From 5af6e5a6f11895c1bb27209c7a8164ae802550d9 Mon Sep 17 00:00:00 2001 From: Floris van Doorn Date: Mon, 29 Apr 2024 16:09:10 +0200 Subject: [PATCH] update blueprint update readme try newcommand --- README.md | 28 +- blueprint/src/chapter/main.tex | 2357 +++++++++++++++++------------ blueprint/src/preamble/common.tex | 44 +- 3 files changed, 1454 insertions(+), 975 deletions(-) diff --git a/README.md b/README.md index fe17cb76..7e94c277 100644 --- a/README.md +++ b/README.md @@ -1,6 +1,10 @@ # Formalization of a generalized Carleson's theorem A (WIP) formalized proof of a generalized Carleson's theorem in Lean +* [Blueprint](http://florisvandoorn.com/carleson/blueprint/) +* [Blueprint as pdf](http://florisvandoorn.com/carleson/blueprint.pdf) +* [Doc pages for this repository](http://florisvandoorn.com/carleson/docs/) + ## Contribute To get the repository, make sure you have [installed Lean](https://leanprover-community.github.io/get_started.html). Then get the repository using `git clone https://github.com/fpvandoorn/carleson.git` and run `lake exe cache get!` inside the repository. Run `lake build` to build all files in this repository. See the README of [my course repository](https://github.com/fpvandoorn/LeanCourse23) for more detailed instructions. @@ -13,25 +17,5 @@ you've worked have no errors (having `sorry`'s is of course fine). ## Build the blueprint -To build the web version of the blueprint, you need a working LaTeX installation. -Furthermore, you need some packages: -``` -sudo apt install graphviz libgraphviz-dev -pip3 install invoke pandoc -cd .. # go to folder where you are happy clone git repos -git clone git@github.com:plastex/plastex -pip3 install ./plastex -git clone git@github.com:PatrickMassot/leanblueprint -pip3 install ./leanblueprint -cd carleson -``` - -To actually build the blueprint, run -``` -lake exe cache get -lake build -inv all -``` - -To view the web-version of the blueprint locally, run `inv serve` and navigate to -`http://localhost:8000/` in your favorite browser. \ No newline at end of file +To test the Blueprint locally, you can compile `print.tex` using XeLaTeX (i.e. `xelatex print.tex` in the folder `blueprint/src`). If you have the Python package `invoke` you can also run `inv bp` which puts the output in `blueprint/print/print.pdf`. +If you feel adventurous and want to build the web version of the blueprint locally, you need to install some packages by following the instructions [here](https://pypi.org/project/leanblueprint/). But if the pdf builds locally, you can just make a pull request and use the online blueprint. \ No newline at end of file diff --git a/blueprint/src/chapter/main.tex b/blueprint/src/chapter/main.tex index 8601c14d..bc449b1f 100644 --- a/blueprint/src/chapter/main.tex +++ b/blueprint/src/chapter/main.tex @@ -22,53 +22,30 @@ \chapter{Introduction} -In \cite{carleson}, L. Carleson -answered a classical question on convergence of -Fourier series of continuous functions by proving that they converge pointwise almost everywhere. -Theorem \ref{classical Carleson} is a version of his result. +In \cite{carleson}, L. Carleson addressed a classical question regarding the convergence of Fourier series of continuous functions by proving their pointwise convergence almost everywhere. Theorem \ref{classical Carleson} represents a version of this result. -Let $f$ be a complex valued, $2\pi $-periodic bounded Borel measurable function on the real line and given an integer $n$, define the Fourier coefficient +Let $f$ be a complex-valued, $2\pi$-periodic bounded Borel measurable function on the real line, and for an integer $n$, define the Fourier coefficient as \begin{equation} \widehat{f}_n:=\frac {1}{2\pi} \int_0^{2\pi} f(x) e^{- i nx} dx . \end{equation} -Define for $N\ge 0$ the partial Fourier sum +Define the partial Fourier sum for $N\ge 0$ as \begin{equation} S_Nf(x):=\sum_{n=-N}^N \widehat{f}_n e^{i nx}\ . \end{equation} -\begin{theorem}\label{classical Carleson} -\uses{metric space Carleson} -Let $f$ be a $2\pi $-periodic complex valued uniformly continuous function an $\R$ that satisfies the bound -$|f(x)|\le 1$ for all $x\in \R$. For all $0<\epsilon\le 2\pi$, -there exists a Borel set $E\subset [0,2\pi]$ with Lebesgue measure -$|E|\le \epsilon$ and a positive integer $N_0$ such that for all -$x\in [0,2\pi]\setminus E$ and all integers $N>N_0$ we have + +\begin{theorem}[classical Carleson] +\label{classical Carleson} +\uses{piecewise constant approximation, convergence for piecewise constant, +control approximation effect,metric space Carleson} +Let $f$ be a $2\pi$-periodic complex-valued uniformly continuous function on $\mathbb{R}$ satisfying the bound $|f(x)|\le 1$ for all $x\in \mathbb{R}$. For all $0<\epsilon\le 2\pi$, there exists a Borel set $E\subset [0,2\pi]$ with Lebesgue measure $|E|\le \epsilon$ and a positive integer $N_0$ such that for all $x\in [0,2\pi]\setminus E$ and all integers $N>N_0$, we have \begin{equation}\label{aeconv} |f(x)-S_N f(x)|\le \epsilon. \end{equation} \end{theorem} -Note that mere continuity implies uniform continuity -in the setting of this theorem. Applying this theorem -with a sequence of $\epsilon_n:= 2^{-n}\delta$ for $n\ge 1$ -and taking the union of corresponding exceptional sets $E_n$, we see that -outside a set of measure $\delta$, the partial Fourier sums -converge pointwise for $N\to \infty$. Applying this with a sequence -of $\delta$ shrinking to zero and taking the intersection of the corresponding exceptional -sets, which has measure zero, we see that the Fourier series converges outside -a set of measure zero. - -The purpose of this paper is twofold. -On the one hand, it prepares computer verification of Theorem \ref{classical Carleson} by presenting a very detailed proof as a blueprint for coding in Lean. -We pass through a bound for a -generalization of the so-called Carlson operator -to doubling metric measure spaces. This generalization is new -and proving these bounds is the second purpose of this paper. -This generalization incorporates several results in the recent literature, most prominently bounds for the polynomial Carleson -operator of V. Lie \cite{lie-polynomial} -as well as its generalization \cite{zk-polynomial}. -A computer verification of our theorem will also be a computer -verification for the bulk of the work in these results. +Note that mere continuity implies uniform continuity in the setting of this theorem. By applying this theorem with a sequence of $\epsilon_n:= 2^{-n}\delta$ for $n\ge 1$ and taking the union of corresponding exceptional sets $E_n$, we see that outside a set of measure $\delta$, the partial Fourier sums converge pointwise for $N\to \infty$. Applying this with a sequence of $\delta$ shrinking to zero and taking the intersection of the corresponding exceptional sets, which has measure zero, we see that the Fourier series converges outside a set of measure zero. +The purpose of this paper is twofold. On the one hand, it prepares computer verification of Theorem \ref{classical Carleson} by presenting a very detailed proof as a blueprint for coding in Lean. We pass through a bound for a generalization of the so-called Carleson operator to doubling metric measure spaces. This generalization is new, and proving these bounds constitutes the second purpose of this paper. This generalization incorporates several results from the recent literature, most prominently bounds for the polynomial Carleson operator of V. Lie \cite{lie-polynomial} as well as its generalization \cite{zk-polynomial}. A computer verification of our theorem will also entail a computer verification for the bulk of the work in these results. We proceed to introduce the setup for our general theorem. @@ -107,11 +84,10 @@ \chapter{Introduction} 2d_{B_1}(\mfa,\mfb) \le d_{B_2}(\mfa,\mfb) . \end{equation} -For every ball $B$ in $X$ and -every $d_B$-ball of radius $2R$ in $\Mf$, there is a collection $\mathcal{B}$ of -at most $2^a$ many $d_B$-balls of radius $R$ covering $B$, that is, +For every ball $B$ in $X$ and every $d_B$-ball $\tilde B$ of radius $2R$ in $\Mf$, there is a collection $\mathcal{B}$ of +at most $2^a$ many $d_B$-balls of radius $R$ covering $\tilde B$, that is, \begin{equation}\label{thirddb} - B\subset \bigcup \mathcal{B}. + \tilde B\subset \bigcup \mathcal{B}. \end{equation} @@ -124,7 +100,7 @@ \chapter{Introduction} (1+d_B(\mfa,\mfb))^{-\frac{1}{a}}, \end{equation} %where $\C^\nu$ denotes the homogeneous H\"older space. -where $\|\cdot\|_{\Lip(B)}$ denotes the inhomogeneous Lipschitz norm on $B$, +where $\|\cdot\|_{\Lip(B)}$ denotes the inhomogeneous Lipschitz norm on $B$: $$ \|\varphi\|_{\Lip(B)} = \sup_{x \in B} |\varphi(x)| + R \sup_{x,y \in B, x \neq y} \frac{|\varphi(x) - \varphi(y)|}{\rho(x,y)}\,. $$ @@ -161,9 +137,9 @@ \chapter{Introduction} Our main result is the following restricted weak type estimate for $T$ in the range $10$. This function vanishes outside $[\frac1{4D},\frac 12]$, is constant one on $[\frac1{2D},\frac 14]$, and is Lipschitz @@ -302,7 +231,7 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv Let a doubling metric measure space $(X,\rho,\mu, a)$ be given. Let a cancellative compatible collection $\Mf$ of functions on $X$ be given. -Let $o\in X$ be a point such that $\mfa(o)=1$ +Let $o\in X$ be a point such that $\mfa(o)=0$ for all $\mfa\in \Mf$. @@ -310,7 +239,8 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv the triangle inequality part, without specific reference. It allows us to call $d_B$ a metric. -\begin{lemma} +\begin{lemma}[ball metric] +\label{ball metric} For any ball $B$ in $X$, the local oscillation $d_{B}$ is a metric on $\Mf$. \end{lemma} @@ -322,12 +252,11 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv d_{B(x,r)}(\mfa,\mfb)=0 \end{equation} implies $\mfa(y)=\mfb(y)$ for all $y\in X$. Assume $\eqref{dvanish}$. -Let +Let $y\in X$ and let \begin{equation} R=1+\rho(x,o)+\rho(x,y)\, . \end{equation} -By an iterated application of -the comparability of the norms \eqref{firstdb} or \eqref{seconddb} +By an iterated application of \eqref{firstdb} or \eqref{seconddb}, \begin{equation} d_{B(x,R)}(\mfa,\mfb)=0. \end{equation} @@ -354,7 +283,8 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv that a function from a measure space to a finite set is measurable if the pre-image of each of the elements in the range is measurable. -\begin{prop}\label{finitary Carleson} +\begin{prop}[finitary Carleson] +\label{finitary Carleson} \uses{linearized truncation, grid existence, tile structure, tile sum operator} Let ${\sigma_1},\sigma_2\colon X\to \mathbb{Z}$ be measurable functions with finite range and ${\sigma_1}\leq \sigma_2$. Let $\tQ\colon X\to \Mf$ be a measurable function with finite range. Let $F,G$ be bounded Borel sets in $X$. Then there is a Borel set $G'$ in $X$ with $2\mu(G')\leq \mu(G)$ such that for all Borel functions $f:X\to \C$ with $|f|\le \mathbf{1}_F$. @@ -468,7 +398,7 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv \end{equation} -\begin{prop} +\begin{prop}[discrete Carleson] \label{discrete Carleson} \uses{tile sum operator, exceptional set, forest union, forest complement} Let $(\mathcal{D}, c, s)$ be a grid structure and \begin{equation*} @@ -559,7 +489,10 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv The following proposition is proved in Section \ref{antichainboundary}. -\begin{prop}\label{antichain operator} +\begin{prop}[antichain operator] +\label{antichain operator} + +\uses{dens2 antichain,dens1 antichain} For any antichain $\mathfrak{A} $ and for all $f:X\to \C$ with $|f|\le \mathbf{1}_F$ and all $g:X\to\C$ with $|g| \le \mathbf{1}_G$ \begin{equation} \label{eq antiprop} |\int \overline{g(x)} \sum_{\fp \in \mathfrak{A}} T_{\fp} f(x)\, d\mu(x)| @@ -587,14 +520,14 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv \begin{equation}\label{forest1} 4\fp\lesssim 1\fu. \end{equation} -For each $\fu, \in \fU$ and each $\fp,\fp''\in \fT(\fu)$ and $\fp'\in \fP$ +For each $\fu \in \fU$ and each $\fp,\fp''\in \fT(\fu)$ and $\fp'\in \fP$ we have \begin{equation}\label{forest2} \fp, \fp'' \in \mathfrak{T}(\fu), \fp \leq \fp' \leq \fp'' \implies \fp' \in \mathfrak{T}(\fu). \end{equation} We have \begin{equation}\label{forest3} - \|\sum_{\fu\in \fU} \mathbf{1}_{\sc(\fu))}\|_\infty \leq 2^n\,. + \|\sum_{\fu\in \fU} \mathbf{1}_{\sc(\fu)}\|_\infty \leq 2^n\,. \end{equation} We have for every $\fu\in \fU$ \begin{equation}\label{forest4} @@ -611,7 +544,8 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv The following proposition is proved in Section \ref{treesection}. -\begin{prop}\label{forest operator} +\begin{prop}[forest operator] +\label{forest operator} For any $n\ge 0$ and any $n$-forest $(\fU,\fT)$ we have for all $f: X \to \mathbb{C}$ with $|f| \le \mathbf{1}_F$ and all bounded $g$ with bounded support $$ | \int \overline{g(x)} \sum_{\fu\in \fU} \sum_{\fp\in \fT(\fu)} T_{\fp} f(x) \, \mathrm{d}\mu(x)| @@ -639,10 +573,10 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv \|\varphi\|_{C^\tau(B)} = \sup_{x \in B} |\varphi(x)| + R^\tau \sup_{x,y \in B, x \neq y} \frac{|\varphi(x) - \varphi(y)|}{\rho(x,y)^\tau}\,. \end{equation} -\lars{fractions in exponents everywhere} -\begin{prop} +\begin{prop}[Holder van der Corput] \label{Holder van der Corput} + \uses{Lipschitz Holder approximation} Let $z\in X$ and $R>0$ and set $B=B(z,R)$. Let $\varphi: X \to \mathbb{C}$ by supported on $B$ and satisfy $\|{\varphi}\|_{C^\tau(B)}<\infty$. @@ -667,8 +601,9 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv \end{equation} Define further $M_{\mathcal{B}}:=M_{\mathcal{B},1}$. -\begin{prop}\label{Hardy Littlewood} -\uses{first exception} +\begin{prop}[Hardy Littlewood] +\label{Hardy Littlewood} +\uses{layer cake representation,covering separable space} Let $\mathcal{B}$ be a finite collection of balls in $X$. If for some $\lambda>0$ and some measurable function $u:X\to [0,\infty)$ we have \begin{equation}\label{eq ball assumption} @@ -689,13 +624,12 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson}), overv \label{eq ball av} \frac{1}{\mu(B)} \int_{B} |w(y)| \, \mathrm{d}\mu(y) \le Mw(x) \end{equation} -and for all $1 < p \le \infty$ +and for all $1 < p_1 \le p_2 \le \infty$ \begin{equation} \label{eq hlm 2} - \|Mw\|_p \le 2^{4a} \frac{p}{p-1}\|w\|_p\,. + \|M(w^{p_1})^{\frac{1}{p_1}}\|_{p_2} \le 2^{4a} \frac{p_2}{p_2-p_1}\|w\|_{p_2}\,. \end{equation} - \end{prop} This completes the overview of the proof of Theorem \ref{metric space Carleson}. @@ -706,25 +640,25 @@ \section{Auxiliary lemmas} First, we record an estimate for the metrical entropy numbers of balls in the space $\Mf$ equipped with any of the metrics $d_B$, following from the doubling property \eqref{thirddb}. -\begin{lemma} - \label{lem entropy Theta} - Let $B \subset X$ be a ball. Let $r > 0$, $\mfa \in \Mf$ and $k \in \mathbb{N}$. Suppose that $\mathcal{Z} \subset B_B(\mfa, r2^k)$ \rs{Double notation} satisfies that for all $z, z' \in \mathcal{Z}$ with $z \ne z'$, we have $d_B(z,z') \ge r$. Then +\begin{lemma}[ball metric entropy] + \label{ball metric entropy} + Let $B' \subset X$ be a ball. Let $r > 0$, $\mfa \in \Mf$ and $k \in \mathbb{N}$. Suppose that $\mathcal{Z} \subset B_{B'}(\mfa, r2^k)$ satisfies that for all $z, z' \in \mathcal{Z}$ with $z \ne z'$, we have $d_{B'}(z,z') \ge r$. Then $$ |\mathcal{Z}| \le 2^{ka}\,. $$ \end{lemma} \begin{proof} - By applying property \eqref{thirddb} $k$ times, we obtain a collection $\mathcal{Z}' \subset \Mf$ with $|\mathcal{Z}'| \le 2^{ka}$ and + By applying property \eqref{thirddb} $k$ times, we obtain a collection $\mathcal{Z}' \subset \Mf$ with $|\mathcal{Z}'| = 2^{ka}$ and $$ - B_{B}(\mfa,r2^k) \subset \bigcup_{z' \in \mathcal{Z}'} B_B(z', \frac{r}{2})\,. + B_{B'}(\mfa,r2^k) \subset \bigcup_{z' \in \mathcal{Z}'} B_{B'}(z', \frac{r}{2})\,. $$ Then each $z \in \mathcal{Z}$ is contained in one of the balls $B(z', \frac{r}{2})$, but by the separation assumption no such ball contains more than one element of $\mathcal{Z}$. Thus $|\mathcal{Z}| \le |\mathcal{Z}'| = 2^{ka}$. \end{proof} The next lemma concerns monotonicity of the metrics $d_{B(c(I), \frac 14 D^{s(I)})}$ with respect to inclusion of cubes $I$ in a grid. -\begin{lemma} +\begin{lemma}[monotone cube metrics] \label{monotone cube metrics} \uses{wiggle order 2, forest separation} Let $(\mathcal{D}, c, s)$ be a grid structure. Denote for cubes $I \in \mathcal{D}$ @@ -770,7 +704,8 @@ \section{Auxiliary lemmas} We also record the following basic estimates for the kernels $K_s$. -\begin{lemma}\label{kernel summand} +\begin{lemma}[kernel summand] +\label{kernel summand} Let $-S\le s\le S$ and $x,y,y'\in X$. If $K_s(x,y)\neq 0$, then we have \begin{equation}\label{supp Ks} @@ -781,7 +716,8 @@ \section{Auxiliary lemmas} \label{eq Ks size} |K_s(x,y)|\le \frac{2^{102 a^3}}{\mu(B(x, D^{s}))}\, \end{equation} - and \begin{equation} + and %if $2\rho(y,y')\leq \rho(x,y)$ + \begin{equation} \label{eq Ks smooth} |K_s(x,y)-K_s(x, y')|\le \frac{2^{150a^3}}{\mu(B(x, D^{s}))} \left(\frac{ \rho(y,y')}{D^s}\right)^{\frac 1a}\,. @@ -808,14 +744,14 @@ \section{Auxiliary lemmas} \end{equation} Using $a\ge 4$ proves \eqref{eq Ks size}. - + \lars{The case where $\rho(y,y') > \rho(x,y)/2$ is missing} Similarly, we obtain with \eqref{eqkernel y smooth} and the lower bound in \eqref{supp Ks} \begin{equation} |K_s(x,y)-K_s(x, y')|\le \frac{2^{a^3}}{\mu(B(x, \frac 14 D^{s-1}))} \left(\frac{ \rho(y,y')}{\frac 14 D^{s-1}}\right)^{\frac 1a}\,. \end{equation} - Using $a \ge 4$, this is estimated by + As above, this is estimated by \begin{equation} \le \frac{4D 2^{2a+101a^3}}{\mu(B(x, D^{s}))} \left(\frac{ \rho(y,y')}{D^{s}}\right)^{\frac 1a} @@ -836,7 +772,8 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})} X=\bigcup_{R>0}B(o,R), \end{equation} because every point of $X$ has finite distance from $o$. -\begin{lemma}\label{R truncation} +\begin{lemma}[R truncation] +\label{R truncation} \uses{S truncation} For all integers $R>0$ $$ @@ -854,25 +791,11 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})} \end{equation} \end{lemma} -We first see how the Lemma \ref{R truncation} implies -Theorem \ref{metric space Carleson}. When $R$ tends to $\infty$, the integrand of the left-hand-side of \eqref{Rcut} -grows monotonically to the integrand of the -left-hand side of \eqref{resweak} for all $x$. -By Lebesgue's monotone convergence theorem, the left-hand side of \eqref{Rcut} converges to the -left-hand side of \eqref{resweak}. This proves Theorem \ref{metric space Carleson}. +We first show how Lemma \ref{R truncation} implies Theorem \ref{metric space Carleson}. As $R$ tends to $\infty$, the integrand of the left-hand side of \eqref{Rcut} grows monotonically toward the integrand of the left-hand side of \eqref{resweak} for all $x$. By Lebesgue's monotone convergence theorem, the left-hand side of \eqref{Rcut} converges to the left-hand side of \eqref{resweak}. This verifies Theorem \ref{metric space Carleson}. -It remains to prove Lemma \ref{R truncation}. -Fix an integer $R>0$. Replacing -$G$ by $G\cap B(o,R)$ if necessary, it suffices to show -\eqref{Rcut} under the assumption that $G$ is contained in $B(o,R)$. We make this assumption. -For every $x\in G$, the domain of integration -in \eqref{TRR} is contained in $B(o,2R)$. -Replacing -$F$ by $F\cap B(o,2R)$ if necessary, it suffices to show -\eqref{Rcut} under the assumption that $F$ is contained in $B(o,2R)$. We make this assumption. +It remains to prove Lemma \ref{R truncation}. Fix an integer $R>0$. By replacing $G$ with $G\cap B(o,R)$ if necessary, it suffices to show \eqref{Rcut} under the assumption that $G$ is contained in $B(o,R)$. We make this assumption. For every $x\in G$, the domain of integration in \eqref{TRR} is contained in $B(o,2R)$. By replacing $F$ with $F\cap B(o,2R)$ if necessary, it suffices to show \eqref{Rcut} under the assumption that $F$ is contained in $B(o,2R)$. We make this assumption. -With the definition \eqref{defks} of $K_s$ -and the partition of unity \eqref{eq psisum}, we write \eqref{TRR} as the sum of +Using the definition \eqref{defks} of $K_s$ and the partition of unity \eqref{eq psisum}, we express \eqref{TRR} as the sum of \begin{equation}\label{middles} {T}_{1,s_1,s_2,\mfa}f(x):=\sum_{s_1 \le s\le s_2} \int K_s(x,y) f(y) e(\mfa(y)) \, \mathrm{d}\mu(y) @@ -883,25 +806,15 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})} \int_{R_1 < \rho(x,y) < R_2} K_s(x,y) f(y) e(\mfa(y)) \, \mathrm{d}\mu(y), \end{equation} -where $s_1$ is the smallest integer such that $D^{s_1-2}R_2>\frac 1{4D}$ and $s_2$ -is the largest integer so that $D^{s_2+2}R_1<\frac 12$. Here we restricted the summation index $s$ -by omitting the summands with $ss_2+2$ because for these summands the function $K_s$ vanishes on the domain of integration, and we have omitted the restriction in the integral -in the summands in \eqref{middles} because in these summands the support of $K_s$ is contained in -the set described by this restriction. +where $s_1$ is the smallest integer such that $D^{s_1-2}R_2>\frac 1{4D}$ and $s_2$ is the largest integer such that $D^{s_2+2}R_1<\frac 12$. We restrict the summation index $s$ by excluding summands with $ss_2+2$ because for these summands, the function $K_s$ vanishes on the domain of integration. We also omit the restriction in the integral for the summands in \eqref{middles} because in these summands, the support of $K_s$ is contained in the set described by this restriction. -We apply the triangle inequality and estimate the -versions of \eqref{Rcut} separately with $T_{R_1,R_2,\mfa}$ replaced by -\eqref{middles} and by each summand of \eqref{boundarys}. -To deal with the case \eqref{middles} we use the following lemma, where we use that if -$\frac 1R\le R_1\le R_2\le R$, then $s_1,s_2$ -as in \eqref{middles} are in an interval $[-S,S]$ for some -sufficiently large $S$ depending on $R$. +We apply the triangle inequality and estimate the versions of \eqref{Rcut} separately with $T_{R_1,R_2,\mfa}$ replaced by \eqref{middles} and by each summand of \eqref{boundarys}. To handle the case \eqref{middles}, we employ the following lemma. Here, we utilize the fact that if $\frac 1R\le R_1\le R_2\le R$, then $s_1$ and $s_2$ as in \eqref{middles} are in an interval $[-S,S]$ for some sufficiently large $S$ depending on $R$. -\begin{lemma}\label{S truncation} +\begin{lemma}[S truncation] +\label{S truncation} \uses{finitary S truncation} For all integers $S>0$ $$ @@ -915,7 +828,7 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})} where $T_{1, s_1,s_2,\mfa}$ is defined in \eqref{middles}. \end{lemma} -To reduce Lemma \ref{R truncation} to Lemma \ref{S truncation}, we need estimates for the summands in \eqref{boundarys}. Using Lemma \ref{kernel summand}, we have for arbitrary $s$ the inequality +To reduce Lemma \ref{R truncation} to Lemma \ref{S truncation}, we need estimates for the summands in \eqref{boundarys}. Using Lemma \ref{kernel summand}, we obtain for arbitrary $s$ the inequality \begin{multline} \left|\int_{R_1 < \rho(x,y) < R_2} K_s(x,y) f(y) e(\mfa(y)) \, \mathrm{d}\mu(y)\right|\\ \leq \frac{2^{102 a^3}}{\mu(B(x, D^{s}))} @@ -923,17 +836,16 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})} \leq 2^{102 a^3} M\mathbf{1}_F(x), \end{multline} where $M\mathbf{1}_F$ is as defined in Proposition \ref{Hardy Littlewood}. -Now the left hand side of \eqref{Rcut}, with $T_{R_1,R_2,\mfa}$ replaced by a summand of \eqref{boundarys}, can be estimated using Hölder's inequality and Proposition \ref{Hardy Littlewood} by +Now, the left-hand side of \eqref{Rcut}, with $T_{R_1,R_2,\mfa}$ replaced by a summand of \eqref{boundarys}, can be estimated using Hölder's inequality and Proposition \ref{Hardy Littlewood} by $$ 2^{102 a^3}\int \mathbf{1}_{G}(x) M\mathbf{1}_F(x)\, d\mu(x) \leq \frac{2^{102 a^3+4a}q}{q-1}\mu(G)^{1-\frac{1}{q}} \mu(F)^{\frac{1}{q}}\,. $$ -Using the triangle inequality to estimate the left hand side of \eqref{Rcut} by contributions from the summands in \eqref{middles} and \eqref{boundarys}, using Lemma \ref{S truncation} to control the first term, and the above to estimate the contribution from the four summands in \eqref{boundarys}, combined with $a>4$ and $q < 2$ completes the reduction of -Lemma \ref{R truncation} to Lemma \ref{S truncation}. +Applying the triangle inequality to estimate the left-hand side of \eqref{Rcut} by contributions from the summands in \eqref{middles} and \eqref{boundarys}, using Lemma \ref{S truncation} to control the first term, and the above to estimate the contribution from the four summands in \eqref{boundarys}, combined with $a\geq 4$ and $q < 2$, completes the reduction of Lemma \ref{R truncation} to Lemma \ref{S truncation}. It remains to prove Lemma \ref{S truncation}. Fix $S>0$. - -\begin{lemma}\label{finitary S truncation} +\begin{lemma}[finitary S truncation] +\label{finitary S truncation} \uses{linearized truncation} For all finite sets $\tilde{\Mf}\subset \Mf$ $$ @@ -949,10 +861,10 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})} We reduce Lemma \ref{S truncation} to Lemma \ref{finitary S truncation}. By the Lebesgue monotone convergence theorem, applied to an increasing sequence of finite sets $\tilde{\Mf}$, inequality \eqref{Sqcut} -continues to hold for countable $\tilde{\Mf}$. +continues to hold for countable $\tilde{\Mf}$. Let $\epsilon=\frac{1}{2S+1}$. Pick some $\mfa_0 \in \Mf$. -Let for $k \ge 0$ the set $\tilde{\Mf}_k$ be a subset of $B_{B(o,2R)}(\mfa, k)$ of maximal size, such that for all $\mfa, \mfb \in \tilde{\Mf}_k$, it holds that $d_{B(o, 2R)}(\mfa, \mfb) \ge \epsilon$. Such a set exists, since by Lemma \ref{lem entropy Theta} there exists an upper bound for the size of such subsets in $B_{B(o, 2R)}(\mfa_0, k)$. Let +For $k \ge 0$, let the set $\tilde{\Mf}_k$ be a subset of $B_{B(o,2R)}(\mfa_0, k)$ of maximal size, such that for all $\mfa, \mfb \in \tilde{\Mf}_k$, it holds that $d_{B(o, 2R)}(\mfa, \mfb) \ge \epsilon$. Such a set exists, since by Lemma \ref{ball metric entropy} there exists an upper bound for the size of such subsets in $B_{B(o, 2R)}(\mfa_0, k)$. Define $$ \tilde{\Mf} := \bigcup_{k \in \mathbb{N}} \tilde{\Mf}_k\,. $$ @@ -961,7 +873,7 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})} d_{B(o, 2R)}(\mfb, \mfa) < \epsilon\,. $$ -For every $\mfa\in \Mf$ we have +For every $\mfa\in \Mf$, we have \begin{equation} \left| T_{1, s_1,s_2,\mfa} \mathbf{1}_{F}(x)\right|= \left|\sum_{s_1 \le s\le s_2} @@ -969,54 +881,46 @@ \chapter{Proof of metric space Carleson (Thm \ref{metric space Carleson})} \end{equation} Moreover, there is a $\tilde{\mfa} \in \tilde{\Mf}$ with $d_{B(o,2R)}(\mfa,\tilde{\mfa})\le \epsilon$. Hence, -$$ - \left| T_{1, s_1,s_2,\mfa} \mathbf{1}_{F}(x)\right|-\left|T_{1, s_1,s_2,\tilde{\mfa}} \mathbf{1}_{F}(x) \right| -$$ -$$ - \leq\sum_{s_1 \le s\le s_2} \int |K_s(x,y)| \mathbf{1}_F(y) |e(\mfa(y)-{\mfa(x)})-e({\tilde{\mfa}(y)}-{\tilde{\mfa}(x)})| \, \mathrm{d}\mu(y) -$$ -$$ - \leq \sum_{s_1 \le s\le s_2} \epsilon \int |K_s(x,y)| \mathbf{1}_F(y) \, \mathrm{d}\mu(y) -$$ +\begin{align*} + &\left| T_{1, s_1,s_2,\mfa} \mathbf{1}_{F}(x)\right|-\left|T_{1, s_1,s_2,\tilde{\mfa}} \mathbf{1}_{F}(x) \right|\\ + &\leq\sum_{s_1 \le s\le s_2} \int |K_s(x,y)| \mathbf{1}_F(y) |e(\mfa(y)-{\mfa(x)})-e({\tilde{\mfa}(y)}-{\tilde{\mfa}(x)})| \, \mathrm{d}\mu(y)\\ + &\leq \sum_{s_1 \le s\le s_2} \epsilon \int |K_s(x,y)| \mathbf{1}_F(y) \, \mathrm{d}\mu(y) +\end{align*} Using Lemma \ref{kernel summand}, we can estimate the above expression by -$$ - \sum_{s_1 \le s\le s_2} \frac{2^{102 a^3}}{\mu(B(x, D^{s}))} -\epsilon \int_{B(x, D^s)} \mathbf{1}_F(y) \, \mathrm{d}\mu(y) -$$ -$$ - \leq (2S+1)\epsilon 2^{102 a^3} M\mathbf{1}_F(x)\le 2^{102 a^3}M\mathbf{1}_F(x) -$$ -% \begin{equation} -% \leq (2S+1) \epsilon M\mathbf{1}_F(x)\lesssim M\mathbf{1}_F(x) -% \end{equation} +\begin{align*} + &\sum_{s_1 \le s\le s_2} \frac{2^{102 a^3}}{\mu(B(x, D^{s}))} +\epsilon \int_{B(x, D^s)} \mathbf{1}_F(y) \, \mathrm{d}\mu(y)\\ + &\leq (2S+1)\epsilon 2^{102 a^3} M\mathbf{1}_F(x)\le 2^{102 a^3}M\mathbf{1}_F(x) +\end{align*} We estimate the left-hand-side of \eqref{Scut} by -the sum of left hand side of \eqref{Sqcut} and - \begin{equation} - \int \mathbf{1}_{G}(x) +the sum of left-hand-side of \eqref{Sqcut} and +\begin{align*} + &\int \mathbf{1}_{G}(x) \max_{-S-S$, -assume $x$ is in $I_1(y_m,-k)$ for $m=1,2$. -Then, for $m=1,2$, there is $z_m\in Y_{k-1}\cap B(y_m,D^k)$ with $x\in I_3(z_m,k-1)$. -Using \eqref{disji} inductively for $j=3$, we -conclude $z_1=z_2$. This implies that the balls -$B(y_1, D^k)$ and $B(y_2, D^k)$ intersect. By construction of $Y_k$, this implies $y_1=y_2$. -This proves \eqref{disji} for $j=1$, - -We next consider \eqref{disji} for $j=3$. -Assume $x$ is in $I_3(y_m,k)$ for $m=1,2$ and $y_m\in Y_k$. If $x$ is in $X_k$, then by definition -\eqref{definei3}, $x\in I_1(y_m,k)$ for $m=1,2$. -As we have already shown \eqref{disji} for $j=1$, -we conclude $y_1=y_2$. This completes the proof in -case $x\in X_k$ and we may assume $x$ is not in $X_k$. By definition \eqref{definei3}, $x$ is not -in $I_3(z,k)$ for any $z$ with $z-S$. Let $x$ be a point of $B(o, 4D^S-2D^k$). -By induction, there is $y'\in Y_{k-1}$ such that -$x\in I_3(y',k-1)$. Using the inductive statement -\eqref{squeezedyadic}, we obtain $x\in B(y',4D^{k-1})$. -As $D>4$, by applying the triangle inequality with -the points, $o$, $x$, and $y'$ we obtain that $y'\in B(o, 4D^S-D^k)$. -By Lemma {\ref{cover big ball}}, $y'$ is in $B(y,2D^k)$ -for some $y\in Y_k$. It follows that $x\in I_2(y,k)$. -This proves \eqref{unioni} for $j=2$. - -We show \eqref{unioni} for $j=3$. -Let $x\in B(o, 4D^S-2D^k)$. In case $x\in X_k$, - then by definition of $X_k$ we have $x\in I_1(y,k)$ for some $y\in Y_k$ and thus $x\in I_3(y,k)$. We may thus assume $x\not\in X_k$. As we have already seen -\eqref{unioni} for $j=2$, - there is $y\in Y_k$ such that $x\in I_2(y,k)$. -We may assume this $y$ is minimal with respect to the order in $Y_k$ -Then $x\in I_3(y,k)$. - This proves \eqref{unioni} for $j=3$. - -Next we show the first inclusion in \eqref{squeezedyadic}. -Let $x\in B(y,\frac 1{2}D^k)$. -As $I_1(y,k)\subset I_3(y,k)$, -it suffices to show $x\in I_1(y,k)$. -If $k=-S$, this follows immediately from -the assumption on $x$ and definition of $I_1$. -Assume $k>-S$. By inductive statement \eqref{unioni} -and $D>4$, there is a -$y'\in Y_{k-1}$ such that $x\in I_3(y',k-1)$. -By inductive statement \eqref{squeezedyadic}, -we conclude $x\in B(y',4D^{k-1})$. -By the triangle inequality with points $x$, $y$, $y'$ and $D>4$, we have -$y'\in B(y,D^k)$. It follows by definition -\eqref{defineij} that -$I_3(y',k-1)\subset I_1(y,k)$ and thus -{$x\in I_3(y,k)$}. This proves the first inclusion -in \eqref{squeezedyadic}. - - -We show the second inclusion in \eqref{squeezedyadic}. -Let $x\in I_3(y,k)$. As $I_1(y,k)\subset I_2(y,k)$ -directly from the definition \eqref{defineij}, -it follows by definition \eqref{definei3} that -$x\in I_2(y,k)$. By definition -\eqref{defineij}, there is $y'\in Y_{k-1}\cap B(y,2D^k)$ -with $x\in I_3(y',k-1)$. By induction, -$x\in B(y', 4D^{k-1})$. By the triangle inequality -applied to the points $x,y',y$ and $D>4$, we conclude -$x\in B(y,4D^k)$. -This shows the second inclusion in \eqref{squeezedyadic} and completes the proof of the lemma. +We prove these statements simultaneously by induction on the ordered set of pairs $(y,k)$. Let $-S\le k\le S$. + +We first consider \eqref{disji} for $j=1$. If $k=-S$, disjointedness of the sets $I_1(y,-S)$ follows by definition of $I_1$ and $Y_k$. If $k>-S$, assume $x$ is in $I_1(y_m,k)$ for $m=1,2$. Then, for $m=1,2$, there is $z_m\in Y_{k-1}\cap B(y_m,D^k)$ with $x\in I_3(z_m,k-1)$. Using \eqref{disji} inductively for $j=3$, we conclude $z_1=z_2$. This implies that the balls $B(y_1, D^k)$ and $B(y_2, D^k)$ intersect. By construction of $Y_k$, this implies $y_1=y_2$. This proves \eqref{disji} for $j=1$. + +We next consider \eqref{disji} for $j=3$. Assume $x$ is in $I_3(y_m,k)$ for $m=1,2$ and $y_m\in Y_k$. If $x$ is in $X_k$, then by definition \eqref{definei3}, $x\in I_1(y_m,k)$ for $m=1,2$. As we have already shown \eqref{disji} for $j=1$, we conclude $y_1=y_2$. This completes the proof in case $x\in X_k$, and we may assume $x$ is not in $X_k$. By definition \eqref{definei3}, $x$ is not in $I_3(z,k)$ for any $z$ with $z-S$. Let $x$ be a point of $B(o, 4D^S-2D^k)$. By induction, there is $y'\in Y_{k-1}$ such that $x\in I_3(y',k-1)$. Using the inductive statement \eqref{squeezedyadic}, we obtain $x\in B(y',4D^{k-1})$. As $D>4$, by applying the triangle inequality with the points, $o$, $x$, and $y'$, we obtain that $y'\in B(o, 4D^S-D^k)$. By Lemma \ref{cover big ball}, $y'$ is in $B(y,2D^k)$ for some $y\in Y_k$. It follows that $x\in I_2(y,k)$. This proves \eqref{unioni} for $j=2$. + +We show \eqref{unioni} for $j=3$. Let $x\in B(o, 4D^S-2D^k)$. In case $x\in X_k$, then by definition of $X_k$ we have $x\in I_1(y,k)$ for some $y\in Y_k$ and thus $x\in I_3(y,k)$. We may thus assume $x\not\in X_k$. As we have already seen \eqref{unioni} for $j=2$, there is $y\in Y_k$ such that $x\in I_2(y,k)$. We may assume this $y$ is minimal with respect to the order in $Y_k$. Then $x\in I_3(y,k)$. This proves \eqref{unioni} for $j=3$. + +Next, we show the first inclusion in \eqref{squeezedyadic}. Let $x\in B(y,\frac 12 D^k)$. As $I_1(y,k)\subset I_3(y,k)$, it suffices to show $x\in I_1(y,k)$. If $k=-S$, this follows immediately from the assumption on $x$ and the definition of $I_1$. Assume $k>-S$. By the inductive statement \eqref{unioni} and $D>4$, there is a $y'\in Y_{k-1}$ such that $x\in I_3(y',k-1)$. By the inductive statement \eqref{squeezedyadic}, we conclude $x\in B(y',4D^{k-1})$. By the triangle inequality with points $x$, $y$, $y'$, and $D>4$, we have $y'\in B(y,D^k)$. It follows by definition \eqref{defineij} that $I_3(y',k-1)\subset I_1(y,k)$, and thus $x\in I_3(y,k)$. This proves the first inclusion in \eqref{squeezedyadic}. + +We show the second inclusion in \eqref{squeezedyadic}. Let $x\in I_3(y,k)$. As $I_1(y,k)\subset I_2(y,k)$ directly from the definition \eqref{defineij}, it follows by definition \eqref{definei3} that $x\in I_2(y,k)$. By definition \eqref{defineij}, there is $y'\in Y_{k-1}\cap B(y,2D^k)$ with $x\in I_3(y',k-1)$. By induction, $x\in B(y', 4D^{k-1})$. By the triangle inequality applied to the points $x,y',y$ and $D>4$, we conclude $x\in B(y,4D^k)$. This shows the second inclusion in \eqref{squeezedyadic} and completes the proof of the lemma. \end{proof} -\begin{lemma} +\begin{lemma}[cover by cubes] \label{cover by cubes} \uses{dyadic property} Let $-S\le l\le k\le S$ and @@ -1421,12 +1262,10 @@ \section{Proof of L.\ref{grid existence}, dyadic structure} We conclude \eqref{3coverdyadic} by induction. \end{proof} -\begin{lemma} +\begin{lemma}[dyadic property] \label{dyadic property} \uses{transitive boundary} -Let $-S\le l\le k\le S$ and -$y\in Y_k$ and $y'\in Y_l$. -with +Let $-S\le l\le k\le S$ and $y\in Y_k$ and $y'\in Y_l$ with $I_3(y',l)\cap I_3(y,k)\neq \emptyset$. Then \begin{equation} \label{3dyadicproperty} @@ -1437,43 +1276,15 @@ \section{Proof of L.\ref{grid existence}, dyadic structure} \end{lemma} \begin{proof} -Let $l,k,y,y'$ as in the Lemma. -Pick $x\in I_3(y',l)\cap I_3(y,k)$. -Assume first $l=k$. By \eqref{disji} of Lemma -\ref{basic grid structure}, we conclude $y'=y$ -and thus \eqref{3dyadicproperty}. -Now assume $l5$ and $k''32$, +\eqref{squeezedyadic} and \eqref{disji} of Lemma \ref{basic grid structure}. Assume then without loss of generality that $l'32$, we conclude from \eqref{bulbul} that +$X\setminus I_3(y,k)$ with $\rho(x,u')<6D^{l'}$. +Using $D>25$, we conclude from the triangle inequality and \eqref{bulbul} that $x\in B(u,\frac 12D^l)$. However, $B(u,\frac 12 D^l)\subset I_3(u,l)$, and $I_3(u,l)\subset I_3(y,k)$, a contradiction to $x\not\in I_3(y,k)$. This proves the lemma. \end{proof} -\begin{lemma} +\begin{lemma}[smaller boundary] \label{smaller boundary} \uses{boundary measure} Let $K = 2^{4a+1}$ @@ -1621,9 +1432,9 @@ \section{Proof of L.\ref{grid existence}, dyadic structure} \begin{proof} We prove this by induction on $n$. If $n=0$, both sides of \eqref{very new small} are equal to - $\mu(I_3(y,k)$. If $n=1$, this follows from Lemma \ref{small boundary}. + $\mu(I_3(y,k))$ by \eqref{disji}. If $n=1$, this follows from Lemma \ref{small boundary}. - Assume $n>1$ and \eqref{very new small} has been proven with $n-1$ in place of $n$. + Assume $n>1$ and \eqref{very new small} has been proven for $n-1$. We write \eqref{very new small} \begin{equation} \sum_{y''\in Y_{k-nK}: (y'',k-nK|y,k)}\mu(I_3(y'',k-nK)) @@ -1636,10 +1447,11 @@ \section{Proof of L.\ref{grid existence}, dyadic structure} = \sum_{y'\in Y_{k-K}:(y',k-K|y,k)} 2^{1-n}\mu(I_3(y',k-K)) \end{equation} Applying \eqref{new small boundary} gives -\eqref{very new small} and proves the lemma. +\eqref{very new small}, and proves the lemma. \end{proof} -\begin{lemma}\label{boundary measure} +\begin{lemma}[boundary measure] +\label{boundary measure} For each $-S\le k\le S$ and $y\in Y_k$ and $0k$. Choose $y\in I\cap J$. By property \eqref{coverdyadic}, there is $K\in \mathcal{D}$ with $s(K)=s(J)-1$ and $y\in K$. By construction -of $J$, and pairwise disjointness of all $I(w,s(J)-1)$ that we have already seen, +of $J$, and pairwise disjointedness of all $I_3(w,s(J)-1)$ that we have already seen, we have $K\subset J$. By minimality of $s(J)$, we have $I\subset K$. This proves $I\subset J$ and thus \eqref{dyadicproperty}. We next establish \eqref{coverball}. Let $-S\leq k\leq S$. Using \eqref{unioni} for $j=3$, we get \begin{equation} x\in B(o, D^S)\subset B(o, 4D^S-2D^k)\subset \bigcup_{y\in Y_k} I_3(y,k)\, . -\end{equation} Thus, there exists a dyadic cube $I=I_3(y', k)$ with $s(I)=k$ and $x\in I$. This proves \eqref{coverball}. +\end{equation} Thus, there exists a dyadic cube $I=I_3(y, k)$ with $s(I)=k$ and $x\in I$. This proves \eqref{coverball}. -\section{Proof of L.\ref{tile structure}, tile structure} +\section{Proof of Lemma \ref{tile structure}, tile structure} \label{subsectiles} Choose a grid structure $(\mathcal{D}, c, s)$. Let $I \in \mathcal{D}$. Suppose that \begin{equation} \label{eq tile Z} - Z \subset \bigcup_{\mfa \in \tQ(X)} B_{I^\circ}(\mfa, 1) + \mathcal{Z} \subset \bigcup_{\mfa \in \tQ(X)} B_{I^\circ}(\mfa, 1) \end{equation} -is such that for any $\mfa, \mfb \in Z$ we have +is such that for any $\mfa, \mfb \in \mathcal{Z}$ we have \begin{equation} \label{eq tile disjoint Z} B_{I^\circ}(\mfa, 0.3) \cap B_{I^\circ}(\mfb, 0.3) = \emptyset\,. \end{equation} -By Lemma \ref{lem entropy Theta} applied to each of the balls $B_{I^\circ}(\mfa, 1)$, $\mfa \in \tQ(X)$, we have +By Lemma \ref{ball metric entropy} applied to each of the balls $B_{I^\circ}(\mfa, 1)$, $\mfa \in \tQ(X)$, we have $$ - |Z| \le 2^{2a} |\tQ(X)|\,. + |\mathcal{Z}| \le 2^{2a} |\tQ(X)|\,. $$ -In particular, there exists a set $Z$ satisfying both \eqref{eq tile Z} and \eqref{eq tile disjoint Z} of maximal cardinality among all such sets. We pick for each $I \in \mathcal{D}$ such a set $Z(I)$. +In particular, there exists a set $\mathcal{Z}$ satisfying both \eqref{eq tile Z} and \eqref{eq tile disjoint Z} of maximal cardinality among all such sets. We pick for each $I \in \mathcal{D}$ such a set $\mathcal{Z}(I)$. -\begin{lemma} +\begin{lemma}[frequency ball cover] \label{frequency ball cover} For each $I \in \mathcal{D}$, we have \begin{equation} \label{eq tile cover} - \tQ(X) \subset \bigcup_{\mfa \in \tQ(X)} B_{I^\circ}(\mfa, 1) \subset \bigcup_{z \in Z(I)} B_{I^\circ}(z, 0.7)\,. + \tQ(X) \subset \bigcup_{\mfa \in \tQ(X)} B_{I^\circ}(\mfa, 1) \subset \bigcup_{z \in \mathcal{Z}(I)} B_{I^\circ}(z, 0.7)\,. \end{equation} \end{lemma} \begin{proof} - To show \eqref{eq tile cover} note that the first inclusion is obvious. For the second inclusion let $\mfb \in \bigcup_{\mfa \in \tQ(X)} B_{I^\circ}(\mfa, 1)$. By maximality of $Z(I)$, there must be a point $z \in Z(I)$ such that $B_{I^\circ}(z, 0.3) \cap B_{I^\circ}(\mfb, 0.3) \ne \emptyset$. Else, $Z(I) \cup \{\mfb\}$ would be a set of larger cardinality than $Z(I)$ satisfying \eqref{eq tile Z} and \eqref{eq tile disjoint Z}. Fix such $z$, and fix a point $z_1 \in B_{I^\circ}(z, 0.3) \cap B_{I^\circ}(\mfb, 0.3)$. By the triangle inequality, we deduce that + To show \eqref{eq tile cover} note that the first inclusion is obvious. For the second inclusion let $\mfb \in \bigcup_{\mfa \in \tQ(X)} B_{I^\circ}(\mfa, 1)$. By maximality of $\mathcal{Z}(I)$, there must be a point $z \in \mathcal{Z}(I)$ such that $B_{I^\circ}(z, 0.3) \cap B_{I^\circ}(\mfb, 0.3) \ne \emptyset$. Else, $\mathcal{Z}(I) \cup \{\mfb\}$ would be a set of larger cardinality than $\mathcal{Z}(I)$ satisfying \eqref{eq tile Z} and \eqref{eq tile disjoint Z}. Fix such $z$, and fix a point $z_1 \in B_{I^\circ}(z, 0.3) \cap B_{I^\circ}(\mfb, 0.3)$. By the triangle inequality, we deduce that $$ d_{I^\circ}(z,\mfb) \le d_{I^\circ}(z,z_1) + d_{I^\circ}(\mfb, z_1) < 0.3 + 0.3 = 0.6\,, $$ @@ -1767,25 +1577,25 @@ \section{Proof of L.\ref{tile structure}, tile structure} We define $$ - \fP = \{(I, z) \ : \ I \in \mathcal{D}, z \in Z(I)\}\,, + \fP = \{(I, z) \ : \ I \in \mathcal{D}, z \in \mathcal{Z}(I)\}\,, $$ $$\sc((I, z)) = I\qquad \text{and} \qquad \fcc((I, z)) = z.$$ We further set $$\ps(\fp) = s(\sc(\fp)),\qquad \qquad \pc(\fp) = c(\sc(\fp)).