Positivity of bounded form $\phi$ on C*-algebra $A$

Question

I am reading Lectures on von Neumann algebras by Stratila and Zsidò. I don't know what to do with exercise 5.1, do you have any tips?

Let $A$ be a $C^*$-algebra and $\phi$ a bounded form on $A$. If there exists an $a \in A$, $a>0$, such that $\phi (a)= ||\phi||\, ||a||$, then $\phi$ is positive.

Thanks for your help.

Answer 1

Rescaling $a$, you may assume that $\|a\|=1$. Then the condition tells us that $\phi$ attains its norm in a positive element of norm $1$.

We may moreover also assume that $A$ is unital (why?).

We show that this automatically implies that also $\phi(1)=\|\phi\|$.

Indeed, all you have to do is choose $\lambda\in \mathbb{C}$ with $|\lambda| = 1$ such that $\lambda\phi(1-a) \in [0, \infty[$. Then, since $\|a + \lambda(1-a)\| \le 1$ (why?), we have $$\|\phi\| = \phi(a) \le \phi(a)+ \lambda \phi(1-a) = \phi(a + \lambda(1-a)) \le \|\phi\|$$ and thus $\|\phi\| = \phi(a) + \lambda\phi(1-a) = \|\phi\| + \lambda\phi(1-a)$ so that $\phi(1-a)=0$, i.e. $\phi(1) = \phi(a) = \|\phi\|$.

Now, you can invoke the well-known fact that if $\phi$ is a bounded functional on a unital $C^*$-algebra that attains its norm in the identity, then it is automatically a positive functional. This is proven in every text on $C^*$-algebras.