The spectral theorem for bounded self-adjoint operators

Originally written by Claude Opus 5.5 (Anthropic), October 2026, and self-checked by that writing AI. GPT-6.1 Sol (OpenAI), at the Ultra setting, read and self-checked the full lesson and all six solutions, completed the measure prerequisites and clarified the normal-operator conditions, October 2026. Public domain (CC0).

This lesson extends the continuous functional calculus of a self-adjoint operator to bounded Borel functions, and proves the spectral theorem. It covers:

bounded Borel functions by closing bounded continuous functions under bounded pointwise limits;
the spectral measure of a vector, obtained from the Riesz representation theorem;
the Borel functional calculus, and its continuity under bounded pointwise convergence;
projection-valued measures and the spectral theorem;
spectral projections, and the approximation of an operator by combinations of them;
the polar decomposition of an operator;
monotone convergence of operators (Vigier's theorem);
Calkin's theorem on the closed ideals of $B(H)$;
the same results for normal operators, whose norm is also the supremum of $|\langle n\xi,\xi\rangle|$ over unit vectors.

The lesson uses the following lessons:

Hilbert spaces and compact operators, cited as the Hilbert-space lesson;
C*-algebras: continuous functional calculus, automatic continuity, positive cones, approximate identities and quotients, cited as the C*-algebra lesson;
the Riesz representation theorem, Theorem 2.2 of Haar measure on locally compact groups;
the density of polynomials in $C(K)$ for compact $K\subseteq\mathbb C$, from the Stone–Weierstrass lesson.

Theorems 2.1–2.2 and 3.1–3.2 of the measure-tools lesson prove monotone and dominated convergence, completeness of $L^2$, and density of simple functions on arbitrary measure spaces. Theorem 2.2 of the Haar lesson proves the positive Riesz theorem on locally compact Hausdorff spaces; here its domain is the compact metric spectrum. The elementary measurability argument used below is included in the conventions.

Conventions

$H$ is a complex Hilbert space, $H\neq\{0\}$. For $H=\{0\}$ every statement below is trivial.
Inner products are linear in the first variable.
$B(H)$ is the C*-algebra of bounded operators. For $\xi,\eta\in H$, $\theta_{\xi,\eta}$ is the rank-one operator $\zeta\mapsto\langle\zeta,\eta\rangle\xi$.
For a self-adjoint $h\in B(H)$, the spectrum $S=\sigma(h)$ is a compact subset of $[-\|h\|,\|h\|]$ (the C*-algebra lesson, Proposition 1.5). It is nonempty because $H\neq\{0\}$ (Banach algebras, spectrum, holomorphic functional calculus and Gelfand theory, Section 5).
For $f\in C(S)$, $f(h)$ is the continuous functional calculus (the C*-algebra lesson, Theorem 5.1). By that theorem, $f\mapsto f(h)$ is an isometric unital $*$-homomorphism, and $\iota(h)=h$ for the identity function $\iota(\lambda)=\lambda$.
For a self-adjoint $a\in B(H)$, $a\geq0$ if and only if $\langle a\xi,\xi\rangle\geq0$ for every $\xi$ (the C*-algebra lesson, Proposition 8.5(11)).
The Borel sets of a metric space $K$ form the $\sigma$-algebra generated by the open sets. A function is Borel if it is measurable for this $\sigma$-algebra.
$B_b(K)$ is the space of bounded complex Borel functions on $K$, with $\|f\|_K=\sup_K|f|$.
A measure is positive and countably additive.
A sequence $(f_n)$ of functions on $K$ converges boundedly to $f$ if $f_n(x)\to f(x)$ at every point and $\sup_n\|f_n\|_K<\infty$.

If $f_n\to f$ boundedly and $\mu$ is a finite measure on the Borel sets, then $\int f_n\,d\mu\to\int f\,d\mu$. This follows from Theorem 2.2 of the measure-tools lesson, applied to real and imaginary parts with the common constant bound. Pointwise limits are Borel: for real $f_n\to f$, $\{f>a\}$ is the union over rational $r>a$ and $N\geq1$ of $\bigcap_{n\geq N}\{f_n>r\}$. Indeed eventual values above $r>a$ force $f\geq r>a$, and $f>a$ allows $r$ strictly between them. Apply this to real and imaginary parts for complex functions.

1. Bounded Borel functions

Lemma 1.1. Let $K$ be a metric space. Let $\mathcal M$ be a set of bounded complex functions on $K$ with two properties:

$\mathcal M$ contains every bounded continuous function;
$\mathcal M$ contains the limit of every boundedly convergent sequence in $\mathcal M$.

Then $\mathcal M$ contains every bounded Borel function on $K$.

Proof. The intersection of all sets with the two properties again has them. Call it $\mathcal B$; then $\mathcal B\subseteq\mathcal M$. Any subset of $\mathcal B$ with the two properties equals $\mathcal B$.

(a) $\mathcal B$ is closed under sums, products, scalar multiples and complex conjugation.

Fix a bounded continuous $f$. The set of $g\in\mathcal B$ with $f+g\in\mathcal B$ and $fg\in\mathcal B$ contains the bounded continuous functions. It is closed under bounded convergence: if $g_n\to g$ boundedly, then $f+g_n\to f+g$ and $fg_n\to fg$ boundedly. So this set is $\mathcal B$.
Now fix $g\in\mathcal B$. The set of $f\in\mathcal B$ with $f+g\in\mathcal B$ and $fg\in\mathcal B$ contains the bounded continuous functions, by the first step. It is closed under bounded convergence for the same reason. So it is $\mathcal B$.
In the same way, the set of $g\in\mathcal B$ with $cg\in\mathcal B$ for all $c\in\mathbb C$ and $\bar g\in\mathcal B$ is $\mathcal B$.

(b) $\Sigma=\{E\subseteq K:1_E\in\mathcal B\}$ is a $\sigma$-algebra.

$K\in\Sigma$, because $1_K$ is continuous.
$1_{K\setminus E}=1-1_E$ and $1_{E\cap F}=1_E1_F$, so $\Sigma$ is closed under complements and finite intersections, hence under finite unions.
If $E_n\in\Sigma$, the indicators of $E_1\cup\dots\cup E_N$ converge boundedly to the indicator of $\bigcup_nE_n$.

(c) $\Sigma$ contains the open sets, hence all Borel sets. Let $U\neq K$ be open. The functions $g_n(x)=\min\big(1,n\operatorname{dist}(x,K\setminus U)\big)$ are continuous with values in $[0,1]$. They converge to $1_U$ at every point, because $\operatorname{dist}(x,K\setminus U)>0$ for $x\in U$. So $U\in\Sigma$, and $\Sigma$ contains the $\sigma$-algebra generated by the open sets.

