Let me be precise about what I mean in the title. Let $A$ be a $C^*$-algebra, which we identify with its image of its universal representation $(\pi, H)$, so the second dual of $A$ is canonically identified with the von Neumann algebra $A''$ generated by $\pi(A) = A$. Denote $\widetilde{A}$ the minimal unitalization of $A$ in $A''$, and $M(A)$ the maximal one, i.e. $M(A) = \{x \in A'' \mid xA \cup Ax \subset A\}$, which is of course a copy of the multiplier algebra of $A$. It is known (see e.g. Pedersen's book, Theorem 3.12.9) that $M(A)_{\mathrm{sa}} = (\widetilde{A}_{\mathrm{sa}})_m \cap (\widetilde{A}_{\mathrm{sa}})^m$, where $(\widetilde{A}_{\mathrm{sa}})_m$ denotes the set of strong limits in $A''$ of bounded decreasing nets in $\widetilde{A}_{\mathrm{sa}}$, and $(\widetilde{A}_{\mathrm{sa}})^m$ the set of strong limits of bounded increasing nets in $\widetilde{A}_{\mathrm{sa}}$. Of course $M(A)_{\mathrm{sa}} \subset A''_{\mathrm{sa}}$ and the inclusion is strict in general. There are a variety of topologies on $M(A)$. Besides the ones that inherited from $B(H)$ as a subspace, one also often considers the strict topology on $M(A)$, which is the locally convex topology defined by the family of semi-norms $\|(\cdot)a\|$ and $\|a(\cdot)\|$ with $a$ running through $A$. It is also known that $M(A)$ is in fact the strict completion of $A$ as a locally convex space.

Now von Neumann's double commutation theorem says that the ultra-strong-$*$ closure of $A$ in $B(H)$ is $A''$. On $M(A)$, denote the ultra-strong-$*$ topology that is inherited from $H$ by $\sigma^*$, and the strict topology by $\beta$. If $\sigma^* = \beta$, then $M(A)$ is ultra-strongly-$*$ complete since it is strictly complete, in particular $M(A)$ is ultra-strongly-$*$ closed in $B(H)$, forcing $M(A) = A''$, but we know this is not always the case, although we do have $M(A) \subset A''$ by the above discussion, i.e. the $\beta$-closure of $A$ is always contained in the ultra-strong-$*$ closure of $A$. Since finer topology gives smaller closure, my question is, is it true that we always have $\beta$ finer than $\sigma^*$?

Here's a special case to get started. Note that if $A = K(H)$, then $A''=B(H)$. In this case, it is fairly easy to see that the strict topology on $M(A)$ is finer than the strong-$*$ topology on $M(A)$. Indeed, take any rank-one projection $p_\xi$ onto $\mathbb{C}\xi$ with $\xi$ being an arbitrary unit vector of $H$, one has, for any $x \in B(H) = M(K(H)) = K(H)''$, that $\|xp_\xi\| = \|xp_\xi x^*\|^{1/2} = \|x\xi\|$ and $\|x^*p_\xi\| = \|x^* p_\xi x\|^{1/2} = \|x^*\xi\|$. As $p_\xi \in K(H)$, and $\|(\cdot)\xi\|$ together with $\|(\cdot)^*\xi\|$ form a generating family of semi-norms of the strong-$*$ topology, we've shown that indeed the strict topology on $M(K(H)) = B(H)$ is finer than the strong-$*$ one. It is known that the strict topology, the strong-$*$ topology, the ultra-strong-$*$ topology and the Arens-Mackey topology on $B(H)$ all agree on bounded parts. So it is reasonable to compare the strict topology to other topologies that are finer than the strong-$*$ one but still agree with it on bounded parts. Two such topologies are already mentioned, the ultra-strong-$*$ topology and the Arens-Mackey topology. We know the dual of $B(H)$ equipped with the strict topology is exactly the predual $B(H)_*$ of $B(H)$, so the Arens-Mackey topology is finer than the strict one. The question remains about the comparison between the ultra-strong-$*$ topology and the strict topology, even in this special case where we do have $M(A)=A''$.