Notes on Lewin Ch4: Spectral Theorem and Functional Calculus

1 Multiplication Operators

Let $𝐵 \subset ℝ^{𝑑}$ be Borel, and let $𝜇$ be a locally finite Borel measure on $𝐵$ . We set $ℌ = 𝐿^{2} (𝐵, d 𝜇; ℂ)$ . Local finiteness implies that $𝐿_{𝑐}^{\infty} \subset ℌ$ .

Example 1. Let $𝐵 = {𝑥_{1}, \dots, 𝑥_{𝑘}} \subset ℝ^{𝑑}$ and $𝜇 = \sum_{𝑖} 𝛿_{𝑥_{𝑖}}$ . Then $𝐿^{2} (𝐵, d 𝜇) ≅ ℂ^{𝑘}$ , so every operator $𝑇$ identifies under the obvious choice of basis $𝑒_{𝑖} = 𝟙_{{𝑥_{𝑖}}}$ with a matrix $𝑀 \in ℂ^{𝑘 \times 𝑘}$ , with $𝑀_{𝑖 𝑗} = ⟨ 𝑒_{𝑖}, 𝑇 𝑒_{𝑗} ⟩ = (𝑇 𝑒_{𝑗}) (𝑥_{𝑖})$ . Then $𝑔 = 𝑇 𝑓$ is the element $𝑔 (𝑥_{𝑖}) = \sum_{𝑗} 𝑀_{𝑖, 𝑗} 𝑓 (𝑥_{𝑗})$ .

A diagonal matrix $𝑀$ corresponds to $𝑓 \mapsto 𝑎 𝑓$ where $𝑎 (𝑥_{𝑖}) = 𝑀_{𝑖 𝑖}$ .
Every Hermitian matrix $𝑀 = 𝑀^{*}$ can be diagonalized (possibly under a different basis).

Let $𝑎 \in 𝐿_{loc}^{2} (𝐵, d 𝜇)$ , then $𝑀_{𝑎}$ is the operator defined by

𝑀_{𝑎} 𝑓 (𝑥) = 𝑎 (𝑥) 𝑓 (𝑥), 𝐷 (𝑀_{𝑎}) = {𝑣 \in 𝐿^{2} (𝐵, d 𝜇) : 𝑎 𝑣 \in 𝐿^{2} (𝐵, d 𝜇)} .

Theorem 1. Let $𝑎 \in 𝐿_{loc}^{2} (𝐵, d 𝜇)$ .

$(𝑀_{𝑎}, 𝐷 (𝑀_{𝑎}))$ is closed.
$𝜎 (𝑀_{𝑎}) = ess ran (𝑎)$ , where the essential range of $𝑎$ is

ess ran (𝑎) = {𝑦 \in ℂ : 𝜇 (| 𝑎 (\cdot) - 𝑦 | \leq 𝜀) > 0 for all 𝜀 > 0} .

The eigenvalues of $𝑀_{𝑎}$ are the $𝜆 \in ess ran (𝑎)$ such that $𝜇 (𝑎 = 𝜆) > 0$ , with the corresponding eigenspace $𝐿^{2} ({𝑎 = 𝜆}, d 𝜇)$ , the space of all square-integrable functions with support in the set ${𝑎 = 𝜆}$ , defined $𝜇$ -a.e.
$(𝑀_{𝑎}, 𝐷 (𝑀_{𝑎}))$ is bounded iff $𝑎 \in 𝐿^{\infty} (𝐵, d 𝜇)$ .
$(𝑀_{𝑎}, 𝐷 (𝑀_{𝑎}))$ is self-adjoint iff $𝑎$ is real-valued (bounded or not).

Bits of the proof. For $𝜆 \notin ess ran (𝑎)$ , there is $𝜀 > 0$ such that $| 𝑎 - 𝜆 | > 𝜀$ $𝜇$ -a.e. Thus, $\frac{1}{𝑎 - 𝜆} \in 𝐿^{\infty} (𝐵, d 𝜇)$ , and the map $𝑣 \mapsto {𝑣 (𝑎 - 𝜆)}^{- 1}$ is bounded $𝐿^{2} (𝐵, d 𝜇) \to 𝐷 (𝑀_{𝑎})$ (using $\frac{𝑎}{𝑎 - 𝜆} = 1 + \frac{𝜆}{𝑎 - 𝜆}$ ) and is an inverse for $𝑀_{𝑎} - 𝜆$ . For $𝜆 \in ess ran (𝑎)$ , there exists $𝑅_{𝑛} > 0$ such that $𝜇 ({| 𝑎 - 𝑦 | < \frac{1}{𝑛}} \cap 𝔹_{𝑅_{𝑛}}) \in (0, \infty)$ for all $𝑛$ . Then with $𝑢_{𝑛} ≔ 𝟙_{{| 𝑎 - 𝑦 | < \frac{1}{𝑛}} \cap 𝔹_{𝑅_{𝑛}}}$ we have
${‖ (𝑀_{𝑎} - 𝜆) 𝑢_{𝑛} ‖}^{2} \leq \frac{1}{𝑛^{2}} {‖ 𝑢_{𝑛} ‖}^{2} .$
So $𝑀_{𝑎} - 𝜆$ cannot be invertible.

