I'm trying to solve the following :
Let $D$ be a simply connected region stricly included in $\mathbb{C}$. Let $\mathbb{D}$ be the open unit disk. Let $f \in \text{Hol}(D)$ be a bounded function with zeroes at $w_1, w_2, ...$. Show that $$ \sum\limits_{n \geq 1}\big| h'(w_n)\big| \text{dist}(w_n, \partial D)$$ converges, where $h : D \rightarrow \mathbb{D}$ denotes the Riemann map, $\text{dist}$ is the euclidean distance and $\partial D$ is the boundary of $D$.
Using the following result on conformal metrics : $$ h^* \tau(z) \leq \frac{2}{\text{dist}(z, \partial D)}$$ where $\tau$ is the hyperbolic (poincaré) metric and $z \in D$, I got
$$ \frac{2}{1- \big| h(z)\big|^2}\cdot \big| h'(z)\big| ~\leq~ \frac{2}{\text{dist}(z, \partial D)}.$$ Hence $$ \text{dist}(z, \partial D)\big| h'(z)\big| ~\leq~ 1- \big| h(z)\big|^2.$$ Therefore $$ \sum\limits_{n \geq 1}\text{dist}(w_n, \partial D)\big| h'(w_n)\big| ~\leq~ \sum\limits_{n \geq 1}\Big(1- \big| h(w_n)\big|^2\Big).$$ In my humble opinion (I have not worked out the details yet the check if it works) using compact exhaustion of $D$ it must be possible to use the fact that there only is a finite number of zeroes in each compact subset of $D$ to show that $\big| h(w_n)\big|^2$ tends to 1. But I don't see how to show that $\big| h(w_n)\big|^2$ tends to 1 fast enough for the series to converge.
So far I have not used the fact that $f$ is bounded, which worries me a little. I don't know how to put this hypothesis to use.
I would be really thankful if you guys could avoid give the answer right away. A hint would be welcome though.
Consider the function $F = f \circ (h^{-1}) \in H^{\infty}(\mathbb{D})$, and recall the conditions on the zeros of a nonconstant bounded holomorphic function on the unit disk. You will also need the fact that the unit disk is bounded.
The zeros of $F$ are the $h(w_n)$. By the Blaschke condition, we have
$$\sum_{n = 1}^{\infty} \bigl(1 - \lvert h(w_n)\rvert\bigr) < +\infty.$$
Since $1 - \lvert h(w_n)\rvert^2 = \bigl(1 + \lvert h(w_n)\rvert\bigr)\bigl(1 - \lvert h(w_n)\rvert\bigr) \leqslant 2\bigl(1 - \lvert h(w_n)\rvert\bigr)$, it follows that
$$\sum_{n = 1}^{\infty} \bigl(1 - \lvert h(w_n)\rvert^2\bigr) < +\infty,$$
and the inequality $\lvert h'(z)\rvert\cdot \operatorname{dist}(z,\partial D) \leqslant 1 - \lvert h(z)\rvert^2$ then yields the assertion.