Pipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Constrained Variational Problems and Variational Inequalities

1 · Prerequisites

2 · Summary

This page develops variational problems whose admissible set is constrained in one of the two ways that dominate the calculus of variations: by a closed convex set, which produces a variational inequality, or by a regular level set of a differentiable map, which produces Lagrange multipliers. It then carries both mechanisms through the obstacle problem and closes with the variational characterisation of the Dirichlet eigenvalues.

The convex-constraint half begins with the geometry of Hilbert space. The characterisation of the metric projection by its variational inequality and the nonexpansiveness of the projection feed the Lions--Stampacchia argument: for a bounded coercive bilinear form on a real Hilbert space and a nonempty closed convex admissible set there is exactly one solution of the variational inequality, obtained as the fixed point of a contraction built from the projection. The solution depends Lipschitz continuously on the data, with constant governed by the coercivity constant. On the Banach-space side the direct method is set up on weakly sequentially closed constraint sets, the strong $L^{2}$ compactness inherited from Rellich's theorem keeps unit normalisation in the limit, and convex norm-closed sets are recognised as weakly closed.

The regular-constraint half proves the finite-dimensional duality lemma on independent functionals, a Banach implicit function theorem for a derivative that is surjective with a complemented kernel, the realisation of every kernel direction by a differentiable level-set curve, and the identification of the tangent space with the kernel of the constraint derivative. The differential of the constrained functional annihilates that kernel, and the multiplier rules for finitely many constraints and for one regular constraint in Hilbert space turn this into $D I (u) = \sum_{i} λ_{i} D G_{i} (u)$ . The multipliers are unique exactly when the constraint gradients are independent; a proportional-constraint counterexample shows how badly this fails otherwise.

The obstacle problem is the model inequality-constrained problem. The admissible set ${v \in H_{0}^{1} (Ω) : v \geq ψ}$ , with the stated trace compatibility $T ψ \leq 0$ , is nonempty, convex, closed and weakly closed; the symmetric energy has a unique minimiser, which is the unique solution of the obstacle variational inequality; the reaction is a nonnegative distribution that vanishes on the open noncontact set when $u$ and $ψ$ have continuous representatives, is supported on the contact set when those representatives are continuous and it is represented by a nonnegative Radon measure, and obeys the Lewy--Stampacchia bound $0 \leq Λ_{u} \leq (L ψ - f)^{+}$ in the stated $L^{2}$ regularity class. The one-dimensional example computes the solution, the contact set and the reaction explicitly, and the companion counterexamples delimit the trace, regularity and product hypotheses.

The page closes with the spectral application: the first Dirichlet eigenfunction minimises the Dirichlet energy on the $L^{2}$ -unit sphere, every minimiser is a weak eigenpair, some minimiser is nonnegative, and the higher eigenvalues are obtained by minimising over the unit sphere intersected with the orthogonal complement of the preceding eigenfunctions. Conventions and choice principles are declared per item: the weak-compactness and direct-method statements assume the ultrafilter lemma, DC and HB; the multiplier, truncation, Hilbert-Sobolev and absolute-value interfaces use the Axiom of Choice; the projection and measure-theoretic sign lemmas use Countable Choice, while Rellich compactness assumes the Axiom of Choice; and the implicit-function, trace and integration-by-parts suppliers declare their own choice footprints.

3 · Logical flowchart

4 · Definitions, theorems and proofs

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The direct method on a weakly closed constraint set

Statement

Assume the ultrafilter lemma, DC and HB (The ultrafilter extension principle (UL/BPI), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The real dominated-extension principle as an additional hypothesis over ZF). Let $X$ be a real reflexive Banach space (Reflexivity is surjectivity of the canonical map), let $A \subseteq X$ be nonempty and weakly sequentially closed (Weak convergence of nets and sequences), and let $I : X \to (- \infty, + \infty]$ be proper, coercive on $A$ and weakly sequentially lower semicontinuous on $A$ (Proper, coercive and weakly lower semicontinuous extended-real functionals). Then $I$ attains its infimum on $A$ : there is $u_{0} \in A$ with $I (u_{0}) = \inf_{A} I$ . In particular, if $C \subseteq X$ is nonempty, convex and norm closed, and the restriction $I ∣_{C}$ is proper, coercive on $C$ and weakly sequentially lower semicontinuous on $C$ , then $C$ is weakly sequentially closed and the conclusion holds for $A = C$ .

Facts & Assumptions

Given: The ultrafilter lemma, DC and HB; a real reflexive Banach space $X$ ; a nonempty weakly sequentially closed set $A \subseteq X$ ; and a proper functional $I : X \to (- \infty, + \infty]$ that is coercive and weakly sequentially lower semicontinuous on $A$ . For the second assertion, a nonempty convex norm-closed set $C \subseteq X$ such that $I ∣_{C}$ is proper, coercive and weakly sequentially lower semicontinuous on $C$ .

[F1]

The direct method in a reflexive Banach space: under the ultrafilter lemma, DC and HB, for every real reflexive Banach space $X$ , every nonempty weakly sequentially closed $A \subseteq X$ and every proper coercive weakly sequentially lower semicontinuous $I : X \to (- \infty, + \infty]$ , there is $u_{0} \in A$ with $I (u_{0}) = \inf_{A} I$ . In particular, a convex norm-closed set $C$ may be used as the admissible set when $I ∣_{C}$ is proper, coercive and weakly sequentially lower semicontinuous on $C$ .

[F2]

Norm closed convex iff weakly closed: under the assumed HB, every convex norm-closed subset of a normed space is weakly closed and therefore weakly sequentially closed, so every weakly convergent sequence in it has its limit in the set.

[A1]

The ultrafilter extension principle (UL/BPI), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The real dominated-extension principle as an additional hypothesis over ZF: the three hypotheses recorded in the statement, exactly as consumed by [F1].

Proof

technique · direct

Given: The hypotheses of the statement, including a real reflexive Banach space $X$ and a nonempty weakly sequentially closed $A \subseteq X$ , with $I$ proper, coercive on $A$ and weakly sequentially lower semicontinuous on $A$ .

1.1givenA1F1

All hypotheses of [F1] are met: $X$ is a real reflexive Banach space, $A$ is nonempty and weakly sequentially closed, and $I$ is proper, coercive on $A$ and weakly sequentially lower semicontinuous on $A$ ; the ultrafilter lemma, DC and HB are assumed [A1]. Hence there is $u_{0} \in A$ with $I (u_{0}) = \inf_{A} I$ .

1.2F1F2

For the second assertion let $C \subseteq X$ be nonempty, convex and norm closed, and assume $I ∣_{C}$ is proper, coercive on $C$ and weakly sequentially lower semicontinuous on $C$ . Then $C$ is weakly sequentially closed by [F2], and all hypotheses of [F1] hold with $A = C$ ; hence there is $u_{0} \in C$ with $I (u_{0}) = \inf_{C} I$ .

2.1step 1.1step 1.2A1∎

Step 1.1 proves the first assertion and step 1.2 the "in particular" clause; no convexity or smoothness of a general admissible set $A$ is claimed beyond what is stated, and the three hypotheses of the statement are used only through [F1].

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Weak H^1 convergence plus Rellich preserves the L^2 unit normalisation

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 1$ , let $Ω \subseteq R^{n}$ be a bounded open set, and let $(u_{j}) \subseteq H_{0}^{1} (Ω)$ be norm bounded with $u_{j} ⇀ u$ weakly in $H_{0}^{1} (Ω)$ (Weak convergence of nets and sequences, Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms) and $∥ u_{j} ∥_{L^{2} (Ω)} = 1$ for every $j$ (The space $L^{p} (μ)$ as the quotient by null functions). Then $u \in H_{0}^{1} (Ω)$ , $∥ u ∥_{L^{2} (Ω)} = 1$ , and some subsequence $(u_{j_{k}})$ converges to $u$ in $L^{2} (Ω)$ .

Facts & Assumptions

Given: A bounded open set $Ω \subseteq R^{n}$ , a norm-bounded sequence $(u_{j})$ in $H_{0}^{1} (Ω) = W_{0}^{1, 2} (Ω)$ converging weakly to $u$ , with $∥ u_{j} ∥_{L^{2} (Ω)} = 1$ for every $j$ , and the Axiom of Choice.

[A1]

AC implies DC implies countable choice, The Axiom of Choice, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The Axiom of Countable Choice ( ${A C}_{ω}$ ): the Axiom of Choice implies Dependent Choice, which implies Countable Choice, so every Countable-Choice hypothesis below is discharged.

[F1]

Compactness of $W_{0}^{1, p} (Ω) ↪ L^{p} (Ω)$ on bounded open sets: for a bounded open $Ω \subseteq R^{n}$ and $1 \leq p < \infty$ the inclusion $W_{0}^{1, p} (Ω) \to L^{p} (Ω)$ is compact: every sequence bounded in $W_{0}^{1, p} (Ω)$ has a subsequence converging in $L^{p} (Ω)$ .

[F2]

Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure: the $W^{1, 2}$ norm dominates the $L^{2}$ norm, so $∥ v ∥_{L^{2} (Ω)} \leq ∥ v ∥_{W_{0}^{1, 2} (Ω)}$ for every $v \in H_{0}^{1} (Ω)$ , and $H_{0}^{1} (Ω) = W_{0}^{1, 2} (Ω)$ is a closed subspace of $W^{1, 2} (Ω)$ contained in $L^{2} (Ω)$ .

[F3]

Weak convergence of nets and sequences: $u_{j} ⇀ u$ in $H_{0}^{1} (Ω)$ means $f (u_{j}) \to f (u)$ for every bounded linear functional $f$ on $H_{0}^{1} (Ω)$ ; in particular the specified limit $u$ lies in $H_{0}^{1} (Ω)$ , and every subsequence inherits the convergence to the same limit.

[F4]

Holder's inequality for integrals, including the endpoint cases: applying real Hölder to $∣ v ∣$ and $∣ w ∣$ gives $\int_{Ω} ∣ v w ∣ \leq ∥ v ∥_{L^{2}} ∥ w ∥_{L^{2}}$ for real or complex $v, w \in L^{2} (Ω)$ .

[F5]

A function with nonnegative test pairings is nonnegative a.e.: under Countable Choice, if $ζ \in L^{2} (O; R)$ and $\int_{O} ζ φ = 0$ for every $φ \in C_{c}^{\infty} (O)$ , then $ζ = 0$ a.e. on $O$ (second clause of that lemma).

[F6]

The reverse triangle inequality in a normed space: $∣ ∥ a ∥ - ∥ b ∥ ∣ \leq ∥ a - b ∥$ in a normed space.

[F7]

The space $L^{p} (μ)$ as the quotient by null functions: elements of $L^{2} (Ω)$ are a.e. classes, and equality of two classes means equality almost everywhere.

Proof

technique · direct

Given: A bounded open set $Ω \subseteq R^{n}$ , a norm-bounded sequence $(u_{j})$ in $H_{0}^{1} (Ω)$ with $u_{j} ⇀ u$ and $∥ u_{j} ∥_{L^{2}} = 1$ for all $j$ , and the Axiom of Choice.

1.1givenF1F2

By [F1] with $p = 2$ and the bounded open set $Ω$ , applied to the norm-bounded sequence $(u_{j}) \subseteq W_{0}^{1, 2} (Ω) = H_{0}^{1} (Ω)$ , there are a subsequence $(u_{j_{k}})$ and a class $w \in L^{2} (Ω)$ with $∥ u_{j_{k}} - w ∥_{L^{2} (Ω)} \to 0$ .

2.1step 1.1F2F3F4F5F7

We claim $w = u$ in $L^{2} (Ω)$ . Fix a real test $φ \in C_{c}^{\infty} (Ω; R)$ and consider the linear functional $f_{φ} (v) : = \int_{Ω} v φ$ , which is bounded on $H_{0}^{1} (Ω)$ because $∣ f_{φ} (v) ∣ \leq ∥ v ∥_{L^{2}} ∥ φ ∥_{L^{2}} \leq ∥ v ∥_{W^{1, 2}} ∥ φ ∥_{L^{2}}$ by [F2, F4]. Since $(u_{j_{k}})$ is a subsequence of a weakly convergent sequence, [F3] gives $f_{φ} (u_{j_{k}}) \to f_{φ} (u)$ , that is $\int_{Ω} u_{j_{k}} φ \to \int_{Ω} u φ$ ; on the other hand [F4] gives $∣ \int_{Ω} (u_{j_{k}} - w) φ ∣ \leq ∥ u_{j_{k}} - w ∥_{L^{2}} ∥ φ ∥_{L^{2}} \to 0$ , so $\int_{Ω} (u - w) φ = 0$ for every $φ \in C_{c}^{\infty} (Ω)$ . Writing $ζ : = u - w \in L^{2} (Ω)$ , the real and imaginary parts belong to $L^{2} (Ω; R)$ since their absolute values are at most $∣ ζ ∣$ . Their pairings with every real test vanish separately. The second clause of [F5], with both open sets equal to $Ω$ , therefore applies to each part: zero pairings in particular satisfy its nonnegative-pairing hypothesis. Both parts vanish a.e., so $ζ = 0$ a.e. and $w = u$ by [F7]; for real scalars the imaginary part is zero already.

3.1step 1.1step 2.1F6algebra

By step 1.1 and step 2.1 the subsequence converges to $u$ in $L^{2} (Ω)$ ; since $∥ u_{j_{k}} ∥_{L^{2}} = 1$ for every $k$ , the reverse triangle inequality [F6] gives $∣ ∥ u ∥_{L^{2}} - ∥ u_{j_{k}} ∥_{L^{2}} ∣ \leq ∥ u - u_{j_{k}} ∥_{L^{2}} \to 0$ , hence $∥ u ∥_{L^{2} (Ω)} = 1$ .

4.1step 1.1step 3.1A1F3∎

The weak limit $u$ lies in $H_{0}^{1} (Ω)$ by [F3]; the subsequence $(u_{j_{k}})$ converges to $u$ in $L^{2} (Ω)$ by steps 1.1 and 2.1; and $∥ u ∥_{L^{2} (Ω)} = 1$ by step 3.1. This proves the assertion; the Countable Choice required by [F5] and by the extraction in [F1] is supplied by the Axiom of Choice through [A1].

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

A split surjective derivative parametrises its level set

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ be a real Banach space (Banach space), let $U \subseteq X$ be open, let $G : U \to R^{m}$ with $m \geq 1$ be of class $C^{1}$ (C k map between Banach spaces, Fréchet derivative between Banach spaces), let $u \in U$ , and suppose $D G (u) : X \to R^{m}$ is surjective. Then there is a finite-dimensional subspace $Y \subseteq X$ such that $X = \ker D G (u) \oplus Y$ is a topological direct sum with $D G (u) ∣_{Y} : Y \to R^{m}$ a bounded linear isomorphism and the coordinate projection onto $Y$ along $\ker D G (u)$ bounded (A complemented closed subspace of a normed space, A closed subspace is complemented exactly when it is the range of a bounded projection). Moreover there are an open $A \subseteq \ker D G (u)$ with $0 \in A$ , an open $B \subseteq Y$ with $0 \in B$ , and a unique $C^{1}$ map $φ : A \to B$ with $φ (0) = 0$ and $D φ (0) = 0$ such that ${z \in u + (A + B) : G (z) = G (u)} = {u + a + φ (a) : a \in A} .$

Facts & Assumptions

Given: A real Banach space $X$ , an open set $U \subseteq X$ , a $C^{1}$ map $G : U \to R^{m}$ with surjective derivative $D G (u)$ at a point $u \in U$ , and the standard unit vectors $e_{0}, \dots, e_{m - 1}$ of $R^{m}$ .

[A1]

The Axiom of Choice: the Axiom of Choice, consumed through the published implicit function theorem, which assumes it; the selections made here itself are finite.

[F1]

Implicit function theorem for Banach spaces: for real Banach spaces $X_{0}, Y, Z$ , an open $W \subseteq X_{0} \times Y$ , a $C^{k}$ map $F : W \to Z$ with $k \geq 1$ , a point $(a_{0}, b_{0}) \in W$ with $F (a_{0}, b_{0}) = 0$ whose partial derivative $D_{Y} F (a_{0}, b_{0}) : Y \to Z$ is a bounded linear isomorphism, there are open $A \subseteq X_{0}$ with $a_{0} \in A$ , $B \subseteq Y$ with $b_{0} \in B$ , and a unique $C^{k}$ map $g : A \to B$ with $g (a_{0}) = b_{0}$ and ${F = 0} \cap (A \times B) = {(a, g (a)) : a \in A}$ ; along the graph $D g (x) = - D_{Y} F (x, g (x))^{- 1} D_{X} F (x, g (x))$ .

[F2]

Fréchet derivative between Banach spaces, C k map between Banach spaces, A bounded linear operator between normed spaces: the Fréchet derivative $D G (u)$ is a bounded linear operator $X \to R^{m}$ , bounded linear operators are continuous, and restrictions of bounded linear operators to subspaces are bounded and linear.

[F3]

The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ , Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ : the standard unit vectors $e_{0}, \dots, e_{m - 1}$ form an ordered basis of the function space $R^{m}$ , so every vector of $R^{m}$ is a unique linear combination $\sum_{i < m} a_{i} e_{i}$ , and $R^{m}$ is finite dimensional; the span of a finite list is the set of its linear combinations.

[F4]

Every natural-number-indexed list of nonempty sets has a choice function on its family of values: a finite family of nonempty sets indexed by a natural number has a choice function.

[F5]

A linear map from a finite-dimensional normed space is bounded: a linear map whose domain admits an ordered basis of finite length is bounded.

[F6]

A finite-dimensional normed subspace is closed, Every finite-dimensional normed space is Banach: a finite-dimensional subspace of a normed space is closed, and a finite-dimensional normed space is complete.

[F7]

A closed subspace of a Banach space is Banach, For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ : a closed subspace of a Banach space is a Banach space, and a continuous map pulls closed sets back to closed sets.

[F8]

A complemented closed subspace of a normed space, A closed subspace is complemented exactly when it is the range of a bounded projection: a closed subspace $M$ is complemented when there is a closed subspace $N$ with unique decomposition $x = m + n$ and both coordinate maps bounded; equivalently there is a bounded linear projection $P$ with $P^{2} = P$ and $ran (P) = M$ .

[F9]

Chain sum product and composition rules for Banach derivatives: sums, scalar multiples, compositions of $C^{1}$ maps and derivatives of affine maps are computed by the chain and sum rules, the derivative of a bounded linear map being the map itself.

[F10]

Finite products of Banach spaces are Banach, The standard product norms on a finite product of normed spaces: a finite product of Banach spaces, with a product norm, is a Banach space.

Proof

technique · direct

Given: A real Banach space $X$ , open $U \subseteq X$ , a $C^{1}$ map $G : U \to R^{m}$ with $D G (u)$ surjective at $u \in U$ .

1.1givenF2F3F4F5F6choosealgebra

(Construction of the complement) For each $i < m$ the set $D G (u)^{- 1} ({e_{i}})$ is nonempty by surjectivity, so finite choice [F4] selects $u_{0}, \dots, u_{m - 1} \in X$ with $D G (u) u_{i} = e_{i}$ ; put $Y : = span {u_{0}, \dots, u_{m - 1}}$ [F3]. The $u_{i}$ are linearly independent: if $\sum_{i < m} c_{i} u_{i} = 0$ , then $\sum_{i < m} c_{i} e_{i} = D G (u) 0 = 0$ and the uniqueness of coordinates in the standard basis forces every $c_{i} = 0$ [F3]. Hence $\dim Y = m$ and $D G (u) ∣_{Y} : Y \to R^{m}$ is a linear bijection, bounded as a restriction of the bounded operator $D G (u)$ [F2]; its inverse is a linear map on the finite-dimensional space $R^{m}$ and is bounded by [F5]. Moreover $Y$ is closed in $X$ and complete by [F6].

2.1step 1.1F2F3F7F8algebra

(Topological splitting and the bounded projection) Since $D G (u) ∣_{Y}$ is injective, $\ker D G (u) \cap Y = {0}$ ; since $R^{m}$ is spanned by the $e_{i}$ [F3], every $x \in X$ has $D G (u) x = \sum_{i < m} a_{i} e_{i}$ for unique $a_{i}$ , and with $y : = \sum_{i < m} a_{i} u_{i} \in Y$ one has $D G (u) (x - y) = 0$ , so $x = (x - y) + y \in \ker D G (u) + Y$ ; the sum is therefore direct. The kernel $\ker D G (u)$ is closed, being the preimage of the closed set ${0}$ under the continuous operator $D G (u)$ [F2, F7], hence is a Banach space by [F7]. Define $P : = (D G (u) ∣_{Y})^{- 1} \circ D G (u) : X \to X$ ; it is linear and bounded by step 1.1, it takes values in $Y$ , restricts to the identity on $Y$ , and $\ker P = \ker D G (u)$ , so $P^{2} = P$ and $ran (P) = Y$ . By [F8], $Y$ is complemented by $\ker D G (u)$ with both coordinate projections bounded, so $X = \ker D G (u) \oplus Y$ is a topological direct sum and the coordinate projection onto $Y$ along $\ker D G (u)$ is bounded.

2.2step 1.1F2F9F10construct

(The auxiliary map) The set $W : = {(a, y) \in \ker D G (u) \times Y : u + a + y \in U}$ is open in the Banach space $\ker D G (u) \times Y$ by [F7, F10] and contains $(0, 0)$ because $u \in U$ . Define $F : W \to R^{m}$ , $F (a, y) : = G (u + a + y) - G (u)$ ; the map $(a, y) \mapsto u + a + y$ is affine and $C^{1}$ with derivative $(h, k) \mapsto h + k$ , so $F$ is $C^{1}$ with $F (0, 0) = 0$ and $D F (0, 0) (h, k) = D G (u) (h + k)$ by the chain and sum rules [F9, F2]; the partial derivative in the $y$ -variable is therefore $D_{Y} F (0, 0) = D G (u) ∣_{Y}$ , a bounded linear isomorphism by step 1.1.

3.1step 2.1step 2.2F1algebra

(Applying the implicit function theorem) Apply [F1] with $X_{0} = \ker D G (u)$ , $Y$ , $Z = R^{m}$ , the map $F$ , and the point $(0, 0)$ : by step 2.2 its hypotheses hold, and it provides open $A \subseteq \ker D G (u)$ with $0 \in A$ , open $B \subseteq Y$ with $0 \in B$ , and a unique $C^{1}$ map $φ : A \to B$ with $φ (0) = 0$ and ${F = 0} \cap (A \times B) = {(a, φ (a)) : a \in A}$ . Since $F (a, y) = 0$ means exactly $G (u + a + y) = G (u)$ , the graph identity reads ${z \in u + (A + B) : G (z) = G (u)} = {u + a + φ (a) : a \in A}$ .

