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Transport of spectral projection-valued measures along isometric bijections

Statement

Assume the Axiom of Choice and the Axiom of Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H, K$ be complex Hilbert spaces (Hilbert space, Real and complex inner product spaces, with the inner product linear in the first argument) and let $U : H \to K$ be a bijective linear map with $⟨ U x, U y ⟩_{K} = ⟨ x, y ⟩_{H} for all x, y \in H .$ Let $T$ be a self-adjoint operator on $H$ with spectral projection-valued measure $E_{T}$ on the Borel $σ$ -algebra of $R$ (Spectral theorem for unbounded self-adjoint operators (PVM form), Projection valued measure, Symmetric, self-adjoint and essentially self-adjoint operators), and define $D (T^{'}) : = U D (T), T^{'} (U x) : = U (T x) (x \in D (T)) .$ Then:

(a) $T^{'}$ is well defined, linear, densely defined and self-adjoint on $K$ ;

(b) the assignment $E (B) : = U E_{T} (B) U^{- 1}$ on Borel sets $B \subseteq R$ is a regular projection-valued measure on the Borel $σ$ -algebra of $R$ , and it is the spectral projection-valued measure of $T^{'}$ ;

(c) for every unit vector $x \in H$ the scalar laws of the two operators agree, $μ_{U x}^{T^{'}} (B) : = ∥ E (B) U x ∥_{K}^{2} = ∥ E_{T} (B) x ∥_{H}^{2} = : μ_{x}^{T} (B) (B \subseteq R Borel);$

(d) $σ (T^{'}) = σ (T)$ .

Facts & Assumptions

Given: complex Hilbert spaces $H, K$ ; a bijective linear isometry $U : H \to K$ ; a self-adjoint operator $T$ on $H$ with spectral projection-valued measure $E_{T}$ ; the operator $T^{'}$ defined in the Statement; a real number $λ$ and Borel sets $B, B_{n} \subseteq R$ .

[F1]

Spectral theorem on a nonzero space. When $H \neq {0}$ , $E_{T}$ is the unique regular projection-valued measure on the Borel $σ$ -algebra of $R$ with $D (T) = {x \in H : \int_{R} λ^{2} d E_{T, x} (λ) < \infty}, T x = \int_{R} λ d E_{T} (λ) x (x \in D (T)),$ where $E_{T, x} (B) = ⟨ E_{T} (B) x, x ⟩$ ; conversely, if a regular projection-valued measure $E$ on $R$ satisfies the two displayed identities for an operator $S$ in place of $T$ , then $E$ is the spectral projection-valued measure of the self-adjoint operator $S$ (Spectral theorem for unbounded self-adjoint operators (PVM form), Integral of a measurable function against a projection-valued measure).

[F2]

Projection-valued measures and their scalar measures. $E_{T} (B)$ is an orthogonal projection for every Borel $B$ , $E_{T} (\emptyset) = 0$ , $E_{T} (R) = I$ , $E_{T} (B \cap C) = E_{T} (B) E_{T} (C)$ , and for pairwise disjoint $B_{n}$ with union $B$ one has $E_{T} (B) x = \sum_{n} E_{T} (B_{n}) x$ in norm for every $x$ ; the scalar pairing $E_{T, x, y} (B) : = ⟨ E_{T} (B) x, y ⟩$ is a finite complex measure and $E_{T, x} (B) : = E_{T, x, x} (B) = ∥ E_{T} (B) x ∥^{2} \geq 0$ ; the regularity of $E_{T}$ means that every $E_{T, x}$ is a regular Borel measure (Projection valued measure, Scalar and complex measures from a pvm).

[F3]

Spectrum as support. On a nonzero complex Hilbert space, for the self-adjoint operator $T$ with spectral projection-valued measure $E_{T}$ one has $E_{T} (R ∖ σ (T)) = 0$ and $E_{T} (V) \neq 0$ for every nonempty relatively open $V \subseteq σ (T)$ (Support of the spectral measure, the spectrum and the values of the scalar law clause (a)).

[F4]

The adjoint. For a densely defined operator $S$ on $H$ , a vector $y \in H$ lies in $D (S^{*})$ exactly when $x \mapsto ⟨ S x, y ⟩$ is bounded on $(D (S), ∥ \cdot ∥)$ ; then $S^{*} y$ is the unique $w \in H$ with $⟨ S x, y ⟩ = ⟨ x, w ⟩$ for every $x \in D (S)$ ; and $S$ is self-adjoint exactly when $S = S^{*}$ , that is $D (S) = D (S^{*})$ and the two actions agree (Adjoint of a densely defined operator, Symmetric, self-adjoint and essentially self-adjoint operators).

