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 $M(A)=A''$.



Is the strict topology on the multiplier algebra of a $C^*$-algebra always finer than the ultrastrong-$*$ topology?
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](https://en.wikipedia.org/wiki/Cohen%E2%80%93Hewitt_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.




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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 $H$.


