Hilbert-type operator induced by radial weight on Hardy spaces

We consider the Hilbert-type operator defined by $H_ω(f)(z)=\int_0^1 f(t)\left(\frac{1}{z}\int_0^z B^ω_t(u)\,du\right)\,ω(t)dt,$ where $\{B^ω_ζ\}_{ζ\in\mathbb{D}}$ are the reproducing kernels of the Bergman space $A^2_ω$ induced by a radial weight $ω$ in the unit disc $\mathbb{D}$. We prove that $H_ω$ is bounded on the Hardy space $H^p$, $1<p<\infty$, if and only if \begin{equation} \label{abs1} \sup_{0\le r<1} \frac{\widehatω(r)}{\widehatω\left( \frac{1+r}{2}\right)}<\infty, \tag† \end{equation} and \begin{equation*} \sup\limits_{0<r<1}\left(\int_0^r \frac{1}{\widehatω(t)^p} dt\right)^{\frac{1}{p}} \left(\int_r^1 \left(\frac{\widehatω(t)}{1-t}\right)^{p'}\,dt\right)^{\frac{1}{p'}} <\infty, \end{equation*} where $\widehatω(r)=\int_r^1 ω(s)\,ds$. We also prove that $H_ω: H^1\to H^1$ is bounded if and only if \eqref{abs1} holds and $\sup\limits_{r \in [0,1)} \frac{\widehatω(r)}{1-r} \left(\int_0^r \frac{ds}{\widehatω(s)}\right)<\infty.$ As for the case $p=\infty$, $H_ω$ is bounded from $H^\infty$ to $BMOA$, or to the Bloch space, if and only if \eqref{abs1} holds. In addition, we prove that there does not exist radial weights $ω$ such that $H_ω: H^p \to H^p$, $1\le p<\infty$, is compact and we consider the action of $H_ω$ on some spaces of analytic functions closely related to Hardy spaces.

No reviews yet.

No comments yet.