Jordan decomposition in Takesaki: theorem III.2.2.1

Solution 1:

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 $\|\omega_1\|=\|\omega_2\|=1$.