10 Proof of The Classical Carleson Theorem
The convergence of partial Fourier sums is proved in Section 10.1 in two steps. In the first step, we establish convergence on a suitable dense subclass of functions. We choose smooth functions as subclass, the convergence is stated in Lemma 10.1.2 and proved in Section 10.2. In the second step, one controls the relevant error of approximating a general function by a function in the subclass. This is stated in Lemma 10.1.3 and proved in Section 10.8. The proof relies on a bound on the real Carleson maximal operator stated in Lemma 10.1.4 and proved in Section 10.9. This latter proof refers to the main Carleson Theorem 1.0.2. Two assumptions in Theorem 1.0.2 require more work. The boundedness of the nontangential maximal operator \(T^*\) defined in 1.0.16 is established in Lemma 10.1.5 using \(L^2\) and weak \(L^1\) bounds for the Hilbert transform, Lemmas 10.1.6 and 10.1.7. These lemmas are proved in Subsections 10.3, 10.4 and 10.5. The cancellative property is verified by Lemma 10.1.8, which is proved in Section 10.6. Several further auxiliary lemmas are stated and proved in Subsection 10.1, the proof of one of these auxiliary lemmas, Lemma 10.1.12, is done in Section 10.7.
All subsections past Section 10.1 are mutually independent.
Subsections 10.3 uses bounds for the Hardy–Littlewood maximal function on the real line. There may be a better path through Lean than we choose here. We refer to Proposition 2.0.6, the assumptions of it, namely that the real line fits into the setting of Chapter 2, is done in Section 10.9.
10.1 The classical Carleson theorem
Let a uniformly continuous \(2\pi \)-periodic function \(f:{\mathbb {R}}\to \mathbb {C}\) and \(0{\lt}\epsilon {\lt}1\) be given. Let
\begin{equation} C_{a,q} := \frac{2^{450a^3}}{(q-1)^6} \end{equation}
10.1.1
denote the constant from Theorem 1.0.2. Define
\begin{equation} \epsilon ' := \frac\epsilon {4 (C_{4,2} \cdot (8 / (\pi \epsilon ))^{\frac12} + \pi )}. \end{equation}
10.1.2
By uniform continuity of \(f\), there is a \(0{\lt}\delta {\lt}\pi \) such that for all \(x,x' \in {\mathbb {R}}\) with \(|x-x'|\le \delta \) we have
\begin{equation} \label{uniconbound} |f(x)-f(x')|\le \epsilon ' \, . \end{equation}
10.1.3
Define
\begin{equation} \label{def-fzero} f_0:=f \ast \phi _\delta , \end{equation}
10.1.4
where \(\phi _\delta \) is a nonnegative smooth bump function with \(\operatorname{\operatorname {supp}}(\phi _\delta ) \subset (-\delta , \delta )\) and \(\int _{\mathbb {R}}\phi _\delta (x) \, dx = 1\).
Lemma
10.1.1
smooth approximation
The function \(f_0\) is \(2\pi \)-periodic. The function \(f_0\) is smooth (and therefore measurable). The function \(f_0\) satisfies for all \(x\in {\mathbb {R}}\):
\begin{equation} \label{eq-ffzero} |f(x)-f_0(x)|\le \epsilon ' \, , \end{equation}
10.1.5
Proof
▶
Most of this is part of the Lean library.
We prove in Section 10.2:
Lemma
10.1.2
convergence for smooth
There exists some \(N_0 \in {\mathbb {N}}\) such that for all \(N{\gt}N_0\) and \(x\in [0,2\pi ]\) we have
\begin{equation} |S_N f_0 (x)- f_0(x)|\le \frac\epsilon 4\, . \end{equation}
10.1.6
We prove in Section 10.8:
Lemma
10.1.3
control approximation effect
There is a set \(E \subset {\mathbb {R}}\) with Lebesgue measure \(|E|\le \epsilon \) such that for all
\begin{equation} x\in [0,2\pi )\setminus E \end{equation}
10.1.7
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}
10.1.8
We are now ready to prove classical Carleson:
Let \(N_0\) be as in Lemma 10.1.2. For every
\begin{equation} x\in [0, 2\pi ) \setminus E\, , \end{equation}
10.1.9
and every \(N{\gt}N_0\) we have by the triangle inequality
\begin{equation*} |f(x)-S_Nf(x)| \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}
10.1.10
Using ??, we estimate 10.1.10 by
\begin{equation} \le \epsilon ' +\frac\epsilon 4 +\frac\epsilon 4\le \epsilon \, . \end{equation}
10.1.11
This shows 1.0.3 for the given \(E\) and \(N_0\).
Let \(\kappa :{\mathbb {R}}\to {\mathbb {R}}\) be the function defined by \(\kappa (0)=0\) and for \(0{\lt}|x|{\lt}1\)
\begin{equation} \label{eq-hilker} \kappa (x)=\frac{ 1-|x|}{1-e^{ix}}\, \end{equation}
10.1.12
and for \(|x|\ge 1\)
\begin{equation} \label{eq-hilker1} \kappa (x)=0\, . \end{equation}
10.1.13
Note that this function is continuous at every point \(x\) with \(|x|{\gt}0\).
The proof of Lemma 10.1.3 will use the following Lemma 10.1.4, which itself is proven in Section 10.9 as an application of Theorem 1.0.2.
Lemma
10.1.4
real Carleson
Let \(F,G\) be Borel subsets of \({\mathbb {R}}\) with finite measure. Let \(f\) be a bounded measurable function on \({\mathbb {R}}\) with \(|f|\le \mathbf{1}_F\). Then
\begin{equation} \left|\int _G Tf(x) \, dx\right| \le C_{4,2} |F|^{\frac12} |G|^{\frac12} \, , \end{equation}
10.1.14
where
\begin{equation} \label{define-T-carleson} T f(x)=\sup _{n\in \mathbb {Z}} \sup _{r{\gt}0}\left|\int _{r{\lt}|x-y|{\lt}1} f(y)\kappa (x-y) e^{iny}\, dy\right|\, . \end{equation}
10.1.15
One of the main assumption of Theorem 1.0.2, concerning the operator \(T_*\) defined in 1.0.16, is verified by the following lemma, which is proved in Section 10.3.
Lemma
10.1.5
nontangential Hilbert
For every bounded measurable function \(g\) with bounded support we have
\begin{equation} \label{concretetstarbound} \| T_*g\| _2\le 2^{43}\| g\| _2, \end{equation}
10.1.16
where
\begin{equation} \label{concretetstar} T_* g(x):=\sup _{0{\lt}r_1{\lt}r_2{\lt}1}\sup _{|x-x'|{\lt}r_1}\frac1{2\pi } \left|\int _{r_1{\lt}|x'-y|{\lt}r_2} g(y) \kappa (x'-y)\, dy\right|\, . \end{equation}
10.1.17
The proof of Lemma 10.1.5 relies on the next two auxiliary Lemmas.
For \(r\in (0,1)\), \(x\in \mathbb {R}\) and a bounded, measurable function \(g\) on \(\mathbb {R}\) with bounded support, we define
\begin{equation} \label{def-H_r} H_r g(x):= \int _{r{\lt}|x-y|{\lt}1} g(y) \kappa (x-y)\, dy. \end{equation}
10.1.18
The following Lemma is proved in Section 10.4
Lemma
10.1.6
Hilbert strong 2 2
Let \(0{\lt}r{\lt}1\). Let \(f\) be a bounded, measurable function on \(\mathbb {R}\) with bounded support. Then
\begin{equation} \label{eq-Hr-L2-bound} \| H_rf\| _{2}\leq 2^{13} \| f\| _2. \end{equation}
10.1.19
The following Lemma is proved in Section 10.5
Lemma
10.1.7
Hilbert weak 1 1
Let \(f\) be a bounded measurable function on \(\mathbb {R}\) with bounded support. Let \(\alpha {\gt}0\). Then for all \(r\in (0, 1)\), we have
\begin{equation} \label{eq-weak-1-1} \mu \left(\{ x\in \mathbb {R}: |H_r f(x)|{\gt}\alpha \} \right)\leq \frac{2^{19}}{\alpha } \int |f(y)|\, dy. \end{equation}
10.1.20
The next lemma will be used to verify that the collection \({\Theta }\) of modulation functions in our application of Theorem 1.0.2 satisfies the condition 1.0.12. It is proved in Section 10.6.
Lemma
10.1.8
van der Corput
Let \(\alpha {\lt}\beta \) be real numbers. Let \(g:{\mathbb {R}}\to {\mathbb {C}}\) be a measurable function and assume
\begin{equation} \| g\| _{Lip(\alpha ,\beta )}:=\sup _{\alpha \le x\le \beta }|g(x)|+|\beta -\alpha | \sup _{\alpha \le x{\lt}y\le \beta } \frac{|g(y)-g(x)|}{|y-x|}{\lt}\infty \, . \end{equation}
10.1.21
Then for any \(0\le \alpha {\lt}\beta \le 2\pi \) we have
\begin{equation} \int _{\alpha }^{\beta } g(x) e^{-inx}\, dx\le 2\pi |\beta -\alpha |\| g\| _{Lip(\alpha ,\beta )}(1+|n||\beta -\alpha |)^{-1}\, . \end{equation}
10.1.22
We close this section with six lemmas that are used across the following subsections.
Lemma
10.1.9
mean zero oscillation
Let \(n\in \mathbb {Z}\) with \(n\neq 0\), then
\begin{equation} \int _0^{2\pi } e^{inx}\, dx=0\, . \end{equation}
10.1.23
Proof
▶
We have
\begin{equation*} \int _0^{2\pi } e^{inx}\, dx=\left[ \frac1{in}e^{inx}\right]_0^{2\pi }=\frac1{in}(e^{2\pi i n}-e^{2\pi i 0})=\frac1{in}(1-1)=0\, . \qedhere \end{equation*}
Lemma
10.1.10
Dirichlet kernel
We have for every \(2\pi \)-periodic bounded measurable \(f\) and every \(N\ge 0\)
\begin{equation} S_Nf(x)=\frac1{2\pi }\int _{0}^{2\pi }f(y) K_N(x-y)\, dy \end{equation}
10.1.24
where \(K_N\) is the \(2\pi \)-periodic continuous function of \({\mathbb {R}}\) given by
\begin{equation} \label{eqksumexp} \sum _{n=-N}^N e^{in x'}\, . \end{equation}
10.1.25
We have for \(e^{ix'}\neq 1\) that
\begin{equation} \label{eqksumhil} K_N(x')=\frac{e^{iNx'}}{1-e^{-ix'}} +\frac{e^{-iNx'}}{1-e^{ix'}} \, . \end{equation}
10.1.26
Proof
▶
We have by definitions and interchanging sum and integral
\begin{equation*} S_Nf(x)=\sum _{n=-N}^N \widehat{f}_n e^{inx} \end{equation*}
\begin{equation*} =\sum _{n=-N}^N \frac1{2\pi }\int _{0}^{2\pi } f(x) e^{in(x-y)}\, dy \end{equation*}
\begin{equation} \label{eq-expsum} =\frac1{2\pi }\int _{0}^{2\pi } f(y) \sum _{n=-N}^N e^{in(x-y)}\, dy\, . \end{equation}
10.1.27
This proves the first statement of the lemma. By a telescoping sum, we have for every \(x'\in {\mathbb {R}}\)
\begin{equation} \left( e^{\frac12 ix'}-e^{-\frac12 ix'}\right) \sum _{n=-N}^N e^{inx'}= e^{(N+\frac12) ix'}-e^{-(N+\frac12) ix'}\, . \end{equation}
10.1.28
If \(e^{ix'}\neq 1\), the first factor on the left-hand side is not \(0\) and we may divide by this factor to obtain
\begin{equation} \sum _{n=-N}^N e^{inx'}= \frac{e^{i(N+\frac12)x'}}{e^{\frac12 ix'}-e^{-\frac12ix'}} -\frac{e^{-i(N+\frac12)x'}}{e^{\frac12 ix'}-e^{-\frac12ix'}} =\frac{e^{iNx'}}{1-e^{-ix'}} +\frac{e^{-iNx'}}{1-e^{ix'}}\, . \end{equation}
10.1.29
This proves the second part of the lemma.
Lemma
10.1.11
lower secant bound
Let \(\eta {\gt}0\) and \(-2\pi +\eta \le x\le 2\pi -\eta \) with \(|x|\ge \eta \). Then
\begin{equation} |1-e^{ix}|\ge \frac{2}{\pi } \eta \end{equation}
10.1.30
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.
The following lemma will be proved in Section 10.7.
