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.

✓ 23 results · all verified · 18 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 5 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

The Direct Method and Euler--Lagrange Equations

1 · Prerequisites

2 · Summary

This page develops the direct method of the calculus of variations and the Euler--Lagrange equation for integral functionals, first on an abstract normed space and then on the Sobolev spaces attached to a bounded $C^{1}$ domain. Extended-real functionals are fixed with their effective domain, properness, coercivity and weak sequential lower semicontinuity, and convex and strictly convex functionals on a real vector space are defined together with the Gateaux and Frechet derivatives of a functional. The direct-method spine then runs: coercivity bounds every finite level set; a norm-bounded sequence in a reflexive Banach space has a weakly convergent subsequence, under the ultrafilter lemma, DC and HB; weak closedness keeps the weak limit admissible; and the liminf passage at a weakly lower semicontinuous functional turns a minimising sequence into a minimiser. The abstract existence theorem combines these lemmas for a functional proper on a nonempty weakly sequentially closed admissible set, and $W^{1, p} (Ω)$ is shown to be reflexive for $1 < p < \infty$ , so the convex integral functional $I (u) = \int_{Ω} f (x, u, D u) d x$ with a Caratheodory integrand attains its infimum on every nonempty affine trace class, under the stated upper growth and coercivity hypotheses and joint convexity in $(s, ξ)$ . Strict convexity gives uniqueness of a minimiser, and for a convex Gateaux differentiable functional stationarity is sufficient for a global minimum. The more general convex variational inequality is proved with finite one-sided derivatives along admissible segments, including boundary points of the convex set. Convex norm sequential lower semicontinuity passes to weak sequential lower semicontinuity under HB and Countable Choice by equality of the ambient norm and weak closures of convex sublevels.

The Euler--Lagrange half of the page proves the first variation formula. The Caratheodory composition lemma makes the integral well defined; the fundamental lemma of the calculus of variations and its boundary form convert the vanishing first variation into the weak Euler--Lagrange equation $\int_{Ω} (f_{ξ} \cdot D φ + f_{s} φ) d x = 0$ on the fixed-trace class, with the boundary fundamental lemma retaining the natural condition on a free boundary. Under $C^{2}$ regularity of the integrand and of the minimiser, integration by parts gives the classical equation $- div f_{ξ} + f_{s} = 0$ ; the free-boundary theorem gives $f_{ξ} (x, u, D u) \cdot ν = 0$ on $\partial Ω$ . A remark records that the Euler--Lagrange equation is necessary but not sufficient without convexity, and the page closes with the Dirichlet principle: the Dirichlet energy $\frac{1}{2} \int_{Ω} ∣ D u ∣^{2} - \int_{Ω} f u$ has a unique minimiser on each admissible affine trace class, that minimiser is the weak solution of $- Δ u = f$ , and it is classical whenever the elliptic regularity theory of the cited suppliers applies. A twice differentiable local minimiser has nonnegative second variation throughout.

Conventions: $Ω \subseteq R^{n}$ is a bounded $C^{1}$ domain, $1 < p < \infty$ , weak lower semicontinuity is sequential, and the Euler--Lagrange equation is written $- div f_{ξ} + f_{s} = 0$ . Choice principles are declared per item: the reflexive-subsequence and full direct-method statements assume the ultrafilter lemma, DC and HB, the weak-closure and trace-class lemmas and fixed-trace Euler--Lagrange theorem assume the Axiom of Choice. Composition, differentiation and the interior fundamental lemma explicitly assume Countable Choice; the boundary lemma explicitly assumes AC and enters through the Lebesgue-point and mollifier interfaces that use the Axiom of Countable Choice.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Proper, coercive and weakly lower semicontinuous extended-real functionals

Definition

Setting. Let $X$ be a real normed space (in particular, a Banach space as in Banach space) and let $A \subseteq X$ be a nonempty subset, the admissible set. An extended-real functional on $A$ is a map $I : A \to (- \infty, + \infty]$ ; its effective domain is $dom I = {u \in A : I (u) < + \infty}$ . Properness. $I$ is proper if $dom I \neq \emptyset$ , equivalently if $\inf_{A} I < + \infty$ ; a point of $dom I$ is a finite competitor. Coercivity. $I$ is coercive on $A$ if for every $M \in R$ there is $R \geq 0$ such that $I (u) > M$ whenever $u \in A$ and $∥ u ∥ \geq R$ ; equivalently (the form used below) every sublevel set ${u \in A : I (u) \leq Λ}$ , $Λ \in R$ , is bounded. Weak lower semicontinuity. $I$ is weakly sequentially lower semicontinuous at $u \in A$ if $I (u) \leq {lim inf}_{j} I (u_{j})$ for every sequence $(u_{j})_{j \in N} \subseteq A$ with $u_{j} ⇀ u$ (Weak convergence of nets and sequences), and weakly sequentially lower semicontinuous on $A$ if this holds at every $u \in A$ . Analogously $I$ is sequentially lower semicontinuous on $A$ if $I (u) \leq {lim inf}_{j} I (u_{j})$ whenever $u_{j} \to u$ in norm; all infima and limits inferior are taken in $\overline{R} = [- \infty, + \infty]$ , using the complete extended order of Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ . For an extended-real sequence $(a_{j})$ , set ${lim inf}_{j} a_{j} : = \sup_{N} \inf_{j \geq N} a_{j}$ ; this extends the tail formula of Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ to sequences that may contain $+ \infty$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined). In particular, the infimum or limit inferior may equal $- \infty$ . Convention. Only the values on $A$ enter these notions, and $I$ is identified with its restriction to $A$ ; a point of $X ∖ A$ is inadmissible, not a point where $I$ equals $+ \infty$ .

Remarks

The two forms of coercivity agree. If $I$ is coercive in the divergence form and $Λ \in R$ , applying the definition with $M = Λ$ gives $R \geq 0$ with $I (u) > Λ$ whenever $u \in A$ and $∥ u ∥ \geq R$ , so the sublevel set ${u \in A : I (u) \leq Λ}$ is contained in the bounded set ${u \in X : ∥ u ∥ < R}$ . Conversely, suppose every sublevel set is bounded and let $M \in R$ be given; the sublevel set $S = {u \in A : I (u) \leq M}$ is bounded, so there is $R \geq 0$ with $∥ u ∥ \leq R$ for all $u \in S$ , and every $u \in A$ with $∥ u ∥ \geq R + 1$ lies outside $S$ , that is, $I (u) > M$ (a value in $(- \infty, + \infty]$ fails $I (u) \leq M$ exactly when it exceeds $M$ ). This is the sense in which the equivalence is asserted.
Properness and a finite infimum. If $u \in dom I$ then $\inf_{A} I \leq I (u) < + \infty$ , and conversely if $\inf_{A} I < + \infty$ then not every value of $I$ on the nonempty set $A$ is $+ \infty$ , so some $u \in A$ satisfies $I (u) < + \infty$ , that is, $u \in dom I$ .
The sublevel-set form is the one used in the compactness step of the direct method, and the divergence form is the one recorded in the sources ([MA] Definition 2.3, [G] Definition 4.1, [T] Section 13.2). No convexity, continuity or topology on $A$ is assumed by these definitions.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Gateaux and Frechet derivatives of a functional

Definition

Let $X$ be a real Banach space, $U \subseteq X$ open and $F : U \to R$ . Frechet differentiability. $F$ is Frechet differentiable at $u \in U$ if there is a bounded linear functional $D F (u) \in X^{*}$ (Fréchet derivative between Banach spaces, A bounded linear operator between normed spaces, The dual space X^* of a normed space and its dual norm) with $F (u + h) = F (u) + D F (u) h + o (∥ h ∥) (h \to 0),$ that is, $\lim_{h \to 0} ∣ F (u + h) - F (u) - D F (u) h ∣ / ∥ h ∥ = 0$ ; such $D F (u)$ is unique. Gateaux differentiability. $F$ is Gateaux differentiable at $u$ in the direction $v \in X$ if the limit $δ F (u; v) : = \lim_{ε \to 0, ε \neq 0} \frac{F (u + ε v) - F (u)}{ε} \in R$ exists; $F$ is Gateaux differentiable at $u$ if that limit exists for every $v \in X$ and the resulting map $v \mapsto δ F (u; v)$ is a bounded linear functional, written $δ F (u) \in X^{*}$ and called the Gateaux (variational) derivative of $F$ at $u$ . Relation and caveat. Frechet differentiability at $u$ implies Gateaux differentiability at $u$ with $δ F (u) = D F (u)$ , because $F (u + ε v) - F (u) = D F (u) (ε v) + o (∣ ε ∣)$ ; the converse fails, and the directional limits $δ F (u; v)$ need not be linear or bounded in $v$ when only the one-dimensional limits exist. For each fixed $v$ the function $φ (ε) : = F (u + ε v)$ satisfies $φ^{'} (0) = δ F (u; v)$ .

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

A norm-closed convex set is weakly sequentially closed

Statement

Assume the Axiom of Choice. Let $X$ be a real or complex normed space and let $K \subseteq X$ be convex and closed in the norm topology. Then $K$ is closed in the weak topology $σ (X, X^{*})$ (Weak topology on a normed space); in particular $K$ is weakly sequentially closed: if $(u_{j}) \subseteq K$ and $u_{j} ⇀ u$ (Weak convergence of nets and sequences), then $u \in K$ .

Facts & Assumptions

Given: The Axiom of Choice (The Axiom of Choice); a real or complex normed space $X$ with dual $X^{*}$ ; a convex set $K \subseteq X$ that is closed in the norm topology. The weak topology is $σ (X, X^{*})$ (Weak topology on a normed space) and weak sequential convergence is as in Weak convergence of nets and sequences.

[F1]

Under the Axiom of Choice, two disjoint nonempty convex sets $C, K \subseteq X$ , with $C$ closed and $K$ compact, are strongly separated by a nonzero functional in $X^{*}$ (Strong separation of a closed and a compact convex set): there are $f \in X^{*}$ , $f \neq 0$ , and a positive gap $\sup_{C} Re f < \inf_{K} Re f$ .

[F2]

The weak topology is the initial topology of the maps $f : X \to K$ , $f \in X^{*}$ , hence every set ${y \in X : Re f (y) > α}$ with $f \in X^{*}$ and $α \in R$ is weakly open, and a subset of $X$ is weakly closed exactly when its complement is weakly open (Weak topology on a normed space).

[F3]

A sequence $u_{j} ⇀ u$ converges weakly in the sense of convergence in $σ (X, X^{*})$ ; a weakly closed set contains the limit of every weakly convergent sequence contained in it (Weak convergence of nets and sequences).

Proof

technique · direct, by separating an exterior point from $K$ with a weak half-space

1.1givenalgebra

Trivial case and set-up. If $K = \emptyset$ then $K$ is closed in every topology, so both assertions hold; assume henceforth $K \neq \emptyset$ and fix a point $x \in X ∖ K$ .

2.1F1step 1.1

Strong separation of $K$ and the singleton ${x}$ . The sets $K$ and ${x}$ are nonempty and convex, $K$ is closed in the norm topology and ${x}$ is compact; they are disjoint because $x \notin K$ . By [F1], applied here, there are $f \in X^{*}$ , $f \neq 0$ , and a real number $α$ with $\sup_{K} Re f \leq α < Re f (x)$ , the gap being the one supplied by the theorem.

3.1F2step 2.1

A weak neighbourhood of $x$ missing $K$ . Put $U : = {y \in X : Re f (y) > α}$ . By [F2] the set $U$ is open in $σ (X, X^{*})$ ; it contains $x$ because $Re f (x) > α$ , and it is disjoint from $K$ because every $y \in K$ satisfies $Re f (y) \leq \sup_{K} Re f \leq α$ .

4.1step 3.1F2

$K$ is weakly closed. Since $x \in X ∖ K$ was arbitrary and step 3.1 produces for it a weak neighbourhood $U \subseteq X ∖ K$ , the complement $X ∖ K$ is weakly open; equivalently $K$ is closed in the weak topology $σ (X, X^{*})$ .

5.1F3step 4.1∎

Weak sequential closedness. Let $(u_{j}) \subseteq K$ with $u_{j} ⇀ u$ . By [F3] the convergence is convergence in $σ (X, X^{*})$ , and a set closed in a topology contains the limit of every convergent sequence in it; hence $u \in K$ .

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

A bounded sequence in a reflexive Banach space has a weakly convergent subsequence

Statement

Assume the ultrafilter lemma, DC and HB. Let $X$ be a real reflexive Banach space (Reflexivity is surjectivity of the canonical map) and let $(u_{j})$ be a norm-bounded sequence in $X$ . Then $(u_{j})$ has a subsequence converging weakly to a point of $X$ (Weak convergence of nets and sequences).

Facts & Assumptions

[F1]

Under the ultrafilter lemma, DC and HB, a real Banach space $X$ is reflexive if and only if every norm-bounded sequence in $X$ has a subsequence that converges weakly to a point of $X$ (Reflexivity is equivalent to weak subsequential compactness of bounded sequences); the convergence is in the sense of Weak convergence of nets and sequences.

Proof

technique · direct, by the forward implication of the reflexivity characterisation

1.1F1given

The three choice principles named in the hypothesis are exactly the ones assumed by [F1], and $X$ is a real reflexive Banach space by hypothesis; the sequence $(u_{j})$ is norm bounded by hypothesis. The forward implication of [F1] therefore applies and produces a strictly increasing sequence of indices $j_{1} < j_{2} < \dots$ and a point $u \in X$ with $u_{j_{k}} ⇀ u$ .

2.1step 1.1∎

The limit $u$ obtained in step 1.1 is a point of $X$ , so $(u_{j})$ has a subsequence converging weakly to a point of $X$ , which is the stated conclusion.

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

W^{1,p}(Omega) is reflexive for 1<p<infinity

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 $Ω \subseteq R^{n}$ be open, $n \geq 1$ , and let $1 < p < \infty$ . Then the real Banach space $W^{1, p} (Ω)$ of Integer-order Sobolev spaces and their norms is reflexive (Reflexivity is surjectivity of the canonical map).

Facts & Assumptions

Given: The ultrafilter lemma, DC and HB; an open set $Ω \subseteq R^{n}$ , $n \geq 1$ ; and $1 < p < \infty$ .

[F1]

The Sobolev norm is the $ℓ^{p}$ -sum norm $∥ u ∥_{W^{1, p} (Ω)} = (\sum_{∣ α ∣ \leq 1} ∥ D^{α} u ∥_{L^{p} (Ω)}^{p})^{1 / p}$ over the finitely many multi-indices $∣ α ∣ \leq 1$ (Integer-order Sobolev spaces and their norms).

[F2]

$W^{1, p} (Ω)$ is a complete normed space (Integer-order Sobolev spaces are Banach); its statement assumes the Axiom of Choice, used there only to derive Countable Choice, which follows from the DC assumed here (Dependent choice implies countable choice).

[F3]

For every measure space and every $1 < p < \infty$ , $L^{p}$ is reflexive under Countable Choice (Reflexivity of Lp for one less p less infinity), and Countable Choice holds here because DC does (Dependent choice implies countable choice).

[F4]

Under the ultrafilter lemma, DC and HB, a real Banach space $X$ is reflexive if and only if every norm-bounded sequence in $X$ has a subsequence converging weakly to a point of $X$ (Reflexivity is equivalent to weak subsequential compactness of bounded sequences, Reflexivity is surjectivity of the canonical map).

[F5]

Under HB, a closed linear subspace of a reflexive Banach space, with the restricted norm, is reflexive (Closed subspaces of reflexive spaces are reflexive).

Proof

technique · direct, by embedding $W^{1,p}(\Omega)$ isometrically into a finite product of reflexive $L^p$ spaces

1.1F1F2given

The gradient embedding. Write $A_{1} = {0, e_{1}, \dots, e_{n}}$ and define $Φ (u) : = (D^{α} u)_{α \in A_{1}}$ for $u \in W^{1, p} (Ω)$ , regarded as an element of the real vector space $Y : = \prod_{α \in A_{1}} L^{p} (Ω)$ equipped with the $ℓ^{p}$ -sum norm $∥ (w_{α}) ∥_{Y} : = (\sum_{α} ∥ w_{α} ∥_{L^{p} (Ω)}^{p})^{1 / p}$ . By [F1] the map $Φ$ is linear and $∥ Φ (u) ∥_{Y} = ∥ u ∥_{W^{1, p} (Ω)}$ for every $u$ ; hence $Φ$ is a linear isometry onto its image $S : = Φ (W^{1, p} (Ω))$ , and $Y$ is a Banach space (a finite $ℓ^{p}$ -sum of the Banach spaces $L^{p} (Ω)$ ).

1.2F2given

$W^{1, p} (Ω)$ is complete. By [F2] the space $W^{1, p} (Ω)$ is a complete normed space, the Countable Choice needed there being supplied by DC.

2.1F3F4step 1.1

Finite sums of reflexive spaces are reflexive. Each factor $L^{p} (Ω)$ is reflexive by [F3]; we show that a finite $ℓ^{p}$ -sum of reflexive Banach spaces is reflexive. For two factors $X, Z$ : a bounded sequence in $X \oplus_{p} Z$ has bounded coordinate sequences, so two successive extractions using [F4] give a subsequence $(x_{k}, z_{k})$ with $x_{k} ⇀ x$ in $X$ and $z_{k} ⇀ z$ in $Z$ ; every bounded linear functional on $X \oplus_{p} Z$ has the form $(x, z) \mapsto f (x) + g (z)$ with $f \in X^{*}$ , $g \in Z^{*}$ bounded by the norm of the functional (restrict the functional to each factor), so $f (x_{k}) + g (z_{k}) \to f (x) + g (z)$ and the subsequence converges weakly in $X \oplus_{p} Z$ ; [F4] then makes $X \oplus_{p} Z$ reflexive. Induction over the finitely many factors gives the reflexivity of $Y$ .

2.2step 1.1step 1.2

$S$ is closed. Since $Φ$ is a surjective isometry from the complete space $W^{1, p} (Ω)$ onto $S$ by steps 1.1 and 1.2, the space $S$ is complete, and a complete subset of the normed space $Y$ is closed.

3.1step 2.1

$Y$ is reflexive. By step 2.1 the finite $ℓ^{p}$ -sum $Y$ of the reflexive spaces $L^{p} (Ω)$ is reflexive.

