# Holomorphic extension on the annulus $A=\{1<|z|<2 \}$

Let $$f$$ be holomorphic in the annulus $$A=\{1<|z|<2 \}$$. Suppose there is a sequence of polynomials that converges locally uniformly in $$A$$ to $$f$$. Prove that there is a function $$F$$ that is holomorphic in $$|z|<2$$ with $$F|_A=f$$.

I really can't think of a way to handle this problem.
I think that the goal should be to find a function holomorphic in $$|z|<1$$ in a way that the limits will agree with $$f$$ on the boundary $$|z|=1$$.
Then the combination of two functions will give the holomorphic extension... Appreciate your help.

Let $$P_n$$ be the polynomials in question. It would suffice to show that $$P_n$$ converges locally uniformly in the disk of radius $$2$$ centered at $$0$$. To do this, take a compact set $$K\subset D_2(0)$$. Pick $$r<2$$ such that $$K\subset D_r(0)$$. By the maximum principle, $$\max_{z\in K}|P_n(z)-P_m(z)|\le \sup_{z\in D_r(0)}|P_n(z)-P_m(z)|=\max_{z\in\partial D_r(0)}|P_n(z)-P_m(z)|\to 0$$ when $$n,m\to\infty$$ by your initial assumption. Can you see how to complete the argument?
• Thank you for the reply. So that means, for any disk of radius $<2$ , we have the sequence converging. Hence by defining $F(z)=\lim\limits_{n\to\infty}P_n(z)$ we have the extension. Am I corrrect? – gune Jan 5 at 20:44
• @gune Yup. Note that we're appealing to the completeness of the space of holomorphic functions with the convergence on compact sets, since the above argument shows that $P_n$ is a Cauchy sequence. – Reveillark Jan 5 at 20:46
• Thank you! And may I know why exactly does $F$ have to be holomorphic? Is it because of it being the uniform limit of holomorphic functions? I'm wondering whether the local uniform convergence would affect.. – gune Jan 5 at 20:48
• Just a quick question. We know the the polynomials in question converge only in the annulus $A$. So what happens if you choose $r$ to be less than 1? Will the last quantity still tend to 0? – Nicholas Roberts Jan 5 at 20:48
• Yeah, I forgot to clarify, we should take $1<r<2$. The fact that the sequence converges uniformly on compact sets is enough to guarantee that the limit is holomorphic. This makes sense, seeing as how being holomorphic is a local property. – Reveillark Jan 5 at 21:02