# Singular integral operators and the Riesz transform

Jordan Bell
November 17, 2017

## 1 Calderón-Zygmund kernels

Let $\omega_{n-1}$ be the measure of $S^{n-1}$. It is

 $\omega_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}.$

Let $v_{n}$ be the measure of the unit ball in $\mathbb{R}^{n}$. It is

 $v_{n}=\frac{\omega_{n-1}}{n}=\frac{2\pi^{n/2}}{n\Gamma(n/2)}.$

For $k,N\geq 0$ and $\phi\in C^{\infty}(\mathbb{R}^{n})$ let

 $p_{k,N}(\phi)=\max_{|\alpha|\leq k}\sup_{x\in\mathbb{R}^{n}}(1+|x|)^{N}|(% \partial^{\alpha}\phi)(x)|.$

A Borel measurable function $K:\mathbb{R}^{n}\setminus\{0\}\to\mathbb{C}$ is called a Calderón-Zygmund kernel if there is some $B$ such that

1. 1.

$|K(x)|\leq B|x|^{-n}$, $x\neq 0$

2. 2.

$\int_{|x|\geq 2|y|}|K(x-y)-K(x)|dx\leq B$, $y\neq 0$

3. 3.

$\int_{R_{1}<|x|, $0.

The following lemma gives a tractable condition under which Condition 2 is satisfied.11 1 Camil Muscalu and Wilhelm Schlag, Classical and Multilinear Harmonic Analysis, volume I, p. 167, Lemma 7.2.

###### Lemma 1.

If $|(\nabla K)(x)|\leq C|x|^{-n-1}$ for all $x\neq 0$ then for $y\neq 0$,

 $\int_{|x|\geq 2|y|}|K(x-y)-K(x)|dx\leq v_{n}2^{n}C.$
###### Proof.

For $|x|>2|y|>0$, if $0\leq t\leq 1$ then

 $|x-ty|\geq|x|-t|y|\geq|x|-|y|>|x|-\frac{|x|}{2}=\frac{|x|}{2}.$

Write $f(t)=K(x-ty)$, for which $f^{\prime}(t)=-(\nabla K)(x-ty)\cdot y$. By the fundamental theorem of calculus,

 $K(x-y)-K(x)=f(1)-f(0)=\int_{0}^{1}f^{\prime}(t)dt=-\int_{0}^{1}(\nabla K)(x-ty% )\cdot ydt,$

thus

 $|K(x-y)-K(x)|\leq\int_{0}^{1}|(\nabla K)(x-ty)||y|dt\leq C|y|\int_{0}^{1}|x-ty% |^{-n-1}dt.$

Then using $|x-ty|>\frac{|x|}{2}$,

 $|K(x-y)-K(x)|\leq C|y|\left(\frac{|x|}{2}\right)^{-n-1}=2^{n+1}C|y||x|^{-n-1}.$

For $|y|>0$, using spherical coordinates,2

 $\displaystyle\int_{|x|\geq 2|y|}|K(x-y)-K(x)|dx$ $\displaystyle\leq\int_{|x|\geq 2|y|}2^{n+1}C|y||x|^{-n-1}dx$ $\displaystyle=2^{n+1}C|y|\int_{2|y|}^{\infty}\left(\int_{S^{n-1}}|r\gamma|^{-n% -1}d\sigma(\gamma)\right)r^{n-1}dr$ $\displaystyle=v_{n}2^{n+1}C|y|\int_{2|y|}^{\infty}r^{-2}dr$ $\displaystyle=v_{n}2^{n+1}C|y|\cdot\frac{1}{2|y|}$ $\displaystyle=v_{n}2^{n}C.$

For a Calderón-Zygmund kernel $K$, for $f\in\mathscr{S}(\mathbb{R}^{n})$, for $x\in\mathbb{R}^{n}$, and for $\epsilon>0$, using Condition 3 with $R_{1}=\epsilon$ and $R_{2}=1$,33 3 https://math.aalto.fi/~parissi1/notes/harmonic.pdf, p. 115, Lemma 6.15.

