Question: Suppose $f$ is analytic in $1<|z|<2$ and that there exists a sequence of polynomials converging to $f$ uniformly on every compact subset of this annulus. Show that $f$ has an analytic extension to all of the disk $|z|<2$.
Thoughts: This feels like it is some sort of variation of Weierstrass'/Hurwitz's/Montel's theorems. Since $f$ is analytic in $1<|z|<2$, and there is some sequence of polynomials, say $\{f_n\}\rightarrow f$ uniformly on every compact subset of this annulus, could we say that $f$ is equicontinuous, and then play around with (maybe) some normal family stuff? Also, would we really be concerned about the annulus that $f$ is analytic on, or is it best to just try and deal with $\mathbb{D}$? I suppose the boundary of $\mathbb{D}$ may pose some problems in this case. Or, would an analytic extension of $f$ in this case even extend $f$ to the reigion $f$ is already analytic? Any help is greatly appreciate! Thank you.
Choose some $R \in (1, 2)$. The uniform convergence of $f_n$ to $f$ (together with the maximum modulus principle) implies that $(f_n)$ is uniformly bounded in $\{ |z| \le R \}$, so that it is a normal family.
Montel's theorem implies that there is a subsequence $f_{n_k}$ converging to a holomorphic function $F$ in $\{ |z| < R \}$. On the other hand, $f_{n_k} \to f$ in $\{ 1 < |z| < R \}$.
It follows that $F(z) = f(z)$ in $\{ 1 < |z| < R \}$, so that $$ \tilde f(z) = \begin{cases} F(z) & \text {for } |z| < R \\ f(z) & \text {for } 1 <|z| < 2 \\ \end{cases} $$ is well-defined and an analytic extension of $f$ to $\{ |z| < 2 \}$.