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A derivative truncation needs its energy band

Physical scope

Pure mathematics; no physical premises or physical claim.

Statement

Assume AC. Let $d p$ denote Lebesgue measure on the momentum line $R$ , and let $H = L^{2} (R, d p; C)$ be the space of equivalence classes modulo Lebesgue-a.e. equality of complex measurable functions $u$ satisfying $\int_{R} ∣ u (p) ∣^{2} d p < \infty$ , with pairing $⟨ u, v ⟩ = \int_{R} u (p) \overline{v (p)} d p$ , linear in its first variable, and induced norm $∥ u ∥_{2} = (\int_{R} ∣ u (p) ∣^{2} d p)^{1 / 2}$ . Define the multiplication operator $A$ by $A = M_{p^{2}}, D (A) = {u \in H : p^{2} u \in H},$ with $(A u) (p) = p^{2} u (p)$ almost everywhere. For each integer $r \geq 1$ , let $A^{r}$ be multiplication by $p^{2 r}$ on $D (A^{r}) : = {u \in H : p^{2 r} u \in H}$ ; set $A^{0} = I$ with $D (A^{0}) = H$ . Fix $M > 0$ , $0 \leq E < M$ , and let $N$ be a nonnegative integer. This is the no-band case for the low-energy remainder estimate: the polynomial approximation to $(M^{2} + p^{2})^{- 1}$ has multiplier remainder $r_{N} (p^{2}) = M^{- 2} \frac{(- p^{2} / M^{2})^{N + 1}}{1 + p^{2} / M^{2}} .$ Let $r_{N} (A)$ denote multiplication by $r_{N} (p^{2})$ on the maximal domain $D (r_{N} (A)) : = {u \in H : r_{N} (p^{2}) u \in H}$ . There is a unit vector $u \in ⋂_{r = 0}^{N + 1} D (A^{r})$ for which $∥ r_{N} (A) u ∥ > M^{- 2} (E^{2} / M^{2})^{N + 1} .$ Thus the low-energy factor in the remainder estimate cannot hold uniformly on the full momentum spectrum. Here $p, M, E$ have mass units, $d p$ has mass units, $A$ has mass-squared units, and $ℏ = c = 1$ . This is a mathematical counterexample to removing the spectral-band hypothesis; it is not a measured physical failure.

Statement references

Low-momentum matching has a controlled remainder

Facts & Assumptions

Given: The complex $L^{2}$ space, the multiplier operator $A$ , $M > 0$ , $0 \leq E < M$ , and $N \geq 0$ .

[A1]

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

[A2]

AC supplies countable selections and prescribed serial paths: AC implies Countable Choice, which is also required by the multiplication and L2 Hilbert suppliers.

[F1]

Low-momentum matching has a controlled remainder: its estimate assumes $J$ belongs to the spectral subspace $E_{A} ([0, E^{2}]) H$ .

[F2]

Spectral measure of a real L2 multiplication operator: for real measurable multiplier $m$ , $M_{m}$ on $D (M_{m}) = {u : m u \in L^{2}}$ is self-adjoint and its spectral projections are multiplication by indicators of $m^{- 1} (B)$ .

[F3]

The space $L^{p} (μ)$ as the quotient by null functions: $L^{2}$ consists of measurable complex functions modulo equality almost everywhere.

[F4]

$L^{2}$ with the integral pairing is a Hilbert space: on complex $L^{2}$ , $⟨ f, g ⟩ = \int f \overline{g}$ is a Hilbert-space inner product with $∥ f ∥_{2}^{2} = \int ∣ f ∣^{2}$ .

Proof

technique · direct multiplier calculation and compactly supported unit vectors

Given: The hypotheses and notation in the Statement.

1.1A1A2F2F3F4givenalgebra

By [A1]–[A2], the AC and Countable Choice assumptions of [F2] and [F4] hold. Apply [F2] with $m (p) = p^{2}$ on the sigma-finite positive-measure space $(R, d p)$ ; sigma-finiteness follows from $R = ⋃_{n \geq 1} [- n, n]$ and $[- 1, 1]$ has positive finite measure. It gives the self-adjoint nonnegative multiplication operator $A$ with the displayed domain and spectral calculus. The multiplier $(M^{2} + p^{2})^{- 1}$ is bounded by $M^{- 2}$ and maps $H$ into $D (A)$ because $p^{2} / (M^{2} + p^{2}) \leq 1$ ; pointwise multiplication then shows it is the two-sided inverse of $M^{2} I + A$ . For $u$ with bounded support, every $A^{r} u$ is in $H$ . Pointwise finite geometric division gives

(M^{2} + p^{2})^{- 1} - M^{- 2} \sum_{r = 0}^{N} (- p^{2} / M^{2})^{r} = r_{N} (p^{2}),

so the remainder on such $u$ is multiplication by $r_{N} (p^{2})$ .

2.1A1F1F2F3F4step 1.1algebra∎

The target coefficient $C_{N} : = M^{- 2} (E^{2} / M^{2})^{N + 1}$ is finite and is strictly less than $M^{- 2}$ . If $N = 0$ , $∣ r_{0} (q) ∣ = M^{- 2} (q / M^{2}) / (1 + q / M^{2}) \to M^{- 2} > C_{0}$ as $q \to \infty$ . If $N \geq 1$ , $∣ r_{N} (q) ∣ = M^{- 2} (q / M^{2})^{N + 1} / (1 + q / M^{2}) \to \infty$ . In either case choose $P > 0$ so that $∣ r_{N} (p^{2}) ∣ > C_{N}$ for every $p \in [P, P + M]$ , and put $u = M^{- 1 / 2} 1_{[P, P + M]}$ . The interval has Lebesgue measure $M$ , so $∥ u ∥_{2}^{2} = M^{- 1} \int_{P}^{P + M} 1 d p = 1$ ; its bounded support puts it in every $D (A^{r})$ , while $∥ r_{N} (A) u ∥_{2}^{2} = M^{- 1} \int_{P}^{P + M} ∣ r_{N} (p^{2}) ∣^{2} d p > C_{N}^{2} .$ Thus the full-space bound fails. The restriction $E < M$ is precisely what makes the proposed low-energy coefficient smaller than the $N = 0$ high-momentum limit; the example does not contradict the band-limited theorem.

A derivative truncation needs its energy band

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources