11 Proof of The Classical Carleson Theorem
The convergence of partial Fourier sums is proved in Section 11.1 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 11.1.2 and proved in Section 11.2. 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 11.1.3 and proved in Section 11.6. The proof relies on a bound on the real Carleson maximal operator stated in Lemma 11.1.5 and proved in Section 11.7, which involves showing that the real line fits into the setting of Chapter 2. This latter proof refers to the two-sided variant of the Carleson Theorem 10.0.1. Two assumptions in Theorem 1.0.2 require more work. The boundedness of the operator \(T_r\) defined in 10.0.2 is established in 11.1.6. This lemma is proved in Section 11.3. The cancellative property is verified by Lemma 11.1.7, which is proved in Section 11.4. Several further auxiliary lemmas are stated and proved in Section 11.1, the proof of one of these auxiliary lemmas, Lemma 11.1.11, is done in Section 11.5.
All subsections past Section 11.1 are mutually independent.
11.1 The classical Carleson theorem
Let a uniformly continuous \(2\pi \)-periodic function \(f:{\mathbb {R}}\to \mathbb {C}\) and \(\epsilon {\gt}0\) be given. Let
\begin{equation} C_{a,q} := \frac{2^{452a^3}}{(q-1)^6} \end{equation}
11.1.1
denote the constant from Theorem 10.0.1. Define
\begin{equation} \epsilon ' := \frac{\epsilon }{4 C_\epsilon } \, , \end{equation}
11.1.2
where
\begin{equation*} C_\epsilon = \left(\frac{8}{\pi \epsilon }\right)^\frac {1}{2} C_{4,2} + \pi \, . \end{equation*}
Since \(f\) is continuous and periodic, \(f\) is uniformly continuous. Thus, 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}
11.1.3
Define
\begin{equation} \label{def-fzero} f_0:=f \ast \phi _\delta , \end{equation}
11.1.4
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
11.1.1
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}
11.1.5
Proof
▶
Periodicity follows directly from the definitions. The other properties are part of the Lean library.
We prove in Section 11.2:
Lemma
11.1.2
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}
11.1.6
We prove in Section 11.6:
Lemma
11.1.3
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}
11.1.7
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}
11.1.8
We are now ready to prove Theorem 1.0.1. We first prove a version with explicit exceptional sets.
Theorem
11.1.4
classical Carleson with exceptional sets
Let \(f\) be a \(2\pi \)-periodic complex-valued continuous function on \(\mathbb {R}\). For all \(\epsilon {\gt}0\), 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 \(N{\gt}N_0\), we have
\begin{equation} \label{aeconv} |f(x)-S_N f(x)|\le \epsilon . \end{equation}
11.1.9
Proof
▶
Let \(N_0\) be as in Lemma 11.1.2. For every
\begin{equation} x\in [0, 2\pi ) \setminus E\, , \end{equation}
11.1.10
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}
11.1.11
Using ??, we estimate 11.1.11 by
\begin{equation} \le \epsilon ' +\frac\epsilon 4 +\frac\epsilon 4\le \epsilon \, . \end{equation}
11.1.12
This shows 11.1.9 for the given \(E\) and \(N_0\).
Now we turn to the proof of the statement of Carleson theorem given in the introduction.
By applying Theorem 11.1.4 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.
Let \(\kappa :{\mathbb {R}}\to {\mathbb {C}}\) 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}
11.1.13
and for \(|x|\ge 1\),
\begin{equation} \label{eq-hilker1} \kappa (x)=0\, . \end{equation}
11.1.14
Note that this function is continuous at every point \(x\) with \(|x|{\gt}0\).
The proof of Lemma 11.1.3 will use the following Lemma 11.1.5, which itself is proven in Section 11.7 as an application of Theorem 1.0.2.
Lemma
11.1.5
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}
11.1.15
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}
11.1.16
One of the main assumptions of Theorem 10.0.1, concerning the operator \(T_r\) defined in 10.0.2, is verified by the following lemma, which is proved in Section 11.3.
