TITLE: Function invariant under Hilbert transform
QUESTION [5 upvotes]: Let $f\in C_0^\infty(\mathbb{R}^n)$ and
$$Hf(x)= \operatorname{p.v.}\int_{\mathbb{R}}\frac{f(x-y)}{y} \, dx$$
the Hilbert transform of $f$. Is it possible that $Hf=f$ (a.e. and possibly after extending $H$ to $L^p$ space) ?

REPLY [8 votes]: Taking the Fourier transform on both sides of $Hf = f$ you get
$$
-i \operatorname{sign}(\xi) \widehat{f}(\xi) = \widehat{f}(\xi).
$$
This is true if and only if $\widehat{f}(\xi) = 0$ almost everywhere, but this implies that $f$ is $0$.