4.1step 2.1step 3.1F9algebra

(The derivative at zero vanishes) The graph identity gives $G (u + a + φ (a)) = G (u)$ for every $a \in A$ . The map $h (a) : = u + a + φ (a)$ is $C^{1}$ with $D h (0) = {i d}_{\ker D G (u)} + D φ (0)$ [F9], so the chain rule [F9] gives $D G (u) \circ (i d + D φ (0)) = D (G \circ h) (0) = 0$ as a map $\ker D G (u) \to R^{m}$ . For $a \in \ker D G (u)$ this reads $D φ (0) a \in \ker D G (u)$ because $D G (u) a = 0$ ; as $D φ (0)$ takes values in $Y$ and $\ker D G (u) \cap Y = {0}$ by step 2.1, $D φ (0) a = 0$ for every $a$ , that is $D φ (0) = 0$ .

5.1step 2.1step 3.1step 4.1F1A1∎

(Conclusion) Step 1.1 and step 2.1 supply the finite-dimensional subspace $Y$ with $X = \ker D G (u) \oplus Y$ a topological direct sum, $D G (u) ∣_{Y}$ a bounded linear isomorphism and the coordinate projection onto $Y$ bounded; step 3.1 supplies the open sets $A$ and $B$ and the map $φ$ together with the parametrisation identity; step 4.1 supplies $D φ (0) = 0$ ; and the uniqueness assertion follows because any other $C^{1}$ map with the same set identity satisfies $F (a, ψ (a)) = 0$ on $A$ and hence equals the unique map $φ$ of [F1]. This proves the statement, the Axiom of Choice having been used only through [F1] [A1].

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Regular constraint directions are realised by level-set curves

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ be a real Banach space, $U \subseteq X$ open, $G : U \to R^{m}$ of class $C^{1}$ (C k map between Banach spaces, Fréchet derivative between Banach spaces), $u \in U$ , and suppose $D G (u)$ is surjective, with $Y$ , $A \subseteq \ker D G (u)$ , $B \subseteq Y$ and $φ : A \to B$ as in A split surjective derivative parametrises its level set. Then for every $h \in \ker D G (u)$ there are $ε > 0$ and a $C^{1}$ curve $c : (- ε, ε) \to X$ with $c (0) = u$ , $c^{'} (0) = h$ and $G (c (t)) = G (u)$ for every $t \in (- ε, ε)$ ; explicitly $c (t) = u + t h + φ (t h)$ .

Facts & Assumptions

Given: The setting of A split surjective derivative parametrises its level set: a split surjective derivative $D G (u)$ at $u \in U$ , and the open sets $A \subseteq \ker D G (u)$ with $0 \in A$ and the $C^{1}$ map $φ : A \to B$ with $φ (0) = 0$ , $D φ (0) = 0$ of that theorem.

[F1]

A split surjective derivative parametrises its level set: $\ker D G (u)$ is a closed subspace of $X$ , and the level set is parametrised as ${z \in u + (A + B) : G (z) = G (u)} = {u + a + φ (a) : a \in A}$ with $A$ open, $0 \in A$ , and $φ$ of class $C^{1}$ with $φ (0) = 0$ and $D φ (0) = 0$ .

[F2]

C k map between Banach spaces, Chain sum product and composition rules for Banach derivatives: sums and scalar multiples of $C^{1}$ maps are $C^{1}$ with the sum rule for derivatives, the derivative of a bounded linear map is the map itself, and composites of $C^{1}$ maps are $C^{1}$ with the chain rule; in particular $t \mapsto t h$ and $a \mapsto u + a + φ (a)$ are $C^{1}$ .

[A1]

The Axiom of Choice: the hypothesis under which the parametrisation of [F1] is available.

Proof

technique · direct

Given: The setting of [F1] and a vector $h \in \ker D G (u)$ .

1.1givenA1F1algebra

Since $A$ is open and $0 \in A$ there is $r > 0$ with $B (0, r) \subseteq A$ ; if $h \neq 0$ , set $ε : = r / (2 ∥ h ∥)$ , and if $h = 0$ take $ε : = 1$ . For $∣ t ∣ < ε$ one has $∥ t h ∥ < r$ , so $t h \in A$ , and $c (t) : = u + t h + φ (t h)$ is a well-defined element of $u + (A + B) \subseteq U$ with $G (c (t)) = G (u)$ by the parametrisation identity of [F1]; moreover $c (0) = u + 0 + φ (0) = u$ .

2.1step 1.1F1F2

The curve $c$ is $C^{1}$ on $(- ε, ε)$ and $c^{'} (t) = h + D φ (t h) h$ for every $t$ , by the sum and chain rules applied to $t \mapsto t h$ and $φ$ [F2]; at $t = 0$ this gives $c^{'} (0) = h + D φ (0) h = h$ because $D φ (0) = 0$ [F1].

2.2step 1.1

In particular $c$ is a $C^{1}$ curve on $(- ε, ε)$ with values in $X$ , $c (0) = u$ , and $G (c (t)) = G (u)$ for every $t$ by step 1.1, which is the asserted realisation of $h$ by a level-set curve.

3.1step 2.1step 2.2A1∎

Steps 1.1, 2.1 and 2.2 prove the claim for an arbitrary $h \in \ker D G (u)$ ; no convexity of the constraint was used, and the Axiom of Choice enters only through the parametrisation supplied by [F1] [A1].

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The tangent space of a regular level set is the kernel of the constraint derivative

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ be a real Banach space, let $U \subseteq X$ be open, let $G : U \to R^{m}$ be of class $C^{1}$ (C k map between Banach spaces, Fréchet derivative between Banach spaces), let $u \in U$ , and suppose $D G (u) : X \to R^{m}$ is surjective. Then the set of derivatives $γ^{'} (0)$ of $C^{1}$ curves $γ : (- ε, ε) \to X$ with $γ (0) = u$ and $G \circ γ$ constant equals $\ker D G (u)$ ; that is, $\ker D G (u)$ is exactly the tangent space of the level set $G^{- 1} (G (u))$ at $u$ .

Facts & Assumptions

Given: A real Banach space $X$ , an open set $U \subseteq X$ , a $C^{1}$ map $G : U \to R^{m}$ with $D G (u)$ surjective at $u \in U$ , under the Axiom of Choice.

[F1]

Chain sum product and composition rules for Banach derivatives, C k map between Banach spaces: if $γ$ is $C^{1}$ near $0$ with $γ (0) = u$ and $G$ is $C^{1}$ near $u$ , then $G \circ γ$ is differentiable at $0$ with $(G \circ γ)^{'} (0) = D G (u) γ^{'} (0)$ ; a constant function has derivative $0$ , and a bounded linear map is its own derivative.

[F2]

Regular constraint directions are realised by level-set curves: under AC, given the chart $Y, A, B, φ$ of A split surjective derivative parametrises its level set (which requires $m \geq 1$ ), every $h \in \ker D G (u)$ is realised by a $C^{1}$ curve $c (t) = u + t h + φ (t h)$ with $c (0) = u$ , $c^{'} (0) = h$ and $G (c (t)) = G (u)$ .

[F3]

Proof

technique · direct

Given: A real Banach space $X$ , an open set $U \subseteq X$ , a $C^{1}$ map $G : U \to R^{m}$ with $D G (u)$ surjective at $u \in U$ .

1.1givenF1

Let $γ : (- ε, ε) \to X$ be $C^{1}$ with $γ (0) = u$ and $G \circ γ$ constant. Then $G \circ γ$ has derivative $0$ at every $t$ , and the chain rule [F1] gives $0 = (G \circ γ)^{'} (0) = D G (u) γ^{'} (0)$ ; hence $γ^{'} (0) \in \ker D G (u)$ .

1.2givenF1F2F3choose

Conversely, let $h \in \ker D G (u)$ . If $m = 0$ , then $G$ is constant by [F3] and $\ker D G (u) = X$ . Choose $r > 0$ with $B (u, r) \subseteq U$ and set $ε = r / (2 (1 + ∥ h ∥))$ ; the affine curve $c (t) = u + t h$ lies in $U$ for $∣ t ∣ < ε$ , is $C^{1}$ , and has $c^{'} (0) = h$ by [F1]. If $m \geq 1$ , surjectivity and AC give the chart $Y, A, B, φ$ of the implicit-function theorem in [F2]; applying the realization lemma to this chart gives a $C^{1}$ curve $c$ with $c (0) = u$ , $c^{'} (0) = h$ and $G \circ c$ constant. In either case $h$ is a derivative of the required kind.

2.1step 1.1step 1.2F2∎

Step 1.1 shows that every such derivative lies in $\ker D G (u)$ and step 1.2 shows that every element of $\ker D G (u)$ occurs, so the two sets are equal; this is the assertion, and the Axiom of Choice was used only through the chart and realization lemma in the case $m \geq 1$ [F2].

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The differential annihilates the tangent kernel at a constrained extremum

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ be a real Banach space, let $U \subseteq X$ be open, let $I : U \to R$ be Fréchet differentiable at $u \in U$ (Fréchet derivative between Banach spaces), and let $G : U \to R^{m}$ be of class $C^{1}$ (C k map between Banach spaces) with $D G (u)$ surjective. If $u$ is a local minimiser or a local maximiser of $I$ on the level set ${G = G (u)}$ , then $D I (u) h = 0$ for every $h \in \ker D G (u)$ .

Facts & Assumptions

Given: A real Banach space $X$ , open $U \subseteq X$ , a map $I$ differentiable at $u$ with differential $D I (u) \in X^{*}$ , a $C^{1}$ map $G$ with $D G (u)$ surjective, and the assumption that $u$ is a local minimiser or local maximiser of $I$ on the level set ${G = G (u)}$ , meaning that for some radius $δ > 0$ one has $I (u) \leq I (z)$ (respectively $I (u) \geq I (z)$ ) for every $z \in U$ with $∥ z - u ∥ < δ$ and $G (z) = G (u)$ .

[F1]

The tangent space of a regular level set is the kernel of the constraint derivative: for every $h \in \ker D G (u)$ there are $ε > 0$ and a $C^{1}$ curve $γ : (- ε, ε) \to X$ with $γ (0) = u$ , $γ^{'} (0) = h$ and $G (γ (t)) = G (u)$ for all $t$ .

[F2]

Chain sum product and composition rules for Banach derivatives, Fréchet derivative between Banach spaces: the composition $t \mapsto I (γ (t))$ is differentiable at $0$ with derivative $D I (u) γ^{'} (0) = D I (u) h$ .

[F3]

Fermat's interior extremum theorem: if $f$ has a local extremum at a point $c$ interior to its domain and is differentiable at $c$ , then $f^{'} (c) = 0$ : a real function on an open interval that is differentiable at an interior point and has a local minimum or local maximum there has derivative $0$ at that point.

[A1]

The Axiom of Choice: the hypothesis under which the level-set parametrisation of [F1] is available.

Proof

technique · direct

Given: The setting above and a vector $h \in \ker D G (u)$ .

1.1givenA1F1

By [F1] choose $ε > 0$ and a $C^{1}$ curve $γ : (- ε, ε) \to X$ with $γ (0) = u$ , $γ^{'} (0) = h$ and $G (γ (t)) = G (u)$ for every $t$ ; by continuity of $γ$ at $0$ and the strict positive radius $δ$ of the local extremum hypothesis, we may shrink $ε$ so that $∥ γ (t) - u ∥ < δ$ for all $t$ .

2.1step 1.1F2

The function $φ (t) : = I (γ (t))$ is defined on the open interval $(- ε, ε)$ , is differentiable at $0$ with $φ^{'} (0) = D I (u) h$ by [F2], and has a local minimum (respectively local maximum) at $t = 0$ : for $∣ t ∣ < ε$ the curve lies in the level set and within distance $δ$ of $u$ , so $φ (0) = I (u) \leq I (γ (t)) = φ (t)$ (respectively $\geq$ ).

3.1step 2.1F3

Fermat's interior extremum theorem [F3] applied to $φ$ at the interior point $0$ gives $φ^{'} (0) = 0$ , that is, $D I (u) h = 0$ .

4.1step 3.1A1∎

Since $h \in \ker D G (u)$ was arbitrary, $D I (u)$ vanishes on all of $\ker D G (u)$ , which is the assertion; the Axiom of Choice was used only through [F1] [A1].

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Functionals vanishing on a common kernel are combinations of an independent family

Statement

Let $X$ be a real vector space (Vector space over a field), let $m \geq 1$ , and let $ψ_{1}, \dots, ψ_{m} \in X^{*}$ be linearly independent linear functionals on $X$ (Linear functionals and the algebraic dual $V^{*} = L (V, F)$ , Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent). Let $φ \in X^{*}$ satisfy $⋂_{i = 1}^{m} \ker ψ_{i} \subseteq \ker φ$ (Kernel and image of a linear map). Then there is a unique $λ \in R^{m}$ with $φ = \sum_{i = 1}^{m} λ_{i} ψ_{i}$ . In particular, for $m = 1$ : if $ψ \neq 0$ and $\ker ψ \subseteq \ker φ$ , then $φ = λ ψ$ for a unique $λ \in R$ .

Facts & Assumptions

Given: A real vector space $X$ , an integer $m \geq 1$ , linearly independent linear functionals $ψ_{1}, \dots, ψ_{m} ⁣ : X \to R$ , and a linear functional $φ ⁣ : X \to R$ with $⋂_{i = 1}^{m} \ker ψ_{i} \subseteq \ker φ$ .

[F1]

Linear functionals and the algebraic dual $V^{*} = L (V, F)$ , The space $L (V, W)$ of linear maps with pointwise addition and scalar multiplication: $X^{*} = L (X, R)$ is the vector space of linear functionals on $X$ with pointwise operations, so a linear combination $x \mapsto \sum_{i} c_{i} ψ_{i} (x)$ of elements of $X^{*}$ is again an element of $X^{*}$ , and the zero of $X^{*}$ is the functional vanishing identically on $X$ .

[F2]

Kernel and image of a linear map, The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial: for a linear map $T$ one has $\ker T = {x : T (x) = 0}$ , and $\ker T$ is a linear subspace of the domain of $T$ .

[F3]

Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent: the list $ψ_{1}, \dots, ψ_{m}$ is linearly independent exactly when $\sum_{i = 1}^{m} c_{i} ψ_{i} = 0$ in $X^{*}$ forces $c_{1} = \dots = c_{m} = 0$ ; in particular every $ψ_{i}$ is nonzero, since otherwise the list would carry the nontrivial relation with coefficient $1$ on the zero term.

[F4]

Linear map between vector spaces over the same field: each $ψ_{i}$ and $φ$ satisfies $ψ_{i} (a x + b y) = a ψ_{i} (x) + b ψ_{i} (y)$ and $φ (a x + b y) = a φ (x) + b φ (y)$ for all $x, y \in X$ and $a, b \in R$ .

Proof

technique · induction on $m$

Given: A real vector space $X$ , an integer $m \geq 1$ , linearly independent functionals $ψ_{1}, \dots, ψ_{m}$ on $X$ , and a functional $φ$ on $X$ with $⋂_{i = 1}^{m} \ker ψ_{i} \subseteq \ker φ$ .

1.1basegivenF1F3F4algebra

Base case $m = 1$ . Let $ψ \neq 0$ and $\ker ψ \subseteq \ker φ$ . As $ψ \neq 0$ there is $y \in X$ with $ψ (y) \neq 0$ , and $x_{1} : = y / ψ (y)$ satisfies $ψ (x_{1}) = 1$ by homogeneity [F4]; for arbitrary $x \in X$ the vector $x - ψ (x) x_{1}$ lies in $\ker ψ \subseteq \ker φ$ , so $0 = φ (x) - ψ (x) φ (x_{1})$ by linearity of $φ$ [F4], that is, $φ = φ (x_{1}) ψ$ ; conversely if $λ ψ = λ^{'} ψ$ with $ψ \neq 0$ , then $(λ - λ^{'}) ψ = 0$ is the zero functional, so evaluating at $y$ gives $λ = λ^{'}$ .

1.2ihassume-hyp

Induction step setup. Let $P (m)$ denote the assertion of the statement for the fixed integer $m$ and an arbitrary real vector space, and suppose $m \geq 2$ while $P (m - 1)$ is known as the induction hypothesis; the goal is to prove $P (m)$ .

2.1step 1.1F1F2F3algebra

Put $V : = \ker ψ_{m}$ , a linear subspace of $X$ [F2], and let $ρ_{i} : = ψ_{i} ∣_{V}$ for $1 \leq i \leq m - 1$ ; each $ρ_{i}$ is a linear functional on $V$ [F1, F2]. The list $ρ_{1}, \dots, ρ_{m - 1}$ is linearly independent: if $\sum_{i < m} c_{i} ρ_{i} = 0$ , then the functional $σ : = \sum_{i < m} c_{i} ψ_{i} \in X^{*}$ vanishes on $V = \ker ψ_{m}$ , so $\ker ψ_{m} \subseteq \ker σ$ and the base case of step 1.1 gives $σ = λ ψ_{m}$ for some $λ \in R$ ; subtracting yields $\sum_{i < m} c_{i} ψ_{i} - λ ψ_{m} = 0$ in $X^{*}$ , whence $c_{1} = \dots = c_{m - 1} = 0$ and $λ = 0$ by independence of $ψ_{1}, \dots, ψ_{m}$ [F1, F3].

3.1step 1.2step 2.1F1F2algebra

The inclusion hypothesis transfers: if $v \in ⋂_{i < m} \ker ρ_{i}$ , then $v \in V = \ker ψ_{m}$ and $ψ_{i} (v) = 0$ for all $i < m$ , so $v \in ⋂_{i \leq m} \ker ψ_{i} \subseteq \ker φ$ ; hence $⋂_{i < m} \ker ρ_{i} \subseteq \ker (φ ∣_{V})$ , and $φ ∣_{V}$ is a linear functional on $V$ [F1, F4]. The induction hypothesis $P (m - 1)$ applied to the real vector space $V$ and the linearly independent list $ρ_{1}, \dots, ρ_{m - 1}$ of step 2.1 therefore provides $μ_{1}, \dots, μ_{m - 1} \in R$ with $φ ∣_{V} = \sum_{i < m} μ_{i} ρ_{i}$ , that is, with $φ - \sum_{i < m} μ_{i} ψ_{i}$ vanishing on $V$ .

4.1step 1.1step 3.1F1F3algebra

Let $τ : = φ - \sum_{i < m} μ_{i} ψ_{i} \in X^{*}$ , a functional vanishing on $V = \ker ψ_{m}$ by step 3.1, so that $\ker ψ_{m} \subseteq \ker τ$ ; since $ψ_{m} \neq 0$ [F3], the base case of step 1.1 yields $τ = μ_{m} ψ_{m}$ for some $μ_{m} \in R$ ; setting $λ_{i} : = μ_{i}$ for $i < m$ gives $φ = \sum_{i = 1}^{m} λ_{i} ψ_{i}$ by [F1].

5.1step 4.1F1F3discharge-induction∎

Uniqueness and discharge of the induction. If $\sum_{i = 1}^{m} λ_{i} ψ_{i} = \sum_{i = 1}^{m} λ_{i}^{'} ψ_{i}$ , then $\sum_{i = 1}^{m} (λ_{i} - λ_{i}^{'}) ψ_{i}$ is the zero element of $X^{*}$ [F1], so $λ_{i} = λ_{i}^{'}$ for every $i$ by linear independence [F3]. Thus $P (m - 1)$ implies $P (m)$ , and with the base case of step 1.1 the principle of induction gives $P (m)$ for every $m \geq 1$ , which is the assertion, including the stated uniqueness.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The Lagrange multiplier rule for one regular constraint in Hilbert space

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $H$ be a real Hilbert space (Hilbert space), let $U \subseteq H$ be open, let $I : U \to R$ be Fréchet differentiable at $u \in U$ , and let $G : U \to R$ be of class $C^{1}$ with $D G (u) \neq 0$ (Fréchet derivative between Banach spaces). If $u$ is a local minimiser or a local maximiser of $I$ on the level set ${G = G (u)}$ , then there is a unique $λ \in R$ with $D I (u) = λ D G (u) .$

Facts & Assumptions

Given: A real Hilbert space $H$ , open $U \subseteq H$ , a functional $I$ Fréchet differentiable at $u$ , a $C^{1}$ function $G$ with $D G (u) \neq 0$ , and the assumption that $u$ is a local minimiser or local maximiser of $I$ on the level set ${G = G (u)}$ .

[F1]

The differential annihilates the tangent kernel at a constrained extremum: under these hypotheses $D I (u) h = 0$ for every $h \in \ker D G (u)$ , and the same conclusion is obtained from the constrained-extremum lemma applied with the single constraint $G$ (the level set and $C^{1}$ hypotheses are exactly those of that lemma with $m = 1$ , whose derivative $D G (u) \neq 0$ is surjective onto $R$ ).

[F2]

Functionals vanishing on a common kernel are combinations of an independent family: if $ψ \neq 0$ is a bounded linear functional on $H$ with $\ker ψ \subseteq \ker φ$ for some bounded linear functional $φ$ , then $φ = λ ψ$ for a unique $λ \in R$ ; this is the $m = 1$ clause of that lemma.

[F3]

Fréchet derivative between Banach spaces, Hilbert space: $D I (u)$ and $D G (u)$ are bounded linear functionals on $H$ (the derivative of a $C^{1}$ function into $R$ ), and $D G (u) \neq 0$ means that $D G (u)$ is not the zero functional.

[A1]

The Axiom of Choice: recorded as in the statement and consumed only through [F1].

Proof

technique · direct

Given: The hypotheses above, including the local extremum at $u$ and $D G (u) \neq 0$ .

1.1givenA1F1F3

Since $D G (u) : H \to R$ is a nonzero bounded linear functional, it is surjective, so the constrained-extremum lemma [F1] applies with the single constraint $G$ : the differential $D I (u)$ vanishes on $\ker D G (u)$ .

2.1step 1.1F2F3

The functionals $ψ : = D G (u) \neq 0$ and $φ : = D I (u)$ satisfy $\ker ψ = \ker D G (u) \subseteq \ker D I (u) = \ker φ$ by step 1.1, so the $m = 1$ clause of [F2] gives a unique $λ \in R$ with $D I (u) = λ D G (u)$ .

3.1step 2.1A1∎

This is the asserted multiplier identity with its uniqueness clause; the Hilbert structure is used only through the standing conventions of the page, the argument being valid in any real Banach space [A1].

