# prove that limit function is analytic for sequence of analytic function on compacts subsets of $\Omega$

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 $\Omega$ ??

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.

• $\Omega$ is not given to be simple connected – Eklavya Aug 3 '17 at 13:44
• To the proposer: If $f$ is analytic on every simply-connected, connected open subset of $\Omega$ then $f$ is analytic on $\Omega$. – DanielWainfleet Aug 3 '17 at 14:24
• 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. – DanielWainfleet Aug 3 '17 at 14:27