The Boundedness Problem for Generalized Hilbert Operators on Hardy Spaces

Abstract

Let $g$ be analytic in the unit disc and consider the generalized Hilbert operator $${\mathcal{H}}_g(f)(z)={\int }_0^{1}f(t)g^{\prime }(tz)dt.$$ The boundedness of ${\mathcal{H}}_g$ on $H^p$ is characterized by the mean Lipschitz condition $g\in \Lambda \left(p,\frac{1}{p}\right)$ when $1<p\le 2$, while the problem remains open for $2<p<\infty$. It has been recently proved that the condition $g\in \Lambda \left(p,\frac{1}{p}\right)$ does not imply the boundedness of ${\mathcal{H}}_g$ on $H^p$, $2<p<\infty$ \cite{GuoTang2026}. We show that this condition is far from sufficient in the latter range: for every $2<p<\infty$, there exists a function $g\in \Lambda \left(p,\frac{1}{p}\right)$ such that ${\mathcal{H}}_g$ is not bounded even from $H^p$ into $H^1$. The main ingredient is an exact characterization of the boundedness of ${\mathcal{H}}_g:H^p\to H^2$ for all $1\le p\le \infty$. In particular, when $2<p<\infty$, this mapping is bounded if and only if $g^{\prime }$ belongs to a certain mixed-norm space. For lacunary symbols, the same mixed-norm condition also characterizes the boundedness of ${\mathcal{H}}_g$ on $H^p$, and hence gives a complete solution of the open problem within this class of symbols. We also show that, for $1\le q\le \infty$, boundedness of ${\mathcal{H}}_g:H^1\to H^q$ is characterized by the condition $g^{\prime }\in H^q$. We also characterize compactness of ${\mathcal{H}}_g$ in the aforementioned cases.