# The Hilbert transform on $\mathbb{R}$

Jordan Bell
April 24, 2016

## 1 The principal value integral

Let $A_{\epsilon}=\{x\in\mathbb{R}:|x|\geq\epsilon\}$. For $f\in\bigcap_{\epsilon>0}L^{1}(A_{\epsilon})$, if $\int_{A_{\epsilon}}f(x)dx$ has a limit as $\epsilon\to 0$, we denote it by

 $PV\int_{\mathbb{R}}f(x)dx=\lim_{\epsilon\to 0}\int_{|x|\geq\epsilon}f(x)dx.$

Let $\mathscr{S}$ be the collection of Schwartz functions $\mathbb{R}\to\mathbb{C}$ and let $\mathscr{S}^{\prime}$ be its dual space, whose elements are called tempered distributions. For $\phi\in\mathscr{S}$, for $k>j$,

 $\displaystyle\left|\int_{A_{1/j}}\frac{\phi(x)}{x}dx-\int_{A_{1/k}}\frac{\phi(% x)}{x}dx\right|$ $\displaystyle=\left|\int_{1/k\leq|x|<1/j}\frac{\phi(x)}{x}dx\right|$ $\displaystyle=\left|\int_{1/k\leq|x|<1/j}\frac{\phi(x)-\phi(0)}{x}dx\right|$ $\displaystyle\leq\int_{1/k\leq|x|<1/j}\left\|\phi^{\prime}\right\|_{b}$ $\displaystyle=\left\|\phi^{\prime}\right\|_{b}\ \cdot\left(\frac{1}{j}-\frac{1% }{k}\right)$ $\displaystyle=\left\|\phi^{\prime}\right\|_{b}\ \cdot\frac{k-j}{kj}$ $\displaystyle\leq\frac{\left\|\phi^{\prime}\right\|_{b}}{j}.$

Therefore $\int_{A_{1/j}}\frac{\phi(x)}{x}dx$ is a Cauchy sequence in $\mathbb{C}$ and hence converges. Then the following limit exists:

 $\left\langle\phi,W\right\rangle=PV\int_{\mathbb{R}}\frac{\phi(x)}{x}dx=\lim_{% \epsilon\to 0}\int_{|x|\geq\epsilon}\frac{\phi(x)}{x}dx.$

It is apparent that $W:\mathscr{S}\to\mathbb{C}$ is linear, and one proves that $W\in\mathscr{S}^{\prime}$.

For $\phi\in\mathscr{S}$, by Hadamard’s lemma, there is a $C^{\infty}$ function $\psi:\mathbb{R}\to\mathbb{C}$ such that $\phi(x)=\phi(0)+x\psi(x)$ for all $x$. For $\epsilon>0$,

 $\displaystyle\int_{\epsilon}^{1}\phi^{\prime}(x)\log xdx$ $\displaystyle=x\psi(x)\log x\bigg{|}_{\epsilon}^{1}-\int_{\epsilon}^{1}(\phi(x% )-\phi(0))\frac{1}{x}dx$ $\displaystyle=-\epsilon\psi(\epsilon)\log\epsilon-\int_{\epsilon}^{1}\frac{% \phi(x)}{x}dx-\phi(0)\log\epsilon$

and

 $\displaystyle\int_{-1}^{-\epsilon}\phi^{\prime}(x)\log|x|dx$ $\displaystyle=x\psi(x)\log(-x)\bigg{|}_{-1}^{-\epsilon}-\int_{-1}^{-\epsilon}(% \phi(x)-\phi(0))\frac{1}{x}dx$ $\displaystyle=-\epsilon\psi(-\epsilon)\log\epsilon-\int_{-1}^{-\epsilon}\frac{% \phi(x)}{x}dx+\phi(0)\log\epsilon.$

Hence

 $\int_{\epsilon\leq|x|\leq 1}\phi^{\prime}(x)\log|x|dx=-\epsilon\cdot(\psi(% \epsilon)+\psi(-\epsilon))\cdot\log\epsilon-\int_{\epsilon\leq|x|\leq 1}\frac{% \phi(x)}{x}dx.$

On the other hand,

 $\int_{|x|\geq 1}\phi^{\prime}(x)\log|x|dx=-\int_{|x|\geq 1}\frac{\phi(x)}{x}dx.$

Therefore

 $PV\int_{\mathbb{R}}\phi^{\prime}(x)\log|x|dx=-PV\int_{\mathbb{R}}\frac{\phi(x)% }{x}dx.$

