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This was a bonus question on an exam:

Suppose $f_n$ is $H(\mathbb{C})$ and $f_n \to f$ almost uniformly. Does that mean $f'(0)$ exists(either always or with some assumptions)?

(Perhaps I'm working on wrong translation - "almost uniformly" means it converges uniformly on every compact set.)

I know that if $f_n \to f$ a.u. then $f_n^{(m)} \to f^{(m)}$ a.u. But $f_n$ is entire, so $f_n'(0)$ exists and the answer is yes, always. Is that correct or some assumptions need to be made? I am worried because a.u. convergence means the $f_n$ converges uniformly to $f$ but only on compact sets and that might break my answer somehow.

blahblah
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  • What does "almost uniform convergence" mean in the exam question? This notion is what I'm sort of familiar with, but it's an unnatural notion of convergence for holomorphic functions. By what you write after the quote, it seems that the natural notion of locally uniform convergence [equivalently, since $\mathbb{C}$ is locally compact, compact convergence] might be referred to as "almost uniform convergence" (unusual). If that is the case: the locally uniform limit of holomorphic functions is holomorphic. – Daniel Fischer Jun 13 '20 at 14:36
  • @DanielFischer I've edited the question – blahblah Jun 14 '20 at 08:04
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    Probably "almost uniform convergence" is the literal translation of the Russian term. It makes sense, but in English, it's not commonly used. – Daniel Fischer Jun 14 '20 at 12:30
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    Does this answer your question? Uniform limit of holomorphic functions The answer in the duplicate even uses your terminology (almost uniformly). – Moishe Kohan Jun 14 '20 at 15:07

1 Answers1

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As you've clarified that the sequence $(f_n)$ is assumed to converge uniformly on every compact subset of $\mathbb{C}$, it means that $f'(0)$ always exists, no additional assumptions needed.

There are many ways to prove that. Perhaps the most appropriate way here uses Cauchy's integral formula.

Fix an arbitrary $r > 0$. On the open disk $D_r(0) = \{ z \in \mathbb{C} : \lvert z\rvert < r\}$, for every $n$ we have $$f_n(z) = \frac{1}{2\pi i} \int\limits_{\lvert w\rvert = r} \frac{f_n(w)}{w - z}\,dw\,.$$ The circle $C_r = \{ w : \lvert w\rvert = r\}$ is compact, therefore $(f_n)$ converges uniformly to $f$ on $C_r$. Hence, for every fixed $z \in D_r(0)$ we have $$f(z) = \lim_{n \to \infty} f_n(z) = \lim_{n \to \infty} \frac{1}{2\pi i} \int\limits_{\lvert w\rvert = r} \frac{f_n(w)}{w-z}\,dw = \frac{1}{2\pi i} \int\limits_{\lvert w\rvert = r} \frac{f(w)}{w-z}\,dw\,, \tag{$\ast$}$$ that is, on $D_r(0)$ we have the representation of $f$ as a Cauchy integral.

Now, on the right hand side of $(\ast)$ differentiation under the integral sign is legitimate, thus we see that the limit function $f$ is holomorphic (on $D_r(0)$, but since $r$ was arbitrary it follows that $f$ is entire), and $$f'(z) = \frac{1}{2\pi i} \int\limits_{\lvert w\rvert = r} \frac{f(w)}{(w-z)^2}\,dw$$ for $\lvert z\rvert < r$. In particular, this holds for $z = 0$.

Daniel Fischer
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