Gelfand-Naimark-Segal Construction

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $f : A \to \C$ be a positive linear functional.
Then there exists a cyclic representation $\tuple {\pi_f, \struct {\HH_f, \innerprod \cdot \cdot_f} }$ with cyclic vector $e_f$ such that:
:$\map f a = \innerprod {\map {\pi_f} a e_f} {e_f}_f$ for each $a \in A$. 
== Proof ==
From Positive Linear Functional on C*-Algebra is Bounded, $f$ is bounded.
Define:
:$\LL_f = \set {x \in A : \map f {x^\ast x} = 0}$
From Positive Linear Functional on C*-Algebra induces Semi-Inner Product, $\LL_f$ is a vector subspace of $A$.
Construct the quotient vector space $A/\LL_f$.
Define $\widetilde {\innerprod \cdot \cdot_f} : A^2 \to \C$ by:
:$\widetilde {\innerprod x y_{f, 0} } = \map f {y^\ast x}$
for each $x, y \in A$.
From Positive Linear Functional on C*-Algebra induces Semi-Inner Product, $\innerprod \cdot \cdot_f^\ast$ is a semi-inner product on $A$.
Note that:
:$\LL_f = \set {x \in A : \widetilde {\innerprod x x_{f, 0} } = 0}$ 
Hence by Semi-Inner Product induces Inner Product on Quotient, we can define $\innerprod \cdot \cdot_{f, 0} : \paren {A/\LL_f}^2 \to \C$ by:
:$\innerprod {x + \LL_f} {y + \LL_f}_{f, 0} = \map f {y^\ast x}$
to obtain an inner product on $A/\LL_f$.
By Completion Theorem (Inner Product Space), there exists a Hilbert space $\struct {\HH_f, \innerprod \cdot \cdot_f}$ and a linear isometry $\phi : A/\LL_f \to \HH_f$ such that $\phi \sqbrk {A/\LL_f}$ is everywhere dense in $\HH_f$ and:
:$\innerprod {\map \phi {x + \LL_f} } {\map \phi {y + \LL_f} }_f = \innerprod {x + \LL_f} {y + \LL_f}_{f, 0}$
for each $x, y \in A$.
We show that $\LL_f$ is a left ideal.
Let $a \in A$ and $x \in \LL_f$.
We then have:
Hence:
:$\map f {\paren {a x}^\ast \paren {a x} } = 0$
giving $a x \in \LL_f$.
Hence, if $x - y \in \LL_f$ for $x, y \in A$ we have:
:$a x - a y \in \LL_f$
and so $a x + \LL_f = a y + \LL_f$.
We can therefore define, for $a \in A$, $\map {\widetilde {\pi_{f, 0} } } a : A/\LL_f \to A/\LL_f$ by:
:$\map {\paren {\map {\widetilde {\pi_{f, 0} } } a} } {x + \LL_f} = a x + \LL_f$
for each $x \in A$. 
For $x, y \in A$ and $\lambda \in \C$ we have:
Hence $\map {\widetilde {\pi_{f, 0} } } a$ is linear. 
We show that $\map {\widetilde {\pi_{f, 0} } } a$ is bounded for each $a \in A$.
From Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum, we have:
:$a^\ast a \le_A \norm {a^\ast a} {\mathbf 1}_A = \norm a^2 {\mathbf 1}_A$
where $\le_A$ be the canonical preordering.
Hence from Conjugation in C*-Algebra preserves Positivity: Corollary, we have:
:$x^\ast a^\ast a x \le_A \norm a^2 x^\ast x$
So from Positive Linear Functional on C*-Algebra is Increasing on Hermitian Elements,w e have:
:$\map f {x^\ast a^\ast a x} \le \norm a^2 \map f {x^\ast x}$
Let $\norm {\, \cdot \,}_{f, 0}$ be the inner product norm induced by $\innerprod \cdot \cdot_{f, 0}$. 
Then we have:
and so $\widetilde {\pi_{f, 0} } a$ is bounded with:
:$\norm {\map {\widetilde {\pi_{f, 0} } } a}_{\map B {A/\LL_f} } \le \norm a$
where $\norm {\, \cdot \,}_{\map B {A/\LL_f} }$ is the norm of a bounded linear operator on $A/\LL_f$.
From Inverse of Linear Isometry is Linear Isometry, $\phi^{-1} : \phi \sqbrk {A/\LL_f} \to A/\LL_f$.
Then set $\map {\pi_{f, 0} } a = \phi \circ \map {\widetilde {\pi_{f, 0} } } a \circ \phi^{-1} : \phi \sqbrk {A/\LL_f} \to \HH_f$.
Since $\phi$ is a linear isometry, $\map {\pi_{f, 0} } a$ is a bounded linear operator with:
:$\norm {\map {\pi_{f, 0} } a}_{\map B {\phi \sqbrk {A/\LL_f} } } \le \norm a$ 
From Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain, there exists a unique extension $\map {\pi_f} a : \HH_f \to \HH_f$ such that:
:$\norm {\map {\pi_f} a}_{\map B {\HH_f} } \le \norm a$ 
We now consider $\pi_f$ as a function $A \to \map B {\HH_f}$.
