, 2D^{s(J)}) \neq \emptyset}  \frac{1}{\mu(\sc(\fp))}\int_{E(\fp)} |g(x)|
+        $$
+        $$
+            \lesssim \rho(y, y')^{\tau'} \sum_{s \geq s(J)} D^{-s\tau'}\inf_{J} Mg
+        $$
+        \begin{equation}
+            \lesssim \left(\frac{\rho(y, y')}{D^{s(J)}}\right)^{\tau'}\inf_{J} Mg\,.  \label{TreeHölder}
+        \end{equation}
+        Then \eqref{TreeHölder} implies that
+        \begin{align}
+            \label{TreeUB}
+            \sup_{B(x_J, 2D^{s(J)})} |T_{\mathfrak{T}}^*g| \leq \inf_{B(x_J, 2^{-3}D^{s(J)})} |T^*_{\mathfrak{T}} g| + C \inf_{J} Mg\,.
+        \end{align}
+    \end{proof}
+
+
+    \begin{lemma}
+        We have for all $J \in \mathcal{J}'$
+        $$
+            \sup_{B(x_J, 2 D^{s(J)})} |T^*_{\mathfrak{T}_2 \cap \mathfrak{S}} g| \le ... \inf_{J} |T^*_{\mathfrak{T}_2} g| + ... \inf_{J} Mg\,.
+        $$
+    \end{lemma}
+
+    \begin{proof}
+        Note that with $\mathfrak{T} = \mathfrak{T}_2 \cap \mathfrak{S}$, condition \eqref{SepTreesProp} is satisfied for all $J \in \mathcal{J}$. Indeed, if $\sc(\fp)^* \cap B(x_J, 2D^{s(J)}) \neq \emptyset $ and $s(\fp) < s(J)$, then $\sc(\fp) \subset B(x_J, 100 D^{s(J) + 1})$, contradicting the definition of $\mathcal{J}$. Thus we can apply \eqref{TreeUB} and obtain for all $J \in \mathcal{J}$:
+        \begin{align}
+            &\quad\sup_{B(x_J, 2 D^{s(J)})} |T^*_{\mathfrak{T}_2 \cap \mathfrak{S}} g|\nonumber \\
+            &\leq \inf_{B(x_J, 2^{-3} D^{s(J)})} |T_{\mathfrak{T}_2 \cap \mathfrak{S}}^* g| + C \inf_{J} Mg\nonumber\\
+            &\leq \inf_{B(x_J, 2^{-3}D^{s(J)})} |T^*_{\mathfrak{T}_2} g| + \sup_{B(x_J, 2^{-3}  D^{s(J)})}|T^*_{\mathfrak{T}_2\setminus \mathfrak{S}} g| + C \inf_{J} Mg\nonumber\\
+            &\leq \inf_{J} |T^*_{\mathfrak{T}_2} g| + C \inf_{J} Mg\,.\label{TreeUB2}
+        \end{align}
+    \end{proof}
+
+    Define
+    \[
+        h_J(y) := \chi_J(y)\cdot(e(-Q_{\mathfrak{T}_1}(y)) T_{\mathfrak{T}_1}^* g_1(y)) \cdot \overline{(e(-Q_{\mathfrak{T}_2}(y)) T_{\mathfrak{T}_2 \cap \mathfrak{S}}^* g_2(y))}\,.
+    \]
+    \begin{lemma}
+        We have for all $J \in \mathcal{J}'$ and al $y, y' \in X$
+        \begin{equation}
+            \label{hHölder}
+            |h_J(y)- h_J(y')| \lesssim  \left(\frac{\rho(y,y')}{D^{s(J)}}\right)^{\tau'} \prod_{j = 1,2} (\inf_{J} |T_{\mathfrak{T}_j}^* g_j| + \inf_J Mg_j)\,.
+        \end{equation}
+    \end{lemma}
+
+    \begin{proof}
+        Since $\mathfrak{T}_1$ is contained in $\mathfrak{S}$, condition \eqref{SepTreesProp} holds also with $\mathfrak{T} = \mathfrak{T}_1$, for all $J \in \mathcal{J}$. Hence \eqref{TreeHölder} and \eqref{TreeUB} hold for $\mathfrak{T} = \mathfrak{T}_1$. Together with \eqref{TreeHölder} for $\mathfrak{T} = \mathfrak{S} \cap \mathfrak{T}_2$ and \eqref{TreeUB2}, as well as the Lipschitz estimate for $\chi_J$, this yields for $y, y' \in B(x_J, 2 D^{s(J)})$:
+
+        Note that $T_{\mathfrak{T}_i}^* g_i$ is continuous and bounded on $X$ for $i=1,2$, and that $T_{\mathfrak{T}_1}^* g_1$ vanishes outside of $I_0$. By definition, $\chi_J$ is continuous and bounded on $I_0$ and vanishes outside of $B(x_J, 2D^{s(J)})$. Thus the functions $h_J$ are continuous on all of $X$ and supported in the ball $B(x_J, 2D^{s(J)})$. Thus \eqref{hHölder} holds for all $y, y' \in X$.
+    \end{proof}
+
+
+    Using that $\mathcal{Q}$ is $\tau$-cancellative, Lemma \ref{lem vdc regularity} and \eqref{DeltaJEqn}, we obtain with $\gamma_{\tau}$ as in Lemma \ref{lem vdc regularity}
+    \begin{align*}
+        \left| \int_{X} T_{\mathfrak{T}_1}^* g_1 \overline{T_{\mathfrak{T}_2 \cap \mathfrak{S}}^* g_2 }\right|
+        &\leq \sum_{J \in \mathcal{J}} \left|\int_{X} e(Q(y)) h_J(y) \, \mathrm{d}\mu(y) \right|\\
+        &\leq \sum_{J \in \mathcal{J}} \Delta_J^{-\gamma_{\tau} } \mu(J) \prod_{j = 1,2}(\inf_{J} |T_{\mathfrak{T}_j}^* g_j| + \inf_J Mg_j)\\
+        &\lesssim  \Delta^{-(1 - \eta)\gamma_{\tau} } \prod_{j = 1,2}\| |T_{\mathfrak{T}_j}^* g_j| + Mg_j\|_{L^2(I_0)}\,.
+    \end{align*}
+    This completes the proof of Lemma \ref{lem big sep tree}.
+\end{proof}
+
+\begin{proof}[Proof of Lemma \ref{lem small sep tree}]
+    Define $\mathcal{J}' = \{J \in \mathcal{J}(\mathfrak{T}_1) \, : \, J \subset I_0\}$, this is a partition of $I_0$.
+
+    \begin{lemma}
+        \label{lem sep aux 3}
+        We claim that for some $s_\Delta$ with $D^{s_\Delta}\sim \Delta^{\eta/(2 \log A)}$, we have
+        \begin{equation}
+            \label{SepTreeClaim}
+            \fp \in \mathfrak{T}_2 \setminus \mathfrak{S}, J \in \mathcal{J}', \sc(\fp)^* \cap J \neq \emptyset \implies s(\fp) \leq s(J) - s_\Delta\,.
+        \end{equation}
+    \end{lemma}
+
+    \begin{proof}
+        Indeed, if $s(\fp) > s(J) - s_\Delta$ and $\sc(\fp)^* \cap J \neq \emptyset$, then $\rho(x_{\sc(\fp)}, x_J) \leq 15 D^{s(\fp) + s_\Delta}$. Since $J \in \mathcal{J}'$, there exists some $\fp' \in \mathfrak{T}_1$ with
+        $$
+            \sc(\fp') \subset B(x_J, 100 D^{s(J) + 2}) \subset B(x_{\sc(\fp)}, 200 D^{s(\fp) + s_\Delta + 2}) =: B\,.
+        $$
+        It follows that
+        \begin{multline*}
+            \Delta \lesssim d_{\sc(\fp')}(Q_{\mathfrak{T}_1}, Q_{\mathfrak{T}_2})
+            \leq d_{B}(Q_{\mathfrak{T}_1}, Q_{\mathfrak{T}_2})\\
+            \lesssim D^{2 s_\Delta  \log A } d_{\sc(\fp)}(Q_{\mathfrak{T}_1}, Q_{\mathfrak{T}_2})\lesssim D^{ 2 s_\Delta \log A} \Delta^{1 - \eta}.
+        \end{multline*}
+        If the constants in $D^{s_\Delta}\sim \Delta^{\eta/(2 \log A)}$ are chosen correctly, this is a contradiction, thus our claim holds.
+    \end{proof}
+
+    Note that $\mathfrak{T}_2 \setminus \mathfrak{S}$ is still a tree, since $\mathfrak{S}$ is an up set.
+    Hence, Lemma \ref{TreeEstimate} implies that
+    \begin{align*}
+        \quad\left| \int T_{\mathfrak{T}_1}^* g_1 \overline{T_{\mathfrak{T}_2 \setminus \mathfrak{S}}^* g_2} \right|\lesssim \|g_1\mathbf{1}_{I_0}\|_2 \|P_{\mathcal{J}'}|T_{\mathfrak{T}_2 \setminus \mathfrak{S}}^* g_2|\|_2\,.
+    \end{align*}
+    By \eqref{SepTreeClaim}, we have
+    \begin{align*}
+        \|P_{\mathcal{J}'}|T_{\mathfrak{T}_2 \setminus \mathfrak{S}}^* g_2|\|_2
+        &\leq \sum_{s \geq s_\Delta} \bigg(\sum_{J \in \mathcal{J}'} \frac{1}{\mu(J)} \bigg|\int_J\sum_{\substack{\fp \in \mathfrak{T}_2 \setminus \mathfrak{S}:\\ s(\fp) = s(J) - s,\, \sc(\fp)^* \cap J \neq \emptyset}} T_{\fp}^* g_2\bigg|^2\bigg)^{1/2}\\
+        &\leq \sum_{s \geq s_\Delta} \bigg(\sum_{J \in \mathcal{J}'} \frac{1}{\mu(J)} \bigg|\int_J Mg_2 \sum_{\substack{I \in \mathcal{D}_{s(J) - s}:\\ I\cap I_0 = \emptyset,\, I^* \cap J \neq \emptyset}}  \mathbf{1}_{I^*}\bigg|^2\bigg)^{1/2}\\
+        &\leq \sum_{s \geq s_\Delta} \bigg(\sum_{J \in \mathcal{J}'}  \int_J (Mg_2)^2  \frac{1}{\mu(J)}\int_J\bigg(\sum_{\substack{I \in \mathcal{D}_{s(J) - s}:\\ I\cap I_0 = \emptyset,\, I^* \cap J \neq \emptyset}}  \mathbf{1}_{I^*}\bigg)^2 \bigg)^{1/2}\,.
+    \end{align*}
+    The sets
+    $$
+        \{I^* \cap J \, : \, I \in \mathcal{D}_{s(J) - s},\, I\cap I_0 = \emptyset,\, I^* \cap J \neq \emptyset\}
+    $$
+    have bounded overlap and are contained in
+    $$
+        \{x \in J \, : \, \rho(x, X \setminus J) \leq 5 D^{s(J)-s}\}\,.