Theorem 2 (Theorem 4.4: Spectral Theorem). Let $(𝐴, 𝐷 (𝐴))$ be self-adjoint on $ℌ$ . Then there exists $𝑑 \geq 1$ , a Borel set $𝐵 \subset ℝ^{𝑑}$ , a locally finite measure $𝜇$ on $𝐵$ , a real-valued locally bounded function $𝑎 \in 𝐿_{loc}^{\infty} (𝐵, d 𝜇)$ , and an isomorphism $𝑈 : ℌ \to 𝐿^{2} (𝐵, d 𝜇)$ such that

𝑈 𝐴 𝑈^{- 1} = 𝑀_{𝑎}, 𝑈 𝐷 (𝐴) = 𝐷 (𝑀_{𝑎}) .

One can take $𝑑 = 2, 𝐵 = 𝜎 (𝐴) \times ℕ \subset ℝ^{2}, 𝑎 (𝑠, 𝑛) = 𝑠$ , and $𝜇$ a finite measure on $𝐵$ .

Corollary (resolvent bound).

‖ {(𝐴 - 𝑧)}^{- 1} ‖ = \frac{1}{𝑑 (𝑧, 𝜎 (𝐴))} .

Corollary (isolated eigenvalues). Every isolated point of the spectrum is an eigenvalue.

Define $𝑓 (𝐴) ≔ 𝑈^{- 1} 𝑀_{𝑓 (𝑎)} 𝑈, 𝐷 (𝑓 (𝐴)) = 𝑈^{- 1} 𝐷 (𝑀_{𝑓 (𝑎)})$ for any such isomorphism. We need to show this is independent of the choice of $𝑈$ .

Theorem 3 (Theorem 4.8: Functional Calculus for bounded Borel functions). Let $(𝐴, 𝐷 (𝐴))$ be self-adjoint. There exists a unique map

𝑓 \in {ℒ︁}^{\infty} (ℝ, ℂ) \mapsto 𝑓 (𝐴) \in ℬ︀ (ℌ)

defined on the $𝐶^{*}$ -algebra ${ℒ︁}^{\infty} (ℝ, ℂ)$ of bounded Borel functions on $ℝ$ , with values in the algebra $ℬ︀ (ℌ)$ of bounded operators on $ℌ$ , such that:

it is a morphism of $𝐶^{*}$ -algebras ( $ℂ$ -linear, preserves product and star operation).
it is continuous, with $‖ 𝑓 (𝐴) ‖ \leq {sup}_{𝑥 \in ℝ} | 𝑓 |$ .
if $𝑓 (𝑥) = {(𝑥 - 𝑧)}^{- 1}$ with $𝑧 \in ℂ \ ℝ$ , then $𝑓 (𝐴) = {(𝐴 - 𝑧)}^{- 1}$ .
if $𝑓 |_{𝜎 (𝐴)} \equiv 0$ then $𝑓 (𝐴) = 0$ .
if $| 𝑓_{𝑛} (𝑥) | \leq 𝐶$ and $𝑓_{𝑛} \to 𝑓$ pointwise on $ℝ$ , then $𝑓_{𝑛} (𝐴) 𝑣 \to 𝑓 (𝐴) 𝑣$ for all $𝑣 \in ℌ$ .

Remark 1 (Remark 4.9: Spectral measure). Let $𝑣$ be a unit vector of $𝐻$ . By Theorem 3, the map

𝑓 \in 𝐶_{𝑏}^{0} (ℝ, ℝ) \mapsto 𝜑_{𝑣} (𝑓) ≔ ⟨ 𝑣, 𝑓 (𝐴) 𝑣 ⟩ \in ℝ

is a continuous linear form. If in addition $𝑓 \geq 0$ we can write

𝑓 (𝐴) = \sqrt{𝑓} {(𝐴)}^{2},

which shows that $𝜑_{𝑣}$ is a positive linear form on $𝐶_{𝑏}^{0}$ . Hence by Riesz–Markov, there is a unique Borel probability measure $𝜇_{𝐴, 𝑣}$ on $ℝ$ such that

⟨ 𝑣, 𝑓 (𝐴) 𝑣 ⟩ = \int_{ℝ} 𝑓 (𝑠) d 𝜇_{𝐴, 𝑣} (𝑠) .

