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Rellich Kondrachov and Sobolev Compactness

1 · Prerequisites

2 · Summary

This page proves the compact embedding theorems behind existence theory for PDEs. It begins with the definition of a compact embedding and the translation estimate for W1,p functions, the two inputs of the Fréchet–Kolmogorov criterion: an Lp-bounded family with vanishing tails and uniform translation control is totally bounded, and relative compactness forces uniform translation continuity. From the criterion follow the Rellich theorems: W01,p(Ω)⋐Lp(Ω) on every bounded open set, and W1,p(Ω)⋐Lp(Ω) on bounded extension domains, with no boundary regularity in the first case. A compactness proof of the Poincaré–Wirtinger inequality and a local Lp-compactness theorem for Wloc1,p-bounded sequences follow, together with the subcritical and critical forms of Rellich–Kondrachov on extension domains, the Morrey compactness into C0,β for p>n, and the higher-order statements.

The second half develops the compactness machinery for fractional Sobolev spaces that the subcritical trace theorem needs: the level-set kernel estimate, a dyadic summability lemma, the Slobodeckij lower bound for dyadic level sets, the critical fractional Sobolev inequality on Rd, and mollification rates that yield fractional Rellich compactness on bounded supports. These feed the compactness of the Sobolev trace into subcritical boundary Lq spaces, the strictly subcritical character of every theorem on the page, and three downstream consequences: the whole-sequence strong convergence corollary for bounded sequences and their weak limits, the compact L2 operator corollary obtained from a bounded map into H01, and the lemma that norm-closed target constraints survive compact extraction.

Notes

Every compactness statement carries the exact choice hypothesis of its proof: the Fréchet–Kolmogorov theorem assumes Countable and Dependent Choice; the Rellich theorems, the Morrey branch and the fractional Rellich theorem assume the Axiom of Choice through their named suppliers, while the bounded-support tail lemma and the compact-operator definition are choice-free. The sequential form of compact embedding is stated under Countable and Dependent Choice.

3 · Logical flowchart

4 · Definitions, theorems and proofs

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

Compactly embedded normed spaces

Definition

Let X and Y be normed spaces over the same field K∈{R,C}, read in the real case from A norm on a real vector space, the induced metric, and the dictionary with the metric axioms and in the complex case from Real and complex scalar conventions for normed spaces, and suppose X⊆Y with continuous inclusion J:X→Y: there is a real C≥0 with ∥x∥Y≤C∥x∥X for every x∈X.

One says that X is compactly embedded in Y, written X⋐Y, when the inclusion operator J is a compact operator in the sense of Compact linear operator: the image under J of every bounded subset of X (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space) has compact closure in Y (Open cover, subcover, compact metric space, and compact subset of a metric space).

The sequential form. Assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)) and the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Then X⋐Y if and only if every bounded sequence (xj) in X has a subsequence (xjk) converging in Y: Indeed, compact closure gives the sequence conclusion by For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice. Conversely, fix bounded B⊆X and let K=J(B)‾. For any sequence (yj) in K, Countable Choice selects bj∈B with ∥yj−bj∥Y<1/j (start at j=1); a convergent subsequence of (bj) gives one of (yj) with the same limit, which belongs to the closed set K. Thus K is sequentially compact and For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice makes it compact. The empty case is immediate. Both readings of X⋐Y are used on this page; the second is the form in which the compactness theorems below are stated.

Continuity of the inclusion is a separate hypothesis and is never inferred from compactness: a compact operator is bounded by Compact linear operator and A bounded linear operator between normed spaces, but the definition above fixes the continuity of J in advance. On this page the continuity of every Sobolev inclusion is verified separately, through the corresponding Sobolev embedding theorem.

Remarks

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

The translation estimate for W1,p functions on Rn

Statement

Assume the Axiom of Choice. Let n≥1, 1≤p<∞, K∈{R,C} and u∈W1,p(Rn;K). For h∈Rn let τhu:=u(⋅−h), read on almost-everywhere classes. Then τhu∈W1,p(Rn) with Dj(τhu)=τhDju, and ∥τhu−u∥Lp(Rn)≤∣h∣ ∥Du∥Lp(Rn), where ∣Du∣=(∑j=1n∣Dju∣2)1/2 and ∣h∣ is the Euclidean norm of h. The estimate is a statement about classes and does not depend on the chosen representatives.

Facts & Assumptions

Given: the Axiom of Choice, n≥1, 1≤p<∞, a scalar field K∈{R,C}, a class u∈W1,p(Rn;K) and a vector h∈Rn.

[F1]

Smooth density in W1,p(Rn). Under Countable Choice, for every u∈W1,p(Rn;K) there are φm∈Cc∞(Rn;K) with ∥φm−u∥W1,p→0; Countable Choice is supplied by the assumed Axiom of Choice. (Compactly supported smooth functions are dense in W^{k,p}(R^n), The Axiom of Countable Choice (ACω))

[F2]

Fundamental theorem of calculus. For a smooth function φ, φ(x−h)−φ(x)=−∫01Dφ(x−th)⋅h dt, with the complex-valued identity read componentwise. (Fundamental theorem of calculus for absolutely continuous functions)

[F3]

Minkowski and translation isometry. Minkowski's integral inequality applies to the t-integral on [0,1], and Lebesgue translation invariance gives ∥w(⋅−th)∥p=∥w∥p for every w∈Lp. (Minkowski's integral inequality, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, The space Lp(μ) as the quotient by null functions)

[F4]

Vector gradient norm. The Lp norm of the Euclidean magnitude ∣Du∣=(∑i∣Diu∣2)1/2 is equivalent, with constants depending only on n,p, to the finite sum of the component Lp norms used in Integer-order Sobolev spaces and their norms. Indeed ∣Diu∣≤∣Du∣≤∑i∣Diu∣, so Minkowski bounds ∥∣Du∣∥p above by ∑i∥Diu∥p. Also ∣∣Dum∣−∣Du∣∣≤∣D(um−u)∣≤∑i∣Di(um−u)∣, which proves convergence in Lp when um→u in W1,p. (Integer-order Sobolev spaces and their norms)

[F5]

Weak derivatives. A locally integrable function v has weak ∂j-derivative w when ∫v ∂jψ=−∫wψ for every ψ∈Cc∞(Rn); if v,w∈Lp, this places v in W1,p in that coordinate. (Weak derivative of a locally integrable function, Integer-order Sobolev spaces and their norms)

Proof

technique · Prove the estimate for smooth approximants along line segments, then use density to pass to the $W^{1,p}$ limit. Identify the translated weak derivatives by testing against compactly supported smooth functions
1.1F1F2F3given

By [F1] and Countable Choice, choose φm∈Cc∞(Rn;K) with ∥φm−u∥W1,p→0. For a smooth φ the fundamental theorem [F2] gives φ(x−h)−φ(x)=−∫01Dφ(x−th)⋅h dt. By Cauchy--Schwarz and [F3], ∥τhφ−φ∥p≤∣h∣∫01∥∣Dφ∣(⋅−th)∥p dt=∣h∣∥Dφ∥p.

1.2F3F5given

For each coordinate j and ψ∈Cc∞(Rn), the change of variables y=x−h and the weak-derivative identity for u give ∫Rn(τhu)(x) ∂jψ(x) dx=∫Rnu(y) ∂jψ(y+h) dy=−∫RnDju(y) ψ(y+h) dy=−∫Rn(τhDju)(x) ψ(x) dx. By [F3], τhDju∈Lp, so [F5] proves τhu∈W1,p and Dj(τhu)=τhDju. Translation acts on almost-everywhere classes because it preserves null sets; hence both the derivative identity and the estimate are representative-independent.

2.1F3F4step 1.1

The convergence φm→u in W1,p gives ∥φm−u∥p→0 and, by [F4], ∥∣Dφm∣−∣Du∣∥p→0. Translation is an Lp isometry by [F3], so ∣∥τhu−u∥p−∥τhφm−φm∥p∣≤2∥u−φm∥p. Letting m→∞ in the inequality of step 1.1 proves ∥τhu−u∥p≤∣h∣∥Du∥p.

3.1step 2.1step 1.2∎

If h=0 the estimate is equality. For every h the preceding argument proves the stated estimate and translated derivative identity, with all expressions depending only on the classes in Lp.

