Suppose \(f\) is entire and has Taylor series \(\sum a_n z^n\) about 0.
Express \(a_n\) as a contour integral along the circle \(\abs{z} = R\).
Apply (a) to show that the above Taylor series converges uniformly on every bounded subset of \(\CC\).
Determine those functions \(f\) for which the above Taylor series converges uniformly on all of \(\CC\).
Solution
Cauchy’s coefficient formula gives, for every \(R>0\), \[ \boxed{a_n=\frac1{2\pi i}\int_{|\zeta|=R} \frac{f(\zeta)}{\zeta^{n+1}}\,d\zeta.} \]
Fix a bounded set \(E\subset\mathbb C\) and choose \(r,R\) with \(E\subset\{|z|\le r\}\) and \(R>r\). If \(M_R=\max_{|\zeta|=R}|f(\zeta)|\), then (a) yields \[ |a_n|\le \frac{M_R}{R^n}. \] Thus for \(z\in E\), \[ |a_nz^n|\le M_R(r/R)^n. \] The geometric majorant is summable, so the Weierstrass \(M\)-test gives uniform convergence on \(E\).
Uniform convergence on all of \(\mathbb C\) occurs exactly when \(f\) is a polynomial. Indeed, if \(\sum a_nz^n\) converges uniformly on \(\mathbb C\), then its terms tend uniformly to zero. Hence for each \(n\) sufficiently large, \[ \sup_{z\in\mathbb C}|a_nz^n|<1. \] If \(a_n\ne0\) and \(n>0\), that supremum is infinite, a contradiction. Therefore all sufficiently large coefficients vanish, so \(f\) is a polynomial. The converse is immediate because the partial sums are eventually equal to \(f\).