Let $\phi: A \to \mathbb{C}$ be a state on a unital $C^*$-algebra $A$ and let $a \in A$ be a positive element with $\phi(a) = 0$. Is it true that $\phi(b^*ab)=0$ for every $b \in B?$
Some observations:
- Since the unitaries of $A$ span $A$, it suffices to check the claim for $b$ a unitary.
- We can try to apply results about completely positive maps/multiplicative maps/Cauchy-Schwarz inequality.
- If $a=1$ and $\phi(a)=0$, then $\phi = 0$ so the statement is trivial.
- An example: Let $\phi: M_n(\mathbb{C}) \to \mathbb{C}: A \mapsto Tr(A)$. If $\phi(A) = 0$. Then (since we may assume that $B$ is a unitary) $\phi(B^*AB) = \phi(A)=0$ so for this functional the claim follows.
- By more or less the same argument, the claim also holds if $A$ is an abelian $C^*$-algebra.
This fails easily. Take $A=\mathbb{B}(\mathbb{C}^2)\cong M_2$ and let $\xi=(1,0)\in\mathbb{C}^2$. Then $\xi$ is a unit vector, so $\phi(x):=\langle x\xi,\xi\rangle$ defines a state on $A$. Note that if $p$ is the orthogonal projection onto the 1 dimensional subspace spanned by $\xi$, then $\phi(p^\bot)=0$. But $p^\bot$ is precisely the projection onto the one-dimensional subspace spanned by $\xi'=(0,1)$. It is well known that, projections of the same rank are unitarily equivalent in Hilbert spaces, so there exists a unitary $u\in\mathbb{B}(\mathbb{C}^2)$ such that $p=u^*p^\bot u$. (if you don't believe me, try the unitary $u=\begin{pmatrix}0&1\\1&0\end{pmatrix}$).
If your claim was true, since $\phi(p^\bot)=0$, we would also have $\phi(u^*p^\bot u)=0$, i.e. $\phi(p)=0$. But $\phi(p)=1$.