With $(𝐵, 𝜇) = (𝜎 (𝐴) \times ℕ, 𝜇)$ and $𝑎 (𝑠, 𝑛) = 𝑠$ from Theorem 2, if $𝑈 : 𝐻 \to 𝐿^{2} (𝐵, 𝜇)$ is the corresponding unitary map, we can write

⟨ 𝑣, 𝑓 (𝐴) 𝑣 ⟩ = {⟨ 𝑈 𝑣, 𝑓 (𝑎) 𝑈 𝑣 ⟩}_{𝐿^{2} (𝐵, 𝜇)} = \int_{𝐵} 𝑓 (𝑎) {| 𝑈 𝑣 |}^{2} d 𝜇 = \int_{𝜎 (𝐴) \times ℕ} 𝑓 (𝑠) {| 𝑈 𝑣 (𝑠, 𝑛) |}^{2} d 𝜇 (𝑠, 𝑛) .

Therefore $𝜇_{𝐴, 𝑣}$ is the pushforward measure¹

𝜇_{𝐴, 𝑣} = 𝑎_{*} ({| 𝑈 𝑣 |}^{2} 𝜇),

i.e. for every Borel set $𝐸 \subset ℝ$ ,

𝜇_{𝐴, 𝑣} (𝐸) = \int_{𝑎^{- 1} (𝐸)} {| 𝑈 𝑣 (𝑥) |}^{2} d 𝜇 (𝑥) = \int_{𝐸 \times ℕ} {| 𝑈 𝑣 (𝑠, 𝑛) |}^{2} d 𝜇 (𝑠, 𝑛) .

In words, $𝜇_{𝐴, 𝑣}$ is the cylindrical projection on $𝜎 (𝐴)$ of the probability measure $| 𝑈 𝑣 (𝑠, 𝑛) |^{2} d 𝜇 (𝑠, 𝑛)$ on $𝜎 (𝐴) \times ℕ$ . We then have $𝑣 \in 𝐷 (𝐴)$ iff $𝜇_{𝐴, 𝑣}$ has a moment of order two, and in this case

\int_{ℝ} 𝑠^{2} d 𝜇_{𝐴, 𝑣} = {‖ 𝐴 𝑣 ‖}^{2} .

We also have

\int_{ℝ} 𝑠 d 𝜇_{𝐴, 𝑣} (𝑠) = ⟨ 𝑣, 𝐴 𝑣 ⟩ .

We will in fact need this construction in the proof of Theorem 2.

For every bounded Borel function $𝑓$ , $𝑓 (𝐴)$ is defined by Theorem 3. By setting $𝐸_{𝐴} (𝐵) ≔ 𝟙_{𝐵} (𝐴)$ we obtain the spectral theorem in terms of projection-valued measures:

Theorem 4 (Projection-valued spectral measure). Let $(𝐴, 𝐷 (𝐴))$ be self-adjoint. There exists a unique map

𝐸_{𝐴} : ℬ︀ (ℝ) \to ℬ︀ (𝐻)

from the Borel subsets of $ℝ$ to the orthogonal projections on $𝐻$ , such that:

$𝐸_{𝐴} (\emptyset) = 0$ and $𝐸_{𝐴} (ℝ) = 𝐼$ .
if ${(𝐵_{𝑛})}_{𝑛}$ are pairwise disjoint Borel subsets of $ℝ$ , then
$𝐸_{𝐴} (\cup_{𝑛} 𝐵_{𝑛}) 𝑣 = \sum_{𝑛} 𝐸_{𝐴} (𝐵_{𝑛}) 𝑣$
for every $𝑣 \in 𝐻$ , where the series converges in $𝐻$ .
for every pair of Borel subsets $𝐵, 𝐶 \subset ℝ$ ,
$𝐸_{𝐴} (𝐵 \cap 𝐶) = 𝐸_{𝐴} (𝐵) 𝐸_{𝐴} (𝐶) .$
for every bounded Borel function $𝑓 \in {ℒ︁}^{\infty} (ℝ, ℂ)$ ,
$𝑓 (𝐴) = \int_{ℝ} 𝑓 (𝑠) d 𝐸_{𝐴} (𝑠) .$
We can recover
$𝐴 = \int_{ℝ} 𝑠 d 𝐸_{𝐴} (𝑠) .$

Definition 1 (Scalar spectral measure). Let $𝑣$ be a unit vector of $𝐻$ . Then $𝜇_{𝐴, 𝑣}$ and $𝐸_{𝐴}$ are related by

𝜇_{𝐴, 𝑣} (𝐵) = ⟨ 𝑣, 𝐸_{𝐴} (𝐵) 𝑣 ⟩ .

Corollary (Corollary 4.10: Functional Calculus for locally bounded Borel functions). Let $(𝐴, 𝐷 (𝐴))$ be self-adjoint and let $𝑓 : ℝ \to ℂ$ be a locally bounded Borel function. Then $𝑓 (𝐴)$ defined above is independent of the isomorphism $𝑈$ used to represent $𝐴$ as a multiplication operator.

