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Relative variation of a cubic spectral jump factor

Statement

Let $A, Ω > 0$ , $0 \leq r < 1$ , and $I = [Ω (1 - r), Ω (1 + r)]$ . Let $J : I \to [0, \infty)$ be a pointwise spectral representative satisfying $J (ω) = A ω^{3}$ for every $ω \in I$ . Define $b (ω) = \sqrt{J (ω)} = \sqrt{A} ω^{3 / 2}$ using the nonnegative square root. Then $b (Ω) > 0$ and, for every real $s$ with $∣ s ∣ \leq r Ω$ , $∣ b (Ω + s) / b (Ω) - 1 ∣ \leq (3 + 3 r + r^{2}) ∣ s ∣ / Ω$ .

For any fixed complex scalar $c$ , define $q_{c} (ω) = c b (ω)$ on $I$ . If $c = 0$ , then $q_{c} (ω) = 0$ throughout $I$ . If $c \neq 0$ , then $q_{c} (Ω) \neq 0$ and $∣ q_{c} (Ω + s) / q_{c} (Ω) - 1 ∣$ obeys the same bound. A consuming physical model may use its fixed dipole matrix element as $c$ ; no dipole operator is assumed here.

For the Lebesgue-measure assertions in this paragraph assume the Axiom of Countable Choice. An identity $J (ω) = A ω^{3}$ only almost everywhere on $I$ , with respect to Lebesgue measure, does not determine the value of a chosen representative at $Ω$ and does not imply the pointwise premise. Likewise, define an integrable regulator here to mean a measurable function $χ : R \to [0, \infty)$ with $\int_{R} χ (ω) d ω < \infty$ , used to form $J_{χ} (ω) = A ω^{3} χ (ω)$ for $ω > 0$ and $J_{χ} (ω) = 0$ for $ω \leq 0$ . Integrability of $χ$ alone implies neither $J_{χ} (ω) = A ω^{3}$ throughout $I$ nor $J_{χ} (Ω) > 0$ .

Facts & Assumptions

[F1]

The spectral-kernel contract uses Lebesgue integration and pointwise positive spectral square roots, with integral regularity imposed separately. Stationary correlation spectra and ULE kernel regularity

[F2]

Write $λ$ for Lebesgue measure on $R$ and $L (R)$ for its measurable sets. Under Countable Choice, $L (R)$ is a sigma-algebra and $λ$ is a complete measure on it. Lebesgue measurable sets, the family L(ℝⁿ), and the restricted set function λₙ Assuming countable choice, L(ℝⁿ) is a sigma-algebra containing every elementary set and λₙ is a complete measure extending elementary volume

[F3]

Measurability of an extended-real-valued function is equivalent to measurability of every strict superlevel set. A property holds almost everywhere if its exceptions lie in a measurable null set. Nonnegative integrals are suprema of simple minorant integrals; a nonnegative real measurable function is integrable exactly when that integral is finite. Extended-real-valued measurable functions Measure-null sets and almost-everywhere statements relative to a measure The nonnegative Lebesgue integral Integrable real and complex functions, and their integrals

[F4]

Under Countable Choice, every bounded interval with endpoints $a \leq b$ is Lebesgue measurable with measure $b - a$ , whatever its endpoint inclusions. In particular $λ ({a}) = 0$ . This is the one-dimensional case of the box theorem, including degenerate boxes. A box in ℝⁿ with parameters aᵢ≤ bᵢ is Lebesgue measurable of measure ∏_i<n(bᵢ-aᵢ), whichever of its faces are included

[F5]

For measurable $E$ , write $1_{E}$ for its indicator. It is a nonnegative simple measurable function; its simple integral is $λ (E)$ and agrees with its nonnegative Lebesgue integral. Nonnegative simple measurable functions The integral of a nonnegative simple function The nonnegative integral agrees with the simple integral on simple functions

[F6]

On any measure space, the nonnegative integral is monotone and homogeneous for positive finite scalars. Monotonicity and nonnegative homogeneity of the nonnegative integral

