The Hilbert transform on $ℝ$

1 The principal value integral

Let $A_{ϵ}=\{x\in ℝ:|x|\ge ϵ\}$. For $f\in {\bigcap }_{ϵ>0}{L}^1(A_{ϵ})$, if ${\int }_{A_{ϵ}}f(x)𝑑x$ has a limit as $ϵ\to 0$, we denote it by

$$PV{\int }_{ℝ}f(x)𝑑x=\underset{ϵ\to 0}{lim}{\int }_{|x|\ge ϵ}f(x)𝑑x.$$

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

$\left|{\displaystyle {\int }_{A_{1/j}}}{\displaystyle \frac{\varphi (x)}{x}}𝑑x-{\displaystyle {\int }_{A_{1/k}}}{\displaystyle \frac{\varphi (x)}{x}}𝑑x\right|$

$={\parallel {\varphi }^{\prime }\parallel }_b\cdot \left({\displaystyle \frac{1}{j}}-{\displaystyle \frac{1}{k}}\right)$

$={\parallel {\varphi }^{\prime }\parallel }_b\cdot {\displaystyle \frac{k-j}{kj}}$

$\le {\displaystyle \frac{{\parallel {\varphi }^{\prime }\parallel }_b}{j}}.$

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

$$⟨\varphi ,W⟩=PV{\int }_{ℝ}\frac{\varphi (x)}{x}𝑑x=\underset{ϵ\to 0}{lim}{\int }_{|x|\ge ϵ}\frac{\varphi (x)}{x}𝑑x.$$

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

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

${\displaystyle {\int }_{ϵ}^1}{\varphi }^{\prime }(x)\log xdx$ $={x\psi (x)\log x|}_{ϵ}^1-{\displaystyle {\int }_{ϵ}^1}(\varphi (x)-\varphi (0)){\displaystyle \frac{1}{x}}𝑑x$

$=-ϵ\psi (ϵ)\log ϵ-{\displaystyle {\int }_{ϵ}^1}{\displaystyle \frac{\varphi (x)}{x}}𝑑x-\varphi (0)\log ϵ$

and

${\displaystyle {\int }_{-1}^{-ϵ}}{\varphi }^{\prime }(x)\log |x|dx$ $={x\psi (x)\log (-x)|}_{-1}^{-ϵ}-{\displaystyle {\int }_{-1}^{-ϵ}}(\varphi (x)-\varphi (0)){\displaystyle \frac{1}{x}}𝑑x$

$=-ϵ\psi (-ϵ)\log ϵ-{\displaystyle {\int }_{-1}^{-ϵ}}{\displaystyle \frac{\varphi (x)}{x}}𝑑x+\varphi (0)\log ϵ.$

Hence

$${\int }_{ϵ\le |x|\le 1}{\varphi }^{\prime }(x)\log |x|dx=-ϵ\cdot (\psi (ϵ)+\psi (-ϵ))\cdot \log ϵ-{\int }_{ϵ\le |x|\le 1}\frac{\varphi (x)}{x}𝑑x.$$

On the other hand,

$${\int }_{|x|\ge 1}{\varphi }^{\prime }(x)\log |x|dx=-{\int }_{|x|\ge 1}\frac{\varphi (x)}{x}𝑑x.$$

Therefore

$$PV{\int }_{ℝ}{\varphi }^{\prime }(x)\log |x|dx=-PV{\int }_{ℝ}\frac{\varphi (x)}{x}𝑑x.$$

Let $\mu \varphi (x)=\varphi (-x)$ and ${\tau }_y\varphi (x)=\varphi (x-y)$. Then

$${\tau }_y\mu \varphi (x)=\mu \varphi (x-y)=\varphi (y-x)={\tau }_x\varphi (y).$$

Write

$\varphi *\psi (x)$ $={\displaystyle {\int }_{ℝ}}\varphi (y)\psi (x-y)𝑑y$

$={\displaystyle {\int }_{ℝ}}\varphi (y)({\tau }_y\psi )(x)𝑑y$

$={\displaystyle {\int }_{ℝ}}\varphi (y)({\tau }_x\mu \psi )(y)𝑑y.$

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

$$(\varphi *u)(x)=⟨{\tau }_x\mu \varphi ,u⟩.$$

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

$$⟨\psi ,\varphi *u⟩=⟨(\mu \varphi )*\psi ,u⟩.$$

Define ${\tau }_{x}u\in {𝒮}^{\prime }$ by

$$⟨\varphi ,{\tau }_{x}u⟩=⟨{\tau }_{-x}\varphi ,u⟩.$$

2 The Hilbert transform

For $ϵ>0$, for $\varphi \in 𝒮$ let

$H_{ϵ}\varphi (x)$ $={\displaystyle \frac{1}{\pi }}{\displaystyle {\int }_{|y|\ge ϵ}}{\displaystyle \frac{\varphi (x-y)}{y}}𝑑y$