We show that $\tuple {\pi_f, \HH_f}$ is a representation. 
First, we show that $\pi_f$ is linear. 
That is:
:$\map {\pi_f} {a + \lambda b} = \map {\pi_f} a + \lambda \map {\pi_f} b$
for $a, b \in A$ and $\lambda \in \C$. 
From Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide, it suffices to show that we have:
:$\map {\pi_f} {a + \lambda b} x = \map {\pi_f} a x + \lambda \map {\pi_f} b x$
for $x \in \phi \sqbrk {A/\LL_f}$.
That is, for $u \in A/\LL_f$, we want to verify that:
:$\map {\pi_f} {a + \lambda b} \map \phi u = \map {\pi_f} a \map \phi u + \lambda \map {\pi_f} b \map \phi u$
We have:
and likewise for $\map {\pi_f} {a + \lambda b} \map \phi u$ and $\map {\pi_f} b \map \phi u$. 
Hence, since $\phi^{-1}$ is a linear isometry, it suffices to show that:
:$\paren {\map {\widetilde {\pi_{f, 0} } } a} u + \lambda \paren {\map {\widetilde {\pi_{f, 0} } } b} u = \paren {\map {\widetilde {\pi_{f, 0} } } {a + \lambda b} } u$
for each $u \in A/\LL_f$. 
This is equivalent to:
:$\paren {a u + \LL_f} + \lambda \paren {b u + \LL_f} = \paren {a + \lambda b} u + \LL_f$ 
which follows from the definition of $A/\LL_f$. 
Hence we have that $\pi_f$ is linear. 
We now show that $\pi_f$ is an algebra homomorphism. 
That is, for each $a, b \in A$:
:$\map {\pi_f} a \map {\pi_f} b = \map {\pi_f} {a b}$ 
Again, from Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide it is enough to verify that:
:$\map {\widetilde {\pi_{f, 0} } } a \map {\widetilde {\pi_{f, 0} } } b u = \map {\widetilde {\pi_{f, 0} } } {a b} u$ 
for each $u \in A/\LL_f$. 
We have:
:$\map {\widetilde {\pi_{f, 0} } } a \map {\widetilde {\pi_{f, 0} } } b u = \map {\paren {\map {\widetilde {\pi_{f, 0} } } a} } {b u + \LL_f} = \paren {a b + \LL_f} = \map {\widetilde {\pi_{f, 0} } } {a b} u$ 
Hence $\pi_f$ is an algebar homomorphism.
We now want to verify that:
:$\paren {\map {\pi_f} a}^\ast = \map {\pi_f} {a^\ast}$ 
From Everywhere Dense Set determines Adjoint Linear Transformation, it is enough to show:
:$\innerprod {\map {\pi_f} a \map \phi {u + \LL_f} } {\map \phi {v + \LL_f} }_f = \innerprod {\map \phi {u + \LL_f} } {\map {\pi_f} {a^\ast} \map \phi {v + \LL_f} }_f$ 
for each $u + \LL_f, v + \LL_f \in A/\LL_f$. 
We have:
 
Hence by Everywhere Dense Set determines Adjoint Linear Transformation, we have:
:$\paren {\map {\pi_f} a}^\ast = \map {\pi_f} {a^\ast}$ 
Finally, we show that $\pi_f$ is unital. 
We have:
:$\paren {\map {\widetilde {\pi_{f, 0} } } { {\mathbf 1}_A} } {u + \LL_f} = u + \LL_f$ 
for each $a \in A$.
Hence we have:
:$\map {\pi_f} { {\mathbf 1}_A} u = u$ 
for all $u \in \phi \sqbrk {A/\LL_f}$. 
From Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide, we therefore have:
:$\map {\pi_f} { {\mathbf 1}_A} = I$ 
Hence $\pi_f$ be a unital $\ast$-algebra homomorphism. 
Further, we have for $a \in A$:
We finally show that $\tuple {\pi_f, \struct {\HH_f, \innerprod \cdot \cdot}_f}$ is a cyclic representation with cyclic vector $e_f = \map \pi { {\mathbf 1}_A + \LL_f}$.
For $a \in A$ we have:
:$\map {\pi_f} a e_f = \map \phi {\map {\widetilde {\pi_{f, 0} } } a \paren { {\mathbf 1}_A + \LL_f} } = \map \phi {a + \LL_f}$
Hence:
:$\map {\pi_f} A e_f = \phi \sqbrk {A/\LL_f}$
By construction, $\phi \sqbrk {A/\LL_f}$ is everywhere dense in $\HH_f$.
That is:
:$\map \cl {\map {\pi_f} A e_f} = \HH_f$
so $\tuple {\pi_f, \struct {\HH_f, \innerprod \cdot \cdot}_f}$ is a cyclic representation with cyclic vector $e_f = \map \pi { {\mathbf 1}_A + \LL_f}$.
== Sources ==
* $\text {VIII}.5.14$
Category:Representations of C*-Algebras