$$ Then \eqref{tilecenter}, \eqref{tilescale} hold by definition. It remains to construct the map $\Omega$, and verify properties \eqref{eq dis freq cover}, \eqref{eq freq dyadic} and -\eqref{eq freq comp ball}. We first construct an auxiliary map $\Omega_1$. For each $I \in \mathcal{D}$, we pick an enumeration of the finite set $Z(I)$ +\eqref{eq freq comp ball}. We first construct an auxiliary map $\Omega_1$. For each $I \in \mathcal{D}$, we pick an enumeration of the finite set $\mathcal{Z}(I)$ $$ - Z(I) = \{z_1, \dotsc, z_M\}\,. + \mathcal{Z}(I) = \{z_1, \dotsc, z_M\}\,. $$ We define {$\Omega_1:\fP \mapsto \mathcal{P}(\Mf) $ as below}. Set $$ - \Omega_1((I, z_1)) = B_{I^\circ}(z_1, 0.7) \setminus \bigcup_{z \in Z(I)\setminus \{z_1\}} B_{I^\circ}(z, 0.3) + \Omega_1((I, z_1)) = B_{I^\circ}(z_1, 0.7) \setminus \bigcup_{z \in \mathcal{Z}(I)\setminus \{z_1\}} B_{I^\circ}(z, 0.3) $$ and then define iteratively \begin{equation} \label{eq def omega1} - \Omega_1((I, z_k)) = B_{I^\circ}(z_k, 0.7) \setminus \bigcup_{z \in Z(I) \setminus \{z_k\}} B_{I^\circ}(z, 0.3) \setminus \bigcup_{i=1}^{k-1} \Omega_1((I, z_i))\,. + \Omega_1((I, z_k)) = B_{I^\circ}(z_k, 0.7) \setminus \bigcup_{z \in \mathcal{Z}(I) \setminus \{z_k\}} B_{I^\circ}(z, 0.3) \setminus \bigcup_{i=1}^{k-1} \Omega_1((I, z_i))\,. \end{equation} -\begin{lemma} +\begin{lemma}[disjoint frequency cubes] \label{disjoint frequency cubes} \uses{dyadic frequency cubes} For each $I \in \mathcal{D}$, and $\fp_1, \fp_2\in \fP(I)$, if $$\Omega_1(\fp_1)\cap \Omega_1(\fp_2)\neq \emptyset,$$ then $\fp_1=\fp_2$. @@ -1794,18 +1604,18 @@ \section{Proof of L.\ref{tile structure}, tile structure} \begin{proof} By the definition of the map $\sc$, we have $$ - \fP(I) = \{(I, z) \, : \, z \in Z(I)\}\,. + \fP(I) = \{(I, z) \, : \, z \in \mathcal{Z}(I)\}\,. $$ By \eqref{eq def omega1}, the set $\Omega_1((I, z_k))$ is disjoint from each $\Omega_1((I, z_i))$ with $i < k$. Thus the sets $\Omega_1(\fp)$, $\fp \in \fP(I)$ are pairwise disjoint. \end{proof} -\begin{lemma} +\begin{lemma}[frequency cube cover] \label{frequency cube cover} \uses{dyadic frequency cubes} For each $I \in \mathcal{D}$, it holds that \begin{equation} \label{eq omega1 cover} - \bigcup_{z \in Z(I)} B_{I^\circ}(z, 0.7)\subset \bigcup_{\fp \in \fP(I)} \Omega_1(\fp)\,. + \bigcup_{z \in \mathcal{Z}(I)} B_{I^\circ}(z, 0.7)\subset \bigcup_{\fp \in \fP(I)} \Omega_1(\fp)\,. \end{equation} For every $\fp \in \fP$, it holds that \begin{equation} @@ -1819,16 +1629,16 @@ \section{Proof of L.\ref{tile structure}, tile structure} The second inclusion in \eqref{eq omega1 incl} then follows from \eqref{eq def omega1} and the equality $B_{\fp}(\fcc(\fp), 0.7) = B_{I^\circ}(z, 0.7)$, which is true by definition. For the first inclusion in \eqref{eq omega1 incl} let $\mfa \in B_{\fp}(\fcc(\fp),0.3)$. Let $k$ be such that $z = z_k$ in the enumeration we chose above. It follows immediately from \eqref{eq def omega1} and \eqref{eq tile disjoint Z} that $\mfa \notin \Omega_1((I, z_i))$ for all $i < k$. Thus, again from \eqref{eq def omega1}, we have - $q \in \Omega_1((I,z_k))$. + $\mfa \in \Omega_1((I,z_k))$. - To show \eqref{eq omega1 cover} let $\mfa \in \bigcup_{z \in Z(I)} B_{I^\circ}(z,0.7)$. - If there exists $z \in Z_1(I)$ with $q \in B_{I^\circ}(z,0.3)$, then + To show \eqref{eq omega1 cover} let $\mfa \in \bigcup_{z \in \mathcal{Z}(I)} B_{I^\circ}(z,0.7)$. + If there exists $z \in \mathcal{Z}(I)$ with $\mfa \in B_{I^\circ}(z,0.3)$, then $$ z \in \Omega_1((I, z)) \subset \bigcup_{\fp \in \fP(I)} \Omega_1(\fp) $$ by the first inclusion in \eqref{eq omega1 incl}. - Now suppose that there exists no $z \in Z(I)$ with $\mfa \in B_{I^\circ}(z, 0.3)$. Let $k$ be minimal such that $\mfa \in B_{I^\circ}(z_k, 0.7)$. Since $\Omega_1((I, z_i)) \subset B_{I^\circ}(z_i, 0.7)$ for each $i$ by \eqref{eq def omega1}, we have that $\mfa \notin \Omega_1((I, z_i))$ for all $i < k$. Hence $\mfa \in \Omega_1((I, z_k))$, again by \eqref{eq def omega1}. + Now suppose that there exists no $z \in \mathcal{Z}(I)$ with $\mfa \in B_{I^\circ}(z, 0.3)$. Let $k$ be minimal such that $\mfa \in B_{I^\circ}(z_k, 0.7)$. Since $\Omega_1((I, z_i)) \subset B_{I^\circ}(z_i, 0.7)$ for each $i$ by \eqref{eq def omega1}, we have that $\mfa \notin \Omega_1((I, z_i))$ for all $i < k$. Hence $\mfa \in \Omega_1((I, z_k))$, again by \eqref{eq def omega1}. \end{proof} Now we are ready to define the function $\Omega$. For all cubes $I \in \mathcal{D}$ such that there exists no $J \in \mathcal{D}$ with $I \subset J$ and $I \ne J$, we define for all $\fp \in \fP(I)$ @@ -1839,23 +1649,23 @@ \section{Proof of L.\ref{tile structure}, tile structure} For cubes $I \in \mathcal{D}$ for which there exists $J \in \mathcal{D}$ with $I \subset J$ and $I \ne J$, we define $\Omega$ by recursion. We can pick an inclusion minimal $J \in \mathcal{D}$ among the finitely many cubes such that $I \subset J$ and $I \ne J$. This $J$ is unique: Suppose that $J'$ is another inclusion minimal cube with $I \subset J'$ and $I \ne J'$. Without loss of generality, we have that $s(J) \le s(J')$. By \eqref{dyadicproperty}, it follows that $J \subset J'$. Since $J'$ is minimal with respect to inclusion, it follows that $J = J'$. Then we define \begin{equation} \label{eq it omega} - \fc(\fp) = \bigcup_{z \in Z(J) \cap \Omega_1(\fp)} \Omega((J, z)) \cup B_{\fp}(\fcc(\fp),0.2)\,. + \fc(\fp) = \bigcup_{z \in \mathcal{Z}(J) \cap \Omega_1(\fp)} \Omega((J, z)) \cup B_{\fp}(\fcc(\fp),0.2)\,. \end{equation} -\begin{lemma} +\begin{lemma}[dyadic frequency cubes] \label{dyadic frequency cubes} With this definition, \eqref{eq dis freq cover}, \eqref{eq freq dyadic} and \eqref{eq freq comp ball} hold. \end{lemma} \begin{proof} - First, we prove \eqref{eq freq comp ball}. If $I \in \mathcal{D}$ is maximal in $\mathcal{D}$ with respect to set inclusion, then \eqref{eq freq comp ball} holds for all $\fp \in \fP(I)$ by \eqref{eq max omega} and\eqref{eq omega1 incl}. Now suppose that $I$ is not maximal in $\mathcal{D}$ with respect to set inclusion. Then we may assume by induction that for all $J \in \mathcal{D}$ with $I \subset J$ and all $\fp' \in \fP(J)$, \eqref{eq freq comp ball} holds. Let $J$ be the unique minimal cube in $\mathcal{D}$ with $I \subsetneq J$. + First, we prove \eqref{eq freq comp ball}. If $I \in \mathcal{D}$ is maximal in $\mathcal{D}$ with respect to set inclusion, then \eqref{eq freq comp ball} holds for all $\fp \in \fP(I)$ by \eqref{eq max omega} and \eqref{eq omega1 incl}. Now suppose that $I$ is not maximal in $\mathcal{D}$ with respect to set inclusion. Then we may assume by induction that for all $J \in \mathcal{D}$ with $I \subset J$ and all $\fp' \in \fP(J)$, \eqref{eq freq comp ball} holds. Let $J$ be the unique minimal cube in $\mathcal{D}$ with $I \subsetneq J$. Suppose that $\mfa \in \Omega(\fp)$. If $\mfa\in B_{\fp}(\mathcal{Q}(\fp), 0.2)$, then since \begin{equation*} B_{\fp}(\mathcal{Q}(\fp), 0.2)\subset B_{\fp}(\mathcal{Q}(\fp), 1)\, , \end{equation*} - we conclude that $\mfa\in B_{\fp}(\mathcal{Q}(\fp), 0.7)$. If not, by \eqref{eq it omega}, there exists $z \in Z(J) \cap \Omega_1(\fp)$ with $\mfa \in \Omega(J,z)$. Using the triangle inequality and \eqref{eq omega1 incl}, we obtain + we conclude that $\mfa\in B_{\fp}(\mathcal{Q}(\fp), 0.7)$. If not, by \eqref{eq it omega}, there exists $z \in \mathcal{Z}(J) \cap \Omega_1(\fp)$ with $\mfa \in \Omega(J,z)$. Using the triangle inequality and \eqref{eq omega1 incl}, we obtain $$ d_{I^\circ}(\fcc(\fp),\mfa) \le d_{I^\circ}(\fcc(\fp), z) + d_{I^\circ}(z, \mfa) \le 0.7 + d_{I^\circ}(z, \mfa)\,. $$ @@ -1867,34 +1677,34 @@ \section{Proof of L.\ref{tile structure}, tile structure} Next, we show \eqref{eq dis freq cover}. Let $I \in \mathcal{D}$. - If $I$ is maximal with respect to inclusion, then disjointness of the sets $\fc(\fp)$ for $\fp \in \fP(I)$ follows from the definition \eqref{eq max omega} and Lemma \ref{disjoint frequency cubes}. To obtain the inclusion in \eqref{eq dis freq cover} one combines the inclusions \eqref{eq tile cover} and \eqref{eq omega1 cover} + If $I$ is maximal with respect to inclusion, then disjointedness of the sets $\fc(\fp)$ for $\fp \in \fP(I)$ follows from the definition \eqref{eq max omega} and Lemma \ref{disjoint frequency cubes}. To obtain the inclusion in \eqref{eq dis freq cover} one combines the inclusions \eqref{eq tile cover} and \eqref{eq omega1 cover} of Lemma \ref{frequency cube cover} with \eqref{eq max omega}. - Now we turn to the case where there exists $J \in \mathcal{D}$ with $I \subset J$ and $I\ne J$. In this case we use induction: It suffices to show \eqref{eq dis freq cover} under the assumption that it holds for all cubes $J \in \mathcal{D}$ with $I \subset J$. As shown before definition \eqref{eq it omega}, we may choose the unique inclusion minimal such $J$. To show disjointness of the sets $\fc(\fp), \fp \in \fP(I)$ we pick two tiles $\fp, \fp' \in \fP(I)$ and $\mfa \in \fc(\fp) \cap \fc(\fp')$. + Now we turn to the case where there exists $J \in \mathcal{D}$ with $I \subset J$ and $I\ne J$. In this case we use induction: It suffices to show \eqref{eq dis freq cover} under the assumption that it holds for all cubes $J \in \mathcal{D}$ with $I \subset J$. As shown before definition \eqref{eq it omega}, we may choose the unique inclusion minimal such $J$. To show disjointedness of the sets $\fc(\fp), \fp \in \fP(I)$ we pick two tiles $\fp, \fp' \in \fP(I)$ and $\mfa \in \fc(\fp) \cap \fc(\fp')$. Then we are by \eqref{eq it omega} in one of the following four cases. - 1. There exist $z \in Z(J) \cap \Omega_1(\fp)$ such that $\mfa \in \Omega(J, z)$, and there exists $z' \in Z(J) \cap \Omega_1(\fp')$ such that $\mfa \in \Omega(J, z')$. By the induction hypothesis, that \eqref{eq dis freq cover} holds for $J$, we must have $z = z'$. By Lemma \ref{disjoint frequency cubes}, we must then have $\fp = \fp'$. + 1. There exist $z \in \mathcal{Z}(J) \cap \Omega_1(\fp)$ such that $\mfa \in \Omega(J, z)$, and there exists $z' \in \mathcal{Z}(J) \cap \Omega_1(\fp')$ such that $\mfa \in \Omega(J, z')$. By the induction hypothesis, that \eqref{eq dis freq cover} holds for $J$, we must have $z = z'$. By Lemma \ref{disjoint frequency cubes}, we must then have $\fp = \fp'$. - 2. There exists $z \in Z(J) \cap \Omega_1(\fp)$ such that $\mfa \in \Omega(J,z)$, and $\mfa \in B_{\fp'}(\fcc(\fp'), 0.2)$. Using the triangle inequality, Lemma \ref{monotone + 2. There exists $z \in \mathcal{Z}(J) \cap \Omega_1(\fp)$ such that $\mfa \in \Omega(J,z)$, and $\mfa \in B_{\fp'}(\fcc(\fp'), 0.2)$. Using the triangle inequality, Lemma \ref{monotone cube metrics} and \eqref{eq freq comp ball}, we obtain $$ d_{\fp'}(\fcc(\fp'),z) \le d_{\fp'}(\fcc(\fp'), \mfa) + d_{\fp'}(z, \mfa) \le 0.2 + 2^{-95a} \cdot 1 < 0.3\,. $$ Thus $z \in \Omega_1(\fp')$ by \eqref{eq omega1 incl}. By Lemma \ref{disjoint frequency cubes}, it follows that $\fp = \fp'$. - 3. There exists $z' \in Z(J) \cap \Omega_1(\fp')$ such that $\mfa \in \Omega(J,z')$, and $\mfa \in B_{\fp}(\fcc(\fp), 0.2)$. This case is the same as case 2., after swapping $\fp$ and $\fp'$. + 3. There exists $z' \in \mathcal{Z}(J) \cap \Omega_1(\fp')$ such that $\mfa \in \Omega(J,z')$, and $\mfa \in B_{\fp}(\fcc(\fp), 0.2)$. This case is the same as case 2., after swapping $\fp$ and $\fp'$. - 4. We have $q \in B_{\fp}(\fcc(\fp), 0.2) \cap B_{\fp'}(\fcc(\fp'), 0.2)$. In this case it follows that $\fp = \fp'$ since the sets $B_{\fp}(\fcc(\fp), 0.2)$ are pairwise disjoint by the inclusion \eqref{eq omega1 incl} and Lemma \ref{disjoint frequency cubes}. + 4. We have $\mfa \in B_{\fp}(\fcc(\fp), 0.2) \cap B_{\fp'}(\fcc(\fp'), 0.2)$. In this case it follows that $\fp = \fp'$ since the sets $B_{\fp}(\fcc(\fp), 0.2)$ are pairwise disjoint by the inclusion \eqref{eq omega1 incl} and Lemma \ref{disjoint frequency cubes}. - To show the inclusion in \eqref{eq dis freq cover}, let $\mfa \in \tQ(X)$. By the induction hypothesis, there exists $\fp \in \fP(J)$ such that $\mfa \in \Omega(\fp)$. By definition of the set $\fP$, we have $\fp = (J, z)$ for some $z \in Z(J)$. By \eqref{eq tile Z}, there exists $x \in X$ with $d_{J^\circ}(\tQ(x), z) \le 1$. By Lemma \ref{monotone cube metrics}, it follows that $d_{I^\circ}(\tQ(x), z) \le 1$. - Thus, by \eqref{eq tile cover}, there exists $z' \in Z(I)$ with $z \in B_{I^\circ}(z', 0.7)$. Then by Lemma \eqref{frequency cube cover} there exists $\fp' \in \fP(I)$ with $z \in Z(J) \cap \Omega_1(\fp')$. Consequently, by \eqref{eq it omega}, $\mfa \in \fc(\fp')$. This completes the proof of \eqref{eq dis freq cover}. + To show the inclusion in \eqref{eq dis freq cover}, let $\mfa \in \tQ(X)$. By the induction hypothesis, there exists $\fp \in \fP(J)$ such that $\mfa \in \Omega(\fp)$. By definition of the set $\fP$, we have $\fp = (J, z)$ for some $z \in \mathcal{Z}(J)$. By \eqref{eq tile Z}, there exists $x \in X$ with $d_{J^\circ}(\tQ(x), z) \le 1$. By Lemma \ref{monotone cube metrics}, it follows that $d_{I^\circ}(\tQ(x), z) \le 1$. + Thus, by \eqref{eq tile cover}, there exists $z' \in \mathcal{Z}(I)$ with $z \in B_{I^\circ}(z', 0.7)$. Then by Lemma \eqref{frequency cube cover} there exists $\fp' \in \fP(I)$ with $z \in \mathcal{Z}(J) \cap \Omega_1(\fp')$. Consequently, by \eqref{eq it omega}, $\mfa \in \fc(\fp')$. This completes the proof of \eqref{eq dis freq cover}. Finally, we show \eqref{eq freq dyadic}. Let $\fp, \fq \in \fP$ with $\sc(\fp) \subset \sc(\fp)$ and $\fc(\fp) \cap \fc(\fq) \ne \emptyset$. If we have $\ps(\fp) \ge \ps(\fq)$, then it follows from \eqref{dyadicproperty} that $I = J$, thus $\fp, \fq \in \fP(I)$. By \eqref{eq dis freq cover} we have then either $\fc(\fp) \cap \fc(\fq) = \emptyset$ or $\fc(\fp) = \fc(\fq)$. By the assumption in \eqref{eq freq dyadic} we have $\fc(\fp) \cap \fc(\fq) \ne \emptyset$, so we must have $\fc(\fp) = \fc(\fq)$ and in particular $\fc(\fq) \subset \fc(\fp)$. So it remains to show \eqref{eq freq dyadic} under the additional assumption that $\ps(\fq) > \ps(\fp)$. In this case, we argue by induction on $\ps(\fq)-\ps(\fp)$. By \eqref{coverdyadic}, there exists a cube $J \in \mathcal{D}$ with $s(J) = \ps(\fq) - 1$ and $J \cap\sc(\fp) \ne \emptyset$. We pick one such $J$. By \eqref{dyadicproperty}, we have $\sc(\fp) \subset J \subset \sc(\fq)$. By \eqref{eq tile Z}, there exists $x \in X$ with $d_{\fq}(\tQ(x), \fcc(\fq)) \le 1$. By Lemma \ref{monotone cube metrics}, it follows that $d_{J^\circ}(\tQ(x), \fcc(\fq)) \le 1$. - Thus, by \eqref{eq tile cover}, there exists $z' \in Z(J)$ with $\fcc(\fq) \in B_{J^\circ}(z', 0.7)$. Then by Lemma \eqref{frequency cube cover} there exists $\fq' \in \fP(J)$ with $\fcc(\fq) \in\Omega_1(\fq')$. + Thus, by \eqref{eq tile cover}, there exists $z' \in \mathcal{Z}(J)$ with $\fcc(\fq) \in B_{J^\circ}(z', 0.7)$. Then by Lemma \eqref{frequency cube cover} there exists $\fq' \in \fP(J)$ with $\fcc(\fq) \in\Omega_1(\fq')$. By \eqref{eq it omega}, it follows that $\Omega(\fq) \subset \Omega(\fq')$. Note that then $\sc(\fp) \subset \sc(\fq')$ and $\fc(\fp) \cap \fc(\fq') \ne \emptyset$ and $\ps(\fq') - \ps(\fp) = \ps(\fq) - \ps(\fp) - 1$. Thus, we have by the induction hypothesis that $\Omega(\fq') \subset \Omega(\fp)$. This completes the proof. \end{proof} @@ -1904,21 +1714,8 @@ \chapter{Proof of discrete Carleson (Prop. \ref{discrete Carleson}) } %\section{Decomposition of grid cubes and first exceptional set} -Let a grid structure $(\mathcal{D}, c, s)$ and a tile structure $(\fP,\sc,\fc,\fcc)$ -for this grid structure be given. -In Subsection \ref{subsectilesorg}, we decompose the -set $\fP$ of tiles into subsets. -Each subset will be controlled by one of three methods. -The guiding principle of the decomposition is -to be able to apply the forest estimate -of Proposition \ref{forest operator} to the final subsets -defined in \eqref{defc5}. This application is done in Subsection \ref{subsecforest}. -The miscellaneous subsets along the construction of the -forests will be either thrown into exceptional sets, -which are defined and controlled in Subsection -\ref{subsetexcset}, or will be controlled by -the antichain estimate of Proposition \ref{antichain operator}, -which is done in Subsection \ref{subsecantichain}. Subsection \ref{subsec lessim aux} contains some auxiliary lemmas needed for the proofs in Subsections \ref{subsecforest}-\ref{subsecantichain}. +Let a grid structure $(\mathcal{D}, c, s)$ and a tile structure $(\fP, \sc, \fc, \fcc)$ for this grid structure be given. In Subsection \ref{subsectilesorg}, we decompose the set $\fP$ of tiles into subsets. Each subset will be controlled by one of three methods. The guiding principle of the decomposition is to be able to apply the forest estimate of Proposition \ref{forest operator} to the final subsets defined in \eqref{defc5}. This application is done in Subsection \ref{subsecforest}. The miscellaneous subsets along the construction of the forests will either be thrown into exceptional sets, which are defined and controlled in Subsection \ref{subsetexcset}, or will be controlled by the antichain estimate of Proposition \ref{antichain operator}, which is done in Subsection \ref{subsecantichain}. Subsection \ref{subsec lessim aux} contains some auxiliary lemmas needed for the proofs in Subsections \ref{subsecforest}-\ref{subsecantichain}. + %The oganisation and summation of these various %estimates is done in Subsection \ref{subsecaddingup}. @@ -1927,16 +1724,16 @@ \chapter{Proof of discrete Carleson (Prop. \ref{discrete Carleson}) } \section{Organisation of the tiles}\label{subsectilesorg} -In the following chain of definitions, $k, n$, and +In the following definitions, $k, n$, and $j$ will be nonnegative integers. Define $\mathcal{C}(G,k)$ to be the set of $I\in \mathcal{D}$ such that there exists a $J\in \mathcal{D}$ with $I\subset J$ and \begin{equation}\label{muhj1} - {\mu(G \cap J)} > 2^{-k-1}{\mu(G)}\, , + {\mu(G \cap J)} > 2^{-k-1}{\mu(J)}\, , \end{equation} -but there does not exists a $J\in \mathcal{D}$ with $I\subset J$ and +but there does not exist a $J\in \mathcal{D}$ with $I\subset J$ and \begin{equation}\label{muhj2} {\mu(G \cap J)} > 2^{-k}{\mu(J)}\,. \end{equation} @@ -1949,7 +1746,7 @@ \section{Organisation of the tiles}\label{subsectilesorg} \begin{equation}\label{ebardense} \mu({E_1}(\fp)) > 2^{-n} \mu(\sc(\fp)) \end{equation} -and there does not exists $\fp'\in \fP(k)$ with +and there does not exist $\fp'\in \fP(k)$ with $\fp'\neq \fp$ and $\fp\le \fp'$ such that \begin{equation}\label{mnkmax} \mu({E_1}(\fp')) > 2^{-n} \mu(\sc(\fp')). @@ -2017,7 +1814,7 @@ \section{Organisation of the tiles}\label{subsectilesorg} \fL_2(k,n,j) \end{equation} to be the set of all $\fp\in \fC_2(k,n,j)$ such that there -does not exists +does not exist $\fu\in \fU_1(k,n,j)$ with $\sc(\fp)\neq \sc(\fu)$ and $2\fp\lesssim \fu$. Define @@ -2087,7 +1884,7 @@ \section{Organisation of the tiles}\label{subsectilesorg} and define \begin{equation}\label{definegone2} G_2:= -\bigcup_{0\le k}\bigcup_{k< n} +\bigcup_{k\ge 0}\bigcup_{k< n} A(2n+6,k,n)\, . \end{equation} Define @@ -2101,7 +1898,8 @@ \section{Organisation of the tiles}\label{subsectilesorg} Define $G'=G_1\cup G_2 \cup G_3$ The following bound of the measure of $G'$ will be proven in Subsection \ref{subsetexcset}. -\begin{lemma}\label{exceptional set} +\begin{lemma}[exceptional set] +\label{exceptional set} \uses{first exception,dense cover,pairwise disjoint,dyadic union,tree count,boundary measure} We have \begin{equation} @@ -2112,7 +1910,8 @@ \section{Organisation of the tiles}\label{subsectilesorg} In Subsection \ref{subsecforest}, we identify each set $\fC_5(k,n,j)$ outside $G'$ as forest and use Proposition \ref{forest operator} to prove the following lemma. -\begin{lemma}\label{forest union} +\begin{lemma}[forest union] +\label{forest union} \uses{relation geometry,C6 forest, C6 convex, forest separation, forest stacking} Let \begin{equation} @@ -2131,12 +1930,13 @@ \section{Organisation of the tiles}\label{subsectilesorg} \ref{forest union} and apply the antichain estimate of Proposition \ref{antichain operator} to prove the following lemma. -\begin{lemma}\label{forest complement} -\uses{antichain decomposition,L0 antichain,l2 antichain,L1 L3 antichain} +\begin{lemma}[forest complement] +\label{forest complement} +\uses{antichain decomposition,L0 antichain,L2 antichain,L1 L3 antichain} Let \begin{equation} - \fP_2 =\fP\setminus \left(\bigcup_{k\ge 0}\bigcup_{n\ge k} - \bigcup_{0\le j\le 2n+3}\fC_5(k,n,j)\right)\,. + \fP_2 =\fP\setminus \fP_1 %\left(\bigcup_{k\ge 0}\bigcup_{n\ge k} + % \bigcup_{0\le j\le 2n+3}\fC_5(k,n,j)\right)\,. \end{equation} For all $f:X\to \C$ with $|f|\le \mathbf{1}_F$ we have \begin{equation} @@ -2147,8 +1947,9 @@ \section{Organisation of the tiles}\label{subsectilesorg} Proposition \ref{discrete Carleson} follows by applying triangle inequality to \eqref{disclesssim} according to the splitting in Lemma \ref{forest union} -and \ref{forest complement} and using both Lemmas as well +and Lemma \ref{forest complement} and using both Lemmas as well as the bound on the set $G'$ given by Lemma \ref{exceptional set}. +\asgar{Perhaps $\frac{2^{260 a^3+1}}{(q-1)^4}$ by triangle inequality.