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Obstacle complementarity in distribution form

Statement

Assume Countable Choice and the Axiom of Choice (The Axiom of Countable Choice (ACω), The Axiom of Choice). Let Ω⊆Rn be a bounded C1 domain (Bounded C^k domains and boundary charts), let L be a uniformly elliptic divergence-form operator with real L∞ coefficients, and let a:H01(Ω;R)×H01(Ω;R)→R be the symmetric, bounded and coercive real restriction of its associated form (Uniformly elliptic divergence-form operators and their sesquilinear forms, Bounded C^k domains and boundary charts); let f∈L2(Ω;R) and let F∈H−1(Ω;R) be its canonical functional F(v)=∫Ωfv on H01(Ω;R), so in particular F(φ)=∫Ωfφ for test functions (The negative Sobolev space H−1(Ω), Locally integrable functions as regular distributions), let ψ∈H1(Ω;R) with Tψ≤0 (The notation Hk and the reserved zero-boundary symbol), and let u∈K be the obstacle solution of Existence and uniqueness for the obstacle problem. For real φ∈Cc∞(Ω;R) put Λu(φ):=a(u,φ)−∫Ωfφ. Extend Λu complex linearly as in The closed convex obstacle set and the obstacle variational inequality. Then:

  1. Λu is a nonnegative distribution: Λu(φ)≥0 for every nonnegative test function φ (Distribution, Test function space d of an open set).
  2. If in addition Λu is represented by a function ζ∈L2(Ω), that is Λu(φ)=∫Ωζφ for every test function, and if u and ψ have continuous representatives on Ω, then ζ≥0 a.e., ζ=0 a.e. on the open set {u>ψ}, and (u−ψ)ζ=0 a.e. on Ω.

Facts & Assumptions

Given: The obstacle setting above, with the obstacle solution u∈K of Existence and uniqueness for the obstacle problem, the reaction Λu(φ)=a(u,φ)−F(φ) on Cc∞(Ω), and, for the second assertion, a representation Λu(φ)=∫Ωζφ with ζ∈L2(Ω) together with continuous representatives of u and ψ on Ω.

[F1]

The closed convex obstacle set and the obstacle variational inequality, Existence and uniqueness for the obstacle problem: K={v∈H01(Ω):v≥ψ a.e.} is nonempty, and u∈K satisfies a(u,v−u)≥F(v−u) for every v∈K (Zero-boundary Sobolev space as a norm closure); the inequality and the membership are almost-everywhere statements about classes.

[F2]

A function with nonnegative test pairings is nonnegative a.e.: if ζ∈L2(O) and ∫Oζφ≥0 for every nonnegative φ∈Cc∞(O), then ζ≥0 a.e. on O.

[F3]

The fundamental lemma of the calculus of variations: if g∈Lloc1(O) and ∫Ogφ=0 for every φ∈Cc∞(O), then g=0 a.e. on O; in particular L2(O)⊆Lloc1(O) by the finite measure of the bounded domain.

[F4]

Uniformly elliptic divergence-form operators and their sesquilinear forms: the real Dirichlet form a is used only on H01(Ω;R)×H01(Ω;R); under the stated hypotheses it is a bounded, coercive symmetric real bilinear form there, and the associated form on H1(Ω) restricts to it. In particular, both the obstacle solution and each test function lie in the form domain.

[F5]

The negative Sobolev space H−1(Ω): in the real convention, every f∈L2(Ω) defines F(v)=∫Ωfv in H−1(Ω), with ∣F(v)∣≤∥f∥2∥v∥2≤∥f∥2∥v∥H01 and hence ∥F∥H−1≤∥f∥2.

Proof

technique · direct

Given: The setting above, in particular the obstacle solution u∈K and the reaction Λu.

1.1givenF1F4F5

(Nonnegativity) Let φ∈Cc∞(Ω) with φ≥0, and put v:=u+φ. Then v∈H01(Ω;R) because u∈H01(Ω;R) and φ∈Cc∞(Ω;R)⊆H01(Ω;R), and v≥u≥ψ a.e. on Ω, so v∈K [F1]. By [F5], F is the continuous H−1 functional used by the variational inequality; testing that inequality at v gives a(u,φ)≥F(φ), that is Λu(φ)≥0; hence Λu is a nonnegative distribution.

1.2givenF1F4

(Vanishing on the noncontact set) Assume now that Λu(φ)=∫Ωζφ for all test functions and that u,ψ have continuous representatives, and put O:={u>ψ}, an open subset of Ω because the difference of the continuous representatives is continuous and positive exactly on O. Let φ∈Cc∞(O) be arbitrary. If φ=0, then Λu(φ)=0. Otherwise its support Kφ is a nonempty compact subset of O; as u−ψ is continuous and positive there, there is m>0 with u−ψ≥m on Kφ. Choose ε>0 with ε∥φ∥∞<m. Then u±εφ≥u−ε∥φ∥∞≥ψ on Kφ, while on Ω∖Kφ one has u±εφ=u≥ψ a.e.; also u±εφ∈H01(Ω;R). Hence both competitors belong to K [F1]. Testing the variational inequality at them gives εΛu(φ)≥0 and −εΛu(φ)≥0, so Λu(φ)=0; that is, ∫Ωζφ=0 for every φ∈Cc∞(O).

2.1step 1.1F2

For the second assertion, the hypothesis of the representation gives ∫Ωζφ=Λu(φ)≥0 for every nonnegative test function by step 1.1, so [F2] applied with O=Ω yields ζ≥0 a.e. on Ω.

2.2step 1.2F3

Step 1.2 gives ∫Ωζφ=0 for every φ∈Cc∞(O) with ζ∈L2(Ω)⊆Lloc1(Ω); consequently [F3] yields ζ=0 a.e. on O={u>ψ}.

3.1step 2.2F1

Finally (u−ψ)ζ=0 a.e. on Ω: on O this follows from ζ=0 a.e. by step 2.2, and on Ω∖O the class u−ψ vanishes a.e. — indeed u≥ψ a.e. by [F1] while on Ω∖{u>ψ} one has u≤ψ pointwise for the continuous representatives, so the set where u<ψ is contained in the complement of O and is null.

4.1step 1.1step 2.1step 2.2step 3.1∎

Step 1.1 proves the nonnegativity of the reaction, step 2.1 the a.e. nonnegativity of its L2 representative, step 2.2 its vanishing on the noncontact set and step 3.1 the complementarity product; all three conclusions of the second assertion use exactly the stated L2-representation and continuity hypotheses, and no product of a distribution with a Sobolev class is formed.

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