avewright/proofwiki-math huggingface dataset:
Theorem: Norm on C*-Algebra is Unique

Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$. 
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be algebra norms on $\struct {A, \ast}$ such that:
:$\struct {A, \ast, \norm {\, \cdot \,}_1}$ and $\struct {A, \ast, \norm {\, \cdot \,}_2}$ are $\text C^\ast$-algebras.
Then:
:$\norm x_1 = \norm x_2$
for each $x \in A$.

Proof:
Note that the spectrum of $x \in A$ is defined independently of the algebra norm on $A$, depending only on the invertible elements of $A$.
With that:
{{begin-eqn}}
{{eqn	| l = \norm x_1^2
		| r = \norm {x x^\ast}_1
		| c = {{Defof|C*-Algebra}}
}}
{{eqn	| r = \map {r_A} {x x^\ast}
		| c = Product of Element in *-Star Algebra with its Star is Hermitian, Spectral Radius of Normal Element of C*-Algebra Equal to Norm
}}
{{eqn	| r = \norm {x x^\ast}_2
		| c = Product of Element in *-Star Algebra with its Star is Hermitian, Spectral Radius of Normal Element of C*-Algebra Equal to Norm
}}
{{eqn	| r = \norm x_2^2
}}
{{end-eqn}}
Since both $\norm x_1$ and $\norm x_2$ are non-negative, it follows that $\norm x_1 = \norm x_2$.
{{qed}}
Category:C*-Algebras
ruh61yjuqsxmfnb22bhl6mfjprjnzfc
Theorem: Norm of Element of C*-Algebra as Supremum over Closed Unit Ball

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $c \in A$.
Then:
:$\ds \norm c = \sup_{\norm b \le 1} \norm {c b} = \sup_{\norm b \le 1} \norm {b c}$

Proof:
Note that if $c = {\mathbf 0}_A$, we have $\norm c = 0$ and:
:$\norm {c b} = \norm {b c} = \norm { {\mathbf 0}_A} = 0$
for each $b \in A$.
Hence we have the theorem in the case $c = {\mathbf 0}_A$.
Now take $c \ne {\mathbf 0}_A$.
From the definition of an algebra norm, we have:
:$\norm {c b} \le \norm c \norm b$
and:
:$\norm {b c} \le \norm b \norm c$
Hence we have:
:$\ds \norm c \le \sup_{\norm b \le 1} \norm {c b}$
and:
:$\ds \norm c \le \sup_{\norm b \le 1} \norm {b c}$
From the definition of a Banach $\ast$-algebra, we have:
:$\norm c = \norm {c^\ast}$
Note that:
:$\ds \norm {\frac {c^\ast} {\norm c} } \le 1$
and:
:$\norm {c b} = \frac {\norm {c c^\ast} } {\norm c}$
By the $\text C^\ast$ identity, we have:
:$\ds \frac {\norm {c c^\ast} } {\norm c} = \frac {\norm c^2} {\norm c} = \norm c$
So we have:
:$\ds \sup_{\norm b \le 1} \norm {c b} \le \norm c$ 
We conclude:
:$\ds \norm c = \sup_{\norm b \le 1} \norm {c b}$
Similarly we have:
:$\ds \frac {\norm {c^\ast c} } {\norm c} = \frac {\norm {c c^\ast} } {\norm c} = \frac {\norm c^2} {\norm c} = \norm c$
by the $\text C^\ast$ identity.
We then also obtain:
:$\ds \sup_{\norm b \le 1} \norm {b c} \le \norm c$
and hence:
:$\ds \norm c = \sup_{\norm b \le 1} \norm {b c}$
{{qed}}
Theorem: Norms of Double Centralizer of C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $\tuple {L, R}$ be a double centralizer of $A$.
Then $\norm L_{\map B A} = \norm R_{\map B A}$, where $\norm L_{\map B A}$ and $\norm R_{\map B A}$ denotes the norm of a bounded linear transformation.