} @@ -2162,7 +1963,9 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} The bound for $G_1$ is follows from the Vitali covering lemma, Proposition \ref{Hardy Littlewood}. -\begin{lemma}\label{first exception} +\begin{lemma} +[first exception]\label{first exception} +\uses{Hardy Littlewood} We have \begin{equation} \mu(G_1)\le 2^{-4}\mu(G)\, . @@ -2183,23 +1986,24 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} $$ \mathcal{B} = \{B(\pc(\fp),r(\fp)) \ : \ \fp \in \fP_{F,G}\} $$ -and the function $h = \mathbf{1}_F$, we obtain +and the function $u = \mathbf{1}_F$, we obtain $$ \mu(\bigcup \mathcal{B}) \le 2^{2a+1} K^{-1} \mu(F)\,. $$ We conclude with \eqref{eq vol sp cube} and $r(\fp)>4D^{\ps(\fp)}$ $$ - \mu(G_1)\le \mu(\bigcup_{\fp\in \fP_{F,G}} \sc(\fp)) + \mu(G_1)= \mu(\bigcup_{\fp\in \fP_{F,G}} \sc(\fp)) \le \mu(\bigcup \mathcal{B})\le 2^{2a+1} K^{-1} \mu (F) = 2^{-4}\mu(G)\,. $$ \end{proof} -We turn to the bound of $G_2$, which relies on the dyadic covering Lemma \ref{dense cover} and the -John-Nirenberg Lemma \ref{john nirenberg} below. +We turn to the bound of $G_2$, which relies on the Dyadic Covering Lemma \ref{dense cover} and the +John-Nirenberg Lemma \ref{John Nirenberg} below. -\begin{lemma}\label{dense cover} -\uses{john nirenberg} +\begin{lemma}[dense cover] +\label{dense cover} +\uses{John Nirenberg} For each $k\ge 0$, the union of all dyadic cubes in $\mathcal{C}(G,k)$ has measure at most $2^{k+1} \mu(G)$ . \end{lemma} @@ -2216,7 +2020,7 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} \sum_{J\in \mathcal{M}^*(k)}\mu(J) \end{equation} Using the definition of $\mathcal{M}(k)$ and then -the pairwise disjointness of elements in +the pairwise disjointedness of elements in $\mathcal{M}^*(k)$, we estimate \eqref{cbymstar} by \begin{equation} @@ -2230,8 +2034,9 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} -\begin{lemma}\label{pairwise disjoint} -\uses{john nirenberg} +\begin{lemma}[pairwise disjoint] +\label{pairwise disjoint} +\uses{John Nirenberg} If $\fp, \fp' \in {\mathfrak{M}}(k,n)$ and \begin{equation}\label{eintersect} {E_1}(\fp)\cap {E_1}(\fp')\neq \emptyset, @@ -2239,7 +2044,7 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} then $\fp=\fp'$. \end{lemma} \begin{proof} -Let $\fp,\fp'$ as in the lemma. As by definition of $E_1$ +Let $\fp,\fp'$ be as in the lemma. By definition of $E_1$, we have $E_1(\fp)\subset \sc(\fp)$ and analogously for $\fp'$, we conclude from \eqref{eintersect} that $\sc(\fp)\cap \sc(\fp')\neq \emptyset$. Let without loss of generality $\sc(\fp)$ be maximal in $\{\sc(\fp),\sc(\fp')\}$, then $\sc(\fp')\subset \sc(\fp)$. @@ -2250,8 +2055,9 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} \end{proof} -\begin{lemma}\label{dyadic union} -\uses{john nirenberg} +\begin{lemma}[dyadic union] +\label{dyadic union} +\uses{John Nirenberg} For each $x\in A(\lambda,k,n)$, there is a dyadic cube $I$ that contains $x$ and is @@ -2260,21 +2066,16 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} \end{lemma} \begin{proof} - -Fix $k,n,\lambda,x$ as in the lemma such that -$x\in A(\lambda,k,n)$. Let -$\mathcal{M}$ be the set of dyadic cubes - $\sc(\fp)$ -with $\fp$ in $\mathfrak{M}(k,n)$ -and $x\in \sc(\fp)$. By definition of -$A(\lambda,k,n)$, the cardinality of $\mathcal{M}$ -is at least $\lambda$. Let $I$ be an cube of -smallest scale in $\mathcal{M}$. Then -$I$ is contained in all cubes of $\mathcal{M}$. +Fix $k,n,\lambda,x$ as in the lemma such that $x\in A(\lambda,k,n)$. +Let $\mathcal{M}$ be the set of dyadic cubes $\sc(\fp)$ with $\fp$ in $\mathfrak{M}(k,n)$ and $x\in \sc(\fp)$. +By definition of $A(\lambda,k,n)$, the cardinality of $\mathcal{M}$ is at least $1+\lambda 2^{n+1}$. +Let $I$ be a cube of smallest scale in $\mathcal{M}$. +Then $I$ is contained in all cubes of $\mathcal{M}$. It follows that $I\subset A(\lambda,k,n)$. \end{proof} -\begin{lemma}\label{john nirenberg} +\begin{lemma}[John Nirenberg] +\label{John Nirenberg} \uses{second exception, top tiles} For all integers $k,n,\lambda\ge 0$, we have \begin{equation}\label{alambdameasure} @@ -2290,7 +2091,7 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} We prove the lemma by induction on $\lambda$. For $\lambda=0$, we use that $A(\lambda)$ by definition of $\mathfrak{M}(k,n)$ is contained in the union of elements in $ \mathcal{C}(G,k)$. Lemma \ref{dense cover} then completes the base of the induction. -Now assume that the statement of Lemma \ref{john nirenberg} +Now assume that the statement of Lemma \ref{John Nirenberg} is proven for some integer $\lambda\ge 0$. The set $A(\lambda+1)$ is contained in the set $A(\lambda)$. Let $\mathcal{M}$ be the set of dyadic cubes which are a subset of $A(\lambda)$. By Lemma \ref{dyadic union}, the union of $\mathcal{M}$ is $A(\lambda)$. @@ -2298,9 +2099,9 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} Let $L\in \mathcal{M}^*$. For each $x\in L$, we have \begin{equation}\label{suminout} - \sum_{\fp \in {\mathfrak{M}}(k,n)} \mathbf{1}_{I(\fp)}(x)= - \sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \subset L} \mathbf{1}_{I(\fp)}(x)+ - \sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \not \subset L} \mathbf{1}_{I(\fp)}(x)\, . + \sum_{\fp \in {\mathfrak{M}}(k,n)} \mathbf{1}_{\sc(\fp)}(x)= + \sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \subset L} \mathbf{1}_{\sc(\fp)}(x)+ + \sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \not \subset L} \mathbf{1}_{\sc(\fp)}(x)\, . \end{equation} If the second sum on the right-hand-side is not zero, there is an element of $\mathcal{D}$ strictly containing $L$. @@ -2315,7 +2116,7 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} the left-hand-side of \eqref{suminout} is at least $1+(\lambda+1) 2^{n+1}$, so we have by the triangle inequality for the first sum on the right-hand side \begin{equation}\label{mnkonl} -\sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \subset L} \mathbf{1}_{I(\fp)}(x)\ge 2^{n+1}\, .\end{equation} +\sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \subset L} \mathbf{1}_{\sc(\fp)}(x)\ge 2^{n+1}\, .\end{equation} By Lemma \ref{pairwise disjoint}, we have \begin{equation} \sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \subset L} \mu({E_1}(\fp)) \leq \mu(L)\, . @@ -2331,7 +2132,7 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} \end{equation} \begin{equation} \le - \int \sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \subset L} \mathbf{1}_{I(\fp)} d\mu + \int \sum_{\fp \in {\mathfrak{M}}(k,n):\sc(\fp) \subset L} \mathbf{1}_{\sc(\fp)} d\mu \le 2^n \mu(L)\, . \end{equation} Hence @@ -2344,7 +2145,8 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} \eqref{alambdameasure} for $\lambda+1$ and completes the proof of the lemma. \end{proof} -\begin{lemma}\label{second exception} +\begin{lemma}[second exception] +\label{second exception} We have \begin{equation} \mu(G_2)\le 2^{-4} \mu(G)\, . @@ -2352,7 +2154,7 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} \end{lemma} \begin{proof} -We use Lemma \ref{john nirenberg} and sum twice a geometric series +We use Lemma \ref{John Nirenberg} and sum twice a geometric series to obtain \begin{equation} \sum_{0\le k}\sum_{k< n} @@ -2367,7 +2169,8 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} We turn to the set $G_3$. -\begin{lemma}\label{top tiles} +\begin{lemma}[top tiles] +\label{top tiles} We have \begin{equation}\label{eq musum} \sum_{\mathfrak{m} \in \mathfrak{M}(k,n)} \mu(\sc(\mathfrak{m}))\le 2^{n+1}2^{k+1}\mu(G). @@ -2379,7 +2182,7 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} \int \sum_{\mathfrak{m} \in \mathfrak{M}(k,n)} \mathbf{1}_{\sc(\mathfrak{m})}(x) \, d\mu(x) \le 2^{n+1} \sum_{\lambda=1}^{|\mathfrak{M}|}\mu(A(\lambda, k,n))\,. \end{equation} -Using Lemma \ref{john nirenberg} +Using Lemma \ref{John Nirenberg} %\ref{alambdameasure} and then summing a geometric series, we estimate this by \begin{equation} @@ -2393,7 +2196,8 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} \end{proof} -\begin{lemma}\label{tree count} +\begin{lemma}[tree count] +\label{tree count} \uses{third exception} Let $k,n,j\ge 0$. We have for every $x\in X$ \begin{equation} @@ -2466,7 +2270,8 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} Inserting this into \eqref{usumbymsum} proves the lemma. \end{proof} -\begin{lemma}\label{boundary exception} +\begin{lemma}[boundary exception] +\label{boundary exception} \uses{third exception} Let $\mathcal{L}(\fu)$ be as defined in \eqref{eq L def}. We have for each $\fu\in \fU_1(k,n,l)$, \begin{equation} @@ -2513,7 +2318,8 @@ \section{Proof of Lemma \ref{exceptional set}, the exceptional sets} - \begin{lemma}\label{third exception} + \begin{lemma}[third exception] + \label{third exception} We have \begin{equation} @@ -2584,7 +2390,7 @@ \section{Auxiliary lemmas} \label{subsec lessim aux} Before proving Lemma \ref{forest union} and Lemma \ref{forest complement}, we collect some useful properties of $\lesssim$. -\begin{lemma} +\begin{lemma}[wiggle order 1] \label{wiggle order 1} \uses{wiggle order 3} If $n\fp \lesssim m\fp'$ and @@ -2595,7 +2401,7 @@ \section{Auxiliary lemmas} This follows immediately from the definition \eqref{wiggleorder} of $\lesssim$ and the two inclusions $B_{\fp}(\fcc(\fp), n) \subset B_{\fp}(\fcc(\fp), n')$ and $B_{\fp'}(\fcc(\fp'), m') \subset B_{\fp'}(\fcc(\fp'), m)$. \end{proof} -\begin{lemma} +\begin{lemma}[wiggle order 2] \label{wiggle order 2} \uses{wiggle order 3} Let $n, m \ge 1$. @@ -2623,7 +2429,7 @@ \section{Auxiliary lemmas} Thus $B_{\fp'}(\fcc(\fp'),n) \subset B_{\fp}(\fcc(\fp),n + 2^{-95a}m)$. Combined with $\sc(\fp) \subset \sc(\fp')$, this yields \eqref{eq wiggle2}. \end{proof} -\begin{lemma} +\begin{lemma}[wiggle order 3] \label{wiggle order 3} \uses{C convex, relation geometry} The following implications hold for all $\fq, \fq' \in \fP$: @@ -2653,7 +2459,7 @@ \section{Auxiliary lemmas} \fp \le \fp' \le \fp'' \ \text{and} \ \fp, \fp'' \in \mathfrak{A} \implies \fp' \in \mathfrak{A}\,. \end{equation} -\begin{lemma} +\begin{lemma}[P convex] \label{P convex} \uses{C convex} For each $k$, the collection $\fP(k)$ is convex. @@ -2667,7 +2473,7 @@ \section{Auxiliary lemmas} and $\mu(G \cap J) > 2^{-k-1} \mu(G)$. Thus \eqref{muhj1} holds for $\sc(\fp')$. On the other hand, by \eqref{muhj2}, there exists no $J \in \mathcal{D}$ with $\sc(\fp) \subset J$ and $\mu(G \cap J) > 2^{-k} \mu(G)$. Since $\sc(\fp) \subset \sc(\fp')$, this implies that \eqref{muhj2} holds for $\sc(\fp')$. Hence $\sc(\fp') \in \mathcal{C}(G,k)$, and therefore by \eqref{eq defPk} $\fp' \in \fP(k)$. \end{proof} -\begin{lemma} +\begin{lemma}[C convex] \label{C convex} \uses{C1 convex} For each $k,n$, the collection $\fC(k,n)$ is convex. @@ -2685,7 +2491,7 @@ \section{Auxiliary lemmas} Thus $\fp' \in \fC(k,n)$. \end{proof} -\begin{lemma} +\begin{lemma}[C1 convex] \label{C1 convex} \uses{C2 convex} For each $k,n,j$, the collection $\fC_1(k,n,j)$ is convex. @@ -2699,7 +2505,7 @@ \section{Auxiliary lemmas} thus $\fp' \in \fC_1(k,n,j)$. \end{proof} -\begin{lemma} +\begin{lemma}[C2 convex] \label{C2 convex} \uses{C3 convex} For each $k,n,j$, the collection $\fC_2(k,n,j)$ is convex. @@ -2712,7 +2518,7 @@ \section{Auxiliary lemmas} \end{equation*} Combined with Lemma \ref{C1 convex}, it follows that $\fp' \in \fC_1(k,n,j)$. Suppose that $\fp' \notin \fC_2(k,n,j)$. By \eqref{eq C2 def}, this implies that there exists $0 \le l' \le Zn$ with $\fp' \in \fL_1(k,n,j,l')$. By the definition \eqref{eq L1 def} of $\fL_1(k,n,j,l')$, this implies that $\fp$ is minimal with respect to $\le$ in $\fC_1(k,n,j) \setminus \bigcup_{l < l'} \fL_1(k,n,j,l)$. Since $\fp \in \fC_2(k,n,j)$ we must have $\fp \ne \fp'$. Thus $\fp \le \fp'$ and $\fp \ne \fp'$. By minimality of $\fp'$ it follows that $\fp \notin \fC_1(k,n,j) \setminus \bigcup_{l < l'} \fL_1(k,n,j,l)$. But by \eqref{eq C2 def} this implies $\fp \notin \fC_2(k,n,j)$, a contradiction. \end{proof} -\begin{lemma} +\begin{lemma}[C3 convex] \label{C3 convex} \uses{C4 convex} For each $k,n,j$, the collection $\fC_3(k,n,j)$ is convex. @@ -2722,7 +2528,7 @@ \section{Auxiliary lemmas} Let $\fp \le \fp' \le \fp''$ with $\fp, \fp'' \in \fC_3(k,n,j)$. By \eqref{eq C3 def} and Lemma \ref{C2 convex} it follows that $\fp' \in \fC_2(k,n,j)$. Suppose that $\fp' \notin \fC_3(k,n,j)$. Then, by \eqref{eq C3 def} and \eqref{eq L2 def}, there exists $\fu \in \fU_1(k,n,j)$ with $2\fp' \lesssim \fu$ and $\sc(\fp') \ne \sc(\fu)$. Together this gives $\sc(\fp') \subsetneq \sc(\fu)$. From $\fp' \le \fp$, \eqref{eq sc1} and transitivity of $\lesssim$ we then have $2\fp \lesssim \fu$. Also, $\sc(\fp) \subset \sc(\fp') \subsetneq \sc(\fu)$, so $\sc(\fp) \ne \sc(\fu)$. But then $\fp \in \fL_2(k,n,j)$, contradicting by \eqref{eq C3 def} the assumption $\fp \in \fC_3(k,n,j)$. \end{proof} -\begin{lemma} +\begin{lemma}[C4 convex] \label{C4 convex} \uses{C5 convex} For each $k,n,j$, the collection $\fC_4(k,n,j)$ is convex. @@ -2732,7 +2538,7 @@ \section{Auxiliary lemmas} Let $\fp \le \fp' \le\fp''$ with $\fp, \fp'' \in \fC_4(k,n,j)$. As before we obtain from the inclusion $\fC_4(k,n,j) \subset \fC_3(k,n,j)$ and Lemma \ref{C3 convex} that $\fp' \in \fC_3(k,n,j)$. Thus, if $\fp' \notin \fC_4(k,n,j)$ then by \eqref{eq L3 def} there exists $l$ such that $\fp' \in \fL_3(k,n,j,l)$. Thus $\fp'$ is maximal with respect to $\le$ in $\fC_3(k,n,j) \setminus \bigcup_{0 \le l' < l} \fL_3(k,n,j,l')$. Since $\fp'' \in \fC_4(k,n,j)$ we must have $\fp' \ne \fp''$. Thus $\fp' \le \fp''$ and $\fp'\ne \fp''$. By minimality of $\fp'$ it follows that $\fp'' \notin \fC_3(k,n,j) \setminus \bigcup_{l < l'} \fL_3(k,n,j,l)$. But by \eqref{eq C4 def} this implies $\fp'' \notin \fC_4(k,n,j)$, a contradiction. \end{proof} -\begin{lemma} +\begin{lemma}[C5 convex] \label{C5 convex} \uses{C6 convex} For each $k,n,j$, the collection $\fC_5(k,n,j)$ is convex. @@ -2743,7 +2549,7 @@ \section{Auxiliary lemmas} By \eqref{eq L4 def}, there exists $\fu \in \fU_1(k,n,j)$ with $\sc(\fp') \subset \bigcup \mathcal{L}(\fu)$. Then also $\sc(\fp) \subset \bigcup \mathcal{L}(\fu)$, a contradiction. \end{proof} -\begin{lemma} +\begin{lemma}[dens compare] \label{dens compare} \uses{C dens1} We have for every $k\ge 0$ and $\fP'\subset \fP(k)$ @@ -2787,7 +2593,7 @@ \section{Auxiliary lemmas} \eqref{muhj2} for $\fp$. \end{proof} -\begin{lemma} +\begin{lemma}[C dens1] \label{C dens1} For each set $\mathfrak{A} \subset \mathfrak{C}(k,n)$, we have $$ @@ -2835,7 +2641,7 @@ \section{Proof of Lemma \ref{forest union}, the forests} if $\fu=\fu'$ or there exists $\fp$ in $\mathfrak{T}_1(\fu)$ with $10 \fp\lesssim \fu'$. -\begin{lemma} +\begin{lemma}[relation geometry] \label{relation geometry} \uses{forest geometry,equivalence relation,forest inner} If $\fu \sim \fu'$, then $\sc(u) = \sc(u')$ and @@ -2866,7 +2672,7 @@ \section{Proof of Lemma \ref{forest union}, the forests} \end{proof} -\begin{lemma} +\begin{lemma}[equivalence relation] \label{equivalence relation} For each $k,n,j$, the relation $\sim$ on $\fU_2(k,n,j)$ is an equivalence relation. @@ -2929,7 +2735,7 @@ \section{Proof of Lemma \ref{forest union}, the forests} \bigcup_{\fu\sim \fu'}\mathfrak{T}_1(\fu')\cap \fC_6(k,n,j)\, . \end{equation} -\begin{lemma} +\begin{lemma}[C6 forest] \label{C6 forest} We have \begin{equation} @@ -2941,7 +2747,7 @@ \section{Proof of Lemma \ref{forest union}, the forests} By \eqref{eq C4 def} and \eqref{defc5}, we have $\fp \in \fC_3(k,n,j)$. By \eqref{eq L2 def} and \eqref{eq C3 def}, there exists $\fu \in \fU_1(k,n,j)$ with $2\fp \lesssim \fu$ and $\sc(\fp) \ne \sc(\fu)$, that is, with $\fp \in \mathfrak{T}_1(\fu)$. Then $\mathfrak{T}_1(\fu)$ is clearly nonempty, so $\fu \in \fU_2(k,n,j)$. By the definition of $\fU_3(k,n,j)$, there exists $\fu' \in \fU_3(k,n,j)$ with $\fu \sim \fu'$. By \eqref{definesv}, we have $\fp \in \mathfrak{T}_2(\fu')$. \end{proof} -\begin{lemma} +\begin{lemma}[C6 convex] \label{C6 convex} \uses{forest convex} Let $\fu \in \fU_3(k,n,j)$. If $\fp \le \fp' \le \fp''$ and $\fp, \fp'' \in \mathfrak{T}_2(\fu)$, then $\fp' \in \mathfrak{T}_2(\fu)$. @@ -2953,7 +2759,7 @@ \section{Proof of Lemma \ref{forest union}, the forests} \end{proof} -\begin{lemma} +\begin{lemma}[forest geometry] \label{forest geometry} For each $\fu\in \fU_3(k,n,j)$, the set $\mathfrak{T}_2(\fu)$ @@ -2976,7 +2782,7 @@ \section{Proof of Lemma \ref{forest union}, the forests} Together with $\sc(\fp) \subset \sc(\fu) = \sc(\fu')$, this gives $4\fp \lesssim 1\fu'$, which is \eqref{forest1}. \end{proof} -\begin{lemma} +\begin{lemma}[forest convex] \label{forest convex} For each $\fu\in \fU_3(k,n,j)$, the set $\mathfrak{T}_2(\fu)$ @@ -2991,7 +2797,7 @@ \section{Proof of Lemma \ref{forest union}, the forests} It follows that $\fp' \in \mathfrak{T}_2(\fu')$, which shows \eqref{forest2}. \end{proof} -\begin{lemma} +\begin{lemma}[forest separation] \label{forest separation} For each $\fu,\fu'\in \fU_3(k,n,j)$ with $\fu\neq \fu'$ and and each $\fp \in \fT_2(\fu)$ @@ -3019,7 +2825,7 @@ \section{Proof of Lemma \ref{forest union}, the forests} The Lemma follows by combining the two displays with the fact that $95 a \ge 1$. \end{proof} -\begin{lemma} +\begin{lemma}[forest inner] \label{forest inner} For each $\fu\in \fU_3(k,n,j)$ and each $\fp \in \mathfrak{T}_2(\fu)$ @@ -3042,7 +2848,7 @@ \section{Proof of Lemma \ref{forest union}, the forests} \end{proof} -\begin{lemma} +\begin{lemma}[forest stacking] \label{forest stacking} It holds that \begin{equation} @@ -3098,7 +2904,7 @@ \section{Proof of Lemma \ref{forest complement}, the antichains} \label{subsecantichain} Define $\fP_{X \setminus G'}$ to be the set of all $\fp \in \fP$ such that $\sc(\fp) \not \subset G'$. -\begin{lemma} +\begin{lemma}[antichain decomposition] \label{antichain decomposition} We have that \begin{align} @@ -3119,7 +2925,7 @@ \section{Proof of Lemma \ref{forest complement}, the antichains} Since $\fp \in \fP_{X \setminus G'}$, we have in particular $\fp \not \subset A(2n + 6, k, n)$, so there exist at most $1 + (4n + 12)2^n < 2^{2n+4}$ tiles $\mathfrak{m} \in \mathfrak{M}(k,n)$ with $\fp \le \mathfrak{m}$. It follows that $\fp \in \fL_0(k,n)$ or $\fp \in \fC_1(k,n,j)$ for some $1 \le j \le 2n + 3$. In the former case we are done, in the latter case the inclusion to be shown follows immediately from the definitions of the collections $\fC_i$ and $\fL_i$. \end{proof} -\begin{lemma} +\begin{lemma}[L0 antichain] \label{L0 antichain} We have that $$ @@ -3171,7 +2977,7 @@ \section{Proof of Lemma \ref{forest complement}, the antichains} Thus, by \eqref{straightorder}, $2\fp_0 \leq 100\mathfrak{m}$, a contradiction to $\fp_0 \notin \fC(k,n)$. \end{proof} -\begin{lemma} +\begin{lemma}[L2 antichain] \label{L2 antichain} Each of the sets $\fL_2(k,n,j)$ is an antichain. \end{lemma} @@ -3181,7 +2987,7 @@ \section{Proof of Lemma \ref{forest complement}, the antichains} If we have $\fp_l \in \fU_1(k,n,j)$, then it follows from $2\fp \lesssim 200 \fp_l \lesssim \fp_l$ and \eqref{eq L2 def} that $\fp \not\in \fL_2(k,n,j)$, a contradiction. Thus, by the definition \eqref{defunkj} of $\fU_1(k,n,j)$, there exists $\fp_{l+1} \in \fC_1(k,n,j)$ with $\sc(\fp_l) \subsetneq \sc(\fp_{l+1}) $ and $\mfa \in B_{\fp_l}(\fcc(\fp_l), 100) \cap B_{\fp_{l+1}}(\fcc(\fp_{l+1}), 100)$. Using the triangle inequality and Lemma \ref{monotone cube metrics}, one deduces that $200 \fp_l \lesssim 200\fp_{l+1}$. This contradicts maximality of $l$. \end{proof} -\begin{lemma} +\begin{lemma}[L1 L3 antichainf] \label{L1 L3 antichain} Each of the sets $\fL_1(k,n,j,l)$ and $\fL_3(k,n,j,l)$ is an antichain. \end{lemma} @@ -3236,13 +3042,13 @@ \chapter{Proof of Proposition \ref{antichain operator}} as the geometric mean of two inequalities, each involving one of the two densities. One of these two inequalities will need a careful estimate formulated in -Lemma \ref{lem basic TT*} of +Lemma \ref{tile correlation} of the $TT^*$ correlation between two tile operators of two tiles. -Lemma \ref{lem basic TT*} will be proven in +Lemma \ref{tile correlation} will be proven in Subsection \ref{sec tile operator}. The summation of the contributions of these individual correlations will require a -geometric Lemma \ref{lem antichain 1} counting the relevant tile pairs. -Lemma \ref{lem antichain 1} will be proven in Subsection +geometric Lemma \ref{antichain tile count} counting the relevant tile pairs. +Lemma \ref{antichain tile count} will be proven in Subsection \ref{subsec geolem}. @@ -3252,9 +3058,9 @@ \chapter{Proof of Proposition \ref{antichain operator}} \section{The density arguments}\label{sec TT* T*T} -We begin with the following crucial disjointness property of the sets $E(\fp)$ with $\fp \in \mathfrak{A}$. -\begin{lemma} -\label{lem antichain -1} +We begin with the following crucial disjointedness property of the sets $E(\fp)$ with $\fp \in \mathfrak{A}$. +\begin{lemma}[tile disjointness] +\label{tile disjointness} Let $\fp,\fp'\in \mathfrak{A}$. If there exists an $x\in X$ with $x\in E(\fp)\cap E(\fp')$, then $\fp= \fp'$. @@ -3277,7 +3083,8 @@ \section{The density arguments}\label{sec TT* T*T} \end{equation} with $\fp\in \mathfrak{A}$ and recall the definition of $M_{\mathcal{B}}$ from Proposition \ref{Hardy Littlewood}. -\begin{lemma}\label{lem hlmbound} +\begin{lemma}[maximal bound antichain]\label{maximal bound antichain} +\uses{tile disjointness} Let $x\in