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Where can I find the proof of this lemma?

$H^\infty$ with the norm $\| \cdot \|_\infty$ is a (complex) Banach space

Where $H^\infty = \{f \in H(U) \mid \|f\|_\infty \lt \infty\}$ with $U$ be the open unit disk in the complex plane, and $H(U)$ is the set of holomorphic functions in $U$.

zhw.
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SamC
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  • @Berci Yes, $H(U)$ is the set of holomorphic functions in $U$. – SamC Sep 06 '15 at 19:17
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    Being a Cauchy sequence $f_n$ in $H^\infty$ means $|f_n-f_m|_\infty\to 0$ as $(m,n)\to (\infty,\infty)$. I guess the proof can go as follows: show that $f_n(z)$ converges pointwise to some $f(z)$ for each $z\in U$, then show that $f\in H(U)$, and finally that $f_n$ converges uniformly to $f$ on $U$. – Berci Sep 06 '15 at 20:19

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You know the proof that $C(X)$ is a Banach space, right? The completeness comes from the fact that a uniform limit of continuous functions is continuous. So do the same thing, except using the fact that a uniform limit of holomorphic functions is holomorphic.

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Nate Eldredge
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