Consider the following fragment from Takesaki's book "Theory of operator algebras I":
Takesaki says that by passing to the unitisation, it suffices to show the proposition for unital $C^*$-algebras. I don't see why this is the case. I understand the proof of the unital case so this can be freely used. I tried the following:
Let $A$ be non-unital. If $\omega$ is a hermitian functional on $A$, then we can define
$$\widetilde{\omega}: \widetilde{A}\to \mathbb{C}: a + \lambda 1 \mapsto \omega(a) + \lambda$$
which is again hermitian (but is it also bounded??).
Then we can apply the unital result and write
$$\widetilde{\omega}= \widetilde{\omega}_+ – \widetilde{\omega}_-, \|\widetilde{\omega}\| = \|\widetilde{\omega}_+\| + \|\widetilde{\omega}_-\|.$$
Defining $\omega_+ = \widetilde{\omega}_+\vert_A$ and $\omega_- = \widetilde{\omega}_-\vert_A$, we get $\omega = \omega_+-\omega_-$ as a difference of two positive linear functionals on $A$. However, it is not clear to me why
$$\|\omega_+\| + \|\omega_-\| = \|\omega\|$$
in this case!
I think if the following result is true, then I can finish the proof:
If $\omega \in S(\widetilde{A})$ is a state and $\omega\vert_A \ne 0$, then $\omega\vert_A$ is also a state. But I don't know why this should be true. Any help/comment/remark is highly appreciated!
Best Answer
You start with a selfadjoint $\omega$ on $A$. Extend by Hahn-Banach to a functional $\omega'$ on $\tilde A$ with $\|\omega'\|=\|\omega\|$. Define $$ \tilde\omega(x)=\tfrac12\,\big(\omega'(x)+\overline{\omega'(x^*)}\big). $$ Then $\tilde\omega$ is selfadjoint. We have $\|\tilde\omega\|\leq\|\omega'\|=\|\omega\|$, but as it extends $\omega$, we get $\|\tilde\omega\|=\|\omega\|$. So we end up with a selfadjoint extension of $\omega$ to $\tilde A$, that preserves the norm.
The argument in Takesaki's book gives you $$ \frac\omega{\|\omega\|}=\lambda\omega_1-\mu\omega_2. $$ These are states in the unitization. But they are also states in $A$. In principle you know that $\|\omega_j\|\leq1$, $j=1,2$. But since the left-hand-side has norm 1, you have $$ 1=\|\lambda\omega_1-\mu\omega_2\|\leq\lambda\|\omega_1\|+\mu\|\omega_2\|\leq\lambda+\mu=1. $$ The only way for this to be an equality is if $\|\omega_1\|=\|\omega_2\|=1$.