Equivalence between trace-class and convergence of the kernel integral.

Question

It is known that when a bounded integral operator $T$ on $L^2(S)$ with kernel $k: S^2\rightarrow \mathbb C$ is trace-class, then we have (See for instance Functional Analysis, Peter Lax, p346) $$\mbox{Tr}(T)=\int_S k(x,x) dx.$$ But what about the reverse? Suppose that the integral above is well defined, can we then say that $T$ is trace-class? I cannot find a proper reference for this.

Answer 1

An easy counterexample is to take $S=\mathbb N$, with the counting measure (so your Hilbert space is $\ell^2(\mathbb N)$) and $$k(x,y)=\begin{cases}1,&\ |x-y|=1\\[0.3cm]0,&\ \text{otherwise}\end{cases}$$ Then $k(x,x)=0$ for all $x$, while $T$ is a unitary.

For non-positive operators, being trace-class is a much stronger condition than just having integrable/summable diagonal.