Let $A,B$ be two $C^*$-algebras. Given two states $\phi\in S(A),\psi\in S(B)$ one defines the elementary tensor state $\phi\odot\psi:A\odot B\to\mathbb{C}$ as the unique linear map satisfying $\phi\odot\psi(a\otimes b)=\phi(a)\psi(b)$ on elementary tensors. One proves with elementary arguments that all such functionals are algebraically positive and continuous with respect to the spatial norm (and by Takesaki's theorem w.r.t any $C^*$-norm on $A\odot B$). Denote the extension of $\phi\odot\psi$ on a state on $A\otimes B$ (the spatial TP) by $\phi\otimes\psi\in S(A\otimes B)$.
Let $0\neq x\in A\otimes B$. It is easily proven that there exist states $\phi\in S(A)$ and $\psi\in S(B)$ so that $\phi\otimes\psi(x^*x)>0$. Here is a simple argument: As seen in Murphy, we have that
$$\|x^*x\|_{\min}=\sup_{\tau\in S(A),\rho\in S(B)}\|\pi_\tau\otimes\pi_\rho(x^*x)\|_{B(H_\tau\otimes H_\rho)}$$
where $(H_\tau,\pi_\tau,\xi_\tau)$ is the GNS triplet. So we find a pair of states $\tau,\rho$ such that $\|\pi_\tau\otimes\pi_\rho((x^*x)^{1/2})\|\neq0$, i.e. $\pi_\tau\otimes\pi_\rho((x^*x)^{1/2})\neq0$. This thing here is an operator on $H_\tau\otimes H_\rho$, i.e. $\pi_\tau\otimes\pi_\rho((x^*x)^{1/2})\in B(H_\tau\otimes H_\rho)$, and as it is non-zero, we can find unit vectors $\xi_1\in H_\tau,\xi_2\in H_\rho$ such that $\pi_\tau\otimes\pi_\rho((x^*x)^{1/2})(\xi_1\otimes\xi_2)\neq0$. Simply set $\phi(a)=\langle\pi_\tau(a)\xi_1,\xi_1\rangle$ and $\psi(b)=\langle\pi_\rho(b)\xi_2,\xi_2\rangle$. These are obviously states on $A$ and $B$, and $\phi\otimes\psi(x^*x)\neq0$.
Here is my question: apparently it is also true that the elementary tensors of pure states separate the positive elements, i.e. given $0\neq x\in (A\otimes B)_+$, we can find pure states $\phi\in\text{PS}(A),\psi\in\text{PS}(B)$ such that $\phi\otimes\psi(x)\neq0$. How can this be proved, or where can I find a proof?
Here are the things I tried
With elementary arguments as those presented in Murphy's book, I was able to show that we also have
$$\|x\|_{\min}=\sup_{\tau\in\text{PS}(A),\rho\in\text{PS}(B)}\|\pi_\tau\otimes\pi_\rho(x)\|_{B(H_\tau\otimes H_\rho)}$$
the proof is the same but one has to observe first that the direct sum of the GNS representations for pure states $\bigoplus_{\tau\in\text{PS}(A)}\pi_\tau:A\to\mathbb{B}(\bigoplus_{\tau\in\text{PS}(A)}H_\tau)$ are faithful.
The thing is that, if I try to mimic the above proof, the rest of the proof gets messed up: If I knew that the vectors $\xi_1,\xi_2$ found earlier are the canonical cyclic vectors, then everything would work out like a charm since the resulting states $\phi,\psi$ would actually be $\tau$ and $\rho$ which would be pure this time. But of course this is too much to expect I guess.
Also, one can show (similarly to the other equations) that
$$\|x^*x\|_{\min}=\sup\bigg\{\frac{\tau\otimes\rho(y^*x^*xy)}{\tau\otimes\rho(y^*y)}:\tau\in\text{PS}(A),\rho\in\text{PS}(B),y\in A\odot B\text{ s.t. }\tau\otimes\rho(y^*y)\neq0\bigg\}.$$
So we could start like this: $x^*x\neq0$, so we get pure states $\tau,\rho$ and some $y\in A\odot B$ such that $\tau\otimes\rho(y^*x^*xy)\neq0$. And everything would be perfect if there wasn't this $y$ appearing here.