Lemma
10.1.12
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\| _{L^2[-\pi , \pi ]} \le \| f\| _{L^2[-\pi , \pi ]}. \end{equation}
10.1.31
Lemma
10.1.13
Hilbert kernel bound
For \(x,y\in {\mathbb {R}}\) with \(x\neq y\) we have
\begin{equation} \label{eqcarl30} |\kappa (x-y)|\le 2^2(2|x-y|)^{-1}\, . \end{equation}
10.1.32
Proof
▶
Fix \(x\neq y\). If \(\kappa (x-y)\) is zero, then 10.1.32 is evident. Assume \(\kappa (x-y)\) is not zero, then \(0{\lt}|x-y|{\lt}1\). We have
\begin{equation} \label{eqcarl31} |\kappa (x-y)|=\left|\frac{1-|x-y|}{1-e^{i(x-y)}}\right|\, . \end{equation}
10.1.33
We estimate with Lemma 10.1.11
\begin{equation} \label{eqcarl311} |\kappa (x-y)|\le \frac{1}{|1-e^{i(x-y)}|}\le \frac2{|x-y|}\, . \end{equation}
10.1.34
This proves 10.1.32 in the given case and completes the proof of the lemma.
Lemma
10.1.14
Hilbert kernel regularity
For \(x,y,y'\in {\mathbb {R}}\) with \(x\neq y,y'\) and
\begin{equation} \label{eq-close-hoelder} 2|y-y'|\le |x-y|\, , \end{equation}
10.1.35
we have
\begin{equation} \label{eqcarl301} |\kappa (x-y) - \kappa (x-y')|\le 2^{8}\frac{1}{|x-y|} \frac{|y-y'|}{|x-y|}\, . \end{equation}
10.1.36
Proof
▶
Upon replacing \(y\) by \(y-x\) and \(y'\) by \(y'-x\) on the left-hand side of 10.1.35, we can assume that \(x = 0\). Then the assumption 10.1.35 implies that \(y\) and \(y'\) have the same sign. Since \(\kappa (y) = \bar\kappa (-y)\) we can assume that they are both positive. Then it follows from 10.1.35 that
\[ \frac{y}{2} \le y' \, . \]
We distinguish four cases. If \(y, y' \le 1\), then we have
\[ |\kappa (-y) - \kappa (-y')| = \left| \frac{1 - y}{1- e^{-iy}} - \frac{1 - y'}{1- e^{-iy'}}\right| \]
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 10.1.11, we bound this by
\[ \le |y - y'| \sup _{\frac{y}{2} \le t \le 1} \frac{3}{|1 - e^{-it}|^2} \le 3 |y-y'| (2 \frac{2}{y})^2 \le 2^{6} \frac{|y-y'|}{|y|^2}\, . \]
If \(y \le 1\) and \(y' {\gt} 1\), then \(\kappa (-y') = 0\) and we have from the first case
\[ |\kappa (-y) - \kappa (-y')| = |\kappa (-y) - \kappa (-1)| \le 2^{6} \frac{|y-1|}{|y|^2} \le 2^{6} \frac{|y-y'|}{|y|^2}\, . \]
Similarly, if \(y {\gt} 1\) and \(y' \le 1\), then \(\kappa (-y) = 0\) and we have from the first case
\[ |\kappa (-y) - \kappa (-y')| = |\kappa (-y') - \kappa (-1)| \le 2^{6} \frac{|y'-1|}{|y'|^2} \le 2^{6} \frac{|y-y'|}{|y'|^2}\, . \]
Using again \(y' \ge \frac{y}{2}\), we bound this by
\[ \le 2^{6} \frac{|y-y'|}{|y / 2|^2} = 2^{8} \frac{|y-y'|}{|y|^2} \]
Finally, if \(y, y' {\gt} 1\) then
\[ |\kappa (-y) - \kappa (-y')| = 0 \le 2^{8} \frac{|y-y'|}{|y|^2}\, . \]
10.2 Smooth functions.
Lemma
10.2.1
Let \(f:{\mathbb {R}}\to {\mathbb {C}}\) be \(2\pi \)-periodic and differentiable, and let \(n \in {\mathbb {Z}}\setminus \{ 0\} \). Then
\begin{equation} \widehat{f}_n = \frac{1}{i n} \widehat{f'}_n. \end{equation}
10.2.1
Proof
▶
This is part of the Lean library.
Lemma
10.2.2
Let \(f:{\mathbb {R}}\to {\mathbb {C}}\) such that
\begin{equation} \sum _{n\in {\mathbb {Z}}} |\widehat{f}_n| {\lt} \infty . \end{equation}
10.2.2
Then
\begin{equation} \sup _{x\in [0,2\pi ]} |f(x) - S_Nf(x)| \rightarrow 0 \end{equation}
10.2.3
as \(N \rightarrow \infty \).
Proof
▶
This is part of the Lean library.
Lemma
10.2.3
Let \(f:{\mathbb {R}}\to {\mathbb {C}}\) be \(2\pi \)-periodic and twice continuously differentiable. Then
\begin{equation} \sup _{x\in [0,2\pi ]} |f(x) - S_Nf(x)| \rightarrow 0 \end{equation}
10.2.4
as \(N \rightarrow \infty \).
Proof
▶
By Lemma 10.2.2, it suffices to show that the Fourier coefficients \(\widehat{f}_n\) are summable. Applying Lemma 10.2.1 twice and using the fact that \(f''\) is continuous and thus bounded on \([0,2\pi ]\) , we compute
\begin{equation*} \sum _{n\in {\mathbb {Z}}} |\widehat{f}_n| = |\widehat{f}_0| + \sum _{n\in {\mathbb {Z}}\setminus \{ 0\} } \frac{1}{n^2} |\widehat{f''}_n| \le |\widehat{f}_0| + \left(\sup _{x\in [0,2\pi ]} |f(x)| \right) \cdot \sum _{n\in {\mathbb {Z}}\setminus \{ 0\} } \frac{1}{n^2} {\lt} \infty . \end{equation*}
10.3 Proof of Cotlar’s Inequality
Lemma
10.3.1
estimate x shift
Let \(0{\lt}r{\lt}1\) and \(x\in \mathbb {R}\). Let \(g\) be a bounded measurable function with bounded support on \(\mathbb {R}\). Let \(Mg\) be the Hardy–Littlewood function defined in Proposition 2.0.6. Then for all \(x'\) with \(|x-x'|{\lt}r\).
\begin{equation} \label{xx'difference} \left|\int _{r{\lt}|x-y|{\lt}1} g(y) \kappa (x-y)\, dy -\int _{r{\lt}|x'-y|{\lt}1} g(y) \kappa (x'-y)\, dy \right|\le 2^{13}Mg(x)\, . \end{equation}
10.3.1
Proof
▶
First note that the conditions \(|x-y|{\lt}1\) and \(|x'-y|{\lt}1\) may be removed as the factor containing \(\kappa \) vanishes without these conditions.
We split the first integral in 10.3.1 into the domains \(r{\lt}|x-y|\le 2r\) and \(2r{\lt}|x-y|\). The integral over the first domain we estimate by \(\eqref{firstxx'}\) below. For the second domain, we observe with \(|x-x'|{\lt}r\) and the triangle inequality that \(r{\lt}|x'-y|\). We therefore combine on this domain with the corresponding part of the second integral in 10.3.1 and estimate that by \(\eqref{secondxx'}\) below. The remaining part of the second integral in 10.3.1 we estimate by \(\eqref{thirdxx'}\). Overall, we have estimated 10.3.1 by
\begin{equation} \label{firstxx'} \int _{r{\lt}|x-y|\le 2r} |g(y)| |\kappa (x-y)|\, dy \end{equation}
10.3.2
\begin{equation} \label{secondxx'} + \left|\int _{2r{\lt}|x-y|} g(y) (\kappa (x'-y)-\kappa (x-y))\, dy \right| \end{equation}
10.3.3
\begin{equation} \label{thirdxx'} +\left|\int _{r{\lt}|x'-y|, r{\lt}|x-y|{\lt}2r} |g(y) \kappa (x'-y)|\, dy \right| \end{equation}
10.3.4
Using the bound on \(\kappa \) in Lemma 10.1.13, we estimate 10.3.2 by
\begin{equation} \label{firstxx'b} \frac8r\int _{|x-y|{\lt}2r} |g(y)|\, dy\, . \end{equation}
10.3.5
Using the definition of \(Mg\), we estimate 10.3.5 by
\begin{equation} \label{firstxx'c} \le 32{Mg(x)}\, . \end{equation}
10.3.6
Similarly, in the domain of 10.3.4 we note by the triangle inequality and assumption on \(x'\) that \(|x'-y|\le 3r\) and thus we estimate 10.3.4 by
\begin{equation} \label{thirdxx'b} \frac8{r}\int _{|x'-y|{\lt}3r} |g(y)| \, dy\le 48 Mg(x) \end{equation}
10.3.7
We turn to the term 10.3.3. Let \(\nu \) be the smallest integer such that \(2^\nu {\gt} 2/r\). Using that the kernel vanishes unless \(|x-y|{\lt}2\), we decompose and estimate 10.3.3 with the triangle inequality by
\begin{equation} \label{secondxx'b} \sum _{j=1}^\nu \int _{2^jr{\lt}|x-y|\le 2^{j+1}r} |g(y)| |\kappa (x'-y)-\kappa (x-y)|\, dy\, . \end{equation}
10.3.8
Using Lemma 10.1.14, we estimate 10.3.8 by
\begin{equation*} 2^{10}\sum _{j=1}^\nu \int _{2^jr{\lt}|x-y|\le 2^{j+1}r} |g(y)| \frac{|x-x'|}{|x-y|^2}\, dy \end{equation*}
\begin{equation*} \le 2^{10}\sum _{j=1}^\nu \frac1{2^{2j}r}\int _{2^jr{\lt}|x-y|\le 2^{j+1}r} |g(y)| \, dy \end{equation*}
\begin{equation} \label{secondxx'c} \le 2^{10}\sum _{j=1}^\nu {2^{1-j}} Mg(x)\, . \end{equation}
10.3.9
Using a geometric series, we estimate 10.3.9 by
\begin{equation} \label{secondxx'd} \le 2^{11} Mg(x)\, . \end{equation}
10.3.10
Summing the estimates for 10.3.2, 10.3.3, and 10.3.4 proves the lemma.
Recall that
\begin{equation*} H_rg(x)=\int _{r{\lt}|x-y|{\lt}1} g(y) \kappa (x-y)\, dy\, . \end{equation*}
Lemma
10.3.2
Cotlar control
Let \(0{\lt}r{\lt}r_1{\lt}1\) and \(x\in \mathbb {R}\). Let \(g\) be a bounded measurable function with bounded support on \(\mathbb {R}\). Let \(Mg\) be the Hardy–Littlewood function defined in Proposition 2.0.6. Then for all \(x'\in {\mathbb {R}}\) with \(|x'-x|{\lt}\frac{r_1}4\) we have
\begin{equation} \label{eq-cotlar-control} \left|H_{r_1} g(x) \right|\le |H_{r}(g-g\mathbf{1}_{[x-\frac{r_1}2,x+\frac{r_1}2]})(x')|+ 2^{15}Mg(x)\, . \end{equation}
10.3.11
Proof
▶
Let \(x\) and \(x'\) be given with \(|x'-x|{\lt}\frac{r_1}4\). By an application of Lemma 10.3.1, we estimate the left-hand-side of 10.3.11 by
\begin{equation} \label{eqcotlar0} |H_{r_1}(g)(x')|+ 2^{13}Mg(x)\, . \end{equation}
10.3.12
We have
\begin{equation} \label{eqcotlar-1} H_{r_1}(g)(x')= \int _{r_1{\lt}|x'-y|{\lt}1} g(y) \kappa (x'-y)\, dy\, . \end{equation}
10.3.13
On the domain \(r_1{\lt}|x'-y|\), we have \(\frac{r_1}2{\lt}|x-y|\). Hence we may write for 10.3.13
\begin{equation*} H_{r_1}(g)(x’)=\int _{r_1{\lt}|x'-y|{\lt}1} (g-g\mathbf{1}_{[x-\frac{r_1}2,x+\frac{r_1}2]})(y) \kappa (x’-y)\, dy \end{equation*}
\begin{equation} \label{eqcotlar1} =H_{r_1}(g-g\mathbf{1}_{[x-\frac{r_1}2,x+\frac{r_1}2]})(x')\, . \end{equation}
10.3.14
Combining the estimate 10.3.12 with the identification 10.3.14, we obtain
\begin{equation} \label{eqcotlar5} \left|H_{r_1} g(x) \right|\le |H_{r_1}(g-g\mathbf{1}_{[x-\frac{r_1}2,x+\frac{r_1}2]})(x')|+ 2^{13}Mg(x)\, . \end{equation}
10.3.15
We have
\begin{equation*} (H_{r}-H_{r_1})(g-g\mathbf{1}_{[x-\frac{r_1}2,x+\frac{r_1}2]})(x’) \end{equation*}
\begin{equation} \label{eqcotlar2} = \int _{ {r}{\lt}|x'-y|{\lt}r_1} (g-g\mathbf{1}_{[x-\frac{r_1}2,x+\frac{r_1}2]})(y)k(x'-y)\, dy\, . \end{equation}
10.3.16
Assume first \(r\ge \frac{r_1}4\). Then we estimate 10.3.16 with Lemma 10.1.13 by
\begin{equation*} \int _{\frac{r_1}4{\lt}|x'-y|{\lt}r_1} |g(y)k(x’-y)|\, dy \end{equation*}
\begin{equation} \label{eqcotlar3} \le \frac{32}{r_1}\int _{|x'-y|{\lt}r_1} |g(y)|\, dy \le 64 Mg(x')\, . \end{equation}
10.3.17
Assume now \(r\le \frac{r_1}4\). As \(|x'-y|\le \frac{r_1}4\) implies \(|x-y|\le \frac{r_1}2\), we see that 10.3.16 equals
\begin{equation*} \int _{ {\frac{r_1}4}{\lt}|x'-y|{\lt}r_1} (g-g\mathbf{1}_{[x-\frac{r_1}2,x+\frac{r_1}2]})(y)k(x’-y)\, dy\, , \end{equation*}
which we again estimate as above by 10.3.17. In both cases, 10.3.11 follows by the triangle inequality from 10.3.15 and the estimate for 10.3.16.
