Uniform limit of analytic functions

Let $$\{f_n(z)\}$$ be a sequence of analytic functions converging uniformly to a function $$f(z)$$ on all compact subsets of a domain $$D$$. Then $$f(z)$$ is analytic in $$D$$.

Suppose we proceed as follows:

It is enough to prove that $$f(z)$$ is analytic at a point $$z_0\in D$$. Let $$D_0$$ be disk with center $$z_0$$ and contained in $$D$$. Clearly $$f(z)$$ is continuous on $$D_0$$. Moreover because of uniform convergence $$\lim_{n\to\infty}\left( \int_C f_n(z)dz\right)=\int_C f(z)dz$$ for every closed contour $$C$$ in $$D_0$$ and hence using Cauchy's theorem, we see that the $$\int_C f_n(z)dz=0$$ for every closed contour $$C$$ in $$D_0$$. Now Morera's theorem finishes the proof.

Question: Where is uniform convergence on compact subsets used?

• What do you want to replace it with? If you remove it completely, the hypothesis becomes "Let $f_n$ be a sequence of analytic functions, and let $f$ be some function not necessarily related to the $f_n$ in any way." Of course in that case $f$ need not be analytic. Commented Feb 3, 2015 at 18:38
• Uniform convergence is used when changing the order of limit and integral. Commented Feb 3, 2015 at 18:43

You need uniform convergence (or something similar) to be able to conclude that $$\lim_{n\to\infty} \int_C f_n(z)\,dz = \int_C f(z)\,dz$$ In paricular, pointwise convergence is not enough for this.
• Of course that and $C$ being compact, sorry i was confused about something and now cant see what. Asked the question too quickly.