Proof:
For each $a, b \in A$, we have:
{{begin-eqn}}
{{eqn	| l = \norm {a \map L b}
		| r = \norm {\map R a b}
		| c = {{Defof|Double Centralizer of C*-Algebra}}
}}
{{eqn	| o = \le
		| r = \norm {\map R a} \norm b
		| c = {{Defof|Norm on Algebra}} 
}}
{{eqn	| o = \le
		| r = \norm R_{\map B A} \norm a \norm b
		| c = Fundamental Property of Norm on Bounded Linear Transformation
}}
{{end-eqn}}
Hence by Norm of Element of C*-Algebra as Supremum over Closed Unit Ball, we have:
:$\ds \norm {\map L b} = \sup_{\norm a \le 1} \norm {a \map L b} \le \norm R_{\map B A} \norm b$ 
By the definition of the norm of a bounded linear transformation, we have:
:$\norm L_{\map B A} \le \norm R_{\map B A}$
For each $a, b \in A$, we also have:
{{begin-eqn}}
{{eqn	| l = \norm {\map R a b}
		| r = \norm {a \map L b}
		| c = {{Defof|Double Centralizer of C*-Algebra}}
}}
{{eqn	| o = \le
		| r = \norm a \norm {\map L b}
		| c = {{Defof|Norm on Algebra}} 
}}
{{eqn	| o = \le
		| r = \norm L_{\map B A} \norm a \norm b
		| c = Fundamental Property of Norm on Bounded Linear Transformation
}}
{{end-eqn}}
Hence by Norm of Element of C*-Algebra as Supremum over Closed Unit Ball, we have:
:$\ds \norm {\map R a} = \sup_{\norm a \le 1} \norm {\map R a b} \le \norm L_{\map B A} \norm b$ 
By the definition of the norm of a bounded linear transformation, we have:
:$\norm R_{\map B A} \le \norm L_{\map B A}$
We conclude:
:$\norm L_{\map B A} = \norm R_{\map B A}$
{{qed}}
Theorem: Double Centralizer Generated by Element of C*-Algebra is Double Centralizer

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $c \in A$.
Define $L_c : A \to A$ and $R_c : A \to A$ by:
:$\map {L_c} a = c a$
and:
:$\map {R_c} a = a c$
for each $a \in A$. 
Then $\tuple {L_c, R_c}$ is a double centralizer of $A$.

Proof:
First, for each $a \in A$ we have:
:$\norm {\map {L_c} a} \le \norm c \norm a$
and:
:$\norm {\map {R_c} a} \le \norm a \norm c$
by the definition of an algebra norm.
Now, for $a, b \in A$ we have:
{{begin-eqn}}
{{eqn	| l = \map {L_c} {a b}
		| r = c a b
}}
{{eqn	| r = \paren {c a} b
}}
{{eqn	| r = \map {L_c} a b
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn	| l = \map {R_c} {a b}
		| r = a b c
}}
{{eqn	| r = a \paren {b c}
}}
{{eqn	| r = a \map {R_c} b
}}
{{end-eqn}}
Finally, we have:
{{begin-eqn}}
{{eqn	| l = \map {R_c} a b
		| r = a c b
}}
{{eqn	| r = a \paren {c b}
}}
{{eqn	| r = a \map {L_c} b
}}
{{end-eqn}}
Hence $\tuple {L_c, R_c}$ is a double centralizer of $A$.
{{qed}}
Theorem: Norm of Double Centralizer Generated by Element of C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $c \in A$.
Let $\tuple {L_c, R_c}$ be the double centralizer generated by $c$.
Then:
:$\norm {L_c}_{\map B A} = \norm {R_c}_{\map B A} = \norm c$
where $\norm {L_c}_{\map B A}$ and $\norm {R_c}_{\map B A}$ denotes the norm of a bounded linear transformation.

