Let $f(z)=\sum _{n=0}^\infty a_nz^n$ be an entire function and let $r$ be a positive real number.Then show that:

  1. $\sum _{n=0}^\infty |a_n|^2r^{2n}\le \sup_{|z|=r} |f(z)|^2$
  2. $\sum _{n=0}^\infty |a_n|^2r^{2n}\le\dfrac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^2 d\theta$

My try:

Since $f(z)=\sum _{n=0}^\infty a_nz^n\implies |f(z)|\le \sum _{n=0}^\infty |a_n||z|^n$

If we take $|z|=r$ then we get;

$f(z)=\sum _{n=0}^\infty a_nz^n\le \sum _{n=0}^\infty |a_n|r^n$

But I am unable to complete the above problem .Please give some hints.


Prove (2) first, then prove (1). Use

$$\lvert f(z)\rvert^2 = f(z)\overline{f(z)} = \sum_{n,m \ge 0} a_n \overline{a_m} z^n \bar{z}^m$$

to obtain

$$\int_0^{2\pi} \lvert f(re^{i\theta})\rvert^2\, d\theta = 2\pi \sum_{n = 0}^\infty \lvert a_n\rvert^2 r^{2n}$$

by term-wise integration.

  • $\begingroup$ can you please explain how did you get the second step after doing term-wise integration ;I am not getting it $\endgroup$ – Learnmore Oct 11 '16 at 3:15
  • $\begingroup$ No problem, @PeterPan. If $z = re^{i\theta}$, $\lvert f(z)\rvert^2$ is a sum of terms $a_n\overline{a_m}r^{n+m}e^{in\theta}e^{-im\theta}$, and $\int_0^{2\pi} e^{in\theta}e^{-im\theta}\, d\theta = \int_0^{2\pi} e^{i(n-m)\theta}\, d\theta$ is $2\pi$ for $m = n$ and $0$ otherwise. Hence, term-wise integration yields $\sum\limits_{n = 0}^\infty a_n \overline{a_n}r^{n+n}(2\pi)$, or $2\pi \sum\limits_{n = 0}^\infty \lvert a_n\rvert^2 r^{2n}$. $\endgroup$ – kobe Oct 11 '16 at 3:24

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