The Spectral Theorem for a Bounded Self-Adjoint Operator

Contents

Statement
The polynomial calculus
The continuous calculus
The spectral measures
The Borel calculus
The projection-valued measure
The multiplication form

This appendix proves Theorem 16.59 of Hilbert Spaces, in both of the forms stated there: the existence and uniqueness of a projection-valued measure $E$ on $\R$, supported on $\sigma(A)$, with $A=\int_{\sigma(A)}\lambda\,\dd E(\lambda)$; and the unitary equivalence of $A$, on a separable Hilbert space, with multiplication by a bounded real function on a finite measure space. It is the theorem three of the quantum postulates of The Postulates of Quantum Mechanics are statements about (Remark 16.63), and it is the replacement, forced by Example 16.56, for the eigenbasis that Theorem 9.79 supplies in finite dimension.

The route is the classical one of von Neumann [vonNeumann:1930] [vonNeumann:1932] in the arrangement of Reed and Simon [Reed:1972]: first the polynomial calculus and the identity $\norm{p(A)}=\sup_{\sigma(A)}\abs{p}$, which is what makes the construction possible at all; then the continuous calculus, obtained by completing the polynomials in the supremum norm; then, for each pair of vectors, a complex measure on $\sigma(A)$; then the extension of the calculus to bounded Borel functions, whose values on indicator functions are the projections $E(\Omega)$; and finally the multiplication form. Two classical theorems are used and are not proved in this treatise — Stone–Weierstrass and Riesz–Markov. Each is stated precisely where it is used, and Remark A24.14 collects them.

Throughout, $A\in\mathcal{B}(\mathcal{H})$ is self-adjoint, $\mathcal{H}\neq\set{0}$, and $\sigma(A)\subseteq\R$ is compact and non-empty (Theorems 16.54 and 16.55). We write $C(\sigma(A))$ for the continuous complex functions on $\sigma(A)$ with the supremum norm $\norm{f}_{\infty}=\sup_{\lambda\in\sigma(A)}\abs{f(\lambda)}$, and $B(\sigma(A))$ for the bounded Borel functions with the same norm.

Statement

Theorem A24.1 (Spectral theorem, both forms).

Let $A\in\mathcal{B}(\mathcal{H})$ be self-adjoint. Then:

there is exactly one projection-valued measure $E$ on $\R$ (Definition 16.58) with $E(\R\setminus\sigma(A))=0$ and

\begin{equation}\tag{A24.1} \braket{x}{Ay}=\int_{\sigma(A)}\lambda\, \braket{x}{\dd E(\lambda)y}\ec\qquad \forall\,x,y\in\mathcal{H}\ec \end{equation}

and every $B\in\mathcal{B}(\mathcal{H})$ commuting with $A$ commutes with every $E(\Omega)$;
if $\mathcal{H}$ is separable there are a measure space $(M,\mu)$ with $\mu(M)<\infty$, a bounded real measurable function $a$ on $M$, and a unitary $U\!:\mathcal{H}\longrightarrow L^{2}(M,\mu)$ with

\begin{equation}\tag{A24.2} \bigl(UAU^{-1}f\bigr)(m)=a(m)\,f(m)\ec\qquad \forall\,f\in L^{2}(M,\mu)\ep \end{equation}

Rests on Definition 16.58, Theorem 16.55 and Proposition 16.43.

The polynomial calculus

For a polynomial $p(z)=\sum_{k=0}^{d}c_{k}z^{k}$ with complex coefficients write $p(A)=\sum_{k}c_{k}A^{k}\in\mathcal{B}(\mathcal{H})$, with $A^{0}=\identity$, and let $\bar p(z)=\sum_{k}c_{k}^{\ast}z^{k}$. Since $A^{\dagger}=A$, Equation (16.24) gives $p(A)^{\dagger}=\bar p(A)$.

Lemma A24.2 (The norm of a self-adjoint operator lies in its spectrum).

Let $B\in\mathcal{B}(\mathcal{H})$ be self-adjoint. Then $\norm{B}\in\sigma(B)$ or $-\norm{B}\in\sigma(B)$; consequently

\begin{equation}\tag{A24.3} \norm{B}=\sup_{\lambda\in\sigma(B)}\abs{\lambda}\ep \end{equation}

Rests on Proposition 16.43, Proposition 16.52 and Definition 16.49.

Proof.

