# The topology of pointwise convergence with the adjoint operator on a von Neumann algebra

Let $H$ be a Hilbert space and ${\mathcal A}$ a von Neumann algebra in $B(H)$. Let us endow ${\mathcal A}$ with the topology of pointwise convergence with the Hermitian adjoint operator, i.e. a net $A_i\in {\mathcal A}$ tends to 0 in ${\mathcal A}$ if $$\forall x\in H\qquad A_ix\overset{H}{\underset{i\to\infty}{\longrightarrow}}0 \quad\&\quad A_i^*x\overset{H}{\underset{i\to\infty}{\longrightarrow}}0.$$ The algebra ${\mathcal A}$ with this topology has totally bounded sets $S\subseteq {\mathcal A}$, which I believe, are described by the property $$\forall x\in H\qquad \text{the sets \{Ax;\ A\in S\} and \{A^*x;\ A\in S\} are totally bounded in H}$$
Let $u:{\mathcal A}\to{\mathbb C}$ be a linear functional, which is continuous on every totally bounded set (equivalently, on each compact set) $S\subseteq {\mathcal A}$ in this sense (with respect to this topology induced from ${\mathcal A}$ to $S$).

Question:

does $u$ belong to ${\mathcal A}_*$ (the predual of ${\mathcal A}$)?

EDIT. I must say, even the case of ${\mathcal A}=B(H)$ is not clear for me. And even if we replace the topology on ${\mathcal A}$ with the usual strong operator topology (i.e. if we remove the requirement $A^*_ix\to 0$). We also can change premises in other different ways, for example, we can think that ${\mathcal A}$ is commutative (for me that would be sufficient now). If anybody could cast some light on any weakened conditions, I would appreciate this very much. I need this for my current work, this is related to the study of envelopes of topological algebras.

[As requested by the asker, I edited this answer to include more background information on the topologies I mention.]

The topology you refer to in your question is the strong$^*$ (operator) topology, defined by the seminorms $$a \mapsto \|a \xi\| \quad \text{ and } \quad a \mapsto \|a^* \xi\|$$ for $\xi \in \mathcal{H}$. The $\sigma$-strong$^*$ (operator) topology is defined by the seminorms $$a \mapsto \sum_{n = 1}^\infty \|a \xi_n\| \quad \text{ and } \quad a \mapsto \sum_{n = 1}^\infty \|a^* \xi_n\|$$ for square-summable sequences $(\xi_n)_{n=1}^\infty$. Similarly, the strong operator topology is defined by the seminorms $$a \mapsto \|a \xi\|$$ and the $\sigma$-strong (operator) topology is defined by the seminorms $$a \mapsto \sum_{n = 1}^\infty \|a \xi_n\|.$$ The weak operator topology is defined by the seminorms $$a \mapsto |\langle a \xi \,\vert\, \eta \rangle|$$ and the $\sigma$-weak (operator) topology, which coincides with the weak$^*$ topology, is defined by the seminorms $$a \mapsto \sum_{n = 1}^\infty |\langle a \xi_n \,\vert\, \eta_n \rangle|.$$ It's a standard result that all of the non-$\sigma$ topologies have the same continuous linear functionals and all of the $\sigma$ topologies have the same continuous linear functionals. Since the $\sigma$ and non-$\sigma$ variants of each topology agree on bounded sets, it follows from the Krein-Smulian Theorem applied to the $\sigma$-weak topology that they all have the same closed convex sets.

The advantage of the $\sigma$ versions of the topologies is that every $*$-isomorphism of von Neumann algebras is a homeomorphism for any of those topologies, which is not true for the non-$\sigma$ versions. Their defining seminorms can also be defined using the predual, with no reference to a Hilbert space, e.g. the $\sigma$-strong$^*$ topology is given by the seminorms $$a \mapsto \varphi(a^* a)^{1/2} \quad \text{ and } \quad a \mapsto \varphi(a a^*)^{1/2}$$ for $\varphi \in \mathcal{M}_*^+$.

