nyuuzyou/wikis huggingface dataset:
Spectrum of Compact Linear Operator on Infinite-Dimensional Banach Space contains Zero

== Theorem ==
Let $X$ be an infinite-dimensional Banach space over $\C$.
Let $T : X \to X$ be a compact linear operator. 
Let $\map \sigma T$ be the spectrum of $T$.
Then $0 \in \map \sigma T$. 
That is, $T$ is not invertible as a bounded linear operator. 
== Proof ==
Suppose $0 \not \in \map \sigma T$. 
Then $T$ is invertible as a bounded linear operator with bounded inverse $T^{-1}$ so that: 
:$T T^{-1} = I$
From Left Composition of Compact Linear Transformation with Bounded Linear Transformation is Compact, this implies that $I$ is compact.
From Identity Operator is Compact iff Finite-Dimensional Normed Vector Space, we have that $X$ is finite-dimensional.
This is a contradiction, so we have $0 \in \map \sigma T$. 
== Sources ==
* $15.3$: Compact Linear Operators
Category:Compact Linear Transformations
Category:Spectra (Bounded Linear Operators)
Spectrum of Element in Maximal Commutative Subalgebra of Unital Banach Algebra

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra.
Let $C$ be a commutative subalgebra of $A$ that is maximal with respect to set inclusion.
Let $x \in C$.
Let $\map {\sigma_A} x$ and $\map {\sigma_C} x$ be the spectra of $x$ in $A$ and $C$ respectively.
Then:
:$\map {\sigma_C} x = \map {\sigma_A} x$
== Proof ==
Let $\map G A$ and $\map G C$ be the groups of units of $A$ and $C$ respectively. 
From Maximal Commutative Subalgebra of Unital Algebra is Unital, $C$ is unital.
From Spectrum of Element in Unital Subalgebra, we have $\map {\sigma_A} x \subseteq \map {\sigma_C} x$.
Now let $\lambda \in \C \setminus \map {\sigma_A} x$. 
Then $\lambda {\mathbf 1}_A - x \in \map G A$.
Let $z = \paren {\lambda {\mathbf 1}_A - x}^{-1}$.
We want to show that $z \in C$.
Let $y \in C$.
Since $y$ commutes with $\lambda {\mathbf 1}_A$ and $z$, we have:
:$z y \paren {\lambda {\mathbf 1}_A - x} z = z \paren {\lambda {\mathbf 1}_A - x} y z$
Since:
:$\paren {\lambda {\mathbf 1}_A - x} z = {\mathbf 1}_A$
and:
:$z \paren {\lambda {\mathbf 1}_A - x} = {\mathbf 1}_A$
we have $z y = y z$.
So $C \cup \set z$ is a set of commuting elements.
Let $\widetilde C$ be the subalgebra generated by $C$.
From Subalgebra Generated by Commuting Elements is Commutative, $\widetilde C$ is commutative.
We also have $C \subseteq \widetilde C$. 
Since $C$ maximal with respect to set inclusion, we have:
:$\widetilde C = C$
Hence $z \in C$.
So, we have $\lambda {\mathbf 1}_A x \in \map G C$.
Hence, we have $\lambda \in \C \setminus \map {\sigma_C} x$. 
So, we have:
:$\C \setminus \map {\sigma_A} x \subseteq \C \setminus \map {\sigma_C} x$
From Set Complement inverts Subsets, we have:
:$\map {\sigma_C} x \subseteq \map {\sigma_A} x$
Hence, we have:
:$\map {\sigma_C} x = \map {\sigma_A} x$
Category:Spectral Theory of Banach Algebras
Category:Spectra (Spectral Theory)
Spectrum of Element in Unital Subalgebra