4.1F5step 2.2step 3.1

$S$ is reflexive. By steps 2.2 and 3.1, $S$ is a closed linear subspace of the reflexive Banach space $Y$ ; [F5] therefore makes $S$ , with the restricted norm, reflexive.

5.1F4step 4.1∎

Reflexivity transfers to $W^{1, p} (Ω)$ . The isometry $Φ : W^{1, p} (Ω) \to S$ satisfies $Φ^{* *} \circ J_{W^{1, p} (Ω)} = J_{S} \circ Φ$ for the canonical maps: both sides send $u$ to the functional $s^{*} \mapsto s^{*} (Φ (u))$ on $S^{*}$ . If $ψ \in W^{1, p} (Ω)^{* *}$ is given, then $Φ^{* *} ψ \in S^{* *}$ and, $S$ being reflexive by step 4.1, $Φ^{* *} ψ = J_{S} (s)$ for some $s \in S$ ; writing $s = Φ (u)$ gives $J_{S} (Φ (u)) = Φ^{* *} (J_{W^{1, p} (Ω)} (u))$ , hence $J_{W^{1, p} (Ω)} (u) = ψ$ because $Φ^{* *}$ is injective. So the canonical map of $W^{1, p} (Ω)$ is surjective, that is, $W^{1, p} (Ω)$ is reflexive.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

A Caratheodory integrand composed with measurable functions is measurable

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $Ω \subseteq R^{n}$ be Lebesgue measurable and let $f : Ω \times R \times R^{n} \to R$ be a Caratheodory integrand: for every $(s, ξ) \in R \times R^{n}$ the map $x \mapsto f (x, s, ξ)$ is measurable (Borel measurable and Lebesgue measurable functions on $R^{n}$ ), and for almost every $x \in Ω$ the map $(s, ξ) \mapsto f (x, s, ξ)$ is continuous. If $u : Ω \to R$ and $w : Ω \to R^{n}$ are measurable, then $x \mapsto f (x, u (x), w (x))$ is measurable.

Facts & Assumptions

Given: Countable Choice; a Lebesgue measurable set $Ω \subseteq R^{n}$ ; a Caratheodory integrand $f : Ω \times R \times R^{n} \to R$ , so that $x \mapsto f (x, s, ξ)$ is measurable for every $(s, ξ) \in R \times R^{n}$ and $(s, ξ) \mapsto f (x, s, ξ)$ is continuous for almost every $x \in Ω$ ; measurable maps $u : Ω \to R$ and $w : Ω \to R^{n}$ . Throughout, $Ω$ carries the trace of the Lebesgue sigma-algebra and the restricted Lebesgue measure, and $N : = {x \in Ω : (s, ξ) \mapsto f (x, s, ξ) is not continuous}$ satisfies $∣ N ∣ = 0$ .

[F1]

Every real-valued measurable function is the pointwise limit everywhere of a sequence of real-valued simple functions (Every measurable function admits simple approximations dominated by its absolute value).

[F2]

Measurable real-valued functions are closed under finite sums, real scalar multiplication, positive and negative parts, and multiplication by measurable indicators. Countable infima and increasing suprema of measurable extended-real functions are measurable; thus ${lim inf}_{k} g_{k} = \sup_{m} \inf_{k \geq m} g_{k}$ is extended-real measurable and need not be finite (Closure properties of measurable functions used by the integral).

[F3]

For a map into $R^{m}$ , measurability means that preimages of Borel sets are measurable in the domain, and when $m = 1$ this is the usual notion of a real-valued measurable function (Borel measurable and Lebesgue measurable functions on $R^{n}$ ); under the ambient Axiom of Countable Choice this is the Lebesgue sigma-algebra framework used throughout.

[F4]

The Lebesgue measure space is complete: every subset of a Lebesgue null set is Lebesgue measurable (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

Proof

technique · direct, by approximation with measurable simple functions and passage to the pointwise limit

1.1F3

Measurability of $u$ and of the coordinates of $w$ . For every $a \in R$ and every coordinate index $i$ the set ${ξ \in R^{n} : ξ_{i} > a}$ is Borel, so $(w^{i})^{- 1} ((a, \infty)) = w^{- 1} ({ξ_{i} > a})$ is measurable in $Ω$ by [F3]; hence each coordinate function $w^{i}$ of $w$ is real-valued measurable, and so is $u$ .

2.1F1step 1.1

Simple approximants. By [F1] applied to $u$ there are simple functions $u_{k} : Ω \to R$ with $u_{k} \to u$ pointwise on $Ω$ , and by [F1] applied to each coordinate $w^{i}$ there are simple functions $s_{k}^{i}$ with $s_{k}^{i} \to w^{i}$ pointwise. Setting $w_{k} : = (s_{k}^{1}, \dots, s_{k}^{n})$ gives, for each $k$ , a map with finitely many values that converges to $w$ pointwise.

3.1F2step 2.1

Measurability of the composed approximations. Fix $k$ and write $u_{k} = \sum_{i = 1}^{I} a_{i} 1_{E_{i}}$ and $w_{k} = \sum_{j = 1}^{J} b_{j} 1_{F_{j}}$ with pairwise disjoint measurable sets $E_{i}, F_{j}$ covering $Ω$ . For each pair $(i, j)$ the map $x \mapsto f (x, a_{i}, b_{j})$ is measurable by the first Caratheodory clause, so $x \mapsto f (x, a_{i}, b_{j}) 1_{E_{i} \cap F_{j}} (x)$ is measurable by the indicator clause of [F2] applied to its positive and negative parts; the finite sum $g_{k} : = \sum_{i, j} f (x, a_{i}, b_{j}) 1_{E_{i} \cap F_{j}}$ is therefore measurable [F2]. Since the $E_{i}$ and the $F_{j}$ partition $Ω$ , one has $g_{k} (x) = f (x, u_{k} (x), w_{k} (x))$ for every $x$ .

4.1F2step 2.1step 3.1

The limit inferior. On $Ω ∖ N$ the map $(s, ξ) \mapsto f (x, s, ξ)$ is continuous, so $g_{k} (x) = f (x, u_{k} (x), w_{k} (x)) \to f (x, u (x), w (x))$ there by step 2.1. Hence the extended-real measurable function $g : = {lim inf}_{k} g_{k}$ , which exists by [F2], satisfies $g (x) = f (x, u (x), w (x))$ for every $x \in Ω ∖ N$ .

5.1F4step 4.1∎

Conclusion. The function $x \mapsto f (x, u (x), w (x))$ differs from the measurable function $g$ only on the null set $N$ . For a Borel set $B \subseteq R$ (also Borel in $\overline{R}$ ) its preimage is the union of ${g \in B} ∖ N$ , which is measurable, and a subset of $N$ , which is measurable by the completeness of Lebesgue measure [F4]. So $x \mapsto f (x, u (x), w (x))$ is measurable.

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

The fundamental lemma of the calculus of variations

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $Ω \subseteq R^{n}$ be open, $n \geq 1$ , and let $g \in L_{l o c}^{1} (Ω; C)$ (Locally integrable functions as regular distributions) satisfy $\int_{Ω} g φ d x = 0 for every φ \in C_{c}^{\infty} (Ω)$ (Test function space d of an open set). Then $g = 0$ almost everywhere on $Ω$ . Moreover, if $g$ is real-valued and $\int_{Ω} g φ d x \geq 0$ for every nonnegative $φ \in C_{c}^{\infty} (Ω)$ , then $g \geq 0$ almost everywhere.

Facts & Assumptions

Given: Countable Choice; an open set $Ω \subseteq R^{n}$ , $n \geq 1$ , and $g \in L_{l o c}^{1} (Ω; C)$ . Part (a) assumes $\int_{Ω} g φ d x = 0$ for every $φ \in C_{c}^{\infty} (Ω)$ ; part (b) assumes $g$ real-valued and $\int_{Ω} g φ d x \geq 0$ for every nonnegative $φ \in C_{c}^{\infty} (Ω)$ . The measure-theoretic suppliers used below are stated under the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), the ambient convention of the Lebesgue framework cited here.

[F1]

Choose a nonnegative smooth bump $b$ equal to one on ${\overline{B}}_{1 / 4} (0)$ and supported inside $B_{1 / 2} (0)$ (Compactly supported scaled Euclidean bumps). Its integral $c$ is finite by boundedness and compact support, and positive because the inner ball contains a box of positive measure (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume). Then $ρ : = b / c$ is nonnegative, smooth, compactly supported in $B_{1} (0)$ and has integral one. It is majorised by the bounded nonincreasing function $Φ (t) : = ∥ ρ ∥_{\infty} 1_{[0, 1]} (t)$ , whose radial integral is finite. Radiality of $ρ$ itself is unnecessary.

[F2]

For $f \in L^{1} (R^{n})$ and a Lebesgue point $x$ of $f$ with value $a = f (x)$ , and for any measurable kernel $K$ with $\int K = 1$ and $∣ K (y) ∣ \leq Φ (∣ y ∣)$ as in [F1], one has $\int ε^{- n} K (y / ε) f (x - y) d y \to a$ as $ε ↓ 0$ (Lebesgue-point convergence for radial-majorized kernels).

[F3]

The Lebesgue set of a class in $L_{l o c}^{1} (R^{n})$ is defined by the averages $λ (B (x, r))^{- 1} \int_{B (x, r)} ∣ f (y) - f (x) ∣ d y \to 0$ , and under the Axiom of Countable Choice it has full Lebesgue measure (Lebesgue points and the Lebesgue set of an $L_{l o c}^{1}$ class, Almost every point is a Lebesgue point of a locally integrable function).

[F4]

A ball is contained in a half-open cube of side $2 r$ centred at the same point, whose Lebesgue measure is $(2 r)^{n}$ ; by monotonicity of the measure, $λ (B (x, r)) \leq 2^{n} r^{n}$ (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

Proof

technique · direct; the sign statement is proved with a mollifier kernel at almost every point, and the vanishing statement follows by applying the sign statement to $g$ and $-g$

1.1givenalgebra

The vanishing statement follows from the sign statement. Assume part (b) proved and first take $g$ real-valued. Applying it to $g$ gives $g \geq 0$ almost everywhere; applying it to $- g$ , whose pairing with every nonnegative $φ$ equals $- \int_{Ω} g φ = 0 \geq 0$ , gives $- g \geq 0$ almost everywhere. Hence $g = 0$ almost everywhere. For complex $g$ , the vanishing pairing with every real test implies vanishing pairings for $Re g$ and $Im g$ ; applying this real argument to each gives part (a). So it suffices to prove the sign statement, and from the next step on we assume $g$ real-valued and $\int_{Ω} g φ \geq 0$ for every nonnegative $φ \in C_{c}^{\infty} (Ω)$ .

1.2algebra

Exhaustion of $Ω$ and localisation. For integers $k \geq 1$ put $Ω_{k} : = {x \in Ω : ∣ x ∣ < k and dist (x, R^{n} ∖ Ω) > 1 / k}$ , with $dist (x, \emptyset) = + \infty$ . Each $Ω_{k}$ is open (both conditions are open or strict), its closure is bounded and contained in $Ω$ , so $\overline{Ω_{k}}$ is a compact subset of $Ω$ ; moreover $Ω_{k} \subseteq Ω_{k + 1}$ and $⋃_{k} Ω_{k} = Ω$ , because for $x \in Ω$ openness gives $dist (x, R^{n} ∖ Ω) > 0$ and one may take $k > \max {∣ x ∣, 1 / dist (x, R^{n} ∖ Ω)}$ .

2.1step 1.2

The localised functions are integrable on $R^{n}$ . Let $g_{k} : = g 1_{Ω_{k}}$ , extended by zero outside $Ω_{k}$ . Since $\overline{Ω_{k}}$ is a compact subset of $Ω$ and $g \in L_{l o c}^{1} (Ω)$ , one has $\int_{R^{n}} ∣ g_{k} ∣ = \int_{Ω_{k}} ∣ g ∣ < \infty$ , so $g_{k} \in L^{1} (R^{n})$ .

3.1F3step 2.1

Almost every point is a Lebesgue point of every $g_{k}$ . By [F3] applied to $g_{k}$ there is a Lebesgue null set $N_{k} \subseteq R^{n}$ such that every $z \notin N_{k}$ is a Lebesgue point of $g_{k}$ ; the union $N : = ⋃_{k} N_{k}$ is again null, being a countable union of null sets.

4.1F3F4step 3.1

At points of $Ω ∖ N$ the Lebesgue averages are small. Fix $z \in Ω ∖ N$ and $k$ with $z \in Ω_{k}$ . Then $z \notin N_{k}$ , so for $A_{k} (r) : = \int_{B (z, r)} ∣ g_{k} (y) - g_{k} (z) ∣ d y$ the Lebesgue point property gives $A_{k} (r) / λ (B (z, r)) \to 0$ ; combined with [F4] this yields $A_{k} (r) / r^{n} \leq 2^{n} A_{k} (r) / λ (B (z, r)) \to 0$ , that is $A_{k} (r) = o (r^{n})$ . Moreover $g_{k} (z) = g (z)$ , since $z \in Ω_{k}$ .

5.1F1F2step 4.1

Kernel convergence. By step 4.1 the point $z$ is a Lebesgue point of $g_{k}$ with $g_{k} (z) = g (z)$ and $A_{k} (r) = o (r^{n})$ , so applying [F2] with $f = g_{k}$ , $x = z$ , $a = g_{k} (z)$ and the kernel $K = ρ$ of [F1] gives, for $r ↓ 0$ , the limit $r^{- n} \int_{R^{n}} ρ (y / r) g_{k} (z - y) d y ⟶ g (z),$ since [F1] realises $ρ$ as a compactly supported kernel with the required bounded nonincreasing majorant.

6.1F1givenstep 4.1

Admissible test functions. Fix $z \in Ω ∖ N$ and $k$ with $z \in Ω_{k}$ as in step 4.1. For $0 < r < dist (z, R^{n} ∖ Ω_{k})$ the function $φ_{r} (w) : = r^{- n} ρ ((z - w) / r)$ lies in $C_{c}^{\infty} (Ω)$ : it is smooth in $w$ , nonnegative, and has support in $B (z, r) \subseteq Ω$ . Changing variables $w = z - y$ gives $\int_{Ω} g (w) φ_{r} (w) d w = \int r^{- n} ρ (y / r) g (z - y) d y$ , and $g = g_{k}$ on the support of this test. Thus the integral equals the mollified $g_{k}$ in step 5.1, and the hypothesis of part (b) gives $\int_{Ω} g φ_{r} \geq 0$ .

7.1step 5.1step 6.1step 1.1∎

Conclusion of the sign statement. For $z \in Ω ∖ N$ and $r ↓ 0$ , step 6.1 keeps the quantities $\int_{Ω} g φ_{r}$ nonnegative while step 5.1 identifies their limit as $g (z)$ ; hence $g (z) \geq 0$ . Since $N$ is null, $g \geq 0$ almost everywhere on $Ω$ , and by step 1.1 this also gives $g = 0$ almost everywhere under the hypotheses of part (a).

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

A twice differentiable local minimiser has nonnegative second variation

Statement

Let $X$ be a real Banach space, $U \subseteq X$ open, $F : U \to R$ of class $C^{2}$ (C k map between Banach spaces) and let $u \in U$ be a local minimiser of $F$ , meaning that there is $δ > 0$ such that $F (w) \geq F (u)$ whenever $w \in U$ and $∥ w - u ∥ < δ$ . Then for every $v \in X$ $D^{2} F (u) [v, v] \geq 0,$ where $D^{2} F (u) \in B (X, B (X, R))$ is the second Frechet derivative of $F$ at $u$ (Fréchet derivative between Banach spaces); that is, the second variation of $F$ at $u$ is nonnegative in every direction.

Facts & Assumptions

Given: A real Banach space $X$ , an open set $U \subseteq X$ , a map $F : U \to R$ of class $C^{2}$ , a local minimiser $u \in U$ of $F$ , and a direction $v \in X$ .

[F1]

By the stated definition of local minimality, there is $δ > 0$ with $F (w) \geq F (u)$ for every $w \in U$ with $∥ w - u ∥ < δ$ ; in the terminology of Local (relative) maximum and minimum of $f : A \to R$ at a point, the strict forms, and what it means for the point to be interior to $A$ , $0$ is then an interior local minimum of the one-variable function $ε \mapsto F (u + ε v)$ (Local (relative) maximum and minimum of $f : A \to R$ at a point, the strict forms, and what it means for the point to be interior to $A$ ).

[F2]

The Banach-space calculus of C k map between Banach spaces together with the chain rule of Chain sum product and composition rules for Banach derivatives gives that $φ (ε) : = F (u + ε v)$ , defined for $∣ ε ∣$ small, is of class $C^{2}$ with $φ^{'} (ε) = D F (u + ε v) v$ and $φ^{''} (ε) = D^{2} F (u + ε v) [v, v]$ ; in particular $φ^{'} (0) = D F (u) v$ and $φ^{''} (0) = D^{2} F (u) [v, v]$ , where $D^{2} F (u) \in B (X, B (X, R))$ is the second Frechet derivative (Fréchet derivative between Banach spaces).

[F3]

If a differentiable function on a real interval has an interior local extremum at a point, then its derivative vanishes there (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 $f^{'} (c) = 0$ ).

[F4]

Taylor's formula with Lagrange remainder at order $1$ : if $φ$ has derivatives through order $2$ on $[0, h]$ , then for some $ξ \in (0, h)$ one has $φ (h) = φ (0) + φ^{'} (0) h + \frac{1}{2} φ^{''} (ξ) h^{2}$ (The Lagrange and Cauchy forms of Taylor's remainder).

Proof

technique · direct, by reduction to a one-variable function along a line

1.1F1algebra

Reduction to one variable. If $v = 0$ then $D^{2} F (u) [0, 0] = 0$ because $D^{2} F (u)$ is linear in each variable, so assume $v \neq 0$ . Since $U$ is open, there is $η > 0$ with $u + ε v \in U$ for $∣ ε ∣ < η$ , and by [F1] there is $δ > 0$ with $F (w) \geq F (u)$ for $∥ w - u ∥ < δ$ . For $∣ ε ∣ < \min (η, δ / ∥ v ∥)$ the point $u + ε v$ lies in $U$ and $∥ (u + ε v) - u ∥ = ∣ ε ∣ ∥ v ∥ < δ$ , so $φ (ε) \geq φ (0)$ : the point $0$ is an interior local minimum of $φ$ .

