Lecture Notes in Functional Analysis

by William L. Paschke

edition 0.9

Contents

Frontmatter

I. Hilbert Space

II. Bounded Operators

III. Compact Operators

IV. The Spectral Theorem

Index and References

## A. Unbounded Operators

A linear operator $$$L$$$ in $$$H$$$ (as distinct from on $$$H$$$) has domain $$$D\subseteq H$$$ and range in $$$H$$$. Write $$$\mathopen{}\left(L, D\right)\mathclose{}$$$ for $$$L : D \to H$$$. Say that $$$\mathopen{}\left({L}_{1}, {D}_{1}\right)\mathclose{}$$$ is an extension of $$$\mathopen{}\left(L, D\right)\mathclose{}$$$ if $$$D\subseteq {D}_{1}$$$ and $$${L}_{1}|D= L$$$.

The graph of $$$L$$$ is $$$\mathop{\mathcal{G}}\mathopen{}\left( L\right)\mathclose{}= \mathop{\mathcal{G}}\mathopen{}\left( \mathopen{}\left(L, D\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left\{\, \mathopen{}\left(x, L\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\,\middle\vert\, x\in D\,\right\}\mathclose{}\subseteq H\oplus H$$$. If $$$D$$$ is closed, then $$$L$$$ is bounded if and only if its graph is closed in $$$H\oplus H$$$ (Closed Graph Theorem). It is easy to give examples of unbounded $$$L$$$ on non-closed $$$D$$$ with closed graph.

Example IV.13

Take $$$S\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ with $$$\operatorname{Ker}\mathopen{}\left( S\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{}$$$ but $$$S$$$ not invertible. Let $$$D= S\mathopen{}\left( H\right)\mathclose{}$$$. Define $$$L : D \to H$$$ by $$$L\mathopen{}\left( S\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}= y$$$ (which is possible because $$$S$$$ is injective). Then $$$\mathop{\mathcal{G}}\mathopen{}\left( \mathopen{}\left(L, D\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left\{\, \mathopen{}\left(x, L\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\,\middle\vert\, x\in D\,\right\}\mathclose{}= \mathopen{}\left\{\, \mathopen{}\left(S\mathopen{}\left( y\right)\mathclose{}, L\mathopen{}\left( S\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}\,\middle\vert\, y\in H\,\right\}\mathclose{}= \mathopen{}\left\{\, \mathopen{}\left(S\mathopen{}\left( y\right)\mathclose{}, y\right)\mathclose{}\,\middle\vert\, y\in H\,\right\}\mathclose{}$$$, which is closed because it is the image of the graph of $$$S$$$ (which is closed) under switching the entries.

Definition IV.14

A linear operator $$$L : D \to H$$$ is closed if its graph is closed in $$$H\oplus H$$$.

Example IV.15

Let $$$H= \mathrm{l}^{0}\mathopen{}\left( {\mathbb{Z}}^{+}\right)\mathclose{}$$$ (as functions on $$${\mathbb{Z}}^{+}$$$). Let $$$α : {\mathbb{Z}}^{+} \to \mathbb{C}$$$ be any function. Let $$$D= \mathopen{}\left\{\, f\in \mathrm{l}^{0}\,\middle\vert\, , α\mathopen{}\left( f\right)\mathclose{}\in \mathrm{l}^{0}, \,\right\}\mathclose{}$$$, $$$L\mathopen{}\left( f\right)\mathclose{}= α\mathopen{}\left( f\right)\mathclose{}$$$. To see $$$L$$$ is closed, suppose $$$\mathopen{}\left({f}_{n}, L\mathopen{}\left( {f}_{n}\right)\mathclose{}\right)\mathclose{} \to \mathopen{}\left(f, g\right)\mathclose{}$$$, $$${f}_{n}\in D$$$. Then $$${f}_{n} \to f$$$, $$$α\mathopen{}\left( {f}_{n}\right)\mathclose{} \to g$$$ in $$$\mathrm{l}^{0}$$$. In particular, $$${f}_{n}\mathopen{}\left( j\right)\mathclose{} \to f\mathopen{}\left( j\right)\mathclose{}$$$ for all $$$j\in {\mathbb{Z}}^{+}$$$, and, using Fatou, $$\sum_{j}{} {\mathopen{}\left\lvert{}g\mathopen{}\left( j\right)\mathclose{}-α\mathopen{}\left( j\right)\mathclose{}f\mathopen{}\left( j\right)\mathclose{}\right\rvert\mathclose{}}^{2} = \sum_{j}{} \lim_{n}{} {\mathopen{}\left\lvert{}g\mathopen{}\left( j\right)\mathclose{}-α\mathopen{}\left( j\right)\mathclose{}{f}_{n}\mathopen{}\left( j\right)\mathclose{}\right\rvert\mathclose{}}^{2} \leq \liminf_{n}{} {\mathopen{}\left\lvert{}g\mathopen{}\left( j\right)\mathclose{}-α\mathopen{}\left( j\right)\mathclose{}{f}_{n}\mathopen{}\left( j\right)\mathclose{}\right\rvert\mathclose{}}^{2} = \liminf_{n}{} {\mathopen{}\left\lVert{}g-α\mathopen{}\left( {f}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} = 0 \text{.}$$ So, $$$α\mathopen{}\left( f\right)\mathclose{}= g\in \mathrm{l}^{0}$$$, $$$f\in D$$$, and $$$\mathopen{}\left(f, g\right)\mathclose{}= \mathopen{}\left(f, L\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\in G\mathopen{}\left( L\right)\mathclose{}$$$.

Definition IV.16

A linear operator $$$L : D \to H$$$ is closable (or pre-closed) if the closure of the graph (in $$$H\oplus H$$$) is the graph of a linear operator.

That is,

1. For all $$$x\in H$$$, $$${\#} \mathopen{}\left(\mathopen{}\left(\mathopen{}\left\{\, x\,\right\}\mathclose{}\oplus H\right)\mathclose{}\cap \overline{\mathop{\mathcal{G}}\mathopen{}\left( L\right)\mathclose{}}\right)\mathclose{} \leq 1$$$.
2. If $$$\mathopen{}\left(x, {y}_{1}\right)\mathclose{}$$$ and $$$\mathopen{}\left(x, {y}_{2}\right)\mathclose{}$$$ are in $$$\overline{\mathop{\mathcal{G}}\mathopen{}\left( L\right)\mathclose{}}$$$ then $$${y}_{1}= {y}_{2}$$$.
3. $$$\mathopen{}\left(0, y\right)\mathclose{}\in \overline{\mathop{\mathcal{G}}\mathopen{}\left( L\right)\mathclose{}}$$$ implies $$$y= 0$$$.
4. $$${x}_{n} \to 0$$$ in $$$D$$$ and $$$L\mathopen{}\left( {x}_{n}\right)\mathclose{} \to y$$$ implies $$$y= 0$$$.

Notation IV.17

For $$$\mathopen{}\left(L, D\right)\mathclose{}$$$ closable, write $$$\overline{L}$$$ for the linear operator with $$$\mathop{\mathcal{G}}\mathopen{}\left( \overline{L}\right)\mathclose{}= \overline{\mathop{\mathcal{G}}\mathopen{}\left( L\right)\mathclose{}}$$$. $$$\overline{L}$$$ is called the closure of $$$L$$$.

Example IV.18

$$$H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}\right)\mathclose{}$$$. $$$D= \mathopen{}\left\{\, f\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}\right)\mathclose{}\cap \mathrm{C}^{1}\mathopen{}\left( \mathbb{R}\right)\mathclose{}\,\middle\vert\, , f'\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathbb{R}\right)\mathclose{}, \,\right\}\mathclose{}$$$. Suppose $$$\mathopen{}\left({f}_{n}\right)\mathclose{}$$$ is a sequence in $$$D$$$ with $$${f}_{n} \to 0$$$ and $$${f}_{n}' \to g$$$ (in $$$\mathrm{L}^{\mathrm{2}}$$$) for some $$$g\in \mathrm{L}^{\mathrm{2}}$$$. Fix $$$a\lt b$$$. Define $$$G$$$ on $$$\mathopen{}\left[a, b\right]\mathclose{}$$$ by $$$G\mathopen{}\left( x\right)\mathclose{}= \int _{a}^{x}{}g$$$. Then $$${f}_{n}'|\mathopen{}\left[a, b\right]\mathclose{} \to g|\mathopen{}\left[a, b\right]\mathclose{}$$$, because $$${f}_{n}|\mathopen{}\left[a, b\right]\mathclose{}-{f}_{n}\mathopen{}\left( a\right)\mathclose{}\mathbf{1} \to G$$$ in $$$\mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{}$$$. But, $$${f}_{n}|\mathopen{}\left[a, b\right]\mathclose{} \to 0$$$, so $$$G$$$ is the $$$\mathrm{L}^{\mathrm{2}}$$$ limit of constant functions, and must therefore be constant. But, $$$G\mathopen{}\left( a\right)\mathclose{}= 0$$$, so $$$G\equiv 0$$$ almost everywhere on $$$\mathopen{}\left[a, b\right]\mathclose{}$$$, and hence on $$$\mathbb{R}$$$.