Derives Lemma A24.2. If $B=0$ then $\sigma(B)=\set{0}$ by Theorem 16.54 and Proposition 16.52 and the claim is trivial, so let $B\neq0$. Exactly as in Equation (A23.4), choose unit vectors $x_{n}$ with $\braket{x_{n}}{Bx_{n}}\longrightarrow\lambda$ where $\lambda\in\R$ and $\abs{\lambda}=\norm{B}$; this is possible by Proposition 16.43 and because $\braket{x_{n}}{Bx_{n}}$ is real (Proposition 16.42). Then, as in Equation (A23.5),

\begin{equation}\tag{A24.4} \norm{(B-\lambda\identity)x_{n}}^{2} =\norm{Bx_{n}}^{2}-2\lambda\braket{x_{n}}{Bx_{n}}+\lambda^{2} \leq2\lambda^{2}-2\lambda\braket{x_{n}}{Bx_{n}} \longrightarrow0\ep \end{equation}

If $B-\lambda\identity$ had a bounded inverse $R$ we would get $1=\norm{x_{n}}\leq\norm{R}\norm{(B-\lambda\identity)x_{n}} \longrightarrow0$, which is false; so $\lambda\in\sigma(B)$ with $\abs{\lambda}=\norm{B}$. Since $\sigma(B)$ is contained in the disc of radius $\norm{B}$ (Proposition 16.52), Equation (A24.3) follows.

∎

Lemma A24.3 (Spectral mapping for polynomials).

For every polynomial $p$ and every $A\in\mathcal{B}(\mathcal{H})$,

\begin{equation}\tag{A24.5} \sigma\bigl(p(A)\bigr)=p\bigl(\sigma(A)\bigr) =\set{p(\lambda)\mid\lambda\in\sigma(A)}\ep \end{equation}

Rests on Definition 16.49 and Proposition 16.37.

Proof.

Derives Lemma A24.3. If $p$ is constant, $p\equiv c$, then $p(A)=c\identity$ and both sides are $\set{c}$ (using $\sigma(A)\neq\varnothing$, Theorem 16.54). Otherwise let $\mu\in\C$ and factor the non-constant polynomial $p(z)-\mu$ over $\C$ (Theorem 12.19):

\begin{equation}\tag{A24.6} p(z)-\mu=c\prod_{j=1}^{d}(z-\alpha_{j})\ec\qquad c\neq0\ec \end{equation}

so that, substituting $A$ and using that polynomials in $A$ commute,

\begin{equation}\tag{A24.7} p(A)-\mu\identity=c\prod_{j=1}^{d}(A-\alpha_{j}\identity)\ep \end{equation}

If every factor is invertible, so is the product, and $\mu\in\rho(p(A))$. Conversely suppose the product $T=\prod_{j}(A-\alpha_{j}\identity)$ is invertible and fix $j$. Its inverse $T^{-1}$ commutes with $A$ — from $TA=AT$ one gets $AT^{-1}=T^{-1}A$ on multiplying by $T^{-1}$ on both sides — hence with every factor. Writing $S_{j}=\prod_{k\neq j}(A-\alpha_{k} \identity)$ we have $T=(A-\alpha_{j}\identity)S_{j}=S_{j} (A-\alpha_{j}\identity)$, so

\begin{equation*} (A-\alpha_{j}\identity)\,S_{j}T^{-1}=\identity=T^{-1}S_{j}\, (A-\alpha_{j}\identity)\ec \end{equation*}

and $A-\alpha_{j}\identity$ is invertible with inverse $S_{j}T^{-1}=T^{-1}S_{j}$. Therefore $\mu\in\sigma(p(A))$ if and only if some $\alpha_{j}\in\sigma(A)$; and by Equation (A24.6) the $\alpha_{j}$ are exactly the roots of $p-\mu$, so this happens exactly when $\mu=p(\lambda)$ for some $\lambda\in\sigma(A)$.

∎

Proposition A24.4 (The polynomial calculus is isometric).

For every polynomial $p$ and every self-adjoint $A\in\mathcal{B}(\mathcal{H})$,

\begin{equation}\tag{A24.8} \norm{p(A)}=\sup_{\lambda\in\sigma(A)}\abs{p(\lambda)} =\norm{p}_{\infty}\ep \end{equation}

Rests on Lemma A24.2, Lemma A24.3 and Proposition 16.39.

Proof.

Derives Proposition A24.4. The operator $p(A)^{\dagger}p(A)=(\bar p\,p)(A)$ is self-adjoint, and $\bar p\,p$ is a polynomial. By the $C^{\ast}$ identity Equation (16.25), then Lemma A24.2 applied to it, then Lemma A24.3,

\begin{equation}\tag{A24.9} \norm{p(A)}^{2}=\norm{(\bar p\,p)(A)} =\sup_{\mu\in\sigma((\bar p\,p)(A))}\abs{\mu} =\sup_{\lambda\in\sigma(A)}\abs{\bar p(\lambda)\,p(\lambda)} =\sup_{\lambda\in\sigma(A)}\abs{p(\lambda)}^{2}\ec \end{equation}

the last step because $\sigma(A)\subseteq\R$ (Theorem 16.55), so that $\bar p(\lambda)=p(\lambda)^{\ast}$ there.

