*-Algebra obtains Banach *-Algebra Structure through *-Algebra Isomorphism
Theorem
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$.
Let $\struct {B, \square, \norm {\, \cdot \,}_B}$ be a Banach $\ast$-algebra.
Let $\phi : A \to B$ be a $\ast$-algebra isomorphism.
Define $\norm {\, \cdot \,}_A : A \to \R$ by:
- $\norm a_A = \norm {\map \phi a}_B$
Then $\struct {A, \ast, \norm {\, \cdot \,}_A}$ is a Banach $\ast$-algebra.
Further if $\struct {B, \square, \norm {\, \cdot \,}_B}$ is a $\text C^\ast$-algebra, then so is $\struct {A, \ast, \norm {\, \cdot \,}_A}$.
Proof
From Algebra obtains Norm Structure through Algebra Isomorphism:
- $\norm {\, \cdot \,}_A$ is an algebra norm.
We want to show that:
- $\norm {a^\ast}_A = \norm a_A$
for each $a \in A$.
Let $a \in A$.
We have:
| \(\ds \norm {a^\ast}_A\) | \(=\) | \(\ds \norm {\map \phi {a^\ast} }_B\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {\map \phi a^\square}_B\) | Definition of *-Algebra Homomorphism | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {\map \phi a}_B\) | Definition of Banach *-Algebra | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm a_A\) |
By definition, $\phi : A \to B$ is an isometric isomorphism.
Hence from Inverse of Isometric Isomorphism between Normed Vector Spaces is Isometric Isomorphism, $\phi^{-1} : B \to A$ is an isometric isomorphism.
From Metric Space Completeness is Preserved by Isometry, the completeness of $B$ implies the completeness of $A$.
So $\struct {A, \norm {\, \cdot \,}_A}$ is a Banach algebra.
Hence $\struct {A, \ast, \norm {\, \cdot \,}_A}$ is a Banach $\ast$-algebra.
Now suppose that $\struct {B, \square, \norm {\, \cdot \,}_B}$ is a $\ast$-algebra so that:
- $\norm {b b^\square}_B = \norm b^2_B$
for each $b \in B$.
We want to show that:
- $\norm {a a^\ast}_A = \norm a^2_A$
for each $a \in A$.
We have:
| \(\ds \norm {a a^\ast}_A\) | \(=\) | \(\ds \norm {\map \phi {a a^\ast} }_B\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {\map \phi a \map \phi {a^\ast} }_B\) | Definition of Algebra Homomorphism | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {\map \phi a \map \phi a^\square}_B\) | Definition of *-Algebra Homomorphism | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {\map \phi a}_B^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \norm a_A^2\) |
$\blacksquare$