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

Relative compactness forces uniform translation continuity in Lp

Statement

Assume the Axiom of Countable Choice. Let 1≤p<∞ and let F⊆Lp(Rn) be relatively compact, that is, its closure in Lp(Rn) is compact. In the displayed nonnegative supremum, take the value 0 if F=∅. Then sup⁡f∈F∥τhf−f∥Lp(Rn)⟶0(∣h∣→0). This is the necessity of the translation hypothesis in the Fr'echet--Kolmogorov criterion.

Facts & Assumptions

Given: the Axiom of Countable Choice, 1≤p<∞, and a relatively compact family F⊆Lp(Rn) with closure F‾.

[F1]

Compact metric spaces are totally bounded. A compact metric space is totally bounded and complete. (A compact metric space is complete and totally bounded, and neither implication uses any choice principle)

[F2]

Total boundedness is finite-net covering. A metric space (X,d) is totally bounded when for every δ>0 there are finitely many points x1,…,xN∈X with X=⋃i=1NB(xi,δ); a subset of a totally bounded space which is itself totally bounded as a subspace has the same property for every δ>0. (Finite ε-net and totally bounded metric space, Open ball, closed ball and sphere in a metric space, A totally bounded metric space is bounded, every subspace of a totally bounded space is totally bounded, and the closure of a totally bounded subset is totally bounded)

[F3]

Continuity of translation in Lp. For every g∈Lp(Rn), ∥τhg−g∥p→0 as ∣h∣→0, where τhg=g(⋅−h) acts on almost-everywhere classes. (∥τhf−f∥p→0 in Lp(Rn) as h→0, for 1≤p<∞, Translation of a function on Rn)

[F4]

Translation is an Lp-isometry. ∥τhg∥p=∥g∥p for every g∈Lp(Rn) and every h, because Lebesgue measure is translation invariant. (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, The space Lp(μ) as the quotient by null functions)

Proof

technique · Cover the compact closure by finitely many small balls, use continuity of translation at the finitely many centres, and propagate to the whole family by the isometry property and the triangle inequality
1.1F1F2given

If F=∅ the supremum is 0 and the claim holds. Otherwise the closure F‾ is a compact subset of Lp(Rn), so (F‾,dp) is a compact metric space and [F1] makes it totally bounded. Given ε>0, [F2] provides finitely many centres f1,…,fN∈F‾ with F‾⊆⋃i=1NB(fi,ε/3).

2.1F3F4step 1.1

For each i the centre fi is an element of Lp(Rn), so [F3] gives δi>0 with ∥τhfi−fi∥p<ε/3 whenever ∣h∣<δi; set δ:=min⁡iδi>0, a minimum over the nonempty finite set of indices. By [F4], ∥τh(f−fi)∥p=∥f−fi∥p for every f∈Lp(Rn) and every h.

3.1F3F4step 1.1step 2.1∎

Fix f∈F and ∣h∣<δ. By step 1.1 there is i with ∥f−fi∥p<ε/3, and the triangle inequality together with steps 2.1 and 1.1 gives ∥τhf−f∥p≤∥τh(f−fi)∥p+∥τhfi−fi∥p+∥fi−f∥p<ε/3+ε/3+ε/3=ε. Hence sup⁡f∈F∥τhf−f∥p≤ε for all ∣h∣<δ, and since ε>0 was arbitrary the supremum tends to 0. Countable Choice enters only through the continuity-of-translation interface [F3].

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

Uniformly supported families have vanishing tails

Statement

Let 1≤p<∞, let E⊆Rn be bounded and let F⊆Lp(Rn) be a family such that every f∈F vanishes Lebesgue-almost everywhere outside E. In the displayed nonnegative supremum, take the value 0 if F=∅. Then for every ε>0 there is R>0 with sup⁡f∈F∫∣x∣>R∣f(x)∣p dx<ε. In particular the tightness hypothesis of the Fr'echet--Kolmogorov criterion is automatic for families supported in one bounded set.

Facts & Assumptions

Given: 1≤p<∞, a bounded set E⊆Rn, and a family F⊆Lp(Rn) whose every member vanishes almost everywhere outside E.

[F1]

Bounded sets lie in balls. E⊆Rn is bounded in the metric sense of Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space if and only if there is a centre x0 and a radius r>0 with E⊆B(x0,r); a ball is contained in the ball about the origin of radius ∣x0∣+r. (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Open ball, closed ball and sphere in a metric space)

[F2]

Restriction to a measurable tail. If A⊆Ec is measurable and f vanishes almost everywhere on Ec, then for any measurable representative g the function ∣g∣p1A is measurable and zero almost everywhere. (The space Lp(μ) as the quotient by null functions, Measure-null sets and almost-everywhere statements relative to a measure)

[F3]

Zero integral and almost-everywhere vanishing. A nonnegative measurable h satisfies ∫h dx=0 if and only if h=0 almost everywhere. (A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere)

Proof

Proof technique: Choose a ball containing E and use the given almost-everywhere vanishing on the measurable tail outside that ball.

1.1F1F2F3given

By [F1] fix R>0 with E⊆B(0,R) and put A:={∣x∣>R}. The set A is measurable and is contained in Ec. If F=∅ then its tail supremum is 0<ε. Otherwise, for each f∈F the hypothesis says that f vanishes almost everywhere on Ec, hence on A; by [F2] the measurable function ∣g∣p1A is zero almost everywhere for any representative g of f, so [F3] gives ∫A∣f∣p dx=0.

2.1step 1.1given∎

Every member of F has tail integral 0 by step 1.1, so sup⁡f∈F∫∣x∣>R∣f∣p dx=0<ε. Since ε>0 was arbitrary, the tightness condition of the Fr'echet--Kolmogorov criterion holds. The argument uses no choice principle: R is obtained from the single bounded set E and the supremum is evaluated at the constant value 0.

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

The Fr'echet--Kolmogorov compactness criterion in Lp(Rn)

Statement

Assume the Axiom of Countable Choice and the Axiom of Dependent Choice. Let 1≤p<∞ and let F⊆Lp(Rn). In each of the three displayed nonnegative suprema, take the value 0 if F=∅. Assume F satisfies: (i) sup⁡f∈F∥f∥Lp(Rn)<∞; (ii) for every ε>0 there is R>0 with sup⁡f∈F(∫∣x∣>R∣f∣p)1/p<ε; and (iii) for every ε>0 there is δ>0 with sup⁡f∈F∥τhf−f∥Lp(Rn)<ε whenever ∣h∣<δ. Then F is totally bounded in Lp(Rn), the closure of F is compact, and every sequence in F has a subsequence converging in Lp(Rn).

Facts & Assumptions

Given: the Axioms of Countable and Dependent Choice, 1≤p<∞, and a family F⊆Lp(Rn) satisfying conditions (i)--(iii) of the statement; write M:=sup⁡f∈F∥f∥p<∞. For R>0 and f∈F put fR:=f1{∣x∣≤R}, and for each ε>0 let ηε be the radial mollifier at scale ε of A radial mollifier family in Rn.

[F1]

Mollification. For fR∈Lp⊆Lloc1, the function fR∗ηδ is smooth and ∂α(fR∗ηδ)=fR∗(∂αηδ). (Convolution with a mollifier is smooth, and derivatives pass under the integral sign)

[F2]

H"older's inequality. For conjugate exponents p,p′, ∫∣uv∣≤∥u∥p∥v∥p′; in particular ∣(fR∗ηδ)(x)∣≤∥fR∥p∥ηδ∥p′ and ∣∇(fR∗ηδ)(x)∣≤∥fR∥p∥∇ηδ∥p′ pointwise. (Holder's inequality for integrals, including the endpoint cases)

[F3]

Minkowski's integral inequality. For measurable F on a product of sigma-finite measure spaces with ∫Y∥F(⋅,y)∥p dν(y)<∞, ∥∫YF(⋅,y) dν(y)∥p≤∫Y∥F(⋅,y)∥p dν(y). (Minkowski's integral inequality)

[F4]

Translations are Lp-isometries. ∥τhg∥p=∥g∥p for every g∈Lp and every h, since Lebesgue measure is translation invariant. (Translation of a function on Rn, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, The space Lp(μ) as the quotient by null functions)

[F5]