Lemma
11.1.6
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}
11.1.17
where
\begin{equation} \label{def-H-r} H_r f(x) := T_r f(x) = \int _{r\le \rho (x,y)} \kappa (x-y) f(y) \, dy \end{equation}
11.1.18
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 11.4.
Lemma
11.1.7
van der Corput
Let \(\alpha \le \beta \) 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)|+\frac{|\beta -\alpha |}{2} \sup _{\alpha \le x{\lt}y\le \beta } \frac{|g(y)-g(x)|}{|y-x|}{\lt}\infty \, . \end{equation}
11.1.19
Then for any \(\alpha \le \beta \) and \(n\in {\mathbb {Z}}\) 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}
11.1.20
We close this section with six lemmas that are used across the following subsections.
Lemma
11.1.8
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}
11.1.21
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
11.1.9
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}
11.1.22
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}
11.1.23
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}
11.1.24
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}
11.1.25
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}
11.1.26
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}
11.1.27
This proves the second part of the lemma.
Lemma
11.1.10
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}
11.1.28
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 11.5.
Lemma
11.1.11
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}
11.1.29
Lemma
11.1.12
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}
11.1.30
Proof
▶
Fix \(x\neq y\). If \(\kappa (x-y)\) is zero, then 11.1.30 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}
11.1.31
We estimate with Lemma 11.1.10
\begin{equation} \label{eqcarl311} |\kappa (x-y)|\le \frac{1}{|1-e^{i(x-y)}|}\le \frac2{|x-y|}\, . \end{equation}
11.1.32
This proves 11.1.30 in the given case and completes the proof of the lemma.
Lemma
11.1.13
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}
11.1.33
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}
11.1.34
Proof
▶
Upon replacing \(y\) by \(y-x\) and \(y'\) by \(y'-x\) on the left-hand side of 11.1.33, we can assume that \(x = 0\). Then the assumption 11.1.33 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 11.1.33 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 11.1.10, 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
\[ |\kappa (-y) - \kappa (-y')| = 0 \le 2^{8} \frac{|y-y'|}{|y|^2}\, . \]
11.2 Smooth functions.
Lemma
11.2.1
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}
11.2.1
Proof
▶
This is part of the Lean library.
Lemma
11.2.2
Let \(f:{\mathbb {R}}\to {\mathbb {C}}\) such that
\begin{equation} \sum _{n\in {\mathbb {Z}}} |\widehat{f}_n| {\lt} \infty . \end{equation}
11.2.2
Then
\begin{equation} \sup _{x\in [0,2\pi ]} |f(x) - S_Nf(x)| \rightarrow 0 \end{equation}
11.2.3
as \(N \rightarrow \infty \).
Proof
▶
This is part of the Lean library.
Lemma
11.2.3
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}
11.2.4
as \(N \rightarrow \infty \).
Proof
▶
By Lemma 11.2.2, it suffices to show that the Fourier coefficients \(\widehat{f}_n\) are summable. Applying Lemma 11.2.1 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*}
11.3 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}
11.3.1
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}
11.3.2
Lemma
11.3.1
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}
11.3.3
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}
11.3.4
We have by the triangle inequality, the square root of the identity in 11.3.4, and Lemma 11.1.11
\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}
11.3.5
This proves 11.3.4 and completes the proof of the lemma.
Lemma
11.3.2
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}
11.3.6
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}
11.3.7
We then have by breaking up the domain of integration and using 11.3.7
\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}
11.3.8
This proves the first identity of the lemma. The second identity follows by substitution of \(y\) by \(y-x\).
Lemma
11.3.3
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}
11.3.9
Proof
▶
Using Fubini and Lemma 11.3.2, 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}
11.3.10
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 11.3.9 with Cauchy-Schwarz and then with 11.3.10 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}
11.3.11
where the minimum is understood to be \(r^{-1}\) in case \(1=e^{ix}\).
