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.



If I multiply the coefficients of a trace-class operator with bounded complex numbers is it still trace class?
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](https://math.stackexchange.com/questions/2907991/hadamard-product-optimal-bound-on-operator-norm) 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.