Thus $$$L$$$ is pre-closed. To see that it is not closed, consider the functions (see figures Figure IV.A and Figure IV.B) $$φ\mathopen{}\left( x\right)\mathclose{}= \begin{cases}0, & \mathopen{}\left\lvert{}x\right\rvert\mathclose{}\leq 1 ; \\ 1-\mathopen{}\left\lvert{}x\right\rvert\mathclose{}, & \mathopen{}\left\lvert{}x\right\rvert\mathclose{}\lt 1\text{.} \end{cases}$$ $$ψ\mathopen{}\left( x\right)\mathclose{}= \begin{cases}0, & \mathopen{}\left\lvert{}x\right\rvert\mathclose{}\geq 1 ; \\ {-} \mathop{\text{sgn}}\mathopen{}\left( x\right)\mathclose{} , & \mathopen{}\left\lvert{}x\right\rvert\mathclose{}\lt 1 \end{cases}$$ Then $$$\mathopen{}\left(φ, ψ\right)\mathclose{}\in \overline{G\mathopen{}\left( L\right)\mathclose{}}\setminus G\mathopen{}\left( L\right)\mathclose{}$$$.

Figure IV.A. $$$φ\mathopen{}\left( x\right)\mathclose{}$$$
Figure IV.B. $$$ψ\mathopen{}\left( x\right)\mathclose{}$$$
Definition IV.19

For any densely defined operator $$$L : D \to H$$$ (i.e. $$$\overline{D}= H$$$), define $$$D^{*}$$$ to be the set of $$$y\in H$$$ such that there exists $$$ŷ$$$ such that $$$\mathopen{}\left\langle{}L\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}= \mathopen{}\left\langle{}x, ŷ\right\rangle\mathclose{}$$$ for all $$$x\in D$$$. Because $$$D$$$ is dense, $$$ŷ$$$ is unique, and we get an operator $$$L^{*} : D^{*} \to H$$$ defined by $$$L^{*}\mathopen{}\left( y\right)\mathclose{}= ŷ$$$, the adjoint of $$$L$$$, with $$$\mathopen{}\left\langle{}L\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}= \mathopen{}\left\langle{}x, L^{*}\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{}$$$ for all $$$x\in D$$$ and $$$y\in D^{*}$$$.

Proposition IV.20

For any densely defined $$$L$$$, the operator $$$L^{*}$$$ is closed.

Proof. Suppose $$$\mathopen{}\left({y}_{n}, L^{*}\mathopen{}\left( {y}_{n}\right)\mathclose{}\right)\mathclose{} \to \mathopen{}\left(y, w\right)\mathclose{}$$$ in $$$H\oplus H$$$ for a sequence $$$\mathopen{}\left({y}_{n}\right)\mathclose{}\in D^{*}$$$. Then for all $$$x\in D$$$, $$$\mathopen{}\left\langle{}L\mathopen{}\left( x\right)\mathclose{}, {y}_{n}\right\rangle\mathclose{} \to \mathopen{}\left\langle{}L\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}$$$ and $$$\mathopen{}\left\langle{}L\mathopen{}\left( x\right)\mathclose{}, {y}_{n}\right\rangle\mathclose{}= \mathopen{}\left\langle{}x, L^{*}\mathopen{}\left( {y}_{n}\right)\mathclose{}\right\rangle\mathclose{} \to \mathopen{}\left\langle{}x, w\right\rangle\mathclose{}$$$. Thus $$$\mathopen{}\left\langle{}L\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}= \mathopen{}\left\langle{}x, w\right\rangle\mathclose{}$$$, making $$$y\in D^{*}$$$, $$$w= L^{*}\mathopen{}\left( y\right)\mathclose{}$$$, and $$$\mathopen{}\left(y, w\right)\mathclose{}\in \mathop{\mathcal{G}}\mathopen{}\left( L^{*}\right)\mathclose{}$$$.

Example IV.21

$$$H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$. Define $$$D= \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$. Consider $$$L : D \to H$$$ defined by $$$L\mathopen{}\left( f\right)\mathclose{}= f'$$$. Then $$$D^{*}\cap \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}= \mathopen{}\left\{\, g\in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}\,\middle\vert\, , g\mathopen{}\left( 0\right)\mathclose{}= 0= g\mathopen{}\left( 1\right)\mathclose{}, \,\right\}\mathclose{}$$$. For $$$f$$$ and $$$g$$$ in $$$\mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$, $$$\int _{0}^{1}{} f'\overline{g} = \mathopen{}\left( f\overline{g} \right|\mathclose{}_{0}^{1}-\int _{0}^{1}{} f \overline{g}'$$$ using integration by parts. If we change the domain by defining $$$D= \mathopen{}\left\{\, f\in \mathrm{C}^{1}\,\middle\vert\, , f\mathopen{}\left( 0\right)\mathclose{}= 0, \,\right\}\mathclose{}$$$, $$$L\mathopen{}\left( f\right)\mathclose{}= f'$$$ makes $$$D^{*}\cap \mathrm{C}^{1}= \mathopen{}\left\{\, g\in \mathrm{C}^{1}\,\middle\vert\, , g\mathopen{}\left( 1\right)\mathclose{}= 0, \,\right\}\mathclose{}$$$. In either case $$$L^{*}|\mathopen{}\left( D^{*}\cap \mathrm{C}^{1}\right)\mathclose{}= {-} \frac{\mathrm{d}}{\mathrm{d}x}$$$.

In what follows, let $$$L$$$ be a linear operator on $$$H$$$ with a dense domain $$$D= \mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}$$$.

Proposition IV.22

For densely defined closable $$$L$$$, we have $$$\overline{L}^{*}= L^{*}$$$.

Proof. Clearly, $$$\mathop{\mathcal{D}}\mathopen{}\left( \overline{L}^{*}\right)\mathclose{}\subseteq \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}$$$ (in general the domain of the adjoint of the extension will be a subset of the domain of the adjoint of the original). Take $$$y\in L^{*}$$$. For $$$x\in \mathop{\mathcal{D}}\mathopen{}\left( \overline{L}\right)\mathclose{}$$$, $$$\mathopen{}\left(x, \overline{L}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\in \mathop{\mathcal{G}}\mathopen{}\left( \overline{L}\right)\mathclose{}= \overline{\mathop{\mathcal{G}}\mathopen{}\left( L\right)\mathclose{}}$$$. We get $$$\mathopen{}\left({x}_{n}\right)\mathclose{}$$$ in $$$\mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}$$$ with $$${x}_{n} \to x$$$ and $$$L\mathopen{}\left( {x}_{n}\right)\mathclose{} \to \overline{L}\mathopen{}\left( x\right)\mathclose{}$$$. Then $$\mathopen{}\left\langle{}\overline{L}\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}= \lim_{n}{} \mathopen{}\left\langle{}L\mathopen{}\left( {x}_{n}\right)\mathclose{}, y\right\rangle\mathclose{} = \lim_{n}{} \mathopen{}\left\langle{}{x}_{n}, L^{*}\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{} = \mathopen{}\left\langle{}x, L^{*}\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{} \text{.}$$ Thus $$$y\in \mathop{\mathcal{D}}\mathopen{}\left( \overline{L}^{*}\right)\mathclose{}$$$ and $$$\overline{L}^{*}\mathopen{}\left( y\right)\mathclose{}= L^{*}\mathopen{}\left( y\right)\mathclose{}$$$.

Remark IV.23

Suppose $$$L$$$ is closed and densely defined. Then $$$\mathop{\mathcal{G}}\mathopen{}\left( L\right)\mathclose{}$$$ is a Hilbert space, and we have the bounded linear map $$$E : \mathop{\mathcal{G}}\mathopen{}\left( L\right)\mathclose{} \to H$$$ defined by $$$E\mathopen{}\left( \mathopen{}\left(x, L\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}= x$$$. Note $$$\mathopen{}\left\lVert{}E\right\rVert\mathclose{}\leq 1$$$. So, we get (bounded) $$$E^{*} : H \to \mathop{\mathcal{G}}\mathopen{}\left( L\right)\mathclose{}$$$.

Proposition IV.24

With $$$L$$$ and $$$E$$$ as above,

1. $$$\operatorname{Ker}\mathopen{}\left( E\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{}$$$, so $$$E^{*}$$$ has dense range in $$$\mathop{\mathcal{G}}\mathopen{}\left( L\right)\mathclose{}$$$.
2. If $$$w\in H$$$ and $$$y= E\mathopen{}\left( E^{*}\mathopen{}\left( w\right)\mathclose{}\right)\mathclose{}$$$ (so that $$$y\in \mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}$$$ and $$$E^{*}\mathopen{}\left( w\right)\mathclose{}= \mathopen{}\left(y, L\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}$$$), then $$$L\mathopen{}\left( y\right)\mathclose{}\in \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}$$$ and $$$y+ L^{*}\mathopen{}\left( L\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}= w$$$.
3. $$$\operatorname{Ran}\mathopen{}\left( L\right)\mathclose{}\equiv L\mathopen{}\left( \mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}\right)\mathclose{}\subseteq \overline{ \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{} }$$$.

