From d044d312d89b583a52e5fd1891d2a3b71d1bdefc Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Sun, 8 Nov 2026 19:29:23 -0500 Subject: [PATCH] Unlimited FC, first draft. --- src/vna/vn/unlimited.tex | 220 +++++++++++++++++++++++++++++++++++++++ 1 file changed, 220 insertions(+) create mode 100644 src/vna/vn/unlimited.tex diff --git a/src/vna/vn/unlimited.tex b/src/vna/vn/unlimited.tex new file mode 100644 index 0000000..5e01f60 --- /dev/null +++ b/src/vna/vn/unlimited.tex @@ -0,0 +1,220 @@ +\section{Unbounded Functional Calculus} +\label{section:unbounded-functional-calculus} + + + +\begin{definition}[Closed] +\label{definition:unbounded-closed} + Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a linear operator, then $T$ is \textbf{closed} if its graph + \[ + \Gamma(T) = \bracsn{(x, y) \in D(T) \times H| Tx = y} + \] + + is a closed subspace of $H \times H$. + + The operator $T$ is \textbf{closable} if it admits a closed extension $S \supset T$. +\end{definition} + + +\begin{definition}[Adjoint] +\label{definition:unbounded-adjoint} + Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a densely defined linear operator, + \[ + D(T^*) = \bracsn{\xi \in H| (\eta \mapsto \dpn{\xi, T\eta}{H}) \text{ is bounded on }D(T)} + \] + + and\footnote{$\eta$ sampled from $D(T)$.} + \[ + T^*: D(T^*) \to H \quad \dpn{T^*\xi, \eta}{H} = \dpn{\xi, T\eta}{H} + \] + + then $T^*$ is a closed operator, known as the \textbf{adjoint} of $T$. +\end{definition} +\begin{proof} + 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 $\theta \in D(T)$, + \[ + \dpn{\eta, \theta}{H} = \limv{n}\dpn{T^*\xi_n, \theta}{H} = \limv{n}\dpn{\xi_n, T\theta}{H} = \dpn{\xi, T\theta}{H} = \dpn{T^*\xi, \theta}{H} + \] + + As $D(T)$ is dense, $\eta = T^*\xi$, so the graph of $T^*$ is closed. +\end{proof} + +\begin{definition}[Self-Adjoint] +\label{definition:unbounded-self-adjoint} + 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 $T = T^*$. +\end{definition} + + + +\begin{lemma} +\label{lemma:cayley-transform-unbounded} + Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a densely defined, self-adjoint operator, then: + \begin{enumerate} + \item For any $\xi \in D(T)$, + \[ + \norm{(T \pm i)\xi}_H^2 = \norm{T\xi}_H^2 + \norm{\xi}_H^2 \ge \norm{\xi}_H + \] + + In particular, $T \pm i$ is injective. + \item $(T \pm i)(H) = H$. + \end{enumerate} + + In other words, $(T \pm i)$ is bijective onto $H$. +\end{lemma} +\begin{proof} + (1): For any $\xi \in D(T)$, + \begin{align*} + \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} \\ + &= \norm{T\xi}_H^2 + \norm{\xi}_H^2 + \dpn{T\xi, i\xi}{H} + \dpn{i\xi, T\xi}{H} \\ + &= \norm{T\xi}_H^2 + \norm{\xi}_H^2 \ge \normn{\xi}_H^2 + \end{align*} + + (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 $\eta = (T + i)\xi \in (T+i)(H)$. + + (2, dense range): Let $\eta \in H$ with $\eta \perp (T + i)(H)$, then for each $\xi \in D(T)$, + \[ + 0 = \dpn{(T + i)\xi, \eta}{H} = \dpn{T\xi, \eta}{H} - \dpn{\xi, i\eta}{H} = \dpn{\xi, (T - i)\eta}{H} + \] + + As $D(T)$ is dense, $(T - 1)\eta = 0$, so $\eta = 0$ by (1). +\end{proof} + +\begin{definition}[Cayley Transform] +\label{definition:unbounded-cayley-transform} + Let $H$ be a complex Hilbert space, $T: D(T) \to T$ be a densely defined, self-adjoint operator, then + \[ + U(T): H \to H \quad U(T) = (T - i)(T + i)^{-1} + \] + + is a unitary operator, known as the \textbf{Cayley transform} of $T$. +\end{definition} +\begin{proof} + 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. +\end{proof} + +\begin{definition}[Unbounded Integration Against Spectral Measure] +\label{definition:unbounded-spectral-integral} + 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 + \[ + D\paren{\int_X f dE} = \bracs{\xi \in H \bigg | \int_X |f|dE_{\xi, \xi} < \infty} + \] + + and + \[ + \int_X f dE: D\paren{\int_X f dE} \to H \quad \angles{\int_X fdE \xi, \eta}_H = \int_X f dE_{\xi, \eta} + \] + + Then, + \begin{enumerate} + \item $\int_X f dE: D(\int_Xf dE) \to H$ is a densely defined linear operator. + \item For any Borel measurable functions $f, g: X \to \complex$, + \[ + D\paren{\int_X f dE \int_X g dE} = D\paren{\int_X g dE} \cap D\paren{\int_X fg dE} + \] + + and $\int_X f dE \int_X g dE = \int_X fg dE$. + + \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. + \end{enumerate} +\end{definition} +\begin{proof} + (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, + \begin{align*} + \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 \\ + \normn{\xi - E(\bracsn{f \le n})\xi}_H^2 &= \normn{E(\bracsn{f > n})\xi}_H^2 + \end{align*} + + 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 $H$. + + 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}, + \[ + \int_X \one_B d|E_{\xi, \eta}| \le E_{\xi, \xi}(B)^{1/2}E_{\eta, \eta}(B)^{1/2} \le E_{\xi, \xi}(B)^{1/2}\norm{\eta}_H + \] + + For any $\phi \in \Sigma(X; \complex)$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, + \[ + \int_X |\phi| d|E_{\xi, \eta}| \le \norm{\eta}_H\sum_{\alpha \in \phi(X)} \alpha E_{\xi, \xi}(\bracsn{\phi = \alpha})^{1/2} \le \norm{\eta}_H\int_X |\phi|^2 dE_{\xi, \xi} + \] + + By the \hyperref[Monotone Convergence Theorem]{theorem:mct}, + \[ + \int_X |f| d|E_{\xi, \eta}| \le \norm{\eta}_H \int_X |f|^2 dE_{\xi, \xi} + \] + + so $\int_X f dE \xi \in H$ is well-defined. + + (2): For each $\xi \in D(\int_X g dE)$, by a monotone class argument, + \[ + \int_X |fg|^2 dE_{\xi, \xi} = \int_X |f|^2 |g|^2 dE_{\xi, \xi} = \int_X |f|^2 dE_{\int_X gdE\xi, \int_X gdE\xi} + \] + + so $\int_X g dE \xi \in D(\int_X f dE)$ if and only if $\xi \in D(\int_X fg dE)$, and + \[ + D\paren{\int_X f dE \int_X g dE} = D\paren{\int_X g dE} \cap D\paren{\int_X fg dE} + \] + + + Let $\xi \in D(\int_X g dE) \cap D(\int_X fg dE)$ and $\eta \in H$, then + \begin{align*} + \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} \\ + &= \angles{\int_X f g dE \xi, \eta}_H + \end{align*} + + (3): For any $\xi, \eta \in D(\int_X f dE) = D(\int_X \ol f dE)$, + \begin{align*} + \angles{\int_X f dE\xi, \eta}_H &= \int_X f dE_{\xi, \eta} = \ol{\int_X \ol f dE_{\eta, \xi}} \\ + &= \ol{\int_X \ol f dE\eta, \xi}_H = \dpn{\xi, \int_X \ol f dE \eta}{H} + \end{align*} + + 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), + \begin{align*} + \angles{\xi, \int_X \frac{\ol f}{1 + |f|} dE\eta}_H &= \angles{\int_X \frac{f}{1 + |f|} dE\xi, \eta}_H \\ + &= \angles{\int_X f dE \int_X (1 + |f|)^{-1}dE\xi, \eta}_H \\ + &= \angles{ \int_X (1 + |f|)^{-1}dE\xi, \paren{\int_X f dE}^{*}\eta}_H \\ + &= \angles{\xi, \int_X (1 + |f|)^{-1}dE\paren{\int_X f dE}^{*}\eta}_H \\ + \end{align*} + + so + \[ + \int_X \frac{\ol f}{1 + |f|} dE = \int_X (1 + |f|)^{-1}dE\paren{\int_X f dE}^{*} + \] + + By (2) and multiplication of $\int_X (1 + |f|) dE$ from the left, $\int_X \ol f dE = \paren{\int_X f dE}^{*}$. +\end{proof} + + +\begin{theorem}[Spectral Theorem I (Unbounded)] +\label{theorem:spectral-theorem-1-unbounded} + 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 $T = \int_{\real} \lambda E(d\lambda)$. +\end{theorem} +\begin{proof} + 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, + \[ + (I - U) = (T+i)(T+i)^{-1} - (T - i)(T+i)^{-1} = 2i(T + i)^{-1}x + \] + + is injective, so $1$ is not an eigenvalue of $U$, and $E(\bracsn{1}) = 0$. Let + \[ + \phi: \partial B_\complex(0, 1) \setminus 1 \to \real \quad \lambda \mapsto i\frac{1 + \lambda}{1 - \lambda} + \] + + then the pushforward $F = \phi_* E$ defined by $F(B) = E(\phi^{-1}(B))$ is also a spectral measure, with + \begin{align*} + \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} \\ + &= i[(T + i)(T+i)^{-1} + (T-i)(T+i)^{-1}] \cdot (2i(T + i)^{-1})^{-1} \\ + &= \frac{1}{2}[(T + i)(T+i)^{-1}(T+i) + (T-i)(T+i)^{-1}(T+i)] = T + \end{align*} + +\end{proof} + + +\begin{theorem}[Polar Decomposition (Unbounded)] +\label{theorem:unbounded-polar-decomposition} + Let $H$ be a complex Hilbert space, $T: H \to H$ be a closed, densely defined linear operator, then: + \begin{enumerate} + \item $T^*T$ is self-adjoint. + \item There exists a unique partial isometry $U \in B(H)$ such that $T = U|T|$, where $|T| = (T^*T)^{1/2}$. + \end{enumerate} +\end{theorem} +% Omitted until further notice. +