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Let $f_1, f_2, f_3, \ldots$ be analytic functions on $D=\{z \in \mathbb{C}:|z|<1\}$, taking values in $D$. If $f_n(z_0) \rightarrow 1$ for some $z_0 \in D$, I would like to prove that $f_n \rightarrow 1$ uniformly on compact subsets of $D$.

I believe I should be using techniques from the study of normal families. We have that $|f_n(z)|<1$ so $f_n$ is a normal family, and hence for any subsequence of $f_n$, there is a further subsequence such that $f_{a_n}\to f:D\to \overline{D}$ uniformly on compact sets of $D$. We can repeat this process with another subsequence to find that $f_{b_n}\to \tilde{f}:D\to \overline{D}$ uniformly on compact sets of $D$. We must have that $f_{b_n},f_{a_n}(z_0)\to 1$ and thus $f(z_0)=\tilde{f}(z_0)$. Now, both $f, \tilde{f}$ are analytic and take their maximum inside of $D$. This means that they are constant, thus $f\equiv \tilde{f}\equiv 1$. Since, we showed that every subsequence of $f_n$ contains a further subsequence converging to the same limit, we have that $f_n\to 1$ uniformly on compact subsets of $D$.

Is the reasoning correct?

Diffusion
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1 Answers1

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Yes, if $(f_n)$ is a sequence of functions holomorphic in a domain $G$ and $f$ is holomorphic in $G$ such that every subsequence of $(f_n)$ has a subsequence $(f_{n_k})$ with $f_{n_k} \to f$ locally uniformly in $G$ then the entire sequence $(f_n)$ converges locally uniformly to $f$ in $G$.

This can easily be proven by contradiction, similar as the corresponding statement for sequences in $\Bbb R$, see for example Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$..

Alternatively one can use the Schwarz-Pick theorem: For $|z| \le r < 1$ is, with $w_n = f_n(z_0)$, $$ \left| \frac{f_n(z) - w_n}{1-\overline{w_n} f_n(z)}\right| \le \left| \frac{z-z_0}{1-\overline{z_0} z}\right| \le K < 1 \\ \implies 1-K^2 \le 1-\left| \frac{f_n(z) - w_n}{1-\overline{w_n} f_n(z)}\right|^2 = \frac{(1-|f_n(z)|^2)(1-|w_n|)^2}{|1-\overline{w_n} f_n(z)|^2} \\ \implies |1-\overline{w_n} f_n(z)|^2 \le \frac{1-|w_n|^2}{1-K^2} \, . $$ It follows that $1-\overline{w_n} f_n(z) \to 0$ uniformly in $|z| \le r$, which implies that $f_n(z) \to 1$ uniformly in $|z| \le r$.

Martin R
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