Proof.

1. $$$\operatorname{Ker}\mathopen{}\left( E\right)\mathclose{}= 0$$$ is clear. $$$\overline{\operatorname{Ran}\mathopen{}\left( E^{*}\right)\mathclose{}}= { \mathopen{}\left(\operatorname{Ker}\mathopen{}\left( E\right)\mathclose{}\right)\mathclose{} }^{\perp}= \mathop{\mathcal{G}}\mathopen{}\left( L\right)\mathclose{}$$$.
2. Take $$$x\in \mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}$$$. Then $$\mathopen{}\left\langle{}x, y\right\rangle\mathclose{}+\mathopen{}\left\langle{}L\mathopen{}\left( x\right)\mathclose{}, L\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left(x, L\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, \mathopen{}\left(y, L\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left(x, L\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, E^{*}\mathopen{}\left( w\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}E\mathopen{}\left( \mathopen{}\left(x, L\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}, w\right\rangle\mathclose{}= \mathopen{}\left\langle{}x, w\right\rangle\mathclose{} \text{.}$$ Thus $$$\mathopen{}\left\langle{}L\mathopen{}\left( x\right)\mathclose{}, L\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}x, w-y\right\rangle\mathclose{}$$$ and $$$L^{*}\mathopen{}\left( L\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}= w-y$$$.
3. Take $$$y\in \mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}$$$. Then $$$\mathopen{}\left(y, L\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}\in \mathop{\mathcal{G}}\mathopen{}\left( L\right)\mathclose{}= \overline{ E^{*}\mathopen{}\left( H\right)\mathclose{}}$$$. So get a sequence $$$\mathopen{}\left( \mathopen{}\left({y}_{n}, L\mathopen{}\left( {y}_{n}\right)\mathclose{}\right)\mathclose{} \right)\mathclose{}\in E^{*}\mathopen{}\left( H\right)\mathclose{}$$$ such that $$$\mathopen{}\left({y}_{n}, L\mathopen{}\left( {y}_{n}\right)\mathclose{}\right)\mathclose{} \to \mathopen{}\left(y, L\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}$$$. Each $$${y}_{n}\in E\mathopen{}\left( E^{*}\mathopen{}\left( H\right)\mathclose{}\right)\mathclose{}\subseteq \mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}$$$, and, by the second part, $$$L\mathopen{}\left( {y}_{n}\right)\mathclose{}\in \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}$$$ and $$$L\mathopen{}\left( y\right)\mathclose{}= \lim_{n}{} L\mathopen{}\left( {y}_{n}\right)\mathclose{} \in \overline{\mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}}$$$.

Proposition IV.25

For densely defined $$$L$$$, we have $$$L$$$ is closable if and only if $$$\mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}$$$ is dense in $$$H$$$.

Proof. (⇒) Apply Proposition IV.24 to $$$\overline{L}$$$. Then $$$\operatorname{Ran}\mathopen{}\left( \overline{L}\right)\mathclose{}\subseteq \overline{ \mathop{\mathcal{D}}\mathopen{}\left( \overline{L}^{*}\right)\mathclose{} }= \overline{ \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{} }$$$. Also for $$$y\in { \mathopen{}\left(\operatorname{Ran}\mathopen{}\left( \overline{L}\right)\mathclose{}\right)\mathclose{} }^{\perp}$$$, $$$\mathopen{}\left\langle{}\overline{L}\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}= 0$$$ for all $$$x\in \mathop{\mathcal{D}}\mathopen{}\left( \overline{L}\right)\mathclose{}$$$. Thus $$$y\in \mathop{\mathcal{D}}\mathopen{}\left( \overline{L}^{*}\right)\mathclose{}$$$ (with $$$\overline{L}^{*}\mathopen{}\left( y\right)\mathclose{}= 0$$$). Thus $$${ \mathopen{}\left(\operatorname{Ran}\mathopen{}\left( \overline{L}\right)\mathclose{}\right)\mathclose{} }^{\perp}\subseteq \mathop{\mathcal{D}}\mathopen{}\left( \overline{L}^{*}\right)\mathclose{}= \overline{\mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}}$$$ and $$$H= \overline{ \operatorname{Ran}\mathopen{}\left( \overline{L}\right)\mathclose{}+{ \mathopen{}\left(\operatorname{Ran}\mathopen{}\left( \overline{L}\right)\mathclose{}\right)\mathclose{} }^{\perp} }\subseteq \overline{\mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}}$$$.

(⇐) Suppose $$${x}_{n} \to 0$$$ in $$$\mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}$$$ and $$$L\mathopen{}\left( {x}_{n}\right)\mathclose{} \to w$$$. Then for all $$$y\in \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}$$$, $$$\mathopen{}\left\langle{}L\mathopen{}\left( {x}_{n}\right)\mathclose{}, y\right\rangle\mathclose{}= \mathopen{}\left\langle{}{x}_{n}, L^{*}\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{} \to 0$$$. Also $$$\mathopen{}\left\langle{}L\mathopen{}\left( {x}_{n}\right)\mathclose{}, y\right\rangle\mathclose{} \to \mathopen{}\left\langle{}w, y\right\rangle\mathclose{}$$$, so $$$\mathopen{}\left\langle{}w, y\right\rangle\mathclose{}= 0$$$. Thus $$$w\in { \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{} }^{\perp}= \mathopen{}\left\{\, 0\,\right\}\mathclose{}$$$.

Back to Proposition IV.24. We have $$$1+ L^{*}\mathopen{}\left( L\right)\mathclose{} : E\mathopen{}\left( E^{*}\mathopen{}\left( H\right)\mathclose{}\right)\mathclose{} \to H$$$. For any $$$y\in \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\mathopen{}\left( L\right)\mathclose{}\right)\mathclose{}$$$ (i.e. $$$y$$$ with $$$L\mathopen{}\left( y\right)\mathclose{}\in \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}$$$), $$\mathopen{}\left\langle{}\mathopen{}\left(1+ L^{*}\mathopen{}\left( L\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( y\right)\mathclose{}, y\right\rangle\mathclose{}= \mathopen{}\left\langle{}y, y\right\rangle\mathclose{}+\mathopen{}\left\langle{} L^{*}\mathopen{}\left( L\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}, y\right\rangle\mathclose{}= {\mathopen{}\left\lVert{}y\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}L\mathopen{}\left( y\right)\mathclose{}\right\rVert\mathclose{}}^{2}\geq {\mathopen{}\left\lVert{}y\right\rVert\mathclose{}}^{2}\geq 0 \text{.}$$ So, $$$1+ L^{*}\mathopen{}\left( L\right)\mathclose{}$$$ is injective on $$$\mathop{\mathcal{D}}\mathopen{}\left( L^{*}L\right)\mathclose{}\subseteq E\mathopen{}\left( E^{*}\mathopen{}\left( H\right)\mathclose{}\right)\mathclose{}$$$. It follows that $$$E\mathopen{}\left( E^{*}\mathopen{}\left( H\right)\mathclose{}\right)\mathclose{}= \mathop{\mathcal{D}}\mathopen{}\left( L^{*}L\right)\mathclose{}$$$, which is dense in $$$H$$$.

Let $$$S : H \to \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\mathopen{}\left( L\right)\mathclose{}\right)\mathclose{}$$$ be the inverse map to $$$1+ L^{*}L$$$. For any $$$w\in H$$$, $$\mathopen{}\left\langle{}\mathopen{}\left(1+ L^{*}L\right)\mathclose{}\mathopen{}\left( S\mathopen{}\left( w\right)\mathclose{}\right)\mathclose{}, w\right\rangle\mathclose{}= \mathopen{}\left\langle{}w, S\mathopen{}\left( w\right)\mathclose{}\right\rangle\mathclose{}\geq {\mathopen{}\left\lVert{}S\mathopen{}\left( w\right)\mathclose{}\right\rVert\mathclose{}}^{2} \text{.}$$ Thus $$${\mathopen{}\left\lVert{}S\mathopen{}\left( w\right)\mathclose{}\right\rVert\mathclose{}}^{2}\leq \mathopen{}\left\lVert{}w\right\rVert\mathclose{}\mathopen{}\left\lVert{}S\mathopen{}\left( w\right)\mathclose{}\right\rVert\mathclose{}$$$, $$$\mathopen{}\left\lVert{}S\mathopen{}\left( w\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}w\right\rVert\mathclose{}$$$ and $$$S\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ with $$$\mathopen{}\left\lVert{}S\right\rVert\mathclose{}\leq 1$$$ and $$$\mathopen{}\left\langle{}w, S\mathopen{}\left( w\right)\mathclose{}\right\rangle\mathclose{}\geq 0$$$ for all $$$w\in H$$$. Therefore $$$S\geq 0$$$. Write $$$S= { \mathopen{}\left(1+ L^{*}L\right)\mathclose{} }^{-1}$$$. Also write $$${\mathopen{}\left(1+ L^{*}L\right)\mathclose{}}^{{-}\frac{1}{2}}= {S}^{\frac{1}{2}}$$$.

