Unlimited FC, first draft.
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Bokuan Li
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\section{Unbounded Functional Calculus}
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\label{section:unbounded-functional-calculus}
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\begin{definition}[Closed]
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\label{definition:unbounded-closed}
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Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a linear operator, then $T$ is \textbf{closed} if its graph
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\[
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\Gamma(T) = \bracsn{(x, y) \in D(T) \times H| Tx = y}
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\]
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is a closed subspace of $H \times H$.
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The operator $T$ is \textbf{closable} if it admits a closed extension $S \supset T$.
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\end{definition}
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\begin{definition}[Adjoint]
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\label{definition:unbounded-adjoint}
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Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a densely defined linear operator,
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\[
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D(T^*) = \bracsn{\xi \in H| (\eta \mapsto \dpn{\xi, T\eta}{H}) \text{ is bounded on }D(T)}
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\]
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and\footnote{$\eta$ sampled from $D(T)$.}
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\[
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T^*: D(T^*) \to H \quad \dpn{T^*\xi, \eta}{H} = \dpn{\xi, T\eta}{H}
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\]
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then $T^*$ is a closed operator, known as the \textbf{adjoint} of $T$.
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\end{definition}
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\begin{proof}
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Let $\seq{\xi_n} \subset D(T^*)$ and $\xi, \eta \in H$ such that $\xi_n \to \xi$ and $T^*\xi_n \to \eta$, then for any $\theta \in D(T)$,
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\[
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\dpn{\eta, \theta}{H} = \limv{n}\dpn{T^*\xi_n, \theta}{H} = \limv{n}\dpn{\xi_n, T\theta}{H} = \dpn{\xi, T\theta}{H} = \dpn{T^*\xi, \theta}{H}
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\]
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As $D(T)$ is dense, $\eta = T^*\xi$, so the graph of $T^*$ is closed.
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\end{proof}
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\begin{definition}[Self-Adjoint]
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\label{definition:unbounded-self-adjoint}
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Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a densely defined linear operator, then $T$ is \textbf{self-adjoint} if $D(T) = D(T^*)$\footnote{Written out to emphasise.} and $T = T^*$.
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\end{definition}
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\begin{lemma}
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\label{lemma:cayley-transform-unbounded}
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Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a densely defined, self-adjoint operator, then:
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\begin{enumerate}
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\item For any $\xi \in D(T)$,
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\[
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\norm{(T \pm i)\xi}_H^2 = \norm{T\xi}_H^2 + \norm{\xi}_H^2 \ge \norm{\xi}_H
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\]
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In particular, $T \pm i$ is injective.
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\item $(T \pm i)(H) = H$.
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\end{enumerate}
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In other words, $(T \pm i)$ is bijective onto $H$.
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\end{lemma}
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\begin{proof}
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(1): For any $\xi \in D(T)$,
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\begin{align*}
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\normn{(T + i)\xi}_H^2 &= \norm{T\xi}_H^2 + \norm{\xi}_H^2 + \dpn{T\xi, i\xi}{H} + \dpn{i\xi, T\xi}{H} \\
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&= \norm{T\xi}_H^2 + \norm{\xi}_H^2 + \dpn{T\xi, i\xi}{H} + \dpn{i\xi, T\xi}{H} \\
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&= \norm{T\xi}_H^2 + \norm{\xi}_H^2 \ge \normn{\xi}_H^2
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\end{align*}
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(2, closed range): For any $\seq{\xi_n} \subset H$ and $\eta \in H$ with $(T + i)\xi_n \to \eta$, (1) implies that $\seq{\xi_n} \subset D(T)$ and $\seq{T\xi_n}$ are both Cauchy. Let $\xi = \limv{n}\xi_n$, then since $T$ is self-adjoint, it is closed, so $T\xi = \limv{n}T\xi_n \in T(H)$, and $\eta = (T + i)\xi \in (T+i)(H)$.
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(2, dense range): Let $\eta \in H$ with $\eta \perp (T + i)(H)$, then for each $\xi \in D(T)$,
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\[
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0 = \dpn{(T + i)\xi, \eta}{H} = \dpn{T\xi, \eta}{H} - \dpn{\xi, i\eta}{H} = \dpn{\xi, (T - i)\eta}{H}
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\]
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As $D(T)$ is dense, $(T - 1)\eta = 0$, so $\eta = 0$ by (1).
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\end{proof}
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\begin{definition}[Cayley Transform]
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\label{definition:unbounded-cayley-transform}
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Let $H$ be a complex Hilbert space, $T: D(T) \to T$ be a densely defined, self-adjoint operator, then
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\[
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U(T): H \to H \quad U(T) = (T - i)(T + i)^{-1}
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\]
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is a unitary operator, known as the \textbf{Cayley transform} of $T$.