 $\begin{split}&\displaystyle\int_{|x-y|\geq\epsilon}K(x-y)f(y)dy\\ \displaystyle=&\displaystyle\int_{\epsilon\leq|x-y|\leq 1}K(x-y)(f(y)-f(x))dy+% \int_{|x-y|\geq 1}K(x-y)f(y)dy.\end{split}$

By Condition 1 there is some $B$ such that $|K(x)|\leq B|x|^{-n}$, and combining this with $|f(y)-f(x)|\leq\left\|\nabla f\right\|_{\infty}|y-x|$,

 $|K(x-y)(f(y)-f(x))|\leq B\left\|\nabla f\right\|_{\infty}|y-x|^{-n+1},$

which is integrable on $\{|x-y|\leq 1\}$. Then by the dominated convergence theorem,

 $\lim_{\epsilon\to 0}\int_{\epsilon\leq|x-y|\leq 1}K(x-y)(f(y)-f(x))dy=\int_{|x% -y|\leq 1}K(x-y)(f(y)-f(x))dy.$
###### Lemma 2.

For a Calderón-Zygmund kernel $K$, for $f\in\mathscr{S}(\mathbb{R}^{n})$, and for $x\in\mathbb{R}^{n}$, the limit

 $\lim_{\epsilon\to 0}\int_{|x-y|\geq\epsilon}K(x-y)f(y)dy$

exists.

## 2 Singular integral operators

For a Calderón-Zygmund kernel $K$ on $\mathbb{R}^{n}$, for $f\in\mathscr{S}(\mathbb{R}^{n})$, and for $x\in\mathbb{R}^{n}$, let

 $(Tf)(x)=\lim_{\epsilon\to 0}\int_{|x-y|\geq\epsilon}K(x-y)f(y)dy.$

We call $T$ a singular integral operator. By Lemma 2 this makes sense.

We prove that singular integral operators are $L^{2}\to L^{2}$ bounded.44 4 Camil Muscalu and Wilhelm Schlag, Classical and Multilinear Harmonic Analysis, volume I, p. 168, Proposition 7.3; Elias M. Stein, Singular Integrals and Differentiability Properties of Functions, p. 35, §3.2, Theorem 2; http://math.uchicago.edu/~may/REU2013/REUPapers/Talbut.pdf

###### Theorem 3.

There is some $C_{n}$ such that for any Calderón-Zygmund kernel $K$ and any $f\in\mathscr{S}(\mathbb{R}^{n})$,

 $\left\|Tf\right\|_{2}\leq C_{n}B\left\|f\right\|_{2}.$
###### Proof.

For $0 and for $\xi\in\mathbb{R}^{n}$ define

 $\displaystyle m_{r,s}(\xi)$ $\displaystyle=\int_{\mathbb{R}^{n}}e^{-2\pi ix\cdot\xi}1_{r<|x|

Take $r<|\xi|^{-1}, for which

 $m_{r,s}(\xi)=\int_{r<|x|<|\xi|^{-1}}e^{-2\pi ix\cdot\xi}K(x)dx+\int_{|\xi|^{-1% }<|x|

For the first integral, using Condition 3 with $R_{1}=r$ and $R_{2}=|\xi|^{-1}$ and then using Condition 1,

 $\displaystyle\left|\int_{r<|x|<|\xi|^{-1}}e^{-2\pi ix\cdot\xi}K(x)dx\right|$ $\displaystyle=\left|\int_{r<|x|<|\xi|^{-1}}(e^{-2\pi ix\cdot\xi}-1)K(x)dx\right|$ $\displaystyle\leq\int_{|x|<|\xi|^{-1}}|e^{-2\pi ix\cdot\xi}-1||K(x)|dx$ $\displaystyle\leq\int_{|x|<|\xi|^{-1}}2\pi|x||\xi||K(x)|dx$ $\displaystyle\leq 2\pi|\xi|\int_{|x|<|\xi|^{-1}}B|x|^{-n+1}dx$ $\displaystyle=2\pi|\xi|\cdot v_{n}|\xi|^{-1}.$

For the second integral, let $z=\frac{\xi}{2|\xi|^{2}}$, and

 $\displaystyle\int_{|\xi|^{-1}<|x| $\displaystyle=-\int_{|\xi|^{-1}<|x| $\displaystyle=-\int_{|\xi|^{-1}<|x-z|

Let

 $\displaystyle R$ $\displaystyle=\int_{|\xi|^{-1}<|x| $\displaystyle=-\int_{|\xi|^{-1}<|x+z|

with which

 $\int_{|\xi|^{-1}<|x|

On the one hand, applying Condition 2, as $|z|=\frac{1}{2|\xi|}$,

 $\displaystyle\left|\int_{|\xi|^{-1}<|x| $\displaystyle\leq\int_{|x|>|\xi|^{-1}}|K(x)-K(x-z)|dx$ $\displaystyle=\int_{|x|>2|z|}|K(x)-K(x-z)|dx$ $\displaystyle\leq B.$

