Question: Let $K \in L^2(R^n\times R^n)$. Are "explicit" necessary and sufficient conditions known such that $K$ is the kernel of some trace-class operator $A \in TC(L^2(R^n))$?
We know that $K$ is the kernel of a trace-class operator if and only if there are $J,L\in L^2(R^n\times R^n)$ such that $K(x,y) = \int J(x,z)L(z,y)\, dz$ almost everywhere. However, constructing the factorization can be hard in general, and I would like to have more explicit conditions preferrably only involving the kernel.
Let $A \in TC(L^2(R^n)) \subset HS(L^2(R^n))$ be a trace-class operator with kernel $K\in HS(L^2(R^n))$. Then it is known (Brislawn, Chris, Kernels of trace class operators, Proc. Am. Math. Soc. 104, No. 4, 1181-1190 (1988). ZBL0695.47017.), that $$ tr A = \int \tilde{K}(x,x)\, dx, $$ where $\tilde{K} : R^n\times R^n \to R$ is given by the limit of the Lebesgue average operator, $$ \tilde{K}(x,y) := \lim_{r\to 0} \frac{1}{|C_r|^2} \iint K(x+s,y+t)\, dsdt. $$ Here, $C_r = [-r,r]^n$. Hence, $K\in L^2(R^n\times R^n)$ the kernel of $A\in TC(L^2(R^n))$ implies integrability of $\tilde{K}(x,x)$. However, integrability is not a necessary condition for $K$ to be the kernel of a trace-class operator.
The motivation comes from defining Sobolev-space analogues of trace-class operators, e.g., the space of trace-class operators $A$ such that $\nabla A \nabla$ is also trace class, which transforms to $\nabla_x\cdot \nabla_y K$ being the kernel of a trace-class operator.