Proposition IV.26

Let $$$L$$$ be a closed, densely defined linear operator on $$$H$$$. Then

1. $$$\mathop{\mathcal{D}}\mathopen{}\left( L^{*}L\right)\mathclose{}$$$ is dense.
2. $$$1+ L^{*}L : \mathop{\mathcal{D}}\mathopen{}\left( L^{*}L\right)\mathclose{} \to H$$$ is bijective.
3. Its inverse $$$S\equiv { \mathopen{}\left(1+ L^{*}L\right)\mathclose{} }^{-1} : H \to \mathop{\mathcal{D}}\mathopen{}\left( L^{*}L\right)\mathclose{}\subseteq H$$$ is a bounded positive operator with $$$\mathopen{}\left\lVert{}S\right\rVert\mathclose{}\leq 1$$$.
4. $$$L{S}^{\frac{1}{2}}= L{\mathopen{}\left(1+ L^{*}L\right)\mathclose{}}^{{-}\frac{1}{2}}$$$ maps $$${S}^{\frac{1}{2}}\mathopen{}\left( H\right)\mathclose{}$$$ (a dense subspace of $$$H$$$) boundedly into $$$H$$$, and $$$\mathopen{}\left\lVert{}L{S}^{\frac{1}{2}}\right\rVert\mathclose{}\leq 1$$$.

Proof. We have already shown all but the last part. Notice $$$\overline{ {S}^{\frac{1}{2}}\mathopen{}\left( H\right)\mathclose{} }= { \mathopen{}\left(\operatorname{Ker}\mathopen{}\left( {S}^{\frac{1}{2}}\right)\mathclose{}\right)\mathclose{} }^{\perp}= { \mathopen{}\left(\operatorname{Ker}\mathopen{}\left( S\right)\mathclose{}\right)\mathclose{} }^{\perp}= H$$$, and $$$S\mathopen{}\left( H\right)\mathclose{}= \mathop{\mathcal{D}}\mathopen{}\left( L^{*}L\right)\mathclose{}\subseteq D\mathopen{}\left( L\right)\mathclose{}$$$. So, $$${S}^{\frac{1}{2}}\mathopen{}\left( H\right)\mathclose{}\subseteq \mathop{\mathcal{D}}\mathopen{}\left( L{S}^{\frac{1}{2}}\right)\mathclose{}$$$. Also, $${\mathopen{}\left\lVert{}L\mathopen{}\left( {S}^{\frac{1}{2}}\mathopen{}\left( {S}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}L\mathopen{}\left( S\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right\rVert\mathclose{}}^{2}= \mathopen{}\left\langle{}L\mathopen{}\left( S\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, L\mathopen{}\left( S\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{} L^{*}\mathopen{}\left( L\mathopen{}\left( S\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}, S\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left(1-S\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}, S\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}\leq {\mathopen{}\left\lVert{}{S}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}$$ because $$$\mathopen{}\left(1+ L^{*}L\right)\mathclose{}\mathopen{}\left( S\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}= x$$$ and $$\mathopen{}\left\langle{}x, S\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}-{\mathopen{}\left\lVert{}S\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2}\leq \mathopen{}\left\langle{}\mathopen{}\left\langle{}x, S\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}{S}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}, {S}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}= {\mathopen{}\left\lVert{}{S}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2} \text{.}$$

Before dealing with self-adjoint operators, we consider a way to construct densely defined operators $$$L$$$ for which $$$\mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}= \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}$$$.

Definition IV.27

A resolution of the identity in $$$H$$$ is an increasing projection-valued map $$$t\mapsto {Q}_{t}$$$ on $$$\mathbb{R}$$$ such that $$$\bigcap_{t}{}{Q}_{t}\mathopen{}\left( H\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{}$$$ and $$$\overline{ \bigcup_{t}{}{Q}_{t}\mathopen{}\left( H\right)\mathclose{} }= H$$$. A spectral resolution of a bounded self-adjoint operator is the special case of this in which $$${Q}_{t}$$$ flattens out at $$$0$$$ once $$$t$$$ moves far enough to the left and at $$$1$$$ once $$$t$$$ moves far enough to the right.

Remark IV.28

In general, we have $$$\lim_{t\to{-}\infty}{} {Q}_{t}\mathopen{}\left( ξ\right)\mathclose{} = 0= \lim_{t\to{+}\infty}{} \mathopen{}\left(1-{Q}_{t}\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}$$$ for all $$$ξ\in H$$$ by Lemma IV.10.

Remark IV.29

For the spectral resolution $$$t\mapsto {Q}_{t}$$$, write $$${E}_{t}= {Q}_{t}-{Q}_{{-}t}$$$ for $$$t\gt 0$$$, and let $$${H}_{t}= {E}_{t}\mathopen{}\left( H\right)\mathclose{}$$$. Notice that $$${E}_{s}\leq {E}_{t}$$$ for $$$0\lt s\leq t$$$ and that $$$\lim_{t\to{+}\infty}{} {E}_{t}\mathopen{}\left( ξ\right)\mathclose{} = ξ$$$ for all $$$ξ$$$, i.e. the union of the $$${H}_{t}$$$'s is dense in $$$H$$$. Now fix a continuous function $$$f : \mathbb{R} \to \mathbb{C}$$$. For each $$$τ\gt 0$$$, define the bounded operator $$${L}_{\mathopen{}\left(f, τ\right)\mathclose{}}= {L}_{τ}= \int _{{-}τ}^{τ}{}f\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}{Q}_{t}$$$. As in our discussion above of continuous functions of bounded self-adjoint operators and Theorem IV.5, the integral is the norm limit of Riemann-Stieltjes sums coming from partitions of $$$\mathopen{}\left[{-}τ, τ\right]\mathclose{}$$$. Thus $$${L}_{τ}$$$ commutes with each $$${Q}_{t}$$$, and $$${L}_{τ}\mathopen{}\left( {E}_{τ}\right)\mathclose{}= {E}_{τ}\mathopen{}\left( {L}_{τ}\right)\mathclose{}= {L}_{τ}$$$. Further, $$${\mathopen{}\left\lVert{}{L}_{τ}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}}^{2}= \int _{{-}τ}^{τ}{} {\mathopen{}\left\lvert{}f\mathopen{}\left( t\right)\mathclose{}\right\rvert\mathclose{}}^{2} \,\mathrm{d} \mathopen{}\left\langle{}{Q}_{t}\mathopen{}\left( ξ\right)\mathclose{}, ξ\right\rangle\mathclose{}$$$ for every $$$ξ$$$, which makes $$$\mathopen{}\left\lVert{}{L}_{τ}\right\rVert\mathclose{}\leq \max_{ \mathopen{}\left\lvert{}t\right\rvert\mathclose{}\leq τ }{}\mathopen{}\left\lvert{}f\mathopen{}\left( t\right)\mathclose{}\right\rvert\mathclose{}$$$.

Proposition IV.30

For $$$ξ\in H$$$, we have $$$\sup_{τ\gt 0}{} \mathopen{}\left\lVert{}{L}_{t}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{} \lt \infty$$$ if and only if $$$\lim_{τ\to{+}\infty}{} {L}_{τ}\mathopen{}\left( ξ\right)\mathclose{}$$$ exists (in norm).

Proof. (⇒) Take $$$0\lt {τ}_{0}\lt {τ}_{1}\lt \dotsb$$$ with $$${τ}_{n} \to \infty$$$ and notice that $${L}_{{τ}_{0}}\mathopen{}\left( ξ\right)\mathclose{}= \sum_{j=1}^{N}{} \mathopen{}\left({L}_{{τ}_{j}}-{L}_{{τ}_{j-1}}\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{} = {L}_{{τ}_{N}}\mathopen{}\left( ξ\right)\mathclose{}$$ for every $$$N\geq 1$$$. The summands are mutually orthogonal, so $${\mathopen{}\left\lVert{}{L}_{{τ}_{0}}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}}^{2}+\sum_{j=1}^{N}{} {\mathopen{}\left\lVert{}\mathopen{}\left({L}_{{τ}_{j}}-{L}_{{τ}_{j-1}}\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}}^{2} = {\mathopen{}\left\lVert{}{L}_{{τ}_{N}}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}}^{2} \text{.}$$ By the assumption, it follows that the series $$${L}_{{τ}_{0}}\mathopen{}\left( ξ\right)\mathclose{}+\sum_{j=1}^{N}{} \mathopen{}\left({L}_{{τ}_{j}}-{L}_{{τ}_{j-1}}\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}$$$ (with partial sums $$${L}_{{τ}_{N}}$$$) converges in norm, say to $$$η\in H$$$. Notice that for $$$τ\gt {τ}_{N}$$$, the vectors $$$η-{L}_{τ}\mathopen{}\left( ξ\right)\mathclose{}$$$ and $$$\mathopen{}\left({L}_{τ}-{L}_{{τ}_{N}}\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}$$$ are orthogonal, so $$$\mathopen{}\left\lVert{}η-{L}_{τ}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}η-{L}_{{τ}_{N}}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}$$$.

The other direction is obvious.