On the other hand, let

 $D=D_{1}\triangle D_{2}=\{x:|\xi|^{-1}<|x+z|

For $x\in D_{1}$ we have

 $|x|\geq|x+z|-|z|>\frac{1}{|\xi|}-\frac{1}{2|\xi|}=\frac{1}{2|\xi|},$

and for $x\in D_{2}$ we have $|x|>\frac{1}{|\xi|}$, so for $x\in D$,

 $|x|>\frac{1}{2|\xi|}.$

Applying Condition 1,

 $|K(x)|\leq B|x|^{-n}<2^{n}B|\xi|^{n}.$

Furthermore, for $x\in D_{1}\setminus D_{2}$ we have $|x|\leq|\xi|^{-1}$, and for $x\in D_{2}\setminus D_{1}$ we have

 $|x|\leq|x+z|+|z|=|x+z|+\frac{1}{2|\xi|}\leq\frac{1}{|\xi|}+\frac{1}{2|\xi|}=% \frac{3}{2|\xi|}.$

Hence

 $D\subset\left\{x:\frac{1}{2|\xi|}<|x|\leq\frac{3}{2|\xi|}\right\},$

so

 $\lambda(D)\leq\left(\frac{3}{2|\xi|}\right)^{n}v_{n}=\left(\frac{3}{2}\right)^% {n}|\xi|^{-n}v_{n}.$

Therefore

 $|R|\leq 2^{n}B|\xi|^{n}\cdot\left(\frac{3}{2}\right)^{n}|\xi|^{-n}v_{n}=3^{n}v% _{n}B$

and then

 $\left|\int_{|\xi|^{-1}<|x|

and finally55 5 The way I organize the argument, I want to use $\left\|m_{r,s}\right\|_{\infty}\leq C_{n}B$, while we have only obtained this bound for $r<|\xi|^{-1}. To make the argument correct I may need to do things in a different order, e.g. apply Fatou’s lemma and then use an inequality instead of using an inequality and then apply Fatou’s lemma.

 $|m_{r,s}(\xi)|\leq 2\pi v_{n}+\frac{1}{2}B+\frac{1}{2}\cdot 3^{n}v_{n}B=C_{n}B.$

Define

 $(T_{r,s}f)(x)=\int_{\mathbb{R}^{n}}1_{r<|y|

Then

 $\displaystyle\widehat{T_{r,s}f}(\xi)$ $\displaystyle=\int_{\mathbb{R}^{n}}e^{-2\pi ix\cdot\xi}\left(\int_{\mathbb{R}^% {n}}1_{r<|y| $\displaystyle=\int_{\mathbb{R}^{n}}1_{r<|y| $\displaystyle=\int_{\mathbb{R}^{n}}1_{r<|y| $\displaystyle=m_{r,s}(\xi)\widehat{f}(\xi),$

and so

 $\left\|\widehat{T_{r,s}f}\right\|_{2}^{2}=\int_{\mathbb{R}^{n}}|\widehat{T_{r,% s}f}(\xi)|^{2}d\xi=\int_{\mathbb{R}^{n}}|m_{r,s}(\xi)\widehat{f}(\xi)|^{2}d\xi% \leq\left\|m_{r,s}\right\|_{\infty}^{2}\left\|\widehat{f}\right\|_{2}^{2},$

by Plancherel’s theorem and the inequality we got for $|m_{r,s}(\xi)|$,

 $\left\|T_{r,s}f\right\|_{2}^{2}\leq\left\|m_{r,s}\right\|_{\infty}^{2}\left\|f% \right\|_{2}^{2}\leq(C_{n}B)^{2}\left\|f\right\|_{2}^{2}.$

For each $x\in\mathbb{R}^{n}$, $(T_{r,s}f)(x)\to(Tf)(x)$ as $r\to 0$ and $s\to\infty$, and thus using Fatou’s lemma,

 $\int_{\mathbb{R}^{n}}|(Tf)(x)|^{2}dx\leq\liminf_{r\to 0,s\to\infty}\int_{% \mathbb{R}^{n}}|(T_{r,s}f)(x)|^{2}dx=(C_{n}B)^{2}\left\|f\right\|_{2}^{2}.$

