Dual of holomorphic functions (with the $L^1$ topology) Let $\Omega$ be a connected domain of the complex plane, and let $E$ be the vector space of integrable holomorphic functions on $\Omega$. Then it can be checked that $E$ is a closed subspace of $L^1(\Omega)$. My question (which I'm sure has a well-known answer, but I couldn't find it) is : what is the dual of $E$ ? More precisely, does it identify with a nice (i.e. explicit) subspace of $L^\infty(\Omega)$ ? 
Note that since we do not endow $E$ with the usual topology of holomorphic functions (uniform convergence on compacts) this has a priori nothing to do with hyperfunctions.
 A: The key term is Bergman space. The space you introduced is usually denoted  $A^1(\Omega)$ or $L^1_a(\Omega)$. When $\Omega$ is a disk, the dual of $A^1(\Omega)$ can be identified with the space of Bloch functions on the disk. This is proved in the book Bergman spaces by Duren and Schuster. (By the way, in the reflexive range $1<p<\infty$ we have $(A^p)^*=A^q$ on the disk, with $p^{-1}+q^{-1}=1$.)
If $\Omega$ is simply-connected and has smooth boundary ($C^{1,\alpha}$ with $\alpha>0$), then the conformal map of $\Omega$ onto the disk has derivative bounded from $0$ and $\infty$. By the change of variable formula, the map induces isomorphism between Bergman spaces on $\Omega$ and on the disk. Thus,   the problem reduces to the case of the disk; the dual is again the Bloch space.
For a general simply-connected domain $\Omega$, the identification of the dual of $A^p(\Omega)$ is difficult; it is related to a major open problem in complex function theory (Brennan's conjecture). As a starting point, see the paper The dual of a Bergman space on simply connected domains by Hedenmalm. 
