Existence of state on a C*-algebra satisfying $|\tau(ab)|=\|ab\|$

Question

(This is a repost of a question from math.SE, https://math.stackexchange.com/questions/1240966/existence-of-state-on-a-c-algebra-satisfying-tauab-ab)

Let $a,b$ be elements of a unital C*-algebra $A$ with $0\leq a,b\leq 1$. Is it the case there is a state $\tau$ on $A$ such that $|\tau(ab)|=\|ab\|$? I'm particularly interested in the case when $a,b$ are projections and $\|ab\|=1$.

If $ab$ is normal (e.g., $a$ and $b$ commute), then this is a standard fact. Also, if $\|a\|=\|b\|=1$ and $\tau$ is a state satisfying $\tau(a)=\tau(b)=1$, then $\tau$ will work, but I don't see how to come up with such a $\tau$ if $a$ and $b$ don't commute.

Answer 1

I think there are $2\times 2$ counterexamples. Take $a = \pmatrix{1&0\cr 0&0}$ and $b = \pmatrix{.5&.5\cr .5&.5}$, so that $c = ab = \pmatrix{.5&.5\cr 0&0}$. If there were a state $\tau$ satisfying $\tau(c) = \lambda$ for some $|\lambda| = \|c\|$ then there would be a face of the state space which does this, so there is an extreme point, i.e., a vector state, with this property. But that would mean that $c$ has an eigenvalue with modulus equal to its norm, which is false (its only eigenvalue is $.5$, but its norm is $1/\sqrt{2}$).

However, if $\|ab\| = 1$ then the result is easy because (fixing some faithful representation) there is a sequence of unit vectors $(v_n)$ with the property that $\langle abv_n, v_n\rangle \to 1$, so you can take a weak* limit of these vector states.

(Key fact: if $a > 0$, $\|a\| = 1$, and $v$ is a unit vector with $\|av\|$ close to $1$ then $av$ is close to $v$. So if $\|ab\| = 1$ then there is a unit vector $v$ with $\|abv\|$ close to $1$, which successively implies that $\|bv\|$ is close to $1$, $bv$ is close to $v$, and $abv$ is close to $v$.)

(I wrote this quickly because I have to run, I hope it's right.)