Prove that $\int_0^{2\pi} \lvert \sum_{n=0}^{\infty} A_n (re^{i\theta})^n \rvert^2=2\pi \sum_{n=0}^{\infty} \lvert A_n\rvert ^2 r^{2n}$

Question

Suppose $f:B_1(0)\to \mathbb{C}$ is a holomorphic function, which is given by its Taylor series $$f(w)=\sum_{n=0}^{\infty} A_n w^n,\,\,A_n\in \mathbb{C}.$$ I want to show that $$\int_0^{2\pi}\lvert f(re^{i\theta})\rvert^2\,d\theta=2\pi\sum_{n=0}^{\infty} \lvert A_n\rvert^2 r^{2n},\,\,r\in (0,1).$$ $\textbf{Proof}$: Because of $$\lvert f(re^{i\theta})\rvert^2=\left\lvert\sum_{n=0}^{\infty} A_n r^n e^{i\theta n}\right\rvert^2=\left(\sum_{n=0}^{\infty} A_n r^n e^{i\theta n}\right)\overline{\left(\sum_{n=0}^{\infty} A_n r^n e^{i\theta n}\right)} \\ =\left(\sum_{n=0}^{\infty} A_n r^n e^{i\theta n}\right)\left( \sum_{n=0}^{\infty} \overline{A_n} r^n e^{-i\theta n}\right),$$ it follows from the Cauchy product formula that $$\lvert f(re^{i\theta})\rvert^2=\sum_{n=0}^{\infty} \sum_{k=0}^{n} A_k r^k e^{i\theta k} \overline{A_{n-k}}r^{n-k} e^{-i\theta(n-k)}=\sum_{n=0}^{\infty} r^n\sum_{k=0}^{n} A_k \overline{A_{n-k}} e^{i\theta(2k-n)}.$$ Since both of the series converge uniformly on the compact set $\{(r,t):t\in [0,2\pi]\}$ and are bounded, their product also converges uniformly and we can interchange the integral and the sum. But the integral from $0$ to $2\pi$ of $e^{i\theta(2k-n)}$ is not equal to zero iff $2k-n=0$ or $2k=n$. But $k$ is a whole number, so for uneven $n$, this equation can't be fulfilled. This means that we can just sum over all even natural numbers: $$\sum_{n=0}^{\infty} r^{2n} \sum_{k=0}^{2n} A_k \overline{A_{2n-k}}e^{i2\theta(k-n)}.$$ Again, the integral won't vanish iff $n=k$, which finally yields $$\int_{0}^{2\pi} \sum_{n=0}^{\infty} r^{2n}\sum_{k=0}^{2n}A_k \overline{A_{2n-k}}e^{i2\theta(k-n)}\,d\theta=\sum_{n=0}^{\infty} r^{2n}\int_{0}^{2\pi} A_n \overline{A_{2n-n}} e^{i2\theta(n-n)}\,d\theta \\ =\sum_{n=0}^{\infty} r^{2n}\int_0^{2\pi} A_n \overline{A_n}\,d\theta=2\pi \sum_{n=0}^{\infty} \lvert{A_n}\rvert^2 r^{2n}.$$

Answer 1

Use $$G=|\sum_{m} Z_m~|^2= \sum_{m} \sum_{n} Z_m \bar Z_n,$$ and $$\int_{0}^{2\pi} e^{i(m-n)\theta} d\theta =2\pi \delta(m-n)$$

So, $$H=\int_{0}^{2\pi}| \sum_m A^m r^m e^{im\theta}|^2 d\theta \implies \int_{0}^{2\pi}| \sum_m A^m r^m \bar A_n r^n e^{i(m-n)\theta}|^2 d\theta $$ $$\implies H= 2\pi \sum_{m} \sum_{n} A_m \bar A_n r^{m+n} \int_{0}^{2\pi} e^{2i(m-n)} d\theta$$ As $\delta(m-n)=1, m=n; 0, ~ if~ m \ne n$, we get $$H=2\pi \sum_{n} |A_n|^2 r^{2n}. $$

Answer 2

So here is my simpler approach by using Cauchy's integral: Since $f(z)$ is a holomorphic function hence: $$ LHS = \int_0^{2\pi}\lvert f(re^{i\theta})\rvert^2\,d\theta =\sum_{n=0}^{\infty} \overline{A_n}\cdot r^n \int_{0}^{2\pi} f\left(re^{i\theta}\right)e^{-in\theta}\mathrm{d}\theta $$ Then we define $C_r: z = re^{i\theta}$ and substitute each integral apply Cauchy integral since $f(z)$ is a holomorphic function then it's continuous on $C_r$. By Cauchy's integral: $$LHS = \sum_{n=0}^{\infty} \overline{A_n}\cdot r^{2n}(-i) \int_{C_r} \frac{f\left(z\right)}{z^{n+1}}\mathrm{d}z = \sum_{n=0}^{\infty} \overline{A_n}\cdot r^{2n}\cdot\frac{2\pi}{n!}\frac{\mathrm{d}^n f(0)}{\mathrm{d}z^{n}}$$ Finally, by Taylor series of $f(z)$, it is trivial that: $$\frac{\mathrm{d}^n f(z)}{\mathrm{d}z^{n}}= n!\cdot A_n + \sum_{k=1}^{\infty}\frac{(n+k)!}{k!}A_nz^k$$ This yeilds our result. $\blacksquare$