$={\displaystyle \frac{1}{\pi }}{\displaystyle {\int }_{|y|\ge ϵ}}{\displaystyle \frac{{\tau }_y\varphi (x)}{y}}𝑑y$

$={\displaystyle \frac{1}{\pi }}{\displaystyle {\int }_{|y|\ge ϵ}}{\displaystyle \frac{{\tau }_x\mu \varphi (y)}{y}}𝑑y.$

Define

$H\varphi (x)$ $=\underset{ϵ\to 0}{lim}{H}_{ϵ}\varphi (x)$

$={\displaystyle \frac{1}{\pi }}\cdot PV{\displaystyle {\int }_{ℝ}}{\displaystyle \frac{{\tau }_x\mu \varphi (y)}{y}}𝑑y$

$={\displaystyle \frac{1}{\pi }}\cdot PV{\displaystyle {\int }_{ℝ}}{\displaystyle \frac{\varphi (x-y)}{y}}𝑑y$

$={\displaystyle \frac{1}{\pi }}\cdot PV{\displaystyle {\int }_{ℝ}}{\displaystyle \frac{\varphi (y)}{x-y}}𝑑y.$

For $x\in ℝ$,

$\varphi *W(x)$ $=⟨{\tau }_x\mu \varphi ,W⟩$

$=PV{\displaystyle {\int }_{ℝ}}{\displaystyle \frac{{\tau }_x\mu \varphi (y)}{y}}𝑑y$

$=PV{\displaystyle {\int }_{ℝ}}{\displaystyle \frac{\varphi (x-y)}{y}}𝑑y$

$=PV{\displaystyle {\int }_{ℝ}}{\displaystyle \frac{\varphi (y)}{x-y}}𝑑y$

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

Thus

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

We calculate

$⟨\varphi ,\widehat{W}⟩$ $=⟨\widehat{\varphi },W⟩$

$=\underset{ϵ\to 0}{lim}{\displaystyle {\int }_{ϵ\le |\xi |\le 1/ϵ}}{\displaystyle \frac{\widehat{\varphi }(\xi )}{\xi }}𝑑\xi$

$=\underset{ϵ\to 0}{lim}{\displaystyle {\int }_{|\xi |\ge ϵ}}\left({\displaystyle {\int }_{ℝ}}\varphi (x)e^{-2\pi ix\xi }𝑑x\right){\displaystyle \frac{1}{\xi }}𝑑\xi$

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

$=\underset{ϵ\to 0}{lim}{\displaystyle {\int }_{ℝ}}\varphi (x)\left({\displaystyle {\int }_{ϵ\le |\xi |\le 1/ϵ}}-i{\displaystyle \frac{\sin 2\pi x\xi }{\xi }}d\xi \right)𝑑x.$

Check that

$$\underset{ϵ\to 0}{lim}{\int }_{ϵ\le |\xi |\le 1/ϵ}\frac{\sin 2\pi x\xi }{\xi }𝑑\xi =\pi \cdot \mathrm{sgn}x.$$

Then, using the dominated convergence theorem,

$$⟨\varphi ,\widehat{W}⟩={\int }_{ℝ}\varphi (x)\cdot -i\pi \cdot \mathrm{sgn}xdx.$$

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

Now,

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

Then

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

Let

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

with which

$$\widehat{H\varphi }=m_H\cdot \widehat{\varphi }.$$

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

$$FH\varphi =m_H\cdot F\varphi .$$

So

$$H\varphi =F^{-1}(m_H\cdot F\varphi ),$$

and hence

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

For $\xi \ne 0$, $m_H{(\xi )}^2=-1$, which yields

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

Thus $H^2=-\mathrm{id}$. Therefore ${\parallel H\varphi \parallel }_{L^2}={\parallel \varphi \parallel }_{L^2}$.

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

$⟨Hf,g⟩$ $=⟨\widehat{Hf},\widehat{g}⟩$

$={\displaystyle {\int }_{ℝ}}\widehat{Hf}(\xi )\cdot \overline{\widehat{g}(\xi )}𝑑\xi$

$={\displaystyle {\int }_{ℝ}}{m}_H(\xi )\cdot \widehat{f}(\xi )\overline{\widehat{g}(\xi )}𝑑\xi$

$=-{\displaystyle {\int }_{ℝ}}\widehat{f}(\xi )\cdot \overline{m_H(\xi )\cdot \widehat{g}(\xi )}𝑑\xi$

$=-{\displaystyle {\int }_{ℝ}}\widehat{f}(\xi )\cdot \overline{\widehat{Hg}(\xi )}𝑑\xi$