X$. Then \begin{equation}\label{hlmbound} @@ -3288,7 +3095,7 @@ \section{The density arguments}\label{sec TT* T*T} \begin{proof} -Fix $x\in X$. By Lemma \ref{lem antichain -1}, there is at most one $\fp \in \mathfrak{A}$ +Fix $x\in X$. By Lemma \ref{tile disjointness}, there is at most one $\fp \in \mathfrak{A}$ such that $T_{\fp} f(x)$ is not zero. If there is no such $\fp$, the estimate \eqref{hlmbound} follows. @@ -3333,8 +3140,9 @@ \section{The density arguments}\label{sec TT* T*T} \tilde{q}=\frac {2q}{1+q}\,. \end{equation} Since $1< q\le 2$, we have $1<\tilde{q} s_1} \psi(D^{-s}x)\,. @@ -4569,14 +4392,14 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate} and the doubling property \eqref{doublingx} $6 +100a^2$ times estimates the last display by \begin{equation} - \label{pf lem nontangential imeq} + \label{pf nontangential operator bound imeq} \le \frac{2^{6a+101a^3}}{\mu(B(x', 16D^{s_1}))}\, . \end{equation} By the triangle inequality and \eqref{eq vol sp cube}, we have $$ B(x', 8D^{s_1}) \subset B(c(I_{s_1}(x)), 16D^{s(I_{s_1}(x))})\,. $$ - Combining this with \eqref{pf lem nontangential imeq}, we conclude that \eqref{eq lower bound term} is at most + Combining this with \eqref{pf nontangential operator bound imeq}, we conclude that \eqref{eq lower bound term} is at most $$ 2^{6a + 101a^3} M_{\mathcal{B},1} f(x)\,. $$ @@ -4609,8 +4432,8 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate} We need the following lemma to prepare the $L^2$-estimate for the auxiliary operators $S_{1, \fu}$. -\begin{lemma} - \label{lem aux overlap} +\begin{lemma}[boundary overlap] + \label{boundary overlap} For every cube $I \in \mathcal{D}$, there exist at most $2^{8a}$ cubes $J \in \mathcal{D}$ with $s(J) = s(I)$ and $B(c(I), 16D^{s(I)}) \cap B(c(J), 16 D^{s(J)}) \ne \emptyset$. \end{lemma} @@ -4624,7 +4447,7 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate} If $\mathcal{C}$ is any finite collection of cubes $J \in \mathcal{D}$ satisfying $s(J) = s(I)$ and \begin{equation*} B(c(I), 16 D^{s(I)}) \cap B(c(J), 16 D^{s(J)}) \ne\emptyset\ , - \end{equation*} then it follows from \eqref{eq vol sp cube} and pairwise disjointness of cubes of the same scale \eqref{dyadicproperty} that the balls $B(c(J), \frac{1}{4} D^{s(J)})$ are pairwise disjoint. Hence + \end{equation*} then it follows from \eqref{eq vol sp cube} and pairwise disjointedness of cubes of the same scale \eqref{dyadicproperty} that the balls $B(c(J), \frac{1}{4} D^{s(J)})$ are pairwise disjoint. Hence \begin{align*} \mu(B(c(I), 32 D^{s(I)})) &\ge \sum_{J \in \mathcal{C}} \mu(B(c(J), \frac{1}{4}D^{s(J)}))\\ &\ge |\mathcal{C}| 2^{-8a} \mu(B(c(I), 32 D^{s(I)}))\,. @@ -4634,7 +4457,7 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate} Now we can bound the operators $S_{1, \fu}$. -\begin{proof}[Proof of Lemma \ref{lem L2 Su estimate}] +\begin{proof}[Proof of Lemma \ref{boundary operator bound}] Note that by definition, $S_{1,\fu}f$ is a finite sum of indicator functions of cubes $I \in \mathcal{D}$ for each locally integrable $f$, and hence is bounded, has bounded support and is integrable. Let $g$ be another function with the same three properties. Then $\bar g S_{1,\fu}f$ is integrable, and we have $$ \Bigg|\int \bar g(y) S_{1,\fu}f(y) \, \mathrm{d}\mu(y)\Bigg| @@ -4648,18 +4471,17 @@ \section{An auxiliary \texorpdfstring{$L^2$}{L2} tree estimate} \end{multline*} Changing the order of summation and using $J \subset B(c(I), 16 D^{s(I)})$ to bound the first average integral by $M_{\mathcal{B},1}|g|(y)$ for any $y \in J$, we obtain \begin{align} - \label{eq lem L2 Su estimate 1} + \label{eq boundary operator bound 1} \le \sum_{J\in\mathcal{J}}\int_J|f(y)| M_{\mathcal{B},1}|g|(y) \, \mathrm{d}\mu(y) \sum_{I \in \mathcal{D} \, : \, J\subset B(c(I),16 D^{s(I)})} D^{(s(J)-s(I))/a}. \end{align} - By \eqref{eq vol sp cube} and \eqref{defineD} the condition $J \subset B(c(I), 16 D^{s(I)})$ implies $s(I) \ge s(J)$. By Lemma \ref{lem aux overlap}, there are at most $2^{8a}$ cubes $I$ at each scale with $J \subset B(c(I), D^{s(I)})$. + By \eqref{eq vol sp cube} and \eqref{defineD} the condition $J \subset B(c(I), 16 D^{s(I)})$ implies $s(I) \ge s(J)$. By Lemma \ref{boundary overlap}, there are at most $2^{8a}$ cubes $I$ at each scale with $J \subset B(c(I), D^{s(I)})$. By convexity of $t \mapsto D^t$ and since $D \ge 2$, we have for all $-1 \le t \le 0$ $$ D^t \le 1 + t\left(1 - \frac{1}{D}\right) \le 1 + \frac{1}{2}t\,, $$ so $(1 - D^{-1/a})^{-1} \le 2a \le 2^a$. - \rs{Ds convex} Using this estimate for the sum of the geometric series, - we conclude that \eqref{eq lem L2 Su estimate 1} is at most + we conclude that \eqref{eq boundary operator bound 1} is at most $$ 2^{9a} \sum_{J\in\mathcal{J}}\int_J|f(y)| M_{\mathcal{B},1}|g|(y) \, \mathrm{d}\mu(y)\,. $$ @@ -4679,8 +4501,9 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate} The main result of this subsection is the following quantitative bound for operators associated to trees, with decay in the the densities $\dens_1$ and $\dens_2$. -\begin{lemma} - \label{lem tree est} +\begin{lemma}[densities tree bound] + \label{densities tree bound} + \uses{tree projection estimate,local dens1 tree bound,local dens2 tree bound} Let $\fu \in \fU$. Then for all $f,g$ bounded with bounded support \begin{equation} \label{eq cor tree est} @@ -4693,10 +4516,11 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate} \end{equation} \end{lemma} -Below, we deduce this lemma from Lemma \ref{TreeEstimate} and the following two estimates controlling the size of support of the operator and its adjoint. +Below, we deduce this lemma from Lemma \ref{tree projection estimate} and the following two estimates controlling the size of support of the operator and its adjoint. -\begin{lemma} - \label{lem 1density estimate tree} +\begin{lemma}[local dens1 tree bound] + \label{local dens1 tree bound} + \uses{monotone cube metrics} Let $\fu \in \fU$ and $L \in \mathcal{L}(\fT(\fu))$. Then \begin{equation} \label{eq 1density estimate tree} @@ -4704,15 +4528,15 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate} \end{equation} \end{lemma} -\begin{lemma} - \label{lem 2density estimate tree} +\begin{lemma}[local dens2 tree bound] + \label{local dens2 tree bound} Let $J \in \mathcal{J}(\fT(\fu))$ be such that there exist $\fq \in \fT(\fu)$ with $J \cap \sc(\fq) \ne \emptyset$. Then $$ \mu(F \cap J) \le 2^{200a^3 + 19} \dens_2(\fT(\fu))\,. $$ \end{lemma} -\begin{proof}[Proof of Lemma \ref{lem tree est}] +\begin{proof}[Proof of Lemma \ref{densities tree bound}] Denote $$ \mathcal{E}(\fu) = \bigcup_{\fp \in \fT(\fu)} E(\fp)\,. @@ -4721,7 +4545,7 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate} $$ \left| \int_X \bar g \sum_{\fp \in \fT(\fu)} T_{\fp} f \, \mathrm{d}\mu \right| = \left| \int_X \overline{ g\mathbf{1}_{\mathcal{E}(\fu)}} \sum_{\fp \in \fT(\fu)} T_{\fp} f \, \mathrm{d}\mu \right|\,. $$ - By Lemma \ref{TreeEstimate}, this is bounded by + By Lemma \ref{tree projection estimate}, this is bounded by \begin{equation} \label{eq both factors tree} \le 2^{104a^3}\|P_{\mathcal{J}(\fT(\fu))}|f|\|_2 \|P_{\mathcal{L}(\fT(\fu))} |\mathbf{1}_{\mathcal{E}(\fu)}g|\|_2\,. @@ -4731,11 +4555,11 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate} $$ \|P_{\mathcal{L}(\fT(\fu))} |\mathbf{1}_{\mathcal{E}(\fu)}g|\|_2 = \left( \sum_{L \in \mathcal{L}(\fT(\fu))} \frac{1}{\mu(L)} \left(\int_{L \cap \mathcal{E}(\fu)} |g(y)| \, \mathrm{d}\mu(y)\right)^2 \right)^{1/2}\,. $$ - By Cauchy-Schwarz and Lemma \ref{lem 1density estimate tree} this is at most + By Cauchy-Schwarz and Lemma \ref{local dens1 tree bound} this is at most $$ \le \left( \sum_{L \in \mathcal{L}(\fT(\fu))} 2^{101a^3} \dens_1(\fT(\fu)) \int_{L \cap \mathcal{E}(\fu)} |g(y)|^2 \, \mathrm{d}\mu(y) \right)^{1/2}\,. $$ - Since cubes $L \in \mathcal{L}(\fT(\fu))$ are pairwise disjoint by Lemma \ref{lem partition}, this is + Since cubes $L \in \mathcal{L}(\fT(\fu))$ are pairwise disjoint by Lemma \ref{dyadic partitions}, this is \begin{equation} \label{eq factor L tree} \le 2^{51 a^3} \dens_1(\fT(\fu))^{1/2} \|g\|_2\,. @@ -4750,7 +4574,7 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate} $$ \le \left( \sum_{J \in \mathcal{J}(\fT(\fu))} \int_J |f(y)|^2 \, \mathrm{d}\mu(y) \right)^{1/2}\,. $$ - Since cubes in $\mathcal{J}(\fT(\fu))$ are pairwise disjoint by Lemma \ref{lem partition}, this at most + Since cubes in $\mathcal{J}(\fT(\fu))$ are pairwise disjoint by Lemma \ref{dyadic partitions}, this at most \begin{equation} \label{eq factor J tree} \|f\|_2\,. @@ -4761,7 +4585,7 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate} $$ \left( \sum_{J \in \mathcal{J}(\fT(\fu))} \int_J |f(y)|^2 \, \mathrm{d}\mu(y) \right)^{1/2} = \left( \sum_{J \in \mathcal{J}(\fT(\fu))} \int_{J \cap F} |f(y)|^2 \, \mathrm{d}\mu(y) \right)^{1/2}\,. $$ - We estimate as before, using now Lemma \ref{lem 2density estimate tree} and Cauchy-Schwarz, that the right hand side is + We estimate as before, using now Lemma \ref{local dens2 tree bound} and Cauchy-Schwarz, that the right hand side is $$ \le 2^{100a^3 + 10} \dens_2(\fT(\fu))^{1/2} \|f\|_2\,. $$ @@ -4770,7 +4594,7 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate} Now we prove the two auxiliary estimates. -\begin{proof}[Proof of Lemma \ref{lem 1density estimate tree}] +\begin{proof}[Proof of Lemma \ref{local dens1 tree bound}] If the set on the right hand side is empty, then \eqref{eq 1density estimate tree} holds. If not, then there exists $\fp \in \fT(\fu)$ with $L \cap \sc(\fp) \ne \emptyset$. Suppose first that there exists such $\fp$ with $\ps(\fp) \le s(L)$. Then by \eqref{dyadicproperty} $\sc(\fp) \subset L$, which gives by the definition of $\mathcal{L}(\fT(\fu))$ that $s(L) = -S$ and hence $L = \sc(\fp)$. Let $\fq \in \fT(\fu)$ with $E(\fq) \cap L \ne \emptyset$. Since $s(L) = -S \le \ps(\fq)$ it follows from \eqref{dyadicproperty} that $\sc(\fp) = L \subset \sc(\fq)$. We have then by Lemma \ref{monotone cube metrics} @@ -4807,7 +4631,7 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate} This completes the proof. \end{proof} -\begin{proof}[Proof of Lemma \ref{lem 2density estimate tree}] +\begin{proof}[Proof of Lemma \ref{local dens2 tree bound}] Suppose first that there exists a tile $\fp \in \fT(\fu)$ with $\sc(\fp) \subset B(c(J), 100 D^{s(J) + 1})$. By the definition of $\mathcal{J}(\fT(\fu))$, this implies that $s(J) = -S$, and in particular $\ps(\fp) \ge s(J)$. Using the triangle inequality and \eqref{eq vol sp cube} it follows that $J \subset B(\pc(\fp), 200 D^{\ps(\fp) + 1})$. From the doubling property \eqref{doublingx}, $D=2^{100a^2}$ and \eqref{eq vol sp cube}, we obtain $$ \mu(\sc(\fp)) \le 2^{100a^3 + 9} \mu(J) @@ -4860,7 +4684,7 @@ \section{The quantitative \texorpdfstring{$L^2$}{L2} tree estimate} \section{Almost orthogonality of separated trees} -The main result of this subsection is the almost orthogonality estimate for operators associated to distinct trees in a forest in Lemma \ref{SeparatedTrees} below. We will deduce this Lemma from Lemmas \ref{lem big sep tree} and \ref{lem small sep tree}, which are proven in Subsections \ref{subsec big tiles} and \ref{subsec rest tiles}, respectively. Before stating the lemma, we introduce some relevant notation. +The main result of this subsection is the almost orthogonality estimate for operators associated to distinct trees in a forest in Lemma \ref{correlation separated trees} below. We will deduce this Lemma from Lemmas \ref{correlation distant tree parts} and \ref{correlation near tree parts}, which are proven in Subsections \ref{subsec big tiles} and \ref{subsec rest tiles}, respectively. Before stating the lemma, we introduce some relevant notation. The adjoint of the operator $T_{\fp}$ defined in \eqref{definetp} is given by \begin{equation} @@ -4868,8 +4692,8 @@ \section{Almost orthogonality of separated trees} T_{\fp}^* g(x) = \int_{E(\fp)} \overline{K_{\ps(\fp)}(y,x)} e(-\tQ(y)(x)+ \tQ(y)(y)) g(y) \, \mathrm{d}\mu(y)\,. \end{equation} -\begin{lemma} - \label{lem Tp support adjoint} +\begin{lemma}[adjoint tile support] + \label{adjoint tile support} For each $\fp \in \fP$, we have $$ T_{\fp}^* g = \mathbf{1}_{B(\pc(\fp), 5D^{\ps(\fp)})} T_{\fp}^* \mathbf{1}_{\sc(\fp)} g\,. @@ -4899,8 +4723,9 @@ \section{Almost orthogonality of separated trees} The second claimed equation follows now since $\sc(\fp) \subset \sc(\fu)$ and by \eqref{forest6} $B(\pc(\fp), 5D^{\ps(\fp)}) \subset \sc(\fu)$. \end{proof} -\begin{lemma} - \label{lem tree adjoint bound} +\begin{lemma}[adjoint tree estimate] + \label{adjoint tree estimate} + \uses{densities tree bound} For all bounded $g$ with bounded support, we have that $$ \left\| \sum_{\fp \in \fT(\fu)} T_{\fp}^* g\right\|_2 \le 2^{155a^3} \dens_1(\fT(\fu))^{1/2} \|g\|_2\,. @@ -4908,7 +4733,7 @@ \section{Almost orthogonality of separated trees} \end{lemma} \begin{proof} - By Cauchy-Schwarz and Lemma \ref{lem tree est}, we have for all bounded $f,g$ with bounded support that + By Cauchy-Schwarz and Lemma \ref{densities tree bound}, we have for all bounded $f,g$ with bounded support that $$ \left| \int_X \overline{\sum_{\fp\in \fT(\fu)} T_{\fp}^* g} f \,\mathrm{d}\mu \right| = \left| \int_X g \sum_{\fp \in \fT(\fu)} T_{\fp} f \,\mathrm{d}\mu \right| $$ @@ -4923,8 +4748,9 @@ \section{Almost orthogonality of separated trees} $$ S_{2,\fu}g := \left|\sum_{\fp \in \fT(\fu)} T_{\fp}^*g \right| + M_{\mathcal{B},1}g + |g|\,. $$ -\begin{lemma} - \label{lem L2 Sg estimate} +\begin{lemma}[adjoint tree control] + \label{adjoint tree control} + \uses{Hardy Littlewood,adjoint tree estimate} We have for all bounded $g$ with bounded support $$ \|S_{2, \fu} g\|_2 \le 2^{156a^3} \|g\|_2\,. @@ -4932,14 +4758,15 @@ \section{Almost orthogonality of separated trees} \end{lemma} \begin{proof} - This follows immediately from Minkowski's inequality, Proposition \ref{Hardy Littlewood} and Lemma \ref{lem tree adjoint bound}, using that $a \ge 4$. + This follows immediately from Minkowski's inequality, Proposition \ref{Hardy Littlewood} and Lemma \ref{adjoint tree estimate}, using that $a \ge 4$. \end{proof} Now we are ready to state the main result of this subsection. -\begin{lemma} - \label{SeparatedTrees} +\begin{lemma}[correlation separated trees] + \label{correlation separated trees} + \uses{adjoint tile support,correlation distant tree parts,correlation near tree parts} For any $\fu_1 \ne \fu_2 \in \fU$ and all bounded $g_1, g_2$ with bounded support, we have \begin{equation} \label{eq lhs sep tree} @@ -4952,17 +4779,17 @@ \section{Almost orthogonality of separated trees} \rs{Constant changed from $2^{550a^3}$, check} \end{lemma} -\begin{proof}[Proof of Lemma \ref{SeparatedTrees}] - By Lemma \ref{lem Tp support adjoint} and \eqref{dyadicproperty}, the left hand side \eqref{eq lhs sep tree} is $0$ unless $\sc(\fu_1) \subset \sc(\fu_2)$ or $\sc(\fu_2) \subset \sc(\fu_1)$. Without loss of generality we assume that $\sc(\fu_1) \subset \sc(\fu_2)$. +\begin{proof}[Proof of Lemma \ref{correlation separated trees}] + By Lemma \ref{adjoint tile support} and \eqref{dyadicproperty}, the left hand side \eqref{eq lhs sep tree} is $0$ unless $\sc(\fu_1) \subset \sc(\fu_2)$ or $\sc(\fu_2) \subset \sc(\fu_1)$. Without loss of generality we assume that $\sc(\fu_1) \subset \sc(\fu_2)$. Define \begin{equation} \label{def Tree S set} \mathfrak{S} := \{\fp \in \fT(\fu_1) \cup \fT(\fu_2) \ : \ d_{\fp}(\fcc(\fu_1), \fcc(\fu_2)) \ge 2^{Zn/2}\,\}. \end{equation} - Lemma \ref{SeparatedTrees} follows by combining the definition \eqref{defineZ} of $Z$ with the following two lemmas. - \begin{lemma} - \label{lem big sep tree} + Lemma \ref{correlation separated trees} follows by combining the definition \eqref{defineZ} of $Z$ with the following two lemmas. + \begin{lemma}[correlation distant tree parts] + \label{correlation distant tree parts} We have for all $\fu_1 \ne \fu_2 \in \fU$ with $\sc(\fu_1) \subset \sc(\fu_2)$ and all bounded $g_1, g_2$ with bounded support \begin{equation} \label{eq lhs big sep tree} @@ -4974,8 +4801,8 @@ \section{Almost orthogonality of separated trees} \end{equation} \rs{Constant changed from $2^{550a^3}$, check} \end{lemma} - \begin{lemma} - \label{lem small sep tree} + \begin{lemma}[correlation near tree parts] + \label{correlation near tree parts} We have for all $\fu_1 \ne \fu_2 \in \fU$ with $\sc(\fu_1) \subset \sc(\fu_2)$ and all bounded $g_1, g_2$ with bounded support \begin{equation} \label{eq lhs small sep tree} @@ -4990,8 +4817,8 @@ \section{Almost orthogonality of separated trees} In the proofs of both lemmas, we will need the following observation. -\begin{lemma} - \label{lem T1 S} +\begin{lemma}[overlap implies distance] + \label{overlap implies distance} Let $\fu_1 \ne \fu_2 \in \fU$ with $\sc(\fu_1) \subset \sc(\fu_2)$. If $\fp \in \fT(\fu_1) \cup \fT(\fu_2)$ with $\sc(\fp) \cap \sc(\fu_1) \ne \emptyset$, then $\fp \in \mathfrak{S}$. In particular, we have $\fT(\fu_1) \subset \mathfrak{S}$. \end{lemma} @@ -5011,15 +4838,15 @@ \section{Almost orthogonality of separated trees} so $\fp \in \mathfrak{S}$. \end{proof} -To simplify the notation, we will write at various places throughout the proof of Lemmas \ref{lem big sep tree} and \ref{lem small sep tree} for a subset $\fC \subset \fP$ +To simplify the notation, we will write at various places throughout the proof of Lemmas \ref{correlation distant tree parts} and \ref{correlation near tree parts} for a subset $\fC \subset \fP$ $$ T_{\fC} f := \sum_{\fp \in \fC} T_{\fp} f\,, \quad\quad T_{\fC}^* g := \sum_{\fp\in\fC} T_{\fp}^* g\,. $$ -\section{Proof of Lemma \ref{lem big sep tree}: Tiles with large separation} +\section{Proof of Lemma \ref{correlation distant tree parts}: Tiles with large separation} \label{subsec big tiles} -Lemma \ref{lem big sep tree} follows from the van der Corput estimate in Proposition \ref{Holder van der Corput}. We apply this proposition in Subsubsection \ref{subsubsec van der corput}. To prepare this application, we first, in Subsubsection \ref{subsubsec pao}, construct a suitable partition of unity, and show then, in Subsubsection \ref{subsubsec holder estimates} the Hölder estimates needed to apply Proposition \ref{Holder van der Corput}. +Lemma \ref{correlation distant tree parts} follows from the van der Corput estimate in Proposition \ref{Holder van der Corput}. We apply this proposition in Subsubsection \ref{subsubsec van der corput}. To prepare this application, we first, in Subsubsection \ref{subsubsec pao}, construct a suitable partition of unity, and show then, in Subsubsection \ref{subsubsec holder estimates} the Hölder estimates needed to apply Proposition \ref{Holder van der Corput}. \subsubsection{A partition of unity} \label{subsubsec pao} @@ -5037,7 +4864,7 @@ \subsubsection{A partition of unity} \end{lemma} \begin{proof} - By Lemma \ref{lem partition}, it remains only to show that each $J \in \mathcal{J}(\mathfrak{S})$ with $J \cap \sc(\fu_1) \ne \emptyset$ is in $\mathcal{J}'$. But if $J \notin \mathcal{J}'$, then by \eqref{dyadicproperty} $\sc(\fu_1) \subsetneq J$. Pick $\fp \in \fT(\fu_1) \subset \mathfrak{S}$. Then $\sc(\fp) \subsetneq J$. This contradicts the definition of $\mathcal{J}(\mathfrak{S})$. + By Lemma \ref{dyadic partitions}, it remains only to show that each $J \in \mathcal{J}(\mathfrak{S})$ with $J \cap \sc(\fu_1) \ne \emptyset$ is in $\mathcal{J}'$. But if $J \notin \mathcal{J}'$, then by \eqref{dyadicproperty} $\sc(\fu_1) \subsetneq J$. Pick $\fp \in \fT(\fu_1) \subset \mathfrak{S}$. Then $\sc(\fp) \subsetneq J$. This contradicts the definition of $\mathcal{J}(\mathfrak{S})$. \end{proof} For cubes $J \in \mathcal{D}$, denote @@ -5248,7 +5075,7 @@ \subsubsection{Hölder estimates for adjoint tree operators} \eqref{T*Hölder1b}+\eqref{T*Hölder1} \le \frac{2^{151a^3}}{\mu(B(\pc(\fp), 4D^{\ps(\fp)}))} \left(\frac{\rho(y,y')}{D^{\ps(\fp)}}\right)^{1/a} \int_{E(\fp)}|g(x)| \, \mathrm{d}\mu(x)\,. $$ - Next, if $y,y' \notin B(\pc(\fp), 5D^{\ps(\fp)})$, then $T_{\fp}^*g(y) = T_{\fp}^*g(y') = 0$, by Lemma \ref{lem Tp support adjoint}. Then \eqref{T*Hölder2} holds. + Next, if $y,y' \notin B(\pc(\fp), 5D^{\ps(\fp)})$, then $T_{\fp}^*g(y) = T_{\fp}^*g(y') = 0$, by Lemma \ref{adjoint tile support}. Then \eqref{T*Hölder2} holds. Finally, if $y \in B(\pc(\fp), 5D^{\ps(\fp)})$ and $y' \notin B(\pc(\fp), 5D^{\ps(\fp)})$, then $$ @@ -5323,7 +5150,7 @@ \subsubsection{Hölder estimates for adjoint tree operators} \end{lemma} \begin{proof} - For the first estimate, assume that $\ps(\fp) < s(J)$, then in particular $\ps(\fp) \le \ps(\fu_1)$. Since $\fp \notin \mathfrak{S}$, we have by Lemma \ref{lem T1 S} that $\sc(\fp) \cap \sc(\fu_1) = \emptyset$. + For the first estimate, assume that $\ps(\fp) < s(J)$, then in particular $\ps(\fp) \le \ps(\fu_1)$. Since $\fp \notin \mathfrak{S}$, we have by Lemma \ref{overlap implies distance} that $\sc(\fp) \cap \sc(\fu_1) = \emptyset$. %by \eqref{dyadicproperty}. Since $B\Big(c(J), \frac{1}{4} D^{s(J)}\Big) \subset \sc(J) \subset \sc(\fu_1)$, this implies $$ @@ -5405,7 +5232,7 @@ \subsubsection{Hölder estimates for adjoint tree operators} \end{lemma} \begin{proof} - By Lemma \ref{lem T1 S}, we have that in both cases, $\fC \subset \mathfrak{S}$. If $\fp \in \fC$ with $B(\sc(\fp)) \cap B(J) \neq \emptyset$ and $\ps(\fp) < s(J)$, then $\sc(\fp) \subset B(c(J), 100 D^{s(J) + 1})$. Since $\fp \in \mathfrak{S}$, it follows from the definition of $\mathcal{J}'$ that $s(J) = -S$, which contradicts $\ps(\fp) < s(J)$. + By Lemma \ref{overlap implies distance}, we have that in both cases, $\fC \subset \mathfrak{S}$. If $\fp \in \fC$ with $B(\sc(\fp)) \cap B(J) \neq \emptyset$ and $\ps(\fp) < s(J)$, then $\sc(\fp) \subset B(c(J), 100 D^{s(J) + 1})$. Since $\fp \in \mathfrak{S}$, it follows from the definition of $\mathcal{J}'$ that $s(J) = -S$, which contradicts $\ps(\fp) < s(J)$. \end{proof} \begin{lemma} @@ -5429,7 +5256,7 @@ \subsubsection{Hölder estimates for adjoint tree operators} Note that \eqref{TreeUB} follows from \eqref{TreeHölder}, since for $y'\in B^\circ{}(J)$, by the triangle inequality, $$\left(\frac{\rho(y,y')}{D^{s(J)}}\right)^{1/a}\le \Big(8 + \frac{1}8\Big)^{1/a}\le 2^{a^3}.