Finite nets for equicontinuous families. An equicontinuous pointwise bounded family in C(K,R), K a compact metric space, is totally bounded for the supremum metric; the same holds after applying the statement to real and imaginary parts of a complex-valued family, and Arzelà--Ascoli for real C(K) under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded records the equivalent compact-closure form. (An equicontinuous pointwise-bounded family in C(K,R) has a finite net in the supremum metric, Arzelà--Ascoli for real C(K) under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded)

[F6]

Total boundedness and its closure. A metric space is totally bounded when it has a finite δ-net for every δ>0; total boundedness passes to subsets and to closures, and for g,h supported in a compact ball K one has ∥g−h∥p≤∥g−h∥∞∣K∣1/p. (Finite ε-net and totally bounded metric space, A totally bounded metric space is bounded, every subspace of a totally bounded space is totally bounded, and the closure of a totally bounded subset is totally bounded, Open ball, closed ball and sphere in a metric space)

Proof

technique · Make the tails small and the translations small, regularise by convolution, use Arzel\`a--Ascoli to get a finite net for the regularised family, and transfer the net back to $\mathcal F$
1.1given

If F=∅ it is totally bounded and its closure is empty. Otherwise fix ε>0. By (ii) choose R>0 with ∥f−fR∥p<ε for every f∈F, and by (iii) choose δ>0 with ∥τhf−f∥p<ε for every f∈F and every ∣h∣<δ; let η:=ηδ/2 be the radial mollifier at scale δ/2, so ∫η=1, η≥0 and supp⁡η⊆B(0,δ/2).

2.1F3F4step 1.1

For every f∈F and every ∣y∣<δ, ∥fR−τyfR∥p≤∥fR−f∥p+∥f−τyf∥p+∥τyf−τyfR∥p<3ε by step 1.1 and [F4]. Since fR−fR∗η=∫η(y)(fR−τyfR)dy, [F3] gives ∥fR−fR∗η∥p≤∫η(y)∥fR−τyfR∥p dy<3ε, and therefore ∥f−fR∗η∥p<4ε.

2.2F1F2step 1.1

By [F1] and [F2], every g=fR∗η satisfies ∥g∥∞≤M∥η∥p′ and ∥∇g∥∞≤M∥∇η∥p′, and g vanishes off the ball of radius R+δ because fR vanishes off the ball of radius R; hence the family G:={fR∗η:f∈F} is uniformly bounded and uniformly Lipschitz, and all its elements are supported in the compact ball K:=B(0,R+δ)‾.

3.1F5F6step 2.2

The real parts {Re⁡g:g∈G} are equicontinuous and pointwise bounded on the compact metric space K, and likewise the imaginary parts; by [F5] both families are totally bounded in the supremum metric, and combining the finitely many real and imaginary sup-balls, G is totally bounded in the supremum metric over K. Since every g is supported in K, the supremum over Rn equals the supremum over K, so for every ϑ>0 the family G has a finite covering by Lp-balls of radius ϑ by [F6].

4.1F6step 2.1step 3.1

Let r>0 and apply steps 1.1--3.1 with ε<r/10, covering G by finitely many Lp-balls of radius r/10. Step 2.1 then covers F by the same centres with radius r/2. Discard empty intersections with F and choose one point of F in each remaining ball. Their radius-r balls cover F by the triangle inequality, so the centres belong to F as required by [F6]. Thus F is totally bounded.

5.1F6F7step 1.1step 4.1∎

By [F6] the closure F‾ is totally bounded, and it is closed in the complete space Lp(Rn), hence complete by [F7]; a complete and totally bounded metric space is compact by [F7], and compactness is equivalent to sequential compactness by [F7], so every sequence in F has a subsequence converging in Lp(Rn). The empty case was disposed of in step 1.1, and no other choice principle is used: Countable Choice covers the completeness-to-compactness step and the finite-net selection, Dependent Choice covers the compactness-sequential equivalence.

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

Compactness of W01,p(Ω)↪Lp(Ω) on bounded open sets

Statement

Assume the Axiom of Choice. Let n≥1, let Ω⊆Rn be a bounded open set and let 1≤p<∞. Then W01,p(Ω) is compactly embedded in Lp(Ω): the inclusion is bounded, and every sequence bounded in W01,p(Ω) has a subsequence converging in Lp(Ω). No regularity of ∂Ω is needed.

Facts & Assumptions

Given: the Axiom of Choice, n≥1, a bounded open Ω⊆Rn, 1≤p<∞, and a sequence (uj) with M:=sup⁡j∥uj∥W01,p(Ω)<∞.

[F1]

Zero extension. For each j the zero extension u~j of uj lies in W1,p(Rn) with Diu~j the zero extension of Diuj and ∥u~j∥W1,p(Rn)=∥uj∥W1,p(Ω); the extension vanishes outside Ω‾. (Zero extension of W_0^{1,p} has no boundary derivative, Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms)

[F2]

Translation estimate. ∥τhu~j−u~j∥Lp(Rn)≤∣h∣ ∥Du~j∥Lp(Rn), under the Axiom of Choice. (The translation estimate for W1,p functions on Rn)

[F3]

Automatic tails. A family in Lp(Rn) whose members all vanish almost everywhere outside the fixed bounded set Ω‾ satisfies the tightness condition of the Fr'echet--Kolmogorov criterion. (Uniformly supported families have vanishing tails, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space)

[F4]

The Fr'echet--Kolmogorov criterion. A bounded family in Lp(Rn) with vanishing tails and uniform translation control is totally bounded, its closure is compact, and every sequence in it has an Lp(Rn)-convergent subsequence. (The Fr'echet--Kolmogorov compactness criterion in Lp(Rn), The Axiom of Countable Choice (ACω), The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain)

[F5]

Restriction is contractive. ∥u∥Lp(Ω)≤∥u∥W01,p(Ω), and the Lp(Ω) norm of the restriction never exceeds the Lp(Rn) norm of an extension. (Integer-order Sobolev spaces and their norms, The space Lp(μ) as the quotient by null functions)

Proof

technique · Extend by zero, verify the three Fr\'echet--Kolmogorov conditions for the extended family, extract an $L^p(\mathbb R^n)$-convergent subsequence, and restrict
1.1F1givenalgebra

By [F1] the extensions satisfy sup⁡j∥u~j∥Lp(Rn)≤M<∞, and the finite-dimensional equivalence of norms gives sup⁡j∥∣Du~j∣∥Lp(Rn)≤CnM<∞ since each component Diu~j is bounded in Lp by M; moreover each u~j vanishes outside the fixed bounded set Ω‾.

2.1F2F3F4step 1.1

The family {u~j} meets the hypotheses of [F4]: it is bounded by step 1.1; its tails vanish by [F3]; and by [F2] ∥τhu~j−u~j∥Lp(Rn)≤∣h∣ ∥∣Du~j∣∥Lp(Rn)≤CnM∣h∣, a bound uniform in j that tends to 0 with ∣h∣.

3.1F1F4F5step 2.1∎

By [F4] there is a subsequence (u~jk) converging in Lp(Rn), say to v; restricting to Ω gives ∥ujk−v∣Ω∥Lp(Ω)≤∥u~jk−v∥Lp(Rn)→0 by [F5], so (ujk) converges in Lp(Ω). Boundedness of the inclusion is the inequality ∥u∥Lp(Ω)≤∥u∥W01,p(Ω) of [F5]; the extraction uses the Countable and Dependent Choice of [F4], and the Axiom of Choice is inherited through the translation estimate [F2].

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

Compactness of W1,p(Ω)↪Lp(Ω) on bounded extension domains

Statement

Assume the Axiom of Choice. Let n≥1, let Ω⊆Rn be a bounded W1,p-extension domain (Sobolev extension domains and extension operators) and let 1≤p<∞. Then W1,p(Ω) is compactly embedded in Lp(Ω): every sequence bounded in W1,p(Ω) has a subsequence converging in Lp(Ω). Every bounded C1 domain is an example.

Facts & Assumptions

Given: the Axiom of Choice, a bounded W1,p-extension domain Ω⊆Rn, 1≤p<∞, and a sequence (uj) with M:=sup⁡j∥uj∥W1,p(Ω)<∞.

