state space is non-empty

In this entry we prove the existence of states for every $C^{*}$-algebra (http://planetmath.org/CALgebra).

Let $\mathcal{A}$ be a $C^{*}$-algebra. For every self-adjoint   (http://planetmath.org/InvolutaryRing) 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 (http://planetmath.org/Ring), 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.

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 $x_{0}$,

 $\displaystyle ev_{x_{0}}:C(X)\longrightarrow\mathbb{C}$ $\displaystyle ev_{x_{0}}(f):=f(x_{0})$

is a multiplicative linear functional of $C(X)$. Hence, $\|ev_{x_{0}}\|=1$ and also $|ev_{x_{0}}(a)|=|a(x_{0})|=\|a\|$.

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 there is a state $\widetilde{\psi}$ on $\widetilde{\mathcal{A}}$ satisfying the required . Now, we just need to take the restriction  (http://planetmath.org/RestrictionOfAFunction) of $\widetilde{\psi}$ to $\mathcal{A}$ and this restriction is a state in $\mathcal{A}$ satisfying the required . $\square$

Title state space is non-empty StateSpaceIsNonempty 2013-03-22 17:45:14 2013-03-22 17:45:14 asteroid (17536) asteroid (17536) 14 asteroid (17536) Theorem msc 46L30 msc 46L05