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,

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

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

Proof.

$\operatorname{Ker}\mathopen{}\left( E\right)\mathclose{}= 0$ is clear. $\overline{\operatorname{Ran}\mathopen{}\left( E^{*}\right)\mathclose{}}= { \mathopen{}\left(\operatorname{Ker}\mathopen{}\left( E\right)\mathclose{}\right)\mathclose{} }^{\perp}= \mathop{\mathcal{G}}\mathopen{}\left( L\right)\mathclose{}$ .
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 $L^{*}\mathopen{}\left( L\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}= w-y$ .
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

$\mathop{\mathcal{D}}\mathopen{}\left( L^{*}L\right)\mathclose{}$ is dense.
$1+ L^{*}L : \mathop{\mathcal{D}}\mathopen{}\left( L^{*}L\right)\mathclose{} \to H$ is bijective.
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 $\mathopen{}\left\lVert{}S\right\rVert\mathclose{}\leq 1$ .
$L{S}^{\frac{1}{2}}= L{\mathopen{}\left(1+ L^{*}L\right)\mathclose{}}^{{-}\frac{1}{2}}$ 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.

Previous: The Spectral Theorem

Next: Additional Exercises