I have encountered a question like this:
Suppose a complex function $f$ is analytic in $|z|<1$, continuous in $|z| \leq 1$, and $|f|=1$ on $|z|=1$. Show that $f$ can be extended as a rational function.
According to the isolation property of zero points, $f$ has only finite zero points in $|z| \leq 1$. But I don’t know how to show that it’s a rational function. Hope someone could help. Thanks!
There are finitely many zeros of $f$ in the unit disc. There is a finite Blaschke product $p(z)$ having the same zeros as $f$ in the unit disc. This is designed so that $|p|=1$ on the unit circle. Then $f=pg$ where $g$ is continuous and non-zero on the closed unit disc, holomorphic on the open unit disc and has $|g|=1$ on the unit circle. By the maximum modulus principle, $g$ is constant. So $f$ is a constant multiple of a finite Blaschke product, which is a rational function on $\Bbb C$.
