Show that a complex function is identically zero.

Question

Show that if $f:\{z : |z| < 1\} \to \mathbb{C}$ is analytic, bounded, and for some $0<a<b<2 \pi$ $$\lim_{r \to 1^-} \sup_{a \leq \theta \leq b} |f(re^{i\theta})| = 0 , $$ then $f=0.$

I tried to use Maximum modulus theorem of Liouville theorem, but we do not know whether $f$ is analytic on unit circle.

Answer 1

Broad hints: There exist $c_n, 1\leq n \leq N$ such that if $g(z)= f(z)f(ze^{ic_1})f(ze^{ic_2})....f(ze^{ic_N})$ then $g(re^{i\theta}) \to 0$ for every $\theta$. Now apply MMP to circle of radius $1-\epsilon$ and let $\epsilon \to 0$.