== Theorem ==
Let $A$ be a unital algebra over $\C$. 
Let $B$ be a unital subalgebra of $A$.
Let $x \in B$.
Let $\map {\sigma_A} x$ and $\map {\sigma_B} x$ be the spectra of $x$ in $A$ and $B$ respectively. 
Then:
:$\map {\sigma_A} x \subseteq \map {\sigma_B} x$
=== Corollary ===
== Proof ==
Let $\map G A$ and $\map G B$ be the group of units of $A$ and $B$ respectively.
From Group of Units of Submonoid is Subgroup, we have:
:$\map G B \subseteq \map G A$
From Set Complement inverts Subsets, we have:
:$A \setminus \map G A \subseteq A \setminus \map G B$
Then, we have:
Category:Spectra (Spectral Theory)
Category:Unital Subalgebras
Category:Spectrum of Element in Unital Subalgebra
Spectrum of Element in Unital Subalgebra/Corollary

== Theorem ==
Let $A$ be a non-unital algebra over $\C$. 
Let $B$ be a subalgebra of $A$.
Let $x \in B$.
Let $\map {\sigma_A} x$ and $\map {\sigma_B} x$ be the spectra of $x$ in $A$ and $B$ respectively. 
Then:
:$\map {\sigma_A} x \subseteq \map {\sigma_B} x$
== Proof ==
Let $A_+$ and $B_+$ be the unitizations of $A$ and $B$ respectively.
From Unitization of Algebra over Field preserves Subalgebra Relation, we have:
:$B_+$ is a unital subalgebra of $A_+$.
From Spectrum of Element in Unital Subalgebra, we have that:
:$\map {\sigma_{A_+} } {\tuple {x, 0} } \subseteq \map {\sigma_{B_+} } {\tuple {x, 0} }$
From the definition of the spectrum in a non-unital algebra, we have:
:$\map {\sigma_{A_+} } {\tuple {x, 0} } = \map {\sigma_A} x$
and:
:$\map {\sigma_{B_+} } {\tuple {x, 0} } = \map {\sigma_B} x$
So, we have:
:$\map {\sigma_A} x \subseteq \map {\sigma_B} x$
Category:Spectrum of Element in Unital Subalgebra
Spectrum of Element in Unitization of Unital Algebra

