TITLE: Complex Analysis: Showing analytic function is zero
QUESTION [5 upvotes]: How I can solve this problem:
Let $f: D \to D$ be an analytic function where $D$ is the unit open disc in $\mathbb C$. Suppose there is a positive number $\delta > 0$ such that ,  $$\lim_{z \to e^{iθ}} ⁡f(z)= 0; \qquad \forall \ |\theta| < \delta.$$ 
Show that $f \equiv 0$ on $D$. 
Thanks
Note: An easier version of this Privalov's Theorem is an exercise problem from Stein and Shakarchi's Complex Analysis textbook which additionally assumes the holomorphic function converges uniformly to $0$ on the portion of the arc. See [1], [2], [3]. The absence of this 'uniform' non-tangential limit on the portion of the arc makes it a slightly more difficult problem.

REPLY [6 votes]: I think the Schwarz reflection principle, applied to the circular arc $\>C\!:  z=e^{i\theta}, \ |\theta|<\delta$, does the trick. 
The function $f$ is analytic on one side of this arc. Extending $f$ with the value $0$ to the points of $C$ makes it continuous on $D\cup C$ and real valued on $C$. It follows that $f$ can be extended analytically  to the outside of $C$ by putting $$\tilde f(z):=\overline{f(1/\bar z)}\qquad\bigl(|z|>1\bigr)\ ,$$
and $\tilde f(z):=f(z)$ otherwise. As $\tilde f$ is now analytic in a neighborhood of $z=1$ and is $\equiv0$ on $C$ it follows that $\tilde f(z)\equiv0$.