# Showing a sequence of analytic functions converges locally uniformly

Let $f_n :U \to \mathbb{C}$ be a sequence of analytic functions on an open and connected set $U$. Suppose that the sequence is locally bounded and that for the set $$D:= \{z \in U : f_n(z) \, \, \mathrm{converges} \}$$ has an accumulation point in $U$.

How would you show that then the whole sequence $f_n$ converges locally uniformaly to an analytic function $f$?

By Montel's theorem, the sequence $(f_n)$ has a subsequence which converges locally uniformly to some analytic function $f: U \to \mathbb{C}$. Assume, $f_n \not \to f$ locally uniformly. Then there exists a compact set $K$, $\varepsilon>0$ and a subsequence $(f_{n_k})_k$ of $(f_n)$ such that $$\forall k: \|f_{n_k}-f\|_K \geq \varepsilon \tag{1}$$
By Montel's theorem, $(f_{n_k})$ has a locally uniformly convergent subsequence $(f_{n_{k_j}})_j$, $$f_{n_{k_j}} \to \tilde{f} \quad \text{locally uniformly}$$ By assumption, $\tilde{f}|_D = f|_D$. Since $f$, $\tilde{f}$ are holomorphic and $D$ has an accumulation point, we conclude $f = \tilde{f}$ (by the identity theorem). Contradiction to (1)!
• Thanks, I understand everything up to the part where $\tilde f |_D = f|_D$. How have we used the fact that $D$ has an accumulation point? – user53076 May 10 '13 at 7:18
• Ok I see how that the identity theorem uses the fact that there is an accumulation point, but how we justify that if $\tilde f = f$ then this is a contradiction to (1)? Even if $f_{n_k}$ and $f_{n_j}$ are completely different sequences? – user53076 May 10 '13 at 7:53
• @user53076 Note that $(f_{n_{k_j}})_j$ is a subsequence of $(f_{n_k})$ and therefore (by assumption, (1)) we have $$\|f_{n_{k_j}}-f\| \stackrel{f=\tilde{f}}{=} \|f_{n_{k_j}}-\tilde{f}\| \geq \varepsilon$$ and this is a contradiction to $f_{n_{k_j}} \to \tilde{f}$. – saz May 10 '13 at 14:56
• @user135520 why do you think that we need that $K$ contains $D$? – saz Jun 13 '18 at 3:12
• @user135520 It's simply the uniqueness of the limit. Since we know that the limit $w:=\lim_{n \to \infty} f_n(z)$ exists ror $z \in D$, we have $w = \lim_{k \to \infty} f_{n_k}(z)$ for any subsequence. In particular, $$\lim_{j \to \infty} f_{n_j}(z) = w = \lim_{k \to \infty} f_{n_k}(z)$$ for any $z \in D$ and any two subsequences $(f_{n_k})$ and $(f_{n_j})$. – saz Jul 14 '18 at 6:48