[F5]

Inner-product calculus. The pairing is linear in the first argument and conjugate-linear in the second, $∥ u ∥^{2} = ⟨ u, u ⟩$ , and $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ (Real and complex inner product spaces, with the inner product linear in the first argument, Hilbert space, Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors).

[F6]

Zero-space spectrum. The resolvent set consists of those $z \in C$ for which $z I - T : D (T) \to H$ is bijective with bounded inverse; its complement is $σ (T)$ . On $H = {0}$ every shift is the unique bijection with bounded inverse, so $σ (T) = \emptyset$ (Resolvent and spectrum of an unbounded operator).

Proof

Given: the data of the Statement and [F1]-[F6].

1.1givenF5

[The inverse is a linear isometry] Since $U$ is bijective, $U^{- 1}$ exists; it is linear because $U$ is linear. For $a = U x$ and $b = U y$ in $K$ , $⟨ U^{- 1} a, U^{- 1} b ⟩_{H} = ⟨ x, y ⟩_{H} = ⟨ U x, U y ⟩_{K} = ⟨ a, b ⟩_{K}$ ; hence $U^{- 1}$ is isometric, and so is $U$ , so $∥ U x ∥_{K} = ∥ x ∥_{H}$ for every $x \in H$ .

1.2givenF2F4F6

[Zero-space case] If $H = {0}$ , bijectivity of $U$ gives $K = {0}$ . The domains of $T$ and $T^{'}$ are the zero space; their actions and adjoints are the unique zero operator by [F4], so clause (a) holds. Every $E_{T} (B)$ and $E (B)$ is that operator; the PVM identities and scalar regularity hold directly, and the integral of the coordinate function has domain ${0}$ and zero action by Integral of a measurable function against a projection-valued measure, so this unique PVM is the spectral PVM of $T^{'}$ , proving (b). There are no unit vectors, so (c) is vacuous; [F6] gives $σ (T^{'}) = σ (T) = \emptyset$ , proving (d). For the remaining steps assume $H \neq {0}$ , hence $K \neq {0}$ , so [F1] and [F3] apply.

2.1givenstep 1.1F2

[ $E$ is a projection-valued measure] For a Borel set $B$ define $E (B) : = U E_{T} (B) U^{- 1} \in B (K)$ . It is an orthogonal projection: $E (B)^{2} = U E_{T} (B) U^{- 1} U E_{T} (B) U^{- 1} = E (B)$ and $⟨ E (B) a, b ⟩_{K} = ⟨ E_{T} (B) U^{- 1} a, U^{- 1} b ⟩_{H} = ⟨ U^{- 1} a, E_{T} (B) U^{- 1} b ⟩_{H} = ⟨ a, E (B) b ⟩_{K}$ for all $a, b \in K$ , using the isometry of $U^{- 1}$ and the self-adjointness of $E_{T} (B)$ . Moreover $E (\emptyset) = 0$ , $E (R) = I$ , $E (B \cap C) = E (B) E (C)$ , and for pairwise disjoint $B_{n}$ with union $B$ , $∥ E (B) a - \sum_{n \leq N} E (B_{n}) a ∥_{K} = ∥ E_{T} (B) U^{- 1} a - \sum_{n \leq N} E_{T} (B_{n}) U^{- 1} a ∥_{H} \to 0$ by [F2]. Hence $E$ is a projection-valued measure on the Borel sets of $R$ .

2.2givenstep 1.1F5

[ $T^{'}$ is well defined, linear and symmetric] If $U x = U x^{'}$ then $x = x^{'}$ by injectivity, so $T^{'} (U x) : = U (T x)$ is well defined; $D (T^{'}) = U D (T)$ is a linear subspace, dense because $D (T)$ is dense and $U$ is bijective, and $T^{'}$ is linear because $T$ and $U$ are. For $x, y \in D (T)$ , using the isometry of $U$ in both directions and the self-adjointness of $T$ , $⟨ T^{'} U x, U y ⟩_{K} = ⟨ U T x, U y ⟩_{K} = ⟨ T x, y ⟩_{H} = ⟨ x, T y ⟩_{H} = ⟨ U x, U T y ⟩_{K} = ⟨ U x, T^{'} U y ⟩_{K}$ .

3.1step 2.1F2

[ $E$ is regular with transported scalar measures] For $y, z \in K$ and Borel $B$ , $E_{y, z} (B) : = ⟨ E (B) y, z ⟩_{K} = ⟨ E_{T} (B) U^{- 1} y, U^{- 1} z ⟩_{H} = E_{T, U^{- 1} y, U^{- 1} z} (B)$ by the isometry of $U^{- 1}$ and the definition of $E$ ; in particular $E_{y} (B) = E_{T, U^{- 1} y} (B)$ with $U^{- 1} y \in H$ . Since $E_{T}$ is regular, each $E_{T, x}$ is a regular Borel measure [F2], so each scalar measure $E_{y}$ of $E$ is regular and $E$ is regular.

