Bilinear series of a symmetric kernel

Remarks

The functions $\varphi_n(x) \overline{\varphi_m(y)}$ form an orthonormal family in $L^2([a, b]^2)$. The coefficient of $K$ on $\varphi_n(x) \overline{\varphi_m(y)}$ is $\int \int K(x, y) \varphi_m(y) \overline{\varphi_n(x)} \, dy \, dx = \mu_m \ip{\varphi_m}{\varphi_n}$, which is $\mu_n$ if $m = n$ and $0$ otherwise. So the series is the orthogonal projection of $K$ onto the span of this family. The difference between $K$ and the series is a kernel with operator $T_K - \sum_n \mu_n \ip{\cdot}{\varphi_n} \varphi_n$. This operator is $0$ by the spectral theorem for compact self-adjoint operators, so the difference is $0$ almost everywhere. The identity for $\sum \mu_n^2$ is then Parseval's identity.

Example: for $K(x, y) = \min(x, y)$ on $[0, 1]$, $$\min(x, y) = \sum_{n \ge 1} \frac{2 \sin\left((n - \tfrac12)\pi x\right) \sin\left((n - \tfrac12)\pi y\right)}{(n - \tfrac12)^2 \pi^2},$$ and the identity gives $\sum_{n \ge 1} \frac{1}{(n - \frac12)^4 \pi^4} = \int_0^1 \int_0^1 \min(x, y)^2 = \frac{1}{6}$.

In general the convergence is only in $L^2$. Under the extra hypotheses of Mercer's theorem, continuity and positivity of the kernel, the series converges absolutely and uniformly.

Going deeper

For a student

The bilinear series diagonalizes the kernel itself, not only the operator. It gives the estimate $\sum \mu_n^2 < \infty$, so the characteristic values grow fast enough that $\sum 1/\lambda_n^2$ converges. For positive continuous kernels it is upgraded to uniform convergence by Mercer's theorem, which also gives the trace formula $\sum \mu_n = \int K(x, x) \, dx$.