Proof:
From Norm of Element of C*-Algebra as Supremum over Closed Unit Ball we have:
:$\ds \norm c = \sup_{\norm b \le 1} \norm {c b} = \sup_{\norm b \le 1} \norm {b c}$
From the definition of the double centralizer generated by $c$, we have:
:$\ds \norm c = \sup_{\norm b \le 1} \norm {\map {L_c} b} = \sup_{\norm b \le 1} \norm {\map {R_c} b}$
By the definition of the norm of a bounded linear transformation, we then have:
:$\norm c = \norm {L_c}_{\map B A} = \norm {R_c}_{\map B A}$
{{qed}}
Theorem: Sufficient Condition for C* Identity

Let $\tuple {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$.
Let $\ast$ be an involution on $A$ such that:
:$\norm a^2 \le \norm {a a^\ast}$
Then $\tuple {A, \ast, \norm {\, \cdot \,} }$ is a $\text C^\ast$-algebra.

Proof:
Let $a \in A$.
From Zero Vector in *-Algebra is Hermitian and {{NormAxiomVector|1}}, we have:
:$\norm a^2 = \norm {a a^\ast}$
for $a = {\mathbf 0}_A$.
Now take $a \ne {\mathbf 0}_A$ so that $\norm a \ne 0$. 
From the definition of an algebra norm:
:$\norm {a a^\ast} \le \norm a \norm {a^\ast}$
Hence:
:$\norm a^2 \le \norm a \norm {a^\ast}$ 
Dividing by $\norm a$ we have:
:$\norm a \le \norm {a^\ast}$
for each $a \in A \setminus \set { {\mathbf 0}_A}$.
Replacing $a$ with $a^\ast$ we get:
:$\norm {a^\ast} \le \norm {a^{\ast \ast} } = \norm a$
Hence we have:
:$\norm a = \norm {a^\ast}$
Hence from:
:$\norm {a a^\ast} \le \norm a \norm {a^\ast}$
we have:
:$\norm {a a^\ast} \le \norm a^2$
Then, we have:
:$\norm a^2 \le \norm {a a^\ast} \le \norm a^2$
giving that:
:$\norm a^2 = \norm {a a^\ast}$
Hence the $C^\ast$ identity holds, so $\tuple {A, \ast, \norm {\, \cdot \,} }$ is a $\text C^\ast$-algebra.
{{qed}}
Category:C*-Algebras
2nxmk58ewbyxj1lhowd3co55rvbmjob
Theorem: Multiplication on Normed Algebra is Continuous

Let $\GF \in \set {\R, \C}$. 
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$.
Let $\struct {A^2, \norm {\, \cdot \,}_{A^2} }$ be the direct product of $A$ with itself equipped with the direct product norm.
Define $m : A^2 \to A$ by:
:$\map m {a, b} = a b$
for each $\tuple {a, b} \in A^2$. 
Then $m$ is continuous.

