Question: Let $\{f_n\}$ be a sequence of analytic functions on $\mathbb{D}$ such that $|f_n(z)|\leq1$ for all $n$ and for all $z\in\mathbb{D}$. Prove that there is a subsequnece $\{f_{n_j}\}$ and an analytic function $f$ on $\mathbb{D}$ satisfying: For every $0<r<1$, $\max_{|z|\leq r}|f(z)-f_{n_j}(z)|\rightarrow 0$ as $n_j\rightarrow\infty$. Show, by example, that this is false in general that $\sup_{z\in\mathbb{D}}|f(z)-f_{n_j}(z)|\rightarrow0$ as $n_j\rightarrow\infty$.
Thoughts I am thinking that the first part is just Montel's Theorem. That is, since $\{f_n\}$ is a sequence of functions in (a bounded set) $\mathbb{D}$, then Montel's implies the existence of a subsequence $\{f_{n_j}\}$, which converges uniformly to a limit function $f$. So, I suppose that is precisely the property that we want to show. Unless I am thinking of it incorrectly? For the next question, finding an example, I was trying to play around with $\frac{1}{z}$, but I wasn't getting anything to come out nicely. Any help would be greatly appreciated! Thakn you.
You are right about the first part: $\{ f_n \} $ is uniformly bounded in $\Bbb D$ so that Montel's theorem implies the existence of a subsequence $(f_{n_j})$ which is locally uniformly convergent: $f_{n_j} \to f$ uniformly on all compact subsets of $\Bbb C$. In particular (actually equivalent) $f_{n_j} \to f$ uniformly on all circles with radius $r < 1$.
For the second part consider the functions $f_n(z) = z^n$ in the unit disk. We have $f_n(z) \to f(z) = 0$ on each compact subset of $\Bbb D$, but $\sup_{z\in\mathbb{D}}|f(z)-f_{n}(z)| = 1$.