Suppose \(f\) is holomorphic in an annulus \(r < |z| < R\), and there exists a sequence of holomorphic polynomials \(p_n\) converging to \(f\) uniformly on compact subsets of the annulus. Show that \(f\) can be extended to the disc \(\{|z| < R\}\) as a holomorphic function.
Solution
Let \(D_R=\{z:|z|<R\}\) and \(A=\{z:r<|z|<R\}\).
1. The sequence \((p_n)\) is uniformly Cauchy on each compact subset of \(D_R\).
Fix a nonempty compact set \(K\subset D_R\). Choose a radius \(\rho\) satisfying \[ \max\{r,\max_{z\in K}|z|\}<\rho<R. \] The circle \(|z|=\rho\) is a compact subset of \(A\). Uniform convergence to \(f\) there implies that \((p_n)\) is uniformly Cauchy on that circle. Each difference \(p_n-p_m\) is holomorphic on a neighborhood of the closed disk \(|z|\leq\rho\). The maximum modulus principle gives \[ \sup_{z\in K}|p_n(z)-p_m(z)| \leq\max_{|z|\leq\rho}|p_n(z)-p_m(z)| =\max_{|z|=\rho}|p_n(z)-p_m(z)|\longrightarrow0 \] as \(n,m\to\infty\) [@SS03]. The assertion for the empty compact set requires no estimate.
2. The limit on the disk is a holomorphic extension of \(f\).
By completeness of \(\mathbb C\), step 1 first gives the pointwise limit \(F(z)=\lim_n p_n(z)\) at each \(z\in D_R\). For a compact \(K\subset D_R\), let \(m\to\infty\) in the uniform Cauchy estimate. It follows that \(p_n\to F\) uniformly on \(K\). Thus convergence is locally uniform on the whole disk. The local uniform limit theorem makes \(F\) holomorphic there [@SS03].
For every \(z\in A\), the hypothesis already gives \(p_n(z)\to f(z)\). Uniqueness of limits in \(\mathbb C\) therefore gives \(F(z)=f(z)\) on \(A\). Hence \(F\) is the required extension to all of \(D_R\).