Proof:
Fix $\tuple {a, b} \in A^2$. 
Let $\tuple {a', b'} \in A^2$. 
Then we have:
{{begin-eqn}}
{{eqn	| l = \norm {\map m {a, b} - \map m {a', b'} }
		| r = \norm {a b - a' b'}
}}
{{eqn	| r = \norm {a b - a' b + a' b - a' b'}
}}
{{eqn	| r = \norm {b \paren {a - a'} + a' \paren {b - b'} }
}}
{{eqn	| o = \le
		| r = \norm {b \paren {a - a'} } + \norm {a' \paren {b - b'} }
		| c = {{NormAxiomVector|3}}
}}
{{eqn	| o = \le
		| r = \norm b \norm {a - a'} + \norm {a'} \norm {b - b'}
}}
{{end-eqn}}
Let $\epsilon > 0$.
Consider first the case of $b = {\mathbf 0}_A$. 
Take $\norm {a - a'} \le 1$. 
Then by the Reverse Triangle Inequality: Normed Vector Space, we have $\norm {a'} \le \norm a + 1$.
Hence, by taking: 
:$\ds \norm {b - b'} < \frac \epsilon {\norm a + 1}$ 
we have:
:$\norm b \norm {a - a'} + \norm {a'} \norm {b - b'} < \epsilon$
So by taking:
:$\ds \norm {\tuple {a, b} - \tuple {a', b'} }_{A^2} < \min \set {1, \frac \epsilon {\norm a + 1} }$
we have:
:$\norm {\map m {a, b} - \map m {a', b'} }_{A^2} < \epsilon$ 
So $m$ is continuous at $\tuple {a, {\mathbf 0}_A}$ for each $a \in A$.
Now consider $b \ne {\mathbf 0}_A$. 
If we take:
:$\ds \norm {a - a'} < \frac \epsilon {2 \norm b}$
then by the Reverse Triangle Inequality: Normed Vector Space, we have:
:$\ds \norm {a'} < \norm a + \frac \epsilon {2 \norm b}$
Then we can take:
:$\ds \norm {b - b'} < \frac \epsilon 2 \paren {\norm a + \frac \epsilon {2 \norm b} }^{-1}$
to obtain:
:$\ds \norm b \norm {a - a'} + \norm {a'} \norm {b - b'} < \norm b \paren {\frac \epsilon {2 \norm b} } + \frac \epsilon 2 \paren {\norm a + \frac \epsilon {2 \norm b} } \paren {\norm a + \frac \epsilon {2 \norm b} }^{-1} = \epsilon$
To conclude, by taking:
:$\ds \norm {\tuple {a, b} - \tuple {a', b'} }_{A^2} < \min \set {\frac \epsilon {2 \norm b}, \frac \epsilon 2 \paren {\norm a + \frac \epsilon {2 \norm b} }^{-1} }$
we have:
:$\norm {\map m {a, b} - \map m {a', b'} }_{A^2} < \epsilon$ 
So $m$ is continuous at $\tuple {a, b}$ for each $a \in A$, $b \ne {\mathbf 0}_A$.
Hence $m$ is continuous at $\tuple {a, b}$ for each $\tuple {a, b} \in A$.
Hence $m$ is continuous.
{{qed}}
Category:Normed Algebras
4rdv0yamy36faphzt5vpb69d1406tqs
Theorem: Multiplier Algebra is Unital C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra. 
Let $\map M A$ be the set of double centralizers of $A$. 
Define:
:$\tuple {L_1, R_1} +_{\map M A} \tuple {L_2, R_2} = \tuple {L_1 +_{\map B A} L_2, R_1 +_{\map B A} R_2}$
:$\lambda \circ_{\map M A} \tuple {L_1, R_1} = \tuple {\lambda \circ_{\map B A} L_1, \lambda \circ_{\map B A} R_1}$
:$\tuple {L_1, R_1} \times_{\map M A} \tuple {L_2, R_2} = \tuple {L_1 L_2, R_2 R_1}$
where $+_{\map B A}$ and $\circ_{\map B A}$ are the respective operations on the space of bounded linear transformations $\map B A$. 
Define $\norm {\, \cdot \,}_{\map M A} : \map M A \to \hointr 0 \infty$ by:
:$\norm {\tuple {L, R} }_{\map M A} = \norm L_{\map B A}$
where $\norm {\, \cdot \,}_{\map B A}$ is the norm on $\map B A$.
For each $\tuple {L, R} \in \map M A$, define $L^\ast : A \to A$ and $R^\ast : A \to A$ by:
:$\map {L^\ast} a = \paren {\map L {a^\ast} }^\ast$ 
and:
:$\map {R^\ast} a = \paren {\map R {a^\ast} }^\ast$
for each $a \in A$. 
Finally, define $\ast : \map M A \to \map M A$ by:
:$\tuple {L, R}^\ast = \tuple {R^\ast, L^\ast}$
for each $\tuple {L, R} \in \map M A$.
Then $\tuple {\map M A, +_{\map B A}, \circ_{\map M A}, \times_{\map M A}, \ast, \norm {\, \cdot \,}_{\map M A} }$ is a unital $\text C^\ast$-algebra.

