43.8 Unbounded Functional Calculus

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 $\theta \in D(T)$,

As $D(T)$ is dense, $T \circ \xi \in D(T)^{*}$, and $\eta = T^{*}\xi$. Thus the graph of $T^{*}$ is closed.$\square$

Proof. (1): For any $\xi \in D(T)$,

(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 $\eta \in H$, then the following are equivalent:

1. (a)

   $\eta \perp (T + i)(H)$.

2. (b)

   $\dpn{T\xi, \eta}{H}= \dpn{\xi, i\eta}{H}$ for all $\xi \in D(T)$.

3. (c)

   $\eta \in D(T^{*})$ with $T^{*}\eta = i\eta$.

4. (d)

   $\eta \in D(T)$ with $(T - i)\eta = 0$.

By (1), $\eta \perp (T+i)(H)$ if and only if $\eta = 0$.$\square$

Proof. By Lemma 43.8.4, $(T+i)^{-1}: H \to D(T)$ 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 Lemma 43.8.4 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.$\square$

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,

where by continuity from above, $E(\bracsn{f > n})\xi \to 0$ 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 Lemma 43.6.2,

For any $\phi \in \Sigma(X; \complex)$, by the Cauchy-Schwarz inequality,

By the Monotone Convergence Theorem,

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,

so $\int_{X} g dE \xi \in D(\int_{X} f dE)$ if and only if $\xi \in D(\int_{X} fg dE)$, and

Let $\xi \in D(\int_{X} g dE) \cap D(\int_{X} fg dE)$ and $\eta \in H$, then

(3): For any $\xi, \eta \in D(\int_{X} f dE) = D(\int_{X} \ol f dE)$,

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 bounded $L^{\infty}$ functional calculus and (2),

so

By (2) and multiplication of $\int_{X} (1 + |f|) dE$ from the left, $\int_{X} \ol f dE = \paren{\int_X f dE}^{*}$.$\square$

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 Spectral Theorem I, 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,

is injective, so $1$ is not an eigenvalue of $U$, and $E(\bracsn{1}) = 0$. Let

then the pushforward $F = \phi_{*} E$ defined by $F(B) = E(\phi^{-1}(B))$ is also a spectral measure, with

$\square$

Proof. (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:

1. (i)

   $(\xi, \eta) \perp \Gamma(T)$.

2. (ii)

   For each $\theta \in D(T)$, $\dpn{\eta, T\theta}{H}= -\dpn{\xi, \theta}{H}$.

3. (iii)

   $\eta \in D(T^{*})$, with $\dpn{T^*\eta, \theta}{H}= -\dpn{\xi, \theta}{H}$ for all $\theta \in H$.

4. (iv)

   $\eta \in D(T^{*})$ with $\xi = -T^{*}\eta$.

so $\Gamma(T)^{\perp} = \bracsn{(-T^*\eta, \eta)| \eta \in D(T^*)}$.

(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

In which case, $T\xi = -\eta \in D(T^{*})$, so $\xi \in D(T^{*}T)$ with

so $I + T^{*}T: D(T^{*}T) \to H$ is surjective.$\square$

Proof. For each $\xi, \eta \in D(T^{*}T)$, by definition of the adjoint,

(Densely Defined): Let $\eta \in D(T^{*}T)^{\perp}$. By (2) of Lemma 43.8.8, there exists $\xi \in D(T^{*}T)$ such that $\eta = \xi + T^{*}T\xi$. In which case,

so $T\xi = \xi = 0$, and $\eta = 0$ as well.

($T^{*}T \subset (T^{*}T)^{*}$): For each $\xi, \eta \in D(T^{*}T)$, $\dpn{T^*T\xi, \eta}{H}= \dpn{\xi, T^*T\eta}{H}$, so $T^{*}T \subset (T^{*}T)^{*}$.

($T^{*}T \supset (T^{*}T)^{*}$): Let $\xi \in D((T^{*}T)^{*})$, then by (2) of Lemma 43.8.8, there exists $\eta \in D(T^{*}T)$ such that $\xi + (T^{*}T)^{*}\xi = \eta + T^{*}T\eta$. In which case, for each $\theta \in D(T^{*}T)$,

As the above holds for all $\theta \in D(T^{*}T)$, (2) of Lemma 43.8.8 implies that $\dpn{\xi, \theta}{H}= \dpn{\eta, \theta}{H}$ for all $\theta \in H$. Therefore $\xi = \eta \in D(T^{*}T)$, and $T^{*}T \supset (T^{*}T)^{*}$.$\square$

Proof. (1): By von Neumann’s theorem, $T^{*}T: D(T^{*}T) \to H$ is a densely defined, self-adjoint, and positive operator. Thus $|T| = (T^{*}T)^{1/2}$ is well-defined, self-adjoint, and positive via the spectral theorem.

(2): By definition, $D(T)$ is isomorphic to the graph of $T$, which is a Hilbert space given that $T$ is closed.

Let $\xi \in D(T)$ with $\xi \perp D(T^{*}T)$, then for each $\eta \in D(T^{*}T)$,

By (2) of Lemma 43.8.8, $(I+T^{*}T): D(T^{*}T) \to H$ is surjective, so $\xi = 0$, and $D(T^{*}T)$ is dense in $D(T)$.

(3): By (3) of Definition 43.8.6, $|T|$ 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 Borel functional calculus^[3],

By (2) of Lemma 43.8.8, $\xi = 0$, and $D(T^{*}T)$ is dense in $D(|T|)$.

(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, $D(T) = D(|T|)$.

(5): For each $\xi \in D(T^{*}T)$,

By the density provided in (2) and (3), $\normn{T\xi}_{H}^{2} = \norm{|T|\xi}_{H}^{2}$ for all $\xi \in D(T)$.$\square$

Proof. By Definition 43.8.10, $|T|: D(T) \to H$ 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

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,

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

and $H = \ker(V) \oplus \ol{|T|(D(T))}$, $U = V$.$\square$