That is,

 $\left\|Tf\right\|_{2}\leq C_{n}B\left\|f\right\|_{2}.$

## 3 The Riesz transform

Let

 $c_{n}=\frac{1}{\pi v_{n-1}}=\frac{\Gamma\left(\frac{n+1}{2}\right)}{\pi^{(n+1)% /2}}.$

For $1\leq j\leq n$, let

 $K_{j}(x)=c_{n}\frac{x_{j}}{|x|^{n+1}}.$

This is a Calderón-Zygmund kernel. For $\phi\in\mathscr{S}(\mathbb{R}^{n})$ define

 $(R_{j}\phi)(x)=\lim_{\epsilon\to 0}\int_{|y-x|\geq\epsilon}K_{j}(x-y)\phi(y)dy% =\lim_{\epsilon\to 0}\int_{|y|\geq\epsilon}K_{j}(y)\phi(x-y)dy.$

We call each $R_{j}$, $1\leq j\leq n$, a Riesz transform.

For $1\leq j\leq n$ define $W_{j}:\mathscr{S}(\mathbb{R}^{n})\to\mathbb{C}$ by

 $\left\langle W_{j},\phi\right\rangle=\frac{\Gamma\left(\frac{n+1}{2}\right)}{% \pi^{\frac{n+1}{2}}}\lim_{\epsilon\to 0}\int_{|y|\geq\epsilon}K_{j}(y)\phi(y)dy.$ (1)

For $\epsilon>0$,

 $\displaystyle\left|\int_{\epsilon\leq|y|\leq 1}K_{j}(y)\phi(y)dy\right|$ $\displaystyle=\left|\int_{\epsilon\leq|y|\leq 1}K_{j}(y)(\phi(y)-\phi(0))dy\right|$ $\displaystyle\leq\int_{\epsilon\leq|y|\leq 1}c_{n}|y|^{-n}\cdot\left\|\nabla% \phi\right\|_{\infty}|y|dy$ $\displaystyle=c_{n}\left\|\nabla\phi\right\|_{\infty}\omega_{n-1}\int_{% \epsilon}^{1}r^{-n+1}\cdot r^{n-1}dr$ $\displaystyle=c_{n}\left\|\nabla\phi\right\|_{\infty}\omega_{n-1}(1-\epsilon).$

For $|y|\geq 1$,

 $\displaystyle\int_{|y|\geq 1}|K_{j}(y)\phi(y)|dy$ $\displaystyle\leq c_{n}\int_{|y|\geq 1}|y|^{-n}(1+|y|^{2})^{-1/2}p_{0,1}(\phi)dy$ $\displaystyle=c_{n}\omega_{n}\int_{1}^{\infty}r^{-n}(1+r)^{-1}\cdot r^{n-1}dr$ $\displaystyle=c_{n}\omega_{n}\log 2.$

It then follows from the dominated convergence theorem that the limit

 $\lim_{\epsilon\to 0}\int_{|y|\geq\epsilon}K_{j}(y)\phi(y)dy$

exists, which shows that the definition (1) makes sense. It is apparent that $W_{j}$ is linear. Then prove that if $\phi_{k}\to\phi$ in $\mathscr{S}(\mathbb{R}^{n})$ then $\left\langle W_{j},\phi_{k}\right\rangle\to\left\langle W_{j},\phi\right\rangle$. This being true means that $W_{j}\in\mathscr{S}^{\prime}(\mathbb{R}^{n})$, namely that each $W_{j}$ is a tempered distribution.

For a function $f:\mathbb{R}^{n}\to\mathbb{C}$, write

 $\widetilde{f}(x)=f(-x),\qquad(\tau_{y}f)(x)=f(x-y).$

For $u\in\mathscr{S}^{\prime}(\mathbb{R}^{n})$ and $h\in\mathscr{S}(\mathbb{R}^{n})$, define

 $\left\langle h*u,\phi\right\rangle=\left\langle u,\widetilde{h}*\phi\right% \rangle,\qquad\phi\in\mathscr{S}(\mathbb{R}^{n}).$

It is a fact that $h*u\in\mathscr{S}^{\prime}(\mathbb{R}^{n})$, and this tempered distribution is induced by the $C^{\infty}$ function $x\mapsto\left\langle u,\tau_{x}\widetilde{h}\right\rangle$.66 6 Loukas Grafakos, Classical Fourier Analysis, second ed., p. 116, Theorem 2.3.20. The Fourier transform of a tempered distribution $u$ is defined by