Proof:
=== Proof that $\map M A$ is a vector space ===
We show that $\map M A$ is a subspace of the direct product $\map B A \times \map B A$. 
We use the One-Step Vector Subspace Test.
From Double Centralizer Generated by Element of C*-Algebra is Double Centralizer, we have $\map M A \ne \O$.
Let $\tuple {L_1, R_1}, \tuple {L_2, R_2} \in \map M A$ and $\lambda \in \C$.
We show that $\tuple {L_1 + \lambda L_2, R_1 + \lambda R_2}$ is a double centralizer. 
For each $a, b \in A$ and $i \in \set {1, 2}$, we have:
:$\map {L_i} {a b} = \map {L_i} a b$
:$\map {R_i} {a b} = a \map {R_i} b$
:$\map {R_i} a b = a \map {L_i} b$
We then have:
:$\map {L_1} {a b} + \lambda \map {L_2} {a b} = \map {L_1} a b + \lambda \map {L_2} a b$
:$\map {R_1} {a b} + \lambda \map {R_2} {a b} = a \map {R_1} b + \lambda a \map {R_2} b$
:$\map {R_1} a b + \lambda \map {R_2} a b = a \map {L_1} b + \lambda a \map {L_2} b$ 
By associativity, we have:
:$\map {\paren {L_1 + \lambda L_2} } {a b} = \paren {\map {L_1} a + \lambda \map {L_2} a} b$
:$\map {\paren {R_1 + \lambda R_2} } {a b} = a \map {\paren {R_1 + \lambda R_2} } b$
:$\map {\paren {R_1 + \lambda R_2} } a b = a \map {\paren {L_1 + \lambda L_2} } b$
Further, $L_1 + \lambda L_2$ and $R_1 + \lambda R_2$ are bounded from Space of Bounded Linear Transformations forms Vector Space.
So $\tuple {L_1 + \lambda L_2, R_1 + \lambda R_2}$ is a double centralizer.
So $\map M A$ is a subspace of the direct product $\map B A \times \map B A$.
Hence $\map M A$ is a vector space.
{{qed|lemma}}
Theorem: Galois Connection is Unique for Given Lower Adjoint

Let $\struct {S, \preceq}$ and $\struct {T, \precsim}$ be ordered sets.
Let $g: S \to T$, $d: T \to S$ be mappings such that $\tuple{g, d}$ is a Galois connection.
Then:
:$\tuple{g, d}$ is the unique Galois connection such that $d$ is the lower adjoint

Proof:
Let $g^\prime: S \to T$ be a mapping such that $\tuple{g^\prime, d}$ is a Galois connection.
We have:
{{begin-eqn}}
{{eqn | q = \forall x \in S 
      | l = \map {g^\prime} x
      | r = \map \max {d^{-1} \sqbrk {x^\preceq} }
      | c = Galois Connection is Expressed by Maximum
}}
{{eqn | r = \map g x
      | c = Galois Connection is Expressed by Maximum
}}
{{end-eqn}}
where:
:$\max$ denotes the maximum
:$d^{-1} \sqbrk {x^\preceq}$ denotes the image of $x^\preceq$ under relation $d^{-1}$
:$x^\preceq$ denotes the lower closure of $x$
From Equality of Mappings:
:$g^\prime = g$
The result follows.
{{qed}}
Category:Galois Connections
c833293tk4xyvzh83a4mlh6au9efrl6
Theorem: Galois Connection is Unique for Given Upper Adjoint

Let $\struct {S, \preceq}$ and $\struct {T, \precsim}$ be ordered sets.
Let $g: S \to T$, $d: T \to S$ be mappings such that $\tuple{g, d}$ is a Galois connection.
Then:
:$\tuple{g, d}$ is the unique Galois connection such that $g$ is the upper adjoint

Proof:
Let $d^\prime: T \to S$ be a mapping such that $\tuple{g, d^\prime}$ is a Galois connection.
We have:
{{begin-eqn}}
{{eqn | q = \forall y \in T 
      | l = \map {d^\prime} y
      | r = \map \min {g^{-1} \sqbrk {t^\succsim} }
      | c = Galois Connection is Expressed by Minimum
}}
{{eqn | r = \map d x
      | c = Galois Connection is Expressed by Minimum
}}
{{end-eqn}}
where:
:$\min$ denotes the minimum
:$g^{-1} \sqbrk {y^\succsim}$ denotes the image of $y^\succsim$ under relation $g^{-1}$
:$y^\succsim$ denotes the upper closure of $y$
From Equality of Mappings:
:$d^\prime = d$
The result follows.
{{qed}}
Category:Galois Connections
jm9bwtrjj3oz3l3gj2xrf4l4eyqy0gx
Theorem: Extension of Continuous Complex-Valued Function Vanishing at Infinity to Alexandroff Extension is Continuous

