If I multiply the coefficients of a trace-class operator with bounded complex numbers is it still trace class?

Question

Suppose that $T \in TC(l^2( \mathbb{Z}))$ is trace class. Consider its kernel $ T(i,j) = \langle e_i, T e_j \rangle $ where $ \{e_i\}_{i \in \mathbb{Z}}$ is an ONB for $l^2( \mathbb{Z})$. Now, consider the operator given by the kernel $T(i,j) K(i,j) $ for some numbers $K(i,j)$ such that $\sup_{i,j} \vert K(i,j)\vert < \infty$.

Is this operator still trace class?

My thoughts: The new operator is the Hadamard/Schur product of $K \circ T$ for some operator $K$ which we do not know is bounded. If $K$ is bounded in operator norm then $\vert \vert K \circ T \vert \vert_1 \leq \vert \vert K \vert \vert_\infty \vert \vert T \vert \vert_1 $. But our condition on $K$ is not enough to ensure that.

We can split $K$ and $T$ up into real and imaginary parts and then their real and imaginary parts into positive and negative parts. By a triangle inequality we can estimate each of the 16 terms $K_i \circ T_j$ where $K_i$ and $T_j$ are positive. By the Schur product theorem then $K_i \circ T_j$ is also positive. Hence we can compute its trace norm by \begin{align*} \vert \vert K_i \circ T_j \vert \vert_1 &= Tr( K_i \circ T_j ) = \sum_n K_i(n,n) T_j(n,n)\\ &\leq \sup_{n} K_i(n,n) \sum_n T_j(n,n) \leq \sup_{n} K_i(n,n) \vert \vert T_j \vert \vert_1 \leq \sup_{n} K_i(n,n) \vert \vert T \vert \vert_1 \end{align*}  where $K_i(n,n), T_j(n,n) \geq 0$ since $K_i$ and $T_j$ are positive operators. We can bound the matrix elements of the real and imaginary parts of $K$, but unfortunately we can't bound the the matrix elements of the positive and negative parts. I suspect one can use this insight to construct a counterexample.

Answer 1

It's a little more complicated than I thought! Frederik Ravn Klausen pointed out an error. Still, I maintain that the product needn't even be bounded.

As the answer to this question shows, in $M_n$ you can find a unitary $U$ and an matrix $A$ whose entries all have modulus 1, whose Hadamard product $A\bullet U$ has operator norm $\sqrt{n}$. Using the duality between operator norm and trace class norm, find a trace class matrix $B$ with trace norm 1 such that ${\rm tr}(B^*(A\bullet U)) = \sqrt{n}$.

We have an identity ${\rm tr}(B^*(A\bullet U)) = {\rm tr}((A\bullet B)^*U)$. Therefore the trace norm of $A\bullet B$ is at least (in fact, exactly) $\sqrt{n}$. Now $\bigoplus 2^{-n} B_{5^n}$ is trace class, and its Hadamard product with $\bigoplus A_{5^n}$ is unbounded.

Answer 2

Nik Weaver's answer gives a nice counter-example. Let me just say a few words of context. Kernels $K$ for which $KT$ is trace-class for all trace-class $T$ are called Schur multipliers. (Not to be confused with an unrelated, and more common, term from group theory). I believe this is because of Schur's 1911 paper (Crelle's journl, in German).

A more modern approach recognises links with completely bounded maps, the Haagerup tensor product, and so forth. One comment is that as the bounded linear operators and the trace-class operators are in duality, we can instead consider kernels which multiply bounded operators to bounded operators, and this is the more common framing of the question. References I know are Pisier's work (JSTOR) and Spronk's paper (PLMS). In particular, we have the following characterisation of such kernels $K$: there must be some Hilbert space $H$ and vectors $\xi_i, \eta_j$ in $H$ with $\sup_i\|\xi_i\|<\infty, \sup_j\|\eta_j\|<\infty$ and $K(i,j) = (\xi_i|\eta_j)$ for all $i,j$. This open up the idea of using the theory of tensor norms on Banach spaces to study such maps, which is (roughly speaking) what Pisier does in the cited paper.

I am not aware of a canonical reference; perhaps others are?