2

Let $f$ be an entire function such that $\left| f \right| \left| Im(z)\right|^2 \leq 1$ then show that $f(z) \equiv 0$.

Liouville's theorem does the job, if I can show that $\left| f \right|$ is bounded. The function is bounded in each strip $\left|Im(z)\right| > C$ but I don't know how that implies that the function is bounded on the whole plane.

sr chunchurria
  • 311
  • 2
    One way of seeing this is by arguing that $f \equiv 0$ on $\mathbb{R}$. Use the fact that $\log |f|$ is subharmonic and apply sub mean-value property for circles with centre on real axis and let the radius go to $\infty$ (if you are not familiar with sub mean-value property then you may also use Jensen's formula) – r9m Jan 08 '21 at 22:49
  • 2
    Have a look at https://math.stackexchange.com/q/377782/42969. That is about $\left| f \right| \left| Im(z)\right| \leq 1$ but the same approach should work. – Martin R Jan 09 '21 at 03:07

0 Answers0