# Show a normal family $\{f_n\}$ converges uniformly on compacts

Let $\{f_n\}$ be a sequence of holomorphic functions on a domain $\Omega\subset \mathbb C$ which is bounded uniformly on compact subsets of $\Omega$. Let $\{z_k\}$ be a sequence of distinct points in $\Omega$. with lim$_{k\to \infty} z_k = z_0 \in \Omega$. Assume that lim$_{n\to\infty} f_n(z_k)$ exists, for all $k$. Prove that $\{f_n\}$ converges uniformly on compact subsets of $\Omega$.

So far I can show that for disk $D$ centered at $z_k$, $\{f_n\}$ converges uniformly on $D$ if $\{f_n^{(i)}(z_k)\}$ converges for all $i$. But I don't know how this will help in proving uniform convergence on arbitrary compacts. Any help will be appreciated.

• Is the sequence $(z_k)$ fixed ? – Youem Apr 15 '18 at 2:42
• Yes. It is fixed. – user136592 Apr 15 '18 at 2:43
• – Daniel Fischer Apr 15 '18 at 11:45

It suffices to show that there exists a holomorphic $f$ with the property that every subsequence of $\{f_n\}$ has a further subsequence which converges to $f$: if this were to hold but $f_n\not\to f$, then by Montel’s theorem we can pass to a subsequence which converges to something that is not $f$, however this is impossible since passing to yet another subsequence shows that this “not $f$” is indeed $f$.

Now by Montel’s theorem yet again, every subsequence of $\{f_n\}$ has a further subsequence which converges to something; but by our condition on $\{z_k\}$, we know that all these convergent subsequences must agree: this is a consequence of the identity principle (the set of points where they agree has an accumulation point).

If $(f_n)_n$ converges uniformly on each closed disk $D\subset \Omega$ then there exists a function $f$ on $\Omega$ such that $(f_n)_n$ converges uniformly to $f$ on any compact subset of $\Omega.$

Proof: (1). For each $p\in\Omega$ there is a closed disk $D$ with $p\in D \subset \Omega,$ and $(f_n)_n$ converges uniformly on $D,$ so $(f_n)_n$ converges point-wise on $\Omega$ to a function $f.$

(2). Let $E$ be a non-empty compact subset of $\Omega.$ For each $p\in E$ let $D_p$ be a closed disk of positive radius, such that $p\in \text {int}(D_p)\subset D_p\subset \Omega.$

Now $C=\{\text {int}(D_p):p\in E\}$ is an open cover of $E,$ and $E$ is compact. So let $F$ be a finite (non-empty) subset of $E$ such that $E\subset \cup_{p\in F}\text {int}(D_p).$ A fortiori, we have $E\subset \cup_{p\in F}D_p.$

For $p\in F$ let $\|f-f_n\|_p=\sup \{|f(x)-f_n(x)|:x\in D_p\}.$ Then $$\lim_{n\to \infty}\|f-f_n\|_p=0$$ which is to say that $(f_n)_n$ converges uniformly to $f$ on $D_p.$

Since $F$ is finite we have $$\sup \{|f(x)-f_n(x)|:x\in E\}\leq \sup \{|f(x)-f_n(x)|: x\in \cup_{p\in F}D_p=$$ $$=\max_{p\in F}(\sup \{|f(x)-f_n(x)|:x\in D_p\})\leq$$ $$\leq \sum_{p\in F}\sup \{|f(x)-f_n(x)|: x\in D_p\}=$$ $$=\sum_{p\in F}\|f-f_n\|_p.$$

Now $F$ is fixed and finite. And each of the terms, in the last summation above, goes to $0$ as $n\to \infty.$ QED.

• I only intended to address this part of the problem, because it was directly asked by the proposer. – DanielWainfleet Apr 15 '18 at 12:13