Lemma
11.3.4
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}
11.3.12
Let
\begin{equation} h(x)= \int _{-\pi }^{\pi } f(y)g(x-y)\, dy \, . \end{equation}
11.3.13
Then
\begin{equation} \| h\| _{L^2[-\pi , \pi ]}\le 2^{5}\| f\| _{L^2[-\pi , \pi ]} \, . \end{equation}
11.3.14
Proof
▶
From monotonicity of the integral and 11.3.12,
\begin{equation} \| g\| _{L^1[-\pi , \pi ]} \le \int _{-\pi }^{\pi }k_r(x)\, dx\, . \end{equation}
11.3.15
Using the symmetry \(k_r(x)=k_r(-x)\), the assumption, and Lemma 11.1.10, 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}
11.3.16
Together with Lemma 11.3.3, this proves the lemma.
Lemma
11.3.5
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}
11.3.17
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}
11.3.18
Proof
▶
We have by definition and Lemma 11.1.9
\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}
11.3.19
This is of the form 11.3.17 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}
11.3.20
With 11.1.23 of Lemma 11.1.9 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}
11.3.21
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}
11.3.22
This proves 11.3.18 for \(|x|\in [0, r)\) since \(k_r(x)=r^{-1}\) in this case.
For \(e^{ix'}\neq 1\) and may use the expression 11.1.24 for \(K_N\) in Lemma 11.1.9 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}
11.3.23
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}
11.3.24
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 11.1.10 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}
11.3.25
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}
11.3.26
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}
11.3.27
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}
11.3.28
As \(e^{-2ix}\neq 1\), we may divide by \(1-e^{-2ix}\) and insert this into 11.3.23 to obtain
\begin{equation} L''(x)= \frac1N \frac{e^{-i2Nx}}{1-e^{ix}} \frac{1-e^{-i2Nx}}{1-e^{-2ix}}\, . \end{equation}
11.3.29
Hence, with Lemma 11.1.10 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}
11.3.30
Inequalities 11.3.21, 11.3.22, 11.3.24, 11.3.25, 11.3.27, and 11.3.30 prove 11.3.18. This completes the proof of the lemma.
We now prove Lemma 11.1.6.
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}
11.3.31
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 11.3.5 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 \(\kappa \) 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 11.3.1 and Lemma 11.3.4, 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}
11.3.32
which gives 11.3.31.
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 11.1.18, we have \(H_r g(x ) = H_r f(c+ 3/2+x)\). Thus, by 11.3.31
\begin{equation} \label{eq-Hr-short-support-2} \| H_rg\| _2 = \| H_r f\| _2 \le 2^{11} \| f\| _2 = \| g\| _2\, . \end{equation}
11.3.33
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 11.3.33, 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.
11.4 The proof of the van der Corput Lemma
Let \(g\) be a Lipschitz continuous function as in the lemma. Assume first that \(n=0\). Then
\begin{equation*} \int _\alpha ^\beta g(x) \, \mathrm{d}x \le |\beta - \alpha | \sup _{\alpha \le x\le \beta }|g(x)| \le |\beta -\alpha |\| g\| _{Lip(\alpha ,\beta )}(1+|n||\beta -\alpha |)^{-1} \end{equation*}
