Theorem: Existence of Unitization of C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra that is not unital as an algebra.
Let $A_+$ be the unitization of $A$ as an algebra.
Define:
:$\tuple {x, \lambda}^\ast = \tuple {x^\ast, \overline \lambda}$
for each $\tuple {x, \lambda} \in A_+$.
Then there exists a unique algebra norm $\norm {\, \cdot \,}_\ast$ such that $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ is a unital $\text C^\ast$-algebra.
Further, $\norm {\, \cdot \,}_\ast$ extends $\norm {\, \cdot \,}$.

Proof:
Let $\map M A$ be the multiplier algebra of $A$. 
Define $T : A_+ \to \map M A$ by:
:$\map T {a, \lambda} = \tuple {L_a + \lambda I_A, R_a + \lambda I_A}$
where $\tuple {L_a, R_a}$ is the double centralizer generated by $a$ and $I_A$ is the identity map.
We show that $T$ is an $\ast$-algebra homomorphism so that $T \sqbrk {A_+}$ is a $\ast$-subalgebra of $\map M A$.
We note that from the proof of C*-Algebra embeds into Multiplier Algebra as C*-Subalgebra, we have:
:$L_a L_b = L_{a b}$
:$R_b R_a = R_{a b}$
:$L_{a + t b} = L_a + t L_b$
:$R_{a + t b} = R_a + t R_b$
:$L_a^\ast = R_{a^\ast}$
:$R_a^\ast = L_{a^\ast}$
for $a, b \in A$ and $t \in \C$.
Now let $\tuple {a, \lambda}, \tuple {b, \mu} \in A_+$ and $t \in \C$.
We then have:
{{begin-eqn}}
{{eqn	| l = \map T {a + t b, \lambda + t \mu}
		| r = \tuple {L_{a + t b} + \paren {\lambda + t \mu} I_A, R_{a + t b} + \paren {\lambda + t \mu} I_A}
}}
{{eqn	| r = \tuple {L_a + \lambda I_A + t \paren {L_b + \mu I_A}, R_a + \lambda I_A + t \paren {R_b + \mu I_A} }
}}
{{eqn	| r = \tuple {L_a + \lambda I_A, R_a + \lambda I_A} + t \tuple {L_b + \mu I_A, R_b + \mu I_A}
}}
{{eqn	| r = \map T {a, \lambda} + t \map T {b, \mu} 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn	| l = \tuple {L_a + \lambda I_A, R_a + \lambda I_A} \tuple {L_b + \mu I_A, R_b + \mu I_A}
		| r = \tuple {\paren {L_a + \lambda I_A} \paren {L_b + \mu I_A}, \paren {R_b + \mu I_A} \paren {R_a + \lambda I_A} }
}}
{{eqn	| r = \tuple {L_a L_b + \lambda L_b + \mu L_a + L_a L_b, R_b R_a + \mu R_a + \lambda R_b + \lambda \mu I_A}
}}
{{eqn	| r = \tuple {L_{a b} + L_{\lambda b} + L_{\mu a} + \lambda \mu I_A, R_{a b} + R_{\lambda b} + R_{\mu a} + \lambda \mu I_A}
}}
{{eqn	| r = \tuple {L_{a b + \lambda b + \mu a} + \lambda \mu I_A, R_{a b + \lambda b + \mu a} + \lambda \mu I_A}
}}
{{eqn	| r = \map \phi {a b + \lambda b + \mu a, \lambda \mu}
}}
{{eqn	| r = \map \phi {\tuple {a, \lambda} \tuple {b, \lambda} }
		| c = {{Defof|Unitization of Algebra over Field}}
}}
{{end-eqn}}
So $\phi$ is an algebra homomorphism.
Further, we have, for each $a \in A$:
{{begin-eqn}}
{{eqn	| l = \map {I_A^\ast} a
		| r = \paren {\map {I_A} {a^\ast} }^\ast 
}}
{{eqn	| r = \paren {a^\ast}^\ast
}}
{{eqn	| r = a
		| c = $(\text C^\ast 1)$ in {{Defof|Involution on Algebra}}
}}
{{end-eqn}}
Also, for $a, b \in A$ and $\lambda \in \C$ we have:
{{begin-eqn}}
{{eqn	| l = \map {\paren {\lambda T}^\ast} b
		| r = \paren {\lambda \map T {b^\ast} }^\ast 
}}
{{eqn	| r = \overline \lambda \map {T^\ast} b
		| c = $(\text C^\ast 4)$ in {{Defof|Involution on Algebra}}
}}
{{end-eqn}}
for any $T$ in a double centralizer.
