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First round of corrections.

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This commit is contained in:
Bokuan Li
2026-11-09 13:43:01 -05:00
parent 1d6d23b30c
commit 8d03304d2d
1 changed files with 14 additions and 14 deletions
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src/vna/vn/unlimited.tex
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@@ -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.