Existence and Uniqueness of Continuous Functional Calculus

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be normal. 
Let $\map {\sigma_A} x$ denote the spectrum of $x$ in $A$.
Let $\iota : \map {\sigma_A} x \to \C$ be the inclusion mapping.
Then there exists a unique unital $\ast$-algebra homomorphism $\Theta_x : \map \CC {\map {\sigma_A} x} \to A$ such that:
:$\map {\Theta_x} \iota = x$ 
where $\map \CC {\map {\sigma_A} x}$ is the space of continuous functions on $\map {\sigma_A} x$.
That is, for each $x \in A$ there exists a unique continuous functional calculus.
We further have:
:$\map {\Theta_x} f \in \map \span {\set {x^n \paren {x^\ast}^m : n, m \ge 0} }^-$
and:
:$\map {\Theta_x} f$ is normal. 
 
== Proof ==
Let $B \subseteq A$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, x, x^\ast}$.
By C*-Algebra Generated by Commutative Self-Adjoint Set is Commutative, $B$ is commutative. 
From Spectrum of Element of Unital C*-Subalgebra of Unital C*-Algebra, we have:
:$\map {\sigma_A} x = \map {\sigma_B} x$
Let $\Phi_B$ be the spectrum of $B$. 
Let $\hat x$ be the Gelfand transform of $x$ in $B$. 
From the Gelfand Representation Theorem, $\hat x : \Phi_B \to \C$ is continuous and has $\hat x \sqbrk {\Phi_B} = \map {\sigma_B} x$. 
We argue that $\hat x$ is a homeomorphism.
From Spectrum of Banach Algebra is Weak-* Locally Compact Hausdorff Space: Corollary, $\Phi_B$ is a compact Hausdorff space.
From Spectrum of Element of Banach Algebra is Compact, $\map {\sigma_B} x$ is compact.
From Continuous Bijection from Compact to Hausdorff is Homeomorphism, it suffices to show that $\hat x$ is an injection. 
Suppose that $\phi_1, \phi_2 \in \Phi_B$ are such that:
:$\map {\phi_1} x = \map {\phi_2} x$
From Character on C*-Algebra is *-Algebra Homomorphism, we have:
:$\map {\phi_1} {x^\ast} = \overline {\map {\phi_1} x} = \overline {\map {\phi_2} x} = \map {\phi_2} {x^\ast}$ 
Since $\phi_1$ and $\phi_2$ are algebra homomorphisms, we have:
:$\map {\phi_i} {x^n \paren {x^\ast}^m} = \paren {\map {\phi_i} x}^n \paren {\map {\phi_i} {x^\ast} }^m$
for $i \in \set {1, 2}$ and $n, m \in \N$. 
Hence we have:
:$\map {\phi_1} {x^n \paren {x^\ast}^m} = \map {\phi_2} {x^n \paren {x^\ast}^m}$
where either $n \ge 1$ or $m \ge 1$.
We also have:
:$\map {\phi_i} { {\mathbf 1}_A} = 1$
for $i \in \set {1, 2}$ from Character on Unital Banach Algebra is Unital Algebra Homomorphism. 
Since $\phi_1$ and $\phi_2$ are linear, we have:
:$\map {\phi_1} u = \map {\phi_2} u$
for:
:$u \in \span \set {x^n \paren {x^m}^\ast : k_1, \ldots, k_n \ge 0}$
From Explicit Form for Generated C*-Algebra:
:$\span \set {x^n \paren {x^m}^\ast : k_1, \ldots, k_n \ge 0}$ is everywhere dense in $B$. 
From Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide, we therefore have:
:$\map {\phi_1} u = \map {\phi_2} u$ for $u \in B$.
Hence $\phi_1 = \phi_2$. 
Hence applying Continuous Bijection from Compact to Hausdorff is Homeomorphism, we have that $\hat x$ is a homeomorphism.
Define $G : B \to \map \CC {\Phi_B}$ by:
:$\map G a = \hat a$
for each $a \in B$. 