Let $X$ be a locally compact Hausdorff space.
Let $X^\ast = X \cup \set p$ be the Alexandroff extension of $X$.
Let $f : X \to \C$ be a continuous complex-valued function vanishing at infinity.
Define $f^\ast : X^\ast \to \C$ by taking:
:$\map {f^\ast} x = \begin{cases}\map f x & x \in X \\ 0 & x = p\end{cases}$
for each $x \in X^\ast$. 
Then $f^\ast$ is continuous.

Proof:
Since $f$ is continuous at each $x \in X$, we have:
:for each $x \in X$:
::for every open neighborhood $U_2$ of $\map f x \in \C$, there exists an open neighborhood $U_1$ of $x \in X$ such that $f \sqbrk {U_1} \subseteq U_2$.
From the definition of the topology on $X^\ast$:
:every open set in $X$ is open in $X^\ast$.
Hence the $U_1$ given above remains an open neighborhood of $x$ in $X^\ast$.
Since $\map {f^\ast} x = \map f x$ for $x \in X$, $f^\ast$ remains continuous at each $x \in X$. 
We therefore only need to concern ourselves with continuity at $p$.
Let $U_2$ be an open neighborhood of $0 \in \C$.
Then there exists $\epsilon > 0$ such that $\set {z \in \C : \cmod z < \epsilon} \subseteq U_2$. 
From the definition of a complex-valued function vanishing at infinity, there exists a compact set $K \subseteq X$ such that:
:$\cmod {\map f x} < \epsilon$ for $x \in X \setminus K$. 
We also have $\cmod {\map {f^\ast} p} = 0 < \epsilon$, so we have:
:$\cmod {\map {f^\ast} x} < \epsilon$ for $x \in \set p \cup \paren {X \setminus K}$.
From the definition of the topology on $X^\ast$:
:$\set p \cup \paren {X \setminus K}$ is open in $X^\ast$.
Hence setting $U_1 = \set p \cup \paren {X \setminus K}$, we have:
:${f^\ast} \sqbrk {U_1} \subseteq \set {z \in \C : \cmod z < \epsilon} \subseteq U_2$
with $p \in U_1$.
So $f^\ast$ is continuous at $p$.
Hence $f^\ast$ is continuous.
{{qed}}
Category:Extension of Continuous Complex-Valued Function Vanishing at Infinity to Alexandroff Extension is Continuous
Category:Alexandroff Extensions
Category:Complex-Valued Functions Vanishing at Infinity
6rr9y3wj1ln6hrpbx4xpnrn7gt1w1ob
Theorem: Complex Numbers form Unital C*-Algebra

Let $\C$ be the complex numbers realised as a normed algebra (this is possible from Complex Numbers form Algebra).
Let $\overline {\, \cdot \,}$ be complex conjugation.
Then $\tuple {\C, \overline {\, \cdot \,} }$ is a unital $\text C^\ast$-algebra.

Proof:
From Complex Plane is Banach Space, $\C$ is a Banach space.
Further, we have:
:$\cmod 1 = 1$
Since $\C$ is also a normed algebra, it therefore a unital Banach algebra.
We prove that complex conjugation satisfies the properties of an involution as well as the $\text C^\ast$ identity.
Theorem: Bounded Complex-Valued Continuous Functions on Topological Space form Unital C*-Algebra

Let $X$ be a topological space. 
Let $\map {\CC_b} {X; \C} := \map {\CC_b} X$ be the space of bounded continuous functions on $X$ valued in $\C$.
Let $\norm {\,\cdot\,}_\infty$ be the supremum norm on $\map {\CC_b} X$.
For each $f \in \map {\CC_b} X$, define $\overline f : X \to \C$ by:
:$\map {\overline f} x = \overline {\map f x}$
for each $x \in X$. 
Equip $\map {\CC_b} X$ with pointwise multiplication. 
Then $\struct {\map {\CC_b} X, \norm {\,\cdot\,}_\infty}$ is a unital $\text C^\ast$-algebra.