(d) Conclusion. By (a)–(c), $\mathcal B$ contains the Borel simple functions, the finite combinations of indicators of Borel sets. Every bounded Borel function $f$ is a uniform limit of such functions. For real $f$ with $0\leq f\leq M$, take $\begin{gathered} f_n\\ =\sum_{0\leq j\leq nM}\frac jn\,1_{\{j/n\leq f<(j+1)/n\}},\\ 0\leq f-f_n<\tfrac1n , \end{gathered}$ and split a complex $f$ into the positive and negative parts of its real and imaginary parts. A uniformly convergent sequence of bounded functions converges boundedly. So $f\in\mathcal B\subseteq\mathcal M$. $\square$

Conversely, every function in $\mathcal B$ is Borel, since the bounded Borel functions have the two properties by the preceding measurability argument. So $\mathcal B=B_b(K)$.

Corollary 1.2 (Measures are determined by continuous functions). Let $K$ be a metric space, let $\mu_1,\dots,\mu_m$ be finite measures on the Borel sets of $K$, and let $c_1,\dots,c_m\in\mathbb C$. Suppose that $\sum_jc_j\int f\,d\mu_j=0$ for every bounded continuous $f$. Then the same holds for every bounded Borel $f$. In particular, two finite Borel measures on $K$ with the same integrals of bounded continuous functions are equal.

Proof. Let $\mathcal M$ be the set of bounded Borel $f$ with $\sum_jc_j\int f\,d\mu_j=0$. It contains the bounded continuous functions. It is closed under bounded convergence, by dominated convergence for each $\mu_j$ and because limits of Borel functions are Borel. Lemma 1.1 applies. For the last statement, take $m=2$, $c=(1,-1)$ and $f=1_E$. $\square$

2. The spectral measure of a vector

Fix a self-adjoint $h\in B(H)$ with spectrum $S$.

Proposition 2.1. For every $\xi\in H$ there is exactly one Radon measure $\mu_\xi$ on $S$ with $\langle f(h)\xi,\xi\rangle=\int_Sf\,d\mu_\xi\qquad(f\in C(S)).$ It is finite, with $\mu_\xi(S)=\|\xi\|^2$. For $c\in\mathbb C$, $\mu_{c\xi}=|c|^2\mu_\xi$.

Proof. The functional $I_\xi(f)=\langle f(h)\xi,\xi\rangle$ on $C(S)$ is linear.

It is positive. If $f\geq0$, put $g=\sqrt f$. Then $g$ is real, so $g(h)$ is self-adjoint, and $f(h)=g(h)^2$. Hence $\langle f(h)\xi,\xi\rangle=\|g(h)\xi\|^2\geq0$.
The measure. $S$ is compact, so $C(S)=C_c(S)$. The Riesz representation theorem (Haar measure on locally compact groups, Theorem 2.2) gives exactly one Radon measure $\mu_\xi$ on $S$ that represents $I_\xi$.
Its total mass. $\mu_\xi(S)$ is finite, since $S$ is compact. It equals $I_\xi(1)=\langle\xi,\xi\rangle$, since $1(h)=1$.
Scaling. $|c|^2\mu_\xi$ is a Radon measure that represents $I_{c\xi}=|c|^2I_\xi$. By uniqueness, it is $\mu_{c\xi}$. $\square$

We also regard $\mu_\xi$ as a measure on the Borel sets of $\mathbb R$, by $\mu_\xi(\Delta)=\mu_\xi(\Delta\cap S)$. It is the spectral measure of $\xi$.

3. The Borel functional calculus

Theorem 3.1. Let $h\in B(H)$ be self-adjoint with spectrum $S$. For every $f\in B_b(S)$ there is exactly one operator $f(h)\in B(H)$ with $\begin{gathered} \langle f(h)\xi,\xi\rangle\\ =\int_Sf\,d\mu_\xi\\ (\xi\in H). \end{gathered} \tag{3.1}$ The map $f\mapsto f(h)$ has the following properties.

For continuous $f$ it is the continuous functional calculus.
It is a unital $*$-homomorphism: it is linear, $(fg)(h)=f(h)g(h)$, $\bar f(h)=f(h)^*$ and $1(h)=1$.
$\|f(h)\xi\|^2=\int_S|f|^2\,d\mu_\xi$. Hence $\|f(h)\|\leq\|f\|_S$. If $f\geq0$, then $f(h)\geq0$. If $f$ is real, then $f(h)$ is self-adjoint.
(Bounded convergence.) If $f_n\to f$ boundedly on $S$, then $f_n(h)\xi\to f(h)\xi$ for every $\xi\in H$.
$f(h)$ commutes with every operator that commutes with $h$.
$\sigma(f(h))$ is contained in the closure of $f(S)$.

For a bounded Borel function $f$ on $\mathbb R$ we write $f(h)$ for $(f|_S)(h)$.

Proof. Uniqueness. Two operators $T,T'$ with $\langle T\xi,\xi\rangle=\langle T'\xi,\xi\rangle$ for every $\xi$ are equal (the Hilbert-space lesson, Corollary 3.2).

A sesquilinear form. For $f\in B_b(S)$ and $\xi,\eta\in H$ put $\beta_f(\xi,\eta)=\frac14\sum_{k=0}^3i^k\int_Sf\,d\mu_{\xi+i^k\eta}.$ For continuous $f$, $\beta_f(\xi,\eta)=\langle f(h)\xi,\eta\rangle$. This is the polarization identity (the Hilbert-space lesson, Proposition 1.1(1)) for the sesquilinear form $(\xi,\eta)\mapsto\langle f(h)\xi,\eta\rangle$.

$\beta_f$ is sesquilinear for every $f\in B_b(S)$.

Fix $\xi,\xi',\eta$. The number $\beta_f(\xi+\xi',\eta)-\beta_f(\xi,\eta)-\beta_f(\xi',\eta)$ has the form $\sum_jc_j\int f\,d\mu_j$ for finitely many finite measures $\mu_j$ on $S$, which do not depend on $f$. It vanishes for continuous $f$, so it vanishes for every $f\in B_b(S)$, by Corollary 1.2.
The same argument gives $\beta_f(c\xi,\eta)=c\beta_f(\xi,\eta)$.
It also gives $\beta_{\bar f}(\eta,\xi)=\overline{\beta_f(\xi,\eta)}$. For continuous $f$ this reads $\langle f(h)^*\eta,\xi\rangle=\overline{\langle f(h)\xi,\eta\rangle}$, and both sides are combinations of integrals of $\bar f$.
Conjugate-linearity in the second variable follows from the last two points.

