For a sequence $\{f_j\}$ of holomorphic functions, what can we say given $\sum_{j=1}^\infty |f_j(0)|$ converges. In particular, let $\{f_j\}$ a sequence of holomorphic functions from $D(0,1)$ to $D(0,1)$ \ $\{0\}$, where $D(0,1)$ denotes the unit disk. I want to show that if $\sum_{j=1}^\infty |f_j(0)|$ converges, then $\sum_{j=1}^\infty f_j(z)^2$ converges absolutely and uniformly on compact sets in $D(0,1/3)$. 
I need a hint to get started. I think there may be a theorem relating to this kind of thing of which I am not aware.  
 A: Suppose $u$ is positive and harmonic in $\mathbb D.$ Then
$$u(z) \ge \frac{1-|z|}{1+|z|}u(0).$$
This is the familiar Harnack inequality from below. Thus, using the minimum principle for harmonic functions, we have
$$\min_{|z|\le r} u(z) = \min_{|z|= r} u(z) \ge \frac{1-r}{1+r}u(0).$$
Now in our problem each $f_j \ne 0$ in $\mathbb D,$ so we can write $f_j= e^{g_j},$ where $g_j$ is holomorphic in the disc. Furthermore, because $|f_j| < 1,$ $g_j$ takes the form $g_j = -(u_j+iv_j),$ where $u_j$ is positive and harmonic on $\mathbb D.$
From the above, $u_j \ge (1-1/3)/(1+ 1/3)u_j(0) = u_j(0)/2$ on the closure of $D(0,1/3).$ Thus, for $|z|\le 1/3,$
$$|f_j(z)| = e^{- u_j(z)} \le e^{-u_j(0)/2}.$$
Thus for such $z,$ $\sum |f_j(z)|^2 \le \sum (e^{-u_j(0))/2})^2 = \sum e^{-u_j(0)} = \sum |f_j(0)| < \infty.$ By Weierstrass M, $\sum f_j^2$ converges uniformly on $\{|z|\le 1/3\}.$
A: Amazing and very neat example ... A diagram with the involved mappings:
 
Let $H_+= \{ \mbox{Im }z>0\}$ and $\psi : H_+ \rightarrow D^*=D\setminus\{0\}$ be the (universal} covering map $\psi(z)=e^{iz}$. Then each $f_j: D\mapsto D^*$ lifts to a map $\hat{f}_j: D \rightarrow H_+$ (unique up to an integer translation). Let $w_j=\hat{f}_j(0)=x_j+i y_j \in H_+$. The assumption on $f_j$ translates into  $\sum_j |\psi(w_j)| = \sum_j e^{-y_j}<+\infty$.
Now, for each $j$ let $h_j: H_+ \rightarrow D$ be given by $h_j(w)=(w-w_j)/(w-\overline{w_j})$. [Reason: We want to get uniform bounds on the image of $\hat {f_j}$]. Then $h_j\circ \hat{f}_j: D\rightarrow D$ is holomorphic and maps $0$ to $0$. By the Schwarz Lemma we have $|h_j(\hat{f}_j(z))|\leq |z|$ for all $z\in D$.
Then for $|z|<r<1$, $v=h_j(\hat{f}_j(z))$  verifies $|v|=r$ and $w=\hat{f}_j(z)= h_j^{-1}(v)=\frac{w_j- v\overline{w_j}}{1-v}=x_j + iy_j \frac{1+v}{1-v}$. But then  $\mbox{im } w \geq y_j \frac{1-r}{1+r}$ which for $r<1/3$ yields $\mbox{im }w \geq y_j \frac{1-1/3}{1+1/3} = \frac12 y_j$. Whence 
$$ \sum_j |f_j(z)|^2 =\sum_j |\psi(\hat{f}_j(z)|^2\leq \sum_j e^{-2 (\frac12 y_j)}= \sum_j e^{-y_j}= \sum_j |f_j(0)| < +\infty$$
Convergence uniform on compacts (just change $r$ a bit). The condition is optimal! Counter-examples for $r>1/3$.
Hopefully didn't do too many mistakes on the way...
