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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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  • $\begingroup$ 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 $f_k(z)=z^k$. $\endgroup$
    – Mike Jury
    Commented May 17, 2013 at 14:50
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    $\begingroup$ @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! $\endgroup$
    – Loïc Teyssier
    Commented May 17, 2013 at 14:54
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    $\begingroup$ 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 $z_0$. $\endgroup$
    – jbc
    Commented May 17, 2013 at 17:19
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    $\begingroup$ @AliTaghavi: Yes, as you can see by considering the partial sums of $\sum n!x^n$. $\endgroup$
    – Loïc Teyssier
    Commented May 14, 2014 at 7:44
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    $\begingroup$ @DuchampGérardH.E. Yes indeed, the proof uses Montel's theoremt. The formal convergence guarantees that the whole series converges, and not merely a subsequence. $\endgroup$
    – Loïc Teyssier
    Commented Dec 20, 2018 at 7:33

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