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 11.1.10, 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 $2 π$ -periodic function $f : R \to C$ and $ϵ > 0$ be given. Let

C_{a, q} := \frac{2^{452 a^{3}}}{(q - 1)^{6}}

11.1.1

denote the constant from Theorem 10.0.1. Define

ϵ^{'} := \frac{ϵ}{4 C_{ϵ}},

11.1.2

where

C_{ϵ} = {(\frac{8}{π ϵ})}^{\frac{1}{2}} C_{4, 2} + π .

Since $f$ is continuous and periodic, $f$ is uniformly continuous. Thus, there is a $0 < δ < π$ such that for all $x, x^{'} \in R$ with $| x - x^{'} | \leq δ$ we have

| f (x) - f (x^{'}) | \leq ϵ^{'} .

11.1.3

Define

f_{0} := f * ϕ_{δ},

11.1.4

where $ϕ_{δ}$ is a nonnegative smooth bump function with $supp (ϕ_{δ}) \subset (- δ, δ)$ and $\int_{R} ϕ_{δ} (x) d x = 1$ .

Lemma 11.1.1 smooth approximation

✓

The function $f_{0}$ is $2 π$ -periodic. The function $f_{0}$ is smooth (and therefore measurable). The function $f_{0}$ satisfies for all $x \in R$ :

| f (x) - f_{0} (x) | \leq ϵ^{'},

11.1.5

Proof ▶

We prove in Section 11.2:

Lemma 11.1.2 convergence for smooth

✓

There exists some $N_{0} \in N$ such that for all $N > N_{0}$ and $x \in [0, 2 π]$ we have

| S_{N} f_{0} (x) - f_{0} (x) | \leq \frac{ϵ}{4} .

11.1.6

We prove in Section 11.6:

Lemma 11.1.3 control approximation effect

✓

There is a set $E \subset R$ with Lebesgue measure $| E | \leq ϵ$ such that for all

x \in [0, 2 π) ∖ E

11.1.7

we have

sup_{N \geq 0} | S_{N} f (x) - S_{N} f_{0} (x) | \leq \frac{ϵ}{4} .

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 $2 π$ -periodic complex-valued continuous function on $R$ . For all $ϵ > 0$ , there exists a Borel set $E \subset [0, 2 π]$ with Lebesgue measure $| E | \leq ϵ$ and a positive integer $N_{0}$ such that for all $x \in [0, 2 π] ∖ E$ and all integers $N > N_{0}$ , we have

| f (x) - S_{N} f (x) | \leq ϵ .

11.1.9

Proof ▶

Let $N_{0}$ be as in Lemma 11.1.2. For every

x \in [0, 2 π) ∖ E,

11.1.10

and every $N > N_{0}$ we have by the triangle inequality

| f (x) - S_{N} f (x) |

\leq | f (x) - f_{0} (x) | + | f_{0} (x) - S_{N} f_{0} (x) | + | S_{N} f_{0} (x) - S_{N} f (x) | .

11.1.11

Using ??, we estimate 11.1.11 by

\leq ϵ^{'} + \frac{ϵ}{4} + \frac{ϵ}{4} \leq ϵ .

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.

Proof of Theorem 1.0.1 ▶

By applying Theorem 11.1.4 with a sequence of $ϵ_{n} := 2^{- n} δ$ for $n \geq 1$ and taking the union of corresponding exceptional sets $E_{n}$ , we see that outside a set of measure $δ$ , the partial Fourier sums converge pointwise for $N \to \infty$ . Applying this with a sequence of $δ$ 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 $κ : R \to C$ be the function defined by $κ (0) = 0$ and for $0 < | x | < 1$

κ (x) = \frac{1 - | x |}{1 - e^{i x}}

11.1.13

and for $| x | \geq 1$ ,

κ (x) = 0 .

11.1.14

Note that this function is continuous at every point $x$ with $| x | > 0$ .

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 $R$ with finite measure. Let $f$ be a bounded measurable function on $R$ with $| f | \leq 1_{F}$ . Then

| \int_{G} T f (x) d x | \leq C_{4, 2} | F |^{\frac{1}{2}} | G |^{\frac{1}{2}},

11.1.15

where

T f (x) = sup_{n \in Z} sup_{r > 0} | \int_{r < | x - y | < 1} f (y) κ (x - y) e^{i n y} d y | .

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 $0 < r < 1$ . Let $f$ be a bounded, measurable function on $R$ with bounded support. Then

‖ H_{r} f ‖_{2} \leq 2^{13} ‖ f ‖_{2},

11.1.17

where

H_{r} f (x) := T_{r} f (x) = \int_{r \leq ρ (x, y)} κ (x - y) f (y) d y

11.1.18

The next lemma will be used to verify that the collection $Θ$ 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 $α \leq β$ be real numbers. Let $g : R \to C$ be a measurable function and assume

‖ g ‖_{L i p (α, β)} := sup_{α \leq x \leq β} | g (x) | + \frac{| β - α |}{2} sup_{α \leq x < y \leq β} \frac{| g (y) - g (x) |}{| y - x |} < \infty .

11.1.19

Then for any $α \leq β$ and $n \in Z$ we have

\int_{α}^{β} g (x) e^{i n x} d x \leq 2 π | β - α | ‖ g ‖_{L i p (α, β)} (1 + | n | | β - α |)^{- 1} .

11.1.20

We close this section with five lemmas that are used across the following subsections.

Lemma 11.1.8 Dirichlet kernel

✓

We have for every $2 π$ -periodic bounded measurable $f$ and every $N \geq 0$

S_{N} f (x) = \frac{1}{2 π} \int_{0}^{2 π} f (y) K_{N} (x - y) d y

11.1.21

where $K_{N}$ is the $2 π$ -periodic continuous function of $R$ given by

\sum_{n = - N}^{N} e^{i n x^{'}} .

11.1.22

We have for $e^{i x^{'}} \neq 1$ that

K_{N} (x^{'}) = \frac{e^{i N x^{'}}}{1 - e^{- i x^{'}}} + \frac{e^{- i N x^{'}}}{1 - e^{i x^{'}}} .

