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 $\{z: |z|<1\}$.
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 $\{z: |z|<1\}$.
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 $|\sum_n f(z^n) | \leq \sum_n |f(z^n)| \leq L_K \sum_n |z^n| \leq L_K \sum_n |z|^n$.