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The Lagrange multiplier rule for finitely many regular constraints

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ be a real Banach space (Banach space), let $U \subseteq X$ be open, let $I : U \to R$ be Fréchet differentiable at $u \in U$ , and let $G = (G_{1}, \dots, G_{m}) : U \to R^{m}$ be of class $C^{1}$ with $D G (u) : X \to R^{m}$ surjective (Fréchet derivative between Banach spaces). If $u$ is a local minimiser or a local maximiser of $I$ on the level set ${G = G (u)}$ , then there is a unique $λ \in R^{m}$ with $D I (u) = \sum_{i = 1}^{m} λ_{i} D G_{i} (u) .$

Facts & Assumptions

Given: A real Banach space $X$ , open $U \subseteq X$ , a functional $I$ Fréchet differentiable at $u$ , a $C^{1}$ map $G = (G_{1}, \dots, G_{m})$ with $D G (u)$ surjective, and the assumption that $u$ is a local minimiser or local maximiser of $I$ on the level set ${G = G (u)}$ .

[F1]

The differential annihilates the tangent kernel at a constrained extremum: under these hypotheses $D I (u) h = 0$ for every $h \in \ker D G (u)$ , that is, $⋂_{i = 1}^{m} \ker D G_{i} (u) \subseteq \ker D I (u)$ .

[F2]

Fréchet derivative between Banach spaces, The dual space X^* of a normed space and its dual norm: $D I (u)$ and each component $D G_{i} (u) = π_{i - 1} \circ D G (u)$ is a bounded linear functional on $X$ , and the kernel of $D G (u)$ is the intersection of the kernels of its components.

[F3]

Every natural-number-indexed list of nonempty sets has a choice function on its family of values: a finite family of nonempty sets indexed by a natural number admits a choice function.

[F4]

Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent: the functionals $ψ_{1}, \dots, ψ_{m} \in X^{*}$ are linearly independent exactly when $\sum c_{i} ψ_{i} = 0$ in $X^{*}$ forces all $c_{i} = 0$ .

[F5]

Functionals vanishing on a common kernel are combinations of an independent family: for $m \geq 1$ , if $ψ_{1}, \dots, ψ_{m} \in X^{*}$ are linearly independent and $⋂_{i = 1}^{m} \ker ψ_{i} \subseteq \ker φ$ for some $φ \in X^{*}$ , then there is a unique $λ \in R^{m}$ with $φ = \sum_{i = 1}^{m} λ_{i} ψ_{i}$ .

[F6]

The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ : $R^{0}$ is the zero space. For $m \geq 1$ , use the one-based labels $e_{i} : = {\hat{e}}_{i - 1}$ for $1 \leq i \leq m$ , where $({\hat{e}}_{k})_{k < m}$ is the supplied standard basis; likewise $G_{i} = π_{i - 1} \circ G$ and $λ_{i} = λ (i - 1)$ . Sums and intersections over $1 \leq i \leq m$ reindex those over $k < m$ ; at $m = 0$ the sum is the zero functional and the intersection of component kernels is $X$ .

[A1]

The Axiom of Choice: recorded as in the statement; the selections below are finite and need no choice principle.

Proof

technique · direct

Given: The hypotheses above, including the local extremum at $u$ .

1.1givenF1F2

By [F1] the differential $D I (u)$ vanishes on $\ker D G (u) = ⋂_{i = 1}^{m} \ker D G_{i} (u)$ [F2].

1.2givenF3F4F6choose

For $m \geq 1$ , the functionals $ψ_{i} : = D G_{i} (u) \in X^{*}$ are linearly independent. Indeed, since $D G (u)$ is surjective, for each $1 \leq i \leq m$ the preimage $D G (u)^{- 1} ({e_{i}})$ of the $i$ -th standard unit vector of [F6] is nonempty, so finite choice [F3], applied to the family indexed by $k < m$ with $i = k + 1$ , selects $x_{1}, \dots, x_{m} \in X$ with $D G (u) x_{i} = e_{i}$ , that is, $ψ_{j} (x_{i}) = δ_{i j}$ ; if $\sum_{i} c_{i} ψ_{i} = 0$ is the zero functional, evaluating at $x_{j}$ gives $c_{j} = 0$ for every $j$ , and independence follows by [F4].

2.1step 1.1step 1.2F5F6

If $m = 0$ , then $\ker D G (u) = X$ by [F6], and step 1.1 gives $D I (u) = 0$ . The unique vector of $R^{0}$ gives the zero empty sum, proving both existence and uniqueness of the multiplier identity. If $m \geq 1$ , apply [F5] with $ψ_{i} = D G_{i} (u)$ and $φ = D I (u)$ : the independence of step 1.2 and the kernel inclusion of step 1.1 are exactly its hypotheses, so there is a unique $λ \in R^{m}$ with $D I (u) = \sum_{i = 1}^{m} λ_{i} D G_{i} (u)$ .

3.1step 2.1A1∎

This is the asserted multiplier identity, with the uniqueness statement included; the constrained-stationarity supplier uses the implicit function theorem under the Axiom of Choice [A1], while the common-kernel argument above uses no additional choice principle.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The multiplier vector is unique when the constraint gradients are independent

Statement

Assume the Axiom of Choice (The Axiom of Choice). In the setting of The Lagrange multiplier rule for finitely many regular constraints — a real Banach space $X$ (Banach space), open $U \subseteq X$ , and a $C^{1}$ map $G = (G_{1}, \dots, G_{m}) : U \to R^{m}$ with $D G (u) : X \to R^{m}$ surjective — if $λ, λ^{'} \in R^{m}$ satisfy $\sum_{i = 1}^{m} λ_{i} D G_{i} (u) = \sum_{i = 1}^{m} λ_{i}^{'} D G_{i} (u)$ in $X^{*}$ , then $λ = λ^{'}$ . Equivalently, the transpose $D G (u)^{*} : R^{m} \to X^{*}$ (The transpose of a bounded operator, Fréchet derivative between Banach spaces) is injective.

Facts & Assumptions

Given: The setting of the multiplier rule: a real Banach space $X$ , open $U \subseteq X$ , a $C^{1}$ map $G$ with $D G (u)$ surjective, and vectors $λ, λ^{'} \in R^{m}$ as in the statement.

[F1]

The Lagrange multiplier rule for finitely many regular constraints: under these hypotheses $D G (u)$ has components $D G_{i} (u) \in X^{*}$ , surjectivity is available exactly as in the multiplier rule, and multiples and sums of the component functionals are formed pointwise.

[F2]

Every natural-number-indexed list of nonempty sets has a choice function on its family of values: a finite family of nonempty sets indexed by a natural number admits a choice function.

[F3]

The transpose of a bounded operator, Fréchet derivative between Banach spaces: for a bounded linear operator $T : X \to R^{m}$ the transpose $T^{*} : (R^{m})^{*} \to X^{*}$ is defined by $T^{*} g = g \circ T$ ; under the identification of $(R^{m})^{*}$ with $R^{m}$ by the standard basis, $(D G (u)^{*} μ) (x) = \sum_{i} μ_{i} D G_{i} (u) x$ , so $D G (u)^{*} μ = 0$ if and only if $\sum_{i} μ_{i} D G_{i} (u)$ is the zero functional.

Proof

technique · direct

Given: The setting above and $λ, λ^{'} \in R^{m}$ with equal associated functionals.

1.1givenF1F2choosealgebra

Put $μ : = λ - λ^{'} \in R^{m}$ . Subtracting the two equal functionals gives $\sum_{i = 1}^{m} μ_{i} D G_{i} (u) = 0$ in $X^{*}$ [F1]. Since $D G (u)$ is surjective, for each $i$ the preimage $D G (u)^{- 1} ({e_{i}})$ is nonempty, and finite choice [F2] selects $x_{1}, \dots, x_{m} \in X$ with $D G (u) x_{i} = e_{i}$ , that is $D G_{j} (u) x_{i} = δ_{i j}$ . Evaluating the vanishing functional at $x_{j}$ gives $0 = \sum_{i} μ_{i} D G_{i} (u) x_{j} = μ_{j}$ , and this holds for every $j$ ; hence $μ = 0$ and $λ = λ^{'}$ .

2.1step 1.1F3

For the equivalent formulation, [F3] identifies $D G (u)^{*} μ \in X^{*}$ with $x \mapsto \sum_{i} μ_{i} D G_{i} (u) x$ . Consequently $D G (u)^{*} μ = 0$ holds exactly when $\sum_{i} μ_{i} D G_{i} (u)$ is the zero functional, which by the evaluation argument of step 1.1 forces $μ = 0$ ; so $D G (u)^{*} : R^{m} \to X^{*}$ is injective.

3.1step 1.1step 2.1∎

Step 1.1 proves the uniqueness of the multiplier vector and step 2.1 the equivalent statement that the transpose of the surjective derivative is injective; this records the independence boundary case in which the surjectivity hypothesis of the multiplier rule cannot be dropped.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The projection onto a closed convex set is characterised by a variational inequality

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a real Hilbert space (Hilbert space), let $K \subseteq H$ be nonempty, closed and convex (Convex sets and continuous real-hyperplane separation in a normed space), let $x \in H$ , and let $P_{K} x$ denote the unique nearest point of $K$ to $x$ , whose existence and uniqueness are supplied by Projection onto a nonempty closed convex set. Then for every $u \in K$ one has $u = P_{K} x ⟺ ⟨ u - x, v - u ⟩ \geq 0 for every v \in K .$

Facts & Assumptions

Given: A real Hilbert space $H$ , a nonempty closed convex set $K \subseteq H$ , a vector $x \in H$ and a point $u \in K$ ; $P_{K} x$ is the unique nearest point of $K$ to $x$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice is the selection principle consumed by the existence-and-uniqueness theorem for nearest points.

[F1]

Projection onto a nonempty closed convex set: under Countable Choice, a nonempty closed convex subset $K$ of a real or complex Hilbert space has exactly one nearest point to $x$ ; hence $P_{K} x$ is well defined and $u = P_{K} x$ holds exactly when $∥ x - u ∥ \leq ∥ x - v ∥$ for every $v \in K$ .

[F2]

Variational characterisation of the nearest point: for $p \in K$ , $p$ is the nearest point of $K$ to $x$ if and only if $Re ⟨ x - p, v - p ⟩ \leq 0$ for every $v \in K$ .

[F3]

Real and complex inner-product spaces and their induced length: the inner product of a real inner product space is real-valued, so $Re ⟨ a, b ⟩ = ⟨ a, b ⟩ = ⟨ b, a ⟩$ and $⟨ a, b ⟩ = - ⟨ - a, b ⟩$ for all vectors $a, b$ ; in particular $u - x = - (x - u)$ .

Proof

technique · direct

Given: A real Hilbert space $H$ , a nonempty closed convex set $K \subseteq H$ , vectors $x \in H$ and $u \in K$ , and the unique nearest point $P_{K} x$ of $K$ to $x$ .

1.1A1F1

By [F1] the point $u$ equals $P_{K} x$ if and only if $u$ is a nearest point of $K$ to $x$ , that is, $∥ x - u ∥ \leq ∥ x - v ∥$ for every $v \in K$ ; here Countable Choice enters through the existence and uniqueness of the nearest point recorded in [A1].

1.2F2

By [F2] the point $u$ is a nearest point of $K$ to $x$ if and only if $Re ⟨ x - u, v - u ⟩ \leq 0$ for every $v \in K$ .

2.1step 1.1step 1.2F3algebra∎

Since the inner product is real-valued [F3], $Re ⟨ x - u, v - u ⟩ = ⟨ x - u, v - u ⟩ = - ⟨ u - x, v - u ⟩$ , so the inequality of step 1.2 is equivalent to $⟨ u - x, v - u ⟩ \geq 0$ for every $v \in K$ ; combining this with step 1.1 gives $u = P_{K} x$ if and only if $⟨ u - x, v - u ⟩ \geq 0$ for every $v \in K$ , which is the asserted characterisation.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The metric projection onto a closed convex set is nonexpansive

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a real Hilbert space (Hilbert space) and let $K \subseteq H$ be nonempty, closed and convex, with metric projection $P_{K} : H \to K$ , the unique nearest point map of Projection onto a nonempty closed convex set. Then for all $x, y \in H$ $∥ P_{K} x - P_{K} y ∥ \leq ∥ x - y ∥,$ and $P_{K}$ is firmly nonexpansive in the equivalent forms $⟨ P_{K} x - P_{K} y, x - y ⟩ \geq ∥ P_{K} x - P_{K} y ∥^{2}, ⟨ (x - P_{K} x) - (y - P_{K} y), P_{K} x - P_{K} y ⟩ \geq 0.$

Facts & Assumptions

Given: A real Hilbert space $H$ , a nonempty closed convex set $K \subseteq H$ , and points $x, y \in H$ ; Countable Choice is available.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice, consumed by the existence-and-uniqueness theorem for nearest points.

[F1]

Projection onto a nonempty closed convex set: under Countable Choice the nearest point $P_{K} z$ of $K$ to $z$ exists and is unique for every $z \in H$ .

[F2]

The projection onto a closed convex set is characterised by a variational inequality: for $u \in K$ one has $u = P_{K} z$ if and only if $⟨ u - z, v - u ⟩ \geq 0$ for every $v \in K$ .

[F3]

Real and complex inner-product spaces and their induced length: the inner product is real-valued on a real inner product space, symmetric and linear in each argument, with $∥ w ∥^{2} = ⟨ w, w ⟩$ .

[F4]

Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ for all vectors $u, v$ of a real or complex inner product space.

Proof

technique · direct

Given: A real Hilbert space $H$ , a nonempty closed convex set $K \subseteq H$ , and points $x, y \in H$ , with Countable Choice available.

1.1A1F1F2

Put $u : = P_{K} x$ and $v : = P_{K} y$ , both well defined by [F1]. By [F2] applied to $u = P_{K} x$ with the admissible point $v \in K$ one has $⟨ u - x, v - u ⟩ \geq 0$ , and applied to $v = P_{K} y$ with the admissible point $u \in K$ one has $⟨ v - y, u - v ⟩ \geq 0$ .

2.1step 1.1F3algebra

Adding the two inequalities of step 1.1 and using $u - v = - (v - u)$ twice gives $0 \leq ⟨ u - x, v - u ⟩ + ⟨ v - y, u - v ⟩ = ⟨ u - x, v - u ⟩ - ⟨ v - y, v - u ⟩ = ⟨ (u - x) - (v - y), v - u ⟩ = ⟨ (u - v) - (x - y), v - u ⟩ = - ∥ u - v ∥^{2} - ⟨ x - y, v - u ⟩ = - ∥ u - v ∥^{2} + ⟨ x - y, u - v ⟩$ , that is $⟨ P_{K} x - P_{K} y, x - y ⟩ = ⟨ u - v, x - y ⟩ \geq ∥ u - v ∥^{2}$ by symmetry of the real inner product, the first firmly nonexpansive form.

3.1step 2.1F3F4algebra

If $u = v$ the nonexpansiveness inequality is trivial; otherwise [F4] gives $∥ u - v ∥^{2} \leq ⟨ u - v, x - y ⟩ \leq ∥ u - v ∥ ∥ x - y ∥$ , and division by the positive number $∥ u - v ∥$ yields $∥ P_{K} x - P_{K} y ∥ = ∥ u - v ∥ \leq ∥ x - y ∥$ .

4.1step 2.1A1F3algebra∎

Finally, $⟨ (x - u) - (y - v), u - v ⟩ = ⟨ x - y, u - v ⟩ - ∥ u - v ∥^{2}$ [F3], so the inequality of step 2.1 is exactly the equivalent form $⟨ (x - P_{K} x) - (y - P_{K} y), P_{K} x - P_{K} y ⟩ \geq 0$ ; this completes the proof of both nonexpansiveness statements for arbitrary $x, y$ and arbitrary admissible $K$ , with Countable Choice used only through the existence of the projections [A1].

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Stampacchia's variational inequality

Statement

Facts & Assumptions

Given: A real Hilbert space $H$ with inner product $⟨ \cdot, \cdot ⟩$ linear in the first argument, a nonempty closed convex $K \subseteq H$ , a bilinear form $a$ bounded by $M$ and coercive with constant $α$ , and a bounded linear functional $F$ , with Countable Choice available.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice, consumed through the Riesz representation theorem and the projection theorem.

[F1]

Riesz representation for Hilbert spaces: there is a unique $f \in H$ with $F (w) = ⟨ w, f ⟩$ for every $w \in H$ .

[F2]

A bounded form is represented by a unique bounded operator: there is a unique bounded linear operator $A \in B (H)$ with $a (u, v) = ⟨ A u, v ⟩$ for all $u, v \in H$ and $∥ A ∥ \leq M$ ; coercivity is equivalent to $⟨ A u, u ⟩ \geq α ∥ u ∥^{2}$ for every $u \in H$ .

[F3]

Bounded, coercive and symmetric sesquilinear forms, Real and complex inner-product spaces and their induced length: $∣ a (u, v) ∣ \leq M ∥ u ∥ ∥ v ∥$ and $a (u, u) \geq α ∥ u ∥^{2}$ , and on a real inner product space $∥ z - ρ A z ∥^{2} = ∥ z ∥^{2} - 2 ρ ⟨ A z, z ⟩ + ρ^{2} ∥ A z ∥^{2}$ .

[F4]

Projection onto a nonempty closed convex set, The metric projection onto a closed convex set is nonexpansive: the metric projection $P_{K} : H \to K$ is well defined and $1$ -Lipschitz on $H$ .

[F5]

Closed subspaces of complete metric spaces are complete; the converse under countable choice, Complete metric space: every Cauchy sequence converges in the space, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric: a closed subset of a complete metric space is complete for the subspace metric; a Hilbert space is complete for its metric.

[F6]

A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point, Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction: a contraction of a nonempty complete metric space has exactly one fixed point.

[F7]

The projection onto a closed convex set is characterised by a variational inequality: for $u \in K$ one has $u = P_{K} x$ if and only if $⟨ u - x, v - u ⟩ \geq 0$ for every $v \in K$ .

[F8]

A bounded linear operator between normed spaces: a bounded linear operator is continuous and $∥ A w ∥ \leq ∥ A ∥ ∥ w ∥$ for every $w$ .

[F9]

Bounded, coercive and symmetric sesquilinear forms: in the real convention, boundedness and coercivity are defined by their inequalities without requiring symmetry; symmetry is an additional property.

Proof

technique · direct

Given: A real Hilbert space $H$ , a nonempty closed convex $K \subseteq H$ , a bounded coercive bilinear form $a$ with constants $M, α > 0$ , a bounded linear functional $F$ , and Countable Choice.

1.1givenA1F1F2F9

By [F1] fix $f \in H$ with $F (w) = ⟨ w, f ⟩$ for all $w$ ; by [F2] fix $A \in B (H)$ with $a (u, v) = ⟨ A u, v ⟩$ , $∥ A ∥ \leq M$ and $⟨ A u, u ⟩ \geq α ∥ u ∥^{2}$ for all $u$ . The real bilinear form need not be symmetric by [F9].

2.1step 1.1F2F3F4F6F8algebra

If $H = {0}$ , then $K = {0}$ and $u = 0$ is the unique solution, since $a (0, 0) = F (0) = 0$ . Otherwise choose $z_{0} \neq 0$ ; boundedness and coercivity give $α ∥ z_{0} ∥^{2} \leq a (z_{0}, z_{0}) \leq M ∥ z_{0} ∥^{2}$ , so $0 < α \leq M$ . Choose $ρ : = α / M^{2}$ and $q : = \sqrt{1 - α^{2} / M^{2}}$ , which satisfies $0 \leq q < 1$ and $q^{2} = 1 - 2 ρ α + ρ^{2} M^{2}$ . For $z \in H$ the expansion of [F3] together with $⟨ A z, z ⟩ \geq α ∥ z ∥^{2}$ and $∥ A z ∥ \leq M ∥ z ∥$ [F2, F8] gives $∥ (I - ρ A) z ∥^{2} = ∥ z ∥^{2} - 2 ρ ⟨ A z, z ⟩ + ρ^{2} ∥ A z ∥^{2} \leq (1 - 2 ρ α + ρ^{2} M^{2}) ∥ z ∥^{2} = q^{2} ∥ z ∥^{2}$ . Hence the map $T (w) : = P_{K} (w - ρ (A w - f))$ satisfies $∥ T (w) - T (w^{'}) ∥ \leq ∥ (I - ρ A) (w - w^{'}) ∥ \leq q ∥ w - w^{'} ∥$ for all $w, w^{'} \in K$ by the nonexpansiveness of $P_{K}$ [F4], that is, $T$ is a contraction of $K$ with constant $q < 1$ [F6].

3.1step 2.1F5F6

The set $K$ is a nonempty closed subset of the complete metric space $H$ , hence complete for the subspace metric [F5]; the contraction $T : K \to K$ of step 2.1 therefore has exactly one fixed point $u \in K$ by [F6], that is, $u = P_{K} (u - ρ (A u - f))$ .

4.1step 1.1step 3.1F7algebra

For $u \in K$ , the fixed point equation $u = P_{K} (u - ρ (A u - f))$ is equivalent, by [F7] applied with $x = u - ρ (A u - f)$ , to $⟨ u - (u - ρ (A u - f)), v - u ⟩ \geq 0$ for every $v \in K$ , that is, to $ρ ⟨ A u - f, v - u ⟩ \geq 0$ for every $v \in K$ ; since $ρ > 0$ this is equivalent to $⟨ A u - f, v - u ⟩ \geq 0$ , hence to $a (u, v - u) \geq F (v - u)$ for every $v \in K$ by step 1.1.

5.1step 4.1A1algebra∎

(Uniqueness and conclusion) Let $u, u^{'} \in K$ both satisfy the variational inequality. Testing the inequality for $u$ at $v = u^{'}$ and the inequality for $u^{'}$ at $v = u$ and adding gives $a (u, u^{'} - u) + a (u^{'}, u - u^{'}) \geq F (u^{'} - u) + F (u - u^{'}) = 0$ ; bilinearity turns the left-hand side into $- a (u - u^{'}, u - u^{'})$ , so $a (u - u^{'}, u - u^{'}) \leq 0$ , and coercivity gives $α ∥ u - u^{'} ∥^{2} \leq a (u - u^{'}, u - u^{'}) \leq 0$ , hence $u = u^{'}$ . The zero-dimensional case was settled in step 2.1; in the remaining case steps 3.1 and 4.1 exhibit the unique fixed point $u \in K$ which solves the inequality, and the uniqueness argument just given makes it the only solution. This proves the statement, Countable Choice having entered only through [F1] and [F4] [A1].

