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.

Solutions Collecting From Web of "Uniform convergence of infinite series"

What you did seems correct.

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 $f$.