Is a bounded function always the Hilbert transform of some other function?

Question

Given $f \in L^\infty(\mathbb R)$, does there always exist a $g$ (in some space) such that \begin{equation*} Hg=f, \end{equation*} where $Hg$ is the Hilbert transform of $g$ ? In other words, is the Hilbert transform defined for $BMO$ functions?

Answer 1

The Hilbert transform is an anti-involution, meaning $H(H(u))=-u$. This is easiest to observe on the Fourier side, where $H$ is the multiplication by $-i\operatorname{sign}\omega$, hence $H^2$ multiplies by $-1$.

Thus, $g=-Hf$ does the job. The function $g$ is generally not in $L^\infty$, but it belongs to BMO.