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?



Convergence of operators in trace-class sense
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 $k>n$.