[F1]

Extension operator. There is a bounded linear E:W1,p(Ω)→W1,p(Rn) with (Eu)∣Ω=u and ∥Eu∥W1,p(Rn)≤∥E∥ ∥u∥W1,p(Ω). (Sobolev extension domains and extension operators)

[F2]

Cutoff. Since Ω‾ is compact and contained in the open set V:=B(0,R), with R>0 large enough to contain Ω‾, there is η∈Cc∞(Rn) with η=1 on Ω‾ and supp⁡η⊆V, a fixed bounded set. This construction also works for the empty domain, choosing any ball V. (A Euclidean bump for a compact set inside an open set, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space)

[F3]

Products with a smooth cutoff. If v∈W1,p(Rn) and η∈Cc∞(Rn), then ηv∈W1,p(Rn) with ∥ηv∥W1,p(Rn)≤Cη∥v∥W1,p(Rn) for a constant depending only on η and p. (Weak Leibniz rule with a smooth factor)

[F4]

Translation estimate and automatic tails. Under the Axiom of Choice, ∥τhv−v∥Lp(Rn)≤∣h∣ ∥∣Dv∣∥Lp(Rn); a family supported in one fixed bounded set has vanishing tails. (The translation estimate for W1,p functions on Rn, Uniformly supported families have vanishing tails)

[F5]

The Fr'echet--Kolmogorov criterion. A bounded family in Lp(Rn) with vanishing tails and uniform translation control has an Lp(Rn)-convergent subsequence. (The Fr'echet--Kolmogorov compactness criterion in Lp(Rn), The Axiom of Countable Choice (ACω), The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain)

[F6]

Restriction and norms. ∥g∣Ω∥Lp(Ω)≤∥g∥Lp(Rn); and (Euj)∣Ω=uj almost everywhere. (Integer-order Sobolev spaces and their norms, The space Lp(μ) as the quotient by null functions)

[F7]

Bounded C1 domains are extension domains. (Bounded C^k domains admit integer-order Sobolev extension)

Proof

technique · Extend, multiply by a fixed cutoff, apply the Fr\'echet--Kolmogorov criterion to the products, and restrict to $\Omega$
1.1F1F2F3F6given

Fix E as in [F1] and η as in [F2], and put vj:=η Euj. By [F3] each vj lies in W1,p(Rn) and is supported in the fixed bounded set supp⁡η; moreover ∥vj∥W1,p(Rn)≤Cη∥Euj∥W1,p(Rn)≤Cη∥E∥M, and vj=(Euj)∣Ω=uj almost everywhere on Ω because η=1 there.

2.1F4F5step 1.1algebra

The family {vj} satisfies the three hypotheses of [F5]: it is bounded in Lp(Rn) by step 1.1; its tails vanish by [F4] because all members are supported in the fixed bounded set supp⁡η; and [F4] gives ∥τhvj−vj∥Lp(Rn)≤∣h∣ ∥∣Dvj∣∥Lp(Rn)≤C′M′∣h∣ with C′ and M′ independent of j, so the translation control is uniform and tends to 0 with ∣h∣.

3.1F1F5F6F7step 1.1step 2.1∎

By [F5] some subsequence (vjk) converges in Lp(Rn), say to v; restricting and using vjk=ujk almost everywhere on Ω together with [F6] gives ∥ujk−v∣Ω∥Lp(Ω)=∥(vjk−v)∣Ω∥Lp(Ω)≤∥vjk−v∥Lp(Rn)→0, so (ujk) converges in Lp(Ω). This proves compactness of the inclusion; its boundedness follows from ∥u∥Lp(Ω)≤∥Eu∥Lp(Rn)≤∥E∥ ∥u∥W1,p(Ω) by [F1] and [F6]. Finally, [F7] says every bounded C1 domain carries such an extension operator, giving the stated examples. The Axiom of Choice is inherited through [F1], [F4] and [F7], while the extraction uses the Countable and Dependent Choice of [F5].

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Poincare-Wirtinger on bounded connected extension domains by Rellich compactness

Statement

Assume the Axiom of Choice. Let n≥1, let Ω⊆Rn be a nonempty bounded connected extension domain, and let 1≤p<∞. Then there is C=C(Ω,p) such that every u∈W1,p(Ω;K) satisfies ∥u−uΩ∥Lp(Ω)≤C∥Du∥Lp(Ω),uΩ:=∣Ω∣−1∫Ωu(x) dx.

Facts & Assumptions

Given: the Axiom of Choice, a nonempty bounded connected extension domain Ω⊆Rn of finite positive measure, 1≤p<∞, and the mean uΩ of u. Nonempty openness supplies a ball inside Ω, and boundedness supplies a containing ball; thus 0<∣Ω∣<∞ by Euclidean balls have positive finite Lebesgue measure.

[F1]

Rellich compactness. Every sequence bounded in W1,p(Ω) has a subsequence converging in Lp(Ω). (Compactness of W1,p(Ω)↪Lp(Ω) on bounded extension domains, Sobolev extension domains and extension operators)

[F2]

The mean is continuous for the Lp norm. ∣uΩ∣≤∣Ω∣−1/p∥u∥Lp and hence ∣∫Ω(uj−u)∣≤∣Ω∣1−1/p∥uj−u∥Lp(Ω), by H"older's inequality on the finite-measure set Ω. (Holder's inequality for integrals, including the endpoint cases, The space Lp(μ) as the quotient by null functions)

[F3]

Weak gradients vanish when tested against convergent subsequences. If uj→u in Lp(Ω) and ∥Duj∥Lp(Ω)→0, then for every φ∈Cc∞(Ω) and every coordinate i, ∫Ωu ∂iφ=lim⁡j∫Ωuj∂iφ=−lim⁡j∫ΩDiuj φ=0. (Integer-order Sobolev spaces and their norms, The space Lp(μ) as the quotient by null functions)

[F4]

Zero gradient implies constancy on components. If u∈Wloc1,p(Ω) and Diu=0 almost everywhere for all i, then u is almost everywhere constant on each connected component of Ω. (Zero weak gradient gives componentwise constants)

Proof

technique · suppose no constant exists, normalise a violating sequence, extract a strongly convergent subsequence by Rellich, and use the vanishing gradient to force the limit to be a constant, contradicting unit norm
1.1F2givenassume-contra

Suppose the assertion fails: for every j≥1 there is vj∈W1,p(Ω) with ∥vj−(vj)Ω∥p>j∥Dvj∥p, and vj is not almost everywhere constant. Put uj:=(vj−(vj)Ω)/∥vj−(vj)Ω∥p; then the mean of uj is 0, ∥uj∥Lp=1, and ∥Duj∥p≤1/j, so ∥uj∥W1,p(Ω)≤(1+n/jp)1/p≤(1+n)1/p for all j.

2.1F1F2step 1.1

By [F1] there is a subsequence ujk→u in Lp(Ω). By [F2] and step 1.1 the means pass to the limit, so ∫Ωu=0; and ∥u∥Lp=1 because ∣∥ujk∥p−∥u∥p∣≤∥ujk−u∥p→0.

3.1F3F4step 2.1discharge-contradiction∎

By [F3] applied to the convergent subsequence of step 2.1 with ∥Dujk∥p≤1/jk→0, ∫Ωu ∂iφ=0 for every test function φ and every i, so Diu=0 almost everywhere and u∈W1,p(Ω); [F4] then makes u an almost everywhere constant on the connected Ω, and since its mean is 0 that constant is 0, contradicting ∥u∥Lp=1 from step 2.1. Hence the constant C exists. Countable Choice selects the violating sequence in step 1.1; the assumed Axiom of Choice also supplies [F1] and [F4].

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Local Lp compactness of Wloc1,p-bounded sequences

Statement

Assume the Axiom of Choice. Let n≥1, let Ω⊆Rn be open and let 1≤p<∞. Let (uj) be a sequence that is bounded in Wloc1,p(Ω): for every Ω′⋐Ω there is CΩ′ with ∥uj∥W1,p(Ω′)≤CΩ′ for all j. Then (uj) has a subsequence converging in Llocp(Ω).

If additionally 1<p<∞, the limit of that subsequence can be chosen in Wloc1,p(Ω), and then it is the limit in Llocp(Ω). For p=1 membership of the limit in Wloc1,1(Ω) is not asserted: the weak-compactness argument below uses the reflexivity of Lp, which fails at p=1, and a weak limit of L1 gradients need not be an L1 function. The compactness conclusion itself is proved for every 1≤p<∞.