We therefore have: 
{{begin-eqn}}
{{eqn	| l = \map \phi {a, \lambda}^\ast
		| r = \paren {L_a + \lambda I_A, R_a + \lambda I_A}^\ast
}}
{{eqn	| r = \paren {R_a^\ast + \overline \lambda I_A^\ast, L_a^\ast + \overline \lambda I_A^\ast} 
}}
{{eqn	| r = \paren {L_{a^\ast} + \overline \lambda I_A, R_{a^\ast} + \overline \lambda I_A}
}}
{{eqn	| r = \map \phi {a^\ast, \overline \lambda}
}}
{{end-eqn}}
Hence $\phi$ is a $\ast$-algebra homomorphism.
So from Image of *-Algebra under *-Algebra Homomorphism is *-Subalgebra, $\phi \sqbrk {A_+}$ is a $\ast$-subalgebra of $\map M A$. 
Now, for $\tuple {a, \lambda} \in A_+$ set:
:$\norm {\tuple {a, \lambda} }_\ast = \norm {\map \phi {a, \lambda} }_{\map M A}$
We have:
:$\norm {\tuple {a, 0} }_\ast = \norm {L_a}_{\map B A} = \norm a$
from Norm of Double Centralizer Generated by Element of C*-Algebra.
Hence $\norm {\, \cdot \,}_\ast$ extends $\norm {\, \cdot \,}$.
We prove that $\norm {\, \cdot \,}_\ast$ is indeed an algebra norm. 
Towards proving {{NormAxiomVector|1}}, suppose that:
:$\norm {\tuple {a, \lambda} }_\ast = \norm {\map \phi {a, \lambda} }_{\map M A} = 0$
Then we have:
:$\norm {L_a + \lambda I_A}_{\map B A} = 0$
Hence:
:$\map {\paren {L_a + \lambda I_A} } b = {\mathbf 0}_A$
and:
:$\norm {\paren {R_a + \lambda I_A} } b = 0$
for each $b \in A$.
That is:
:$a b + \lambda b = {\mathbf 0}_A$
:$b a + \lambda b = {\mathbf 0}_A$
for each $b \in A$.
Suppose that $\lambda \ne 0$.
Then we have:
:$\paren {-\lambda^{-1} a} b = b$
and:
:$b \paren {-\lambda^{-1} a} = b$ 
Hence $-\lambda^{-1} a$ would be an identity element for $A$, contrary to our assumption that $A$ is not unital as an algebra.
So $\lambda = 0$.
We then have:
:$a b = {\mathbf 0}_A$
for each $b \in A$.
In particular, setting $b = a^\ast$, we have $a a^\ast = {\mathbf 0}_A$.
Then $\norm {a a^\ast} = 0$.
Hence the $\text C^\ast$ identity, we have $\norm a^2 = 0$.
This gives $a = {\mathbf 0}_A$ from {{NormAxiomVector|1}} on ${\, \norm \,}$.
Hence $\tuple {a, \lambda} = \tuple { {\mathbf 0}_A, 0} = {\mathbf 0}_{A_+}$.
Hence {{NormAxiomVector|1}} is proved.
We move onto {{NormAxiomVector|2}}.
Let $\tuple {a, \lambda} \in A_+$ and $t \in \C$.
We then have:
{{begin-eqn}}
{{eqn	| l = \norm {t \tuple {a, \lambda} }_\ast
		| r = \norm {\tuple {t a, t \lambda} }_\ast 
}}
{{eqn	| r = \norm {\map \phi {t a, t \lambda} }_{\map M A}
}}
{{eqn	| r = \norm {t \map \phi {a, \lambda} }_{\map M A}
		| c = $\phi$ is linear
}}
{{eqn	| r = \cmod t \norm {\map \phi {a, \lambda} }_{\map M A}
		| c = {{NormAxiomVector|2}} for $\norm {\, \cdot \,}_{\map M A}$
}}
{{eqn	| r = \cmod t \norm {\tuple {a, \lambda} }_\ast
}}
{{end-eqn}}
This proves {{NormAxiomVector|2}}.
We now prove {{NormAxiomVector|3}}.
Let $\tuple {a, \lambda}, \tuple {b, \mu} \in A_+$.
Then we have:
{{begin-eqn}}
{{eqn	| l = \norm {\tuple {a, \lambda} + \tuple {b, \mu} }_\ast
		| r = \norm {\tuple {a + b, \lambda + \mu} }_\ast
}}
{{eqn	| r = \norm {\map \phi {a + b, \lambda + \mu} }_{\map M A}
}}
{{eqn	| r = \norm {\map \phi {a, \lambda} + \map \phi {b, \mu} }_{\map M A}
		| c = since $\phi$ is linear
}}
{{eqn	| o = \le
		| r = \norm {\map \phi {a, \lambda} }_{\map M A} + \norm {\map \phi {b, \mu} }_{\map M A}
		| c = {{NormAxiomVector|3}} for $\norm {\, \cdot \,}_{\map M A}$
}}
{{eqn	| r = \norm {\tuple {a, \lambda} }_\ast + \norm {\tuple {b, \mu} }_\ast 
}}
{{end-eqn}}
Hence {{NormAxiomVector|3}} is also proven.
