Let $A_n\to 0$ strongly and each $A_n$ be trace-class.
It would have been nice if $A_n\to0$ in trace-class norm, but this is false.
Here is a possible counter-example:
Let the Hilbert space be $\ell^2(\mathbb{N})$ with eigenbasis $(e_i)_i$ and define $A_n := e_n \otimes e_n^\ast$. Then $A_n \to 0$ strongly because given any $\psi\in\ell^2(\mathbb{N})$, $\|A_n \psi\| = |\psi(n)|\to0$ because $\psi$ is square-summable. On the other hand, $\|A_n\|_1 = \|e_n\|^2=1$ for all $n$ so that there's no way that $\|A_n\|_1\to0$.
My question: is this all correct? Is there any way to salvage some statement out of this? What I mean is this: If $S_n\to0$ strongly and $T$ is trace-class then $T S_n\to0$ in trace-class norm. So it seems like if one adds some sort of condition on $A_n$, e.g., that it may be written as $A_n = B_n C_n$ with some conditions on $B_n$, $C_n$ then one could maybe conclude $A_n\to0$ in trace-class norm. Any thoughts on possible lemmas?
Think of the trace as the analogue of the measure/integral in a infinite-measure space and think of the strong operator topology as "bounded and pointwise" convergence. Lebesgue's Dominated Convergence Theorem tells you that you can exchange limits with integrals if you have a uniform bound which is integrable. Therefore if there is an $A_\infty$ of trace-class such that $A_n \leq A_\infty$ and $A_n \to 0$ strongly I would expect to have $A_n \to 0$ in trace-norm.