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Positive unbounded spectral calculus from a bounded resolvent

Statement

Assume the Axiom of Choice. Let $H \neq {0}$ be a complex Hilbert space, with inner product linear in its first variable, and let $A : D (A) \subseteq H \to H$ be densely defined, self-adjoint, and nonnegative, meaning $⟨ A u, u ⟩ \geq 0$ for every $u \in D (A)$ . Here self-adjoint means equality with the Hilbert-space adjoint, including equality of domains. A projection-valued measure (PVM) on a measurable space $(X, Σ)$ is a map $E : Σ \to B (H)$ such that every $E (S)$ is a bounded orthogonal projection ( $E (S)^{2} = E (S) = E (S)^{*}$ ), $E (\emptyset) = 0$ , $E (X) = I$ , $E (S \cap T) = E (S) E (T)$ , and $E (⋃_{n} S_{n}) u = \sum_{n} E (S_{n}) u$ in norm for each disjoint sequence $(S_{n})$ and each $u \in H$ . On a locally compact Hausdorff Borel space, such a PVM is regular when each finite scalar measure $μ_{u} (S) : = ⟨ E (S) u, u ⟩$ is inner regular on Borel sets and outer regular by open supersets. Then $I + A : D (A) \to H$ is bijective. Its inverse $B : = (I + A)^{- 1}$ is a bounded positive (meaning $⟨ B x, x ⟩ \geq 0$ for all $x \in H$ ) self-adjoint injection with $∥ B ∥ \leq 1$ and dense range $D (A)$ . For a bounded operator, its spectrum $σ (B)$ is the set of complex $z$ for which $B - z I$ has no bounded two-sided inverse. Let $E_{B}$ be the regular spectral PVM on the Borel subsets of the nonempty compact set $σ (B)$ satisfying $B = \int_{σ (B)} b d E_{B} (b)$ . For every ambient Borel set $S \subseteq C$ , also write $E_{B} (S) : = E_{B} (S \cap σ (B))$ , the extension by zero outside the spectrum. Then $σ (B) \subseteq [0, 1]$ and $E_{B} ({0}) = 0$ . Define $ϕ (b) = b^{- 1} - 1$ for $b \in σ (B) ∖ {0}$ and, for every Borel $S \subseteq [0, \infty)$ , define

$E_{A} (S) : = E_{B} ({b \in σ (B) ∖ {0} : ϕ (b) \in S}) .$

This is a PVM on the Borel subsets of $[0, \infty)$ . For each $u \in H$ put $μ_{u} (S) : = ⟨ E_{A} (S) u, u ⟩$ . The unbounded integral $\int_{[0, \infty)} λ d E_{A} (λ)$ denotes the operator $T_{A}$ with domain

$D (T_{A}) : = {u \in H : \int_{[0, \infty)} λ^{2} d μ_{u} (λ) < \infty}$

and action

$T_{A} u : = \lim_{N \to \infty} Φ_{E_{A}} (λ 1_{[0, N]}) (u),$

where $Φ_{E_{A}} (g)$ is the bounded Borel PVM integral of $g$ . Then $A = T_{A}$ and hence

$A = \int_{[0, \infty)} λ d E_{A} (λ), D (A) = {u \in H : \int_{[0, \infty)} λ^{2} d ⟨ E_{A} (λ) u, u ⟩ < \infty} .$

For every bounded Borel function $f : [0, \infty) \to C$ , $f (A) : = Φ_{E_{A}} (f)$ is bounded, satisfies $∥ f (A) ∥ \leq \sup_{λ \geq 0} ∣ f (λ) ∣$ , and obeys $∥ f (A) u ∥^{2} = \int ∣ f ∣^{2} d μ_{u}$ . The condition $E_{B} ({0}) = 0$ does not assert that $0 \notin σ (B)$ .

