Singular integral operators and the Riesz transform

1 Calderón-Zygmund kernels

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

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

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

$$v_n=\frac{{\omega }_{n-1}}{n}=\frac{2{\pi }^{n/2}}{n\mathrm{\Gamma }(n/2)}.$$

For $k,N\ge 0$ and $\varphi \in C^{\mathrm{\infty }}({ℝ}^n)$ let

$$p_{k,N}(\varphi )=\underset{|\alpha |\le k}{\max }\underset{x\in {ℝ}^n}{sup}{(1+|x|)}^N|({\partial }^{\alpha }\varphi )(x)|.$$

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

1. $|K(x)|\le B{|x|}^{-n}$, $x\ne 0$

2. ${\int }_{|x|\ge 2|y|}|K(x-y)-K(x)|𝑑x\le B$, $y\ne 0$

3. , .

The following lemma gives a tractable condition under which Condition 2 is satisfied.

Lemma 1.

If $\mathrm{|}\mathrm{(}\mathrm{\nabla }\mathit{}K\mathrm{)}\mathit{}\mathrm{(}x\mathrm{)}\mathrm{|}\mathrm{\le }C\mathit{}{\mathrm{|}x\mathrm{|}}^{\mathrm{-}n\mathrm{-}\mathrm{1}}$ for all $x\mathrm{\ne }\mathrm{0}$ then for $y\mathrm{\ne }\mathrm{0}$,

$${\int }_{|x|\ge 2|y|}|K(x-y)-K(x)|𝑑x\le v_n{2}^{n}C.$$

Proof.

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

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

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

$$|K(x-y)-K(x)|\le {\int }_0^1|(\nabla K)(x-ty)||y|𝑑t\le C|y|{\int }_0^1{|x-ty|}^{-n-1}𝑑t.$$

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

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

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

${\displaystyle {\int }_{|x|\ge 2|y|}}|K(x-y)-K(x)|𝑑x$ $\le {\displaystyle {\int }_{|x|\ge 2|y|}}{2}^{n+1}C|y|{|x|}^{-n-1}𝑑x$

$=2^{n+1}C|y|{\displaystyle {\int }_{2|y|}^{\mathrm{\infty }}}\left({\displaystyle {\int }_{S^{n-1}}}{|r\gamma |}^{-n-1}𝑑\sigma (\gamma )\right)r^{n-1}𝑑r$

$=v_n{2}^{n+1}C|y|{\displaystyle {\int }_{2|y|}^{\mathrm{\infty }}}{r}^{-2}𝑑r$

$=v_n{2}^{n+1}C|y|\cdot {\displaystyle \frac{1}{2|y|}}$

$=v_n{2}^{n}C.$

∎

For a Calderón-Zygmund kernel $K$, for $f\in 𝒮({ℝ}^n)$, for $x\in {ℝ}^n$, and for $ϵ>0$, using Condition 3 with $R_1=ϵ$ and $R_2=1$,

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

$$|K(x-y)(f(y)-f(x))|\le B{\parallel \nabla f\parallel }_{\mathrm{\infty }}{|y-x|}^{-n+1},$$

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

$$\underset{ϵ\to 0}{lim}{\int }_{ϵ\le |x-y|\le 1}K(x-y)(f(y)-f(x))𝑑y={\int }_{|x-y|\le 1}K(x-y)(f(y)-f(x))𝑑y.$$

Lemma 2.

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

$$\underset{ϵ\to 0}{lim}{\int }_{|x-y|\ge ϵ}K(x-y)f(y)𝑑y$$

exists.

2 Singular integral operators

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

$$(Tf)(x)=\underset{ϵ\to 0}{lim}{\int }_{|x-y|\ge ϵ}K(x-y)f(y)𝑑y.$$

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.

Theorem 3.

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

$${\parallel Tf\parallel }_2\le C_{n}B{\parallel f\parallel }_2.$$

Proof.

For and for $\xi \in {ℝ}^n$ define

$m_{r,s}(\xi )$

Take , for which

For the first integral, using Condition 3 with $R_1=r$ and $R_2={|\xi |}^{-1}$ and then using Condition 1,

$=2\pi |\xi |\cdot v_n{|\xi |}^{-1}.$

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

Let

$R$

with which

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

$\le {\displaystyle {\int }_{|x|>{|\xi |}^{-1}}}|K(x)-K(x-z)|𝑑x$

$={\displaystyle {\int }_{|x|>2|z|}}|K(x)-K(x-z)|𝑑x$

$\le B.$

On the other hand, let

For $x\in D_1$ we have

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

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,

Furthermore, for $x\in D_1\setminus D_2$ we have $|x|\le {|\xi |}^{-1}$, and for $x\in D_2\setminus D_1$ we have

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

Hence

so

$$\lambda (D)\le {\left(\frac{3}{2|\xi |}\right)}^n{v}_n={\left(\frac{3}{2}\right)}^n{|\xi |}^{-n}{v}_n.$$

Therefore

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

and then

and finally

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

Define

Then

$\widehat{T_{r,s}f}(\xi )$

$=m_{r,s}(\xi )\widehat{f}(\xi ),$

and so

$${\parallel \widehat{T_{r,s}f}\parallel }_2^2={\int }_{{ℝ}^n}{|\widehat{T_{r,s}f}(\xi )|}^2𝑑\xi ={\int }_{{ℝ}^n}{|m_{r,s}(\xi )\widehat{f}(\xi )|}^2𝑑\xi \le {\parallel m_{r,s}\parallel }_{\mathrm{\infty }}^2{\parallel \widehat{f}\parallel }_2^2,$$

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

$${\parallel T_{r,s}f\parallel }_2^2\le {\parallel m_{r,s}\parallel }_{\mathrm{\infty }}^2{\parallel f\parallel }_2^2\le {(C_{n}B)}^2{\parallel f\parallel }_2^2.$$

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

$${\int }_{{ℝ}^n}{|(Tf)(x)|}^2𝑑x\le \underset{r\to 0,s\to \mathrm{\infty }}{lim\; inf}{\int }_{{ℝ}^n}{|(T_{r,s}f)(x)|}^2𝑑x={(C_{n}B)}^2{\parallel f\parallel }_2^2.$$

That is,

$${\parallel Tf\parallel }_2\le C_{n}B{\parallel f\parallel }_2.$$

∎

3 The Riesz transform

Let

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

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

$$K_j(x)=c_n\frac{x_j}{{|x|}^{n+1}}.$$

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

$$(R_j\varphi )(x)=\underset{ϵ\to 0}{lim}{\int }_{|y-x|\ge ϵ}{K}_j(x-y)\varphi (y)𝑑y=\underset{ϵ\to 0}{lim}{\int }_{|y|\ge ϵ}{K}_j(y)\varphi (x-y)𝑑y.$$

We call each $R_j$, $1\le j\le n$, a Riesz transform.

For $1\le j\le n$ define $W_j:𝒮({ℝ}^n)\to ℂ$ by

$$⟨W_j,\varphi ⟩=\frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{{\pi }^{\frac{n+1}{2}}}\underset{ϵ\to 0}{lim}{\int }_{|y|\ge ϵ}{K}_j(y)\varphi (y)𝑑y.$$ (1)

For $ϵ>0$,

$\left|{\displaystyle {\int }_{ϵ\le |y|\le 1}}{K}_j(y)\varphi (y)𝑑y\right|$ $=\left|{\displaystyle {\int }_{ϵ\le |y|\le 1}}{K}_j(y)(\varphi (y)-\varphi (0))𝑑y\right|$

$\le {\displaystyle {\int }_{ϵ\le |y|\le 1}}{c}_n{|y|}^{-n}\cdot {\parallel \nabla \varphi \parallel }_{\mathrm{\infty }}|y|𝑑y$

$=c_n{\parallel \nabla \varphi \parallel }_{\mathrm{\infty }}{\omega }_{n-1}{\displaystyle {\int }_{ϵ}^1}{r}^{-n+1}\cdot r^{n-1}𝑑r$