11.1.23

Proof ▶

We have by definitions and interchanging sum and integral

S_{N} f (x) = \sum_{n = - N}^{N} {\hat{f}}_{n} e^{i n x}

= \sum_{n = - N}^{N} \frac{1}{2 π} \int_{0}^{2 π} f (x) e^{i n (x - y)} d y

= \frac{1}{2 π} \int_{0}^{2 π} f (y) \sum_{n = - N}^{N} e^{i n (x - y)} d y .

11.1.24

This proves the first statement of the lemma. By a telescoping sum, we have for every $x^{'} \in R$

(e^{\frac{1}{2} i x^{'}} - e^{- \frac{1}{2} i x^{'}}) \sum_{n = - N}^{N} e^{i n x^{'}} = e^{(N + \frac{1}{2}) i x^{'}} - e^{- (N + \frac{1}{2}) i x^{'}} .

11.1.25

If $e^{i x^{'}} \neq 1$ , the first factor on the left-hand side is not $0$ and we may divide by this factor to obtain

\sum_{n = - N}^{N} e^{i n x^{'}} = \frac{e^{i (N + \frac{1}{2}) x^{'}}}{e^{\frac{1}{2} i x^{'}} - e^{- \frac{1}{2} i x^{'}}} - \frac{e^{- i (N + \frac{1}{2}) x^{'}}}{e^{\frac{1}{2} i x^{'}} - e^{- \frac{1}{2} i x^{'}}} = \frac{e^{i N x^{'}}}{1 - e^{- i x^{'}}} + \frac{e^{- i N x^{'}}}{1 - e^{i x^{'}}} .

11.1.26

This proves the second part of the lemma.

Lemma 11.1.9 lower secant bound

✓

Let $η > 0$ and $- 2 π + η \leq x \leq 2 π - η$ with $| x | \geq η$ . Then

| 1 - e^{i x} | \geq \frac{2}{π} η

11.1.27

Proof ▶

We have

| 1 - e^{i x} | = \sqrt{(1 - \cos (x))^{2} + \sin^{2} (x)} \geq | \sin (x) | .

If $0 \leq x \leq \frac{π}{2}$ , then we have from concavity of $\sin$ on $[0, π]$ and $\sin (0) = 0$ and $\sin (\frac{π}{2}) = 1$

| \sin (x) | \geq \frac{2}{π} x \geq \frac{2}{π} η .

When $x \in \frac{m π}{2} + [0, \frac{π}{2}]$ 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 11.1.10 spectral projection bound

✓

Let $f$ be a bounded $2 π$ -periodic measurable function. Then, for all $N \geq 0$

‖ S_{N} f ‖_{L^{2} [- π, π]} \leq ‖ f ‖_{L^{2} [- π, π]} .

11.1.28

Lemma 11.1.11 Hilbert kernel bound

✓

For $x, y \in R$ with $x \neq y$ we have

| κ (x - y) | \leq 2^{2} (2 | x - y |)^{- 1} .

11.1.29

Proof ▶

Fix $x \neq y$ . If $κ (x - y)$ is zero, then 11.1.29 is evident. Assume $κ (x - y)$ is not zero, then $0 < | x - y | < 1$ . We have

| κ (x - y) | = | \frac{1 - | x - y |}{1 - e^{i (x - y)}} | .

11.1.30

We estimate with Lemma 11.1.9

| κ (x - y) | \leq \frac{1}{| 1 - e^{i (x - y)} |} \leq \frac{2}{| x - y |} .

11.1.31

This proves 11.1.29 in the given case and completes the proof of the lemma.

Lemma 11.1.12 Hilbert kernel regularity

✓

For $x, y, y^{'} \in R$ with $x \neq y, y^{'}$ and

2 | y - y^{'} | \leq | x - y |,

11.1.32

we have

| κ (x - y) - κ (x - y^{'}) | \leq 2^{8} \frac{1}{| x - y |} \frac{| y - y^{'} |}{| x - y |} .

11.1.33

Proof ▶

Upon replacing $y$ by $y - x$ and $y^{'}$ by $y^{'} - x$ on the left-hand side of 11.1.32, we can assume that $x = 0$ . Then the assumption 11.1.32 implies that $y$ and $y^{'}$ have the same sign. Since $κ (y) = \bar{κ} (- y)$ we can assume that they are both positive. Then it follows from 11.1.32 that

\frac{y}{2} \leq y^{'} .

We distinguish four cases. If $y, y^{'} \leq 1$ , then we have

| κ (- y) - κ (- y^{'}) | = | \frac{1 - y}{1 - e^{- i y}} - \frac{1 - y^{'}}{1 - e^{- i y^{'}}} |

and by the fundamental theorem of calculus

= | \int_{y^{'}}^{y} \frac{- 1 + e^{- i t} + i (1 - t) e^{i t}}{(1 - e^{- i t})^{2}} d t | .

Using $y^{'} \geq \frac{y}{2}$ and Lemma 11.1.9, we bound this by

\leq | y - y^{'} | sup_{\frac{y}{2} \leq t \leq 1} \frac{3}{| 1 - e^{- i t} |^{2}} \leq 3 | y - y^{'} | (2 \frac{2}{y})^{2} \leq 2^{6} \frac{| y - y^{'} |}{| y |^{2}} .

If $y \leq 1$ and $y^{'} > 1$ , then $κ (- y^{'}) = 0$ and we have from the first case

| κ (- y) - κ (- y^{'}) | = | κ (- y) - κ (- 1) | \leq 2^{6} \frac{| y - 1 |}{| y |^{2}} \leq 2^{6} \frac{| y - y^{'} |}{| y |^{2}} .

Similarly, if $y > 1$ and $y^{'} \leq 1$ , then $κ (- y) = 0$ and we have from the first case

| κ (- y) - κ (- y^{'}) | = | κ (- y^{'}) - κ (- 1) | \leq 2^{6} \frac{| y^{'} - 1 |}{| y^{'} |^{2}} \leq 2^{6} \frac{| y - y^{'} |}{| y^{'} |^{2}} .

