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Books.ConvexAnalysis_Rockafellar_1970.Chap06.section31_part14

theorem helperForLemma_31_0_12_counterexample_dualSup_eq_zero :
⨆ (xStar : Fin 1), fenchelConjugate 1 (fun (x : Fin 1) => ↑(x 0 ^ 2)) xStar - fenchelConjugate 1 (fun (x : Fin 1) => 0) xStar = 0

Helper for Lemma 31.0.12: in the one-dimensional counterexample, the left-hand dual supremum is exactly 0. The origin contributes 0, while every nonzero dual vector contributes because the conjugate of the constant-zero function is the singleton indicator at the origin.

Helper for Lemma 31.0.12: any proof of the current theorem schema specializes to the explicit counterexample and forces the impossible identity 0 = ⊥.

theorem helperForLemma_31_0_12_counterexampleLeadingEqualityFalse :
¬⨆ (xStar : Fin 1), fenchelConjugate 1 (fun (x : Fin 1) => ↑(x 0 ^ 2)) xStar - fenchelConjugate 1 (fun (x : Fin 1) => 0) xStar = Filter.liminf (translatedDifferenceValueFunction (fun (x : Fin 1) => 0) fun (x : Fin 1) => ↑(x 0 ^ 2)) (nhds 0)

Helper for Lemma 31.0.12: the one-dimensional quadratic/zero counterexample already falsifies the leading equality claimed in the target theorem.

theorem helperForLemma_31_0_12_counterexampleTailRelationsHold :
Filter.liminf (translatedDifferenceValueFunction (fun (x : Fin 1) => 0) fun (x : Fin 1) => ↑(x 0 ^ 2)) (nhds 0) translatedDifferenceValueFunction (fun (x : Fin 1) => 0) (fun (x : Fin 1) => ↑(x 0 ^ 2)) 0 translatedDifferenceValueFunction (fun (x : Fin 1) => 0) (fun (x : Fin 1) => ↑(x 0 ^ 2)) 0 = functionInfimumEReal fun (x : Fin 1) => 0 - ↑(x 0 ^ 2)

Helper for Lemma 31.0.12: the one-dimensional quadratic/zero counterexample still satisfies the valid tail inequalities liminf p ≤ p(0) and p(0) = inf_x (f x - g x), so the obstruction is isolated to the leading equality.

theorem helperForLemma_31_0_12_counterexampleFullConclusionFalse :
¬(⨆ (xStar : Fin 1), fenchelConjugate 1 (fun (x : Fin 1) => ↑(x 0 ^ 2)) xStar - fenchelConjugate 1 (fun (x : Fin 1) => 0) xStar = Filter.liminf (translatedDifferenceValueFunction (fun (x : Fin 1) => 0) fun (x : Fin 1) => ↑(x 0 ^ 2)) (nhds 0) Filter.liminf (translatedDifferenceValueFunction (fun (x : Fin 1) => 0) fun (x : Fin 1) => ↑(x 0 ^ 2)) (nhds 0) translatedDifferenceValueFunction (fun (x : Fin 1) => 0) (fun (x : Fin 1) => ↑(x 0 ^ 2)) 0 translatedDifferenceValueFunction (fun (x : Fin 1) => 0) (fun (x : Fin 1) => ↑(x 0 ^ 2)) 0 = functionInfimumEReal fun (x : Fin 1) => 0 - ↑(x 0 ^ 2))

Helper for Lemma 31.0.12: the one-dimensional quadratic/zero counterexample falsifies the entire displayed conclusion, even though the tail inequalities remain valid there.

Helper for Lemma 31.0.12: the explicit quadratic/zero counterexample refutes the current universal theorem schema, so no local proof can exist until the leading equality is repaired.

Helper for Lemma 31.0.12: the direct counterexample to the current theorem header uses the singleton indicator at the origin as the convex function f.

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      Helper for Lemma 31.0.12: the direct counterexample to the current theorem header uses the quadratic function x ↦ (x 0)^2 as g.

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          Helper for Lemma 31.0.12: for every dual vector x⋆, the affine perturbation x ↦ ⟪x, x⋆⟫ - (x 0)^2 is unbounded below on ℝ^1.

          Helper for Lemma 31.0.12: the concave conjugate of the quadratic counterexample is everywhere, because each affine-minus-quadratic slice is unbounded below.

          Helper for Lemma 31.0.12: the Fenchel conjugate of the singleton indicator counterexample is the constant-zero function.

          Helper for Lemma 31.0.12: in the actual current theorem header, the direct indicator/quadratic counterexample has dual supremum .

          Helper for Lemma 31.0.12: for the direct indicator/quadratic counterexample, the translated value function is the negative square u ↦ -(u 0)^2.

          Helper for Lemma 31.0.12: the liminf side of the actual current theorem header evaluates to 0 on the direct indicator/quadratic counterexample.

          Helper for Lemma 31.0.12: the actual current theorem schema is already refuted by the direct indicator/quadratic counterexample, so no proof can exist without repairing the statement upstream.

          Lemma 31.0.12 (Inequality Between the Dual Objective and p(u)): let f : ℝ^n → ℝ ∪ {+∞} be proper convex and let g : ℝ^n → ℝ ∪ {-∞} be proper concave. Assume f and g are closed, and either dom f ∩ dom g ≠ ∅ or dom f⋆ ∩ dom g⋆ ≠ ∅, encoded here by the nonemptiness of effectiveDomainconcaveEffectiveDomain or of effectiveDomain(f⋆) ∩ concaveConjugateEffectiveDomain(g). For the translated value function p(u) = inf_x (f x - g (x + u)), represented by translatedDifferenceValueFunction f g, one has sup_xStar φ(xStar) = liminf_{u → 0} p(u) ≤ p(0) = inf_x (f x - g x), where φ = fenchelDualObjective f g.