Let $A$ be a $C^\*$-algebra. Let $M(A)$ be its multiplier $C^\*$-algebras. The strict topology on $M(A)$ is given by
$$x\_\lambda \to x \iff \forall a\in A: (\|x\_\lambda a-xa\| + \|ax\_\lambda - ax\| \to 0 ).$$


We can identify $M(A) \cong \mathcal{L}(A)$ where $\mathcal{L}(A)$ are the adjoinable operators of the right Hilbert $A$-module $A$ with respect to the inner product $\langle a,b \rangle:= a^\*b.$ The strict topology on $\mathcal{L}(A)$ is given as follows:


$$t\_\lambda \to t \iff \forall a \in A: (\|t\_i(a)-t(a)\| + \|t\_i^\*(a) -t^\*(a)\| \to 0).$$


When we identify $A$ as a subset of $\mathcal{L}(A)$ via the mapping $ab^\* \mapsto \theta\_{a,b}$ where $\theta\_{a,b}(x) = a\langle b,x\rangle$, we obtain two notions of strict topologies on $M(A)$. I guess on bounded sets these topologies agree, but are these topologies in general equal?



Strict topology on the multiplier algebra
Yes, these topologies agree basically by definition once you understand the isomorphism $M(A)\cong \mathcal L(A)$ (since $tθ\_{a,b}=θ\_{t(a),b}$ the canonical isomorphism $A≅ \mathcal K(A)$ takes this element to $t(a)b^\*=t(ab^\*)$). The isomorphism $A \to \mathcal K(A)$ is given by $a \mapsto \theta\_{a\_1,a\_2}$ where $a=a\_1a\_2^\ast$ is any way of writing $a$ as a product of two elements (any element
in a $C^\ast$-algebra can be written as a product of two elements, e.g. if a $a= u|a|$ is the polar decomposition in the enveloping von Neumann algebra $A^{\ast\ast}$, then $a\_1 := u|a|^{1/2} \in A$ (this can be seen by approximating $t\mapsto t^{1/2}$ by polynomials) and $a = a\_1 |a|^{1/2}$).