$=c_n{\parallel \nabla \varphi \parallel }_{\mathrm{\infty }}{\omega }_{n-1}(1-ϵ).$

For $|y|\ge 1$,

${\displaystyle {\int }_{|y|\ge 1}}|K_j(y)\varphi (y)|𝑑y$ $\le c_n{\displaystyle {\int }_{|y|\ge 1}}{|y|}^{-n}{(1+{|y|}^2)}^{-1/2}{p}_{0,1}(\varphi )𝑑y$

$=c_n{\omega }_n{\displaystyle {\int }_1^{\mathrm{\infty }}}{r}^{-n}{(1+r)}^{-1}\cdot r^{n-1}𝑑r$

$=c_n{\omega }_n\log 2.$

It then follows from the dominated convergence theorem that the limit

$$\underset{ϵ\to 0}{lim}{\int }_{|y|\ge ϵ}{K}_j(y)\varphi (y)𝑑y$$

exists, which shows that the definition (1) makes sense. It is apparent that $W_j$ is linear. Then prove that if ${\varphi }_k\to \varphi$ in $𝒮({ℝ}^n)$ then $⟨W_j,{\varphi }_k⟩\to ⟨W_j,\varphi ⟩$. This being true means that $W_j\in {𝒮}^{\prime }({ℝ}^n)$, namely that each $W_j$ is a tempered distribution.

For a function $f:{ℝ}^n\to ℂ$, write

$$\tilde{f}(x)=f(-x),({\tau }_{y}f)(x)=f(x-y).$$

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

$$⟨h*u,\varphi ⟩=⟨u,\tilde{h}*\varphi ⟩,\varphi \in 𝒮({ℝ}^n).$$

It is a fact that $h*u\in {𝒮}^{\prime }({ℝ}^n)$, and this tempered distribution is induced by the $C^{\mathrm{\infty }}$ function $x\mapsto ⟨u,{\tau }_x\tilde{h}⟩$. The Fourier transform of a tempered distribution $u$ is defined by

$$⟨\widehat{u},\varphi ⟩=⟨u,\widehat{\varphi }⟩,\varphi \in 𝒮({ℝ}^n),$$

where

$$\widehat{\varphi }(\xi )={\int }_{{ℝ}^n}{e}^{-2\pi ix\cdot \xi }\varphi (x)𝑑x,\xi \in {ℝ}^n.$$

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

$$⟨hu,\varphi ⟩=⟨u,h\varphi ⟩,\varphi \in 𝒮({ℝ}^n).$$

It is a fact that $hu$ is itself a tempered distribution. It is proved that

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

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

Lemma 4.

For $\mathrm{1}\mathrm{\le }j\mathrm{\le }n$, for $\varphi \mathrm{\in }\mathrm{S}\mathit{}\mathrm{(}{\mathrm{R}}^n\mathrm{)}$, and for $x\mathrm{\in }{\mathrm{R}}^n$,

$$(R_j\varphi )(x)=(\varphi *W_j)(x).$$

We will use the following identity for integrals over $S^{n-1}$.

Lemma 5.

For $\xi \mathrm{\ne }\mathrm{0}$ and for $\mathrm{1}\mathrm{\le }j\mathrm{\le }n$,

$${\int }_{S^{n-1}}\mathrm{sgn}(\xi \cdot \theta ){\theta }_j𝑑\sigma (\theta )=\frac{2{\omega }_{n-2}}{n-1}\frac{{\xi }_j}{|\xi |}.$$

Proof.

It is a fact that

(2)

It suffices to prove the claim when $\xi \in S^{n-1}$. For $1\le j\le n$ there is $A_j={(a_{i,k})}_{i,k}\in S{O}_n(ℝ)$ such that

$$A_j{e}_j=\xi ,$$

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

${\displaystyle {\int }_{S^{n-1}}}\mathrm{sgn}(\xi \cdot \theta ){\theta }_j𝑑\sigma (\theta )$ $={\displaystyle {\int }_{S^{n-1}}}\mathrm{sgn}(A_j{e}_j\cdot \theta ){(A{A}^{-1}\theta )}_j𝑑\sigma (\theta )$

$={\displaystyle {\int }_{S^{n-1}}}\mathrm{sgn}(e_j\cdot A_j^{-1}\theta ){(A{A}^{-1}\theta )}_j𝑑\sigma (\theta )$

$={\displaystyle {\int }_{S^{n-1}}}\mathrm{sgn}(e_j\cdot \theta ){(A\theta )}_{j}d(A_j^{-1}{}_{*}\sigma )(\theta )$

$={\displaystyle {\int }_{S^{n-1}}}\mathrm{sgn}(e_j\cdot \theta ){(A\theta )}_j𝑑\sigma (\theta )$

$={\displaystyle {\int }_{S^{n-1}}}\mathrm{sgn}({\theta }_j){\displaystyle \sum _{k=1}^n}{a}_{j,k}{\theta }_{k}d\theta .$

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

$${\int }_{S^{n-1}}\mathrm{sgn}(\xi \cdot \theta ){\theta }_j𝑑\sigma (\theta )={\int }_{S^{n-1}}{\xi }_j|{\theta }_j|𝑑\sigma (\theta )=\frac{{\xi }_j}{|\xi |}{\int }_{S^{n-1}}|{\theta }_j|𝑑\sigma (\theta ).$$

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

$${\int }_{S^{n-1}}\mathrm{sgn}(\xi \cdot \theta ){\theta }_j𝑑\sigma (\theta )=\frac{{\xi }_j}{|\xi |}{\int }_{S^{n-1}}|{\theta }_1|𝑑\sigma (\theta ).$$

It is a fact that

$${\int }_{R{S}^{n-1}}f(\theta )𝑑\sigma (\theta )={\int }_{-R}^R{\int }_{\sqrt{R^2-s^2}{S}^{n-2}}f(s,\varphi )𝑑\varphi \frac{Rds}{\sqrt{R^2-s^2}}.$$

Using this with $f(\theta )=f({\theta }_1,\mathrm{\dots },{\theta }_n)=|{\theta }_1|$ and using that the measure of $R{S}^{n-2}$ is $R^{n-2}{\omega }_{n-1}$, we calculate

${\displaystyle {\int }_{S^{n-1}}}|{\theta }_1|𝑑\sigma (\theta )$ $={\displaystyle {\int }_{-1}^1}{\displaystyle {\int }_{\sqrt{1-s^2}{S}^{n-2}}}|s|𝑑\varphi {\displaystyle \frac{ds}{\sqrt{1-s^2}}}$

$={\displaystyle {\int }_{-1}^1}{(1-s^2)}^{\frac{n-2}{2}-\frac{1}{2}}{\omega }_{n-2}|s|𝑑\varphi$

$=2{\omega }_{n-2}{\displaystyle {\int }_0^1}{(1-s^2)}^{\frac{n-3}{2}}s𝑑s$

$={\omega }_{n-2}{\displaystyle {\int }_0^1}{u}^{\frac{n-3}{2}}𝑑u$

$={\displaystyle \frac{2{\omega }_{n-2}}{n-1}}.$

∎

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 $\xi \mapsto -i\frac{{\xi }_j}{|\xi |}$.

Theorem 6.

For $\mathrm{1}\mathrm{\le }j\mathrm{\le }n$ and for $\varphi \mathrm{\in }\mathrm{S}\mathit{}\mathrm{(}{\mathrm{R}}^n\mathrm{)}$,

$$⟨{\widehat{W}}_j,\varphi ⟩={\int }_{{ℝ}^n}-i\varphi (x)\frac{x_j}{|x|}dx.$$

Proof.

We calculate

$⟨W_j,\widehat{\varphi }⟩$ $={\displaystyle \frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{{\pi }^{\frac{n+1}{2}}}}\underset{ϵ\to 0}{lim}{\displaystyle {\int }_{|\xi |\ge ϵ}}{K}_j(\xi )\widehat{\varphi }(\xi )𝑑\xi$

$={\displaystyle \frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{{\pi }^{\frac{n+1}{2}}}}\underset{ϵ\to 0}{lim}{\displaystyle {\int }_{ϵ\le |\xi |\le 1/ϵ}}{K}_j(\xi )\widehat{\varphi }(\xi )𝑑\xi$

$={\displaystyle \frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{{\pi }^{\frac{n+1}{2}}}}\underset{ϵ\to 0}{lim}{\displaystyle {\int }_{ϵ\le |\xi |\le 1/ϵ}}\left({\displaystyle {\int }_{{ℝ}^n}}{e}^{-2\pi ix\cdot \xi }\varphi (x)𝑑x\right){\displaystyle \frac{{\xi }_j}{{|\xi |}^{n+1}}}𝑑\xi$

$={\displaystyle \frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{{\pi }^{\frac{n+1}{2}}}}\underset{ϵ\to 0}{lim}{\displaystyle {\int }_{{ℝ}^n}}\varphi (x)\left({\displaystyle {\int }_{ϵ\le |\xi |\le 1/ϵ}}{e}^{-2\pi ix\cdot \xi }{\displaystyle \frac{{\xi }_j}{{|\xi |}^{n+1}}}𝑑\xi \right)𝑑x.$

For the inside integral, because $\theta \mapsto \cos (-2\pi r{x}_j{\theta }_j){\theta }_j$ is an odd function,

${\displaystyle {\int }_{ϵ\le |\xi |\le 1/ϵ}}{e}^{-2\pi ix\cdot \xi }{\displaystyle \frac{{\xi }_j}{{|\xi |}^{n+1}}}𝑑\xi$ $={\displaystyle {\int }_{ϵ\le r\le 1/ϵ}}\left({\displaystyle {\int }_{S^{n-1}}}{e}^{-2\pi ix\cdot (r\theta )}{\displaystyle \frac{r{\theta }_j}{r^{n+1}}}𝑑\sigma (\theta )\right)r^{n-1}𝑑r$

$={\displaystyle {\int }_{ϵ\le r\le 1/ϵ}}\left({\displaystyle {\int }_{S^{n-1}}}{e}^{-2\pi irx\cdot \theta }{\theta }_j𝑑\sigma (\theta )\right)r^{-1}𝑑r$

$={\displaystyle {\int }_{ϵ\le r\le 1/ϵ}}\left({\displaystyle {\int }_{S^{n-1}}}i\sin (-2\pi rx\cdot \theta ){\theta }_j𝑑\sigma (\theta )\right)r^{-1}𝑑r$

$=-i{\displaystyle {\int }_{ϵ\le r\le 1/ϵ}}\left({\displaystyle {\int }_{S^{n-1}}}\sin (2\pi rx\cdot \theta ){\theta }_j𝑑\sigma (\theta )\right)r^{-1}𝑑r$

$=-i{\displaystyle {\int }_{S^{n-1}}}\left({\displaystyle {\int }_{ϵ\le r\le 1/ϵ}}\sin (2\pi rx\cdot \theta )r^{-1}𝑑r\right){\theta }_j𝑑\sigma (\theta ).$

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

$$\left|{\int }_a^b\frac{\sin t}{t}𝑑t\right|\le 4,$$

thus for $x\ne 0$,

$$|f_{ϵ}(x)|\le 4{\omega }_{n-1}.$$

As

$$\underset{ϵ\to 0}{lim}{f}_{ϵ}(x)=-i{\int }_{S^{n-1}}\mathrm{sgn}(x\cdot \theta )\frac{\pi }{2}{\theta }_j𝑑\sigma (\theta ),$$

applying the dominated convergence theorem yields

Then using Lemma 5 and putting the above together we get

$⟨W_j,\widehat{\varphi }⟩$ $={\displaystyle \frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{{\pi }^{\frac{n+1}{2}}}}\cdot -i{\displaystyle \frac{\pi }{2}}{\displaystyle {\int }_{{ℝ}^n}}\varphi (x)\left({\displaystyle {\int }_{S^{n-1}}}\mathrm{sgn}(x\cdot \theta ){\theta }_{j}d\sigma (\theta )\right)dx$