Lemma
10.3.3
Cotlar sets
Let \(0{\lt}r{\lt}r_1{\lt}1\) and \(x\in \mathbb {R}\). Let \(g\) be a bounded measurable function with bounded support on \(\mathbb {R}\). Let \(Mg\) be the Hardy–Littlewood maximal function defined in Proposition 2.0.6. Let \(x\in {\mathbb {R}}\). Then the measure \(|F_1|\) of the set \(F_1\) of all \(x'\in [x-\frac{r_1}4,x+\frac{r_1}4]\) such that
\begin{equation} \label{first-cotlar-exception} |H_{r}(g)(x')|{\gt}4M(H_{r}g)(x) \end{equation}
10.3.18
is less than or equal to \(r_1/8\). Moreover, the measure \(|F_2|\) of the set \(F_2\) of all \(x'\in [x-\frac{r_1}4,x+\frac{r_1}4]\) such that
\begin{equation} \label{second-cotlar-exception} |H_{r}(g\mathbf{1}_{[x-\frac{r_1}2,x+\frac{r_1}2]})|{\gt}2^{22}Mg(x) \end{equation}
10.3.19
is less than \(r_1/8\).
Proof
▶
Let \(r\), \(r_1\), \(x\) and \(g\) be given. If \(M(H_{r}g)(x)=0\), then \(H_{r}g\) is constant zero and \(F_1\) is empty and the estimate on \(F_1\) trivial. Assume \(M(H_{r}g)(x){\gt}0\). We have with 10.3.18
\begin{equation} M(H_{r}g)(x)\ge \frac2{r_1}\int _{|x'-x|{\lt}\frac{r_1}4}|H_{r}g(x')|\, dx' \end{equation}
10.3.20
\begin{equation} \ge \frac2{r_1}\int _{F_1} 4 M(H_{r}g)(x)\, dx' \end{equation}
10.3.21
Dividing by \(M(H_{r}g)(x)\) gives
\begin{equation} 1\ge \frac8{r_1} |F_1|\, . \end{equation}
10.3.22
This gives the desired bound for the measure of \(F_1\).
We turn to the set \(F_2\). Similarly as above we may assume \(Mg(x){\gt}0\). The set \(F_2\) is then estimated with Lemma 10.1.7 by
\begin{equation} \frac{2^{19}}{2^{22}Mg(x)}\int |g\mathbf{1}_{[x-\frac{r_1}2, x+\frac{r_1}2]}|(y)\, dy \end{equation}
10.3.23
\begin{equation} \le \frac{2^{19}}{2^{22}Mg(x)}r_1Mg(x)= \frac{r_1}8\, . \end{equation}
10.3.24
This gives the desired bound for the measure of \(F_2\).
Lemma
10.3.4
Cotlar estimate
Let \(0{\lt}r{\lt}r_1{\lt}1\) and \(x\in \mathbb {R}\). Let \(g\) be a bounded measurable function with bounded support on \(\mathbb {R}\). Let \(Mg\) and \(M(H_r g)\) be the respective Hardy–Littlewood maximal functions defined in Proposition 2.0.6. Then for all \(x\in {\mathbb {R}}\)
\begin{equation} \label{eq-cotlar-estimate} \left| \int _{r_1{\lt}|x-y|{\lt}1} g(y) \kappa (x-y)\, dy \right| \end{equation}
10.3.25
\begin{equation} \le 2^{2}M(H_rg)(x)+ 2^{23} Mg(x) \, . \end{equation}
10.3.26
Proof
▶
By Lemma 10.3.2, the measure of the set of all \(x'\in [x-\frac{r_1}4,x+\frac{r_1}4]\) such that at least one of the conditions 10.3.18 and 10.3.19 is satisfied is at most \(r_1/4\) and hence not all of \(x'\in [x-\frac{r_1}4,x+\frac{r_1}4]\). Pick an \(x'\in [x-\frac{r_1}4,x+\frac{r_1}4]\) such that both conditions are not satisfied. Applying Lemma 10.3.2 for this \(x'\) and using the triangle inequality estimates the left-hand side of 10.3.25 by
\begin{equation} 4M(H_rg)(x)+2^{22}Mg(x)+2^{15}Mg(x)\, . \end{equation}
10.3.27
This proves the lemma.
Lemma
10.3.5
simple nontangential Hilbert
For every \(0{\lt}r{\lt}1\) and every bounded measurable function \(g\) with bounded support we have
\begin{equation} \label{trzerobound} \| T_{r}g\| _2\le 2^{42}\| g\| _2, \end{equation}
10.3.28
where
\begin{equation} \label{eq-simple--nontangential} T_{r} g(x):=\sup _{r{\lt}r_1{\lt}1}\sup _{|x-x'|{\lt}r_1}\frac1{2\pi } \left|\int _{r_1{\lt}|x'-y|{\lt}1} g(y) \kappa (x'-y)\, dy\right|\, . \end{equation}
10.3.29
Proof
▶
With Lemma 10.3.1 and the triangle inequality, we estimate for every \(x\in {\mathbb {R}}\)
\begin{equation} T_{r} g(x) \le 2^{13} Mg(x)+\sup _{r{\lt}r_1{\lt}1} \frac1{2\pi } \left|\int _{r_1{\lt}|x-y|{\lt}1} g(y) \kappa (x-y)\, dy\right|\, . \end{equation}
10.3.30
Using further Lemma 10.3.4, we estimate
\begin{equation} T_{r} g(x) \le 2^{13} Mg(x)+2^{23} Mg(x)+2^{2}M(H_{r}g)(x)\, . \end{equation}
10.3.31
Taking the \(L^2\) norm and using Proposition 2.0.6with \(a=4\) and \(p_2=2\) and \(p_1=1\) , we obtain
\begin{equation} \| T_{r} g\| _2 \le 2^{24} \| Mg\| _2+2^{2}\| M(H_{r}g)\| _2 \end{equation}
10.3.32
\begin{equation} \le 2^{41} \| g\| _2+2^{19}\| H_{r}(g)\| _2\, . \end{equation}
10.3.33
Applying Lemma 10.1.6, gives
\begin{equation} \| T_{r} g\| _2\le 2^{41} \| g\| _2+2^{32}\| g\| _2\, . \end{equation}
10.3.34
This shows 10.3.28 and completes the proof of the lemma.
Fix \(g\) as in the Lemma. Applying Lemma 10.3.5 with a sequence of \(r\) tending to \(0\) and using Lebesgue monotone convergence shows
\begin{equation} \label{tzerobound} \| T_{0}g\| _2\le 2^{42}\| g\| _2, \end{equation}
10.3.35
where
\begin{equation} \label{eq-simpler--nontangential} T_{0} g(x):=\sup _{0{\lt}r_1{\lt}1}\sup _{|x-x'|{\lt}r_1}\frac1{2\pi } \left|\int _{r_1{\lt}|x'-y|{\lt}1} g(y) \kappa (x'-y)\, dy\right|\, . \end{equation}
10.3.36
We now write by the triangle inequality
\begin{equation} \label{concretetstartriangle} T_* g(x)\le \sup _{0{\lt}r_1{\lt}r_2{\lt}1}\sup _{|x-x'|{\lt}r_1}\frac1{2\pi } \left|\int _{r_1{\lt}|x'-y|{\lt}1}g(y) \kappa (x'-y)\, dy\right| \end{equation}
10.3.37
\begin{equation*} +\sup _{0{\lt}r_1{\lt} r_2{\lt}1}\sup _{|x-x'|{\lt}r_1}\frac1{2\pi } \left|\int _{r_2{\lt}|x'-y|{\lt}1} g(y) \kappa (x’-y)\, dy\right|\, . \end{equation*}
Noting that the first integral does not depend on \(r_2\) and estimating the second integral by the larger supremum over all \(|x-x'|{\lt}r_2\), at which time the integral does not depend on \(r_1\), we estimate 10.3.37 by
\begin{equation} \label{concretetstartriangle2} \sup _{0{\lt}r_1{\lt}1}\sup _{|x-x'|{\lt}r_1}\frac1{2\pi } \left|\int _{r_1{\lt}|x'-y|{\lt}1}g(y) \kappa (x'-y)\, dy\right| \end{equation}
10.3.38
\begin{equation*} +\sup _{0{\lt} r_2{\lt}1}\sup _{|x-x'|{\lt}r_2}\frac1{2\pi } \left|\int _{r_2{\lt}|x'-y|{\lt}1} g(y) \kappa (x’-y)\, dy\right|\, . \end{equation*}
Applying the triangle inequality on the left-hand side of 10.1.16 and applying 10.3.35 twice proves 10.1.16. This completes the proof of Lemma 10.1.5.
10.4 The truncated Hilbert transform
Let \(M_n\) be the modulation operator acting on measurable \(2\pi \)-periodic functions defined by
\begin{equation} M_ng(x)=g(x) e^{inx}\, . \end{equation}
10.4.1
Define the approximate Hilbert transform by
\begin{equation} L_N g=\frac1N\sum _{n=0}^{N-1} M_{-n-N} S_{N+n}M_{N+n}g\, . \end{equation}
10.4.2
Lemma
10.4.1
modulated averaged projection
We have for every bounded measurable \(2\pi \)-periodic function \(g\)
\begin{equation} \label{lnbound} \| L_Ng\| _{L^2[-\pi , \pi ]}\le \| g\| _{L^2[-\pi , \pi ]}\, . \end{equation}
10.4.3
Proof
▶
We have
\begin{equation} \label{mnbound} \| M_ng\| _{L^2[-\pi , \pi ]}^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}
10.4.4
We have by the triangle inequality, the square root of the identity in 10.4.4, and Lemma 10.1.12
\begin{equation*} \| L_ng\| _{L^2[-\pi , \pi ]}=\| \frac1N\sum _{n=0}^{N-1} M_{-n-N} S_{N+n}M_{N+n}g\| _{L^2[-\pi , \pi ]} \end{equation*}
\begin{equation*} \le \frac1N\sum _{n=0}^{N-1} \| M_{-n-N} S_{N+n}M_{N+n}g\| _{L^2[-\pi , \pi ]} = \frac1N\sum _{n=0}^{N-1} \| S_{N+n}M_{N+n}g\| _{L^2[-\pi , \pi ]} \end{equation*}
\begin{equation} \le \frac1N\sum _{n=0}^{N-1} \| M_{N+n}g\| _{L^2[-\pi , \pi ]} = \frac1N\sum _{n=0}^{N-1} \| g\| _{L^2[-\pi , \pi ]} =\| g\| _{L^2[-\pi , \pi ]}\, . \end{equation}
10.4.5
This proves 10.4.4 and completes the proof of the lemma.
Lemma
10.4.2
periodic domain shift
Let \(f\) be a bounded \(2\pi \)-periodic function. We have for any \(0 \le x\le 2\pi \) that
\begin{equation} \int _0^{2\pi } f(y)\, dy= \int _{-x}^{2\pi -x} f(y)\, dy =\int _{-\pi }^{\pi } f(y-x)\, dy\, . \end{equation}
10.4.6
Proof
▶
We have by periodicity and change of variables
\begin{equation} \label{eqhil9} \int _{-x}^{0} f(y)\, dy=\int _{-x}^{0} f(y+2\pi )\, dy= \int _{2\pi -x}^{2\pi } f(y)\, dy\, . \end{equation}
10.4.7
We then have by breaking up the domain of integration and using 10.4.7
\begin{equation*} \int _0^{2\pi } f(y)\, dy= \int _0^{2\pi -x} f(y)\, dy+ \int _{2\pi -x}^{2\pi } f(y)\, dy \end{equation*}
\begin{equation} = \int _0^{2\pi -x} f(y)\, dy+ \int _{ -x}^{0} f(y)\, dy = \int _{-x}^{2\pi -x} f(y)\, dy\, . \end{equation}
10.4.8
This proves the first identity of the lemma. The second identity follows by substitution of \(y\) by \(y-x\).