Using the proposition, we define the operator $$$L= {L}_{f}$$$ on the domain $$$\mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}= \mathopen{}\left\{\, ξ\in H\,\middle\vert\, , \sup_{τ\gt 0}{} \mathopen{}\left\lVert{}{L}_{τ}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{} \lt \infty, \,\right\}\mathclose{}$$$ by $$$L\mathopen{}\left( ξ\right)\mathclose{}= \lim_{τ\to{+}\infty}{} {L}_{τ}\mathopen{}\left( ξ\right)\mathclose{}$$$, and write $$$L= \int _{{-}\infty}^{\infty}{}f\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d} {Q}_{t}$$$. Notice that $$$\bigcup_{τ}{}{H}_{τ}\subseteq \mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}$$$ (so $$$L$$$ is densely defined), and that $$$L$$$ coincides with $$${L}_{τ}$$$ on $$${H}_{τ}$$$. If the given continuous function $$$f$$$ is bounded, then $$$\mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}= H$$$ and $$$L$$$ is bounded with $$$\mathopen{}\left\lVert{}L\right\rVert\mathclose{}\leq \sup_{t\in \mathbb{R}}{}\mathopen{}\left\lvert{}f\mathopen{}\left( t\right)\mathclose{}\right\rvert\mathclose{}$$$.

Proposition IV.31

$$$\mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}= \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}$$$ and $$$L^{*}= {L}_{\overline{f}}$$$.

Proof. Calculating with Riemann-Stieltjes sums, we see that $$${L}_{τ}^{*}= {L}_{{\mathopen{}\left(\overline{f}, τ\right)\mathclose{}}_{}}$$$ for $$$τ\gt 0$$$, and $${\mathopen{}\left\lVert{} {L}_{τ}^{*}\mathopen{}\left( η\right)\mathclose{}\right\rVert\mathclose{}}^{2}= \int _{{-}τ}^{τ}{} {\mathopen{}\left\lvert{}f\mathopen{}\left( t\right)\mathclose{}\right\rvert\mathclose{}}^{2} \,\mathrm{d} \mathopen{}\left\langle{}{Q}_{t}\mathopen{}\left( η\right)\mathclose{}, η\right\rangle\mathclose{} = {\mathopen{}\left\lVert{}{L}_{τ}\mathopen{}\left( η\right)\mathclose{}\right\rVert\mathclose{}}^{2}$$ for all $$$η\in H$$$. Notice this norm equality implies $$$\mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}= \mathop{\mathcal{D}}\mathopen{}\left( {L}_{\overline{f}}\right)\mathclose{}$$$.

To show that $$$\mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}\subseteq \mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}$$$, fix $$$η\in \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}$$$, and let $$$φ$$$ be the bounded linear functional on $$$\mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}$$$ defined by $$$φ\mathopen{}\left( ξ\right)\mathclose{}= \mathopen{}\left\langle{}L\mathopen{}\left( ξ\right)\mathclose{}, η\right\rangle\mathclose{}$$$. For $$$ξ\in {H}_{τ}$$$, we have $$$φ\mathopen{}\left( ξ\right)\mathclose{}= \mathopen{}\left\langle{}{L}_{τ}\mathopen{}\left( ξ\right)\mathclose{}, η\right\rangle\mathclose{}= \mathopen{}\left\langle{}ξ, {L}_{τ}^{*}\mathopen{}\left( η\right)\mathclose{}\right\rangle\mathclose{}$$$, so $$\mathopen{}\left\lVert{}φ|{H}_{τ}\right\rVert\mathclose{}= \mathopen{}\left\lVert{} {L}_{τ}^{*}\mathopen{}\left( η\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}{L}_{τ}\mathopen{}\left( η\right)\mathclose{}\right\rVert\mathclose{} \text{.}$$ This makes $$$\mathopen{}\left\lVert{}{L}_{τ}\mathopen{}\left( η\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}φ\right\rVert\mathclose{}$$$ for all $$$τ\gt 0$$$, which puts $$$η\in \mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}$$$.

On the other hand, for $$$ξ$$$ and $$$η$$$ in $$$\mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}$$$, we have $$\mathopen{}\left\lvert{}\mathopen{}\left\langle{}L\mathopen{}\left( ξ\right)\mathclose{}, η\right\rangle\mathclose{}\right\rvert\mathclose{}= \lim_{τ\to\infty}{} \mathopen{}\left\lvert{}\mathopen{}\left\langle{}{L}_{τ}\mathopen{}\left( ξ\right)\mathclose{}, η\right\rangle\mathclose{}\right\rvert\mathclose{} = \lim_{τ\to\infty}{} \mathopen{}\left\lvert{}\mathopen{}\left\langle{}ξ, {L}_{τ}^{*}\mathopen{}\left( η\right)\mathclose{}\right\rangle\mathclose{}\right\rvert\mathclose{} \leq \mathopen{}\left\lVert{}ξ\right\rVert\mathclose{}\sup_{τ\gt 0}{} \mathopen{}\left\lVert{}{L}_{τ}\mathopen{}\left( η\right)\mathclose{}\right\rVert\mathclose{} = \mathopen{}\left\lVert{}ξ\right\rVert\mathclose{}\mathopen{}\left\lVert{}L\mathopen{}\left( η\right)\mathclose{}\right\rVert\mathclose{} \text{.}$$ It follows that $$$\mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}\subseteq \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}$$$, and hence $$$\mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}= \mathop{\mathcal{D}}\mathopen{}\left( L^{*}\right)\mathclose{}$$$.

Finally, take $$$ξ\in \mathop{\mathcal{D}}\mathopen{}\left( L\right)\mathclose{}$$$ and $$$η\in {H}_{τ}$$$. For all $$$t\geq τ$$$, we have $$\mathopen{}\left\langle{} L^{*}\mathopen{}\left( ξ\right)\mathclose{}, η\right\rangle\mathclose{}= \mathopen{}\left\langle{}ξ, L\mathopen{}\left( η\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}ξ, {L}_{t}\mathopen{}\left( η\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}{L}_{{\mathopen{}\left(\overline{f}, t\right)\mathclose{}}_{}}\mathopen{}\left( ξ\right)\mathclose{}, η\right\rangle\mathclose{} \text{.}$$ Letting $$$t$$$ approach $$$\infty$$$, we get $$$\mathopen{}\left\langle{} L^{*}\mathopen{}\left( ξ\right)\mathclose{}, η\right\rangle\mathclose{}= \mathopen{}\left\langle{}{L}_{\overline{f}}\mathopen{}\left( ξ\right)\mathclose{}, η\right\rangle\mathclose{}$$$. Since the union of the $$${H}_{τ}$$$'s is dense in $$$H$$$, we conclude that $$$L^{*}\mathopen{}\left( ξ\right)\mathclose{}= {L}_{\overline{f}}\mathopen{}\left( ξ\right)\mathclose{}$$$.

Definition IV.32

An operator $$$T$$$ in $$$H$$$ is self-adjoint provided $$$\mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}= \mathop{\mathcal{D}}\mathopen{}\left( T^{*}\right)\mathclose{}$$$ and that $$$\mathopen{}\left\langle{}T\mathopen{}\left( ξ\right)\mathclose{}, η\right\rangle\mathclose{}= \mathopen{}\left\langle{}ξ, T\mathopen{}\left( η\right)\mathclose{}\right\rangle\mathclose{}$$$ for all $$$ξ$$$ and $$$η$$$ in $$$\mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$.

Remark IV.33

Such an operator is necessarily closed because the adjoint of any operator is closed.

Example IV.34

For any resolution of the identity $$$t\mapsto {Q}_{t}$$$, the operator $$${L}_{f}$$$ is self-adjoint for real $$$f$$$ by the proposition above.

Example IV.35

Let $$$H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( X, μ\right)\mathclose{}$$$. For any measurable function $$$α : X \to \mathopen{}\left({-}\infty, \infty\right)\mathclose{}$$$, the operator of multiplication by $$$α$$$ on the domain $$$\mathopen{}\left\{\, ξ\in \mathrm{L}^{\mathrm{2}}\,\middle\vert\, , α\mathopen{}\left( ξ\right)\mathclose{}\in \mathrm{L}^{\mathrm{2}}, \,\right\}\mathclose{}$$$ is densely defined and self-adjoint. The given domain is dense because it contains $$$\mathrm{L}^{\mathrm{2}}\mathopen{}\left( {\mathopen{}\left\lvert{}α\right\rvert\mathclose{}}^{-1}\mathopen{}\left( \mathopen{}\left[n, {-}n\right]\mathclose{}\right)\mathclose{}\right)\mathclose{}$$$ for every $$$n$$$. The argument for self-adjointness resembles the proof of the proposition above. Observe that letting $$${Q}_{t}$$$ equal the projection on $$$\mathrm{L}^{\mathrm{2}}\mathopen{}\left( {α}^{-1}\mathopen{}\left( \mathopen{}\left({-}\infty, t\right]\mathclose{}\right)\mathclose{}\right)\mathclose{}$$$ gives a resolution of the identity.

Remark IV.36

Our version of the spectral theorem, coming up, says that Example IV.34 with $$$f\mathopen{}\left( t\right)\mathclose{}= t$$$ is universal. Another version of the spectral theorem, which we will not prove, says that Example IV.35 is also universal.