+    $$
+    Thus, by the small boundary property, the inner integral is bounded by $\mu(J) D^{-\kappa s}$. We further estimate:
+    \begin{align*}
+            \sum_{s \geq s_\Delta} D^{-\kappa s/2} \|\mathbf{1}_{I_0}Mg_2\|_2 \lesssim D^{-\kappa s_\Delta/2} \|\mathbf{1}_{I_0}Mg_2\|_2 \lesssim \Delta^{-\kappa \eta/(4 \log A)} \|\mathbf{1}_{I_0}Mg_2\|_2\,.
+    \end{align*}
+    This completes the estimate of the contribution of $\mathfrak{T}_2 \setminus \mathfrak{S}$. Finally, we can optimize $\eta$ to obtain the claimed estimate with $\delta = \frac{4 \gamma_{\tau} \log A }{\kappa + 4 \gamma_{\tau} \log A }$ .
+\end{proof}
+
+
+
+
+
+
+\section{Forests}
+An $n$-row is an $n$-forest $(\fU, \fT)$, i.e. satisfying \eqref{forest1} -\eqref{forest6}, such that in addition the sets $\sc(\fu), \fu \in \fU$ are pairwise disjoint.
+
+\begin{lemma}
+    \label{lem row decomposition}
+    Let $(\fU, \fT)$ be an $n$-forest. Then there exists a decomposition
+    $$
+        \fU = \dot{\bigcup_{1 \le j \le 2^n}} \fU_j
+    $$
+    such that for all $j = 1, \dotsc, 2^n$ the pair $(\fU_j, \fT|_{\fU_j})$ is an $n$-row.
+\end{lemma}
+
+\begin{proof}
+    Define recursively $\fU_j$ to be a maximal disjoint set of tiles $\fu$ in
+    $$
+        \fU \setminus \bigcup_{j' < j} \fU_{j'}
+    $$
+    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 $2^n - j$ 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 at most one less than the number of $\fu \in \fU \setminus \bigcup_{j' \le j-1} \fU_{j'}$ with $x \in \sc(\fu)$ or zero.
+\end{proof}
+
+We pick a decomposition of the forest $(\fU, \fT)$ into $2^n$ $n$-rows $(\fU_j, \fT_j) := (\fU_j, \fT|_{\fU_j})$ as in Lemma \ref{lem row decomposition}.
+
+\begin{lemma}
+    \label{lem row bound}
+    For each $1 \le j \le 2^n$, we have
+    $$
+        \left\| \sum_{\fu \in \fU_j} \sum_{\fp \in \fT(\fu)}  T_{\fp}^* g\right\|_2 \le ...  2^{-n/2} \|g\|_2
+    $$
+    and
+    $$
+        \left\| \sum_{\fu \in \fU_j} \sum_{\fp \in \fT(\fu)} \mathbf{1}_F T_{\fp}^* g\right\|_2 \le ... \dens_2(\fT(\fu))^{1/2} 2^{-n/2} \|g\|_2\,.
+    $$
+\end{lemma}
+
+\begin{proof}
+    By Corollary \ref{cor tree est}, we have for each $\fu \in \fU$ and all $f$  that
+    \begin{equation}
+        \label{eq explicit tree bound 1}
+        \left\|\sum_{\fp \in \fT(\fu)} T_{\fp} f \right\|_{2}  \le ... 2^{-n/2}  \|f\|_2\,
+    \end{equation}
+    and
+    \begin{equation}
+        \label{eq explicit tree bound 2}
+        \left\|\sum_{\fp \in \fT(\fu)} T_{\fp} \mathbf{1}_F f \right\|_{2} \le ... 2^{-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 $g$ \lars{conditions}:
+    $$
+        \left\|\mathbf{1}_F \sum_{\fu \in \fU_j} \sum_{\fp \in \fT(\fu)} T_{\fp}^* g\right\|_2^2 = \left\|\mathbf{1}_F \sum_{\fu \in \fU_j} \sum_{\fp \in \fT(\fu)} \mathbf{1}_{\sc(\fu)} T_{\fp}^* \mathbf{1}_{\sc(\fu)} g\right\|_2^2
+    $$
+    $$
+        = \sum_{\fu \in \fU_j} \int_{\sc(\fu)} \left| \mathbf{1}_F \sum_{\fp \in \fT(\fu)} T_{\fp}^* \mathbf{1}_{\sc(\fu)} g\right|^2 \, \mathrm{d}\mu
+        \le \sum_{\fu \in \fU_j} \left\|\sum_{\fp \in \fT(\fu)} \mathbf{1}_F T_{\fp}^* \mathbf{1}_{\sc(\fu)} g\right\|_2^2\,.
+    $$
+    Applying the estimate for the adjoint operator following from equation \eqref{eq explicit tree bound 2}, we obtain
+    $$
+        \le ...2^{-n/2} \dens_2(\fT(\fu))^{1/2} \sum_{\fu \in \fU_j} \left\| \mathbf{1}_{\sc(\fu)} g\right\|_2^2\,.
+    $$
+    Again by disjointness of the cubes $\sc(\fu)$, this is estimated by
+    $$
+        ...2^{-n/2} \dens_2(\fT(\fu))^{1/2} \|g\|_2^2\,.
+    $$
+    The proof of the first estimate from \eqref{eq explicit tree bound 1} is exactly the same.
+\end{proof}
+
+\begin{lemma}
+    \label{lem sep row bound}
+    For all $1 \le j < j' \le 2^n$ and for all $g_1, g_2 \in L^2(X)$ \lars{write correct conditions} it holds that
+    $$
+        \left| \int \sum_{\fu \in \fU_j} \sum_{\fu' \in \fU_{j'}} \sum_{\fp \in \fT_j(\fu)} \sum_{\fp' \in \fT_{j'}(\fu')} T^*_{\fp} g_1 \overline{T^*_{\fp'} g_2} \, \mathrm{d}\mu \right| \le ... \Delta^{-\delta} \|g_1\|_2 \|g_2\|_2\,,
+    $$
+    where $\delta$ is as in Lemma \ref{SeparatedTrees}.
+\end{lemma}
+
+\begin{proof}
+    To save some space we will write for subsets $\fC \subset \fP$
+    $$
+        T_{\fC}^* = \sum_{\fp \in \fC} T_{\fp}^*\,.
+    $$
+    We have by Lemma \ref{lem Tp support adjoint} 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 \lars{Give the operators on right a name, for now $S$}
+    $$
+        ... \Delta^{-\delta} \sum_{\fu \in \fU_j} \sum_{\fu' \in \fU_{j'}} \|S g_1\|_{L^2(\sc(\fu')\cap \sc(\fu)} \|Sg_2\|_{L^2(\sc(\fu')\cap\sc(\fu))}\,.
+    $$
+    We apply the Cauchy-Schwarz inequality in the form $\sum_{i \in M} a_i b_i \le (\sum_{i \in M} a_i^2 )^{1/2}(\sum_{i \in M} b_i^2 )^{1/2}$ to the outer two sums:
+    $$
+        \le ... \Delta^{-\delta} \left(\sum_{\fu \in \fU_j} \sum_{\fu' \in \fU_{j'}} \|S g_1\|_{L^2(\sc(\fu')\cap \sc(\fu))}^2 \right)^{1/2} \left(\sum_{\fu \in \fU_j} \sum_{\fu' \in \fU_{j'}} \|S g_2\|_{L^2(\sc(\fu')\cap\sc(\fu))}^2 \right)^{1/2}\,.
+    $$
+    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
+    $$
+        \sum_{\fu \in \fU_j}\sum_{\fu' \in \fU_{j'}} \|S g_1\|_{L^2(\sc(\fu')\cap \sc(\fu))}^2
+        = \sum_{\fu \in \fU_j}\sum_{\fu' \in \fU_{j'}} \int_{\sc(\fu) \cap \sc(\fu')} |Sg_1(y)|^2 \, \mathrm{d}\mu(y)
+    $$
+    $$
+        \le \int_X |Sg_1(y)|^2 \, \mathrm{d}\mu(y) = \|Sg_1\|_2^2\,.
+    $$
+    Arguing similar for $g_2$, we can  continue our estimate with
+    $$
+        \le ... \Delta^{-\delta} \|Sg_1\|_2 \|Sg_2\|_2\,.
+    $$
+    By Lemma \ref{lem L2 Sg estimate}, the lemma follows.
+\end{proof}
+
+Define for $1 \le j \le 2^n$
+$$
+    E_j := \bigcup_{\fu \in \fU_j} \bigcup_{\fp \in \fT(\fu)} E(\fp)\,.
+$$
+
+\begin{lemma}
+    \label{lem disjoint support}
+    The sets $E_j$, $1 \le j \le 2^n$ are pairwise disjoint.
+\end{lemma}
+
+\begin{proof}
+    Suppose that $\fp \in \fT(\fu)$ and $\fp' \in \fT(\fu')$ with $\fu \ne \fu'$ and $x \in E(\fp) \cap E(\fp')$. Suppose without loss of generality that $s(\fp) \le s(\fp')$. Then $x \in \sc(\fp) \cap \sc(\fp') \subset \sc(\fu')$. By \eqref{dyadicproperty} it follows that $\sc(\fp) \subset \sc(\fu')$. By \eqref{forest5}, it follows that
+    $$
+        d_{\fp}(Q(\fp), Q(\fu')) > 2^{Z(n+1)}\,.
+    $$
+    By the triangle inequality. Lemma \ref{lem cube monotone} and \eqref{forest1} it follows that
+    \begin{align*}
+        d_{\fp}(Q(\fp), Q(\fp')) &\ge d_{\fp}(Q(\fp), Q(\fu')) - d_{\fp}(Q(\fp'), Q(\fu'))\\
+        &> 2^{Z(n+1)} - d_{\fp'}(Q(\fp'), Q(\fu'))\\
+        &\ge 2^{Z(n+1)} - 4\,.
+    \end{align*}
+    Since $Z \ge 3$ \lars{assumption on $Z$}, it follows that $Q(\fp') \notin B_{\fp}(Q(\fp), 1)$, so $\Omega(\fp') \not\subset \Omega(\fp)$ by \eqref{eq freq comp ball}. Hence, by \eqref{eq freq dyadic}, $\Omega(\fp) \cap \Omega(\fp') = \emptyset$. But if $x \in E(\fp) \cap E(\fp')$ then $\Theta(x) \in \Omega(\fp) \cap \Omega(\fp')$. This is a contradiction, the lemma follows.
+\end{proof}
+
+Now we prove Proposition \ref{forestprop}.
+
+\begin{proof}[Proof of Proposition \ref{forestprop}]
+    To save some space, we will write
+    $$
+        T_{\mathfrak{R}_j}^* = \sum_{\fu \in \fU_j} \sum_{\fp \in \fT(\fu)} T_{\fp}^*\,.
+    $$
+    By \eqref{eq adjoint}, we have for each $j$ and each $g$
+    $$
+        T_{\mathfrak{R}_j}^*g = \sum_{\fu \in \fU_j} \sum_{\fp \in \fT(\fu)} T_{\fp}^* g = \sum_{\fu \in \fU_j} \sum_{\fp \in \fT(\fu)} T_{\fp}^* \mathbf{1}_{E_j} g = T_{\mathfrak{R}_j}^* \mathbf{1}_{E_j} g\,.
