Characterization of Character on Banach Algebra

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Theorem

Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$.

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

Let $\phi : A_+ \to \C$ be a character.


Then either:

$(1) \quad \map \phi {x, \lambda} = \lambda$ for each $\tuple {x, \lambda} \in A_+$
$(2) \quad $ there exists a character $\widetilde \phi$ on $A$ such that $\map \phi {x, \lambda} = \map {\widetilde \phi} x + \lambda$ for each $\tuple {x, \lambda} \in A_+$.

Conversely, every map of the form given in $(2)$ is a character.


Proof

Note that since $\phi$ is linear, we have:

$\map \phi {x, \lambda} = \map \phi {x, 0} + \lambda \map \phi {0, 1}$

Then from Character on Unital Banach Algebra is Unital Algebra Homomorphism, we have:

$\map \phi {0, 1} = 1$

so that:

$\map \phi {x, \lambda} = \map \phi {x, 0} + \lambda$


We first show that:

$\map \phi {x, \lambda} = \lambda$

defines a character.

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

Then we have:

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

hence $\phi$ is linear.

We also have:

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

hence $\phi$ is a character.

In this case $\map \phi {x, 0} = 0$ for each $x \in A$.


Now suppose that $\map \phi {x, 0} \ne 0$ for each $x \in A$.

We show that:

$\map {\widetilde \phi} x = \map \phi {x, 0}$

is a character on $A$.

For $\tuple {x, \lambda}, \tuple {y, \mu} \in A_+$ and $t \in \C$, we have:

\(\ds \map {\widetilde \phi} x + t \map {\widetilde \phi} y\) \(=\) \(\ds \map \phi {x, 0} + t \map \phi {y, 0}\)
\(\ds \) \(=\) \(\ds \map \phi {x + t y, 0}\)
\(\ds \) \(=\) \(\ds \map {\widetilde \phi} {x + t y}\)

and:

\(\ds \map {\widetilde \phi} x \map {\widetilde \phi} y\) \(=\) \(\ds \map \phi {x, 0} \map \phi {y, 0}\)
\(\ds \) \(=\) \(\ds \map \phi {x y, 0}\)
\(\ds \) \(=\) \(\ds \map {\widetilde \phi} {x y}\)

So $\widetilde \phi$ is a character on $A$ and $\phi$ has the form:

$\map \phi {x, \lambda} = \map {\widetilde \phi} x + \lambda$

where $\widetilde \phi$ is a character on $A$.

So every character on $A_+$ has the form in $(2)$.


Conversely, let $\widetilde \phi$ be a character on $A$.

Define:

$\map \phi {x, \lambda} = \map {\widetilde \phi} x + \lambda$

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

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

Then we have:

\(\ds \map \phi {x + t y, \lambda + t \mu}\) \(=\) \(\ds \map {\widetilde \phi} {x + t y} + \paren {\lambda + t \mu}\)
\(\ds \) \(=\) \(\ds \paren {\map {\widetilde \phi} x + \lambda} + t \paren {\map {\widetilde \phi} y + \mu}\) since $\widetilde \phi$ is linear
\(\ds \) \(=\) \(\ds \map \phi {x, \lambda} + t \map \phi {y, \mu}\)

and:

\(\ds \map \phi {x, \lambda} \map \phi {y, \mu}\) \(=\) \(\ds \paren {\map {\widetilde \phi} x + \lambda} \paren {\map {\widetilde \phi} y + \mu}\)
\(\ds \) \(=\) \(\ds \map {\widetilde \phi} x \map {\widetilde \phi} y + \lambda \map {\widetilde \phi} y + \mu \map {\widetilde \phi} x + \lambda \mu\)
\(\ds \) \(=\) \(\ds \map {\widetilde \phi} {x y + \lambda y + \mu x} + \lambda \mu\) Definition of Character (Banach Algebra)
\(\ds \) \(=\) \(\ds \map \phi {x y + \lambda y + \mu x, \lambda \mu}\)
\(\ds \) \(=\) \(\ds \map \phi {\tuple {x, \lambda} \tuple {y, \mu} }\) Definition of Unitization of Algebra over Field

So $\phi$ is a character.

$\blacksquare$

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