if $\sup_{0<r<1} \int_0^{2\pi} |f(re^{i t})|^2 dt < \infty$ then $\sum{|a_n|^2} <\infty$

Question

Let $f(z) = \sum_0^\infty a_n z^n$ analytic for $|z|<1$. I need to show that if $$\sup_{0<r<1} \frac{1}{2 \pi}\int_0^{2\pi}|f(re^{i t})|^2 dt < \infty$$ then

$$\sum_0^\infty |a_n|^2 <\infty$$

My attempt: If $\gamma$ is the circle centered at $0$ an radius $0<r<1$ then $$ \frac{1}{2 \pi}\int_0^{2\pi}|f(re^{i t})|^2 dt =\frac{1}{2 \pi i}\int_{\gamma} \frac{|f(z)|^2}{z} dz$$ The numerator in the integrand isn't analytic, so I don't know how to handle it. If instead we consider $$ \left|\frac{1}{2 \pi}\int_0^{2\pi}f(re^{i t})^2 dt \right|=\left|\frac{1}{2 \pi i}\int_{\gamma} \frac{f(z)^2}{z} dz\right|$$ By Cauchy's formula we get that the previous integral is equal to $|{f(0)}^2|=|{a_0}^2|$. A result which doesn't look useful.

Answer 1

Hint: Note that $f(re^{it})=\sum\limits_{n=0}^\infty a_n r^n e^{int}.$ Plug this into your bound and use the orthogonality properties of the family of functions $\{e^{int}\}_{n\in\mathbb{Z}}$. (I can expand more, if needed)

Answer 2

If $0<r<1$, computing the integral (and noting that we can exchange integration with summation by uniform converge on the circle of radius $r$), we get that $\Sigma{|a_n|^{2}r^{2n}} < M < \infty$

Fix $m>0$ and pick $r_m=1-\frac{1}{2m+2}$, so $r_m^{2n} \ge \frac{1}{e}, 0\le n \le m$ which immediately gives that the partial sums of $\Sigma{|a_n|^2}$ are bounded by $eM$