This follows from the functional calculus for bounded functions because we can describe $𝐷 (𝑓 (𝐴))$ and $𝑓 (𝐴) 𝑣$ in terms of the corresponding functional calculus for $𝑓_{𝑛} = 𝑓 𝟙_{{| 𝑓 | < 𝑛}}$ ,

\begin{matrix} 𝑣 \in 𝐷 (𝑓 (𝐴)) ⟺ \underset{𝑛 \to \infty}{lim sup} ‖ 𝑓_{𝑛} (𝐴) 𝑣 ‖ < \infty, \\ 𝑓 (𝐴) 𝑣 = lim_{𝑛 \to \infty} 𝑓_{𝑛} (𝐴) 𝑣 . \end{matrix}

2 Proof of Theorems 4.4 and 4.8

Structure of the proof:
1. First we prove the functional calculus for the “resolvent algebra” $𝒜︀$ of resolvents. This gives Theorem 5 using the Stone–Weierstrass theorem. This calculus is for functions in $𝐶_{lim}^{0} ≔ ℂ + 𝐶_{0}^{0} (ℝ, ℂ)$ (continuous functions with equal limits at $\pm \infty$ ).
2. Deduce Theorem 2.
3. Use the monotone class theorem to deduce uniqueness and hence Theorem 3 (the other properties of Theorem 3 follow directly from the definition).
$𝐶_{lim}^{0}$ is a unital $𝐶^{*}$ -algebra.

Theorem 5 (Theorem 4.11: Continuous functional calculus). Let $(𝐴, 𝐷 (𝐴))$ be self-adjoint. There exists a unique map

𝑓 \in 𝐶_{lim}^{0} (ℝ, ℂ) \mapsto 𝑓 (𝐴) \in ℬ︀ (ℌ)

such that:

it is a morphism of $𝐶^{*}$ -algebras ( $ℂ$ -linear, preserves product and star operation).
it is continuous, with $‖ 𝑓 (𝐴) ‖ \leq {sup}_{𝑥 \in ℝ} | 𝑓 |$ .
if $𝑓 (𝑥) = {(𝑥 - 𝑧)}^{- 1}$ with $𝑧 \in ℂ \ ℝ$ , then $𝑓 (𝐴) = {(𝐴 - 𝑧)}^{- 1}$ .

Proof. Define the resolvent algebra $𝒜︀$ as the algebra generated by constant functions and the rational maps $𝑥 \mapsto {(𝑥 - 𝑧)}^{- 1}$ for $𝑧 \in ℂ \ ℝ$ . By the Stone–Weierstrass theorem, $𝒜︀$ is dense in $𝐶_{lim}^{0}$ . It is clearly a $𝐶^{*}$ -algebra.

We first show that there is a unique morphism of $𝐶^{*}$ -algebras $𝒜︀ \to ℬ︀ (ℌ)$ sending ${(𝑥 - 𝑧)}^{- 1}$ to ${(𝐴 - 𝑧)}^{- 1}$ . We are forced to directly map ${(𝑥 - 𝑧)}^{- 1}$ to ${(𝐴 - 𝑧)}^{- 1}$ by (iii) and similarly for linear combinations of products by (i). That this assignment is unique and well-defined follows from

${({(𝐴 - 𝑧)}^{- 1})}^{*} = {(𝐴 - \bar{𝑧})}^{- 1}$ , and
the resolvent identity
${(𝐴 - 𝑧)}^{- 1} - {(𝐴 - 𝑤)}^{- 1} = (𝑤 - 𝑧) {(𝐴 - 𝑧)}^{- 1} {(𝐴 - 𝑤)}^{- 1},$
which demonstrates commutativity.

For (ii) we will use Lemma 1.

Lemma 1 gives us (ii) by using

{‖ 𝑓 ‖}_{\infty}^{2} - | 𝑓 (𝑥) |^{2} \in 𝒜︀ .

This continuity allows us to well-define $𝑓 (𝐴)$ for all $𝑓 \in 𝐶_{lim}^{0}$ by approximating by elements of $𝒜︀$ (via Stone–Weierstrass) and using the continuity to show that the limit is independent of the choice of approximating sequence. Uniqueness follows from the density of $𝒜︀$ in $𝐶_{lim}^{0}$ .∎

Lemma 1 (Lemma: Stability under the square root). Let $𝑓 \in 𝒜︀$ be non-negative on $ℝ$ . Then there is a unique $𝑔 \in 𝒜︀$ such that $𝑔^{2} = 𝑓$ and $𝑔 \geq 0$ on $ℝ$ . In particular,

𝑓 (𝐴) = {𝑔 (𝐴)}^{2} \geq 0 .

Proof. Let us write bold letters $𝜶, 𝒑, 𝒒, \dots$ to denote vectors / multiindices of some length and set ${(𝑓 (𝜶))}^{𝜷}$ to mean $\prod_{𝑖} {(𝑓 (𝛼_{𝑖}))}^{𝛽_{𝑖}}$ when $𝜶$ and $𝜷$ are the same length.

Note that every $𝑓 \in 𝒜︀$ can be written as a reduced rational function $𝑓 = \frac{𝑃}{𝑄}$ where $𝑃$ and $𝑄$ are polynomials with no common roots, and $𝑄$ has no real roots. Conversely every such rational function is in $𝒜︀$ by decomposing as partial fractions.