Both the strong$^*$ and $\sigma$-strong$^*$ topologies are quasicomplete, meaning that all closed and bounded sets are complete, and hence total boundedness is equivalent to precompactness. Also, the notion of boundedness (in the topological vector space sense) agrees with the ordinary notion of norm boundedness for all of these topologies. For the $\sigma$ topologies for whom the continuous linear functionals are $\mathcal{M}_*$, this follows from the Banach-Mackey Theorem. For the non-$\sigma$ variants it also follows because their continuous linear functionals are dense in $\mathcal{M}_*$.

Since a net $(a_\lambda)$ is Cauchy (resp. converges to $a$) in the strong$^*$-topology if and only if $(a_\lambda)$ and $(a_\lambda^*)$ are Cauchy (resp. converge to $a$ and $a^*$) in the strong operator topology, we just need to show quasicompleteness for the strong operator topology. But a bounded Cauchy net in the strong operator topology has a limit by the Uniform Boundedness Principle. This also shows that the $\sigma$-strong and $\sigma$-strong$^*$ topologies are quasicomplete, because they are finer than their corresponding non-$\sigma$ versions but agree on bounded sets. Hence your question is a well-defined question about the $\sigma$-strong$^*$ topology of an abstract von Neumann algebra.

If $\mathcal{M}$ is a von Neumann algebra, then any linear functional continuous on compact subsets of the $\sigma$-strong$^*$ topology is sequentially $\sigma$-strong$^*$ continuous. Let $\varphi$ be a linear functional on $\mathcal{M}$ that is $\sigma$-strong$^*$-continuous on compact sets. If $(x_n) \to 0$ is $\sigma$-strong$^*$-convergent, then $K = \overline{\{ x_n : n \in \mathbb{N} \}}$ is $\sigma$-strong$^*$-compact. Continuity of $\varphi$ on $K$ implies that $(\varphi(x_n)) \to \varphi(0)$. Therefore $\varphi$ is sequentially continuous.

If $\mathcal{M}_*$ is separable, then the $\sigma$-strong$^*$ topology is metrizable on bounded parts of $\mathcal{M}$. Choose a dense subset $(\psi_n)_{n=1}^\infty$ of the unit ball of $\mathcal{M}_*$. Then $$d(x, y) = \sum_{n = 1}^\infty \psi_n((x - y)^* (x - y))^{1/2} + \sum_{n = 1}^\infty \psi_n((x - y) (x - y)^*)^{1/2}$$ defines a metric for the strong$^*$ topology on the unit ball of $\mathcal{M}$. More generally, if $\mathcal{M}$ is countably decomposable and $\psi$ is a faithful normal state on $\mathcal{M}$, then $$d(x, y) = \psi((x - y)^* (x - y))^{1/2} + \psi((x - y) (x - y)^*)^{1/2}$$ defines a metric for the strong$^*$ topology on the unit ball of $\mathcal{M}$.

Since sequential continuity is equivalent to continuity on metric spaces, this gives an affirmative answer to your question when $\mathcal{M}$ is countably decomposable. Similar results hold for the other operator topologies.

Matthias Neufang has shown that sequential $\sigma$-weak continuity is equivalent to $\sigma$-weak continuity if and only if the decomposability number of $\mathcal{M}$ is not a real-valued measurable cardinal, with $\ell^\infty(\kappa)$ for a real-valued measurable cardinal $\kappa$ being a counterexample. The arguments should generalize to the $\sigma$-strong$^*$ topology, because they reduce to the commutative case, and then reduce to a direct sum of finite measure spaces, which give countably decomposable von Neumann algebras.