Using again $y^{'} \geq \frac{y}{2}$ , we bound this by

\leq 2^{6} \frac{| y - y^{'} |}{| y / 2 |^{2}} = 2^{8} \frac{| y - y^{'} |}{| y |^{2}}

Finally, if $y, y^{'} > 1$ then

| κ (- y) - κ (- y^{'}) | = 0 \leq 2^{8} \frac{| y - y^{'} |}{| y |^{2}} .

11.2 Smooth functions.

Lemma 11.2.1

✓

Let $f : R \to C$ be $2 π$ -periodic and continuously differentiable, and let $n \in Z ∖ {0}$ . Then

{\hat{f}}_{n} = \frac{1}{i n} {\hat{f^{'}}}_{n} .

11.2.1

Proof ▶

Lemma 11.2.2

✓

Let $f : R \to C$ such that

\sum_{n \in Z} | {\hat{f}}_{n} | < \infty .

11.2.2

Then

sup_{x \in [0, 2 π]} | f (x) - S_{N} f (x) | \to 0

11.2.3

as $N \to \infty$ .

Proof ▶

Lemma 11.2.3

✓

Let $f : R \to C$ be $2 π$ -periodic and twice continuously differentiable. Then

sup_{x \in [0, 2 π]} | f (x) - S_{N} f (x) | \to 0

11.2.4

as $N \to \infty$ .

Proof ▶

By Lemma 11.2.2, it suffices to show that the Fourier coefficients ${\hat{f}}_{n}$ are summable. Applying Lemma 11.2.1 twice and using the fact that $f^{″}$ is continuous and thus bounded on $[0, 2 π]$ , we compute

\sum_{n \in Z} | {\hat{f}}_{n} | = | {\hat{f}}_{0} | + \sum_{n \in Z ∖ {0}} \frac{1}{n^{2}} | {\hat{f^{″}}}_{n} | \leq | {\hat{f}}_{0} | + (sup_{x \in [0, 2 π]} | f (x) |) \cdot \sum_{n \in Z ∖ {0}} \frac{1}{n^{2}} < \infty .

Proof of Lemma 11.1.2 ▶

11.3 The truncated Hilbert transform

Let $M_{n}$ be the modulation operator acting on measurable $2 π$ -periodic functions defined by

M_{n} g (x) = g (x) e^{i n x} .

11.3.1

Define the approximate Hilbert transform by

L_{N} g = \frac{1}{N} \sum_{n = 0}^{N - 1} M_{- n - N} S_{N + n} M_{N + n} g .

11.3.2

Lemma 11.3.1 modulated averaged projection

We have for every bounded measurable $2 π$ -periodic function $g$

‖ L_{N} g ‖_{L^{2} [- π, π]} \leq ‖ g ‖_{L^{2} [- π, π]} .

11.3.3

Proof ▶

We have

‖ M_{n} g ‖_{L^{2} [- π, π]}^{2} = \int_{- π}^{π} | e^{i n x} g (x) |^{2} d x = \int_{- π}^{π} | g (x) |^{2} d x = ‖ g ‖_{L^{2} [- π, π]}^{2} .

11.3.4

We have by the triangle inequality, the square root of the identity in 11.3.4, and Lemma 11.1.10

‖ L_{n} g ‖_{L^{2} [- π, π]} = ‖ \frac{1}{N} \sum_{n = 0}^{N - 1} M_{- n - N} S_{N + n} M_{N + n} g ‖_{L^{2} [- π, π]}

\leq \frac{1}{N} \sum_{n = 0}^{N - 1} ‖ M_{- n - N} S_{N + n} M_{N + n} g ‖_{L^{2} [- π, π]} = \frac{1}{N} \sum_{n = 0}^{N - 1} ‖ S_{N + n} M_{N + n} g ‖_{L^{2} [- π, π]}

\leq \frac{1}{N} \sum_{n = 0}^{N - 1} ‖ M_{N + n} g ‖_{L^{2} [- π, π]} = \frac{1}{N} \sum_{n = 0}^{N - 1} ‖ g ‖_{L^{2} [- π, π]} = ‖ g ‖_{L^{2} [- π, π]} .

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 $2 π$ -periodic function. We have for any $0 \leq x \leq 2 π$ that

\int_{0}^{2 π} f (y) d y = \int_{- x}^{2 π - x} f (y) d y = \int_{- π}^{π} f (y - x) d y .

11.3.6

Proof ▶

We have by periodicity and change of variables

\int_{- x}^{0} f (y) d y = \int_{- x}^{0} f (y + 2 π) d y = \int_{2 π - x}^{2 π} f (y) d y .

11.3.7

We then have by breaking up the domain of integration and using 11.3.7

\int_{0}^{2 π} f (y) d y = \int_{0}^{2 π - x} f (y) d y + \int_{2 π - x}^{2 π} f (y) d y

= \int_{0}^{2 π - x} f (y) d y + \int_{- x}^{0} f (y) d y = \int_{- x}^{2 π - x} f (y) d y .

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 $2 π$ -periodic functions on $R$ . Then

{(\int_{- π}^{π} {(\int_{- π}^{π} f (y) g (x - y) d y)}^{2} d x)}^{\frac{1}{2}} \leq ‖ f ‖_{L^{2} [- π, π]} ‖ g ‖_{L^{1} [- π, π]} .

11.3.9

Proof ▶

Using Fubini and Lemma 11.3.2, we observe

\int_{- π}^{π} \int_{- π}^{π} f (y)^{2} g (x - y) d y d x = \int_{- π}^{π} f (y)^{2} \int_{- π}^{π} g (x - y) d x d y

= \int_{- π}^{π} f (y)^{2} \int_{- π}^{π} g (x) d x d y = ‖ f ‖_{L^{2} [- π, π]}^{2} ‖ g ‖_{L^{1} [- π, π]} .