∎

The continuous calculus

Theorem A24.5 (Stone–Weierstrass; quoted).

Let $K$ be a compact Hausdorff space and let $\mathcal{A}\subseteq C(K)$ be a subalgebra that contains the constants, separates the points of $K$, and is closed under complex conjugation. Then $\mathcal{A}$ is dense in $C(K)$ for the supremum norm. For $K\subseteq\R$ compact the polynomials form such a subalgebra, and the statement reduces to the classical Weierstrass approximation theorem quoted in Remark 16.1. Rests on Definition 10.9.

Proposition A24.6 (Continuous functional calculus).

There is exactly one map $\Phi\!:C(\sigma(A))\longrightarrow\mathcal{B}(\mathcal{H})$ that is linear, multiplicative, sends $1\longmapsto\identity$ and $\lambda\longmapsto A$, and is continuous for $\norm{\cdot}_{\infty}$ and the operator norm. It satisfies, writing $f(A)=\Phi(f)$,

\begin{equation}\tag{A24.10} \norm{f(A)}=\norm{f}_{\infty}\ec\qquad f(A)^{\dagger}=\bar f(A)\ec\qquad f\geq0\implies f(A)\geq0\ec \end{equation}

and $f(A)$ commutes with every bounded operator commuting with $A$. Rests on Proposition A24.4, Theorem A24.5 and Proposition 16.61.

Proof.

Derives Proposition A24.6. Well defined on polynomials. If two polynomials agree on $\sigma(A)$ their difference $q$ has $\norm{q}_{\infty}=0$, so $q(A)=0$ by Equation (A24.8): the map $p|_{\sigma(A)}\longmapsto p(A)$ depends only on the restriction, and it is isometric.

Extension. The polynomials restricted to the compact set $\sigma(A)\subseteq\R$ are dense in $C(\sigma(A))$ by Theorem A24.5. An isometric linear map defined on a dense subspace of a normed space, with values in the complete space $\mathcal{B}(\mathcal{H})$ (Proposition 16.37), extends uniquely to an isometry of the closure: given $f$, choose polynomials $p_{n}$ with $\norm{p_{n}-f}_{\infty}\longrightarrow0$; then $\norm{p_{n}(A)-p_{m}(A)}=\norm{p_{n}-p_{m}}_{\infty}$ makes $(p_{n}(A))$ Cauchy, and its limit $\Phi(f)$ is independent of the approximating sequence by the same identity. Linearity and multiplicativity pass to the limit because addition and multiplication are continuous in a Banach algebra (Proposition 16.37), and the first identity of Equation (A24.10) is the limit of Equation (A24.8). Taking adjoints term by term — legitimate because the adjoint is isometric (Proposition 16.39) — gives $\Phi(f)^{\dagger}=\Phi(\bar f)$.

Positivity. If $f\geq0$ on $\sigma(A)$ then $g=\sqrt{f}$ is continuous and real, so $\Phi(f)=\Phi(g)^{2}=\Phi(g)^{\dagger}\Phi(g)$ is positive by Proposition 16.42(iv).

Commutation. If $BA=AB$ then $B$ commutes with every polynomial in $A$, hence with the norm limits $\Phi(f)$.

Uniqueness is Proposition 16.61, already proved in the chapter.

∎

The spectral measures

Theorem A24.7 (Riesz–Markov; quoted).

Let $K$ be a compact metric space. For every bounded linear functional $\ell$ on $C(K)$ there is exactly one regular complex Borel measure $\nu$ on $K$ with

\begin{equation}\tag{A24.11} \ell(f)=\int_{K}f\,\dd\nu\ec\qquad\forall\,f\in C(K)\ec \end{equation}

and $\norm{\ell}$ equals the total variation $\abs{\nu}(K)$. If $\ell(f)\geq0$ whenever $f\geq0$, then $\nu$ is a positive measure and $\norm{\ell}=\nu(K)=\ell(1)$. Regularity carries with it that $C(K)$ is dense in $L^{2}(K,\nu)$ for every positive finite Borel measure $\nu$. Rests on Definition 10.9.

Proposition A24.8 (The measures $\mu_{x,y}$).

