Lidskii's trace formula for integral operators

Remarks

The proof has two steps. First, for a trace class integral operator with continuous kernel, the operator trace equals the trace of the kernel $\int_a^b K(s, s) \, ds$. Second, Lidskii's theorem says that the operator trace equals the sum of the eigenvalues, counted with algebraic multiplicity.

Example: on $[0, 1]$ let $K(s, t) = \min(s, t)$. The eigenfunctions are $\sin((n - \tfrac12)\pi s)$ with eigenvalues $\lambda_n = 1/((n - \tfrac12)^2 \pi^2)$, $n \ge 1$. Since $K$ is a positive kernel and $\sum_n \lambda_n < \infty$, the operator is trace class. The formula reads $\sum_{n \ge 1} \frac{4}{(2n - 1)^2 \pi^2} = \int_0^1 s \, ds = \frac12$, which is the identity $\sum_{n \ge 1} (2n-1)^{-2} = \pi^2/8$.

The trace class hypothesis cannot be dropped. There are continuous kernels whose operators are not trace class, and for them the sum of the eigenvalues need not converge absolutely. Take the Volterra kernel $K(s, t) = 1$ for $t \le s$ and $K(s, t) = 0$ otherwise. Its operator has no nonzero eigenvalue, while the diagonal integral is $b - a$. There is no contradiction: this kernel is not continuous, and the operator is not trace class.

Going deeper

For a student

For a continuous Hermitian kernel of positive type, the formula follows from Mercer's theorem by integrating the expansion $K(s, s) = \sum_n \lambda_n \abs{\varphi_n(s)}^2$. Lidskii's theorem extends it to kernels that are not self-adjoint, where the eigenvalues may be complex and the eigenfunctions need not form a basis. The formula is the first case of the trace identities $\sum_n \lambda_n^m = \int_a^b K_m(s, s) \, ds$ for the iterated kernels $K_m$, which hold for every continuous kernel when $m \ge 2$, since then $T^m$ is trace class.