Let $\{f_n : \mathbb{D} → \mathbb{D}\}_{n=1}^\infty$ be a sequence of analytic functions such that the sequences $\{f_n(1/k)\}_{n=1}^\infty$ converge for any $k ∈ N, k > 1.$ Prove that the sequence ${f_n(z)}_{n=1}^\infty$ converges for all $z ∈ \mathbb{D}.$
This is a qual problem. I thought about maybe somehow using the identity principle or Schwarz' lemma but so far am having no luck.
The family $\mathscr{F} = \{ f_n : n \in \mathbb{Z}^+\}$ is a normal family, since $\mathbb{D}$ is bounded. Thus every sequence in $\mathscr{F}$ has a compactly convergent subsequence.
Now, let $\sigma, \tau \colon \mathbb{Z}^+ \to \mathbb{Z}^+$ two strictly increasing maps such that the subsequences $(f_{\sigma(n)})$ and $(f_{\tau(n)})$ of the original sequence converge compactly. Denote their respective limit functions by $f^{\sigma}$ and $f^{\tau}$.
Since for all $k \in \mathbb{N}\setminus\{0,1\}$ the sequence $f_n(1/k)$ is assumed to converge, it follows that
$$f^\sigma\left(\tfrac{1}{k}\right) = f^\tau\left(\tfrac{1}{k}\right)$$
for all $k \in \mathbb{N}\setminus\{0,1\}$, and since $(1/k)$ has a limit point in $\mathbb{D}$ (namely $0$), the identity theorem asserts that $f^\sigma \equiv f^\tau$.
Thus all convergent subsequences converge to the same limit.
This is a general criterion for the convergence of a sequence:
A sequence $(x_n)$ in a topological space $X$ converges to a point $x\in X$ if and only if every subsequence $(x_{n_k})$ of $(x_n)$ has a further subsequence $\left(x_{n_{k_m}}\right)$ that converges to $x$.
So we not only have the pointwise convergence of $f_n$, we even have the locally uniform convergence.
