strict topology on multiplier algebras

Question

Suppose $A$ is a $C^*$ algebra,$M(A)$ is the multiplier algebra.If $S$ is a subset of $M(A)$ which is compact for the strict topology on $M(A)$,is $S$ also a subset of $M(M(A))$ which is compact for the strict topology on $M(M(A))$?

Answer 1

So, if I understand the question, you are asking this: Let $S$ be a subset of $M(A)$ which is compact for the strict topology. (The strict topology is such that a net $(x_i)$ converges to $x$ exactly when $x_ia\rightarrow xa, ax_i\rightarrow ax$ in norm, for each $a\in A$). Now set $B=M(A)$, a unital algebra, and consider $M(B)$, which is just $B$. We also consider the strict on $M(B)=B$, which is just the norm topology (let $a=1$ in the above description). Is $S \subseteq B$ also compact in the strict (=norm) topology.

So, in short:

Let $S$ be a subset of $M(A)$ which is compact for the strict topology. Is $S$ also compact for the norm topology on $M(A)$?

As you might expect, the answer is "no". For a counter-example, let $A=K(H)$ the compact operators on a separable Hilbert space with basis $(e_n)$, and let $t_n(\xi) = (\xi | e_n) e_n$ be the rank-one projection onto the span of $(e_n)$. Then $t_n =t_n^*$ and $t_n(\xi)\rightarrow 0$ for each $\xi\in H$, and $(t_n)$ is bounded. It follows that $t_n\rightarrow 0$ in the strict topology on $M(A) = B(H)$. Set $$ S = \{ t_n : n\geq 1 \} \cup \{0\}. $$ It follows that $S$ is compact in $M(A)$. However, as a subset of $B(H)$ with the norm topology, $S$ is not compact.