Lemma
10.4.3
Young convolution
Let \(f\) and \(g\) be two bounded non-negative measurable \(2\pi \)-periodic functions on \({\mathbb {R}}\). Then
\begin{equation} \label{eqyoung} \left(\int _{-\pi }^{\pi } \left(\int _{-\pi }^{\pi } f(y)g(x-y)\, dy\right)^2\, dx\right)^{\frac12}\le \| f\| _{L^2[-\pi , \pi ]} \| g\| _{L^1[-\pi , \pi ]}\, . \end{equation}
10.4.9
Proof
▶
Using Fubini and Lemma 10.4.2, we observe
\begin{equation*} \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 _{-\pi }^{\pi }f(y)^2\int _{-\pi }^{\pi }g(x) \, dx dy =\| f\| _{L^2[-\pi , \pi ]}^2\| g\| _{L^1[-\pi , \pi ]}\, . \end{equation}
10.4.10
Let \(h\) be the nonnegative square root of \(g\), then \(h\) is bounded and \(2\pi \)-periodic with \(h^2=g\). We estimate the square of the left-hand side of 10.4.9 with Cauchy-Schwarz and then with 10.4.10 by
\begin{equation*} \int _{-\pi }^{\pi } (\int _{-\pi }^{\pi }f(y)h(x-y)h(x-y)\, dy)^2\, dx \end{equation*}
\begin{equation*} \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 \end{equation*}
\begin{equation*} = \| f\| _{L^2[-\pi , \pi ]}^2\| g\| _{L^1[-\pi , \pi ]}^2\, . \end{equation*}
Taking square roots, this proves the lemma.
For \(0{\lt}r{\lt}1\), Define the kernel \(k_r\) to be the \(2\pi \)-periodic function
\begin{equation} |k_r(x)|:=\min \left(r^{-1}, 1+\frac r{|1-e^{ix}|^2}\right)\, , \end{equation}
10.4.11
where the minimum is understood to be \(r^{-1}\) in case \(1=e^{ix}\).
Lemma
10.4.4
integrable bump convolution
Let \(g,f\) be bounded measurable \(2\pi \)-periodic functions. Let \(0{\lt}r{\lt}\pi \). Assume we have for all \(0\le x\le 2\pi \)
\begin{equation} \label{ebump1} |g(x)|\le k_r(x)\, . \end{equation}
10.4.12
Let
\begin{equation} h(x)= \int _{-\pi }^{\pi } f(y)g(x-y)\, dy \, . \end{equation}
10.4.13
Then
\begin{equation} \| h\| _{L^2[-\pi , \pi ]}\le 2^{5}\| f\| _{L^2[-\pi , \pi ]} \, . \end{equation}
10.4.14
Proof
▶
From monotonicity of the integral and 10.4.12,
\begin{equation} \| g\| _{L^1[-\pi , \pi ]} \le \int _{-\pi }^{\pi }k_r(x)\, dx\, . \end{equation}
10.4.15
Using the symmetry \(k_r(x)=k_r(-x)\), the assumption, and Lemma 10.1.11, the last display is equal to
\begin{equation*} = 2 \int _0^\pi \min \left(\frac1r, 1+\frac r{|1-e^{ix}|^2}\right)\, dx \end{equation*}
\begin{equation*} \le 2\int _0^{r} \frac1r \, dx+2\int _r^{\pi }1+\frac{64r}{x^2}\, dx \end{equation*}
\begin{equation} \le 2+2\pi + 2\left(\frac{64r}r-\frac{64r}{\pi }\right) \le 2^{5}\, . \end{equation}
10.4.16
Together with Lemma 10.4.3, this proves the lemma.
Lemma
10.4.5
dirichlet approximation
Let \(0{\lt}r{\lt}1\). Let \(N\) be the smallest integer larger than \(\frac1r\). There is a \(2\pi \)-periodic continuous function \({L’}\) on \({\mathbb {R}}\) that satisfies for all \(-\pi \le x\le \pi \) and all \(2\pi \)-periodic bounded measurable functions \(f\) on \({\mathbb {R}}\)
\begin{equation} \label{lthroughlprime} L_Nf(x)=\frac1{2\pi }\int _{-\pi }^{\pi }f(y) {L’}(x-y)\, dy \end{equation}
10.4.17
and
\begin{equation} \label{eqdifflhil} \left|L'(x)-\mathbf{1}_{\{ y:\, r{\lt}|y|{\lt}1\} } \kappa (x)\right|\le 2^{5}k_r(x)\, . \end{equation}
10.4.18
Proof
▶
We have by definition and Lemma 10.1.10
\begin{equation} L_Ng(x)= \frac1N\sum _{n=0}^{N-1} \int _{-\pi }^{\pi } e^{-i(N+n)x} K_{N+n}(x-y) e^{i(N+n)y}g(y) \, dy \, .\end{equation}
10.4.19
This is of the form 10.4.17 with the continuous function
\begin{equation} {L’}(x)= \frac1N\sum _{n=0}^{N-1} K_{N+n}(x) e^{-i(N+n)x}\, . \end{equation}
10.4.20
With 10.1.25 of Lemma 10.1.10 we have \(|K_N(x)|\le N\) for every \(x\) and thus
\begin{equation} \label{eqhil13} |{L’}(x)|\le \frac1N\sum _{n=0}^{N-1} (N+n) \le 2N\le 2^2 r^{-1}\, . \end{equation}
10.4.21
Therefore, for \(|x|\in [0, r)\cup (1, \pi ]\), we have
\begin{equation} \label{eqdiffzero} \left|L'(x)-\mathbf{1}_{\{ y:\, r{\lt}|y|{\lt}1\} }(x)\kappa (x)\right|=|L'(x)|\leq 2^{2} r^{-1}. \end{equation}
10.4.22
This proves 10.4.18 for \(|x|\in [0, r)\) since \(k_r(x)=r^{-1}\) in this case.
For \(e^{ix'}\neq 1\) and may use the expression 10.1.26 for \(K_N\) in Lemma 10.1.10 to obtain
\begin{equation*} {L’}(x)= \frac1N\sum _{n=0}^{N-1} \left(\frac{e^{i(N+n)x}}{1-e^{-ix}} +\frac{e^{-i(N+n)x}}{1-e^{ix}}\right) e^{-i(N+n)x} \end{equation*}
\begin{equation*} = \frac1N\sum _{n=0}^{N-1} \left(\frac{1}{1-e^{-ix}} +\frac{e^{-i2(N+n)x}}{1-e^{ix}}\right) \end{equation*}
\begin{equation} \label{eqhil3} = \frac{1}{1-e^{-ix}} + \frac1N \frac{e^{-i2Nx}}{1-e^{ix}} \sum _{n=0}^{N-1} {e^{-i2nx}} \end{equation}
10.4.23
and thus
\begin{equation} \label{eq-L'L''} {L’}(x) -\mathbf{1}_{\{ y:\, r{\lt}|y|{\lt}1\} }\kappa (x)=L''(x)+ \frac{1-\mathbf{1}_{{\{ y:\, r{\lt}|y|{\lt}1\} }}(x)(1-|x|)}{1-e^{-ix}}, \end{equation}
10.4.24
where
\[ L''(x):=\frac1N \frac{e^{-i2Nx}}{1-e^{ix}} \sum _{n=0}^{N-1} {e^{-i2nx}}. \]
For \(x\in [-\pi , r]\cup [r, \pi ]\), we have using Lemma 10.1.11 that
\begin{equation*} \left|\frac{1-\mathbf{1}_{{\{ y:\, r{\lt}|y|{\lt}1\} }}(x)(1-|x|)}{1-e^{-ix}} \right|=\left|\frac{\min (|x|, 1)}{1-e^{-ix}} \right|\leq \frac{8\min (|x|, 1)}{|x|} \end{equation*}
\begin{equation} \label{eq-diffzero2} \leq 2^3\cdot 1\leq 2^{3} k_r(x). \end{equation}
10.4.25
Next, we need to estimate \(L''(x)\). If the real part of \(e^{ix}\) is negative, we have
\begin{equation} 1\le |1-e^{ix}|\le 2\, . \end{equation}
10.4.26
and hence
\begin{equation} \label{eqhil12} |L''(x)|\le \frac1N \sum _{n=0}^{N-1} 1=1\le 1+\frac r{|1-e^{ix}|^2}\, . \end{equation}
10.4.27
If the real part of \(e^{ix}\) is positive and in particular while still \(e^{ix}\neq \pm 1\), then we have by telescoping
\begin{equation} (1-e^{-2ix}) \sum _{n=0}^{N-1} {e^{-i2nx}}=1-e^{-i2Nx}\, . \end{equation}
10.4.28
As \(e^{-2ix}\neq 1\), we may divide by \(1-e^{-2ix}\) and insert this into 10.4.23 to obtain
\begin{equation} L''(x)= \frac1N \frac{e^{-i2Nx}}{1-e^{ix}} \frac{1-e^{-i2Nx}}{1-e^{-2ix}}\, . \end{equation}
10.4.29
Hence, with Lemma 10.1.11 and nonnegativity of the real part of \(e^{ix}\)
\begin{equation*} |L”(x)| \le \frac2N \frac{1}{|1-e^{ix}|} \frac{1}{|1-e^{-2ix}|} \end{equation*}
\begin{equation} \label{eqhil11} = \frac2N \frac{1}{|1-e^{ix}|^2} \frac{1}{|1+e^{ix}|}\le \frac{4r}{|1-e^{ix}|^2}\le 2^{2} \left(1+\frac{r}{|1-e^{ix}|^2}\right) \end{equation}
10.4.30
Inequalities 10.4.21, 10.4.22, 10.4.24, 10.4.25, 10.4.27, and 10.4.30 prove 10.4.18. This completes the proof of the lemma.
We now prove Lemma 10.1.6.
We first show that if \(f\) is supported in \([-3/2, 3/2]\), then
\begin{equation} \label{eq-Hr-short-support} \| H_r f\| _{L^2({\mathbb {R}})} \le 2^{16} \| f\| _{L^2({\mathbb {R}})}\, . \end{equation}
10.4.31
Let \(\tilde{f}\) be the \(2\pi \)-periodic extension of \(f\) to \(\mathbb {R}\). Let \(N\) be the smallest integer larger than \(\frac1r\). Then, by Lemma 10.4.5 and the triangle inequality, for \(x\in [-\pi , \pi ]\) we have
\begin{equation*} |H_r \tilde{f}(x)|\leq 2\pi |L_N \tilde{f}(x)|+2^{5}\left|\int _{-\pi }^{\pi }k_r(x-y)\tilde{f}(y)\, dy\right|. \end{equation*}
Taking \(L^2\) norm over the interval \([-\pi , \pi ]\) and using its sub-additivity, we get
\[ \| H_r \tilde{f}\| _{L^2([-\pi , \pi ])} \]
\begin{equation*} \leq 2\pi \| L_N \tilde{f}\| _{L^2([-\pi , \pi ])}\, + 2^{5}\left(\int _{-\pi }^{\pi } \left|\int _{-\pi }^{\pi }k_r(x-y)\tilde{f}(y)\, dy\right|^2\, dx\right)^{\frac{1}{2}}. \end{equation*}
Since \(k_r\) is supported in \([-1,1]\), we have that \(H_rf\) is supported in \([-5/2, 5/2]\) and agrees there with \(H_r \tilde f(x)\). Using Lemma 10.4.1 and Lemma 10.4.4, we conclude
\begin{equation} \| H_r f\| _{L^2({\mathbb {R}})} \le \| H_r \tilde f\| _{L^2([-\pi , \pi ])} \leq 2\pi \| f\| _{L^2({\mathbb {R}})} + 2^{10}\| f\| _{L^2({\mathbb {R}})}\, , \end{equation}
10.4.32
which gives 10.4.31.
Suppose now that \(f\) is supported in \([c, c+3]\) for some \(c \in {\mathbb {R}}\). Then the function \(g(x) = f(c+ \frac{3}{2} +x)\) is supported in \([-3/2,3/2]\). By a change of variables in ??, we have \(H_r g(x ) = H_r f(c+ 3/2+x)\). Thus, by 10.4.31
\begin{equation} \label{eq-Hr-short-support-2} \| H_rg\| _2 = \| H_r f\| _2 \le 2^{11} \| f\| _2 = \| g\| _2\, . \end{equation}
10.4.33
Let now \(f\) be arbitrary. Since \(\kappa (x) = 0\) for \(|x| {\gt} 1\), we have for all \(x \in [c+1, c+2]\)
\[ H_rf(x) = H_r(f \mathbf{1}_{[c, c+3]})(x)\, . \]
Thus
\[ \int _{c+1}^{c+2} |H_r f(x)|^2 \, \mathrm{d}x \le \int _{{\mathbb {R}}} |H_r(f \mathbf{1}_{[c, c+3]})(x)|^2 \, \mathrm{d}x\, . \]
Applying the bound 10.4.33, this is
\[ \le 2^{11} \int _{c}^{c+3} |f(x)|^2 \, \mathrm{d}x\, . \]
Summing over all \(c \in \mathbb {Z}\), we obtain
\[ \int _{{\mathbb {R}}} |H_rf(x)|^2 \, \mathrm{d}x \le 3 \cdot 2^{11} \int _{{\mathbb {R}}} |f(x)|^2 \, \mathrm{d}x\, . \]
This completes the proof.