 $\left\langle\widehat{u},\phi\right\rangle=\left\langle u,\widehat{\phi}\right% \rangle,\qquad\phi\in\mathscr{S}(\mathbb{R}^{n}),$

where

 $\widehat{\phi}(\xi)=\int_{\mathbb{R}^{n}}e^{-2\pi ix\cdot\xi}\phi(x)dx,\qquad% \xi\in\mathbb{R}^{n}.$

It is a fact that $\widehat{u}$ is itself a tempered distribution. Finally, for a tempered distribution $u$ and a Schwartz function $h$, we define

 $\left\langle hu,\phi\right\rangle=\left\langle u,h\phi\right\rangle,\qquad\phi% \in\mathscr{S}(\mathbb{R}^{n}).$

It is a fact that $hu$ is itself a tempered distribution. It is proved that77 7 Loukas Grafakos, Classical Fourier Analysis, second ed., p. 120, Proposition 2.3.22.

 $\widehat{\phi*u}=\widehat{\phi}\widehat{u}.$

The left-hand side is the Fourier transform of the tempered distribution $\phi*u$, and the right-hand side is the product of the Schwartz function $\widehat{\phi}$ and the tempered distribution $\widehat{u}$.

###### Lemma 4.

For $1\leq j\leq n$, for $\phi\in\mathscr{S}(\mathbb{R}^{n})$, and for $x\in\mathbb{R}^{n}$,

 $(R_{j}\phi)(x)=(\phi*W_{j})(x).$

We will use the following identity for integrals over $S^{n-1}$.88 8 Loukas Grafakos, Classical Fourier Analysis, second ed., p. 261, Lemma 4.1.15.

###### Lemma 5.

For $\xi\neq 0$ and for $1\leq j\leq n$,

 $\int_{S^{n-1}}\mathrm{sgn}\,(\xi\cdot\theta)\theta_{j}d\sigma(\theta)=\frac{2% \omega_{n-2}}{n-1}\frac{\xi_{j}}{|\xi|}.$
###### Proof.

It is a fact that

 $\int_{S^{n-1}}\mathrm{sgn}\,(\theta_{k})\theta_{j}d\sigma(\theta)=\begin{cases% }0&k\neq j\\ \int_{S^{n-1}}|\theta_{j}|d\sigma(\theta)&k=j.\end{cases}$ (2)

It suffices to prove the claim when $\xi\in S^{n-1}$. For $1\leq j\leq n$ there is $A_{j}=(a_{i,k})_{i,k}\in SO_{n}(\mathbb{R})$ such that9

 $A_{j}e_{j}=\xi,$

for which $a_{i,j}=\xi_{i}$. Using that $A_{j}^{T}=A_{j}^{-1}$ and that $\sigma$ is invariant under $O(n)$ we calculate

 $\displaystyle\int_{S^{n-1}}\mathrm{sgn}\,(\xi\cdot\theta)\theta_{j}d\sigma(\theta)$ $\displaystyle=\int_{S^{n-1}}\mathrm{sgn}\,(A_{j}e_{j}\cdot\theta)(AA^{-1}% \theta)_{j}d\sigma(\theta)$ $\displaystyle=\int_{S^{n-1}}\mathrm{sgn}\,(e_{j}\cdot A_{j}^{-1}\theta)(AA^{-1% }\theta)_{j}d\sigma(\theta)$ $\displaystyle=\int_{S^{n-1}}\mathrm{sgn}\,(e_{j}\cdot\theta)(A\theta)_{j}d({A_% {j}^{-1}}_{*}\sigma)(\theta)$ $\displaystyle=\int_{S^{n-1}}\mathrm{sgn}\,(e_{j}\cdot\theta)(A\theta)_{j}d% \sigma(\theta)$ $\displaystyle=\int_{S^{n-1}}\mathrm{sgn}\,(\theta_{j})\sum_{k=1}^{n}a_{j,k}% \theta_{k}d\theta.$

Applying Lemma 2 and $a_{j,j}=\xi_{j}$, this becomes

 $\int_{S^{n-1}}\mathrm{sgn}\,(\xi\cdot\theta)\theta_{j}d\sigma(\theta)=\int_{S^% {n-1}}\xi_{j}|\theta_{j}|d\sigma(\theta)=\frac{\xi_{j}}{|\xi|}\int_{S^{n-1}}|% \theta_{j}|d\sigma(\theta).$

Hence for each $1\leq j\leq n$,

 $\int_{S^{n-1}}\mathrm{sgn}\,(\xi\cdot\theta)\theta_{j}d\sigma(\theta)=\frac{% \xi_{j}}{|\xi|}\int_{S^{n-1}}|\theta_{1}|d\sigma(\theta).$

It is a fact that1010 10 Loukas Grafakos, Classical Fourier Analysis, second ed., p. 441, Appendix D.2.