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\end{definition}
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\begin{proof}
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By \autoref{lemma:cayley-transform-unbounded}, $(T+i)^{-1}: H \to D(T)$ and $(T - i): D(T) \to H$ are well-defined linear bijections, so $U(T)$ is a well-defined linear isomorphism. For any $\xi \in D(T)$, (1) of \autoref{lemma:cayley-transform-unbounded} implies that $\norm{(T + i)\xi}_H = \norm{(T - i)\xi}_H$. As $(T+i)^{-1}$ and $(T - i)$ are both bijections, $U(T)$ is an isometry. Therefore $U(T)$ is unitary.
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\end{proof}
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\begin{definition}[Unbounded Integration Against Spectral Measure]
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\label{definition:unbounded-spectral-integral}
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Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, $E: \cb_X \to B(H)$ be a spectral measure relative to $H$. For each Borel measurable function $f: X \to \complex$, let
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\[
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D\paren{\int_X f dE} = \bracs{\xi \in H \bigg | \int_X |f|dE_{\xi, \xi} < \infty}
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\]
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and
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\[
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\int_X f dE: D\paren{\int_X f dE} \to H \quad \angles{\int_X fdE \xi, \eta}_H = \int_X f dE_{\xi, \eta}
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\]
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Then,
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\begin{enumerate}
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\item $\int_X f dE: D(\int_Xf dE) \to H$ is a densely defined linear operator.
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\item For any Borel measurable functions $f, g: X \to \complex$,
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\[
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D\paren{\int_X f dE \int_X g dE} = D\paren{\int_X g dE} \cap D\paren{\int_X fg dE}
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\]
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and $\int_X f dE \int_X g dE = \int_X fg dE$.
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\item For any Borel measurable functions $f: X \to \complex$, $(\int_X f dE)^* = \int_X \ol f dE$. In particular, $\int_X f dE$ is always closed.
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\end{enumerate}
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\end{definition}
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\begin{proof}
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(1): For each $n \in \natp$, $\one_{\bracsn{f \le n}}f \in B^\infty(X; \complex)$. Thus for each $\xi \in H$, by a monotone class argument,
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\begin{align*}
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\int_X |f|^2 dE_{E(\bracsn{f \le n}\xi, \bracsn{f \le n}\xi)} &= \int_{\bracsn{f \le n}}|f|^2 dE_{\xi, \xi} < \infty \\
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\normn{\xi - E(\bracsn{f \le n})\xi}_H^2 &= \normn{E(\bracsn{f > n})\xi}_H^2
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\end{align*}
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where by \hyperref[continuity from above]{proposition:measure-properties}, $E(\bracsn{f > n})\xi \to 0$ as $n \to \infty$. Therefore $D(\int_X f dE)$ is dense in $H$.
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It remains to show that $\int_X f dE$ is defined on the claimed domain. Let $\xi \in D(\int_X f dE)$, $\eta \in H$, and $B \in \cb_X$, then by (1) and (3) of \autoref{lemma:spectral-measure-properties},
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\[
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\int_X \one_B d|E_{\xi, \eta}| \le E_{\xi, \xi}(B)^{1/2}E_{\eta, \eta}(B)^{1/2} \le E_{\xi, \xi}(B)^{1/2}\norm{\eta}_H
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\]
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For any $\phi \in \Sigma(X; \complex)$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz},
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\[
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\int_X |\phi| d|E_{\xi, \eta}| \le \norm{\eta}_H\sum_{\alpha \in \phi(X)} \alpha E_{\xi, \xi}(\bracsn{\phi = \alpha})^{1/2} \le \norm{\eta}_H\int_X |\phi|^2 dE_{\xi, \xi}
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\]
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By the \hyperref[Monotone Convergence Theorem]{theorem:mct},
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\[
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\int_X |f| d|E_{\xi, \eta}| \le \norm{\eta}_H \int_X |f|^2 dE_{\xi, \xi}
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\]
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so $\int_X f dE \xi \in H$ is well-defined.
