Second draft.
This commit is contained in:
Bokuan Li
+129
-6
@@ -163,7 +163,7 @@
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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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(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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\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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\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{\dpn{\int_X \ol f dE\eta, \xi}}_H = \dpn{\xi, \int_X \ol f dE \eta}{H}
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&= \ol{\dpn{\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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\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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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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@@ -207,14 +207,137 @@
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\end{proof}
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\end{proof}
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\begin{lemma}
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\label{lemma:closed-unbonuded-adjoint}
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Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a densely defined linear operator, and
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\[
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\Gamma(T) = \bracsn{(\xi, T\xi)|\xi \in D(T)} \subset H \oplus H
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\]
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be the graph of $T$, then:
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\begin{enumerate}
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\item $\Gamma(T)^\perp = \bracsn{(-T^*\eta, \eta)| \eta \in D(T^*)}$.
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\item If $T$ is closed, then the mapping $I + T^*T: D(T^*T) \to H$ is surjective.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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(1): Let $\Gamma(T) = \bracsn{(\xi, T\xi)| \xi \in D(T)}$ be the graph of $T$, then for any $(\xi, \eta) \in H \oplus H$, the following are equivalent:
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\begin{enumerate}[label=(\roman*)]
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\item $(\xi, \eta) \perp \Gamma(T)$.
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\item For each $\theta \in D(T)$, $\dpn{\eta, T\theta}{H} = -\dpn{\xi, \theta}{H}$.
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\item $\eta \in D(T^*)$, with $\dpn{T^*\eta, \theta}{H} = -\dpn{\xi, \theta}{H}$ for all $\theta \in H$.
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\item $\eta \in D(T^*)$ with $\xi = -T^*\eta$.
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\end{enumerate}
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so $\Gamma(T)^\perp = \bracsn{(-T^*\eta, \eta)| \eta \in D(T^*)}$.
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(2): Given that $T$ is closed, $H \oplus H = \Gamma(T) \oplus \Gamma(T)^\perp$. In particular, for each $\theta \in H$, there exists $\xi \in D(T)$ and $\eta \in D(T^*)$ such that
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\[
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(\theta, 0) = (\xi, T\xi) + (-T^*\eta, \eta)
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\]
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In which case, $T\xi = -\eta \in D(T^*)$, so $\xi \in D(T^*T)$ with
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\[
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\theta = \xi - T^*\eta = \xi + T^*T\xi
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\]
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so $I + T^*T: D(T^*T) \to H$ is surjective.
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\end{proof}
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\begin{theorem}[von Neumann]
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\label{theorem:von-neumann-closed-operator}
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Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a closed, densely defined linear operator, then $T^*T: D(T^*T) \to H$ is a densely defined, self-adjoint, and positive operator.
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\end{theorem}
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\begin{proof}
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For each $\xi, \eta \in D(T^*T)$, by definition of the adjoint,
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\[
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\dpn{T^*T\xi, \eta}{H} = \dpn{T\xi, T\eta}{H} = \dpn{\xi, T^*T\eta}{H}
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\]
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(Densely Defined): Let $\eta \in D(T^*T)^\perp$. By (2) of \autoref{lemma:closed-unbonuded-adjoint}, there exists $\xi \in D(T^*T)$ such that $\eta = \xi + T^*T\xi$. In which case,
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\[
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0 = \dpn{\xi, \eta}{H} = \dpn{\xi, \xi}{H} + \dpn{\xi, T^*T\xi}{H} = \norm{\xi}_H^2 + \norm{T\xi}_H^2
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\]
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so $T\xi = \xi = 0$, and $\eta = 0$ as well.
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($T^*T \subset (T^*T)^*$): For each $\xi, \eta \in D(T^*T)$, $\dpn{T^*T\xi, \eta}{H} = \dpn{\xi, T^*T\eta}{H}$, so $T^*T \subset (T^*T)^*$.
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($T^*T \supset (T^*T)^*$): Let $\xi \in D((T^*T)^*)$, then by (2) of \autoref{lemma:closed-unbonuded-adjoint}, there exists $\eta \in D(T^*T)$ such that $\xi + (T^*T)^*\xi = \eta + T^*T\eta$. In which case, for each $\theta \in D(T^*T)$,
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\begin{align*}
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\dpn{(I+T^*T)^*\xi, \theta}{H} &= \dpn{(I+T^*T)\eta, \theta}{H} \\
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\dpn{\xi, (I + T^*T)\theta}{H} &= \dpn{\eta, (I+T^*T)^*\theta}{H}
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\end{align*}
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As the above holds for all $\theta \in D(T^*T)$, (2) of \autoref{lemma:closed-unbonuded-adjoint} implies that $\dpn{\xi, \theta}{H} = \dpn{\eta, \theta}{H}$ for all $\theta \in H$. Therefore $\xi = \eta \in D(T^*T)$, and $T^*T \supset (T^*T)^*$.
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\end{proof}
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\begin{definition}[Absolute Value]
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\label{definition:unbounded-absolute-value}
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Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a closed, densely defined linear operator, and $|T| = (T^*T)^{1/2}: D(|T|) \to H$, then:
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\begin{enumerate}
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\item $|T|$ is a densely defined, self-adjoint, and positive operator.
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\item Equip $D(T)$ with the inner product $\dpn{\xi, \eta}{D(T)} = \dpn{\xi, \eta}{H} + \dpn{T\xi, T\eta}{H}$, then $D(T)$ is a Hilbert space with $D(T^*T)$ being dense in $D(T)$.
