If a sequence of holomorphic functions converge to a continuous function, is the limit holomorphic?

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)

• 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 $[0,1]$) Jun 19, 2019 at 2:21
• I didn't realize that. I am only reading parts of this book so I am not sure how Conway defines the symbol $\to$ Jun 19, 2019 at 2:39
• 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... Jun 19, 2019 at 3:02
• 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. Jun 19, 2019 at 3:08
• 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. Jun 19, 2019 at 3:08

The above Theorem 2.1 is a famous theorem of Weierstraß. In this theorem $$\to$$ means uniform convergence on compact subsets.
• are you sure that's what it means? wouldn't saying $f \in C(G,\mathbb{C})$ be redundant? Jun 19, 2019 at 3:28
• Yes, I am sure. And yes, $f \in C(G,\mathbb{C})$ is reundant.
• I know how to prove this if "$\to$" means uniform convergence. But can you tell me exactly where in the book this is defined? Jun 19, 2019 at 7:01