1.2F2

The derivatives of the reduced function. By [F2] the function $φ$ is of class $C^{2}$ near $0$ , its second derivative is continuous there, and $φ^{'} (0) = D F (u) v$ , $φ^{''} (0) = D^{2} F (u) [v, v]$ .

2.1F3step 1.2

Fermat's theorem. In the case $v \neq 0$ of step 1.1 the point $0$ lies in the interior of the interval on which $φ$ is defined and is an interior local minimum of the differentiable function $φ$ , so $φ^{'} (0) = 0$ by [F3]; combined with step 1.2 this gives $D F (u) v = 0$ .

3.1F4step 2.1algebra

Taylor expansion at order one. Let $h > 0$ be small enough that $φ$ is of class $C^{2}$ on $[0, h]$ and $φ (h) \geq φ (0)$ . By [F4] there is $ξ_{h} \in (0, h)$ with $φ (h) - φ (0) = φ^{'} (0) h + \frac{1}{2} φ^{''} (ξ_{h}) h^{2}$ , and $φ^{'} (0) = 0$ by step 2.1, so $φ (h) - φ (0) = \frac{1}{2} φ^{''} (ξ_{h}) h^{2}$ . Since $φ (h) - φ (0) \geq 0$ and $h^{2} > 0$ , it follows that $φ^{''} (ξ_{h}) \geq 0$ .

4.1step 3.1step 1.2step 1.1∎

Passage to the limit. As $h ↓ 0$ one has $ξ_{h} \to 0$ because $0 < ξ_{h} < h$ , and $φ^{''}$ is continuous at $0$ by step 1.2, so $φ^{''} (0) = \lim_{h ↓ 0} φ^{''} (ξ_{h}) \geq 0$ . In the case $v \neq 0$ , $φ^{''} (0) = D^{2} F (u) [v, v]$ by step 1.2 and hence $D^{2} F (u) [v, v] \geq 0$ ; the case $v = 0$ was settled in step 1.1. As $v$ was arbitrary, the second variation of $F$ at $u$ is nonnegative in every direction.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Convex and strictly convex functionals on a convex subset of a real vector space

Definition

Let $X$ be a real vector space and $K \subseteq X$ . The set $K$ is convex if $λ u + (1 - λ) v \in K$ for all $u, v \in K$ and $λ \in [0, 1]$ . Fix a nonempty convex $K$ and an extended-real functional $I : K \to (- \infty, + \infty]$ (Proper, coercive and weakly lower semicontinuous extended-real functionals), with the sums and positive-weight products of The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined. For convex combinations only, additionally define $0 \cdot (+ \infty) = 0$ ; this is a local convention, since that product is left undefined in the general extended-real arithmetic. Then $I$ is convex if $I (λ u + (1 - λ) v) \leq λ I (u) + (1 - λ) I (v), u, v \in K, λ \in [0, 1],$ and strictly convex if the inequality is strict whenever $u \neq v$ , $I (u), I (v) < + \infty$ and $λ \in (0, 1)$ . A convex functional has convex sublevel sets: for every $t \in R$ the set ${u \in K : I (u) \leq t}$ is convex. Only real coefficients are used: on a complex vector space these notions are read on the underlying real structure.

Remarks

Sublevel sets. Let $I$ be convex and let $t \in R$ . If $u, v \in K$ satisfy $I (u) \leq t$ and $I (v) \leq t$ , then $I (u), I (v) < + \infty$ , and for $λ \in [0, 1]$ convexity and the extended-real conventions give $I (λ u + (1 - λ) v) \leq λ I (u) + (1 - λ) I (v) \leq \max {I (u), I (v)} \leq t$ ; hence ${u \in K : I (u) \leq t}$ is convex. This is the property used when a sublevel set is intersected with a weakly closed admissible set.
Endpoint coefficients. At $λ = 0$ and $λ = 1$ the defining inequality reads $I (v) \leq I (v)$ and $I (u) \leq I (u)$ , using $0 \cdot (+ \infty) = 0$ for the extended value $+ \infty$ ; the strict form is therefore imposed only for $0 < λ < 1$ , as stated.
Finite competitors. For $0 < λ < 1$ , if either $I (u)$ or $I (v)$ is $+ \infty$ , the right-hand side of the convexity inequality is $+ \infty$ , so it carries no information at such a pair; the strict form is correspondingly restricted to pairs in the effective domain $dom I$ (Proper, coercive and weakly lower semicontinuous extended-real functionals).

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

Coercivity bounds every finite-level sequence

Statement

Let $I : A \to (- \infty, + \infty]$ be coercive on the nonempty set $A \subseteq X$ (Proper, coercive and weakly lower semicontinuous extended-real functionals). Then for every $Λ \in R$ the sublevel set ${u \in A : I (u) \leq Λ}$ is norm bounded; consequently every sequence $(u_{j}) \subseteq A$ with $\sup_{j} I (u_{j}) < + \infty$ is norm bounded. In particular every minimising sequence $(u_{j})$ with $I (u_{j}) \to \inf_{A} I < + \infty$ is norm bounded.

Facts & Assumptions

Given: A nonempty set $A$ in a real Banach space, an extended-real functional $I : A \to (- \infty, + \infty]$ that is coercive on $A$ , and real numbers $Λ$ .

[F1]

Coercivity of $I$ on $A$ is equivalent to the boundedness of every sublevel set ${u \in A : I (u) \leq Λ}$ , $Λ \in R$ (Proper, coercive and weakly lower semicontinuous extended-real functionals).

[F2]

The number $\inf_{A} I$ is the greatest lower bound of the values of $I$ on $A$ (Greatest lower bound (infimum)). If a sequence $(a_{j})$ in $(- \infty, + \infty]$ converges to a finite real $L$ , then its tail is bounded above by $L + 1$ ; if $a_{j} \to - \infty$ , its tail is bounded above by $0$ . A finite initial segment need not be bounded above as a sequence of values when it contains $+ \infty$ .

Proof

technique · direct, by placing the values in a sublevel set and quoting the sublevel form of coercivity

1.1F1given

Bounded sublevels. Let $Λ \in R$ . By [F1] the sublevel set ${u \in A : I (u) \leq Λ}$ is bounded in the norm of $X$ ; that is, there is $R \geq 0$ with $∥ u ∥ \leq R$ for every $u \in A$ with $I (u) \leq Λ$ .

2.1step 1.1

Sequences with finite sup of values are bounded. Let $(u_{j}) \subseteq A$ satisfy $\sup_{j} I (u_{j}) = : Λ_{0} < + \infty$ . Then $Λ_{0} \in R$ and $I (u_{j}) \leq Λ_{0}$ for every $j$ , so the whole sequence lies in the sublevel set ${I \leq Λ_{0}}$ , which is bounded by step 1.1; hence $(u_{j})$ is norm bounded.

3.1F2step 2.1∎

Minimising sequences with finite infimum. Let $(u_{j}) \subseteq A$ be a minimising sequence, $I (u_{j}) \to \inf_{A} I$ , with $\inf_{A} I < + \infty$ . By [F2] there are $N$ and a real $M$ with $I (u_{j}) \leq M$ for every $j \geq N$ . Step 2.1 shows that the tail $(u_{j})_{j \geq N}$ is norm bounded. The finite set of initial vectors $u_{1}, \dots, u_{N - 1}$ is also norm bounded, so the entire sequence is norm bounded, even if some initial functional values equal $+ \infty$ .

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

Weak closedness keeps the direct-method limit admissible

Statement

Let $X$ be a normed space and let $K \subseteq X$ be weakly sequentially closed (Weak convergence of nets and sequences). If $(u_{j}) \subseteq K$ and $u_{j} ⇀ u$ , then $u \in K$ . In particular, under the Axiom of Choice every nonempty convex norm-closed $K$ has this property, by A norm-closed convex set is weakly sequentially closed.

Facts & Assumptions

Given: A normed space $X$ and a subset $K \subseteq X$ that is weakly sequentially closed: every sequence $(u_{j}) \subseteq K$ with $u_{j} ⇀ u$ satisfies $u \in K$ (Weak convergence of nets and sequences).

[F1]

A set $K$ is weakly sequentially closed when it contains the weak limit of every weakly convergent sequence contained in it; the relation $u_{j} ⇀ u$ denotes convergence in the weak topology $σ (X, X^{*})$ (Weak convergence of nets and sequences).

[F2]

Under the Axiom of Choice, a convex subset of a real or complex normed space that is closed in the norm topology is closed in the weak topology $σ (X, X^{*})$ , hence weakly sequentially closed (A norm-closed convex set is weakly sequentially closed).

Proof

technique · direct, unpacking the definition and quoting the closed-convex lemma

1.1F1given

First assertion. Let $(u_{j}) \subseteq K$ with $u_{j} ⇀ u$ . By the definition [F1] of weak sequential closedness of $K$ recorded in the hypothesis, $u \in K$ ; this is exactly the first sentence of the statement.

2.1F1F2step 1.1∎

Second assertion. Assume additionally that the Axiom of Choice holds and that $K$ is nonempty, convex and norm closed. By [F2] the set $K$ is weakly closed, and a weakly closed set is in particular weakly sequentially closed: if $(u_{j}) \subseteq K$ and $u_{j} ⇀ u$ , then $u$ lies in the weak closure of $K$ , which is $K$ . Hence such a $K$ satisfies the hypothesis of step 1.1 and contains every weak limit of its weakly convergent sequences.

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

The liminf passage makes the weak limit a minimiser

Statement

Let $A \subseteq X$ and $I : A \to (- \infty, + \infty]$ be proper, and let $(u_{j}) \subseteq A$ be a minimising sequence with $u_{j} ⇀ u \in A$ (Weak convergence of nets and sequences). If $I$ is weakly sequentially lower semicontinuous at $u$ (Proper, coercive and weakly lower semicontinuous extended-real functionals), then $I (u) = \inf_{A} I$ , so $u$ is a minimiser of $I$ on $A$ .

Facts & Assumptions

Given: A set $A \subseteq X$ , a proper extended-real functional $I : A \to (- \infty, + \infty]$ (Proper, coercive and weakly lower semicontinuous extended-real functionals), a minimising sequence $(u_{j}) \subseteq A$ with $u_{j} ⇀ u$ and $u \in A$ , and weak sequential lower semicontinuity of $I$ at $u$ .

[F1]

Weak sequential lower semicontinuity of $I$ at $u$ means $I (u) \leq {lim inf}_{j} I (u_{j})$ for every sequence $(u_{j}) \subseteq A$ with $u_{j} ⇀ u$ (Proper, coercive and weakly lower semicontinuous extended-real functionals, Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ ).

[F2]

Infimum and limit inferior: $\inf_{A} I$ is the greatest lower bound of $I$ on $A$ in $[- \infty, + \infty]$ and $\inf_{A} I \leq I (w)$ for every $w \in A$ (Proper, coercive and weakly lower semicontinuous extended-real functionals).

[F3]

If a sequence of extended reals converges to a limit, its limit inferior equals that limit (Proper, coercive and weakly lower semicontinuous extended-real functionals).

Proof

technique · direct

1.1F3F2given

The limit inferior of the values. Since $(u_{j})$ is minimising, $I (u_{j}) \to \inf_{A} I$ ; by [F3] therefore ${lim inf}_{j} I (u_{j}) = \inf_{A} I$ .

2.1F1step 1.1

Lower semicontinuity. Applying [F1] to the sequence $(u_{j})$ , which lies in $A$ and converges weakly to $u \in A$ , gives $I (u) \leq {lim inf}_{j} I (u_{j}) = \inf_{A} I$ .

2.2F2step 1.1

The reverse inequality. Since $u \in A$ , the defining property of the infimum [F2] gives $\inf_{A} I \leq I (u)$ .

3.1step 2.1step 2.2given∎

Conclusion. If $\inf_{A} I = - \infty$ , step 2.1 would give $I (u) \leq - \infty$ , impossible because $I$ takes values in $(- \infty, + \infty]$ ; hence under the hypotheses this case cannot occur. Otherwise $\inf_{A} I$ is finite, and steps 2.1 and 2.2 combine to $I (u) = \inf_{A} I$ , so $u$ attains the infimum and is a minimiser.

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

Differentiation of an integral functional under growth domination

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $Ω \subseteq R^{n}$ be a bounded $C^{1}$ domain (Bounded C^k domains and boundary charts), $1 < p < \infty$ , and let $f : Ω \times R \times R^{n} \to R$ be a Caratheodory integrand (A Caratheodory integrand composed with measurable functions is measurable) whose classical partial derivatives $f_{s}$ , $f_{ξ}$ exist and are continuous in $(s, ξ)$ for almost every $x$ , with a constant $C \geq 0$ and functions $G \in L^{1} (Ω)$ , $g \in L^{p^{'}} (Ω)$ , where $p^{'} = p / (p - 1)$ , such that for almost every $x$ and all $(s, ξ)$ $∣ f (x, s, ξ) ∣ \leq C (1 + ∣ s ∣^{p} + ∣ ξ ∣^{p}) + G (x), ∣ f_{s} (x, s, ξ) ∣ + ∣ f_{ξ} (x, s, ξ) ∣ \leq C (1 + ∣ s ∣^{p - 1} + ∣ ξ ∣^{p - 1}) + g (x) .$ Then $I (u) : = \int_{Ω} f (x, u (x), D u (x)) d x$ is well defined and finite on $W^{1, p} (Ω)$ (Integer-order Sobolev spaces and their norms), and for all $u, v \in W^{1, p} (Ω)$ $δ I (u; v) = \int_{Ω} (f_{s} (x, u, D u) v + f_{ξ} (x, u, D u) \cdot D v) d x .$ In particular $I$ is Gateaux differentiable at every $u$ , with bounded Gateaux derivative $δ I (u) \in W^{1, p} (Ω)^{*}$ (Gateaux and Frechet derivatives of a functional).

Facts & Assumptions

Given: Countable Choice; a bounded $C^{1}$ domain $Ω \subseteq R^{n}$ , $1 < p < \infty$ with Holder conjugate $p^{'} = p / (p - 1)$ , and a Caratheodory integrand $f : Ω \times R \times R^{n} \to R$ whose classical partials $f_{s}, f_{ξ}$ exist and are continuous in $(s, ξ)$ for almost every $x$ , with $C \geq 0$ , $G \in L^{1} (Ω)$ , $g \in L^{p^{'}} (Ω)$ and, for almost every $x$ and all $(s, ξ)$ , $∣ f (x, s, ξ) ∣ \leq C (1 + ∣ s ∣^{p} + ∣ ξ ∣^{p}) + G (x), ∣ f_{s} (x, s, ξ) ∣ + ∣ f_{ξ} (x, s, ξ) ∣ \leq C (1 + ∣ s ∣^{p - 1} + ∣ ξ ∣^{p - 1}) + g (x) .$

[F1]

For a Caratheodory integrand and measurable $u, w$ , the composition $x \mapsto f (x, u (x), w (x))$ is measurable (A Caratheodory integrand composed with measurable functions is measurable).

[F2]

The proof of Integer-order Sobolev spaces are Banach uses only Countable Choice after AC supplies it, so the Countable Choice assumed here supplies that same completeness argument and makes $W^{1, p}$ a Banach space. On a bounded domain, the class $u \in W^{1, p} (Ω)$ has $u, D u \in L^{p} (Ω)$ and finite norm $∥ u ∥_{W^{1, p} (Ω)}$ , with $∥ u ∥_{p} \leq ∥ u ∥_{W^{1, p}}$ and $∥ ∣ D u ∣ ∥_{p} \leq n ∥ u ∥_{W^{1, p}}$ , since $∣ D u ∣ \leq \sum_{i} ∣ D_{i} u ∣$ ; the constant $1$ lies in $L^{1} (Ω) \cap L^{p^{'}} (Ω)$ , $∣ G ∣ \in L^{1} (Ω)$ , and $∣ u ∣^{p - 1}, ∣ D u ∣^{p - 1}$ (and the analogous monomials in $v, D v$ ) lie in $L^{p^{'}} (Ω)$ (Integer-order Sobolev spaces and their norms, Bounded C^k domains and boundary charts).

[F3]

Holder's inequality: $\int ∣ h w ∣ \leq ∥ h ∥_{p^{'}} ∥ w ∥_{p}$ for $h \in L^{p^{'}} (Ω)$ and $w \in L^{p} (Ω)$ (Holder's inequality for integrals, including the endpoint cases).

[F4]

Dominated convergence: if measurable $q_{k} \to q$ almost everywhere and $∣ q_{k} ∣ \leq H$ almost everywhere for a single $H \in L^{1} (Ω)$ , then $\int q_{k} \to \int q$ (Dominated convergence).

[F5]

If the classical partial derivatives of $(s, ξ) \mapsto f (x, s, ξ)$ are continuous at a point, then that map is totally differentiable there (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative); the chain rule identifies the total derivative of $t \mapsto f (x, s + t v, ξ + t w)$ as $f_{s} (x, s + t v, ξ + t w) v + f_{ξ} (x, s + t v, ξ + t w) \cdot w$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ); the one-variable mean value theorem applies to $t \mapsto f (x, s + t v, ξ + t w)$ on a compact interval (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[F6]

$I$ is Gateaux differentiable at $u$ with Gâteaux derivative $δ I (u) \in W^{1, p} (Ω)^{*}$ precisely when the limits $δ I (u; v)$ exist for all $v$ and $v \mapsto δ I (u; v)$ is a bounded linear functional (Gateaux and Frechet derivatives of a functional).

Proof

technique · direct, by difference quotients, the mean value theorem, a uniform integrable dominator and dominated convergence

1.1F1F2given

Well-definedness and finiteness. For $u \in W^{1, p} (Ω)$ the map $x \mapsto f (x, u (x), D u (x))$ is measurable by [F1], and the first bound of the hypothesis, together with $G \leq ∣ G ∣$ , gives $∣ f (x, u (x), D u (x)) ∣ \leq C (1 + ∣ u (x) ∣^{p} + ∣ D u (x) ∣^{p}) + ∣ G (x) ∣$ , whose integral is finite by [F2]. Hence $I (u) = \int_{Ω} f (x, u (x), D u (x)) d x$ is a well-defined real number for every $u \in W^{1, p} (Ω)$ .