For $x,y\in\mathcal{H}$ there is a unique regular complex Borel measure $\mu_{x,y}$ on $\sigma(A)$ with

\begin{equation}\tag{A24.12} \braket{x}{f(A)\,y}=\int_{\sigma(A)}f\,\dd\mu_{x,y}\ec\qquad \forall\,f\in C(\sigma(A))\ep \end{equation}

The assignment is linear in $y$, antilinear in $x$, obeys $\mu_{y,x}=\mu_{x,y}^{\ast}$ and $\abs{\mu_{x,y}}(\sigma(A))\leq\norm{x}\norm{y}$; and $\mu_{x} :=\mu_{x,x}$ is a positive measure with $\mu_{x}(\sigma(A))=\norm{x}^{2}$. Rests on Proposition A24.6, Theorem A24.7 and Theorem 16.46.

Proof.

Derives Proposition A24.8. The map $f\longmapsto\braket{x}{f(A)y}$ is linear on $C(\sigma(A))$ and bounded, since by Equation (16.1) and Equation (A24.10)

\begin{equation}\tag{A24.13} \abs{\braket{x}{f(A)y}}\leq\norm{x}\,\norm{f(A)}\,\norm{y} =\norm{f}_{\infty}\norm{x}\norm{y}\ep \end{equation}

Theorem A24.7 supplies a unique regular complex Borel measure $\mu_{x,y}$ satisfying Equation (A24.12), with $\abs{\mu_{x,y}}(\sigma(A))\leq\norm{x}\norm{y}$ by the norm statement there. Linearity in $y$ and antilinearity in $x$ hold for the functionals and therefore, by the uniqueness in Theorem A24.7, for the measures. For the conjugation rule,

\begin{equation*} \left(\int f\,\dd\mu_{x,y}\right)^{\ast} =\braket{f(A)y}{x}=\braket{y}{f(A)^{\dagger}x} =\braket{y}{\bar f(A)x}=\int\bar f\,\dd\mu_{y,x}\ec \end{equation*}

and since $f\longmapsto\bar f$ exhausts $C(\sigma(A))$, uniqueness gives $\mu_{y,x}=\mu_{x,y}^{\ast}$. Finally, if $f\geq0$ then $\braket{x}{f(A)x}\geq0$ by the positivity in Equation (A24.10), so $\mu_{x}$ is a positive measure, and taking $f=1$ in Equation (A24.12) gives $\mu_{x}(\sigma(A))=\braket{x}{x}=\norm{x}^{2}$.

∎

The Borel calculus

Proposition A24.9 (Bounded Borel functional calculus).

There is a unique map $g\longmapsto g(A)$ from $B(\sigma(A))$ to $\mathcal{B}(\mathcal{H})$ extending $\Phi$ and satisfying

\begin{equation}\tag{A24.14} \braket{x}{g(A)\,y}=\int_{\sigma(A)}g\,\dd\mu_{x,y}\ec\qquad \forall\,x,y\in\mathcal{H}\ep \end{equation}

It is linear, multiplicative, satisfies $g(A)^{\dagger}=\bar g(A)$ and $\norm{g(A)}\leq\norm{g}_{\infty}$, and $g(A)$ commutes with every bounded operator commuting with $A$. Rests on Proposition A24.8, Theorem 16.46 and Proposition A24.6.

Proof.

Derives Proposition A24.9. Existence of the operator. Fix $g\in B(\sigma(A))$ and consider

\begin{equation*} b_{g}(x,y)=\int_{\sigma(A)}g\,\dd\mu_{x,y}\ec\qquad \abs{b_{g}(x,y)}\leq\norm{g}_{\infty}\, \abs{\mu_{x,y}}(\sigma(A))\leq\norm{g}_{\infty}\norm{x}\norm{y}\ec \end{equation*}

which by Proposition A24.8 is antilinear in $x$ and linear in $y$. For fixed $x$ the map $y\longmapsto b_{g}(x,y)$ is a bounded linear functional, so by Theorem 16.46 there is a unique vector $w(x)$ with $b_{g}(x,y)=\braket{w(x)}{y}$; the map $x\longmapsto w(x)$ is linear — antilinearity in the first slot on both sides — and bounded by $\norm{g}_{\infty}$. Its adjoint in the sense of Theorem 16.38 is the operator we call $g(A)$, so that Equation (A24.14) holds with $\norm{g(A)}\leq\norm{g}_{\infty}$. For continuous $g$ the formula reproduces Equation (A24.12), so the map extends $\Phi$, and it is unique because Equation (A24.14) determines all matrix elements. Linearity in $g$ and $g(A)^{\dagger}=\bar g(A)$ follow from Equation (A24.14) and $\mu_{y,x}=\mu_{x,y}^{\ast}$.

