# Let $A$ be a unital $C^*$-algebra, $a\in A,\ x,y\in A_{sa}$. Does there exist a state $\phi$ on $A$ such that $\phi(xa^*ay)=\lVert a\rVert^2\phi(xy)$?

If $$A$$ is a commutative, unital $$C^*$$-algebra, then $$A=C(X)$$ for some compact, $$T_2$$ space $$X$$. Then $$a=f,x=g$$ and $$y=h$$ are continuous functions on $$X$$. Then there is $$x_0\in X$$ such that $$\lVert f\rVert=|f(x_0)|$$. Take $$\phi=\hat{x_0}$$. Then $$\phi(xa^*ay)=g(x_0)|f(x_0)|^2h(x_0)=\lVert f\rVert^2\phi(gh)=\lVert a\rVert^2\phi(xy)$$.

I'm doubtful about the non-commutative case here. If $$x,y=1$$, then the we consider the (commutative) $$C^*$$-algebra generated by $$a^*a$$, and then extend to whole of $$A$$ by Hahn-Banach. But here $$x,y,a^*a$$ may not commute with each other, so the $$C^*$$-algebra generated by $$x,y,a^*a$$ may not be commutative. Therefore, I cannot apply the above argument.

Can anyone help me with the non-commutative case? Thanks for your help in advance.

Let $$A=B(H)$$. Fix an orthonormal basis $$\{e_n\}$$, and let $$\{E_{kj}\}$$ the corresponding matrix units. Let $$x=\sum_kE_{k,k+1},\qquad y=\sum_kE_{k+1,k},\qquad a=E_{11}.$$ Then $$\|a\|=1$$, and $$x(1-a^*a)y=\sum_{h,k,j}E_{h,h+1}E_{k+1,k+1}E_{j+1,j}=\sum_kE_{kk}=I.$$ This means that $$\phi(x(1-a^*a)y)=1$$ for any state $$\phi$$, which means that $$1+\phi(xa^*ay)=\|a\|^2\,\phi(xy)$$ for any state $$\phi$$.
• I'm a bit curious why the computation $x(1-a^*a)y=1$ fails if $H$ is finite dimensional Oct 25, 2023 at 21:30
• In finite dimension the equality $x(1-a^*a)y=1$ implies that $1-a^*a$ is invertible. This forces $\|a\|<1$ (because $1$ cannot be an eigenvalues of $a^*a$), contrary to what was assumed. Oct 25, 2023 at 21:54