1.2F5given

Difference quotients and their pointwise limit. Fix $u, v \in W^{1, p} (Ω)$ and put $q_{ε} (x) : = ε^{- 1} (f (x, u (x) + ε v (x), D u (x) + ε D v (x)) - f (x, u (x), D u (x)))$ for $ε \neq 0$ . For almost every $x$ the map $(s, ξ) \mapsto f (x, s, ξ)$ is $C^{1}$ , so [F5] applies to $g_{x} (t) : = f (x, u (x) + t v (x), D u (x) + t D v (x))$ : by the mean value theorem there is $θ = θ (x, ε) \in (0, 1)$ with $q_{ε} (x) = g_{x}^{'} (θ ε)$ , and $g_{x}^{'} (t) = f_{s} (x, u + t v, D u + t D v) v (x) + f_{ξ} (x, u + t v, D u + t D v) \cdot D v (x)$ . As $ε \to 0$ the arguments $(u (x) + θ ε v (x), D u (x) + θ ε D v (x))$ tend to $(u (x), D u (x))$ , so continuity of the partials gives the pointwise limit $q_{ε} (x) \to f_{s} (x, u (x), D u (x)) v (x) + f_{ξ} (x, u (x), D u (x)) \cdot D v (x)$ for almost every $x$ .

2.1F2F3step 1.2

A single integrable dominator. For almost every $x$ and every $0 < ∣ ε ∣ \leq 1$ , the representation of step 1.2 and the second bound of the hypothesis give, with $θ = θ (x, ε)$ and the elementary estimate $(a + b)^{p - 1} \leq 2^{p - 1} (a^{p - 1} + b^{p - 1})$ , $∣ q_{ε} (x) ∣ \leq h (x) (∣ v (x) ∣ + ∣ D v (x) ∣), h : = C^{'} (1 + ∣ u ∣^{p - 1} + ∣ v ∣^{p - 1} + ∣ D u ∣^{p - 1} + ∣ D v ∣^{p - 1}) + ∣ g ∣,$ for a constant $C^{'}$ depending only on $C$ and $p$ . Since $u, D u, v, D v \in L^{p} (Ω)$ , the monomials $∣ u ∣^{p - 1}, \dots, ∣ D v ∣^{p - 1}$ lie in $L^{p^{'}} (Ω)$ , and $∣ g ∣ \in L^{p^{'}} (Ω)$ , while the constant term is integrable on the bounded domain; hence $h \in L^{p^{'}} (Ω)$ , and Hölder's inequality [F3] gives $h (∣ v ∣ + ∣ D v ∣) \in L^{1} (Ω)$ , uniformly in $ε$ .

3.1F4step 2.1

Dominated convergence identifies the limit. Let $ε_{k} \to 0$ , $ε_{k} \neq 0$ , and choose $K$ such that $∣ ε_{k} ∣ \leq 1$ for every $k \geq K$ . By step 1.2 the functions $q_{ε_{k}}$ converge pointwise almost everywhere to $L (x) : = f_{s} (x, u, D u) v + f_{ξ} (x, u, D u) \cdot D v$ , and by step 2.1 the tail $(q_{ε_{k}})_{k \geq K}$ is dominated by the single $L^{1}$ function $h (∣ v ∣ + ∣ D v ∣)$ . Applying [F4] to this tail gives $\int_{Ω} q_{ε_{k}} \to \int_{Ω} L$ ; removing a finite prefix does not change the limit. Since the sequence $ε_{k} \to 0$ was arbitrary, $\lim_{ε \to 0} ε^{- 1} (I (u + ε v) - I (u)) = \int_{Ω} (f_{s} (x, u, D u) v + f_{ξ} (x, u, D u) \cdot D v) d x$ .

4.1F3F6step 3.1∎

Boundedness of the derivative, and conclusion. The map $v \mapsto δ I (u; v) : = \int_{Ω} (f_{s} (x, u, D u) v + f_{ξ} (x, u, D u) \cdot D v) d x$ is linear in $v$ by linearity of the integral, and the pointwise estimate $∣ f_{s} (x, u, D u) v + f_{ξ} (x, u, D u) \cdot D v ∣ \leq h_{u} (x) (∣ v ∣ + ∣ D v ∣)$ with $h_{u} : = C (1 + ∣ u ∣^{p - 1} + ∣ D u ∣^{p - 1}) + ∣ g ∣ \in L^{p^{'}} (Ω)$ gives, by [F3], $∣ δ I (u; v) ∣ \leq (1 + n) ∥ h_{u} ∥_{p^{'}} ∥ v ∥_{W^{1, p} (Ω)}$ ; hence $δ I (u) \in W^{1, p} (Ω)^{*}$ . By [F6] the functional $I$ is Gateaux differentiable at $u$ with derivative $δ I (u)$ and the displayed formula, as claimed.

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

The first variation vanishes at an interior minimiser

Statement

Let $X$ be a real Banach space, $U \subseteq X$ open, $F : U \to R$ Gateaux differentiable at $u \in U$ (Gateaux and Frechet derivatives of a functional) and suppose $u$ is a local minimiser of $F$ : $F (u) \leq F (w)$ for all $w \in U$ with $∥ w - u ∥$ small (Local (relative) maximum and minimum of $f : A \to R$ at a point, the strict forms, and what it means for the point to be interior to $A$ ). Then $δ F (u; v) = 0 for every v \in X .$ More generally, if $V \subseteq X$ is a linear subspace and $F (u) \leq F (w)$ for all $w \in (u + V) \cap U$ with $∥ w - u ∥$ small, then $δ F (u; v) = 0$ for every $v \in V$ .

Facts & Assumptions

Given: A real Banach space $X$ , an open set $U \subseteq X$ , a map $F : U \to R$ that is Gateaux differentiable at $u \in U$ , and the assumption that $u$ is a local minimiser: $F (u) \leq F (w)$ for all $w \in U$ with $∥ w - u ∥$ small. For the general form, a linear subspace $V \subseteq X$ with $F (u) \leq F (w)$ for all $w \in (u + V) \cap U$ with $∥ w - u ∥$ small.

[F1]

Gateaux differentiability of $F$ at $u$ means that $δ F (u; v) = \lim_{ε \to 0, ε \neq 0} ε^{- 1} (F (u + ε v) - F (u))$ exists for every $v \in X$ and that $v \mapsto δ F (u; v)$ is a bounded linear functional; for each fixed $v$ the function $φ (ε) : = F (u + ε v)$ satisfies $φ^{'} (0) = δ F (u; v)$ (Gateaux and Frechet derivatives of a functional).

[F2]

The point $0$ is an interior local minimum of $φ$ when $φ (0) \leq φ (ε)$ for all $ε$ with $∣ ε ∣$ small, the one-variable notion of Local (relative) maximum and minimum of $f : A \to R$ at a point, the strict forms, and what it means for the point to be interior to $A$ (Local (relative) maximum and minimum of $f : A \to R$ at a point, the strict forms, and what it means for the point to be interior to $A$ ).

[F3]

If a function on a real interval is differentiable at an interior local extremum, then its derivative vanishes there (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 $f^{'} (c) = 0$ ).

Proof

technique · direct, reducing to the one-variable Fermat theorem along each admissible line

1.1F1F2given

Reduction to one variable. Fix $v \in X$ . Since $U$ is open and $u \in U$ , there is $ε_{0} > 0$ with $u + ε v \in U$ for every $∣ ε ∣ < ε_{0}$ ; define $φ (ε) : = F (u + ε v)$ for those $ε$ . By [F1] $φ^{'} (0)$ exists and equals $δ F (u; v)$ . The local minimality of $u$ gives $ε_{1} \in (0, ε_{0}]$ with $F (u) \leq F (u + ε v)$ , that is $φ (0) \leq φ (ε)$ , whenever $∣ ε ∣ < ε_{1}$ ; in the terminology of [F2], $0$ is an interior local minimum of $φ$ .

2.1F3step 1.1

Fermat's theorem applied to $φ$ . The function $φ$ is differentiable at its interior point $0$ and has a local minimum there, so by [F3] $φ^{'} (0) = 0$ ; by step 1.1 this reads $δ F (u; v) = 0$ .

3.1F1step 2.1∎

The general admissible-affine form. Let $V \subseteq X$ be a linear subspace and suppose $F (u) \leq F (w)$ for all $w \in (u + V) \cap U$ with $∥ w - u ∥$ small. Fix $v \in V$ . For $∣ ε ∣$ small the point $u + ε v$ belongs to $(u + V) \cap U$ , because $V$ is a linear subspace and $U$ is open; the argument of steps 1.1 and 2.1 therefore applies verbatim to this $v$ and yields $δ F (u; v) = 0$ . As $v \in V$ was arbitrary, the first variation vanishes on the whole subspace $V$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

The affine Dirichlet trace class is nonempty, convex and weakly closed

Statement

Assume the Axiom of Choice. Let $n \geq 2$ and let $Ω \subseteq R^{n}$ be a bounded $C^{1}$ domain, $1 < p < \infty$ , and let $g \in W^{1 - 1 / p, p} (\partial Ω)$ lie in the trace range of $T : W^{1, p} (Ω) \to W^{1 - 1 / p, p} (\partial Ω)$ (The sharp trace theorem: boundedness and range in the fractional space, The $L^{p}$ trace operator on a bounded $C^{1}$ domain). Then the affine trace class $K_{g} : = {v \in W^{1, p} (Ω) : T v = g}$ is nonempty, convex, norm closed in $W^{1, p} (Ω)$ and weakly closed; moreover, for every right inverse $R$ of $T$ with $T R g = g$ , $K_{g} = R g + W_{0}^{1, p} (Ω)$ (A bounded right inverse of the trace, supported in a prescribed collar, The kernel of the trace is the closure of the test functions, Zero-boundary Sobolev space as a norm closure).

Facts & Assumptions

Given: The Axiom of Choice; $n \geq 2$ ; a bounded $C^{1}$ domain $Ω \subseteq R^{n}$ , $1 < p < \infty$ , and $g \in W^{1 - 1 / p, p} (\partial Ω)$ lying in the range of the trace operator $T : W^{1, p} (Ω) \to W^{1 - 1 / p, p} (\partial Ω)$ of The $L^{p}$ trace operator on a bounded $C^{1}$ domain; the affine class $K_{g} : = {v \in W^{1, p} (Ω) : T v = g}$ .

[F1]

For the stated $n \geq 2$ and $1 < p < \infty$ , the trace operator $T : W^{1, p} (Ω) \to W^{θ, p} (\partial Ω)$ , $θ = 1 - 1 / p$ , is bounded and surjective onto the fractional Sobolev space (The sharp trace theorem: boundedness and range in the fractional space, The $L^{p}$ trace operator on a bounded $C^{1}$ domain).

[F2]

There is a bounded right inverse $R : W^{θ, p} (\partial Ω) \to W^{1, p} (Ω)$ with $T \circ R = i d$ ; it is not unique (A bounded right inverse of the trace, supported in a prescribed collar).

[F3]

$\ker T = W_{0}^{1, p} (Ω)$ , the $W^{1, p}$ -closure of $C_{c}^{\infty} (Ω)$ (The kernel of the trace is the closure of the test functions, Zero-boundary Sobolev space as a norm closure); in particular $W_{0}^{1, p} (Ω)$ is a linear subspace of $W^{1, p} (Ω)$ , which is a normed space for $∥ \cdot ∥_{W^{1, p} (Ω)}$ (Integer-order Sobolev spaces and their norms).

[F4]

Under the Axiom of Choice, every convex subset of a real or complex normed space that is closed in the norm topology is weakly closed (A norm-closed convex set is weakly sequentially closed).

Proof

technique · direct, by writing $K_g$ as a translate of the kernel of the trace and applying the closed-convex weak-closure lemma

1.1F1F2given

Nonemptiness. Since $g$ lies in the range of $T$ there is $u \in W^{1, p} (Ω)$ with $T u = g$ , so $K_{g} \neq \emptyset$ ; alternatively [F2] gives $R g \in K_{g}$ because $T R g = g$ .

1.2F2F3given

The class is a translate of the kernel. Fix a right inverse $R$ of $T$ , which exists by [F2] and satisfies $T R g = g$ . For $u \in W^{1, p} (Ω)$ one has $u \in K_{g} ⟺ T u = g ⟺ T (u - R g) = 0 ⟺ u - R g \in \ker T = W_{0}^{1, p} (Ω)$ , using linearity of $T$ and [F3]. Hence $K_{g} = R g + W_{0}^{1, p} (Ω)$ .

2.1F3step 1.2

Convexity. Let $u, v \in K_{g}$ and $λ \in [0, 1]$ . By step 1.2 the elements $u - R g$ and $v - R g$ lie in the linear subspace $W_{0}^{1, p} (Ω)$ , so $λ u + (1 - λ) v - R g = λ (u - R g) + (1 - λ) (v - R g) \in W_{0}^{1, p} (Ω)$ and therefore $λ u + (1 - λ) v \in K_{g}$ . Thus $K_{g}$ is convex.

2.2F1F3step 1.2

Norm closedness. The operator $T$ is bounded by [F1], hence continuous, and $K_{g} = T^{- 1} ({g})$ is the preimage of the singleton ${g}$ , which is closed in the normed space $W^{1 - 1 / p, p} (\partial Ω)$ ; a continuous preimage of a closed set is closed. So $K_{g}$ is closed in the norm topology of $W^{1, p} (Ω)$ .

3.1F4step 2.1step 2.2

Weak closedness. By steps 2.1 and 2.2 the set $K_{g}$ is convex and closed in the norm topology, so [F4] applies under the Axiom of Choice and $K_{g}$ is weakly closed.

4.1F2F3step 1.2step 3.1∎

The identity for an arbitrary right inverse. Let $R$ be any bounded right inverse of $T$ , so that $T R g = g$ . The argument of step 1.2 used only this identity and the kernel description [F3], so it gives $K_{g} = R g + W_{0}^{1, p} (Ω)$ for this $R$ as well. This, together with steps 1.1, 2.1 and 3.1, establishes every clause of the statement.

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

The boundary fundamental lemma of the calculus of variations

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $Ω \subseteq R^{n}$ , $n \geq 2$ , be a bounded $C^{1}$ domain with surface measure $σ$ on $\partial Ω$ (Surface integration on compact C1 hypersurfaces, Bounded C1 domains and their outward normals). If $g \in C (\overline{Ω})$ satisfies $\int_{\partial Ω} g φ d σ = 0 for every φ \in C^{\infty} (\overline{Ω}),$ then $g = 0$ on $\partial Ω$ . Equivalently, if $h \in C (\overline{Ω}; R^{n})$ satisfies $\int_{\partial Ω} (h \cdot ν) φ d σ = 0$ for every $φ \in C^{\infty} (\overline{Ω})$ , then $h \cdot ν = 0$ on $\partial Ω$ , where $ν$ is the outward unit normal.

Facts & Assumptions

Given: The Axiom of Choice; a bounded $C^{1}$ domain $Ω \subseteq R^{n}$ with surface measure $σ$ on $\partial Ω$ ; a function $g \in C (\overline{Ω})$ with $\int_{\partial Ω} g φ d σ = 0$ for every $φ \in C^{\infty} (\overline{Ω})$ . For the equivalent formulation, $h \in C (\overline{Ω}; R^{n})$ with $\int_{\partial Ω} (h \cdot ν) φ d σ = 0$ for every $φ \in C^{\infty} (\overline{Ω})$ .

[F0]

Under the Axiom of Choice, the Axiom of Countable Choice holds (AC supplies the countable and dependent choices used in Banach integration), which is the measure convention under which the boundary charts, the ambient partitions and the surface integral are set up.

[F1]

A bounded $C^{1}$ domain is locally a graph: near each boundary point, after a rigid change of coordinates, $\partial Ω$ is ${z = (y, t) : t = h (y)}$ for a $C^{1}$ function $h$ on a ball, $Ω$ is locally the subgraph, and the outward normal is $ν = (- D h, 1) / \sqrt{1 + ∣ D h ∣^{2}}$ ; the surface integral over a compact face contained in a regular patch is computed by the chart $X (y) = (y, h (y))$ with Gram factor $J (y) = \sqrt{1 + ∣ D h (y) ∣^{2}} \geq 1$ (Bounded C1 domains and their outward normals, Surface integration on compact C1 hypersurfaces), the definition being assembled from finitely many charts with an ambient smooth partition of unity (Finite ambient partitions near compact sets).

[F2]

For every $0 < r < R$ and every centre $a$ there is a smooth bump equal to one on ${\overline{B}}_{r} (a)$ and supported strictly inside $B_{R} (a)$ (Compactly supported scaled Euclidean bumps).

[F3]

If $G \in L_{l o c}^{1} (U)$ on an open set $U \subseteq R^{m}$ satisfies $\int_{U} G ψ = 0$ for every $ψ \in C_{c}^{\infty} (U)$ , then $G = 0$ almost everywhere on $U$ (The fundamental lemma of the calculus of variations).

Proof

technique · direct, by flattening the boundary at an arbitrary boundary point and applying the fundamental lemma to the charted integrand

1.1F0F1given

Local chart at a boundary point. Fix $x_{0} \in \partial Ω$ . By [F1] we may, after translating and applying a rigid motion, assume $x_{0} = 0$ and find $ρ > 0$ , $h \in C^{1} (B (0, ρ))$ and a neighbourhood $U ∋ 0$ with the boundary in $U$ is the graph of $h$ and the domain in $U$ is its subgraph, with both sets intersected with $U$ ; write $X (y) : = (y, h (y))$ and $J : = \sqrt{1 + ∣ D h ∣^{2}} \geq 1$ . Any function on $\partial Ω$ whose support lies in this patch has surface integral equal to the chart integral against $J$ , by [F1].