Multiplicativity, first step. Let $f\in C(\sigma(A))$ and $x,y\in\mathcal{H}$. For every $h\in C(\sigma(A))$,

\begin{equation*} \int h\,\dd\mu_{x,f(A)y}=\braket{x}{h(A)f(A)y} =\braket{x}{(hf)(A)y}=\int hf\,\dd\mu_{x,y}\ec \end{equation*}

using multiplicativity of the continuous calculus. The two regular measures $\mu_{x,f(A)y}$ and $f\,\mu_{x,y}$ therefore integrate every continuous function alike, so they are equal by the uniqueness in Theorem A24.7. Consequently, for $g\in B(\sigma(A))$,

\begin{equation}\tag{A24.15} \braket{x}{g(A)f(A)y}=\int g\,\dd\mu_{x,f(A)y} =\int gf\,\dd\mu_{x,y}=\braket{x}{(gf)(A)y}\ec \end{equation}

i.e. $g(A)f(A)=(gf)(A)$ for $g$ Borel and $f$ continuous. Taking adjoints in Equation (A24.15) and replacing $(f,g)$ by $(\bar f,\bar g)$ gives $f(A)g(A)=(fg)(A)$ as well.

Multiplicativity, second step. Now let $g\in B(\sigma(A))$ be fixed. For every $h\in C(\sigma(A))$, by the step just proved,

\begin{equation*} \int h\,\dd\mu_{x,g(A)y}=\braket{x}{h(A)g(A)y} =\braket{x}{(hg)(A)y}=\int hg\,\dd\mu_{x,y}\ec \end{equation*}

so $\mu_{x,g(A)y}=g\,\mu_{x,y}$ by uniqueness again. Hence for any two bounded Borel $g,g'$,

\begin{equation}\tag{A24.16} \braket{x}{g'(A)g(A)y}=\int g'\,\dd\mu_{x,g(A)y} =\int g'g\,\dd\mu_{x,y}=\braket{x}{(g'g)(A)y}\ep \end{equation}

Commutation. Let $BA=AB$. By Proposition A24.6, $B$ commutes with $f(A)$ for every continuous $f$, so for such $f$

\begin{equation*} \int f\,\dd\mu_{B^{\dagger}x,y}=\braket{B^{\dagger}x}{f(A)y} =\braket{x}{Bf(A)y}=\braket{x}{f(A)By} =\int f\,\dd\mu_{x,By}\ec \end{equation*}

whence $\mu_{B^{\dagger}x,y}=\mu_{x,By}$, and evaluating Equation (A24.14) on both sides gives $\braket{x}{Bg(A)y}=\braket{x}{g(A)By}$ for every Borel $g$.

∎

The projection-valued measure

Proof of Theorem A24.1, part (1): existence. Derives Theorem A24.1. For a Borel set $\Omega\subseteq\R$ let $\chi_{\Omega}$ be its indicator function, restricted to $\sigma(A)$, and put

\begin{equation}\tag{A24.17} E(\Omega)=\chi_{\Omega}(A)\ep \end{equation}

This is a bounded operator by Proposition A24.9.

Projections. $\chi_{\Omega}^{2}=\chi_{\Omega}$ and $\bar\chi_{\Omega}=\chi_{\Omega}$, so $E(\Omega)^{2}=E(\Omega)=E(\Omega)^{\dagger}$: each $E(\Omega)$ is a projection in the sense of Proposition 16.21.

Normalisation and multiplicativity. $\chi_{\varnothing}=0$ and $\chi_{\R}=1$ on $\sigma(A)$ give $E(\varnothing)=0$ and $E(\R)=\identity$; also $E(\R\setminus\sigma(A))=0$, since that indicator vanishes identically on $\sigma(A)$, which is the support statement. And $\chi_{\Omega_{1}\cap\Omega_{2}}=\chi_{\Omega_{1}}\chi_{\Omega_{2}}$ gives $E(\Omega_{1}\cap\Omega_{2})=E(\Omega_{1})E(\Omega_{2})$ by Equation (A24.16).

Countable additivity. Finite additivity is linearity of the calculus: for disjoint $\Omega_{1},\Omega_{2}$, $\chi_{\Omega_{1}\cup\Omega_{2}}=\chi_{\Omega_{1}}+\chi_{\Omega_{2}}$. For a disjoint sequence $\set{\Omega_{n}}$ with union $\Omega$, write $S_{N}=\sum_{n\leq N}E(\Omega_{n})=E(\bigcup_{n\leq N}\Omega_{n})$, so that $E(\Omega)-S_{N}=E(\Omega_{>N})$ with $\Omega_{>N}=\bigcup_{n>N}\Omega_{n}$. Since $E(\Omega_{>N})$ is a projection,

\begin{equation}\tag{A24.18} \norm{E(\Omega)x-S_{N}x}^{2} =\braket{x}{E(\Omega_{>N})x} =\mu_{x}(\Omega_{>N})=\sum_{n>N}\mu_{x}(\Omega_{n})\ec \end{equation}

using Equation (A24.14) with $g=\chi_{\Omega_{>N}}$ and the countable additivity of the ordinary positive measure $\mu_{x}$. The series $\sum_{n}\mu_{x}(\Omega_{n})=\mu_{x}(\Omega)\leq \norm{x}^{2}$ converges, so its tail tends to $0$, and Equation (16.36) holds. Thus $E$ is a projection-valued measure in the sense of Definition 16.58.