+    $$
+    Hence, by Lemma \ref{lem row decomposition},
+    $$
+        \left\|\sum_{\fu \in \fU} \sum_{\fp \in \fT(\fu)} T^*_{\fp} g\right\|_2^2 = \left\|\sum_{j = 1}^{2^n} T^*_{\mathfrak{R}_{j}} g\right\|_2^2  =  \left\|\sum_{j=1}^{2^n} T^*_{\mathfrak{R}_{j}} \mathbf{1}_{E_j} g\right\|_2^2
+    $$
+    $$
+        = \int_X \left|\sum_{j=1}^{2^n} T^*_{\mathfrak{R}_{j}} \mathbf{1}_{E_j} g\right|^2 \, \mathrm{d}\mu
+    $$
+    $$
+        = \sum_{j=1}^{2^n} \int_X |T_{\mathfrak{R}_j}^* \mathbf{1}_{E_j} g|^2 + \sum_{j =1}^{2^n} \sum_{\substack{j' = 1\\j' \ne j}}^{2^n} \int_X \overline{ T_{\mathfrak{R}_j}^* \mathbf{1}_{E_j} g} T_{\mathfrak{R}_{j'}}^* \mathbf{1}_{E_{j'}} g \, \mathrm{d}\mu\,.
+    $$
+    We use Lemma \ref{lem row bound} to estimate each term in the first sum, and Lemma \ref{lem sep row bound} to bound each term in the second sum
+    $$
+        \le ...  2^{-n} \sum_{j = 1}^{2^n} \|\mathbf{1}_{E_j} g\|_2^2 + ... 2^{-Z\delta n}\sum_{j=1}^{2^n}\sum_{j' = 1}^{2^n} \|\mathbf{1}_{E_j} g\|_2 \|\mathbf{1}_{E_{j'}}g\|_2\,.
+    $$
+    By Cauchy-Schwarz in the second two sums, this is at most
+    $$
+        (... 2^{-n} + ... 2^{n}2^{-Z\delta n}) \sum_{j = 1}^n \|\mathbf{1}_{E_j} g\|_2^2\,,
+    $$
+    and by disjointness of the sets $E_j$ and the choice $Z = 2/\delta$, this is finally bounded by
+    $$
+        ... 2^{-n} \|g\|_2^2\,.
+    $$
+    Taking adjoints, it follows that for all $f$
+    \begin{equation}
+        \label{eq forest bound 1}
+        \left\|\sum_{\fu \in \fU} \sum_{\fp \in \fT(\fu)} T_{\fp} f\right\|_2 \le ... 2^{-n/2} \|f\|_2\,.
+    \end{equation}
+    On the other hand, we have by disjointness of the sets $E_j$
+    $$
+        \left\|\sum_{\fu \in \fU} \sum_{\fp \in \fT(\fu)} T_{\fp} f\right\|_2^2 =  \left\|\sum_{j=1}^{2^n} \mathbf{1}_{E_j} T_{\mathfrak{R}_{j}} f\right\|_2^2 = \sum_{j = 1}^{2^n} \|\mathbf{1}_{E_j} T_{\mathfrak{R}_{j}} f\|_2^2\,.
+    $$
+    If $f \le \mathbf{1}_F$ then we obtain from Lemma \ref{lem row bound}
+    \begin{equation}
+        \label{eq forest bound 2}
+        \le ... \dens_2(?) 2^{-n} \sum_{j = 1}^{2^n} \|f\|_2^2
+        = ... \dens_2(?) \|f_2\|^2\,.
+    \end{equation}
+    The Proposition follows by taking a geometric average of \eqref{eq forest bound 1} and \eqref{eq forest bound 2}.
+\end{proof}
+
+\chapter{Proof of P. \ref{lem vdc regularity}, the H\"older cancellative condition}
+\label{liphoel}
+
+\rs{Checked, see comments}
+We need the following auxiliary lemma.
+
+\begin{lemma}
+    \label{lem regularity aux}
+    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 $\|\varphi\|_{C^\tau}$. 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}
+        |\varphi(x) - \tilde \varphi(x)| \leq t^{\tau} \|\varphi\|_{C^\tau(B)}
+    \end{equation}and
+    \begin{equation}\label{eq secondt}
+        \|\tilde \varphi\|_{\Lip(B(z,2R))} \leq 2^{4a}t^{-1-a} \|\varphi\|_{C^{\tau}(B)}\, .
+    \end{equation}
+\end{lemma}
+
+
+\begin{proof}
+    %Let $s = R/t$.
+    Define for $x,y\in X$ the Lipschitz \rs{With respect to which ball?} and thus measurable function
+        \begin{equation}
+            L(x,y) := \max\{0, 1 - \frac{\rho(x,y)}{tR}\}\, .
+    \end{equation}
+We  have that $L(x,y)\neq 0$ implies
+\begin{equation}\label{eql01}
+    y\in B(x, tR)\, .
+\end{equation}
+We have for $y\in B(x, 2^{-1}tR)$ that
+\begin{equation}\label{eql30}
+            |L(x,y)|\ge 2^{-1}  \ .
+    \end{equation}
+Hence
+\begin{equation}
+        \int L(x,y) \, \mathrm{d}\mu(y)\ge  2^{-1}\mu(B(x, 2^{-1}Rt)\, .
+    \end{equation}
+    Let $n$ be the smallest integer so that $2^nt\ge 1$. Iterating $n+2$ times the doubling condition \eqref{doublingx}, we obtain
+        \begin{equation}\label{eql32}
+        \int L(x,y) \, \mathrm{d}\mu(y)\ge  2^{-1-a(n+2)}\mu(B(x, 2R))\, .
+    \end{equation}
+
+Now define
+    $$
+        \tilde \varphi(x) := \left(\int L(x,y) \, \mathrm{d}\mu(y)\right)^{-1}\int L(x,y) \varphi(y) \, \mathrm{d}\mu(y)\, .
+    $$
+Using that $\varphi$ is supported in $B(z,R)$ and
+\eqref{eql01}, we have that $\tilde{\varphi}$ is supported in $B(z,2R)$.
+
+We prove \eqref{eq firstt}.
+    For any $x\in X$, using
+    that $L$ is nonnegative,
+    \begin{equation}\label{eql1}
+    \left(\int L(x,y) \, \mathrm{d}\mu(y)\right)
+        |\varphi(x) - \tilde \varphi(x)|
+    \end{equation}
+    \begin{equation}\label{eql2}
+    = \left| \int L(x,y)(\varphi(x) - \varphi(y)) \, \mathrm{d}\mu(y)\right|\, .
+    \end{equation}
+Using \eqref{eql01}, we estimate the last display by
+    \begin{equation}\label{eql3}
+            \le   \int_{B(x, tR)} L(x,y)|\varphi(x) - \varphi(y)| \, \mathrm{d}\mu(y)\, .\end{equation}
+    Using the definition of $\|\varphi\|_{C^\tau(B)}$, we estimate the last display further by
+        \begin{equation}\label{eql4}
+            \le   \left(\int_{B(x, tR)} L(x,y)
+            \rho(x,y)^\tau \, \mathrm{d}\mu(y) \right)\|\varphi\|_{C^\tau(B)}R^{-\tau}\, .
+    \end{equation}
+    Using the condition on the domain of integration to estimate $\rho(x,y)$ by $tR$ and then expanding the domain by positivity of the integrand, we estimate this further by
+
+    \begin{equation}\label{eql5}
+            \le   \left(\int L(x,y) \, \mathrm{d}\mu(y)\right)
+            \|\varphi\|_{C^\tau(B)} t^{\tau} \, .
+    \end{equation}
+    Dividing the string of inequalities from \eqref{eql1} to
+\eqref{eql5} by the positive integral of $L$ proves \eqref{eq firstt} .
+
+
+We turn to \eqref{eq secondt}. For every $x\in X$, we have
+\begin{equation}
+    \left|\int L(x,y) \, \mathrm{d}\mu(y)\right||\tilde{\varphi}(x)|
+    =\left|\int L(x,y) {\varphi}(y)\, \mathrm{d}\mu(y)\right|
+\end{equation}
+    \begin{equation}
+    \le \left|\int L(x,y) \, \mathrm{d}\mu(y)\right| \sup_{x'\in X}
+    |{\varphi}(x')|\ .
+\end{equation}
+Dividing by the integral of $L$, we obtain
+\begin{equation}\label{eql42}
+    |\tilde{\varphi}(x)|\le \sup_{x'\in X}
+    |{\varphi}(x')|\le \|\varphi\|_{C^\tau}\ .
+\end{equation}
+
+If $\rho(x,x')\ge R$, then we have
+    \begin{equation}\label{eql52}
+    R\frac{|\tilde{\varphi}(x') - \tilde \varphi(x)|}{\rho(x,x')} \le
+    2\sup_{x''\in X} |\tilde{\varphi}(x'')|\le 2\|\varphi\|_{C^\tau}\, .
+\end{equation}
+
+Now assume  $\rho(x,x')< R$. For $y\in X$ we have by the triangle inequality and a two fold case distinction
+for the maximum in the definition of $L$,
+\begin{equation}\label{eql10}
+    |L(x,y) - L(x',y)| \le \frac{\rho(x,x')}{tR}.
+\end{equation}
+We compute with \eqref{eql10}, first adding and subtracting a term in the integral,
+    \begin{equation}\left(\int L(x,y) \, \mathrm{d}\mu(y)\right)
+        |\tilde{\varphi}(x') - \tilde \varphi(x)|=
+    \end{equation}
+    \begin{equation}\left|\int L(x,y) \tilde{\varphi}(x')
+    -L(x,y) \tilde{\varphi}(x)
+    +L(x',y) \tilde{\varphi}(x')-
+        L(x',y) \tilde{\varphi}(x')
+    \, \mathrm{d}\mu(y)\right|
+    \end{equation}
+    \begin{equation}\label{eql21}\le  \left| \int (L(x',y)-L(x,y)) \varphi(y) \, \mathrm{d}\mu(y)\right|
+    \end{equation}
+    \begin{equation}\label{eql22}+  \left| \int L(x,y)  \, \mathrm{d}\mu(y)-\int L(x',y)  \, \mathrm{d}\mu(y)\right||\tilde{\varphi}(x')|
+    \end{equation}
+\begin{equation}\label{eql23}\le 2  \int |L(x,y)  -L(x',y)|  \, \mathrm{d}\mu(y)
+\|\varphi\|_{C^\tau}\, ,
+    \end{equation}
+where in the last inequality we have used  \eqref{eql42}.
+Using further \eqref{eql10} and the support of $L$, we estimate the last display by
+\begin{equation}\label{eql224}\le 2  \frac{\rho(x,x')} {tR}\mu(B(x,tR)\cup B(x',tR))
+\|\varphi\|_{C^\tau}\, .
+    \end{equation}
+    Using $\rho(x,x')<R$, we estimate the last display by
+\begin{equation}\label{eql225}\le 2
+\frac{\rho(x,x')} {tR}
+\mu(B(x,2R))
+\|\varphi\|_{C^\tau}\, .
+    \end{equation}
+
+Dividing by the integral over $L$ and using \eqref{eql32}, we obtain
+\begin{equation}\label{eql226}
+    \frac {R |\tilde{\varphi}(x') - \tilde \varphi(x)|}{\rho(x,x')}
+    \le   2^{2+a(n+2)}t^{-1}\|\varphi\|_{C^\tau} \le
+    2^{2+3a} t^{-1-a} \|\varphi\|_{C^\tau}\, .
