• $H$ be a separable $\mathbb R$-Hilbert space
  • $L\in\mathfrak L(H,\mathfrak L(H,\mathbb R))$
  • $T\in\mathfrak L(H)$ be nonnegative, self-adjoint and nuclear (trace-class)

Note that$^1$ $$\operatorname{tr}\left(\left(L\otimes_\pi\operatorname{id}_H\right)T\right)=LT,\tag1$$ where on the left-hand side $L$ is considered as being an element of $\mathfrak L(H)$ and on the right-hand side $L$ is considered as being an element of $\left(H\:\hat\otimes_\pi\:H\right)'$.

Now, assume $L\in\mathfrak L(H,\mathfrak L(H))$. $L$ can be considered as being an element of $\mathfrak L(H\:\hat\otimes_\pi\:H,H)$ and hence the right-hand side of $(1)$ is still well-defined. Is there a generalization of the trace functional such that (with a suitable identification of $L$ for the left-hand side) we still have the identity $(1)$?

$^1$ If $E,F,X,Y$ are $\mathbb R$-Banach spaces, $S\in\mathfrak L(X,E)$ and $T\in\mathfrak L(Y,F)$, let $S\otimes_\pi T$ denote the unique bounded linear operator from $X\:\hat\otimes_\pi\:Y$ to $E\:\hat\otimes_\pi\:F$ with $$(S\otimes_\pi T)(x\otimes y)=Sx\otimes Ty\;\;\;\text{for all }(x,y)\in X\times Y.\tag2$$


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