# 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](hilbert-spaces-and-compact-operators.md), cited as *the Hilbert-space lesson*;
- [C\*-algebras: continuous functional calculus, automatic continuity, positive cones, approximate identities and quotients](c-star-algebras-continuous-functional-calculus-automatic-continuity-positive-cones.md), 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](banach-algebras-spectrum-holomorphic-functional-calculus-and-gelfand-theory.md), 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.
1. For continuous \(f\) it is the continuous functional calculus.
2. It is a unital \(*\)-homomorphism: it is linear, \((fg)(h)=f(h)g(h)\), \(\bar f(h)=f(h)^*\) and \(1(h)=1\).
3. \(\|f(h)\xi\|^2=\int_S|f|^2\,d\mu_\xi\). Hence \(\|f(h)\|\leq\|f\|_S\). If \(f\geq0\), then \(f(h)\geq0\). If \(f\) is real, then \(f(h)\) is self-adjoint.
4. (*Bounded convergence.*) If \(f_n\to f\) boundedly on \(S\), then \(f_n(h)\xi\to f(h)\xi\) for every \(\xi\in H\).
5. \(f(h)\) commutes with every operator that commutes with \(h\).
6. \(\sigma(f(h))\) 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.
1. \(E(\varnothing)=0\). If \(\Delta\cap\Delta'=\varnothing\), then \(E(\Delta\cup\Delta')=E(\Delta)+E(\Delta')\) and \(E(\Delta)E(\Delta')=0\).
2. \(E(\Delta\cap\Delta')=E(\Delta)E(\Delta')=E(\Delta')E(\Delta)\) for all Borel sets \(\Delta,\Delta'\).
3. 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 \(S\).
1. \(E_h(\Delta)=1_\Delta(h)\), 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 \(h=\int\lambda\,dE_h(\lambda)\).
2. (*Uniqueness.*) Let \(E\) be a projection-valued measure on \(\mathbb R\), supported by a compact set \(L\), with \(h=\int\iota_L\,dE\). Then \(E=E_h\).
3. 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 \(\langle f(h)\xi,\xi\rangle=\int f\,d\mu_\xi\).
4. 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 \(E=E_h\).
1. 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.
2. 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 \(0\).
3. \(\ker h=E(\{0\})H\).
4. 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.
1. Every \(\lambda\in\sigma(n)\) is an approximate eigenvalue: there are unit vectors \(\xi_k\) with \(\|(n-\lambda)\xi_k\|\to0\).
2. \(\|n\|=\sup_{\|\xi\|=1}|\langle n\xi,\xi\rangle|\).

**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](the-double-commutant-theorem.md) 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](compact-and-trace-class-operators-the-predual-of-b-h-and-the-operator-topologies.md). Calkin's theorem is used in [Representations and positive functionals: the GNS construction and the Gelfand–Naimark theorem](representations-and-positive-functionals-the-gns-construction-and-the-gelfand-naimark.md).

## 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"](https://gdz.sub.uni-goettingen.de/download/pdf/PPN252457811_1906/LOG_0028.pdf), *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*](https://bruceblackadar.com/Mathematics/Cycr.pdf), 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*](https://bruceblackadar.com/Mathematics/Cycr.pdf), I.3.2.5.

The proofs are written here in our own words.

*Freely accessible reading:* [Jesse Peterson, *Notes on operator algebras*, §3.7](https://math.vanderbilt.edu/peters10/teaching/spring2015/OperatorAlgebras.pdf) 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.