let $\Omega$ be a bounded domain in $\mathbb{C}$ and suppose that $(f_n)$ be a sequence in holomorphic functions on $\Omega$. suppose that $(f_n)$ converge to $f:\Omega \to\mathbb{C}$ uniformly on compacts sets of $\Omega$.show that $f $ is holomoric on $\Omega$ .

i know that limit of sequence of analytic functions is analytic on some domain. but how can i use it for compacts subsets of $\Omega$ to conclude that it is analtyic on $\Omega$ ??


1 Answer 1


The kind of convergence you refer to is (compact) normal convergence.

This can be shown using Morera's theorem in conjunction with Cauchy's integral theorem, which states:

A continuous function $f$ is analytic in an open domain $\Omega$ iff for every closed, zero-homological, rectifiable curve $\gamma$ the line integral $\oint_\gamma f(z) d z = 0$. (A zero-homological loop is a loop with winding number zero.)

We can use this result to prove the desired property:

Continuity of the limit follows immediately from the uniform convergence. It remains to verify the key element of Cauchy's integral theorem, namely that all line integrals $\oint_\gamma f(z) dz = 0$ for the limit. Provided that $f_n\to f$ uniformly on every compact set, you have in particular $f_n\to f$ uniformly on every closed loop $\gamma$ (of finite length). Hence, for any zero-homological loop $\gamma$, and every $\epsilon>0$ there exists some $N(\epsilon)\in \mathbb N$ such that $|f_n(z)-f(z)|<\epsilon$ for all $n>N(\epsilon)$. If $l(\gamma)$ denotes the length of $\gamma$, we then obtain $$\left|\oint_\gamma f_n(z) dz - \oint_\gamma f(z) dz\right| \le l(\gamma)\epsilon,\quad\forall n>N(\epsilon).$$ Since for every $n$, the integral $\oint_\gamma f_n(z) d z = 0$, this proves that $\oint_\gamma f(z) d z = 0$. Hence $f$ is analytic.

  • $\begingroup$ $\Omega$ is not given to be simple connected $\endgroup$
    – Eklavya
    Aug 3, 2017 at 13:44
  • 1
    $\begingroup$ To the proposer: If $f$ is analytic on every simply-connected, connected open subset of $\Omega$ then $f$ is analytic on $\Omega$. $\endgroup$ Aug 3, 2017 at 14:24
  • $\begingroup$ There was a comment that has been deleted concerning the continuity of $f$. ... If $X,Y$ are metric spaces and $f_n:X\to Y$ are continuous and $f_n\to f$ uniformly then $ f$ is continuous. $\endgroup$ Aug 3, 2017 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.