10.5 Calderón-Zygmund Decomposition
For \(I=[s, t)\subset \mathbb {R}\), we define
\begin{equation} \label{eq-bisec} \textrm{bi} (I):= \left\{ \left[s, \frac{s+t}{2}\right), \left[\frac{s+t}{2}, t\right)\right\} . \end{equation}
10.5.1
In what follows, we write \(\mu \) for the Lebesgue measure on \({\mathbb {R}}\).
Lemma
10.5.1
interval bisection
Let \(I=[s, t)\subset \mathbb {R}\) be a bounded, right-open interval. Let \(I_1, I_2\in \textrm{bi} (I)\) with \(I_1\neq I_2\). Then
\begin{equation} \label{eq-bi-size} \mu (I_1)=\mu (I_2)=\frac{\mu (I)}{2}. \end{equation}
10.5.2
Further,
\begin{equation} \label{eq-bi-union} I=I_1\cup I_2, \end{equation}
10.5.3
and the intervals \(I_1\) and \(I_2\) are disjoint.
Proof
▶
This in some form can be taken from the Lean library.
For a bounded interval \(I=[a, b)\subset \mathbb {R}\) and a non-negative integer \(n\), we inductively define
\begin{equation} \label{eq-ch-I} \textrm{ch}_0(I):=\{ I\} , \qquad \textrm{ch}_n(I)=\cup _{J\in \textrm{ch}_{n-1}(J)} \textrm{bi} (J). \end{equation}
10.5.4
Lemma
10.5.2
bisection children
Let \(I=[s, t)\) and let \(n\) be a non-negative integer. Then the intervals in \(\textrm{ch}_n(I)\) are pairwise disjoint. For any \(J\in \textrm{ch}_n(I)\),
\begin{equation} \label{eq-ch-size} \mu (J)=\frac{\mu (I)}{2^{n}}=\frac{t-s}{2^n}. \end{equation}
10.5.5
Further,
\begin{equation} \label{eq-ch-cover} I= \cup _{J\in \textrm{ch}_n(I)} J \end{equation}
10.5.6
and
\begin{equation} \label{eq-ch-card} |\textrm{ch}_n(I)|= 2^{n+1}. \end{equation}
10.5.7
Proof
▶
This follows by induction on \(n\), using Lemma 10.5.1.
Lemma
10.5.3
Lebesgue differentiation
Let \(f\) be a bounded measurable function with bounded support. Then for \(\mu \) almost every \(x\), we have
\[ \frac{1}{\mu (I_n)}\int _{I_n} f(y)\, dy= f(x), \]
where \(\{ I_n\} _{n\geq 1}\) is a sequence of intervals such that \(x\in I_n\) for each \(n\geq 1\) and
\[ \lim _{n\to \infty } \mu (I_n)=0. \]
Proof
▶
This follows from the Lebesgue differentiation theorem, which is already formalized in Lean.
Lemma
10.5.4
stopping time
Let \(f\) be a bounded, measurable function with bounded support on \(\mathbb {R}\). Let \(\alpha {\gt}0\). Then there exists \(A\subset \mathbb {R}\) such that the following properties 10.5.8, 10.5.9, 10.5.10, 10.5.11, and 10.5.12 are satisfied. For all \(x\in A\)
\begin{equation} \label{eq-CZ-good-set} |f(x)|\leq \alpha . \end{equation}
10.5.8
The set \(\mathbb {R}\setminus A\) can be decomposed into a countable union of intervals
\begin{equation} \label{eq-CZ-good-set-comp} \mathbb {R}\setminus A= \cup _{j} I_j= \cup _{j} [s_j, t_j), \end{equation}
10.5.9
such that
\begin{equation} \label{eq-CZ-bad-sets-1} [s_j, t_j)\cap [s_{j'}, t_{j'})=\emptyset \text{ for } j\neq j' \, . \end{equation}
10.5.10
For each \(j\),
\begin{equation} \label{eq-CZ-bad-sets-2} \frac{\alpha }{2}\leq \frac{1}{t_j-s_j}\int _{s_j}^{t_j} |f(y)|\, dy\leq \alpha . \end{equation}
10.5.11
Further,
\begin{equation} \label{eq-CZ-meas-b-set} \sum _{j}(t_j-s_j)\leq \frac{2}{\alpha }\int |f(y)|\, dy\, . \end{equation}
10.5.12
Proof
▶
Since \(f\) is bounded with bounded support, there exists a non-negative integer \(\ell \) such that
\begin{equation} \label{eq-f-supp} f(x)=0 \textrm{ for } x\not\in [-2^{\ell -1}, 2^{\ell -1})\, , \end{equation}
10.5.13
and
\[ 2^{-\ell }\int _{-2^{\ell -1}}^{2^{\ell -1}} |f(y)|\, dy{\lt} \alpha \, . \]
Let
\[ I_0:=[-2^{\ell }, 2^{\ell }),\qquad \mathcal{Q}_0:=\{ I_0\} \, , \]
and
\[ \Tilde {\mathcal{Q}}_1:=\textrm{bi}(I_0). \]
For \(n\geq 1\), we inductively define
\begin{equation*} {\mathcal{Q}}_n:=\left\{ I\in \Tilde {\mathcal{Q}}_n: \frac{1}{\mu (I)}\int _{I} |f(y)|\, dy{\gt} \frac{\alpha }{2}\right\} \, , \end{equation*}
and
\begin{equation*} \Tilde {\mathcal{Q}}_{n+1}:=\cup _{I\in {\Tilde {\mathcal{Q}}}_n\setminus \mathcal{Q}_n} \textrm{bi} (I)\, . \end{equation*}
Finally, let
\begin{equation*} \mathcal{Q}:=\cup _{n\geq 1} \mathcal{Q}_n\, . \end{equation*}
For each \(n\) we have
\[ \mathcal{Q}_n\subseteq \textrm{ch}_n(I_0). \]
Therefore, by Lemma 10.5.2,
\[ |\mathcal{Q}_n|\leq \frac{\mu (I_0)}{2^n}= 2^{\ell +1-n}. \]
It follows that \(\mathcal{Q}\) is a countable union of finite sets and hence, is countable itself. Let \(\{ I_j\} _{j\geq 1}\) be an enumeration of this set, with
\[ I_j:=[s_j, t_j), \]
for \(s_j, t_j\in I_0\) and \(s_j{\lt}t_j\). We set
\begin{equation*} A:=\mathbb {R}\setminus \cup _{j\geq 1} [s_j, t_j). \end{equation*}
Then 10.5.9 holds by definition. We next show 10.5.10. Let \(I_j, I_{j'}\in \mathcal{Q}\). Then, there exist \(1\leq n, n'\) such that \(I_j\in \mathcal{Q}_n\) and \(I_{j'}\in \mathcal{Q}_{n'}\). If \(n=n'\), 10.5.10 follows from Lemma 10.5.2 since \(\mathcal{Q}_n\subseteq \textrm{ch}_n(I_0).\) Otherwise, assume without loss of generality that \(n{\lt}n'\). Then by construction, there exists \(J\in \Tilde {\mathcal{Q}}_{n}\setminus {\mathcal{Q}_n}\) such that \(I_{j'} \subset J\). Since \(I_{j}\in \mathcal{Q}_n\), it follows that \(I_j\neq J\). Since \(I_j, J\in \Tilde {\mathcal{Q}}_{n}\subset \textrm{ch}_n(I_0)\), by Lemma 10.5.2, we deduce that they are disjoint. Since \(I_{j'}\subset J\), we conclude that \(I_j\) and \(I_{j'}\) are disjoint as well. This proves 10.5.10.
To see 10.5.11, let \(I_j=(s_j, t_j)\in \mathcal{Q}\). Then \(I_j\in \mathcal{Q}_n\) for some \(n\geq 1\). By definition of \(\mathcal{Q}_n\), we have
\[ \frac{\alpha }{2}{\lt}\frac{1}{\mu (I)}\int _{I} |f(y)|\, dy=\frac{1}{t_j-s_j}\int _{s_j}^{t_j} |f(y)|\, dy. \]
As \(I_j\in \mathcal{Q}_n\subseteq \Tilde {\mathcal{Q}_{n}}\), by definition of the latter set, there exists \(J\in \Tilde {\mathcal{Q}}_{n-1}\setminus \mathcal{Q}_{n-1}\) with \(I\in \textrm{bi} (J)\). Since \(J\not\in \mathcal{Q}_{n-1}\), we conclude that
\[ \frac{1}{\mu (J)}\int _J |f(y)|\, dy\leq \frac{\alpha }{2}. \]
Using Lemma 10.5.1 and the above, we get
\[ \frac{1}{t_j-s_j}\int _{s_j}^{t_j} |f(y)|\, dy=\frac{1}{\mu (I)}\int _I |f(y)|\, dy\leq \frac{2}{\mu (J)} \int _J |f(y)|\, dy\leq \alpha . \]
This establishes 10.5.11. Using this, we see that for each \(j\), we have
\[ t_j-s_j\leq \frac{2}{\alpha }\int _{s_j}^{t_j} |f(y)|\, dy. \]
Summing up in \(j\) and using the disjointedness property 10.5.10, we get
\[ \sum _{j} (t_j-s_j)\leq \frac{2}{\alpha }\int _{s_j}^{t_j} |f(y)|\, dy= \frac{2}{\alpha }\int |f(y)|\, dy. \]
Finally, we show \(\eqref{eq-CZ-good-set}\). Let \(x\in A\). If \(x\not\in [-2^{\ell }, 2^{\ell }]\), then by 10.5.13 that \(f(x)=0\). Thus 10.5.8 is true in this case. Alternately, let \(x\in I_0\cap A\). Then for each \(n\), there exists \(I(n)\in \Tilde {\mathcal{Q}}_n\) such that \(x\in I(n)\) and
\begin{equation} \label{eq-avg-In} \frac{1}{\mu (I(n))}\int _{I(n)} |f(y)| dy\leq {\alpha }. \end{equation}
10.5.14
Since \(\mu (I(n))=2^{\ell +1-n}\), we have
\[ \lim _{n\to \infty } \mu (I(n))=0. \]
By Lemma 10.5.3, we also have
\begin{equation*} \lim _{n\to \infty }\frac{1}{\mu (I(n))}\int _{I(n)} |f(y)| dy= |f(x)| \end{equation*}
for almost every \(x\in A\). Combining the above with 10.5.14, we conclude that
\[ |f(x)|\leq \alpha . \]
This finishes the proof of 10.5.8, and hence the lemma.
Calderón-Zygmund decomposition is a tool to extend \(L^2\) bounds to \(L^p\) bounds with \(p{\lt}2\) or to the so-called weak \((1, 1)\) type endpoint bound. It is classical and can be found in
[
.