By Gelfand-Naimark Theorem: Commutative Case, $G$ is an isometric unital $\ast$-algebra isomorphism.
From Inverse of Isometric *-Algebra Isomorphism is Isometric *-Algebra Isomorphism, $G^{-1} : \map \CC {\Phi_B} \to B$ is an isometric unital $\ast$-algebra isomorphism.
Define $T : \map \CC {\map {\sigma_B} x} \to \map \CC {\Phi_B}$ by: 
:$T \psi = \psi \circ \hat x$ 
for each $\psi \in \map \CC {\map {\sigma_B} x}$.
Further, $\map T 1 = 1 \circ \hat x = 1$, so $T$ is unital.
From Spaces of Continuous Functions on Homeomorphic Compact Hausdorff Spaces are Isometrically *-Algebra Isomorphic, $T$ is an isometric $\ast$-algebra isomorphism.
From Composition of Isometric *-Algebra Isomorphisms is Isometric *-Algebra Isomorphism, $G^{-1} \circ T : \map \CC {\map {\sigma_B} x} \to B$ is an isometric unital $\ast$-algebra isomorphism.
Set $\Theta_x = G^{-1} \circ T$. 
We prove that $\Theta_x$ is the desired map.
We now only need to prove that $\map {\Theta_x} \iota = x$.
We have: 
Hence, recalling $\map {\sigma_B} x = \map {\sigma_A} x$ and $B \subseteq A$, $\Theta_x : \map \CC {\map {\sigma_A} x} \to A$ is a continuous functional calculus for $x$. 
Since:
:$B = \span \set {x^n \paren {x^\ast}^m : n, m \ge 0}^-$
we have:
:$\map {\Theta_x} f \in \span \set {x^n \paren {x^\ast}^m : n, m \ge 0}^-$ for each $f \in \map \CC {\map {\sigma_A} x}$.
That $\map {\Theta_x} f$ is normal follows immediately from the fact that $B$ is commutative.
To conclude, we only need to show that $\Theta_x$ is unique.
Suppose that $\Gamma_x : \map \CC {\map {\sigma_A} x} \to A$ is a unital $\ast$-algebra homomorphism with:
:$\map {\Gamma_x} \iota = x = \map {\Theta_x} \iota$
Since $\Gamma_x$ and $\Theta_x$ are $\ast$-algebra homomorphisms, we have:
:$\map {\Gamma_x} {\overline \iota} = x^\ast = \map {\Theta_x} {\overline \iota}$ 
Then, since $\Gamma_x$ and $\Theta_x$ are algebra homomorphisms, we have:
:$\map {\Gamma_x} p = \map p {x, x^\ast} = \map {\Theta_x} p$
where:
:$\ds \map p z = \sum_{j = 1}^n \alpha_j z^{n_j} \overline z^{m_j}$
for $z \in \map \CC {\map {\sigma_A} x}$ and:
:$\ds \map p {x, x^\ast} = \sum_{j = 1}^n \alpha_j x^{n_j} \paren {x^\ast}^{m_j}$
where $\alpha_j \in \C$ and $n_j, m_j \in \Z_{\ge 0}$ and $n \in \N$.
Write:
:$\AA = \span \set {p_{n, m} : n, m \ge 0}$ 
where:
:$\map {p_{n, m} } z = z^n \overline z^m$ 
We have therefore shown that $\map {\Gamma_x} f = \map {\Theta_x} f$ for all $f \in \AA$. 
From Subalgebra Generated by Inclusion and Conjugate is Everywhere Dense in Space of Continuous Functions on Compact Subset of Complex Numbers, $\AA$ is dense in $\map \CC {\map {\sigma_A} x}$.
Hence from Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide, we have:
:$\map {\Theta_x} f = \map {\Gamma_x} f$ 
for each $f \in \map \CC {\map {\sigma_A} x}$. 
So $\Theta_x = \Gamma_x$. 
Hence the continuous functional calculus is unique.
Category:Continuous Functional Calculus
Category:C*-Algebras