7.6 Uniform convergence of analytic functions

Part (c) of [link] gives an example showing that the uniform limit of a sequence of differentiable functions of a real variable need notbe differentiable. Indeed, when thinking about uniform convergence of functions, the fundamental result to remember isthat the uniform limit of continuous functions is continuous ( [link] ). The functions in [link] were differentiable functions of a real variable. The fact is that, for functions of a complex variable, things are as usual much more simple.The following theorem is yet another masterpiece of Weierstrass.

Suppose $U$ is an open subset of $C,$ and that $\left\{f_n\right\}$ is a sequence of analytic functions on $U$ that converges uniformly to a function $f.$ Then $f$ is analytic on $U.$ That is, the uniform limit of differentiable functions on an open set $U$ in the complex plane is also differentiable on $U.$

Though this theorem sounds impressive and perhaps unexpected, it is really just a combination of [link] and the Cauchy Integral Formula. Indeed, let $c$ be a point in $U,$ and let $r>0$ be such that ${\overline{B}}_r\left(c\right)\subseteq U.$ Then the sequence $\left\{f_n\right\}$ converges uniformly to $f$ on the boundary $C_r$ of this closed disk. Moreover, for any $z\in B_r\left(c\right),$ the sequence $\{f_n\left(\zeta \right)/(\zeta -z)\}$ converges uniformly to $f\left(\zeta \right)/(\zeta -z)$ on $C_r.$ Hence, by [link] , we have

Hence, by part (a) of [link] , $f$ is expandable in a Taylor series around $c,$ i.e., $f$ is analytic on $U.$