Statement references

Baez, Quantum Theory and Analysis, §5

Princeton MAT520, §11.5

Facts & Assumptions

Given: The Axiom of Choice, $H \neq {0}$ , and the densely defined nonnegative self-adjoint operator $A$ on $H$ from the Statement.

[A1]

The Axiom of Choice: every family of nonempty sets has a choice function.

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): every countable family of nonempty sets has a choice function; this is the countable-family instance of [A1].

[F1]

Spectral theorem for bounded normal operators pvm form: under the Axiom of Choice, every bounded normal operator on a nonzero complex Hilbert space has a unique regular spectral projection-valued measure $E$ on its spectrum with operator integral equal to that operator.

[F2]

Projection valued measure: a PVM assigns an orthogonal projection to each measurable set, is multiplicative on intersections, is normalized by $E (X) = I$ , and is strongly countably additive; regularity means each finite scalar measure $E_{u} (S) = ⟨ E (S) u, u ⟩$ is inner regular on Borel sets and outer regular by open supersets.

[F3]

Bounded borel pvm integral: for a PVM $E$ and bounded Borel $h$ , the integral $Φ_{E} (h)$ is bounded, $∥ Φ_{E} (h) ∥ \leq ∥ h ∥_{\infty}$ , and $∥ Φ_{E} (h) x ∥^{2} = \int ∣ h ∣^{2} d E_{x}$ , where $E_{x} (S) = ⟨ E (S) x, x ⟩$ .

[F4]

Pvm integral is a star homomorphism: the bounded PVM integral is multiplicative, unital, and preserves complex conjugation as the operator adjoint.

[F5]

Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ for vectors in a complex Hilbert space.

[F6]

Adjoint of a densely defined operator: under Countable Choice, for a densely defined $T$ and the first-variable-linear inner product, $y \in D (T^{*})$ exactly when $x \mapsto ⟨ T x, y ⟩$ is bounded on $D (T)$ in the ambient norm $∥ x ∥_{H}$ . The vector $T^{*} y$ is the unique representing vector satisfying $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ for all $x \in D (T)$ . In particular, if this identity holds with a vector $v \in H$ , [F5] bounds the functional by $∥ v ∥ ∥ x ∥$ , so $y \in D (T^{*})$ and $T^{*} y = v$ . The required Countable Choice follows from [A1]–[A2].

Proof

technique · direct

Given: The hypotheses in [A1]–[A2], a nonzero complex Hilbert space $H$ , and a densely defined nonnegative self-adjoint operator $A$ .

