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Let $N$ be a von Neumann algebra and let $(\phi_\lambda)$ be a net of normal states on $N$ so that $\phi_\lambda\to\phi$ in the weak-* topology, i.e. $\phi_\lambda(x)\to\phi(x)$ for all $x\in N$. What can we infer about $\phi$ in this case? Of course, $\phi$ is going to be a state on $N$, since $\phi(x^*x)=\lim_\lambda\phi_\lambda(x^*x)\geq0$ and $\phi(1_N)=\lim_\lambda\phi_\lambda(1_N)=1$.

But can we say anything more about $\phi$? If not, can we prove that the normal states on $N$ are dense in the weak-* topology in the state space $S(N)$?

I was thinking that, if this density is true, it suffices to prove it for the case of $\mathcal{B(H)}$. Indeed, if this is true for those von Neumann algebras, then we take an arbitrary von Neumann algebra $N$ and concretely represent it on some Hilbert space, i.e. we have an inclusion of von Neumann algebras $N\subset\mathcal{B(H)}$. Now normal states of von Neumann subalgebras extend to normal states and, in general, states extend to states (in the $C^*$-algebraic setting). So we take a state $\phi$ on $N$, we extend it on a state $\Phi$ on $\mathcal{B(H)}$ and then we approximate $\Phi$ in the weak-* sense by normal states on $\mathcal{B(H)}$. We restrict those to $N$ and these are normal states on $N$ approximating $\phi$ weak-* on $N$.

I have no idea on how to prove this though, any help is appreciated.

Just dropped in
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1 Answers1

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Every state on $N$ is the weak$^*$-limit of normal states.

In order to prove it, let $C$ be the weak$^*$-closure of the set of normal states and suppose by contradiction that the (non-normal) state $\varphi _0$ is not in $C$. Since $C$ is clearly weak$^*$-closed and convex, the Hahn-Banach separation Theorem produces a weak$^*$-continuous linear functional $\Lambda $ on $N^*$, and a real number $\alpha $, such that $$ \Re(\Lambda (\varphi ))\leq \alpha < \Re(\Lambda (\varphi _0)), \quad\forall \varphi \in C. $$

It is well known that weak$^*$-continuous functionals must be given by evaluation at some element of $N$, so there is $x$ in $N$ such that $\Lambda (\varphi ) = \varphi (x)$, for every $\varphi $ in $N^*$, and hence $$ \Re(\varphi (x))\leq \alpha < \Re(\varphi _0(x)), \quad\forall \varphi \in C. \tag 1 $$

Letting $a=(x+x^*)/2$, notice that for every self-adjoint linear functional $\psi $ on $N$ one has $$ \Re(\psi (x)) = (\psi (x)+\overline{\psi (x)})/2 = (\psi (x)+\psi (x^*))/2 = \psi (a), $$ so (1) gives $$ \varphi (a)\leq \alpha < \varphi _0(a), \quad\forall \varphi \in C. $$

From the first inequality one deduces that $a\leq \alpha \cdot 1$, but this clearly contradicts the second inequality.

Ruy
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  • Very nice proof. Thanks Ruy, for some reason I got confused and thought that Hahn-Banach applied for compact, convex sets (probably got confused with Krein-Milman) and I turned it down. I guess I must insist more! – Just dropped in Feb 23 '21 at 18:24
  • You are welcome! Regarding compactness, notice that $C$ is indeed compact in the weak*-topology by Alaoglu's Theorem. But of course Hahn-Banach applies to all closed convex sets. – Ruy Feb 23 '21 at 18:28
  • Shoot, of course. I must focus! – Just dropped in Feb 23 '21 at 18:32
  • @Ruy What is $A$? Also, I think at some point an $x$ shoudl be replaced by an $a$. – J. De Ro Feb 23 '21 at 23:15
  • @QuantumSpace, sorry, that $A$ should be an $N$ (already corrected). Besides that I could not find any other typos. Could you please point out the precise place you think $x$ should be $a$? – Ruy Feb 23 '21 at 23:32