Orsdxy

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,
$$||f||_{C^{alpha}(Omega)}=sup_{xneq yinOmega}frac{|f(x)-f(y)}{|x-y|^{alpha}}.$$
Is it true that
$$||mathcal{H}{f}||_{C^{alpha}}leq C||f||_{C^{alpha}}?$$

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 $widehat{f_n}$ is supported in the annulus $$A_n={xi:2^{n-1}<|xi|<2^{n+1}}$$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.

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 $$widehat{Hf}(xi)=sgn(xi)hat f(xi).$$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 $$widehat{phi_n}(xi)=sgn(xi)quad(xiin A_n)$$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 $$||Hf||_{Lip_1}=||(Hf)'||_infty=||H(f')||_infty
notle c||f'||_infty=c||f||_{Lip_1}.$$

This proves in turn that $L^infty$ and $Lip_1$ are not Besov spaces, because any Besov space has a characterization analogous to the above, with a different defining inequality, and it follows exactly as above that $H$ is bounded on every Besov space.

It is not true.

We take the function
$$
f(x) =
begin{cases}
1 &:& xgeq 1,
\ x &:& -1<x<1,
\ -1 &:& xleq -1.
end{cases}
$$
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
$$
frac{|f(x)-f(y)|}{|x-y|^alpha}
= |x-y|^{1-alpha} leq 1+|x-y| leq 3.
$$
Thus $fin C^alpha$, i.e. $|f|_{C^alpha}$ is finite.

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