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

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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 5 '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. –  user2468 Apr 5 '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 5 '12 at 17:35
    
Is $\alpha$ a complex or real number? –  user2468 Apr 5 '12 at 21:41
    
@J.D.: $\alpha$ is real, positive, and finite. –  Monica Apr 5 '12 at 21:54
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