11.3.10

Let $h$ be the nonnegative square root of $g$ , then $h$ is bounded and $2 π$ -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

\int_{- π}^{π} (\int_{- π}^{π} f (y) h (x - y) h (x - y) d y)^{2} d x

\leq \int_{- π}^{π} (\int_{- π}^{π} f (y)^{2} g (x - y) d y) (\int_{- π}^{π} g (x - y) d y) d x

= ‖ f ‖_{L^{2} [- π, π]}^{2} ‖ g ‖_{L^{1} [- π, π]}^{2} .

Taking square roots, this proves the lemma.

For $0 < r < 1$ , Define the kernel $k_{r}$ to be the $2 π$ -periodic function

k_{r} (x) := min (r^{- 1}, 1 + \frac{r}{| 1 - e^{i x} |^{2}}),

11.3.11

where the minimum is understood to be $r^{- 1}$ in case $1 = e^{i x}$ .

Lemma 11.3.4 integrable bump convolution

Let $g, f$ be bounded measurable $2 π$ -periodic functions. Let $0 < r < π$ . Assume we have for all $0 \leq x \leq 2 π$

| g (x) | \leq k_{r} (x) .

11.3.12

Let

h (x) = \int_{- π}^{π} f (y) g (x - y) d y .

11.3.13

Then

‖ h ‖_{L^{2} [- π, π]} \leq 2^{5} ‖ f ‖_{L^{2} [- π, π]} .

11.3.14

Proof ▶

From monotonicity of the integral and 11.3.12,

‖ g ‖_{L^{1} [- π, π]} \leq \int_{- π}^{π} k_{r} (x) d x .

11.3.15

Using the symmetry $k_{r} (x) = k_{r} (- x)$ , the assumption, and Lemma 11.1.9, the last display is equal to

= 2 \int_{0}^{π} min (\frac{1}{r}, 1 + \frac{r}{| 1 - e^{i x} |^{2}}) d x

\leq 2 \int_{0}^{r} \frac{1}{r} d x + 2 \int_{r}^{π} 1 + \frac{64 r}{x^{2}} d x

\leq 2 + 2 π + 2 (\frac{64 r}{r} - \frac{64 r}{π}) \leq 2^{5} .

11.3.16

Together with Lemma 11.3.3, this proves the lemma.

Lemma 11.3.5 Dirichlet approximation

Let $0 < r < 1$ . Let $N$ be the smallest integer larger than $\frac{1}{r}$ . There is a $2 π$ -periodic continuous function $L^{'}$ on $R$ that satisfies for all $- π \leq x \leq π$ and all $2 π$ -periodic bounded measurable functions $f$ on $R$

L_{N} f (x) = \frac{1}{2 π} \int_{- π}^{π} f (y) L^{'} (x - y) d y

11.3.17

and

| L^{'} (x) - 1_{{y : r < | y | < 1}} κ (x) | \leq 2^{5} k_{r} (x) .

11.3.18

Proof ▶

We have by definition and Lemma 11.1.8

L_{N} g (x) = \frac{1}{N} \sum_{n = 0}^{N - 1} \int_{- π}^{π} e^{- i (N + n) x} K_{N + n} (x - y) e^{i (N + n) y} g (y) d y .

11.3.19

This is of the form 11.3.17 with the continuous function

L^{'} (x) = \frac{1}{N} \sum_{n = 0}^{N - 1} K_{N + n} (x) e^{- i (N + n) x} .

11.3.20

With 11.1.22 of Lemma 11.1.8 we have $| K_{N} (x) | \leq N$ for every $x$ and thus

| L^{'} (x) | \leq \frac{1}{N} \sum_{n = 0}^{N - 1} (N + n) \leq 2 N \leq 2^{2} r^{- 1} .

11.3.21

Therefore, for $| x | \in [0, r) \cup (1, π]$ , we have

| L^{'} (x) - 1_{{y : r < | y | < 1}} (x) κ (x) | = | L^{'} (x) | \leq 2^{2} r^{- 1} .

11.3.22

This proves 11.3.18 for $| x | \in [0, r)$ since $k_{r} (x) = r^{- 1}$ in this case.

For $e^{i x^{'}} \neq 1$ and may use the expression 11.1.23 for $K_{N}$ in Lemma 11.1.8 to obtain

L^{'} (x) = \frac{1}{N} \sum_{n = 0}^{N - 1} (\frac{e^{i (N + n) x}}{1 - e^{- i x}} + \frac{e^{- i (N + n) x}}{1 - e^{i x}}) e^{- i (N + n) x}

= \frac{1}{N} \sum_{n = 0}^{N - 1} (\frac{1}{1 - e^{- i x}} + \frac{e^{- i 2 (N + n) x}}{1 - e^{i x}})

= \frac{1}{1 - e^{- i x}} + \frac{1}{N} \frac{e^{- i 2 N x}}{1 - e^{i x}} \sum_{n = 0}^{N - 1} e^{- i 2 n x}

11.3.23

and thus

L^{'} (x) - 1_{{y : r < | y | < 1}} κ (x) = L^{″} (x) + \frac{1 - 1_{{y : r < | y | < 1}} (x) (1 - | x |)}{1 - e^{- i x}},

11.3.24

where

L^{″} (x) := \frac{1}{N} \frac{e^{- i 2 N x}}{1 - e^{i x}} \sum_{n = 0}^{N - 1} e^{- i 2 n x} .

For $x \in [- π, r] \cup [r, π]$ , we have using Lemma 11.1.9 that

| \frac{1 - 1_{{y : r < | y | < 1}} (x) (1 - | x |)}{1 - e^{- i x}} | = | \frac{min (| x |, 1)}{1 - e^{- i x}} | \leq \frac{8 min (| x |, 1)}{| x |}

\leq 2^{3} \cdot 1 \leq 2^{3} k_{r} (x) .

11.3.25

Next, we need to estimate $L^{″} (x)$ . If the real part of $e^{i x}$ is negative, we have

1 \leq | 1 - e^{i x} | \leq 2 .

11.3.26

and hence

| L^{″} (x) | \leq \frac{1}{N} \sum_{n = 0}^{N - 1} 1 = 1 \leq 1 + \frac{r}{| 1 - e^{i x} |^{2}} .