Proof:
From Bounded Continuous Functions on Topological Space form Banach Space, $\struct {\map {\CC_b} X, \norm {\,\cdot\,}_\infty}$ is a Banach space over $\C$.
From Product of Continuous Functions on Topological Ring is Continuous, $f g$ is continuous for each $f, g \in \map {\CC_b} X$.
Further, from Product of Bounded Mappings on Normed Algebra is Bounded, $f g$ is bounded.
For each $f, g \in \map {\CC_b} X$ and $x \in X$ we have:
:$\cmod {\map f x \map g x} = \cmod {\map f x} \cmod {\map g x}$
so that:
:$\norm {f g}_\infty \le \norm f_\infty \norm g_\infty$
Hence $\struct {\map {\CC_b} X, \norm {\,\cdot\,}_\infty}$ is a Banach algebra with pointwise multiplication.
Further, the function ${\mathbf 1}_{\map {\CC_b} X} : X \to \C$ defined by:
:$\map { {\mathbf 1}_{\map {\CC_b} X} } x = 1$
for each $x \in X$, satisfies:
:$\map f x \map { {\mathbf 1}_{\map {\CC_b} X} } x = \map { {\mathbf 1}_{\map {\CC_b} X} } x \map f x = 1$
From Constant Mapping is Continuous, ${\mathbf 1}_{\map {\CC_b} X}$ is continuous.
Further:
:$\cmod { \map { {\mathbf 1}_{\map {\CC_b} X} } x} = 1$
for each $x \in X$.
Hence ${\mathbf 1}_A \in \map {\CC_b} X$ with:
:$\norm { {\mathbf 1}_{\map {\CC_b} X} }_\infty = 1$
So ${\mathbf 1}_{\map {\CC_b} X}$ is an identity element for $\map {\CC_b} X$, making $\map {\CC_b} X$ a unital Banach algebra.
That $\overline {\, \cdot \,}$ is an involution on $\map {\CC_b} X$ follows from Complex Numbers form Unital C*-Algebra.
Finally, from Complex Numbers form Unital C*-Algebra we have:
:$\cmod {\map f x \map {\overline f} x} = \cmod {\map f x}^2$
Taking supremums we have:
:$\norm {f \overline f}_\infty = \norm f^2$ 
for each $f \in \map {\CC_b} X$.
So $\overline {\, \cdot \,}$ satisfies the $\text C^\ast$ identity.
Hence $\struct {\map {\CC_b} X, \norm {\,\cdot\,}_\infty}$ is a unital $\text C^\ast$-algebra.
{{qed}}
Category:C*-Algebras
jo75rfvyaa0hv0e93u9zoxbw5lpgcaf
Theorem: Closed *-Subalgebra of C*-Algebra is C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $B \subseteq A$ be a closed $\ast$-subalgebra of $A$. 
Then $\struct {B, \ast, \norm {\, \cdot \,} }$ is a $\text C^\ast$-algebra.

Proof:
Since $\struct {B, \ast}$ is a $\ast$-subalgebra of $A$, it is in particular a $\ast$-algebra.
From Closed Subspace of Banach Space forms Banach Space, $\struct {B, \ast, \norm {\, \cdot \,} }$ is additionally a Banach space.
We have:
:$\norm {x y} \le \norm x \norm y$
and:
:$\norm {x x^\ast} = \norm x^2$
for each $x \in B$, since these properties hold for all $x \in A$.
Hence $\struct {B, \ast, \norm {\, \cdot \,} }$ is a $\text C^\ast$-algebra.
{{qed}}
Category:C*-Algebras
i1dfctsbrlkd5307vhuu7q4k3r3utdv
Theorem: Complex Conjugate of Complex-Valued Function Vanishing at Infinity Vanishes at Infinity

Let $X$ be a locally compact Hausdorff space. 
Let $f : X \to \C$ be a complex-valued function vanishing at infinity.
Define $\overline f : X \to \C$ by:
:$\forall x \in \C: \map {\overline f} x = \overline {\map f x}$
where $\overline {\map f x}$ denotes the conjugate of $\overline {\map f x}$.
Then $\overline f$ is a complex-valued function vanishing at infinity.