Let $\mu\phi(x)=\phi(-x)$ and $\tau_{y}\phi(x)=\phi(x-y)$. Then

 $\tau_{y}\mu\phi(x)=\mu\phi(x-y)=\phi(y-x)=\tau_{x}\phi(y).$

Write

 $\displaystyle\phi*\psi(x)$ $\displaystyle=\int_{\mathbb{R}}\phi(y)\psi(x-y)dy$ $\displaystyle=\int_{\mathbb{R}}\phi(y)(\tau_{y}\psi)(x)dy$ $\displaystyle=\int_{\mathbb{R}}\phi(y)(\tau_{x}\mu\psi)(y)dy.$

For $u\in\mathscr{S}^{\prime}$ and for $\phi\in\mathscr{S}$, define $\phi*u:\mathbb{R}\to\mathbb{C}$ by

 $(\phi*u)(x)=\left\langle\tau_{x}\mu\phi,u\right\rangle.$

One proves that $\phi*u$ is a tempered distribution, and satisfies

 $\left\langle\psi,\phi*u\right\rangle=\left\langle(\mu\phi)*\psi,u\right\rangle.$

Define $\tau_{x}u\in\mathscr{S}^{\prime}$ by

 $\left\langle\phi,\tau_{x}u\right\rangle=\left\langle\tau_{-x}\phi,u\right\rangle.$

## 2 The Hilbert transform

For $\epsilon>0$, for $\phi\in\mathscr{S}$ let

 $\displaystyle H_{\epsilon}\phi(x)$ $\displaystyle=\frac{1}{\pi}\int_{|y|\geq\epsilon}\frac{\phi(x-y)}{y}dy$ $\displaystyle=\frac{1}{\pi}\int_{|y|\geq\epsilon}\frac{\tau_{y}\phi(x)}{y}dy$ $\displaystyle=\frac{1}{\pi}\int_{|y|\geq\epsilon}\frac{\tau_{x}\mu\phi(y)}{y}dy.$

Define

 $\displaystyle H\phi(x)$ $\displaystyle=\lim_{\epsilon\to 0}H_{\epsilon}\phi(x)$ $\displaystyle=\frac{1}{\pi}\cdot PV\int_{\mathbb{R}}\frac{\tau_{x}\mu\phi(y)}{% y}dy$ $\displaystyle=\frac{1}{\pi}\cdot PV\int_{\mathbb{R}}\frac{\phi(x-y)}{y}dy$ $\displaystyle=\frac{1}{\pi}\cdot PV\int_{\mathbb{R}}\frac{\phi(y)}{x-y}dy.$

For $x\in\mathbb{R}$,

 $\displaystyle\phi*W(x)$ $\displaystyle=\left\langle\tau_{x}\mu\phi,W\right\rangle$ $\displaystyle=PV\int_{\mathbb{R}}\frac{\tau_{x}\mu\phi(y)}{y}dy$ $\displaystyle=PV\int_{\mathbb{R}}\frac{\phi(x-y)}{y}dy$ $\displaystyle=PV\int_{\mathbb{R}}\frac{\phi(y)}{x-y}dy$ $\displaystyle=\lim_{\epsilon\to 0}\int_{|x|\geq\epsilon}\frac{\phi(y)}{x-y}dy.$

Thus

 $H\phi=\frac{1}{\pi}\phi*W=\phi*\left(\frac{W}{\pi}\right).$

We calculate

 $\displaystyle\left\langle\phi,\widehat{W}\right\rangle$ $\displaystyle=\left\langle\widehat{\phi},W\right\rangle$ $\displaystyle=\lim_{\epsilon\to 0}\int_{\epsilon\leq|\xi|\leq 1/\epsilon}\frac% {\widehat{\phi}(\xi)}{\xi}d\xi$ $\displaystyle=\lim_{\epsilon\to 0}\int_{|\xi|\geq\epsilon}\left(\int_{\mathbb{% R}}\phi(x)e^{-2\pi ix\xi}dx\right)\frac{1}{\xi}d\xi$ $\displaystyle=\lim_{\epsilon\to 0}\int_{\mathbb{R}}\phi(x)\left(\int_{\epsilon% \leq|\xi|\leq 1/\epsilon}\frac{e^{-2\pi ix\xi}}{\xi}d\xi\right)dx$ $\displaystyle=\lim_{\epsilon\to 0}\int_{\mathbb{R}}\phi(x)\left(\int_{\epsilon% \leq|\xi|\leq 1/\epsilon}-i\frac{\sin 2\pi x\xi}{\xi}d\xi\right)dx.$

Check that

 $\lim_{\epsilon\to 0}\int_{\epsilon\leq|\xi|\leq 1/\epsilon}\frac{\sin 2\pi x% \xi}{\xi}d\xi=\pi\cdot\mathrm{sgn}\,x.$

Then, using the dominated convergence theorem,

 $\left\langle\phi,\widehat{W}\right\rangle=\int_{\mathbb{R}}\phi(x)\cdot-i\pi% \cdot\mathrm{sgn}\,xdx.$

Thus, $\widehat{W}=-\pi i\cdot\mathrm{sgn}\,$.