11.3.27

If the real part of $e^{i x}$ is positive and in particular while still $e^{i x} \neq \pm 1$ , then we have by telescoping

(1 - e^{- 2 i x}) \sum_{n = 0}^{N - 1} e^{- i 2 n x} = 1 - e^{- i 2 N x} .

11.3.28

As $e^{- 2 i x} \neq 1$ , we may divide by $1 - e^{- 2 i x}$ and insert this into 11.3.23 to obtain

L^{″} (x) = \frac{1}{N} \frac{e^{- i 2 N x}}{1 - e^{i x}} \frac{1 - e^{- i 2 N x}}{1 - e^{- 2 i x}} .

11.3.29

Hence, with Lemma 11.1.9 and nonnegativity of the real part of $e^{i x}$

| L ” (x) | \leq \frac{2}{N} \frac{1}{| 1 - e^{i x} |} \frac{1}{| 1 - e^{- 2 i x} |}

= \frac{2}{N} \frac{1}{| 1 - e^{i x} |^{2}} \frac{1}{| 1 + e^{i x} |} \leq \frac{4 r}{| 1 - e^{i x} |^{2}} \leq 2^{2} (1 + \frac{r}{| 1 - e^{i x} |^{2}})

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.

Proof of Lemma 11.1.6 ▶

We first show that if $f$ is supported in $[- 3 / 2, 3 / 2]$ , then

‖ H_{r} f ‖_{L^{2} (R)} \leq 2^{16} ‖ f ‖_{L^{2} (R)} .

11.3.31

Let $\tilde{f}$ be the $2 π$ -periodic extension of $f$ to $R$ . Let $N$ be the smallest integer larger than $\frac{1}{r}$ . Then, by Lemma 11.3.5 and the triangle inequality, for $x \in [- π, π]$ we have

| H_{r} \tilde{f} (x) | \leq 2 π | L_{N} \tilde{f} (x) | + 2^{5} | \int_{- π}^{π} k_{r} (x - y) \tilde{f} (y) d y | .

Taking $L^{2}$ norm over the interval $[- π, π]$ and using its sub-additivity, we get

‖ H_{r} \tilde{f} ‖_{L^{2} ([- π, π])}

\leq 2 π ‖ L_{N} \tilde{f} ‖_{L^{2} ([- π, π])} + 2^{5} {(\int_{- π}^{π} {| \int_{- π}^{π} k_{r} (x - y) \tilde{f} (y) d y |}^{2} d x)}^{\frac{1}{2}} .

Since $κ$ is supported in $[- 1, 1]$ , we have that $H_{r} f$ 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

‖ H_{r} f ‖_{L^{2} (R)} \leq ‖ H_{r} \tilde{f} ‖_{L^{2} ([- π, π])} \leq 2 π ‖ f ‖_{L^{2} (R)} + 2^{10} ‖ f ‖_{L^{2} (R)},

11.3.32

which gives 11.3.31.

Suppose now that $f$ is supported in $[c, c + 3]$ for some $c \in R$ . 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

‖ H_{r} g ‖_{2} = ‖ H_{r} f ‖_{2} \leq 2^{11} ‖ f ‖_{2} = ‖ g ‖_{2} .

11.3.33

Let now $f$ be arbitrary. Since $κ (x) = 0$ for $| x | > 1$ , we have for all $x \in [c + 1, c + 2]$

H_{r} f (x) = H_{r} (f 1_{[c, c + 3]}) (x) .

Thus

\int_{c + 1}^{c + 2} | H_{r} f (x) |^{2} d x \leq \int_{R} | H_{r} (f 1_{[c, c + 3]}) (x) |^{2} d x .

Applying the bound 11.3.33, this is

\leq 2^{11} \int_{c}^{c + 3} | f (x) |^{2} d x .

Summing over all $c \in Z$ , we obtain

\int_{R} | H_{r} f (x) |^{2} d x \leq 3 \cdot 2^{11} \int_{R} | f (x) |^{2} d x .

This completes the proof.

11.4 The proof of the van der Corput Lemma

Proof of Lemma 11.1.7 ▶

Let $g$ be a Lipschitz continuous function as in the lemma. Assume first that $n = 0$ . Then

\int_{α}^{β} g (x) d x \leq | β - α | sup_{α \leq x \leq β} | g (x) | \leq | β - α | ‖ g ‖_{L i p (α, β)} (1 + | n | | β - α |)^{- 1}

Assume now $n \neq 0$ . Without loss of generality, we may assume $n > 0$ . We distinguish two cases. If $β - α < \frac{π}{n}$ , we have by the triangle inequality

| \int_{α}^{β} g (x) e^{i n x} d x | \leq | β - α | sup_{x \in [α, β]} | g (x) | \leq 2 π | β - α | ‖ g ‖_{L i p (α, β)} (1 + | n | | β - α |)^{- 1} .

We turn to the case $\frac{π}{n} \leq β - α$ . We have

e^{i n (x + π / n)} = - e^{i n x} .

Using this, we write

\int_{α}^{β} g (x) e^{i n x} d x = \frac{1}{2} \int_{α}^{β} g (x) e^{i n x} d x - \frac{1}{2} \int_{α}^{β} g (x) e^{i n (x + π / n)}) d x .

We split the first integral at $α + \frac{π}{n}$ and the second one at $β - \frac{π}{n}$ , and make a change of variables in the second part of the first integral to obtain

= \frac{1}{2} \int_{α}^{α + \frac{π}{n}} g (x) e^{i n x} d x - \frac{1}{2} \int_{β - \frac{π}{n}}^{β} g (x) e^{i n (x + π / n)} d x

+ \frac{1}{2} \int_{α + \frac{π}{n}}^{β} (g (x) - g (x - \frac{π}{n})) e^{i n x} d x .

The sum of the first two terms is by the triangle inequality bounded by

\frac{π}{n} sup_{x \in [α, β]} | g (x) | .