+\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}.
+\end{proof}
+
+
+We turn to the proof of Proposition \ref{lem vdc regularity}.
+Let $z\in X$ and $R>0$ and set $B=B(z,R)$. Let $\varphi$
+be given as in Proposition \ref{lem vdc regularity}.
+Set
+\begin{equation}\label{eql69}
+    t:=(1+d_B(\mfa,\mfb))^{-\tau/(2+a)}
+\end{equation}
+and define $\tilde{\varphi}$ as in Lemma \ref{lem regularity aux}. 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|
+    \end{equation}
+        \begin{equation}\label{eql61}
+    \le     \left|\int e(\mfa(x)-{\mfb(x)}) \tilde{\varphi} (x)\, \mathrm{d}\mu(x)\right|
+    \end{equation}
+            \begin{equation}\label{eql62}
+        +  \left|\int e(\mfa(x)-{\mfb(x)}) (\varphi (x)-\tilde{\varphi}(x))\, \mathrm{d}\mu(x)\right|
+    \end{equation}
+Using the $\tau$ cancellative condition \eqref{eq vdc cond}  of $\Mf$, the term \eqref{eql61} is bounded above by
+    \begin{equation}\label{eql63}
+        2^a \mu(B(z,2R)) \|\tilde{\varphi}\|_{\Lip(B(z,2R))} (1 + d_{B(z,2R)}(\mfa,\mfb))^{-\tau}  \, .
+    \end{equation}
+
+
+
+Using   the doubling condition \eqref{doublingx},
+the inequality \eqref{eq secondt}, and the estimate
+$d_B\le d_{B(z,2R)}$ from the definition,
+we estimate \eqref{eql63} from above by
+\begin{equation}\label{eql64}
+        2^{3+4a}t^{-1-a} \mu(B) \|{\varphi}\|_{C^\tau(B)}
+        (1 + d_{B}(\mfa,\mfb))^{-\tau}  \, .
+    \end{equation}
+    \rs{Should the constant above be $2^{1+5a}$?}
+The term \eqref{eql62} we estimate using
+\eqref{eq firstt} and that
+$\mfa$ and $\mfb$ are real and thus $e(\mfa)$ and
+$e(\mfb)$ bounded in absolute value by $1$.
+We obtain for \eqref{eql62} with \eqref{doublingx}
+the upper bound
+    \begin{equation}\label{eql65}
+        \mu(B(z,2R)) t^{\tau} \|\varphi\|_{C^\tau(B)}
+        \le  2^a \mu(B) t^{\tau} \|\varphi\|_{C^\tau(B)}
+        \,.
+    \end{equation}
+Using the definition \eqref{eql69} of $t$ and adding
+\eqref{eql64} and \eqref{eql65} estimates
+\eqref{eql60} from above by
+\begin{equation}
+        2^{3+4a} \mu(B) \|{\varphi}\|_{C^\tau(B)}
+        (1 + d_{B}(\mfa,\mfb))^{-\tau/(2+a)}
+        \end{equation}
+\begin{equation} +
+        2^a \mu(B) \|{\varphi}\|_{C^\tau(B)}
+        (1 + d_{B}(\mfa,\mfb))^{-\tau^2/(2+a)}\, .
+    \end{equation}
+\begin{equation}\label{eql66}
+        \le  2^{4+4a} \mu(B) \|{\varphi}\|_{C^\tau(B)}
+        (1 + d_{B}(\mfa,\mfb))^{-\tau^2/(2+a)}  \, ,
+    \end{equation}
+where we used $\tau\le 1$. \rs{Should the constant above be $2^{1+5a}$?}
+This completes the proof of Proposition \ref{lem vdc regularity}.
+
+%\begin{proof}
+%    Define
+%    $$
+%        L(x,y) = \max\{0, 1 - \rho(x,y)/(Cs)\}\,.
+%    $$
+%    For $C \leq 1$ sufficiently small, this satisfies the third property. Note further that for each $x$,
+%    $$
+%        \int L(x,y) \, \mathrm{d}\mu(y) = \int_{0}^1 \mu(B(x,Cst)) \, \mathrm{d}t\,.
+%    $$
+%    This implies that
+%    $$
+%        \mu(B(x,s)) \geq \mu(B(x, Cs)) \geq \int L(x,y) \, \mathrm{d}\mu(y) \geq \frac{1}{2A} \mu(B(x,Cs)) \gtrsim \mu(B(x,s))\,.
+%    $$
+%    We define
+%    $$
+%        K(x,y) = \left(\int L(x,y) \, \mathrm{d}\mu(y)\right)^{-1} L(x,y)\,.
+%    $$
+%    This also satisfies the condition 1) and the upper bound of condition 2). To check the Hölder estimate in condition 2), we fix $y$ and first assume that $x, x' \in B(y, 2As)$. Then, by condition 3) and the doubling property, we have that $\mu(B(x,s)) \sim \mu(B(x',s)) \sim \mu(B(y,s))$. This implies:
+%    \begin{align*}
+%        |K(x,y) - K(x',y)| &\leq L(x,y)  \left|\bigg(\int L(x,y) \, \mathrm{d}\mu(y)\bigg)^{-1} - \bigg(\int L(x',y) \, \mathrm{d}\mu(y)\bigg)^{-1}\right|\\
+%        & + \bigg(\int L(x',y) \, \mathrm{d}\mu(y)\bigg)^{-1} |L(x,y) - L(x',y)|\\
+%        &\lesssim \frac{1}{\mu(B(y,s))^{-2}}  \int |L(x,y) - L(x',y)| \, \mathrm{d}\mu(y)  + \frac{\rho(x,x')}{\mu(B(x,s))s}\\
+%        &\lesssim \frac{\rho(x,x')}{\mu(B(y,s))s}\,.
+%    \end{align*}
+%    Now assume that $x \in B(y,s)$ and $x' \notin B(y, 2As)$. Then $K(x',y) = 0$ and $\rho(x,x') \geq s$, so the required estimate also holds. In the final case $x,x' \notin B(y, 2As)$ the left hand side vanishes, so the inequality also holds.
+%\end{proof}
+
+
+
+\chapter{Proof of P. \ref{prop hlm}, Besicovitch covering and Hardy Littlewood}
+\label{sec hlm}
+\rs{Checked, see comments}
+
+\lars{I also wrote this in section 2, but should this not be called `Vitali covering and Hardy Littlewood'?}
+\rs{Agree that it should be called Vitali covering instead}
+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}
+Let $1\le q$. \rs{$1\le q< \infty?$} Then for any measurable function $h:X\to [0,\infty)$ on the measure space $X$
+relative to the measure $\mu$
+we have
+\begin{equation}\label{eq layercake}
+    \|h\|_q^q=q\int_0^\infty \lambda^{q-1}\mu(\{x: h(x)\ge \lambda\})\, d\lambda\, .
+\end{equation}
+\end{lemma}
+\begin{proof}
+    The left-hand side of \eqref{eq hlm} is by definition
+\begin{equation}
+    \int_X h(x)^q \, d\mu(x)\, .\end{equation}
+    Writing $h(x)$ as an elementary integral in $\lambda$ and then using Fubini, we write for the last display
+    \begin{equation}
+    =\int_X \int _0^{h(x)}
+    q \lambda^{q-1} d\lambda\, d\mu(x)
+\end{equation}
+\begin{equation}
+    =q\int _0^{\infty}
+    \lambda^{q-1} \mu(\{x: h(x)\ge \lambda\}) d\lambda\, .
+\end{equation}
+This proves the lemma.
+\end{proof}
+
+We turn to the proof of Proposition \ref{prop hlm}. \rs{Change $q'$ notation}\rs{Look at other comments in the proposition statement}
+Let the collection $\mathcal{B}$ be given.
+We first show \eqref{eq besico}.
+
+
+
+We recursively choose a finite sequence $B_i\in \mathcal{B}$
+for $i\ge 0$ as follows. Assume $B_{i'}$
+is already chosen for $0\le i'<i$.
+If there exists a ball $B_{i}\in \mathcal{B}$ so that $B_{i}$
+is disjoint from all $B_{i'}$
+with $0\le i'<i$, then choose
+such a ball $B_i=B(x_i,r_i)$ with maximal $r_i$.
+
+If there is no such ball, stop the selection and set
+$i'':=i$.
+
+By  disjointness of the chosen balls, we have
+\begin{equation}
+\sum_{0\le i<i''}\int_{B_i} h(x)\, d\mu(x) \le \int_X  h(x)\, d\mu(x)\, .
+\end{equation}
+By \eqref{eq ball assumption}, we conclude
+\begin{equation}\label{eqbes1}
+\lambda \sum_{0\le i<i''}\mu(B_i)
+\le  \int_X  h(x)\, d\mu(x)\, .
+\end{equation}
+Let $x\in \bigcup \mathcal{B}$.
+\rs{Should we write let $x\in \bigcup_{B\in \mathcal{B}} B$?}
+Choose a ball $B'=B(x',r')\in \mathcal{B}$
+such that $x\in B'$.
+If $B'$ is one of the selected balls, then
+\begin{equation}\label{3rone}
+    x\in \bigcup _{0\le i< i''}B_i\subset \bigcup _{0\le i< i''}B(x_i,3r_i)\, .
+\end{equation}
+If $B'$ is not one of the selected balls, then as it is not selected at time $i''$ there is a selected ball $B_i$  with
+$B'\cap B_i\neq \emptyset$.
+Choose such $B_i$ with minimal index $i$. As $B'$ is therefore disjoint from all
+balls $B_{i'}$ with $i'<i$ and
+as it was not selected in place of $B_i$, we have $r_i\ge r'$.
+
+Using a point $y$ in the intersection of $B_i$ and $B'$,
+we conclude by the triangle inequality
+\begin{equation}
+    \rho(x_i,x')\le \rho(x_i,y)+\rho(x',y)\le r_i+r'\le 2r_i \, .
+\end{equation}
+By the triangle inequality again, we further conclude
+\begin{equation}
+    \rho(x_i,x)\le \rho(x_i,x')+\rho(x',x)\le 2r_i+r'\le 3r_i \, .
+\end{equation}
+It follows that
+\begin{equation}\label{3rtwo}
+    x\in  \bigcup _{0\le i< i''}B(x_i,3r_i)\, .
+\end{equation}
+With \eqref{3rone} and \eqref{3rtwo}, we conclude
+\begin{equation}
+\bigcup \mathcal{B}\subset
+\bigcup _{0\le i< i''}B(x_i,3r_i)\, .
+\end{equation}
+With the doubling property
+\eqref{doublingx} applied twice, we conclude
+\begin{equation}\label{eqbes2}
+    \mu(\bigcup{\mathcal{B}})
+    \le \sum _{0\le i< i''}\mu (B(x_i,3r_i))
+    \le 2^{2a}\sum _{0\le i< i''}\mu (B_i)\, .
+\end{equation}
+With \eqref{eqbes1} and \eqref{eqbes2} we conclude
+\eqref{eq besico}.
+
+
+We turn to the proof of \eqref{eq hlm}. We first consider the case $q'=1$ and recall $M_{\mathcal{B}}=M_{\mathcal{B},1}$.