== Theorem ==
Let $A$ be a unital algebra over $\C$.
Let $A_+$ be the unitization of $A$.
Let $a \in A$. 
Then:
:$\map {\sigma_{A_+} } {\tuple {a, 0} } = \map {\sigma_A} a \cup \set 0$
where $\sigma_{A_+}$ and $\sigma_A$ denote the spectrum in $A_+$ and $A$ respectively. 
== Proof ==
First, we note that $0 \in \map {\sigma_{A_+} } {\tuple {a, 0} }$.
Note that for each $\tuple {b, \lambda} \in A_+$, we have $\tuple {a, 0} \tuple {b, \lambda} = \tuple {a b + \lambda a, 0}$.
This cannot be equal to $\tuple { {\mathbf 0}_A, 1}$.
Hence $\tuple {a, 0}$ is not invertible and $0 \in \map {\sigma_{A_+} } {\tuple {a, 0} }$.
Now take $\lambda \in \C \setminus \set 0$. 
We show that $\lambda \tuple {0, 1} - \tuple {a, 0}$ is invertible in $A_+$ $\lambda {\mathbf 1}_A - a$ is invertible in $A$ for $\lambda \ne 0$. 
We have:
:$\lambda \tuple {0, 1} - \tuple {a, 0} = \tuple {-a, \lambda}$
Suppose that $\tuple {-a, \lambda}$ is invertible in $A_+$ with inverse $\tuple {b, \mu}$. 
Then:
:$\tuple {-a, \lambda} \tuple {b, \mu} = \tuple {-a b - \mu a + \lambda b, \lambda \mu} = \tuple { {\mathbf 0}_A, 1}$
and:
:$\tuple {b, \mu} \tuple {-a, \lambda} = \tuple {-b a - \mu a + \lambda b, \lambda \mu} = \tuple { {\mathbf 0}_A, 1}$
Hence $\lambda \mu = 1$ and $-a b - \mu a + \lambda b = -b a - \mu a + \lambda b = {\mathbf 0}_A$.
We then have:
:$\paren {\lambda {\mathbf 1}_A - a} \paren {b + \mu {\mathbf 1}_A} = \lambda b - a b + \lambda \mu {\mathbf 1}_A - \mu a = \paren {-a c - \mu a + \lambda b} + \paren {\lambda \mu} {\mathbf 1}_A$
as above we have $-a b - \mu a + \lambda b = {\mathbf 0}_A$ and $\lambda \mu = 1$ so:
:$\paren {\lambda {\mathbf 1}_A - a} \paren {b + \mu {\mathbf 1}_A} = {\mathbf 1}_A$
Similarly we have:
:$\paren {b + \mu {\mathbf 1}_A} \paren {\lambda {\mathbf 1}_A - a} = \lambda b - b a + \lambda \mu {\mathbf 1}_A - \mu a = \paren {-b a - \mu a + \lambda b} + \paren {\lambda \mu} {\mathbf 1}_A = {\mathbf 1}_A$
Hence if $\tuple {-a, \lambda}$ is invertible in $A_+$ then $\lambda {\mathbf 1}_A - a$ is invertible in $A$.
Now suppose that $\lambda {\mathbf 1}_A - a$ is invertible in $A$ with inverse $c$. 
We show that $\tuple {-a, \lambda}$ is invertible in $A_+$.
So:
:$\lambda c - a c = \lambda c - c a = {\mathbf 1}_A$
Hence:
and:
So $\tuple {-a, \lambda}$ is invertible with inverse $\tuple {c - \lambda^{-1} {\mathbf 1}_A, \lambda^{-1} }$. 
Hence for $\lambda \ne 0$, $\lambda \tuple {0, 1} - \tuple {a, 0}$ is invertible in $A_+$ $\lambda {\mathbf 1}_A - a$ is invertible in $A$.
Hence we have:
:$\map {\sigma_{A_+} } {\tuple {a, 0} } \setminus \set 0 = \map {\sigma_A} a$
Hence:
:$\paren {\map {\sigma_{A_+} } {\tuple {a, 0} } \setminus \set 0} \cup \set 0 = \map {\sigma_A} a \cup \set 0$ 
Since $0 \in \map {\sigma_{A_+} } {\tuple {a, 0} }$ as already established, we have:
:$\map {\sigma_{A_+} } {\tuple {a, 0} } = \map {\sigma_A} a \cup \set 0$
Category:Unitizations of Algebras over Fields
Category:Spectra (Spectral Theory)
Spectrum of Element of Banach Algebra is Bounded

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$. 
Let $x \in A$.
Let $\map {\sigma_A} x$ be the spectrum of $x$ in $A$.
Then $\map {\sigma_A} x$ is bounded, and in particular:
:$\cmod \lambda \le \norm x$ for all $\lambda \in \map {\sigma_A} x$
=== Corollary ===
== Proof ==
Suppose first that $\struct {A, \norm {\, \cdot \,} }$ is unital.
Let $\map G A$ be the group of units.
Let $\lambda \in \C$ be such that $\cmod \lambda > \norm x$.
Then from , we have:
:$\ds \norm {\frac x \lambda} < 1$
From Element of Unital Banach Algebra Close to Identity is Invertible:
:$\ds {\mathbf 1}_A - \frac x \lambda \in \map G A$
Hence:
:$\lambda {\mathbf 1}_A - x \in \map G A$
So, we have:
:$\set {\lambda \in \C : \cmod \lambda > \norm x} \subseteq \set {\lambda \in \C : \lambda {\mathbf 1}_A - x \in \map G A}$
so that:
:$\set {\lambda \in \C : \lambda {\mathbf 1}_A - x \not \in \map G A} \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm x}$
So we have:
:$\map {\sigma_A} x \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm x}$
in the case that $\struct {A, \norm {\, \cdot \,} }$ is unital.
Now suppose that $\struct {A, \norm {\, \cdot \,} }$ is not unital.
Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the normed unitization of $\struct {A, \norm {\, \cdot \,} }$.
From the definition of the spectrum in a non-unital algebra, we have:
:$\map {\sigma_A} x = \map {\sigma_{A_+} } {\tuple {x, 0} }$
From the unital case, we have that $\map {\sigma_{A_+} } {\tuple {x, 0} }$ is bounded with:
:$\cmod \lambda \le \norm {\tuple {x, 0} }_{A_+}$ for all $\lambda \in \map {\sigma_A} x$.
We have:
:$\norm {\tuple {x, 0} }_{A_+} = \norm x$
and so:
:$\cmod \lambda \le \norm x$ for all $\lambda \in \map {\sigma_A} x$.
Category:Spectra (Spectral Theory)
Category:Banach Algebras
Category:Spectrum of Element of Banach Algebra is Bounded
Spectrum of Element of Banach Algebra is Bounded/Corollary