Facts & Assumptions

Given: the Axiom of Choice, an open set Ω⊆Rn, 1≤p<∞, and a sequence (uj) bounded in Wloc1,p(Ω).

[F1]

Countable relatively compact ball cover. The rational balls B(q,r) with q∈Qn, r∈Q>0 and B(q,r)‾⊆Ω form a countable cover of Ω: given x∈Ω, openness and density of Qn and Q give such a ball containing x. Countability follows from countability of Q and finite products, and each closed ball is compact by Heine--Borel. (The rationals embed densely in the reals, Q is countably infinite, A product of two at most countable sets is at most countable, Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line)

[F2]

Rellich on smooth balls. Every bounded ball is a bounded smooth extension domain, and every sequence bounded in W1,p on such a ball has a subsequence converging in Lp. (Bounded C^k domains admit integer-order Sobolev extension, Compactness of W1,p(Ω)↪Lp(Ω) on bounded extension domains)

[F3]

Reflexivity of Lp. Assume Countable Choice. For 1<p<∞ and every measure space, Lp is reflexive, so every norm-bounded sequence in Lp has a weakly convergent subsequence under the ultrafilter lemma, Dependent Choice and Hahn--Banach, all supplied by the Axiom of Choice. (Reflexivity of Lp for one less p less infinity, Reflexivity is equivalent to weak subsequential compactness of bounded sequences, Weak convergence of nets and sequences, The Axiom of Countable Choice (ACω), The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain)

[F4]

Weak derivatives pass to weak limits. If uk⇀u and Diuk⇀vi in Lp(B), 1<p<∞, then u∈W1,p(B) with Diu=vi: test against φ∈Cc∞(B) and pass to the limit in ∫uk∂iφ=−∫Diukφ. (Integer-order Sobolev spaces and their norms, Weak derivative of a locally integrable function, Weak convergence of nets and sequences)

Proof

technique · Extract successively on a countable cover by relatively compact rational balls and diagonalise. For $p>1$, use weak compactness on each ball to identify the local limit's weak derivatives
1.1F1F2given

If Ω=∅, the assertions are immediate. Otherwise enumerate the countable cover in [F1] as (Bm)m≥1. The local boundedness hypothesis makes (uj) bounded in W1,p(B1), so [F2] gives a subsequence converging in Lp(B1). Recursively, after obtaining a subsequence converging on B1,…,Bm, apply [F2] to that subsequence on Bm+1 and retain a further subsequence converging there. Countable and Dependent Choice select these nested subsequences. The diagonal sequence (vj), taking the j-th term of the j-th subsequence, is eventually a subsequence of each stage; hence it converges in Lp(Bm) for every m.

2.1F1F5step 1.1

Let um be the Lp(Bm) limit of (vj). On every overlap Bm∩Bℓ the limits um and uℓ agree almost everywhere, by uniqueness of limits of the same sequence in Lp(Bm∩Bℓ). Since the cover is countable, these compatible classes patch to a class u∈Llocp(Ω). If Ω′⋐Ω, compactness and [F5] give a finite subcover Ω′⊆⋃m∈JBm; therefore ∥vj−u∥Lp(Ω′)p≤∑m∈J∥vj−u∥Lp(Bm)p⟶0. Thus vj→u in Llocp(Ω).

3.1F2F3F4step 1.1step 2.1∎

Suppose 1<p<∞ and fix m. The sequence (vj) is bounded in W1,p(Bm). By [F3], after finitely many further subsequence extractions, its function and each of its n weak derivatives converge weakly in Lp(Bm), say vjk⇀w and Divjk⇀wi. Step 2.1 gives strong convergence to um in Lp(Bm), so the weak limit is w=um. Passing to the limit in the weak-derivative identities against each φ∈Cc∞(Bm) and using [F4] gives um∈W1,p(Bm) with Dium=wi. This argument may use a further subsequence depending on m: it identifies the already fixed strong limit um, so it identifies the derivatives of the already fixed limit on every ball without changing the diagonal sequence of step 1.1. On overlaps these derivative classes agree by the weak test identity. For any Ω′⋐Ω, choose a finite ball subcover of Ω′‾ and an ambient smooth partition equal to one near that compact set (Finite ambient partitions near compact sets). Testing after multiplication by the partition pieces proves the weak derivative identity on Ω′; the partition-gradient terms sum to zero, and the finitely many local Lp bounds give global Lp(Ω′) bounds. Thus u∈Wloc1,p(Ω). For p=1 this weak-compactness step is unavailable, and no membership of the limit in Wloc1,1 is asserted. The assumed Axiom of Choice supplies the countable selections in step 1.1 and the Countable and Dependent Choice interfaces of [F2] and [F3].

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Subcritical compactness for W01,p on arbitrary bounded open sets

Statement

Assume the Axiom of Choice. Let n≥2, let Ω⊆Rn be a bounded open set with no boundary regularity assumed, let 1≤p<n and p∗=npn−p. For every 1≤q<p∗ the space W01,p(Ω) is compactly embedded in Lq(Ω): every sequence bounded in W01,p(Ω) has a subsequence converging in Lq(Ω).

Facts & Assumptions

Given: the Axiom of Choice, a bounded open set Ω⊆Rn, 1≤p<n, p∗=np/(n−p), a target exponent 1≤q<p∗, and a sequence (uj) with M:=sup⁡j∥uj∥W01,p(Ω)<∞.

[F2]

Sobolev inequality for W01,p of an arbitrary bounded open set. ∥v∥Lp∗(Ω)≤C∥Dv∥Lp(Ω) for all v∈W01,p(Ω), with C=C(n,p); the zero extensions of the uj therefore satisfy ∥uj∥Lp∗(Ω)≤CM. (The Sobolev inequality for zero-boundary Sobolev closures on open sets, Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms, The Sobolev conjugate exponent and the scaling identity)

[F5]

For p=1, supply the endpoint separately: the zero extension v of a zero-boundary class belongs to W1,1(Rn). Choose smooth compactly supported vj→v in W1,1 by Compactly supported smooth functions are dense in W^{k,p}(R^n). The endpoint The p=1 Gagliardo-Nirenberg-Sobolev inequality on differences makes (vj) Cauchy in Ln/(n−1); completeness and the almost-everywhere subsequence theorem identify this limit with v, since vj→v in L1 as well. Passing to the limit in the endpoint inequality gives ∥v∥n/(n−1)≤C∥Dv∥1, and restriction supplies the bound asserted in [F2]. (Riesz-Fischer completeness of Lp for 1≤p≤∞, Assuming Countable Choice, Lp-convergent sequences have almost-everywhere convergent subsequences)

[F3]

Lyapunov interpolation and H"older. For p≤b<p∗ and 1/b=θ/p+(1−θ)/p∗, ∥g∥b≤∥g∥pθ∥g∥p∗1−θ; for b≤p, ∥g∥Lb(Ω)≤∣Ω∣1/b−1/p∥g∥Lp(Ω). (Lyapunov interpolation inequality for Lp norms, Holder's inequality for integrals, including the endpoint cases, The space Lp(μ) as the quotient by null functions)

[F4]

Under Countable Choice, each Lq(Ω), 1≤q<∞, is complete. (Riesz-Fischer completeness of Lp for 1≤p≤∞)

Proof

technique · direct
1.1F1F2F5given

If Ω=∅, all classes are zero and the claim is immediate. Otherwise, by [F1] extract a subsequence converging in Lp(Ω); write wk:=ujk. By [F2] the differences satisfy ∥wk−wℓ∥Lp∗(Ω)≤2CM for all k,ℓ.

2.1F1F3F4step 1.1

Fix 1≤q<p∗. If q≤p, then [F3] gives ∥wk−wℓ∥Lq(Ω)≤∣Ω∣1/q−1/p∥wk−wℓ∥Lp(Ω)→0 by step 1.1; at q=p the factor is 1. If p<q<p∗, choose θ∈(0,1) with 1/q=θ/p+(1−θ)/p∗; [F3] gives ∥wk−wℓ∥Lq(Ω)≤∥wk−wℓ∥Lp(Ω)θ(2CM)1−θ→0 by step 1.1. In both cases (wk) is Cauchy in Lq(Ω), hence converges there by [F4].