The diagonal. By Proposition 2.1, $\mu_{(1+i^k)\xi}=|1+i^k|^2\mu_\xi$, and $\sum_ki^k|1+i^k|^2=4+2i+0-2i=4$. So $\beta_f(\xi,\xi)=\int_Sf\,d\mu_\xi$.

A bound. $|\int f\,d\mu_\zeta|\leq\|f\|_S\|\zeta\|^2$, and $\sum_k\|\xi+i^k\eta\|^2=4(\|\xi\|^2+\|\eta\|^2)$, because the cross terms cancel. So $|\beta_f(\xi,\eta)|\leq\|f\|_S\big(\|\xi\|^2+\|\eta\|^2\big).$ For $\xi,\eta\neq0$, replace $\xi$ by $t\xi$ and $\eta$ by $t^{-1}\eta$ with $t=(\|\eta\|/\|\xi\|)^{1/2}$. This does not change $\beta_f(\xi,\eta)$, and gives $|\beta_f(\xi,\eta)|\leq2\|f\|_S\|\xi\|\|\eta\|$.

Existence. By the Hilbert-space lesson, Theorem 3.1, there is $f(h)\in B(H)$ with $\langle f(h)\xi,\eta\rangle=\beta_f(\xi,\eta)$. It satisfies (3.1). For continuous $f$, the continuous calculus satisfies (3.1) by the definition of $\mu_\xi$. By uniqueness, the two agree, which is (1).

Weak continuity. Let $f_n\to f$ boundedly. Then $\int f_n\,d\mu_\zeta\to\int f\,d\mu_\zeta$ for every $\zeta$, so $\begin{gathered} \langle f_n(h)\xi,\eta\rangle\\ =\beta_{f_n}(\xi,\eta)\to\beta_f(\xi,\eta)\\ =\langle f(h)\xi,\eta\rangle . \end{gathered} \tag{3.2}$

(2) Linearity, $1(h)=1$ and $\bar f(h)=f(h)^*$ follow from (3.1) and uniqueness. For example, $\begin{gathered} \langle\bar f(h)\xi,\xi\rangle\\ =\int\bar f\,d\mu_\xi\\ =\overline{\langle f(h)\xi,\xi\rangle}\\ =\langle f(h)^*\xi,\xi\rangle . \end{gathered}$ Multiplicativity, first step. Let $\mathcal M_1$ be the set of $f\in B_b(S)$ with $(fg)(h)=f(h)g(h)$ for every $g\in C(S)$.

It contains $C(S)$, by the C*-algebra lesson, Theorem 5.1(2).
It is closed under bounded convergence. Let $f_n\to f$ boundedly with $f_n\in\mathcal M_1$, and let $g\in C(S)$. Then $f_ng\to fg$ boundedly. By (3.2), applied to $f_ng$ and to $f_n$ at the vector $g(h)\xi$, $\begin{gathered} \langle(fg)(h)\xi,\eta\rangle\\ =\lim_n\langle(f_ng)(h)\xi,\eta\rangle\\ =\lim_n\langle f_n(h)g(h)\xi,\eta\rangle\\ =\langle f(h)g(h)\xi,\eta\rangle . \end{gathered}$ By Lemma 1.1, $\mathcal M_1=B_b(S)$.

Multiplicativity, second step. Let $\mathcal M_2$ be the set of $g\in B_b(S)$ with $(fg)(h)=f(h)g(h)$ for every $f\in B_b(S)$.

It contains $C(S)$, by the first step.
It is closed under bounded convergence, because $\langle f(h)g_n(h)\xi,\eta\rangle=\langle g_n(h)\xi,f(h)^*\eta\rangle$ and (3.2) applies to the $g_n$.

So $\mathcal M_2=B_b(S)$.

(3) By (2) and (3.1), $\begin{gathered} \|f(h)\xi\|^2\\ =\langle f(h)^*f(h)\xi,\xi\rangle\\ =\langle(\bar ff)(h)\xi,\xi\rangle\\ =\int|f|^2\,d\mu_\xi\\ \leq\|f\|_S^2\,\mu_\xi(S)\\ =\|f\|_S^2\|\xi\|^2 . \end{gathered}$ If $f\geq0$, then $\langle f(h)\xi,\xi\rangle=\int f\,d\mu_\xi\geq0$ for every $\xi$, so $f(h)\geq0$. If $f$ is real, then $f(h)^*=\bar f(h)=f(h)$.

(4) By (2) and (3), $\|(f_n(h)-f(h))\xi\|^2=\int|f_n-f|^2\,d\mu_\xi$. This tends to $0$ by dominated convergence.

(5) Let $Th=hT$. Since $h=h^*$, $T$ commutes with every $f(h)$, $f\in C(S)$ (the C*-algebra lesson, Theorem 5.1(6)). The set of $f\in B_b(S)$ with $Tf(h)=f(h)T$ is closed under bounded convergence, by (4). By Lemma 1.1 it is $B_b(S)$.

(6) Let $\lambda$ lie outside the closure of $f(S)$. Then $g=(f-\lambda)^{-1}$ is a bounded Borel function on $S$. By (2), $g(h)$ is an inverse of $f(h)-\lambda$. $\square$

In (6) the inclusion can be strict (Exercise 2).

4. Projection-valued measures and the spectral theorem

Definition 4.1. Let $X$ be a metric space; we use $X=\mathbb R$ and $X=\mathbb C$. A projection-valued measure on $X$, acting on $H$, is a map $E$ from the Borel sets of $X$ to the projections of $B(H)$ with two properties:

$E(X)=1$;
for every $\xi\in H$, the function $\mu^E_\xi(\Delta)=\langle E(\Delta)\xi,\xi\rangle=\|E(\Delta)\xi\|^2$ is a measure.

$E$ is supported by a Borel set $Y$ if $E(Y)=1$.

Lemma 4.2. Let $E$ be a projection-valued measure.

$E(\varnothing)=0$. If $\Delta\cap\Delta'=\varnothing$, then $E(\Delta\cup\Delta')=E(\Delta)+E(\Delta')$ and $E(\Delta)E(\Delta')=0$.
$E(\Delta\cap\Delta')=E(\Delta)E(\Delta')=E(\Delta')E(\Delta)$ for all Borel sets $\Delta,\Delta'$.
If $\Delta$ is the union of disjoint Borel sets $\Delta_n$, then $E(\Delta)\xi=\sum_nE(\Delta_n)\xi$ for every $\xi\in H$.

Proof. (1) $\|E(\varnothing)\xi\|^2=\mu^E_\xi(\varnothing)=0$ for every $\xi$.