== Corollary ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$. 
Let $x \in A$.
Let $\map {r_A} x$ be the spectral radius of $x$ in $A$.
Then:
:$\map {r_A} x \le \norm x$
== Proof ==
Let $\map {\sigma_A} x$ be the spectrum of $x$ in $A$.
From the definition of the spectral radius we have:
:$\ds \map {r_A} x = \sup_{\lambda \in \map {\sigma_A} x} \cmod \lambda$ 
From Spectrum of Element of Banach Algebra is Bounded, we have:
:$\cmod \lambda \le \norm x$ for each $\lambda \in \map {\sigma_A} x$. 
Hence we have:
:$\ds \map {r_A} x \le \norm x$
from the definition of supremum. 
Category:Spectrum of Element of Banach Algebra is Bounded
Spectrum of Element of Banach Algebra is Closed

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$. 
Let $x \in A$.
Let $\map {\sigma_A} x$ be the spectrum of $x$ in $A$.
Then $\map {\sigma_A} x$ is closed.
== Proof ==
 suppose that $A$ is unital, swapping $A$ for its unitization if necessary.
Let $\map {\rho_A} x$ be the resolvent set of $x$ in $A$.
From Resolvent Set of Element of Banach Algebra is Open, $\map {\rho_A} x$ is open.
From the definition of the spectrum, we have $\map {\sigma_A} x = \C \setminus \map {\rho_A} x$.
From the definition of a closed set, $\map {\sigma_A} x$ is therefore closed.
Category:Spectra (Spectral Theory)
Category:Spectral Theory of Banach Algebras
Spectrum of Element of Banach Algebra is Compact

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$. 
Let $x \in A$.
Let $\map {\sigma_A} x$ be the spectrum of $x$ in $A$.
Then $\map {\sigma_A} x$ is compact.
== Proof ==
The result follows immediately from:
:Spectrum of Element of Banach Algebra is Bounded
:Spectrum of Element of Banach Algebra is Closed
Category:Spectra (Spectral Theory)
Category:Banach Algebras
Spectrum of Element of Banach Algebra is Non-Empty

