If $\varphi$ is a positive linear functional then the following are equivalent.

Question

$\begin{align} &\text{(1) } \varphi(x^*x)=\varphi(xx^*)\text{ for all }x\in A.\\&\text{(2) } \varphi(u^*au)=\varphi(a)\text{ for all }a\in A \text{ and unitary } u\in \tilde A.\end{align}$

$\tilde A$ is the unitalization of $A$ by adding a $1$ ($A$ may have its own $1_A$).

my trying:

Extend $\varphi$ by defining $\varphi(a+k)=\varphi(a)+k\|\varphi\|$. Then $(1)\Leftrightarrow\varphi((a+k)^*(a+k))=\varphi((a+k)(a+k)^*)$ and $(2)\Leftrightarrow\varphi(u^*(a+k)u)=\varphi(a+k)$ .

Thus we may assume that $A$ is unital and prove that

$\begin{align} &\text{(1) } \varphi(x^*x)=\varphi(xx^*)\text{ for all }x\in A.\\&\text{(2) } \varphi(u^*au)=\varphi(a)\text{ for all }a\in A \text{ and unitary } u\in A.\end{align}$

are equivalent.

Since every $a$ can be decomposed in to $a=\text{Re}(a)^+-\text{Re}(a)^-+i\text{ Im}(a)^+-i\text{ Im}(a)^-$, and $\varphi(u^*xu)=\varphi(\sqrt x u^*u\sqrt x)=\varphi(x)$ for positive $x$, that (1) implies (2) is trivial.

However, I have no idea on how to show (2) implies (1). If $|x|$ is invertible, that there is a unitary $u$ such that $x=u|x|$ follows from polar decomposition. Then $ux^*xu^*=xx^*$ follows. Similarly, if $x$ can be approximated by invertible elements, $\varphi(x^*x)=\varphi(xx^*)$ still holds. But this is true only for certain kinds of C*-algebras. Do I need to use Gelfand-Naimark representation on this?

Answer 1

Use the following statement:

If $A$ is a unital $C^*$-algebra and $x\in A$ there are unitaries $u_1,...,u_4$ and complex numbers $a_1,...,a_4$ with $x=\sum_i a_i u_i$.

So:

$$\varphi(x^*x)=\varphi\left(\sum_i(a_iu_i)^*\sum_j(a_ju_j)\right)= \sum_{ij} \overline{a_i}a_j\varphi(u_i^*u_j)$$

now note that $\varphi(u_i^*u_j)= \varphi\left(u_i(u_i^*u_j)u_i^*\right) = \varphi(u_ju_i^*)$ by (2). Hence: $$\varphi(x^*x)=\sum_{ij}\overline{a_i}a_j\varphi(u_ju_i^*) =\varphi\left(\sum_j(a_ju_j)\sum_i(a_iu_i)^*\right)=\varphi(xx^*)$$