TITLE: an equality of a series and an integral involing squared norms
QUESTION [1 upvotes]: I was reading the solution of a problem, and I did not understand one step. Let's consider the power series expansion of a holomorphic function $ f: D\to D$ (where D is the unit open disk) and $f(0)=0$.
$
$$ f(z)=
\sum\limits_{k = 1}^\infty  {a_k z^k } 
$$
Well I don't understand why:
$$
\sum\limits_{k = 1}^\infty  {\left| {a_k } \right|^2 }  = \mathop {\lim }\limits_{r \to 1} \int\limits_0^{2\pi } {\left| {f\left( {re^{it} } \right)} \right|^2 dt} 
$$
I don't know if this result holds only in this kind of functions, or it's more general (other holomorphic functions)

REPLY [3 votes]: The identity you have written is not quite correct. It is off by a factor of $2\pi$. Also, the function $f$ does not have to map into $D$, and it doesn't have to map $0$ to $0$. It does need to be holomorphic on $D$, though.
For $z\in D$, compute $|f(z)|^2$ using the power series expansion for $f$: $$|f(z)|^2 = \left(\sum a_nz^n\right)\left(\sum \overline{a}_k \overline{z}^k\right) = \sum_{(n,k)} a_n\overline{a}_kz^n\overline{z}^k.$$ You can then use this expression to compute $$\int_0^{2\pi}|f(re^{it})|^2\,dt = \sum_{(n,k)}a_n\overline{a}_k\int_0^{2\pi}r^{n+k}e^{i(n-k)t}\,dt.$$ The integrals on the right hand side of this expression are $0$ unless $n - k = 0.$ The equality then simplifies to $$\int_0^{2\pi}|f(re^{it})|^2\,dt = 2\pi \sum_k a_k\overline{a}_k r^{2k} = 2\pi\sum_k |a_k|^2r^{2k}.$$ Letting $r\to 1$ gives $$\lim_{r\to 1^-}\int_0^{2\pi}|f(re^{it})|^2\,dt = 2\pi\sum_k |a_k|^2.$$
{"set_name": "stack_exchange", "score": 1, "question_id": 217231}