Let $f$ be an analytic function on $D(0, 1)$ and continuous on $\overline{D(0, 1)}$. We assume that $f \equiv 0$ on the $arc$ defined by {$e^{i\theta}, 0\le \theta_1 \lt \theta \lt \theta_2 \le 2\pi$}. $\quad$ prove that $f= 0$ on $D(0, 1)$.
Asked
Active
Viewed 85 times
2 Answers
1
Schwarz reflection principle across the arc. For $|z|>1$ define $f(z) = \overline{f(1/\overline{z})}$. This is analytic on $D(0,1)$ and $|z|>1$, and continuous on the union of these and the arc. Using Morera's theorem, it is analytic there. Then by the identity theorem, it is $0$ everywhere.
Robert Israel
- 448,999
1
Another way: if you multiply together a bunch of functions of the form $f(cz)$, where $c$ is a complex number with $|c|=1$, you can get a function that is $0$ on the entire boundary. By the maximum principle, the function is $0$ everywhere. $f$ and the rotations can each only have countably many zeroes unless they are constant, but this product has uncountably many. Thus, $f$ is $0$ everywhere.
user387394
- 594