I have recently discovered the notion of the multiplier algebra $M(A)$ of a (non-unital) $C^*$-algebra $A$ and its strict topology, and have a basic question which I cannot find an answer to but which is probably easily answered.
Suppose that $(a_n)_{n \in N} \subset A$ is a net in $A$ which converges strictly to $a \in A$ (the key point being that everything remains in $A$, not its multiplier algebra). Does it follow that $(a_n)_{n \in N}$ actually converges to $a$ in norm?
My intuition suggests that this should be true via some approximate identity argument, but when I try to work it out I seem to require an approximate identity which acts similarly to the identity "uniformly" over the net $(a_n)_{n \in N}$.
If anyone knows the answer or has a good reference which might help me figure things out myself, please let me know.
It is not true. Let $A=C_0(\mathbb R)$, and let $$ a_n(t)=\max\{0, 1-|x-n|\}. $$ For any $g\in A$ you have $\|g\,a_n\|\to0$, since $g\,a_n$ is supported on $[n-1,n+1]$ and $|g|$ is arbitrarily small on said interval for $n$ big enough. So $a_n\to 0$ strictly. But $\|a_n\|=1$ for all $n$.
Analogously you can take $A=K (H)$ for an infinite-dimensional Hilbert space. Let $\{P_n\}$ be a pairwise orthogonal sequence of rank-one projections. Then $TP_n\to0$ and $P_nT\to 0$ for all $T\in A$, but $\|P_n\|=1$ for all $n$.