+We write for the $q$-th power of left-hand side of \eqref{eq hlm}
+with Lemma \ref{lem layercake}
+and a change of variables
+\begin{equation}
+    \|M_{\mathcal{B}}h(x)\|_q^q
+    =q\int _0^{\infty}
+    \lambda^{q-1} \mu(\{x: M_{\mathcal{B}}(x)\ge \lambda\}) d\lambda\,
+\end{equation}
+\begin{equation} \label{eqbesi11}
+    =2^q q\int _0^{\infty}
+    \lambda^{q-1} \mu(\{x: M_{\mathcal{B}}(x)\ge 2\lambda\}) d\lambda\, .
+\end{equation}
+Fix $\lambda\ge 0$ and let $x\in X$ satisfy $M_{\mathcal{B}}(x)\ge 2\lambda$. By definition of $M_{\mathcal{B}}$, there is a ball
+$B'\in \mathcal{B}$ such that
+$x\in B'$ and
+\begin{equation}\label{eqbesi10}
+\int_{B'} h(y)\, d\mu(y)\ge 2\lambda \mu({B'})   \, .
+\end{equation}
+Define
+$h_\lambda(y):=0$ if $|h(y)|<\lambda$ and $h_\lambda(y):=h(y)$ if $|h(y)|\ge \lambda$.
+Then with \eqref{eqbesi10}
+\begin{equation}
+\int_{B'} h_\lambda (y)\, d\mu(y)
+=\int_{B'} h (y)\, d\mu(y)-
+\int_{B'} (h-h_\lambda) (y) d\mu(y)\,
+\end{equation}
+\begin{equation}
+\ge 2\lambda \mu({B'})-
+\int_{B'} (h-h_\lambda) (y) d\mu(y)\, .
+\end{equation}
+As $(h-h_\lambda)(y)\le \lambda$
+by definition, we can estimate the last display by
+\begin{equation}
+\ge 2\lambda \mu({B'})-
+\int_{B'} \lambda  (y) d\mu(y)
+=\lambda \mu({B'})\, .
+\end{equation}
+Hence $x$ is contained in
+$\bigcup(\mathcal{B}_\lambda)$, \rs{Union notation}
+where $\mathcal{B}_\lambda$
+is the collection of balls $B''$ in $\mathcal{B}$ such that
+\begin{equation}
+    \int_{B''} h_\lambda (y)\, d\mu(y)\ge \lambda \mu(B'')\, .
+\end{equation}
+We have thus seen
+\begin{equation}
+    \{x: M_{\mathcal{B}}(x)\ge 2\lambda\}\subset
+    \bigcup \mathcal{B}_\lambda
+\, .
+\end{equation}
+Applying  \eqref{eq besico} to the  collection $\mathcal{B}_\lambda$
+gives
+\begin{equation}
+    \lambda \mu(\{x: M_{\mathcal{B}}(x)\ge 2\lambda\})\le
+    2^{2a}
+    \int h_\lambda (x)\, dx\, .
+\end{equation}
+With Lemma \ref{lem layercake},
+\begin{equation}\label{eqbesi12}
+    \lambda \mu(\{x: M_{\mathcal{B}}(x)\ge 2\lambda\})\le
+    2^{2a}
+    \int_0^\infty  \mu (\{x: |h_\lambda (x)|\ge \lambda'\})\, d\lambda'\, .
+\end{equation}
+By definition of $h_\lambda$, making a case distinction between $\lambda\ge \lambda'$ and $\lambda <\lambda'$, we see that
+\begin{equation}\label{eqbesi13}
+    \mu (\{x: |h_\lambda (x)|\ge \lambda'\})
+    \le
+    \mu (\{x: |h (x)|\ge \max(\lambda,\lambda')\})\, .
+\end{equation}
+We obtain with \eqref{eqbesi11},
+\eqref{eqbesi12}, and \eqref{eqbesi13}
+\begin{equation}
+    \|M_{\mathcal{B}}h(x)\|_q^q
+    \le 2^{q+2a} q
+    \int_0^\infty \lambda^{q-2}
+    \int_0^\infty
+    \mu (\{x: |h (x)|\ge \max(\lambda,\lambda')\})
+    \, d\lambda'd\lambda\, .
+\end{equation}
+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}
+\begin{equation}
+    \int_0^\infty \lambda^{q-2}
+    \int_0^\lambda
+    \mu (\{x: |h (x)|\ge \lambda\})
+    \, d\lambda'd\lambda
+\end{equation}
+\begin{equation}
+    =\int_0^\infty \lambda^{q-1}
+        \mu (\{x: |h (x)|\ge \lambda\})
+d\lambda.
+\end{equation}
+\begin{equation}\label{eqbesi14}
+    =q^{-1} \|h\|_q^q\, .
+\end{equation}
+We have for $\lambda< \lambda'$
+with Fubini and Lemma \ref{lem layercake}
+\begin{equation}
+    \int_0^\infty \lambda^{q-2}
+    \int_\lambda^\infty
+    \mu (\{x: |h (x)|\ge \lambda'\})
+    \, d\lambda'd\lambda.
+\end{equation}
+\begin{equation}
+    =\int_0^\infty \int_0^{\lambda'}\lambda^{q-2}
+        \mu (\{x: |h (x)|\ge \lambda'\})
+d\lambda d\lambda'.
+\end{equation}
+\begin{equation}
+    =(q-1)^{-1}\int_0^\infty (\lambda')^{q-1}
+        \mu (\{x: |h (x)|\ge \lambda'\})
+d\lambda'.
+\end{equation}
+\begin{equation}\label{eqbesi15}
+    =(q-1)^{-1} q^{-1}\|h\|_q^q\, .
+\end{equation}
+Adding the two estimates
+\eqref{eqbesi14} and \eqref{eqbesi15} gives
+\begin{equation}
+    \|M_{\mathcal{B}}h(x)\|_q^q
+    \le 2^{q+2a} (1+(q-1)^{-1})\|h\|_q^q
+    = 2^{q+2a} q(q-1)^{-1}\|h\|_q^q
+    \, .
+    \end{equation}
+With $a\ge 1$ and $q>1$, taking the $q$-th root, we obtain \eqref{eq hlm}.
+We turn to the case of general
+$1\le q'<q$.
+We have
+\begin{equation}
+    M_{\mathcal{B},q'}h=(M_{\mathcal{B}} (|h|^{q'}))^{\frac 1{q'}}\, .
+\end{equation}
+Applying the special case of \eqref{eq hlm} for $M_{\mathcal{B}}$ gives
+\begin{equation}
+    \|M_{\mathcal{B},q'}h\|_q=
+    \|M_{\mathcal{B}} (|h|^{q'})\|_{q/q'}^{\frac 1{q'}}
+\end{equation}
+\begin{equation}
+    \le 2^{2a} (q/q') (q/q'-1)^{-1}
+    \|(|h|^{q'})\|_{q/q'}^{\frac 1{q'}}
+    =2^{2a} q(q-q')^{-1}\|h\|_q\, .
+\end{equation}
+This proves \eqref{eq hlm} in general.
+and completes the proof of Propsoition \ref{prop hlm}.
+
+\chapter{Proof of Theorem \ref{classical}}
+
+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}.
+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{thm main 1}. Two assumptions in Theorem \ref{thm main 1} require more work, this is prepared in Lemmas \ref{lem hilbert} and \ref{lem vdc} and proved in Subsections \ref{10hilbert}
+and \ref{10vandercorput}. Subsections \ref{10piecewise},
+\ref{10hilbert}, \ref{10vandercorput}, and \ref{10difference}
+are mutually independent.
+
+
+\section{The classical Carleson theorem}
+\label{10classical}
+
+Let a uniformly continuous $2\pi$ periodic function $f:\R\to \mathbb{C}$ be given. Let $\epsilon>0$ be given.
+
+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\, .
+\end{equation}
+Let $K$ be the Gaussian bracket of $2\pi/\delta$, that is the unique integer with
+\begin{equation}
+    K-1\le  \frac{2\pi}{\delta} < K\, .
+\end{equation}
+and note that $K>2$ by assumption on $\delta$.
+For each $x\in \R$, let $k(x)$ be the Gaussian bracket of $Kx/2\pi$, that is the unique integer such that
+\begin{equation}\label{definekx}
+k(x)\le \frac{Kx}{2\pi}<k(x)+1\, .
+\end{equation}
+Define
+\begin{equation}\label{def fzero}
+f_0(x):=f\left(\frac{2\pi k(x)}{K}\right)\, .
+\end{equation}
+\begin{lemma}
+The function $f_0$ is measurable.
+The function $f_0$ is $2\pi $- periodic.
+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\, ,
+\end{equation}
+\begin{equation}\label{eq ffzero}
+|f_0(x)|\le  1\, .
+\end{equation}
+\end{lemma}
+
+\begin{proof}
+Let $F$ be any set in $\mathbb{C}$. We show that
+$f_0^{-1}(F)$ is measurable. As
+$Z:=\{2\pi k/K, k\in \mathbb{Z}\}$ is countable, it suffcies to show
+that $f_0^{-1}(F)\setminus Z$ is measurable. It then
+suffices to show that $f_0^{-1}(F)\setminus Z$ is open.
+Let $x\in f_0^{-1}(F)\setminus Z$. There is a $k$
+such that $x\in [2\pi k/K, 2\pi (k+1))$. As $x\not\in Z$,
+we have that $x\in (2\pi k/K, 2\pi (k+1))$
+but $f_0$ is constant on $(2\pi k/K, 2\pi (k+1))$,
+hence $(2\pi k/K, 2\pi (k+1))$ is a subset of
+$f_0^{-1}(F)\setminus Z$. This proves that
+$f_0^{-1}(F)\setminus Z$ is open and completes the proof of measurability of $f_0$.
+
+
+
+
+To see that $f_0$ is $2\pi$ periodic, we observe by
+adding $2\pi$ to \eqref{definekx} that
+\begin{equation}
+    k(x+2\pi)=k(x)+K
+\end{equation}
+and hence by Definition \eqref{def fzero}
+\begin{equation}
+    f_0(x+2\pi)=f(\frac {2\pi k(x+2\pi)}K)=f(\frac{2\pi (k(x)+K)}{K})
+        =f(\frac{2\pi k(x)}K+2\pi)\ ,
+\end{equation}
+which by $2\pi$-periodicity of $f$ is equal to
+\begin{equation}
+    =f(\frac{2\pi k(x)}K)=f_0(x)\, .
+\end{equation}
+    This proves that $f_0$ is $2\pi$- periodic
+
+
+
+
+To see \eqref{eq ffzero}, observe that by \eqref{definekx} and 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\, .
+\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 $1$.
+This completes the proof of the lemma.
+\end{proof}
+
+Define the set $E_1$ to be
+\begin{equation}
+E_1=\bigcup_{k=0}^{K}[\frac{2\pi}K (k-\frac {\epsilon}{16\pi }),
+\frac{2\pi}K (k-\frac {\epsilon}{16\pi})]\, .
+\end{equation}
+Then we have for the Lebesgue measure of $E_1$, using $K>1$,
+\begin{equation}
+    |E_1|\le \sum_{k=0}^{K}\frac {\epsilon}{4K}= \frac{\epsilon}{4}\left(1+\frac 1K\right)\le \frac \epsilon 2\, \, .
