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?

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  • $\begingroup$ 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? $\endgroup$ – gune Jan 5 at 20:44
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    $\begingroup$ @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. $\endgroup$ – Reveillark Jan 5 at 20:46
  • $\begingroup$ 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.. $\endgroup$ – gune Jan 5 at 20:48
  • $\begingroup$ 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? $\endgroup$ – Nicholas Roberts Jan 5 at 20:48
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    $\begingroup$ 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. $\endgroup$ – Reveillark Jan 5 at 21:02

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