34.9 Positive Linear Functionals
Definition 34.9.1 (Positive Linear Functional).label Let $A$ be a unital $C^{*}$-algebra and $\phi \in A^{*}$, then $\phi$ is positive if $\dpn{x, \phi}{A}\ge 0$ for all positive elements $x \in A$.
Theorem 34.9.2.label Let $A$ be a unital $C^{*}$-algebra and $\phi \in \hom(A; \complex)$, then the following are equivalent:
- (1)
$\phi$ is a positive linear functional.
- (2)
$\phi \in A^{*}$ with $\normn{\phi}_{A^*}= \dpn{1, \phi}{A}$.
Proof. (1) $\Rightarrow$ (2): For any $x \in A_{sa}$ with $\norm{x}_{A} \le 1$, $\sigma_{A}(1 - x) \subset 1 - [-1, 1] = [0, 2]$ by Proposition 33.5.11. Thus $1 - x \ge 0$ by Proposition 34.8.2, and $\dpn{x, \phi}{A}\le \dpn{1, \phi}{A}$. By Proposition 11.2.7, $\norm{\phi}_{A^*}\le \dpn{1, \phi}{A}$, so $\norm{\phi}_{A^*}= \dpn{1, \phi}{A}$.
(2) $\Rightarrow$ (1): For each $x \in A_{sa}$, $\sigma_{A}(x) = \sigma_{A[x]}(x)$ by Corollary 34.2.4, so $x \ge 0$ in $A$ if and only if $x \ge 0$ in $A[x]$ by Proposition 34.8.2. In addition, the Gelfand-Naimark Theorem implies that $x \ge 0$ if and only if $\Gamma_{A[x]}(x) \ge 0$. Moreover, since $1 \in A[x]$, the norm of $\phi$ alongside (2) is preserved by restricting to $A[x]$. By considering each commutative subalgebra, assume without loss of generality that $A$ is commutative.
By the Riesz Representation Theorem, $\phi$ takes the form of a complex Radon measure $\mu$ on $\Omega(A)$, and $\norm{\phi}_{A^*}= \norm{\mu}_{\text{var}}$. For each Borel set $E \in \cb_{\Omega(A)}$,
which is only possible if $\mu(E), \mu(\Omega(A) \setminus E) \ge 0$. As this holds for all $E \in \cb_{\Omega(A)}$, $\mu$ is positive, and $\phi$ then is a positive linear functional.$\square$
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