2.1F2step 1.1

A cutoff and suitable test functions. Choose $0 < r < R$ with $B (0, R) \subseteq U$ and let $η$ be the smooth bump of [F2] with $η = 1$ on ${\overline{B}}_{r} (0)$ and $supp η \subseteq B_{R} (0)$ . Choose $δ > 0$ with $δ < ρ$ and $∣ (y, h (y)) ∣ < r$ for $∣ y ∣ < δ$ . For every $ψ \in C_{c}^{\infty} ({∣ y ∣ < δ})$ define $φ (z) : = ψ (z^{'}) η (z)$ , where $z^{'} = (z_{1}, \dots, z_{n - 1})$ ; then $φ \in C_{c}^{\infty} (R^{n})$ , hence $φ \in C^{\infty} (\overline{Ω})$ , and for $∣ y ∣ < δ$ one has $φ (X (y)) = ψ (y) η (X (y)) = ψ (y)$ because $X (y) \in {\overline{B}}_{r} (0)$ there.

3.1F1step 1.1step 2.1

The local integral identity. The hypothesis gives $\int_{\partial Ω} g φ d σ = 0$ for the test function $φ$ of step 2.1, whose boundary support lies in the patch of step 1.1; the chart formula therefore yields $0 = \int_{B (0, ρ)} g (X (y)) φ (X (y)) J (y) d y = \int_{B (0, ρ)} G (y) ψ (y) d y$ , where $G : = g \circ X \cdot J$ is continuous because $g$ is continuous on $\overline{Ω}$ and $h$ is $C^{1}$ . As $ψ \in C_{c}^{\infty} ({∣ y ∣ < δ})$ was arbitrary, $G \in L_{l o c}^{1}$ satisfies $\int G ψ = 0$ for every test function supported in that ball.

4.1F3step 3.1

The fundamental lemma at $x_{0}$ . Applying [F3] to $G$ on the ball ${∣ y ∣ < δ}$ gives $G = 0$ almost everywhere; since $J \geq 1$ , this implies $g \circ X = 0$ almost everywhere, and since $y \mapsto g (X (y))$ is continuous, $g (X (y)) = 0$ for every $∣ y ∣ < δ$ . In particular $g (x_{0}) = g (X (0)) = 0$ .

5.1step 4.1

Conclusion on the boundary. The point $x_{0} \in \partial Ω$ was arbitrary, so $g = 0$ on $\partial Ω$ .

6.1F1step 5.1∎

The vector-valued formulation. Let $h \in C (\overline{Ω}; R^{n})$ satisfy $\int_{\partial Ω} (h \cdot ν) φ d σ = 0$ for every $φ \in C^{\infty} (\overline{Ω})$ . The boundary function $γ : = (h \cdot ν) ∣_{\partial Ω}$ is continuous, because $h$ is continuous on $\overline{Ω}$ and the normal field $ν$ is continuous on the $C^{1}$ boundary [F1]; the argument of steps 1.1–5.1 uses only the boundary values of the continuous integrand and the linearity of the integral in it, so it applies with $g$ replaced by $γ$ and gives $γ = 0$ on $\partial Ω$ , that is $h \cdot ν = 0$ on $\partial Ω$ .

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

A convex norm-lower-semicontinuous functional is weakly lower semicontinuous

Statement

Assume HB (The real dominated-extension principle as an additional hypothesis over ZF) and Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $X$ be a real normed space, let $K \subseteq X$ be nonempty and convex (Convex and strictly convex functionals on a convex subset of a real vector space) and let $I : K \to (- \infty, + \infty]$ be convex and sequentially lower semicontinuous in the norm topology (Proper, coercive and weakly lower semicontinuous extended-real functionals). Then $I$ is weakly sequentially lower semicontinuous on $K$ : for every $(u_{j}) \subseteq K$ with $u_{j} ⇀ u \in K$ (Weak convergence of nets and sequences), $I (u) \leq {lim inf}_{j} I (u_{j}) .$

Facts & Assumptions

Given: HB and Countable Choice; a real normed space $X$ , a nonempty convex set $K \subseteq X$ , and a convex functional $I : K \to (- \infty, + \infty]$ that is sequentially lower semicontinuous in the norm topology.

[F1]

A convex functional has convex sublevel sets: for every $t \in R$ the set ${v \in K : I (v) \leq t}$ is convex (Convex and strictly convex functionals on a convex subset of a real vector space).

[F2]

Norm sequential lower semicontinuity means $I (v) \leq {lim inf}_{j} I (v_{j})$ whenever $v_{j} \to v \in K$ in norm with $v_{j} \in K$ (Proper, coercive and weakly lower semicontinuous extended-real functionals). It gives closedness of sublevels relative to $K$ , not necessarily in $X$ .

[F3]

Under HB the norm and weak closures in $X$ of any convex subset coincide (Norm closed convex iff weakly closed).

[F6]

Countable Choice selects a point from each nonempty set $S \cap B (u, 1 / m)$ , $m \geq 1$ , whenever $u$ lies in the norm closure of $S$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F4]

Weak convergence $u_{j} ⇀ u$ is convergence in $σ (X, X^{*})$ ; in particular every subsequence of a weakly convergent sequence converges weakly to the same limit (Weak convergence of nets and sequences).

[F5]

Limit inferior: if ${lim inf}_{j} a_{j} < t$ for a sequence in $(- \infty, + \infty]$ and a real $t$ , then $a_{j} \leq t$ for infinitely many $j$ , so a strictly increasing sequence of indices $j_{k}$ with $a_{j_{k}} \leq t$ for all $k$ exists (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ ).

Proof

technique · direct, by comparing the norm and weak closures of a convex sublevel set

1.1F5givenalgebra

Suppose $u_{j} ⇀ u$ with $(u_{j}) \subseteq K$ and $u \in K$ , and put $ℓ : = {lim inf}_{j} I (u_{j})$ . If $I (u) > ℓ$ , then since $I (u) \in (- \infty, + \infty]$ there is a real $t$ with $ℓ < t < I (u)$ (if $I (u)$ is finite take $t$ between; if $I (u) = + \infty$ take any real $t > ℓ$ ).

2.1F4F5step 1.1

A subsequence in the sublevel set. By [F5], applied to the sequence $(I (u_{j}))$ and this $t$ , there is a strictly increasing sequence of indices $j_{k}$ with $I (u_{j_{k}}) \leq t$ for every $k$ . By [F4] the subsequence still satisfies $u_{j_{k}} ⇀ u$ .

3.1F1F3step 2.1

Use the ambient closures. Put $S_{t} : = {v \in K : I (v) \leq t}$ . It is nonempty by step 2.1 and convex by [F1]. Since $u_{j_{k}} \in S_{t}$ and $u_{j_{k}} ⇀ u$ , the point $u$ lies in the weak closure of $S_{t}$ in $X$ . By [F3] it therefore lies in its norm closure. No ambient closedness of $K$ or $S_{t}$ is required.

4.1F2F6step 3.1given

Recover the relative sublevel inequality. For each integer $m \geq 1$ , choose $v_{m} \in S_{t}$ with $∥ v_{m} - u ∥ < 1 / m$ , using [F6]. Then $v_{m} \to u$ in norm, $v_{m} \in K$ and $u \in K$ . Thus [F2] gives $I (u) \leq {lim inf}_{m} I (v_{m}) \leq t$ .

5.1step 4.1step 1.1∎

Conclusion. Step 4.1 gives $I (u) \leq t < I (u)$ by the choice of $t$ in step 1.1, a contradiction; hence $I (u) \leq ℓ = {lim inf}_{j} I (u_{j})$ . As $(u_{j})$ and $u$ were arbitrary, $I$ is weakly sequentially lower semicontinuous on $K$ .

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

Strict convexity gives uniqueness of a minimiser

Statement

Let $K$ be a convex subset of a real vector space and let $I : K \to (- \infty, + \infty]$ be proper and strictly convex (Convex and strictly convex functionals on a convex subset of a real vector space, Proper, coercive and weakly lower semicontinuous extended-real functionals). If $u, v \in K$ both minimise $I$ on $K$ , then $u = v$ .

Facts & Assumptions

Given: A convex subset $K$ of a real vector space and a proper, strictly convex extended-real functional $I : K \to (- \infty, + \infty]$ (Convex and strictly convex functionals on a convex subset of a real vector space, Proper, coercive and weakly lower semicontinuous extended-real functionals); points $u, v \in K$ that both minimise $I$ on $K$ , in the sense that $I (u) = I (v) = \inf_{K} I$ .

[F1]

Strict convexity: for $u \neq v$ with $I (u), I (v) < + \infty$ and $λ \in (0, 1)$ one has $I (λ u + (1 - λ) v) < λ I (u) + (1 - λ) I (v)$ ; convexity gives $λ u + (1 - λ) v \in K$ (Convex and strictly convex functionals on a convex subset of a real vector space).

[F2]

The infimum is a lower bound: $\inf_{K} I \leq I (w)$ for every $w \in K$ (Greatest lower bound (infimum)).

Proof

technique · direct, by evaluating strict convexity at the midpoint of two minimisers

1.1F1F2given

Set-up. Let $u, v \in K$ both minimise $I$ and suppose for contradiction that $u \neq v$ . Properness gives $\inf_{K} I < + \infty$ , so $t : = \inf_{K} I = I (u) = I (v)$ is finite.

2.1F1step 1.1algebra

Strict convexity at the midpoint. The midpoint $w : = \frac{1}{2} u + \frac{1}{2} v$ lies in the convex set $K$ , and strict convexity with $λ = \frac{1}{2}$ applies because $u \neq v$ and $I (u) = I (v) = t < + \infty$ : hence $I (w) < \frac{1}{2} I (u) + \frac{1}{2} I (v) = t$ .

3.1F2step 2.1∎

Contradiction. Step 2.1 gives $I (w) < t = \inf_{K} I$ , while [F2] gives $\inf_{K} I \leq I (w)$ since $w \in K$ . This is impossible, so $u = v$ ; two distinct minimisers cannot exist.

Remarks

Properness is necessary. Without it the statement is false: on $K = [0, 1] \subseteq R$ the functional $I \equiv + \infty$ is convex and vacuously strictly convex, and $0$ and $1$ are two distinct points at which $I$ equals $\inf_{K} I = + \infty$ . Properness, equivalently the existence of a finite competitor, is what excludes this degenerate case, and it holds in the finite-valued integral-functional applications (Proper, coercive and weakly lower semicontinuous extended-real functionals).

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

The weak Euler-Lagrange equation for integral functionals with fixed trace

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $Ω \subseteq R^{n}$ , $n \geq 2$ , be a bounded $C^{1}$ domain, $1 < p < \infty$ , let $f$ and $I$ satisfy the hypotheses of Differentiation of an integral functional under growth domination, and let $g \in W^{1 - 1 / p, p} (\partial Ω)$ lie in the trace range of $T : W^{1, p} (Ω) \to W^{1 - 1 / p, p} (\partial Ω)$ (The sharp trace theorem: boundedness and range in the fractional space, The fractional Sobolev space on a compact $C^{1}$ boundary, The $L^{p}$ trace operator on a bounded $C^{1}$ domain). Let $u \in W^{1, p} (Ω)$ with $T u = g$ be a local minimiser of $I$ among the functions with trace $g$ : $I (u) \leq I (w)$ for all $w \in W^{1, p} (Ω)$ with $T w = g$ and $∥ w - u ∥_{W^{1, p}}$ small. Then $\int_{Ω} (f_{ξ} (x, u, D u) \cdot D φ + f_{s} (x, u, D u) φ) d x = 0$ for every $φ \in W_{0}^{1, p} (Ω)$ (Zero-boundary Sobolev space as a norm closure); equivalently, for every $φ \in C_{c}^{\infty} (Ω)$ (Test function space d of an open set).

Facts & Assumptions

Given: The Axiom of Choice; a bounded $C^{1}$ domain $Ω \subseteq R^{n}$ , $1 < p < \infty$ , an integrand $f$ and functional $I (u) = \int_{Ω} f (x, u, D u) d x$ satisfying the hypotheses of Differentiation of an integral functional under growth domination, and $g \in W^{1 - 1 / p, p} (\partial Ω)$ in the trace range of $T$ ; a local minimiser $u \in W^{1, p} (Ω)$ of $I$ among the functions of trace $g$ . The Sobolev and trace framework is set up under the Axiom of Choice, used through Countable Choice (The Axiom of Choice), and $W^{1, p} (Ω)$ is a Banach space (Integer-order Sobolev spaces are Banach).

[F1]

$T : W^{1, p} (Ω) \to W^{1 - 1 / p, p} (\partial Ω)$ is linear and $\ker T = W_{0}^{1, p} (Ω)$ , the $W^{1, p}$ -closure of $C_{c}^{\infty} (Ω)$ (The kernel of the trace is the closure of the test functions, Zero-boundary Sobolev space as a norm closure, The $L^{p}$ trace operator on a bounded $C^{1}$ domain).

[F2]

Affine form of the first-variation theorem: if $F : U \to R$ on the open $U \subseteq X$ is Gateaux differentiable at $u$ and $F (u) \leq F (w)$ for all $w \in (u + V) \cap U$ with $∥ w - u ∥$ small, $V \subseteq X$ a linear subspace of the Banach space $X$ , then $δ F (u; v) = 0$ for every $v \in V$ (The first variation vanishes at an interior minimiser, Integer-order Sobolev spaces are Banach).

[F3]

The differentiation lemma: $I$ is Gateaux differentiable at $u$ with $δ I (u; v) = \int_{Ω} (f_{s} (x, u, D u) v + f_{ξ} (x, u, D u) \cdot D v) d x$ for every $v \in W^{1, p} (Ω)$ (Differentiation of an integral functional under growth domination).

[F4]

$C_{c}^{\infty} (Ω) \subseteq W_{0}^{1, p} (Ω)$ because $W_{0}^{1, p} (Ω)$ is defined as the closure of $C_{c}^{\infty} (Ω)$ in $W^{1, p} (Ω)$ (Zero-boundary Sobolev space as a norm closure, Test function space d of an open set).

Proof

technique · direct, by testing the affine first-variation theorem against the kernel of the trace

1.1F1given

Variations preserving the trace. Fix $φ \in W_{0}^{1, p} (Ω)$ . Then $T φ = 0$ by [F1], and linearity of $T$ gives $T (u + ε φ) = T u + ε T φ = g$ for every $ε \in R$ ; thus every point of the affine line $u + R φ$ has trace $g$ .

2.1givenstep 1.1

Local minimality along the line. For $ε$ with $∣ ε ∣$ small, the point $u + ε φ$ lies in the local admissible neighbourhood of $u$ among the functions of trace $g$ and has norm distance $∣ ε ∣ ∥ φ ∥_{W^{1, p}}$ from $u$ ; hence $I (u) \leq I (u + ε φ)$ . Therefore $u$ is a local minimiser of $I$ on the affine set $(u + W_{0}^{1, p} (Ω)) \cap W^{1, p} (Ω)$ .

3.1F2step 2.1

The first variation vanishes. By [F2] applied with $X = W^{1, p} (Ω)$ , $V = W_{0}^{1, p} (Ω)$ and the local minimality of step 2.1, $δ I (u; φ) = 0$ .

4.1F1F3F4step 3.1∎

Computing the derivative. By [F3] the Gateaux derivative is $δ I (u; φ) = \int_{Ω} (f_{s} (x, u, D u) φ + f_{ξ} (x, u, D u) \cdot D φ) d x$ ; together with step 3.1 this gives the displayed identity for the arbitrary element $φ \in W_{0}^{1, p} (Ω)$ . Finally, if $φ \in C_{c}^{\infty} (Ω)$ then $φ \in W_{0}^{1, p} (Ω)$ by [F4], so the identity holds in particular for every such test function. Conversely, the derivative in [F3] is bounded on $W^{1, p}$ , and every $φ \in W_{0}^{1, p}$ is a norm limit of compactly supported smooth functions by [F1]; continuity passes the identity from those tests to $φ$ .

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

The direct method in a reflexive Banach space

Statement

Assume the ultrafilter lemma, DC and HB. 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 $I : A \to (- \infty, + \infty]$ 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 exists $u_{0} \in A$ with $I (u_{0}) = \inf_{A} I$ . The admissible set may be taken convex and norm closed in $X$ , by Norm closed convex iff weakly closed.

Facts & Assumptions

Given: The ultrafilter lemma (The ultrafilter extension principle (UL/BPI)), the principle of dependent choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain) and HB (The real dominated-extension principle as an additional hypothesis over ZF); a real reflexive Banach space $X$ (Reflexivity is surjectivity of the canonical map); a nonempty set $A \subseteq X$ that is weakly sequentially closed; and a proper extended-real functional $I : A \to (- \infty, + \infty]$ , coercive on $A$ and weakly sequentially lower semicontinuous on $A$ (Proper, coercive and weakly lower semicontinuous extended-real functionals). Write $α : = \inf_{A} I$ in the complete extended order specified by Proper, coercive and weakly lower semicontinuous extended-real functionals.

[F1]

Properness gives $α < + \infty$ ; moreover for every real $ε > 0$ there is $u \in A$ with $I (u) < α + ε$ when $α \in R$ , and when $α = - \infty$ there is for every $M \in R$ a point $u \in A$ with $I (u) \leq M$ (Proper, coercive and weakly lower semicontinuous extended-real functionals, Greatest lower bound (infimum)).

[F2]

Coercivity bounds finite-level sequences: every sequence $(u_{j}) \subseteq A$ with $\sup_{j} I (u_{j}) < + \infty$ is norm bounded, and in particular every minimising sequence with $I (u_{j}) \to \inf_{A} I < + \infty$ is norm bounded (Coercivity bounds every finite-level sequence).

[F3]

Under the ultrafilter lemma, DC and HB, every norm-bounded sequence in a real reflexive Banach space has a subsequence converging weakly to a point of $X$ (A bounded sequence in a reflexive Banach space has a weakly convergent subsequence).

[F4]

If $A$ is weakly sequentially closed, $(u_{j}) \subseteq A$ and $u_{j} ⇀ u$ , then $u \in A$ (Weak closedness keeps the direct-method limit admissible).

[F5]

If $(u_{j}) \subseteq A$ is minimising, $u_{j} ⇀ u \in A$ and $I$ is weakly sequentially lower semicontinuous at $u$ , then $I (u) = \inf_{A} I$ (The liminf passage makes the weak limit a minimiser).

[F6]

Dependent choice implies countable choice, so a countable sequence of independent nonempty selections can be made along $N$ (Dependent choice implies countable choice).

