# state space is non-empty

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

Theorem - 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)|=\parallel a\parallel $.

$$

*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.

Thus, by the Gelfand-Naimark theorem^{}, $\mathcal{B}$ is isomorphic^{} to $C(X)$, the space of continuous functions^{} $X\u27f6\u2102$ 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, $\parallel a\parallel =|a({x}_{0})|$.

The evaluation function^{} at ${x}_{0}$,

$e{v}_{{x}_{0}}:C(X)\u27f6\u2102$ | ||

$e{v}_{{x}_{0}}(f):=f({x}_{0})$ |

is a multiplicative linear functional of $C(X)$. Hence, $\parallel e{v}_{{x}_{0}}\parallel =1$ and also $|e{v}_{{x}_{0}}(a)|=|a({x}_{0})|=\parallel a\parallel $.

We can now extend $e{v}_{{x}_{0}}$ to a linear functional^{} $\psi $ on $\mathcal{A}$ such that $\parallel \psi \parallel =\parallel e{v}_{{x}_{0}}\parallel =1$, using the Hahn-Banach theorem^{}.

Also, $\psi (e)=e{v}_{{x}_{0}}(e)=1$ 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)|=|e{v}_{{x}_{0}}(a)|=\parallel a\parallel $.

In case $\mathcal{A}$ does not have an identity element we can consider its minimal unitization $\stackrel{~}{\mathcal{A}}$. By the preceding there is a state $\stackrel{~}{\psi}$ on $\stackrel{~}{\mathcal{A}}$ satisfying the required . Now, we just need to take the restriction^{} (http://planetmath.org/RestrictionOfAFunction) of $\stackrel{~}{\psi}$ to $\mathcal{A}$ and this restriction is a state in $\mathcal{A}$ satisfying the required . $\mathrm{\square}$

Title | state space is non-empty |
---|---|

Canonical name | StateSpaceIsNonempty |

Date of creation | 2013-03-22 17:45:14 |

Last modified on | 2013-03-22 17:45:14 |

Owner | asteroid (17536) |

Last modified by | asteroid (17536) |

Numerical id | 14 |

Author | asteroid (17536) |

Entry type | Theorem |

Classification | msc 46L30 |

Classification | msc 46L05 |