I have been struggling on the following problem.
Suppose $f$ is an entire analytic function such that $|f(z)|>1$ if $|z|>1$. Show that $f$ is a polynomial.
My idea is as followed: all zeros of $|f(z)|$ lie inside $|z|\leq 1$. Applying Argument Principle, we can show that number of zeros of $f$ is bounded. So we can assume
$f(z)=(z-z_1)…(z-z_M)g(z)$
where g is entire analytic without any zeros.
Then I would like to apply Liouville's Theorem: the point is that it isn't too clear to me why $|\dfrac{1}{g(z)}|$ is a bounded function.
Best Answer
If the Taylor series about 0 does not terminate, $f(1/z)$ has an essential singularity at $0$ (why?)
Then from the Casorati–Weierstrass theorem (have you learnt this?) you know $f(1/z)$ cannot be bounded from below near 0. Contradiction!