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Let $D$ be a bounded domain in $\mathbb{C}^n$ and $\varphi$ be a non-positive plurisubharmonic function on $D$. The weighted Bergman space $A^2(D,e^{-\varphi})$ is the space of holomorphic functions in $L^2(D,e^{-\varphi})$: $$ A^2(D,e^{-\varphi}):=\{f\in\mathcal{O}(D):\int_D|f(z)|^2 e^{-\varphi(z)}dV(z)<\infty\}. $$ The weighted Bergman space $A^2(D,e^{-\varphi})$ is a Hilbert space with the following inner product: $$ <f,g>=\int_D f(z)\overline{g(z)}e^{-\varphi(z)}dV(z), $$ where $dV(z)=dx_1\dots dx_ndy_1\dots dy_n$.

When $\varphi(z)\equiv 0$, it is clearly that the Bergman space is separable.

But when $\varphi$ is a nontrivial plurisubharmonic function, how to prove that the weighted Bergman space $A^2(D,e^{-\varphi})$ is separable?

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  • $\begingroup$ Am I missing something, or isn't this immediate from the fact that $L^2(D, e^{-\varphi })$ is separable? Indeed, $L^2(X,\mu)$ is separable for any $\sigma$-finite measure $\mu$ on a standard Borel space $X$. $\endgroup$
    – Nate Eldredge
    Commented Mar 29, 2018 at 2:44
  • $\begingroup$ The measure of $L^2(D,e^{-\varphi})$ seems not to be $\sigma-$finite. In fact, there might exist $x_0\in D$ such that $\int_U e^{-\varphi}=+\infty$, where $U\ni x_0$ is any open neighborhood of $x_0$. This phenomenon depends on the singularity of the plurisubharmonic function. The set $\{x\in D|\varphi(x)=-\infty\}$ may not be empty set. $\endgroup$
    – quotient
    Commented Mar 29, 2018 at 7:37
  • $\begingroup$ Okay, but any function in $L^2(D,e^{-\varphi})$ is necessarily zero almost everywhere on $\{\varphi = -\infty\}$, else it would not be square-integrable. So the space $L^2(D, e^{-\varphi})$ is equivalent to $L^2(D \cap \{\varphi > -\infty\}, e^{-\varphi})$, and the latter is $\sigma$-finite. $\endgroup$
    – Nate Eldredge
    Commented Mar 29, 2018 at 13:34
  • $\begingroup$ In fact, if $\varphi = -\infty$ on some open set $U$, then any $f \in A^2(D, e^{-\varphi})$ must vanish a.e. on that set, so it must vanish everywhere on that set (by continuity). And by analyticity it therefore must vanish everywhere on every component of $D$ that meets $U$. So you might as well just exclude those components from your domain. $\endgroup$
    – Nate Eldredge
    Commented Mar 29, 2018 at 13:38

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Let $B\subset D$ be dense and countable. Let $B'\subset A^{2}(D,e^{-\varphi})^*$ be the set of point evaluations on $A^{2}(D,e^{-\varphi})$ at the points of $B$. Any $f\in A^{2}(D,e^{-\varphi})$ is continuous, and so if $f(b)=0$ for any $b\in B$, then $f\equiv0$. Hence, $B'$ is a countable total set. Since $A^{2}(D,e^{-\varphi})$ is reflexive it follows that its dual is separable, and so it is separable itself.

This proof in fact shows that any reflexive space of continuous functions over a separable topological space, such that the point evaluations are continuous functionals is reflexive.

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