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Suppose that I have an infinite matrix $K(i,j)$ with the promise that there is some $C$ such that $\vert K(i,j) \vert \leq C$ for all $i,j \in \mathbb{Z}$.

Consider the Hilbert space $l^2( \mathbb{Z})$ and the trace-class operators on that Hilbert space. For operators $ \rho = \sum_{i,j} \rho(i,j) \vert i \rangle \langle j \vert $ which is trace class. Then consider the operation $S_K(\rho) = \sum_{i,j} K(i,j) \rho(i,j) \vert i \rangle \langle j \vert $. Is $S_K( \rho)$ nescessarily trace class? If yes is there a bound on $\vert \vert S_K \vert \vert_{1,1}$?

The answer and question here is very related: https://mathoverflow.net/questions/406705/if-i-multiply-the-coefficients-of-a-trace-class-operator-with-bounded-complex-nu

Frederik Ravn Klausen
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  • What is your work on this question ? – Jean Marie Oct 20 '21 at 16:37
  • Well, tried to use the counterexample here: https://math.stackexchange.com/questions/2907991/hadamard-product-optimal-bound-on-operator-norm without luck. I tried to split both K and rho into real and imaginary, and then further to positive and negative parts using the bound on the trace norms for each of these parts. Using the Schur product theorem we know that the Hadamard product of two positive operators is positive. So if $K \circ \rho$ is trace-class we can compute the trace norm using the trace. The problem is that one needs a bound on the operator norm of either the positive or neg K. – Frederik Ravn Klausen Oct 20 '21 at 17:03

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