# What do weak-star limits of normal states on von Neumann algebras look like?

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.

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.

• 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! – JustDroppedIn Feb 23 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 at 18:28
• Shoot, of course. I must focus! – JustDroppedIn Feb 23 at 18:32
• @Ruy What is $A$? Also, I think at some point an $x$ shoudl be replaced by an $a$. – QuantumSpace Feb 23 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 at 23:32