What is $\sum_{n=0}^{\infty}|a_nz^n|^2=\frac{1}{2 \pi}\int_{-\pi}^{\pi}|f(ze^{it})|^2dt$ for?

Question

Let $f : \mathbb C\rightarrow \mathbb C$ be an analytic function : $f(z)= \sum a_n z^n$

It holds that

$$a_n z^n= \frac{1}{2 \pi}\int_{-\pi}^{\pi}f(ze^{it})e^{-int}dt$$

and $$\sum_{n=0}^{\infty}|a_nz^n|^2=\frac{1}{2 \pi}\int_{-\pi}^{\pi}|f(ze^{it})|^2dt$$

I wonder what the last formula is useful for (the first gives the coefficients in the power series of $f$ knowing only $f$). Can you give some examples where this identity proves useful ?

Answer 1

It's useful for polynomial..like:

• Let $f(z)=\sum_{n= 0}^{+\infty}a_n z^n$, the radius of convergence $R\ge 1$.

$\forall n,\quad a_n\in \mathbb{Z}$ and $f$ is bounded on the open unit disk.

Then one can prove that $f$ is a polynomial.

• (Oral X) Let $f(z)=z^{n}+a_{n-1}z^{n-1}+...+a_{0}$ be a polynomial of degree $n\geq1$.

If $\lvert f(z)\rvert\le 1$, for all $\lvert z\rvert=1$, then one can show that $f=z^n$.

• I forgot this famous theorem : Due to Liouville

The only developable functions in power series and bounded on $\mathbb{C}$ are constant functions.

Answer 2

By considering $a_0$ and $f(z)=\underbrace{ \exp( \exp(... \exp(}_{n-1}z)))$, we have the very nice $$\int_0^{2\pi}\underbrace{\exp(\exp(...\exp(}_{n}it)))dt=2\pi\underbrace{ \exp( \exp(... \exp(}_{n-2}1))).$$