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 $1<p<\infty$, 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
$$||\mathcal{H}{f}||_{C^{\alpha}}\leq C||f||_{C^{\alpha}}?$$
Edit: I will accept answer of the David as correct.
To the other readers, I ask you to read all the answers and comments to understand such a choice.
 A: 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 $$H_{\epsilon, A}f(x)=\int_{\epsilon<|y|<A}f(x-y)\frac{dy}y,$$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.
A: It is not true.
We take the function
$$
f(x) = 
\begin{cases}
 1 &:& x\geq 1,
\\ x &:& -1<x<1,
\\ -1 &:& x\leq -1.
\end{cases}
$$
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 $\|\mathcal Hf\|=\infty$.
A: 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 $$||Hf||_{Lip_1}=||(Hf)'||_\infty=||H(f')||_\infty
\not\le c||f'||_\infty=c||f||_{Lip_1}.$$
