Convergence of operators in trace-class sense

Question

I will be very brief.

Let $A, C$ be two bounded operators on a Hilbert space $\mathcal{H}$, such that both $AC$ and $CA$ are trace-class operators. Let $B_{n}$ be a sequence of bounded operators such that:

$i)$ $B_{n}\to I$ as $n\to \infty$ strongly, where $I$ is the identity operator;

$ii)$ $AB_{n}C$ is a trace-class operator for any $n \in \mathbb{N}$.

Could someone tell me if it holds that $AB_{n}C\to AC$ in trace-class sense? If we replace strong convergence, by norm convergence, does the statement hold?

Answer 1

No. Let $P$ be an orthogonal projection on $H$ whose range and kernel are both infinite dimensional, and set $C=P$, $A=I-P$. Let $(e_n)$ be an ON basis for $PH$ and $(f_n)$ an ON basis for $(I-P)H$. Let $B_n x_k$ be $x_k$ if $x$ is $e$ or $f$ and $k\le n$, let $B_ne_{n+1}=f_{n+1}$, and let $B_nx_k = 0$ for other values of $x = e,\ f$ and $k$. This answers your first question with each $B_n$ finite rank.

For your second question, let $A$ and $C$ be the same, but let $B_n=I+D_n$, where $D_n e_k = n^{-1/2} f_k$ for $k\le n$, $D_n f_k=0$ for all $k$, and $D_ne_k=0$ for $k>n$.