Restriction of state on unitisation

Let $A$ be a $C^*$-algebra with unitisation $\widetilde{A}$. Assume $\omega$ is a state on $\widetilde{A}$ such that $\omega(A)\ne 0$. Is it true that $\omega\vert_A$ is a state on $A$?

I can show that $\omega\vert_A$ is positive and satisfies $\|\omega\vert_A\|\le \|\omega\|=1$ but I don't see the other inequality. I tried playing with an approximate unit but it didn't work out.

Solution 1:

No, it's not true. Consider for instance $A=c_0$, and $\omega$ be given by $$ \omega(x,\lambda)=\tfrac{x_1}2+\lambda. $$ This is a state, but $\|\omega|_A\|=\tfrac12$. One can easily see this, by thinking of $c_0+\mathbb C1$ as multiplication operators on $\ell^2$. Then $$ \omega(x,\lambda)=\tfrac12\,\big[\langle (x,\lambda)e_1,e_1\rangle+\langle (x,\lambda)e_2,e_2\rangle\big], $$ a convex combination of states.

More generally, any $\varphi\in A^*_+$ with $\|\varphi\|\leq1$ admits a extension to a state in $\tilde A$. Indeed, we define $$ \tilde\varphi(a,\lambda)=\varphi(a)+\lambda. $$ This is obviously linear and unital, so it is enough to show that it is positive. For this, \begin{align} \big|\tilde\varphi\big((a,\lambda)^*(a,\lambda)\big)\big| &=\varphi(a^*a+2\operatorname{Re}\overline\lambda a)+|\lambda|^2\\[0.3cm] &=\varphi(a^*a)+2\operatorname{Re}\overline\lambda\,\varphi(a)+|\lambda|^2\\[0.3cm] &\geq|\varphi(a)|^2+2\operatorname{Re}\overline\lambda\,\varphi(a)+|\lambda|^2\\[0.3cm] &=|\varphi(a)+\lambda|^2\geq0. \end{align} The key inequality is Kadison's Schwarz inequality, that together with $\|\varphi\|\leq1$ gives us $$|\varphi(a)|^2\leq\|\varphi\|\,\varphi(a^*a)\leq\varphi(a^*a).$$