$$ - By the triangle inequality, Lemma \ref{lem Tp support adjoint} and Lemma \ref{lem tile Hölder}, we have for all $y, y' \in B(J)$ + By the triangle inequality, Lemma \ref{adjoint tile support} and Lemma \ref{lem tile Hölder}, we have for all $y, y' \in B(J)$ \begin{equation} \label{eq C Lip} |e(\fcc(\fu)(y)) T_{\fC}^* g(y) - e(\fcc(\fu)(y')) T_{\fC}^* g(y')| @@ -5447,7 +5274,6 @@ \subsubsection{Hölder estimates for adjoint tree operators} $$ \le 2^{152a^3} \left(\frac{\rho(y,y')}{D^s}\right)^{1/a} \sum_{S \ge s \ge s(J)} D^{(s(J) - s)/a} \inf_J M_{\mathcal{B},1} |g|\,. $$ - \rs{Ds convex} By convexity of $t \mapsto D^t$ and since $D \ge 2$, we have for all $-1 \le t \le 0$ $$ D^t \le 1 + t(1 - \frac{1}{D}) \le 1 + \frac{1}{2}t\,. @@ -5487,7 +5313,7 @@ \subsubsection{Hölder estimates for adjoint tree operators} \begin{proof}[Proof of Lemma \ref{lem hHölder}] Let $P$ be the product on the right hand side of \eqref{hHölder}, and $h_J$ as defined in \eqref{def hj}. - By \eqref{eq pao 2} and Lemma \ref{lem Tp support adjoint}, the function $h_J$ is supported in $B(J) \cap \sc(\fu_1)$. + By \eqref{eq pao 2} and Lemma \ref{adjoint tile support}, the function $h_J$ is supported in $B(J) \cap \sc(\fu_1)$. By \eqref{eq pao 2} and Lemma \ref{lem sep tree aux 1}, we have for all $y \in B(J)$: $$ |h_J(y)| \le 2^{308a^3} P\,. @@ -5536,7 +5362,7 @@ \subsubsection{The van der Corput estimate} \end{lemma} \begin{proof} - Since $\emptyset \ne \fT(\fu_1) \subset \mathfrak{S}$ by Lemma \ref{lem T1 S}, there exists at least one tile $\fp \in \mathcal{S}$ with $\sc(\fp) \subsetneq \sc(\fu_1)$. Thus $\sc(\fu_1) \notin \mathcal{J}'$, so $J \subsetneq \sc(\fu_1)$. Thus there exists a cube $J' \in \mathcal{D}$ with $J \subset J'$ and $s(J') = s(J) + 1$, by \eqref{coverdyadic} and \eqref{dyadicproperty}. By definition of $\mathcal{J'}$ and the triangle inequality, there exists $\fp \in \mathfrak{S}$ such that + Since $\emptyset \ne \fT(\fu_1) \subset \mathfrak{S}$ by Lemma \ref{overlap implies distance}, there exists at least one tile $\fp \in \mathcal{S}$ with $\sc(\fp) \subsetneq \sc(\fu_1)$. Thus $\sc(\fu_1) \notin \mathcal{J}'$, so $J \subsetneq \sc(\fu_1)$. Thus there exists a cube $J' \in \mathcal{D}$ with $J \subset J'$ and $s(J') = s(J) + 1$, by \eqref{coverdyadic} and \eqref{dyadicproperty}. By definition of $\mathcal{J'}$ and the triangle inequality, there exists $\fp \in \mathfrak{S}$ such that $$ \sc(\fp) \subset B(c(J'), 100 D^{s(J') + 1}) \subset B(c(J), 128 D^{s(J)+2})\,. $$ @@ -5551,13 +5377,13 @@ \subsubsection{The van der Corput estimate} which gives the lemma using $a \ge 4$. \end{proof} - Now we are ready to prove Lemma \ref{lem big sep tree}. - \begin{proof}[Proof of Lemma \ref{lem big sep tree}] + Now we are ready to prove Lemma \ref{correlation distant tree parts}. + \begin{proof}[Proof of Lemma \ref{correlation distant tree parts}] We have $$ \eqref{eq lhs big sep tree} = \left| \int_{X} T_{\fT(\fu_1)}^* g_1 \overline{T_{\fT(\fu_2) \cap \mathfrak{S}}^* g_2 }\right|\,. $$ - By Lemma \ref{lem Tp support adjoint}, the right hand side is supported in $\sc(\fu_1)$. Using \eqref{eq pao 1} and the definition \eqref{def hj} of $h_J$, we thus have + By Lemma \ref{adjoint tile support}, the right hand side is supported in $\sc(\fu_1)$. Using \eqref{eq pao 1} and the definition \eqref{def hj} of $h_J$, we thus have $$ \le \sum_{J \in \mathcal{J}'} \left|\int_{B(J)} e(\fcc(\fu_2)(y) - \fcc(\fu_1)(y)) h_J(y) \, \mathrm{d}\mu(y) \right|\,. $$ @@ -5590,16 +5416,16 @@ \subsubsection{The van der Corput estimate} $$ \eqref{eq big sep 1} \le 2^{650a^3} 2^{-Zn/(4a^2 + 2a^3)} \int_X \prod_{j=1}^2 ( |T_{\fT(\fu_j)}^* g_j|(x) + M_{\mathcal{B},1} g_j(x)) \, \mathrm{d}\mu(x)\,. $$ - Applying the Cauchy-Schwarz inequality, Lemma \ref{lem big sep tree} follows. + Applying the Cauchy-Schwarz inequality, Lemma \ref{correlation distant tree parts} follows. \end{proof} -\section{Proof of Lemma \ref{lem small sep tree}: The remaining tiles} +\section{Proof of Lemma \ref{correlation near tree parts}: The remaining tiles} \label{subsec rest tiles} - Recall that + We define $$ - \mathcal{J}' = \{J \in \mathcal{J}(\fT(\fu_1)) \, : \, J \subset \sc(\fu_1)\}\,. + \mathcal{J}' = \{J \in \mathcal{J}(\fT(\fu_1)) \, : \, J \subset \sc(\fu_1)\}\,, $$ - \rs{This lemma has been proven before} + note that this is different from the $\mathcal{J}'$ defined in the previous subsection. \begin{lemma} \label{lem J' partition 2} We have @@ -5609,10 +5435,10 @@ \section{Proof of Lemma \ref{lem small sep tree}: The remaining tiles} \end{lemma} \begin{proof} - By Lemma \ref{lem partition}, it remains only to show that each $J \in \mathcal{J}(\fT(\fu_1))$ with $J \cap \sc(\fu_1) \ne \emptyset$ is in $\mathcal{J}'$. But if $J \notin \mathcal{J}'$, then by \eqref{dyadicproperty} $\sc(\fu_1) \subsetneq J$. Pick $\fp \in \fT(\fu_1)$. Then $\sc(\fp) \subsetneq J$. This contradicts the definition of $\mathcal{J}(\fT(\fu_1))$. + By Lemma \ref{dyadic partitions}, it remains only to show that each $J \in \mathcal{J}(\fT(\fu_1))$ with $J \cap \sc(\fu_1) \ne \emptyset$ is in $\mathcal{J}'$. But if $J \notin \mathcal{J}'$, then by \eqref{dyadicproperty} $\sc(\fu_1) \subsetneq J$. Pick $\fp \in \fT(\fu_1)$. Then $\sc(\fp) \subsetneq J$. This contradicts the definition of $\mathcal{J}(\fT(\fu_1))$. \end{proof} - Lemma \ref{lem small sep tree} follows from the following key estimate. + Lemma \ref{correlation near tree parts} follows from the following key estimate. \begin{lemma} \label{lem sep tree small 1} @@ -5623,10 +5449,10 @@ \section{Proof of Lemma \ref{lem small sep tree}: The remaining tiles} $$ \end{lemma} - We prove this lemma below. First, we deduce Lemma \ref{lem small sep tree}. + We prove this lemma below. First, we deduce Lemma \ref{correlation near tree parts}. - \begin{proof}[Proof of Lemma \ref{lem small sep tree}] - By Lemma \ref{TreeEstimate} and Lemma \ref{lem Tp support adjoint}, we have + \begin{proof}[Proof of Lemma \ref{correlation near tree parts}] + By Lemma \ref{tree projection estimate} and Lemma \ref{adjoint tile support}, we have \begin{align*} \eqref{eq lhs small sep tree} \le 2^{104a^3} \|P_{\mathcal{L}(\fT(\fu_1))} |g_1\mathbf{1}_{\sc(\fu_1)}| \|_2 \|P_{\mathcal{J}(\fT(\fu_1) )}|T_{\fT(\fu_2) \setminus \mathfrak{S}}^* g_2|\|_2\,. \end{align*} @@ -5733,13 +5559,13 @@ \section{Proof of Lemma \ref{lem small sep tree}: The remaining tiles} $$ \le \sum_{s = -S}^S \left( \sum_{J \in \mathcal{J}'} \frac{1}{\mu(J)} \left|\int_J \sum_{\substack{\fp \in \fT(\fu_2) \setminus \mathfrak{S}\\ \ps(\fp) = s}} T_{\fp}^* g_2 \, \mathrm{d}\mu(y) \right|^2\right)^{1/2}\,. $$ - By Lemma \ref{lem Tp support adjoint}, the integral in the last display is $0$ if $J \cap B(\sc(\fp)) = \emptyset$. By Lemma \ref{lem sep aux 3}, it follows with $s_1 := \frac{Zn}{202a^3} - 2$: + By Lemma \ref{adjoint tile support}, the integral in the last display is $0$ if $J \cap B(\sc(\fp)) = \emptyset$. By Lemma \ref{lem sep aux 3}, it follows with $s_1 := \frac{Zn}{202a^3} - 2$: \begin{equation} \label{eq sep tree small 1} = \sum_{s = s_1}^{s_1 + 2S} \Bigg( \sum_{J \in \mathcal{J}'} \frac{1}{\mu(J)} \Bigg|\int_J \sum_{\substack{\fp \in \fT(\fu_2) \setminus \mathfrak{S}\\ \ps(\fp) = s(J) - s\\ J \cap B(\sc(\fp)) \ne \emptyset}} T_{\fp}^* g_2 \, \mathrm{d}\mu(y) \Bigg|^2\Bigg)^{1/2}\,. \end{equation} - We have by Lemma \ref{lem Tp support adjoint} and \eqref{eq Ks size} + We have by Lemma \ref{adjoint tile support} and \eqref{eq Ks size} $$ \int |T_{\fp}^* g_2|(y) \, \mathrm{d}\mu(y) $$ @@ -5772,7 +5598,7 @@ \section{Proof of Lemma \ref{lem small sep tree}: The remaining tiles} $$ \le 2^{103a^3} \int_J M_{\mathcal{B},1} |g_2|(y) \mathbf{1}_{B(I)}(y) \, \mathrm{d}\mu(y)\,. $$ - By Lemma \ref{lem T1 S}, we have $\sc(\fp) \cap \sc(\fu_1) = \emptyset$ for all $\fp \in \fT(\fu_2) \setminus \mathfrak{S}$. + By Lemma \ref{overlap implies distance}, we have $\sc(\fp) \cap \sc(\fu_1) = \emptyset$ for all $\fp \in \fT(\fu_2) \setminus \mathfrak{S}$. Thus we can estimate \eqref{eq sep tree small 1} by $$ 2^{103a^3} \sum_{s = s_1}^{s_1 + 2S} \Bigg( \sum_{J \in \mathcal{J}'} \frac{1}{\mu(J)} \Bigg|\int_J \sum_{\substack{I \in \mathcal{D}, s(I) = s(J) - s\\ I \cap \sc(\fu_1) = \emptyset\\ @@ -5795,7 +5621,6 @@ \section{Proof of Lemma \ref{lem small sep tree}: The remaining tiles} $$ \le 2^{116a^3} D^{-s_1 \kappa /2} \frac{1}{1 - D^{-\kappa/2}} \|\mathbf{1}_{\sc(\fu_1)} M_{\mathcal{B},1} |g_2|\|_2\,. $$ - \rs{Ds convex} By convexity of $t \mapsto D^t$ and since $D \ge 2$, we have for all $-1 \le t \le 0$ $$ D^t \le 1 + t(1 - \frac{1}{D}) \le 1 + \frac{1}{2}t\,. @@ -5833,7 +5658,7 @@ \section{Forests} $$ with inclusion maximal $\sc(\fu)$. Properties \eqref{forest1}, -\eqref{forest6} for $(\fU_j, \fT|_{\fU_k})$ follow immediately from the corresponding properties for $(\fU, \fT)$, and the cubes $\sc(\fu), \fu \in \fU_j$ are disjoint by definition. The collections $\fU_j$ are also disjoint by definition. - Now we show by induction on $j$ that each point is contained in at most \rs{at least?} $2^n - j$ cubes $\sc(\fu)$ with $\fu \in \fU \setminus \bigcup_{j' \le j} \fU_{j'}$. This implies that $\bigcup_{j = 1}^{2^n} \fU_j = \fU$, which completes the proof of the Lemma. For $j = 0$ each point is contained in at most $2^n$ cubes by \eqref{forest3}. For larger $j$, if $x$ is contained in any cube $\sc(\fu)$ with $\fu \in \fU \setminus \bigcup_{j' < j} \fU_{j'}$, then it is contained in a maximal such cube. Thus it is contained in a cube in $\sc(\fu)$ with $\fu \in \fU_j$. Thus the number $\fu \in \fU \setminus \bigcup_{j' \le j} \fU_{j'}$ with $x\in \sc(\fu)$ is zero, or is less than the number of $\fu \in \fU \setminus \bigcup_{j' \le j-1} \fU_{j'}$ with $x \in \sc(\fu)$ by at least one. + Now we show by induction on $j$ that each point is contained in at most $2^n - j$ cubes $\sc(\fu)$ with $\fu \in \fU \setminus \bigcup_{j' \le j} \fU_{j'}$. This implies that $\bigcup_{j = 1}^{2^n} \fU_j = \fU$, which completes the proof of the Lemma. For $j = 0$ each point is contained in at most $2^n$ cubes by \eqref{forest3}. For larger $j$, if $x$ is contained in any cube $\sc(\fu)$ with $\fu \in \fU \setminus \bigcup_{j' < j} \fU_{j'}$, then it is contained in a maximal such cube. Thus it is contained in a cube in $\sc(\fu)$ with $\fu \in \fU_j$. Thus the number $\fu \in \fU \setminus \bigcup_{j' \le j} \fU_{j'}$ with $x\in \sc(\fu)$ is zero, or is less than the number of $\fu \in \fU \setminus \bigcup_{j' \le j-1} \fU_{j'}$ with $x \in \sc(\fu)$ by at least one. \end{proof} We pick a decomposition of the forest $(\fU, \fT)$ into $2^n$ $n$-rows @@ -5857,7 +5682,7 @@ \section{Forests} \end{lemma} \begin{proof} - By Corollary \ref{lem tree est} and the density assumption \eqref{forest4}, we have for each $\fu \in \fU$ and all bounded $f$ of bounded support that + By Corollary \ref{densities tree bound} and the density assumption \eqref{forest4}, we have for each $\fu \in \fU$ and all bounded $f$ of bounded support that \begin{equation} \label{eq explicit tree bound 1} \left\|\sum_{\fp \in \fT(\fu)} T_{\fp} f \right\|_{2} \le 2^{155a^3} 2^{(4a+1-n)/2} \|f\|_2\, @@ -5867,7 +5692,7 @@ \section{Forests} \label{eq explicit tree bound 2} \left\|\sum_{\fp \in \fT(\fu)} T_{\fp} \mathbf{1}_F f \right\|_{2} \le 2^{256a^3} 2^{(4a + 1-n)/2} \dens_2(\fT(\fu))^{1/2} \|f\|_2\,. \end{equation} - Since for each $j$ the top cubes $\sc(\fu)$, $\fu \in \fU_j$ are disjoint, we further have for all bounded $g$ of bounded support by Lemma \ref{lem Tp support adjoint} + Since for each $j$ the top cubes $\sc(\fu)$, $\fu \in \fU_j$ are disjoint, we further have for all bounded $g$ of bounded support by Lemma \ref{adjoint tile support} $$ \left\|\mathbf{1}_F \sum_{\fu \in \fU_j} \sum_{\fp \in \fT(\fu)} T_{\fp}^* g\right\|_2^2 = \left\|\mathbf{1}_F \sum_{\fu \in \fU_j} \sum_{\fp \in \fT(\fu)} \mathbf{1}_{\sc(\fu)} T_{\fp}^* \mathbf{1}_{\sc(\fu)} g\right\|_2^2 $$ @@ -5879,7 +5704,7 @@ \section{Forests} $$ \le 2^{256a^3} 2^{(4a+1-n)/2} \max_{\fu \in \fU_j}\dens_2(\fT(\fu))^{1/2} \sum_{\fu \in \fU_j} \left\| \mathbf{1}_{\sc(\fu)} g\right\|_2^2\,. $$ - Again by disjointness of the cubes $\sc(\fu)$, this is estimated by + Again by disjointedness of the cubes $\sc(\fu)$, this is estimated by $$ 2^{256a^3} 2^{(4a+1-n)/2} \max_{\fu \in \fU_j}\dens_2(\fT(\fu))^{1/2} \|g\|_2^2\,. $$ @@ -5903,14 +5728,14 @@ \section{Forests} $$ T_{\fC}^* = \sum_{\fp \in \fC} T_{\fp}^*\,. $$ - We have by Lemma \ref{lem Tp support adjoint} and the triangle inequality that + We have by Lemma \ref{adjoint tile support} and the triangle inequality that $$ \left| \int \sum_{\fu \in \fU_j} \sum_{\fu' \in \fU_{j'}} \sum_{\fp \in \fT_j(\fu)} \sum_{\fp' \in \fT_{j'}(\fu')} T^*_{\fp} g_1 \overline{T^*_{\fp'} g_2} \, \mathrm{d}\mu \right| $$ $$ \le \sum_{\fu \in \fU_j} \sum_{\fu' \in \fU_{j'}} \left| \int T^*_{\fT_j(\fu)} (\mathbf{1}_{\sc(\fu)} g_1) \overline{T^*_{\fT_{j'}(\fu')} (\mathbf{1}_{\sc(\fu')} g_2)} \, \mathrm{d}\mu \right|\,. $$ - By Lemma \ref{SeparatedTrees}, this is bounded by + By Lemma \ref{correlation separated trees}, this is bounded by \begin{equation} \label{eq S2uu'} 2^{650a^3-3n} \sum_{\fu \in \fU_j} \sum_{\fu' \in \fU_{j'}} \|S_{2,\fu} g_1\|_{L^2(\sc(\fu')\cap \sc(\fu)} \|S_{2, \fu'} g_2\|_{L^2(\sc(\fu')\cap\sc(\fu))}\,. @@ -5926,7 +5751,7 @@ \section{Forests} \left(\sum_{\fu \in \fU_j} \sum_{\fu' \in \fU_{j'}} \|S_{2,\fu'} g_2\|_{L^2(\sc(\fu')\cap\sc(\fu))}^2 \right)^{1/2}\,. $$ \rs{Constant changed form 550 to 650, check} - By pairwise disjointness of the sets $\sc(\fu)$ for $\fu \in \fU_j$ and of the sets $\sc(\fu')$ for $\fu' \in \fU_{j'}$, we have + By pairwise disjointedness of the sets $\sc(\fu)$ for $\fu \in \fU_j$ and of the sets $\sc(\fu')$ for $\fu' \in \fU_{j'}$, we have $$ \sum_{\fu \in \fU_j}\sum_{\fu' \in \fU_{j'}} \|S_{2,\fu} g_1\|_{L^2(\sc(\fu')\cap \sc(\fu))}^2 = \sum_{\fu \in \fU_j}\sum_{\fu' \in \fU_{j'}} \int_{\sc(\fu) \cap \sc(\fu')} |S_{2,\fu} g_1(y)|^2 \, \mathrm{d}\mu(y) @@ -5939,7 +5764,7 @@ \section{Forests} \le 2^{650a^3-3n} \|S_{2,\fu}g_1\|_2 \|S_{2,\fu'} g_2\|_2\,. $$ \rs{Constant changed from 550 to 650, check} - The lemma now follows from Lemma \ref{lem L2 Sg estimate}. + The lemma now follows from Lemma \ref{adjoint tree control}. \rs{Constant changed, check.} \end{proof} @@ -5997,7 +5822,7 @@ \section{Forests} $$ 2^{862a^3} (2^{-n} + 2^{n}2^{-3 n}) \sum_{j = 1}^n \|\mathbf{1}_{E_j} g\|_2^2\,, $$ - and by disjointness of the sets $E_j$, this is at most + and by disjointedness of the sets $E_j$, this is at most $$ 2^{862a^3 - n} \|g\|_2^2\,. $$ @@ -6006,7 +5831,7 @@ \section{Forests} \label{eq forest bound 1} \left\|\sum_{\fu \in \fU} \sum_{\fp \in \fT(\fu)} T_{\fp} f\right\|_2 \le 2^{431a^3-\frac{n}{2}} \|f\|_2\,. \end{equation} - On the other hand, we have by disjointness of the sets $E_j$ + On the other hand, we have by disjointedness of the sets $E_j$ $$ \left\|\sum_{\fu \in \fU} \sum_{\fp \in \fT(\fu)} T_{\fp} f\right\|_2^2 = \left\|\sum_{j=1}^{2^n} \mathbf{1}_{E_j} T_{\mathfrak{R}_{j}} f\right\|_2^2 = \sum_{j = 1}^{2^n} \|\mathbf{1}_{E_j} T_{\mathfrak{R}_{j}} f\|_2^2\,. $$ @@ -6021,14 +5846,14 @@ \section{Forests} Proposition \ref{Holder van der Corput} follows by taking the product of the $(2 - \frac{2}{q})$-th power of \eqref{eq forest bound 1} and the $(\frac{2}{q} - 1)$-st power of \eqref{eq forest bound 2}. \end{proof} -\chapter{Proof of P. \ref{Holder van der Corput}, the H\"older cancellative condition} +\chapter{Proof of Prop. \ref{Holder van der Corput}, the H\"older cancellative condition} \label{liphoel} We need the following auxiliary lemma. Recall that $\tau = 1/a$. -\begin{lemma} - \label{lem regularity aux} +\begin{lemma}[Lipschitz Holder approximation] + \label{Lipschitz Holder approximation} Let $z\in X$ and $R>0$. Let $\varphi: X \to \mathbb{C}$ be a function supported in the ball $B:=B(z,R)$ with finite norm $\|\varphi\|_{C^\tau(B)}$. Let $0 0$, there exists a countable collection $C(r) \subset X$ of points such that $$ X \subset \bigcup_{c \in C(r)} B(c, r)\,. @@ -6543,7 +6368,7 @@ \chapter{Proof of P. \ref{Hardy Littlewood}, Vitali covering and Hardy Littlewoo $$ \mathcal{B}_\infty = \{B(c, 2^k) \ : \ c \in C(2^k), k \in \mathbb{N}\}\,. $$ -By Lemma \ref{lem separable}, this is a countable collection of balls. We choose an enumeration $\mathcal{B}_\infty = \{B_1, \dotsc\}$ and define +By Lemma \ref{covering separable space}, this is a countable collection of balls. We choose an enumeration $\mathcal{B}_\infty = \{B_1, \dotsc\}$ and define $$ \mathcal{B}_n = \{B_1, \dotsc, B_n\}\,. $$ @@ -6567,21 +6392,15 @@ \chapter{Proof of P. \ref{Hardy Littlewood}, Vitali covering and Hardy Littlewoo \chapter{Proof of Theorem \ref{classical Carleson}} -\lars{TODO: Subsection 4 has to be rewritten to prove the bound for the correct maximal operator. - -Subsection 6 is not written at all. -Subsection 7 is not complete. - -I have checked all other subsections, and I have checked Subsectino 7 as far as it is written.