Proof of Young's convolution inequality #231
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YaelDillies
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Feb 10, 2025
fpvandoorn
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Feb 10, 2025
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When I tried to write the
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YaelDillies
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Feb 13, 2025
| [μ.IsAddLeftInvariant] {p q : ENNReal} (hpq : p.IsConjExponent q) | ||
| (hL : ∀ (x y : G), ‖L (f x) (g y)‖ ≤ ‖f x‖ * ‖g y‖) | ||
| (hf : AEStronglyMeasurable f μ) (hg : AEStronglyMeasurable g μ) (x₀ : G) : | ||
| ∫⁻ (a : G), ‖(L (f a)) (g (x₀ - a))‖ₑ ∂μ ≤ eLpNorm f p μ * eLpNorm g q μ := by |
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| ∫⁻ (a : G), ‖(L (f a)) (g (x₀ - a))‖ₑ ∂μ ≤ eLpNorm f p μ * eLpNorm g q μ := by | |
| ∫⁻ a, ‖L (f a) (g (x₀ - a))‖ₑ ∂μ ≤ eLpNorm f p μ * eLpNorm g q μ := by |
| /-- A generalization of Young's convolution inequality for the `ℒr` seminorm of a convolution | ||
| `(f ⋆[L, μ] g)`, which applies for any `L`. -/ | ||
| theorem eLpNorm_convolution_le'' {p q r : ℝ≥0∞} | ||
| (hp : p ≥ 1) (hq : q ≥ 1) (hr : r ≥ 1) (hpqr : 1 / p + 1 / q = 1 / r + 1) |
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| (hp : p ≥ 1) (hq : q ≥ 1) (hr : r ≥ 1) (hpqr : 1 / p + 1 / q = 1 / r + 1) | |
| (hp : 1 ≤ p) (hq : 1 ≤ q) (hr : 1 ≤ r) (hpqr : p⁻¹ + q⁻¹ = r⁻¹ + 1) |
and everywhere else
| replace hL : ∀ (x y : G), ‖L (f x) (g y)‖ ≤ 1 * ‖f x‖ * ‖g y‖ := by simpa using hL | ||
| simpa using eLpNorm_convolution_le hp hq hr hpqr hf hg hL | ||
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| /-- A generalization of Young's convolution inequality for the `ℒr` seminorm of a convolution |
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| /-- A generalization of Young's convolution inequality for the `ℒr` seminorm of a convolution | |
| /-- A generalization of **Young's convolution** inequality for the `ℒr` seminorm of a convolution |
| /-- A generalization of Young's convolution inequality that allows an arbitrary `L` as long as | ||
| a bound on the size of `L` (on the ranges of `f` and `g`) is known. See also | ||
| `eLpNorm_convolution_le''`, which is stated similarly in terms of `‖L‖ₑ`. -/ | ||
| theorem eLpNorm_convolution_le {p q r : ℝ≥0∞} |
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| theorem eLpNorm_convolution_le {p q r : ℝ≥0∞} | |
| theorem eLpNorm_convolution_le_of_norm_le_mul {p q r : ℝ≥0∞} |
| /-- **Young's convolution inequality**: the `ℒr` seminorm of a convolution `(f ⋆[L, μ] g)` is | ||
| bounded by the product of the `ℒp` and `ℒq` seminorms, where `1 / p + 1 / q = 1 / r + 1` and | ||
| `‖L‖ₑ ≤ 1`. This includes the standard form of the inequality, in which `L` is multiplication. -/ | ||
| theorem eLpNorm_convolution_le' {p q r : ℝ≥0∞} |
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| theorem eLpNorm_convolution_le' {p q r : ℝ≥0∞} | |
| theorem eLpNorm_convolution_le_of_norm_le {p q r : ℝ≥0∞} |
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| /-- A generalization of Young's convolution inequality for the `ℒr` seminorm of a convolution | ||
| `(f ⋆[L, μ] g)`, which applies for any `L`. -/ | ||
| theorem eLpNorm_convolution_le'' {p q r : ℝ≥0∞} |
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| theorem eLpNorm_convolution_le'' {p q r : ℝ≥0∞} | |
| theorem eLpNorm_convolution_le_enorm_mul {p q r : ℝ≥0∞} |
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Proof of Young's convolution inequality.
AEMeasurable.lintegral_prod_right'by analogy toMeasureTheory.AEStronglyMeasurable.integral_prod_right'. I noticed that the latter theorem has noleftversion. Was that an intentional decision, or should we add some left theorems?