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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 $\dfrac{2}{\pi}$?

Monica
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    Why all the downvotes? I am not sure, but I think your formula may be incorrect? I think the quantity you are estimating should just be less than $\alpha r$ – Keaton Apr 05 '12 at 17:19
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    OP is a new user in math.SE. Downvoters should be nice & provide explanations, and/or suggestions for OP to improve her question. –  Apr 05 '12 at 17:30
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    Seems like a perfectly reasonable question to me. OP has left the imperative mode, shown some thought about the subject, and made clear where the problem is. – Ross Millikan Apr 05 '12 at 17:35
  • Is $\alpha$ a complex or real number? –  Apr 05 '12 at 21:41
  • @J.D.: $\alpha$ is real, positive, and finite. – Monica Apr 05 '12 at 21:54
  • Are you sure it's not $|f(z)|\le A|e^{\alpha z}|$ and $|\log |f(z)||$? – anon Apr 05 '12 at 22:24
  • As @Keaton suggested, I think the quantity should be just inferior to αr: for example just take $f(z) = e^{α|z|}$, which is an entire function and you'll see the majoration cannot be $2αr/π$. – revers Apr 10 '12 at 14:28
  • @revers, your function is not analytic, so it is not entire. It is true that the attempt here does not work, but for $f(z) = e^{\alpha z}$ the inequality is correct, since the left-hand side in that case is the average of a harmonic function, so it is $\log |f(0)| = 0$ by the mean value property, for all $r$. – Lukas Geyer Oct 28 '12 at 17:40
  • The inequality you wrote is not correct. Why are you trying to prove it? What is the reason of your conjecture? – Alexandre Eremenko Mar 29 '15 at 16:01

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