Now,

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

Then

 $\widehat{H\phi}(\xi)=\frac{1}{\pi}\widehat{\phi*W}(\xi)=\frac{1}{\pi}\widehat{% \phi}(\xi)\cdot\widehat{W}(\xi)=\widehat{\phi}(\xi)\cdot-i\cdot\mathrm{sgn}\,(% \xi).$

Let

 $m_{H}(\xi)=-i\cdot\mathrm{sgn}\,(\xi),$

with which

 $\widehat{H\phi}=m_{H}\cdot\widehat{\phi}.$

Writing $F\phi=\widehat{\phi}$,

 $FH\phi=m_{H}\cdot F\phi.$

So

 $H\phi=F^{-1}(m_{H}\cdot F\phi),$

and hence

 $H^{2}\phi=F^{-1}(m_{H}\cdot FH\phi)=F^{-1}(m_{H}\cdot m_{H}F\phi).$

For $\xi\neq 0$, $m_{H}(\xi)^{2}=-1$, which yields

 $H^{2}\phi=F^{-1}(-F\phi)=-\phi.$

Thus $H^{2}=-\mathrm{id}$. Therefore $\left\|H\phi\right\|_{L^{2}}=\left\|\phi\right\|_{L^{2}}$.

Thus it makes sense to define $H:L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})$. For $f,g\in L^{2}(\mathbb{R})$, by Plancherel’s theorem, and as $\overline{m_{H}}=-m_{H}$,

 $\displaystyle\left\langle Hf,g\right\rangle$ $\displaystyle=\left\langle\widehat{Hf},\widehat{g}\right\rangle$ $\displaystyle=\int_{\mathbb{R}}\widehat{Hf}(\xi)\cdot\overline{\widehat{g}(\xi% )}d\xi$ $\displaystyle=\int_{\mathbb{R}}m_{H}(\xi)\cdot\widehat{f}(\xi)\overline{% \widehat{g}(\xi)}d\xi$ $\displaystyle=-\int_{\mathbb{R}}\widehat{f}(\xi)\cdot\overline{m_{H}(\xi)\cdot% \widehat{g}(\xi)}d\xi$ $\displaystyle=-\int_{\mathbb{R}}\widehat{f}(\xi)\cdot\overline{\widehat{Hg}(% \xi)}d\xi$ $\displaystyle=-\left\langle f,Hg\right\rangle.$

But $\left\langle Hf,g\right\rangle=\left\langle f,H^{*}g\right\rangle$, so

 $\left\langle f,H^{*}g\right\rangle=\left\langle f,-Hg\right\rangle,$

which implies that $H^{*}g=-Hg$ and thus $H^{*}=-H$. Furthermore,

 $\widehat{H^{*}g}(\xi)=-\widehat{Hg}(\xi)=-m_{H}(\xi)\cdot\widehat{g}(\xi)=i% \cdot\mathrm{sgn}\,(\xi)\cdot\widehat{g}(\xi).$

## 3 The Poisson kernel

For $y>0$, calculate

 $\displaystyle\int_{\mathbb{R}}e^{2\pi i\xi x}e^{-2\pi y|\xi|}d\xi$ $\displaystyle=\frac{1}{2\pi ix+2\pi y}-\frac{1}{2\pi ix-2\pi y}$ $\displaystyle=\frac{2\pi ix-2\pi y-2\pi ix-2\pi y}{-4\pi^{2}x^{2}-4\pi^{2}y^{2}}$ $\displaystyle=\frac{y}{\pi(x^{2}+y^{2})}$

and

 $\displaystyle-i\int_{\mathbb{R}}e^{2\pi i\xi x}\mathrm{sgn}\,(\xi)e^{-2\pi y|% \xi|}d\xi$ $\displaystyle=\frac{i}{2\pi ix+2\pi y}+\frac{i}{2\pi ix-2\pi y}$ $\displaystyle=\frac{i(2\pi ix-2\pi y+2\pi ix+2\pi y)}{-4\pi^{2}x^{2}-4\pi^{2}y% ^{2}}$ $\displaystyle=\frac{-4\pi x}{-4\pi^{2}x^{2}-4\pi^{2}y^{2}}$ $\displaystyle=\frac{x}{\pi(x^{2}+y^{2})}.$

