In this entry we prove the existence of states for every $C^*$ -algebra.

Theorem - Let $\mathcal{A}$ be a $C^*$ -algebra. For every self-adjoint element $a \in \mathcal{A}$ there exists a state $\psi$ on $\mathcal{A}$ such that $|\psi(a)|=\|a\|$ .

$\;$

Proof : We first consider the case where $\mathcal{A}$ is unital, with identity element $e$ .

Let $\mathcal{B}$ be the $C^*$ -subalgebra generated by $a$ and $e$ . Since $a$ is self-adjoint, $\mathcal{B}$ is a comutative $C^*$ -algebra with identity element.

Thus, by the Gelfand-Naimark theorem, $\mathcal{B}$ is isomorphic to $C(X)$ , the space of continuous functions $X \longrightarrow \mathbb{C}$ for some compact set $X$ .

Regarding $a$ as an element of $C(X)$ , $a$ attains a maximum at a point $x_0 \in X$ , since $X$ is compact. Hence, $\|a\| = |a(x_0)|$ .

The evaluation function at $x_0$ ,

We can now extend $ev_{x_0}$ to a linear functional $\psi$ on $\mathcal{A}$ such that $\|\psi\| = \|ev_{x_0}\|=1$ , using the Hahn-Banach theorem.

Also, $\psi(e) = ev_{x_0}(e)=1$ and so $\psi$ is a norm one positive linear functional, i.e. $\psi$ is a state on $\mathcal{A}$ .

Of course, $\psi$ is such that $|\psi(a)| = |ev_{x_0}(a)| = \|a\|$ .

In case $\mathcal{A}$ does not have an identity element we can consider its minimal unitization $\widetilde{\mathcal{A}}$ . By the preceding argument there is a state $\widetilde{\psi}$ on $\widetilde{\mathcal{A}}$ satisfying the required property. Now, we just need to take the restriction of $\widetilde{\psi}$ to $\mathcal{A}$ and this restriction is a state in $\mathcal{A}$ satisfying the required property. $\square$