Character on Banach Algebra is Continuous

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

Let $\phi : A \to \C$ be a character on $A$.

Then $\phi$ is continuous and further:
 * $\norm \phi_{A^\ast} \le 1$

Proof
Let $\map G A$ be the group of units of $A$.

Suppose first that $\struct {A, \norm {\, \cdot \,} }$ is unital.

We show that:
 * $\cmod {\map \phi x} \le \norm x$ for each $x \in A$.

From Continuity of Linear Functionals, we will then have that $\phi$ is continuous with $\norm \phi_{A^\ast} \le 1$.

that $x \in A$ is such that:
 * $\cmod {\map \phi x} > \norm x$

Since then $\map \phi x \ne 0$, we have $x \ne {\mathbf 0}_A$.

From, we have:
 * $\ds \norm {\frac x {\map \phi x} } < 1$

From Element of Unital Banach Algebra Close to Identity is Invertible, we have that:
 * $\ds {\mathbf 1}_A - \frac x {\map \phi x} \in \map G A$

So there exists $z \in A$ such that:
 * $\ds z \paren { {\mathbf 1}_A - \frac x {\map \phi x} } = {\mathbf 1}_A$

From Character on Unital Banach Algebra is Unital Algebra Homomorphism, we have $\map \phi { {\mathbf 1}_A} = 1$.

So, we have:

This is a contradiction since $A \ne \set { {\mathbf 0}_A}$.

Hence we have:
 * $\norm {\map \phi x} \le \norm x$ for all $x \in A$.

So we have that $\phi$ is continuous with:
 * $\norm \phi_{A^\ast} \le 1$

Now suppose that $\struct {A, \norm {\, \cdot \,} }$ is not unital.

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

Define $\phi_+ : A_+ \to \C$ by:
 * $\map {\phi_+} {\tuple {x, \lambda} } = \map \phi x + \lambda$

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

From Character on Non-Unital Banach Algebra induces Character on Unitization, $\phi_+$ is a character on $A_+$.

Further:
 * $\map {\phi_+} {\tuple {x, 0} } = \map \phi x$ for each $x \in A$.

From the unital case, we have that $\phi_+$ is continuous with $\norm {\phi_+}_{A_+^\ast} \le 1$.

Then, we have:

So $\phi$ is continuous with $\norm \phi_{A^\ast} \le 1$.