Let $\mathcal{A}$ be an infinite-dimensional W*-algebra, that is, an infinite-dimensional $C^{*}$-algebra which is the Banach dual of a Banach space $\mathcal{A}_{*}$ (equivalently, $\mathcal{A}$ is an abstract von Neumann algebra).
Consider the so-called normal positive, linear functionals on $\mathcal{A}$.
According to the general theory of W*-algebras, these are all those positive linear functionals in the dual space $\mathcal{A}^{*}$ that are in the image $i(\mathcal{A}_{*})\subset\mathcal{A}^{*}$ of $\mathcal{A}_{*}$ in its bidual $(\mathcal{A}_{*})^{**}=\mathcal{A}^{*}$.
Now, consider the set $\mathcal{P}$ composed of all those normal,positive, linear functionals on $\mathcal{A}$ that are faithful (assuming, of course, that $\mathcal{A}$ admits normal, positive, linear functionals that are normal), that is, $\omega\in\mathcal{P}$ is such that $\omega(\mathbf{a}\mathbf{a}^{\dagger})=0$ for some $\mathbf{a}\in\mathcal{A}$ implies $\mathbf{a}=\mathbf{0}$.
Is the set $\mathcal{P}$ open in $\mathcal{A}^{*}$ and/or $\mathcal{A}_{*}$?
Best Answer
If you are considering the weak$^*$-topology, the answer is "no". Take $\varphi$ be a faithful, normal, state, and $\{p_j\}$ and increasing net of projections with $p_j\nearrow 1$. Then the net $\{\varphi(p_j\cdot )\}_j$ converges weak$^*$ (i.e., pointwise) to $\varphi$. None of the functionals $\varphi(p_j\cdot)$ is faithful. We can also make the approximating functionals non-normal by using the trick I mention below.
When you consider the norm topology, normal functionals are norm-limits of non-normal ones: take $\varphi$ normal, $\psi$ non-normal, and form the sequence $\varphi+\tfrac1n\,\psi$.
And regarding things purely in the predual. Let $\mathcal A=B(H)$ with $H$ separable. Then $\mathcal A_*$ is $L^1(H)$, the trace-class operators. Fix an orthonormal basis $\{e_n\}_{n\in\mathbb N}$. Let $\varphi\in L^1(H)$ be $$ \varphi = \sum_n\frac1{n^2}\,\langle \cdot,e_n\rangle\,e_n. $$ We may see it as a linear functional $$ \varphi(T)=\sum_n \frac1{n^2}\,\langle Te_n,e_n\rangle. $$ Then $\varphi$ is linear, positive, faithful. Let $\varphi_m\in L^1(T)$ be given by $$ \varphi_m(T)=\sum_{n\leq m}\frac1{n^2}\,\langle Te_n,e_n\rangle. $$ Then $\varphi_m\to\varphi$ in norm (the trace-norm, which is the norm in the predual), and $\varphi_m$ is not faithful for any $m$.