L2 convergence of Fourier series

Remarks

Both claims follow from the Fourier basis theorem, which says that the monomials $e_n$ form an orthonormal basis of $L^2(\mathbb{T})$. $S_N f$ is the orthogonal projection of $f$ onto the span of $e_{-N}, \dots, e_N$, so $\norm{f - S_N f}_2^2 = \norm{f}_2^2 - \sum_{\abs{n} \le N} \abs{\hat f(n)}^2$. Completeness of the basis makes the right side tend to $0$; the second formula is Parseval's identity for this basis. Example: for $f(x) = x$ on $(-\pi, \pi)$, $\hat f(0) = 0$ and $\abs{\hat f(n)} = 1/\abs{n}$ for $n \ne 0$, and Parseval gives $\frac{\pi^2}{3} = 2 \sum_{n \ge 1} \frac{1}{n^2}$, which is the value $\frac{\pi^2}{6}$ of the Basel problem. Convergence in $L^2$ does not give convergence at each point; pointwise convergence almost everywhere for $L^2$ functions is the much deeper Carleson's theorem. The converse direction is the Riesz–Fischer theorem: every sequence $(c_n) \in \ell^2(\Z)$ is the sequence of Fourier coefficients of some $f \in L^2(\mathbb{T})$.

Going deeper

For a student

This is the simplest and most robust convergence theorem for Fourier series. It needs no smoothness, only square integrability, and it holds for every function in $L^2(\mathbb{T})$. The analogous statement in $L^p(\mathbb{T})$ for $1 < p < \infty$ is true but harder; it rests on the boundedness of the conjugate function, a theorem of M. Riesz. For $p = 1$ and in $C(\mathbb{T})$ with the supremum norm the analogous statement is false, because the Lebesgue constants of the Dirichlet kernel are unbounded.