Assume now \(n\ne 0\). Without loss of generality, we may assume \(n{\gt}0\). We distinguish two cases. If \(\beta -\alpha {\lt} \frac{\pi }{n}\), we have by the triangle inequality
\begin{equation*} \left|\int _\alpha ^\beta g(x) e^{inx} \, \mathrm{d}x\right| \le |\beta -\alpha | \sup _{x \in [\alpha ,\beta ]} |g(x)| \le 2\pi |\beta -\alpha |\| g\| _{Lip(\alpha ,\beta )}(1+|n||\beta -\alpha |)^{-1} \, . \end{equation*}
We turn to the case \(\frac{\pi }{n} \le \beta -\alpha \). 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
\[ \frac{1}{2} \int _{\alpha + \frac{\pi }{n}}^\beta |g(x) - g(x - \frac{\pi }{n})| \, \mathrm{d}x \]
\[ \le \frac{|\beta -\alpha |}{2} \frac{\pi }{n} \sup _{\alpha \le x {\lt} y \le \beta } \frac{|g(x) - g(y)|}{|x-y|}\, . \]
Adding the two terms, we obtain
\[ \left|\int _\alpha ^\beta g(x) e^{-inx} \, \mathrm{d}x\right| \le \frac{\pi }{n} \| g\| _{\mathrm{Lip}(\alpha ,\beta )}\, . \]
This completes the proof of the lemma, using that with \(\frac{\pi }{n} \le \beta -\alpha \),
\[ \frac{\pi }{n} = \frac{2 \pi |\beta -\alpha |}{2n|\beta -\alpha |} \le 2 \pi |\beta -\alpha |(1 + n|\beta -\alpha |)^{-1}\, . \]
11.5 Partial sums as orthogonal projections
This subsection proves Lemma 11.1.11
Lemma
11.5.1
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}
11.5.1
Proof
▶
Let \(N{\gt}0\) be given. With \(K_N\) as in Lemma 11.1.9,
\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}
11.5.2
We have by Lemma 11.1.9
\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}
11.5.3
By Lemma 11.1.8, the summands for \(n\neq n'\) vanish. We obtain for 11.5.3
\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}
11.5.4
Applying Fubini in 11.5.2 and using 11.5.4 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}
11.5.5
This proves the lemma.
Lemma
11.5.2
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}
11.5.6
Proof
▶
We have with \(K_N\) as in Lemma 11.1.9 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}
11.5.7
Further, with Lemma 11.1.9 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}
11.5.8
This proves the lemma.
We turn to the proof of Lemma 11.1.11.
We have with Lemma 11.5.2, then Lemma 11.5.1 and the Lemma 11.5.2 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}
11.5.9
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}
11.5.10
\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 11.5.9, 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}
11.5.11
As 11.5.10 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}
11.5.12
As both sides are positive, we may take the square root of this inequality. This completes the proof of the lemma.
11.6 The error bound
Lemma
11.6.1
Dirichlet kernel - Hilbert kernel relation
For all \(N\in {\mathbb {Z}}\) and \(x\in [-\pi ,\pi ] \setminus \{ 0\} \),
\begin{equation*} \left|K_N(x) - (e^{-iNx}\kappa (x) + \overline{e^{-iNx}\kappa (x)})\right| \le \pi \, . \end{equation*}
Proof
▶
Let \(N\in {\mathbb {Z}}\) and \(x\in [-\pi ,\pi ] \setminus \{ 0\} \). With Lemma 11.1.9, we obtain
\begin{equation*} K_N(x) - (e^{-iNx}\kappa (x) + \overline{e^{-iNx}\kappa (x)}) = e^{-iNx} \frac{\min (|x|, 1) }{1 - e^{ix}} + e^{iNx} \frac{\min (|x|, 1) }{1 - e^{-ix}} \, . \end{equation*}
