Multiplier Algebra is Unital C*-Algebra

== Theorem ==
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.
=== $\map M A$ is an algebra ===
From Direct Product of Algebras is Algebra, $\map B A \times \map B A$ is an algebra.
It then suffices to show that $\map M A$ is an subalgebra of $\map M A$. 
Let $\tuple {L_1, R_1}, \tuple {L_2, R_2} \in \map M A$.
We show that $\tuple {L_1 L_2, R_2 R_1}$ is a double centralizer.
From Composition of Bounded Linear Transformations is Bounded Linear Transformation, $L_1 L_2$ and $R_2 R_1$ are both bounded.
For $a, b \in A$ we have:
and:
finally:
Hence $\tuple {L_1 L_2, R_2 R_1}$ is a double centralizer.
=== $\map M A$ is a Banach algebra ===
From Norms of Double Centralizer of C*-Algebra, we have:
:$\norm {\tuple {L, R} }_{\map M A} = \norm L_{\map B A} = \norm R_{\map B A}$
Then we have:
:$\norm {\tuple {L, R} }_{\map B A \times \map B A} = \max \set {\norm L_{\map B A}, \norm R_{\map B A} } = \norm {\tuple {L, R} }_{\map M A}$
Hence from Norm on Space of Bounded Linear Transformations is Norm and Norm on Bounded Linear Transformation is Submultiplicative, we have that $\norm {\, \cdot \,}_{\map B A}$ is a algebra norm. 
Now from Direct Product of Banach Spaces is Banach Space, $\map B A \times \map B A$ is a Banach space.
From Closed Subspace of Banach Space forms Banach Space, it is enough to show that $\map M A$ is closed in $\map B A \times \map B A$. 
Let $\sequence {\tuple {L_n, R_n} }_{n \mathop \in \N}$ be a sequence of double centralizers on $A$ converging to $\tuple {L, R}$. 
From Convergence in Direct Product Norm, we have $L_n \to L$ and $R_n \to R$ in $\map B A$. 
For $a, b \in A$ and $n \in \N$, we then have: 
:$\map {L_n} {a b} = \map {L_n} a b$
:$\map {R_n} {a b} = a \map {R_n} b$
:$\map {R_n} a b = a \map {L_n} b$
From Multiplication on Normed Algebra is Continuous, by taking $n \to \infty$ we have:
:$\map L {a b} = \map L a b$
:$\map R {a b} = a \map R b$
:$\map R a b = a \map L b$
for each $a, b \in A$. 
So $\map M A$ is closed in $\map B A \times \map B A$ and hence a Banach space.
Hence $\map M A$ is a Banach algebra.
=== $\map M A$ is a $\text C^\ast$-algebra ===
Let $\tuple {L, R} \in \map M A$. 
From the definition of $\ast$, we have:
:$\tuple {L, R}^\ast = \tuple {R^\ast, L^\ast}$
where:
:$\map {L^\ast} a = \paren {\map L {a^\ast} }^\ast$
and:
:$\map {R^\ast} a = \paren {\map R {a^\ast} }^\ast$
for each $a \in A$. 
We show that:
:$\tuple {R^\ast, L^\ast}$ is a double centralizer. 
We have: 
:$\map L {b^\ast a^\ast} = \map L {b^\ast} a^\ast$
:$\map R {b^\ast a^\ast} = b^\ast \map R {a^\ast}$
:$\map R {b^\ast} a^\ast = a^\ast \map L {b^\ast}$
Taking $\ast$s, we have:
:$\paren {\map L {\paren {a b}^\ast} }^\ast = a \paren {\map L {b^\ast} }^\ast$
:$\paren {\map R {\paren {a b}^\ast} }^\ast = \paren {\map R {a^\ast} }^\ast b$
:$a \paren {\map R {b^\ast} }^\ast = \paren {\map L {b^\ast} }^\ast a$
So that:
:$\map {L^\ast} {a b} = a \map {L^\ast} b$
:$\map {R^\ast} {a b} = \map {R^\ast} a b$
:$a \map {R^\ast} b = \map {L^\ast} b a$ 
To finish we just show that $L^\ast$ and $R^\ast$ are bounded.
Since $L$ and $R$ are bounded, there exists $M_1, M_2 > 0$ such that:
:$\norm {\map L a}_{\map B A} \le M_1 \norm a$
and:
:$\norm {\map R a}_{\map B A} \le M_2 \norm a$ 
By C* Identity implies Involution is Isometry, we have $\norm {x^\ast} = \norm x$ for each $x \in A$.
Now, for each $a \in A$ we have:
and:
Hence $\tuple {L^\ast, R^\ast}$ is a double centralizer.
==== Proof of $(\text C^\ast 1)$ ====
We show $L^{\ast \ast} = L$ and $R^{\ast \ast} = R$.
Let $a \in A$.
Then: 
Swapping $L$ for $R$ in the above we also obtain:
:$\map {R^{\ast \ast} } a = \map R a$
for each $a \in A$.
Hence we have:
:$\tuple {L, R}^{\ast \ast} = \tuple {R^\ast, L^\ast}^\ast = \tuple {L^{\ast \ast}, R^{\ast \ast} } = \tuple {L, R}$
==== Proof of $(\text C^\ast 2)$ ====
Let $\tuple {L_1, R_1}, \tuple {L_2, R_2} \in \map M A$. 
Firstly we have:
and similarly:
:$\paren {R_1 + R_2}^\ast = R_1^\ast + R_2^\ast$
Now, we have:
==== Proof of $(\text C^\ast 3)$ ====
Let $\tuple {L_1, R_1}, \tuple {L_2, R_2} \in \map M A$. 
For $a \in A$, we have: 
hence:
:$\paren {L_1 L_2}^\ast = L_1^\ast L_2^\ast$ 
Similarly we have:
:$\paren {R_2 R_1}^\ast = R_2^\ast R_1^\ast$
We deduce that:
==== Proof of $(\text C^\ast 4)$ ====
Let $\tuple {L, R} \in \map M A$ and $\lambda \in \C$.
For $a \in A$ we have: 
so:
:$\paren {\lambda L}^\ast = \overline \lambda L^\ast$
Similarly, we have:
:$\paren {\lambda R}^\ast = \overline \lambda R^\ast$ 
Now, we have:
==== Proof of $(\text C^\ast 5)$ ====
Let:
:$T = \tuple {L, R} \in \map M A$
We now prove that:
:$\norm {T T^\ast}_{\map M A} = \norm T^2_{\map M A}$
For each $a \in A$, we have:
Hence:
So from Sufficient Condition for C* Identity, we have that $\map M A$ is a $\text C^\ast$-algebra. 
=== $\map M A$ is unital ===
Let $I_A$ be the identity mapping on $A$. 
Set:
:${\mathbf 1}_{\map M A} = \tuple {I_A, I_A}$
We then have:
:${\mathbf 1}_{\map M A} \tuple {L, R} = \tuple {I L, I R} = \tuple {L, R} = \tuple {L I, R I} = \tuple {L, R} {\mathbf 1}_{\map M A}$
Finally, we have:
:$\norm {\tuple {I_A, I_A} }_{\map M A} = \norm {I_A}_{\map B A} = 1$
from Norm of Identity Linear Transformation.
Hence $\map M A$ is a unital Banach algebra with identity element ${\mathbf 1}_{\map M A}$.
Category:Multiplier Algebras
Category:C*-Algebras