$=-⟨f,Hg⟩.$

But $⟨Hf,g⟩=⟨f,H^{*}g⟩$, so

$$⟨f,H^{*}g⟩=⟨f,-Hg⟩,$$

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

${\displaystyle {\int }_{ℝ}}{e}^{2\pi i\xi x}{e}^{-2\pi y|\xi |}𝑑\xi$ $={\displaystyle \frac{1}{2\pi ix+2\pi y}}-{\displaystyle \frac{1}{2\pi ix-2\pi y}}$

$={\displaystyle \frac{2\pi ix-2\pi y-2\pi ix-2\pi y}{-4{\pi }^2{x}^2-4{\pi }^2{y}^2}}$

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

and

$-i{\displaystyle {\int }_{ℝ}}{e}^{2\pi i\xi x}\mathrm{sgn}(\xi )e^{-2\pi y|\xi |}𝑑\xi$ $={\displaystyle \frac{i}{2\pi ix+2\pi y}}+{\displaystyle \frac{i}{2\pi ix-2\pi y}}$

$={\displaystyle \frac{i(2\pi ix-2\pi y+2\pi ix+2\pi y)}{-4{\pi }^2{x}^2-4{\pi }^2{y}^2}}$

$={\displaystyle \frac{-4\pi x}{-4{\pi }^2{x}^2-4{\pi }^2{y}^2}}$

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

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,

$$P_y(x)+i{Q}_y(x)=\frac{1}{\pi }\frac{y+ix}{x^2+y^2}=\frac{1}{\pi }\frac{1}{y-ix}.$$

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

$(P_y*f)(x)$ $={\displaystyle {\int }_{ℝ}}{P}_y(x-t)f(t)𝑑t$

$={\displaystyle {\int }_{ℝ}}f(t){\displaystyle \frac{y}{\pi ({(x-t)}^2+y^2)}}𝑑t$

$={\displaystyle {\int }_{ℝ}}f(x-t){\displaystyle \frac{y}{\pi (t^2+y^2)}}𝑑t$

and

$(Q_y*f)(x)$ $={\displaystyle {\int }_{ℝ}}{Q}_y(x-t)f(t)𝑑t$

$={\displaystyle {\int }_{ℝ}}f(t){\displaystyle \frac{x-t}{\pi ({(x-t)}^2+y^2)}}𝑑t$

$={\displaystyle {\int }_{ℝ}}f(x-t){\displaystyle \frac{t}{\pi (t^2+y^2)}}𝑑t.$

Then

$P_y*f(x)+i{Q}_y*f(x)$ $={\displaystyle {\int }_{ℝ}}f(x-t){\displaystyle \frac{1}{\pi }}{\displaystyle \frac{1}{y-it}}𝑑t$

$={\displaystyle {\int }_{ℝ}}f(t){\displaystyle \frac{1}{\pi }}{\displaystyle \frac{1}{y-ix+it}}𝑑t$

$={\displaystyle \frac{i}{\pi }}{\displaystyle {\int }_{ℝ}}{\displaystyle \frac{f(t)}{x+iy-t}}𝑑t.$

For $y_1,y_2>0$,

$\widehat{P_{y_1}*P_{y_2}}(\xi )$ $=\widehat{P_{y_1}}(\xi )\cdot \widehat{P_{y_2}}(\xi )$

$=e^{-2\pi y_1|\xi |}\cdot e^{-2\pi y_2|\xi |}$

$=e^{-2\pi (y_1+y_2)|\xi |}$

$=\widehat{P_{y_1+y_2}}(\xi ).$

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 $ℍ=\{z\in ℂ:\mathrm{Im}z>0\}$ and for $\varphi \in 𝒮$ let

$$F_{\varphi }(z)=\frac{i}{\pi }{\int }_{ℝ}\frac{\varphi (t)}{z-t}𝑑t,z\in ℍ,$$

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

$$P_y*\varphi (x)+i{Q}_y*\varphi (x)=\frac{i}{\pi }{\int }_{ℝ}\frac{\varphi (t)}{x+iy-t}𝑑t=F_{\varphi }(z).$$

It is proved that for and $f\in L^p(ℝ)$, $Q_{ϵ}*f-H_{ϵ}f\to 0$ in $L^p$ as $ϵ\to 0$, and that for almost all $x\in ℝ$, $Q_{ϵ}*f(x)-H_{ϵ}f(x)\to 0$ as $ϵ\to 0$.

For , it can be proved that there is some $C_p$ such that

$${\parallel H\varphi \parallel }_{L^p}\le C_p{\parallel \varphi \parallel }_{L^p}$$

for all $\varphi \in 𝒮$, with $C_p\le 2p$ for and $C_p\le \frac{2p}{p-1}$ for .