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\item Equip $D(|T|)$ with the inner product $\dpn{\xi, \eta}{D(|T|)} = \dpn{\xi, \eta}{H} + \dpn{|T|\xi, |T|\eta}{H}$, then $D(|T|)$ is a Hilbert space with $D(T^*T)$ being dense in $D(|T|)$.
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\item $D(T) = D(|T|)$.
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\item For each $\xi \in D(T) = D(|T|)$, $\normn{T\xi}_H = \norm{|T|\xi}_H$. In particular, $\ker T = \ker |T|$.
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\end{enumerate}
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The operator $|T|$ is the \textbf{absolute value} of $T$.
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\end{definition}
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\begin{proof}
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(1): By \hyperref[von Neumann's theorem]{theorem:von-neumann-closed-operator}, $T^*T: D(T^*T) \to H$ is a densely defined, self-adjoint, and positive operator. Thus $|T| = (T^*T)^{1/2}$ is well-defined, self-adjoint, and positive via the \hyperref[spectral theorem]{theorem:spectral-theorem-1-unbounded}.
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(2): By definition, $D(T)$ is isomorphic to the graph of $T$, which is a Hilbert space given that $T$ is closed.
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Let $\xi \in D(T)$ with $\xi \perp D(T^*T)$, then for each $\eta \in D(T^*T)$,
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\[
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0 = \dpn{\xi, \eta}{H} + \dpn{T\xi, T\eta}{H} = \dpn{\xi, (I + T^*T)\eta}{H}
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\]
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By (2) of \autoref{lemma:closed-unbonuded-adjoint}, $(I+T^*T): D(T^*T) \to H$ is surjective, so $\xi = 0$, and $D(T^*T)$ is dense in $D(T)$.
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(3): By (3) of \autoref{definition:unbounded-spectral-integral}, $|T|$ is self-adjoint, and in particular closed. Since $D(T)$ is isomorphic to the graph of $|T|$, $D(T)$ is a Hilbert space. Similarly, for any $\xi \in D(T)$ with $\xi \perp D(T^*T)$ and $\eta \in D(T^*T)$, by the \hyperref[Borel functional calculus]{definition:unbounded-spectral-integral}\footnote{The fact that the spectral measure of a positive element is supported in $[0, \infty)$ is needed here.},
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\[
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0 = \dpn{\xi, \eta}{H} + \dpn{|T|\xi, |T|\eta}{H} = \dpn{\xi, (I + T^*T)\eta}{H}
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\]
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By (2) of \autoref{lemma:closed-unbonuded-adjoint}, $\xi = 0$, and $D(T^*T)$ is dense in $D(|T|)$.
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(4): For any $\xi \in D(T^*T)$, $\norm{\xi}_{D(T)} = \norm{\xi}_{D(|T|)}$. By (2) and (3), $D(T)$ and $D(|T|)$ are both the completions of $D(T^*T)$ under the same norm, so, $D(T) = D(|T|)$.
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(5): For each $\xi \in D(T^*T)$,
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\[
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\normn{T\xi}_H^2 = \dpn{T\xi, T\xi}{H} = \dpn{\xi, T^*T\xi}{H} = \dpn{|T|\xi, |T|\xi}{H} = \norm{|T|\xi}_H^2
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\]
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By the density provided in (2) and (3), $\normn{T\xi}_H^2 = \norm{|T|\xi}_H^2$ for all $\xi \in D(T^*T)$.
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\end{proof}
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\begin{theorem}[Polar Decomposition (Unbounded)]
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\begin{theorem}[Polar Decomposition (Unbounded)]
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\label{theorem:unbounded-polar-decomposition}
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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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Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a closed, densely defined linear operator, then there exists a unique partial isometry $U \in B(H)$ with $\ker U = \ker T$ such that $T = U|T|$, where $|T|$ is the \hyperref[absolute value]{definition:unbounded-absolute-value} of $T$.
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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 $U \in B(H)$ with $\ker U = \ker T$ 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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\end{theorem}
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% Omitted until further notice.
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% Omitted until further notice.
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\begin{proof}
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By \autoref{definition:unbounded-absolute-value}, $|T|: D(T) \to H$ is a densely defined, self-adjoint, and positive operator with $\normn{T\xi}_H = \norm{|T|\xi}_H$ for all $\xi \in D(T)$ and $\ker(T) = \ker(|T|)$. Let
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\[
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U: |T|(D(T)) \to T(D(T)) \quad |T|\xi \mapsto T\xi
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\]
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then $U$ extends into a well-defined isometry $\ol{|T|(D(T))} \to \ol{T(D(T))}$. Further extend $U$ to $H$ by setting $U|_{|T|(D(T))^\perp} = 0$, then $U: H \to \ol{T(D(T))} \subset H$ is a partial isometry with $T = U|T|$. As $|T|$ is self-adjoint,
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\[
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\ker(U) = |T|(D(T))^\perp = \ker(|T|) = \ker(T)
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\]
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Finally, let $V: H \to H$ be a partial isometry with $T = V|T|$ and $\ker(V) = \ker(T)$, then for each $\xi \in D(T)$, $V|T|\xi = T\xi = U|T|\xi$, so $U = V$ on $|T|(D(T))$. By continuity, $U = V$ on $\ol{T(D(T))}$. As
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\[
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\ker(V) = \ker(T) = \ker(U) = |T|(D(T))^\perp
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\]
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and $H = \ker(V) \oplus \ol{|T|(D(T))}$, $U = V$.
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\end{proof}
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