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While reading Conway's GTM11, I encountered the following theorem. In the theorem, $G\subset\mathbb{C}$ is a domain and $H(G)$ means the holomorphic functions on $G$, and $C(G,\mathbb{C})$ means the continuous functions on $G$.

(Chap VII.) 2.1 Theorem. If $\{f_n\}$ is a sequence in $H(G)$ and $f$ belongs to $C(G,\mathbb{C})$ such that $f_n\to f$ then $f$ is analytic and $f_n^{(k)}\to f^{(k)}$ for each integer $k\geq1$.

And the proof says

Proof. We will show that $f$ is analytic by Morera's theorem. Let $T$ be a triangle contained inside a disk $D\subset G$. Since $T$ is compact, $\{f_n\}$ converges to $f$ uniformly over $T$. Hence...

My question: how does it follow from $T$ being compact that $f_n\to f$ is uniform?

From $T$ being compact I can conclude that there exists $0\leq\delta_n<\infty$ such that $|f_n(z)-f(z)|\leq\delta_n$ for all $z\in T$. If I can show that $\delta_n\to0$, then the convergence is uniform. So I tried to do this by contradiction. Suppose $\delta_n\not\to0$, and there exists $\delta>0$ and $\{z_{n_k}\}\subset T$ such that

$$|f_{n_k}(z_{n_k})-f(z_{n_k})|>\delta>0$$

Since $T$ is compact and $z_{n_k}\in T$, $\{z_{n_k}\}$ has a convergent subsequence, hence we may assume $\{z_{n_k}\}$ is convergent in the first place and denote the putative limit by $z_0$. Then we can write

$$0<\delta<|f_{n_k}(z_{n_k})-f(z_{n_k})|\\ \leq|f_{n_k}(z_{n_k})-f_{n_k}(z_0)|+|f_{n_k}(z_0)-f(z_0)|+|f(z_0)-f(z_{n_k})|$$

Obviously the second and the third term can be made arbitrarily small. And if I could show the first term can also be arbitrarily small, I would obtain a contradiction and finish the proof. But I don't how whether or how this can be done. (I do know that if $\{f_n\}$ is equicontinuous over $T$ then we are done, but it is not in the assumption of the theorem)

Any suggestions? Thanks in advance.

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    $\begingroup$ What does "$f_n \to f$" mean? If it means uniformly on compact sets (which I think it does), then $T$ being compact obviously means $f_n \to f$ uniformly on $T$. I think you're interpreting it as pointwise convergence and trying to show pointwise convergence on a compact set implies uniform convergence, which it most certainly does not (e.g. $x^n$ on $[0,1]$) $\endgroup$ Jun 19, 2019 at 2:21
  • $\begingroup$ I didn't realize that. I am only reading parts of this book so I am not sure how Conway defines the symbol $\to$ $\endgroup$
    – trisct
    Jun 19, 2019 at 2:39
  • $\begingroup$ actually, I'm not sure that it means uniformly on compact sets. it might mean pointwise convergence. the Morera proof goes by establishing $\int_T f(z)dz = 0$ for each triangle $T$ (which is sufficient for showing $f$ is analytic, by Morera). it establishes $\int_T f(z)dz = 0$ by $\int_T f_n(z)dz = 0$ for each $n$ (by Cauchy) and some sort of dominated convergence or uniform convergence. but I don't see why you necessarily have uniform convergence... $\endgroup$ Jun 19, 2019 at 3:02
  • $\begingroup$ I think I know how to show equicontinuity if the $f_n$'s are uniformly bounded. To show equicontinuity, it suffices to show locally a uniform bound on $f_n'$, which can be done via Cauchy's integral representation: $f_n'(z) = \int_T \frac{f_n(\zeta)}{(z-\zeta)^2}d\zeta$ if the $f_n$'s are uniformly bounded. $\endgroup$ Jun 19, 2019 at 3:08
  • $\begingroup$ But also, if the $f_n$'s are uniformly bounded, then you can use dominated convergence theorem together with Morera, as I explained in my second comment. $\endgroup$ Jun 19, 2019 at 3:08

1 Answer 1

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The above Theorem 2.1 is a famous theorem of Weierstraß. In this theorem $\to$ means uniform convergence on compact subsets.

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  • $\begingroup$ are you sure that's what it means? wouldn't saying $f \in C(G,\mathbb{C})$ be redundant? $\endgroup$ Jun 19, 2019 at 3:28
  • $\begingroup$ Yes, I am sure. And yes, $f \in C(G,\mathbb{C})$ is reundant. $\endgroup$
    – Fred
    Jun 19, 2019 at 5:28
  • $\begingroup$ I know how to prove this if "$\to$" means uniform convergence. But can you tell me exactly where in the book this is defined? $\endgroup$
    – trisct
    Jun 19, 2019 at 7:01

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