Compact convergence of bivariate complex series

Question

Let $b_n\in H(D(0,R))$ be a sequence of holomorphic functions on a disc of radius $R$ centered at $0$, $\Omega=D(0,R)\times D(0,S)$, $|w_0|<S$. Take a compactly convergent (that is, uniformly convergent on every compact $K\in\Omega$) series $$f(z,w)=\sum_{n=0}^\infty b_n(z)(w-w_0)^n.$$ In $\Omega$ we can also write $$f(z,w)=\sum^\infty_{n=0}a_n(z)w^n$$ for some $a_n\in H(D(0,R))$. This series is (absolutely) convergent on $\Omega$, because the first one is, but is it compactly convergent?

Answer 1

By the first statement, the $f(z,w)$ is holomorphic. The functions $a_n$ are uniquely determined by $f$: it is $\frac{1}{n!} \frac{\partial^n f}{\partial w^n}(z,0)$. The function $f$ has a power series that converges absolutely uniformly on compact subsets $$ f(z,w) = \sum_{j,k} c_{j,k} z^jw^k $$ Now that means we can group the terms in any way we want, including as $$ f(z,w) = \sum_{k} \left( \sum_j c_{j,k} z^j \right) w^k $$ The limit is uniform here if it was uniform above (that is not hard to show). The sum in the parentheses is your $a_k(z)$ since that was the unique function there.