+\end{equation}
+
+We prove in Subsection \ref{10piecewise}
+\begin{lemma}
+\label{lem piece}
+    For all $N>\frac {2^{25} K^2}{\epsilon ^3}$ and
+    \begin{equation}
+        x\in [0,2\pi)\setminus E_1
+    \end{equation}
+we have
+    \begin{equation}
+        |S_N f_0 (x)- f_0(x)|\le \frac \epsilon 4\, .
+    \end{equation}
+\end{lemma}
+
+We prove in Subsection \ref{10difference}
+\begin{lemma}\label{lem diff}
+    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
+    \end{equation}
+    we have
+    \begin{equation}
+        \sup_{N\ge 0} |S_Nf(x)-S_Nf_0(x)|
+        \le \frac \epsilon 4
+    \end{equation}
+\end{lemma}
+
+Define
+\begin{equation}
+    E:=E_1\cup E_2\, .
+\end{equation}
+Then
+\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}.
+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}
+    |f(x)-S_Nf(x)|
+    \end{equation}
+    \begin{equation}
+    \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
+\begin{equation}
+    \le 2^{-2^{50}} \epsilon +\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}.
+
+The proof of Lemma \ref{lem diff} will essentially be
+an application of Lemma \ref{lem rcarleson}
+the following variant of Theorem \ref{thm main 1}.
+Define for $0<r<\frac \pi2$ the function $k:\R\to \R$
+by $k(0)=0$ and for $x\neq 0$
+\begin{equation}\label{eq hilker}
+k(x)=\frac {\max (1-|x|, 0)}{1-e^{ix}}\, .
+\end{equation}
+The following Lemma is proved in Subsection \ref{10rcarleson}.
+\begin{lemma}\label{lem rcarleson}
+    Let $F,G$ be Borel subsets of $\R$ with finite measure. Let $f$ be a bounded measurable function on $\R$ with $|f|\le 1_F$. Then
+\begin{equation}
+    |\int _G Tf(x) \, dx| \le 2^{2^{40}}\|f\|_2 |G|^{\frac 12} \, ,
+\end{equation}
+where
+\begin{equation}
+    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|\, .
+\end{equation}
+\end{lemma}
+
+
+The following Lemma is proved in Subsection \ref{10hilbert}.
+\begin{lemma}\label{lem hilbert}
+    For every $0<r<\pi$ and every  bounded measurable function $g$ with bounded support we have
+\begin{equation}
+    \|T_rg\|_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|\, .
+\end{equation}
+\end{lemma}
+The following Lemma is proved in Subsection \ref{10vandercorput}.
+\begin{lemma}\label{lem vdc}
+    Let $0\le a<b\le 2\pi$. 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\, .
+    \end{equation}
+    Then for any $0\le a<b\le 2\pi$ we have
+    \begin{equation}
+        \int _{a}^{b} g(x) e^{-inx}\, dx\le 2^{10} |b-a|\|g\|_{Lip(a,b)}(1+|n||b-a|)^{-1}\, .
+    \end{equation}
+
+\end{lemma}
+
+
+
+
+
+
+We close this section with three Lemmas that are used
+across the following subsections.
+
+\begin{lemma}\label{lem expintegral}
+Let $n\in \mathbb{Z}$ with   $n\neq 0$, then
+\begin{equation}
+\int_0^{2\pi} e^{inx}\, dx=0
+\end{equation}
+\end{lemma}
+\begin{proof}
+We have
+\begin{equation}
+\int_0^{2\pi} e^{inx}\, dx=\left[ \frac 1{in}e^{inx}\right]_0^{2\pi}=\frac  1{in}(e^{2\pi i n}-e^{2\pi i 0})=\frac 1{in}(1-1)=0\, .
+\end{equation}
+
+\end{proof}
+
+\begin{lemma}\label{dirichlet}
+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
+\end{equation}
+where $K_N$ is the $2\pi$ periodic continuous function of
+$\R$ given by
+\begin{equation}\label{eqksumexp}
+\sum_{n=-N}^N e^{in x'}\, .
+\end{equation}
+We have for $e^{ix'}\neq 1$ that
+\begin{equation}\label{eqksumhil}
+    K_N(x')=\frac{e^{iNx'}}{1-e^{-ix'}}
+        +\frac {e^{-iNx'}}{1-e^{ix'}} \, .
+\end{equation}
+
+
+\end{lemma}
+
+
+\begin{proof}
+We have by definitions and interchanging sum and integral
+    \begin{equation*}
+        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}
+    f(x) e^{in(x-y)}\, dy
+    \end{equation*}
+    \begin{equation}\label{eq expsum}
+        \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.
+By a telescoping sum, we have for every $x'\in \R$
+\begin{equation}
+    \left( e^{\frac 12 ix'}-e^{-\frac 12 ix'}\right)  \sum_{n=-N}^N e^{inx'}= e^{(N+\frac 12) ix'}-e^{-(N+\frac 12) ix'}\, .
+\end{equation}
+
+If $e^{ix'}\neq 1$, the first factor on the left-hand side is not $0$ and we may divide by this factor to obtain
+\begin{equation}
+        \sum_{n=-N}^N e^{inx'}= \frac{e^{i(N+\frac 1 2)x'}}{e^{\frac 12 ix'}-e^{-\frac 12ix'}}
+        -\frac{e^{-i(N+\frac 1 2)x'}}{e^{\frac 12 ix'}-e^{-\frac 12ix'}}
+        =\frac{e^{iNx'}}{1-e^{-ix'}}
+        +\frac {e^{-iNx'}}{1-e^{ix'}}\, .
+\end{equation}
+This  proves the second part of the lemma.
+\end{proof}
+
+The following lemma will be proved in Subsection \ref{10projection}.
+
+\begin{lemma}\label{lem l2sn}
+    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.
+    \end{equation}
+\end{lemma}
+
+
+
+
+
+\section{Piecewise constant functions.}
+\label{10piecewise}
+
+
+
+
+
+
+
+
+
+We first compute the partial Fourier sumes for constant functions.
+
+
+\begin{lemma}\label{constant}
+If $f$ satisfies $f(x)=f(0)$ for all $x\in \R$, then for all $N\ge 0$ we have $S_Nf=f$.
+    \end{lemma}
+\begin{proof}
+    We compute with Lemma \ref{lem expintegral} for $n\in \mathbb{Z}$ with $n\neq 0$,
+    \begin{equation}
+        \widehat{f}_n=\frac 1{2\pi}\int_0^{2\pi}f(y)e^{-iny}\, dy=\frac {f(0)}{2\pi}\int_0^{2\pi}e^{-iny}\, dy=0\, .
+    \end{equation}
+hence for $N\ge 0$
+\begin{equation}
+    S_nf(x)=\sum_{n=-N}^N \widehat{f}_n e^{inx}=\widehat{f_0}=\frac{1}{2\pi}\int_0^{2\pi}f(y)\, dy
+    =\frac{f(0)}{2\pi}\int_0^{2\pi}\, dy=f(x)\, .
+\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}
+    \end{proof}
+
+\begin{lemma}\label{lem diri}
+Let $\eta>0$. Let
+\begin{equation}
+    -2\pi +\eta \le a<b \le 2\pi-\eta
+\end{equation}
+and assume
+\begin{equation}\label{eq abdelta}
+    [a,b]\cap [-\eta,\eta]=\emptyset \, .
+\end{equation}
+Then
+    \begin{equation}\label{dirichletint}
+    \left|\int_a^b
+\frac {e^{iNx}} {1-e^{-ix}}
+dx\right| \le \frac{64}N(\frac{|b-a|}{\eta^2}+\frac 2{\eta}).
+\end{equation}
+\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}
+    and thus the integrand is continuous.
+    We have with partial integration
+\begin{equation*}
+    \int_a^b
+\frac {e^{-iNx}} {1-e^{ix}}
+dx
+\end{equation*}
+\begin{equation}\label{eq part int}
+=\frac 1{-iN}\int_a^b
+{e^{-iNx}}\frac {ie^{ix}} {(1-e^{ix})^2}\, dx
++\left[ \frac 1{-iN}
+\frac {e^{-iNx}} {1-e^{ix}}\right]_a^b\, .
+\end{equation}
+
+
+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},
+\begin{equation*}
+\left|   \frac 1 {-iN}\int_a^b
+{e^{-iNx}}\frac{ie^{ix}}
+{(1-e^{ix})^2}\,
+d\psi\right|
+\end{equation*}
+\begin{equation}\label{firstpi}
+    \le \frac 1N \int_a^b|1-e^{-ix}|^{-2}\le \frac {|b-a|}N\sup_{x\in [a,b]}|1-e^{-ix}|^{-2}
+\end{equation}
+\begin{equation}
+    \le  \frac {64|b-a|}{\delta ^2N}
+\end{equation}
+We have
+for the second summand in \eqref{eq part int}
+with Lemma \ref{expbound} again
+
+
+\begin{equation*}
+    \left|\left[ \frac 1{-iN}
+\frac {e^{-iNx}} {1-e^{ix}}\right]_a^b\right|
+\end{equation*}
+\begin{equation}\label{secondpi}
+    \le \frac 1N (|1-e^{ia}|^{-1}+|1-e^{ib}|^{-1})
+\le \frac {16}{\delta N}\, .
+\end{equation}
+Using the triangle inequality in \eqref{eq part int}
+and applying the estimates \eqref{firstpi} and \eqref{secondpi}  proves the lemma.
+\end{proof}
+
+
+
+Let $\mathcal{G}$ be the class of $2\pi$-periodic measurable functions $g$ which satisfy for all $x\in \R$
+\begin{equation}
+    g(\frac{2\pi k(x)}K)=g(x)
+\end{equation}
+and
+\begin{equation}
+    |g(x)|\le 2\, .
+\end{equation}
+
+
+
+\begin{lemma}\label{lem g with zero}
+Let $g\in \mathcal{G}$. Assume
+$x\in [0,2\pi)\setminus E_1 $ and $g(x)=0$.
+Then for all $N>\frac {2^25 K^2}{\epsilon ^3}$
+\begin{equation}\label{single char f}
+|S_Ng(x)|\le \frac \epsilon 4\, ,
+\end{equation}
+\end{lemma}
+\begin{proof}
+
+
+
+
+
+
+    With Lemma \ref{dirichlet}, breaking up the domain of integration into a
+    partition of subintervals,
+    \begin{equation*}
+        |S_N(g)|=|\int_{0}^{2\pi} g(y)K_N(x-y)\, dy|
+    \end{equation*}
+    \begin{equation}\label{eqcf1}
+    =  |\sum_{k=0}^{K-1}\int_{\frac{2\pi k}K}^{\frac{2\pi (k+1)}K} g(y)K_N(x-y)\, dy|\, .
+    \end{equation}
+Using $g\in \mathcal{G}$ and that $ k(y)=k$ for each
+\begin{equation}
+    y\in [\frac {2\pi k}K, \frac{2\pi (k+1)}K)\, ,
+\end{equation}
+and then applying the triangle inequality with the upper bound on $|g|$ and the identity $g(x)=g(2\pi k(x)/K)=0$,  we estimate \eqref{eqcf1} by
+        \begin{equation*}
+    =  |\sum_{k=0}^{K-1} g(\frac{2\pi k}K) \int_{\frac{2\pi k}K}^{\frac{2\pi (k+1)}K} K_N(x-y)\, dy|
+    \end{equation*}
+        \begin{equation}\label{eqcf3}
+    \le 2 \sum_{0\le k<K, k\neq k(x)} |\int_{\frac{2\pi k}K}^{\frac{2\pi (k+1)}K} K_N(x-y)\, dy|\, .
