Weak-star convergence implies trace-norm convergence

Question

By definition, if bounded operators $a_i$ converge to $0$ in the weak*-star topology, then $\operatorname{tr} a_it \to 0$ for any trace-class $t$.

Does this also hold for the trace-norm instead of the trace? I.e., $\lVert a_it\rVert_1 \to 0$ for any trace-class $t$.

Answer 1

No. Let $(e_n)$ be an orthonormal basis of a Hilbert space. Define $T: v \mapsto \langle v, e_1\rangle e_1$, i.e., $T$ is orthogonal projection onto the span of $e_1$. For each $n$ define $A_n: v \mapsto \langle v, e_1\rangle e_n$. Then $A_nT = A_n$ and $|A_nT| = |A_n| = T$. So $\|A_nT\|_1 = 1$ for all $n$, but $A_n \to 0$ weak*.