[F7]

Under HB, which is assumed here, a convex subset of a real or complex normed space is norm closed if and only if it is weakly closed (Norm closed convex iff weakly closed). A weakly closed set is weakly sequentially closed, so an admissible set that is convex and norm closed satisfies the theorem hypothesis under the stated choice principles.

[F8]

Under HB and Countable Choice (supplied here by DC), in the convex case the weak-lower-semicontinuity hypothesis of the theorem is verified by convexity plus norm lower semicontinuity: a convex norm-lower-semicontinuous functional on a convex set is weakly sequentially lower semicontinuous (A convex norm-lower-semicontinuous functional is weakly lower semicontinuous). This is the role of that lemma for the present theorem and for its convex applications.

Proof

technique · direct, by selecting a minimising sequence, extracting a weakly convergent subsequence and passing to the limit

1.1F1F6given

A minimising sequence. If $α \in R$ , [F1] supplies for each $j \in N$ a point $u_{j} \in A$ with $I (u_{j}) < α + 2^{- j}$ ; if $α = - \infty$ , [F1] supplies $u_{j} \in A$ with $I (u_{j}) \leq - j$ . In both cases $(u_{j}) \subseteq A$ satisfies $I (u_{j}) \to α$ , so it is a minimising sequence. The countably many selections are licensed by [F6].

2.1F2step 1.1

Boundedness. In the finite case $I (u_{j}) < α + 1 < + \infty$ for all $j$ ; in the case $α = - \infty$ one has $I (u_{j}) \leq 0$ for all $j$ . Hence $\sup_{j} I (u_{j}) < + \infty$ , and [F2] makes $(u_{j})$ norm bounded.

3.1F3F4step 2.1

A weakly convergent subsequence with admissible limit. By [F3] there are a strictly increasing sequence $j_{k}$ and a point $u_{0} \in X$ with $u_{j_{k}} ⇀ u_{0}$ ; the subsequence lies in $A$ , which is weakly sequentially closed, so [F4] gives $u_{0} \in A$ .

4.1F5F7F8step 3.1∎

The limit is a minimiser, and the convex special case. The subsequence $(u_{j_{k}})$ is still minimising, $I (u_{j_{k}}) \to α$ , and $u_{j_{k}} ⇀ u_{0} \in A$ , so the weak lower semicontinuity hypothesis and [F5] give $I (u_{0}) = \inf_{A} I$ : the infimum is attained on $A$ . If in addition $A$ is convex and closed in the norm topology, the HB-form of [F7] and the weak-closed-to-weakly-sequentially-closed passage make $A$ weakly sequentially closed, so the theorem applies to that admissible set; and in the convex case the weak lower semicontinuity hypothesis itself is supplied by [F8] whenever $I$ is convex and norm lower semicontinuous. No stronger choice principle than the HB assumed here is needed for these clauses.

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

Stationarity is sufficient for a global minimum of a convex differentiable functional

Statement

Let $K$ be a convex subset of a real Banach space and let $I : K \to (- \infty, + \infty]$ be convex (Convex and strictly convex functionals on a convex subset of a real vector space). Fix $u \in K$ with $I (u) < + \infty$ , and assume that for every $v \in K$ the finite one-sided admissible directional derivative $δ_{+} I (u; v - u) : = \lim_{t ↓ 0} \frac{I (u + t (v - u)) - I (u)}{t}$ exists and is nonnegative. Then $I (u) = \inf_{K} I$ . If $I$ is strictly convex, $u$ is the unique minimiser. In particular, for a real-valued Gateaux differentiable functional on an open neighbourhood of $K$ , the condition $δ I (u; v - u) \geq 0$ for every $v \in K$ suffices, since the one-sided derivative agrees with that of Gateaux and Frechet derivatives of a functional.

Facts & Assumptions

Given: A convex $K$ in a real Banach space; a convex extended-real functional $I$ ; a finite competitor $u \in K$ ; and finite nonnegative one-sided derivatives $δ_{+} I (u; v - u)$ for every $v \in K$ . Only $0 < t < 1$ is used, so the segment is admissible even when $u$ lies on the boundary of $K$ .

[F1]

Convexity of $I$ : for $w, z \in K$ and $λ \in [0, 1]$ one has $I (λ w + (1 - λ) z) \leq λ I (w) + (1 - λ) I (z)$ , with the extended-real conventions; in particular the segment ${u + ε (v - u) : ε \in [0, 1]}$ lies in $K$ for $v \in K$ (Convex and strictly convex functionals on a convex subset of a real vector space).

[F2]

Three-slope inequality: if $φ : I_{0} \to R$ is convex on an interval and $x < y < z$ lie in $I_{0}$ , then the secant slopes satisfy $s (x, y) \leq s (x, z) \leq s (y, z)$ , where $s (a, b) = (f (b) - f (a)) / (b - a)$ (For a convex function and $x < y < z$ , the three secant slopes satisfy $s (x, y) \leq s (x, z) \leq s (y, z)$ ). Equivalently, the supporting-line form of convexity applies at every interior point with a slope between the one-sided derivatives (Every slope between the left and right derivatives of a convex function gives a supporting line).

[F3]

The admissible derivative is the limit of the secant slopes $(φ_{w} (t) - φ_{w} (0)) / t$ as $t ↓ 0$ , where $φ_{w} (t) : = I (u + t w)$ . When an ordinary Gateaux derivative exists on an open neighbourhood, this is its one-sided restriction (Gateaux and Frechet derivatives of a functional).

[F4]

A proper, strictly convex functional has at most one minimiser on a convex set (Strict convexity gives uniqueness of a minimiser); in step 4.1 the functional is proper because $I (u)$ is finite.

Proof

technique · direct, by monotonicity of the secant slopes along the admissible segment

1.1F1F3given

Reduction to a segment. Fix $v \in K$ . If $I (v) = + \infty$ then $I (u) \leq I (v)$ is automatic because $I (u)$ is finite by hypothesis; so assume $I (v) < + \infty$ . Define $φ (ε) : = I (u + ε (v - u))$ for $ε \in [0, 1]$ . By [F1] the segment lies in $K$ and $φ (ε) \leq (1 - ε) I (u) + ε I (v) < + \infty$ , while $φ (ε) > - \infty$ by the codomain of $I$ ; hence $φ : [0, 1] \to R$ is a finite convex function.

1.2F3given

The one-sided derivative. By the differentiability hypothesis the secant slope $s (0, ε) = ε^{- 1} (φ (ε) - φ (0))$ has the finite limit $δ_{+} I (u; v - u) \geq 0$ as $ε ↓ 0$ .

2.1F2step 1.1

Secant comparison. By [F2], applied on the interval $[0, 1]$ to the convex function $φ$ and the points $0 < ε < 1$ , one has $s (0, ε) \leq s (0, 1)$ .

3.1step 1.2step 2.1algebra

Passing to the limit. Letting $ε ↓ 0$ in the inequality of step 2.1 gives $δ_{+} I (u; v - u) \leq s (0, 1) = φ (1) - φ (0) = I (v) - I (u)$ ; since $δ_{+} I (u; v - u) \geq 0$ by hypothesis, it follows that $I (v) \geq I (u)$ .

4.1F4step 3.1∎

Conclusion and uniqueness. As $v \in K$ was arbitrary, $I (u) \leq I (v)$ for every $v \in K$ , so $I (u)$ is a lower bound for $I$ on $K$ ; since $u \in K$ , it is the greatest lower bound, $I (u) = \inf_{K} I$ . If $I$ is moreover strictly convex and $v \in K$ is any other minimiser, then both $u$ and $v$ are finite minimisers and [F4] gives $u = v$ , so $u$ is the unique minimiser.

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

The classical Euler-Lagrange equation under regularity

Statement

Let the hypotheses of The weak Euler-Lagrange equation for integral functionals with fixed trace hold, and assume in addition that $f \in C^{2} (\overline{Ω} \times R \times R^{n})$ and $u \in C^{2} (\overline{Ω})$ ( $C^{k}$ maps and multi-index derivative notation in Euclidean space). Then $w (x) : = f_{ξ} (x, u (x), D u (x)) \in C^{1} (Ω; R^{n})$ and $- div w + f_{s} (x, u (x), D u (x)) = 0 (x \in Ω),$ the classical Euler-Lagrange equation (Divergence and curl of a $C^{1}$ vector field), with the first variation supplied by Differentiation of an integral functional under growth domination.

Facts & Assumptions

Given: The hypotheses of The weak Euler-Lagrange equation for integral functionals with fixed trace (a bounded $C^{1}$ domain $Ω$ , $1 < p < \infty$ , the Caratheodory integrand $f$ with the stated growth bounds, a local minimiser $u \in W^{1, p} (Ω)$ of the integral functional among the functions of trace $g$ with $g$ in the trace range), together with $f \in C^{2} (\overline{Ω} \times R \times R^{n})$ and $u \in C^{2} (\overline{Ω})$ . The measure-theoretic background is the Axiom-of-Countable-Choice framework of the published surface and divergence theory (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For every $φ \in W_{0}^{1, p} (Ω)$ , and in particular for every $φ \in C_{c}^{\infty} (Ω)$ , the weak Euler-Lagrange identity holds: $\int_{Ω} (f_{ξ} (x, u, D u) \cdot D φ + f_{s} (x, u, D u) φ) d x = 0$ (The weak Euler-Lagrange equation for integral functionals with fixed trace).

[F2]

Composites of $C^{k}$ Euclidean maps are $C^{k}$ ( $C^{k}$ Euclidean maps are closed under componentwise algebra and composition): since $f \in C^{2}$ gives $f_{s}, f_{ξ} \in C^{1}$ and $x \mapsto (x, u (x), D u (x))$ is $C^{1}$ for $u \in C^{2} (\overline{Ω})$ , the functions $x \mapsto f_{s} (x, u (x), D u (x))$ and $w (x) : = f_{ξ} (x, u (x), D u (x))$ are of class $C^{1}$ on $Ω$ ; consequently $div w$ is continuous and $f_{s} - div w$ is continuous on $Ω$ (Divergence and curl of a $C^{1}$ vector field, $C^{k}$ maps and multi-index derivative notation in Euclidean space).

[F3]

Divergence theorem: for a bounded $C^{1}$ domain and $F \in C^{1} (\overline{Ω}; R^{n})$ , $\int_{Ω} div F d x = \int_{\partial Ω} F \cdot ν d σ$ (Divergence on a bounded C1 Euclidean domain); the first Green identity is the special case $F = v D u$ of this identity (First Green identity).

[F4]

Fundamental lemma: if $g \in L_{l o c}^{1} (Ω)$ satisfies $\int_{Ω} g φ = 0$ for every $φ \in C_{c}^{\infty} (Ω)$ , then $g = 0$ almost everywhere; a continuous such $g$ vanishes everywhere (The fundamental lemma of the calculus of variations).

Proof

technique · direct, by testing the weak equation with compactly supported functions and integrating by parts

1.1F2given

Regularity of the coefficients. By [F2] the vector field $w (x) = f_{ξ} (x, u (x), D u (x))$ is of class $C^{1}$ on the open set $Ω$ , and the function $x \mapsto f_{s} (x, u (x), D u (x))$ is continuous; hence $div w$ is continuous and so is $f_{s} - div w$ .

2.1F1step 1.1

The weak identity. Let $φ \in C_{c}^{\infty} (Ω)$ . Then $φ \in W_{0}^{1, p} (Ω)$ , so [F1] gives $\int_{Ω} (f_{ξ} (x, u, D u) \cdot D φ + f_{s} (x, u, D u) φ) d x = 0$ , that is $\int_{Ω} w \cdot D φ d x = - \int_{Ω} f_{s} φ d x$ .

2.2F3step 1.1

Integration by parts with compact support. The field $F : = φ w$ is $C^{1}$ and compactly supported in $Ω$ ; in particular $F$ extends by zero to a $C^{1}$ field on $\overline{Ω}$ , so [F3] may be applied to it. Since $φ = 0$ on $\partial Ω$ , the boundary term vanishes and $\int_{Ω} div (φ w) d x = 0$ . By the product rule $div (φ w) = D φ \cdot w + φ div w$ , hence $\int_{Ω} w \cdot D φ d x = - \int_{Ω} (div w) φ d x$ .

3.1step 2.1step 2.2

The combined identity. Substituting step 2.2 into step 2.1 gives $\int_{Ω} (f_{s} (x, u, D u) - div w) φ d x = 0$ for every $φ \in C_{c}^{\infty} (Ω)$ .

4.1F4step 3.1∎

The fundamental lemma. The function $g : = f_{s} (x, u (x), D u (x)) - div w (x)$ is continuous on $Ω$ by step 1.1 and is orthogonal to every test function by step 3.1; [F4] gives $g = 0$ almost everywhere, and continuity upgrades this to $g = 0$ everywhere on $Ω$ . Hence $- div w + f_{s} (x, u (x), D u (x)) = 0$ on $Ω$ , the classical Euler-Lagrange equation.

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

The direct method for convex integral functionals

Statement

Assume the ultrafilter lemma, DC and HB. Let $n \geq 2$ , $1 < p < \infty$ , let $Ω \subseteq R^{n}$ be a bounded $C^{1}$ domain, and let $u_{b} \in W^{1, p} (Ω)$ . Set $K : = u_{b} + W_{0}^{1, p} (Ω),$ where $W_{0}^{1, p} (Ω)$ is the closure of $C_{c}^{\infty} (Ω)$ in $W^{1, p} (Ω)$ (Zero-boundary Sobolev space as a norm closure). Let $f : Ω \times R \times R^{n} \to R$ be a Caratheodory integrand (A Caratheodory integrand composed with measurable functions is measurable) such that for almost every $x$ the map $(s, ξ) \mapsto f (x, s, ξ)$ is convex and lower semicontinuous (Convex and strictly convex functionals on a convex subset of a real vector space). Assume the upper growth bound $f (x, s, ξ) \leq C (1 + ∣ s ∣^{p} + ∣ ξ ∣^{p}) + G (x) for almost every x and all (s, ξ),$ where $G \in L^{1} (Ω)$ , together with the coercivity hypothesis: there are $ν > 0$ , $c \geq 0$ , $q \in [1, p]$ and $h \in L^{1} (Ω)$ , $h \geq 0$ , with $f (x, s, ξ) \geq ν ∣ ξ ∣^{p} - c ∣ s ∣^{q} - h (x)$ for almost every $x$ and all $(s, ξ)$ , where, in the case $q = p$ , the smallness condition $2^{p - 1} c C_{P}^{p} \leq ν 2^{- p}$ holds for a Poincare constant $C_{P}$ of $W_{0}^{1, p} (Ω)$ (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction). Then $I (u) = \int_{Ω} f (x, u (x), D u (x)) d x$ is finite on $W^{1, p} (Ω)$ and attains its infimum on $K$ . If in addition $f$ satisfies the hypotheses of Differentiation of an integral functional under growth domination, then every minimiser $u$ satisfies $\int_{Ω} (f_{s} (x, u, D u) v + f_{ξ} (x, u, D u) \cdot D v) d x = 0 (v \in W_{0}^{1, p} (Ω)),$ the weak Euler-Lagrange equation for zero-boundary variations. If $f (x, \cdot, \cdot)$ is strictly convex for almost every $x$ , the minimiser is unique.

Facts & Assumptions

Given: The ultrafilter lemma, DC (which implies Countable Choice by Dependent choice implies countable choice) and HB; a bounded $C^{1}$ domain $Ω \subseteq R^{n}$ , $n \geq 2$ , $1 < p < \infty$ ; a lift $u_{b} \in W^{1, p} (Ω)$ and the nonempty affine class $K = u_{b} + W_{0}^{1, p} (Ω)$ ; and a Caratheodory integrand $f : Ω \times R \times R^{n} \to R$ with $(s, ξ) \mapsto f (x, s, ξ)$ convex and lower semicontinuous for almost every $x$ , satisfying the upper growth bound $f (x, s, ξ) \leq C (1 + ∣ s ∣^{p} + ∣ ξ ∣^{p}) + G (x)$ with $G \in L^{1} (Ω)$ , and the coercivity bound $f (x, s, ξ) \geq ν ∣ ξ ∣^{p} - c ∣ s ∣^{q} - h (x)$ with $ν > 0$ , $c \geq 0$ , $q \in [1, p]$ , $0 \leq h \in L^{1} (Ω)$ , where $2^{p - 1} c C_{P}^{p} \leq ν 2^{- p}$ holds in the case $q = p$ for a Poincare constant $C_{P}$ of $W_{0}^{1, p} (Ω)$ (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).

[F1]

For measurable $u, w$ the composition $x \mapsto f (x, u (x), w (x))$ is measurable (A Caratheodory integrand composed with measurable functions is measurable).

[F2]

$W_{0}^{1, p} (Ω)$ is a closed linear subspace by its definition as the closure of $C_{c}^{\infty} (Ω)$ ; hence $K = u_{b} + W_{0}^{1, p} (Ω)$ is nonempty, convex and norm closed. Under HB, norm-closed convex sets are weakly closed (Zero-boundary Sobolev space as a norm closure, Norm closed convex iff weakly closed).

[F3]

On the bounded domain, $u \in W^{1, p} (Ω)$ has $u, D u \in L^{p} (Ω)$ , and $L^{p} (Ω) \subseteq L^{q} (Ω)$ for $q \leq p$ with $∥ w ∥_{q} \leq ∣ Ω ∣^{1 / q - 1 / p} ∥ w ∥_{p}$ ; for $q < p$ this follows by applying Holder to $∣ w ∣^{q}$ and $1$ with conjugate exponents $p / q$ and $p / (p - q)$ , while $q = p$ is equality (Holder's inequality for integrals, including the endpoint cases); the space $W^{1, p} (Ω)$ is a real Banach space and its classes are $L^{p}$ classes (Integer-order Sobolev spaces and their norms, The space $L^{p} (μ)$ as the quotient by null functions).

[F4]

Fatou's lemma for nonnegative measurable functions (Fatou's lemma).

[F5]

If a sequence converges in $L^{p}$ , $1 \leq p < \infty$ , then a subsequence converges almost everywhere (Assuming Countable Choice, $L^{p}$ -convergent sequences have almost-everywhere convergent subsequences).