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(2): For each $\xi \in D(\int_X g dE)$, by a monotone class argument,
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\[
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\int_X |fg|^2 dE_{\xi, \xi} = \int_X |f|^2 |g|^2 dE_{\xi, \xi} = \int_X |f|^2 dE_{\int_X gdE\xi, \int_X gdE\xi}
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\]
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so $\int_X g dE \xi \in D(\int_X f dE)$ if and only if $\xi \in D(\int_X fg dE)$, and
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\[
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D\paren{\int_X f dE \int_X g dE} = D\paren{\int_X g dE} \cap D\paren{\int_X fg dE}
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\]
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Let $\xi \in D(\int_X g dE) \cap D(\int_X fg dE)$ and $\eta \in H$, then
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\begin{align*}
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\angles{\int_X f dE \int_X g dE \xi, \eta}_H &= \int_X f dE_{\int_X g dE \xi, \eta} = \int_X fg dE_{\xi, \eta} \\
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&= \angles{\int_X f g dE \xi, \eta}_H
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\end{align*}
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(3): For any $\xi, \eta \in D(\int_X f dE) = D(\int_X \ol f dE)$,
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\begin{align*}
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\angles{\int_X f dE\xi, \eta}_H &= \int_X f dE_{\xi, \eta} = \ol{\int_X \ol f dE_{\eta, \xi}} \\
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&= \ol{\int_X \ol f dE\eta, \xi}_H = \dpn{\xi, \int_X \ol f dE \eta}{H}
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\end{align*}
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so $\int_X \ol f dE \subset (\int_X f dE)^*$. On the other hand, for each $\xi \in H$ and $\eta \in D[(\int_X f dE)^*]$, by the \hyperref[bounded $L^\infty$ functional calculus]{definition:linfty-functional-calculus} and (2),
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\begin{align*}
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\angles{\xi, \int_X \frac{\ol f}{1 + |f|} dE\eta}_H &= \angles{\int_X \frac{f}{1 + |f|} dE\xi, \eta}_H \\
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&= \angles{\int_X f dE \int_X (1 + |f|)^{-1}dE\xi, \eta}_H \\
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&= \angles{ \int_X (1 + |f|)^{-1}dE\xi, \paren{\int_X f dE}^{*}\eta}_H \\
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&= \angles{\xi, \int_X (1 + |f|)^{-1}dE\paren{\int_X f dE}^{*}\eta}_H \\
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\end{align*}
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so
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\[
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\int_X \frac{\ol f}{1 + |f|} dE = \int_X (1 + |f|)^{-1}dE\paren{\int_X f dE}^{*}
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\]
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By (2) and multiplication of $\int_X (1 + |f|) dE$ from the left, $\int_X \ol f dE = \paren{\int_X f dE}^{*}$.
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\end{proof}
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\begin{theorem}[Spectral Theorem I (Unbounded)]
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\label{theorem:spectral-theorem-1-unbounded}
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Let $H$ be a complex Hilbert space and $T: D(T) \to H$ be a densely defined, self-adjoint operator, then there exists a unique spectral measure $E: \cb_\real \to B(H)$ such that $T = \int_{\real} \lambda E(d\lambda)$.
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\end{theorem}
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\begin{proof}
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Let $U(T) = (T - i)(T + i)^{-1} \in U(B(H))$ be the Cayley transform of $T$, then $U(T)$ is unitary. By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1}, there exists a unique spectral measure $E: \cb_{\partial B_\complex(0, 1)} \to B(H)$ such that $U = \int_{\partial B_\complex(0, 1)} \lambda E(d\lambda)$. Moreover,
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\[
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(I - U) = (T+i)(T+i)^{-1} - (T - i)(T+i)^{-1} = 2i(T + i)^{-1}x
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\]
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is injective, so $1$ is not an eigenvalue of $U$, and $E(\bracsn{1}) = 0$. Let
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\[
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\phi: \partial B_\complex(0, 1) \setminus 1 \to \real \quad \lambda \mapsto i\frac{1 + \lambda}{1 - \lambda}
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\]
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then the pushforward $F = \phi_* E$ defined by $F(B) = E(\phi^{-1}(B))$ is also a spectral measure, with
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\begin{align*}
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\int_\real \lambda F(d\lambda) &= \int_{\partial B_\complex(0, 1) \setminus \bracsn{1}} i\frac{1 + \lambda}{1 - \lambda} E(d\lambda) = i(1 + U)(1 - U)^{-1} \\
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&= i[(T + i)(T+i)^{-1} + (T-i)(T+i)^{-1}] \cdot (2i(T + i)^{-1})^{-1} \\
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&= \frac{1}{2}[(T + i)(T+i)^{-1}(T+i) + (T-i)(T+i)^{-1}(T+i)] = T
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\end{align*}
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\end{proof}
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\begin{theorem}[Polar Decomposition (Unbounded)]
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\label{theorem:unbounded-polar-decomposition}
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Let $H$ be a complex Hilbert space, $T: H \to H$ be a closed, densely defined linear operator, then:
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\begin{enumerate}
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\item $T^*T$ is self-adjoint.
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\item There exists a unique partial isometry $U \in B(H)$ such that $T = U|T|$, where $|T| = (T^*T)^{1/2}$.
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\end{enumerate}
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\end{theorem}
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% Omitted until further notice.
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