# Hölder norm of the Hilbert Transform

Let $$\mathcal{H}$$ the Hilbert transform defined by $$\mathcal{H}f(x)= p.v.\int_{-\infty}^{+\infty}\frac{f(x-y)}{y}dy.$$ We know that, for each $$1, it is true that $$||\mathcal{H}f||_{L^p}\leq C_p||f||_{L^p}$$ for some positive constant depending only of $$p$$.

My question is: Consider $$0<\alpha<1$$ and $$||f||_{C^{\alpha}}$$ is the $$\alpha^{th}$$-Holder norm, e.g, $$||f||_{C^{\alpha}(\Omega)}=\sup_{x\neq y\in\Omega}\frac{|f(x)-f(y)|}{|x-y|^{\alpha}}.$$ Is it true that $$||\mathcal{H}{f}||_{C^{\alpha}}\leq C||f||_{C^{\alpha}}?$$

• Yes, take $C=1$. – supinf Feb 11 at 12:03
• adjusted, @supinf. – VVCM Feb 11 at 12:08
• The $L^\infty\to L^\infty$ estimate fails, so there must be some explicit example of a sequence $f_n$ such that $\|Hf_n\|_{L^\infty}/\|f_n\|_{L^\infty}\to \infty$. I guess that the same counterexample would make the $C^\alpha \to C^\alpha$ estimate fail as well. Can you please check? I am interested. – Giuseppe Negro Feb 11 at 12:23
• @GiuseppeNegro Your Idea does not work if the function $f_n$ is not continuous. And i think the heaviside function could be a counterexample for $L^\infty$, but i did not check it. – supinf Feb 11 at 14:18
• @orange, $p.v.$ is the principal value. – VVCM Feb 11 at 19:53

Adding another answer because this really has little to do with my previous answer; instead it's a comment on how one might "interpret" things to reconcile the contradiction between my "yes" and supinf's "no".

We need to be a little careful. Let $$H_{\epsilon, A}f(x)=\int_{\epsilon<|y|so $$H=\lim_{\epsilon\to0,\\A\to\infty}H_{\epsilon, A}.$$

supinf gave a simple example of $$f\in C^\alpha$$ such that $$H_{\epsilon, A}f(0)\to-\infty$$. In fact his example has $$Hf(x)=-\infty$$ for every $$x$$, so if we want to talk about the Hilbert transform on $$C^\alpha$$ we need to modify the definition. Look at it this way:

Of course the $$C^\alpha$$ norm is just a seminorm. It's clear that $$||f||=0$$ if and only if $$f$$ is constant, so we do have a norm on the quotient space $$X_\alpha=C^\alpha/\Bbb C$$, consisting of $$C^\alpha$$ modulo constants.

When I said that $$H$$ was bounded on $$C^\alpha$$ I should have said it was bounded on $$X_\alpha$$. In that context we shouldn't expect pointwise convergence; instead we have this:

True Fact. If $$f\in C^\alpha$$ there exist $$g\in C^\alpha$$ and constants $$c_{\epsilon,A}$$ such that $$H_{\epsilon, A}f(x)-c_{\epsilon, A}\to g(x)$$ for every $$x$$.

I'm not going to show that $$g\in C^\alpha$$ here; that's contained in my previous answer. But I will show that the limit $$g(x)$$ exists (and is finite) for every $$x$$; this resolves the contradiction given by supinf: If he'd defined $$Hf=g$$ then he would not have obtained $$Hf=-\infty$$.

Define $$H=\int f(x-y)\frac{dy}y=\int_{-\infty}^{-1}+\int_{-1}^1+\int_1^\infty=H^{-}+H^0+H^+.$$

First, $$H^0$$ is no problem. If we say $$H_\epsilon^0=\int_{\epsilon<|y|<1}$$ then $$H^0f(x)-H_\epsilon^0f(x)=\int_0^\epsilon (f(x-y)-f(x+y)\frac{dy}y;$$since $$f(x-y)-f(x+y)=O(y^\alpha)$$ this shows that in fact $$H_\epsilon^0 f\to H^0f$$ uniformly.

Now say $$H^+_A=\int_1^A$$. We do need to subtract a constant $$c_A^+$$ to get this to converge. The obvious choice is $$c_A^+=H_A^+f(0)$$, since that certainly gives convergence for $$x=0$$. If I did the calculus correctly we have \begin{align}H_A^+f(x)-H_A^+(0)&=\int_{1-x}^1f(-y)\frac1{y-x}dy \\&+\int_{1}^{A-x}f(-y)\left(\frac1{y-x}-\frac1y\right)dy \\&-\int_{A-x}^Af(-y)\frac{dy}y.\end{align}The first integral on the RHS is independent of $$A$$ , while the second integral tends to somethhing finite as $$A\to\infty$$, since $$f(y)=O(y^\alpha)$$ and $$1/(y+x)=1/y=O(1/y^2)$$; similarly the tird integral tends to $$0$$.

Similarly if $$H_A^-=\int_{-A}^{-1}$$ there exists $$c^-_A$$ such that $$H_A^--c_A^-$$ is pointwise convergent; hence $$H_{\epsilon,A}-(c^+_A+c^-_A)$$ is pointwise convergent.

• This is an issue that also appears for other function spaces; for example, I read in the book of Tao on nonlinear wave equations that negative order Sobolev spaces should be interpreted as "modulo polynomials". I never understood what he meant but now that you write this I think I get it. Thanks! – Giuseppe Negro Feb 12 at 18:15
• Indeed. In any space where $f=\sum f_n$ as "above" the convergence is at best modulo polynomials. Because $\sum\widehat f_n$ can't see what $\hat f$ does at the origin; a distribution with Fourier transform supported at the origin is precisely a polynomial. Here the only polynomials in $C^\alpha$ are the constants. – David C. Ullrich Feb 12 at 19:53

I believe the answer is yes. Going to cheat, pulling out a big gun.

