Schmidt's singular value decomposition of a kernel

Remarks

The operator $T_K^* T_K$ is compact, self-adjoint and has nonnegative eigenvalues; it is the integral operator with the kernel $\int_a^b \overline{K(t, x)} K(t, y) \, dt$. The spectral theorem for compact self-adjoint operators gives orthonormal eigenfunctions $v_n$ with eigenvalues $s_n^2 > 0$. Then $u_n = T_K v_n / s_n$ is an orthonormal family, since $\ip{T_K v_n}{T_K v_m} = \ip{T_K^* T_K v_n}{v_m} = s_n^2 \delta_{nm}$. Parts 3 and 4 follow as for the bilinear series of a symmetric kernel, using the orthonormal family $u_n(x) \overline{v_m(y)}$.

Example: the Volterra kernel on $[0, 1]$, equal to $1$ for $y \le x$ and $0$ otherwise, has no eigenvalues, but it has singular values $s_n = \frac{1}{(n - \frac12)\pi}$, with $v_n(y) = \sqrt{2} \cos\left((n - \tfrac12)\pi y\right)$ and $u_n(x) = \sqrt{2} \sin\left((n - \tfrac12)\pi x\right)$. Part 4 gives $\sum_n \frac{1}{(n - \frac12)^2 \pi^2} = \frac12$, the area of the triangle where the kernel is $1$.

For a Hermitian kernel, the $s_n$ are the absolute values of the nonzero eigenvalues, and one may take $u_n = \pm v_n$.

Going deeper

For a student

The decomposition is the right tool for non-symmetric kernels, where eigenfunctions may fail to exist or to be orthogonal. Keeping the first $m$ terms gives the best approximation of $K$ in the $L^2$ norm by a degenerate kernel of rank at most $m$, with error $\left(\sum_{n > m} s_n^2\right)^{1/2}$. For a Fredholm equation of the first kind $T_K \varphi = f$, it gives the formal solution $\varphi = \sum_n s_n^{-1} \ip{f}{u_n} v_n$, and the growth of $1/s_n$ shows why such equations are ill-posed.