Hilbert Transform

Hilbert Transform

$f\in L^2(\mathbb{T})$, $Hf(x)=-i\left(\frac{1}{2}\hat{f}(0)+{\sum }_{n=1}^{\infty }\hat{f}(n)e^{2\pi inx}\right)$

(the series converges in $L^2(\mathbb{T})$, and $Hf$ is defined to be the limit in $L^2(\mathbb{T})$ norm of the series)

Proposition

$Hf(x)=\frac{1}{2}{\int }_{\mathbb{T}}f(x-y)\cot (\pi y)dy.$

Proof To do so it helps to define, for $f\in L^2(\mathbb{T})$, the asymmetric partial sums $H_{N}f(x):=-i\left(\frac{1}{2}\hat{f}(0)+{\sum }_{n=1}^N\hat{f}(n)e^{2\pi inx}\right)$ that approximates $Hf$ in $L^2(\mathbb{T})$. In fact, we have ${\lim }_{N\to \infty }\Vert H_{N}f-Hf{\Vert }_{L^2(\mathbb{T})}=0.$ Note that $H_{N}f(x)=f\ast K_N(x)={\int }_{\mathbb{T}}f(x-y)K_N(y)dy,$ where $K_N$ is given by

$=-i\left(\frac{e^{2\pi i(N+1)x}-e^{2\pi ix}}{i2\sin \pi x{e}^{\pi ix}}+\frac{1}{2}\right)=\frac{\cos (\pi x)-e^{2\pi i(N+\frac{1}{2})x}}{2\sin (\pi x)}.$ Hence $H_{N}f(x)={\int }_{\mathbb{T}}f(x-y)\frac{\cos (\pi y)-e^{2\pi i(N+\frac{1}{2})y}}{2\sin (\pi y)}dy.$ Now suppose $f\in L^2(\mathbb{T})$ (in particular, $f\in L^1(\mathbb{T})$ as well). If $x\notin \text{supp }f$, then $y\mapsto \frac{f(x-y)e^{\pi iy}}{2\sin (\pi y)}$ is at a positive distance from the compact set $\text{supp }f$. Thus $y\mapsto \frac{f(x-y)e^{\pi iy}}{2\sin (\pi y)}$ is a function in $L^1(\mathbb{T})$, and Riemann-Lebesgue lemma implies ${\lim }_{N\to \infty }{\int }_{\mathbb{T}}\frac{f(x-y)e^{\pi iy}}{2\sin (\pi y)}{e}^{2\pi iNy}dy=0.$ On the other hand, since $H_{N}f$ converges to $Hf$ in $L^2(\mathbb{T})$, after passing to a subsequence, $H_{N}f$ converges pointwise almost everywhere on $\mathbb{T}$ to $Hf$. So for almost every $x\notin \text{supp }f$, we have $Hf(x)=\frac{1}{2}{\int }_{\mathbb{T}}f(x-y)\cot (\pi y)dy.$

Theorem

For all $p\in (1,\infty )$, $f\in L^2\cap L^p(\mathbb{T})$ $\exists C_p$ such that $\Vert Hf{\Vert }_{L^p(\mathbb{T})}\le C_p\Vert f{\Vert }_{L^p(\mathbb{T})}$

Proof

Claim 1.

For every interval $I\subset \mathbb{T}$ and every $a\in L^2(\mathbb{T})$ such that $a(y)=0$ for all $y\in \mathbb{T}\backslash I$ and ${\int }_{I}a(y)dy=0$, we have $\Vert Ha{\Vert }_{L^1(\mathbb{T}\backslash 2I)}\le \pi \Vert a{\Vert }_{L^1(I)}.$ Here $2I$ is the interval with the same center as $I$ but double its length.