+    \end{equation}
+Doing a variable substitution, we obtain for \eqref{eqcf3}
+\begin{equation}
+    2 \sum_{0\le k<K, k\neq k(x)}
+    |\int_{x-2\pi (k+1)/K}^{x-2\pi k/K} K_N(y)\, dy|\, .
+\end{equation}
+Fix $0\le k<K$ with $k\neq k(x)$ and set
+\begin{equation}
+    a=x-2\pi (k+1)/K\, ,
+\end{equation}
+\begin{equation}
+    b=x-2\pi k/K\, .
+\end{equation}
+
+
+
+As $x\in [0,2\pi)\setminus E_1$, we have
+with $\eta=\epsilon/(8K)$
+\begin{equation}
+    a\ge x-2\pi\ge -2\pi+\eta\, ,
+\end{equation}
+\begin{equation}
+    b\le x \le 2\pi -\eta\, .
+\end{equation}
+If $b< 0$, then $2\pi k/K> x$ and as $x\not \in E_1$ also
+$2\pi k-\eta \ge x$. Hence $b \le -\eta $. If $b\ge 0$, then $2\pi k/K\le x$. As $k\neq k(x)$, we also have
+$2\pi (k+1)/K\le x$. As $x\not \in E_1$, we have
+$2\pi (k+1)/K+\eta \le x$. It follows that
+$0<a$. In both cases, we have seen
+\eqref{eq abdelta}.
+
+
+
+With Lemma \ref{dirichlet} and the triangle inequality,
+it follows that
+\begin{equation}
+    |\int_{x-2\pi (k+1)/K}^{x-2\pi k/K} K_N(y)\, dy|
+    \le |\int_{a}^{b}
+    \frac{e^{iNy}}{1-e^{-iy}}\, dy|+
+    |\int_{a}^{b}
+    \frac{e^{-iNy}}{1-e^{iy}}\, dy|
+\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
+\begin{equation}
+    \le 2|\int_{a}^{b}
+    \frac{e^{iNy}}{1-e^{-iy}}\, dy|
+    \le \frac{128}N(\frac{|b-a|}{\eta^2}+\frac 2{\eta})
+    \le \frac{2^{20}K}{N\epsilon^2 }\le 2^{-3}\epsilon K^{-1}\, .
+\end{equation}
+Inserting this in \eqref{eqcf3} and adding up proves
+the lemma.
+
+
+
+\end{proof}
+
+We now prove  Lemma \ref{lem piece}
+
+    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 $g=f-h$.
+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
+\begin{equation}
+    S_Nf_0(x)- f_0(x)=SN(g+h)(x)- S_Nh(x)=S_Ng(x)\, .
+\end{equation}
+Lemma \ref{lem piece}
+now follows by Lemma \ref{lem g with zero}.
+
+
+
+
+\section{The truncated Hilbert transform}
+\label{10hilbert}
+
+
+
+
+
+
+
+
+
+
+
+Let $M_n$ be the modulation operator
+acting on measurable $2\pi$ periodic functions
+defined by
+\begin{equation}
+    M_ng(x)=g(x) e^{inx}\, .
+\end{equation}
+Define the approximate Hilbert transform by
+\begin{equation}
+    L_N g=\frac 1N\sum_{n=0}^{N-1}
+        M_{-n-N} S_{N+n}M_{N+n}g\, .
+\end{equation}
+
+
+\begin{lemma}
+We have for every bounded measurable $2\pi$ periodic function $g$
+\begin{equation}\label{lnbound}
+    \|L_Ng\|_2\le \|g\|_2
+\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\, .
+    \end{equation}
+        We have by the triangle inequality, the square root of Identity \eqref{mnbound}, and Lemma \ref{lem l2sn}
+    \begin{equation*}
+        \|L_ng\|_2=\|\frac 1N\sum_{n=0}^{N-1}
+        M_{-n-N} S_{N+n}M_{N+n}g\|_2
+    \end{equation*}
+    \begin{equation*}
+        \le \frac 1N\sum_{n=0}^{N-1} \|
+        M_{-n-N} S_{N+n}M_{N+n}g\|_2
+            = \frac 1N\sum_{n=0}^{N-1} \|
+    S_{N+n}M_{N+n}g\|_2
+    \end{equation*}
+        \begin{equation}
+        \le \frac 1N\sum_{n=0}^{N-1} \|
+    M_{N+n}g\|_2  = \frac 1N\sum_{n=0}^{N-1} \|
+g\|_2 =\|g\|_2\, .
+    \end{equation}
+This proves \eqref{mnbound} and completes the proof of the lemma.
+\end{proof}
+
+\begin{lemma}\label{lem shift}
+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
+\end{equation}
+\end{lemma}
+\begin{proof}
+    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\, .
+\end{equation}
+We then have by breaking up the domain of integration
+and using \eqref{eqhil9}
+\begin{equation*}
+    \int_0^{2\pi} f(y)\, dy= \int_0^{2\pi -x}   f(y)\, dy+
+    \int_{2\pi -x}^{2\pi}   f(y)\, dy
+    \end{equation*}
+\begin{equation}
+= \int_0^{2\pi -x}   f(y)\, dy+
+    \int_{ -x}^{0}   f(y)\, dy
+    = \int_{-x}^{2\pi-x} f(y)\, dy\, .
+    \end{equation}
+This proves the first identity of the lemma. The second identity follows by substitution of $y$ by $y-x$.
+\end{proof}
+
+
+
+\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}
+        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
+    \, dy
+\end{equation*}
+\begin{equation}\label{eqhil4}
+=\int_0^{2\pi}f(y)^2\int_0^{2\pi}g(x) \, dx
+        dy
+=\|f\|_2^2\|g\|_1\, .
+\end{equation}
+
+    Let $h$ be the  nonnegative square root of $g$, then
+    $h$ is bounded and $2\pi$ periodic with $h^2=g$.
+    We estimate the square of the left-hand side of
+    \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
+    \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
+    = \|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
+\begin{equation}
+    |k(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}
+Let $g,f$ be bounded measurable $2\pi$-
+periodic functions. Let $0<r<\pi$.
+Assume we have for all $0\le x\le 2\pi$
+\begin{equation}
+    |g(x)|\le k_r(x)\, .
+\end{equation}
+Let
+\begin{equation}
+    h(x)= \int_0^{2\pi} f(y)g(x-y)\, dy \, .
+\end{equation}
+Then
+\begin{equation}
+    \|h\|_2\le 2^{5}\|f\|_2  \, .
+\end{equation}
+
+\end{lemma}
+
+\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
+\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
+\end{equation*}
+\begin{equation*}
+    \le 2\int_0^{r} \frac 1r \, dy+2\int_r^{\pi}1+\frac {8r}{x^2}\, dr
+\end{equation*}
+\begin{equation}
+    \le 2+2\pi + 2\left(\frac {8r}r-\frac {8r}{\pi}\right)
+    \le 2^{5}\, .
+\end{equation}
+    Together with Lemma \ref{young}, this proves the lemma.
+\end{proof}
+
+\begin{lemma}\label{lem dirichlet2}
+Let $0<r<\pi$. 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 $0\le x\le 2\pi$
+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
+\end{equation}
+and
+\begin{equation}\label{eqdifflhil}
+    |L'(x)-\frac{1_{[r, 2\pi -r]}(x)}{1-e^{-ix}}|\le 2^{10}k_r(x)\, .
+\end{equation}
+\end{lemma}
+
+
+\begin{proof}
+We have by definition and Lemma \ref{dirichlet}
+\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}
+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}
+
+
+For $e^{ix'}\neq 1$
+and may use the expression
+\label{eqksumhil} for $K_N$
+in Lemma \ref{dirichlet} 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}
+    L''(x):={L'}(x)  -\frac{1}{1-e^{-ix}}=\frac 1N \frac {e^{-i2Nx}}{1-e^{ix}}
+        \sum_{n=0}^{N-1}
+    {e^{-i2nx}}
+\end{equation}
+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\, .
+\end{equation}
+and hence
+\begin{equation}\label{eqhil12}
+    |L''(x)|\le
+        \frac 1N
+        \sum_{n=0}^{N-1}
+    1=1\le 2^{10}\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{expbound} 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^{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.
+\end{proof}
+
+
+\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{1_{[r, 2\pi -r]}(x)}{1-e^{-ix}}-\frac{1_{[r, 2\pi -r]}(x')}{1-e^{-ix'}}|\le 2^{10}k_r(x)\, .
+\end{equation}
+\end{lemma}
+
+\begin{proof}
+We have by Lemma \ref{expbound} for each $0<x\le \pi$
+\begin{equation}
+    |\frac{1_{[r, 2\pi -r]}(x)}{1-e^{-ix}}|\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 analoguous inequality holds for $-\pi\le x\le 3\pi$.
+Hence
+\begin{equation}\label{eqhil16}
+    |\frac{1_{[r, 2\pi -r]}(x)}{1-e^{-ix}}-\frac{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{1_{[r, 2\pi -r]}(x)}{1-e^{-ix}}-\frac{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 thos $x$, 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}
+
+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{dirichlet3}, 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{Partial sums as $L^2$ projections}
+\label{10projection}
+
+
+
+
+
+
+
+
+
+\begin{lemma}\label{lem 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},
+\begin{equation*}
+S_N (S_Nf) (x)=
+\int_0^{2\pi} S_Nf(y)K_N(x-y)\, dy
+\end{equation*}
+\begin{equation}\label{eqhil1}
+=
+\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}
+\begin{equation*}
+\int_0^{2\pi} K_N(y-y') K_N(x-y)\,  dy
+\end{equation*}
+\begin{equation*}
+=\sum_{n=-N}^N\sum_{n'=-N}^N
+\int_0^{2\pi} e^{in(y-y')}e^{in'(x-y)}\,  dy
+\end{equation*}
+\begin{equation}\label{eqhil6}
+=\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.
+We obtain for \eqref{eqhil6}
+\begin{equation}\label{eqhil2}
+=\sum_{n=-N}^N
+e^{in(x-y')}\int_0^{2\pi} \,  dy=K_N(x-y')\, .
+\end{equation}
+Applying Fubini in  \eqref{eqhil1} and using
+\eqref{eqhil2} gives
+\begin{equation}
+S_N(S_Nf)(x)=
+\int_0^{2\pi}  f(y')K(x-y')  \, dy'=S_N f(x)
+\end{equation}
+This proves the lemma.
+\end{proof}
+\begin{lemma}\label{selfadjoint}
+    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 $x$
+    \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
+\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
+    \end{equation*}
+    \begin{equation}
+=
+\frac 1{2\pi} \int_0^{2\pi} \int_{0}^{2\pi}\overline{f(y)} K_N(y-x)
+g(x)\, dx dy
+=\int_0^{2\pi} \overline{f(x)} S_Ng(x)\, dx
+\, .