Another approach would be this product states on the tensor product *-algebra
using the so-called slice map, but I want to consider elements of the entire $C^*$-tensor product and not just the algebraic TP. That means that (among many other things) I'd have to extend the slice map to the spatial TP and I don't see this happening at the moment.
edit: of course slice maps extend to the spatial tp, this is a simple corollary of the fact that the TP of two completely positive maps is again c.p.
Best Answer
Apparently this is not as hard as I thought. I will assume that $A$ is unital for ease.
First of all, note that, given a state $\phi\in s(A)$, the slice map $\eta_\phi$ is simply the tensor product $\phi\otimes\text{id}_B:A\otimes B\to\mathbb{C}\otimes B\cong B$, so, since states are c.p.c maps (i.e. completely positive and contractive), the tensor product of two c.p.c maps extends to a c.p.c map to the spatial tensor product, i.e. the slice map extends to a c.p.c map $\eta_\phi:A\otimes B\to B$.
Now let $x\in(A\otimes B)_+$ be non-zero. I claim that we can find a pure state $\phi\in\text{PS}(A)$ such that $\eta_\phi(x)\neq0$.
Proof of the claim: assume that $\eta_\phi(x)=0$ for all pure states $\phi$.
Let $n\ge1$ and $\phi_1,\dots,\phi_n\in\text{PS}(A)$ and $t_1,\dots,t_n\in[0,1]$ with $\sum_{i=1}^nt_i=1$. consider the convex combination $\omega=\sum_{i=1}^nt_i\phi_i$. Then one easily sees that $$\eta_\omega=\sum_{i=1}^nt_i\eta_{\phi_i}$$ so $\eta_\omega(x)=0$, i.e. $\eta_\phi(x)=0$ for all $\phi$ convex combination of pure states. But the convex combinations of pure states are weak star dense in the state space $S(A)$. Now, given a state $\phi\in S(A)$, we find a net $\{\phi_\lambda\}$ of convex combinations of pure states such that $\phi_\lambda\to\phi$ in the weak star topology. Let $\varepsilon>0$ and find $z=\sum_{i=1}^na_i\otimes b_i\in A\odot B$ with $\|x-z\|<\varepsilon$. Then $$\|\eta_\phi(x)\|=\|\eta_\phi(x)-\eta_{\phi_\lambda}(x)\|\leq\|\eta_\phi(x)-\eta_\phi(z)\|+\|\eta_\phi(z)-\eta_{\phi_\lambda}(z)\|+\|\eta_{\phi_\lambda}(z)-\eta_{\phi_\lambda}(x)\|\leq$$ $$\leq2\varepsilon+\|\sum_{i=1}^n(\phi(a_i)-\phi_\lambda(a_i))b_i\|<3\varepsilon$$ for large $\lambda$. Since $\varepsilon>0$ was arbitrary, we get $\eta_\phi(x)=0$, so $\eta_\phi(x)=0$ for all $\phi\in S(A)$. Composing with any state $\psi\in S(B)$, we get that $\psi\circ\eta_\phi(x)=0$ for all $\phi\in S(A),\psi\in S(B)$. But $\psi\circ\eta_\phi=\phi\otimes\psi$, as this is easily verified on elementary tensors, so we conclude that $\psi\otimes\phi(x)=0$ for all states $\phi\in S(A)$ and $\psi\in S(B)$. But we already showed in the question that the states separate the points of the spatial tensor product, so we get a contradiction.
Continuing the proof: Now that we have such a pure state, we have that $\eta_\phi(x)\in B_+$ is non-zero, so, by elementary theory (i.e. this can be found in Murphy) we can find a pure state $\psi\in\text{PS}(B)$ such that $|\psi(\eta_\phi(x))|=\|\eta_\phi(x)\|\neq0$. But note that $\psi\circ\eta_\phi(a\otimes b)=\psi(\phi(a)b)=\phi(a)\psi(b)$ on elementary tensors, i.e. $\psi\circ\eta_\phi=\phi\otimes\psi$ and thus $\phi\otimes\psi(x)\neq0$, as we wanted.