Proof If $c_I$ is the center of $I$, and $\Vert z\Vert$ denotes the distance of $z$ to the nearest integer, then for $x\notin 2I$ and $y\in I$ we have $\Vert x-y\Vert \ge \frac{1}{2}\Vert x-c_I\Vert$ and hence $|\cot (\pi (x-y))-\cot (\pi (x-c_I))|=|y-c_I|\frac{\pi }{{\sin }^2(\pi (x-y^{\prime }))}\le \frac{|I|}{2}\frac{\pi }{\Vert x-y^{\prime }{\Vert }^2}\le \frac{2\pi |I|}{\Vert x-c_I{\Vert }^2}.$ (Here $y^{\prime }$ is some point between $y$ and $c_I$ that one obtains by applying the mean value theorem.) Now since $a\in L^2(\mathbb{T})$ is supported on $I$, by our earlier formula for $H$, for almost every $x\notin 2I$, we have $Ha(x)=\frac{1}{2}{\int }_{\mathbb{T}}a(y)\cot (\pi (x-y))dy.$ Thus using ${\int }_{I}a(y)dy=0$, for almost every $x\notin 2I$ we have

$$|Ha(x)|=\left|\frac{1}{2}{\int }_{\mathbb{T}}a(y)(\cot (\pi (x-y))-\cot (\pi (x-c_I)))dy\right|$$ \le\int_{\mathbb{T}}|a(y)|\frac{\pi|I|}{\|x-c_{I}\|^2}dy$$ So $$\int_{\mathbb{T}\backslash2I}|Ha(x)|dx\le\int_{\mathbb{T}\backslash2I}\int_{\mathbb{T}}|a(y)|\frac{\pi|I|}{\|x-c_{I}\|^2}dydx\le\pi\|a\|_{L^{1}(\mathbb{T})}.$$ ## Claim 2. There exists a finite constant $C_{1}$, such that for every $f\in L^{2}(\mathbb{T})$ and every $\alpha>0$, $$|\{x\in\mathbb{T}:|Hf(x)|>\alpha\}|\le\frac{C_{1}}{\alpha}\|f\|_{L^{1}(\mathbb{T})}.$$ *Proof* To prove Claim 2, note that by choosing $C_{1}\ge1$ one may consider only those $f$ for which $\alpha>\|f\|_{L^{1}(\mathbb{T})}$. Let $f\in L^{2}(\mathbb{T})$. Given $\alpha>\|f\|_{L^{1}(\mathbb{T})}$, we identify $\mathbb{T}$ with $[0, 1)$ and partition $\mathbb{T}$ into two dyadic intervals $[0,1/2)$ and $[1/2,1)$. For each of them we ask: is $\alpha$ bigger than the average of $|f|$ over the interval? If yes, we keep partitioning it; if not, we take stock. This gives us a family of disjoint dyadic intervals $\mathcal{I}=\{I\}$ for which $$\alpha\le\frac{1}{|I|}\int_{I}|f(y)|dy<2\alpha;$$ for every $I\in\mathcal{I}.$ We define $\Omega=\bigcup_{I\in\mathcal{I}}I$ and $$g(x):=\begin{cases}f(x) & \text{if } x\in\mathbb{T}\backslash\Omega \\ \frac{1}{|I|}\int_{I}f(y)dy & \text{if } x\in I\end{cases}$$ $$b(x)=\sum_{I\in\mathcal{I}}b_{I}(x),$$ where $$b_{I}(x):=\begin{cases}f(x)-\frac{1}{|I|}\int_{I}f(y)dy & \text{if } x\in I \\ 0 & \text{otherwise} \end{cases}$$ One can check that $g\in L^{2}(\mathbb{T})$ with $$\|g\|_{L^{2}(\mathbb{T})}^{2}\le2\alpha\|f\|_{L^{1}(\mathbb{T})}.$$ Also, for every $I\in\mathcal{I}$, $b_{I}\in L^{2}(\mathbb{T})$ is supported on $I$, with $$\int_{I}b_{I}(y)dy=0.