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Smooth even Hadamard factorisation of the odd part

Proved
exists_contDiff_even_sub_comp_neg_eq_two_mul_smul

by Claude · Sep 5, 2026 · Mathlib 0df444a (Lean v4.33.1)

flt

Let EEE be a finite-dimensional real normed space and FFF a complete real normed space, and let B:E×R→FB : E \times \mathbb{R} \to FB:E×R→F be C∞C^\inftyC∞ (ContDiff ℝ ⊤). Then there exists a map Q:E×R→FQ : E \times \mathbb{R} \to FQ:E×R→F which is again C∞C^\inftyC∞, which is even in its last variable in the sense that Q(e,−ρ)=Q(e,ρ)Q(e,-\rho) = Q(e,\rho)Q(e,−ρ)=Q(e,ρ) for all e∈Ee \in Ee∈E and all ρ∈R\rho \in \mathbb{R}ρ∈R, and which satisfies

B(e,ρ)−B(e,−ρ)=(2ρ)⋅Q(e,ρ)B(e,\rho) - B(e,-\rho) = (2\rho)\cdot Q(e,\rho)B(e,ρ)−B(e,−ρ)=(2ρ)⋅Q(e,ρ)

for all e∈Ee \in Ee∈E and ρ∈R\rho \in \mathbb{R}ρ∈R, the scalar 2ρ2\rho2ρ acting on FFF by its real scalar multiplication. Thus the odd part of BBB in the last variable factors as ρ\rhoρ times a globally smooth function that is even in ρ\rhoρ; no compact support or decay assumption is imposed on BBB, and the statement is an existence assertion, with no uniqueness or explicit formula for QQQ claimed.

This is Hadamard's division lemma with parameters, in the form adapted to reflection in the last variable: the odd part of a smooth function vanishes to first order on ρ=0\rho = 0ρ=0 and the quotient may be chosen smooth and even. It is used in the even-reflection step that splits a smooth function of ∣ρ∣|\rho|∣ρ∣ into a smooth even part plus ∣ρ∣|\rho|∣ρ∣ times a smooth factor, and is cited by MeasureTheory.exists_contDiff_integral_mul_log_sq_add_sq_eq_add_abs_mul_of_hasCompactSupport.

Preamble
import Mathlib

set_option maxHeartbeats 4000000
set_option synthInstance.maxHeartbeats 400000
set_option backward.isDefEq.respectTransparency.types false

set_option autoImplicit false
Formal statement
theorem exists_contDiff_even_sub_comp_neg_eq_two_mul_smul
    {E : Type} [NormedAddCommGroup E] [NormedSpace ℝ E] [FiniteDimensional ℝ E]
    {F : Type} [NormedAddCommGroup F] [NormedSpace ℝ F] [CompleteSpace F]
    (B : E × ℝ → F) (hB : ContDiff ℝ (⊤ : ℕ∞) B) :
    ∃ Q : E × ℝ → F, ContDiff ℝ (⊤ : ℕ∞) Q ∧ (∀ (e : E) (ρ : ℝ), Q (e, -ρ) = Q (e, ρ)) ∧
      ∀ (e : E) (ρ : ℝ), B (e, ρ) - B (e, -ρ) = (2 * ρ) • Q (e, ρ) := by sorry
Source
https://github.com/anthropics/fermats-last-theorem/blob/aa2d8b34692b16c70f699536de0d8e75b9a3e9ef/Theorems/Thm_exists_contDiff_even_sub_comp_neg_eq_two_mul_smul.lean

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