Let for $𝑓 = \frac{𝑃}{𝑄} \in 𝒜︀$ ,

𝑃 = {𝑐 (𝑥 - 𝜶)}^{𝒑} {(𝑥 - 𝒛)}^{𝒑^{'}}, 𝑄 = {(𝑥 - 𝝃)}^{𝒒},

with $𝛼_{𝑖} \in ℝ$ , $𝑧_{𝑖} \in ℂ$ , $𝜉_{𝑖} \in ℂ$ .

Next suppose that $𝑓 \geq 0$ on $ℝ$ . In particular then $𝑓 (𝑥) \in ℝ$ when $𝑥 \in ℝ$ , so

𝑃 \bar{𝑄} = 𝑄 \bar{𝑃},

which implies that $𝑐 \in ℝ$ by comparing the leading term, and also that

{(𝑥 - 𝒛)}^{𝒑^{'}} {(𝑥 - \bar{𝝃})}^{𝒒} = {(𝑥 - 𝝃)}^{𝒒} {(𝑥 - \bar{𝒛})}^{𝒑^{'}} .

Since the fraction $\frac{𝑃}{𝑄}$ is reduced, none of the $𝑧_{𝑖}$ can equal the $𝜉_{𝑖}$ , so each ${(𝑥 - 𝑧_{𝑖})}^{𝑝_{𝑖}^{'}}$ must be a factor of ${(𝑥 - \bar{𝒛})}^{𝒒}$ . Similarly, each ${(𝑥 - \bar{𝜉_{𝑖}})}^{𝑞_{𝑖}}$ must be a factor of ${(𝑥 - 𝝃)}^{𝒒}$ . It follows that the complex numbers appear with their conjugates with equal multiplicity, so

𝑓 (𝑥) = {𝑐 (𝑥 - 𝜶)}^{𝒑} \frac{{| 𝑥 - 𝒛 |}^{2 𝒑^{'}}}{{| 𝑥 - 𝝃 |}^{2 𝒒}} .

Now, using that $𝑓 \geq 0$ on $ℝ$ , we have $𝑐 \geq 0$ and $𝑝_{𝑖}$ are even. So we can take

𝑔 (𝑥) = \sqrt{𝑐} {(𝑥 - 𝜶)}^{\frac{𝒑}{2}} \frac{{| 𝑥 - 𝒛 |}^{𝒑^{'}}}{{| 𝑥 - 𝝃 |}^{𝒒}},

as required.∎

Proof (of the Spectral Theorem, Theorem 2). As in Remark 1, we can construct the scalar spectral measure $𝜇_{𝐴, 𝑣}$ for some unit vector $𝑣$ , as follows. By the Riesz–Markov representation theorem, there is a unique Borel probability measure $𝜇_{𝐴, 𝑣}$ on $ℝ$ such that

⟨ 𝑣, 𝑓 (𝐴) 𝑣 ⟩ = \int_{ℝ} 𝑓 (𝑠) d 𝜇_{𝐴, 𝑣} (𝑠)

for every bounded continuous function $𝑓$ . Now observe that

⟨ 𝑔 (𝐴) 𝑣, 𝑓 (𝐴) 𝑣 ⟩ = \int_{ℝ} \bar{𝑔 (𝑠)} 𝑓 (𝑠) d 𝜇_{𝐴, 𝑣} (𝑠)

so that the map $𝑓 \mapsto 𝑓 (𝐴) 𝑣$ is an isometry $𝐶_{lim}^{0} (ℝ, ℂ) \to ℌ$ with respect to the inner product. By taking the closure, this gives an isometry

𝑈 : 𝐿^{2} (ℝ, d 𝜇_{𝐴, 𝑣}) \to {𝒳︀}_{𝑣} ≔ \bar{{𝑓 (𝐴) 𝑣 : 𝑓 \in 𝐶_{lim}^{0} (ℝ, ℂ)}}, 𝑓 \mapsto 𝑓 (𝐴) 𝑣 .

Note that

⟨ 𝑔 (𝐴) 𝑣, {(𝐴 - 𝑧)}^{- 1} 𝑓 (𝐴) 𝑣 ⟩ = ⟨ 𝑣, (\bar{𝑔} {(• - 𝑧)}^{- 1} 𝑓) (𝐴) 𝑣 ⟩ = \frac{\int_{ℝ (\bar{𝑔 (𝑠)} 𝑓 (𝑠))}}{𝑠 - 𝑧} d 𝜇_{𝐴, 𝑣} (𝑠)

Thus, extending by continuity, we see that ${(𝐴 - 𝑧)}^{- 1}$ restricted to ${𝒳︀}_{𝑣}$ is unitarily equivalent to multiplication by ${(𝑠 - 𝑧)}^{- 1}$ . It follows that $𝐴$ is unitarily equivalent to multiplication by $𝑠$ . Indeed, let