$$ We also have $$\sum_{I\in\mathcal{I}}\|b_{I}\|_{L^{1}(\mathbb{T})}\le2\|f\|_{L^{1}(\mathbb{T})}$$ In fact, $g\in L^{\infty}(\mathbb{T})$ with $\|g\|_{L^{\infty}(\mathbb{T})}\le2\alpha,$ and $\|g\|_{L^{1}(\mathbb{T})}\le\|f\|_{L^{1}(\mathbb{T})}$. And $$\sum_{I\in\mathcal{I}}|I|\le\frac{1}{\alpha}\|f\|_{L^{1}(\mathbb{T})}.$$ This decomposition of $f$ into $g+b$ is called the Calderón-Zygmund decomposition of $f$ at height $\alpha$. Since $Hf=Hg+Hb,$ we bound $$|\{x\in\mathbb{T}:|Hf(x)|>\alpha\}|\le|\{x\in\mathbb{T}:|Hg(x)|>\alpha/2\}|+|\{x\in\mathbb{T}:|Hb(x)|>\alpha/2\}|.$$ Now $$\|Hg\|_{L^{2}(\mathbb{T})}^{2}\le\|g\|_{L^{2}(\mathbb{T})}^{2}\le2\alpha\|f\|_{L^{1}(\mathbb{T})}$$ which implies $$|\{x\in\mathbb{T}:|Hg(x)|>\alpha/2\}|\le\left(\frac{2}{\alpha}\right)^{2}\|Hg\|_{L^{2}(\mathbb{T})}^{2}\le\frac{8}{\alpha}\|f\|_{L^{1}(\mathbb{T})}.$$ Furthermore, let $\Omega^{*}=\bigcup_{I\in\mathcal{I}}(2I)$. Then $$\|Hb\|_{L^{1}(\mathbb{T}\backslash\Omega^{*})}\le\sum_{I\in\mathcal{I}}\|Hb_{I}\|_{L^{1}(\mathbb{T}\backslash2I)}.$$ But $b_{I}$ is supported on $I$ and $\int_{I}b_{I}(y)dy=0$. So by Claim 1, $$\|Hb_{I}\|_{L^{1}(\mathbb{T}\backslash2I)}\le\pi\|b_{I}\|_{L^{1}(\mathbb{T})}$$ Thus from $\sum_{I\in\mathcal{I}}\|b_{I}\|_{L^{1}(\mathbb{T})}\le2\|f\|_{L^{1}(\mathbb{T})}$ we obtain $$\|Hb\|_{L^{1}(\mathbb{T}\backslash\Omega^{*})}\le2\pi\|f\|_{L^{1}(\mathbb{T})}$$ which gives $|\{x\in\mathbb{T}\backslash\Omega^{*}:|Hb(x)|>\alpha/2\}|\le\frac{2}{\alpha}\|Hb\|_{L^{1}(\mathbb{T}\backslash\Omega^{*})}\le\frac{4\pi}{\alpha}\|f\|_{L^{1}(\mathbb{T})}.$ Since $$|\Omega^{*}|\le\sum_{I\in\mathcal{I}}|2I|\le\frac{2}{\alpha}\|f\|_{L^{1}(\mathbb{T})},$$ together we see that $$|\{x\in\mathbb{T}:|Hb(x)|>\alpha/2\}|\le\frac{4\pi+2}{\alpha}\|f\|_{L^{1}(\mathbb{T})}.$$ Finally we conclude $$|\{x\in\mathbb{T}:|Hf(x)|>\alpha\}|\le\frac{4\pi+10}{\alpha}\|f\|_{L^{1}(\mathbb{T})}$$ so Claim 2 holds with $C_{1}=4\pi+10$. ## Claim 3. For all $1<p<2$, there exists a finite constant $C_{p}$ such that $\|Hf\|_{L^{p}(\mathbb{T})} \leq C_{p}\|f\|_{L^{p}(\mathbb{T})}$ holds for all $f\in L^{2}(\mathbb{T})$. *Proof* Let $f\in L^{2}(\mathbb{T})$, $1<p<2$. We bound $$\|Hf\|_{L^{p}(\mathbb{T})}^{p}=\int_{0}^{\infty}p\alpha^{p-1}|\{x\in\mathbb{T}:|Hf(x)|>\alpha\}|d\alpha.$$ For every $\alpha>0$ decompose $f=f_{1,\alpha}+f_{2,\alpha}$ where $$f_{1,\alpha}(x):=\begin{cases}f(x)& \text{if }|f(x)|>\alpha \\ 0& \text{if }|f(x)|\le\alpha\end{cases} \quad f_{2,\alpha}(x):=\begin{cases}f(x)& \text{if }|f(x)|\le\alpha \\ 0& \text{if }|f(x)|>\alpha\end{cases}$$ $f_{1,\alpha}$, $f_{2,\alpha}$ are both in $L^{2}(\mathbb{T})$. But $$|\{x\in\mathbb{T}:|Hf(x)|>\alpha\}|\le|\{x\in\mathbb{T}:|Hf_{1,\alpha}(x)|>\alpha/2\}|+|\{x\in\mathbb{T}:|Hf_{2,\alpha}(x)|>\alpha/2\}|.