Let $T$ be a trace class operator on $\ell^2 (\mathbb{N})$. Let $A$ be a multiplier on $\ell^2 (\mathbb{N})$ defined by a sequence $a=(a_n)_{n\in\mathbb{N}}$ in $\ell^{\infty} (\mathbb{N})$. That is, for $u=(u_n)_{n\in\mathbb{N}}$ in $\ell^{2} (\mathbb{N})$, $(Au)_n=a_nu_n$ for all $n$ in $\mathbb{N}$. I would like to control the Schatten 1-norm of the compositions $AT$ and $TA$.
I know that $$\|TA\|_1\le \|T\|_1\|A\|$$ and that $$\|AT\|_1\le \|T\|_1\|A\|,$$ where $\|\cdot\|_1$ denotes the Schatten 1-norm and $\|\cdot\|$ denotes the operator norm. Now clearly $\|A\|=\|a\|_{\infty}$, so $$\|AT\|_1,\|TA\|_1\le \|T\|_1\|a\|_\infty.$$
So to make the Schatten 1-norms of the compositions small, it suffices to make the terms $a_n$ uniformly small. My question is whether it is possible to make $\|AT\|_1$ and $\|TA\|_1$ small by controlling only finitely many $a_n$. If you think about the entries of the matrix of, say, $TA$ with respect to the standard orthonormal basis of $\ell^{2} (\mathbb{N})$, $$\langle TAe_m,e_n\rangle=a_m \langle Te_m,e_n\rangle.$$ Suppose that $\|a\|_{\infty}=1$, say. So (with a lot of hand waving here) I only need to make $a_m$ small when $\langle Te_m,e_n\rangle$ is "large" and I need not worry about the other $a_m$ when $\langle Te_m,e_n\rangle$ is small, for $|a_m \langle Te_m,e_n\rangle|$ is at worst $|\langle Te_m,e_n\rangle|$.
The context of this question is quantum mechanics. $T$ is a density operator and $A$ is a potential. So in addition $T$ is positive semidefinite and $a_n\in\mathbb{R}$, although I do not see how this additional information changes anything.