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Endpoint trace commutes with Sobolev truncation on an interval

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $I = (a, b)$ with $- \infty < a < b < \infty$ , let $1 \leq p < \infty$ , and let $u \in W^{1, p} (I; R)$ (Integer-order Sobolev spaces and their norms). Let $u^{*}$ be its unique absolutely continuous representative on $[a, b]$ (One-dimensional $W^{1, p}$ functions have unique absolutely continuous representatives) and put $T u : = (u^{*} (a), u^{*} (b)) \in R^{2}$ , ordered componentwise. Then:

$T$ is well defined, linear and bounded, and $\ker T = W_{0}^{1, p} (I)$ (Zero-boundary Sobolev space as a norm closure).
For every $k \in R$ the function $(u - k)^{+}$ belongs to $W^{1, p} (I)$ , and $T ((u - k)^{+}) = (T u - k)^{+}$ componentwise; moreover $(u - k)^{+} \in W_{0}^{1, p} (I)$ if and only if $T u \leq (k, k)$ componentwise. In particular $T (u^{+}) = (T u)^{+}$ , $T (u^{-}) = (T u)^{-}$ and $T ∣ u ∣ = ∣ T u ∣$ .

Facts & Assumptions

Given: An interval $I = (a, b)$ with finite endpoints, an exponent $1 \leq p < \infty$ , a class $u \in W^{1, p} (I; R)$ with weak derivative $u^{'}$ , its unique absolutely continuous representative $u^{*}$ on $[a, b]$ , the trace pair $T u = (u^{*} (a), u^{*} (b))$ , and a real number $k$ .

[A1]

The Axiom of Choice: the Axiom of Choice, consumed only through the cited suppliers that assume it or Countable Choice.

[F1]

One-dimensional $W^{1, p}$ functions have unique absolutely continuous representatives: for $u \in W^{1, p} (I)$ there is exactly one continuous locally absolutely continuous representative $u^{*}$ of the class, and on a bounded interval it extends uniquely to an absolutely continuous function on $[a, b]$ with $u^{*} (x) = u^{*} (a) + \int_{a}^{x} u^{'}$ for $x \in [a, b]$ ; in particular $u^{*} (x) - u^{*} (y) = \int_{y}^{x} u^{'}$ for all $x, y \in [a, b]$ .

[F2]

The one-dimensional endpoint estimate on a bounded interval: with $ε = (b - a) / 2$ one has $∣ u^{*} (a) ∣ \leq C_{p} (a, b) ∥ u ∥_{W^{1, p}}$ and $∣ u^{*} (b) ∣ \leq C_{p} (a, b) ∥ u ∥_{W^{1, p}}$ , where $C_{p} (a, b) < \infty$ is the explicit constant of that estimate (for $1 < p < \infty$ the $p$ -th powers obey the displayed bound with $2^{p - 1} (ε^{- 1} \int_{a}^{a + ε} ∣ u^{*} ∣^{p} + ε^{p - 1} \int_{a}^{a + ε} ∣ u^{'} ∣^{p})$ , for $p = 1$ the unsquared bound).

[F3]

Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure: $W^{1, p} (I)$ consists of $L^{p}$ classes with $L^{p}$ weak derivatives, and $W_{0}^{1, p} (I)$ is the closure of $C_{c}^{\infty} (I)$ in the $W^{1, p}$ norm; in particular $W_{0}^{1, p} (I)$ is closed in $W^{1, p} (I)$ .

[F4]

Linearity, locality, and commutation of weak derivatives: weak differentiation is linear.

[F5]

Classical derivatives agree with weak derivatives, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included: a $C^{1}$ function on $I$ has its classical derivative as weak derivative; the interval $I = (a, b)$ is a box with finite Lebesgue measure, so the constant function $1$ lies in $L^{p} (I)$ and hence in $W^{1, p} (I)$ with weak derivative $0$ , the classical derivative of the constant.

[F6]

Positive, negative, and truncated Sobolev functions: for a real class $w \in W^{1, p} (I)$ the classes $w^{+} = \max {w, 0}$ and $w^{-} = \max {- w, 0}$ lie in $W^{1, p} (I)$ with $D w^{+} = 1_{{w > 0}} D w$ and $D w^{-} = - 1_{{w < 0}} D w$ a.e.

[F7]

Compactly supported scaled Euclidean bumps, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ : in dimension one there is a fixed smooth bump $b_{0}$ with $0 \leq b_{0} \leq 1$ , $b_{0} = 1$ on $[- 1, 1]$ and $supp b_{0} \subseteq (- 2, 2)$ ; by the chain rule the rescalings $x \mapsto b_{0} ((x - c) / r)$ and $x \mapsto b_{0} ((c - x) / r)$ are smooth with derivatives $\pm r^{- 1} b_{0}^{'} ((x - c) / r)$ and $\mp r^{- 1} b_{0}^{'} ((c - x) / r)$ , so these derivatives are bounded in absolute value by $∥ b_{0}^{'} ∥_{\infty} / r$ , and products of such rescalings obey the product rule.

[F8]

Holder's inequality for integrals, including the endpoint cases: for $1 < p < \infty$ , $\int_{a}^{x} ∣ u^{'} ∣ \leq (x - a)^{1 - 1 / p} (\int_{a}^{x} ∣ u^{'} ∣^{p})^{1 / p}$ for $x \in (a, b)$ , and for $p = 1$ the left-hand side is itself $\int_{a}^{x} ∣ u^{'} ∣^{1}$ ; the reflected estimate holds on $(x, b)$ .

[F9]

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Dominated convergence: iterated integrals of nonnegative measurable functions on the finite-measure product $(a, b) \times (a, b)$ may be interchanged, and pointwise convergent dominated families have convergent integrals.

[F10]

Weak Leibniz rule with a smooth factor: for $u \in W^{1, p} (I)$ and $η \in C_{c}^{\infty} (I)$ one has $η u \in W^{1, p} (I)$ with $D (η u) = η^{'} u + η u^{'}$ ; also $∣ b_{0}^{'} ∣$ is bounded on the compact support of $b_{0}$ .

[F11]

Compactly supported Sobolev functions extend by zero in every integer order, Compactly supported smooth functions are dense in W^{k,p}(R^n): a class in $W^{1, p} (I)$ vanishing a.e. outside a compact subset of $I$ extends by zero to a class in $W^{1, p} (R)$ with the same norm, and $C_{c}^{\infty} (R)$ is dense in $W^{1, p} (R)$ for $p < \infty$ .

[F12]

Bounded restriction and cutoff localisation in Sobolev spaces: restriction to an open subset is a contraction on $W^{1, p}$ , and multiplication by a fixed $ψ \in C_{c}^{\infty} (I)$ is bounded on $W^{1, p} (I)$ with a constant depending only on finitely many sup norms of derivatives of $ψ$ .

[F13]

Test function cutoffs and euclidean localization: for compact $K \subseteq I$ there is $ψ \in C_{c}^{\infty} (I)$ with $0 \leq ψ \leq 1$ and $ψ = 1$ on a neighbourhood of $K$ ; the construction is choice-free.

Proof

technique · direct

Given: An interval $I = (a, b)$ with finite endpoints, $1 \leq p < \infty$ , a class $u \in W^{1, p} (I; R)$ with weak derivative $u^{'}$ , its absolutely continuous representative $u^{*}$ on $[a, b]$ , the pair $T u = (u^{*} (a), u^{*} (b))$ , and $k \in R$ .

1.1givenF1A1

(Well-definedness of $T$ ) By [F1] the class $u$ has exactly one continuous locally absolutely continuous representative, it extends uniquely to an absolutely continuous function $u^{*}$ on $[a, b]$ , and its endpoint values are determined by the class; hence $T$ is well defined, and [F1] also gives $u^{*} (x) = u^{*} (a) + \int_{a}^{x} u^{'}$ for every $x \in [a, b]$ .

1.2givenF2

(Norm estimate) With $ε = (b - a) / 2$ , [F2] bounds $∣ u^{*} (a) ∣$ and $∣ u^{*} (b) ∣$ by $C_{p} (a, b) ∥ u ∥_{W^{1, p}}$ , hence $∥ T u ∥_{R^{2}} \leq \sqrt{2} C_{p} (a, b) ∥ u ∥_{W^{1, p}}$ for every $u$ .

1.3givenF1F4

(Linearity) For $u, v \in W^{1, p} (I)$ the class $u + v$ has weak derivative $u^{'} + v^{'}$ by [F4], and $u^{*} + v^{*}$ is continuous and absolutely continuous; by the uniqueness clause of [F1] it is the representative of $u + v$ , so $(u + v)^{*} = u^{*} + v^{*}$ and $T (u + v) = T u + T v$ ; for $c \in R$ the class $c u$ has weak derivative $c u^{'}$ and representative $c u^{*}$ by [F1] and [F4], so $T (c u) = c T u$ ; hence $T$ is linear.

1.4givenF4F5F6

(Truncation membership) The constant function $k$ lies in $W^{1, p} (I)$ with weak derivative $0$ by [F5], so $w : = u - k$ lies in $W^{1, p} (I)$ with weak derivative $u^{'}$ by [F4]; by [F6] the classes $w^{+} = \max {w, 0}$ and $w^{-} = \max {- w, 0}$ lie in $W^{1, p} (I)$ with $D w^{+} = 1_{{w > 0}} u^{'}$ and $D w^{-} = - 1_{{w < 0}} u^{'}$ a.e.; in particular $(u - k)^{+} \in W^{1, p} (I)$ .

2.1step 1.4F1F6

(Endpoint values of the truncated class) Apply [F1] to the class $w^{+}$ of step 1.4: it has a unique continuous representative $w^{*}$ on $[a, b]$ , absolutely continuous, with $w^{*} (x) = w^{*} (a) + \int_{a}^{x} D w^{+}$ . The continuous function $x \mapsto (u^{*} (x) - k)^{+}$ on $[a, b]$ is a representative of $w^{+}$ because $u^{*} - k$ represents $u - k$ and the positive part is well defined on a.e. classes [F6]; two continuous representatives of one class agree on the dense interval $(a, b)$ , hence by continuity on all of $[a, b]$ . Therefore $T (w^{+}) = ((u^{*} (a) - k)^{+}, (u^{*} (b) - k)^{+}) = ((T u - k)^{+})$ componentwise.

2.2step 1.2F1F3

( $W_{0}^{1, p} \subseteq \ker T$ ) Each $φ \in C_{c}^{\infty} (I)$ , extended by zero, is its own absolutely continuous representative with $φ (a) = φ (b) = 0$ , so $T φ = 0$ by [F1]; by step 1.2 the map $T$ is bounded, hence continuous, and by [F3] the space $W_{0}^{1, p} (I)$ is the closure of $C_{c}^{\infty} (I)$ , so $T$ vanishes on all of $W_{0}^{1, p} (I)$ .

2.3step 1.1F1F8algebra

(Endpoint decay when $T u = 0$ ) Assume $T u = 0$ , so $u^{*} (a) = u^{*} (b) = 0$ . For $x \in (a, b)$ , [F1] gives $u^{*} (x) = u^{*} (a) + \int_{a}^{x} u^{'} = \int_{a}^{x} u^{'}$ and, symmetrically, $u^{*} (x) = u^{*} (b) - \int_{x}^{b} u^{'} = - \int_{x}^{b} u^{'}$ ; hence $∣ u^{*} (x) ∣ \leq \int_{a}^{x} ∣ u^{'} ∣$ and $∣ u^{*} (x) ∣ \leq \int_{x}^{b} ∣ u^{'} ∣$ . For $p > 1$ [F8] turns this into $∣ u^{*} (x) ∣^{p} \leq (x - a)^{p - 1} \int_{a}^{x} ∣ u^{'} ∣^{p}$ and $∣ u^{*} (x) ∣^{p} \leq (b - x)^{p - 1} \int_{x}^{b} ∣ u^{'} ∣^{p}$ , with the same inequalities for $p = 1$ since $(x - a)^{0} = (b - x)^{0} = 1$ .

3.1step 2.3F7F8F9F10algebra

(Cutoffs and convergence to $u$ ) Assume $T u = 0$ and fix $0 < δ < (b - a) / 4$ . Let $b_{0}$ be the bump of [F7] and put $η_{δ} (x) : = 1 - b_{0} ((x - a) / δ)$ , $ρ_{δ} (x) : = 1 - b_{0} ((b - x) / δ)$ , $χ_{δ} : = η_{δ} ρ_{δ}$ . Then $0 \leq χ_{δ} \leq 1$ , $χ_{δ} = 1$ on $[a + 2 δ, b - 2 δ]$ , $χ_{δ} = 0$ on $[a, a + δ] \cup [b - δ, b]$ , so $supp χ_{δ} \subseteq [a + δ, b - δ] \subset I$ , and the chain and product rules give $∣ χ_{δ}^{'} ∣ \leq 2 M / δ$ with $M : = ∥ b_{0}^{'} ∥_{\infty} < \infty$ [F7]. By [F10] $χ_{δ} u \in W^{1, p} (I)$ with $D (χ_{δ} u) = χ_{δ}^{'} u + χ_{δ} u^{'}$ ; moreover $χ_{δ} \to 1$ pointwise on $I$ , so dominated convergence [F9] gives $∥ (1 - χ_{δ}) u ∥_{p} \to 0$ and $∥ (1 - χ_{δ}) u^{'} ∥_{p} \to 0$ . For the remaining term, [F8] and step 2.3 give pointwise a.e. $∣ u^{*} ∣^{p} \leq (x - a)^{p - 1} \int_{a}^{x} ∣ u^{'} ∣^{p}$ on $(a, a + 2 δ)$ , so Tonelli [F9] yields $\int_{a}^{a + 2 δ} ∣ u^{*} ∣^{p} \leq \frac{(2 δ)^{p}}{p} \int_{a}^{a + 2 δ} ∣ u^{'} ∣^{p}$ , and reflecting at $b$ the same bound holds on $(b - 2 δ, b)$ ; consequently $\int_{I} ∣ χ_{δ}^{'} u ∣^{p} \leq (2 M / δ)^{p} \int_{(a, a + 2 δ) \cup (b - 2 δ, b)} ∣ u^{*} ∣^{p} \leq \frac{(4 M)^{p}}{p} \int_{(a, a + 2 δ) \cup (b - 2 δ, b)} ∣ u^{'} ∣^{p} \to 0$ because the endpoint regions shrink to null sets and $∣ u^{'} ∣^{p}$ is integrable [F9]. Hence $D (χ_{δ} u) \to u^{'}$ in $L^{p}$ and $χ_{δ} u \to u$ in $L^{p}$ , that is, $χ_{δ} u \to u$ in $W^{1, p} (I)$ .

4.1step 3.1F3F11F12F13

(Each $χ_{δ} u$ lies in $W_{0}^{1, p}$ ) Fix $δ$ as in step 3.1. Since $χ_{δ} u$ vanishes a.e. outside the compact set $[a + δ, b - δ] \subset I$ , its zero extension $E_{0} (χ_{δ} u)$ lies in $W^{1, p} (R)$ with equal norm by [F11], and by the density corollary there are $φ_{j} \in C_{c}^{\infty} (R)$ with $φ_{j} \to E_{0} (χ_{δ} u)$ in $W^{1, p} (R)$ . By [F13] choose $ψ \in C_{c}^{\infty} (I)$ with $ψ = 1$ on a neighbourhood of $[a + δ, b - δ]$ ; then $ψ φ_{j} \in C_{c}^{\infty} (I)$ and $ψ φ_{j} - χ_{δ} u$ is the restriction to $I$ of $ψ (φ_{j} - E_{0} (χ_{δ} u))$ , so the multiplication and restriction bounds of [F12] give $∥ ψ φ_{j} - χ_{δ} u ∥_{W^{1, p} (I)} \leq C_{ψ} ∥ φ_{j} - E_{0} (χ_{δ} u) ∥_{W^{1, p} (R)} \to 0$ . Thus $χ_{δ} u$ lies in the closure of $C_{c}^{\infty} (I)$ , which is $W_{0}^{1, p} (I)$ by [F3].

5.1step 1.2step 2.2step 4.1F3

((i) concluded: $\ker T = W_{0}^{1, p}$ ) Steps 1.1-1.3 show that $T$ is well defined, linear and bounded (with the estimate of step 1.2); step 2.2 gives $W_{0}^{1, p} (I) \subseteq \ker T$ . Conversely, if $T u = 0$ , then step 3.1 exhibits $χ_{δ} u \to u$ in $W^{1, p} (I)$ with $χ_{δ} u \in W_{0}^{1, p} (I)$ for each $δ$ by step 4.1; since $W_{0}^{1, p} (I)$ is closed in $W^{1, p} (I)$ [F3], $u \in W_{0}^{1, p} (I)$ . Hence $\ker T = W_{0}^{1, p} (I)$ .

6.1step 1.4step 2.1step 5.1

(The truncation iff) Fix $k \in R$ . By step 2.1, $T ((u - k)^{+}) = ((T u - k)^{+})$ ; by step 5.1, $(u - k)^{+} \in W_{0}^{1, p} (I)$ exactly when this trace pair vanishes, that is, exactly when $(u^{*} (a) - k)^{+} = 0$ and $(u^{*} (b) - k)^{+} = 0$ , equivalently $u^{*} (a) \leq k$ and $u^{*} (b) \leq k$ , equivalently $T u \leq (k, k)$ componentwise; membership $(u - k)^{+} \in W^{1, p} (I)$ is step 1.4.

7.1step 1.3step 5.1step 6.1A1∎

(The particular identities and discharge) Taking $k = 0$ in step 6.1 gives $T (u^{+}) = (T u)^{+}$ and $u^{+} \in W_{0}^{1, p} (I)$ exactly when $T u \leq (0, 0)$ . Applying step 6.1 to the class $- u$ , whose representative is $- u^{*}$ and whose trace pair is $- T u$ , gives $T ((- u)^{+}) = (- T u)^{+} = (T u)^{-}$ , that is $T (u^{-}) = (T u)^{-}$ ; since $∣ u ∣ = u^{+} + u^{-}$ , linearity of $T$ from step 1.3 gives $T ∣ u ∣ = T (u^{+}) + T (u^{-}) = (T u)^{+} + (T u)^{-} = ∣ T u ∣$ componentwise. Steps 5.1, 6.1 and the present step prove all the assertions; the Axiom of Choice was used only through the representative interfaces [F1], the endpoint estimate [F2] and the truncation, extension, density and multiplication interfaces [F6, F11, F12], which assume it or Countable Choice [A1].

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The closed convex obstacle set and the obstacle variational inequality

Definition

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 1$ , and let $Ω \subseteq R^{n}$ be a bounded open set of one of the following two kinds (Bounded C^k domains and boundary charts).

Trace conventions. If $n = 1$ , then $Ω = (a, b)$ is a bounded interval and $T$ is the endpoint-pair trace of Endpoint trace commutes with Sobolev truncation on an interval, ordered componentwise. If $n \geq 2$ , then $Ω \subset R^{n}$ is a bounded $C^{1}$ domain and $T$ is the trace operator of The $L^{p}$ trace operator on a bounded $C^{1}$ domain. An obstacle is a real function $ψ$ with $ψ \in H^{1} (Ω; R)$ (The notation $H^{k}$ and the reserved zero-boundary symbol, Integer-order Sobolev spaces and their norms) such that $T ψ \leq 0$ — componentwise at the two endpoints when $n = 1$ , and almost everywhere on $\partial Ω$ when $n \geq 2$ . By Endpoint trace commutes with Sobolev truncation on an interval for $n = 1$ and A function whose trace is at most a level has positive part in the zero-boundary space for $n \geq 2$ , this boundary order condition is equivalent to $ψ^{+} \in H_{0}^{1} (Ω)$ .

The obstacle admissible set is $K : = {v \in H_{0}^{1} (Ω; R) : v \geq ψ a.e. on Ω},$ where the inequality $v \geq ψ$ is imposed on almost-everywhere classes and is therefore independent of the chosen representatives (The space $L^{p} (μ)$ as the quotient by null functions, Zero-boundary Sobolev space as a norm closure).

Data. Let $a : H_{0}^{1} (Ω) \times H_{0}^{1} (Ω) \to R$ be a bounded coercive bilinear form with constants $M, α > 0$ (Bounded, coercive and symmetric sesquilinear forms) and let $F \in H^{- 1} (Ω)$ (The negative Sobolev space $H^{- 1} (Ω)$ ).

The obstacle variational inequality is the problem: find $u \in K$ with $a (u, v - u) \geq F (v - u) for every v \in K .$

Energy and reaction. In the symmetric case the associated energy is $J (v) : = \frac{1}{2} a (v, v) - F (v)$ for $v \in H_{0}^{1} (Ω)$ ; the reaction distribution of a solution $u$ is the distribution on $Ω$ defined by $Λ_{u} (φ) : = a (u, φ) - F (φ) (φ \in C_{c}^{\infty} (Ω)),$ the right-hand side being well defined on real test functions because $C_{c}^{\infty} (Ω; R) \subseteq H_{0}^{1} (Ω; R)$ (Test function space d of an open set). It is linear and continuous: boundedness gives $∣ Λ_{u} (φ) ∣ \leq (M ∥ u ∥_{H^{1}} + ∥ F ∥_{H^{- 1}}) ∥ φ ∥_{H^{1}}$ , and for tests supported in a compact $S \subseteq Ω$ , $∥ φ ∥_{H^{1}} \leq \sqrt{(n + 1) ∣ S ∣} \max_{∣ α ∣ \leq 1} ∥ D^{α} φ ∥_{\infty}$ . Thus $∣ Λ_{u} ∣$ is a seminorm continuous on every fixed-support test space and hence continuous for Test function topology. Its complex-linear extension is $Λ_{u} (φ_{1} + i φ_{2}) = Λ_{u} (φ_{1}) + i Λ_{u} (φ_{2})$ for real tests $φ_{1}, φ_{2}$ , giving a distribution in the convention of Distribution; nonnegativity is tested on real nonnegative tests.

