= Solution
For a real $2\pi$-periodic square-integrable <function>, use the <Fourier series> convention
$$
f(x)\sim\frac{a_0}{2}+\sum_{n=1}^\infty[a_n\cos nx+b_n\sin nx],
$$
with <Fourier coefficients>
$$
\boxed{a_n=\frac1\pi\int_{-\pi}^{\pi}f(x)\cos nx\,dx\quad(n\geq0),\qquad
b_n=\frac1\pi\int_{-\pi}^{\pi}f(x)\sin nx\,dx\quad(n\geq1)}.
$$
The trigonometric <orthogonality> relations are $\int_{-\pi}^{\pi}\cos nx\cos mx\,dx=\pi\delta_{nm}$ for positive $n,m$, with the same formula for sines; sine-cosine products integrate to zero. The constant has squared integral $2\pi$ and is <orthogonal> to every positive harmonic. Thus the squared norm of the $N$th partial sum is
$$
\frac1\pi\int_{-\pi}^{\pi}f_N(x)^2\,dx
=\frac{a_0^2}{2}+\sum_{n=1}^N(a_n^2+b_n^2).
$$
Taking the limit in the <L2 space> gives the Fourier-series <Parseval identity>:
$$
\boxed{\frac1\pi\int_{-\pi}^{\pi}f(x)^2\,dx
=\frac{a_0^2}{2}+\sum_{n=1}^\infty(a_n^2+b_n^2)}.
$$
Square integrability is the natural hypothesis for this finite norm identity; additional smoothness can be used when pointwise convergence is required.
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