Let $f$ be an entire function such that $\int_0^{2\pi} |f(r e^{it})| dt\leq r^{\frac{41}{3}}$ for all $r>0$. Prove that $f$ is constant.
I thought of using Liouville's theorem, combined with Cauchy's estimates. To do so, I tried to use Cauchy's integral formula, but it is not possible to bound the term $|w-z|$ which appears below in the integral.
Anyone can give me a hint? Thanks!
Best Answer
Take $\gamma$ as the circle of radius $r$ around the origin. Cauchy's Integral Formula gives us
$$ |f^{(n)}(0)| = \frac{n!}{2\pi} \left|\int_{\gamma} \frac{f(\xi)}{\xi^{n+1}}d \xi \right| \leq \frac{n!}{2\pi}\int_0^{2\pi}\frac{|f(re^{it})|}{|r|^{n}}dt. $$ Can you finish from here?