Using Lemma 11.1.10 with \(\eta = \min (|x|, 1)\), we bound
\begin{equation*} \left|K_N(x) - (e^{-iNx}\kappa (x) + \overline{e^{-iNx}\kappa (x)})\right| \le \frac{\min (|x|, 1) }{|1 - e^{ix}|} + \frac{\min (|x|, 1)}{|1 - e^{-ix}|} \le \frac{\pi }{2} + \frac{\pi }{2} = \pi \, . \end{equation*}
Lemma
11.6.2
partial Fourier sum bound
Let \(g:{\mathbb {R}}\to {\mathbb {C}}\) be a measurable \(2\pi \)-periodic function such that for some \(\delta {\gt}0\) and every \(x\in {\mathbb {R}}\),
\begin{equation} |g(x)|\le \delta \, . \end{equation}
11.6.1
Then for every \(x\in [0,2\pi ]\) and \(N{\gt}0\),
\begin{equation*} |S_N g(x)| \le \frac{1}{2\pi } (Tg(x) + T\bar{g}(x)) + \pi \delta . \end{equation*}
Proof
▶
Let \(x\in [0,2\pi ]\) and \(N{\gt}0\). We have with Lemma 11.1.9
\begin{equation*} |S_N g(x)| = \frac{1}{2\pi } \left| \int _0^{2\pi } g(y) K_N(x-y) \, dy\right|\, . \end{equation*}
We use \(2\pi \)-periodicity of \(g\) and \(K_N\) to shift the domain of integration to obtain
\begin{equation*} = \frac{1}{2\pi } \left|\int _{x-\pi }^{x+\pi } g(y) K_N(x-y) \, dy\right|\, . \end{equation*}
Using the triangle inequality, we split this as
\begin{equation} \label{eq-diff-integrable} \le \frac{1}{2\pi } \left|\int _{x-\pi }^{x+\pi } g(y) \left(K_N(x-y) - \max (|x-y|,0) K_N(x-y)\right) \, dy \right| \end{equation}
11.6.2
\begin{equation} \label{eq-diff-singular} + \frac{1}{2\pi } \left|\int _{x-\pi }^{x+\pi } g(y) \max (|x-y|,0) K_N(x-y) \, dy\right|\, . \end{equation}
11.6.3
Note that all integrals are well defined, since \(K_N\) is by 11.1.23 bounded by \(2N+1\). Using that
\begin{equation} \label{eq-Dirichlet-Hilbert} \max (|x-y|,0) K_N(x-y) = e^{-iN(x-y)}\kappa (x-y) + \overline{e^{-iN(x-y)}\kappa (x-y)} \, , \end{equation}
11.6.4
Lemma 11.6.1 and 11.6.5, we bound 11.6.2 by
\begin{equation*} \frac{1}{2\pi } \int _{x-\pi }^{x+\pi } |g(y)| \left|K_N(x-y) - e^{-iN(x-y)}\kappa (x-y) + \overline{e^{-iN(x-y)}\kappa (x-y)}\right|\, dy \le \pi \delta \, . \end{equation*}
By dominated convergence and since \(\kappa (x-y) = 0\) for \(|x-y| {\gt} 1\), 11.6.3 equals
\begin{equation*} \frac{1}{2\pi } \lim _{r \to 0^+} \left| \int _{r {\lt} |x-y| {\lt} 1} g(y) \max (|x-y|,0) K_N(x-y) \, dy\right|\, . \end{equation*}
We bound the limit by a supremum and rewrite using 11.6.4,
\begin{equation*} \le \frac{1}{2\pi } \sup _{r {\gt} 0} \left| \int _{r {\lt} |x-y| {\lt} 1} g(y) \left(e^{-iN(x-y)}\kappa (x-y) + \overline{e^{-iN(x-y)}\kappa (x-y)}\right) \, dy\right| \end{equation*}
Using the triangle inequality, we further bound this by
\begin{alignat*}{3} \le & & & \frac{1}{2\pi } \sup _{r {\gt} 0} \left| \int _{r {\lt} |x-y| {\lt} 1} g(y) e^{-iNy} \kappa (x-y) \, dy\right| \\ & + & & \frac{1}{2\pi } \sup _{r {\gt} 0} \left| \int _{r {\lt} |x-y| {\lt} 1} \overline{g}(y) e^{-iNy} \kappa (x-y) \, dy\right|\, . \end{alignat*}
By the definition 11.1.16 of \(T\), this is
\begin{equation*} \le \frac{1}{2\pi } (Tg(x) + T\bar{g}(x))\, . \end{equation*}
Lemma
11.6.3
real Carleson operator measurable
Let \(f\) be a bounded measurable function on \({\mathbb {R}}\). Then \(Tf\) as defined in 11.1.16 is measurable.