 $\int_{RS^{n-1}}f(\theta)d\sigma(\theta)=\int_{-R}^{R}\int_{\sqrt{R^{2}-s^{2}}S% ^{n-2}}f(s,\phi)d\phi\frac{Rds}{\sqrt{R^{2}-s^{2}}}.$

Using this with $f(\theta)=f(\theta_{1},\ldots,\theta_{n})=|\theta_{1}|$ and using that the measure of $RS^{n-2}$ is $R^{n-2}\omega_{n-1}$, we calculate

 $\displaystyle\int_{S^{n-1}}|\theta_{1}|d\sigma(\theta)$ $\displaystyle=\int_{-1}^{1}\int_{\sqrt{1-s^{2}}S^{n-2}}|s|d\phi\frac{ds}{\sqrt% {1-s^{2}}}$ $\displaystyle=\int_{-1}^{1}(1-s^{2})^{\frac{n-2}{2}-\frac{1}{2}}\omega_{n-2}|s% |d\phi$ $\displaystyle=2\omega_{n-2}\int_{0}^{1}(1-s^{2})^{\frac{n-3}{2}}sds$ $\displaystyle=\omega_{n-2}\int_{0}^{1}u^{\frac{n-3}{2}}du$ $\displaystyle=\frac{2\omega_{n-2}}{n-1}.$

We now calculate the Fourier transform of the $W_{j}$. We show that the Fourier transform of the tempered distribution $W_{j}$ is induced by the function $\xi\mapsto-i\frac{\xi_{j}}{|\xi|}$.1111 11 Loukas Grafakos, Classical Fourier Analysis, second ed., p. 260, Proposition 4.1.14.

###### Theorem 6.

For $1\leq j\leq n$ and for $\phi\in\mathscr{S}(\mathbb{R}^{n})$,

 $\left\langle\widehat{W}_{j},\phi\right\rangle=\int_{\mathbb{R}^{n}}-i\phi(x)% \frac{x_{j}}{|x|}dx.$
###### Proof.

We calculate

 $\displaystyle\left\langle W_{j},\widehat{\phi}\right\rangle$ $\displaystyle=\frac{\Gamma\left(\frac{n+1}{2}\right)}{\pi^{\frac{n+1}{2}}}\lim% _{\epsilon\to 0}\int_{|\xi|\geq\epsilon}K_{j}(\xi)\widehat{\phi}(\xi)d\xi$ $\displaystyle=\frac{\Gamma\left(\frac{n+1}{2}\right)}{\pi^{\frac{n+1}{2}}}\lim% _{\epsilon\to 0}\int_{\epsilon\leq|\xi|\leq 1/\epsilon}K_{j}(\xi)\widehat{\phi% }(\xi)d\xi$ $\displaystyle=\frac{\Gamma\left(\frac{n+1}{2}\right)}{\pi^{\frac{n+1}{2}}}\lim% _{\epsilon\to 0}\int_{\epsilon\leq|\xi|\leq 1/\epsilon}\left(\int_{\mathbb{R}^% {n}}e^{-2\pi ix\cdot\xi}\phi(x)dx\right)\frac{\xi_{j}}{|\xi|^{n+1}}d\xi$ $\displaystyle=\frac{\Gamma\left(\frac{n+1}{2}\right)}{\pi^{\frac{n+1}{2}}}\lim% _{\epsilon\to 0}\int_{\mathbb{R}^{n}}\phi(x)\left(\int_{\epsilon\leq|\xi|\leq 1% /\epsilon}e^{-2\pi ix\cdot\xi}\frac{\xi_{j}}{|\xi|^{n+1}}d\xi\right)dx.$

For the inside integral, because $\theta\mapsto\cos(-2\pi rx_{j}\theta_{j})\theta_{j}$ is an odd function,