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$. 
Let $x \in A$.
Let $\map {\sigma_A} x$ be the spectrum of $x$ in $A$.
Then $\map {\sigma_A} x \ne \O$.
=== Corollary ===
== Proof ==
Suppose first that $\struct {A, \norm {\, \cdot \,} }$ is a unital Banach algebra.
Let $\map {\rho_A} x$ be the resolvent set of $x$ in $A$.
 that $\map {\sigma_A} x = \O$.
Then $\map {\rho_A} x = \C$.
Define $R : \C \to A$ by:
:$\map R \lambda = \paren {\lambda {\mathbf 1}_A - x}^{-1}$
for each $\lambda \in \C$.
From Resolvent Mapping is Analytic: Banach Algebra, $R$ is analytic.
With view to apply Liouville's Theorem (Complex Analysis): Banach Space, we show that $R$ is bounded.
Let $\lambda \in \C$ be such that $\cmod \lambda > \norm x + 1$.
Then $\cmod \lambda > \norm x$, and so:
:$\ds \norm {\frac x \lambda} < 1$
from .
From Element of Unital Banach Algebra Close to Identity is Invertible, we have:
:$\ds \norm {\paren { {\mathbf 1}_A - \frac x \lambda}^{-1} } \le \frac 1 {1 - \norm {\frac x \lambda} } = \frac {\cmod \lambda} {\cmod \lambda - \norm x}$
by .
Hence, we have:
:$\ds \norm {\paren {\lambda {\mathbf 1}_A - x}^{-1} } = \cmod \lambda^{-1} \norm {\paren { {\mathbf 1}_A - \frac x \lambda}^{-1} } \le \frac 1 {\cmod \lambda - \norm x} < 1$
From Resolvent Mapping is Continuous: Banach Algebra, $R$ is continuous.
From Continuous Function on Compact Space is Bounded, there exists $M > 0$ such that:
:$\norm {\paren {\lambda {\mathbf 1}_A - x}^{-1} } \le M$ for $\lambda \in \C$ with $\cmod \lambda \le \norm x + 1$.
Then:
:$\norm {\paren {\lambda {\mathbf 1}_A - x}^{-1} } \le \max \set {1, M}$ for all $\lambda \in \C$. 
From Liouville's Theorem (Complex Analysis): Banach Space, $R$ is constant. 
Then:
:$\paren {2 {\mathbf 1}_A - x}^{-1} = \paren { {\mathbf 1}_A - x}^{-1}$.
Note that for $y, z \in A$ we have:
:$y^{-1} = z^{-1}$ $y = z$.
So, we have:
:$2 {\mathbf 1}_A - x = {\mathbf 1}_A - x$
giving:
:${\mathbf 1}_A = {\mathbf 0}_A$
This is impossible since $\struct {A, \norm {\, \cdot \,} }$ is unital. 
Hence we have reached a contradiction and we have $\map {\sigma_A} x \ne \O$.
Suppose now that $\struct {A, \norm {\, \cdot \,} }$ is not unital.
Then, we have:
:$\map {\sigma_A} x = \map {\sigma_{A_+} } {\tuple {x, 0} }$
where $A_+$ is the normed unitization of $A$.
From the previous case, we have $\map {\sigma_{A_+} } {\tuple {x, 0} } \ne \O$ and hence $\map {\sigma_A} x \ne \O$.
Category:Spectral Theory of Banach Algebras
Category:Spectrum of Element of Banach Algebra is Non-Empty
Spectrum of Element of Banach Algebra is Non-Empty/Corollary

