If $D$ is the unit disk in the complex plane, $\mu$ is the Lebesgue measure on $D$ and $H(D)$ is the space of holomorphic functions on $D$, define the Bergman projection by: $$B:L^1(\mu)\rightarrow H(D), f\mapsto\left(z\mapsto\frac{1}{\pi}\int_D\frac{f(w)}{(1-z\bar{w})^2}\operatorname{d}\mu(w)\right),$$
I know from the book Duren and Schuster - Bergman spaces, page 37, proof of theorem 6, that $B$ cannot map continuously $L^1(\mu)$ into itself.
I'm wondering if from this result we can infer that $\exists f \in L^1(\mu), B(f)\notin L^1(\mu)$.
The reason why I'm asking myself this question is that in the corresponding set of Hardy space $h^1(D)$ we can in fact deduce the analogous result for Cauchy-Szego projection, thanks to the injectivity and continuity of Fourier transform and the closed graph theorem. However, here we don't have anything analogous to Fourier transform, so I'm wondering if we could deduce the analogous result for $L^1(\mu)$ in some manner.