 $\displaystyle\int_{\epsilon\leq|\xi|\leq 1/\epsilon}e^{-2\pi ix\cdot\xi}\frac{% \xi_{j}}{|\xi|^{n+1}}d\xi$ $\displaystyle=\int_{\epsilon\leq r\leq 1/\epsilon}\left(\int_{S^{n-1}}e^{-2\pi ix% \cdot(r\theta)}\frac{r\theta_{j}}{r^{n+1}}d\sigma(\theta)\right)r^{n-1}dr$ $\displaystyle=\int_{\epsilon\leq r\leq 1/\epsilon}\left(\int_{S^{n-1}}e^{-2\pi irx% \cdot\theta}\theta_{j}d\sigma(\theta)\right)r^{-1}dr$ $\displaystyle=\int_{\epsilon\leq r\leq 1/\epsilon}\left(\int_{S^{n-1}}i\sin(-2% \pi rx\cdot\theta)\theta_{j}d\sigma(\theta)\right)r^{-1}dr$ $\displaystyle=-i\int_{\epsilon\leq r\leq 1/\epsilon}\left(\int_{S^{n-1}}\sin(2% \pi rx\cdot\theta)\theta_{j}d\sigma(\theta)\right)r^{-1}dr$ $\displaystyle=-i\int_{S^{n-1}}\left(\int_{\epsilon\leq r\leq 1/\epsilon}\sin(2% \pi rx\cdot\theta)r^{-1}dr\right)\theta_{j}d\sigma(\theta).$

Call the whole last expression $f_{\epsilon}(x)$. It is a fact that for $0,

 $\left|\int_{a}^{b}\frac{\sin t}{t}dt\right|\leq 4,$

thus for $x\neq 0$,

 $|f_{\epsilon}(x)|\leq 4\omega_{n-1}.$

As1212 12 Loukas Grafakos, Classical Fourier Analysis, second ed., p. 263, Exercise 4.1.1.

 $\lim_{\epsilon\to 0}f_{\epsilon}(x)=-i\int_{S^{n-1}}\mathrm{sgn}\,(x\cdot% \theta)\frac{\pi}{2}\theta_{j}d\sigma(\theta),$

applying the dominated convergence theorem yields

 $\begin{split}&\displaystyle\lim_{\epsilon\to 0}\int_{\mathbb{R}^{n}}\phi(x)% \left(\int_{\epsilon\leq|\xi|\leq 1/\epsilon}e^{-2\pi ix\cdot\xi}\frac{\xi_{j}% }{|\xi|^{n+1}}d\xi\right)dx\\ \displaystyle=&\displaystyle\int_{\mathbb{R}^{n}}\phi(x)\left(-i\int_{S^{n-1}}% \mathrm{sgn}\,(x\cdot\theta)\frac{\pi}{2}\theta_{j}d\sigma(\theta)\right)dx\\ \displaystyle=&\displaystyle-i\frac{\pi}{2}\int_{\mathbb{R}^{n}}\phi(x)\left(% \int_{S^{n-1}}\mathrm{sgn}\,(x\cdot\theta)\theta_{j}d\sigma(\theta)\right)dx.% \end{split}$

Then using Lemma 5 and putting the above together we get

 $\displaystyle\left\langle W_{j},\widehat{\phi}\right\rangle$ $\displaystyle=\frac{\Gamma\left(\frac{n+1}{2}\right)}{\pi^{\frac{n+1}{2}}}% \cdot-i\frac{\pi}{2}\int_{\mathbb{R}^{n}}\phi(x)\left(\int_{S^{n-1}}\mathrm{% sgn}\,(x\cdot\theta)\theta_{j}d\sigma(\theta)\right)dx$ $\displaystyle=-i\frac{\pi}{2}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\pi^{% \frac{n+1}{2}}}\int_{\mathbb{R}^{n}}\phi(x)\frac{2\omega_{n-2}}{n-1}\frac{x_{j% }}{|x|}dx.$

We work out that

 $\frac{\pi}{2}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\pi^{\frac{n+1}{2}}}\cdot% \frac{2\omega_{n-2}}{n-1}=1,$

and therefore

 $\left\langle W_{j},\widehat{\phi}\right\rangle=-i\int_{\mathbb{R}^{n}}\phi(x)% \frac{x_{j}}{|x|}dx,$

completing the proof. ∎

Because $R_{j}h=h*W_{j}$,

 $\left\langle\widehat{R_{j}h},\phi\right\rangle=\left\langle\widehat{h}\widehat% {W}_{j},\phi\right\rangle=\left\langle\widehat{W}_{j},\widehat{h}\phi\right% \rangle=\int_{\mathbb{R}^{n}}-i\widehat{h}(\xi)\phi(\xi)\frac{\xi_{j}}{|\xi|}d\xi.$
###### Theorem 7.