For disjoint $\Delta,\Delta'$, additivity of $\mu^E_\xi$ gives $\langle E(\Delta\cup\Delta')\xi,\xi\rangle=\langle(E(\Delta)+E(\Delta'))\xi,\xi\rangle$ for every $\xi$. So $E(\Delta\cup\Delta')=E(\Delta)+E(\Delta')$ (the Hilbert-space lesson, Corollary 3.2).
If $P$, $Q$ and $P+Q$ are projections, then $(P+Q)^2=P+Q$ gives $PQ+QP=0$. Multiplying by $P$ on the left and on the right gives $PQ+PQP=0=PQP+QP$. So $PQ=QP$, and then $2PQ=0$.

(2) The sets $\Delta\cap\Delta'$, $\Delta\setminus\Delta'$ and $\Delta'\setminus\Delta$ are disjoint. By (1), $\begin{gathered} E(\Delta)\\ =E(\Delta\cap\Delta')+E(\Delta\setminus\Delta'),\\ E(\Delta')\\ =E(\Delta\cap\Delta')+E(\Delta'\setminus\Delta), \end{gathered}$ and the products of the projections of two disjoint sets vanish. Multiplying out gives $E(\Delta)E(\Delta')=E(\Delta\cap\Delta')$, and likewise in the other order.

(3) The vectors $E(\Delta_n)\xi$ are mutually orthogonal, by (1). Also by (1), $\begin{gathered} \Big\|E(\Delta)\xi-\sum_{n\leq N}E(\Delta_n)\xi\Big\|^2\\ =\Big\|E\Big(\Delta\setminus\bigcup_{n\leq N}\Delta_n\Big)\xi\Big\|^2\\ =\mu^E_\xi\Big(\Delta\setminus\bigcup_{n\leq N}\Delta_n\Big). \end{gathered}$ This tends to $0$, because $\mu^E_\xi$ is a finite measure. $\square$

Proposition 4.3 (Integration against a projection-valued measure). Let $E$ be a projection-valued measure on $X$. There is exactly one linear map $f\mapsto\int f\,dE$ from $B_b(X)$ to $B(H)$ with $\begin{gathered} \int1_\Delta\,dE\\ =E(\Delta)\\ \text{for every Borel set }\Delta,\\ \Big\|\int f\,dE\Big\|\\ \leq\|f\|_X . \end{gathered}$ It is a unital $*$-homomorphism, and $\big\langle\big(\int f\,dE\big)\xi,\xi\big\rangle=\int f\,d\mu^E_\xi$.

Proof. Simple functions. For a Borel simple function $f=\sum_jc_j1_{\Delta_j}$ with disjoint $\Delta_j$, put $\int f\,dE=\sum_jc_jE(\Delta_j)$.

This does not depend on the representation: two representations with disjoint sets have a common refinement, and Lemma 4.2(1) shows that refining does not change the sum.
The vectors $E(\Delta_j)\xi$ are orthogonal, and $\sum_j\|E(\Delta_j)\xi\|^2=\|E(\bigcup_j\Delta_j)\xi\|^2\leq\|\xi\|^2$. So $\begin{gathered} \Big\|\Big(\int f\,dE\Big)\xi\Big\|^2\\ =\sum_j|c_j|^2\|E(\Delta_j)\xi\|^2\\ \leq\|f\|_X^2\|\xi\|^2 . \end{gathered}$
On simple functions the map is linear and multiplicative (Lemma 4.2(2)), it preserves the involution (each $E(\Delta)$ is self-adjoint), it is unital, and $\langle(\int f\,dE)\xi,\xi\rangle=\sum_jc_j\mu^E_\xi(\Delta_j)=\int f\,d\mu^E_\xi$.

Extension. The simple functions are dense in $B_b(X)$ for $\|\cdot\|_X$ (Lemma 1.1, step (d)). A linear contraction defined on them extends uniquely to a linear contraction on $B_b(X)$. The identities above pass to uniform limits, since $\mu^E_\xi$ is finite. Any linear contraction with $\int1_\Delta\,dE=E(\Delta)$ agrees with this one on simple functions, hence everywhere. $\square$

Theorem 4.4 (Spectral theorem). Let $h\in B(H)$ be self-adjoint with spectrum $S$.

$E_h(\Delta)=1_\Delta(h)$, for Borel sets $\Delta\subseteq\mathbb R$, is a projection-valued measure on $\mathbb R$, supported by $S\subseteq[-\|h\|,\|h\|]$. For every bounded Borel function $f$ on $\mathbb R$, $\int f\,dE_h=f(h)$. In particular $h=\int\iota_S\,dE_h ,$ where $\iota_S(\lambda)=\lambda$ on $S$ and $\iota_S=0$ off $S$. This is usually written $h=\int\lambda\,dE_h(\lambda)$.
(Uniqueness.) Let $E$ be a projection-valued measure on $\mathbb R$, supported by a compact set $L$, with $h=\int\iota_L\,dE$. Then $E=E_h$.
For $\xi\in H$, the spectral measure satisfies $\mu_\xi(\Delta)=\langle E_h(\Delta)\xi,\xi\rangle=\|E_h(\Delta)\xi\|^2$, $\mu_\xi(\mathbb R)=\|\xi\|^2$, and $\langle f(h)\xi,\xi\rangle=\int f\,d\mu_\xi$.
Each $E_h(\Delta)$ commutes with every operator that commutes with $h$.

Proof. (1) $1_\Delta$ is real and $1_\Delta^2=1_\Delta$, so $E_h(\Delta)$ is a projection, by Theorem 3.1(2). Also:

$E_h(\mathbb R)=1(h)=1$;
$\langle E_h(\Delta)\xi,\xi\rangle=\mu_\xi(\Delta\cap S)$ is a measure in $\Delta$;
$E_h(S)=1$;
$S\subseteq[-\|h\|,\|h\|]$ (Conventions).

The maps $f\mapsto\int f\,dE_h$ and $f\mapsto f(h)$ are linear contractions on the bounded Borel functions on $\mathbb R$ (Proposition 4.3 and Theorem 3.1(3)). They agree on indicators, hence on simple functions, hence everywhere. For $f=\iota_S$, $f|_S$ is the identity function of $S$, and the continuous calculus sends it to $h$.

(2) Let $L'=L\cup S$, a compact set.

$E(\mathbb R\setminus L)=0$, so $\mu^E_\xi$ is carried by $L$.
$\int\cdot\,dE$ is a unital homomorphism, so $\int(p\circ\iota_L)\,dE=p(h)$ for every polynomial $p$. Since $p\circ\iota_L=p$ on $L$, $\int_Lp\,d\mu^E_\xi=\langle p(h)\xi,\xi\rangle=\int_Sp\,d\mu_\xi .$
So the finite measures $\mu^E_\xi$ and $\mu_\xi$, restricted to the Borel subsets of $L'$, have the same integrals of polynomials. The polynomials are uniformly dense in $C(L')$ (the Stone–Weierstrass lesson), so the two measures have the same integrals of continuous functions on $L'$. By Corollary 1.2 they are equal.
Both vanish off $L'$. So $\langle E(\Delta)\xi,\xi\rangle=\mu_\xi(\Delta)=\langle E_h(\Delta)\xi,\xi\rangle$ for every $\xi$, and $E(\Delta)=E_h(\Delta)$.

(3) restates the definitions with Proposition 2.1 and (3.1). (4) is Theorem 3.1(5). $\square$

Remark 4.5 (Exponentials). For $t\in\mathbb R$, $e^{ith}$ denotes $f(h)$ with $f(\lambda)=e^{it\lambda}$. Since $f$ is continuous, this is the continuous calculus. The partial sums of $\sum_n(it\lambda)^n/n!$ converge to $f$ uniformly on $S$, and the calculus is isometric. So $e^{ith}$ is the sum of the norm-convergent series $\sum_n(ith)^n/n!$. It is unitary, because $\bar ff=1$.

5. Spectral projections and polar decomposition

The projections $E_h(\Delta)$ are the spectral projections of $h$.

Proposition 5.1. Let $h\in B(H)$ be self-adjoint, with spectrum $S$ and $E=E_h$.

For every $n\geq1$ there are disjoint Borel sets $\Delta_1,\dots,\Delta_m$ and real numbers $c_1,\dots,c_m$ with $\|h-\sum_jc_jE(\Delta_j)\|\leq1/n$. So $h$ is a norm limit of real linear combinations of its spectral projections.
If $h\geq0$, the $c_j$ can be taken $\geq0$ and the $\Delta_j$ inside $[1/n,\infty)$. So $h$ is a norm limit of nonnegative combinations of spectral projections of Borel sets bounded away from $0$.
$\ker h=E(\{0\})H$.
Let $h\geq0$. Then $E((0,\infty))$ is the projection onto the closure of $hH$. As $\varepsilon\downarrow0$, the operators $h(h+\varepsilon)^{-1}$ increase and converge strongly to $E((0,\infty))$.

Proof. (1) Let $s_n(\lambda)=\lfloor n\lambda\rfloor/n$. Then $0\leq\lambda-s_n(\lambda)<1/n$, and on the bounded set $S$ the function $s_n$ takes finitely many values. So $s_n(h)=\sum_k\frac kn\,E\big([k/n,(k+1)/n)\big),$ a finite sum. By Theorem 3.1(3), $\|h-s_n(h)\|\leq\sup_S|\lambda-s_n(\lambda)|\leq1/n$.

(2) If $h\geq0$, then $S\subseteq[0,\infty)$ (the C*-algebra lesson, Theorem 8.2). The term with $k=0$ vanishes, so only the sets $[k/n,(k+1)/n)$ with $k\geq1$ occur.

(3) $hE(\{0\})=(\iota1_{\{0\}})(h)=0$, since $\lambda1_{\{0\}}(\lambda)=0$. Conversely, let $h\xi=0$.

Then $\int\lambda^2\,d\mu_\xi(\lambda)=\|h\xi\|^2=0$, by Theorem 3.1(3).
So $\mu_\xi(S\setminus\{0\})=0$.
Hence $\|E(\{0\})\xi\|^2=\mu_\xi(\{0\})=\|\xi\|^2$, and $E(\{0\})\xi=\xi$.

(4) A vector $\eta$ is orthogonal to $hH$ exactly when $h^*\eta=h\eta=0$.

By the Hilbert-space lesson, Theorem 2.2(3), the closure of $hH$ is $(\ker h)^\perp$.
By (3), $\begin{gathered} (\ker h)^\perp\\ =(1-E(\{0\}))H\\ =E(S\setminus\{0\})H\\ =E((0,\infty))H \end{gathered}$ , because $S\subseteq[0,\infty)$.
Let $f_\varepsilon(\lambda)=\lambda/(\lambda+\varepsilon)$ on $[0,\infty)$. By the continuous calculus, $f_\varepsilon(h)=h(h+\varepsilon)^{-1}$.
As $\varepsilon$ decreases, $f_\varepsilon$ increases, so $f_\varepsilon(h)$ increases (Theorem 3.1(3)).
$\begin{gathered} \|(E((0,\infty))-f_\varepsilon(h))\xi\|^2\\ =\int(1_{(0,\infty)}-f_\varepsilon)^2\,d\mu_\xi \end{gathered}$ . The integrand decreases to $0$ at every point of $[0,\infty)$ as $\varepsilon\downarrow0$. So the integral decreases, and it tends to $0$ along $\varepsilon=1/n$ by dominated convergence. $\square$

A partial isometry is an operator $u$ that is isometric on a closed subspace $L$, its initial space, and zero on $L^\perp$. Its range $uH=u(L)$ is closed, the final space.

Proposition 5.2 (Polar decomposition). Let $x\in B(H)$ and $|x|=(x^*x)^{1/2}$. There is exactly one operator $u$ with $x=u|x|$ and $\ker u=\ker x$. It is a partial isometry with initial space $(\ker x)^\perp$, the closure of $|x|H$, and final space the closure of $xH$. Moreover $u^*u$ and $uu^*$ are the projections onto these two spaces, and $u^*x=|x|$.

Proof.

An isometry. $\||x|\xi\|^2=\langle x^*x\xi,\xi\rangle=\|x\xi\|^2$. So $|x|\xi\mapsto x\xi$ is a well-defined isometry from $|x|H$ onto $xH$. It extends to an isometry from the closure $L$ of $|x|H$ onto a closed subspace, which contains $xH$ and lies in its closure, so it is the closure of $xH$.
The kernel. By the norm identity, $\ker|x|=\ker x$. Since $|x|$ is self-adjoint, $L^\perp=\ker|x|$ (the Hilbert-space lesson, Theorem 2.2). So $L=(\ker x)^\perp$.
The operator. Let $u$ be the isometry on $L$ and $0$ on $L^\perp$. Then $u|x|=x$ and $\ker u=L^\perp=\ker x$.
Uniqueness. If $x=v|x|$ and $\ker v=\ker x$, then $v=u$ on $|x|H$, hence on $L$ by continuity, and $v=0=u$ on $L^\perp=\ker x$.
The projections. $\langle u^*u\xi,\xi\rangle=\|u\xi\|^2=\|P_L\xi\|^2=\langle P_L\xi,\xi\rangle$, where $P_L$ is the projection onto $L$. So $u^*u=P_L$ (the Hilbert-space lesson, Corollary 3.2). Then $uP_L=u$, so $uu^*$ is a self-adjoint idempotent, and its range is $uH$, the final space. Finally $u^*x=u^*u|x|=P_L|x|=|x|$. $\square$

6. Monotone convergence of operators

Theorem 6.1 (Vigier). Let $(a_i)_{i\in I}$ be an increasing net of self-adjoint operators on $H$ with $C=\sup_i\|a_i\|<\infty$. Then:

there is a self-adjoint $a\in B(H)$ with $a_i\xi\to a\xi$ for every $\xi$;
$a$ is the least upper bound of the $a_i$ among self-adjoint operators.

