Toeplitz operator $T_u$ defined on the Bergman space $A^2 (\mathbb D)$ corresponding to the symbol $u \in L^{\infty} (\mathbb D)$ is defined as $$T_u : f \mapsto P (uf),\ f \in A^2 (\mathbb D)$$ where $P : L^2(\mathbb D) \longrightarrow A^2(\mathbb D)$ is the orthogonal projection.
In one of the seminars I came to know that Toeplitz operators are integral operators on the Bergman space but I don't know how the conclusion is drawn. Can someone shed some light on it?
Thanks for your kind attention.
Note that the Bergman space $A^2(\mathbb D)$ is a reproducing kernel Hilbert space with the reproducing kernel given by $$B_{\mathbb D} (z,w) = \frac {1} {\left (1 - z \overline w \right )^2},\ z, w \in \mathbb D$$ with respect to the normalized area measure $A$ of $\mathbb D.$ This reproducing kernel is known as the Bergman kernel. Note that the complex valued function $B_{z} = B_{\mathbb D} (z, \cdot)$ defined on $\mathbb D$ is in $A^2(\mathbb D),$ for all $z \in \mathbb D$ and hence $P(B_{z}) = B_{z},$ where $P : L^2(\mathbb D) \longrightarrow A^2(\mathbb D)$ is the orthogonal projection. So for all $f \in A^2(\mathbb D)$ we have $f(z) = \left \langle f, B_{z} \right \rangle_{A^2(\mathbb D)}.$ Now $$\begin{align*} T_u (f) (z) & = P(uf)(z) \\ & = \left \langle P(uf), B_z \right \rangle_{A^2(\mathbb D)} \\ & = \left \langle uf, P(B_z) \right \rangle_{L^2(\mathbb D)} \\ & = \left \langle uf, B_z \right \rangle_{L^2 (\mathbb D)} \\ & = \int_{\mathbb D} u(\zeta) f(\zeta) \overline {B_z (\zeta})\ dA (\zeta) \\ & = \int_{\mathbb D} \frac {u(\zeta) f(\zeta)} {(1 - \overline z \zeta)^2}\ dA(\zeta) \end{align*}$$
Hence the Toeplitz operator $T_u$ corresponding to the symbol $u \in L^{\infty} (\mathbb D)$ is an integral operator on $A^2(\mathbb D).$