Character on Non-Unital Banach Algebra induces Character on Unitization

Theorem

Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$ that is not unital.

Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the normed unitization of $\struct {A, \norm {\, \cdot \,} }$.

Let $\phi$ be a character on $A$.

Define:

$\map {\phi_+} {\tuple {x, \lambda} } = \map \phi x + \lambda$

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

Then $\phi_+$ is a character on $A_+$.

Proof

Let $\tuple {x, \lambda}, \tuple {y, \mu} \in A_+$ and $t \in \C$.

We have:

\(\ds \map {\phi_+} {\tuple {x, \lambda} + t \tuple {y, \mu} }\)

\(=\)

\(\ds \map {\phi_+} {\tuple {x + t y, \lambda + t \mu} }\)

\(\ds \)

\(=\)

\(\ds \map \phi {x + t y} + \lambda + t \mu\)

\(\ds \)

\(=\)

\(\ds \paren {\map \phi x + \lambda} + t \paren {\map \phi y + \mu}\)

Definition of Linear Functional

\(\ds \)

\(=\)

\(\ds \map {\phi_+} {\tuple {x, \lambda} } + t \map {\phi_+} {\tuple {y, \mu} }\)

and so $\phi_+$ is linear.

To show that $\phi_+$ is a character, it remains to show that:

$\map {\phi_+} {\tuple {x, \lambda} \tuple {y, \mu} } = \map {\phi_+} {\tuple {x, \lambda} } \map {\phi_+} {\tuple {y, \lambda} }$

We have:

\(\ds \map {\phi_+} {\tuple {x, \lambda} \tuple {y, \mu} }\)

\(=\)

\(\ds \map {\phi_+} {\tuple {x y + \lambda y + \mu x, \lambda \mu} }\)

Definition of Unitization of Algebra over Field

\(\ds \)

\(=\)

\(\ds \map \phi {x y + \lambda y + \mu x} + \lambda \mu\)

\(\ds \)

\(=\)

\(\ds \map \phi x \map \phi y + \lambda \map \phi y + \mu \map \phi x + \lambda \mu\)

Definition of Character (Banach Algebra)

\(\ds \)

\(=\)

\(\ds \paren {\map \phi x + \lambda} \paren {\map \phi y + \mu}\)

\(\ds \)

\(=\)

\(\ds \map {\phi_+} {\tuple {x, \lambda} } \map {\phi_+} {\tuple {y, \mu} }\)

So $\phi_+$ is a character.

$\blacksquare$