+\end{equation}
+    This proves the lemma.
+\end{proof}
+
+
+
+We have with Lemma \ref{selfadjoint}, then Lemma \ref{lem projection} and the Lemma\ref{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
+\end{equation*}
+\begin{equation}\label{eqhil7}
+    =\int_0^{2\pi}  f(x)\overline{S_Nf(x)}\, dx=
+    \int_0^{2\pi}  S_N f(x)\overline{f(x)}\, dx\, .
+\end{equation}
+
+We have by the distributive law
+\begin{equation}\label{diffnorm}
+    \int_0^{2\pi} (f(x)-S_Nf(x))(\overline{f(x)-S_Nf(x)})\, dx=
+\end{equation}
+\begin{equation*}
+    \int_0^{2\pi} f(x)\overline{f(x)}
+    -S_Nf(x)\overline{f(x)}
+    -f(x)\overline{S_Nf(x)}
+        + S_Nf(x)\overline{S_Nf(x)}\, dx
+\end{equation*}
+Using the various identities expressed in \eqref{eqhil7}, this becomes
+\begin{equation}
+    =\int_0^{2\pi} f(x)\overline{f(x)})\, dx
+    -
+    \int_0^{2\pi} S_Nf(x)\overline{S_Nf(x)}\, dx\, .
+\end{equation}
+As \eqref{diffnorm} has is nonnegative integrand and is thus nonnegative, we conclude
+\begin{equation}
+    \int_0^{2\pi} S_Nf(x)\overline{S_Nf(x)}\, dx\le
+    \int_0^{2\pi} f(x)\overline{f(x)})\, dx\, .
+\end{equation}
+As both sides are positive, we may take the square root of this inequality.
+This completes the proof of the lemma.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+\section{The error bound}
+\label{10difference}
+
+\section{Carleson on the real line}
+\label{10carleson}
+
+We prove Lemma \ref{lem rcarleson}.
+
+Consider the standard distance function
+\begin{equation}
+    \rho(x,y)=|x-y|
+\end{equation}
+on the real line $\R$.
+\begin{lemma}
+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}
+    For $x\in R$ and $R>0$, the ball $B(x,R)$ is the interval $(x-R,x+R)$
+\end{lemma}
+\begin{proof}
+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 $|x'-x|<R$.
+hence $x'\in B(x,R)$.
+This implies the lemma.
+\end{proof}
+We consider the Lebesgue measure $\mu$ on $\R$.
+\begin{lemma}
+    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}
+    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}
+\begin{equation}
+    \mu(B(x,R))=\mu((x-R,x+R))=2R\, .
+\end{equation}
+Where the last identity is taken from the Lean library.
+\end{proof}
+
+\begin{lemma}\label{lem r 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}
+\begin{equation}
+    \mu(B(x,2R)=4R=2\mu(B(x,R)\, .
+\end{equation}
+This proves the lemma
+\end{proof}
+
+For each $n\in \mathbb{Z}$ define
+$\mfa_n:\R\to \R$ by
+\begin{equation}
+    \mfa_n(x)=nx\, .
+\end{equation}
+
+
+Let $\Mf$ be the collection $\{\theta_n, n\in \mathbb{Z}\}$.
+Note that for every $n\in \mathbb{Z}$ we have $\theta_n(0)=0$.
+Define $d$ as in \eqref{definedE}. Note that
+with for
+$n,m\in \mathbb{Z}$ and $x\in \R$ and $x>0$
+\begin{equation}
+    d_{B(x,R)}(\mfa_n,\mfa_m)=\sup_{y,y'\in B(x,R)}|ny-ny'-my+my'|
+\end{equation}
+\begin{equation}\label{eqcarl1}
+        =\sup_{y,y'\in B(x,R)}|(n-m)(y-x)-(n-m)(y'-x)|
+    \end{equation}
+By the triangle inequality, we have for \eqref{eqcarl1} the upper bound
+\begin{equation}\label{eqcarl2}
+        \le 2|n-m|R\, .
+    \end{equation}
+On the other hand, for any $0<R'<R$, we have
+by choosing suitable $y,y'\in \{x-R', x+R'\}$,
+we have for \eqref{eqcarl1} the lower bound
+\begin{equation}\label{eqcarl3}
+        \ge 2|n-m|R'\, .
+    \end{equation}
+As $R'<R$ is arbitrary, we have with \eqref{eqcarl2}
+and \eqref{eqcarl3}
+\begin{equation}\label{eqcarl4}
+    d_{B(x,R)}(\mfa_n,\mfa_m)=2R|n-m|\, .
+\end{equation}
+
+
+
+\begin{lemma}\label{lem fdb1}
+    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}
+    d_{B(x',2R)}(\mfa_n,\mfa_m)\le 2 d_{B(x,R)}(\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}
+
+\begin{lemma}\label{lem sdb1}
+    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) \, .
+\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}
+
+
+\begin{lemma}\label{lem tdb1}
+    For every $x\in \R$ and $R>0$ and every
+    $n\in \mathbb{Z}$ and $R'>0$,
+    there exist $m_1, m_2, m_3\in \mathbb{Z}$
+    such that
+    \begin{equation}\label{eqcarl5}
+        B'\subset B_1\cup B_2\cup B_3\, ,
+    \end{equation}
+where
+\begin{equation}
+B'=     \{ \mfa \in \Mf: d_{B(x,R)}(\mfa, \mfa_n)<2R'\}
+\end{equation}
+and for $j=1,2,3$
+\begin{equation}
+    B_j=
+        \{ \mfa \in \Mf: d_{B(x,R)}(\mfa, \mfa_{m_j})<R'\}
+        \, .
+\end{equation}
+
+\end{lemma}
+\begin{proof}
+Let $m_1$ be the largest integer smaller than
+or equal to
+$n-R'/2$.
+Let $m_2=2$.
+Let $m_3$ be the smallest integer larger than
+or equal to $n+R'/2$.
+
+Let $\mfa_{n'}\in B'$, then with Lemma \ref{lem ball int},
+we have
+\begin{equation}\label{eqcarl6}
+    R|n-n'|\le R'\, .
+\end{equation}
+Assume first $n'\le n-R'/2$. By definition of $m_1$,
+we have $n'\le m_1$.  With \eqref{eqcarl6}
+we have
+\begin{equation*}
+    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\, .
+\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 $n+R'/2<n'$
+By definition of $m_3$,
+we have $m_3\le n'$.  With \eqref{eqcarl6}
+we have
+\begin{equation*}
+    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\, .
+\end{equation}
+We conclude $\mfa_{n'}\in B_1$.
+This completes the proof of the lemma in all cases.
+\end{proof}
+
+\begin{lemma}
+    For any $x\in \R$ and $R>0$ and any
+    function $\varphi: X\to \C$ supported on $B'=B(x,R)$
+    such that
+\begin{equation}
+    \|\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)}\,.
+\end{equation}
+is finite and for any $n,m\in \mathbb{Z}$,
+\begin{equation}
+    \label{eq vdc cond1}
+    |\int_{B'} e(\mfa_n(x)-{\mfa_m(x)}) \varphi(x) d\mu(x)|\le 2^6 \mu(B')\frac{\|\varphi\|_{\Lip(B')}}{1+d_{B'}(\mfa_n,\mfa_m)}
+\, .
+\end{equation}
+
+
+\end{lemma}
+\begin{proof}
+Set $n'=n-m$. Then we have to prove
+\begin{equation}
+    \label{eq vdc cond2}
+    |\int_{x-R}^{x+R} e^{in'y}\varphi(y) dy|\le R\|\varphi\|_{\Lip(B)}
+(1+2R|n'|)^{-1}\, .
+\end{equation}
+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
+\begin{equation}
+    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}
+    2\int_{x-R}^{x+R} e^{in'y}\varphi(y) dy
+\end{equation}
+\begin{equation*}
+=\int_{x-R}^{x-R+\pi/n'} e^{in'y}\varphi(y) dy
++\int_{x-R+\pi/n'}^{x+R} e^{in'y}\varphi(y) dy
+\end{equation*}
+\begin{equation*}
++\int_{x-R}^{x+R-\pi/n'} e^{in'y}\varphi(y) dy
++\int_{x+R-\pi/n'}^{x+R} e^{in'y}\varphi(y) dy\, .
+\end{equation*}
+We estimate the sum of the second and third term
+on the right-hand side of  \eqref{eqcarl10} using variable substitution and $e^{i\pi}=-1$ by
+\begin{equation*}
+    \left|\int_{x-R+\pi/n'}^{x+R} e^{in'y}\varphi(y) dy
+    +\int_{x-R}^{x+R-\pi/n'} e^{in'y}\varphi(y) dy\right|
+\end{equation*}
+\begin{equation*}
+    =\left|\int_{x-R}^{x+R-\pi/n' }e^{in'(y+\pi/n')}\varphi(y+\pi/n')
+    + e^{in'y}\varphi(y) dy\right|
+\end{equation*}
+\begin{equation*}
+    =\left|\int_{x-R}^{x+R-\pi/n' }e^{in'y}(-\varphi(y+\pi/n')
+    + \varphi(y)) dy\right|
+\end{equation*}
+\begin{equation*}
+    \le 2R\sup_{y\in (x-R,x+R-\pi/n')}
+    |\varphi(y+\pi/n')
+    - \varphi(y)|
+\end{equation*}
+\begin{equation*}
+    \le 2R \|\varphi\|_{Lip(B')}\frac{\pi}{Rn'}
+\end{equation*}
+\begin{equation}\label{eqcarl21}
+    \le 2^5 R \|\varphi\|_{Lip(B')}\frac{1}{(1+2Rn')},
+\end{equation}
+where in the last line we have used $Rn'\ge \pi$.
+We estimate the first and fourth term on the right hand side of \eqref{eqcarl10} by
+\begin{equation*}
+    \left|\int_{x-R'}^{x-R+\pi/n'} e^{in'y}\varphi(y) dy
+    +\int_{x+R-\pi/n'}^{x+R} e^{in'y}\varphi(y) dy\right|
+\end{equation*}
+\begin{equation}\label{eqcarl22}
+    \le \frac{2\pi}{n'}\sup_{y\in B'}|\varphi(y)|
+        \le 2^4 R\|\varphi(y)\|_{Lip(B')}\frac 1 {1+Rn'}\, ,
+\end{equation}
+where in the last line we have used $Rn'\ge \pi$.
+Adding \eqref{eqcarl21} and \eqref{eqcarl22}
+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 function $K:\R\times \R\to \mathbb{C}$ by
+\begin{equation}
+    K(x,y):=k(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}
+    For $x,y\in \R$ with $x\neq y$ we have
+    \begin{equation}\label{eqcarl30}
+        |K(x,y)|\le 2^6(2|x-y|)^{-1}\, .
+    \end{equation}
+\end{lemma}
+\begin{proof}
+    Fix $x\neq y$. 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}
+\begin{equation}\label{eqcarl31}
+|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}
+    For $x,y,y'\in \R$ with $x\neq y,y'$ and
+    \begin{equation}
+        2|y-y'|\le |x-y|\, ,
+    \end{equation}
+    we have
+    \begin{equation}\label{eqcarl30}
+        |K(x,y)|\le 2^6(2|x-y|)^{-1}\, .
+    \end{equation}
+\end{lemma}
 
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