Let $A$ be a $C^*$-algebra and $\tau:A\to\mathbb{C}$ linear. Claim: the following conditions are equivalent:
$\tau(ab)=\tau(ba)$ for all $a,b\in A$
$\tau(x^*x)=\tau(xx^*)$ for all $x\in A$
$\tau(uau^*)=\tau(a)$ for all positive $a\in A$ und for all unitaries $u$ in the unitization $A^1$ of $A$.
The direction 1=>2 is trivial. But I'm stuck to prove 2=> 3 and 3=>1.
One idea for the direction $3\Rightarrow 1$ is: If it's possible to prove that $\tau(ux)=\tau( xu)$ for all unitaries $u\in A^1$ and for all $x\in A$, I can conclude that $\tau(ax)=\tau(xa)$ for all $a\in A^1$ and for all $x\in A$ (because $a\in A^1$ can be written as a linear combination of unitaries), so I can conclude 1. But all my tries to prove $\tau(ux)=\tau( xu)$ for all unitaries $u\in A^1$ and for all $x\in A$ were pointless, can you help me?
I also have no idea how to do $2\Rightarrow 3$ . I would appreciate hints.
Best
for $2\implies 3$: you have, since $a\geq0$, $$ \tau(uau^*)=\tau(ua^{1/2}a^{1/2}u^*)=\tau((a^{1/2}u^*)^*a^{1/2}u^*) =\tau(a^{1/2}u^*(a^{1/2}u^*)^*)=\tau(a^{1/2}u^*ua^{1/2})=\tau(a^{1/2}a^{1/2}) =\tau(a) $$
For $3\implies 1$: since the condition holds for positive $a$ and every operator is a linear combination of positives, the equality holds for arbitrary $a$: $$\tau(uau^*)=\tau(a).$$ In particular it holds for $au$, and we obtain $$\tau(ua)=\tau(au)$$ for $a\in A$ and $u\in A$ a unitary. Now, since $b$ can be written as a linear combination of unitaries, $\tau(ab)=\tau(ab)$.