Character on Banach Algebra is Surjective

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 surjective.

Proof 1

From Image of Submodule under Linear Transformation is Submodule, $\phi \sqbrk A$ is a vector subspace of $\C$.

From Dimension of Proper Subspace is Less Than its Superspace, we have:

$\dim \phi \sqbrk A \le \dim \C = 1$

and so we either have $\phi \sqbrk A = \set 0$ or $\phi \sqbrk A = \C$.

Since $\phi \ne 0$ by the definition of a character, we have $\phi \sqbrk A = \C$.

$\blacksquare$

Proof 2

As $\phi$ is non-zero, there exists an $x_0 \in A$ such that:

$\map \phi {x_0} \in \C \setminus \set 0$

Thus, for each $a \in \C$:

\(\ds \frac a {\map \phi {x_0} } x_0\)

\(\in\)

\(\ds A\)

as $A$ is a $\C$-algebra

and:

\(\ds \map \phi {\frac a {\map \phi {x_0} } x_0}\)

\(=\)

\(\ds \frac a {\map \phi {x_0} } \map \phi {x_0}\)

as $\phi$ is a $\C$-algebra homomorphism

\(\ds \)

\(=\)

\(\ds a\)

$\blacksquare$