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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.

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    $\begingroup$ If the kernel is continuously differentiable often enough and its derivatives are still L2, then the operator is trace class, since a high enough power of the resolvent of the Laplacian is Hilbert-Schmidt. $\endgroup$
    – user130903
    May 7, 2021 at 11:23
  • $\begingroup$ Thank you! I have been thinking that the Lebesgue differentiation theorem applied to the kernel (one must have $\nabla_1\cdot\nabla_2K\in L^2(R^3\times R^3)$) could just possibly imply traceability of the undifferentiated kernel. $\endgroup$
    – Nemis L.
    May 7, 2021 at 11:37

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