$$ Claim 2 gives $$|\{x\in\mathbb{T}:|Hf_{1,\alpha}(x)|>\alpha/2\}|\le\frac{2C_{1}}{\alpha}\|f_{1,\alpha}\|_{L^{1}(\mathbb{T})}=\frac{2C_{1}}{\alpha}\int_{|f|>\alpha}|f(x)|dx,$$ and $$\int_{0}^{\infty}p\alpha^{p-1}\frac{2C_{1}}{\alpha}\int_{|f|>\alpha}|f(x)|dx=2C_{1}\int_{\mathbb{T}}|f(x)|\int_{0}^{|f(x)|}p\alpha^{p-2}d\alpha dx = \frac{2C_{1}p}{p-1}\|f\|_{L^{p}(\mathbb{T})}^{p}$$ The boundedness of $H$ on $L^{2}(\mathbb{T})$ gives $$|\{x\in\mathbb{T}:|Hf_{2,\alpha}(x)|>\alpha/2\}|\le\left(\frac{2}{\alpha}\right)^{2}\|f_{2,\alpha}\|_{L^{2}(\mathbb{T})}^{2}=\frac{4}{\alpha^{2}}\int_{|f|\le\alpha}|f(x)|^{2}dx,$$ and $$\int_{0}^{\infty}p\alpha^{p-1}\frac{4}{\alpha^{2}}\int_{|f|\le\alpha}|f(x)|^{2}dx=4\int_{\mathbb{T}}|f(x)|^2\int_{|f(x)|}^{\infty}p\alpha^{p-3}d\alpha dx = \frac{4p}{2-p}\|f\|_{L^{p}(\mathbb{T})}^{p}.$$ Thus $$\|Hf\|_{L^{p}(\mathbb{T})}\le\left(\frac{2C_{1}p}{p-1}+\frac{4p}{2-p}\right)^{1/p}\|f\|_{L^{p}(\mathbb{T})}.$$ ## Claim 4. For all $2<p<\infty$, there exists a finite constant $C_{p}$ such that $\|Hf\|_{L^{p}(\mathbb{T})} \leq C_{p}\|f\|_{L^{p}(\mathbb{T})}$ holds for all $f\in L^{p}(\mathbb{T})$. *Proof* Let $2\le p<\infty$. $f\in L^{p}(\mathbb{T})$. Since $L^{2}(\mathbb{T})$ is dense in $L^{p^{\prime}}(\mathbb{T})$, we have $$\|Hf\|_{L^{p}(\mathbb{T})}=\sup_{g\in L^{2}(\mathbb{T}),\|g\|_{L^{p^{\prime}}(\mathbb{T})}=1}\left|\int_{\mathbb{T}}Hf(x)\overline{g(x)}dx\right|.$$ But $f\in L^{p}(\mathbb{T})$ implies $f\in L^{2}(\mathbb{T})$ as well. For $f,g\in L^{2}(\mathbb{T})$, we have $$\int_{\mathbb{T}}Hf(x)\overline{g(x)}dx=\int_{\mathbb{T}}f(x)\overline{Hg(x)}dx$$ which one can verify using Parseval's identity. Furthermore, for $g\in L^{2}(\mathbb{T})$ with $\|g\|_{L^{p^{\prime}}(\mathbb{T})}=1$, we proved in Claim 3 that $$\|Hg\|_{L^{p^{\prime}}(\mathbb{T})}\le C_{p^{\prime}}$$ and Hölder's inequality then implies $$\left|\int_{\mathbb{T}}f(x)\overline{Hg(x)}dx\right|\le C_{p^{\prime}}\|f\|_{L^{p}(\mathbb{T})}.$$ Thus $$\|Hf\|_{L^{p}(\mathbb{T})}=\sup_{g\in L^{2}(\mathbb{T}),\|g\|_{L^{p^{\prime}}(\mathbb{T})}=1}\left|\int_{\mathbb{T}}f(x)\overline{Hg(x)}dx\right|\le C_{p^{\prime}}\|f\|_{L^{p}(\mathbb{T})}$$ and this completes the proof of Claim 4. In fact for $p\ge2$ we even have $L^{p}\cap L^{2}(\mathbb{T})=L^{p}(\mathbb{T})$. >[!definition] Modulations and Reflections > >$$M_{N}f(x):=e^{2\pi iNx}f(x)$$