When is an integral transform trace class?

It may be worth noting the phenomena that can appear in Hilbert spaces, where study of the things is more decisive, both positive and negative.

First, I like the "definition" of "trace class" $T:X\rightarrow Y$ with Hilbert spaces $X,Y$ to be that $T$ is a composition of two Hilbert-Schmidt operators (which are defined as being in the HS-norm completion of the algebraic tensor product $X^*\otimes_{\mathrm {alg}}Y$. This gives an intrinsic definition... which, if desired, is provably equivalent to the (ugly) requirement that $\sum |\langle Tx_i,y_i\rangle| <\infty$ for every pair of orthonormal bases.

The reason I recall this cliche is that, in many applications of interest (to me!), natural operators are visibly Hilbert-Schmidt (if compact at all), and the issue becomes to prove trace-class. In practice (for me) it often happens that we know that every one of these integral operators is a finite sum of compositions of two such, proving trace-class.

Sometimes proof of the latter is highly non-trivial, as in the Cartier/Dixmier-Malliavin proof that test functions on Lie groups are finite linear combinations of convolutions of pairs of such. The totally-disconnected group analogue is trivial.

That summing or integrating down the diagonal fails is easy to illustrate with not-normal operators: the shift operator on one-sided or two-sided $\ell^2$ might seem to have trace absolutely summing to $0$, but it is not trace class at all. Integral analogues of this are clear.

Edit: in response to question about reference, etc.: in Lang's "SL(2,R)" the equivalence of the coordinate-dependent definition of "trace class", and the definition as composition of two Hilbert-Schmidt, are carefully compared. Further, in that same source, various conditions on a kernel assuring that its trace is equal to its integral over the diagonal are carefully treated. (I must say "... in contrast to dangerously glib treatments elsewhere").

Further edit: in response to Yemon Choi's comments: yes, the space of trace-class operators is also the closure of finite-rank operators with respect to the "trace norm"... At the moment, verification of the equivalence seems straightforward.

There are many results of the kind you ask about in the book

I. C. Gohberg and M. G. Krein, Introduction to the theory of linear nonselfadjoint operators. Providence, RI: American Mathematical Society, 1969.

It contains both necessary and sufficient conditions, and counter-examples.

A remark on (Q3): There is this famous example of T.Carleman (1916 Acta Math link) where he constructs a (normal ) operator with a continuous kernel such that it belongs to all Schatten p-classes if and only if $p\geq 2.$

More precisely it's possible to construct $k(x)=\sum_n c_ne^{2\pi i n x}$ continuous and periodic with $\sum_n|c_n|^p=\infty$ for $p<2$. Then $Tf=f\ast k$ acting on $L^2(\mathbb T)$ yields the desired result.

Provided some extra regularity on the kernel, the trace formula works fine (there are a lot of results in the literature)

Regarding (Q4) I personally find C. Brislawn's result very interesting but rather difficult to implement in practice.