} The convergence of partial Fourier sums is proved in Subsection \ref{10classical} in two steps. In the first step, -one establishes convergence on a suitable dense subclass of functions. We choose piece-wise constant functions as subclass, the convergence is stated in Lemma \ref{lem piece} and proved in Subsection \ref{10piecewise}. +one establishes convergence on a suitable dense subclass of functions. We choose piece-wise constant functions as subclass, the convergence is stated in Lemma \ref{convergence for piecewise constant} and proved in Subsection \ref{10piecewise}. In the second step, one controls the relevant error of approximating a general function by a function in the subclass by a bound of the Carleson maximal operator. This approximation result is stated -in Lemma \ref{lem diff} and proved in Subsection \ref{10difference}. -This latter proof refers to the main Theorem \ref{metric space Carleson}. Two assumptions in Theorem \ref{metric space Carleson} require more work, this is prepared in Lemmas \ref{lem hilbert} and \ref{lem vdc} and proved in Subsections \ref{10hilbert} +in Lemma \ref{control approximation effect} and proved in Subsection \ref{10difference}. +This latter proof refers to the main Theorem \ref{metric space Carleson}. Two assumptions in Theorem \ref{metric space Carleson} require more work, this is prepared in Lemmas \ref{lem hilbert} and \ref{van der Corput} and proved in Subsections \ref{10hilbert} and \ref{10vandercorput}. Subsections \ref{10piecewise}, \ref{10hilbert}, \ref{10vandercorput}, and \ref{10difference} are mutually independent. @@ -6590,15 +6409,15 @@ \chapter{Proof of Theorem \ref{classical Carleson}} \section{The classical Carleson theorem} \label{10classical} -Let a uniformly continuous $2\pi$-periodic function $f:\R\to \mathbb{C}$ bounded in absolute value by $1$ be given. Let $\epsilon>0$ be given. +Let a uniformly continuous $2\pi$-periodic function $f:\R\to \mathbb{C}$ bounded in absolute value by $1$ be given. Let $\epsilon>0$ be given, without loss of generality we also assume $\epsilon \le 1$. By uniform continuity of $f$, there is a $0<\delta<\pi$ such that for all $x,x' \in \R$ with $|x-x'|\le \delta$ we have \begin{equation}\label{uniconbound} -|f(x)-f(x')|\le 2^{-2^{50}}\epsilon\, . +|f(x)-f(x')|\le 2^{-2^{50}}\epsilon^2\, . \end{equation} -Let $K$ be the Gaussian bracket of $2\pi/\delta$, that is the unique integer with +Let $K$ be the Gaussian bracket of $\frac{2\pi}\delta+1$, that is the unique integer with \begin{equation} K-1\le \frac{2\pi}{\delta} < K\, . \end{equation} @@ -6611,12 +6430,15 @@ \section{The classical Carleson theorem} \begin{equation}\label{def fzero} f_0(x):=f\left(\frac{2\pi k(x)}{K}\right)\, . \end{equation} -\begin{lemma} +\begin{lemma}[piecewise constant +approximation] +\label{piecewise constant +approximation} The function $f_0$ is measurable. The function $f_0$ is $2\pi $- periodic. -The function $f_0$ satisfies for all $x\in \R$, +The function $f_0$ satisfies for all $x\in \R$: \begin{equation}\label{eq ffzero} -|f(x)-f_0(x)|\le 2^{-2^{50}} \epsilon\, , +|f(x)-f_0(x)|\le 2^{-2^{50}} \epsilon^2\, , \end{equation} \begin{equation}\label{eq ffzero1} |f_0(x)|\le 1\, . @@ -6641,9 +6463,14 @@ \section{The classical Carleson theorem} To see that $f_0$ is $2\pi$-periodic, we observe by -adding $2\pi$ to \eqref{definekx} that +applying \eqref{definekx} to $x+2\pi$ and subtracting $K$ +that +\begin{equation}\label{definekxshifted} +k(x+2\pi)-K\le \frac{Kx}{2\pi}\frac {2^{25} K^2}{\epsilon ^3}$ and \begin{equation} x\in [0,2\pi)\setminus E_1 @@ -6693,15 +6522,18 @@ \section{The classical Carleson theorem} \end{lemma} -We prove in Subsection \ref{10difference}: \lars{The proof of this Lemma is still missing completely} -\begin{lemma}\label{lem diff} +We prove in Subsection \ref{10difference}: +\begin{lemma}[control approximation effect] +\label{control approximation effect} +\uses{real Carleson, dirichlet kernel,lower secant bound} There is a set $E_2 \subset \R$ with Lebesgue measure $|E_2|\le \frac \epsilon 2$ such that for all \begin{equation} - x\in [0,1)\setminus E_2 + x\in [0,2\pi)\setminus E_2 \end{equation} we have \begin{equation} + \label{eq max partial sum diff} \sup_{N\ge 0} |S_Nf(x)-S_Nf_0(x)| \le \frac \epsilon 4\,. \end{equation} @@ -6715,70 +6547,84 @@ \section{The classical Carleson theorem} \begin{equation} |E|\le |E_1|+|E_2|\le \frac \epsilon 2 +\frac \epsilon 2 \le \epsilon\, . \end{equation} -Let $N_0$ be as in Lemma \ref{lem piece}. +Let $N_0$ be the unique integer such that +\begin{equation} + N_0- 1\le {2^{25} K^2}{\epsilon ^{-3}}N_0$ we have by the triangle inequality -\begin{equation} +\begin{equation*} |f(x)-S_Nf(x)| - \end{equation} - \begin{equation} + \end{equation*} + \begin{equation}\label{epsilonthird} \le |f(x)-f_0(x)|+ |f_0(x)-S_Nf_0(x)|+|S_Nf_0(x)-S_N f(x)|\, . \end{equation} -Using \eqref{uniconbound} and Lemmas \ref{lem piece} -and \ref{lem diff}, we estimate the last display by +Using \eqref{uniconbound} and Lemmas \ref{convergence for piecewise constant} +and \ref{control approximation effect}, we estimate \eqref{epsilonthird} by \begin{equation} - \le 2^{-2^{50}} \epsilon +\frac \epsilon 4 +\frac \epsilon 4\le \epsilon\, . + \le 2^{-2^{50}} \epsilon^2 +\frac \epsilon 4 +\frac \epsilon 4\le \epsilon\, . \end{equation} This shows \eqref{aeconv} for the given $E$ and $N_0$ and completes the proof of Theorem \ref{classical Carleson}. -The proof of Lemma \ref{lem diff} will essentially be +The proof of Lemma \ref{control approximation effect} will essentially be an application of the following variant of Theorem \ref{metric space Carleson}. -Let $k:\R\to \R$ be the function defined by -by $k(0)=0$ and for $x\neq 0$ +Let $\kappa:\R\to \R$ be the function defined by +$\kappa(0)=0$ and for $0<|x|<1$ \begin{equation}\label{eq hilker} -k(x)=\frac {\max (1-|x|, 0)}{1-e^{ix}}\, . +\kappa(x)=\frac { 1-|x|}{1-e^{ix}}\, +\end{equation} +\ct{make sure you never divide anything by zero... dangerous if $x=2\pi m$} +and for $|x|\ge 1$ +\begin{equation}\label{eq hilker1} +\kappa(x)=0\, . \end{equation} +Note that this function is continuous at every point $x$ with $|x|>0$. The following lemma is proved in Subsection \ref{10carleson}. -\begin{lemma}\label{lem rcarleson} +\begin{lemma}\label{real Carleson} +\uses{metric space Carleson,real line metric,real line measure,real line doubling,frequency ball doubling,frequency ball growth,frequency ball cover,real van der Corput,Hilbert kernel bound,Hilbert kernel regularity} Let $F,G$ be Borel subsets of $\R$ with finite measure. Let $f$ be a bounded measurable function on $\R$ with $|f|\le \mathbf{1}_F$. Then \begin{equation} |\int _G Tf(x) \, dx| \le 2^{2^{40}}|F|^{\frac 12} |G|^{\frac 12} \, , \end{equation} where \begin{equation} + \label{define T carleson} T f(x)=\sup_{n\in \mathbb{Z}} - \sup_{r>0}\left|\int_{r<|x-y|<1} f(y)k(x-y) e^{iny}\, dy\right|\, . + \sup_{r>0}\left|\int_{r<|x-y|<1} f(y)\kappa(x-y) e^{iny}\, dy\right|\, . \end{equation} \end{lemma} -The following lemma is proved in Subsection \ref{10hilbert}. \lars{This used to be the assumption of our main theorem. Now that we changed it, we should change this Lemma to prove the bound for the full maximal operator, i.e. with another sup over $r$} +The following lemma is proved in Subsection \ref{10hilbert}. +\ct{Lars was right. I changed the following so that it exactly fits what we need. This makes 10.3 mildly more laborous, but that's where one can get synergies.... +The $2^8$ can be a bit larger if needed} \begin{lemma}\label{lem hilbert} - For every $00$ and $-2\pi +\eta \le x\le 2\pi-\eta$ with $|x|\ge \eta$. Then +\begin{equation} + |1-e^{ix}|\ge \frac {\eta} 8 +\end{equation} +\end{lemma} +\begin{proof} + +We have +$$ + |1 - e^{ix}| = \sqrt{(1 - \cos(x))^2 + \sin^2(x)} \ge |\sin(x)|\,. +$$ +If $0 \le x \le \frac{\pi}{2}$, then we have from concavity of $\sin$ on $[0, \pi]$ and $\sin(0) = 0$ and $\sin(\frac{\pi}{2}) = 1$ +$$ + |\sin(x)| \ge \frac{2}{\pi} x \ge \frac{2}{\pi} \eta\,. +$$ +When $x\in \frac{m\pi}{2} + [0, \frac{\pi}{2}]$ for $m \in \{-4, -3, -2, -1, 1, 2, 3\}$ one can argue similarly. + \end{proof} + The following lemma will be proved in Subsection \ref{10projection}. -\begin{lemma}\label{lem l2sn} +\begin{lemma}[spectral projection bound] +\label{spectral projection bound} Let $f$ be a bounded $2\pi$-periodic measurable function. Then, for all $N\ge 0$ \begin{equation}\label{snbound} \|S_Nf\|_2\le \|f\|_2. @@ -6880,11 +6749,12 @@ \section{Piecewise constant functions.} We first compute the partial Fourier sums for constant functions. -\begin{lemma}\label{constant} +\begin{lemma}[constant function] +\label{constant function} If $h$ satisfies $h(x)=h(0)$ for all $x\in \R$, then for all $N\ge 0$ we have $S_Nh=h$. \end{lemma} \begin{proof} - We compute with Lemma \ref{lem expintegral} for $n\in \mathbb{Z}$ with $n\neq 0$, + We compute with Lemma \ref{mean zero oscillation} for $n\in \mathbb{Z}$ with $n\neq 0$, \begin{equation} \widehat{h}_n=\frac 1{2\pi}\int_0^{2\pi}h(y)e^{-iny}\, dy=\frac {h(0)}{2\pi}\int_0^{2\pi}e^{-iny}\, dy=0\, . \end{equation} @@ -6895,27 +6765,10 @@ \section{Piecewise constant functions.} \end{equation} This proves the lemma. \end{proof} -\begin{lemma}\label{expbound} -Let $\eta>0$ and $-2\pi +\eta \le x\le 2\pi-\eta$ with $|x|\ge \eta$. Then -\begin{equation} - |1-e^{ix}|\ge \frac {\eta} 8 -\end{equation} -\end{lemma} -\begin{proof} -\ct{check with Floris what the library has here} \lars{I wrote something that Lean can definitely do.} -\lars{(And to not forget it, the sharp constant is really $\frac{2}{\pi}$, because $|1 - e^{ix}|$ is concave on $[0, \pi]$)} -We have -$$ - |1 - e^{ix}| = \sqrt{(1 - \cos(x))^2 + \sin^2(x)} \ge |\sin(x)|\,. -$$ -If $0 \le x \le \frac{\pi}{2}$, then we have from concavity of $\sin$ on $[0, \pi]$ and $\sin(0) = 0$ and $\sin(\frac{\pi}{2}) = 1$ -$$ - |\sin(x)| \ge \frac{2}{\pi} x \ge \frac{2}{\pi} \eta\,. -$$ -When $x\in \frac{m\pi}{2} + [0, \frac{\pi}{2}]$ for $m \in \{-4, -3, -2, -1, 1, 2, 3\}$ one can argue similarly. - \end{proof} -\begin{lemma}\label{lem diri} +\begin{lemma}[Dirichlet kernel bound] +\label{dirichlet kernel bound} +\uses{lower secant bound} Let $\eta>0$. Let \begin{equation} -2\pi +\eta \le a\frac {2^{25} K^2}{\epsilon^3}$ @@ -7005,10 +6860,10 @@ \section{Piecewise constant functions.} - With Lemma \ref{dirichlet}, breaking up the domain of integration into a + With Lemma \ref{dirichlet kernel}, breaking up the domain of integration into a partition of subintervals, \begin{equation*} - |S_N(g)|=\left|\int_{0}^{2\pi} g(y)K_N(x-y)\, dy\right| + |S_N(g)|=\left|\int_{-\pi}^{\pi} g(y)K_N(x-y)\, dy\right| \end{equation*} \begin{equation}\label{eqcf1} = \left|\sum_{k=0}^{K-1}\int_{\frac{2\pi k}K}^{\frac{2\pi (k+1)}K} g(y)K_N(x-y)\, dy\right|\, . @@ -7053,7 +6908,7 @@ \section{Piecewise constant functions.} -With Lemma \ref{dirichlet} and the triangle inequality, +With Lemma \ref{dirichlet kernel} and the triangle inequality, it follows that \begin{equation} \left|\int_{x-2\pi (k+1)/K}^{x-2\pi k/K} K_N(y)\, dy\right| @@ -7062,7 +6917,7 @@ \section{Piecewise constant functions.} \left|\int_{a}^{b} \frac{e^{-iNy}}{1-e^{iy}}\, dy\right| \end{equation} -Using that the two integrals on the right-hand side are complex conjugates of each other, and using Lemma \ref{lem diri} and $\epsilon <2\pi$, this is bounded by +Using that the two integrals on the right-hand side are complex conjugates of each other, and using Lemma \ref{dirichlet kernel bound} and $\epsilon <2\pi$, this is bounded by \begin{equation} \le 2\left|\int_{a}^{b} \frac{e^{iNy}}{1-e^{-iy}}\, dy\right| @@ -7073,18 +6928,18 @@ \section{Piecewise constant functions.} the lemma. \end{proof} -We now prove Lemma \ref{lem piece} +We now prove Lemma \ref{convergence for piecewise constant} Let $x\in [0,2\pi)\setminus E_1 $ and $N>\frac {2^{25} K^2}{\epsilon ^3}$. Let $h$ be the function which is constant equal to $f_0(x)$ and let $g=f_0-h$. By the bound on $f$ and the triangle inequality, $|g|$ is bounded by $2$. Hence $g$ is in the class $\mathcal{G}$. We also have $g(x)=0$. -Using Lemma \ref{constant}, we obtain +Using Lemma \ref{constant function}, we obtain \begin{equation} S_Nf_0(x)- f_0(x)=S_N(g+h)(x)- S_Nh(x)=S_Ng(x)\, . \end{equation} -Lemma \ref{lem piece} -now follows by Lemma \ref{lem g with zero}. +Lemma \ref{convergence for piecewise constant} +now follows by Lemma \ref{flat interval partial sum}. @@ -7092,13 +6947,11 @@ \section{The truncated Hilbert transform} \label{10hilbert} \lars{I am skipping this subsection because it has to be rewritten anyway} +\rs{To be discussed: 1. Using $L^2$ norms in Cotlar leads to a weak $(2,2)$ estimates for $T_*$} +\rs{2. Obtaining $L^2$ estimate for $T_0$ using uniform $L^2$ bound for truncated kernels} +\rs{3. Passing to $\mathbb{R}$ from periodic functions} - - - - - - +\rs{Change norms} Let $M_n$ be the modulation operator @@ -7117,16 +6970,16 @@ \section{The truncated Hilbert transform} \begin{lemma} We have for every bounded measurable $2\pi$-periodic function $g$ \begin{equation}\label{lnbound} - \|L_Ng\|_2\le \|g\|_2 + \|L_Ng\|_{L^2[-\pi, \pi]}\le \|g\|_{L^2[-\pi, \pi]} \end{equation} \end{lemma} \begin{proof} We have \begin{equation}\label{mnbound} - \|M_ng\|_2^2=\int _0^{2\pi} |e^{inx}g(x)|^2\, dx - =\int _0^{2\pi} |g(x)|^2\, dx=\|g\|_2^2\, . + \|M_ng\|_2^2=\int_{-\pi}^{\pi} |e^{inx}g(x)|^2\, dx + =\int_{-\pi}^{\pi} |g(x)|^2\, dx=\|g\|_{L^2[-\pi, \pi]}^2\, . \end{equation} - We have by the triangle inequality, the square root of Identity \eqref{mnbound}, and Lemma \ref{lem l2sn} + We have by the triangle inequality, the square root of the identity in\eqref{mnbound}, and Lemma \ref{spectral projection bound} \begin{equation*} \|L_ng\|_2=\|\frac 1N\sum_{n=0}^{N-1} M_{-n-N} S_{N+n}M_{N+n}g\|_2 @@ -7147,14 +7000,16 @@ \section{The truncated Hilbert transform} \begin{lemma}\label{lem shift} \lean{Function.Periodic.intervalIntegral_add_eq, intervalIntegral.integral_comp_sub_right} +\leanok Let $f$ be a bounded $2\pi$-periodic function. We have for any $0 \le x\le 2\pi$ that \begin{equation} \int_0^{2\pi} f(y)\, dy= \int_{-x}^{2\pi -x} f(y)\, dy - =\int_{0}^{2\pi} f(y-x)\, dy + =\int_{-\pi}^{\pi} f(y-x)\, dy \end{equation} \end{lemma} \begin{proof} +\leanok We have by periodicity and change of variables \begin{equation}\label{eqhil9} \int_{-x}^{0} f(y)\, dy=\int_{-x}^{0} f(y+2\pi)\, dy= \int_{2\pi -x}^{2\pi} f(y)\, dy\, . @@ -7178,19 +7033,19 @@ \section{The truncated Hilbert transform} \begin{lemma}\label{young} Let $f$ and $g$ be two bounded non-negative measurable $2\pi$-periodic functions on $\R$. Then \begin{equation}\label{eqyoung} - \left(\int_0^{2\pi} \left(\int_0^{2\pi} + \left(\int_{-\pi}^{\pi} \left(\int_{-\pi}^{\pi} f(y)g(x-y)\, dy\right)^2\, dx\right)^{\frac 12}\le \|f\|_2 \|g\|_1\, . \end{equation} \end{lemma} \begin{proof} Using Fubini and Lemma \ref{lem shift}, we observe \begin{equation*} - \int_0^{2\pi}\int_0^{2\pi}f(y)^2g(x-y)\, dy - \, dx=\int_0^{2\pi}f(y)^2\int_0^{2\pi}g(x-y)\, dx + \int_{-\pi}^{\pi}\int_{-\pi}^{\pi}f(y)^2g(x-y)\, dy + \, dx=\int_{-\pi}^{\pi}f(y)^2\int_{-\pi}^{\pi}g(x-y)\, dx \, dy \end{equation*} \begin{equation}\label{eqhil4} -=\int_0^{2\pi}f(y)^2\int_0^{2\pi}g(x) \, dx +=\int_{-\pi}^{\pi}f(y)^2\int_{-\pi}^{\pi}g(x) \, dx dy =\|f\|_2^2\|g\|_1\, . \end{equation} @@ -7201,19 +7056,19 @@ \section{The truncated Hilbert transform} \eqref{eqyoung} with Cauchy-Schwarz and then with \eqref{eqhil4} by \begin{equation*} - \int_0^{2\pi} (\int_0^{2\pi}f(y)h(x-y)h(x-y)\, dy)^2\, dx + \int_{-\pi}^{\pi} (\int_{-\pi}^{\pi}f(y)h(x-y)h(x-y)\, dy)^2\, dx \end{equation*} \begin{equation} - \le \int_0^{2\pi}\left(\int_0^{2\pi}f(y)^2g(x-y)\, dy\right) - \left(\int_0^{2\pi}g(x-y)\, dy\right)\, dx + \le \int_{-\pi}^{\pi}\left(\int_{-\pi}^{\pi}f(y)^2g(x-y)\, dy\right) + \left(\int_{-\pi}^{\pi}g(x-y)\, dy\right)\, dx = \|f\|_2^2\|g\|_1^2\, . \end{equation} Taking square roots, this proves the lemma. \end{proof} -For $0r$, we have +$$|x-y|\geq |x'-y|-|x-x'|>r-\frac{r}{2}=\frac{r}{2}.$$ +Therefore the domain of integration above is contained in the set +$\{y:\, |x-y|<\frac{r}{2}\}.$ Decomposing this set into dyadic ranges, we see that the integral above is estimated by +\begin{equation} + \label{eq rKer dyadic sum} + \sum_{j=0}^{\lfloor \log_2(\frac{1}{r})\rfloor+1}\int_{2^{j-1}r\leq |x-y|<2^{j}r}|\kappa(x-y)-\kappa(x'-y)||g(y)|\, dy. +\end{equation} +% Since $|x-x'|<\frac{r}{2}<\frac{|x-y|}{2}$, we can argue as in Lemma \ref{Hilbert kernel regularity} to get the estimate +We claim that +\begin{equation} +\label{eq kappa reg Tr} + |\kappa(x-y)-\kappa(x'-y)|\leq 2^{10}\frac{1}{|x-y|} \frac{|x-x'|}{|x-y|}\,, +\end{equation} +%Detailed proof +\rs{This proof is repeated several times so can be a lemma at the beginning of the section} +To see the above, let $z=x-y$ and $z'=x-y'$. Since $|z-z'|<\frac{r}{2}<\frac{|z|}{2}$, we conclude that $z, z'$ have the same sign. Since $\kappa(z)=\overline{\kappa(-z)}$, we may assume without loss of generality that they are both positive. Then our assumption implies that + $$ + \frac{z}{2} \le z' \,. + $$ + We now consider four cases. If $z, z' \le 1$, then we have + $$ + |\kappa(z)-\kappa(z')| = \left| \frac{1 - z}{1- e^{-iz}} - \frac{1 - z'}{1- e^{-iz'}}\right| + $$ + and by the fundamental theorem of calculus + $$ + = \left| \int_{z'}^{z} \frac{-1 + e^{-it} + i(1-t)e^{it}}{(1 - e^{-it})^2} \,dt \right|\,. + $$ + Using $z' \ge \frac{z}{2}$ and Lemma \ref{lower secant bound}, we bound this by + $$ + \le |z - z'| \sup_{\frac{z}{2} \le t \le 1} \frac{3}{|1 - e^{-it}|^2} \le 3 |z-z'| (8 \frac{2}{z})^2 \le 2^{10} \frac{|z-z'|}{|z|^2}\,. + $$ + If $z \le 1$ and $z' > 1$, then $\kappa(z')= 0$ and we have from the first case + $$ + |\kappa(z)-\kappa(z')| = |\kappa(z)| \le 2^{10} \frac{|z-1|}{|z|^2} \le 2^{10} \frac{|z-z'|}{|z|^2}\,. + $$ + The case when $z > 1$ and $z' \le 1$ can be handled similarly. + Finally, if $z, z' > 1$ then + $$ + |\kappa(z)-\kappa(z')| = 0 \le 2^{10} \frac{|z-z'|}{|z|^2}\,. + $$ + +Plugging \eqref{eq kappa reg Tr} into \eqref{eq rKer dyadic sum}, we get +\begin{equation} + \label{eq rker HL} + |H_rg(x)-H_rg(x')|\leq 2^{10} \sum_{j=0}^{\lfloor \log_2(\frac{1}{r})\rfloor+1}\frac{r}{2}(2^{j-1}r)^{-2}\int_{|x-y|<2^{j}r}|g(y)|\, dy. +\end{equation} +By Proposition \ref{Hardy Littlewood}, there exists a measurable function $Mg: \mathbb{R}\to [0, \infty)$ such that +for each $j$ in the above sum, +$$\frac{1}{2^{j}r}\int_{|x-y|<2^{j}r}|g(y)|\, dy\leq Mg(x),$$ +and +$$\|Mg\|_2\leq 2^{17}\|g\|_2.$$ +In combination with \eqref{eq rker HL}, this yields +\begin{equation*} + |H_rg(x)-H_rg(x')|\leq 2^{10} \sum_{j=0}^{\lfloor \log_2(\frac{1}{r})\rfloor+1}2^{-j+1}Mg(x)\leq 2^{12} Mg(x). +\end{equation*} \end{proof} -We now prove Lemma \ref{lem hilbert}. -We have with $r$ and $N$ as above for every $x$ + + + +% \begin{lemma}\label{lem dirichlet3} +% Let $00$, we have \begin{equation} - \|T_rg\|_2\le \|L_N\|_2+2^{11}\|g*k_r\|_2\ . + \mu\left(\{x'\in \mathbb{R}: |x'-x|<\frac,\, |T_0g(x')|>\alpha\}\right)\leq \alpha^{-\frac{3}{2}}\int_{|x-y|<\frac{r}{2}} |T_0g(y)|^2\, dy. \end{equation} +Using Proposition \ref{Hardy Littlewood} (more precisely, \eqref{eq hlm 2} with $w=|T_0g|^{\frac{3}{2}}$), we conclude that there exists a measurable function $M(|T_0g|^{\frac{3}{2}})$ such that the right hand side above can be estimated by +\begin{equation} + \label{eq HL1} + \alpha^{-{\frac{3}{2}}} M(|T_0g|^{\frac{3}{2}})(x)\mu(\{x'\in \mathbb{R}: |x-x'|<\frac{r}{2}\}). + %= 2\frac{r}{2} \alpha^{-2} M(|L_Ng|^2)(x). +\end{equation} +Moreover, this function satisfies the estimate +$$\|M(|T_0g|^{\frac{3}{2}})^{\frac{2}{3}}\|_2\leq 2^{\frac{2.16}{3}}\left(\frac{2}{2-\frac{3}{2}}\right)^{\frac{2}{3}}\|T_0g\|_2\leq 2^{12} \|T_0g\|_2.$$ +Similarly, we have +\begin{equation*} + \mu\left(\{x'\in \mathbb{R}: |x'-x|<\frac{r}{2},\, |T_0g_{x,1}(x')|>\alpha\}\right)\leq \alpha^{-2}\int_{|x-y|<\frac{r}{2}} |T_0g_{x,1}(y)|^2\, dy. +\end{equation*} +This time, we use Lemma \ref{lem kappa op L2 bound} to estimate the above by +\begin{equation*} + 2\alpha^{-2}\int |T_0g_{x,1}(y)|^2\, dy \leq 2^2 \alpha^{-2}\int |g_{x,1}(y)|^2\, dy + = 2^2 \alpha^{-2}\int_{|y-x'|<\frac{r}{2}} |g(y)|^2 \,dy. +\end{equation*} +Using Proposition \ref{Hardy Littlewood} again (more precisely, \eqref{eq hlm 2} with $w=|g|^2$), we obtain another measurable function $M(|g|^2)$ such that the above can be estimated by +\begin{equation} + \label{eq HL2} + 2^2\alpha^{-2} M(|g|^2)(x)\mu(\{x'\in \mathbb{R}: |x-x'|<\frac{r}{2}\}). + %= 2^3 \frac{r}{2} \alpha^{-2} M(|g|^2)(x). +\end{equation} +From \eqref{eq HL1} and \eqref{eq HL2}, we deduce that for +\begin{equation*} + \label{eq alph choice} + \alpha=\max\left(4 (M(|T_0g|^{\frac{3}{2}})(x))^{\frac{2}{3}}, 4 (M(|g|^2)(x))^{\frac{1}{2}} \right), +\end{equation*} +we have the estimate +\begin{multline*} + \mu(\{x'\in \mathbb{R}: |x-x'|<\frac{r}{2}, |T_0g_{x,1}(x')|>\alpha \text{ or }|T_0g(x')|>\alpha\}\\\leq \frac{1}{2} \mu(\{x'\in \mathbb{R}: |x-x'|<\frac{r}{2}\}. +\end{multline*} +Making this choice of $\alpha$, we conclude that there exists an $x'$ with $|x-x'|<\frac{r}{2}$ such that \eqref{eq cotlar} is true. +\end{proof} -By Lemma \ref{lnbound} and Lemma\ref{krbound}, we obtain - +\begin{lemma} +\label{lem xx' g2 diff} +Let $0 1$, then $\kappa(z')= 0$ and we have from the first case + $$ + |\kappa(z)-\kappa(z')| = |\kappa(z)| \le 2^{10} \frac{|z-1|}{|z|^2} \le 2^{10} \frac{|z-z'|}{|z|^2}\,. + $$ + The case when $z > 1$ and $z' \le 1$ can be handled similarly. + Finally, if $z, z' > 1$ then + $$ + |\kappa(z)-\kappa(z')| = 0 \le 2^{10} \frac{|z-z'|}{|z|^2}\,. + $$ + +Plugging \eqref{eq kappa reg} into \eqref{eq hilbker dyadic sum}, we get +\begin{equation*} + \label{eq hilbker HL} + |T_0g_{x,2}(x)-T_0g_{x,2}(x')|\leq 2^{10} \sum_{j=0}^{\lfloor \log_2(\frac{1}{r})\rfloor}\frac{r}{2}(2^jr)^{-2}\int_{|x-y|<2^{j+1}r}|g(y)|\, dy. +\end{equation*} + +By Proposition \ref{Hardy Littlewood}, there exists a measurable function $Mg: \mathbb{R}\to [0, \infty)$ such that +for each $j$ in the above sum, +$$\frac{1}{2^{j+1}r}\int_{|x-y|<2^{j+1}r}|g(y)|\, dy\leq Mg(x),$$ +and +$$\|Mg\|_2\leq 2^{17}\|g\|_2.