For $1\leq j\leq n$ and for $h\in\mathscr{S}(\mathbb{R}^{n})$,

 $\widehat{R_{j}h}(\xi)=-i\frac{\xi_{j}}{|\xi|}\widehat{h}(\xi),\qquad\xi\in% \mathbb{R}^{n}.$

In other words, the multiplier of the Riesz transform $R_{j}$ is $m_{j}(\xi)=-i\frac{\xi_{j}}{|\xi|}$.

## 4 Properties of the Riesz transform

###### Theorem 8.
 $-I=\sum_{j=1}^{n}R_{j}^{2},$

where $I(h)=h$ for $h\in\mathscr{S}(\mathbb{R}^{n})$.

###### Proof.

For $h\in\mathscr{S}(\mathbb{R}^{n})$,

 $\displaystyle\widehat{R_{j}^{2}h}(\xi)=-i\frac{\xi_{j}}{|\xi|}\widehat{R_{j}h}% (\xi)=-i\frac{\xi_{j}}{|\xi|}\cdot-i\frac{\xi_{j}}{|\xi|}\widehat{h}(\xi)=-% \frac{\xi_{j}^{2}}{|\xi|^{2}}\widehat{h}(\xi),$

hence

 $\sum_{j=1}^{n}\widehat{R_{j}^{2}h}=-\widehat{h}.$

Taking the inverse Fourier transform,

 $\sum_{j=1}^{n}R_{j}^{2}h=-h,$

i.e.

 $\sum_{j=1}^{n}R_{j}^{2}=-I.$

For a tempered distribution $u$, for $1\leq j\leq n$, we define

 $\left\langle\partial_{j}u,\phi\right\rangle=(-1)\left\langle u,\partial_{j}% \phi\right\rangle,\qquad\phi\in\mathscr{S}(\mathbb{R}^{n}).$

It is a fact that $\partial_{j}u$ is itself a tempered distribution. One proves that

 $\widehat{\partial_{j}u}=(2\pi i\xi_{j})\widehat{u}.$

Each side of the above equation is a tempered distribution. Then

 $\widehat{\Delta u}=\sum_{j=1}^{n}\widehat{\partial_{j}^{2}u}=\sum_{j=1}^{n}(2% \pi i\xi_{j})^{2}\widehat{u}=-4\pi^{2}\sum_{j=1}^{n}\xi_{j}^{2}\widehat{u}=-4% \pi^{2}|\xi|^{2}\widehat{u}.$

Suppose that $f$ is a Schwartz function and that $u$ is a tempered distribution satisfying

 $\Delta u=f,$

called Poisson’s equation. Then

 $-4\pi^{2}|\xi|^{2}\widehat{u}=\widehat{f}.$

For $1\leq j,k\leq n$,

 $\displaystyle\partial_{j}\partial_{k}u$ $\displaystyle=\mathscr{F}^{-1}(\mathscr{F}(\partial_{j}\partial_{k}u))$ $\displaystyle=\mathscr{F}^{-1}((2\pi i\xi_{j})(2\pi i\xi_{k})\widehat{u})$ $\displaystyle=\mathscr{F}^{-1}\left(-4\pi^{2}\xi_{j}\xi_{k}\cdot\frac{\widehat% {f}}{-4\pi^{2}|\xi|^{2}}\right)$ $\displaystyle=\mathscr{F}^{-1}\left(\frac{\xi_{j}\xi_{k}}{|\xi|^{2}}\widehat{f% }\right).$

Using Theorem 7,

 $\displaystyle R_{j}R_{k}f$ $\displaystyle=\mathscr{F}^{-1}\mathscr{F}(R_{j}R_{k}f)$ $\displaystyle=\mathscr{F}^{-1}\left(-i\frac{\xi_{j}}{|\xi|}\widehat{R_{k}f}\right)$ $\displaystyle=\mathscr{F}^{-1}\left(-i\frac{\xi_{j}}{|\xi|}\cdot-i\frac{\xi_{k% }}{|\xi|}\widehat{f}\right)$ $\displaystyle=\mathscr{F}^{-1}\left(-\frac{\xi_{j}\xi_{k}}{|\xi|^{2}}\widehat{% f}\right).$

Therefore

 $\partial_{j}\partial_{k}u=-R_{j}R_{k}f.$
###### Theorem 9.

If $f$ is a Schwartz function and $u$ is a tempered distribution satisfying

 $\Delta u=f,$

then for $1\leq j,k\leq n$,

 $\partial_{j}\partial_{k}u=-R_{j}R_{k}f.$