Solution (source code)

= Solution

Expanding the square and using orthogonality of Fourier modes,
$$
\begin{aligned}
\frac1{2\pi}\int_{-\pi}^{\pi}|\widehat v(\theta)|^2d\theta
&=\sum_{m,j}v_m\overline{v_j}
\frac1{2\pi}\int_{-\pi}^{\pi}e^{-i(m-j)\theta}d\theta\\
&=\sum_m|v_m|^2.
\end{aligned}
$$
Finite truncations justify the calculation directly, and passage to the $\ell^2$ <limit> proves <Parseval identity> for every square-summable <sequence>.