The Vitali covering estimate to Young's inequality for integral…
TheoremThe Vitali covering estimate#
If every point of S is the centre of a ball of radius at
most R on which ∫⁻ g is at least t times the measure of the ball, then
t * μ S ≤ 4 ^ n * ∫⁻ g, with n the dimension of the space.
The factor 4 ^ n is the cost of passing from a pairwise disjoint subfamily of those balls to its
fourfold enlargement, which still covers S.
TauCeti.mul_measure_le_of_forall_exists_ball.{u_1} {E : Type u_1} [NormedAddCommGroup E]
[NormedSpace ℝ E] [MeasurableSpace E] [BorelSpace E] [FiniteDimensional ℝ E]
{μ : MeasureTheory.Measure E} [μ.IsAddHaarMeasure] {t : ENNReal} {S : Set E} {R : ℝ}
{g : E → ENNReal}
(hS :
∀ x ∈ S, ∃ r ∈ Set.Ioc 0 R, t * μ (Metric.ball x r) ≤ ∫⁻ (y : E) in Metric.ball x r, g y ∂μ) :
t * μ S ≤ 4 ^ Module.finrank ℝ E * ∫⁻ (y : E), g y ∂μ
Used by 1
TheoremWolff's lemma (TauCeti.exists_circleLIntegral_sq_lt)#
If 2 π times the plane integral of g ^ 2 is smaller than
c * log (R / r), then some circle of radius ρ strictly between r and R has
circleLIntegral g ζ ρ ^ 2 < c.
Since log (R / r) grows without bound as the annulus is made longer while the plane integral
stays fixed, this makes the circle integral arbitrarily small at arbitrarily small radii: a weight
of finite square integral cannot have a large circle integral on every circle about ζ.
TauCeti.exists_circleLIntegral_sq_lt {g : ℂ → ENNReal} (hg : Measurable g) (ζ : ℂ) {r R : ℝ}
(hr : 0 < r) (hrR : r < R) {c : ENNReal}
(hc :
ENNReal.ofReal (2 * Real.pi) * ∫⁻ (z : ℂ), g z ^ 2 < c * ENNReal.ofReal (Real.log (R / r))) :
∃ ρ ∈ Set.Ioo r R, TauCeti.circleLIntegral g ζ ρ ^ 2 < c
Lean
Formalized in Tau Ceti as TauCeti.exists_circleLIntegral_sq_lt.
Other forms
Wolff's lemma in the limit: a weight of finite square integral has small circle integrals at arbitrarily small radii. Formalized in Tau Ceti as TauCeti.exists_circleLIntegral_lt_of_lintegral_sq_ne_top.
TauCeti.exists_circleLIntegral_lt_of_lintegral_sq_ne_top {g : ℂ → ENNReal} (hg : Measurable g)
(hfin : ∫⁻ (z : ℂ), g z ^ 2 ≠ ⊤) (ζ : ℂ) {c : ENNReal} (hc : c ≠ 0) {R : ℝ} (hR : 0 < R) :
∃ ρ ∈ Set.Ioo 0 R, TauCeti.circleLIntegral g ζ ρ < c
Sources
TheoremYoung's inequality for integral kernels#
Let 1 ≤ p ≤ q and let r be the exponent with
1/p + 1/r = 1 + 1/q. If the r-th powers of the kernel k have row integrals
∫⁻ y, k x y ^ r ∂ν at most A for almost every x and column integrals ∫⁻ x, k x y ^ r ∂μ
at most B for almost every y, then the integral operator with kernel k satisfies
∫⁻ x, (∫⁻ y, k x y * g y ∂ν) ^ q ∂μ ≤ A ^ (q (1 - 1/p)) * B * (∫⁻ y, g y ^ p ∂ν) ^ (q / p).
The kernel need only be a.e. measurable: the proof applies the measurable case to a measurable
modification of k, which agrees with k on almost every row and on almost every column.
For p = q this is Schur's test, TauCeti.lintegral_rpow_lintegral_mul_le.
TauCeti.lintegral_rpow_lintegral_mul_le_of_inv_add_inv_eq.{u_1, u_2} {α : Type u_1} {β : Type u_2}
[MeasurableSpace α] [MeasurableSpace β] {μ : MeasureTheory.Measure α}
{ν : MeasureTheory.Measure β} [MeasureTheory.SFinite μ] [MeasureTheory.SFinite ν]
{k : α → β → ENNReal} (hk : AEMeasurable (Function.uncurry k) (μ.prod ν)) {g : β → ENNReal}
(hg : AEMeasurable g ν) {p q r : ℝ} (hp : 1 ≤ p) (hpq : p ≤ q) (hr : p⁻¹ + r⁻¹ = 1 + q⁻¹)
{A B : ENNReal} (hA : ∀ᵐ (x : α) ∂μ, ∫⁻ (y : β), k x y ^ r ∂ν ≤ A)
(hB : ∀ᵐ (y : β) ∂ν, ∫⁻ (x : α), k x y ^ r ∂μ ≤ B) :
∫⁻ (x : α), (∫⁻ (y : β), k x y * g y ∂ν) ^ q ∂μ ≤
A ^ (q * (1 - p⁻¹)) * B * (∫⁻ (y : β), g y ^ p ∂ν) ^ (q / p)
Lean
Formalized in Tau Ceti as TauCeti.lintegral_rpow_lintegral_mul_le_of_inv_add_inv_eq.
Sources
Uses 1 · Used by 2
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