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The obstacle admissible set is nonempty, convex, closed and weakly closed

Statement

Assume Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice), inherited through the truncation and trace suppliers named below. Let $n \geq 1$ ; when $n = 1$ use a bounded interval and the endpoint trace of Endpoint trace commutes with Sobolev truncation on an interval, and when $n \geq 2$ use a bounded $C^{1}$ domain with the published trace definitions (Bounded C^k domains and boundary charts). Let $ψ \in H^{1} (Ω; R)$ satisfy $T ψ \leq 0$ (componentwise at the endpoints for $n = 1$ , almost everywhere on $\partial Ω$ for $n \geq 2$ ), so that $K = {v \in H_{0}^{1} (Ω) : v \geq ψ a.e.}$ is the admissible set of The closed convex obstacle set and the obstacle variational inequality (Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms). Then $K$ is nonempty, convex, closed in the norm of $H_{0}^{1} (Ω)$ and weakly sequentially closed: if $(v_{j}) \subseteq K$ and $v_{j} ⇀ v$ in $H_{0}^{1} (Ω)$ , then $v \in K$ (Weak convergence of nets and sequences).

Facts & Assumptions

Given: The setting above and an obstacle $ψ \in H^{1} (Ω; R)$ with $T ψ \leq 0$ ; the admissible set $K = {v \in H_{0}^{1} (Ω) : v \geq ψ a.e.}$ .

[F1]

The closed convex obstacle set and the obstacle variational inequality: the boundary condition $T ψ \leq 0$ is equivalent to $ψ^{+} \in H_{0}^{1} (Ω)$ (through the interval lemma for $n = 1$ and A function whose trace is at most a level has positive part in the zero-boundary space for $n \geq 2$ ), and $K$ is defined by the representative-independent almost-everywhere inequality $v \geq ψ$ .

[F2]

Positive, negative, and truncated Sobolev functions: $ψ^{+} = \max {ψ, 0}$ is the class of the pointwise maximum, and on representatives $ψ^{+} \geq ψ$ pointwise.

[F3]

Assuming Countable Choice, $L^{p}$ -convergent sequences have almost-everywhere convergent subsequences: a sequence converging in $L^{2} (Ω)$ has a subsequence converging almost everywhere on $Ω$ ; a sequence converging in $H_{0}^{1} (Ω)$ therefore has such a subsequence, since the $H_{0}^{1}$ norm dominates the $L^{2}$ norm (Integer-order Sobolev spaces and their norms).

[F4]

Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms: $H_{0}^{1} (Ω)$ is a closed linear subspace of $H^{1} (Ω)$ , and the almost-everywhere inequality $v \geq ψ$ is a condition on classes, independent of representatives.

[F5]

A norm-closed convex set is weakly sequentially closed: under the Axiom of Choice a convex norm closed subset of a normed space is weakly closed, hence weakly sequentially closed.

Proof

technique · direct

Given: The setting above, with $T ψ \leq 0$ and $K$ as defined.

1.1givenF1F2

(Nonemptiness) By [F1] the boundary condition gives $ψ^{+} \in H_{0}^{1} (Ω)$ , and by [F2] one has $ψ^{+} \geq ψ$ almost everywhere, so $ψ^{+} \in K$ .

1.2givenF4

(Convexity) Let $v, w \in K$ and $0 \leq t \leq 1$ . Then $t v + (1 - t) w \in H_{0}^{1} (Ω)$ because $H_{0}^{1} (Ω)$ is a linear subspace [F4], and $t v + (1 - t) w \geq ψ$ almost everywhere because this holds for $v$ and $w$ ; hence $t v + (1 - t) w \in K$ and $K$ is convex.

1.3givenF3F4

(Norm closedness) Let $(v_{j}) \subseteq K$ with $v_{j} \to v$ in $H_{0}^{1} (Ω)$ . By [F3] a subsequence converges to $v$ almost everywhere, and $v_{j} \geq ψ$ almost everywhere for every $j$ , so $v \geq ψ$ almost everywhere; since $H_{0}^{1} (Ω)$ is closed in $H^{1} (Ω)$ [F4], $v \in H_{0}^{1} (Ω)$ and hence $v \in K$ by [F4].

2.1step 1.1step 1.2step 1.3F5∎

(Weak sequential closedness and conclusion) By steps 1.1-1.3 the set $K$ is nonempty, convex and norm closed; [F5] therefore makes $K$ weakly closed, so in particular weakly sequentially closed: every weakly convergent sequence in $K$ has its limit in $K$ . This proves all the assertions; the choice principles enter only through the truncation interface of [F1] and the weak-closedness criterion [F5].

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Existence and uniqueness for the obstacle problem

Statement

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )), inherited from the obstacle setting and Hilbert-space supplier. Let $Ω$ , $ψ$ , $K$ , $a$ and $F$ be as in The closed convex obstacle set and the obstacle variational inequality, with $a$ symmetric as well as bounded and coercive (Bounded, coercive and symmetric sesquilinear forms) and $F \in H^{- 1} (Ω)$ (The negative Sobolev space $H^{- 1} (Ω)$ ), and let $J (v) = \frac{1}{2} a (v, v) - F (v)$ be the energy. Then $J$ attains its infimum on $K$ at exactly one $u \in K$ , and $u$ is the unique solution of the obstacle variational inequality $a (u, v - u) \geq F (v - u) (v \in K) .$ Equivalently: $u \in K$ minimises $J$ on $K$ if and only if $u$ solves the variational inequality.

Facts & Assumptions

Given: The obstacle setting of The closed convex obstacle set and the obstacle variational inequality: the admissible set $K \subseteq H_{0}^{1} (Ω)$ (Zero-boundary Sobolev space as a norm closure, The notation $H^{k}$ and the reserved zero-boundary symbol, Integer-order Sobolev spaces and their norms), a symmetric bounded coercive bilinear form $a$ with constants $M, α > 0$ , a functional $F \in H^{- 1} (Ω)$ , and the energy $J (v) = \frac{1}{2} a (v, v) - F (v)$ .

[A1]

The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ ): the Axiom of Choice and Countable Choice are available, as required by the obstacle setting and Hilbert-space supplier.

[F1]

The obstacle admissible set is nonempty, convex, closed and weakly closed: $K$ is nonempty, convex and closed in the norm of $H_{0}^{1} (Ω)$ , hence weakly sequentially closed.

[F2]

Stampacchia's variational inequality: for a nonempty closed convex $K \subseteq H_{0}^{1} (Ω)$ and a bounded coercive bilinear form $a$ (no symmetry needed) there is exactly one $u \in K$ with $a (u, v - u) \geq F (v - u)$ for every $v \in K$ .

[F3]

Bounded, coercive and symmetric sesquilinear forms: $a$ is bilinear and symmetric with $a (w, w) \geq α ∥ w ∥^{2} \geq 0$ and $∣ a (w_{1}, w_{2}) ∣ \leq M ∥ w_{1} ∥ ∥ w_{2} ∥$ for all $w, w_{1}, w_{2}$ ; consequently $a (u + t w, u + t w) = a (u, u) + 2 t a (u, w) + t^{2} a (w, w)$ for all real $t$ .

[F4]

The negative Sobolev space $H^{- 1} (Ω)$ : $F$ is a bounded linear functional on $H_{0}^{1} (Ω)$ , so $F$ is linear and $J$ is a real-valued function on $H_{0}^{1} (Ω)$ .

[F5]

$H^{k}$ is a Hilbert space under the derivative-sum inner product, Zero-boundary Sobolev space as a norm closure, A closed subspace of a Banach space is Banach: under the Axiom of Choice, $H^{1} (Ω; R)$ is a real Hilbert space; its closed linear subspace $H_{0}^{1} (Ω; R)$ is complete with the restricted inner product and hence is a real Hilbert space.

Proof

technique · direct

Given: The setting above, with $K$ nonempty closed convex by [F1] and $a$ symmetric bounded coercive.

1.1givenA1F1F2F5

By [F5] the space $H_{0}^{1} (Ω; R)$ is a real Hilbert space. By [F2] applied to the admissible set $K$ of [F1] there is exactly one $u^{*} \in K$ with $a (u^{*}, v - u^{*}) \geq F (v - u^{*})$ for every $v \in K$ ; call it the variational solution.

1.2givenF3F4

Every variational solution minimises $J$ on $K$ : if $u \in K$ satisfies the inequality and $v \in K$ with $w : = v - u$ , then $J (v) - J (u) = \frac{1}{2} a (v, v) - \frac{1}{2} a (u, u) - F (w) = \frac{1}{2} a (w, w) + a (u, w) - F (w)$ by the symmetry and bilinearity of [F3], and this is at least $\frac{α}{2} ∥ w ∥^{2} \geq 0$ because $a (w, w) \geq α ∥ w ∥^{2}$ and $a (u, w) - F (w) \geq 0$ by the inequality and the linearity of $F$ [F4].

1.3givenF1F3F4algebra

Every minimiser solves the variational inequality: let $u$ minimise $J$ on $K$ , let $v \in K$ , put $w : = v - u$ , and for $0 < t \leq 1$ let $v_{t} : = u + t w \in K$ , which lies in $K$ by convexity [F1]. Then $0 \leq J (v_{t}) - J (u) = t (a (u, w) - F (w)) + \frac{t^{2}}{2} a (w, w)$ by [F3, F4]. If $c : = a (u, w) - F (w)$ were negative, then $a (w, w) \geq 0$ would give $0 \leq c + \frac{t}{2} a (w, w)$ for all $t$ ; choosing $t \leq 1$ with $t a (w, w) \leq ∣ c ∣$ when $a (w, w) > 0$ , and any $t$ when $a (w, w) = 0$ , yields $c + \frac{t}{2} a (w, w) \leq c + \frac{∣ c ∣}{2} = \frac{c}{2} < 0$ , a contradiction; hence $c \geq 0$ , that is $a (u, w) \geq F (w)$ .

2.1step 1.1step 1.2step 1.3A1∎

By step 1.1 there is exactly one variational solution $u^{*}$ , and it minimises $J$ by step 1.2. Conversely, if $u \in K$ minimises $J$ , then step 1.3 makes $u$ a variational solution, hence $u = u^{*}$ by the uniqueness in step 1.1. Therefore $J$ attains its infimum on $K$ at exactly one point, namely the variational solution $u^{*}$ , and $u$ minimises $J$ if and only if $u$ solves the variational inequality; the quantitative form $J (v) - J (u^{*}) \geq \frac{α}{2} ∥ v - u^{*} ∥^{2}$ of step 1.2 makes the minimiser unique as well. Countable Choice is consumed by [F2]; the Axiom of Choice supplies the obstacle trace conventions of [F1] and the Hilbert-space prerequisite [F5].

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Lipschitz stability of strongly monotone variational inequalities

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a real Hilbert space (Hilbert space), let $K \subseteq H$ be nonempty, closed and convex, let $a$ be a bounded coercive bilinear form with constants $M, α > 0$ (Bounded, coercive and symmetric sesquilinear forms), and let $F_{1}, F_{2} \in H^{*}$ with the dual norm $∥ \cdot ∥_{*}$ (The dual space X^* of a normed space and its dual norm, The operator norm as the least bound and as the unit-sphere or unit-ball supremum). Let $u_{i} \in K$ be the unique solution of the variational inequality with data $F_{i}$ , that is, $a (u_{i}, v - u_{i}) \geq F_{i} (v - u_{i})$ for every $v \in K$ (Stampacchia's variational inequality). Then $α ∥ u_{1} - u_{2} ∥ \leq ∥ F_{1} - F_{2} ∥_{*} .$

Facts & Assumptions

Given: A real Hilbert space $H$ , a nonempty closed convex $K \subseteq H$ , a bounded coercive bilinear form $a$ with constants $M, α > 0$ , bounded linear functionals $F_{1}, F_{2}$ on $H$ , and their unique variational solutions $u_{1}, u_{2} \in K$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice, consumed through the existence-and-uniqueness theorem for the variational inequality.

[F1]

Stampacchia's variational inequality: for each bounded linear functional $F$ on $H$ there is exactly one $u \in K$ with $a (u, v - u) \geq F (v - u)$ for every $v \in K$ ; in particular $u_{1}$ and $u_{2}$ are well defined and satisfy $a (u_{1}, v - u_{1}) \geq F_{1} (v - u_{1})$ and $a (u_{2}, v - u_{2}) \geq F_{2} (v - u_{2})$ for all $v \in K$ .

[F2]

Bounded, coercive and symmetric sesquilinear forms: $a$ is bilinear with $a (u, u) \geq α ∥ u ∥^{2}$ for every $u \in H$ .

[F3]

The dual space X^* of a normed space and its dual norm, The operator norm as the least bound and as the unit-sphere or unit-ball supremum: $H^{*}$ is the normed space of bounded linear functionals with dual norm $∥ G ∥_{*} = \sup {∣ G (w) ∣ : ∥ w ∥ \leq 1}$ , so $∣ G (w) ∣ \leq ∥ G ∥_{*} ∥ w ∥$ for every $w \in H$ ; in particular $F_{1} - F_{2} \in H^{*}$ .

Proof

technique · direct

Given: The setting above, with the unique solutions $u_{1}, u_{2} \in K$ of the two variational inequalities.

1.1givenA1F1

Testing the inequality for $u_{1}$ at the admissible point $v = u_{2}$ and the inequality for $u_{2}$ at $v = u_{1}$ [F1] gives $a (u_{1}, u_{2} - u_{1}) \geq F_{1} (u_{2} - u_{1})$ and $a (u_{2}, u_{1} - u_{2}) \geq F_{2} (u_{1} - u_{2})$ .

2.1step 1.1F2algebra

Adding the two inequalities of step 1.1 and using bilinearity [F2] gives $a (u_{1} - u_{2}, u_{2} - u_{1}) \geq (F_{1} - F_{2}) (u_{2} - u_{1})$ , that is, $- a (u_{1} - u_{2}, u_{1} - u_{2}) \geq (F_{1} - F_{2}) (u_{2} - u_{1})$ ; coercivity [F2] bounds the left-hand side by $- α ∥ u_{1} - u_{2} ∥^{2}$ , so $α ∥ u_{1} - u_{2} ∥^{2} \leq - (F_{1} - F_{2}) (u_{2} - u_{1}) = (F_{1} - F_{2}) (u_{1} - u_{2})$ .

3.1step 2.1A1F3algebra∎

If $u_{1} = u_{2}$ the asserted inequality is trivial. Otherwise the dual-norm estimate [F3] gives $(F_{1} - F_{2}) (u_{1} - u_{2}) \leq ∥ F_{1} - F_{2} ∥_{*} ∥ u_{1} - u_{2} ∥$ , so step 2.1 yields $α ∥ u_{1} - u_{2} ∥^{2} \leq ∥ F_{1} - F_{2} ∥_{*} ∥ u_{1} - u_{2} ∥$ ; dividing by the positive number $∥ u_{1} - u_{2} ∥$ gives $α ∥ u_{1} - u_{2} ∥ \leq ∥ F_{1} - F_{2} ∥_{*}$ , which is the assertion; Countable Choice was used only through the existence and uniqueness theorem [A1].

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

A function with nonnegative test pairings is nonnegative a.e.

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $Ω \subseteq R^{n}$ be open and let $ζ \in L^{2} (Ω; R)$ satisfy $\int_{Ω} ζ φ d x \geq 0 for every φ \in C_{c}^{\infty} (Ω) with φ \geq 0.$ Then $ζ \geq 0$ almost everywhere on $Ω$ . Moreover, if $O \subseteq Ω$ is open and $\int_{Ω} ζ φ d x = 0$ for every $φ \in C_{c}^{\infty} (O)$ , then $ζ = 0$ almost everywhere on $O$ .

Facts & Assumptions

Given: An open set $Ω \subseteq R^{n}$ , a real $L^{2}$ class $ζ$ on $Ω$ , the nonnegative-pairing hypothesis, and an open set $O \subseteq Ω$ for the second claim.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice selects one element from each member of a natural-number-indexed family of nonempty sets.

[F1]

The mollifier family generated by a unit-mass smooth bump, A unit-mass smooth bump generates an $L^{1}$ approximate identity, Explicit compactly supported smooth cutoffs: there is a nonnegative unit-mass bump $ρ \in C_{c}^{\infty} (R^{n})$ , obtained by normalising the explicit nonnegative cutoff that equals $1$ on $∣ x ∣ \leq 1$ and vanishes for $∣ x ∣ \geq 2$ , and the rescalings $ρ_{ε} (x) = ε^{- n} ρ (x / ε)$ satisfy $ρ_{ε} \geq 0$ , $\int ρ_{ε} = 1$ and $supp ρ_{ε} \subseteq {\overline{B}}_{2 ε} (0)$ .

[F2]

Complex translation, convolution, approximate identities, and mollification: for real $g \in L^{2} (Ω)$ with a representative vanishing outside a compact set $S \subseteq R^{n}$ and for $0 < ε$ small, the class $ρ_{ε} * g$ has a smooth representative $φ_{ε}$ with $supp φ_{ε} \subseteq \overline{S + supp ρ_{ε}}$ , and $ρ_{ε} * g \to g$ in $L^{2} (R^{n})$ as $ε ↓ 0$ ; the assertions are choice-dependent only through the approximate-identity interface.

[F3]

Monotonicity and nonnegative homogeneity of the nonnegative integral: the nonnegative integral is monotone; in particular, the integral of a nonnegative measurable function is nonnegative.

[F4]

Holder's inequality for integrals, including the endpoint cases: for measurable real functions, $\int ∣ f g ∣ \leq ∥ f ∥_{2} ∥ g ∥_{2}$ whenever $f, g \in L^{2}$ .

[F5]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere: a nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere.

[F6]

A compact set and a disjoint closed set have a positive norm-distance gap: a nonempty compact set and a disjoint nonempty closed set in a normed space have positive distance.

[F7]

Test function cutoffs and euclidean localization: for compact $K \subseteq O^{'} \subseteq R^{n}$ with $O^{'}$ open there is $χ \in C_{c}^{\infty} (O^{'})$ with $0 \leq χ \leq 1$ and $χ = 1$ on a neighbourhood of $K$ ; this construction is choice-free.

[F8]

Subsets and countable unions of null subsets of $R^{m}$ are null: every subset of a null set is null, and under Countable Choice every countable union of null sets is null.

[F9]

Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line: a subset of $R^{n}$ that is closed and bounded is compact.

[F10]

The positive and negative parts of a function: $ζ = ζ^{+} - ζ^{-}$ with $ζ^{+}, ζ^{-} \geq 0$ and $ζ^{+} ζ^{-} = 0$ pointwise; hence $ζ \geq 0$ a.e. exactly when $ζ^{-} = 0$ a.e.

[F11]

The space $L^{p} (μ)$ as the quotient by null functions, Test function space d of an open set: elements of $L^{2} (Ω)$ are a.e. classes of measurable functions, while elements of $C_{c}^{\infty} (Ω)$ are actual smooth compactly supported functions, so the pairing $φ \mapsto \int_{Ω} ζ φ$ depends only on the class of $ζ$ and on the function $φ$ ; for an open $O \subseteq Ω$ the restriction of the class $ζ$ is a class in $L^{2} (O)$ with $\int_{O} ∣ ζ ∣^{2} \leq \int_{Ω} ∣ ζ ∣^{2}$ by monotonicity of the nonnegative integral, and $\int_{O} ζ φ = \int_{Ω} ζ φ$ for every $φ \in C_{c}^{\infty} (O)$ .

Proof

technique · direct

Given: An open set $Ω \subseteq R^{n}$ , a real class $ζ \in L^{2} (Ω)$ with $\int_{Ω} ζ φ \geq 0$ for every nonnegative $φ \in C_{c}^{\infty} (Ω)$ , and an open subset $O \subseteq Ω$ .

1.1givenF1F2F3F4F6F10F11algebra

(Local vanishing, uniformly in the open set and the class) Let $ω \subseteq R^{n}$ be open, let $z \in L^{2} (ω)$ satisfy $\int_{ω} z φ \geq 0$ for every nonnegative $φ \in C_{c}^{\infty} (ω)$ , let $O^{'} \subseteq ω$ be open, and let $η \in C_{c}^{\infty} (O^{'})$ with $η \geq 0$ ; we claim $\int_{ω} (z^{-})^{2} η = 0$ . If $η = 0$ this is trivial, so assume $η \neq 0$ . Choose a real representative $f$ of the class $z$ and set $g : = z^{-} η$ , the class of the measurable function $f^{-} η$ ; since $η$ is bounded with compact support and $f^{-} \leq ∣ f ∣$ , this class lies in $L^{2} (ω)$ and has a representative vanishing outside the compact set $S : = supp η \subseteq O^{'}$ [F10, F11]. If $O^{'} = R^{n}$ , choose any $ε > 0$ ; otherwise [F6] applies to $S$ and the nonempty closed set $R^{n} ∖ O^{'}$ , giving $d : = dist (S, R^{n} ∖ O^{'}) > 0$ , and choose $0 < ε < d / 2$ . Let $φ_{ε}$ be the smooth representative of $ρ_{ε} * g$ from [F2]. It is compactly supported; when $O^{'} \neq R^{n}$ , its support lies in $\overline{S + supp ρ_{ε}} \subseteq O^{'}$ because $supp ρ_{ε} \subseteq {\overline{B}}_{2 ε} (0)$ and $2 ε < d$ [F1]. Since $ρ_{ε} \geq 0$ and $f^{-} η \geq 0$ , monotonicity of the integral gives $φ_{ε} \geq 0$ throughout [F3], so $φ_{ε} \in C_{c}^{\infty} (ω)$ is a nonnegative test function and the hypothesis yields $\int_{ω} z φ_{ε} \geq 0$ ; on the other hand $∣ \int_{ω} z (φ_{ε} - g) ∣ \leq ∥ φ_{ε} - g ∥_{2} ∥ z ∥_{2} \to 0$ as $ε ↓ 0$ by [F2] and [F4], so $\int_{ω} z g \geq 0$ . Finally $z g = z z^{-} η = - z^{+} z^{-} η - (z^{-})^{2} η = - (z^{-})^{2} η$ because $z^{+} z^{-} = 0$ pointwise [F10], so $\int_{ω} (z^{-})^{2} η \leq 0$ ; as the integrand is nonnegative this forces $\int_{ω} (z^{-})^{2} η = 0$ . The argument uses only the pairing hypothesis on the open set $ω$ .