It represents $A$. Taking $g=\chi_{\Omega}$ in Equation (A24.14) identifies the complex measure of Definition 16.58 with the measure of Proposition A24.8: $\braket{x}{E(\Omega)y}=\mu_{x,y}(\Omega)$. Hence, taking $f(\lambda)=\lambda$ in Equation (A24.12) — the continuous calculus assigns $A$ to that function — we obtain Equation (A24.1). The commutation statement is the last part of Proposition A24.9.

∎

Lemma A24.10 (Integration against a projection-valued measure).

Let $F$ be any projection-valued measure on $\R$ supported on a compact set $K$, and for $g\in B(K)$ define $\int g\,\dd F$ by $\braket{x}{(\int g\,\dd F)y}=\int g\,\dd\braket{x}{F y}$. Then $g\longmapsto\int g\,\dd F$ is linear and multiplicative. Rests on Definition 16.58 and Theorem 16.46.

Proof.

Derives Lemma A24.10. Exactly as in Proposition A24.9, the sesquilinear form is bounded by $\norm{g}_{\infty}\norm{x}\norm{y}$ — the total variation of $\braket{x}{F(\cdot)y}$ is at most $\norm{x}\norm{y}$, by the Cauchy–Schwarz inequality applied to $\braket{x}{F(\Omega)y}=\braket{F(\Omega)x}{F(\Omega)y}$ over a partition — so the operator exists, and linearity is clear. For multiplicativity, note first that for simple functions $g=\sum_{i}c_{i}\chi_{\Omega_{i}}$ and $g'=\sum_{j}c'_{j}\chi_{\Omega'_{j}}$ the property $F(\Omega)F(\Omega')=F(\Omega\cap\Omega')$ gives

\begin{equation*} \left(\int g\,\dd F\right)\left(\int g'\,\dd F\right) =\sum_{i,j}c_{i}c'_{j}\,F(\Omega_{i}\cap\Omega'_{j}) =\int gg'\,\dd F\ep \end{equation*}

A bounded Borel function is a uniform limit of simple Borel functions — partition the disc of radius $\norm{g}_{\infty}$ into finitely many Borel pieces of diameter $<\varepsilon$ and take preimages — and $\norm{\int g\,\dd F}\leq\norm{g}_{\infty}$, so both sides pass to the uniform limit.

∎

Proof of Theorem A24.1, part (1): uniqueness. Derives Theorem A24.1. Let $E$ and $E'$ both satisfy Equation (A24.1) and be supported on $\sigma(A)$, and write $\nu_{x,y},\nu'_{x,y}$ for the associated complex measures. By Lemma A24.10, $\int\lambda^{n}\,\dd E=(\int\lambda\,\dd E)^{n}=A^{n}$ for every $n\geq0$, and likewise for $E'$; hence

\begin{equation*} \int_{\sigma(A)}p\,\dd\nu_{x,y}=\braket{x}{p(A)y} =\int_{\sigma(A)}p\,\dd\nu'_{x,y} \end{equation*}

for every polynomial $p$. The polynomials are dense in $C(\sigma(A))$ (Theorem A24.5) and both measures are finite, so the two integrate every continuous function alike; by the uniqueness in Theorem A24.7, $\nu_{x,y}=\nu'_{x,y}$. Taking $\Omega$ Borel and $x,y$ arbitrary gives $\braket{x}{E(\Omega)y}=\braket{x}{E'(\Omega)y}$, i.e. $E=E'$.

∎

The multiplication form

Definition A24.11 (Cyclic vector and cyclic subspace).

For $x\in\mathcal{H}$ the cyclic subspace generated by $x$ is

\begin{equation}\tag{A24.19} \mathcal{H}_{x}=\overline{\set{f(A)\,x\mid f\in C(\sigma(A))}}\ec \end{equation}

a closed $A$-invariant subspace containing $x=1(A)x$. The vector $x$ is cyclic for $A$ if $\mathcal{H}_{x}=\mathcal{H}$. Rests on Proposition A24.6 and Theorem 16.18.