𝐵 ≔ 𝑈 𝐴 𝑈^{- 1}

Then ${(𝐵 - 𝑧)}^{- 1} = {𝑈 (𝐴 - 𝑧)}^{- 1} 𝑈^{- 1} = 𝑀_{{(𝑠 - 𝑧)}^{- 1}}$ . At the same time, from

(𝑠 - 𝑧) {(𝑀_{𝑠} - 𝑧)}^{- 1} 𝑓 = (𝑀_{𝑠} - 𝑧) {(𝑀_{𝑠} - 𝑧)}^{- 1} 𝑓 = 𝑓

we see that ${(𝑀_{𝑠} - 𝑧)}^{- 1} = 𝑀_{{(𝑠 - 𝑧)}^{- 1}}$ as well. In particular the resolvents’ ranges are the same, so $𝐷 (𝐵) = 𝐷 (𝑀_{𝑠})$ , and $𝐼 = (𝐵 - 𝑧) {(𝑀_{𝑠} - 𝑧)}^{- 1}$ gives that $𝐵 = 𝑀_{𝑠}$ .

If there exists $𝑣$ such that ${𝒳︀}_{𝑣} = ℌ$ , then we are done. If not, we need to iterate the argument.

Lemma 2 (Invariance of ${𝒳︀}_{𝑣}$ ). The subspace

{𝒳︀}_{𝑣}

is invariant under

{(𝐴 - 𝑧)}^{- 1}

, and

{𝒳︀}_{𝑣}^{⟂}

is also invariant under

{(𝐴 - 𝑧)}^{- 1}

, for every

𝑧 \in ℂ \ ℝ

Proof (of lemma). ${(𝐴 - 𝑧)}^{- 1} {𝒳︀}_{𝑣} \subset {𝒳︀}_{𝑣}$ is because ${(𝐴 - 𝑧)}^{- 1} 𝑓 \in 𝐶_{lim}^{0} (ℝ, ℂ)$ for every $𝑓 \in 𝐶_{lim}^{0} (ℝ, ℂ)$ , and ${(𝐴 - 𝑧)}^{- 1}$ is continuous.

And for a similar reason, ${(𝐴 - 𝑧)}^{- 1} {𝒳︀}_{𝑣}^{⟂} \subset {𝒳︀}_{𝑣}^{⟂}$ , as $⟨ 𝑓 (𝐴) 𝑣, {(𝐴 - 𝑧)}^{- 1} 𝑤 ⟩ = ⟨ {(𝐴 - \bar{𝑧})}^{- 1} 𝑓 (𝐴) 𝑣, 𝑤 ⟩ = ⟨ ({(• - \bar{𝑧})}^{- 1} 𝑓) (𝐴) 𝑣, 𝑤 ⟩ = 0 .$ ∎

Now, we can write

ℌ = ⨁_{𝑛 \in ℕ} {𝒳︀}_{𝑣_{𝑛}}

as follows - let $𝑒_{𝑛}$ be an orthonormal basis for $ℌ$ , and put $𝑣_{1} = 𝑒_{1}$ . Then, let $𝑣_{2} = 𝑃_{{𝒳︀}_{𝑣_{1}}}^{⟂} 𝑒_{𝑗}$ (orthogonal projection) where $𝑒_{𝑗}$ is the first basis vector not in ${𝒳︀}_{𝑣_{1}}$ . Note that ${𝒳︀}_{𝑣_{2}} ⟂ {𝒳︀}_{𝑣_{1}}$ , since $𝑣_{2} ⟂ {𝒳︀}_{𝑣_{1}}$ , and

⟨ 𝑓 (𝐴) 𝑣_{1}, 𝑔 (𝐴) 𝑣_{2} ⟩ = ⟨ 𝑣_{1}, (\bar{𝑓} 𝑔) (𝐴) 𝑣_{2} ⟩ = 0 .

Continuing in this way gives the desired decomposition of $ℌ$ into a direct sum of invariant subspaces.

We now have that ${(𝐴 - 𝑧)}^{- 1}$ on each invariant subspace ${𝒳︀}_{𝑣_{𝑛}}$ is unitarily equivalent to multiplication by ${(𝑠 - 𝑧)}^{- 1}$ on $𝐿^{2} (ℝ, d 𝜇_{𝐴, 𝑣_{𝑛}})$ . We can combine them into a single isomorphism to $𝐿^{2} (𝐵, d 𝜇)$ with $𝐵 = ℝ \times ℕ$ and $𝜇 (𝑉 \times {𝑛}) = 2^{- 𝑛} 𝜇_{𝐴, 𝑣_{𝑛}} (𝑉)$ , and define $𝑎 (𝑠, 𝑛) = 𝑠$ . It follows as before that $𝐴$ is unitarily equivalent to multiplication by $𝑠$ . This completes the proof of Theorem 2, apart from the special form claimed that we can take in fact $𝐵 = 𝜎 (𝐴) \times ℕ$ . This is covered in the next lemma.∎

Lemma 3 (Support of the spectral measure). Let

𝑣 \in 𝐻

with

‖ 𝑣 ‖ = 1

. Then

𝜇_{𝐴, 𝑣} (ℝ \ 𝜎 (𝐴)) = 0

Proof. Suppose $ℝ \ 𝜎 (𝐴) \neq \emptyset$ and let $𝜆_{0} \in ℝ \ 𝜎 (𝐴)$ . Then ${(𝐴 - 𝑧)}^{- 1}$ is bounded on a small ball (in $ℂ$ ) around $𝜆_{0}$ . For concreteness we take

𝑟 = \frac{1}{2 ‖ {(𝐴 - 𝜆_{0})}^{- 1} ‖} .

