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Spectral analyticity on the modular strip

Statement

Assume the Axiom of Choice. Let $H$ be a complex Hilbert space and let $A$ be a self-adjoint operator whose spectral projection valued measure $E$ is supported on $(0, \infty)$ . For $ξ \in H$ , put $μ_{ξ} (B) = ⟨ E (B) ξ, ξ ⟩$ . Define $D (A^{1 / 2}) = {ξ \in H : \int_{(0, \infty)} λ d μ_{ξ} (λ) < \infty} .$ For $z \in S = {z \in C : 0 \leq Im z \leq 1}$ and $ξ \in D (A^{1 / 2})$ , define $A^{- i z / 2} ξ$ by the unbounded Borel functional calculus for the function $λ \mapsto \exp (- i z \log (λ) / 2)$ . This vector is defined, strongly continuous in $z$ on $S$ , and satisfies $∥ A^{- i z / 2} ξ ∥^{2} \leq ∥ ξ ∥^{2} + ∥ A^{1 / 2} ξ ∥^{2} .$ The map $z \mapsto A^{- i z / 2} ξ$ is norm-holomorphic on the open strip $S^{\circ}$ .

Statement references

Unbounded linear operators: domain, graph and extension

Symmetric, self-adjoint and essentially self-adjoint operators

Spectral theorem for unbounded self-adjoint operators (PVM form)

Unbounded Borel functional calculus: domains, products, spectral mapping

Dominated convergence

Facts & Assumptions

Given: The Axiom of Choice, a complex Hilbert space $H$ , a self-adjoint $A$ with spectral measure $E$ supported on $(0, \infty)$ , and $ξ \in D (A^{1 / 2})$ .

[A1]

Assume the Axiom of Choice (The Axiom of Choice).

[F1]

On nonzero $H$ , the unbounded self-adjoint spectral theorem supplies $E$ ; on $H = {0}$ use the zero PVM. The scalar spectral measure is $μ_{ξ} (B) = ⟨ E (B) ξ, ξ ⟩$ , with total mass $∥ ξ ∥^{2}$ (Spectral theorem for unbounded self-adjoint operators (PVM form), Integral of a measurable function against a projection-valued measure).

[F2]

Under Countable Choice, the truncation-defined PVM integral has domain $D (f (E)) = {η : \int ∣ f ∣^{2} d μ_{η} < \infty}$ and satisfies $∥ f (E) η ∥^{2} = \int ∣ f ∣^{2} d μ_{η}$ on this domain, including on the zero space (The unbounded PVM integral is densely defined, closed and normal). The assumed Choice supplies Countable Choice by restricting the families to countable ones. Finite linear combinations of multipliers agree with the corresponding combinations of their integrals on the common domain, by the sum and product clauses of Unbounded Borel functional calculus: domains, products, spectral mapping (apply the product clause to constants for scalar multiples). Multipliers on $(0, \infty)$ are extended by zero on its $E$ -null complement.

[F3]

Dominated convergence applies on an arbitrary measure space when one integrable majorant dominates the sequence (Dominated convergence).

Proof

Given: [A1], [F1]–[F3] and the Statement data.

1.1A1F1F2given

Write $z = t + i s$ with $0 \leq s \leq 1$ . The spectral multiplier $f_{z} (λ) = \exp (- i z \log (λ) / 2)$ has $∣ f_{z} (λ) ∣^{2} = λ^{s}$ . Since $λ^{s} \leq 1 + λ$ for every $λ > 0$ and $s \in [0, 1]$ , [F2] gives $∥ A^{- i z / 2} ξ ∥^{2} = \int λ^{s} d μ_{ξ} \leq \int (1 + λ) d μ_{ξ} = ∥ ξ ∥^{2} + ∥ A^{1 / 2} ξ ∥^{2} .$ This also proves that every vector named in the Statement is in the domain of the indicated multiplier.

2.1F1F2F3step 1.1

If $z_{n} = t_{n} + i s_{n} \to z = t + i s$ in $S$ , then $f_{z_{n}} (λ) \to f_{z} (λ)$ pointwise and $∣ f_{z_{n}} (λ) - f_{z} (λ) ∣^{2} \leq 2 λ^{s_{n}} + 2 λ^{s} \leq 4 (1 + λ) .$ The majorant is integrable by $ξ \in D (A^{1 / 2})$ . Dominated convergence applied to the scalar spectral measure therefore gives $∥ A^{- i z_{n} / 2} ξ - A^{- i z / 2} ξ ∥^{2} \to 0$ , proving strong continuity on the closed strip.

2.2F1F2F3step 1.1

Fix $z_{0}$ with $s_{0} = Im z_{0} \in (0, 1)$ . Choose $δ > 0$ so that $2 δ < s_{0} < 1 - 2 δ$ . For $∣ h ∣ \leq δ$ , the scalar difference quotient of $f_{z} (λ)$ along the segment from $z_{0}$ to $z_{0} + h$ is bounded by $\frac{1}{2} ∣ \log λ ∣ \sup_{0 \leq r \leq 1} λ^{Im (z_{0} + r h) / 2} .$ When $0 < λ \leq 1$ this is bounded by $\frac{1}{2} ∣ \log λ ∣ λ^{δ / 2}$ , and when $λ \geq 1$ it is bounded by $\frac{1}{2} ∣ \log λ ∣ λ^{(1 - δ) / 2}$ . The first expression has a finite supremum and the second is bounded by a constant depending on $δ$ times $λ^{1 / 2}$ . Thus the difference quotients are dominated in $L^{2} (μ_{ξ})$ by a constant times $1 + λ^{1 / 2}$ . Their pointwise limit is $(- i / 2) (\log λ) f_{z_{0}} (λ)$ . Dominated convergence in the spectral norm proves that the vector difference quotient converges in $H$ to the spectral integral of this limit. Hence the vector map is complex-differentiable at $z_{0}$ .

3.1step 1.1step 2.1step 2.2∎

Step 2.2 applies at every point of $S^{\circ}$ , so the vector map is norm-holomorphic throughout the open strip. The estimate and closed-strip continuity were proved in steps 1.1 and 2.1.

Spectral analyticity on the modular strip

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources