Theorem: Spectrum of Element of Space of Continuous Functions Vanishing at Infinity

Let $X$ be a locally compact Hausdorff space.
Let $A = \map {\CC_0} X$ be the Banach algebra of continuous complex-valued functions vanishing at infinity.
Let $f \in \map {\CC_0} X$.
Then:
:$\map {\sigma_A} f = f \sqbrk X \cup \set 0$ 
where $\sigma_A$ denotes the spectrum in $A$.

Proof:
Let $\map {\CC_b} X$ be the Banach algebra of bounded continuous functions on $X$ valued in $\C$.
Define $\mathbf 1 : X \to \C$ by:
:$\map {\mathbf 1} x = 1$ for each $x \in X$.
Then $\mathbf 1$ is the identity element for $\map {\CC_b} X$. 
Let $A_+$ be the unitization of $A$.
Let $A \oplus \C \mathbf 1$ be the internal direct sum of $\C$ and $\C \mathbf 1$ in $\map {\CC_b} X$.
Define $\phi : A_+ \to A \oplus \C \mathbf 1$ by:
:$\map \phi {f, \lambda} = f + \lambda$
for each $\tuple {f, \lambda} \in A_+$, where:
:$\map {\paren {f + \lambda} } x = \map f x + \lambda$
for each $x \in X$. 
From Unitization of Non-Unital Subalgebra of Unital Algebra is Isomorphic to Unital Subalgebra, $\phi$ is a unital algebra isomorphism.
We then have:
{{begin-eqn}}
{{eqn	| l = \map {\sigma_A} f
		| r = \map {\sigma_{A_+} } {\tuple {f, 0} }
		| c = {{Defof|Spectrum (Spectral Theory)|Non-Unital Algebra}}
}}
{{eqn	| r = \map {\sigma_{A \oplus \C \mathbf 1} } {\map \phi {f, 0} }
}}
{{eqn	| r = \map {\sigma_{A \oplus \C \mathbf 1} } f
}}
{{end-eqn}}
We want to show:
:$\map {\sigma_{A \oplus \C \mathbf 1} } f = f \sqbrk X \cup \set 0$
We first show that:
:$f \sqbrk X \cup \set 0 \subseteq \map {\sigma_{A \oplus \C \mathbf 1} } f$
From Reciprocal of Continuous Complex-Valued Function Vanishing at Infinity does not Vanish at Infinity, we have $0 \in \map {\sigma_{A \oplus \C \mathbf 1} } f$.
If $\lambda \in f \sqbrk X$, then there exists $x \in X$ such that $\lambda = \map f x$. 
Then $\map {\paren {\lambda - f} } x = 0$. 
So $\lambda - f$ cannot have an inverse, and $\lambda \in \map {\sigma_{A \oplus \C \mathbf 1} } f$.
Hence we have:
:$f \sqbrk X \cup \set 0 \subseteq \map {\sigma_{A \oplus \C \mathbf 1} } f$
We now show that:
:$\map {\sigma_{A \oplus \C \mathbf 1} } f \subseteq f \sqbrk X \cup \set 0$
For this, we show that if $\lambda \in \C \setminus \paren {f \sqbrk X \cup \set 0}$, then $\lambda - f$ has an inverse in $A \oplus \C \mathbf 1$.
Let $\lambda \in \C \setminus \paren {f \sqbrk X \cup \set 0}$.
We show that:
:$\ds \frac 1 {\lambda - f} - \frac 1 \lambda$ vanishes at infinity.
Let $\epsilon > 0$. 
From Quotient Rule for Continuous Complex Functions, there exists $\delta > 0$ such that whenever $\cmod w < \delta$, we have:
:$\ds \cmod {\frac 1 {\lambda - w} - \frac 1 \lambda} < \epsilon$ 
Since $f$ vanishes at infinity, there exists a compact $K \subseteq X$ such that:
:$\cmod {\map f x} < \delta$ for each $x \in X \setminus K$.
Then:
:$\ds \cmod {\frac 1 {\lambda - \map f x} - \frac 1 \lambda} < \epsilon$ for $x \in X \setminus K$. 
Hence:
:$\ds \frac 1 {\lambda - f} - \frac 1 \lambda \in \map {\CC_0} X$
Then, we have:
:$\ds \frac 1 {\lambda - f} = \paren {\frac 1 {\lambda - f} - \frac 1 \lambda} + \frac 1 \lambda \in A \oplus \C \mathbf 1$
Hence $\lambda - f$ has an inverse in $A \oplus \C \mathbf 1$.
So we have $\lambda \not \in \map {\sigma_A} f$.
We conclude:
:$\map {\sigma_A} f = f \sqbrk X \cup \set 0$ 
{{qed}}
Category:Banach Algebras
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