Lemma
10.5.5
Calderon Zygmund decomposition
Let \(f\) be a bounded, measurable function with bounded support. Let \(\alpha {\gt}0\) and \(\gamma \in (0, 1)\). Then there exists a measurable functions \(g\), a countable family of disjoint intervals \(I_j = [s_j, t_j)\), and a countable family of measurable functions \(\{ g_j\} _{j\geq 1}\) such that for almost every \(x \in {\mathbb {R}}\)
\begin{equation} \label{eq-gb-dec} f(x)= g(x)+ \sum _{j\geq 1} b_j(x) \end{equation}
10.5.15
and such that the following holds. For almost every \(x\in \mathbb {R}\),
\begin{equation} \label{eq-g-max} |g(x)|\leq \gamma \alpha \, . \end{equation}
10.5.16
We have
\begin{equation} \label{eq-g-L1-norm} \int |g(y)|\, dy\leq \int |f(y)|\, dy. \end{equation}
10.5.17
For every \(j\)
\begin{equation} \label{eq-supp-bj} \operatorname {supp} b_j \subset I_j\, . \end{equation}
10.5.18
For every \(j\)
\begin{equation} \label{eq-bad-mean-zero} \int _{I_j} b_j(x)\, dx=0, \end{equation}
10.5.19
and
\begin{equation} \label{eq-bj-L1} \int _{I_j} |b_j(x)|\, dx \leq 2\gamma \alpha . \end{equation}
10.5.20
We have
\begin{equation} \label{eq-bset-length-sum} \sum _j (t_j-s_j)\leq \frac{2}{\gamma \alpha }\int |f(y)|\, dy \end{equation}
10.5.21
and
\begin{equation} \label{eq-b-L1} \sum _{j}\int _{I_j} |b_j(y)|\, dy\leq 2 \int |f(y)|\, dy\, . \end{equation}
10.5.22
Proof
▶
Applying Lemma 10.5.4 to \(f\) and \(\gamma \alpha \), we obtain a collection \(I_j\) of intervals such that the conditions 10.5.8-10.5.12 are satisfied. We set \(A = {\mathbb {R}}\setminus \cup _j I_j\) and
\begin{equation} \label{eq-g-def} g(x):=\begin{cases} f(x), & x\in A,\\ \frac{1}{\mu (I_j)}\int _{I_j} f(y)\, dy, & x\in (s_j, t_j),\\ 0, & x\in [s_j, t_j]\setminus (s_j, t_j), \end{cases} \end{equation}
10.5.23
and, for each \(j\),
\begin{equation} b_j(x):=\begin{cases} f(x)-\frac{1}{\mu (I_j)}\int _{I_j} f(y)\, dy, & x\in (s_j, t_j),\\ 0, & x\not\in (s_j, t_j). \end{cases} \end{equation}
10.5.27
Then 10.5.18 and 10.5.21 are true by construction and Lemma 10.5.4. Further, let \(b(x)=\sum _{j} b_j(x).\) Then
\begin{equation*} f(x)=g(x)+b(x)=g(x)+\sum _{j} b_j(x) , \end{equation*}
for all \(x\) not in the measure zero set \(\cup _j \{ s_j, t_j\} .\)
For almost every \(x\in A\), we get using 10.5.8
\[ |g(x)|\leq \gamma \alpha \, . \]
In the case when neither of the above is true, we have \(g(x)=0\) by definition. Thus, we obtain 10.5.16.
To prove 10.5.17, we estimate
\[ \int |g(y)|\, dy\leq \int _A |f(y)|\, dy+ \sum _{j} \int _{I_j}\frac{1}{\mu (I_j)}\int _{I_j}|f(y)|\, dy. \]
Since the intervals \(I_j\) are disjoint, and \(A=\mathbb {R}\setminus \cup _j I_j\), we conclude
\[ \int |g(y)|\, dy\leq \int _A |f(y)|\, dy+\sum _{j} \int _{I_j}|f(y)|\, dy= \int |f(y)|\, dy. \]
This establishes 10.5.17. If \(x\in I_j\) for some \(j\), 10.5.11 yields
\[ |g(x)|\leq \frac{1}{\mu (I_j)}\int _{I_j}|f(y)|\, dy\leq \gamma \alpha . \]
Further, for each \(j\), it follows from the definition of \(t_j\) that
\[ \int _{I_j} b_j(x)\, dx= \int _{I_j} f(x)\, dx-\int _{I_j} \frac{1}{\mu (I_j)}\int _{I_j}f(y)\, dy\, dx \]
\begin{equation*} =\int _{I_j} f(x)\, dx- \int _{I_j}f(y)\, dy=0. \end{equation*}
This establishes 10.5.19.
Using the triangle inequality, we have that
\begin{equation*} \int _{I_j} |b_j(y)|\, dy\leq \int _{I_j} |f(y)|\, dy + \int _{I_j} \frac{1}{\mu (I_j)}\int _{I_j} |f(x)|\, dx\, dy. \end{equation*}
\begin{equation} \label{eq-bj-int} =2 \int _{I_j} |f(y)|\, dy. \end{equation}
10.5.30
Dividing both sides by \(\mu (I_j)\) and using 10.5.11, we obtain 10.5.20.
Further, summing up 10.5.30 in \(j\) yields
\begin{equation*} \sum _{j}\int _{I_j} |b_j(y)|\, dy\leq \sum _{j}\int _{I_j} |f(y)|\, dy + \sum _{j}\int _{I_j} \frac{1}{\mu (I_j)}\int _{I_j} |f(x)|\, dx\, dy. \end{equation*}
Using the disjointedness of \((s_j, t_j)\), we get
\begin{equation*} \sum _{j}\int _{I_j} |b_j(y)|\, dy\leq 2 \int |f(y)|\, dy. \end{equation*}
This proves 10.5.22, and completes the proof.
Using Lemma 10.5.5 for \(f\) and \(2^{-10}\alpha \), we obtain the decomposition
\[ f=g+b=g+\sum _j b_j \]
such that the properties 10.5.15-10.5.22 are satisfied with \(\gamma =2^{-10}\). For each \(j\), let
\begin{equation} \label{eq-Ij-cj} I_j=[s_j, t_j), \qquad c_j:=\frac{s_j+t_j}{2}, \end{equation}
10.5.31
and
\begin{equation} \label{eq-Ij*} I_j{^*}=\left[s_j-{2(t_j-s_j)}, t_j+2(t_j-s_j)\right]. \end{equation}
10.5.32
Then \(I_j^*\) is an interval with the same center as \(I_j\) but with
\begin{equation} \label{eq-Ij*-dim} \mu (I_j^*)=5(t_j-s_j)=5\mu (I_j). \end{equation}
10.5.33
Let
\begin{equation} \label{eq-omega} \Omega :=\cup _j I_{j}^*. \end{equation}
10.5.34
By definition, for each \(x\in \mathbb {R}\setminus \Omega \) and \(y\in I_j\),
\begin{equation} \label{eq-Om-cj} |x-c_j|{\gt}\frac{5(t_j-s_j)}{2}\geq 5|y-c_j|, \end{equation}
10.5.35
and
\begin{equation} \label{eq-Om-y} |x-y|{\gt}{2(t_j-s_j)}. \end{equation}
10.5.36
It follows by the triangle inequality and subadditivity of \(\mu \) that
\[ \mu \left(\{ x\in \mathbb {R}: |H_r f(x)|{\gt}\alpha \} \right) \]
\begin{equation} \label{eq-set-dec-1} \le \mu \left(\{ x\in \mathbb {R}: |H_r g(x)|{\gt}{\alpha }/2\} \right)+ \mu \left(\{ x\in \mathbb {R}: |H_r b(x)|{\gt}{\alpha }/2\} \right). \end{equation}
10.5.37
We estimate using monotonicity of the integral
\[ \mu \left(\{ x\in \mathbb {R}: |H_r g(x)|{\gt}{\alpha }/2\} \right)\leq \frac{4}{\alpha ^2} \int |H_r g(y)|^2\, dy. \]
Using Lemma 10.1.6 followed by 10.5.16 and 10.5.17, we estimate the right hand side above by
\[ \leq 2^{26}\frac{4}{\alpha ^2} \int |g(y)|^2\, dy\leq \frac{2^{18}}{\alpha } \int |g(y)|\, dy\leq \frac{2^{18}}{\alpha } \int |f(y)|\, dy. \]
Thus, we conclude
\begin{equation} \label{eq-g-func-bd} \mu \left(\{ x\in \mathbb {R}: |H_r g(x)|{\gt}{\alpha }/2\} \right)\leq \frac{2^{18}}{\alpha } \int |f(y)|\, dy. \end{equation}
10.5.38
Next, we estimate
\[ \mu \left(\{ x\in \mathbb {R}: |H_r b(x)|{\gt}\alpha /2\} \right) \]
\[ \le \mu (\Omega ) + \mu \left(\{ x\in \mathbb {R}\setminus \Omega : |H_r b(x)|{\gt}{\alpha }/2\} \right)\, . \]
Using 10.5.33 and 10.5.21, we conclude that
\begin{equation} \label{eq-omega-bd} \mu (\Omega ) \le \sum _{j} \mu (I_j^*) = 5 \sum _{j} (t_j-s_j)\leq \frac{2^{13}}{\alpha } \int |f(y)|\, dy\, . \end{equation}
10.5.39
We now focus on estimating the remaining term
\[ \mu \left(\{ x\in \mathbb {R}\setminus \Omega : |H_r b(x)|{\gt}{\alpha }/2\} \right)\, . \]
For \(x\in \mathbb {R}\setminus \Omega \), define
\begin{align*} \mathcal{J}_1(x)& :=\{ j\, :[s_j, t_j)\cap [x-r, x+r]=\emptyset \} ,\\ \mathcal{J}_2(x)& :=\{ j\, : |y-x|=r \text{ for some } y \in [s_j, t_j)\} ,\\ \mathcal{J}_3(x)& :=\{ j\, : [s_j, t_j)\subset [x-r, x+r]\} . \end{align*}
Since \(H_rb_j(x)=0\) for all \(j\in \mathcal{J}_3(x)\), using the triangle inequality and the decomposition above, we get
\begin{equation} \label{eq-b-dec} |H_r b(x)|\leq \sum _{j\in \mathcal{J}_1(x)} |H_rb_j(x)|+\sum _{j\in \mathcal{J}_2(x)} |H_rb_j(x)|. \end{equation}
10.5.40
Further, for \(j\in \mathcal{J}_1(x)\), we have
\[ H_rb_j(x)=\int _{I_j} \kappa (x-y)\mathbf{1}_{\{ z:\, r{\lt}|z|{\lt}1\} }(x-y) b_j(y)\, dy=\int _{I_j} \kappa (x-y) b_j(y)\, dy\, . \]
Using 10.5.19, the above is equal to
\[ \int _{I_j} (\kappa (x-y)-\kappa (x-c_j)) b_j(y). \]
Thus, using the triangle inequality, 10.5.35, 10.5.36 and Lemma 10.1.14, we can estimate
\[ \sum _{j\in \mathcal{J}_1(x)} |H_rb_j(x)|\leq \sum _{j\in \mathcal{J}_1(x)}2^{10}\int _{I_j}\frac{|y-c_j|}{|x-c_j|^2} |b_j(y)|\, dy \]
\begin{equation} \label{eq-J1-diff-est} \leq 2^{10}\sum _{j} \frac{(t_j-s_j)}{|x-c_j|^2}\int _{I_j} |b_j(y)|\, dy:=F_1(x). \end{equation}
10.5.41
Next, we estimate the second sum in 10.5.40. For each \(j\in \mathcal{J}_2(x)\), set
\[ d_j:=\frac{1}{t_j-s_j}\int _{I_j} \mathbf{1}_{\{ z:\, r{\lt}|z|{\lt}1\} }(x-y) b_j(y)\, dy. \]
Then by 10.5.20
\begin{equation} \label{eq-dj-est} |d_j|\leq 2\cdot 2^{-10} \alpha =2^{-9}\alpha . \end{equation}
10.5.42
For each \(j\in \mathcal{J}_2(x)\), we have
\[ H_r b_j(x)=\int _{I_j} \kappa (x-y) (\mathbf{1}_{\{ z:\, r{\lt}|z|{\lt}1\} }(x-y)b_j(y)-d_j)\, dy \]
\[ + \int _{I_j} d_j \kappa (x-y) \, dy. \]
\[ =\int _{I_j} (\kappa (x-y)-\kappa (x-c_j)) (\mathbf{1}_{\{ z:\, r{\lt}|z|{\lt}1\} }(x-y)b_j(y)-d_j)\, dy \]
\[ + \int _{I_j} d_j \kappa (x-y) \, dy. \]
Thus, using the triangle inequality, the estimate above and 10.5.42, we obtain
\[ |H_r b_j(x)|\leq \]
\begin{equation} \label{eq-J2-diff-est} \int _{I_j} |(\kappa (x-y)-\kappa (x-c_j)) \left(|b_j(y)|+2^{-9}\alpha \right)\, dy +2^{-9}\alpha \int _{I_j} |\kappa (x-y)| \, dy. \end{equation}
10.5.43
Using 10.5.35, 10.5.36 and Lemma 10.1.14, and arguing as in 10.5.41, we get the the first term above can be estimated by
\begin{equation} \label{eq-J2-diff-est-2} 2^{10}\sum _{j} \frac{(t_j-s_j)}{|x-c_j|^2}\left(\int _{I_j} |b_j(y)|+2^{-9} \alpha (t_j-s_j)\right)=F_1(x)+F_2(x), \end{equation}
10.5.44
with \(F_1\) as in 10.5.41 and
\begin{equation} \label{eq-def-F2} F_2(x):= 2\sum _{j} \frac{(t_j-s_j)}{|x-c_j|^2} \alpha (t_j-s_j). \end{equation}
10.5.45
For each \(j\in \mathcal{J}_2(x)\), let \(y_j\in [s_j, t_j)\) be such that
\[ |x-y_j|=r. \]
Using 10.5.36, we have
\[ r=|x-y_j|{\gt}2(t_j-s_j). \]
Further, using the triangle inequality, for each \(y\in I_j\), we obtain
\[ |x-y|\leq |x-y_j|+|y-y_j|\leq r+(t_j-s_j){\lt}\frac{3r}{2}, \]
and
\[ |x-y|\geq |x-y_j|-|y-y_j|\geq r-(t_j-s_j){\gt}\frac{r}{2}. \]
We conclude that
\[ [s_j, t_j)\subset \left\{ y\, : \frac{r}{2}{\lt}|x-y|{\lt}\frac{3r}{2}\right\} . \]
Using this and Lemma 10.1.13, we get
\begin{equation} \label{eq-J2-diff-est-4} \int _{I_j}|\kappa (x-y)|\, dy \leq \int _{\{ y\, : \frac{r}{2}{\lt}|x-y|{\lt}\frac{3r}{2}\} }\frac{8}{|x-y|}\, dy\leq 32\, . \end{equation}
10.5.46
Combining 10.5.40, 10.5.41, 10.5.43, 10.5.44 and 10.5.46, we get
\begin{equation*} |H_rb(x)|\leq 2F_1(x)+F_2(x)+2^{-4}\alpha . \end{equation*}
Using the triangle inequality, we deduce that
\[ \mu ({\{ x\in \Omega : |H_r b(x)|{\gt}\alpha /4\} })\leq \]
\begin{equation} \label{eq-set-dec-3} \mu (\{ x\in \Omega : |F_1(x)|{\gt} 2^{-4} \alpha \} )+\mu (\{ x\in \Omega : | F_2(x)|{\gt} 2^{-4}\alpha \} ). \end{equation}
10.5.47
We estimate
\begin{equation} \label{eq-F1-est-1} \mu (\{ x\in \Omega : |F_1(x)|{\gt} 2^{-4}\alpha \} )\leq \frac{2^{14}}{\alpha }\sum _{j} \int _{I_j} |b_j(y)|\, dy\int _{\Omega } \frac{(t_j-s_j)}{|x-c_j|^2}\, dx. \end{equation}
10.5.48
Using 10.5.35, we can bound
\begin{equation} \label{eq-Om-int} \int _{\Omega } \frac{(t_j-s_j)}{|x-c_j|^2}\, dx \le 2(t_j - s_j) \int _{5|t_j - s_j|}^\infty \frac{1}{t^2} \, \mathrm{d}t = \frac{2}{5}\, . \end{equation}
10.5.49
Plugging 10.5.49 into 10.5.48 and using 10.5.22, we conclude that
\begin{equation} \label{eq-F1-est-2} \mu (\{ x\in \Omega : |F_1(x)|{\gt}2^{-4}\alpha \} )\leq \frac{2^{14}}{\alpha }\int |f(y)|\, dy\, . \end{equation}
10.5.50
Similarly, we estimate
\[ \mu (\{ x\in \Omega : |F_2(x)|{\gt}2^{-4} \alpha \} )\leq \frac{2^{5}}{\alpha }\sum _{j} \alpha (t_j-s_j)\, \int _{\Omega } \frac{(t_j-s_j)}{|x-c_j|^2}\, dx. \]
Using 10.5.49 and 10.5.21, this is bounded by
\begin{equation} \label{eq-F2-est} 2^{4}\sum _{j} (t_j-s_j)\leq \frac{2^{15}}{\alpha }\int |f(y)|\, dy. \end{equation}
10.5.51
Combining estimates 10.5.37, 10.5.38, 10.5.39, 10.5.47, 10.5.50 and 10.5.51 yields 10.1.20.
