This question came up on a past qualifying exam. I am not sure whether the question is stated wrong or if my solution is wrong.
Here is the question:
Suppose $f(z) = \sum_{n=0}^{\infty}a_{k}z^{k}$ for $|z| < 1$, where $a_{k} \in \mathbb{C}$ for all $k \in \mathbb{N}$. Prove that $$ \sup_{0 < r < 1}\int_{t=0}^{2\pi}|f(re^{it})|^{2}dt = \sum_{k=0}^{\infty}|a_{k}|^{2}.$$
Here is my solution:
Let $0 < r < 1$. Then for $|z| = r$, $$|f(z)|^{2} = \left( \sum_{n=0}^{\infty}a_{n} z^{n}\right)\left(\sum_{n=0}^{\infty}\bar{a}_{n}\bar{z}^{n}\right) = \sum_{k=0}^{\infty}\sum_{j=0}^{k}a_{j}\bar{a}_{k-j}z^{j}\bar{z}^{k-j},$$ where the convergence is absolute. Hence, \begin{align} \int_{0}^{2\pi}|f(re^{it})|^{2}dt & = \int_{0}^{2\pi}\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}a_{j}\bar{a}_{k-j}r^{k}e^{it(2j-k)}dt \\ & = \sum_{k=0}^{\infty}\int_{0}^{2\pi}\sum_{j=0}^{k}a_{j}\bar{a}_{k-j}r^{k}e^{it(2j-k)}dt \\ & = \sum_{k=0}^{\infty}r^{k}\sum_{j=0}^{k}a_{j}\bar{a}_{k-j}\underbrace{ \int_{0}^{2\pi}e^{it(2j-k)}dt}_{= 0\ \text{ unless }\ 2j = k} \\ & = \sum_{k=0}^{\infty}r^{2k}a_{k}\bar{a}_{k}2\pi \\ & = 2\pi \sum_{k=0}^{\infty}r^{2k}|a_{k}|^{2}. \end{align} We can justify bringing the sum outside of the integral by Fubini's theorem and the fact that the series is absolutely convergent. Now take the limit as $r \rightarrow 1$ and we get $$ \sup_{0 < r < 1}\int_{t=0}^{2\pi}|f(re^{it})|^{2}dt = 2\pi \sum_{k=0}^{\infty}|a_{k}|^{2}.$$
Is my solution correct? My solution differs from the what the integral is supposed to be by a factor of $2\pi$.