# Normal convergence of sum of analytic function [duplicate]

Possible Duplicate:
Uniform convergence of infinite series

Suppose that $f(x)$ is analytic in $\{z: |z|<1\}$ and $f(0) =0$. Prove that $\sum f(z^n)$ converges uniformly on compact subsets of $\{z: |z|<1\}$.

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I know this somehow uses the Schwartz lemma. Form $f(0)=0$ I can have $f(z^n) = z^n g(z^n)$ with g(0) not equal to 0. Then I don't know if my g(z^n) will be bounded on D. If that is the case I will be able to apply Weierstrass Approximation Theorem to conclude that the series is bounded on D. I want to see how the rigorous proof of this goes. –  Deepak Dec 19 '12 at 16:42
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## marked as duplicate by Davide Giraudo, Matt Pressland, rschwieb, tomasz, Hagen von EitzenDec 19 '12 at 17:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

Here is a proof that just uses the fact that $f$ is $C^1$: Let $K$ be a compact subset of the unit ball $B$. Since $f$ is analytic, we have $L_K = \sup_{z \in K} |f'(z)| < \infty$. Hence $f$ is Lipschitz on $K$, and since $f(0) = 0$, we have $|f(z)| \leq L_K |z|$ for $x \in B$.

Hence $|\sum_n f(z^n) | \leq \sum_n |f(z^n)| \leq L_K \sum_n |z^n| \leq L_K \sum_n |z|^n$.

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thanks a lot. This gives another idea of proof. –  Deepak Dec 19 '12 at 17:28
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