2.1step 1.1F5F7F8F9F10

(From local test functions to a.e. vanishing) Let $ω$ and $z$ be as in step 1.1. Let $O^{'} \subseteq ω$ be open and let $K \subseteq O^{'}$ be compact; by [F7] there is $χ \in C_{c}^{\infty} (O^{'})$ with $0 \leq χ \leq 1$ and $χ = 1$ on a neighbourhood $N$ of $K$ , and step 1.1 with $η = χ$ gives $\int_{ω} (z^{-})^{2} χ = 0$ ; as this integrand is nonnegative and measurable, [F5] gives $(z^{-})^{2} χ = 0$ a.e. on $ω$ , hence $(z^{-})^{2} = 0$ a.e. on $K$ . Now for arbitrary open $O^{'} \subseteq ω$ for $j \in N$ take $K_{j} : = {x : ∣ x ∣ \leq j + 1}$ if $O^{'} = R^{n}$ and $K_{j} : = {x \in O^{'} : ∣ x ∣ \leq j + 1 and dist (x, R^{n} ∖ O^{'}) \geq 1 / (j + 1)}$ otherwise; each $K_{j}$ is closed, bounded and contained in $O^{'}$ , hence compact by [F9], and the $K_{j}$ cover $O^{'}$ (a point $x \in O^{'}$ has a ball $B (x, r) \subseteq O^{'}$ , so $x \in K_{j}$ for every $j + 1 \geq \max {∣ x ∣, 1 / r}$ ). Since $(z^{-})^{2}$ vanishes a.e. on each $K_{j}$ , it vanishes a.e. on the union $O^{'}$ by [F8]. Taking $O^{'} = ω$ , $(z^{-})^{2} = 0$ a.e. on $ω$ , so $z^{-} = 0$ a.e. on $ω$ and $z \geq 0$ a.e. on $ω$ by [F10].

3.1step 2.1F10

(First assertion) Apply step 2.1 with $(ω, z) = (Ω, ζ)$ : the nonnegative-pairing hypothesis holds by assumption, so $(ζ^{-})^{2} = 0$ a.e. on $Ω$ , hence $ζ^{-} = 0$ a.e. on $Ω$ and $ζ \geq 0$ a.e. on $Ω$ by [F10].

3.2step 2.1F10F11

(Second assertion) Assume additionally that $\int_{Ω} ζ φ = 0$ for every $φ \in C_{c}^{\infty} (O)$ , and put $ω : = O$ and $z : = ζ ∣_{O}$ , the restriction of the class, which lies in $L^{2} (O)$ with $\int_{O} ∣ ζ ∣^{2} \leq \int_{Ω} ∣ ζ ∣^{2}$ and $\int_{O} ζ φ = \int_{Ω} ζ φ$ for every $φ \in C_{c}^{\infty} (O)$ [F11]. For every nonnegative $φ \in C_{c}^{\infty} (O)$ one has $\int_{O} ζ φ = 0 \geq 0$ and also $\int_{O} (- ζ) φ = 0 \geq 0$ , so step 2.1 applies with $(ω, z) = (O, ζ ∣_{O})$ and with $(ω, z) = (O, - ζ ∣_{O})$ , whose negative parts are $ζ^{-}$ and $ζ^{+}$ respectively; hence $(ζ^{-})^{2} = 0$ and $(ζ^{+})^{2} = 0$ a.e. on $O$ , so $ζ^{+} = ζ^{-} = 0$ a.e. on $O$ and $ζ = 0$ a.e. on $O$ by [F10].

4.1step 3.1step 3.2A1F2F8∎

Step 3.1 proves the first assertion and step 3.2 the second for arbitrary open $Ω$ , class $ζ$ and open $O \subseteq Ω$ ; the argument of steps 1.1-2.1 is uniform in the pair $(ω, z)$ , so the second assertion needed no re-run of the estimates. Countable Choice was used in the approximate-identity interface of step 1.1 and in the countable-union step of step 2.1 [A1, F2, F8], while the localisation itself is choice-free.

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Obstacle complementarity in distribution form

Statement

Assume Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice). Let $Ω \subseteq R^{n}$ be a bounded $C^{1}$ domain (Bounded C^k domains and boundary charts), let $L$ be a uniformly elliptic divergence-form operator with real $L^{\infty}$ coefficients, and let $a : H_{0}^{1} (Ω; R) \times H_{0}^{1} (Ω; R) \to R$ 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 $f \in L^{2} (Ω; R)$ and let $F \in H^{- 1} (Ω; R)$ be its canonical functional $F (v) = \int_{Ω} f v$ on $H_{0}^{1} (Ω; R)$ , so in particular $F (φ) = \int_{Ω} f φ$ for test functions (The negative Sobolev space $H^{- 1} (Ω)$ , Locally integrable functions as regular distributions), let $ψ \in H^{1} (Ω; R)$ with $T ψ \leq 0$ (The notation $H^{k}$ and the reserved zero-boundary symbol), and let $u \in K$ be the obstacle solution of Existence and uniqueness for the obstacle problem. For real $φ \in C_{c}^{\infty} (Ω; R)$ put $Λ_{u} (φ) : = a (u, φ) - \int_{Ω} f φ .$ Extend $Λ_{u}$ complex linearly as in The closed convex obstacle set and the obstacle variational inequality. Then:

$Λ_{u}$ is a nonnegative distribution: $Λ_{u} (φ) \geq 0$ for every nonnegative test function $φ$ (Distribution, Test function space d of an open set).
If in addition $Λ_{u}$ is represented by a function $ζ \in L^{2} (Ω)$ , that is $Λ_{u} (φ) = \int_{Ω} ζ φ$ for every test function, and if $u$ and $ψ$ have continuous representatives on $Ω$ , then $ζ \geq 0$ 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 $u \in K$ of Existence and uniqueness for the obstacle problem, the reaction $Λ_{u} (φ) = a (u, φ) - F (φ)$ on $C_{c}^{\infty} (Ω)$ , and, for the second assertion, a representation $Λ_{u} (φ) = \int_{Ω} ζ φ$ with $ζ \in L^{2} (Ω)$ 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: $K = {v \in H_{0}^{1} (Ω) : v \geq ψ a.e.}$ is nonempty, and $u \in K$ satisfies $a (u, v - u) \geq F (v - u)$ for every $v \in K$ (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 $ζ \in L^{2} (O)$ and $\int_{O} ζ φ \geq 0$ for every nonnegative $φ \in C_{c}^{\infty} (O)$ , then $ζ \geq 0$ a.e. on $O$ .

[F3]

The fundamental lemma of the calculus of variations: if $g \in L_{l o c}^{1} (O)$ and $\int_{O} g φ = 0$ for every $φ \in C_{c}^{\infty} (O)$ , then $g = 0$ a.e. on $O$ ; in particular $L^{2} (O) \subseteq L_{l o c}^{1} (O)$ 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 $H_{0}^{1} (Ω; R) \times H_{0}^{1} (Ω; R)$ ; under the stated hypotheses it is a bounded, coercive symmetric real bilinear form there, and the associated form on $H^{1} (Ω)$ 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 $f \in L^{2} (Ω)$ defines $F (v) = \int_{Ω} f v$ in $H^{- 1} (Ω)$ , with $∣ F (v) ∣ \leq ∥ f ∥_{2} ∥ v ∥_{2} \leq ∥ f ∥_{2} ∥ v ∥_{H_{0}^{1}}$ and hence $∥ F ∥_{H^{- 1}} \leq ∥ f ∥_{2}$ .

Proof

technique · direct

Given: The setting above, in particular the obstacle solution $u \in K$ and the reaction $Λ_{u}$ .

1.1givenF1F4F5

(Nonnegativity) Let $φ \in C_{c}^{\infty} (Ω)$ with $φ \geq 0$ , and put $v : = u + φ$ . Then $v \in H_{0}^{1} (Ω; R)$ because $u \in H_{0}^{1} (Ω; R)$ and $φ \in C_{c}^{\infty} (Ω; R) \subseteq H_{0}^{1} (Ω; R)$ , and $v \geq u \geq ψ$ a.e. on $Ω$ , so $v \in K$ [F1]. By [F5], $F$ is the continuous $H^{- 1}$ functional used by the variational inequality; testing that inequality at $v$ gives $a (u, φ) \geq F (φ)$ , that is $Λ_{u} (φ) \geq 0$ ; hence $Λ_{u}$ is a nonnegative distribution.

1.2givenF1F4

(Vanishing on the noncontact set) Assume now that $Λ_{u} (φ) = \int_{Ω} ζ φ$ 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 $φ \in C_{c}^{\infty} (O)$ be arbitrary. If $φ = 0$ , then $Λ_{u} (φ) = 0$ . Otherwise its support $K_{φ}$ is a nonempty compact subset of $O$ ; as $u - ψ$ is continuous and positive there, there is $m > 0$ with $u - ψ \geq m$ on $K_{φ}$ . Choose $ε > 0$ with $ε ∥ φ ∥_{\infty} < m$ . Then $u \pm ε φ \geq u - ε ∥ φ ∥_{\infty} \geq ψ$ on $K_{φ}$ , while on $Ω ∖ K_{φ}$ one has $u \pm ε φ = u \geq ψ$ a.e.; also $u \pm ε φ \in H_{0}^{1} (Ω; R)$ . Hence both competitors belong to $K$ [F1]. Testing the variational inequality at them gives $ε Λ_{u} (φ) \geq 0$ and $- ε Λ_{u} (φ) \geq 0$ , so $Λ_{u} (φ) = 0$ ; that is, $\int_{Ω} ζ φ = 0$ for every $φ \in C_{c}^{\infty} (O)$ .

2.1step 1.1F2

For the second assertion, the hypothesis of the representation gives $\int_{Ω} ζ φ = Λ_{u} (φ) \geq 0$ for every nonnegative test function by step 1.1, so [F2] applied with $O = Ω$ yields $ζ \geq 0$ a.e. on $Ω$ .

2.2step 1.2F3

Step 1.2 gives $\int_{Ω} ζ φ = 0$ for every $φ \in C_{c}^{\infty} (O)$ with $ζ \in L^{2} (Ω) \subseteq L_{l o c}^{1} (Ω)$ ; 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 $u \geq ψ$ a.e. by [F1] while on $Ω ∖ {u > ψ}$ one has $u \leq ψ$ 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 $L^{2}$ 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 $L^{2}$ -representation and continuity hypotheses, and no product of a distribution with a Sobolev class is formed.

CorollaryStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The obstacle reaction is supported on the contact set under measure regularity

Statement

Assume Countable Choice and the Axiom of Choice, in the setting of Obstacle complementarity in distribution form, and suppose additionally that the reaction agrees on $C_{c}^{\infty} (Ω)$ with a nonnegative Radon measure $μ$ (Radon measure on an LCH space): $Λ_{u} (φ) = \int_{Ω} φ d μ$ for every test function. Assume $u$ and $ψ$ have continuous representatives. Then $μ ({u > ψ}) = 0$ , so $μ$ is concentrated on the contact set ${u = ψ}$ , and $\int_{Ω} (u - ψ) d μ = 0$ .

Facts & Assumptions

Given: The obstacle setting of Obstacle complementarity in distribution form: a bounded $C^{1}$ domain $Ω \subseteq R^{n}$ (a bounded interval when $n = 1$ ), a uniformly elliptic divergence-form operator $L$ with symmetric bounded coercive form $a$ , $f \in L^{2} (Ω)$ , $F (φ) = \int_{Ω} f φ$ , an obstacle $ψ \in H^{1} (Ω)$ with $T ψ \leq 0$ , the admissible set $K = {v \in H_{0}^{1} (Ω) : v \geq ψ a.e.}$ (Zero-boundary Sobolev space as a norm closure), the obstacle solution $u \in K$ , and the reaction $Λ_{u} (φ) = a (u, φ) - F (φ)$ for real tests, extended complex linearly (The closed convex obstacle set and the obstacle variational inequality). The functions $u$ and $ψ$ are represented by continuous functions on $Ω$ , again written $u$ and $ψ$ . A nonnegative Radon measure $μ$ on the locally compact space $Ω$ satisfies $Λ_{u} (φ) = \int_{Ω} φ d μ$ for every $φ \in C_{c}^{\infty} (Ω)$ (Test function space d of an open set).

[F1]

The closed convex obstacle set and the obstacle variational inequality, Existence and uniqueness for the obstacle problem: $u \in K$ satisfies $a (u, v - u) \geq F (v - u)$ for every $v \in K$ , so in particular $u \geq ψ$ almost everywhere on $Ω$ ; and $C_{c}^{\infty} (Ω; R) \subseteq H_{0}^{1} (Ω; R)$ , so a compactly supported smooth function belongs to the zero-boundary test space.

[F2]

Obstacle complementarity in distribution form: $Λ_{u} (φ) = a (u, φ) - F (φ)$ on real tests, extended complex linearly, defines the reaction distribution of the solution $u$ on $C_{c}^{\infty} (Ω)$ (Distribution), and the additional product conclusions recorded there require the separate hypothesis that $Λ_{u}$ be represented by an $L^{2}$ function.

[F3]

Test function cutoffs and euclidean localization: in ZF, for compact $K_{0} \subseteq O \subseteq R^{n}$ with $O$ open there is $χ \in C_{c}^{\infty} (O)$ with $0 \leq χ \leq 1$ and $χ = 1$ on a neighborhood of $K_{0}$ .

[F4]

Radon measure on an LCH space: $μ$ is a Borel measure on the locally compact Hausdorff space $Ω$ with $μ (K_{0}) < \infty$ for every compact $K_{0}$ , outer regular on Borel sets and inner regular on open sets: $μ (O) = \sup {μ (K_{0}) : K_{0} \subseteq O compact}$ for every open $O$ .

[F5]

Euclidean balls have positive finite Lebesgue measure: every Euclidean ball has positive finite Lebesgue measure, so a nonempty open subset of $R^{n}$ is not Lebesgue-null (Measure-null sets and almost-everywhere statements relative to a measure).

[F6]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere, A nonnegative integral over a null set vanishes: a nonnegative measurable function has integral $0$ if and only if it vanishes almost everywhere, and the integral of a nonnegative measurable function over a null set vanishes.

[F7]

Measures are monotone: a measure is monotone under inclusion of measurable sets.

[F8]

A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value: a continuous real function on a nonempty compact metric space attains its minimum and maximum, so a positive continuous function has a positive minimum there and a continuous test function has finite sup norm.

Proof

technique · direct

Given: The setting above, with the continuous representatives of $u$ and $ψ$ , the nonnegative Radon measure $μ$ representing the reaction on test functions, and the open set $O : = {u > ψ} \subseteq Ω$ .

1.1givenF1F2F8

( $O$ is open and the two-sided test vanishes there) Because $u - ψ$ is continuous, $O$ is open in $Ω$ , hence in $R^{n}$ ; and for every real $χ \in C_{c}^{\infty} (O; R)$ one has $Λ_{u} (χ) = 0$ . Indeed, if $χ = 0$ this is trivial, so let $χ \neq 0$ , put $S : = supp χ \subseteq O$ , a nonempty compact set, and let $m : = \min_{S} (u - ψ) > 0$ , which is positive because $u - ψ$ is continuous and positive on $S$ . For every $0 < ε < m / ∥ χ ∥_{\infty}$ define $v_{\pm} : = u \pm ε χ$ ; then $v_{\pm} \in H_{0}^{1} (Ω)$ by [F1]. On the set ${χ \neq 0} \subseteq S$ one has $v_{\pm} - ψ \geq (u - ψ) - ε ∣ χ ∣ \geq m - ε ∥ χ ∥_{\infty} > 0$ , while off ${χ \neq 0}$ one has $v_{\pm} = u \geq ψ$ almost everywhere by [F1]; hence $v_{\pm} \in K$ . Testing the variational inequality [F1] at $v_{\pm}$ gives $a (u, \pm ε χ) \geq F (\pm ε χ)$ , that is $\pm ε Λ_{u} (χ) \geq 0$ , so $Λ_{u} (χ) = 0$ . For a complex test $χ = χ_{1} + i χ_{2}$ on $O$ , apply this argument to its real and imaginary parts; complex linearity [F2] gives $Λ_{u} (χ) = 0$ as well.

2.1givenstep 1.1F3F6F7

(Compact subsets of the noncontact set are $μ$ -null) Let $K_{0} \subseteq O$ be compact. By [F3] there is $χ \in C_{c}^{\infty} (O)$ with $0 \leq χ \leq 1$ and $χ = 1$ on a neighborhood $U$ of $K_{0}$ ; then $K_{0} \subseteq U \subseteq {χ = 1} \subseteq {χ \neq 0}$ . Step 1.1 and the measure representation give $\int_{Ω} χ d μ = Λ_{u} (χ) = 0$ , and $χ \geq 0$ is continuous, hence $μ$ -measurable; by [F6] $χ = 0$ $μ$ -almost everywhere, that is $μ ({χ \neq 0}) = 0$ . Monotonicity [F7] applied to $K_{0} \subseteq {χ \neq 0}$ gives $μ (K_{0}) = 0$ .

3.1step 2.1F4

( $μ (O) = 0$ by inner regularity) By [F4] inner regularity on the open set $O$ gives $μ (O) = \sup {μ (K_{0}) : K_{0} \subseteq O compact}$ , and every term of this supremum is $0$ by step 2.1, so $μ (O) = 0$ ; this includes the case $O = \emptyset$ , where the only compact subset is the empty set, whose measure is $0$ .

4.1givenstep 3.1F1F5

(The complementary open set is empty) The set ${u < ψ}$ is open in $Ω$ and has Lebesgue measure zero, because $u \geq ψ$ almost everywhere by [F1]; were it nonempty it would contain a Euclidean ball of positive measure by [F5]. Hence ${u < ψ} = \emptyset$ and therefore $Ω ∖ {u = ψ} = {u > ψ} \cup {u < ψ} = O$ with $μ (Ω ∖ {u = ψ}) = μ (O) = 0$ ; equivalently, $μ$ is concentrated on the contact set ${u = ψ}$ .

5.1step 3.1step 4.1F1F3F4F5F6∎

(The integral vanishes) The functions $(u - ψ)^{+}$ and $(u - ψ)^{-}$ are nonnegative and continuous, hence $μ$ -measurable; $(u - ψ)^{+}$ vanishes outside $O$ , which is $μ$ -null by step 3.1, and $(u - ψ)^{-}$ vanishes outside ${u < ψ} = \emptyset$ by step 4.1, so [F6] gives $\int_{Ω} (u - ψ)^{+} d μ = 0 = \int_{Ω} (u - ψ)^{-} d μ$ ; both integrals being finite, $\int_{Ω} (u - ψ) d μ = 0$ . This proves all the asserted conclusions: $μ ({u > ψ}) = 0$ , concentration on ${u = ψ}$ and vanishing of the integral. The positive-measure ball supplier [F5] uses Countable Choice; the cutoff [F3] is constructed in ZF and inner regularity is part of [F4], so the local argument makes no further selections. The assumed Axiom of Choice and Countable Choice also cover the inherited obstacle setting and existence of $u$ [F1].

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Lewy–Stampacchia distribution bound for bounded-coefficient obstacle forms

Statement

Assume Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice), inherited through the existence, density, truncation and sign suppliers cited below. Let $n \geq 1$ and let $Ω \subseteq R^{n}$ be a bounded open set of the two kinds of The closed convex obstacle set and the obstacle variational inequality: a bounded interval when $n = 1$ , and a bounded $C^{1}$ domain when $n \geq 2$ (Bounded C^k domains and boundary charts). Let $L$ be a real symmetric uniformly elliptic divergence-form operator with bounded measurable real coefficients and coercive form $a$ (Uniformly elliptic divergence-form operators and their sesquilinear forms), let $f \in L^{2} (Ω; R)$ , and let the real obstacle $ψ \in H^{2} (Ω; R) \cap H_{0}^{1} (Ω; R)$ (The notation $H^{k}$ and the reserved zero-boundary symbol, Integer-order Sobolev spaces and their norms) satisfy the additional hypothesis that its distributional image $L ψ$ is represented by an $L^{2} (Ω; R)$ function; put $h : = L ψ - f \in L^{2} (Ω; R)$ . Let $u \in K = {v \in H_{0}^{1} (Ω; R) : v \geq ψ}$ solve the obstacle problem (Existence and uniqueness for the obstacle problem, Zero-boundary Sobolev space as a norm closure) with reaction $Λ_{u} (φ) = a (u, φ) - \int_{Ω} f φ$ on test functions (Obstacle complementarity in distribution form, Distribution, Test function space d of an open set). Then, in the sense of distributions, $0 \leq Λ_{u} \leq (L ψ - f)^{+},$ that is, $Λ_{u}$ is a nonnegative distribution bounded above by the $L^{2}$ function $h^{+}$ .

Facts & Assumptions

Given: The bounded open set $Ω$ of the two kinds above with $n \geq 1$ ; a real symmetric uniformly elliptic divergence-form operator $L = - D_{i} (a^{i j} D_{j} \cdot) + b^{i} D_{i} \cdot + c \cdot$ with bounded measurable real coefficients and coercive form $a (u, v) = \int_{Ω} (a^{i j} D_{j} u D_{i} v + b^{i} D_{i} u v + c u v) d x$ of Uniformly elliptic divergence-form operators and their sesquilinear forms; $f \in L^{2} (Ω; R)$ and the functional $F (v) = \int_{Ω} f v$ ; the obstacle $ψ \in H^{2} (Ω; R) \cap H_{0}^{1} (Ω; R)$ with $L ψ$ represented by an $L^{2}$ function; $h = L ψ - f$ ; the admissible set $K = {v \in H_{0}^{1} (Ω; R) : v \geq ψ a.e.}$ and the unique obstacle solution $u \in K$ with reaction $Λ_{u} (φ) = a (u, φ) - F (φ)$ .

[F1]

The closed convex obstacle set and the obstacle variational inequality, Existence and uniqueness for the obstacle problem: $K$ is nonempty and $u \in K$ satisfies $a (u, v - u) \geq F (v - u)$ for every $v \in K$ ; in particular $u \geq ψ$ almost everywhere on $Ω$ . The form $a$ is symmetric, bounded and coercive.

[F2]

Uniformly elliptic divergence-form operators and their sesquilinear forms, The elliptic form is well defined and bounded on $H^{1}$ : the form $a$ is a well-defined bounded bilinear form on $H^{1} (Ω)$ with $∣ a (u, v) ∣ \leq M ∥ u ∥_{H^{1}} ∥ v ∥_{H^{1}}$ , and for every $w \in H^{1} (Ω)$ and every $φ \in C_{c}^{\infty} (Ω)$ the distribution $L w$ pairs as $⟨ L w, φ ⟩ = a (w, φ)$ ; the calculation below retains the bounded drift term, and does not require it to vanish.

[F3]

Zero-boundary Sobolev space as a norm closure, Test function space d of an open set: $H_{0}^{1} (Ω)$ is the closure of $C_{c}^{\infty} (Ω)$ in the $H^{1}$ norm, so $C_{c}^{\infty} (Ω)$ is dense in $H_{0}^{1} (Ω)$ ; every $H^{1}$ class vanishing almost everywhere outside a compact subset of $Ω$ lies in $H_{0}^{1} (Ω)$ : extend it by zero using Compactly supported Sobolev functions extend by zero in every integer order, approximate the extension by Compactly supported smooth functions are dense in W^{k,p}(R^n), and multiply the approximants by a fixed interior smooth cutoff equal to one near its support (Test function cutoffs and euclidean localization, Bounded restriction and cutoff localisation in Sobolev spaces). The resulting interior tests converge in $H^{1}$ .

