Another proof of Liouville's theorem.

Question

Let $f(z)=\sum_{n} a_n z^n$ has radius of convergence $R>0$ and $0<r<R$. Show:

$$\frac{1}{2\pi} \int_0^{2\pi} |f(re^{it})|^2 dt= \sum_{n}|a_n|^2r^{2n}$$

Use this equality to prove Liouvelle's Theorem.

Can anyone give me a hint to how to proceed with this?

Answer 1

Note that by multiplying out and squaring the series for $f$, and using the relation $|w|^2=w\bar{w}$ we see that $$\frac{1}{2\pi} \int_0^{2\pi} |f(re^{it})|^2 dt = \frac{1}{2\pi} \int_0^{2\pi} \big(\sum_{j=0}^{\infty} c_jr^je^{ijt} \big)\big(\sum_{k=0}^{\infty} \bar{c_k} r^k e^{-ikt} \big) dt $$$$=\frac{1}{2\pi} \sum_{j,k=0}^{\infty}c_j\bar{c_k} r^{j+k} \int_0^{2\pi} e^{i(j-k)t} dt $$and in this last expression note that the integrand is zero unless $j=k$. I was purposefully informal with my infinite sums (there should be some limits in there).

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For the second part of your question, suppose $f$ is a bounded entire function, say $|f| \leq M$.

Then the left hand side of the above equation is bounded above by $M^2$ for all $r>0$.

Now, for contradiction, assume that some $c_n \neq 0$. Then the right hand side is greater than or equal to $|c_n|^2r^{2n}$ which approaches to $\infty$ as $r \to \infty$.

Do you see a contradiction?

Answer 2

We have $f(z)=\sum_n c_nz^n$.

Substituting $z=re^{it}$ and evaluating,

$\left|f(re^{it})\right|^2=\sum_n\left|c_n\right|^2r^{2n}+\sum_n\sum_mg(n,m,r)\times e^{i(n-m)t}$

Integrating both sides from $t=0$ to $t=2\pi$ should do it.