[F6]

Under HB and Countable Choice, a convex functional that is sequentially lower semicontinuous in the norm topology on a nonempty convex set is weakly sequentially lower semicontinuous (A convex norm-lower-semicontinuous functional is weakly lower semicontinuous).

[F7]

$W^{1, p} (Ω)$ is a real reflexive Banach space under the ultrafilter lemma, DC and HB (W^{1,p}(Omega) is reflexive for 1<p<infinity, Reflexivity is surjectivity of the canonical map), and the direct method in a reflexive Banach space yields a minimiser (The direct method in a reflexive Banach space); a nonempty convex norm-closed set is admissible by Norm closed convex iff weakly closed.

[F8]

If $f$ satisfies the hypotheses of Differentiation of an integral functional under growth domination, then $I$ is Gateaux differentiable with the displayed integral derivative. A minimiser on $u_{b} + W_{0}^{1, p} (Ω)$ is a local minimiser along every direction in $W_{0}^{1, p} (Ω)$ , so the first-variation theorem gives vanishing derivative on that space (The first variation vanishes at an interior minimiser).

[F9]

A proper, strictly convex functional has at most one minimiser on a convex set (Strict convexity gives uniqueness of a minimiser, Convex and strictly convex functionals on a convex subset of a real vector space); in the application $I$ is finite on the nonempty class $K$ , hence proper.

Proof

technique · direct; finiteness, convexity and norm lower semicontinuity of $I$, then the coercivity estimate and the direct method

1.1F1F2F3given

Finiteness of $I$ . For $u \in W^{1, p} (Ω)$ the integrand $x \mapsto f (x, u (x), D u (x))$ is measurable by [F1]. Its positive part is bounded by $C (1 + ∣ u ∣^{p} + ∣ D u ∣^{p}) + ∣ G ∣$ because $G \leq ∣ G ∣$ , and this has finite integral because $Ω$ is bounded, $u, D u \in L^{p}$ and $∣ G ∣ \in L^{1}$ ; its negative part is bounded by $c ∣ u ∣^{q} + h$ , whose integral is finite because $q \leq p$ , $u \in L^{q}$ by [F3] and $h \in L^{1}$ . Hence $I (u) = \int_{Ω} f (x, u (x), D u (x)) d x$ is a well-defined real number, and $I$ is proper as $K \neq \emptyset$ .

2.1F3step 1.1algebra

Convexity of $I$ . For $u, v \in W^{1, p} (Ω)$ and $λ \in [0, 1]$ the pair $(λ u + (1 - λ) v, λ D u + (1 - λ) D v)$ equals $λ (u, D u) + (1 - λ) (v, D v)$ because the weak gradient is linear, and for almost every $x$ the convexity of $f (x, \cdot, \cdot)$ gives $f (x, λ u + (1 - λ) v, λ D u + (1 - λ) D v) \leq λ f (x, u, D u) + (1 - λ) f (x, v, D v)$ . All three functions are integrable by step 1.1, so integrating gives $I (λ u + (1 - λ) v) \leq λ I (u) + (1 - λ) I (v)$ : $I$ is convex on the real vector space $W^{1, p} (Ω)$ .

2.2F4F5step 1.1given

Norm lower semicontinuity of $I$ . Let $u_{j} \to u$ in $W^{1, p}$ . Suppose, for contradiction, that $I (u) > {lim inf}_{j} I (u_{j}) = : ℓ$ ; since $I$ is real-valued, choose a real $t$ with $ℓ < t < I (u)$ . Then $I (u_{j}) \leq t$ for infinitely many $j$ , so passing to that subsequence (and relabelling) we may assume $I (u_{j}) \leq t$ for all $j$ and still $u_{j} \to u$ in $W^{1, p}$ ; in particular $u_{j} \to u$ and $D u_{j} \to D u$ in $L^{p}$ . By [F5] pass to a further subsequence with $u_{j} \to u$ and $D u_{j} \to D u$ almost everywhere. Since $(s, ξ) \mapsto f (x, s, ξ)$ is lower semicontinuous at $(u (x), D u (x))$ for almost every $x$ , the pointwise limit inferior satisfies $f (x, u, D u) + c ∣ u ∣^{q} + h \leq {lim inf}_{j} (f (x, u_{j}, D u_{j}) + c ∣ u_{j} ∣^{q} + h)$ . The shifted integrands $φ_{j} : = f (x, u_{j}, D u_{j}) + c ∣ u_{j} ∣^{q} + h$ are nonnegative by the coercivity bound and measurable by [F1], so Fatou's lemma [F4] gives $\int_{Ω} (f (x, u, D u) + c ∣ u ∣^{q} + h) \leq {lim inf}_{j} \int_{Ω} φ_{j} = {lim inf}_{j} I (u_{j}) + c ∥ u ∥_{q}^{q} + ∥ h ∥_{1}$ , where $∥ u_{j} ∥_{q} \to ∥ u ∥_{q}$ because $u_{j} \to u$ in $L^{p}$ and $q \leq p$ . Cancelling the common finite terms yields $I (u) \leq {lim inf}_{j} I (u_{j}) \leq t$ , contradicting $t < I (u)$ . Hence $I$ is sequentially lower semicontinuous in the norm topology.

2.3F2F3step 1.1givenalgebra

Coercivity of $I$ on $K$ . Write each $u \in K$ as $u = u_{b} + v$ with $v \in W_{0}^{1, p} (Ω)$ . The lower growth bound gives $I (u) \geq ν ∥ D u ∥_{p}^{p} - c ∥ u ∥_{q}^{q} - ∥ h ∥_{1}$ , while the triangle inequality and $(a + b)^{p} \leq 2^{p - 1} (a^{p} + b^{p})$ give $∥ D u ∥_{p}^{p} \geq 2^{1 - p} ∥ D v ∥_{p}^{p} - ∥ D u_{b} ∥_{p}^{p}$ . By Poincare, $∥ v ∥_{p} \leq C_{P} ∥ D v ∥_{p}$ , and with $C_{Ω} : = ∣ Ω ∣^{1 / q - 1 / p}$ (equal to $1$ when $q = p$ ), Holder and the triangle inequality imply $∥ u ∥_{q}^{q} \leq C_{Ω}^{q} 2^{q - 1} (∥ u_{b} ∥_{p}^{q} + C_{P}^{q} ∥ D v ∥_{p}^{q}) .$ If $q < p$ , these estimates yield $I (u) \geq ν 2^{1 - p} ∥ D v ∥_{p}^{p} - c^{'} ∥ D v ∥_{p}^{q} - C^{''}$ , which tends to $+ \infty$ as $∥ D v ∥_{p} \to \infty$ . If $q = p$ , they yield $I (u) \geq (ν 2^{1 - p} - c 2^{p - 1} C_{P}^{p}) ∥ D v ∥_{p}^{p} - C^{''} \geq ν 2^{- p} ∥ D v ∥_{p}^{p} - C^{''}$ by the smallness assumption. Finally, $u = u_{b} + v$ and Poincare give $∥ u ∥_{W^{1, p}} \leq C_{b} + C_{1} ∥ D v ∥_{p}$ for fixed finite constants $C_{b}, C_{1}$ , so $∥ u ∥_{W^{1, p}} \to \infty$ forces $∥ D v ∥_{p} \to \infty$ . Thus $I$ is coercive on $K$ .

3.1F6step 2.1step 2.2

Weak sequential lower semicontinuity on $K$ . By steps 2.1 and 2.2 the functional $I$ is convex and norm lower semicontinuous on the convex set $W^{1, p} (Ω)$ ; [F6] therefore makes $I$ weakly sequentially lower semicontinuous on $W^{1, p} (Ω)$ , hence on the subset $K$ .

4.1F2F7step 1.1step 2.3step 3.1

Existence of a minimiser. By [F7] the space $W^{1, p} (Ω)$ is a real reflexive Banach space under the present choice principles, and $K$ is nonempty, convex and weakly closed by [F2], in particular weakly sequentially closed. The functional $I$ is proper by step 1.1, coercive on $K$ by step 2.3 and weakly sequentially lower semicontinuous on $K$ by step 3.1, so the direct method [F7] provides $u_{0} \in K$ with $I (u_{0}) = \inf_{K} I$ .

5.1F8step 4.1

The Euler-Lagrange clause. Suppose in addition that $f$ satisfies the hypotheses of the differentiation lemma. Then [F8] gives the Gateaux derivative formula. Since $K = u_{b} + W_{0}^{1, p} (Ω)$ , a minimiser $u_{0}$ is a local minimiser along every direction in $W_{0}^{1, p} (Ω)$ ; applying the first-variation theorem with $V = W_{0}^{1, p} (Ω)$ yields $\int_{Ω} (f_{s} (x, u_{0}, D u_{0}) v + f_{ξ} (x, u_{0}, D u_{0}) \cdot D v) d x = 0 (v \in W_{0}^{1, p} (Ω)) .$ This is the weak Euler-Lagrange equation for the affine zero-boundary variation class.

6.1F9step 1.1step 4.1∎

Strict convexity and uniqueness. Assume finally that $f (x, \cdot, \cdot)$ is strictly convex for almost every $x$ . For distinct $u \neq v$ in $K$ the set $S : = {x : (u (x), D u (x)) \neq (v (x), D v (x))}$ has positive measure, because otherwise $u = v$ and $D u = D v$ almost everywhere, that is, $u = v$ as elements of $W^{1, p} (Ω)$ . For almost every $x \in S$ and every $λ \in (0, 1)$ the strict convexity of $f (x, \cdot, \cdot)$ gives $f (x, λ u + (1 - λ) v, λ D u + (1 - λ) D v) < λ f (x, u, D u) + (1 - λ) f (x, v, D v)$ , the values being finite by step 1.1; integrating over $S$ and using the convex inequality elsewhere gives $I (λ u + (1 - λ) v) < λ I (u) + (1 - λ) I (v)$ , so $I$ is strictly convex on the convex set $K$ . By [F9] the minimiser of step 4.1 is then unique.

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

The natural boundary condition for free boundary variations

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let the hypotheses of The weak Euler-Lagrange equation for integral functionals with fixed trace hold, but with no prescribed trace: $u \in W^{1, p} (Ω)$ is a local minimiser of $I$ on the whole of $W^{1, p} (Ω)$ . Assume in addition that $f \in C^{2} (\overline{Ω} \times R \times R^{n})$ and $u \in C^{2} (\overline{Ω})$ , and let $ν$ be the outward unit normal (Bounded C1 domains and their outward normals). Then $- div (f_{ξ} (x, u, D u)) + f_{s} (x, u, D u) = 0 in Ω, f_{ξ} (x, u (x), D u (x)) \cdot ν (x) = 0 on \partial Ω .$ The second identity is the natural (Neumann-type) boundary condition attached to free boundary variations; no such condition appears when the trace is fixed.

Facts & Assumptions

Given: A bounded $C^{1}$ domain $Ω$ , $1 < p < \infty$ , an integrand $f$ and functional $I$ as in the hypotheses of The weak Euler-Lagrange equation for integral functionals with fixed trace, and a local minimiser $u \in W^{1, p} (Ω)$ of $I$ on the whole of $W^{1, p} (Ω)$ with no prescribed trace. In addition $f \in C^{2} (\overline{Ω} \times R \times R^{n})$ and $u \in C^{2} (\overline{Ω})$ . The boundary theory and the separation used below are set up under the Axiom of Choice (The Axiom of Choice), and $ν$ is the outward unit normal (Bounded C1 domains and their outward normals).

[F1]

The classical Euler-Lagrange equation holds in the interior: with $w (x) : = f_{ξ} (x, u (x), D u (x))$ one has $w \in C^{1} (Ω)$ , and $- div w + f_{s} (x, u (x), D u (x)) = 0$ on $Ω$ (The classical Euler-Lagrange equation under regularity).

[F2]

First variation vanishes: for every $v$ in the Banach space $W^{1, p} (Ω)$ (Integer-order Sobolev spaces and their norms), including every $φ \in C^{\infty} (\overline{Ω})$ , one has $δ I (u; v) = 0$ , because $u$ is a local minimiser on the whole space and $I$ is Gateaux differentiable there (The first variation vanishes at an interior minimiser, Differentiation of an integral functional under growth domination); explicitly $δ I (u; φ) = \int_{Ω} (f_{ξ} (x, u, D u) \cdot D φ + f_{s} (x, u, D u) φ) d x$ .

[F3]

Since $f \in C^{2}$ and $u \in C^{2} (\overline{Ω})$ , the composition $x \mapsto f_{ξ} (x, u (x), D u (x))$ is of class $C^{1}$ on $\overline{Ω}$ ( $C^{k}$ Euclidean maps are closed under componentwise algebra and composition).

[F4]

Divergence theorem: for $F \in C^{1} (\overline{Ω}; R^{n})$ , $\int_{Ω} div F d x = \int_{\partial Ω} F \cdot ν d σ$ (Divergence on a bounded C1 Euclidean domain, First Green identity).

[F5]

Boundary fundamental lemma: if $h \in C (\overline{Ω}; R^{n})$ satisfies $\int_{\partial Ω} (h \cdot ν) φ d σ = 0$ for every $φ \in C^{\infty} (\overline{Ω})$ , then $h \cdot ν = 0$ on $\partial Ω$ (The boundary fundamental lemma of the calculus of variations).

Proof

technique · direct, combining the fixed-trace interior equation with the free boundary variation

1.1F1F3given

The interior equation. Since $u$ is a local minimiser of $I$ on the whole of $W^{1, p} (Ω)$ , it is in particular a local minimiser among the functions with the fixed trace $g : = T u$ , which lies in the trace range by definition; the hypotheses of the fixed-trace case hold, so [F1] gives the interior equation $- div w + f_{s} (x, u, D u) = 0$ on $Ω$ , where $w = f_{ξ} (x, u, D u)$ . By [F3] the field $w$ extends to a $C^{1}$ field on $\overline{Ω}$ .

2.1F2step 1.1

The free variation. Let $φ \in C^{\infty} (\overline{Ω})$ . Then $φ \in W^{1, p} (Ω)$ , and by [F2] the first variation vanishes: $\int_{Ω} (w \cdot D φ + f_{s} (x, u, D u) φ) d x = 0$ .

3.1step 1.1step 2.1algebra

Substituting the interior equation. Replacing $f_{s} (x, u, D u)$ by $div w$ in step 2.1, which is legitimate pointwise on $Ω$ by step 1.1, and using the product rule $div (φ w) = D φ \cdot w + φ div w$ , gives $\int_{Ω} div (φ w) d x = 0$ for every $φ \in C^{\infty} (\overline{Ω})$ .

4.1F4step 3.1

The boundary term. The field $F : = φ w$ lies in $C^{1} (\overline{Ω}; R^{n})$ , so the divergence theorem [F4] applies and $0 = \int_{Ω} div (φ w) d x = \int_{\partial Ω} φ (w \cdot ν) d σ$ for every $φ \in C^{\infty} (\overline{Ω})$ .

5.1F5step 4.1∎

The natural boundary condition. Step 4.1 says that $h : = w$ satisfies $\int_{\partial Ω} (h \cdot ν) φ d σ = 0$ for every $φ \in C^{\infty} (\overline{Ω})$ ; since $w$ is continuous on $\overline{Ω}$ by [F3], the boundary fundamental lemma [F5] gives $h \cdot ν = w \cdot ν = 0$ on $\partial Ω$ . Together with step 1.1 this is the interior equation and the natural boundary condition, and no boundary condition of this kind appears in the fixed-trace case handled by [F1].

RemarkRemark: Literature-sourcedProof: Not applicableOpen item page →

Euler-Lagrange is necessary but not sufficient without convexity

Remark

For a Gateaux differentiable functional on an open set, the first variation vanishes at an interior local minimiser; on an affine admissible class $u + V$ it vanishes in the directions $v \in V$ (The first variation vanishes at an interior minimiser). For integral functionals satisfying its differentiation and fixed-trace hypotheses, The weak Euler-Lagrange equation for integral functionals with fixed trace gives the weak Euler-Lagrange equation for zero-boundary variations. Conversely, for a convex functional Gateaux differentiable on an open neighbourhood of a convex admissible set $K$ , the condition $δ I (u; v - u) \geq 0$ for every $v \in K$ suffices for a global minimum (Stationarity is sufficient for a global minimum of a convex differentiable functional). Without convexity the three notions must be kept apart: a stationary point solves the Euler-Lagrange equation, a local minimiser minimises among nearby admissible competitors, and a global minimiser minimises on the whole admissible set. In general none of the implications "stationary $\Rightarrow$ local minimiser", "local minimiser $\Rightarrow$ global minimiser" or "global minimiser $\Rightarrow$ unique" holds, and the Euler-Lagrange equation alone therefore cannot be used as an existence criterion. Convexity upgrades the variational inequality to global minimality, whereas coercivity and weak lower semicontinuity enter the separate existence argument; a concave quadratic functional is the standard illustration, and the companion page records explicit counterexamples. For the other failed implications already mentioned, $F (t) = t^{2} - 3 t^{3} + t^{4}$ has a local minimum at $0$ (the coefficient $1 - 3 t + t^{2}$ is positive near $0$ ) but $F (1) = - 1 < F (0) = 0$ , while $(t^{2} - 1)^{2}$ has the two global minimisers $\pm 1$ .

