How to prove the following inequalities

Question

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.

Answer 1

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.