1.1givenA1A2F5F6algebra

For $u \in D (A)$ , [F5] and nonnegativity give $∥ (I + A) u ∥ ∥ u ∥ \geq ∣ ⟨ (I + A) u, u ⟩ ∣ = ∥ u ∥^{2} + ⟨ A u, u ⟩ \geq ∥ u ∥^{2}$ . Hence $I + A$ is injective and bounded below by $1$ . To prove $A$ is closed, suppose $u_{n} \in D (A)$ , $u_{n} \to u$ , and $A u_{n} \to v$ . For every $w \in D (A)$ , self-adjointness gives $⟨ A w, u ⟩ = \lim_{n} ⟨ A w, u_{n} ⟩ = \lim_{n} ⟨ w, A u_{n} ⟩ = ⟨ w, v ⟩$ . In particular, $∣ ⟨ A w, u ⟩ ∣ = ∣ ⟨ w, v ⟩ ∣ \leq ∥ w ∥ ∥ v ∥$ by [F5], so this functional is bounded in the ambient $H$ norm on $D (A)$ . By [F6], $u \in D (A^{*})$ and its unique representing vector is $A^{*} u = v$ . Self-adjointness, including equality of domains, yields $u \in D (A)$ and $A u = v$ . Thus $A$ is closed. For $w \in D ((I + A)^{*})$ , write $z = (I + A)^{*} w$ . Subtracting the identity term gives $⟨ A u, w ⟩ = ⟨ u, z - w ⟩$ for every $u \in D (A)$ . By [F5] this is bounded by $∥ u ∥ ∥ z - w ∥$ , so [F6] gives $w \in D (A^{*})$ and $A^{*} w = z - w$ , or $(I + A)^{*} w = w + A^{*} w$ . Conversely, if $w \in D (A^{*})$ , then $⟨ (I + A) u, w ⟩ = ⟨ u, w + A^{*} w ⟩$ for every $u \in D (A)$ . The functional in $u$ is bounded by $∥ u ∥ ∥ w + A^{*} w ∥$ by [F5], so [F6] gives $w \in D ((I + A)^{*})$ with $(I + A)^{*} w = w + A^{*} w$ . Thus $(I + A)^{*} = I + A^{*} = I + A$ with domain $D (A)$ , so $I + A$ is self-adjoint. If $u_{n} \in D (A)$ and $(I + A) u_{n}$ converges, the lower bound makes $(u_{n})$ Cauchy; closedness then puts the limit in the range, so the range of $I + A$ is closed. If $y$ is orthogonal to that range, then $⟨ (I + A) x, y ⟩ = 0 = ⟨ x, 0 ⟩$ for all $x \in D (A)$ ; [F6] with representing vector zero gives $y \in \ker (I + A)^{*} = \ker (I + A)$ , which is zero by the same lower bound; the range is therefore dense and hence all of $H$ . Thus $B = (I + A)^{- 1}$ exists everywhere, is bounded with norm at most $1$ , is injective, and has range $D (A)$ , which is dense.

2.1givenstep 1.1F5algebra

If $x = (I + A) u$ and $y = (I + A) v$ , then $⟨ B x, y ⟩ = ⟨ u, (I + A) v ⟩ = ⟨ (I + A) u, v ⟩ = ⟨ x, B y ⟩$ , so $B$ is self-adjoint. Also $⟨ B x, x ⟩ = ⟨ u, (I + A) u ⟩ = ∥ u ∥^{2} + ⟨ A u, u ⟩ \geq 0$ , so $B$ is positive. For nonreal $z$ , Cauchy–Schwarz [F5] and self-adjointness give $∥ (B - z I) w ∥ \geq ∣ Im z ∣ ∥ w ∥$ and the same lower bound for its adjoint, so its range is closed and dense. For real $z < 0$ , positivity and [F5] give $∥ (B - z I) w ∥ \geq (- z) ∥ w ∥$ , again giving closed and dense range. For real $z > 1$ , the Neumann series $(z I - B)^{- 1} = z^{- 1} \sum_{n \geq 0} (B / z)^{n}$ converges in operator norm, so $B - z I$ is invertible as well. Hence $σ (B) \subseteq [0, 1]$ .

3.1A1A2F1F2F3F4step 2.1

By [A1] and [F1], $B$ has its regular spectral PVM $E_{B}$ on $σ (B)$ , extended to ambient Borel sets by intersection with $σ (B)$ as in the Statement. If $x = E_{B} ({0}) y$ , then [F4] gives $B x = Φ_{E_{B}} (b 1_{{0}}) y = 0$ , so $E_{B} ({0}) H \subseteq \ker B$ . Conversely, if $B x = 0$ , [F3] gives $0 = ∥ B x ∥^{2} = \int b^{2} d E_{x} (b)$ . Within $[0, 1]$ , ${b > 0} = ⋃_{n \geq 1} [1 / n, 1]$ and $b^{2} \geq n^{- 2}$ on each such interval, $E_{x} ([1 / n, 1]) = 0$ for every $n$ . Strong countable additivity and $E_{B} (σ (B)) = I$ then imply $x = E_{B} ({0}) x$ . Since $B$ is injective, $E_{B} ({0}) = 0$ . This is a zero spectral projection; it does not remove $0$ from the operator spectrum.