The same holds for decreasing nets, with the greatest lower bound.

Proof. The limit.

For each $\xi$, the number $\langle a_i\xi,\xi\rangle$ is real, increases with $i$, and is at most $C\|\xi\|^2$. So it converges.
By polarization (the Hilbert-space lesson, Proposition 1.1(1)), $\beta(\xi,\eta)=\lim_i\langle a_i\xi,\eta\rangle$ exists for all $\xi,\eta$.
$\beta$ is sesquilinear, $|\beta(\xi,\eta)|\leq C\|\xi\|\|\eta\|$, and $\beta(\xi,\xi)$ is real.
By the Hilbert-space lesson, Theorem 3.1, there is $a\in B(H)$ with $\langle a\xi,\eta\rangle=\beta(\xi,\eta)$. It is self-adjoint by Corollary 3.2 of that lesson.

The least upper bound. $\langle a\xi,\xi\rangle=\sup_i\langle a_i\xi,\xi\rangle$, so $a\geq a_i$ for every $i$. If $b$ is self-adjoint and $b\geq a_i$ for every $i$, then $\langle b\xi,\xi\rangle\geq\langle a\xi,\xi\rangle$, so $b\geq a$.

Strong convergence. Put $b_i=a-a_i\geq0$.

$\|a\|\leq C$, by the Hilbert-space lesson, Corollary 3.2, so $\|b_i\|\leq2C$.
For a positive operator $b$, $b^2\leq\|b\|b$, because $\lambda^2\leq\|b\|\lambda$ on $\sigma(b)\subseteq[0,\|b\|]$ (Theorem 3.1(3)).
Hence $\begin{gathered} \|b_i\xi\|^2\\ =\langle b_i^2\xi,\xi\rangle\\ \leq2C\langle b_i\xi,\xi\rangle\\ =2C\big(\langle a\xi,\xi\rangle-\langle a_i\xi,\xi\rangle\big)\to0 . \end{gathered}$

For a decreasing net, apply this to $(-a_i)$. $\square$

7. Calkin's theorem

Theorem 7.1 (Calkin). Let $H$ be a separable infinite-dimensional Hilbert space. The closed two-sided ideals of $B(H)$ are $\{0\}$, $K(H)$ and $B(H)$.

Proof. $K(H)$ is a closed ideal (the Hilbert-space lesson, Theorem 5.1). It is not $B(H)$, since an orthonormal sequence has no convergent subsequence, so the identity is not compact. Let $J\neq\{0\}$ be a closed two-sided ideal.

$J$ contains $K(H)$. Take $x\in J$ and $\zeta\in H$ with $x\zeta\neq0$. For $\alpha,\beta\in H$, $\theta_{\alpha,x\zeta}\,x\,\theta_{\zeta,\beta}=\|x\zeta\|^2\,\theta_{\alpha,\beta}.$ Indeed, both sides send $\gamma$ to $\langle\gamma,\beta\rangle\|x\zeta\|^2\alpha$. So $J$ contains every rank-one operator, hence every finite-rank operator. Since $J$ is closed, $K(H)\subseteq J$ (the Hilbert-space lesson, Theorem 5.1(3)).

If $J\neq K(H)$, then $J=B(H)$. Take $x\in J\setminus K(H)$.

$a=x^*x\in J$ is positive and not compact. Suppose $a$ were compact. Let $(\xi_n)$ be bounded, with $\|\xi_n\|\leq M$. Some subsequence has $a\xi_n$ convergent, and along it $\begin{gathered} \|x(\xi_n-\xi_m)\|^2\\ =\langle a(\xi_n-\xi_m),\xi_n-\xi_m\rangle\\ \leq2M\|a(\xi_n-\xi_m)\|\to0 . \end{gathered}$ So $x$ would be compact, a contradiction.
A spectral projection of infinite rank. For $\varepsilon>0$ let $E_\varepsilon=1_{[\varepsilon,\infty)}(a)$. The function $\lambda1_{[0,\varepsilon)}(\lambda)$ is at most $\varepsilon$ on $[0,\infty)$, so $\|a-aE_\varepsilon\|\leq\varepsilon$ by Theorem 3.1. If every $E_\varepsilon$ had finite rank, then every $aE_\varepsilon$ would have finite rank, and $a$ would be compact. So some $E=E_\varepsilon$ has infinite-dimensional range.
$E\in J$. Let $g(\lambda)=\lambda^{-1}1_{[\varepsilon,\infty)}(\lambda)$, a bounded Borel function on $[0,\infty)$. Then $g(a)a=(g\iota)(a)=E$. So $E\in J$.
The identity is in $J$. $EH$ is a closed infinite-dimensional subspace of the separable space $H$, so it is separable. So $EH$ and $H$ both have countably infinite orthonormal bases $(f_n)$ and $(e_n)$ (the Hilbert-space lesson, Theorem 4.1(4)). The operator $v$ with $ve_n=f_n$ is an isometry (Theorem 4.1(2) of that lesson). Since $Ev=v$, we get $1=v^*v=v^*Ev\in J$. So $J=B(H)$. $\square$

Separability is used only to find the isometry $v$. Exercise 5 shows that the conclusion fails without it.

8. Normal operators

Theorem 8.1. Let $n\in B(H)$ be normal, with spectrum $S\subseteq\mathbb C$. Sections 2–5 hold with $h$ replaced by $n$ and $\mathbb R$ by $\mathbb C$, with the following changes. These conditions give the normal version without invoking Fuglede’s theorem:

in Theorem 3.1(5) and Theorem 4.4(4), the operator must commute with $n$ and $n^*$;
in Proposition 5.1(1), the coefficients $c_j$ are complex, and a grid of mesh $1/(2n)$ gives the same $1/n$ error bound;
the positive-coefficient and positive-support assertions, Proposition 5.1(2) and (4), still assume that the operator is positive; such an operator is self-adjoint. The kernel formula in (3) holds for every normal operator, using $|z|^2$ in its proof.

