81 lines
6.0 KiB
TeX
81 lines
6.0 KiB
TeX
\section{The $L^\infty$ Functional Calculus}
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\label{section:linfty-functional-calculus}
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\begin{definition}[$L^\infty$ Functional Calculus]
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\label{definition:linfty-functional-calculus}
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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
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\begin{enumerate}
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\item There exists a unique spectral measure $E: \cb_{\sigma_{B(H)}(T)} \to A$ such that
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\[
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T = \int_{\sigma_{B(H)}(T)}\lambda E(d\lambda) \quad T^* = \int_{\sigma_{B(H)}(T)}\ol \lambda E(d\lambda)
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\]
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\item 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
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\[
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I_E: \mathscr{E}^* \to A \quad \phi \mapsto \phi(T) := \int_{\sigma_{B(H)}(T)}\phi dE
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\]
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is a *-isomorphism.
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\item $I_E: \mathscr{E}^* \to A$ is the unique weak* to weak-operator continuous unital *-homomorphism such that $I_E(\text{Id}) = T$.
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\end{enumerate}
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The spectral measure $E$ is the \textbf{resolution of the identity} for $T$, and the mapping $f \mapsto f(T)$ on $\mathscr{E}^*$ is the \textbf{$L^\infty$-functional calculus} of $T$.
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\end{definition}
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\begin{proof}
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(1), (2): By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1} applied to $B(H)[T]$, there exists a unique spectral measure $E$ on $\sigma_{B(H)}(T)$ such that the mapping
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\[
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I_E: \mathscr{E}^{*} \to A \quad \phi \mapsto \int_{\sigma_{B(H)}(T)} \phi dE
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\]
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is a *-isomorphism that extends the inverse Gelfand transform $\Gamma_{B(H)[T]}^{-1}: C(\sigma_{B(H)}(T); \complex) \to B(H)[T]$.
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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 \autoref{definition:spectral-measure-integral}.
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(3): By uniqueness of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, and (2), the mapping $\phi \mapsto \phi(T)$ is unique.
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\end{proof}
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\begin{theorem}
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\label{theorem:vn-projection-norm-dense}
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Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then:
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\begin{enumerate}
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\item For each normal operator $T \in A$ and $f \in B^\infty(\sigma_{B(H)}(T); \complex)$ with $f(0) = 0$, $f(T) \in A$.
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\item The linear span of projections in $A$ is norm dense in $A$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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(1): Let $B \subset \sigma_{B(H)}(T) \setminus \bracs{0}$ be a Borel set. First suppose that $0 \not\in \ol{B}$. By \hyperref[Urysohn's Lemma]{lemma:urysohn}, there exists $f \in C(\sigma_{B(H)}(T); [0, 1])$ with $f(0) = 0$ and $f|_{\ol B} = 1$. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, 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.
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By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, $h(T) \in A$ 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 \hyperref[$L^\infty$ functional calculus]{definition:linfty-functional-calculus} implies that $\one_B(T) \in A$ as well.
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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.
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By linearity, $g(T) \in A$ for all $g \in \Sigma(\sigma_{B(H)}(T); \complex)$ with $g(0) = 0$. By \autoref{lemma:separable-metric-space-approx-identity}, $\bracsn{g \in \Sigma(\sigma_{B(H)}(T); \complex)|g(0) = 0}$ 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 $f(0) = 0$.
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\end{proof}
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\begin{theorem}
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\label{theorem:von-neumann-group-connected}
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Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then:
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\begin{enumerate}
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\item $G(A)$ is path-connected in the norm topology.
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\item The unitary group of $A$ is path-connected in the norm topology.
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\item $I(A)$ is trivial.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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Using \autoref{theorem:existence-of-projections-vna} and after possibly shrinking $H$, assume without loss of generality that $I \in A$.
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After choosing and extending a branch of the complex logarithm, let $\phi: \complex \to \complex$ be a Borel measurable function such that:
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\begin{enumerate}[label=(\roman*)]
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\item $e^{\phi(z)} = z$ for all $z \in \complex \setminus \bracs{0}$.
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\item For each $0 < r < R$, $\phi$ is bounded on the annulus $\ol{B(0, R)} \setminus B(0, r)$.
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\end{enumerate}
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(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 \hyperref[Borel functional calculus]{definition:linfty-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 $T$.
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(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 \autoref{corollary:spectrum-characterisation-iff}.
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\end{proof}
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