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At some point I needed to prove that some formal (iterative) construction yielded an actual convergent power series. To do so I was led to prove the following lemma, where $\mathcal{B}(\Delta)$ is the Banach space of bounded, holomorphic functions on a domain $\Delta$.

Lemma. Let $\Delta$ be a domain in $\mathbb C^{m}$ and consider a bounded sequence $(f_k)_k$ of $\mathcal{B}(\Delta)$ satisfying the additional property that there exists some point $z_{0}\in\Delta$ such that the corresponding sequence of Taylor series $(\sum_n a_n ^k (z-z_0)^n)_k$ at $z_{0}$ is convergent in $\mathbb C [[z-z_0]]$ equipped with the projective topology (that is, each $(a_n ^k)_k$ converges in $\mathbb C$). Then $(f_{k})_{k}$ converges uniformly on compact sets of $\Delta$ towards some $f\in\mathcal{B}(\Delta)$.

This lemma is elementary to derive (I can include the proof if needed), so I doubt very much it is original. Yet I've never come across this statement in classical references or research articles I've been reading. Do you know a reference/name associated to it?

Thanks in advance!

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I don't know a name or reference, but for your side question, if you mean the topology given by the supremum norm over $\Delta$, the answer is certainly no--take $\Delta$ to be the unit disk and $f_k(z)=z^k$. – Mike Jury May 17 '13 at 14:50
@Mike: yyou're right. I will remove my side question, which was whether the convergence always happens in $\mathbf B (\Delta)$. That was a very stupid question indeed! – Loïc Teyssier May 17 '13 at 14:54
I can't give a reference but there is a large number of related results which follow from the following general considerations: the unit ball of $H^\infty$ is compact for the topology of compact convergence and so the latter coincides there with any weaker Hausdorff topology. In your case, this would be the weak topology induced by evaluation of the derivatives at $z_0$. – jbc May 17 '13 at 17:19
@jbc: that's somehow how the proof runs. I understand this statement does not deserve a special place in the realm of complex analysis, but I wanted to check whether that was the case or not. Thanks anyway for your answer. – Loïc Teyssier May 18 '13 at 7:42
@AliTaghavi: Yes, as you can see by considering the partial sums of $\sum n!x^n$. – Loïc Teyssier May 14 '14 at 7:44

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