Then $‖ {(𝐴 - 𝑧)}^{- 1} ‖ \leq 2$ for $𝑧 \in 𝔹_{𝜆_{0}, 𝑟}$ .

Let $𝜆 \in [𝜆_{0} - \frac{𝑟}{2}, 𝜆_{0} + \frac{𝑟}{2}]$ . Put $𝑧_{𝑛} = 𝜆 + \frac{𝑖}{𝑛}$ for $𝑛 \geq \frac{2}{𝑟}$ so that $𝑧_{𝑛} \in 𝔹_{𝜆_{0}, 𝑟}$ . Then

4 \geq {‖ {(𝐴 - 𝑧_{𝑛})}^{- 1} ‖}^{2} \geq ⟨ 𝑣, {(𝐴 - 𝑧_{𝑛})}^{- 1} 𝑣 ⟩ = \int_{ℝ} \frac{1}{{| 𝑠 - 𝜆 |}^{2} + \frac{1}{𝑛^{2}}} d 𝜇_{𝐴, 𝑣} (𝑠) .

Integrating over $𝜆 \in [𝜆_{0} - \frac{𝑟}{2}, 𝜆_{0} + \frac{𝑟}{2}]$ and using Tonelli’s theorem gives

\begin{aligned} 4 𝑟 & \geq \int_{𝜆_{0} - \frac{𝑟}{2}}^{𝜆_{0} + \frac{𝑟}{2}} \int_{ℝ} \frac{1}{{| 𝑠 - 𝜆 |}^{2} + \frac{1}{𝑛^{2}}} d 𝜇_{𝐴, 𝑣} (𝑠) d 𝜆 \\ = \int_{ℝ} \int_{𝜆_{0} - \frac{𝑟}{2}}^{𝜆_{0} + \frac{𝑟}{2}} \frac{1}{{| 𝑠 - 𝜆 |}^{2} + \frac{1}{𝑛^{2}}} d 𝜆 d 𝜇_{𝐴, 𝑣} (𝑠) \\ \geq \int_{𝜆_{0} - \frac{𝑟}{4}}^{𝜆_{0} + \frac{𝑟}{4}} \int_{𝑠 - \frac{𝑟}{4}}^{𝑠 + \frac{𝑟}{4}} \frac{1}{{| 𝑠 - 𝜆 |}^{2} + \frac{1}{𝑛^{2}}} d 𝜆 d 𝜇_{𝐴, 𝑣} (𝑠) \\ = 𝜇_{𝐴, 𝑣} ([𝜆_{0} - \frac{𝑟}{4}, 𝜆_{0} + \frac{𝑟}{4}]) \int_{- \frac{𝑟}{4}}^{\frac{𝑟}{4}} \frac{1}{𝑡^{2} + \frac{1}{𝑛^{2}}} d 𝑡 \\ = 2 𝑛 arctan (\frac{𝑟 𝑛}{4}) 𝜇_{𝐴, 𝑣} ([𝜆_{0} - \frac{𝑟}{4}, 𝜆_{0} + \frac{𝑟}{4}]) . \end{aligned}

It follows from $𝑛 arctan (\frac{𝑟 𝑛}{4}) \underset{𝑛 \to \infty}{\to} \infty$ that $𝜇_{𝐴, 𝑣} ([𝜆_{0} - \frac{𝑟}{4}, 𝜆_{0} + \frac{𝑟}{4}]) = 0$ . Since $𝜆_{0}$ was arbitrary, we have $𝜇_{𝐴, 𝑣} (ℝ \ 𝜎 (𝐴)) = 0$ as claimed.∎

Proof (of Theorem 3). Given the spectral theorem we have already above the construction of the functional calculus for measurable functions satisfying the required properties, save uniqueness.

So consider a second functional calculus $𝑓 \mapsto {𝑓 (𝐴)}^{'}$ satisfying the same properties. Then since they agree for $𝑓 (𝑥) = {(𝑥 - 𝑧)}^{- 1}$ , they agree on the resolvent algebra $𝒜︀$ , and by continuity they agree on $𝐶_{lim}^{0}$ .