Lemma A24.12 (The cyclic case).

Let $x\neq0$ be cyclic for $A$. Then there is a unitary $U\!:\mathcal{H}\longrightarrow L^{2}(\sigma(A),\mu_{x})$ with $Uf(A)x=f$ for every $f\in C(\sigma(A))$, and

\begin{equation}\tag{A24.20} \bigl(UAU^{-1}h\bigr)(\lambda)=\lambda\,h(\lambda)\ec\qquad \forall\,h\in L^{2}(\sigma(A),\mu_{x})\ep \end{equation}

Rests on Definition A24.11, Proposition A24.8 and Theorem A24.7.

Proof.

Derives Lemma A24.12. For $f\in C(\sigma(A))$, using Equation (A24.10) and Equation (A24.12),

\begin{equation}\tag{A24.21} \norm{f(A)x}^{2}=\braket{x}{\bar f(A)f(A)x} =\braket{x}{(\abs{f}^{2})(A)x} =\int_{\sigma(A)}\abs{f}^{2}\,\dd\mu_{x} =\norm{f}_{L^{2}(\mu_{x})}^{2}\ec \end{equation}

so $f(A)x\longmapsto f$ is a well-defined linear isometry from the dense subspace $\set{f(A)x}$ of $\mathcal{H}$ onto the subspace $C(\sigma(A))$ of $L^{2}(\sigma(A),\mu_{x})$ — well defined because $f(A)x=g(A)x$ forces $\norm{f-g}_{L^{2}(\mu_{x})}=0$ by Equation (A24.21). Its target is dense by the regularity statement of Theorem A24.7, and both spaces are complete, so the isometry extends to a unitary $U$ (a densely defined isometry with dense range extends to a surjective isometry by taking limits, and Equation (16.3) then gives preservation of the inner product). Finally, for $f$ continuous, $UA f(A)x=U(\lambda f)(A)x=\lambda f=\lambda\cdot(Uf(A)x)$, which is Equation (A24.20) on a dense subspace; both sides are bounded, so it holds everywhere.

∎

Lemma A24.13 (Decomposition into cyclic subspaces).

Let $\mathcal{H}$ be separable and $A$ self-adjoint. Then there are finitely or countably many non-zero vectors $x_{n}$ such that the cyclic subspaces $\mathcal{H}_{x_{n}}$ are mutually orthogonal and

\begin{equation}\tag{A24.22} \mathcal{H}=\bigoplus_{n}\mathcal{H}_{x_{n}}\ep \end{equation}

Rests on Definition A24.11, Proposition 16.87 and Definition 16.32.

Proof.

Derives Lemma A24.13. Let $\set{y_{j}}_{j\in\N}$ be dense in $\mathcal{H}$ (Definition 16.32). Construct the $x_{n}$ recursively. Put $K_{0}=\mathcal{H}$. Given mutually orthogonal cyclic subspaces $\mathcal{H}_{x_{1}},\dots,\mathcal{H}_{x_{n-1}}$, let $K_{n-1}$ be the orthogonal complement of their sum. Each $\mathcal{H}_{x_{i}}$ is $A$-invariant, and $A$ is self-adjoint, so $K_{n-1}$ is $A$-invariant too (Lemma A23.3), and it is closed. If $K_{n-1}=\set{0}$, stop. Otherwise let $j(n)$ be the least index with $P_{K_{n-1}}y_{j}\neq0$ — such an index exists, since otherwise every $y_{j}$ would lie in $K_{n-1}^{\perp}$ and, by density and closedness, so would every vector, forcing $K_{n-1}=\set{0}$ — and set $x_{n}=P_{K_{n-1}}y_{j(n)}$, $\mathcal{H}_{x_{n}}\subseteq K_{n-1}$.

By construction $j(1)<j(2)<\cdots$, because $P_{K_{n-1}}y_{j}=0$ for every $j<j(n)$ not already used means $y_{j}\in \mathcal{H}_{x_{1}}\oplus\cdots\oplus\mathcal{H}_{x_{n-1}}$, while $y_{j(n)}=(\text{its part in that sum})+x_{n}$ also lies in $\mathcal{H}_{x_{1}}\oplus\cdots\oplus\mathcal{H}_{x_{n}}$. Hence every $y_{j}$ lies in $\bigoplus_{n}\mathcal{H}_{x_{n}}$, whose closure is therefore all of $\mathcal{H}$; that closure is the internal direct sum of Proposition 16.87, which is Equation (A24.22). If instead the recursion stops because $K_{n-1}=\set{0}$, the sum is already $\mathcal{H}$.

