Character on Banach Algebra is Surjective
Jump to navigation
Jump to search
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds a\) |
$\blacksquare$