The third term is by the triangle inequality at most

\frac{1}{2} \int_{α + \frac{π}{n}}^{β} | g (x) - g (x - \frac{π}{n}) | d x

\leq \frac{| β - α |}{2} \frac{π}{n} sup_{α \leq x < y \leq β} \frac{| g (x) - g (y) |}{| x - y |} .

Adding the two terms, we obtain

| \int_{α}^{β} g (x) e^{- i n x} d x | \leq \frac{π}{n} ‖ g ‖_{Lip (α, β)} .

This completes the proof of the lemma, using that with $\frac{π}{n} \leq β - α$ ,

\frac{π}{n} = \frac{2 π | β - α |}{2 n | β - α |} \leq 2 π | β - α | (1 + n | β - α |)^{- 1} .

11.5 Partial sums as orthogonal projections

This subsection proves Lemma 11.1.10

Proof of Lemma 11.1.10 ▶

The functions $e_{n} : x \mapsto e^{i n x}$ form an orthonormal basis in $L^{2} [- π, π]$ (this is already in Mathlib). Therefore we have

\begin{aligned} ‖ S_{N} f ‖_{L^{2} [- π, π]}^{2} & = ‖ \sum_{n = - N}^{N} ⟨ {\hat{f}}_{n}, e_{n} ⟩_{L^{2} [- π, π]} ‖_{L^{2} [- π, π]}^{2} \\ = \sum_{n = - N}^{N} | {\hat{f}}_{n} | \\ \leq \sum_{n \in Z} | {\hat{f}}_{n} | \\ = ‖ f ‖_{L^{2} [- π, π]} . \end{aligned}

This completes the proof of the lemma.

11.6 The error bound

Lemma 11.6.1 Dirichlet kernel - Hilbert kernel relation

✓

For all $N \in Z$ and $x \in [- π, π] ∖ {0}$ ,

| K_{N} (x) - (e^{- i N x} κ (x) + \overset{―}{e^{- i N x} κ (x)}) | \leq π .

Proof ▶

Let $N \in Z$ and $x \in [- π, π] ∖ {0}$ . With Lemma 11.1.8, we obtain

K_{N} (x) - (e^{- i N x} κ (x) + \overset{―}{e^{- i N x} κ (x)}) = e^{- i N x} \frac{min (| x |, 1)}{1 - e^{i x}} + e^{i N x} \frac{min (| x |, 1)}{1 - e^{- i x}} .

Using Lemma 11.1.9 with $η = min (| x |, 1)$ , we bound

| K_{N} (x) - (e^{- i N x} κ (x) + \overset{―}{e^{- i N x} κ (x)}) | \leq \frac{min (| x |, 1)}{| 1 - e^{i x} |} + \frac{min (| x |, 1)}{| 1 - e^{- i x} |} \leq \frac{π}{2} + \frac{π}{2} = π .

Lemma 11.6.2 partial Fourier sum bound

✓

Let $g : R \to C$ be a measurable $2 π$ -periodic function such that for some $δ > 0$ and every $x \in R$ ,

| g (x) | \leq δ .

11.6.1

Then for every $x \in [0, 2 π]$ and $N > 0$ ,

| S_{N} g (x) | \leq \frac{1}{2 π} (T g (x) + T \bar{g} (x)) + π δ .

Proof ▶

Let $x \in [0, 2 π]$ and $N > 0$ . We have with Lemma 11.1.8

| S_{N} g (x) | = \frac{1}{2 π} | \int_{0}^{2 π} g (y) K_{N} (x - y) d y | .

We use $2 π$ -periodicity of $g$ and $K_{N}$ to shift the domain of integration to obtain

= \frac{1}{2 π} | \int_{x - π}^{x + π} g (y) K_{N} (x - y) d y | .

Using the triangle inequality, we split this as

\leq \frac{1}{2 π} | \int_{x - π}^{x + π} g (y) (K_{N} (x - y) - max (| x - y |, 0) K_{N} (x - y)) d y |

11.6.2

+ \frac{1}{2 π} | \int_{x - π}^{x + π} g (y) max (| x - y |, 0) K_{N} (x - y) d y | .

11.6.3

Note that all integrals are well defined, since $K_{N}$ is by 11.1.22 bounded by $2 N + 1$ . Using that

max (| x - y |, 0) K_{N} (x - y) = e^{- i N (x - y)} κ (x - y) + \overset{―}{e^{- i N (x - y)} κ (x - y)},

11.6.4

Lemma 11.6.1 and 11.6.5, we bound 11.6.2 by

\frac{1}{2 π} \int_{x - π}^{x + π} | g (y) | | K_{N} (x - y) - e^{- i N (x - y)} κ (x - y) + \overset{―}{e^{- i N (x - y)} κ (x - y)} | d y \leq π δ .

By dominated convergence and since $κ (x - y) = 0$ for $| x - y | > 1$ , 11.6.3 equals

\frac{1}{2 π} lim_{r \to 0^{+}} | \int_{r < | x - y | < 1} g (y) max (| x - y |, 0) K_{N} (x - y) d y | .

We bound the limit by a supremum and rewrite using 11.6.4,

\leq \frac{1}{2 π} sup_{r > 0} | \int_{r < | x - y | < 1} g (y) (e^{- i N (x - y)} κ (x - y) + \overset{―}{e^{- i N (x - y)} κ (x - y)}) d y |

Using the triangle inequality, we further bound this by

\begin{aligned} \leq & \frac{1}{2 π} sup_{r > 0} | \int_{r < | x - y | < 1} g (y) e^{- i N y} κ (x - y) d y | \\ + & \frac{1}{2 π} sup_{r > 0} | \int_{r < | x - y | < 1} \overset{―}{g} (y) e^{- i N y} κ (x - y) d y | . \end{aligned}

By the definition 11.1.16 of $T$ , this is

\leq \frac{1}{2 π} (T g (x) + T \bar{g} (x)) .

