If the Toeplitz operator $T_u = 0$ then $u = 0$ almost everywhere.

Question

Let $u \in L^{\infty} (\mathbb D).$ Assume that $T_u = 0,$ where $T_u$ is the Toeplitz operator on $A^2 (\mathbb D)$ corresponding to the symbol $u.$ Then $T_u (f) = 0$ for every $f \in A^2 (\mathbb D)$ and hence $P (uf) = 0,$ where $P : L^2 (\mathbb D) \longrightarrow A^2 (\mathbb D)$ is the orthogonal projection. This shows that $uf \in \left (A^2 (\mathbb D) \right )^{\perp}$ for all $f \in A^2 (\mathbb D)$ and therefore $\left \langle uf, f \right \rangle_{A^2 (\mathbb D)} = 0$ for all $f \in A^2 (\mathbb D).$ Hence $$\int_{\mathbb D} u (\zeta) \left \lvert f (\zeta) \right \rvert^2\ dA(\zeta) = 0$$ for all $f \in A^2 (\mathbb D).$ If $f$ is any non-zero constant then it implies that $u = 0$ almost everywhere. Am I right?