Trace of an integral operator

Remarks

Example: for $K(s, t) = \min(s, t)$ on $[0, 1]$, the kernel is continuous and $\operatorname{tr} K = \int_0^1 s \, ds = 1/2$. Non-example: the weakly singular kernel $K(s, t) = \abs{s - t}^{-1/2}$ on $[0, 1]$ is not continuous, and it is infinite on the diagonal, so the formula gives no number. For a degenerate kernel $K(s, t) = \sum_{i=1}^n u_i(s) v_i(t)$, the trace is $\sum_{i=1}^n \int_a^b u_i(s) v_i(s) \, ds$. This is the trace of the $n \times n$ matrix with entries $\int_a^b v_i(s) u_j(s) \, ds$, which represents $T$ on the span of $u_1, \dots, u_n$. The traces $\int_a^b K_m(s, s) \, ds$ of the iterated kernels $K_m$, $m \ge 2$, appear in the power series of the Fredholm determinant. A continuous kernel need not define a trace class operator on $L^2(a, b)$; when it does, the number above equals the operator trace.