Lemma 11.6.3 real Carleson operator measurable

✓

Let $f$ be a bounded measurable function on $R$ . Then $T f$ 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 $n \in Z$ ,

x \mapsto sup_{r > 0} | \int_{r < | x - y | < 1} f (y) κ (x - y) e^{i n y} d y |

is measurable. So let $n \in Z$ . Note that for each $x \in R$ , the function

r \mapsto | \int_{r < | x - y | < 1} f (y) κ (x - y) e^{i n y} d y |

is continuous on $(0, \infty)$ since the integrand is locally bounded on the domain $0 < | x - y | < 1$ by the assumptions on $f$ and Lemma 11.1.11. Thus, for each $x \in R$ ,

sup_{r > 0} | \int_{r < | x - y | < 1} f (y) κ (x - y) e^{i n y} d y | = sup_{r \in Q_{> 0}} | \int_{r < | x - y | < 1} f (y) κ (x - y) e^{i n y} d y |

The right hand side is again a countable supremum so it remains to show that for every $r \in Q_{> 0}$ ,

x \mapsto | \int_{r < | x - y | < 1} f (y) κ (x - y) e^{i n y} d y | = | \int 1_{{r < | x - \cdot | < 1}} (y) f (y) κ (x - y) e^{i n y} d y |

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 $g : R \to C$ be a measurable $2 π$ -periodic function such that for some $δ > 0$ and every $x \in R$ ,

| g (x) | \leq δ .

11.6.5

Then for every $ϵ > 0$ , there exists a measurable set $E \subset [0, 2 π]$ with $| E | < ϵ$ such that for every $x \in [0, 2 π] ∖ E$ and $N > 0$ ,

| S_{N} g (x) | \leq C_{ϵ} δ,

11.6.6

where

C_{ϵ} = {(\frac{8}{π ϵ})}^{\frac{1}{2}} C_{4, 2} + π .

11.6.7

Proof ▶

Define

E := {x \in [0, 2 π] : sup_{N > 0} | S_{N} g (x) | > C_{ϵ} δ} .

Then 11.6.6 clearly holds, and it remains to show that $| E | \leq ϵ$ . Using Lemma 11.6.2, we obtain

E \subset {x \in [0, 2 π] : C_{ϵ} δ < \frac{1}{2 π} (T g (x) + T \bar{g} (x)) + π δ} \subset E_{1} \cup E_{2},

where

\begin{aligned} E_{1} := & {x \in [0, 2 π] : π (C_{ϵ} - π) δ < T g (x)} \\ E_{2} := & {x \in [0, 2 π] : π (C_{ϵ} - π) δ < T \bar{g} (x)} . \end{aligned}

By Lemma 11.6.3, $E_{1}$ and $E_{2}$ are measurable. Thus,

π (C_{ϵ} - π) δ | E_{1} | \leq \int_{E_{1}} T g (x) d x = δ \int_{E_{1}} T (δ^{- 1} g 1_{[- π, 3 π]}) (x) d x .

Applying Lemma 11.1.5 with $F = [- π, 3 π]$ and $G = E^{'}$ , it follows that this is

\leq δ \cdot C_{4, 2} | F |^{\frac{1}{2}} | E_{1} |^{\frac{1}{2}} \leq (4 π)^{\frac{1}{2}} C_{4, 2} δ \cdot | E^{'} |^{\frac{1}{2}} .

Rearranging, we obtain

| E_{1} | \leq {(\frac{(4 π)^{\frac{1}{2}} C_{4, 2}}{π (C_{ϵ} - π)})}^{2} = \frac{ϵ}{2} .

Analogously, we get the same estimate for $| E_{2} |$ . This completes the proof using $| E | \leq | E_{1} | + | E_{2} |$ .

Proof of Lemma 11.1.3 ▶

Lemma 11.1.3 now follows directly from Lemma 11.6.4 with $δ := ϵ^{'}$ .

11.7 Carleson on the real line

We prove Lemma 11.1.5.

Consider the standard distance function

ρ (x, y) = | x - y |

11.7.1

on the real line $R$ .

Lemma 11.7.1 real line metric

✓

The space $(R, ρ)$ is a complete locally compact metric space.

Proof ▶

Lemma 11.7.2 real line ball

✓

For $x \in R$ and $R > 0$ , the ball $B (x, R)$ is the interval $(x - R, x + R)$

Proof ▶

Let $x^{'} \in B (x, R)$ . By definition of the ball, $| x^{'} - x | < R$ . It follows that $x^{'} - x < R$ and $x - x^{'} < R$ . It follows $x^{'} < x + R$ and $x^{'} > x - R$ . This implies $x^{'} \in (x - R, x + R)$ . Conversely, let $x^{'} \in (x - R, x + R)$ . Then $x^{'} < x + R$ and $x^{'} > x - R$ . It follows that $x^{'} - x < R$ and $x - x^{'} < R$ . It follows that $| x^{'} - x | < R$ , hence $x^{'} \in B (x, R)$ . This proves the lemma.

We consider the Lebesgue measure $μ$ on $R$ .

Lemma 11.7.3 real line measure

✓

The measure $μ$ is a sigma-finite non-zero Radon-Borel measure on $R$ .

Proof ▶

Lemma 11.7.4 real line ball measure

✓

We have for every $x \in R$ and $R > 0$

μ (B (x, R)) = 2 R .

11.7.2

Proof ▶

We have with Lemma 11.7.2

μ (B (x, R)) = μ ((x - R, x + R)) = 2 R .

11.7.3

Lemma 11.7.5 real line doubling

✓

We have for every $x \in R$ and $R > 0$

μ (B (x, 2 R)) = 2 μ (B (x, R)) .

11.7.4

Proof ▶

We have with Lemma 11.7.4

μ (B (x, 2 R) = 4 R = 2 μ (B (x, R)) .

11.7.5

This proves the lemma.

The preceding four lemmas show that $(R, ρ, μ, 4)$ is a doubling metric measure space. Indeed, we even show that $(R, ρ, μ, 1)$ is a doubling metric measure space, but we may relax the estimate in Lemma 11.7.5 to conclude that $(R, ρ, μ, 4)$ is a doubling metric measure space.

