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!