3.2step 2.2F4F5

[ $T^{'}$ is self-adjoint] Let $y^{'} \in D ((T^{'})^{*})$ , put $z : = U^{- 1} y^{'} \in H$ , $w : = (T^{'})^{*} y^{'} \in K$ and $v : = U^{- 1} w \in H$ . For every $x \in D (T)$ , $⟨ T x, z ⟩_{H} = ⟨ U T x, y^{'} ⟩_{K} = ⟨ T^{'} U x, y^{'} ⟩_{K} = ⟨ U x, w ⟩_{K} = ⟨ x, v ⟩_{H}$ by the isometry of $U^{- 1}$ ; hence the functional $x \mapsto ⟨ T x, z ⟩_{H}$ is bounded on $(D (T), ∥ \cdot ∥_{H})$ by $∥ v ∥_{H} ∥ x ∥_{H}$ , so $z \in D (T^{*}) = D (T)$ and $T z = v$ by [F4]. Therefore $y^{'} = U z \in U D (T) = D (T^{'})$ and $(T^{'})^{*} y^{'} = w = U v = U T z = T^{'} y^{'}$ ; thus $D ((T^{'})^{*}) \subseteq D (T^{'})$ , and with step 2.2 and [F4] the operator $T^{'}$ is self-adjoint.

4.1step 3.1F1F5

[ $E$ represents $T^{'}$ ] Let $y \in D (T^{'})$ , that is $x : = U^{- 1} y \in D (T)$ . By [F1] applied to $T$ at $x$ , $\int λ^{2} d E_{T, x} < \infty$ and $\int λ d E_{T} (λ) x = T x$ ; by step 3.1 with $y = z$ , $E_{y} = E_{T, x}$ , so $\int λ^{2} d E_{y} < \infty$ and $\int λ d E_{y} = \int λ d E_{T, x}$ . Conversely, if $\int λ^{2} d E_{y} < \infty$ for some $y \in K$ , then step 3.1 gives $\int λ^{2} d E_{T, U^{- 1} y} < \infty$ , so $U^{- 1} y \in D (T)$ by [F1] and $y \in U D (T) = D (T^{'})$ . Thus $D (T^{'}) = {y \in K : \int λ^{2} d E_{y} (λ) < \infty}$ , the exact domain of the unbounded integral $S : = \int λ d E (λ)$ against the projection-valued measure $E$ . For $y \in D (T^{'})$ and every $z \in K$ , $⟨ S y, z ⟩_{K} = \int λ d E_{y, z} (λ) = \int λ d E_{T, U^{- 1} y, U^{- 1} z} (λ) = ⟨ T U^{- 1} y, U^{- 1} z ⟩_{H} = ⟨ T^{'} y, z ⟩_{K}$ , the middle equality by [F1] applied to $T$ at $U^{- 1} y \in D (T)$ ; since the pairings separate points of $K$ , $S y = T^{'} y$ . By the converse clause of [F1] applied to $S = T^{'}$ and the regular projection-valued measure $E$ , the measure $E$ is the spectral projection-valued measure of $T^{'}$ .

4.2step 2.1step 3.1F2F5

[Clause (c)] For $x \in H$ , $E (B) U x = U E_{T} (B) U^{- 1} U x = U E_{T} (B) x$ , so $∥ E (B) U x ∥_{K}^{2} = ∥ E_{T} (B) x ∥_{H}^{2}$ by the isometry of $U$ ; for unit $x$ this is the asserted identity of scalar laws, both sides being the values at $B$ of the positive measures $E_{U x}$ and $E_{T, x}$ of [F2].

5.1step 4.1F3∎

[Clause (d)] By [F3] applied to $T$ , $σ (T)$ is the support of $E_{T}$ ; by step 4.1 the measure $E$ is the spectral projection-valued measure of $T^{'}$ , so [F3] applied to $T^{'}$ shows that $σ (T^{'})$ is the support of $E$ . For every Borel $B$ , $E (B) = 0$ if and only if $E_{T} (B) = 0$ , because $U$ and $U^{- 1}$ are bijective; hence $E$ and $E_{T}$ have the same null sets and the same nonempty relatively open subsets on which they fail to vanish, so their supports -- and therefore $σ (T^{'})$ and $σ (T)$ -- coincide.

Transport of spectral projection-valued measures along isometric bijections

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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