For each $n \in Z$ define $ϑ_{n} : R \to R$ by

ϑ_{n} (x) = n x .

11.7.6

Let $Θ$ be the collection ${ϑ_{n}, n \in Z}$ . Note that for every $n \in Z$ we have $ϑ_{n} (0) = 0$ . Define

d_{B (x, R)} (ϑ_{n}, ϑ_{m}) := 2 R | n - m | .

11.7.7

Lemma 11.7.6 frequency metric

✓

For every $R > 0$ and $x \in X$ , the function $d_{B (x, R)}$ is a metric on $Θ$ .

Proof ▶

This follows immediately from the fact that the standard metric on $Z$ is a metric.

Lemma 11.7.7 oscillation control

✓

For every $R > 0$ and $x \in X$ , and for all $n, m \in Z$ , we have

sup_{y, y^{'} \in B (x, R)} | n y - n y^{'} - m y + m y^{'} | \leq 2 | n - m | R .

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 $R, R^{'} > 0$ with $B (x, R) \subset B (x, R^{'})$ , and for any $n, m \in Z$

d_{B (x, R)} (ϑ_{n}, ϑ_{m}) \leq d_{B (x^{'}, R^{'})} (ϑ_{n}, ϑ_{m}) .

Proof ▶

This follows immediately from the definition 11.7.7 and $R \leq R^{'}$ .

Lemma 11.7.9 frequency ball doubling

✓

For any $x, x^{'} \in R$ and $R > 0$ with $x \in B (x^{'}, 2 R)$ and any $n, m \in Z$ , we have

d_{B (x^{'}, 2 R)} (ϑ_{n}, ϑ_{m}) \leq 2 d_{B (x, R)} (ϑ_{n}, ϑ_{m}) .

11.7.9

Proof ▶

With 11.7.7, both sides of 11.7.9 are equal to $4 R | n - m |$ . This proves the lemma.

Lemma 11.7.10 frequency ball growth

✓

For any $x, x^{'} \in R$ and $R > 0$ with $B (x, R) \subset B (x^{'}, 2 R)$ and any $n, m \in Z$ , we have

2 d_{B (x, R)} (ϑ_{n}, ϑ_{m}) \leq d_{B (x^{'}, 2 R)} (ϑ_{n}, ϑ_{m}) .

11.7.10

Proof ▶

With 11.7.7, both sides of 11.7.9 are equal to $4 R | n - m |$ . This proves the lemma.

Lemma 11.7.11 integer ball cover

✓

For every $x \in R$ and $R > 0$ and every $n \in Z$ and $R^{'} > 0$ , there exist $m_{1}, m_{2}, m_{3} \in Z$ such that

B^{'} \subset B_{1} \cup B_{2} \cup B_{3},

11.7.11

where

B^{'} = {ϑ \in Θ : d_{B (x, R)} (ϑ, ϑ_{n}) < 2 R^{'}}

11.7.12

and for $j = 1, 2, 3$

B_{j} = {ϑ \in Θ : d_{B (x, R)} (ϑ, ϑ_{m_{j}}) < R^{'}} .

11.7.13

Proof ▶

Let $m_{1}$ be the largest integer smaller than or equal to $n - \frac{R^{'}}{2 R}$ . Let $m_{2} = n$ . Let $m_{3}$ be the smallest integer larger than or equal to $n + \frac{R^{'}}{2 R}$ .

Let $ϑ_{n^{'}} \in B^{'}$ , then with 11.7.7, we have

2 R | n - n^{'} | < 2 R^{'} .

11.7.14

Assume first $n^{'} \leq m_{1}$ . With 11.7.14 we have

R | m_{1} - n^{'} | = R (m_{1} - n^{'}) = R (m_{1} - n) + R (n - n^{'})

< - \frac{R^{'}}{2} + R^{'} = - \frac{R^{'}}{2} .

11.7.15

We conclude $ϑ_{n^{'}} \in B_{1}$ .

Assume next $m_{1} < n^{'} < m_{3}$ . Then $ϑ_{n^{'}} \in B_{2}$ .

Assume finally that $m_{3} \leq n^{'}$ . With 11.7.14 we have

R | m_{3} - n^{'} | = R (n^{'} - m_{3}) = R (n^{'} - n) + R (n - m_{3})

< R^{'} - \frac{R^{'}}{2} = - \frac{R^{'}}{2} .

11.7.16

We conclude $ϑ_{n^{'}} \in B_{1}$ . This completes the proof of the lemma.

Lemma 11.7.12 real van der Corput

✓

For any $x \in R$ and $R > 0$ and any function $φ : X \to C$ supported on $B^{'} = B (x, R)$ such that

‖ φ ‖_{Lip (B^{'})} = sup_{x \in B^{'}} | φ (x) | + R sup_{x, y \in B^{'}, x \neq y} \frac{| φ (x) - φ (y) |}{ρ (x, y)}

11.7.17

is finite and for any $n, m \in Z$ , we have

| \int_{B^{'}} e (ϑ_{n} (x) - ϑ_{m} (x)) φ (x) d μ (x) | \leq 2 π μ (B^{'}) \frac{‖ φ ‖_{Lip (B^{'})}}{1 + d_{B^{'}} (ϑ_{n}, ϑ_{m})} .

11.7.18

Proof ▶

Set $n^{'} = n - m$ . Then we have to prove

| \int_{x - R}^{x + R} e^{i n^{'} y} φ (y) d y | \leq 4 π R ‖ φ ‖_{Lip (B^{'})} (1 + 2 R | n^{'} |)^{- 1} .

11.7.19

This follows from Lemma 11.1.7 with $α = x - R$ and $β = x + R$ .

Proof of Lemma 11.1.5 ▶

The preceding chain of lemmas establishes that $Θ$ is a cancellative, compatible collection of functions on $(R, ρ, μ, 4)$ . 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 $κ$ as near 11.1.13, define the function $K : R \times R \to C$ as in Theorem 1.0.2 by

K (x, y) := κ (x - y) .

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 $(R, ρ, μ, 4)$ . 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. □