∎

Proof of Theorem A24.1, part (2). Derives Theorem A24.1. Take the decomposition Equation (A24.22) and write $A_{n}$ for the restriction of $A$ to $\mathcal{H}_{x_{n}}$, a bounded self-adjoint operator on that Hilbert space for which $x_{n}$ is cyclic, with $\sigma(A_{n})\subseteq\sigma(A)$ and with the same functional calculus (the calculus of $A$ leaves $\mathcal{H}_{x_{n}}$ invariant). Lemma A24.12 gives unitaries $U_{n}\!:\mathcal{H}_{x_{n}}\longrightarrow L^{2}(\sigma(A),\mu_{n})$, $\mu_{n}=\mu_{x_{n}}$, carrying $A$ into multiplication by $\lambda$.

Let $M$ be the disjoint union of countably many copies of $\sigma(A)$, the $n$th copy written $\sigma(A)\times\set{n}$, and define on it the measure

\begin{equation}\tag{A24.23} \mu\bigl(\Omega\times\set{n}\bigr)=c_{n}\,\mu_{n}(\Omega)\ec \qquad c_{n}=\frac{2^{-n}}{\norm{x_{n}}^{2}}\ec \end{equation}

so that, by Proposition A24.8, $\mu(M)=\sum_{n}c_{n}\norm{x_{n}}^{2}=\sum_{n}2^{-n}\leq1<\infty$. Define $a(\lambda,n)=\lambda$, a bounded real measurable function on $M$ because $\sigma(A)$ is compact and real. The map

\begin{equation}\tag{A24.24} U\Bigl(\sum_{n}v_{n}\Bigr)\Big|_{\sigma(A)\times\set{n}} =c_{n}^{-1/2}\,U_{n}v_{n}\ec\qquad v_{n}\in\mathcal{H}_{x_{n}}\ec \end{equation}

is unitary from $\mathcal{H}$ onto $L^{2}(M,\mu)$: on the $n$th block it is an isometry from $\mathcal{H}_{x_{n}}$ onto $L^{2}(\sigma(A),c_{n}\mu_{n})$, because $\int\abs{c_{n}^{-1/2}h}^{2}\,c_{n}\dd\mu_{n} =\int\abs{h}^{2}\dd\mu_{n}$, and $L^{2}(M,\mu)$ is by construction the orthogonal direct sum of those blocks (Proposition 16.85). Since multiplication by $\lambda$ is unaffected by the constant $c_{n}^{-1/2}$, Equation (A24.20) gives Equation (A24.2), which completes the proof of Theorem A24.1 and, with it, of Theorem 16.59.

∎

Remark A24.14 (What is quoted here).

Two classical theorems enter the proof and are not derived in this treatise; every other step above is carried out from Hilbert Spaces and its predecessors.

Theorem A24.5 (Stone–Weierstrass) is used twice: to extend the polynomial calculus to $C(\sigma(A))$, and in the uniqueness argument. Only the special case of a compact subset of $\R$, where it is the Weierstrass approximation theorem, is ever needed; that theorem is already declared as quoted in Remark 16.1 and is used there for Proposition 16.61.
Theorem A24.7 (Riesz–Markov) is the representation of a bounded functional on $C(K)$ by a regular complex Borel measure, together with the uniqueness of that measure and the density of $C(K)$ in $L^{2}(K,\nu)$. It is a theorem of measure theory, and this treatise develops no measure theory (Remark 16.1); it is what converts the operator inequality Equation (A24.13) into a measure, and there is no route to a projection-valued measure that avoids it. Statement and proof: [Reed:1972], chapter IV.

Nothing else is assumed. In particular the polynomial identity Equation (A24.8) — the step that makes the whole construction possible, and the one a reader should check first — rests only on the $C^{\ast}$ identity Equation (16.25) and the numerical-radius formula Equation (16.26), both proved in the chapter.

Remark A24.15.

The Spectral Theorem for a Bounded Self-Adjoint Operator discharges the proof obligation of Theorem 16.59 (Section 16.4.2). The projection-valued measure it constructs is what Definition 16.60 integrates against, so the functional calculus $f(A)$ of the chapter — unique by Proposition 16.61 — now exists as well as being unique; Proposition A24.9 is that existence statement for bounded Borel $f$. The theorem is used again in Stone's Theorem on One-Parameter Unitary Groups, where the unbounded case is derived from it and the exponential $\exp(-\ii Ht/\hbar)$ is defined by Equation (A24.14), and in The Nuclear Spectral Theorem of Gelfand and Maurin, whose direct-integral decomposition is Lemma A24.13 read fibrewise. Physically it is Remark 16.63: the support of $E$ is the set of possible measured values, and $\norm{E(\Omega)\psi}^{2}$ is the Born probability.