# Uniform convergence of infinite series

Suppose $f$ is a holomorphic function (not necessarily bounded) on $\mathbb{D}$ such that $f(0) = 0$. Prove the the infinite series $\sum_{n=1}^\infty f(z^n)$ converges uniformly on compact subsets of $\mathbb{D}$.

I met this problem on today's qual. Here is what I have so far, since $f(0) = 0$, we can write $f(z) = z^m h(z)$ for some integer $m$. Then $f(z^n) = z^{nm}h(z^n)$. We might then use Cauchy's criterion for uniform convergence to finish the proof.

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What you did seems correct. – Davide Giraudo Aug 10 '12 at 20:27

An alternative way is the following: let $M_R:=\sup_{|z|\leq R}|f(z)|$. For a fixed $R<1$, define $g(z):=\frac 1{1+M_R}f(Rz)$, from the open unit disk to itself. Then by Schwarz lemma, we have $|g(z)|\leq |z|$ for all $z\in D$, hence $|f(Rz)|\leq (1+M_R)|z|$ for all $z\in D$. We get that $|f(z)|\leq \frac{1+M_R}R|z|$ for all $z$ in the closed ball of center $0$ and radius $R$, which shows that the series $\sum_nf(z^n)$ is normally convergent on this set.
More simply, for a fixed $R\in (0,1)$, we have $$\sup_{|z|\leq R}|f(z^n)|\leq \sup_{|z|\leq R^n}|f(z)|.$$ Then we use continuity at $0$ of $f$.
Thanks Davide, this is a really nice idea. Although $f$ is not necessarily bounded on $D$, it is bounded on every compact subsets of $D$. – Hongshan Li Aug 11 '12 at 3:26