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An entire function $f(z)$ is of exponential type $\alpha$ if there exists $A$ such that $|f(z)|\leq Ae^{\alpha|z|}$ for all $z\in \mathbb C$. Given that $A=1$: how to prove that $\frac{1}{2\pi}\int_{0}^{2\pi}\log|f(re^{i\theta})|\,d\theta\leq \frac{2\alpha r}{\pi}$ for all $r>0$.

I did the following:

$\frac{1}{2\pi}\int_{0}^{2\pi}\log|f(re^{i\theta})|\,d\theta\leq \log(A)+\alpha |r|=0+r\alpha $

but I don't know how to get the $\dfrac{2}{\pi}$?

  • 0
    The inequality you wrote is not correct. Why are you trying to prove it? What is the reason of your conjecture?2015-03-29

0 Answers 0