For $y>0$ let

 $P_{y}(x)=\frac{1}{\pi}\frac{y}{x^{2}+y^{2}}$

and

 $Q_{y}(x)=\frac{1}{\pi}\frac{x}{x^{2}+y^{2}}.$

Then

 $\widehat{P}_{y}(\xi)=e^{-2\pi y|\xi|}$

and

 $\widehat{Q}_{y}(\xi)=-i\cdot\mathrm{sgn}\,(\xi)e^{-2\pi y|\xi|}.$

Also,

 $P_{y}(x)+iQ_{y}(x)=\frac{1}{\pi}\frac{y+ix}{x^{2}+y^{2}}=\frac{1}{\pi}\frac{1}% {y-ix}.$

For a Borel measurable function $f:\mathbb{R}\to\mathbb{C}$ for which the integral exists,

 $\displaystyle(P_{y}*f)(x)$ $\displaystyle=\int_{\mathbb{R}}P_{y}(x-t)f(t)dt$ $\displaystyle=\int_{\mathbb{R}}f(t)\frac{y}{\pi((x-t)^{2}+y^{2})}dt$ $\displaystyle=\int_{\mathbb{R}}f(x-t)\frac{y}{\pi(t^{2}+y^{2})}dt$

and

 $\displaystyle(Q_{y}*f)(x)$ $\displaystyle=\int_{\mathbb{R}}Q_{y}(x-t)f(t)dt$ $\displaystyle=\int_{\mathbb{R}}f(t)\frac{x-t}{\pi((x-t)^{2}+y^{2})}dt$ $\displaystyle=\int_{\mathbb{R}}f(x-t)\frac{t}{\pi(t^{2}+y^{2})}dt.$

Then

 $\displaystyle P_{y}*f(x)+iQ_{y}*f(x)$ $\displaystyle=\int_{\mathbb{R}}f(x-t)\frac{1}{\pi}\frac{1}{y-it}dt$ $\displaystyle=\int_{\mathbb{R}}f(t)\frac{1}{\pi}\frac{1}{y-ix+it}dt$ $\displaystyle=\frac{i}{\pi}\int_{\mathbb{R}}\frac{f(t)}{x+iy-t}dt.$

For $y_{1},y_{2}>0$,

 $\displaystyle\widehat{P_{y_{1}}*P_{y_{2}}}(\xi)$ $\displaystyle=\widehat{P_{y_{1}}}(\xi)\cdot\widehat{P_{y_{2}}}(\xi)$ $\displaystyle=e^{-2\pi y_{1}|\xi|}\cdot e^{-2\pi y_{2}|\xi|}$ $\displaystyle=e^{-2\pi(y_{1}+y_{2})|\xi|}$ $\displaystyle=\widehat{P_{y_{1}+y_{2}}}(\xi).$

Therefore $(P_{y})_{y>0}$ is a semigroup using convolution: for $y_{1},y_{2}>0$,

 $P_{y_{1}}*P_{y_{2}}=P_{y_{1}+y_{2}}.$

Let $\mathbb{H}=\{z\in\mathbb{C}:\mathrm{Im}\,z>0\}$ and for $\phi\in\mathscr{S}$ let

 $F_{\phi}(z)=\frac{i}{\pi}\int_{\mathbb{R}}\frac{\phi(t)}{z-t}dt,\qquad z\in% \mathbb{H},$

which is a complex analytic function. For $z=x+iy\in\mathbb{H}$,

 $P_{y}*\phi(x)+iQ_{y}*\phi(x)=\frac{i}{\pi}\int_{\mathbb{R}}\frac{\phi(t)}{x+iy% -t}dt=F_{\phi}(z).$

It is proved that for $1\leq p<\infty$ and $f\in L^{p}(\mathbb{R})$, $Q_{\epsilon}*f-H_{\epsilon}f\to 0$ in $L^{p}$ as $\epsilon\to 0$, and that for almost all $x\in\mathbb{R}$, $Q_{\epsilon}*f(x)-H_{\epsilon}f(x)\to 0$ as $\epsilon\to 0$.11 1 Loukas Grafakos, Classical Fourier Analysis, second ed., p. 254, Theorem 4.1.5.

For $1, it can be proved that there is some $C_{p}$ such that

 $\left\|H\phi\right\|_{L^{p}}\leq C_{p}\left\|\phi\right\|_{L^{p}}$

for all $\phi\in\mathscr{S}$, with $C_{p}\leq 2p$ for $2\leq p<\infty$ and $C_{p}\leq\frac{2p}{p-1}$ for $1.22 2 Loukas Grafakos, Classical Fourier Analysis, second ed., p. 255, Theorem 4.1.7.