Norm Preserves Ordering on Positive Elements of C*-Algebra

Theorem

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.

Let $\le_A$ be the canonical preordering on $A$.

Let $a, b \in A$ be such that:

${\mathbf 0}_A \le_A a \le_A b$

Then:

$\norm a \le \norm b$

Proof

First take $A$ unital.

From Identity Element is Order Unit on Set of Hermitian Elements of Unital C*-Algebra, we obtain:

$b \le_A \norm b {\mathbf 1}_A$

Since $\le_A$ is a preordering and $a \le_A b$, we have:

${\mathbf 0}_A \le_A a \le_A \norm b {\mathbf 1}_A$

Let $B$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, a}$.

From Subalgebra Generated by Commuting Elements is Commutative, $B$ is commutative.

Let $\Phi_B$ be the spectrum of $B$.

Let $\struct {\map \CC {\Phi_B}, \overline \cdot, \norm {\, \cdot \,}_\infty}$ be the $\text C^\ast$-algebra of continuous complex-valued functions on $\Phi_B$ vanishing at infinity.

From Gelfand-Naimark Theorem: Commutative Case: Unital, there exists an isometric unital $\ast$-algebra isomorphism $G : B \to \map \CC {\Phi_B}$.

Write $f = \map G a$.

From *-Algebra Homomorphism preserves Canonical Preordering of C*-Algebra and Canonical Preordering of C*-Algebra of Continuous Functions Vanishing at Infinity, we have:

${\mathbf 0}_A \le_B a \le_B \beta {\mathbf 1}_A$

for some $\beta \in \R$ if and only if:

$0 \le \map f \phi \le \beta$ for all $\phi \in \Phi_B$.

From Continuous Function on Compact Space is Bounded, $f$ is bounded and hence such a $\beta$ exists.

We have for all such $\beta$:

$\ds \beta \ge \sup_{\phi \in \Phi_B} \map f \phi$

where the supremum is finite.

Since $\map f \phi \ge 0$ for each $\phi \in \Phi_B$, we have:

$\ds \sup_{\phi \in \Phi_B} \map f \phi = \sup_{\phi \in \Phi_B} \cmod {\map f \phi} = \norm f_\infty$

Since $G$ is isometric, we have $\norm f_\infty = \norm a$.

So we have $\beta \ge \norm a$ whenever:

$a \le_B \beta {\mathbf 1}_A$

From Spectrum of Element of Unital C*-Subalgebra of Unital C*-Algebra we have:

$\map {\sigma_B} x = \map {\sigma_A} x$ for each $x \in B$.

Hence $x$ is positive in $B$ if and only if it is positive in $A$.

So we have $a \le_B \beta {\mathbf 1}_A$ if and only if $a \le_A \beta {\mathbf 1}_A$.

We have established that the latter holds for $\beta = \norm b$, so we obtain $\norm b \ge \norm a$.

This completes the proof in the unital case.

Now suppose that $A$ is not unital.

Let $A_+$ be the unitization of $A$.

Let $\le_+$ be the canonical preordering on $A_+$.

Define:

$\tuple {u, \lambda}^\ast = \tuple {u^\ast, \overline \lambda}$

for each $\tuple {u, \lambda} \in A_+$.

From Existence of Unitization of C*-Algebra, there exists a unique algebra norm $\norm {\, \cdot \,}_\ast$ extending $\norm {\, \cdot \,}$ such that $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ is a unital $\text C^\ast$-algebra.

By definition we have:

$\map {\sigma_{A_+} } {\tuple {x, 0} } = \map {\sigma_A} x$ for each $x \in A$.

Hence we have that $\tuple {x, 0}$ is positive in $A_+$ if and only if $x$ is positive in $A$.

Hence we obtain:

${\mathbf 0}_{A_+} \le_+ \tuple {a, 0} \le_+ \tuple {b, 0}$

Hence by the unital case, we have:

$\norm {\tuple {a, 0} }_\ast \le \norm {\tuple {b, 0} }_\ast$

Since $\norm {\, \cdot \,}_\ast$ extends $\norm {\, \cdot \,}$, we have:

$\norm a \le \norm b$

$\blacksquare$

Sources

• 1990: Gerard J. Murphy: C*-Algebras and Operator Theory ... (previous) ... (next): $2.2$: Positive Elements of $C^\ast$-Algebras