34.7 The Continuous Functional Calculus
Theorem 34.7.1 (Spectral Theorem for $C^{*}$-Algebras).label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be normal, then the mapping
is a homeomorphism.
Proof. Firstly, $A[x]$ is commutative by Proposition 34.2.2. Thus Corollary 34.2.4 and (3) of Proposition 33.8.2 imply that
and $\Phi$ is a surjection onto $\sigma_{A}(x)$.
On the other hand, the Gelfand-Naimark Theorem implies that $\Phi(x^{*}) = \ol{\Phi(x)}$, so since $A[x]$ is the smallest $C^{*}$-algebra containing $x$, any element $\psi \in \Omega(A[x])$ is uniquely determined by $\psi(x)$. Therefore $\Phi$ is injective.
Finally, since $\Omega(A[x])$ is equipped with the weak* topology and $\Phi$ is the evaluation map at $x$, it is continuous.
By Proposition 5.16.5, $\Phi$ is a homeomorphism.$\square$
Definition 34.7.2 (Continuous Functional Calculus).label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be normal, then there exists a unique continuous unital *-homomorphism
such that:
- (1)
$\one(x) = 1_{A}$.
- (2)
$\text{Id}(x) = x$.
- (3)
$\overline{\text{Id}}(x) = x^{*}$.
Moreover,
- (4)
Under the identification $\Omega(A[x]) = \sigma_{A}(x)$, for each $f \in C(\sigma_{A}(x); \complex)$, $f(x) = \Gamma^{-1}_{A[x]}(f)$.
- (5)
The mapping $f \mapsto f(x)$ is a unital *-isomorphism.
- (6)
For each $U \subset \complex$ open and $f \in C(U; \complex)$, the mapping
\[\bracs{x \in A|\sigma_A(x) \subset U}\to A \quad x \mapsto f(x)\]is continuous.
- (7)
Let $B$ be a unital $C^{*}$-algebra and $\Phi: A \to B$ be a unital *-homomorphism, then for any $x \in A$ and $f \in C(\sigma_{A}(x); \complex)$, $\Phi(f(x)) = f(\Phi(x))$.
Proof. (1), (4), (5): By the Spectral Theorem, $\Omega(A[x])$ and $\sigma_{A}(x)$ may be identified. For each $f \in C(\sigma_{A}(x); \complex)$, define $f(x) = \Gamma^{-1}_{A[x]}(f)$, then the mapping $f \mapsto f(x)$ is a unital *-isomorphism by the Gelfand-Naimark Theorem.
(2): The identification
given by the Spectral Theorem implies that $\Gamma_{A[x]}(x) = \text{Id}$.
(3): The mapping $f \mapsto f(x)$ is a *-homomorphism.
(6): Fix $x \in A$ and let $K \in \cn_{U}(\sigma_{A}(x); \complex)$ be compact. By Proposition 33.5.10, there exists $r > 0$ such that $\sigma_{A}(y) \subset K$ for all $y \in B_{A}(x, r)$.
Let $\eps > 0$. By the Stone-Weierstrass Theorem, there exists $p \in \complex[z, \ol z]$ such that $\sup_{\lambda \in K}|p(\lambda) - f(\lambda)| < \eps$. In which case,
for all $y \in B_{A}(x, r)$. Since $p$ is a continuous map on $A$, there exists $\delta \in (0, r)$ such that $\norm{p(x) - p(y)}_{A} < \eps$ for all $y \in B_{A}(x, \delta)$. Therefore $\norm{f(x) - f(y)}_{A} < 3\eps$ for all $y \in B_{A}(x, \delta)$.
(Uniqueness): By the Stone-Weierstrass Theorem.$\square$
Theorem 34.7.3 (Spectral Mapping Theorem (Continuous)).label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be normal, then:
- (1)
For every $f \in C(\sigma_{A}(x); \complex)$, $\sigma_{A}(f(x)) = f(\sigma_{A}(x))$.
- (2)
For every $f \in C(\sigma_{A}(x); \complex)$ and $g \in C(\sigma_{A}(f(x)); \complex)$, $(f \circ g)(x) = f(g(x))$.
Proof. (1): Since $f(x) \in A[x]$, by Proposition 33.8.2 and definition of the continuous functional calculus,
(2): Firstly, the theorem holds directly if $f, g \in \complex[z, \ol z]$ are polynomials in $z$ and $\ol z$.
Suppose that $g \in \complex[z, \ol z]$ but $f$ is arbitrary. By the Stone-Weierstrass Theorem, there exists $\seq{p_n}\subset \complex[z, \ol z]$ such that $p_{n} \to f$ uniformly on $\sigma_{A}(x)$. By property (6) of the continuous functional calculus,
Finally, suppose that both $f$ and $g$ are arbitary. By the Stone-Weierstrass Theorem, there exists $\seq{p_n}\subset \complex[z, \ol z]$ such that $p_{n} \to g$ uniformly on $\sigma_{A}(f(x))$. By continuity of the continuous functional calculus,
$\square$
Corollary 34.7.4.label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be normal, then:
- (1)
$x$ is self-adjoint if and only if $\sigma(x) \subset \real$.
- (2)
$x$ is unitary if and only if $\sigma(x) \subset \partial B_{\complex}(0, 1)$.
- (3)
$x$ is a projection if and only if $\sigma(x) \subset \bracs{0, 1}$.
Corollary 34.7.5.label Let $A$ be a unital $C^{*}$-algebra, then
- (1)
For any $x \in A_{sa}$, there exists a unitary element $u \in A$ such that $x = (u + u^{*})/(2\norm{x}_{A})$.
- (2)
For any $x \in A$, there exists unitary elements $u, v \in A$ such that $x = (u + u^{*})/(\norm{x}_{A}) + i(v + v^{*})/(2\norm{x}_{A})$.
Proof. (1): Assume without loss of generality that $\norm{x}_{A} \le 1$. In which case, $\sigma(x) \subset [-1, 1]$, and $f(\lambda) = \lambda + i\sqrt{1 - \lambda^{2}}$ is defined and continuous on $[-1, 1]$. Furthermore, $|f(\lambda)| = 1$ for all $\lambda \in [-1, 1]$. Thus Corollary 34.7.4 implies that $f(x)$ is unitary. Finally, since $f + \ol f = 2\text{Id}$, $x = (f(x) + \ol{f(x)})/(2\norm{x}_{A})$.$\square$
Corollary 34.7.6.label Let $A, B$ be unital $C^{*}$-algebras and $\Phi: A \to B$ be an injective unital *-homomorphism, then for each $x \in A$,
- (1)
$\sigma_{B}(\Phi(x)) = \sigma_{A}(x)$.
- (2)
$\norm{\Phi(x)}_{B} = \norm{x}_{A}$.
Proof, [II.10.7, Zhu93]. (1): Since $\Phi(G(A)) \subset G(B)$, $\sigma_{B}(\Phi(x)) \subset \sigma_{A}(x)$. If $\sigma_{B}(\Phi(x)) \subsetneq \sigma_{A}(x)$, then Urysohn’s Lemma implies that there exists $C(\sigma_{A}(x); \complex)$ such that $f|_{\sigma_B(\Phi(x))}= 0$ but $f \ne 0$. In which case, by (7) of the continuous functional calculus, $\Phi(f(x)) = f(\Phi(x)) = 0$, which contradicts the fact that $\Phi$ is injective.
(2): By Corollary 34.4.5, $\Phi$ is isometric.$\square$
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