Proof
▶
Since a countable supremum of measurable functions is measurable, it suffices to show that for every \(n\in {\mathbb {Z}}\),
\begin{equation*} x \mapsto \sup _{r{\gt}0}\left|\int _{r{\lt}|x-y|{\lt}1} f(y)\kappa (x-y) e^{iny}\, dy\right| \end{equation*}
is measurable. So let \(n\in {\mathbb {Z}}\). Note that for each \(x\in {\mathbb {R}}\), the function
\begin{equation*} r \mapsto \left|\int _{r{\lt}|x-y|{\lt}1} f(y)\kappa (x-y) e^{iny}\, dy\right| \end{equation*}
is continuous on \((0,\infty )\) since the integrand is locally bounded on the domain \(0{\lt}|x-y|{\lt}1\) by the assumptions on \(f\) and Lemma 11.1.12. Thus, for each \(x\in {\mathbb {R}}\),
\begin{equation*} \sup _{r{\gt}0}\left|\int _{r{\lt}|x-y|{\lt}1} f(y)\kappa (x-y) e^{iny}\, dy\right| =\sup _{r\in {\mathbb {Q}}_{{\gt}0}}\left|\int _{r{\lt}|x-y|{\lt}1} f(y)\kappa (x-y) e^{iny}\, dy\right| \end{equation*}
The right hand side is again a countable supremum so it remains to show that for every \(r\in {\mathbb {Q}}_{{\gt}0}\),
\begin{equation*} x \mapsto \left|\int _{r{\lt}|x-y|{\lt}1} f(y)\kappa (x-y) e^{iny}\, dy\right| = \left|\int \mathbf{1}_{\{ r{\lt}|x-\cdot |{\lt}1\} }(y) f(y)\kappa (x-y) e^{iny}\, dy\right| \end{equation*}
is measurable, which follows from the fact that the integrand is measurable in \((x,y)\).
Lemma
11.6.4
partial Fourier sums of small
Let \(g:{\mathbb {R}}\to {\mathbb {C}}\) be a measurable \(2\pi \)-periodic function such that for some \(\delta {\gt}0\) and every \(x\in {\mathbb {R}}\),
\begin{equation} \label{g-small} |g(x)|\le \delta \, . \end{equation}
11.6.5
Then for every \(\epsilon {\gt}0\), there exists a measurable set \(E\subset [0,2\pi ]\) with \(|E|{\lt}\epsilon \) such that for every \(x\in [0,2\pi ]\setminus E\) and \(N{\gt}0\),
\begin{equation} \label{S-Ng-small} |S_N g(x)|\le C_\epsilon \delta , \end{equation}
11.6.6
where
\begin{equation} \label{C-epsilon-def} C_\epsilon = \left(\frac{8}{\pi \epsilon }\right)^\frac {1}{2} C_{4,2} + \pi \, . \end{equation}
11.6.7
Proof
▶
Define
\begin{equation*} E := \{ x \in [0, 2\pi ] \ : \ \sup _{N {\gt} 0} |S_N g (x)| {\gt} C_\epsilon \delta \} \, . \end{equation*}
Then 11.6.6 clearly holds, and it remains to show that \(|E| \le \epsilon \). Using Lemma 11.6.2, we obtain
\begin{equation*} E \subset \{ x\in [0,2\pi ] \ : \ C_\epsilon \delta {\lt} \frac{1}{2\pi } (Tg(x) + T\bar{g}(x)) + \pi \delta \} \subset E_1 \cup E_2, \end{equation*}
where
\begin{align*} E_1 :=& \{ x\in [0,2\pi ] \ : \ \pi (C_\epsilon - \pi ) \delta {\lt} Tg(x)\} \\ E_2 :=& \{ x\in [0,2\pi ] \ : \ \pi (C_\epsilon - \pi ) \delta {\lt} T\bar{g}(x)\} . \end{align*}
By Lemma 11.6.3, \(E_1\) and \(E_2\) are measurable. Thus,
\begin{equation*} \pi (C_\epsilon - \pi ) \delta |E_1| \le \int _{E_1} Tg(x) \, dx = \delta \int _{E_1} T(\delta ^{-1} g\mathbf{1}_{[-\pi ,3\pi ]})(x) \, dx \, . \end{equation*}
Applying Lemma 11.1.5 with \(F = [-\pi , 3\pi ]\) and \(G = E'\), it follows that this is
\begin{equation*} \le \delta \cdot C_{4,2} |F|^{\frac{1}{2}} |E_1|^{\frac{1}{2}} \le (4\pi )^\frac {1}{2} C_{4,2} \delta \cdot |E’|^{\frac{1}{2}}\, . \end{equation*}
Rearranging, we obtain
\begin{equation*} |E_1| \le \left(\frac{(4\pi )^\frac {1}{2} C_{4,2}}{\pi (C_\epsilon - \pi )}\right)^2 = \frac{\epsilon }{2}\, . \end{equation*}
Analogously, we get the same estimate for \(|E_2|\). This completes the proof using \(|E| \le |E_1| + |E_2|\).
