How to prove the density relation $\overline{\text{Span}\{\varphi_{x,y}:x ,y\in H\}}^{\Vert\cdot\Vert} = N_*$ for any von Newmann algebra $N$? Let $N \subset B(H)$ (for some Hilbert space $H$) be a von Neumann algebra. For every $x,y \in H$ we define $\varphi_{x,y}: N\rightarrow \mathbb{C}$ by $\phi_{x,y}(T) = \langle Tx,y\rangle $. Clearly, $\varphi_{x,y} \in N_*$, the Banach space of all normal linear functionals on $N$. The problem is to show that the spanning set of all such functionals $\{\varphi_{x,y}:x ,y\in H\}$ is norm-dense in $N_*$, i.e., $$\overline{\text{Span}\{\varphi_{x,y}:x ,y\in H\}}^{\Vert\cdot\Vert} = N_*.$$
I think this is not an easy problem (for me). I just, unfortunately, have no idea how to find a sequence in the spanning set converging to a previously given functional in $N_*$.
Any help is appreciated.
Edit: Here I restate the definition of a normal functional on $N$: A normal linear functional on $N$ is a linear map $\varphi: N \rightarrow \mathbb{C}$ such that for all increasing, self-adjoint nets $(a_\lambda)_{\lambda \in \Lambda}$ with $a_\lambda \rightarrow a$ (also self-adjoint), we have $\varphi(a_\lambda) = \lim_{\lambda} \varphi(a_\lambda)$.
 A: $\def\tr{\operatorname{Tr}}$
It is known that all normal functionals are of the form $T\longmapsto \tr(AT)$ for $A$ trace-class. "Trace-class" means that $\tr(|A|)<\infty$. The functionals you are proposing are of the form $T\longmapsto\tr(ET)$, where $E$ is rank-one; linear combinations of those will give maps $T\longmapsto\tr(AT)$ with $A$ finite-rank, and norm limits of these will be $T\longmapsto\tr(AT)$ with $A$ trace-class.
The above is usually proven in the form that there exist countable sequence $\{\xi_n\}$, $\{\eta_n\}$ such that the normal functional is of the form
$$
T\longmapsto \sum_n\langle T\xi_n,\eta_n\rangle. 
$$
This result appears for example as Theorem 7.1.12 in Kadison-Ringrose II.
The span and the closure are always necessary, as there are normal functionals at distance 1 from the ones of the form $T\longmapsto\tr(ET)$, where $E$ is rank-one. We want to fix $A$, and calculate the norm of $\psi_{A,E}:T\longmapsto\tr((A-E)T)$. It is known that
$$
\|\psi_{A,E}\|=\|A-E\|_1=\tr(|A-E|). 
$$
Hence the question is the same as asking whether the rank-one operators are dense in the trace-class operators (in the trace norm). The trace norm dominates the operator norm so this would imply that that every trace-class operator is a limit of rank-one operators. And this is not true. Indeed, fix an orthonormal basis $\{e_n\}$ and let $A$ be the rank-two operator
$$
Ax=\langle x,e_1\rangle\,e_1+\langle x,e_2\rangle\,e_2.
$$
Now fix a rank-one operator $E$. Then $E$ is of the form
$$
Ex=\langle x,y\rangle z
$$
for certain $y,z\in H$; we may assume $\|y\|=1$. Let $x$ be a unit vector in $\{y\}^\perp$. Then
$$
\|A-E\|\geq\|Ax-Ex\|=\|Ax\|.
$$
Write $y=\sum_ny_ne_n$. If $y_1=0$, then we may take $x=e_1$. Otherwise, let $$x=\frac1{\sqrt{|y_1|^2+|y_2|^2}}\,\big(-y_2e_1+y_1e_1\big).$$ In both cases we get $x\perp y$ and $Ax=x$. It follows that $\|A-E\|\geq1$. Translated to your problem, if $N=B(H)$ then the normal functional $T\longmapsto \tr(AT)$ is at distance 1 or more from the sets of states $\{\phi_{x,y}:\ x,y\}$.
