Application of Maximum Principle or Identity Theorem.

Question

Let $f$ be holomorphic and bounded. in the unit disk $\{z\in \mathbb{C} : |z|\leq 1 \}.$

If $f$ converges uniformly to zero in the sector $\frac{\pi}{4}<argz<\frac{\pi}{3}$ as $|z|$ tends to 1,

then $f$ is identically $0$ in the closed unit disk.

I think this problem is related to Maximum Principle or Identity Theorem.

I want you guys to help me.

Answer 1

I think that your hypothesis are that $f$ is holomorphic and bounded in the open unit disk. With these hypothesis, the result follows from theorem 11.22 (chapter Harmonic functions) of Rudin Real and complex analysis. Here is a proof following Rudin ( but not completely). Let $A$ the subarc $\{\exp(i\theta), \frac{\pi}{4}<\theta<\frac{\pi}{3}\}$. Suppose that $|f(z)|\leq M$ for all $z$.

Put $M(r)=\sup\{|f(z)|, |z|=r, z/|z|\in A\}$. Then we know that $M(r)\to 0$ if $r\to 1$. Let $m$ large, such that with $\eta=\exp(2i\pi/m)$, we have $S=\{z, |z|=1\}=\cup_{k=0}^{m-1}\eta^k A$. Put $g(z)=\prod_{k=0}^{m-1}f(\eta^k z)$. Then we get easily that $|g(z)|\leq M^{m-1}M(r)$ for all $z$ such that $|z|=r$. By the maximum principe, we get that $|g(z)\leq M^{m-1}M(r)$ for all $z$, $|z|\leq r$. As $M(r)\to 0$ if $r\to 1$, it is easy to conclude that $g=0$, hence $f=0$. .

Answer 2

Hint: Define $f^*$ an auxiliarry function on all $z$ such that $\pi/4 < \arg(z)< \pi /3$ by :

$f^*=f$ inside the open disk, and

$f^*=0$ outside.

Using Morera's theorem, show that $f^*$ is holomorphic. Conclude using the (generalized) identity theorem [the arc is a set of points of accumulation].