Every uniformly convergent sequence of polynomials converges on a simply connected domain

Question

Here is a really weird problem.

Suppose $P_i$ is a sequence of polynomials which converges to the analytic $f$ on a domain $U$ in $\mathbb{C}$, uniformly on compact subsets. Then there is a simply connected domain $V$ containing $U$ on which this also holds.

I am totally stuck. I assume $V$ is the union of $U$ with the bounded components of its complement. I can show that $V$ is open using some small tricks, but I have no idea how to proceed otherwise. A classmate of mine said something about showing the polynomials form a uniformly Cauchy sequence, but I know he also ran into problems so I don't know if this is a good approach.

Does anyone see a way to proceed? Maybe the case for an annulus is instructive, but I can't solve that either haha

Thanks so much!

Answer 1

Just to get you started: Suppose $U = \{1/2 < |z| < 1\}.$ Suppose $p_n$ converges uniformly on compact subsets of $U.$ Then $p_n$ converges uniformly on compact subsets of the open unit disc $\mathbb D.$ Proof: The sequence $p_n$ is uniformly Cauchy on any $\{|z|=r\}$ if $r$ is close to $1.$ Hence it will be uniformly Cauchy on any $\{|z|\le r\}$ by the maximum modulus theorem.