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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?

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  • $\begingroup$ 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. $\endgroup$ Commented Feb 3, 2015 at 18:38
  • $\begingroup$ Uniform convergence is used when changing the order of limit and integral. $\endgroup$ Commented Feb 3, 2015 at 18:43

2 Answers 2

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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.

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  • $\begingroup$ Of course that and $C$ being compact, sorry i was confused about something and now cant see what. Asked the question too quickly. $\endgroup$
    – user114539
    Commented Feb 3, 2015 at 18:51
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If you replace the condition with uniform convergence on the entire plane, the conclusion would still be true but not as useful since you replacing the hypothesis with a stronger condition. In fact the only power series which converge uniformly on the entire plane are polynomials.

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