[F4]

Positive-part truncation calculus and admissible cut-off weak tests: truncations and compactly supported cutoff products have the stated $H^{1}$ / $H_{0}^{1}$ membership and gradient formulas. Its weak-test interface extends a local inequality to a cutoff test when the local source is in $L^{2}$ , and permits a global $H_{0}^{1}$ test when the source functional is continuous on $H_{0}^{1}$ ; it does not pair a general $L_{l o c}^{1}$ source with an arbitrary $H_{0}^{1}$ function.

[F5]

Weak Leibniz rule with a smooth factor: for $η \in C_{c}^{\infty} (Ω)$ and $z \in H^{1} (Ω)$ the product $η z$ lies in $H^{1} (Ω)$ with $D (η z) = η D z + z D η$ a.e.

[F6]

Uniformly elliptic divergence-form operators and their sesquilinear forms: uniform ellipticity gives $a^{i j} D_{j} w D_{i} w \geq θ ∣ D w ∣^{2} \geq 0$ a.e. for every real $w \in H^{1} (Ω)$ .

[F7]

Dominated convergence: if measurable functions converge pointwise a.e. and are dominated in absolute value by one integrable function, their integrals converge.

[F8]

A function with nonnegative test pairings is nonnegative a.e.: an $L^{2}$ class whose pairings with all nonnegative test functions are nonnegative is itself nonnegative almost everywhere.

[F9]

The space $L^{p} (μ)$ as the quotient by null functions, Holder's inequality for integrals, including the endpoint cases: for $h \in L^{2} (Ω)$ and $φ \in C_{c}^{\infty} (Ω)$ the product $h φ$ is in $L^{1} (Ω)$ , and $h 1_{E} \leq h^{+}$ pointwise for every measurable set $E$ , with both classes in $L^{2} (Ω)$ ; all pointwise statements are read on representatives and hold a.e. independently of the representative.

[F10]

The reaction $Λ_{u} (v) = a (u, v) - \int_{Ω} f v$ is a bounded functional on $H_{0}^{1} (Ω)$ : boundedness of $a$ is [F2], and $f \in L^{2}$ acts continuously by Cauchy--Schwarz and $∥ v ∥_{L^{2}} \leq ∥ v ∥_{H^{1}}$ . Under the real specialization of The negative Sobolev space $H^{- 1} (Ω)$ , this is exactly an $H^{- 1}$ source functional, so the global test-extension clause of [F4] applies.

Proof

technique · direct

Given: The setting above, the obstacle solution $u \in K$ , the class $w : = u - ψ \in H_{0}^{1} (Ω; R)$ , and the reaction functional $Λ (v) : = a (u, v) - F (v)$ defined on $H_{0}^{1} (Ω)$ .

1.1givenF1

For every $q \in H_{0}^{1} (Ω; R)$ with $q \geq 0$ a.e. one has $u + q \in K$ , because $u + q \in H_{0}^{1} (Ω)$ and $(u + q) - ψ = w + q \geq 0$ a.e. by [F1]; testing the variational inequality [F1] at $v = u + q$ gives $Λ (q) \geq 0$ .

1.2givenF1F2F3F9

The classes $w = u - ψ$ and $q = w$ satisfy $w \in H_{0}^{1} (Ω)$ and $w \geq 0$ a.e. by [F1]. For every $v \in C_{c}^{\infty} (Ω)$ the definition of $L ψ$ gives $a (ψ, v) = ⟨ L ψ, v ⟩ = \int_{Ω} (L ψ) v$ [F2]; both $v \mapsto a (ψ, v)$ and $v \mapsto \int_{Ω} (L ψ) v$ are bounded linear functionals on $H_{0}^{1} (Ω)$ — the first by [F2], the second because $∥ v ∥_{L^{2}} \leq ∥ v ∥_{H^{1}}$ and $L ψ \in L^{2}$ — and they agree on the dense subspace $C_{c}^{\infty} (Ω)$ [F3], so they agree on all of $H_{0}^{1} (Ω)$ . Hence, for every $v \in H_{0}^{1} (Ω)$ , $Λ (v) = a (w, v) + \int_{Ω} h v$ by bilinearity of $a$ and $u = w + ψ$ .

1.3givenF1

Testing the variational inequality at $v = ψ \in K$ gives $a (u, ψ - u) \geq F (ψ - u)$ , that is $- Λ (w) \geq 0$ or $Λ (w) \leq 0$ ; testing at $v = u + w = 2 u - ψ \in K$ , which is admissible because $(2 u - ψ) - ψ = 2 w \geq 0$ a.e., gives $Λ (w) \geq 0$ . Hence $Λ (w) = 0$ .

1.4givenF4

For $δ > 0$ put $θ_{δ} : = (1 - w / δ)^{+} = δ^{- 1} (w - δ)^{-}$ and $m_{δ} : = 1 - θ_{δ} = \min {w / δ, 1}$ ; these are the truncations of [F4] at the level $δ$ , so $θ_{δ}, m_{δ} \in H^{1} (Ω; R)$ with $0 \leq θ_{δ}, m_{δ} \leq 1$ , $m_{δ} = w / δ$ on ${w \leq δ}$ , and $D θ_{δ} = - δ^{- 1} 1_{{0 < w < δ}} D w, D m_{δ} = δ^{- 1} 1_{{0 < w < δ}} D w a.e.,$ the indicator being restricted to ${0 < w < δ}$ because $D w = 0$ a.e. on the level set ${w = 0}$ [F4]; moreover $D w = 0$ a.e. on $E : = {w = 0}$ .

2.1step 1.1step 1.3F1F3F4F5F10

Let $φ \in C_{c}^{\infty} (Ω)$ with $φ \geq 0$ and let $δ > 0$ . The products $φ m_{δ}$ and $(∥ φ ∥_{\infty} / δ) w - φ m_{δ}$ lie in $H_{0}^{1} (Ω)$ by [F3] and [F5], and both are nonnegative a.e.: the first because $φ \geq 0$ and $m_{δ} \geq 0$ , the second because $φ m_{δ} \leq ∥ φ ∥_{\infty} m_{δ} \leq (∥ φ ∥_{\infty} / δ) w$ . The global H $^{- 1}$ test clause of [F4] applies by [F10]; independently, [F1] states the obstacle variational inequality for every such $H_{0}^{1}$ competitor. Thus step 1.1 gives $Λ (φ m_{δ}) \geq 0$ and $Λ ((∥ φ ∥_{\infty} / δ) w - φ m_{δ}) \geq 0$ ; by linearity and $Λ (w) = 0$ of step 1.3 the second inequality reads $- Λ (φ m_{δ}) \geq 0$ . Hence $Λ (φ m_{δ}) = 0$ , and since $φ θ_{δ} = φ - φ m_{δ}$ , linearity gives $Λ (φ θ_{δ}) = Λ (φ)$ .

2.2step 1.2step 1.4F5

Expanding the form and using $D (φ θ_{δ}) = θ_{δ} D φ + φ D θ_{δ}$ [F5] and the decomposition of step 1.2, for every $φ \in C_{c}^{\infty} (Ω)$ and $δ > 0$ one has $Λ (φ θ_{δ}) = \int_{Ω} θ_{δ} a^{i j} D_{j} w D_{i} φ - δ^{- 1} \int_{{0 < w < δ}} φ a^{i j} D_{j} w D_{i} w + \int_{Ω} b^{i} D_{i} w φ θ_{δ} + \int_{Ω} c w φ θ_{δ} + \int_{Ω} h φ θ_{δ} .$

3.1step 1.1step 2.1step 2.2F6F7F9

Fix $φ \in C_{c}^{\infty} (Ω)$ with $φ \geq 0$ and abbreviate the five terms of step 2.2 as $A_{δ}$ , $- T_{δ}$ , $B_{δ}$ , $C_{δ}$ , $H_{δ}$ , so that $Λ (φ) = Λ (φ θ_{δ}) = A_{δ} - T_{δ} + B_{δ} + C_{δ} + H_{δ}$ by steps 2.1 and 2.2, with $T_{δ} \geq 0$ because $φ a^{i j} D_{j} w D_{i} w \geq θ φ ∣ D w ∣^{2} \geq 0$ a.e. by [F6]. As $δ ↓ 0$ : $A_{δ} \to 0$ by [F7], since $θ_{δ} \to 1_{E}$ a.e., $∣ θ_{δ} a^{i j} D_{j} w D_{i} φ ∣ \leq n M_{a} ∣ D w ∣ ∣ D φ ∣ \in L^{1} (Ω)$ , and $D w = 0$ a.e. on $E$ ; $B_{δ} \to 0$ by [F7], since $∣ b^{i} D_{i} w φ θ_{δ} ∣ \leq n M_{b} ∣ D w ∣ ∣ φ ∣$ is integrable and the pointwise limit vanishes on $E$ via $D w = 0$ a.e. there; $C_{δ} \to 0$ by [F7], since $∣ c w φ θ_{δ} ∣ \leq M_{c} ∣ φ ∣ δ / 4 \leq M_{c} ∣ φ ∣ / 4$ for $δ \leq 1$ and $w θ_{δ} \leq δ / 4$ ; and $H_{δ} \to \int_{E} h φ$ by [F7], since $h φ$ is integrable [F9] and $θ_{δ} \to 1_{E}$ a.e. Therefore $T_{δ} = A_{δ} + B_{δ} + C_{δ} + H_{δ} - Λ (φ)$ converges to $\int_{E} h φ - Λ (φ)$ , and $T_{δ} \geq 0$ gives $\int_{E} h φ \geq Λ (φ)$ ; with $Λ (φ) \geq 0$ from step 1.1, $0 \leq Λ (φ) \leq \int_{E} h φ$ .

4.1step 3.1F8F9

The class $g : = h 1_{E}$ lies in $L^{2} (Ω)$ by [F9], and step 3.1 gives $\int_{Ω} g φ = \int_{E} h φ \geq Λ (φ) \geq 0$ for every $φ \in C_{c}^{\infty} (Ω)$ with $φ \geq 0$ ; by [F8] therefore $g \geq 0$ a.e. on $Ω$ , that is $h \geq 0$ a.e. on $E$ .

5.1step 1.1step 3.1step 4.1F1F3F4F8F9∎

For every $φ \in C_{c}^{\infty} (Ω)$ with $φ \geq 0$ one has $\int_{E} h φ = \int_{Ω} (h 1_{E}) φ \leq \int_{Ω} h^{+} φ$ because $h 1_{E} \leq h^{+}$ pointwise [F9], while steps 1.1 and 3.1 give $0 \leq Λ (φ) \leq \int_{E} h φ$ ; hence $0 \leq Λ (φ) \leq \int_{Ω} h^{+} φ$ for every nonnegative test function, which is precisely the distributional statement $0 \leq Λ_{u} \leq (L ψ - f)^{+}$ . Countable Choice is consumed through the existence theorem [F1], the density of $C_{c}^{\infty} (Ω)$ in $H_{0}^{1} (Ω)$ [F3] and the sign lemma [F8], and the Axiom of Choice through the ACL-based truncation calculus [F4] and the trace conventions of the obstacle setting [F1]; no further choice principle is used.

Source note

The cited article [OU] proves the Lewy–Stampacchia inequality in the entropy-solution class under a hypothesis that the obstacle’s positive part is bounded and without lower-order terms; it corroborates the principal-part case but is not used to justify the bounded drift and potential terms, which are handled here by the explicit truncation calculation above. The additional hypothesis that $L ψ$ be represented by an $L^{2}$ function is what makes $h = L ψ - f$ an honest $L^{2}$ object and the upper bound an $L^{2}$ function.

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The absolute value preserves the L^2 norm and the Dirichlet energy on H^1_0

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 1$ , let $Ω \subseteq R^{n}$ be open, and let $u \in H_{0}^{1} (Ω; R)$ (Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms). Then $∣ u ∣ \in H_{0}^{1} (Ω)$ , $∥ ∣ u ∣ ∥_{L^{2} (Ω)} = ∥ u ∥_{L^{2} (Ω)}$ (The space $L^{p} (μ)$ as the quotient by null functions), and $\int_{Ω} ∣ D ∣ u ∣ ∣^{2} d x = \int_{Ω} ∣ D u ∣^{2} d x .$ Consequently the constrained minimisation of the Dirichlet energy on the $L^{2}$ -unit sphere may be restricted to nonnegative competitors.

Facts & Assumptions

Given: An open set $Ω \subseteq R^{n}$ , $n \geq 1$ , and a real class $u \in H_{0}^{1} (Ω)$ .

[A1]

The Axiom of Choice: the Axiom of Choice, inherited through the Sobolev composition and subsequence suppliers.

[A2]

AC implies DC implies countable choice: the Axiom of Choice implies Countable Choice, so [F4] applies under the Statement's assumption.

[F1]

Positive, negative, and truncated Sobolev functions: for a real Sobolev class $v$ , $∣ v ∣ \in H^{1} (Ω)$ and $D ∣ v ∣ = sgn (v) D v$ a.e., with $D v = 0$ a.e. on ${v = 0}$ ; also $∣ v ∣ = v^{+} + v^{-}$ .

[F2]

Integer-order Sobolev spaces and their norms: $H^{1} (Ω)$ is the space of $L^{2}$ classes with all first weak derivatives in $L^{2}$ , with norm controlling both $∥ v ∥_{L^{2}}$ and $∥ D v ∥_{L^{2}}$ .

[F3]

Zero-boundary Sobolev space as a norm closure: $H_{0}^{1} (Ω)$ is the closure of $C_{c}^{\infty} (Ω)$ in $H^{1} (Ω)$ and is closed in that norm.

[F4]

Assuming Countable Choice, $L^{p}$ -convergent sequences have almost-everywhere convergent subsequences: if a sequence converges in $L^{2} (Ω)$ , it has a subsequence converging almost everywhere.

[F5]

The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ : for smooth $v$ and a smooth scalar function $Φ$ , $D (Φ \circ v) = Φ^{'} (v) D v$ .

[F6]

Dominated convergence: an almost-everywhere convergent sequence dominated by an integrable function has convergent integrals.

[F7]

The space $L^{p} (μ)$ as the quotient by null functions: $L^{2}$ norms and pointwise compositions are well defined on almost-everywhere classes.

Proof

technique · direct

Given: The open set $Ω$ and real class $u \in H_{0}^{1} (Ω)$ above.

1.1F1F3F5F6

(Smooth compactly supported case). Fix $v \in C_{c}^{\infty} (Ω; R)$ and, for $ε > 0$ , set $Φ_{ε} (s) = \sqrt{s^{2} + ε^{2}} - ε$ . Then $Φ_{ε}$ is smooth, $Φ_{ε} (0) = 0$ , $0 \leq Φ_{ε} (s) \leq ∣ s ∣$ , and $∣ Φ_{ε}^{'} (s) ∣ \leq 1$ . Thus $Φ_{ε} \circ v \in C_{c}^{\infty} (Ω)$ . As $ε ↓ 0$ , the squared value error is bounded by $∣ v ∣^{2} \in L^{1}$ and tends pointwise to zero, so $Φ_{ε} (v) \to ∣ v ∣$ in $L^{2}$ by [F6]. By [F5], $D (Φ_{ε} (v)) = Φ_{ε}^{'} (v) D v$ ; the squared gradient error tends pointwise to zero and is bounded by $4 ∣ D v ∣^{2} \in L^{1}$ , so [F1] and [F6] give convergence to $D ∣ v ∣$ in $L^{2}$ . Hence $Φ_{ε} (v) \to ∣ v ∣$ in $H^{1}$ , so $∣ v ∣ \in H_{0}^{1} (Ω)$ by [F3].

2.1A2F2F3F4

(Approximation and almost-everywhere convergence). By [F3] choose $v_{j} \in C_{c}^{\infty} (Ω)$ with $v_{j} \to u$ in $H^{1}$ . In particular $v_{j} \to u$ in $L^{2}$ ; by [A2], Countable Choice is available, so [F4] lets us pass to a subsequence, still denoted $v_{j}$ , with $v_{j} \to u$ almost everywhere. Step 1.1 gives $∣ v_{j} ∣ \in H_{0}^{1} (Ω)$ for every $j$ . The pointwise inequality $∣ ∣ v_{j} ∣ - ∣ u ∣ ∣ \leq ∣ v_{j} - u ∣$ shows $∣ v_{j} ∣ \to ∣ u ∣$ in $L^{2}$ .

3.1step 2.1F1F6

(Convergence of the gradients). By [F1], $D ∣ v_{j} ∣ - D ∣ u ∣ = sgn (v_{j}) (D v_{j} - D u) + (sgn (v_{j}) - sgn (u)) D u .$ The first term tends to zero in $L^{2}$ because $∣ sgn (v_{j}) ∣ \leq 1$ and $D v_{j} \to D u$ in $L^{2}$ . For the second, at almost every point where $u \neq 0$ the signs converge by the pointwise convergence in step 2.1; on ${u = 0}$ one has $D u = 0$ a.e. by [F1]. Thus its squared magnitude tends to zero a.e. and is bounded by $4 ∣ D u ∣^{2} \in L^{1}$ , so [F6] gives convergence to zero in $L^{2}$ . Therefore $D ∣ v_{j} ∣ \to D ∣ u ∣$ in $L^{2}$ .

4.1step 2.1step 3.1F1F3F7A1A2∎

Since $∣ v_{j} ∣ \to ∣ u ∣$ in $H^{1}$ by steps 2.1-3.1 and $H_{0}^{1} (Ω)$ is closed by [F3], $∣ u ∣ \in H_{0}^{1} (Ω)$ . Pointwise $∣ ∣ u ∣ ∣ = ∣ u ∣$ , so the $L^{2}$ norms agree by [F7]; and [F1], including $D u = 0$ a.e. on ${u = 0}$ , gives $∣ D ∣ u ∣ ∣ = ∣ D u ∣$ a.e., hence equality of the Dirichlet energies. Consequently every unit-sphere competitor $u$ is replaced by the nonnegative competitor $∣ u ∣$ with the same energy, so the infimum is unchanged when minimisation is restricted to nonnegative competitors. The Axiom of Choice enters only through the declared composition and subsequence suppliers.

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The first Dirichlet eigenfunction by constrained minimisation

Statement

Assume the Axiom of Choice, the ultrafilter lemma, DC and HB (The Axiom of Choice, The ultrafilter extension principle (UL/BPI), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The real dominated-extension principle as an additional hypothesis over ZF). Let $n \geq 1$ , let $Ω \subseteq R^{n}$ be a nonempty bounded open set, let $E (u) = \int_{Ω} ∣ D u ∣^{2} d x$ on $H_{0}^{1} (Ω; R)$ (The notation $H^{k}$ and the reserved zero-boundary symbol, Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure) and let $S = {u \in H_{0}^{1} (Ω) : ∥ u ∥_{L^{2}} = 1}$ (The space $L^{p} (μ)$ as the quotient by null functions). Then $S$ is nonempty and $E$ attains its infimum $λ_{1}$ on $S$ . Every minimiser $u_{0}$ is a weak eigenpair of the Dirichlet Laplacian, $\int_{Ω} D u_{0} \cdot D h d x = λ_{1} \int_{Ω} u_{0} h d x (h \in H_{0}^{1} (Ω)),$ with $λ_{1} = E (u_{0}) > 0$ (Symmetric elliptic weak eigenpairs); some minimiser is nonnegative, and $λ_{1}$ coincides with the first Dirichlet eigenvalue listed in Discrete spectrum of a symmetric elliptic Dirichlet operator for the principal Dirichlet form (The $L^{2}$ operator associated with a symmetric elliptic form), the minimisers being exactly the elements of $S \cap E_{λ_{1}}$ .

Facts & Assumptions

Given: A nonempty bounded open set $Ω \subseteq R^{n}$ , the Dirichlet energy $E (u) = \int_{Ω} ∣ D u ∣^{2} d x$ on $H_{0}^{1} (Ω; R)$ , and the $L^{2}$ -unit sphere $S = {u \in H_{0}^{1} (Ω) : ∥ u ∥_{L^{2}} = 1}$ .

[F1]

Zero-boundary Sobolev space as a norm closure, $H^{k}$ is a Hilbert space under the derivative-sum inner product, A closed subspace of a Banach space is Banach: $H_{0}^{1} (Ω)$ is the norm closure of $C_{c}^{\infty} (Ω)$ in $H^{1} (Ω) = W^{1, 2} (Ω)$ , hence a closed subspace of the Banach space $H^{1} (Ω)$ and itself a real Banach space with the $H^{1}$ norm.

[F2]

W^{1,p}(Omega) is reflexive for 1<p<infinity, Closed subspaces of reflexive spaces are reflexive, Reflexivity is surjectivity of the canonical map: under the ultrafilter lemma, DC and HB the space $W^{1, 2} (Ω)$ is reflexive, and under HB its closed subspace $H_{0}^{1} (Ω)$ is reflexive, hence a real reflexive Banach space.

[F3]

Convex and strictly convex functionals on a convex subset of a real vector space, A convex norm-lower-semicontinuous functional is weakly lower semicontinuous: $E$ is convex, being the squared norm of the bounded linear map $u \mapsto D u$ composed with the convex square; it is continuous because $∣ E (u) - E (v) ∣ \leq ∥ D u - D v ∥_{L^{2}} (∥ D u ∥_{L^{2}} + ∥ D v ∥_{L^{2}})$ . By the convex-lower-semicontinuity lemma (Axiom of Choice) $E$ is weakly sequentially lower semicontinuous on every nonempty convex subset of $H_{0}^{1} (Ω)$ .

[F4]

Test function cutoffs and euclidean localization: since $Ω$ is nonempty and open there is a nonzero $φ \in C_{c}^{\infty} (Ω)$ , for instance a cutoff equal to one on a neighbourhood of a chosen point; then $u_{1} : = φ / ∥ φ ∥_{L^{2}}$ lies in $S$ , so $S \neq \emptyset$ and $λ_{1} : = \inf_{S} E \leq E (u_{1}) < \infty$ .

