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

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.

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.

• Great example! Do you need $8^n$ instead of $4^n$ to make sure that the operator norm is infinite? Oct 21, 2021 at 12:18
• Oh right, I'll correct it. Oct 21, 2021 at 14:20
• Wait, the trace norm of a unitary $U$ is $Tr( \sqrt{ U^* U}) = Tr( 1) = n$. So the operator $\oplus 2^{-n} U_{8^n}$ is not trace class? Oct 22, 2021 at 10:21
• Right again! But still fixable ... Oct 22, 2021 at 11:43
• Thanks, if you just want to get that the trace norm is not bounded one can look at $A_n = \sqrt{n} U_n$ where $U_n$ is the unitary and then let $B_n$ be the matrix of all $\frac{1}{n}$. Then $\oplus 2^{-n} B_n$ is trace class with trace norm 1 and $\oplus \sqrt{n} U_n \circ B_n = \oplus \frac{1}{\sqrt{n}} U_n$ which has trace norm $\sum_n \sqrt{n} = \infty$. Oct 22, 2021 at 12:11

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?

• The reference I like to give is Pisier's book Similarity problems and completely bounded maps. Oct 21, 2021 at 12:04
• Thanks for providing this context, so if I have a concrete kernel and I want to check whether it is a Schur multiplier can the theory help me? Oct 23, 2021 at 18:03
• Well, you need to prove if the kernel can be written in the way stated. I guess I would have my doubts about whether this is always a practical test... Oct 23, 2021 at 19:49