Now fix $𝑣 \in ℌ$ and consider the two linear forms

ℓ (𝑓) = ⟨ 𝑣, 𝑓 (𝐴) 𝑣 ⟩ = \int_{𝜎 (𝐴)} 𝑓 (𝑠) d 𝜇_{𝐴, 𝑣} (𝑠), ℓ^{'} (𝑓) = ⟨ 𝑣, {𝑓 (𝐴)}^{'} 𝑣 ⟩

$ℓ = ℓ^{'}$ would imply $𝑓 (𝐴) = {𝑓 (𝐴)}^{'}$ by polarization. Riesz–Markov gives us uniqueness of the corresponding Borel measure, but this is not the same as uniqueness of the functional (for instance if the measure was $𝛿_{0}$ the linear functional could a Banach limit given by some ultrafilter.). For this, we invoke

Theorem 6 (Functional Monotone Class Theorem). Let $𝒜︀$ be a unital algebra of bounded real-valued functions on X. Let $ℋ︀$ be a vector space of bounded functions such that $𝒜︀ \subseteq ℋ︀$ , and suppose $ℋ︀$ is closed under bounded monotone pointwise limits:

0 \leq 𝑓_{𝑛} ↑ 𝑓, sup_{𝑥} 𝑓 (𝑥) < \infty, 𝑓_{𝑛} \in ℋ︀ ⟹ 𝑓 \in ℋ︀ .

Then $ℋ︀$ contains every bounded function measurable with respect to $𝜎 (𝒜︀)$ .

With this theorem and (v) of Theorem 3, we have that $ℓ = ℓ^{'}$ on all bounded Borel functions, and hence $𝑓 (𝐴) = {𝑓 (𝐴)}^{'}$ for all bounded Borel functions, as needed.∎

3 Spectral Projections

As always let $𝐴$ be self-adjoint on $ℌ$ . To each Borel $𝐹 \subset ℝ$ , the functional calculus gives us the associated spectral projection $𝟙_{𝐹} (𝐴)$ .

Proposition 1.

We have the following properties:

$𝟙_{𝐹} (𝐴) = 𝟙_{𝐹} {(𝐴)}^{*} = 𝟙_{𝐹} {(𝐴)}^{2}$
$𝟙_{\emptyset} (𝐴) = 0, 𝟙_{ℝ} (𝐴) = 1 (𝐴) = {Id}_{ℌ}$ ,
If $𝐹 = ⋃_{𝑛 \geq 1} 𝐹_{𝑛}$ , then $𝟙_{𝐹} (𝐴) = \sum_{𝑛 \geq 1} 𝟙_{𝐹_{𝑛}} (𝐴)$ ,
$𝟙_{𝐹_{1} \cap 𝐹_{2}} (𝐴) = 𝟙_{𝐹_{1}} (𝐴) 𝟙_{𝐹_{2}} (𝐴)$ ,

In addition, using the specific representation of $𝐴$ as a multiplication operator $𝑀_{𝑎}$ by $𝑎 (𝑠, 𝑛) = 𝑠$ on $𝐿^{2} (𝜎 (𝐴) \times ℕ, d 𝜇)$ , we have

𝟙_{𝐹} (𝐴) = 𝑈^{- 1} 𝑀_{𝟙_{𝐹} (𝑎)} 𝑈 = 𝑈^{- 1} 𝑀_{𝟙_{𝐹 \times ℕ}} 𝑈,

so $ran 𝟙_{𝐹} (𝐴) = 𝑈^{- 1} (𝐿^{2} (𝐹 \times ℕ, d 𝜇)$ , and $rank (𝟙_{𝐹} (𝐴)) = dim 𝐿^{2} (𝐹 \times ℕ, d 𝜇)$ .

Lemma 4.

$𝜆 \in 𝜎 (𝐴)$ iff $𝟙_{𝜆 - 𝜀, 𝜆 + 𝜀} (𝐴) \neq 0$ for all $𝜀 > 0$ .
$𝜆$ is an eigenvalue of $𝐴$ iff $𝟙_{{𝜆}} (𝐴) \neq 0$ , in which case $𝟙_{{𝜆}} (𝐴)$ is the orthogonal projection to the corresponding eigenspace $ker (𝐴 - 𝜆)$ .

¹ The book writes
$d 𝜇_{𝐴, 𝑣} (𝑠) = \sum_{𝑛 \in ℕ} | 𝑈 𝑣 (𝑠, 𝑛) |^{2} d 𝜇 (𝑠, 𝑛)$
but this is a little fast and loose with notation; the sum is a partial integration of $d 𝜇 (𝑠, 𝑛)$ that comes from the pushforward. The proper way is to define the slice measures $𝜇_{𝑛} (𝐸) = 𝜇 (𝐸 \times {𝑛})$ , then we can write $d 𝜇_{𝐴, 𝑣} (𝑠) = \sum_{𝑛 \in ℕ} | 𝑈 𝑣 (𝑠, 𝑛) |^{2} d 𝜇_{𝑛} (𝑠)$ .

1 Multiplication Operators

2 Proof of Theorems 4.4 and 4.8

3 Spectral Projections

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