I have an idea for a possible counterexample to the noncommutative Stone-Weierstrass problem. A good answer to the following question would really help.


Let $\mathcal{A}$ be the C\*-algebra of $2\times 2$ complex matrices, let $\mathcal{B}$ be the C\*-subalgebra of $2\times 2$ diagonal matrices, and let $v$ and $w$ be the unit vectors $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$, respectively. Then $f: A \mapsto \langle Av,v\rangle$ and $g: A \mapsto \langle Aw,w\rangle$ are pure states on $\mathcal{A}$. They are distinct, but $f(A) = g(A)$ for all $A \in \mathcal{B}$: we say that $\mathcal{B}$ fails to *separate* $f$ and $g$.


I would like to find a C\*-algebra $\mathcal{A}'$ which (unitally) contains $\mathcal{A}$ together with a C\*-subalgebra $\mathcal{B}'$ which contains $\mathcal{B}$, such that $\mathcal{B}'$ separates any extensions $f'$ and $g'$ of $f$ and $g$ to states on $\mathcal{A}'$. This could be trivially accomplished by setting $\mathcal{A}' = \mathcal{B}' = \mathcal{A}$, so let me add one more condition. Let $P = \left[\matrix{1&1\cr1&1}\right]$ and observe that the distance from $P$ to $\mathcal{B}$ is $1$. (Evaluation in the $(1,2)$ matrix entry is a norm one linear functional on $\mathcal{A}$ which takes the value $1$ on $P$ and is zero on $\mathcal{B}$. Conversely, the distance from $P$ to the identity matrix is $1$.) To prevent trivial solutions to my problem, I impose the additional requirement that the distance from $P$ to $\mathcal{B}'$ must still be $1$.



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> Can we find a C\*-algebra $\mathcal{A}'$ which unitally contains $\mathcal{A}$ together with a C\*-subalgebra $\mathcal{B}'$ which contains $\mathcal{B}$, such that (1) the distance from $P$ to $\mathcal{B}'$ is $1$ and (2) $\mathcal{B}'$ separates any two states on $\mathcal{A}'$ which extend $f$ and $g$, respectively?
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Separating pure states on the $2\times 2$ matrix algebra
The answer to my question is no. Let $\mathcal{A} = M\_2$ be the algebra of $2\times 2$ complex matrices, let $\mathcal{B}$ be the subalgebra of diagonal matrices, let $\mathcal{A}'$ be a C\*-algebra which unitally contains $\mathcal{A}$, and let $\mathcal{B}'$ be a C\*-subalgebra of $\mathcal{A}'$ which contains $\mathcal{B}$.


$\mathcal{A}$ contains the matrix units $e\_{11}$, $e\_{12}$, $e\_{21}$, $e\_{22}$, and hence so does $\mathcal{A}'$. So we have $\mathcal{A}' \cong M\_2\otimes \mathcal{C}$ for some other unital C\*-algebra $\mathcal{C}$. Also, since $\mathcal{B}$ contains $e\_{11}$ and $e\_{22}$, we have $$\mathcal{B}' = e\_{11}\mathcal{B}'e\_{11} + e\_{11}\mathcal{B}'e\_{22} + e\_{22}\mathcal{B}'e\_{11} + e\_{22}\mathcal{B}'e\_{22} = \mathcal{B}'\_{11} + \mathcal{B}'\_{12} + \mathcal{B}'\_{21} + \mathcal{B}'\_{22}.$$ We have $P = e\_{11} + e\_{12} + e\_{21} + e\_{22}$. There are two cases.


Case 1. $\mathcal{B}'\_{12}$ contains an element $x$ satisfying $\|x - e\_{12}\| < 1$. Then $\mathcal{B}'$ contains the element $y = e\_{11} + x + x^\* + e\_{22}$, and we have
$$\|y - P\| = \|(x - e\_{12}) + (x^\* - e\_{21})\| = \|x - e\_{12}\| < 1.$$ So in this case the distance from $\mathcal{B}'$ to $P$ is strictly less than 1.


Case 2. We have $\|x - e\_{12}\| \geq 1$ for all $x \in \mathcal{B}'\_{12}$. By the Hahn-Banach theorem, we can find a norm one linear functional on $e\_{11}\mathcal{A}'e\_{22}$ which takes $e\_{12}$ to $1$ and vanishes on $\mathcal{B}'\_{12}$. By the isomorphism $e\_{11}\mathcal{A}'e\_{22} \cong \mathcal{C}$, this yields a state $h$ on $\mathcal{C}$. Then $f\otimes h$ and $g\otimes h$ are two states on $\mathcal{A}'$ which respectively extend $f$ and $g$, but $\mathcal{B}'$ does not separate them, because they agree on $\mathcal{B}'\_{11} \subseteq e\_{11}\mathcal{A}'e\_{11}$ and $\mathcal{B}'\_{22} \subseteq e\_{22}\mathcal{A}'e\_{22}$, and they both vanish on $\mathcal{B}'\_{12}$ and $\mathcal{B}'\_{21}$. So in this case $\mathcal{B}'$ fails to separate all extensions of $f$ and $g$.