$=-i{\displaystyle \frac{\pi }{2}}{\displaystyle \frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{{\pi }^{\frac{n+1}{2}}}}{\displaystyle {\int }_{{ℝ}^n}}\varphi (x){\displaystyle \frac{2{\omega }_{n-2}}{n-1}}{\displaystyle \frac{x_j}{|x|}}𝑑x.$

We work out that

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

and therefore

$$⟨W_j,\widehat{\varphi }⟩=-i{\int }_{{ℝ}^n}\varphi (x)\frac{x_j}{|x|}𝑑x,$$

completing the proof. ∎

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

$$⟨\widehat{R_{j}h},\varphi ⟩=⟨\widehat{h}{\widehat{W}}_j,\varphi ⟩=⟨{\widehat{W}}_j,\widehat{h}\varphi ⟩={\int }_{{ℝ}^n}-i\widehat{h}(\xi )\varphi (\xi )\frac{{\xi }_j}{|\xi |}d\xi .$$

Theorem 7.

For $\mathrm{1}\mathrm{\le }j\mathrm{\le }n$ and for $h\mathrm{\in }\mathrm{S}\mathit{}\mathrm{(}{\mathrm{R}}^n\mathrm{)}$,

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

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 $I\mathit{}\mathrm{(}h\mathrm{)}\mathrm{=}h$ for $h\mathrm{\in }\mathrm{S}\mathit{}\mathrm{(}{\mathrm{R}}^n\mathrm{)}$.

Proof.

For $h\in 𝒮({ℝ}^n)$,

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

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 $1\le j\le n$, we define

$$⟨{\partial }_{j}u,\varphi ⟩=(-1)⟨u,{\partial }_j\varphi ⟩,\varphi \in 𝒮({ℝ}^n).$$

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

$$\widehat{\mathrm{\Delta }u}=\sum _{j=1}^n\widehat{{\partial }_j^{2}u}=\sum _{j=1}^n{(2\pi i{\xi }_j)}^2\widehat{u}=-4{\pi }^2\sum _{j=1}^n{\xi }_j^2\widehat{u}=-4{\pi }^2{|\xi |}^2\widehat{u}.$$

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

$$\mathrm{\Delta }u=f,$$

called Poisson’s equation. Then

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

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

${\partial }_j{\partial }_{k}u$ $={ℱ}^{-1}(ℱ({\partial }_j{\partial }_{k}u))$

$={ℱ}^{-1}((2\pi i{\xi }_j)(2\pi i{\xi }_k)\widehat{u})$

$={ℱ}^{-1}\left(-4{\pi }^2{\xi }_j{\xi }_k\cdot {\displaystyle \frac{\widehat{f}}{-4{\pi }^2{|\xi |}^2}}\right)$

$={ℱ}^{-1}\left({\displaystyle \frac{{\xi }_j{\xi }_k}{{|\xi |}^2}}\widehat{f}\right).$

Using Theorem 7,

$R_j{R}_{k}f$ $={ℱ}^{-1}ℱ(R_j{R}_{k}f)$

$={ℱ}^{-1}\left(-i{\displaystyle \frac{{\xi }_j}{|\xi |}}\widehat{R_{k}f}\right)$

$={ℱ}^{-1}(-i{\displaystyle \frac{{\xi }_j}{|\xi |}}\cdot -i{\displaystyle \frac{{\xi }_k}{|\xi |}}\widehat{f})$

$={ℱ}^{-1}\left(-{\displaystyle \frac{{\xi }_j{\xi }_k}{{|\xi |}^2}}\widehat{f}\right).$

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

$$\mathrm{\Delta }u=f,$$

then for $\mathrm{1}\mathrm{\le }j\mathrm{,}k\mathrm{\le }n$,

$${\partial }_j{\partial }_{k}u=-R_j{R}_{k}f.$$