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

  • 2
    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$2012-04-05
  • 5
    **OP is a new user in math.SE. Downvoters should be nice & provide explanations, and/or suggestions for OP to improve her question.**2012-04-05
  • 4
    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.2012-04-05
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    Is $\alpha$ a complex or real number?2012-04-05
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    @J.D.: $\alpha$ is real, positive, and finite.2012-04-05
  • 0
    Are you sure it's not $|f(z)|\le A|e^{\alpha z}|$ and $|\log |f(z)||$?2012-04-05
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    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 $2αr/π$.2012-04-10
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    @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 $r$.2012-10-28
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    The inequality you wrote is not correct. Why are you trying to prove it? What is the reason of your conjecture?2015-03-29

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