We now proceed to show that for any densely defined self-adjoint operator $$$T$$$, there is a resolution of the identity $$$t\mapsto {Q}_{t}$$$ such that $$$T= \int _{{-}\infty}^{\infty}{}t\,\mathrm{d} {Q}_{t}$$$ (including coincidence of the domains). The main idea is to change variables in the spectral resolution of the bounded self-adjoint operator $$$A= T{\mathopen{}\left(1+{T}^{2}\right)\mathclose{}}^{{-}\frac{1}{2}}$$$ to obtain the desired spectral resolution of the identity in $$$H$$$.

From Proposition IV.26, with $$$L= L^{*}= T$$$, we have the bounded positive operator $$$S= { \mathopen{}\left(1+{T}^{2}\right)\mathclose{} }^{-1}$$$ mapping $$$H$$$ injectively onto $$$\mathop{\mathcal{D}}\mathopen{}\left( {T}^{2}\right)\mathclose{}= \mathopen{}\left\{\, ξ\in \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}\,\middle\vert\, , T\mathopen{}\left( ξ\right)\mathclose{}\in \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}, \,\right\}\mathclose{}$$$. Notice that this makes $$$S\mathopen{}\left( H\right)\mathclose{}\subseteq \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$.

Lemma IV.37

$$$T\mathopen{}\left( S\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}= S\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}$$$ for all $$$ξ\in \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$.

Proof. Notice that $$${T}^{2}\mathopen{}\left( S\mathopen{}\left( \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left(1-S\right)\mathclose{}\mathopen{}\left( \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{}\subseteq \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$, so the following calculation makes sense for $$$ξ\in \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$: $$\mathopen{}\left(1+{T}^{2}\right)\mathclose{}\mathopen{}\left( T\mathopen{}\left( S\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}= T\mathopen{}\left( \mathopen{}\left(I+{T}^{2}\right)\mathclose{}\mathopen{}\left( S\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}= T\mathopen{}\left( ξ\right)\mathclose{}= \mathopen{}\left(1+{T}^{2}\right)\mathclose{}\mathopen{}\left( S\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right)\mathclose{} \text{.}$$ Since $$$T\mathopen{}\left( S\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}$$$ and $$$S\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}$$$ are both in $$$\mathop{\mathcal{D}}\mathopen{}\left( {T}^{2}\right)\mathclose{}$$$ and $$$1+{T}^{2}$$$ is injective on $$$\mathop{\mathcal{D}}\mathopen{}\left( {T}^{2}\right)\mathclose{}$$$, we get $$$T\mathopen{}\left( S\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}= S\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}$$$ as promised.

The following lemma improves slightly on part 4 of Proposition IV.26.

Lemma IV.38

$$${S}^{\frac{1}{2}}T\subseteq \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$.

Proof. For $$$η\in H$$$ and $$$ξ\in \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$, we have $$\mathopen{}\left\lvert{}\mathopen{}\left\langle{}T\mathopen{}\left( ξ\right)\mathclose{}, {S}^{\frac{1}{2}}\mathopen{}\left( η\right)\mathclose{}\right\rangle\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lvert{}\mathopen{}\left\langle{}{S}^{\frac{1}{2}}\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}, η\right\rangle\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}{S}^{\frac{1}{2}}\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right\rVert\mathclose{}\mathopen{}\left\lVert{}η\right\rVert\mathclose{} \text{.}$$ Furthermore, using $$$T\mathopen{}\left( S\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\in \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}= \mathop{\mathcal{D}}\mathopen{}\left( T^{*}\right)\mathclose{}$$$, $${\mathopen{}\left\lVert{}{S}^{\frac{1}{2}}\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right\rVert\mathclose{}}^{2}= \mathopen{}\left\langle{}{S}^{\frac{1}{2}}\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}, {S}^{\frac{1}{2}}\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}S\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}, T\mathopen{}\left( ξ\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}T\mathopen{}\left( S\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}, T\mathopen{}\left( ξ\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}{T}^{2}\mathopen{}\left( S\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}, ξ\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left(1-S\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}, ξ\right\rangle\mathclose{}\leq {\mathopen{}\left\lVert{}ξ\right\rVert\mathclose{}}^{2} \text{,}$$ so $$$\mathopen{}\left\lvert{}\mathopen{}\left\langle{}T\mathopen{}\left( ξ\right)\mathclose{}, {S}^{\frac{1}{2}}\mathopen{}\left( η\right)\mathclose{}\right\rangle\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}ξ\right\rVert\mathclose{}\mathopen{}\left\lVert{}η\right\rVert\mathclose{} \text{.}$$$ This means $$${S}^{\frac{1}{2}}\mathopen{}\left( η\right)\mathclose{}\in \mathop{\mathcal{D}}\mathopen{}\left( T^{*}\right)\mathclose{}= \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$.

We know from Proposition IV.26 that $$$0\leq S\leq 1$$$, so functional calculus gives us $$$f\mathopen{}\left( S\right)\mathclose{}$$$ for $$$f\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$.

Lemma IV.39

For every $$$f\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$, the operator $$$f\mathopen{}\left( S\right)\mathclose{}$$$ maps $$$\mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$ into itself and commutes with $$$T$$$ on $$$\mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$.

Proof. We know this is true for $$$f\mathopen{}\left( S\right)\mathclose{}= S$$$. Fix $$$ξ\in \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$. Then $$${S}^{2}\mathopen{}\left( ξ\right)\mathclose{}\in S\mathopen{}\left( T\right)\mathclose{}\subseteq \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$, and $$${S}^{2}\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}= S\mathopen{}\left( T\mathopen{}\left( S\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}= T\mathopen{}\left( {S}^{2}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}$$$. And so on: the assertion of the lemma holds when $$$f$$$ is a polynomial. For general continuous $$$f$$$, get a sequence $$$\mathopen{}\left({p}_{n}\right)\mathclose{}$$$ of polynomials converging uniformly to $$$f$$$ on $$$\mathopen{}\left[0, 1\right]\mathclose{}$$$, so $$$\mathopen{}\left\lVert{}{p}_{n}\mathopen{}\left( S\right)\mathclose{}-f\mathopen{}\left( S\right)\mathclose{}\right\rVert\mathclose{} \to 0$$$. In $$$\mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{}$$$, we have $$\mathopen{}\left({p}_{n}\mathopen{}\left( S\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}, T\mathopen{}\left( {p}_{n}\mathopen{}\left( S\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left({p}_{n}\mathopen{}\left( S\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}, {p}_{n}\mathopen{}\left( S\right)\mathclose{}\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right)\mathclose{} \to \mathopen{}\left(f\mathopen{}\left( S\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}, f\mathopen{}\left( S\right)\mathclose{}\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right)\mathclose{} \text{.}$$ This puts $$$\mathopen{}\left(f\mathopen{}\left( S\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}, f\mathopen{}\left( S\right)\mathclose{}\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}\in \mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{}$$$ because $$$T$$$ is a closed operator, so $$$f\mathopen{}\left( S\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}\in \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$ and $$$T\mathopen{}\left( f\mathopen{}\left( S\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}= f\mathopen{}\left( S\right)\mathclose{}\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}$$$.

Let $$$A= T{S}^{\frac{1}{2}}$$$. By Lemma IV.38, this defines $$$A$$$ on all of $$$H$$$. From Proposition IV.26 we know that $$$A$$$ is bounded, with $$$\mathopen{}\left\lVert{}A\right\rVert\mathclose{}\leq 1$$$.

Lemma IV.40

$$$A= A^{*}$$$ and $$${A}^{2}= 1-S$$$.

Proof. For $$$ξ\in \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$ and $$$η\in H$$$, we use Lemma IV.38 and Lemma IV.39 to obtain $$\mathopen{}\left\langle{}A\mathopen{}\left( ξ\right)\mathclose{}, η\right\rangle\mathclose{}= \mathopen{}\left\langle{}T\mathopen{}\left( {S}^{\frac{1}{2}}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}, η\right\rangle\mathclose{}= \mathopen{}\left\langle{}{S}^{\frac{1}{2}}\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}, η\right\rangle\mathclose{}= \mathopen{}\left\langle{}T\mathopen{}\left( ξ\right)\mathclose{}, {S}^{\frac{1}{2}}\mathopen{}\left( η\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}ξ, T\mathopen{}\left( {S}^{\frac{1}{2}}\mathopen{}\left( η\right)\mathclose{}\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}ξ, A\mathopen{}\left( η\right)\mathclose{}\right\rangle\mathclose{} \text{.}$$ Since $$$A$$$ is bounded and $$$\overline{\mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}}= H$$$, this works for any $$$ξ\in H$$$.

For the second claim, for $$$ξ\in \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$, we again use Lemma IV.39 to obtain $${A}^{2}\mathopen{}\left( ξ\right)\mathclose{}= A\mathopen{}\left( T\mathopen{}\left( {S}^{\frac{1}{2}}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}= A\mathopen{}\left( {S}^{\frac{1}{2}}\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}= T\mathopen{}\left( S\mathopen{}\left( T\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}= {T}^{2}\mathopen{}\left( S\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left(1-S\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{} \text{,}$$ so the bounded operators $$${A}^{2}$$$ and $$$1-S$$$ agree on a dense subspace.

