mjhjmtu

Let $mathcalH$ the Hilbert transform defined by
$$mathcalHf(x)= p.v.int_-infty^+inftyfracf(x-y)ydy.$$
We know that, for each $1<p<infty$, it is true that
$$||mathcalHf||_L^pleq C_p||f||_L^p$$
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,
$$||f||_C^alpha(Omega)=sup_xneq yinOmegafracx-y.$$
Is it true that
$$||mathcalHf||_C^alphaleq C||f||_C^alpha?$$

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.

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 $$H_epsilon, Af(x)=int_epsilon<f(x-y)fracdyy,$$so $$H=lim_epsilonto0,\AtoinftyH_epsilon, A.$$

supinf gave a simple example of $fin C^alpha$ such that $H_epsilon, Af(0)to-infty$. 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 $fin C^alpha$ there exist $gin C^alpha$ and constants $c_epsilon,A$ such that $H_epsilon, Af(x)-c_epsilon, Ato g(x)$ for every $x$.

I'm not going to show that $gin C^alpha$ 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 $$H=int f(x-y)fracdyy=int_-infty^-1+int_-1^1+int_1^infty=H^-+H^0+H^+.$$

First, $H^0$ is no problem. If we say $H_epsilon^0=int_epsilon<$ then $$H^0f(x)-H_epsilon^0f(x)=int_0^epsilon (f(x-y)-f(x+y)fracdyy;$$since $f(x-y)-f(x+y)=O(y^alpha)$ this shows that in fact $H_epsilon^0 fto H^0f$ 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
$$beginalignH_A^+f(x)-H_A^+(0)&=int_1-x^1f(-y)frac1y-xdy
\&+int_1^A-xf(-y)left(frac1y-x-frac1yright)dy
\&-int_A-x^Af(-y)fracdyy.endalign$$The first integral on the RHS is independent of $A$ , while the second integral tends to somethhing finite as $Atoinfty$, 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.

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 $fin C^alpha$ if and only if $$f=sum_ninBbb Z f_n,$$where $widehatf_n$ is supported in the annulus $$A_n=xi$$and $$2^nalpha||f_n||_inftyle c.$$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 $$||Hf_n||_inftyle c||f_n||_infty,$$and that's actually true, even though $H$ is not bounded on $L^infty$.

Recall that up to an irrelevant constant $$widehatHf(xi)=sgn(xi)hat f(xi).$$Choose a Schwarz function $phi_0$ with $$widehatphi_0(xi)=sgn(xi)quad(xiin A_0).$$If $phi_n$ is an appropriate dilate of $phi_0$ we have $$widehatphi_n(xi)=sgn(xi)quad(xiin A_n)$$and $$||phi_n||_1=||phi_0||_1.$$Hence $$Hf_n=phi_n*f_n,$$and hence $$||Hf_n||_inftyle||phi_n||_1||f_n||_infty=||phi_0||_1||f_n||_infty.$$

(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, $fin Lip_1$ if and only if $f'in L^infty$. It's clear, say from the Fourier transform, that $H$ commutes with the derivative. So $$||Hf||_Lip_1=||(Hf)'||_infty=||H(f')||_infty
notle c||f'||_infty=c||f||_Lip_1.$$

It is not true.

We take the function
$$
f(x) =
begincases
1 &:& xgeq 1,
\ x &:& -1<x<1,
\ -1 &:& xleq -1.
endcases
$$
First, let us verify that $|f|_C^alpha leq 3$.
Let $x,yinmathbb [-1,1]$ 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
$$
fracx-y
= |x-y|^1-alpha leq 1+|x-y| leq 3.
$$
Thus $fin C^alpha$, i.e. $|f|_C^alpha$ is finite.

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