Function invariant under Hilbert transform

Question

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) ?

Answer 1

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 $0$.