Given a sequence of functionals $(\omega_n)_{n \in \mathcal{N}}$ on a Von Neumann algebra $\mathcal{W}$ converging to $\omega$, I have the following doubts:
- Is it true that $\omega_n$ is pure $\forall n \in \mathcal{N} \iff \omega$ is pure?
- Given G group, is it true that $\omega_n$ is G-invariant $\forall n \in \mathcal{N} \iff \omega$ is G-invariant?
The leftward directions of both questions are easily seen to be false, regardless of topology. For example, set $\mathcal{W} = \mathbb{M}_2$. Then the states represented by density matrices $\begin{pmatrix} 1-\frac{1}{n} & 0\\ 0 & \frac{1}{n}\end{pmatrix}$ are not pure, but they converge in norm to the pure state represented by $\begin{pmatrix} 1 & 0\\ 0 & 0\end{pmatrix}$. Similarly, $G = \mathbb{Z}/2\mathbb{Z}$ acts on $\mathbb{M}_2$ by sending the generator to the automorphism given by conjugation by $\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}$. Then the states represented by $\begin{pmatrix} \frac{1}{2}-\frac{1}{n} & 0\\ 0 & \frac{1}{2}+\frac{1}{n}\end{pmatrix}$ are not $G$-invariant, but they converge in norm to the state represented by $\begin{pmatrix} \frac{1}{2} & 0\\ 0 & \frac{1}{2}\end{pmatrix}$, which is $G$-invariant.
The forward direction of the second question is true, regardless of topology. This is because $\omega \mapsto \omega(g^{-1}x) - \omega(x) \in \mathbb{C}$ is continuous in the weak$^\ast$ topology (and therefore also norm topology), so the collection of functionals such that $\omega(g^{-1}x) = \omega(x)$ for each fixed $g \in G$ and $x \in \mathcal{W}$ is closed. Since $\omega$ being $G$-invariant simply means the above holds for all $g$ and $x$, the collection of $G$-invariant functionals is an intersection of closed sets, whence closed itself.
I don’t know the full answer to the forward direction of the first question, but here are some partial answers and thoughts: the claim is true, regardless of topology, if $\mathcal{W}$ is finite-dimensional, since then pure states are simply vector spaces on a finite-dimensional Hilbert space, so the space of pure states is compact. It is also true, regardless of topology, if $\mathcal{W}$ is abelian, since then pure states are multiplicative states and the space of pure states is the Gelfand spectrum of $\mathcal{W}$, which is weak$^\ast$ compact (and therefore also norm closed).
In case $\mathcal{W}$ is neither abelian nor finite-dimensional, the following is “sort of” a counterexample for the statement with the weak$^\ast$ topology. Let $\mathcal{W} = B(\ell^2)$. Recall that a pure state on $B(\ell^2)$ must be either a vector state or evaluates to zero on compact operators $K(\ell^2)$. Let $\omega_n$ be the vector states associated to the unit vector $\frac{1}{\sqrt{2}}(e_1 + e_n)$ and let $\omega$ be a weak$^\ast$ cluster point of $\omega_n$. Then one easily observes that $\omega$ restricts to the functional $\frac{1}{2} e_1 \otimes e_1$ on $K(\ell^2)$, which is not a vector state, nor zero, whence $\omega$ cannot be pure.
This is only “sort of” a counterexample because $\omega$ is only a weak$^\ast$ cluster point, and therefore an ultralimit of the sequence $(\omega_n)$. I’m not sure this issue can be resolved (or that it should), since the space of states under weak$^\ast$ topology is not first-countable for infinite-dimensional vNa’s, so the topology is not really captured by limits of sequences.
I don’t know about the case for norm topology, but there’s probably not many nontrivial limits of sequences of pure states in the norm topology to begin with, so I wouldn’t be surprised if the answer is true in that case.