In the case of $\ell^\infty(I)$, your question has an affirmative answer. First, some preliminaries on various topologies on $\ell^\infty(I)$. This has some overlap with the last question I answered. The Mackey topology is the finest vector topology on $\ell^\infty(I)$ such that the continuous linear functionals are $\ell^1(I)$, and it is given by convergence on the absolutely convex weakly compact subsets of $\ell^1(I)$. The bounded weak$^*$ topology (or the equicontinuous weak$^*$ topology in the duals of non-normed spaces) is the finest vector topology that agrees with the weak$^*$ topology on bounded sets, and it is given by convergence on the absolutely convex norm compact sets of $\ell^1(I)$. Its continuous linear functionals are also $\ell^1(I)$. Since $\ell^1(I)$ has the Schur property, these topologies coincide on $\ell^1(I)$.

By a theorem of Akemann, the strong$^*$ and Mackey topologies agree on bounded sets, so the strong$^*$ topology agrees with the weak$^*$ topology on bounded subsets of $\ell^\infty(I)$. The Grothendieck Completeness Theorem applied to $\ell^1(I)$ says that the completeness of $\ell^1(I)$ implies that every linear functional on $\ell^\infty(I)$ that is weak$^*$-continuous on the absolutely convex weak$^*$-compact subsets of $\ell^\infty(I)$ (or just the unit ball) is in $\ell^1(I)$.

Since every commutative von Neumann algebra is a product of copies of $\mathbb{C}$ and $\mathrm{L}^\infty([0, 1])$, whose strong$^*$ topologies are all sequential, one might expect this to generalize by analyzing weak compactness in the $\ell^1$ sum of the preduals with respect to weak compactness in the individual components, but I can't see how to make it follow easily from any other theorem.

• Cameron, could you, please, specify the terms? The $\sigma$-strong* topology, the strong* topology, what are they exactly? – Sergei Akbarov Nov 12 '16 at 10:33
• @SergeiAkbarov I added some more background info to my answer. Sorry about that. Let me know if anything else isn't clear. – Cameron Zwarich Nov 12 '16 at 12:40
• OK, thank you, Cameron, I look at this. – Sergei Akbarov Nov 12 '16 at 12:43
• @CameronZwarich, could you clarify the statement about the spaces of continuous linear functionals being the same for all the topologies mentioned in your answer? For example, not every $\sigma$-weakly continuous functional is weakly continuous. – Mateusz Wasilewski Nov 12 '16 at 13:49
• @MateuszWasilewski I meant that all of the $\sigma$ topologies have the same continuous linear functionals and so do all of the non-$\sigma$ topologies, so by the Krein-Smulian Theorem applied to the $\sigma$-weak topology they all have the same closed convex sets. I edited it to fix that. – Cameron Zwarich Nov 12 '16 at 15:12

This is a comment on the above answer but it will be too long and in any case I am not entitled. The questions you pose are answered in much more detail in a Comptes Rendus article "Topologies sur les espaces des opérateurs dans l'espace hilbertien" (A276 (1973) 1509). This discusses three locally convex topologies on a von Neumann algebra---the finest locally convex ones which coincide with the weak operator, the strong operator and the symmetric strong operator topology on the unit ball. They are all complete, have the same convergent sequences as the above-mentioned ones, have the same dual space (the normal functionals) and have the property that a linear functional thereon is continuous if and only it is continuous on the unit ball for each or any of the above operator topologies. This can be found in the secondary literature (in english) as chapter IV of "Saks spaces and applications to functional analysis" which is readily available online. Particularly relevant is Prop. 2.3 on p. 204. The commutative case is treated in Ch. 3.

• I should add that this answers the OP in the case where the underlying Hilbert space is separable since the above topologies are metrisable on the unit ball. I am not sure about what happens in the general case. – clyde Nov 12 '16 at 17:57
• Clyde, it will take me some time to find these papers. To avoid misunderstandings I want to ask you this: the continuity of a functional $f$ with respect to which of these topologies is equivalent to the continuity of $f$ on totally bounded sets with respect to this topology? – Sergei Akbarov Nov 12 '16 at 18:45
• In the separable case, this is true for any of the topologies. This is because continuity is determined on the unit ball which is in each case – clyde Nov 13 '16 at 13:36
• metrisable. Thus continuity is determined on convergent sequences and these are totally bounded. – clyde Nov 13 '16 at 13:38