Norm in the minimal tensor product of C*-algebras

Question

Let $A$ and $B$ be two $C^*$-algebras, and let $A \otimes B$ denote their minimal tensor product. Given positive, linear functionals $\varphi$ on $A$ and $\psi$ on $B$, we obtain a positive, linear functional $\varphi\otimes\psi\colon A\otimes B \to \mathbb{C}$, which on elementary tensors is given by $$ \varphi\otimes\psi(a\otimes b) = \varphi(a)\psi(b). $$ Using that bounded, linear functionals on $C^*$-algebras are linear combinations of positive, linear functionals, it also follows that $\varphi\otimes\psi\colon A\otimes B \to \mathbb{C}$ is well-defined for $\varphi \in A^*$ and $\psi \in B^*$.

Question 1: Given a non-zero element $t \in A \otimes B$, do there exist states $\varphi$ on $A$ and $\psi$ on $B$ such that $\varphi\otimes\psi(t) \neq 0$?

Question 2: More generally, given $t \in A \otimes B$, do we have $$ \|t\| = \sup\{ |\varphi\otimes\psi(t)| : \varphi \in A^*, \psi \in B^*, \|\varphi\| \leq 1, \|\psi\| \leq 1 \}? $$

Answer 1

1. Yes. Since positive linear functionals span all bounded linear functionals, the span of $\varphi \otimes \psi$ for states $\varphi \in A^\ast$ and $\psi \in B^\ast$ include $\phi \otimes \omega$ for all $\phi \in A^\ast$ and $\omega \in B^\ast$. Fix faithful representations $\pi_1: A \to B(H)$ and $\pi_2: B \to B(K)$. Then $A \otimes B$ is faithfully represented on $B(H \otimes K)$ in the standard way, and vector state $\langle h_1 \otimes k_1 | \cdot h_2 \otimes k_2 \rangle \in (A \otimes B)^\ast$ for any $h_1, h_2 \in H$, $k_1, k_2 \in K$ is of the form $\phi \otimes \omega$ where $\phi \in A^\ast$ and $\omega \in B^\ast$. Vector states of the above form clearly densely span all vector states. Thus, for any $t \in A \otimes B$, if $\varphi \otimes \psi(t) = 0$ for all states $\varphi \in A^\ast$ and $\psi \in B^\ast$, then $\langle \xi | t\eta \rangle = 0$ for all $\xi, \eta \in H \otimes K$, so $t = 0$.
2. No, this is not true even for $A = B = M_2(\mathbb{C})$. Indeed, let $t$ be the orthogonal projection onto $\text{span}\{\frac{1}{\sqrt{2}}(e_1 \otimes e_1 + e_2 \otimes e_2)\}$, which has norm $1$. $\varphi \in A^\ast$, $\psi \in B^\ast$ of norms at most $1$ correspond to matrices $x, y \in M_2(\mathbb{C})$ of trace norms at most $1$, so,

$$\begin{split} |\varphi \otimes \psi(t)| &= \frac{1}{2}|\langle e_1 \otimes e_1 + e_2 \otimes e_2 | (x \otimes y)(e_1 \otimes e_1 + e_2 \otimes e_2) \rangle|\\ &= \frac{1}{2}|\sum_{i=1}^2 \sum_{j=1}^2 x_{ij}y_{ij}|\\ &\leq \frac{1}{2}(\sum_{i=1}^2 \sum_{j=1}^2 |x_{ij}|^2)^{1/2}(\sum_{i=1}^2 \sum_{j=1}^2 |y_{ij}|^2)^{1/2}\\ &= \frac{1}{2}\|x\|_2\|y\|_2\\ &\leq \frac{1}{2}\|x\|_1\|y\|_1\\ &\leq \frac{1}{2} \end{split}$$

So $\|t\| = 1 > \frac{1}{2} \geq \sup \{|\varphi \otimes \psi(t)|: \varphi \in A^\ast, \psi \in B^\ast, \|\varphi\| \leq 1, \|\psi\| \leq 1\}$.