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\C$. 
Let $x \in A$.
Let $\map {\sigma_A} x$ be the spectrum of $x$ in $A$.
Then $\map {\sigma_A} x \ne \O$.
== Proof ==
Suppose first that $\struct {A, \norm {\, \cdot \,} }$ is unital. 
From Completion Theorem (Normed Algebra), there exists a unital Banach algebra $\struct {\widetilde A, \widetilde {\norm {\, \cdot \,} } }$ and an isometric algebra homomorphism $\phi : A \to \widetilde A$ where:
:${\mathbf 1}_{\widetilde A} = \map \phi { {\mathbf 1}_A}$
From Spectrum of Element of Banach Algebra is Non-Empty, we have:
:$\map {\sigma_{\widetilde A} } {\map \phi x} \ne \O$
From Spectrum of Element in Unital Subalgebra, we have:
:$\map {\sigma_{\widetilde A} } {\map \phi x} \subseteq \map {\sigma_{\phi \sqbrk A} } {\map \phi x}$
so:
:$\map {\sigma_{\phi \sqbrk A} } {\map \phi x} \ne \O$
To finish we show that:
:$\map {\sigma_{\phi \sqbrk A} } {\map \phi x} = \map {\sigma_A} x$
We have $\lambda \in \map {\sigma_{\phi \sqbrk A} } {\map \phi x}$ $\lambda \map \phi { {\mathbf 1}_A} - \map \phi x \in \map G {\phi \sqbrk A}$ where $\map G {\phi \sqbrk A}$ is the group of units of $\phi \sqbrk A$.
Since $\phi$ is linear, we have $\lambda \in \map {\sigma_{\phi \sqbrk A} } {\map \phi x}$ $\map \phi {\lambda {\mathbf 1}_A - x} \in \map G {\phi \sqbrk A}$.
This is equivalent to the existence of $z \in A$ such that $\map \phi {\lambda {\mathbf 1}_A - x} \map \phi z = \map \phi z \map \phi {\lambda {\mathbf 1}_A - x} = \map \phi { {\mathbf 1}_A}$. 
Since $\phi$ is an algebra homomorphism, this is equivalent to:
:$\map \phi {\paren {\lambda {\mathbf 1}_A - x} z} = \map \phi {z \paren {\lambda {\mathbf 1}_A - x} } = \map \phi { {\mathbf 1}_A}$
Since $\phi$ is a bijection, this is equivalent to:
:$\paren {\lambda {\mathbf 1}_A - x} z = z \paren {\lambda {\mathbf 1}_A - x} = {\mathbf 1}_A$
for $z \in A$.
That is, equivalent to $\lambda {\mathbf 1}_A - x \in \map G A$.
Hence equivalent to $\lambda \in \map {\sigma_A} x$. 
So:
:$\map {\sigma_A} x = \map {\sigma_{\phi \sqbrk A} } {\map \phi x} \ne \O$
Suppose now that $\struct {A, \norm {\, \cdot \,} }$ is not unital.
Then, we have:
:$\map {\sigma_A} x = \map {\sigma_{A_+} } {\tuple {x, 0} }$
where $A_+$ is the normed unitization of $A$.
From the previous case, we have $\map {\sigma_{A_+} } {\tuple {x, 0} } \ne \O$ and hence $\map {\sigma_A} x \ne \O$.
Category:Spectrum of Element of Banach Algebra is Non-Empty
Spectrum of Element of Space of Continuous Functions Vanishing at Infinity

== Theorem ==
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:
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:
:$\map {\sigma_A} f = f \sqbrk X \cup \set 0$ 
Category:Banach Algebras
Spectrum of Element of Space of Continuous Functions on Compact Hausdorff Space