11.7 Carleson on the real line
We prove Lemma 11.1.5.
Consider the standard distance function
\begin{equation} \rho (x,y)=|x-y| \end{equation}
11.7.1
on the real line \({\mathbb {R}}\).
Lemma
11.7.1
real line metric
The space \(({\mathbb {R}},\rho )\) is a complete locally compact metric space.
Proof
▶
This is part of the Lean library.
Lemma
11.7.2
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
11.7.3
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
11.7.4
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}
11.7.2
Proof
▶
We have with Lemma 11.7.2
\begin{equation} \mu (B(x,R))=\mu ((x-R,x+R))=2R\, . \end{equation}
11.7.3
Lemma
11.7.5
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}
11.7.4
Proof
▶
We have with Lemma 11.7.4
\begin{equation} \mu (B(x,2R)=4R=2\mu (B(x,R))\, . \end{equation}
11.7.5
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 11.7.5 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}
11.7.6
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}
11.7.7
Lemma
11.7.6
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
11.7.7
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}
11.7.8
Proof
▶
The right hand side of 11.7.8 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
11.7.8
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 11.7.7 and \(R \le R'\).
Lemma
11.7.9
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}
11.7.9
Proof
▶
With 11.7.7, both sides of 11.7.9 are equal to \(4R|n-m|\). This proves the lemma.
Lemma
11.7.10
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}
11.7.10
Proof
▶
With 11.7.7, both sides of 11.7.9 are equal to \(4R|n-m|\). This proves the lemma.
Lemma
11.7.11
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}
11.7.11
where
\begin{equation} B'= \{ {\vartheta }\in {\Theta }: d_{B(x,R)}({\vartheta }, {\vartheta }_n){\lt}2R'\} \end{equation}
11.7.12
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}
11.7.13
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 11.7.7, we have
\begin{equation} \label{eqcarl6} 2R|n-n'|{\lt} 2R'\, . \end{equation}
11.7.14
Assume first \(n'\le m_1\). With 11.7.14 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}
11.7.15
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 11.7.14 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}
11.7.16
We conclude \({\vartheta }_{n'}\in B_1\). This completes the proof of the lemma.
Lemma
11.7.12
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}
11.7.17
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}
11.7.18
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}
11.7.19
This follows from Lemma 11.1.7 with \(\alpha = x - R\) and \(\beta = x + R\).
The preceding chain of lemmas establishes 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 11.1.13, 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}
11.7.20
The function \(K\) is continuous outside the diagonal \(x=y\) and vanishes on the diagonal. Hence it is measurable.
By ??, it follows that \(K\) is a two-sided Calderón–Zygmund kernel on \(({\mathbb {R}},\rho ,\mu ,4)\). Lemma 11.1.6 verifies 10.0.3. Thus the assumptions of Theorem 10.0.1 are all satisfied. Applying the Theorem, Lemma 11.1.5 follows.