[F5]

The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction, Coercivity of the principal Dirichlet form, Dependent choice implies countable choice: since the bounded set $Ω$ is bounded in every direction, Poincaré at $p = 2$ gives a constant $C_{P}$ with $∥ v ∥_{L^{2}} \leq C_{P} ∥ D v ∥_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ ; by the coercivity lemma (Axiom of Choice and Countable Choice, the latter from DC) the model principal form satisfies $E (v) = ∥ D v ∥_{L^{2}}^{2} \geq (1 + C_{P}^{2})^{- 1} ∥ v ∥_{H_{0}^{1}}^{2}$ .

[F6]

A bounded sequence in a reflexive Banach space has a weakly convergent subsequence: under the ultrafilter lemma, DC and HB every norm-bounded sequence in a real reflexive Banach space has a weakly convergent subsequence.

[F7]

Weak H^1 convergence plus Rellich preserves the L^2 unit normalisation: if $Ω$ is bounded and open, $(v_{j}) \subseteq H_{0}^{1} (Ω)$ is norm bounded with $v_{j} ⇀ v$ in $H_{0}^{1} (Ω)$ and $∥ v_{j} ∥_{L^{2}} = 1$ , then $v \in H_{0}^{1} (Ω)$ and $∥ v ∥_{L^{2}} = 1$ .

[F8]

The Lagrange multiplier rule for finitely many regular constraints, Fréchet derivative between Banach spaces: let $X$ be a real Banach space, $U \subseteq X$ open, $I : U \to R$ Fréchet differentiable at $u$ , and $G : U \to R^{m}$ of class $C^{1}$ with $D G (u)$ surjective; if $u$ is a local minimiser or maximiser of $I$ on the level set ${G = G (u)}$ , then there is a unique $λ \in R^{m}$ with $D I (u) = \sum_{i} λ_{i} D G_{i} (u)$ .

[F9]

The absolute value preserves the L^2 norm and the Dirichlet energy on H^1_0: for every open $Ω \subseteq R^{n}$ and real $v \in H_{0}^{1} (Ω)$ , one has $∣ v ∣ \in H_{0}^{1} (Ω)$ , $∥ ∣ v ∣ ∥_{L^{2}} = ∥ v ∥_{L^{2}}$ and $\int_{Ω} ∣ D ∣ v ∣ ∣^{2} = \int_{Ω} ∣ D v ∣^{2}$ .

[F10]

The Rayleigh principle for the first Dirichlet eigenvalue, Discrete spectrum of a symmetric elliptic Dirichlet operator, The $L^{2}$ operator associated with a symmetric elliptic form, Symmetric elliptic weak eigenpairs: in the symmetric case over a bounded open set the discrete spectral theorem provides the nondecreasing eigenvalue list $λ_{1} \leq λ_{2} \leq \dots$ of the principal Dirichlet form and an orthonormal basis ${e_{j}}$ of $L^{2} (Ω)$ of weak eigenfunctions; the Rayleigh principle states that its first eigenvalue equals $\min_{v \neq 0} a_{0} (v, v) / ∥ v ∥_{L^{2}}^{2}$ with $a_{0} (v, w) = \int_{Ω} D v \cdot D w$ , the minimum being attained exactly on $E_{λ_{1}} ∖ {0}$ , and that it is positive whenever the form is coercive, as it is for the principal form by [F5].

Proof

technique · direct

Given: The set $Ω$ , the energy $E$ and the unit sphere $S$ above.

1.1givenF1F2F3F4

By [F4] the set $S$ is nonempty and $λ_{1} = \inf_{S} E \leq E (u_{1}) < \infty$ ; by [F1] and [F2] the space $H_{0}^{1} (Ω)$ is a real reflexive Banach space with the $H^{1}$ norm, and by [F3] the functional $E$ is weakly sequentially lower semicontinuous on the convex set $H_{0}^{1} (Ω)$ .

2.1step 1.1F6

Since $0 \leq λ_{1} < \infty$ , DC supplies a sequence $v_{j} \in S$ with $E (v_{j}) < λ_{1} + 1 / j$ for $j \geq 1$ , so $E (v_{j}) \to λ_{1}$ . Since $∥ v_{j} ∥_{L^{2}} = 1$ and $E (v_{j}) \leq E (u_{1}) + 1$ for all large $j$ , one has $∥ v_{j} ∥_{H_{0}^{1}}^{2} = 1 + E (v_{j}) \leq E (u_{1}) + 2$ , so $(v_{j})$ is norm bounded; by [F6] some subsequence satisfies $v_{j_{l}} ⇀ v_{0}$ in $H_{0}^{1} (Ω)$ .

3.1step 2.1F7

Since $Ω$ is bounded and open, $∥ v_{j_{l}} ∥_{L^{2}} = 1$ and $v_{j_{l}} ⇀ v_{0}$ , [F7] gives $v_{0} \in H_{0}^{1} (Ω)$ and $∥ v_{0} ∥_{L^{2}} = 1$ , that is $v_{0} \in S$ .

4.1step 1.1step 2.1step 3.1F3

By weak lower semicontinuity [F3] and step 2.1, $E (v_{0}) \leq {lim inf}_{l} E (v_{j_{l}}) = λ_{1}$ ; since $v_{0} \in S$ by step 3.1, also $E (v_{0}) \geq λ_{1}$ . Hence $E (v_{0}) = λ_{1}$ : the infimum is attained on $S$ .

5.1step 4.1F8

Let $v_{0} \in S$ be any minimiser and put $G (u) : = ∥ u ∥_{L^{2}}^{2}$ ; then $S = {G = 1} = {G = G (v_{0})}$ , and $E$ , $G$ are Fréchet differentiable at $v_{0}$ with $D E (v_{0}) h = 2 \int_{Ω} D v_{0} \cdot D h d x$ and $D G (v_{0}) h = 2 \int_{Ω} v_{0} h d x$ , because the remainders $\int_{Ω} ∣ D h ∣^{2}$ and $(\int_{Ω} h^{2})$ are $o (∥ h ∥_{H_{0}^{1}})$ . The same derivative formula holds at every $u \in H_{0}^{1}$ , and $∥ D G (u) - D G (v) ∥ \leq 2 ∥ u - v ∥_{L^{2}} \leq 2 ∥ u - v ∥_{H_{0}^{1}}$ by Cauchy--Schwarz, so $G$ is $C^{1}$ . Since $D G (v_{0}) v_{0} = 2 ∥ v_{0} ∥_{L^{2}}^{2} = 2 \neq 0$ , the functional $D G (v_{0})$ is surjective onto $R$ , so the multiplier rule [F8] with $m = 1$ gives a unique $λ \in R$ with $D E (v_{0}) = λ D G (v_{0})$ , that is $\int_{Ω} D v_{0} \cdot D h = λ \int_{Ω} v_{0} h$ for every $h \in H_{0}^{1} (Ω)$ . Testing $h = v_{0}$ gives $λ = E (v_{0}) = λ_{1}$ ; hence every minimiser is a weak eigenpair with eigenvalue $λ_{1}$ .

5.2step 4.1F9

A nonnegative minimiser exists: by [F9] the class $∣ v_{0} ∣$ lies in $H_{0}^{1} (Ω)$ with the same $L^{2}$ norm and the same energy, so $∣ v_{0} ∣ \in S$ and $E (∣ v_{0} ∣) = E (v_{0}) = λ_{1}$ ; thus $∣ v_{0} ∣$ is a minimiser and it is nonnegative.

6.1step 5.1F5

The eigenvalue is positive: by [F5], $1 = ∥ v_{0} ∥_{L^{2}} \leq C_{P} ∥ D v_{0} ∥_{L^{2}}$ , so $λ_{1} = E (v_{0}) = ∥ D v_{0} ∥_{L^{2}}^{2} \geq C_{P}^{- 2} > 0$ .

7.1step 4.1step 5.1step 6.1step 5.2F5F10∎

Finally, apply the Rayleigh principle [F10] to the principal Dirichlet form $a_{0} (v, w) = \int_{Ω} D v \cdot D w$ : its first listed eigenvalue equals $\min_{v \neq 0} a_{0} (v, v) / ∥ v ∥_{L^{2}}^{2} = \min_{S} E = λ_{1}$ , the minimum being attained exactly on the eigenspace $E_{λ_{1}}$ minus the origin, and positivity holds since the principal form is coercive by [F5]. Hence the listed first Dirichlet eigenvalue is $λ_{1}$ , and the minimisers of $E$ on $S$ are exactly the elements of $S \cap E_{λ_{1}}$ ; steps 5.1 and 6.1 show that every such minimiser is a weak eigenpair with eigenvalue $λ_{1} = E (v_{0}) > 0$ , and step 5.2 supplies a nonnegative minimiser.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Higher eigenvalues by orthogonality-constrained minimisation

Statement

Assume the Axiom of Choice, the ultrafilter lemma, DC and HB (The Axiom of Choice, The ultrafilter extension principle (UL/BPI), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The real dominated-extension principle as an additional hypothesis over ZF). Let $n \geq 1$ , let $Ω \subseteq R^{n}$ be a nonempty bounded open set, let $λ_{1} \leq λ_{2} \leq \dots$ and ${e_{j}}$ be the eigenvalues and orthonormal eigenbasis of the real Dirichlet Laplacian, supplied by Discrete spectrum of a symmetric elliptic Dirichlet operator with $a_{0} (v, w) = \int_{Ω} D v \cdot D w$ (The $L^{2}$ operator associated with a symmetric elliptic form, Symmetric elliptic weak eigenpairs, The notation $H^{k}$ and the reserved zero-boundary symbol), and fix $k \geq 2$ . Put $S_{k} = {u \in H_{0}^{1} (Ω; R) : ∥ u ∥_{L^{2}} = 1, (u, e_{j})_{L^{2}} = 0 for j < k}$ and $E (u) = \int_{Ω} ∣ D u ∣^{2}$ (Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms, The space $L^{p} (μ)$ as the quotient by null functions, Orthogonality and the orthogonal complement). Then $S_{k}$ is nonempty, $E$ attains its infimum $μ_{k}$ on $S_{k}$ , every minimiser is a weak eigenpair with eigenvalue $μ_{k}$ , and $μ_{k} = λ_{k}$ (eigenvalues counted with multiplicity); moreover every minimiser lies in the eigenspace $E_{λ_{k}}$ and is orthogonal in $L^{2}$ to $e_{1}, \dots, e_{k - 1}$ .

Facts & Assumptions

Given: A nonempty bounded open set $Ω \subseteq R^{n}$ , the principal Dirichlet form $a_{0} (v, w) = \int_{Ω} D v \cdot D w$ with its eigenvalue list $λ_{1} \leq λ_{2} \leq \dots$ and orthonormal eigenbasis ${e_{j}}_{j \geq 1}$ of $L^{2} (Ω)$ , an index $k \geq 2$ , the set $S_{k}$ and the energy $E (v) = a_{0} (v, v)$ .

[F1]

Specialise the spectral theorem to the real form $a_{0} (v, w) = \int_{Ω} D v \cdot D w$ (identity principal coefficients, zero drift and potential). Discrete spectrum of a symmetric elliptic Dirichlet operator, Symmetric elliptic weak eigenpairs, The $L^{2}$ operator associated with a symmetric elliptic form: each $e_{j}$ lies in $H_{0}^{1} (Ω)$ with $∥ e_{j} ∥_{L^{2}} = 1$ , $a_{0} (e_{j}, v) = λ_{j} (e_{j}, v)_{L^{2}}$ for every $v \in H_{0}^{1} (Ω)$ , the list is nondecreasing, and ${e_{j}}$ is a Hilbert basis of $L^{2} (Ω)$ (Orthonormal families, complete orthonormal systems and Hilbert bases).

[F2]

Zero-boundary Sobolev space as a norm closure, $H^{k}$ is a Hilbert space under the derivative-sum inner product, A closed subspace of a Banach space is Banach: $H_{0}^{1} (Ω)$ is a closed subspace of $H^{1} (Ω) = W^{1, 2} (Ω)$ , hence a real Banach space, and under HB its closed subspace of the reflexive space $W^{1, 2} (Ω)$ is reflexive (W^{1,p}(Omega) is reflexive for 1<p<infinity, Closed subspaces of reflexive spaces are reflexive, Reflexivity is surjectivity of the canonical map).

[F3]

Convex and strictly convex functionals on a convex subset of a real vector space, A convex norm-lower-semicontinuous functional is weakly lower semicontinuous: $E$ is convex and continuous on $H_{0}^{1} (Ω)$ and therefore weakly sequentially lower semicontinuous on every nonempty convex subset (Axiom of Choice through the convex closedness lemma).

[F4]

[F5]

Weak H^1 convergence plus Rellich preserves the L^2 unit normalisation: for bounded open $Ω$ , if $(v_{j}) \subseteq H_{0}^{1} (Ω)$ is norm bounded with $v_{j} ⇀ v$ and $∥ v_{j} ∥_{L^{2}} = 1$ , then $∥ v ∥_{L^{2}} = 1$ .

[F6]

Smooth compactly supported functions of an open set are dense in $L^{2}$ , The space $L^{p} (μ)$ as the quotient by null functions: $H_{0}^{1} (Ω)$ is dense in $L^{2} (Ω)$ ; hence an $L^{2}$ class orthogonal to $H_{0}^{1} (Ω)$ is zero. For each $j$ the functional $L_{j} (v) : = (v, e_{j})_{L^{2}}$ is bounded on $H_{0}^{1} (Ω)$ because $∥ v ∥_{L^{2}} \leq ∥ v ∥_{H_{0}^{1}}$ , so $v_{j} ⇀ v$ in $H_{0}^{1} (Ω)$ implies $L_{j} (v_{j}) \to L_{j} (v)$ .

[F7]

The Lagrange multiplier rule for finitely many regular constraints, Fréchet derivative between Banach spaces: for a real Banach space $X$ , open $U \subseteq X$ , $I : U \to R$ Fréchet differentiable at $u$ and $G : U \to R^{m}$ of class $C^{1}$ with $D G (u)$ surjective, a local extremum of $I$ on the level set ${G = G (u)}$ admits a unique $λ \in R^{m}$ with $D I (u) = \sum_{i} λ_{i} D G_{i} (u)$ ; the Axiom of Choice is consumed here.

[F8]

Eigenfunctions for distinct symmetric elliptic eigenvalues are $L^{2}$ -orthogonal: weak eigenfunctions of the symmetric case with distinct eigenvalues are $L^{2}$ -orthogonal.

Proof

technique · direct

Given: The spectral data and the set $S_{k}$ above.

1.1givenF1

By [F1] the function $e_{k}$ has $∥ e_{k} ∥_{L^{2}} = 1$ and $(e_{k}, e_{j})_{L^{2}} = 0$ for $j < k$ , so $e_{k} \in S_{k}$ : the set $S_{k}$ is nonempty and $μ_{k} : = \inf_{S_{k}} E \leq E (e_{k}) = a_{0} (e_{k}, e_{k}) = λ_{k} ∥ e_{k} ∥_{L^{2}}^{2} = λ_{k} < \infty$ .

1.2F2F3F6

By [F2] the space $H_{0}^{1} (Ω)$ is a real reflexive Banach space, by [F3] the energy $E$ is weakly sequentially lower semicontinuous on $H_{0}^{1} (Ω)$ , and by [F6] each constraint functional $L_{j}$ , $j < k$ , is bounded on $H_{0}^{1} (Ω)$ .

2.1step 1.1step 1.2F4

Since $0 \leq μ_{k} < \infty$ , DC supplies a sequence $v_{j} \in S_{k}$ with $E (v_{j}) < μ_{k} + 1 / j$ for $j \geq 1$ , so $E (v_{j}) \to μ_{k}$ . Since $∥ v_{j} ∥_{L^{2}} = 1$ and $E (v_{j}) \leq λ_{k} + 1$ for all large $j$ , one has $∥ v_{j} ∥_{H_{0}^{1}}^{2} = 1 + E (v_{j}) \leq λ_{k} + 2$ ; by [F4] some subsequence satisfies $v_{j_{l}} ⇀ v_{0}$ in $H_{0}^{1} (Ω)$ .

3.1step 2.1F5F6

The limit stays constrained: [F5] applied to the subsequence gives $v_{0} \in H_{0}^{1} (Ω)$ and $∥ v_{0} ∥_{L^{2}} = 1$ , while for every $j < k$ the bounded functional $L_{j}$ of [F6] gives $(v_{0}, e_{j})_{L^{2}} = \lim_{l} (v_{j_{l}}, e_{j})_{L^{2}} = 0$ . Hence $v_{0} \in S_{k}$ .

4.1step 1.2step 2.1step 3.1F3

By weak lower semicontinuity [F3] we get $E (v_{0}) \leq {lim inf}_{l} E (v_{j_{l}}) = μ_{k}$ , and $v_{0} \in S_{k}$ gives $E (v_{0}) \geq μ_{k}$ ; hence $E (v_{0}) = μ_{k}$ , so $E$ attains its infimum on $S_{k}$ .

5.1step 4.1F1F6F7

Let $v_{0} \in S_{k}$ be any minimiser and define $G : X \to R^{k}$ on $X = H_{0}^{1} (Ω)$ by $G (u) = (∥ u ∥_{L^{2}}^{2} - 1, (u, e_{1})_{L^{2}}, \dots, (u, e_{k - 1})_{L^{2}})$ ; then ${G = 0} = S_{k} = {G = G (v_{0})}$ , $E$ and $G$ are Fréchet differentiable at $v_{0}$ with $D E (v_{0}) h = 2 \int_{Ω} D v_{0} \cdot D h$ and $D G (v_{0}) h = (2 (v_{0}, h)_{L^{2}}, (h, e_{1})_{L^{2}}, \dots, (h, e_{k - 1})_{L^{2}})$ , because the remainders are $\int_{Ω} ∣ D h ∣^{2}$ and the pairings $\int_{Ω} h^{2}$ . Surjectivity is explicit: $D G (v_{0}) (v_{0} / 2) = (1, 0, \dots, 0)$ and $D G (v_{0}) e_{j}$ is the standard coordinate vector with its $1$ in position $j + 1$ , for $j < k$ , using orthonormality and the constraints. The derivative formula holds at every $u \in X$ , its first component varies by at most $2 ∥ u - v ∥_{L^{2}} ∥ h ∥_{L^{2}} \leq 2 ∥ u - v ∥_{H^{1}} ∥ h ∥_{H^{1}}$ , and its other components are constant bounded functionals. Hence $G$ is $C^{1}$ with surjective derivative at $v_{0}$ , and [F7] gives unique multipliers $α, η_{1}, \dots, η_{k - 1} \in R$ with $2 a_{0} (v_{0}, h) = 2 α (v_{0}, h)_{L^{2}} + \sum_{j < k} η_{j} (h, e_{j})_{L^{2}}$ for every $h \in H_{0}^{1} (Ω)$ .

6.1step 4.1step 5.1F1

Test the identity of step 5.1 at $h = e_{j}$ with $j < k$ . Symmetry and [F1] give $2 a_{0} (v_{0}, e_{j}) = 2 λ_{j} (v_{0}, e_{j})_{L^{2}} = 0$ , whereas its right-hand side is $η_{j}$ by the constraints and orthonormality, so $η_{j} = 0$ . Testing at $h = v_{0}$ gives $2 E (v_{0}) = 2 α ∥ v_{0} ∥_{L^{2}}^{2}$ , so $α = μ_{k}$ . Therefore $a_{0} (v_{0}, h) = μ_{k} (v_{0}, h)_{L^{2}}$ for every $h \in H_{0}^{1} (Ω)$ , and every minimiser is a weak eigenpair with eigenvalue $μ_{k}$ .

7.1step 3.1step 6.1F1F8

It remains to identify $μ_{k}$ with $λ_{k}$ ; already $μ_{k} \leq λ_{k}$ by step 1.1. Suppose $μ_{k} < λ_{k}$ : for every $j \geq k$ one has $λ_{j} \geq λ_{k} > μ_{k}$ , so the eigenfunctions $v_{0}$ and $e_{j}$ have distinct eigenvalues and [F8] gives $(v_{0}, e_{j})_{L^{2}} = 0$ ; for $j < k$ the same holds because $v_{0} \in S_{k}$ . Thus $v_{0}$ is $L^{2}$ -orthogonal to every element of the Hilbert basis ${e_{j}}$ [F1], so $v_{0} = 0$ in $L^{2}$ , contradicting $∥ v_{0} ∥_{L^{2}} = 1$ . Hence $μ_{k} = λ_{k}$ .

8.1step 3.1step 4.1step 6.1step 7.1F1F2F4F3F7F8∎

Consequently $μ_{k} = λ_{k}$ , every minimiser is a weak eigenpair with eigenvalue $λ_{k}$ by step 6.1 and therefore lies in the eigenspace $E_{λ_{k}}$ , and by membership in $S_{k}$ it is $L^{2}$ -orthogonal to $e_{1}, \dots, e_{k - 1}$ . The Axiom of Choice enters through the convex-lower-semicontinuity and multiplier suppliers [F3, F7], Countable Choice through the orthogonality corollary [F8] (it follows from the assumed DC via Dependent choice implies countable choice), and the ultrafilter lemma, DC and HB through the weak-compactness and reflexivity suppliers [F2, F4].

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Pointwise and integral constraints have different regularity tests

Remark

The multiplier rules of this page apply to equality constraints given by a $C^{1}$ map with surjective derivative and produce a multiplier equation (The Lagrange multiplier rule for finitely many regular constraints, The Lagrange multiplier rule for one regular constraint in Hilbert space), while an obstacle constraint $u \geq ψ$ is a closed convex inequality constraint that does not by itself provide a differentiable multiplier field: its first-order information is the variational inequality $a (u, v - u) \geq F (v - u)$ (The closed convex obstacle set and the obstacle variational inequality), and complementarity is expressed through the reaction distribution, not through a pointwise product (Obstacle complementarity in distribution form).

Two different regularity tests are therefore in force. The equality rule needs surjectivity of the constraint derivative at the extremum; when that test fails the multiplier equation can fail outright, as The degenerate constraint $x^{2} + y^{2} = 0$ defeats the multiplier conclusion records for the degenerate constraint $x^{2} + y^{2} = 0$ . The obstacle set has no derivative to test and instead needs regularity of the reaction if one wants more than the distributional inequality: the function version is stated with an $L^{2}$ reaction density and continuous representatives, while the measure version uses a Radon representation and continuous representatives (Obstacle complementarity in distribution form, The obstacle reaction is supported on the contact set under measure regularity). Minimality alone supplies neither an $L^{2}$ density nor continuous representatives; the conditional measure formulation does not assert that a nonnegative distribution can fail to admit a Radon representation. In particular the smooth finite-dimensional Lagrange multiplier theorem must not be applied to the obstacle set.

5 · Examples, counterexamples and false statements

None yet.