$$ +In combination with \eqref{eq hilbker HL}, this yields +\begin{equation*} + |T_0g_{x,2}(x)-T_0g_{x,2}(x')|\leq 2^{10} \sum_{j=0}^{\lfloor \log_2(\frac{1}{r})\rfloor}2^{-j}Mg(x)\leq 2^{11} Mg(x). +\end{equation*} +\end{proof} + + +\begin{proof}[Proof of Lemma \ref{lem hilbert}] +We now prove Lemma \ref{lem hilbert}. \rs{Write} +\end{proof} + +% We now prove Lemma \ref{lem hilbert}. +% We have with $r$ and $N$ as above for every $x$ +% \begin{equation} +% |T_rg(x)|\le |L_Ng(x)|+ |H_rg(x)-L_Ng(x)|+ |T_rg(x)-H_rg(x)|\, . +% \end{equation} +% With Lemma \ref{lem dirichlet2} and Lemma \ref{dirichlet kernel}, we estimate this by + +% \begin{equation} +% \le |L_Ng(x)|+ 2^{11}|\int_0^{2\pi} g(y) k_r(x-y)\, dy|\, . +% \end{equation} + +% By monotonicity and sub-additivity of the $L^2$ norm, we obtain +% \begin{equation} +% \|T_rg\|_2\le \|L_N\|_2+2^{11}\|g*k_r\|_2\ . +% \end{equation} + +% By Lemma \ref{lnbound} and Lemma \ref{krbound}, we obtain + +% \begin{equation} +% \|T_rg\|_2\le 2^{30}\|g\|_2\ . +% \end{equation} -\section{The proof of the van der Corput Lemma \ref{lem vdc}} +\section{The proof of the van der Corput Lemma \ref{van der Corput}} \label{10vandercorput} Let $g$ be a Lipschitz continuous function as in the lemma. We have @@ -7434,38 +7781,38 @@ \section{The proof of the van der Corput Lemma \ref{lem vdc}} $$ Using this, we write $$ - \int_a^b g(x) e^{-inx} \, \mathrm{d}x - = \frac{1}{2} \int_a^b g(x) e^{inx} \, \mathrm{d}x - \frac{1}{2} \int_a^b g(x) e^{in(x + \pi/n)}) \, \mathrm{d}x\,. + \int_\alpha^\beta g(x) e^{-inx} \, \mathrm{d}x + = \frac{1}{2} \int_\alpha^\beta g(x) e^{inx} \, \mathrm{d}x - \frac{1}{2} \int_\alpha^\beta g(x) e^{in(x + \pi/n)}) \, \mathrm{d}x\,. $$ -We split the the first integral at $a + \frac{\pi}{n}$ and the second one at $b - \frac{\pi}{n}$, and make a change of variables in the first part of the second integral to obtain +We split the the first integral at $\alpha + \frac{\pi}{n}$ and the second one at $\beta - \frac{\pi}{n}$, and make a change of variables in the first part of the second integral to obtain $$ - = \frac{1}{2} \int_{a}^{a + \frac{\pi}{n}} g(x) e^{inx} \, \mathrm{d}x - \frac{1}{2} \int_{b - \frac{\pi}{n}}^{b} g(x) e^{in(x + \pi/n)} \, \mathrm{d}x + = \frac{1}{2} \int_{\alpha}^{\alpha + \frac{\pi}{n}} g(x) e^{inx} \, \mathrm{d}x - \frac{1}{2} \int_{\beta - \frac{\pi}{n}}^{\beta} g(x) e^{in(x + \pi/n)} \, \mathrm{d}x $$ $$ - + \frac{1}{2} \int_{a + \frac{\pi}{n}}^{b} (g(x) - g(x - \frac{\pi}{n}) e^{inx} \, \mathrm{d}x\,. + + \frac{1}{2} \int_{\alpha + \frac{\pi}{n}}^{\beta} (g(x) - g(x - \frac{\pi}{n}) e^{inx} \, \mathrm{d}x\,. $$ The sum of the first two terms is by the triangle inequality bounded by $$ - \frac{\pi}{n} \sup_{x \in [a,b]} |g(x)|\,. + \frac{\pi}{n} \sup_{x \in [\alpha,\beta]} |g(x)|\,. $$ The third term is by the triangle inequality at most $$ - \frac{1}{2} \int_{a + \frac{\pi}{n}}^b |g(x) - g(x - \frac{\pi}{n})| \, \mathrm{d}x + \frac{1}{2} \int_{\alpha + \frac{\pi}{n}}^\beta |g(x) - g(x - \frac{\pi}{n})| \, \mathrm{d}x $$ $$ - \le \frac{\pi}{2n} \sup_{a \le x < y \le b} \frac{|g(x) - g(y)|}{|x-y|} |b-a|\,. + \le \frac{\pi}{2n} \sup_{\alpha \le x < y \le \beta} \frac{|g(x) - g(y)|}{|x-y|} |\beta-\alpha|\,. $$ Adding the two terms, we obtain $$ - \left|\int_a^b g(x) e^{-inx} \, \mathrm{d}x\right| \le \frac{\pi}{n} \|g\|_{\mathrm{Lip}(a,b)}\,. + \left|\int_\alpha^\beta g(x) e^{-inx} \, \mathrm{d}x\right| \le \frac{\pi}{n} \|g\|_{\mathrm{Lip}(\alpha,\beta)}\,. $$ By the triangle inequality, we also have $$ - \left|\int_a^b g(x) e^{-inx} \, \mathrm{d}x\right| \le |b -a| \sup_{x \in [a,b]} |g(x)| \le |b-a| \|g\|_{\mathrm{Lip}(a,b)}\,. + \left|\int_\alpha^\beta g(x) e^{-inx} \, \mathrm{d}x\right| \le |\beta -\alpha| \sup_{x \in [\alpha,\beta]} |g(x)| \le |\beta-\alpha| \|g\|_{\mathrm{Lip}(\alpha,\beta)}\,. $$ This completes the proof of the lemma, using that $$ - \min\{|b-a|, \frac{\pi}{n}\} \le 2 \pi |b-a|(1 + n|b-a|)^{-1}\,. + \min\{|\beta-\alpha|, \frac{\pi}{n}\} \le 2 \pi |\beta-\alpha|(1 + n|\beta-\alpha|)^{-1}\,. $$ @@ -7475,21 +7822,22 @@ \section{The proof of the van der Corput Lemma \ref{lem vdc}} \section{Partial sums as orthogonal projections} \label{10projection} +This subsection proves Lemma \ref{spectral projection bound} - -\begin{lemma}\label{lem projection} +\begin{lemma}[partial sum projection] +\label{partial sum projection} Let $f$ be a bounded $2\pi$-periodic measurable function. Then, for all $N\ge 0$ \begin{equation}\label{projection} S_N(S_N f)=S_Nf\, . \end{equation} \end{lemma} \begin{proof} -Let $N>0$ be given. With $K_N$ as in Lemma \ref{dirichlet}, +Let $N>0$ be given. With $K_N$ as in Lemma \ref{dirichlet kernel}, \begin{equation*} S_N (S_Nf) (x)= \int_0^{2\pi} S_Nf(y)K_N(x-y)\, dy @@ -7498,7 +7846,7 @@ \section{Partial sums as orthogonal projections} = \int_0^{2\pi} \int_0^{2\pi} f(y')K_N(y-y') K_N(x-y)\, \, dy' dy\, . \end{equation} -We have by Lemma \ref{dirichlet} +We have by Lemma \ref{dirichlet kernel} \begin{equation*} \int_0^{2\pi} K_N(y-y') K_N(x-y)\, dy \end{equation*} @@ -7510,7 +7858,7 @@ \section{Partial sums as orthogonal projections} =\sum_{n=-N}^N\sum_{n'=-N}^N e^{i(n'x-ny')}\int_0^{2\pi} e^{i(n-n')y}\, dy\, . \end{equation} -By Lemma \ref{lem expintegral}, the summands for $n\neq n'$ vanish. +By Lemma \ref{mean zero oscillation}, the summands for $n\neq n'$ vanish. We obtain for \eqref{eqhil6} \begin{equation}\label{eqhil2} =\sum_{n=-N}^N @@ -7524,26 +7872,28 @@ \section{Partial sums as orthogonal projections} \end{equation} This proves the lemma. \end{proof} -\begin{lemma}\label{selfadjoint} +\begin{lemma}[partial sum selfadjoint] +\label{partial sum selfadjoint} +\uses{dirichlet kernel} We have for any $2\pi$-periodic bounded measurable $g,f$ that \begin{equation} \int_0^{2\pi} \overline{S_Nf(x)} g(x)=\int_0^{2\pi} \overline{f(x)} S_Ng(x)\, dx\, . \end{equation} \end{lemma} \begin{proof} - We have with $K_N$ as in Lemma \ref{dirichlet} for every $x$ + We have with $K_N$ as in Lemma \ref{dirichlet kernel} for every $x$ \begin{equation} \overline{K_N(x)}=\sum_{n=-N}^N\overline{ e^{in x}}= {\sum_{n=-N}^N e^{-in x}}=K_N(-x)\, . \end{equation} - Further, with Lemma \ref{dirichlet} and Fubini + Further, with Lemma \ref{dirichlet kernel} and Fubini \begin{equation*} \int_0^{2\pi} \overline{S_Nf(x)} g(x) -= \frac 1{2\pi} \int_0^{2\pi} \int_{0}^{2\pi}\overline{f(y) K_N(x-y)} g(x)\, dy dx += \frac 1{2\pi} \int_0^{2\pi} \int_{-\pi}^{\pi}\overline{f(y) K_N(x-y)} g(x)\, dy dx \end{equation*} \begin{equation} = -\frac 1{2\pi} \int_0^{2\pi} \int_{0}^{2\pi}\overline{f(y)} K_N(y-x) +\frac 1{2\pi} \int_0^{2\pi} \int_{-\pi}^{\pi}\overline{f(y)} K_N(y-x) g(x)\, dx dy =\int_0^{2\pi} \overline{f(x)} S_Ng(x)\, dx \, . @@ -7552,8 +7902,10 @@ \section{Partial sums as orthogonal projections} \end{proof} +We turn to the proof of Lemma +\ref{spectral projection bound}. -We have with Lemma \ref{selfadjoint}, then Lemma \ref{lem projection} and the Lemma\ref{selfadjoint} again +We have with Lemma \ref{partial sum selfadjoint}, then Lemma \ref{partial sum projection} and the Lemma \ref{partial sum selfadjoint} again \begin{equation*} \int_0^{2\pi} S_Nf(x)\overline{S_Nf(x)}\, dx \int_0^{2\pi} f(x)\overline{S_N(S_Nf)(x)}\, dx @@ -7602,75 +7954,173 @@ \section{Partial sums as orthogonal projections} \section{The error bound} \label{10difference} +We prove Lemma \ref{control approximation effect}. Define +$$ + E_2 := \{x \in [0, 2\pi) \ : \ \sup_{N > 0} |S_N f(x) - S_N f_0(x)| > \frac{\epsilon}{4} \}\,. +$$ +Then \eqref{eq max partial sum diff} clearly holds, and it remains to show that $|E_2| \le \frac{\epsilon}{2}$. This will follow from Lemma \ref{real Carleson}. + +Let $x \in E_2$. Then there exists $N > 0$ with +$$ + |S_N f(x) - S_N f_0(x)| > \frac{\epsilon}{4}\,, +$$ +pick such $N$. We have with Lemma \ref{dirichlet kernel} +$$ + \frac{\epsilon}{4} < |S_N f(x) - S_N f_0(x)| = \frac{1}{2\pi} \left| \int_0^{2\pi} (f(y) - f_0(y)) K_N(x-y) \, dy\right|\,. +$$ +We make a change of variables, replacing $y$ by $x -y$. Then we use $2\pi$-periodicity of $f$, $f_0$ and $K_N$ to shift the domain of integration to obtain +$$ + = \frac{1}{2\pi} \left|\int_{-\pi}^{\pi} (f(x-y) - f_0(x-y)) K_N(y) \, dy\right|\,. +$$ +Using the triangle inequality, we split this as +$$ + \le \frac{1}{2\pi} \left|\int_{-\pi}^{\pi} (f(x-y) - f_0(x-y)) \max(1 - |y|, 0) K_N(y) \, dy\right| +$$ +$$ + + \frac{1}{2\pi} \left|\int_{-\pi}^{\pi} (f(x-y) - f_0(x-y)) \min(|y|, 1) K_N(y) \, dy\right|\,. +$$ +Note that all integrals are well defined, since $K_N$ is by \eqref{eqksumexp} bounded by $2N+1$. +Using \eqref{eqksumhil} and the definition \eqref{eq hilker} of $k$, we rewrite the two terms and obtain +\begin{equation} + \label{eq diff singular} + \frac{\epsilon}{4} < \frac{1}{2\pi} \left| \int_{-\pi}^{\pi} (f(x-y) - f_0(x-y)) (e^{iNy} \overline{k}(y) + e^{-iNy}k(y)) \, dy\right| +\end{equation} +\begin{equation} + \label{eq diff integrable} + + \frac{1}{2\pi} \left|\int_{-\pi}^{\pi} (f(x-y) - f_0(x-y)) ( e^{iNy} \frac{\min\{|y|, 1\} }{1 - e^{-iy}} + e^{-iNy} \frac{\min\{|y|, 1\} }{1 - e^{iy}}) \, dy \right|\,. +\end{equation} +By Lemma \ref{lower secant bound} with $\eta = |y|$, we have for $-1 \le y \le 1$ +$$ + |e^{iNy}\frac{\min\{|y|, 1\} }{1 - e^{-iy}}| = \frac{|y|}{|1 - e^{iy}|} \le 8\,. +$$ +By Lemma \ref{lower secant bound} with $\eta = 1$, we have for $1 \le |y| \le \pi$ +$$ + |e^{iNy}\frac{\min\{|y|, 1\} }{1 - e^{-iy}}| = \frac{1}{|1 - e^{iy}|} \le 8\,. +$$ +Thus we obtain using the triangle inequality and \eqref{eq ffzero} +$$ + \eqref{eq diff integrable} \le \frac{16}{2\pi} \int_{-\pi}^{\pi} |f(x-y) - f_0(x-y)| \, dy \le 2^{4-2^{50}} \epsilon^2\,. +$$ +Consequently, we have that +$$ + \frac{\epsilon}{8} \le \frac{1}{2\pi} \left| \int_{-\pi}^{\pi} (f(x-y) - f_0(x-y)) (e^{iNy} \overline{k}(y) + e^{-iNy}k(y)) \, dy\right|\,. +$$ +By dominated convergence and since $k(y) = 0$ for $|y| > 1$, this equals +$$ + = \frac{1}{2\pi} \lim_{r \to 0^+} \left| \int_{r < |y| < 1} (f(x-y) - f_0(x-y)) (e^{iNy} \overline{k}(y) + e^{-iNy}k(y)) \, dy\right|\,. +$$ +Let $h = (f - f_0) \mathbf{1}_{[-\pi, 3\pi]}$. Since $x \in [0, 2\pi]$, the above integral does not change if we replace $(f - f_0)$ by $h$. We do that, apply the triangle inequality and bound the limits by suprema +$$ + \le \frac{1}{2\pi} \sup_{r > 0} \left| \int_{r < |y| < 1} h(x-y) e^{-iNy} k(y) \, dy\right| +$$ +$$ + + \frac{1}{2\pi} \sup_{r > 0} \left| \int_{r < |y| < 1} \overline{h}(x-y) e^{-iNy} k(y) \, dy\right|\,. +$$ +By the definition \eqref{define T carleson} of $T$, this is +$$ + \le \frac{1}{2\pi} (Th(x) + T\bar{h}(x))\,. +$$ +Thus for each $x \in E_2$, at least one of $Th(x)$ and $T\bar h(x)$ is larger than $\frac{\epsilon}{16}$. +Thus at least one of $Th$ and $T\bar h$ is $\ge \frac{\epsilon}{16}$ on a subset $E_2'$ of $E_2$ with $2|E_2'| \ge |E_2|$. Without loss of generality this is $Th$. By assumption \eqref{eq ffzero}, we have $|2^{2^{50}}\epsilon^{-2} h| \le \mathbf{1}_{[-\pi, 3\pi]}$. Applying Lemma \ref{real Carleson} with $F = [-\pi, 3\pi]$ and $G = E_2'$, it follows that +$$ + \frac{\epsilon}{16} |E_2'| \le \int_{E_2'} Th(x) \, dx = 2^{-2^{50}}\epsilon^2 \int_{E_2'} T(2^{2^{50}}\epsilon^{-2} h)(x) \, dx +$$ +$$ + \le 2^{-2^{50}} \epsilon^2 \cdot 2^{2^{40}} |F|^{\frac{1}{2}} |E_2'|^{\frac{1}{2}} \le 2^{-2^{40}} \frac{\epsilon^2}{16} |E_2'|^{\frac{1}{2}}\,. +$$ +Rearranging, we obtain +$$ + |E_2'| \le 2^{-2^{41}}\epsilon^2 \le \frac{\epsilon}{4}\,. +$$ +This completes the proof using $|E_2| \le 2|E_2'|$. + + + + \section{Carleson on the real line} \label{10carleson} -We prove Lemma \ref{lem rcarleson}. +We prove Lemma \ref{real Carleson}. Consider the standard distance function \begin{equation} \rho(x,y)=|x-y| \end{equation} on the real line $\R$. -\begin{lemma} +\begin{lemma}[real line metric] +\label{real line metric} The space $(\R,\rho)$ is a complete locally compact metric space. \end{lemma} \begin{proof} This is part of the Lean library. \end{proof} -\begin{lemma}\label{lem ball int} +\begin{lemma} +\label{real line ball} \lean{Real.ball_eq_Ioo} + \leanok For $x\in R$ and $R>0$, the ball $B(x,R)$ is the interval $(x-R,x+R)$ \end{lemma} \begin{proof} +\leanok Let $x'\in B(x,R)$. By definition of the ball, $|x'-x|x-R$. This implies $x'\in (x-R,x+R)$. Conversely, let $x'\in (x-R,x+R)$. Then $x'x-R$. It follows that -$x'-x0$ \begin{equation} \mu(B(x,R))=2R\, . \end{equation} \end{lemma} \begin{proof} -We have with lemma \ref{lem ball int} +\leanok +We have with Lemma \ref{real line ball} \begin{equation} \mu(B(x,R))=\mu((x-R,x+R))=2R\, . \end{equation} -Where the last identity is taken from the Lean library. \lars{I dont think we have to write this, we dont do it for other trivial manipulations either.} \end{proof} -\begin{lemma}\label{lem r doubling} +\begin{lemma}[real line doubling] +\label{real line doubling} We have for every $x\in \R$ and $R>0$ \begin{equation} \mu(B(x,2R))=2\mu(B(x,R))\, . \end{equation} \end{lemma} \begin{proof} - We have with Lemma \ref{lem ball meas} + We have with Lemma \ref{real line ball measure} \begin{equation} \mu(B(x,2R)=4R=2\mu(B(x,R)\, . \end{equation} This proves the lemma. \end{proof} + +The preceeding four lemmas show that $(\R, \rho, \mu, 4)$ is a doubling metric measure space. Indeed, we even show +that $(\R, \rho, \mu, 1)$ is a doubling metric measure space, but we may relax the estimate in Lemma \ref{real line doubling} to conclude that $(\R, \rho, \mu, 4)$ +is a doubling metric measure space. + + For each $n\in \mathbb{Z}$ define $\mfa_n:\R\to \R$ by \begin{equation} @@ -7680,9 +8130,11 @@ \section{Carleson on the real line} Let $\Mf$ be the collection $\{\mfa_n, n\in \mathbb{Z}\}$. Note that for every $n\in \mathbb{Z}$ we have $\mfa_n(0)=0$. -Define $d$ as in \eqref{definedE}. Note that -with for +Define $d_E$ as in \eqref{definedE}. Note that +for $n,m\in \mathbb{Z}$ and $x\in \R$ and $x>0$ +we have + \begin{equation} d_{B(x,R)}(\mfa_n,\mfa_m)=\sup_{y,y'\in B(x,R)}|ny-ny'-my+my'| \end{equation} @@ -7707,7 +8159,8 @@ \section{Carleson on the real line} -\begin{lemma}\label{lem fdb1} +\begin{lemma}[frequency ball doubling] +\label{frequency ball doubling} For any $x,x'\in \R$ and $R>0$ with $x\in B(x',2R)$ and any $n,m\in \mathbb{Z}$, we have \begin{equation}\label{firstdb1} @@ -7718,19 +8171,20 @@ \section{Carleson on the real line} With \eqref{eqcarl4}, both sides of \eqref{firstdb1} are equal to $4R|n-m|$. This proves the lemma. \end{proof} -\begin{lemma}\label{lem sdb1} +\begin{lemma}[frequency ball growth] +\label{frequency ball growth} For any $x,x'\in \R$ and $R>0$ with $B(x,R)\subset B(x',2R)$ and any $n,m\in \mathbb{Z}$, we have \begin{equation}\label{seconddb1} - 2d_{B(x,R)}(\mfa_n,\mfa_m)\le 2 d_{B(x',2R)}(\mfa_n,\mfa_m) \, . + 2d_{B(x,R)}(\mfa_n,\mfa_m)\le d_{B(x',2R)}(\mfa_n,\mfa_m) \, . \end{equation} \end{lemma} \begin{proof} With \eqref{eqcarl4}, both sides of \eqref{firstdb1} are equal to $4R|n-m|$. This proves the lemma. \end{proof} -\lars{Here, all the doubling properties are checked with $a = 1$. It is obvious that they are then also satisfied for larger $a$, but we do not prove that. In the end we will have to choose $a \ge 4$} -\begin{lemma}\label{lem tdb1} +\begin{lemma}[frequency ball cover] +\label{frequency ball cover} For every $x\in \R$ and $R>0$ and every $n\in \mathbb{Z}$ and $R'>0$, there exist $m_1, m_2, m_3\in \mathbb{Z}$ @@ -7761,7 +8215,7 @@ \section{Carleson on the real line} Let $\mfa_{n'}\in B'$, then with \eqref{eqcarl4}, we have \begin{equation}\label{eqcarl6} - R|n-n'|\le R'\, . + 2R|n-n'|< 2R'\, . \end{equation} Assume first $n'\le n-R'/2$. By definition of $m_1$, @@ -7771,14 +8225,14 @@ \section{Carleson on the real line} R|m_1-n'|=R(m_1-n')=R(m_1-n)+R(n-n') \end{equation*} \begin{equation} - \le -\frac{R'}2+R'=-\frac{R'}2\, . + < -\frac{R'}2+R'=-\frac{R'}2\, . \end{equation} We conclude $\mfa_{n'}\in B_1$. Assume next $n-R'/20$ and any function $\varphi: X\to \C$ supported on $B'=B(x,R)$ such that @@ -7813,10 +8269,10 @@ \section{Carleson on the real line} Set $n'=n-m$. Then we have to prove \begin{equation} \label{eq vdc cond2} - \left|\int_{x-R}^{x+R} e^{in'y}\varphi(y) dy\right|\le 4\pi R\|\varphi\|_{\Lip(B)} + \left|\int_{x-R}^{x+R} e^{in'y}\varphi(y) dy\right|\le 4\pi R\|\varphi\|_{\Lip(B')} (1+2R|n'|)^{-1}\, . \end{equation} -This follows from Lemma \ref{lem vdc} with $a = x - R$ and $b = x + 2R$. +This follows from Lemma \ref{van der Corput} with $\alpha = x - R$ and $\beta = x + 2R$. %We do a case distinction whether $Rn'\le \pi$ %or $Rn'> \pi$. %Assume first $Rn'\le \pi$. We estimate the left-hand side of \eqref{eq vdc cond2} by @@ -7824,7 +8280,7 @@ \section{Carleson on the real line} % 2R\sup_{y\in B'}|\varphi(y)|\le 2^4R\|\varphi\|_{\Lip(B)}(1+2R|n'|)^{-1}\, . %\end{equation} %This proves \eqref{eq vdc cond2} in this case. - +% %Assume now $Rn'> \pi$. %We write %\begin{equation}\label{eqcarl10} @@ -7878,52 +8334,91 @@ \section{Carleson on the real line} %with the triangle inequality proves \eqref{eq vdc cond2} in the given case and completes the proof of the lemma. \end{proof} -With $k$ as in \eqref{eq hilker}, define +The preceding three lemmas establish that $\Mf$ is a cancellative, compatible collection of functions on $(\R, \rho, \mu, 4)$. Again, some of the statements in these lemmas +are stronger than what is needed for $a=4$, but can be relaxed to give the desired conclusion for $a=4$. + + + + +With $\kappa$ as near \eqref{eq hilker}, define the function $K:\R\times \R\to \mathbb{C}$ by \begin{equation} - K(x,y):=k(x-y)\, . + K(x,y):=\kappa(x-y)\, . \end{equation} The function $K$ is continuous outside the diagonal $x=y$ and vanishes on the diagonal. Hence it is measurable. -\begin{lemma} +\begin{lemma}[Hilbert kernel bound] +\label{Hilbert kernel bound} For $x,y\in \R$ with $x\neq y$ we have \begin{equation}\label{eqcarl30} |K(x,y)|\le 2^4(2|x-y|)^{-1}\, . \end{equation} \end{lemma} \begin{proof} - Fix $x\neq y$. We have + Fix $x\neq y$. If $K(x,y)$ is zero, then \eqref{eqcarl30} is evident. Assume $K(x,y)$ is not zero, then $0<|x-y|<1$. + We have \begin{equation}\label{eqcarl31} -|K(x,y)|=\frac {\max (1-|x-y|, 0)}{1-e^{i(x-y)}}\, . -\end{equation} -We make the case distinction whether -$|x-y|>1$ or $|x-y|\le 1$. -Assume first $|x-y|>1$. Then \eqref{eqcarl31} -vanishes and \eqref{eqcarl30} holds. -Now assume $|x-y|\le 1$. The we estimate -with Lemma \ref{expbound} +|K(x,y)|=|\kappa(x-y)|=\left|\frac {1-|x-y|}{1-e^{i(x-y)}}\right|\, . +\end{equation} +We estimate +with Lemma \ref{lower secant bound} \begin{equation}\label{eqcarl311} |K(x,y)|\le \frac {1}{|1-e^{i(x-y)}|}\le \frac 8{|x-y|}\, . \end{equation} This proves \eqref{eqcarl30} in the given case and completes the proof of the lemma. \end{proof} -\begin{lemma} +\begin{lemma}[Hilbert kernel regularity] +\label{Hilbert kernel regularity} +\uses{lower secant bound} For $x,y,y'\in \R$ with $x\neq y,y'$ and \begin{equation} + \label{eq close hoelder} 2|y-y'|\le |x-y|\, , \end{equation} we have \begin{equation}\label{eqcarl301} - |K(x,y) - K(x, y')|\le 2^6\frac{|y-y'|}{2|x-y|}\, . + |K(x,y) - K(x, y')|\le 2^{10}\frac{1}{|x-y|} \frac{|y-y'|}{|x-y|}\, . \end{equation} \end{lemma} -\lars{I checked until here, this subsection is not complete either.} - \begin{proof} - ... + Since $K(x,y) = k(x-y) = K(0, y-x)$ we can assume that $x = 0$. Then the assumption \eqref{eq close hoelder} implies that $y$ and $y'$ have the same sign. Since $K(0,y) = k(y) = \bar k(-y) = \bar K(0, -y)$ we can assume that they are both positive. Then it follows from \eqref{eq close hoelder} that + $$ + \frac{y}{2} \le y' \,. + $$ + We distinguish four cases. If $y, y' \le 1$, then we have + $$ + |K(0,y) - K(0,y')| = \left| \frac{1 - y}{1- e^{-iy}} - \frac{1 - y'}{1- e^{-iy'}}\right| + $$ + and by the fundamental theorem of calculus + $$ + = \left| \int_{y'}^{y} \frac{-1 + e^{-it} + i(1-t)e^{it}}{(1 - e^{-it})^2} \,dt \right|\,. + $$ + Using $y' \ge \frac{y}{2}$ and Lemma \ref{lower secant bound}, we bound this by + $$ + \le |y - y'| \sup_{\frac{y}{2} \le t \le 1} \frac{3}{|1 - e^{-it}|^2} \le 3 |y-y'| (8 \frac{2}{y})^2 \le 2^{10} \frac{|y-y'|}{|y|^2}\,. + $$ + If $y \le 1$ and $y' > 1$, then $K(0, y') = 0$ and we have from the first case + $$ + |K(0,y) - K(0,y')| = |K(0,y) - K(0,1)| \le 2^{10} \frac{|y-1|}{|y|^2} \le 2^{10} \frac{|y-y'|}{|y|^2}\,. + $$ + Similarly, if $y > 1$ and $y' \le 1$, then $K(0, y) = 0$ and we have by the same computation as for the first case + $$ + |K(0,y) - K(0,y')| = |K(0,y') - K(0,1)| \le 2^{10} \frac{|y-1|}{|y|^2} \le 2^{10} \frac{|y-y'|}{|y|^2}\,. + $$ + Finally, if $y, y' > 1$ then + $$ + |K(0,y) - K(0,y')| = 0 \le 2^{10} \frac{|y-y'|}{|y|^2}\,. + $$ \end{proof} +By the previous two lemmas, $K$ is a one-sided Calder\'on--Zygmund kernel on $(\R,\rho,\mu,4)$. + + +The operator $T^*$ defined in \eqref{def tang unm op} coincides in our setting with the operator $T^*$ defined in \eqref{concretetstar}. By Lemma \ref{lem hilbert}, this operator satisfies the bound \eqref{nontanbound}. + +Thus the assumptions of Theorem \ref{metric space Carleson} are all satisfied. Applying the Theorem, Lemma \ref{real Carleson} follows. + \printbibliography diff --git a/blueprint/src/preamble/common.tex b/blueprint/src/preamble/common.tex index 0cc74e74..3beff4e1 100644 --- a/blueprint/src/preamble/common.tex +++ b/blueprint/src/preamble/common.tex @@ -46,25 +46,25 @@ \DeclareMathOperator{\bd}{\operatorname{bd}} \DeclareMathOperator*{\esssup}{\operatorname{ess\,sup}} -\def \fp {\mathfrak p} -\def \fP {\mathfrak P} -\def \fu {\mathfrak u} -\def \fU {\mathfrak U} -\def \fv {\mathfrak v} -\def \fq {\mathfrak q} -\def \fQ {\mathfrak Q} -\def\fT{\mathfrak T} -\def\fL{\mathfrak L} -\def\fC{\mathfrak C} -\def\pc{\mathrm{c}} -\def\ps{\mathrm{s}} -\def \AD{{\bf s}} -\def \fc{\Omega} -\def\borel{\mathcal{E}} -\def\borelb{\mathcal{G}} -\def\sc{\mathcal{I}} -\def \tQ{{Q}} -\def\mfa{\vartheta} -\def\mfb{\theta} -\def\Mf{\Theta} -\def\fcc{\mathcal{Q}} +\newcommand{\fp}{\mathfrak p} +\newcommand{\fP}{\mathfrak P} +\newcommand{\fu}{\mathfrak u} +\newcommand{\fU}{\mathfrak U} +\newcommand{\fv}{\mathfrak v} +\newcommand{\fq}{\mathfrak q} +\newcommand{\fQ}{\mathfrak Q} +\newcommand{\fT}{\mathfrak T} +\newcommand{\fL}{\mathfrak L} +\newcommand{\fC}{\mathfrak C} +\newcommand{\pc}{\mathrm{c}} +\newcommand{\ps}{\mathrm{s}} +\newcommand{\AD}{{\bf s}} +\newcommand{\fc}{\Omega} +\newcommand{\borel}{\mathcal{E}} +\newcommand{\borelb}{\mathcal{G}} +\newcommand{\sc}{\mathcal{I}} +\newcommand{\tQ}{{Q}} +\newcommand{\mfa}{\vartheta} +\newcommand{\mfb}{\theta} +\newcommand{\Mf}{\Theta} +\newcommand{\fcc}{\mathcal{Q}}