What kind of $\mathcal{L}^1$ function has bounded Hilbert transform?

Question

It is known that if a function $f \in \mathcal{L}^1 $, then its Hilbert transform may not map it to the $\mathcal{L}^1$ space. So my question is under what kind of condition, the Hilbert transform of $f$ can still be in the $\mathcal{L}^1$ space?

Answer 1

The space of integrable functions whose Hilbert transform is also integrable is the Hardy space $H^1$ (not to be confused with the Sobolev space $H^1$.) A necessary condition (but not sufficient) is that its integral be $0$.