Product of Powers of Positive Element of Unital C*-Algebra

Theorem

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

Let $a \in A$ be positive elements.

Let $\alpha, \beta > 0$ be real numbers.

Let $a^\alpha$, $a^\beta$ and $a^{\alpha + \beta}$ be given by the continuous functional calculus.

Then:

$a^{\alpha + \beta} = a^\alpha a^\beta$

Proof

Let $\map {\sigma_A} a \subseteq \hointr 0 \infty$ be the spectrum of $a$.

Define $e_\alpha : \map {\sigma_A} a \to \hointr 0 \infty$, $e_\beta : \map {\sigma_A} a \to \hointr 0 \infty$ and $e_{\alpha + \beta} : \map {\sigma_A} a \to \hointr 0 \infty$ by:

$\map {e_\alpha} t = t^\alpha$

$\map {e_\beta} t = t^\beta$

$\map {e_{\alpha + \beta} } t = t^{\alpha + \beta}$

for each $t \in \map {\sigma_A} a$.

For $t \in \map {\sigma_A} a$ we have:

$\map {e_\alpha} t \map {e_\beta} t = t^\alpha t^\beta = t^{\alpha + \beta} = \map {e_{\alpha + \beta} } t$

by Product of Powers.

Hence since the continuous functional calculus is an algebra homomorphism, we obtain:

$\map {e_\alpha} a \map {e_\beta} a = \map {e_{\alpha + \beta} } a$

That is:

$a^\alpha a^\beta = a^{\alpha + \beta}$

$\blacksquare$