In particular, $E_n(\Delta)=1_\Delta(n)$ is the only projection-valued measure on $\mathbb C$ that is supported by a compact set $L$ and has $n=\int\iota_L\,dE_n$.

Proof. The proofs in Sections 2–5 use only the following facts.

$S$ is a compact metric space.
$f\mapsto f(n)$ is a unital $*$-homomorphism from $C(S)$ to $B(H)$ that maps real functions to self-adjoint operators. This holds for normal $n$ by the C*-algebra lesson, Theorem 5.1(1), (2) and (9).
Theorem 3.1(5) uses the C*-algebra lesson, Theorem 5.1(6). That theorem needs commutation with $n$ and $n^*$.
The uniqueness proof uses the polynomials $p(\lambda,\bar\lambda)$, which are dense in $C(L')$ for compact $L'\subseteq\mathbb C$ (the Stone–Weierstrass lesson), and $\int p(\iota_L,\bar\iota_L)\,dE=p(n,n^*)$.
In Proposition 5.1(1), use $s_k(z)=(\lfloor k\operatorname{Re}z\rfloor+i\lfloor k\operatorname{Im}z\rfloor)/k$, with $|z-s_k(z)|<\sqrt2/k$. Taking $k=2n$ gives the $1/n$ estimate in Proposition 5.1(1). Assertions about a positive operator reduce to the self-adjoint case. For a general normal operator the kernel proof uses $\int|z|^2\,d\mu_\xi=\|n\xi\|^2$ and the projection of $\{0\}$. $\square$

Proposition 8.2. Let $n\in B(H)$ be normal.

Every $\lambda\in\sigma(n)$ is an approximate eigenvalue: there are unit vectors $\xi_k$ with $\|(n-\lambda)\xi_k\|\to0$.
$\|n\|=\sup_{\|\xi\|=1}|\langle n\xi,\xi\rangle|$.

Proof. (1) $m=n-\lambda$ is normal, so $\|m^*\xi\|^2=\langle mm^*\xi,\xi\rangle=\langle m^*m\xi,\xi\rangle=\|m\xi\|^2$ for every $\xi$. Suppose $\|m\xi\|\geq c\|\xi\|$ for some $c>0$ and all $\xi$.

Then $m$ is injective, and its range is closed: if $m\xi_j$ is Cauchy, so is $\xi_j$.
The orthogonal complement of the range is $\ker m^*$, which equals $\ker m=\{0\}$. So $m$ is bijective, and its inverse is bounded by $1/c$.
This contradicts $\lambda\in\sigma(n)$. So no such $c$ exists, which gives the vectors $\xi_k$.

(2) The supremum is at most $\|n\|$, by Cauchy–Schwarz. For $\xi_k$ as in (1), $\begin{gathered} |\langle n\xi_k,\xi_k\rangle-\lambda|\\ =|\langle(n-\lambda)\xi_k,\xi_k\rangle|\\ \leq\|(n-\lambda)\xi_k\|\to0 \end{gathered}$ . So the supremum is at least $|\lambda|$ for every $\lambda\in\sigma(n)$. By the C*-algebra lesson, Theorem 1.3, some $\lambda\in\sigma(n)$ has $|\lambda|=r(n)=\|n\|$. $\square$

Exercises

Exercise 1 (easy; Multiplication operators). Let $\mu$ be a finite Borel measure on a compact set $K\subseteq\mathbb R$, and let $h$ be multiplication by $\lambda$ on $L^2(K,\mu)$. Let $\operatorname{supp}\mu$ be the set of points all of whose neighbourhoods have positive measure. Show:

$\sigma(h)=\operatorname{supp}\mu$;
$f(h)$ is multiplication by $f$, for every bounded Borel $f$;
$E_h(\Delta)$ is multiplication by $1_\Delta$;
the spectral measure of the constant function $1$ is $\mu$.

Solution. If the measure is zero, the space and all its operators are zero, the spectrum and support are empty, and all assertions hold in that sense. Otherwise:

The support. $K\setminus\operatorname{supp}\mu$ is the union of the open sets of measure $0$. Since $\mathbb R$ has a countable base of open intervals, it is a countable union of null sets, hence null.
$c\notin\operatorname{supp}\mu$. Then $\delta=\operatorname{dist}(c,\operatorname{supp}\mu)>0$. Almost everywhere $|\lambda-c|\geq\delta$, so multiplication by $(\lambda-c)^{-1}$ is a bounded inverse of $h-c$.
$c\in\operatorname{supp}\mu$. Let $U_k=(c-1/k,c+1/k)$. Then $\mu(U_k)>0$. The unit vectors $\xi_k=\mu(U_k)^{-1/2}1_{U_k}$ have $\|(h-c)\xi_k\|\leq1/k$. So $h-c$ is not bounded below, and not invertible. This proves $\sigma(h)=\operatorname{supp}\mu=:S$.
Continuous $f$. For $f\in C(S)$, multiplication by $f$ makes sense on $L^2(\mu)$, because $\mu(K\setminus S)=0$. The map $f\mapsto M_f$ is a unital $*$-homomorphism into $B(H)$ that sends $\iota$ to $h$. By the uniqueness in the C*-algebra lesson, Theorem 5.1(1), $f(h)=M_f$.
Spectral measures. Hence $\langle f(h)\xi,\xi\rangle=\int f|\xi|^2\,d\mu$ for $f\in C(S)$. By Corollary 1.2, $\mu_\xi=|\xi|^2\mu$ on $S$.
Borel $f$. For bounded Borel $f$, $\langle M_f\xi,\xi\rangle=\int f|\xi|^2\,d\mu=\int f\,d\mu_\xi$. By uniqueness in Theorem 3.1, $f(h)=M_f$.
The last two claims are the cases $f=1_\Delta$ and $\xi=1$. $\square$

Exercise 2 (medium; A dense set of eigenvalues). Let $(q_k)$ enumerate $\mathbb Q\cap[0,1]$ without repetitions, and let $h$ be the operator on $\ell^2(\mathbb N)$ with $he_k=q_ke_k$. Show:

$\sigma(h)=[0,1]$;
$E_h(\{q_k\})$ is the projection onto $\mathbb Ce_k$;
$E_h([0,1]\setminus\mathbb Q)=0$.

Conclude that $f=1_{[0,1]\setminus\mathbb Q}$ has $f(h)=0$, although the closure of $f(\sigma(h))$ is $\{0,1\}$.

Solution.