10.6 The proof of the van der Corput Lemma
Let \(g\) be a Lipschitz continuous function as in the lemma. We have
\[ e^{in(x + \pi /n)} = -e^{inx}\, . \]
Using this, we write
\[ \int _\alpha ^\beta g(x) e^{-inx} \, \mathrm{d}x = \frac{1}{2} \int _\alpha ^\beta g(x) e^{inx} \, \mathrm{d}x - \frac{1}{2} \int _\alpha ^\beta g(x) e^{in(x + \pi /n)}) \, \mathrm{d}x\, . \]
We split the the first integral at \(\alpha + \frac{\pi }{n}\) and the second one at \(\beta - \frac{\pi }{n}\), and make a change of variables in the second part of the first integral to obtain
\[ = \frac{1}{2} \int _{\alpha }^{\alpha + \frac{\pi }{n}} g(x) e^{inx} \, \mathrm{d}x - \frac{1}{2} \int _{\beta - \frac{\pi }{n}}^{\beta } g(x) e^{in(x + \pi /n)} \, \mathrm{d}x \]
\[ + \frac{1}{2} \int _{\alpha + \frac{\pi }{n}}^{\beta } (g(x) - g(x - \frac{\pi }{n})) e^{inx} \, \mathrm{d}x\, . \]
The sum of the first two terms is by the triangle inequality bounded by
\[ \frac{\pi }{n} \sup _{x \in [\alpha ,\beta ]} |g(x)|\, . \]
The third term is by the triangle inequality at most
\[ \frac{1}{2} \int _{\alpha + \frac{\pi }{n}}^\beta |g(x) - g(x - \frac{\pi }{n})| \, \mathrm{d}x \]
\[ \le \frac{\pi }{2n} \sup _{\alpha \le x {\lt} y \le \beta } \frac{|g(x) - g(y)|}{|x-y|} |\beta -\alpha |\, . \]
Adding the two terms, we obtain
\[ \left|\int _\alpha ^\beta g(x) e^{-inx} \, \mathrm{d}x\right| \le \frac{\pi }{n} \| g\| _{\mathrm{Lip}(\alpha ,\beta )}\, . \]
By the triangle inequality, we also have
\[ \left|\int _\alpha ^\beta g(x) e^{-inx} \, \mathrm{d}x\right| \le |\beta -\alpha | \sup _{x \in [\alpha ,\beta ]} |g(x)| \le |\beta -\alpha | \| g\| _{\mathrm{Lip}(\alpha ,\beta )}\, . \]
This completes the proof of the lemma, using that
\[ \min \{ |\beta -\alpha |, \frac{\pi }{n}\} \le 2 \pi |\beta -\alpha |(1 + n|\beta -\alpha |)^{-1}\, . \]
10.7 Partial sums as orthogonal projections
This subsection proves Lemma 10.1.12
Lemma
10.7.1
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}
10.7.1
Proof
▶
Let \(N{\gt}0\) be given. With \(K_N\) as in Lemma 10.1.10,
\begin{equation*} S_N (S_Nf) (x)= \frac{1}{2\pi } \int _0^{2\pi } S_Nf(y)K_N(x-y)\, dy \end{equation*}
\begin{equation} \label{eqhil1} = \frac{1}{(2\pi )^2}\int _0^{2\pi } \int _0^{2\pi } f(y')K_N(y-y') K_N(x-y)\, \, dy' dy\, . \end{equation}
10.7.2
We have by Lemma 10.1.10
\begin{equation*} \frac{1}{2\pi }\int _0^{2\pi } K_N(y-y’) K_N(x-y)\, dy \end{equation*}
\begin{equation*} =\frac{1}{2\pi }\sum _{n=-N}^N\sum _{n'=-N}^N \int _0^{2\pi } e^{in(y-y')}e^{in'(x-y)}\, dy \end{equation*}
\begin{equation} \label{eqhil6} =\frac{1}{2\pi }\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}
10.7.3
By Lemma 10.1.9, the summands for \(n\neq n'\) vanish. We obtain for 10.7.3
\begin{equation} \label{eqhil2} =\frac{1}{2\pi }\sum _{n=-N}^N e^{in(x-y')}\int _0^{2\pi } \, dy=K_N(x-y')\, . \end{equation}
10.7.4
Applying Fubini in 10.7.2 and using 10.7.4 gives
\begin{equation} S_N(S_Nf)(x)= \frac{1}{2\pi } \int _0^{2\pi } f(y')K(x-y') \, dy'=S_N f(x) \end{equation}
10.7.5
This proves the lemma.
Lemma
10.7.2
partial sum selfadjoint
We have for any \(2\pi \)-periodic bounded measurable \(g,f\) that
\begin{equation} \int _0^{2\pi } \overline{S_Nf(x)} g(x)=\int _0^{2\pi } \overline{f(x)} S_Ng(x)\, dx\, . \end{equation}
10.7.6
Proof
▶
We have with \(K_N\) as in Lemma 10.1.10 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}
10.7.7
Further, with Lemma 10.1.10 and Fubini
\begin{equation*} \int _0^{2\pi } \overline{S_Nf(x)} g(x) = \frac1{2\pi } \int _0^{2\pi } \int _{-\pi }^{\pi }\overline{f(y) K_N(x-y)} g(x)\, dy dx \end{equation*}
\begin{equation} = \frac1{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 \, . \end{equation}
10.7.8
This proves the lemma.
We turn to the proof of Lemma 10.1.12.
We have with Lemma 10.7.2, then Lemma 10.7.1 and the Lemma 10.7.2 again
\begin{equation*} \int _0^{2\pi } S_Nf(x)\overline{S_Nf(x)}\, dx \int _0^{2\pi } f(x)\overline{S_N(S_Nf)(x)}\, dx \end{equation*}
\begin{equation} \label{eqhil7} =\int _0^{2\pi } f(x)\overline{S_Nf(x)}\, dx= \int _0^{2\pi } S_N f(x)\overline{f(x)}\, dx\, . \end{equation}
10.7.9
We have by the distributive law
\begin{equation} \label{diffnorm} \int _0^{2\pi } (f(x)-S_Nf(x))(\overline{f(x)-S_Nf(x)})\, dx= \end{equation}
10.7.10
\begin{equation*} \int _0^{2\pi } f(x)\overline{f(x)} -S_Nf(x)\overline{f(x)} -f(x)\overline{S_Nf(x)} + S_Nf(x)\overline{S_Nf(x)}\, dx \end{equation*}
Using the various identities expressed in 10.7.9, this becomes
\begin{equation} =\int _0^{2\pi } f(x)\overline{f(x)}\, dx - \int _0^{2\pi } S_Nf(x)\overline{S_Nf(x)}\, dx\, . \end{equation}
10.7.11
As 10.7.10 has nonnegative integrand and is thus nonnegative, we conclude
\begin{equation} \int _0^{2\pi } S_Nf(x)\overline{S_Nf(x)}\, dx\le \int _0^{2\pi } f(x)\overline{f(x)})\, dx\, . \end{equation}
10.7.12
As both sides are positive, we may take the square root of this inequality. This completes the proof of the lemma.
10.8 The error bound
Define
\[ E := \{ x \in [0, 2\pi ) \ : \ \sup _{N {\gt} 0} |S_N f(x) - S_N f_0(x)| {\gt} \frac{\epsilon }{4} \} \, . \]
Then 10.1.8 clearly holds, and it remains to show that \(|E| \le \frac{\epsilon }{2}\). This will follow from Lemma 10.1.4.
