$\sum_{n=0}^\infty|f_n(z)|^3$ converges uniformly on $|z|\le 1/2$ I have been thinking about this problem for some time but have no progress on it.
Let $\Bbb D$ be the unit disc and $f_n:\Bbb D\to\Bbb D\backslash\{0\}$ holomorphic $\sum_{n=0}^\infty|f_n(0)|<\infty$
(a) Show that $\sum_{n=0}^\infty|f_n(z)|^3$ converges uniformly on $|z|\le 1/2$
(b) give an example of $f_n$ such that $\sum_{n=0}^\infty|f_n(z)|^3$ diverges for all $|z|>1/2$
I think there should be some way to relate $f_n(0)$ and $f_n(z)$ but I cannot figure out. Could you help?
 A: By considering $\alpha_nf_n, |\alpha_n|=1$ we can assume $f_n(0)=a_n >0$ as nothing else changes in the problem; the hypothesis then implies $f_n=e^{g_n}, \Re g_n <0, g_n(0)=b_n<0$
Since $\Re (g_n/b_n) >0$ the Herglotz inequalities give $\frac{1-r}{1+r} \le |g_n(z)/b_n| \le \frac{1+r}{1-r}, |z|=r<1$
(apply Schwartz lemma to $h_n=\frac{g_n/b_n-1}{g_n/b_n+1}, h_n(0)=0, |h_n(z)|<1, |z|<1$)
In particular for $|z|=r \le 1/2, \Re g_n/b_n \ge 1/3$ or since $b_n<0$ gives $\Re g_n \le b_n/3$ so $|f_n(z)|^3 =e^{3\Re g_n(z)} \le e^{b_n}=a_n$
Since $\sum a_n < \infty$ it folows $\sum |f_n(z)|^3 < \infty$ uniformly in $|z| \le 1/2$
For point b, pick $0<a_n<1$, such that $\sum {a_n}<\infty$, and $\sum a_n^{1-\epsilon}=\infty$, for all $1>\epsilon >0$ (for example $a_n=1/(n \log^2 n), n \ge 2$ would do) and $b_n =\log a_n<0, f_n(z)=e^{b_n\frac{z+1}{1-z}}, |f_n| <1$ so for $r > 1/2$ we have:
$f_n(-r)=e^{b_nc_r}, 0<c_r<1/3, \sum f_n(-r)^3=\sum a_n^{3c_r}= \infty$ since $0<3c_r<1$ so we are done!
A: I would like to supplement @Conrad's answer for point (b) as he/she has shown that $\sum |f_n(z)|^3=\infty$ for $z=-r$ while the problem requires $\sum |f_n(z)|^3$ diverges for all $|z|>\frac12$.
This gap is not hard to fill. Let $\lambda$ be an irrational number. We define
$$f_n(z)=\exp [b_n\frac{1+e^{2\pi i\lambda n}\cdot z}{1-e^{2\pi i\lambda n}\cdot z}]$$
It's well known that the sequence $(e^{2\pi i\lambda n})$ is dense in $\{|z|=1\}$. Therefore for any $|z|=r$, there exists a subsequence $n_k$ such that $e^{2\pi i\lambda n_k}\cdot z\to -r$ as $k\to\infty$. Hence for $k$ large, we have
$$\Re \frac{1+e^{2\pi i\lambda n_k}\cdot z}{1-e^{2\pi i\lambda n_k}\cdot z}<c_r+\epsilon<\frac13$$
$$\sum_n |f_n(z)|^3\ge \sum_k|f_{n_k}(z)|^3\ge \sum_k\exp [3b_{n_k}(c_r+\epsilon)]=\sum_ka_{n_k}^{3(c_r+\epsilon)}=\infty$$
This finishes the proof.