Proof:
{{explain|What is the significance of $X$ being a locally compact Hausdorff space? What would happen to this proof if it were not?}}
Let $\epsilon > 0$.
Then since $f$ vanishes at infinity, there exists a compact set $K \subseteq X$ such that:
:$\cmod {\map f x} < \epsilon$ for all $x \in X \setminus K$. 
From Complex Modulus equals Complex Modulus of Conjugate, we then have:
:$\cmod {\overline {\map f x} } < \epsilon$ for all $x \in X \setminus K$. 
Since $\epsilon$ was arbitrary: 
:$\overline f$ is a complex-valued function vanishing at infinity.
{{qed}}
Category:Complex-Valued Functions Vanishing at Infinity
p8rfhzwy573hsichu5ucg001yz44s3l
Theorem: Space of Continuous Functions Vanishing at Infinity is C*-Algebra

Let $X$ be a locally compact Hausdorff space. 
Let $\map {\CC_0} X$ be the set of continuous complex-valued functions vanishing at infinity.
Equip $\map {\CC_0} X$ with pointwise addition, pointwise scalar multiplication and pointwise multiplication.
For each $f \in \map {\CC_0} X$, define:
:$\map {\overline f} x = \overline {\map f x}$
for each $x \in X$.
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $\map {\CC_0} X$.
Then $\tuple {\map {\CC_0} X, \overline {\, \cdot \,}, \norm {\, \cdot \,}_\infty}$ is a $\text C^\ast$-algebra.

Proof:
From Bounded Complex-Valued Continuous Functions on Topological Space form Unital C*-Algebra, $\tuple {\map {\CC_b} X, \overline {\, \cdot \,}, \norm {\, \cdot \,}_\infty}$ is a $\text C^\ast$-algebra where $\map {\CC_b} X$ is the space of bounded continuous functions on $X$ valued in $\C$.
From Closed *-Subalgebra of C*-Algebra is C*-Algebra, it suffices to show that $\map {\CC_0} X$ is a closed $\ast$-subalgebra of $\map {\CC_b} X$.
In Space of Continuous Functions on Locally Compact Hausdorff Space Vanishing at Infinity is Banach Space, $\map {\CC_0} X$ is a Banach space. 
Hence from Subspace of Complete Metric Space is Closed iff Complete, $\map {\CC_0} X$ is closed in $\map {\CC_b} X$. 
From Product of Complex-Valued Functions Vanishing at Infinity Vanishes at Infinity, we have $f g \in \map {\CC_0} X$ for each $f, g \in \map {\CC_0} X$. 
Hence $\map {\CC_0} X$ is a subalgebra of $\map {\CC_b} X$. 
Finally, from Complex Conjugate of Complex-Valued Function Vanishing at Infinity Vanishes at Infinity we have $\overline f \in \map {\CC_b} X$ whenever $f \in \map {\CC_B} X$.
Hence $\map {\CC_0} X$ is a $\ast$-subalgebra of $\map {\CC_b} X$.
Hence applying Closed *-Subalgebra of C*-Algebra is C*-Algebra, 
From Linear Combination of Complex-Valued Functions Vanishing at Infinity Vanishes at Infinity, $\tuple {\map {\CC_0} X, \overline {\, \cdot \,}, \norm {\, \cdot \,}_\infty}$ is a $\text C^\ast$-algebra.
{{qed}}
Category:C*-Algebras
Category:Complex-Valued Functions Vanishing at Infinity
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