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.