determining complex function Problem
Let $f$ be holomorphic in $D=\{|x+iy|< 1\}$, with


*

*$|f|\le |y|^{-1/2}$

*$\lim_{r\rightarrow 1} f(re^{i\theta}) = 0$, for any $\theta\in[0,2\pi]$


Prove $f = 0$.

It is an old qualify problem, I think it seems enough if I apply maximum principle according to second condition[not sure...:)], I am not very sure how to use the first condition.
Thanks.
 A: Thanks to @YoniRozenshein for pointing out an initial error, and to @Yimin for pointing out a solution.
Apply the DCT to Cauchy's integral formula.
Suppose $|z_0| < s <r <1$. Then $f(z_0) = \frac{1}{2 \pi i} \oint_{\gamma_r} \frac{f(z)}{z-z_0} dz = \frac{1}{2 \pi } \int_0^{2 \pi} \frac{f(r e^{i \theta})}{r e^{i \theta}-z_0} r e^{i \theta} d\theta$, from which we get $|f(z_0)| \le \frac{r}{2 \pi } \int_0^{2 \pi} \frac{|f(r e^{i \theta})|}{|r e^{i \theta}-z_0|}   d\theta \le \frac{1}{2 \pi } \int_0^{2 \pi} \frac{|f(r e^{i \theta})|}{|r e^{i \theta}-z_0|}   d\theta$.
Let $g_r (\theta) = \frac{|f(r e^{i \theta})|}{|r e^{i \theta}-z_0|}$, for $r \in (s,1)$. Since $|r e^{i \theta}-z_0|\ge s -|z_0|$, we have $g_r (\theta) \le \frac{|f(r e^{i \theta})|}{s -|z_0|}$, and hence $\lim_{r \uparrow 1} g_r (\theta) = 0$ for all $\theta$. To apply the DCT, we need to bound $g_r$ by some integrable function $g_1$.
We have $|f(r e^{i \theta})| \le \frac{1}{\sqrt{r |\sin \theta|}}$ by assumption, and $|\sin(x)| \ge \frac{1}{2} \min_{k \in \mathbb{Z}} |x-k\pi|$, which gives
$g_r (\theta) \le g_1(\theta)=  (\frac{1}{s-|z_0|} \frac{1}{\sqrt{s}}) \frac{\sqrt{2}}{\sqrt{\min_{k \in \mathbb{Z}} |\theta-k\pi|}} $. A small amount of work shows that $\int_0^{2 \pi} g_1 = 8\sqrt{\pi} (\frac{1}{s-|z_0|} \frac{1}{\sqrt{s}})  $, hence integrable.
It follows that $|f(z_0)| \le \frac{1}{2 \pi} \lim_{r \uparrow 1} \int_0^{2 \pi} g_r = 0$. Since $z_0 \in D$ was arbitrary, we are finished.