If $$0<\alpha<1$$ then $$C^\alpha$$ is in fact a Besov space; we have $$f\in C^\alpha$$ if and only if $$f=\sum_{n\in\Bbb Z} f_n,$$where $$\widehat{f_n}$$ is supported in the annulus $$A_n=\{\xi:2^{n-1}<|\xi|<2^{n+1}\}$$and $$2^{n\alpha}||f_n||_\infty\le c.$$This sort of decomposition of a Besov space often makes it trivial to determine whether a convolution operator is bounded.

Edit: No, it's not quite right to say $$C^\alpha$$ is a Besov space. What's actually a Besov space is the quotient $$X_\alpha=C^\alpha/\Bbb C$$, the space of $$C^\alpha$$ functions modulo constants. I was trying to avoid technical details, but given the other answer showing that taken literally $$H$$ does not map $$C^\alpha$$ to $$C^\alpha$$ it seems we cannot ignore the issue.

Here for example we're done if we can show that $$||Hf_n||_\infty\le c||f_n||_\infty,$$and that's actually true, even though $$H$$ is not bounded on $$L^\infty$$.

Recall that up to an irrelevant constant $$\widehat{Hf}(\xi)=sgn(\xi)\hat f(\xi).$$Choose a Schwarz function $$\phi_0$$ with $$\widehat\phi_0(\xi)=sgn(\xi)\quad(\xi\in A_0).$$If $$\phi_n$$ is an appropriate dilate of $$\phi_0$$ we have $$\widehat{\phi_n}(\xi)=sgn(\xi)\quad(\xi\in A_n)$$and $$||\phi_n||_1=||\phi_0||_1.$$Hence $$Hf_n=\phi_n*f_n,$$and hence $$||Hf_n||_\infty\le||\phi_n||_1||f_n||_\infty=||\phi_0||_1||f_n||_\infty.$$

(The inequality fails for $$\alpha=1$$; one explanation for why is that $$Lip_1$$ is not a Besov space...)

Why is that, I've been asked. First, $$H$$ is certainly not bounded on $$L^\infty$$; if $$f=\chi_{(0,\infty)}$$ then $$Hf$$ is not bounded.

Hence $$H$$ is not bounded on $$Lip_1$$. Because, in the sense of distributions, $$f\in Lip_1$$ if and only if $$f'\in L^\infty$$. It's clear, say from the Fourier transform, that $$H$$ commutes with the derivative. So $$||Hf||_{Lip_1}=||(Hf)'||_\infty=||H(f')||_\infty \not\le c||f'||_\infty=c||f||_{Lip_1}.$$

• This is a great answer, very interesting. It would be even greater if you could give some details on why the estimate fails for $\alpha=0$ (the $L^\infty$ case) and $\alpha=1$. – Giuseppe Negro Feb 11 at 14:25
• @orange Why yes. – David C. Ullrich Feb 12 at 14:15
• @DavidC.Ullrich, can I make this Palley-Littlewood characterization for the $L^p$ spaces? – VVCM Feb 13 at 13:05
• Yes, There is a Littlewood-Paley theory for $L^p$, but only for $1 < p < \infty$. Nice answer. The fact that the Hilbert transform is bounded between Hölder classes can also be proved without using that the Hilbert transform is a Fourier multiplier. It holds that Calderón-Zygmund operators are bounded over those classes. – Adrián González-Pérez Feb 13 at 21:23

It is not true.

We take the function $$f(x) = \begin{cases} 1 &:& x\geq 1, \\ x &:& -1 First, let us verify that $$\|f\|_{C^\alpha} \leq 3$$. Let $$x,y\in\mathbb [-1,1]$$ be given (the other cases for $$x,y$$ are not that interesting, since $$f$$ has the same value as $$f(1)$$ or $$f(-1)$$). Then $$\frac{|f(x)-f(y)|}{|x-y|^\alpha} = |x-y|^{1-\alpha} \leq 1+|x-y| \leq 3.$$ Thus $$f\in C^\alpha$$, i.e. $$\|f\|_{C^\alpha}$$ is finite.

Calculating $$\mathcal Hf$$ at $$x=0$$ yields $$\mathcal (Hf)(0) = p.v.\int_{-\infty}^\infty \frac{f(-y)}{y} \mathrm dy = \int_{-\infty}^{-1} \frac1y + p.v.\int_{-1}^1 (-1) \mathrm dy + \int_1^\infty \frac{-1}y \mathrm dy = -\infty -2 -\infty = -\infty.$$ Hence $$\mathcal Hf$$ is not in $$C^\alpha$$ because it is not continuous. Therefore $$\|\mathcal Hf\|=\infty$$.

• I have a question, in what sense is $\mathcal H f$ defined, since $f$ is not a usual test function? – Calvin Khor Feb 11 at 14:55
• The function $f$ must also decay at infinity for the inequality to hold. This decay is implicit in the Besov decomposition of David. – Giuseppe Negro Feb 11 at 14:57
• @CalvinKhor I used the definition given by VVCM in the original question. – supinf Feb 11 at 15:02
• Oh, the PV includes a cutoff at infinity. thanks – Calvin Khor Feb 11 at 15:04
• @GiuseppeNegro It was not a requirement that the function decays at infinity. Even if the functions vanish at infinity, it might be possible to find a sequence that leads to a contradiction to the inequality conjectured in the original post. – supinf Feb 11 at 15:07