The spectrum. For $c\notin[0,1]$, the diagonal operator with entries $(q_k-c)^{-1}$ is bounded by $\operatorname{dist}(c,[0,1])^{-1}$ and inverts $h-c$. Each $q_k$ is an eigenvalue. The spectrum is closed, so it contains the closure of $\{q_k\}$, which is $[0,1]$.
Spectral measures. For a polynomial $p$, $\langle p(h)\xi,\xi\rangle=\sum_kp(q_k)|\xi_k|^2$. By uniform approximation, the same holds for continuous $p$. By Corollary 1.2, $\mu_\xi=\sum_k|\xi_k|^2\delta_{q_k}$.
The projections. $\langle E_h(\{q_k\})\xi,\xi\rangle=|\xi_k|^2=\langle\theta_{e_k,e_k}\xi,\xi\rangle$. And $\langle E_h([0,1]\setminus\mathbb Q)\xi,\xi\rangle=0$ for every $\xi$. $\square$

Exercise 3 (medium; Unitaries are exponentials). Let $u\in B(H)$ be unitary. Show that $u=e^{ih}$ for a self-adjoint $h$ with $\|h\|\leq\pi$ that commutes with every operator commuting with $u$ and $u^*$. Deduce that the unitary group of $B(H)$ is path-connected in the norm topology.

Solution.

The logarithm. $\sigma(u)$ lies in the unit circle $\mathbb T$ (the C*-algebra lesson, Proposition 1.5). Let $\theta(e^{it})=t$ for $t\in(-\pi,\pi]$. This is a Borel function on $\mathbb T$, continuous off $-1$. Put $h=\theta(u)$, the calculus of the normal operator $u$ (Theorem 8.1). It is self-adjoint, because $\theta$ is real, and $\sigma(h)\subseteq[-\pi,\pi]$ by Theorem 3.1(6). It commutes with every operator that commutes with $u$ and $u^*$.
A composition rule. For a polynomial $p$, $p(h)=(p\circ\theta)(u)$, by the homomorphism property. Let $g$ be continuous on $[-\pi,\pi]$, and let $p_j\to g$ uniformly there. Then $p_j(h)\to g(h)$, by the continuous calculus of $h$. Also $(p_j\circ\theta)(u)\to(g\circ\theta)(u)$, by Theorem 3.1(3). So $g(h)=(g\circ\theta)(u)$.
The exponential. With $g(t)=e^{it}$, $g\circ\theta$ is the identity function on $\mathbb T$. So $e^{ih}=u$.
A path. $u_t=e^{ith}$, $t\in[0,1]$, is a path of unitaries from $1$ to $u$. It is norm-continuous: $\begin{gathered} \|e^{ith}-e^{ish}\|\\ \leq\sup_{|\lambda|\leq\pi}|e^{it\lambda}-e^{is\lambda}|\\ \leq\pi|t-s| \end{gathered}$ . $\square$

Exercise 4 (easy; The Calkin algebra). Let $H$ be separable and infinite-dimensional. Show that $B(H)/K(H)$ has no closed two-sided ideals other than $\{0\}$ and itself.

Solution. Let $q$ be the quotient map and $I$ a closed two-sided ideal of the quotient.

$q^{-1}(I)$ is a closed two-sided ideal of $B(H)$ that contains $K(H)$. By Calkin's theorem it is $K(H)$ or $B(H)$.
Since $q$ is surjective, $I=q(q^{-1}(I))$, which is $\{0\}$ or the whole quotient. $\square$

Exercise 5 (hard; Without separability). Let $H$ be a Hilbert space that is not separable. Let $J$ be the norm closure of the set of operators whose range is separable. Show that $J$ is a closed two-sided ideal with $K(H)\subsetneq J\subsetneq B(H)$.

Solution.

An ideal. If $x$ has separable range, so do $yx$ (a continuous image of a separable set) and $xy$ (its range lies in that of $x$). If $x$ and $x'$ have separable ranges, so does $x+x'$. So these operators form a two-sided ideal, and its closure $J$ is a closed two-sided ideal.
$K(H)\subseteq J$. A compact operator has separable range (the proof of the Hilbert-space lesson, Theorem 5.1(3)).
$K(H)\neq J$. The projection onto the closed span of a countably infinite orthonormal family has separable range, and it is not compact.
$J\neq B(H)$. Suppose $\|1-x\|<1$ for some $x$ with separable range. Then $x$ is invertible (Neumann series), so its range is $H$, which is not separable. So $1\notin J$. $\square$

Exercise 6 (easy; The support of a positive operator). Let $h\geq0$. Show that $E_h((0,\infty))$ is the smallest projection $p$ with $ph=h$.

Solution.

On $S\subseteq[0,\infty)$, $\lambda1_{(0,\infty)}(\lambda)=\lambda$. So $E_h((0,\infty))h=h$.
If $ph=h$, then $p$ is the identity on $hH$, hence on its closure, which is $E_h((0,\infty))H$ by Proposition 5.1(4).
So $pE_h((0,\infty))=E_h((0,\infty))$, which says $E_h((0,\infty))\leq p$. $\square$

Where this leads

Every operator that commutes with $h$ commutes with all spectral projections of $h$. So a von Neumann algebra contains the spectral projections and the bounded Borel functions of each of its self-adjoint elements. This is how the projections enter The double commutant theorem and the lessons after it. The spectral measures $\mu_\xi$ and the unitary groups $e^{ith}$ are used in Compact and trace-class operators, the predual of B(H), and the operator topologies. Calkin's theorem is used in Representations and positive functionals: the GNS construction and the Gelfand–Naimark theorem.

References

The spectral theorem for bounded self-adjoint operators goes back to Hilbert's work on quadratic forms in infinitely many variables: D. Hilbert, "Grundzüge einer allgemeinen Theorie der linearen Integralgleichungen. Vierte Mitteilung", Nachrichten von der Gesellschaft der Wissenschaften zu Göttingen, Mathematisch-Physikalische Klasse (1906). The proof given here, through the continuous functional calculus and the Riesz representation theorem, is standard.
Calkin's theorem on the closed two-sided ideals of the algebra of bounded operators on a separable Hilbert space is due to J. W. Calkin (1941); B. Blackadar, Operator Algebras: Theory of C*-Algebras and von Neumann Algebras, I.8.7.2, states it together with the nonseparable case.
The monotone convergence theorem for operators is due to J.-P. Vigier (1946); see B. Blackadar, Operator Algebras, I.3.2.5.

The proofs are written here in our own words.

Freely accessible reading: Jesse Peterson, Notes on operator algebras, §3.7 gives a route through spectral measures and bounded Borel calculus; the exact null-set and essential-range statements are proved here. The lesson includes its own complete proofs at the stated hypotheses; references to human sources do not imply permission to adapt their expression.

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