For inequality constraints, two-sided variations need not be admissible. Local minimality gives only a nonnegative one-sided derivative along an admissible segment, since $[I (u + t v) - I (u)] / t \geq 0$ for sufficiently small feasible $t > 0$ ; it need not give stationarity in arbitrary directions.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

The Dirichlet principle for the Poisson equation

Statement

Assume the Axiom of Choice (The Axiom of Choice), the ultrafilter lemma, DC and HB. Let $Ω \subseteq R^{n}$ , $n \geq 2$ , be a bounded $C^{1}$ domain, let $f \in L^{2} (Ω)$ and let $g \in H^{1 / 2} (\partial Ω) = W^{1 / 2, 2} (\partial Ω)$ lie in the trace range of $T : H^{1} (Ω) \to H^{1 / 2} (\partial Ω)$ (The sharp trace theorem: boundedness and range in the fractional space, The fractional Sobolev space on a compact $C^{1}$ boundary). Put $I (u) = \frac{1}{2} \int_{Ω} ∣ D u ∣^{2} d x - \int_{Ω} f u d x, K_{g} = {u \in H^{1} (Ω) : T u = g} .$ Then: (i) $I$ is strictly convex, coercive and weakly sequentially lower semicontinuous on $K_{g}$ (Convex and strictly convex functionals on a convex subset of a real vector space, Proper, coercive and weakly lower semicontinuous extended-real functionals), and attains its infimum at exactly one $u_{0} \in K_{g}$ (The direct method for convex integral functionals, Strict convexity gives uniqueness of a minimiser); (ii) $u_{0}$ is the unique weak solution of the Poisson problem $- Δ u = f$ with trace $g$ in the sense of Weak Dirichlet solutions for a divergence-form operator, so that $\int_{Ω} D u_{0} \cdot D φ d x = \int_{Ω} f φ d x$ for every $φ \in H_{0}^{1} (Ω)$ (The weak Euler-Lagrange equation for integral functionals with fixed trace, Existence and uniqueness for the weak Dirichlet Poisson problem, The inhomogeneous weak Dirichlet problem by a trace lifting); (iii) the classical one-directional Dirichlet principle holds: if $v \in C^{2} (\overline{Ω})$ satisfies $- Δ v = f$ in $Ω$ and $v ∣_{\partial Ω} = g$ , then $I (v) \leq I (w)$ for every $w \in K_{g}$ (First Green identity, Classical solutions satisfy the weak formulation).

Facts & Assumptions

Given: The Axiom of Choice (The Axiom of Choice), the ultrafilter lemma, DC and HB; a bounded $C^{1}$ domain $Ω \subseteq R^{n}$ , $n \geq 2$ ; $f \in L^{2} (Ω)$ ; $g \in H^{1 / 2} (\partial Ω)$ in the trace range of $T : H^{1} (Ω) \to H^{1 / 2} (\partial Ω)$ with affine class $K_{g}$ ; and the energy $I (u) = \frac{1}{2} \int_{Ω} ∣ D u ∣^{2} d x - \int_{Ω} f u d x$ .

[F0]

The Axiom of Choice is explicitly assumed here because the trace, trace-kernel and weak-Poisson suppliers used below state their conclusions under AC (The Axiom of Choice).

[F1]

The trace operator $T$ is bounded with $T u = u ∣_{\partial Ω}$ for continuous $u$ , and $H^{1 / 2} (\partial Ω) = W^{1 / 2, 2} (\partial Ω)$ (The $L^{p}$ trace operator on a bounded $C^{1}$ domain, The sharp trace theorem: boundedness and range in the fractional space, The fractional Sobolev space on a compact $C^{1}$ boundary).

[F2]

$K_{g}$ is nonempty, convex and weakly closed, and equals $R g + H_{0}^{1} (Ω)$ for any right inverse $R$ (The affine Dirichlet trace class is nonempty, convex and weakly closed); $\ker T = H_{0}^{1} (Ω)$ (The kernel of the trace is the closure of the test functions, Zero-boundary Sobolev space as a norm closure).

[F3]

The direct method for convex integral functionals: with $p = 2$ , a Caratheodory integrand convex and lower semicontinuous in $(s, ξ)$ satisfying the upper growth bound with $G \in L^{1} (Ω)$ and the coercivity bound holds, the functional attains its infimum on $K_{g}$ ; if the integrand satisfies the differentiation hypotheses, every minimiser solves the weak Euler-Lagrange equation, and strict convexity of the integrand in $(s, ξ)$ makes the minimiser unique (The direct method for convex integral functionals, The weak Euler-Lagrange equation for integral functionals with fixed trace, Strict convexity gives uniqueness of a minimiser).

[F4]

The weak Dirichlet solution of $- Δ u = f$ with trace $g$ is a class $u \in H^{1} (Ω)$ with $T u = g$ and $\int_{Ω} D u \cdot D φ = \int_{Ω} f φ$ for every $φ \in H_{0}^{1} (Ω)$ (Weak Dirichlet solutions for a divergence-form operator); such a solution exists and is unique, and agrees with the lifting construction (Existence and uniqueness for the weak Dirichlet Poisson problem, The inhomogeneous weak Dirichlet problem by a trace lifting).

[F5]

Poincare's inequality on $H_{0}^{1} (Ω)$ with constant $C_{P}$ (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction); $H^{1} (Ω) = W^{1, 2} (Ω)$ carries the Sobolev norm and inner product (The notation $H^{k}$ and the reserved zero-boundary symbol, Integer-order Sobolev spaces and their norms, The Sobolev space $H^{1}$ is a Hilbert space).

[F6]

If $v \in C^{2} (\overline{Ω})$ satisfies $- Δ v = f$ almost everywhere, first Green identity with $φ \in C_{c}^{\infty} (Ω)$ gives $\int_{Ω} D v \cdot D φ = \int_{Ω} f φ$ because the boundary test vanishes (First Green identity). Holder bounds both pairings by a constant times $∥ φ ∥_{H^{1}}$ , so density extends this identity to $H_{0}^{1} (Ω)$ (Holder's inequality for integrals, including the endpoint cases, Zero-boundary Sobolev space as a norm closure). This does not require the classical solution itself to have zero trace; the zero-trace-only supplier Classical solutions satisfy the weak formulation is therefore not applied to $v$ .

[F7]

The basic definitions: convex and strictly convex functionals, proper coercive weakly lower semicontinuous functionals (Convex and strictly convex functionals on a convex subset of a real vector space, Proper, coercive and weakly lower semicontinuous extended-real functionals).

Proof

technique · direct, by checking the hypotheses of the convex direct method for the Dirichlet integrand

1.1F5F7givenalgebra

The integrand and its bounds. Put $f_{0} (x, s, ξ) : = \frac{1}{2} ∣ ξ ∣^{2} - f (x) s$ . It is a Caratheodory integrand, jointly convex and continuous in $(s, ξ)$ , and the elementary inequality $∣ f (x) s ∣ \leq \frac{1}{2} ∣ f (x) ∣^{2} + \frac{1}{2} ∣ s ∣^{2}$ gives the upper bound $f_{0} \leq \frac{1}{2} (1 + ∣ s ∣^{2} + ∣ ξ ∣^{2}) + \frac{1}{2} ∣ f (x) ∣^{2}$ , admissible with $p = 2$ , $C = \frac{1}{2}$ and $G = \frac{1}{2} ∣ f ∣^{2} \in L^{1} (Ω)$ .

1.2F5givenalgebra

The coercivity bound with the smallness condition. Fix $ε > 0$ with $2 ε C_{P}^{2} \leq \frac{1}{8}$ ; Cauchy's inequality $∣ f (x) s ∣ \leq ε ∣ s ∣^{2} + \frac{1}{4 ε} ∣ f (x) ∣^{2}$ gives $f_{0} (x, s, ξ) \geq \frac{1}{2} ∣ ξ ∣^{2} - ε ∣ s ∣^{2} - \frac{1}{4 ε} ∣ f (x) ∣^{2}$ , which is the coercivity bound with $ν = \frac{1}{2}$ , $c = ε$ , $q = p = 2$ and $h = \frac{1}{4 ε} ∣ f ∣^{2} \geq 0$ , $h \in L^{1} (Ω)$ ; the smallness condition $2^{p - 1} c C_{P}^{p} = 2 ε C_{P}^{2} \leq \frac{1}{8} = ν \cdot 2^{- p}$ holds by the choice of $ε$ .

2.1F3F2F0step 1.1step 1.2

The direct method applies. By steps 1.1 and 1.2 the integrand satisfies all hypotheses of [F3] with $p = 2$ ; the class $K_{g}$ is nonempty, convex and weakly closed by [F2]; hence $I$ attains its infimum at some $u_{0} \in K_{g}$ , is coercive and weakly sequentially lower semicontinuous on $K_{g}$ .

3.1F2F3F5F7F0step 2.1algebra

Strict convexity of $I$ on $K_{g}$ . The functional is $I = Q - L$ with $Q (u) = \frac{1}{2} ∥ D u ∥_{2}^{2}$ and $L (u) = \int f u$ . The term $L$ is affine. The quadratic term $Q$ is strictly convex on $K_{g}$ : if $u \neq v$ in $K_{g}$ then $D (u - v)$ does not vanish almost everywhere, because $u - v \in H_{0}^{1} (Ω)$ by [F2], and Poincare [F5] would force $u - v = 0$ if $D (u - v) = 0$ ; consequently $Q (λ u + (1 - λ) v) < λ Q (u) + (1 - λ) Q (v)$ for $0 < λ < 1$ by the parallelogram identity. Hence $I$ is strictly convex on the convex set $K_{g}$ , and the minimiser $u_{0}$ of step 2.1 is unique by the strict-convexity uniqueness corollary Strict convexity gives uniqueness of a minimiser.

3.2F3F5F0step 2.1

The weak Euler-Lagrange equation. The integrand $f_{0}$ satisfies the differentiation hypotheses with $p = 2$ : $f_{0, s} = - f$ and $f_{0, ξ} = ξ$ are continuous in $(s, ξ)$ , $∣ f_{0} ∣ \leq \frac{1}{2} (1 + ∣ s ∣^{2} + ∣ ξ ∣^{2}) + \frac{1}{2} ∣ f ∣^{2}$ and $∣ f_{0, s} ∣ + ∣ f_{0, ξ} ∣ \leq (1 + ∣ s ∣ + ∣ ξ ∣) + ∣ f (x) ∣$ with $∣ f ∣ \in L^{2} (Ω) = L^{p^{'}} (Ω)$ . Hence the conditional clause of [F3] applies to the minimiser $u_{0}$ : $\int_{Ω} (D u_{0} \cdot D φ - f (x) φ) d x = 0$ for every $φ \in H_{0}^{1} (Ω)$ , that is $\int_{Ω} D u_{0} \cdot D φ = \int_{Ω} f φ$ .

4.1F4F0step 2.1step 3.1step 3.2

Identification with the weak Dirichlet solution. By steps 2.1 and 3.2 the minimiser $u_{0} \in H^{1} (Ω)$ satisfies $T u_{0} = g$ and $\int_{Ω} D u_{0} \cdot D φ = \int_{Ω} f φ$ for every $φ \in H_{0}^{1} (Ω)$ ; this is exactly the weak Dirichlet solution of $- Δ u = f$ with trace $g$ in the sense of [F4], and by the uniqueness statement of [F4] it is the unique such solution. This proves (i) and (ii).

5.1F1F2F5F6F0step 4.1∎

The classical one-directional principle. Let $v \in C^{2} (\overline{Ω})$ satisfy $- Δ v = f$ in $Ω$ and $v ∣_{\partial Ω} = g$ . Then $T v = g$ by [F1] (the trace restricts continuous functions pointwise), so for every $w \in K_{g}$ the difference $η : = w - v$ has $T η = 0$ , that is $η \in H_{0}^{1} (Ω)$ by [F2]. By [F6], applied to the classical solution $v$ , one has $\int_{Ω} D v \cdot D η = \int_{Ω} f η$ . Expanding the energy, $I (w) - I (v) = \frac{1}{2} \int_{Ω} (∣ D w ∣^{2} - ∣ D v ∣^{2}) d x - \int_{Ω} f η d x = \frac{1}{2} \int_{Ω} ∣ D η ∣^{2} d x + \int_{Ω} D v \cdot D η d x - \int_{Ω} f η d x = \frac{1}{2} \int_{Ω} ∣ D η ∣^{2} d x \geq 0$ , with equality if and only if $D η = 0$ almost everywhere, that is $w = v$ by [F5]. Hence $I (v) \leq I (w)$ for every $w \in K_{g}$ , the classical Dirichlet principle.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Minimisers are classical when elliptic regularity applies

Statement

Assume the Axiom of Choice and Countable Choice. Let $Ω \subseteq R^{n}$ , $n \geq 2$ , be a bounded domain, let $f$ and $g$ be given and let $u_{0} \in K_{g}$ be the minimiser of the Dirichlet energy of The Dirichlet principle for the Poisson equation. Then: (i) if $Ω$ is a bounded $C^{\infty}$ domain, $f$ extends to a $C^{\infty}$ function on a neighbourhood of $\overline{Ω}$ , and there is $G \in C^{\infty} (U)$ on a neighbourhood $U$ of $\overline{Ω}$ with $G ∣_{\partial Ω} = g$ , then $u_{0}$ agrees almost everywhere with a function $\tilde{u} \in C^{\infty} (\overline{Ω})$ satisfying $- Δ \tilde{u} = f$ pointwise in $Ω$ and $\tilde{u} = g$ on $\partial Ω$ (Smooth weak Dirichlet solutions are classical); (ii) if $0 < α < 1$ , $Ω$ is a bounded $C^{2, α}$ domain, $f \in C^{0, α} (\overline{Ω})$ and $g \in C^{2, α} (\overline{Ω})$ , then $u_{0} \in C^{2, α} (\overline{Ω})$ , $- Δ u_{0} = f$ pointwise and $u_{0} = g$ on $\partial Ω$ (Global Schauder regularity for the weak Dirichlet Laplacian). Variational existence alone gives only $u_{0} \in H^{1} (Ω)$ ; the smoothness asserted here is a consequence of elliptic regularity and fails without the corresponding hypotheses on the domain, coefficients and data.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; a bounded domain $Ω \subseteq R^{n}$ , $n \geq 2$ ; data $f$ and $g$ ; and the minimiser $u_{0} \in K_{g}$ of the Dirichlet energy of The Dirichlet principle for the Poisson equation. In case (i), $Ω$ is a bounded $C^{\infty}$ domain, $f$ extends smoothly to a neighbourhood of $\overline{Ω}$ , and $G \in C^{\infty} (U)$ on a neighbourhood $U$ of $\overline{Ω}$ satisfies $G ∣_{\partial Ω} = g$ ; in case (ii) $0 < α < 1$ , $Ω$ is a bounded $C^{2, α}$ domain, $f \in C^{0, α} (\overline{Ω})$ and $g \in C^{2, α} (\overline{Ω})$ .

[F1]

The minimiser is the unique weak solution of $- Δ u = f$ with trace $g$ in the sense of Weak Dirichlet solutions for a divergence-form operator (The Dirichlet principle for the Poisson equation, The weak Euler-Lagrange equation for integral functionals with fixed trace).

[F2]

The trace of a smooth function is its boundary restriction, the kernel of the trace on a bounded $C^{1}$ domain is $H_{0}^{1}$ , and $H_{0}^{1}$ is the closure of $C_{c}^{\infty}$ (The trace agrees with classical restriction for continuous Sobolev functions, The kernel of the trace is the closure of the test functions, Zero-boundary Sobolev space as a norm closure).

[F3]

For $G \in C^{\infty} (U)$ and every $φ \in C_{c}^{\infty} (Ω)$ , classical integration by parts gives $\int_{Ω} D G \cdot D \overline{φ} d x = - \int_{Ω} (Δ G) \overline{φ} d x$ . Both functionals extend continuously to $H_{0}^{1} (Ω)$ because $D G \in L^{2}$ and $Δ G \in L^{2}$ on the bounded domain (Holder's inequality for integrals, including the endpoint cases).

[F4]

Smooth zero-boundary elliptic regularity: on a bounded $C^{\infty}$ domain, a zero-trace weak solution with smooth coefficients and forcing agrees almost everywhere with a $C^{\infty} (\overline{Ω})$ solution, satisfies the equation pointwise, and vanishes on the boundary (Smooth weak Dirichlet solutions are classical).

[F5]

Schauder regularity: if $0 < α < 1$ , $Ω$ is a bounded $C^{2, α}$ domain and the data are Holder, then the unique weak Dirichlet solution of $- Δ u = f$ lies in $C^{2, α} (\overline{Ω})$ , solves the equation pointwise and attains $g$ classically on $\partial Ω$ (Global Schauder regularity for the weak Dirichlet Laplacian).

Proof

technique · subtract a smooth boundary lift in the smooth case, then apply zero-boundary regularity; use the Schauder supplier directly in case (ii)

1.1F1given

The variational starting point. By [F1] the minimiser $u_{0}$ is the unique weak solution of $- Δ u = f$ with trace $g$ ; the variational analysis alone gives only $u_{0} \in H^{1} (Ω)$ , and no higher regularity is asserted by it.

1.2F1F2F3

Case (i): lift and zero trace. Let $G$ be the smooth extension in the hypothesis and set $v : = u_{0} - G$ . By [F1], $T u_{0} = g$ ; by [F2], $T G = G ∣_{\partial Ω} = g$ , so $T v = 0$ and the trace-kernel theorem gives $v \in H_{0}^{1} (Ω)$ . For every $φ \in C_{c}^{\infty} (Ω)$ , [F1] and [F3] give $\int_{Ω} D v \cdot D \overline{φ} d x = \int_{Ω} (f + Δ G) \overline{φ} d x .$ Both sides are continuous in the $H^{1}$ norm; density of $C_{c}^{\infty} (Ω)$ in $H_{0}^{1} (Ω)$ extends the identity to all $H_{0}^{1}$ tests. Thus $v$ is the zero-trace weak solution with smooth forcing $f + Δ G$ .

2.1F4step 1.2

Apply smooth regularity and restore the lift. The coefficients of $- Δ$ are smooth, and $f + Δ G$ extends smoothly to a neighbourhood of $\overline{Ω}$ . Supplier [F4] applies to $v$ , giving a smooth representative $\tilde{v} \in C^{\infty} (\overline{Ω})$ with $- Δ \tilde{v} = f + Δ G$ and $\tilde{v} = 0$ on $\partial Ω$ . Then $\tilde{u} : = \tilde{v} + G$ represents $u_{0}$ , satisfies $- Δ \tilde{u} = f$ pointwise and has boundary values $g$ .

2.2F1F5step 1.1

Case (ii). Under the Holder hypotheses of case (ii), [F5] applies to the same weak solution and yields $u_{0} \in C^{2, α} (\overline{Ω})$ with $- Δ u_{0} = f$ pointwise and $u_{0} = g$ on $\partial Ω$ .

3.1step 1.1step 2.1step 2.2∎

The warning. Both conclusions are consequences of elliptic regularity under the stated hypotheses on the domain, the coefficients and the data; without them variational existence alone gives only $u_{0} \in H^{1} (Ω)$ , as the companion counterexamples on weak solutions without higher regularity record.

5 · Examples, counterexamples and false statements

None yet.