Show that if \(f\) is holomorphic on \(\DD\) then \(f\) has a power series expansion that converges uniformly on every compact \(K\subset \DD\).
Solution
Since \(f\) is holomorphic on \(\mathbb D\), its Taylor series at \(0\) has radius of convergence at least \(1\): \[ f(z)=\sum_{n=0}^\infty a_nz^n, \qquad |z|<1. \] Let \(K\subset\mathbb D\) be compact. Then \[ r:=\max_{z\in K}|z|<1. \] Choose \(R\) with \(r<R<1\) and put \[ M_R:=\max_{|\zeta|=R}|f(\zeta)|. \] Cauchy’s estimate gives \(|a_n|\le M_R/R^n\). Hence for \(z\in K\), \[ |a_nz^n|\le M_R(r/R)^n. \] The geometric series on the right converges, so the Weierstrass \(M\)-test shows that the Taylor series converges uniformly on \(K\).