== Theorem ==
Let $K$ be a locally compact Hausdorff space.
Let $A = \map \CC K$ be the space of continuous functions on $K$ considered as a Banach algebra.
Let $f \in \map \CC K$.
Then:
:$\map {\sigma_A} f = f \sqbrk K$
where $\sigma_A$ denotes spectrum in $A$. 
== Proof ==
We first show that:
:$f \sqbrk K \subseteq \map {\sigma_A} f$
Let $\lambda \in f \sqbrk K$. 
Then there exists $x \in K$ such that $\lambda = \map f x$.
Then $\map {\paren {\lambda - f} } x = 0$.
Hence there cannot exist $g \in \map \CC K$ such that:
:$\map g x \map {\paren {\lambda - f} } x = 1$
So $\lambda - f$ is not invertible and $\lambda \in \map {\sigma_A} f$.
We now show:
:$\C \setminus f \sqbrk K \subseteq \map {\rho_A} f$
where $\rho_A$ denotes the resolvent set in $A$. 
Let $\lambda \in \C \setminus f \sqbrk K$. 
Then $\map f x \ne \lambda$ for all $x \in X$.
So we have $\map {\paren {\lambda - f} } x \ne 0$ for each $x \in X$. 
Hence from Quotient Rule for Continuous Complex Functions, we have that:
:$\ds \frac 1 {\lambda - f} \in \map \CC K$
Hence we have $\lambda \in \map {\rho_A} f$. 
We conclude that:
:$\map {\sigma_A} f = f \sqbrk K$
Category:Banach Algebras
Spectrum of Element of Unital C*-Subalgebra of Unital C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra with identity element ${\mathbf 1}_A$.
Let $B \subseteq A$ be a unital $\text C^\ast$-subalgebra of $A$. 
Let $\sigma_A$ and $\sigma_B$ be the spectrum in $A$ and $B$ respectively. 
Let $x \in B$. 
Then we have:
:$\map {\sigma_A} x = \map {\sigma_B} x$ 
== Proof ==
First take $x$ to be Hermitian. 
Let $B' \subseteq B$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, x}$. 
By C*-Algebra Generated by Commutative Self-Adjoint Set is Commutative, $B'$ is commutative.
From Spectrum of Hermitian Element in Unital C*-Algebra is Real, we have $\map {\sigma_{B'} } x \subseteq \R$. 
From Spectrum of Element in Unital Subalgebra, we therefore have $\map {\sigma_B} x \subseteq \map {\sigma_{B'} } x \subseteq \R$. 
From Interior of Set of Real Numbers in Complex Numbers is Empty: Corollary, we have $\map {\sigma_B} x^\circ = \O$ where $\map {\sigma_B} x^\circ$ denotes the interior of $\map {\sigma_B} x$. 
From the definition of boundary, we have that $\partial \map {\sigma_B} x = \map {\sigma_B} x^- \setminus \map {\sigma_B} x^\circ$.
Since $\map {\sigma_B} x$ is closed from Spectrum of Element of Banach Algebra is Closed and $\map {\sigma_B} x^\circ = \O$, we then have that $\map {\sigma_B} x = \partial \map {\sigma_B} x$.
Then, from Boundary of Spectrum of Element in Subalgebra of Unital Banach Algebra we have:
:$\map {\sigma_B} x = \partial \map {\sigma_B} x \subseteq \partial \map {\sigma_A} x$
From Set is Closed iff it Contains its Boundary, we have $\partial \map {\sigma_A} x \subseteq \map {\sigma_A} x$. 
So $\map {\sigma_B} x \subseteq \map {\sigma_A} x$. 
From Spectrum of Element in Unital Subalgebra we have $\map {\sigma_A} x \subseteq \map {\sigma_B} x$. 
Hence we conclude $\map {\sigma_A} x = \map {\sigma_B} x$ in this case. 
In particular, $0 \in \map {\sigma_A} x$ $0 \in \map {\sigma_B} x$. 
Hence if $x$ is Hermitian, it is invertible in $B$ it is invertible in $A$.
Let $b \in B$. 
Suppose that $b$ is invertible in $A$. 
We show that it is invertible in $B$. 
Take $a \in A$ such that $a b = b a = {\mathbf 1}_A$.
From Inverse of Star of Element in Unital *-Algebra, we have $b^\ast a^\ast = a^\ast b^\ast = {\mathbf 1}_A$. 
We therefore have:
and:
Hence $b b^\ast$ is invertible in $A$.
From Product of Element in *-Star Algebra with its Star is Hermitian, $b b^\ast$ is Hermitian.
Hence from what we have already shown, $b b^\ast$ is invertible in $B$. 
So there exists $c \in B$ such that $b b^\ast c = {\mathbf 1}_A$. 
Then:
Since $b^\ast \in B$ and $c \in B$, we have $a \in B$. 
So $b$ is invertible in $B$. 
Clearly conversely if $b$ is invertible in $B$, then it is invertible in $A$.
Hence we have shown that if $b \in B$, then $b$ is invertible in $B$ it is invertible in $A$.
In particular, if $x \in B$ then ${\mathbf 1}_A - x$ is invertible in $B$ it is invertible in $A$.
This is precisely the statement $\map {\sigma_A} x = \map {\sigma_B} x$. 
== Sources ==
* : $2.1.11$: Theorem
Category:C*-Algebras