TITLE: When is an integral transform trace class?
QUESTION [29 upvotes]: Given a measure space $(X, \mu)$ and a measurable integral kernel $k : X \times X \rightarrow \mathbb{C}$, the operator 
$$ K f(\xi) =\int_{X} f(x) k(x,\xi) d \mu(x),$$
the operator $K$ is Hilbert Schmidt iff $k \in L^2(X \times X, \mu \otimes\mu)$! 
Q1:The main point of this questions, what are necessary and sufficient conditions for it to be trace class? 
I know various instances, where
$$ \mathrm{tr} K = \int_X k(x,x) d \mu(x).$$
Q2:What are counterexamples, where $x \mapsto k(x,x)$ is integrable, but the operator is not trace class?
Q3:What are counterexamples for a $\sigma$ finite measure space, where $k$ is compactly supported and continuous, but the kernel transformation is not trace class and the above formula fails?
Q4: Is there a good survey/reference for these questions.

REPLY [6 votes]: 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.