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This is from wikipedia. I tried to prove it.

We have

$$f(0)= \frac{1}{2\pi}\int_0^{2\pi} f(e^{i\theta })d \theta . $$

Taking the absolute value, we have

$$ |f(0)| \leq \frac{1}{2\pi}\int_0^{2\pi} |f(e^{i\theta }) | d \theta . $$

Then we would like to the take the logarithm of both sides. However, the $\log$-function is convex upwards, and we have

$$ \log\left(\frac{1}{2\pi}\int_0^{2\pi} |f(e^{i\theta }) | d \theta \right) \geq \frac{1}{2\pi}\int_0^{2\pi} \log( |f(e^{i\theta }) |) d \theta .$$

Wrong direction!

Can anyone give a proof?

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Beamer
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1 Answers1

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The proof follows from the fact that if $f(z_0) \ne 0$, there is an analytic logarithm of $\log f$ near $z_0$ and $\log|f|=\Re \log f$ is harmonic so we have equality in the subharmonic property by the mean value property of harmonic functions, otherwise, LHS is $-\infty$ so we obviously have the required inequality.

Also whenever we pass through zeroes, Jensen's theorem shows that the integral of $\log |f|$ increases so we get the sub-mean property on larger neighborhoods of $z_0$ that may contain zeroes even if $f(z_0) \ne 0$ but that is not required since it is not hard to prove that subharmonicity is a local property (in other words if the function is usc and if the sub-mean property holds in an open neighborhood of any point, it holds in the whole domain)

Conrad
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