4.1A2F2F4step 3.1construct

The map $ϕ (b) = b^{- 1} - 1$ on $σ (B) ∖ {0}$ is Borel. For a Borel set $S \subseteq [0, \infty)$ set $E_{A} (S) : = E_{B} ({b \in σ (B) ∖ {0} : ϕ (b) \in S})$ . Inverse images preserve disjoint unions and intersections, so [F2] makes $E_{A}$ a PVM; its total value is $E_{B} (σ (B) ∖ {0}) = I$ because $E_{B} ({0}) = 0$ . The defining identity on indicator functions, extended first to simple functions and then by uniform approximation, gives $Φ_{E_{A}} (f) = Φ_{E_{B}} (g)$ for every bounded Borel $f$ on $[0, \infty)$ , where $g (b) = f (ϕ (b))$ for $b > 0$ and $g (0) = 0$ .

5.1A2F3F4step 3.1step 4.1algebra

In integrals set $ϕ (0) = 0$ , and interpret each inverse-power multiplier with a cutoff as zero where its cutoff vanishes. Write $μ_{x}^{B} (S) = ⟨ E_{B} (S) x, x ⟩$ . From the definition in step 4.1, $μ_{x} (S) = μ_{x}^{B} ({b > 0 : ϕ (b) \in S})$ , so $\int λ^{2} d μ_{x} (λ) = \int ϕ (b)^{2} d μ_{x}^{B} (b)$ . If $x = B y$ , multiplicativity and the star-homomorphism identity [F4] give $d μ_{x}^{B} (b) = b^{2} d μ_{y}^{B} (b)$ ; hence $\int ϕ (b)^{2} d μ_{x}^{B} (b) = \int_{(0, 1]} (1 - b)^{2} d μ_{y}^{B} (b) \leq ∥ y ∥^{2}$ . Thus $x$ lies in the domain defining $T_{A}$ . Conversely, if $\int ϕ^{2} d μ_{x}^{B} < \infty$ , then $\int_{(0, 1]} b^{- 2} d μ_{x}^{B} = \int (ϕ + 1)^{2} d μ_{x}^{B} \leq 2 \int ϕ^{2} d μ_{x}^{B} + 2 ∥ x ∥^{2} < \infty$ . The bounded truncations $y_{N} : = Φ_{E_{B}} (b^{- 1} 1_{[1 / N, 1]}) (x)$ are Cauchy by the quadratic identity [F3], since their squared tails are tails of this finite integral, and converge to some $y \in H$ . Multiplying by $B$ gives $B y_{N} = E_{B} ([1 / N, 1]) x \to x$ by strong countable additivity and $E_{B} ({0}) = 0$ , so $B y = x$ . Hence $D (T_{A}) = Ran B = D (A)$ . For $x = B y$ , the truncation in the definition of $T_{A}$ is, by step 4.1, $Φ_{E_{B}} ((1 - b) 1_{[1 / (N + 1), 1]}) (y)$ . These bounded multipliers converge pointwise off ${0}$ to $1 - b$ and are uniformly bounded by $1$ ; [F4] gives strong convergence to $Φ_{E_{B}} (1 - b) y = (I - B) y = y - x = A x$ . Thus $T_{A} = A$ on the common domain, proving the operator identity.

6.1A2F2F3step 4.1algebra∎

If $f$ is bounded Borel on $[0, \infty)$ , step 4.1 gives $f (A) = Φ_{E_{B}} (g)$ for $g (b) = f (ϕ (b))$ on $b > 0$ and $g (0) = 0$ . By [F3], $∥ f (A) ∥ \leq ∥ g ∥_{\infty} \leq \sup_{λ \geq 0} ∣ f (λ) ∣$ , and $∥ f (A) u ∥^{2} = \int ∣ f ∣^{2} d μ_{u}$ . This proves the bounded Borel calculus claims.

Positive unbounded spectral calculus from a bounded resolvent

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources