Hilbert–Schmidt expansion theorem

Remarks

For fixed $x$, the numbers $\mu_n \overline{\varphi_n(x)}$ are the Fourier coefficients of the function $y \mapsto \overline{K(x, y)}$, so Bessel's inequality gives $\sum_n \mu_n^2 \abs{\varphi_n(x)}^2 \le \int_a^b \abs{K(x, y)}^2 \, dy \le C$ for a constant $C$ independent of $x$. The Cauchy–Schwarz inequality then bounds the tail $\sum_{n \ge m} \abs{\mu_n \ip{h}{\varphi_n} \varphi_n(x)}$ by $\sqrt{C} \left(\sum_{n \ge m} \abs{\ip{h}{\varphi_n}}^2\right)^{1/2}$, which tends to $0$ uniformly in $x$. By the spectral theorem for compact self-adjoint operators, the series converges to $f$ in $L^2$; the uniform limit is continuous and so is $f$, hence they agree everywhere.

Example: for $K(x, y) = \min(x, y)$ on $[0, 1]$, the eigenfunctions are $\sqrt{2} \sin\left((n - \tfrac12) \pi x\right)$. Every function $f = T_K h$ with $h \in L^2[0, 1]$ is continuous with $f(0) = 0$, and its expansion in these sines converges uniformly on $[0, 1]$.

The hypothesis $f = T_K h$ matters. For a general function $f \in L^2[a, b]$, the eigenfunction series converges only in $L^2$. Its sum is the orthogonal projection of $f$ onto the closed span of the $\varphi_n$, which may differ from $f$.

Going deeper

For a student

This is the classical form of the spectral theorem for compact self-adjoint operators, adapted to continuous kernels. It is used to solve the Fredholm equation of the second kind with a symmetric kernel: the solution of $\varphi - \lambda T_K \varphi = f$, for $\lambda$ not a characteristic value, is $\varphi = f + \lambda \sum_n \frac{\mu_n \ip{f}{\varphi_n}}{1 - \lambda \mu_n} \varphi_n$. Through Green's functions it gives the uniform convergence of eigenfunction expansions for Sturm–Liouville problems.