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It is fairly well known that if $T_\varphi$ is a Toeplitz operator on the Hardy-Hilbert space, then $\lVert T_\varphi \rVert = \lVert \varphi \rVert _{\infty}$.

Now, if $\varphi \in L^\infty (\mathbb D)$ and $T_{\varphi}$ is a Toeplitz operator on the Bergman space then it is easy to see from the definition that $\lVert T_\varphi \rVert \le \lVert {\varphi} \rVert_{\infty}$. Is it also the case that bounded of the operator implies boundedness of the symbol? That is, if $\varphi \in L^2 (\mathbb D)$ is some symbol such that $T_\varphi$ is densely defined and bounded operator, do we have that $\lVert \varphi \rVert_{\infty} \le M \lVert T_{\varphi} \rVert$ for some $M > 0$?

I presume that people would have already studied these topics but I am unable to find so. References will be appreciated!

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It is not neccesary in general that $\varphi \in L^\infty(\mathbb{D})$, but it is necessary and sufficient that in a certain sense $\varphi$ must be bounded ``on average in the hyperbolic sense''. The right condition for positive $ \varphi$ is that there exists a constant $C>0$ such that

\begin{equation*} \frac{1}{(1-|w|)^2}\int_{D(z,(1-|z|)/2)}\varphi(w) dxdy \leq C,\,\, \text{for all}\,\, z \in \mathbb{D} \end{equation*} where $D(z,(1-|z|)/2)$ is the Euclidean disc of center $z$ and radius $(1-|z|)/2$.

You can find this theorem in the Book of Zhu "Operator Theory in function spaces", Theorem 6.2.3. The idea for sufficiency is that $T_\varphi(f) = P(\varphi f)$, where $P$ is the Bergman projection. \begin{equation*} |\langle P(\varphi f), g \rangle |\leq \int_\mathbb{D} |\varphi(z) f(z) g(z) | dxdy \le \Big( \int_\mathbb{D} |f(z)|^2 \varphi(z) dxdy \Big)^{\frac{1}{2}} \Big( \int |g(z)|^2\varphi(z) dxdy \Big)^{\frac{1}{2}}. \end{equation*} So if the measure $\varphi(z)dxdy $ is a Carleson measure for the Bergman space we have that \begin{equation*} |\langle T_\varphi(f), g \rangle | \leq C_\varphi \Vert f\Vert \Vert g\Vert. \end{equation*} In particular the function $\varphi(z) = \log\frac{1}{|z|}$ gives a bounded Toeplitz operator but it is not a bounded function.

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