Let $$$A$$$ have spectral resolution $$$λ\mapsto {P}_{λ}$$$. Then $$${P}_{λ}= 0$$$ for $$$λ\lt {-}1$$$ and $$${P}_{λ}= 1$$$ for $$$λ\gt 1$$$. Further, $$${\pm}1$$$ are not eigenvalues of $$$A$$$, because $$$1-{A}^{2}= B$$$, whose kernel is $$$\mathopen{}\left\{\, 0\,\right\}\mathclose{}$$$. By Proposition IV.12, this means $$$\bigcap_{t\leq {-}1}{}{P}_{t}\mathopen{}\left( H\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{}$$$ and $$$\overline{ \bigcup_{t\leq {-}1}{}{P}_{t}\mathopen{}\left( H\right)\mathclose{} }= H$$$. Consider the strictly increasing function $$$Φ : \mathbb{R} \to \mathopen{}\left({-}1, 1\right)\mathclose{}$$$ defined by $$$Φ\mathopen{}\left( t\right)\mathclose{}= t{\mathopen{}\left(1+{t}^{2}\right)\mathclose{}}^{{-}\frac{1}{2}}$$$. Let $$${Q}_{t}= {P}_{Φ\mathopen{}\left( t\right)\mathclose{}}$$$. This makes $$$\bigcap_{t}{}{Q}_{t}\mathopen{}\left( H\right)\mathclose{}= \bigcap_{t\leq {-}1}{}{P}_{t}\mathopen{}\left( H\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{}$$$ and $$$\overline{ \bigcup_{t}{}{Q}_{t}\mathopen{}\left( H\right)\mathclose{} }= \overline{ \bigcup_{t\leq {-}1}{}{P}_{t}\mathopen{}\left( H\right)\mathclose{} }= H$$$, so $$$t\mapsto {Q}_{t}$$$ is a resolution of the identity. Define the self-adjoint operator $$$B$$$ by $$B= \int _{{-}\infty}^{\infty}{}t\,\mathrm{d} {Q}_{t} \text{.}$$

Our mission will be accomplished once we show that $$$\mathop{\mathcal{D}}\mathopen{}\left( B\right)\mathclose{}= \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$ and $$$B= T$$$, that is, that $$$\mathop{\mathcal{G}}\mathopen{}\left( K\right)\mathclose{}= \mathop{\mathcal{G}}\mathopen{}\left( G, T\right)\mathclose{}$$$.

As before, write $$${E}_{τ}= {Q}_{τ}-{Q}_{{-}τ}= {P}_{Φ\mathopen{}\left( τ\right)\mathclose{}}-{P}_{{-}Φ\mathopen{}\left( τ\right)\mathclose{}}$$$ for $$$τ\gt 0$$$, and $$${H}_{τ}= {E}_{τ}\mathopen{}\left( T\right)\mathclose{}$$$. What happens when we compress $$$B$$$, $$$A$$$, and $$$T$$$ to $$${H}_{τ}$$$? Since $$${E}_{τ}$$$ is a difference of spectral projections for $$$B$$$ and for $$$A$$$, the subspace $$${H}_{τ}$$$ is invariant for both of these operators, and the compressions $$${B}_{τ}= B\mathopen{}\left( {E}_{τ}\right)\mathclose{}= {E}_{τ}\mathopen{}\left( B\right)\mathclose{}$$$ and $$${A}_{τ}= A\mathopen{}\left( {E}_{τ}\right)\mathclose{}= {E}_{τ}\mathopen{}\left( A\right)\mathclose{}$$$ are given as spectral integrals by $${B}_{τ}= \int _{{-}τ}^{τ}{}t\,\mathrm{d}{Q}_{t}$$ and $${A}_{τ}= \int _{{-}Φ\mathopen{}\left( τ\right)\mathclose{}}^{Φ\mathopen{}\left( τ\right)\mathclose{}}{}λ\,\mathrm{d}{P}_{λ} \text{.}$$ They satisfy $${-}\infty\lt {-}τ\leq {B}_{τ}\leq τ\lt \infty$$ and $${-}1\lt {-}Φ\mathopen{}\left( τ\right)\mathclose{}\leq {A}_{τ}\leq Φ\mathopen{}\left( τ\right)\mathclose{}\lt 1 \text{.}$$ Notice this makes $$$1-{{A}_{τ}}^{2}$$$ invertible.

Lemma IV.41

For $$$τ\gt 0$$$: (i) $$${H}_{τ}\subseteq \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{}$$$; and (ii) for $$$ξ\in {H}_{τ}$$$, we have $$$T\mathopen{}\left( ξ\right)\mathclose{}= {A}_{τ}{\mathopen{}\left(1-{{A}_{τ}}^{2}\right)\mathclose{}}^{{-}\frac{1}{2}}ξ$$$.

Proof. Take $$$ξ\in {H}_{τ}$$$. For (i), observe $$$\mathopen{}\left(1-{A}^{2}\right)\mathclose{}ξ= \mathopen{}\left(1-{{A}_{τ}}^{2}\right)\mathclose{}ξ$$$, and thus $$ξ= {\mathopen{}\left(1-{{A}_{τ}}^{2}\right)\mathclose{}}^{{-}1}\mathopen{}\left(1-{A}^{2}\right)\mathclose{}ξ= \mathopen{}\left(1-{A}^{2}\right)\mathclose{}{\mathopen{}\left(1-{{A}_{τ}}^{2}\right)\mathclose{}}^{{-}1}ξ= S{\mathopen{}\left(1-{{A}_{τ}}^{2}\right)\mathclose{}}^{{-}1}ξ\in S\mathopen{}\left( T\right)\mathclose{}\subseteq \mathop{\mathcal{D}}\mathopen{}\left( T\right)\mathclose{} \text{.}$$

For (ii), recall $$$A= T{S}^{\frac{1}{2}}= T{\mathopen{}\left(1-{A}^{2}\right)\mathclose{}}^{\frac{1}{2}}$$$, so $$${A}_{τ}= T{\mathopen{}\left(1-{{A}_{τ}}^{2}\right)\mathclose{}}^{\frac{1}{2}}$$$. Evaluating at $$$ξ$$$, we get $$Tξ= T{\mathopen{}\left(1-{{A}_{τ}}^{2}\right)\mathclose{}}^{\frac{1}{2}}{\mathopen{}\left(1-{{A}_{τ}}^{2}\right)\mathclose{}}^{{-}\frac{1}{2}}ξ= {A}_{τ}{\mathopen{}\left(1-{{A}_{τ}}^{2}\right)\mathclose{}}^{{-}\frac{1}{2}}ξ \text{.}$$

Thus $$$T$$$ maps $$${H}_{τ}$$$ boundedly into itself, and commutes with $$${E}_{τ}$$$. The compression $$${F}_{τ}$$$ given by $$${F}_{τ}= T{E}_{τ}= {E}_{τ}T$$$ is, by part (ii) of the lemma, $$${A}_{τ}{\mathopen{}\left(1-{{A}_{τ}}^{2}\right)\mathclose{}}^{{-}\frac{1}{2}}$$$. This means $$${F}_{τ}$$$ is given by the spectral integral $${F}_{τ}= \int _{{-}Φ\mathopen{}\left( τ\right)\mathclose{}}^{Φ\mathopen{}\left( τ\right)\mathclose{}}{}\frac{λ}{\sqrt{1-{λ}^{2}}}\,\mathrm{d}{P}_{λ} \text{.}$$

Notice that the integrand is $$${Φ}^{-1}\mathopen{}\left( λ\right)\mathclose{}$$$.

Lemma IV.42

$$${F}_{τ}= {B}_{τ}$$$ for all $$$τ\gt 0$$$.

Proof. Both integrals are norm limits of Riemann-Stieltjes sums. Just as in the scalar case, the change of variables $$$t= {Φ}^{-1}\mathopen{}\left( λ\right)\mathclose{}$$$ transforms Riemann-Stieltjes sums for the integral in $$$λ$$$ for $$${F}_{τ}$$$ into Riemann-Stieltjes sums for the integral in $$$t$$$ for $$${B}_{τ}$$$. The reverse change of variables $$$λ= Φ\mathopen{}\left( t\right)\mathclose{}$$$ inverts this transformation.

Now we can prove the theorem.

Theorem IV.43

For every densely defined self-adjoint operator $$$T$$$ in $$$H$$$, there is a resolution of the identity $$$t\mapsto {Q}_{t}$$$ such that $$$T= \int _{{-}\infty}^{\infty}{}t\,\mathrm{d}{Q}_{t}$$$.