Let \(x \in E\). Then there exists \(N {\gt} 0\) with
\[ |S_N f(x) - S_N f_0(x)| {\gt} \frac{\epsilon }{4}\, , \]
pick such \(N\). We have with Lemma 10.1.10
\[ \frac{\epsilon }{4} {\lt} |S_N f(x) - S_N f_0(x)| = \frac{1}{2\pi } \left| \int _0^{2\pi } (f(y) - f_0(y)) K_N(x-y) \, dy\right|\, . \]
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 10.1.25 bounded by \(2N+1\). Using 10.1.26 and the definition 10.1.12 of \(\kappa \), we rewrite the two terms and obtain
\begin{equation} \label{eq-diff-singular} \frac{\epsilon }{4} {\lt} \frac{1}{2\pi } \left| \int _{-\pi }^{\pi } (f(x-y) - f_0(x-y)) (e^{iNy} \overline{\kappa }(y) + e^{-iNy}\kappa (y)) \, dy\right| \end{equation}
10.8.1
\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}
10.8.2
By Lemma 10.1.11 with \(\eta = |y|\), we have for \(-1 \le y \le 1\)
\[ |e^{iNy}\frac{\min \{ |y|, 1\} }{1 - e^{-iy}}| = \frac{|y|}{|1 - e^{iy}|} \le \frac\pi 2 \, . \]
By Lemma 10.1.11 with \(\eta = 1\), we have for \(1 \le |y| \le \pi \)
\[ |e^{iNy}\frac{\min \{ |y|, 1\} }{1 - e^{-iy}}| = \frac{1}{|1 - e^{iy}|} \le \frac\pi 2 \, . \]
Thus we obtain using the triangle inequality and 10.1.5
\[ \eqref{eq-diff-integrable} \le \frac{1}{2} \int _{-\pi }^{\pi } |f(x-y) - f_0(x-y)| \, dy \le \pi \epsilon '. \]
Consequently, we have that
\[ \frac{\epsilon }{4} - \pi \epsilon ' \le \frac{1}{2\pi } \left| \int _{-\pi }^{\pi } (f(x-y) - f_0(x-y)) (e^{iNy} \overline{\kappa }(y) + e^{-iNy}\kappa (y)) \, dy\right|\, . \]
By dominated convergence and since \(\kappa (y) = 0\) for \(|y| {\gt} 1\), this equals
\[ = \frac{1}{2\pi } \lim _{r \to 0^+} \left| \int _{r {\lt} |y| {\lt} 1} (f(x-y) - f_0(x-y)) (e^{iNy} \overline{\kappa }(y) + e^{-iNy}\kappa (y)) \, dy\right|\, . \]
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
\[ \le \frac{1}{2\pi } \sup _{r {\gt} 0} \left| \int _{r {\lt} |y| {\lt} 1} h(x-y) e^{-iNy} \kappa (y) \, dy\right| \]
\[ + \frac{1}{2\pi } \sup _{r {\gt} 0} \left| \int _{r {\lt} |y| {\lt} 1} \overline{h}(x-y) e^{-iNy} \kappa (y) \, dy\right|\, . \]
By the definition 10.1.15 of \(T\), this is
\[ \le \frac{1}{2\pi } (Th(x) + T\bar{h}(x))\, . \]
Thus for each \(x \in E\), at least one of \(Th(x)\) and \(T\bar h(x)\) is larger than \(\frac{\pi (\epsilon - 4 \pi \epsilon ')}{4}\). Thus at least one of \(Th\) and \(T\bar h\) is \(\ge \frac{\pi (\epsilon - 4 \pi \epsilon ')}{4}\) on a subset \(E'\) of \(E\) with \(2|E'| \ge |E|\). Without loss of generality this is \(Th\). By assumption 10.1.5, we have \(|2^{2^{50}}\epsilon ^{-2} h| \le \mathbf{1}_{[-\pi , 3\pi ]}\). Applying Lemma 10.1.4 with \(F = [-\pi , 3\pi ]\) and \(G = E'\), it follows that
\[ \frac{\pi (\epsilon - 4 \pi \epsilon ')}{4} |E'| \le \int _{E'} Th(x) \, dx = \epsilon ' \int _{E'} T(\epsilon '^{-1} h)(x) \, dx \]
\[ \le \epsilon ' \cdot C_{4,2} |F|^{\frac{1}{2}} |E'|^{\frac{1}{2}} \le (4\pi )^\frac {1}{2} C_{4,2} \epsilon ' \cdot |E'|^{\frac{1}{2}}\, . \]
Rearranging, we obtain
\[ |E'| \le \left(\frac{4 (4\pi )^\frac {1}{2} C_{4,2} \epsilon '}{\pi (\epsilon - 4 \pi \epsilon ')}\right)^2 = \frac{\epsilon }{2}\, . \]
This completes the proof using \(|E| \le 2|E'|\).
10.9 Carleson on the real line
We prove Lemma 10.1.4.
Consider the standard distance function
\begin{equation} \rho (x,y)=|x-y| \end{equation}
10.9.1
on the real line \({\mathbb {R}}\).
Lemma
10.9.1
real line metric
The space \(({\mathbb {R}},\rho )\) is a complete locally compact metric space.
Proof
▶
This is part of the Lean library.
Lemma
10.9.2
real line ball
For \(x\in R\) and \(R{\gt}0\), the ball \(B(x,R)\) is the interval \((x-R,x+R)\)
Proof
▶
Let \(x'\in B(x,R)\). By definition of the ball, \(|x'-x|{\lt}R\). It follows that \(x'-x{\lt}R\) and \(x-x'{\lt}R\). It follows \(x'{\lt}x+R\) and \(x'{\gt}x-R\). This implies \(x'\in (x-R,x+R)\). Conversely, let \(x'\in (x-R,x+R)\). Then \(x'{\lt}x+R\) and \(x'{\gt}x-R\). It follows that \(x'-x{\lt}R\) and \(x-x'{\lt}R\). It follows that \(|x'-x|{\lt}R\), hence \(x'\in B(x,R)\). This proves the lemma.
We consider the Lebesgue measure \(\mu \) on \({\mathbb {R}}\).
Lemma
10.9.3
real line measure
The measure \(\mu \) is a sigma-finite non-zero Radon-Borel measure on \({\mathbb {R}}\).
Proof
▶
This is part of the Lean library.
Lemma
10.9.4
real line ball measure
We have for every \(x\in {\mathbb {R}}\) and \(R{\gt}0\)
\begin{equation} \mu (B(x,R))=2R\, . \end{equation}
10.9.2
Proof
▶
We have with Lemma 10.9.2
\begin{equation} \mu (B(x,R))=\mu ((x-R,x+R))=2R\, . \end{equation}
10.9.3
Lemma
10.9.5
real line doubling
We have for every \(x\in {\mathbb {R}}\) and \(R{\gt}0\)
\begin{equation} \mu (B(x,2R))=2\mu (B(x,R))\, . \end{equation}
10.9.4
Proof
▶
We have with Lemma 10.9.4
\begin{equation} \mu (B(x,2R)=4R=2\mu (B(x,R)\, . \end{equation}
10.9.5
This proves the lemma.
The preceding four lemmas show that \(({\mathbb {R}}, \rho , \mu , 4)\) is a doubling metric measure space. Indeed, we even show that \(({\mathbb {R}}, \rho , \mu , 1)\) is a doubling metric measure space, but we may relax the estimate in Lemma 10.9.5 to conclude that \(({\mathbb {R}}, \rho , \mu , 4)\) is a doubling metric measure space.
For each \(n\in \mathbb {Z}\) define \({\vartheta }_n:{\mathbb {R}}\to {\mathbb {R}}\) by
\begin{equation} {\vartheta }_n(x)=nx\, . \end{equation}
10.9.6
Let \({\Theta }\) be the collection \(\{ {\vartheta }_n, n\in \mathbb {Z}\} \). Note that for every \(n\in \mathbb {Z}\) we have \({\vartheta }_n(0)=0\). Define
\begin{equation} \label{eqcarl4} d_{B(x,R)}({\vartheta }_n, {\vartheta }_m) := 2R|n-m|\, . \end{equation}
10.9.7
Lemma
10.9.6
frequency metric
For every \(R {\gt} 0\) and \(x \in X\), the function \(d_{B(x,R)}\) is a metric on \({\Theta }\).
Proof
▶
This follows immediately from the fact that the standard metric on \(\mathbb {Z}\) is a metric.
Lemma
10.9.7
oscillation control
For every \(R {\gt} 0\) and \(x \in X\), and for all \(n, m \in \mathbb {Z}\), we have
\begin{equation} \label{eqcarl2} \sup _{y,y'\in B(x,R)}|ny-ny'-my+my'|\le 2|n-m|R\, . \end{equation}
10.9.8
Proof
▶
The right hand side of 10.9.8 equals
\[ \sup _{y,y'\in B(x,R)}|(n-m)(y-x)-(n-m)(y'-x)|\, . \]
The lemma then follows from the triangle inequality.
Lemma
10.9.8
frequency monotone
For any \(x, x' \in X\) and \(R, R' {\gt} 0\) with \(B(x,R) \subset B(x, R')\), and for any \(n, m \in \mathbb {Z}\)
\[ d_{B(x,R)}({\vartheta }_n, {\vartheta }_m) \le d_{B(x',R')}({\vartheta }_n, {\vartheta }_m)\, . \]
Proof
▶
This follows immediately from the definition 10.9.7 and \(R \le R'\).
Lemma
10.9.9
frequency ball doubling
For any \(x,x'\in {\mathbb {R}}\) and \(R{\gt}0\) with \(x\in B(x',2R)\) and any \(n,m\in \mathbb {Z}\), we have
\begin{equation} \label{firstdb1} d_{B(x',2R)}({\vartheta }_n,{\vartheta }_m)\le 2 d_{B(x,R)}({\vartheta }_n,{\vartheta }_m) \, . \end{equation}
10.9.9
Proof
▶
With 10.9.7, both sides of 10.9.9 are equal to \(4R|n-m|\). This proves the lemma.
Lemma
10.9.10
frequency ball growth
For any \(x,x'\in {\mathbb {R}}\) and \(R{\gt}0\) 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)}({\vartheta }_n,{\vartheta }_m)\le d_{B(x',2R)}({\vartheta }_n,{\vartheta }_m) \, . \end{equation}
10.9.10
Proof
▶
With 10.9.7, both sides of 10.9.9 are equal to \(4R|n-m|\). This proves the lemma.
Lemma
10.9.11
integer ball cover
For every \(x\in {\mathbb {R}}\) and \(R{\gt}0\) and every \(n\in \mathbb {Z}\) and \(R'{\gt}0\), there exist \(m_1, m_2, m_3\in \mathbb {Z}\) such that
\begin{equation} \label{eqcarl5} B'\subset B_1\cup B_2\cup B_3\, , \end{equation}
10.9.11
where
\begin{equation} B'= \{ {\vartheta }\in {\Theta }: d_{B(x,R)}({\vartheta }, {\vartheta }_n){\lt}2R'\} \end{equation}
10.9.12
and for \(j=1,2,3\)
\begin{equation} B_j= \{ {\vartheta }\in {\Theta }: d_{B(x,R)}({\vartheta }, {\vartheta }_{m_j}){\lt}R'\} \, . \end{equation}
10.9.13
Proof
▶
Let \(m_1\) be the largest integer smaller than or equal to \(n- \frac{R'}{2R}\). Let \(m_2=n\). Let \(m_3\) be the smallest integer larger than or equal to \(n+ \frac{R'}{2R}\).
Let \({\vartheta }_{n'}\in B'\), then with 10.9.7, we have
\begin{equation} \label{eqcarl6} 2R|n-n'|{\lt} 2R'\, . \end{equation}
10.9.14
Assume first \(n'\le m_1\). With 10.9.14 we have
\begin{equation*} R|m_1-n’|=R(m_1-n’)=R(m_1-n)+R(n-n’) \end{equation*}
\begin{equation} {\lt} -\frac{R'}2+R'=-\frac{R'}2\, . \end{equation}
10.9.15
We conclude \({\vartheta }_{n'}\in B_1\).
Assume next \(m_1{\lt}n'{\lt}m_3\). Then \({\vartheta }_{n'}\in B_2\).
Assume finally that \(m_3\le n'\). With 10.9.14 we have
\begin{equation*} R|m_3-n’|=R(n’-m_3)=R(n’-n)+R(n-m_3) \end{equation*}
\begin{equation} {\lt} R' -\frac{R'}2=-\frac{R'}2\, . \end{equation}
10.9.16
We conclude \({\vartheta }_{n'}\in B_1\). This completes the proof of the lemma.
Lemma
10.9.12
real van der Corput
For any \(x\in {\mathbb {R}}\) and \(R{\gt}0\) and any function \(\varphi : X\to {\mathbb {C}}\) supported on \(B'=B(x,R)\) such that
\begin{equation} \| \varphi \| _{\operatorname{\operatorname {Lip}}(B')} = \sup _{x \in B'} |\varphi (x)| + R \sup _{x,y \in B', x \neq y} \frac{|\varphi (x) - \varphi (y)|}{\rho (x,y)} \end{equation}
10.9.17
is finite and for any \(n,m\in \mathbb {Z}\), we have
\begin{equation} \label{eq-vdc-cond1} \left|\int _{B'} e({\vartheta }_n(x)-{{\vartheta }_m(x)}) \varphi (x) d\mu (x)\right|\le 2\pi \mu (B')\frac{\| \varphi \| _{\operatorname{\operatorname {Lip}}(B')}}{1+d_{B'}({\vartheta }_n,{\vartheta }_m)} \, . \end{equation}
10.9.18
Proof
▶
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 \| _{\operatorname{\operatorname {Lip}}(B')} (1+2R|n'|)^{-1}\, . \end{equation}
10.9.19
This follows from Lemma 10.1.8 with \(\alpha = x - R\) and \(\beta = x + R\).
The preceding chain of lemmas establish that \({\Theta }\) is a cancellative, compatible collection of functions on \(({\mathbb {R}}, \rho , \mu , 4)\). 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 10.1.12, define the function \(K:{\mathbb {R}}\times {\mathbb {R}}\to \mathbb {C}\) as in Theorem 1.0.2 by
\begin{equation} K(x,y):=\kappa (x-y)\, . \end{equation}
10.9.20
The function \(K\) is continuous outside the diagonal \(x=y\) and vanishes on the diagonal. Hence it is measurable.
By the Lemmas 10.1.13 and 10.1.14, it follows that \(K\) is a one-sided Calderón–Zygmund kernel on \(({\mathbb {R}},\rho ,\mu ,4)\).
The operator \(T^*\) defined in 1.0.16 coincides in our setting with the operator \(T^*\) defined in 10.1.17. By Lemma 10.1.5, this operator satisfies the bound 1.0.18.
Thus the assumptions of Theorem 1.0.2 are all satisfied. Applying the Theorem, Lemma 10.1.4 follows.