37.5 The $L^{\infty}$ Functional Calculus
Definition 37.5.1 ($L^{\infty}$ Functional Calculus).label Let $H$ be a complex Hilbert space, $T \in B(H)$ be normal, and $A \subset B(H)$ be the smallest von Neumann algebra acting on $H$ containing $T$ and $I$, then
- (1)
There exists a unique spectral measure $E: \cb_{\sigma_{B(H)}(T)}\to A$ such that
\[T = \int_{\sigma_{B(H)}(T)}\lambda E(d\lambda) \quad T^{*} = \int_{\sigma_{B(H)}(T)}\ol \lambda E(d\lambda)\] - (2)
Let $\mathscr{E}\subset M_{R}(\sigma_{B(H)}(T); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and $A \subset B(H)$ be the von Neumann algebra generated by $I$ and $T$, then
\[I_{E}: \mathscr{E}^{*} \to A \quad \phi \mapsto \phi(T) := \int_{\sigma_{B(H)}(T)}\phi dE\]is a *-isomorphism.
- (3)
$I_{E}: \mathscr{E}^{*} \to A$ is the unique weak* to weak-operator continuous unital *-homomorphism such that $I_{E}(\text{Id}) = T$.
The spectral measure $E$ is the resolution of the identity for $T$, and the mapping $f \mapsto f(T)$ on $\mathscr{E}^{*}$ is the $L^{\infty}$-functional calculus of $T$.
Proof. (1), (2): By Spectral Theorem I applied to $B(H)[T]$, there exists a unique spectral measure $E$ on $\sigma_{B(H)}(T)$ such that the mapping
is a *-isomorphism that extends the inverse Gelfand transform $\Gamma_{B(H)[T]}^{-1}: C(\sigma_{B(H)}(T); \complex) \to B(H)[T]$.
For each $\phi \in \mathscr{E}^{*}$, let $\phi(T) = \int_{\sigma_{B(H)}(T)}\phi dE$, then the mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $A$ by Definition 37.4.3.
(3): By uniqueness of the continuous functional calculus, Goldstine’s Theorem, and (2), the mapping $\phi \mapsto \phi(T)$ is unique.$\square$
Theorem 37.5.2.label Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then:
- (1)
For each normal operator $T \in A$ and $f \in B^{\infty}(\sigma_{B(H)}(T); \complex)$ with $f(0) = 0$, $f(T) \in A$.
- (2)
The linear span of projections in $A$ is norm dense in $A$.
Proof. (1): Let $B \subset \sigma_{B(H)}(T) \setminus \bracs{0}$ be a Borel set. First suppose that $0 \not\in \ol{B}$. By Urysohn’s Lemma, there exists $f \in C(\sigma_{B(H)}(T); [0, 1])$ with $f(0) = 0$ and $f|_{\ol B}= 1$. By Goldstine’s Theorem, there exists a net $\angles{g_\gamma}_{\gamma \in C}\subset C(\sigma_{B(H)}(T); \complex)$ such that $g_{\gamma} \to \one_{B}$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$. As $f\one_{B} = \one_{B}$, $fg_{\gamma} \to \one_{B}$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$ as well.
By the Stone-Weierstrass Theorem, $h(T) \in A$ for all $h \in C(\sigma_{B(H)}(T); \complex)$ with $h(0) = 0$. In particular, $fg_{\gamma}(T) \in A$ for all $\gamma \in C$. Thus the $L^{\infty}$ functional calculus implies that $\one_{B}(T) \in A$ as well.
If $B$ is arbitrary, then $\one_{B \setminus B_\complex(0, r)}\to \one_{B}$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$ as $r \downto 0$. As $\one_{B \setminus B_{\complex}(0, r)}(T) \in A$ for all $r > 0$, $\one_{B}(T) \in A$ as well.
By linearity, $g(T) \in A$ for all $g \in \Sigma(\sigma_{B(H)}(T); \complex)$ with $g(0) = 0$. By Lemma 26.6.4, $\bracsn{g \in \Sigma(\sigma_{B(H)}(T); \complex)|g(0) = 0}$ is uniformly dense in $\bracsn{f \in B^\infty(\sigma_{B(H)(T)}; \complex)|f(0) = 0}$. Therefore $f(T) \in A$ for all $f \in B^{\infty}(\sigma_{B(H)}(T); \complex)$ with $f(0) = 0$.$\square$
Theorem 37.5.3.label Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then:
- (1)
$G(A)$ is path-connected in the norm topology.
- (2)
The unitary group of $A$ is path-connected in the norm topology.
- (3)
$I(A)$ is trivial.
Proof. Using Theorem 37.3.1 and after possibly shrinking $H$, assume without loss of generality that $I \in A$.
After choosing and extending a branch of the complex logarithm, let $\phi: \complex \to \complex$ be a Borel measurable function such that:
- (i)
$e^{\phi(z)}= z$ for all $z \in \complex \setminus \bracs{0}$.
- (ii)
For each $0 < r < R$, $\phi$ is bounded on the annulus $\ol{B(0, R)}\setminus B(0, r)$.
(1): Let $T \in G(A)$, then there exists $0 < r < R$ such that $\sigma_{A}(T) \subset \ol{B(0, R)}\setminus B(0, r)$. In which case, $\phi$ is a bounded Borel measurable function on $\sigma_{A}(T)$. By the Borel functional calculus, $\phi(T) \in A$ with $T = e^{\phi(T)}$. In which case, the path $t \mapsto e^{t\phi(T)}$ is a norm-continuous path in $G(A)$ from $I$ to $T$.
(2): In particular, as $e^{t\phi}(\partial B(0, 1)) \subset \partial B(0, 1)$, the spectrum of $e^{t\phi}$ as an element in the domain of the $L^{\infty}$ functional calculus, is contained in $\partial B(0, 1)$. Thus if $T$ is unitary, then the path $t \mapsto e^{t\phi(T)}$ lies in the unitary group of $A$ by Corollary 36.6.3.$\square$
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