We now prove that:
:$\norm {\tuple {a, \lambda} \tuple {b, \mu} }_\ast \le \norm {\tuple {a, \lambda} }_\ast \norm {\tuple {b, \mu} }_\ast$
for each $\tuple {a, \lambda}, \tuple {b, \mu} \in A_+$.
We have, for $\tuple {a, \lambda}, \tuple {b, \mu} \in A_+$:
{{begin-eqn}}
{{eqn	| l = \norm {\tuple {a, \lambda} \tuple {b, \mu} }_\ast
		| r = \norm {\tuple {a b + \lambda b + \mu a, \lambda \mu} }_\ast
}}
{{eqn	| r = \norm {\map \phi {a b + \lambda b + \mu a, \lambda \mu} }_{\map M A}
}}
{{eqn	| r = \norm {\map \phi {\tuple {a, \lambda} \tuple {b, \mu} } }_{\map M A}
		| c = $\phi$ is an algebra homomorphism
}}
{{eqn	| o = \le
		| r = \norm {\map \phi {a, \lambda} }_{\map M A} \norm {\map \phi {b, \mu} }_{\map M A}
		| c = $\norm {\, \cdot \,}_{\map M A}$ is a algebra norm
}}
{{eqn	| r = \norm {\tuple {a, \lambda} }_\ast \norm {\tuple {b, \mu} }_\ast 
}}
{{end-eqn}}
By definition, we now have that $\phi : \struct {A_+, \norm {\, \cdot \,}_\ast} \to \struct {\map M A, \norm {\, \cdot \,}_{\map M A} }$ is a linear isometry.
Further, $\map M A$ is a Banach algebra.
By Image of Closed Set under Linear Isometry from Banach Space to Normed Vector Space is Closed, we have that $\phi \sqbrk {A_+}$ is closed in $\map M A$. 
Now let $\sequence {\tuple {a_n, \lambda_n} }_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {A_+, \norm {\, \cdot \,}_\ast}$. 
Then $\sequence {\map \phi {a_n, \lambda_n} }_{n \mathop \in \N}$ is Cauchy sequence in $\struct {\map M A, \norm {\, \cdot \,}_{\map M A} }$.
Since $\phi \sqbrk {A_+}$ is a closed in $\map M A$, there exists $\tuple {a, \lambda} \in A_+$ such that $\map \phi {a_n, \lambda_n} \to \map \phi {a, \lambda}$.
Then:
:$\norm {\tuple {a, \lambda} - \tuple {a_n, \lambda_n} }_\ast = \norm {\map \phi {a, \lambda} - \map \phi {a_n, \lambda_n} }_{\map M A} \to 0$
Hence $\tuple {a_n, \lambda_n} \to \tuple {a, \lambda}$. 
So every Cauchy sequence in $\struct {A_+, \norm {\, \cdot \,}_\ast}$ converges.
So $\struct {A_+, \norm {\, \cdot \,}_\ast}$ is a Banach algebra which is also a $\ast$-algebra. 
It remains to show the $\text C^\ast$ identity for $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$.
We have, for $\tuple {a, \lambda} \in A_+$:
{{begin-eqn}}
{{eqn	| l = \norm {\tuple {a, \lambda} \tuple {a, \lambda}^\ast}_\ast
		| r = \norm {\tuple {a, \lambda} \tuple {a^\ast, \overline \lambda} }_\ast 
}}
{{eqn	| r = \norm {\tuple {a a^\ast + \lambda a^\ast + \overline \lambda a, \lambda \overline \lambda} }_\ast
}}
{{eqn	| r = \norm {\map \phi {a a^\ast + \lambda a^\ast + \overline \lambda a, \lambda \overline \lambda} }_{\map M A}
}}
{{eqn	| r = \norm {\map \phi {\tuple {a, \lambda} \tuple {a^\ast, \overline \lambda} } }_{\map M A}
}}
{{eqn	| r = \norm {\map \phi {a, \lambda} \paren {\map \phi {a, \lambda} }^\ast}_{\map M A}
}}
{{eqn	| r = \norm {\map \phi {a, \lambda} }_{\map M A}^2
		| c = {{Defof|C* Identity}}
}}
{{eqn	| r = \norm {\tuple {a, \lambda} }_\ast^2 
}}
{{end-eqn}}
Hence we have the $\text C^\ast$ identity for $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$.
Hence $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ is a unital $\text C^\ast$-algebra.
Uniqueness of $\norm {\, \cdot \,}_\ast$ follows from Norm on C*-Algebra is Unique.
{{qed}}
Category:Multiplier Algebras
Category:C*-Algebras
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