Existence of Unitization of C*-Algebra

== Theorem ==
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:
and:
So $\phi$ is an algebra homomorphism.
Further, we have, for each $a \in A$:
Also, for $a, b \in A$ and $\lambda \in \C$ we have:
for any $T$ in a double centralizer.
We therefore have: 
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 , 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 on ${\, \norm \,}$.
Hence $\tuple {a, \lambda} = \tuple { {\mathbf 0}_A, 0} = {\mathbf 0}_{A_+}$.
Hence is proved.
We move onto .
Let $\tuple {a, \lambda} \in A_+$ and $t \in \C$.
We then have:
This proves .
We now prove .
Let $\tuple {a, \lambda}, \tuple {b, \mu} \in A_+$.
Then we have:
Hence 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 $\tuple {a, \lambda}, \tuple {b, \mu} \in A_+$:
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 $\tuple {a, \lambda} \in A_+$:
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.
Category:Multiplier Algebras
Category:C*-Algebras