Proof. What remains to be shown is that $$$\mathop{\mathcal{G}}\mathopen{}\left( B\right)\mathclose{}= \mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{}$$$. Take $$$\mathopen{}\left(ξ, B\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\in \mathop{\mathcal{G}}\mathopen{}\left( B\right)\mathclose{}$$$. Then by Lemma IV.42 we have $$\mathopen{}\left(ξ, {B}_{ξ}\right)\mathclose{}= \lim_{τ\to{+}\infty}{} \mathopen{}\left({E}_{τ}\mathopen{}\left( ξ\right)\mathclose{}, {B}_{τ}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{} = \lim_{τ\to{+}\infty}{} \mathopen{}\left({E}_{τ}\mathopen{}\left( ξ\right)\mathclose{}, {F}_{τ}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{} = \lim_{τ\to{+}\infty}{} \mathopen{}\left({E}_{τ}\mathopen{}\left( ξ\right)\mathclose{}, T\mathopen{}\left( {E}_{τ}\mathopen{}\left( ξ\right)\mathclose{}\right)\mathclose{}\right)\mathclose{} \in \overline{\mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{}}= \mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{} \text{.}$$ Hence $$$\mathop{\mathcal{G}}\mathopen{}\left( B\right)\mathclose{}\subseteq \mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{}$$$. For the reverse inclusion, suppose $$$\mathopen{}\left(η, T\mathopen{}\left( η\right)\mathclose{}\right)\mathclose{}\in \mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{}\ominus \mathop{\mathcal{G}}\mathopen{}\left( B\right)\mathclose{}$$$. For $$$ξ\in {H}_{τ}$$$ we have $$0= \mathopen{}\left\langle{}η, ξ\right\rangle\mathclose{}+\mathopen{}\left\langle{}T\mathopen{}\left( η\right)\mathclose{}, B\mathopen{}\left( ξ\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}η, ξ\right\rangle\mathclose{}+\mathopen{}\left\langle{}{F}_{τ}\mathopen{}\left( η\right)\mathclose{}, {B}_{τ}\mathopen{}\left( ξ\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}η, \mathopen{}\left(1+{{B}_{τ}}^{2}\right)\mathclose{}ξ\right\rangle\mathclose{} \text{.}$$ But $$${B}_{τ}$$$ is a bounded self-adjoint operator for which $$${B}_{τ}{H}_{τ}\subseteq {H}_{τ}$$$, whence it follows that the invertible operator $$$1+{{B}_{τ}}^{2}$$$ maps $$${H}_{τ}$$$ onto itself. Thus $$$η$$$ is orthogonal to the dense subspace $$$\bigcup_{τ}{} {H}_{τ}$$$, which makes $$$η= 0$$$.

As discussed earlier, we have for $$$T$$$ a reasonably well-behaved functional calculus for continuous (complex-valued) functions on $$$\mathbb{R}$$$ which yields a densely defined operator $$$f\mathopen{}\left( T\right)\mathclose{}$$$ with $$$\mathop{\mathcal{D}}\mathopen{}\left( \mathopen{}\left(f\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{}^{*}\right)\mathclose{}= \mathop{\mathcal{D}}\mathopen{}\left( \overline{f}\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{}$$$ for every such function $$$f$$$. Explicitly, in the ambient notation, $$f\mathopen{}\left( T\right)\mathclose{}= \int _{{-}\infty}^{\infty}{}f\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}{Q}_{t} \text{,}$$ which means $$$f\mathopen{}\left( T\right)\mathclose{}ξ= \lim_{τ\to{+}\infty}{} {f\mathopen{}\left( T\right)\mathclose{}}_{τ}ξ$$$ for $$$ξ\in \mathop{\mathcal{D}}\mathopen{}\left( f\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{}$$$, which is the set of all vectors for which the limit exists, where $$${f\mathopen{}\left( T\right)\mathclose{}}_{τ}= \int _{{-}τ}^{τ}{}f\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}{Q}_{t}$$$. Arithmetic with functions works as is should provided one restricts to the dense subspace $$$\bigcup_{τ}{}{H}_{τ}$$$, which is contained in the domain of every continuous function of $$$T$$$. Ambiguities vanish when $$$f$$$ is bounded, which as we have seen makes $$$f\mathopen{}\left( T\right)\mathclose{}$$$ bounded, with norm at most $$$\sup_{t\in \mathbb{R}}{}\mathopen{}\left\lvert{}f\mathopen{}\left( t\right)\mathclose{}\right\rvert\mathclose{}$$$. For bounded functions, we have $$$\mathopen{}\left(f\mathopen{}\left( g\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( T\right)\mathclose{}= f\mathopen{}\left( T\right)\mathclose{}g\mathopen{}\left( T\right)\mathclose{}$$$, $$$\mathopen{}\left(f+g\right)\mathclose{}\mathopen{}\left( T\right)\mathclose{}= f\mathopen{}\left( T\right)\mathclose{}+g\mathopen{}\left( T\right)\mathclose{}$$$ and $$$\overline{f}\mathopen{}\left( T\right)\mathclose{}= f\mathopen{}\left( T\right)\mathclose{}^{*}$$$.

In particular, we can form the bounded operator $$${\mathrm{e}}^{\mathrm{i}T}$$$, defined as $$$\exp\mathopen{}\left( \mathrm{i}T\right)\mathclose{}$$$. Its adjoint $$${\mathrm{e}}^{{-}\mathrm{i}T}$$$ is also its inverse, that is, $$${\mathrm{e}}^{\mathrm{i}T}$$$ is unitary. Parlay this into a unitary representation $$$t\mapsto {U}_{t}$$$ of $$$\mathbb{R}$$$ on $$$T$$$ by setting $$${U}_{t}= {\mathrm{e}}^{{-}\mathrm{i}tT}$$$ for real $$$t$$$. These unitary operators obey $$${U}_{{-}t}= {U}_{t}^{*}$$$ and $$${U}_{s+t}= {U}_{s}{U}_{t}$$$. It is not hard to check that the representation is strongly continuous in the sense that $$$\lim_{t\to0}{}{U}_{t}ξ= ξ$$$ for all $$$ξ\in T$$$. A famous theorem of M. Stone asserts that every strongly continuous unitary representation of $$$\mathbb{R}$$$ on $$$T$$$ arises in this way.

We conclude with a look at how these notions play out in quantum mechanics. In the classical framework, the states of a mechanical system with $$$d$$$ degrees of freedom are vectors in $$${\mathbb{R}}^{d}$$$. To keep track of a quantum mechanical system, we must move to the Hilbert space $$$H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( {\mathbb{R}}^{d}\right)\mathclose{}$$$. The state of the system is described by a unit vector $$$ψ$$$ in $$$H$$$, which imposes the probability density $$${\mathopen{}\left\lvert{}ψ\right\rvert\mathclose{}}^{2}$$$ on $$${\mathbb{R}}^{d}$$$. The observables of the system (things that we can imagine measuring in a way that yields a real number at each observation, but in a random way governed by $$${\mathopen{}\left\lvert{}ψ\right\rvert\mathclose{}}^{2}$$$) are bounded self-adjoint operators on $$$H$$$. For an observable $$$A$$$, the expected value of $$$A$$$ when the system is in the state $$$ψ$$$ is $$$\mathopen{}\left\langle{}A\mathopen{}\left( ψ\right)\mathclose{}, ψ\right\rangle\mathclose{}$$$. Evolution of the system over time is determined by a self-adjoint operator $$$B$$$ called the Hamiltonian, which is unbounded in almost all cases of interest. The way $$$B$$$ moves the system is by the unitary representation $$${U}_{t}= {\mathrm{e}}^{{-}\mathrm{i}tB}$$$. From one point of view, the state $$$ψ$$$ remains fixed, while the observable $$$A$$$ at time $$$0$$$ becomes $$${U}_{t} ^{*}A{U}_{t}$$$ at time $$$t$$$. The expected value of $$$A$$$ at time $$$t$$$ in the state $$$ψ$$$ is then $$$\mathopen{}\left\langle{} {U}_{t} ^{*}A{U}_{t}ξ, ξ\right\rangle\mathclose{}$$$. This is of course the same as $$$\mathopen{}\left\langle{}A{U}_{t}ξ, {U}_{t}ξ\right\rangle\mathclose{}$$$, so in terms of what can usefully be measured in the laboratory, we can equally well regard $$$A$$$ as fixed and the state moving from $$${ψ}_{0}$$$ at time $$$0$$$ to $$${ψ}_{t}= {U}_{t}{ψ}_{0}$$$ at time $$$t$$$. This means $$$ψ= ψ\mathopen{}\left( \mathbf{x}, t\right)\mathclose{}= {ψ}_{t}\mathopen{}\left( \mathbf{x}\right)\mathclose{}= \exp\mathopen{}\left( {-}\mathrm{i}tT\right)\mathclose{}{ψ}_{0}$$$, so $$$\frac{\partial }{\partial t}\mathopen{}\left( ψ\mathopen{}\left( \mathbf{x}, t\right)\mathclose{}\right)\mathclose{}= {-}\mathrm{i}B\exp\mathopen{}\left( {-}\mathrm{i}tB\right)\mathclose{}{ψ}_{0}= {-}\mathrm{i}Bψ\mathopen{}\left( \mathbf{x}, t\right)\mathclose{}$$$. This is Schrödinger's equation: $$$\mathrm{i}\frac{\partial ψ}{\partial t}= Bψ$$$.

To go beyond mere epistemology, one has to grab hold of physically meaningful Hamiltonians—which would, as they say, take us beyond the scope of this course.