Is the strict topology on the multiplier algebra of a $C^*$-algebra always finer than the ultrastrong-$*$ topology? 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 $M(A)=A''$.
 A: The answer is "yes".
Use the standard technique: if necessary, replace $H$ by $H\otimes\ell_2$ to ensure that for each $\omega\in B(H)_*$ (the predual of $B(H)$, the trace-class operators on $H$) there is $\xi\in H$ so that $\omega(y) = (y(\xi)|\xi)$ for each $y\in A''$.  As $A$ acts non-degenerately on $H$ (because $A''$ must be unital) and as replacing $H$ by $H\otimes\ell_2$ does not change this, we may apply the Cohen–Hewitt factorization theorem.  This allows us to find $a\in A, \xi'\in H$ with $a(\xi') = \xi$.
Now let $(x_i)$ be a net in $M(A)$ which converges strictly (that is, for the $\beta$ topology) to $x\in M(A)\subseteq A''$.  This means that $x_i a \rightarrow xa$ and $a^*x_i\rightarrow a^*x$ in norm, so also $x_i^*a\rightarrow x^*a$, in norm.  In particular,
$$ \lim_i \omega((x_i-x)^*(x_i-x))
= \lim_i \|(x_i-x)\xi\|^2
= \lim_i \|(x_i-x)a(\xi')\|^2 = 0, $$
and similarly $\omega((x_i-x)(x_i-x)^*) \rightarrow 0$.  As $\omega$ was arbitrary, this shows that $x_i\rightarrow x$ in the $\sigma$-strong$^*$-topology, that is, for the $\sigma^*$ topology.

In fact, we can avoid the "standard technique".  Restriction of functionals defines a quotient map $B(H)_* \rightarrow M_*$ from the trace-class operators to the (unique) predual of our von Neumann algebra $M$.  Then the ultra-strong$^*$-topology is given by the seminorms
$$ M\ni x \mapsto ( \omega(x^*x) + \omega(xx^*) )^{1/2} $$
as $\omega$ varies through the positive part of $M_*$.  In this way, we see that there is no dependence on $H$.  (I guess there is an extra argument needed here: that a positive member of $M_*$ lifts to a positive trace-class operator).
In our case, $M=A''\cong A^{**}$ the bidual, as the representation of $A$ is universal.  So $M_* \cong A^*$ the dual of $A$.  For $\mu\in A^*$ and $a\in A$ define $a\cdot\mu\in A^*$ to be the functional $A\ni b\mapsto \mu(ba)$.  There are various ways to show that the linear span of $\{ a\cdot\mu : a\in A, \mu\in A^* \}$ is norm-dense in $A^*$.  (For example, polar decomposition of functionals and the GNS construction.)  We can hence directly apply Cohen--Hewitt to $A^*$ to show that for each $\mu\in A^*$ there is $a\in A, \mu'\in A^*$ with $\mu = a\cdot\mu'$.  Similarly we can regard $A^*$ as a right $A$-module, and then the same argument shows that we can find $a'\in A,\mu''\in A^*$ with $\mu' = \mu''\cdot a'$, so that $\mu = a\cdot\mu''\cdot a'$ (this being a bimodule, so the order of left/right actions doesn't matter).
Now the same argument works: if $(x_i)$ in $M(A)\subseteq A^{**}$ converges strictly to $x\in M(A)$ then $\|(x_i-x)a\| \rightarrow 0$ and $\|a'(x_i-x)^*\| = \|(x_i-x)a'^*\| \rightarrow 0$, and so
$$ \mu((x_i-x)^*(x_i-x)) = \mu''(a'(x_i-x)^*(x_i-x)a) \rightarrow 0. $$
Thus we have converted the argument to not depend on $H$.
