From 8d03304d2d0991f3cfe2d46a9e3ecbfe4fd8b5c3 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Mon, 9 Nov 2026 13:43:01 -0500 Subject: [PATCH] First round of corrections. --- src/vna/vn/unlimited.tex | 28 ++++++++++++++-------------- 1 file changed, 14 insertions(+), 14 deletions(-) diff --git a/src/vna/vn/unlimited.tex b/src/vna/vn/unlimited.tex index 5e01f60..7f5d9eb 100644 --- a/src/vna/vn/unlimited.tex +++ b/src/vna/vn/unlimited.tex @@ -52,7 +52,7 @@ \begin{enumerate} \item For any $\xi \in D(T)$, \[ - \norm{(T \pm i)\xi}_H^2 = \norm{T\xi}_H^2 + \norm{\xi}_H^2 \ge \norm{\xi}_H + \norm{(T \pm i)\xi}_H^2 = \norm{T\xi}_H^2 + \norm{\xi}_H^2 \ge \norm{\xi}_H^2 \] In particular, $T \pm i$ is injective. @@ -65,23 +65,23 @@ (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 + i\dpn{T\xi, \xi}{H} -i \dpn{\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, closed range): For any $\seq{\xi_n} \subset D(T)$ 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). + As $D(T)$ is dense, $(T - i)\eta = 0$, 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 + Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a densely defined, self-adjoint operator, then \[ U(T): H \to H \quad U(T) = (T - i)(T + i)^{-1} \] @@ -96,7 +96,7 @@ \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 \[ - D\paren{\int_X f dE} = \bracs{\xi \in H \bigg | \int_X |f|dE_{\xi, \xi} < \infty} + D\paren{\int_X f dE} = \bracs{\xi \in H \bigg | \int_X |f|^2 dE_{\xi, \xi} < \infty} \] and @@ -114,7 +114,7 @@ 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. + \item For any Borel measurable function $f: X \to \complex$, $(\int_X f dE)^* = \int_X \ol f dE$. In particular, $\int_X f dE$ is always closed. \end{enumerate} \end{definition} \begin{proof} @@ -133,12 +133,12 @@ For any $\phi \in \Sigma(X; \complex)$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, \[ - \int_X |\phi| d|E_{\xi, \eta}| \le \norm{\eta}_H\sum_{\alpha \in \phi(X)} \alpha E_{\xi, \xi}(\bracsn{\phi = \alpha})^{1/2} \le \norm{\eta}_H\int_X |\phi|^2 dE_{\xi, \xi} + \int_X |\phi| d|E_{\xi, \eta}| \le \norm{\eta}_H\sum_{\alpha \in \phi(X)} |\alpha| E_{\xi, \xi}(\bracsn{\phi = \alpha})^{1/2} \le \norm{\eta}_H\braks{\int_X |\phi|^2 dE_{\xi, \xi}}^{1/2} \] By the \hyperref[Monotone Convergence Theorem]{theorem:mct}, \[ - \int_X |f| d|E_{\xi, \eta}| \le \norm{\eta}_H \int_X |f|^2 dE_{\xi, \xi} + \int_X |f| d|E_{\xi, \eta}| \le \norm{\eta}_H \braks{\int_X |f|^2 dE_{\xi, \xi}}^{1/2} \] so $\int_X f dE \xi \in H$ is well-defined. @@ -163,7 +163,7 @@ (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} + &= \ol{\dpn{\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), @@ -188,14 +188,14 @@ 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, + 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 $U(T) = \int_{\partial B_\complex(0, 1)} \lambda E(d\lambda)$. Moreover, \[ - (I - U) = (T+i)(T+i)^{-1} - (T - i)(T+i)^{-1} = 2i(T + i)^{-1}x + (I - U) = (T+i)(T+i)^{-1} - (T - i)(T+i)^{-1} = 2i(T + i)^{-1} \] is injective, so $1$ is not an eigenvalue of $U$, and $E(\bracsn{1}) = 0$. Let \[ - \phi: \partial B_\complex(0, 1) \setminus 1 \to \real \quad \lambda \mapsto i\frac{1 + \lambda}{1 - \lambda} + \phi: \partial B_\complex(0, 1) \setminus \bracsn{1} \to \real \quad \lambda \mapsto i\frac{1 + \lambda}{1 - \lambda} \] then the pushforward $F = \phi_* E$ defined by $F(B) = E(\phi^{-1}(B))$ is also a spectral measure, with @@ -213,7 +213,7 @@ 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 $|T| = (T^*T)^{1/2}$. + \item There exists a unique partial isometry $U \in B(H)$ with $\ker U = \ker T$ such that $T = U|T|$, where $|T| = (T^*T)^{1/2}$. \end{enumerate} \end{theorem} % Omitted until further notice.