[F7]

If two integrable real or complex functions agree almost everywhere, their integrals over every measurable set agree. Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree

[A1]

Assume Countable Choice for the Lebesgue-measure qualifications and counterexamples, as required by [F2] and [F4]. It asserts that for every sequence of nonempty sets there is a function choosing one member of each set. No choice assumption is used in the pointwise cubic bound or scalar cancellation. The Axiom of Countable Choice (AC_ω)

Proof

Given: The parameters, pointwise cubic law, scalar coefficient and regulator convention in the Statement; Countable Choice for the Lebesgue-measure qualifications only. Goal: Prove the variation bound and zero preservation, and show why almost-everywhere data or regulator integrability alone do not supply the pointwise premise.

1.1givenF1

Put $x = s / Ω$ . The frequency stays positive because $∣ x ∣ \leq r < 1$ , and $b (Ω) = \sqrt{A} Ω^{3 / 2} > 0$ . Dividing positive square roots gives $b (Ω + s) / b (Ω) = (1 + x)^{3 / 2}$ . Rationalizing gives $(1 + x)^{3 / 2} - 1 = x (3 + 3 x + x^{2}) / ((1 + x)^{3 / 2} + 1)$ .

1.2givenA1F2F3F4F5F6F7

For the weaker almost-everywhere premise, use [A1] and let $J_{0} : R \to [0, \infty)$ equal $A ω^{3}$ on $I$ and zero off $I$ . Let $J_{1}$ equal $J_{0}$ except that $J_{1} (Ω) = 0$ . For a negative threshold both strict superlevel sets are $R$ . For a nonnegative threshold the superlevel set of $J_{0}$ is a bounded interval or empty, because the cubic is strictly increasing on the positive interval $I$ ; the superlevel set of $J_{1}$ is obtained by removing at most ${Ω}$ . Thus [F2]–[F4] establish measurability of both functions. Put $M = A [Ω (1 + r)]^{3} > 0$ . Since $0 \leq J_{i} \leq M 1_{I}$ , [F4]–[F6] give $\int_{R} J_{i} d λ \leq M λ (I) = 2 M r Ω < \infty$ for $i = 0, 1$ , so both functions are integrable by [F3]. Their exceptional set is contained in the measurable null singleton ${Ω}$ by [F4], so they agree almost everywhere by [F3]. Consequently [F7] gives identical integrals over every Lebesgue measurable set. Yet $J_{1}$ is cubic almost everywhere on $I$ and $J_{1} (Ω) = 0 \neq A Ω^{3}$ . This also covers $r = 0$ , when $I = {Ω}$ and both integrals vanish. Integral data therefore cannot determine the differing reference values.

2.1givenstep 1.1

The denominator is at least one, and $∣ 3 + 3 x + x^{2} ∣ \leq 3 + 3 r + r^{2}$ . Taking absolute values proves the claimed bound. If $c = 0$ , the defining product $q_{c} = c b$ vanishes at every point. If $c \neq 0$ , positivity of $b (Ω)$ gives $q_{c} (Ω) \neq 0$ , and cancellation gives $q_{c} (Ω + s) / q_{c} (Ω) = b (Ω + s) / b (Ω)$ .

3.1givenA1F3F4F5step 2.1step 1.2∎

For the regulator claim, use [A1] and take $χ = 1_{[0, Ω / 2]}$ . The interval is measurable by [F4], and [F5] makes $χ$ a nonnegative measurable function with $\int_{R} χ d λ = λ ([0, Ω / 2]) = Ω / 2 < \infty$ . Thus it is an integrable regulator by [F3]. Since $Ω > 0$ , we have $χ (Ω) = 0$ . The defined regulated spectrum therefore satisfies $J_{χ} (Ω) = 0 \neq A Ω^{3}$ , disproving both proposed implications from regulator integrability alone. These counterexamples concern only the weaker data, not the pointwise hypothesis used for the bound.

Depends on

Used by · 1

Small-lift amplitude approximation and energy-scale scope

Dependency tree · 2 levels

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