3.1F1F2step 1.1step 2.1∎

Every bounded sequence in W01,p(Ω) therefore has a subsequence convergent in Lq(Ω), which is the asserted compact embedding; no property of ∂Ω was used. The Axiom of Choice is inherited through [F1] and the supplier [F2]; the extraction uses the Countable and Dependent Choice of [F1].

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The Rellich--Kondrachov theorem for 1≤p<n on bounded extension domains

Statement

Assume the Axiom of Choice. Let n≥2, let Ω⊆Rn be a bounded extension domain, let 1≤p<n and p∗=npn−p. Then for every 1≤q<p∗ the inclusion W1,p(Ω)↪Lq(Ω) is bounded and compact: every sequence bounded in W1,p(Ω) has a subsequence converging in Lq(Ω).

Facts & Assumptions

Given: the Axiom of Choice, a bounded extension domain Ω⊆Rn, 1≤p<n, p∗=np/(n−p), a target exponent 1≤q<p∗, and a sequence (uj) with M:=sup⁡j∥uj∥W1,p(Ω)<∞.

[F2]

Sobolev embedding on bounded extension domains. ∥v∥Lp∗(Ω)≤C∥v∥W1,p(Ω) for all v∈W1,p(Ω), C=C(n,p,Ω), hence ∥uj∥Lp∗(Ω)≤CM and ∥uk−uℓ∥Lp∗(Ω)≤2CM. (Sobolev embedding on bounded extension domains for p<n, The Sobolev conjugate exponent and the scaling identity, Integer-order Sobolev spaces and their norms)

[F5]

For p=1, supply the endpoint separately: take the given extension v=Eu∈W1,1(Rn) and smooth compactly supported vj→v in W1,1 by Compactly supported smooth functions are dense in W^{k,p}(R^n). The endpoint The p=1 Gagliardo-Nirenberg-Sobolev inequality on differences makes (vj) Cauchy in Ln/(n−1); completeness and almost-everywhere subsequences identify this limit with v, since vj→v in L1. Passing to the limit gives ∥v∥n/(n−1)≤C∥Dv∥1≤C∥E∥∥u∥W1,1, and restriction gives [F2]. (Riesz-Fischer completeness of Lp for 1≤p≤∞, Assuming Countable Choice, Lp-convergent sequences have almost-everywhere convergent subsequences)

[F3]

Interpolation and H"older. For p≤b<p∗, ∥g∥b≤∥g∥pθ∥g∥p∗1−θ with 1/b=θ/p+(1−θ)/p∗; for b≤p, ∥g∥Lb(Ω)≤∣Ω∣1/b−1/p∥g∥Lp(Ω). (Lyapunov interpolation inequality for Lp norms, Holder's inequality for integrals, including the endpoint cases, The space Lp(μ) as the quotient by null functions)

[F4]

Under Countable Choice, each Lq(Ω), 1≤q<∞, is complete. (Riesz-Fischer completeness of Lp for 1≤p≤∞)

Proof

technique · direct
1.1F1F2F5given

If Ω=∅, all classes are zero and the claim is immediate. Otherwise, by [F1] extract a subsequence converging in Lp(Ω) and write wk:=ujk; by [F2] the differences satisfy ∥wk−wℓ∥Lp∗(Ω)≤2CM.

2.1F1F3F4step 1.1

Fix 1≤q<p∗. If q≤p then ∥wk−wℓ∥Lq≤∣Ω∣1/q−1/p∥wk−wℓ∥Lp→0 by [F3] and step 1.1; if p<q<p∗ then [F3] gives ∥wk−wℓ∥Lq≤∥wk−wℓ∥Lpθ(2CM)1−θ→0. Completeness [F4] therefore makes (wk) converge in Lq(Ω).

3.1F1F2F3step 1.1step 2.1∎

Boundedness of the inclusion holds for every q≥p by [F2] and the interpolation bound of [F3], and for q<p by H"older's inequality of [F3]; compactness is the extraction just proved, so W1,p(Ω)⋐Lq(Ω) in the sense of Compactly embedded normed spaces. The Axiom of Choice is inherited through [F1] and the supplier [F2].

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Rellich--Kondrachov at the critical source exponent p=n

Statement

Assume the Axiom of Choice. Let n≥2 and let Ω⊆Rn be a bounded extension domain. Then W1,n(Ω) is compactly embedded in Lq(Ω) for every finite q: for each fixed 1≤q<∞, every sequence bounded in W1,n(Ω) has a subsequence converging in Lq(Ω). There is no claim of compactness into L∞.

Facts & Assumptions

Given: the Axiom of Choice, a bounded extension domain Ω⊆Rn, n≥2, and a sequence (uj) with M:=sup⁡j∥uj∥W1,n(Ω)<∞.

[F2]

Critical embedding into every finite Lq′. For every finite q′>1 there is C(q′) with ∥v∥Lq′(Ω)≤C(q′)∥v∥W1,n(Ω) for all v∈W1,n(Ω); the sequence is therefore uniformly bounded in Lq′(Ω) for each fixed finite q′. (Higher-order Sobolev embedding (case k=1, p=n), Integer-order Sobolev spaces and their norms)

[F3]

Interpolation and H"older. For n≤b<q′, ∥g∥b≤∥g∥nθ∥g∥q′1−θ with 1/b=θ/n+(1−θ)/q′; for b≤n, ∥g∥Lb(Ω)≤∣Ω∣1/b−1/n∥g∥Ln(Ω). (Lyapunov interpolation inequality for Lp norms, Holder's inequality for integrals, including the endpoint cases, The space Lp(μ) as the quotient by null functions)

[F4]

Under Countable Choice, each Lq(Ω), 1≤q<∞, is complete. (Riesz-Fischer completeness of Lp for 1≤p≤∞)

Proof

technique · direct
1.1F1F2given

If Ω=∅, all classes are zero and the claim is immediate. Otherwise fix 1≤q<∞ and choose q′>max⁡{q,n}. By [F1] extract a subsequence converging in Ln(Ω) and write wk:=ujk; by [F2] the differences satisfy ∥wk−wℓ∥Lq′(Ω)≤2C(q′)M.

2.1F1F3F4step 1.1

If q≤n, then ∥wk−wℓ∥Lq≤∣Ω∣1/q−1/n∥wk−wℓ∥Ln→0 by [F3] and step 1.1. If q>n, then n<q<q′ and [F3] gives ∥wk−wℓ∥Lq≤∥wk−wℓ∥Lnθ(2C(q′)M)1−θ→0 for the corresponding θ∈(0,1). In both cases (wk) is Cauchy, hence convergent by [F4], in Lq(Ω).

3.1F1F2step 1.1step 2.1∎

Every bounded sequence in W1,n(Ω) therefore has a subsequence converging in Lq(Ω) for each fixed finite q, so W1,n(Ω)⋐Lq(Ω) in the sense of Compactly embedded normed spaces. No compactness into L∞ is asserted. The proof uses only finite target exponents. The Axiom of Choice is inherited through [F1] and the supplier [F2].

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Morrey--Rellich compactness for p>n

Statement

Assume the Axiom of Choice. Let n≥2, let Ω⊆Rn be a bounded extension domain, let n<p<∞ and α=1−np. Replace every u∈W1,p(Ω) by its continuous Morrey representative u∗ (Morrey's inequality for p>n). Then every sequence bounded in W1,p(Ω) has a subsequence whose representatives converge in C0,β(Ω‾) for every 0≤β<α; in particular W1,p(Ω) is compactly embedded in every C0,β(Ω‾), 0≤β<α, and in Lq(Ω) for every 1≤q<∞.

Facts & Assumptions

Given: the Axiom of Choice, a bounded extension domain Ω⊆Rn, n<p<∞, α=1−n/p, and a sequence (uj) with M:=sup⁡j∥uj∥W1,p(Ω)<∞.

[F1]

Extension and cutoff. There are a bounded extension operator E and a fixed η∈Cc∞(Rn) with η=1 on Ω; the products vj:=η Euj lie in W1,p(Rn), are supported in the fixed compact set supp⁡η, satisfy ∥vj∥W1,p(Rn)≤C1M, and equal uj almost everywhere on Ω. (Sobolev extension domains and extension operators, A Euclidean bump for a compact set inside an open set, Weak Leibniz rule with a smooth factor, Integer-order Sobolev spaces and their norms)

[F2]

Morrey's inequality on the compact set used here. Each vj in [F1] is supported in a fixed compact set K. Choose one ball B(x,R) containing K∪Ω‾. The supplier's local estimate on B(x,R), with B(x,2R)⊂Rn, gives [vj∗]C0,α(B(x,R))≤C2∥Dvj∥Lp(B(x,2R)) for the continuous representative. Its average on B(x,R) has modulus at most ∣B(x,R)∣−1/p∥vj∥Lp(B(x,R)), so the same oscillation estimate also bounds ∥vj∗∥L∞(B(x,R)) by C3∥vj∥W1,p(Rn). Continuous representatives are unique because continuous functions equal almost everywhere on an open ball are equal everywhere there. Thus both norms on Ω‾ are bounded by C4∥vj∥W1,p(Rn), with constants depending only on n,p,R. (Morrey's inequality for p>n, Local Hölder and scaled C-two-alpha norms on balls, The space Lp(μ) as the quotient by null functions)

[F3]

Arzel`a--Ascoli. A uniformly bounded, equicontinuous family of real functions on a compact metric space has a uniformly convergent subsequence; for complex-valued families apply this to real and imaginary parts. (Arzelà--Ascoli for real C(K) under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded)

[F4]

H"older interpolation. For a bounded function g:K→K on any set K⊆Rn, write [g]C0,γ(K):=sup⁡x≠y∈K∣g(x)−g(y)∣/∣x−y∣γ and ∥g∥C0,γ(K):=∥g∥L∞(K)+[g]C0,γ(K). For 0<β<α and θ=β/α, the bound ∣g(x)−g(y)∣≤min⁡{2∥g∥L∞(K),[g]C0,α(K)∣x−y∣α} gives [g]C0,β(K)≤21−θ∥g∥L∞(K)1−θ[g]C0,α(K)θ, and hence ∥g∥C0,β(K)≤∥g∥L∞(K)+21−θ∥g∥L∞(K)1−θ[g]C0,α(K)θ. These follow from min⁡{A,B}≤A1−θBθ for A,B≥0; the seminorm and supremum conventions agree with Local Hölder and scaled C-two-alpha norms on balls.

Proof

technique · extend, cut off, apply Morrey's inequality for uniform $C^{0,\alpha}$ bounds, extract uniformly convergent representatives by Arzel\`a--Ascoli, and interpolate down to $C^{0,\beta}$
1.1F1F2given

If Ω=∅ the claim is immediate. Otherwise, by [F1] and [F2] each vj∗ satisfies ∥vj∗∥L∞(Ω‾)≤C3C1M and [vj∗]C0,α(Ω‾)≤C2C1M; hence the family {vj∗∣Ω‾} is uniformly bounded and α-H"older, in particular equicontinuous, on the compact set Ω‾.

2.1F3step 1.1

By [F3] applied to the real and imaginary parts on the compact metric space Ω‾, a subsequence of (vj∗∣Ω‾) converges uniformly, that is, in C0,0(Ω‾); along it the C0,α seminorms stay bounded by step 1.1.

3.1F1F2F4step 1.1step 2.1∎

Fix 0<β<α and put θ=β/α. For the differences g=vk∗−vℓ∗ of the uniformly convergent subsequence, step 1.1 gives [g]C0,α(Ω‾)≤2C2C1M, while ∥g∥L∞(Ω‾)→0. By [F4], ∥g∥C0,β(Ω‾)≤∥g∥L∞(Ω‾)+21−θ∥g∥L∞(Ω‾)1−θ(2C2C1M)θ⟶0, If v is the uniform limit, passing to the limit in each difference quotient shows [v]C0,α≤C2C1M. Apply the same estimate to g=vk∗−v to obtain convergence in C0,β(Ω‾); the case β=0 is step 2.1. Since vj∗=uj almost everywhere on Ω by [F1] and [F2], this is the convergence of the Morrey representatives of the uj, and uniform convergence on the bounded Ω‾ implies convergence in Lq(Ω) for every 1≤q<∞, so W1,p(Ω) is compactly embedded in each C0,β(Ω‾), β<α, and in each Lq(Ω), 1≤q<∞. The Axiom of Choice is inherited through the extension operator and Morrey's inequality.

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

Higher-order Rellich--Kondrachov compactness

Statement

Assume the Axiom of Choice. Let n≥2, let Ω⊆Rn be a bounded extension domain, let k>m≥0 be integers and 1≤p<∞; write pk−m∗=npn−(k−m)p when (k−m)p<n.

  • (a) If (k−m)p<n, then Wk,p(Ω) is compactly embedded in Wm,q(Ω) for every 1≤q<pk−m∗.
  • (b) If (k−m)p≥n, then Wk,p(Ω) is compactly embedded in Wm,q(Ω) for every finite q.
  • (c) If (k−m)p>n and 0≤β<1 with β<k−m−np, then every sequence bounded in Wk,p(Ω) has a subsequence whose representatives converge in Cm,β(Ω‾).

Facts & Assumptions

Given: the Axiom of Choice, a bounded extension domain Ω⊆Rn, integers k>m≥0, 1≤p<∞, and a sequence (uj) with M:=sup⁡j∥uj∥Wk,p(Ω)<∞.

[F1]

Lower-order derivatives. For every ∣α∣≤m, the sequence (Dαuj) is bounded in W1,p(Ω), with norm at most CM, because k−∣α∣≥1. (Weak partial derivatives lower the Sobolev order, Integer-order Sobolev spaces and their norms, The notation Hk and the reserved zero-boundary symbol)

[F2]

First-order Lp compactness. On a bounded extension domain, every sequence bounded in W1,p(Ω) has a subsequence converging in Lp(Ω). (Compactness of W1,p(Ω)↪Lp(Ω) on bounded extension domains)

[F3]

Higher-order continuous embeddings. Applying the higher-order Sobolev embedding on the support ball B to each Dαvj, where vj=ηEuj is the compactly supported extension in step 1.1, gives, uniformly in j and ∣α∣≤m, an Lpk−m∗ bound when (k−m)p<n, a bound in every finite Lr when (k−m)p≥n, and a Cm,β′(Ω‾) bound for (uj) by restriction when (k−m)p>n and 0<β′<min⁡{1,k−m−np}. For derivatives with ∣α∣<m, the remaining Sobolev order is larger; finite-measure inclusion handles any stronger resulting integrability. (Weak partial derivatives lower the Sobolev order, Higher-order Sobolev embedding, Sobolev extension domains and extension operators, Compactly embedded normed spaces)

[F4]

Finite-measure inclusion, interpolation, and completeness. If 1≤q≤p, then ∥g∥Lq(Ω)≤∣Ω∣1/q−1/p∥g∥Lp(Ω); if p<q<r<∞ and 1/q=λ/p+(1−λ)/r, then ∥g∥Lq≤∥g∥Lpλ∥g∥Lr1−λ. Each Lq(Ω) is complete for 1≤q<∞. (Holder's inequality for integrals, including the endpoint cases, Lyapunov interpolation inequality for Lp norms, Riesz-Fischer completeness of Lp for 1≤p≤∞, The space Lp(μ) as the quotient by null functions)

[F5]

Weak derivatives pass to strong limits. If fj→f and Difj→gi in Lq(Ω) for 1≤q<∞, passing to the limit in the test identity shows Dif=gi weakly. Iterating gives the same conclusion for all derivatives of order at most m. (Weak derivative of a locally integrable function, Integer-order Sobolev spaces and their norms)

[F7]

Uniform limits preserve classical derivatives. If C1 functions and their first derivatives converge uniformly on compact balls, the limit is C1 there and its derivatives are the corresponding limits; apply the fundamental theorem of calculus on line segments, coordinate by coordinate. (Fundamental theorem of calculus for absolutely continuous functions, Ck maps and multi-index derivative notation in Euclidean space)

[F8]

H"older interpolation. For 0<β<β′<1, a uniformly convergent sequence with uniformly bounded C0,β′ seminorms converges in C0,β. Indeed, the difference quotient is bounded by the minimum of 2∥g∥∞∣x−y∣−β and [g]C0,β′∣x−y∣β′−β, yielding the usual interpolation estimate with a constant depending on β,β′. (Local Hölder and scaled C-two-alpha norms on balls)

Proof

technique · extract strong $L^p$ convergence for the finite derivative family, interpolate against higher-order Sobolev bounds, and use Arzel\`a--Ascoli for the supercritical Hölder conclusion
1.1F1F2F3given

If Ω=∅, all target spaces are trivial and the assertions hold. Otherwise choose the bounded extension E at (k,p) and a smooth cutoff η equal to one near Ω‾, with compact support in a ball B. By the weak Leibniz rule, vj=ηEuj is bounded in Wk,p(Rn) and equals uj on Ω (A Euclidean bump for a compact set inside an open set, Weak Leibniz rule with a smooth factor). For each ∣α∣≤m, Dαvj is bounded in W1,p(B) by [F1]. The smooth ball is a W1,p-extension domain (Bounded C^k domains admit integer-order Sobolev extension), so [F2] applied on B, followed by restriction to Ω, gives a common subsequence on which all Dαuj converge in Lp(Ω). Only finitely many derivative sequences are extracted.

2.1F3F4F5step 1.1

Consider cases (a) and (b). By [F3], for every ∣α∣≤m the sequence (Dαujk) is bounded in Lpk−m∗(Ω) in case (a), and in every finite Lr(Ω) in case (b). If q≤p, finite- measure inclusion [F4] and step 1.1 make each derivative sequence Cauchy in Lq. If p<q<pk−m∗ in case (a), choose r=pk−m∗; if p<q<∞ in case (b), choose any finite r>q. Lyapunov interpolation [F4] applied to differences, whose Lp norms tend to zero by step 1.1 and whose Lr norms are uniformly bounded by [F3], makes every derivative sequence Cauchy in Lq. Completeness of Lq gives limits vα. Passing to the limit in the weak derivative test identities by [F5] shows that vα=Dαv0 for all ∣α∣≤m, so ujk→v0 in Wm,q(Ω). The higher-order embedding [F3] also gives boundedness of the inclusion into each stated target, so this is compact embedding.

3.1F3F6F7F8step 1.1∎

Consider case (c), and fix 0≤β<min⁡{1,k−m−np}. Choose β′ with β<β′<min⁡{1,k−m−np}. By [F3] the sequence is bounded in Cm,β′(Ω‾), so each of its finitely many derivative families of orders at most m is uniformly bounded and equicontinuous on the compact set Ω‾. By [F6], applying Arzel`a--Ascoli successively to these derivative families gives a common subsequence on which every Dαujk converges uniformly to a continuous function vα on Ω‾. On each ball compactly contained in Ω, [F7] applied to line segments shows that vα+ei is the classical i-th derivative of vα whenever ∣α∣<m; hence v0 is a representative in Cm(Ω) whose derivatives through order m extend continuously to Ω‾. For β=0 the uniform convergence is the desired Cm,0 convergence. For β>0, [F8] applied to each difference Dα(ujk−ujℓ), using uniform convergence and the uniform C0,β′ bounds, gives convergence in Cm,β(Ω‾) to v0: each uniform limit vα retains the bounded β′ seminorm by passage to the limit in the pointwise difference quotients, so [F8] applies directly to Dαujk−vα. Thus the asserted compact embedding holds, and the Axiom of Choice supplies the subsequence and the choice interfaces of [F2] and [F6].

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

Bounded Sobolev sequences have strongly convergent subsequences with the weak limit as limit

Statement

Assume the Axiom of Choice. Let n≥2, let Ω⊆Rn be a bounded extension domain, let 1≤p<n, p∗=npn−p, and let (uj) be bounded in W1,p(Ω) with uj⇀u weakly in W1,p(Ω). Then uj→u in Lq(Ω) for every 1≤q<p∗; the convergence is of the whole sequence, not merely of a subsequence.

Facts & Assumptions

Given: the Axiom of Choice, a bounded extension domain Ω⊆Rn, 1≤p<n, a bounded sequence (uj) with uj⇀u weakly in W1,p(Ω), and 1≤q<p∗.

[F1]

Rellich--Kondrachov. Every bounded sequence in W1,p(Ω) has a subsequence converging in Lq(Ω). (The Rellich--Kondrachov theorem for 1≤p<n on bounded extension domains, Sobolev extension domains and extension operators)

[F2]

Weak convergence tested against Lp′. For g∈Lp′(Ω) with 1/p+1/p′=1, the functional v↦∫Ωgv is bounded on W1,p(Ω), so ∫Ωguj→∫Ωgu. In particular ∫Auj→∫Au for every measurable A of finite measure. For complex-valued classes, these integral identities are read componentwise. (Weak convergence of nets and sequences, Holder's inequality for integrals, including the endpoint cases, The space Lp(μ) as the quotient by null functions)

[F3]

Strong convergence tested against Lq′. If vk→v in Lq(Ω) and g∈Lq′(Ω), then ∫gvk→∫gv by H"older's inequality. (Holder's inequality for integrals, including the endpoint cases)

Proof

Proof technique: every subsequence has a further Lq-convergent subsequence; identify its limit with the weak limit u by testing against finite-measure set indicators; conclude that the whole sequence converges.

1.1F1given

Let (ujk) be any subsequence. It is bounded in W1,p(Ω), so by [F1] it has a further subsequence (ujkr) converging in Lq(Ω) to some v.

2.1F2F3step 1.1algebra

For every measurable A⊆Ω of finite measure, [F2] applied to g=1A gives ∫Aujkr→∫Au, while [F3] applied to g=1A∈Lq′(Ω) gives ∫Aujkr→∫Av; hence ∫A(u−v)=0. For each m≥1, apply this to the sets where the real or imaginary part of u−v is greater than 1/m or less than −1/m, intersected with B(0,m). Each such set has measure zero, since the corresponding signed part of the integral has magnitude at least its measure divided by m. As Ω is bounded, these sets cover the nonzero real and imaginary parts, so u=v almost everywhere on Ω.

3.1F1step 2.1∎

Thus every subsequence of (uj) has a further subsequence converging in Lq(Ω) to the same limit u; in a metric space this forces the whole sequence to converge to u, because otherwise some ε>0 would admit a subsequence staying ε-away from u, and that subsequence would in turn have a further subsequence converging to u. The Axiom of Choice is inherited through [F1], and the weak topology is Hausdorff as recorded in Weak topology is hausdorff.

Remarks

The identification of the strong limit with the weak limit does not use the density of test functions in Lq′ for q=1: the finite-measure indicator test functions lie in Lp′∩Lq′ and separate almost-everywhere classes.

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Weak H1 convergence plus compactness gives strong L2 convergence

Statement

Assume the Axiom of Choice. Let n≥1, let Ω⊆Rn be a bounded extension domain and let (uj) be bounded in H1(Ω)=W1,2(Ω) with uj⇀u weakly in H1(Ω). Then uj→u in L2(Ω).

Facts & Assumptions

Given: the Axiom of Choice, a bounded extension domain Ω⊆Rn, n≥1, and a bounded sequence (uj) with uj⇀u weakly in H1(Ω).

[F1]

First-order Rellich compactness. Since H1(Ω)=W1,2(Ω), every bounded sequence in H1(Ω) has a subsequence converging in L2(Ω) on this bounded extension domain. (Compactness of W1,p(Ω)↪Lp(Ω) on bounded extension domains, The notation Hk and the reserved zero-boundary symbol)

[F2]

Finite-measure indicators test both limits. For each measurable A⊆Ω, the functional v↦∫Av is bounded on H1(Ω) and on L2(Ω) by H"older's inequality. Weak H1 convergence and strong L2 convergence therefore give the same limit for these integrals. Since Ω is bounded, u−v∈L1(Ω); if its integral over every measurable set is zero, then its real and imaginary parts vanish almost everywhere. (Weak convergence of nets and sequences, Holder's inequality for integrals, including the endpoint cases, The space Lp(μ) as the quotient by null functions)

Proof

technique · every subsequence has a further $L^2$-convergent subsequence; indicator tests identify its limit with the weak limit, forcing convergence of the whole sequence
1.1F1given

Let (ujk) be any subsequence. It is bounded in H1(Ω), so by [F1] it has a further subsequence (ujkr) converging in L2(Ω) to some v.

2.1F2step 1.1

For each measurable A⊆Ω, [F2] gives ∫Aujkr→∫Au by weak convergence and ∫Aujkr→∫Av by strong L2 convergence. Thus ∫A(u−v)=0 for every such A. Applying this to the sets where the real or imaginary part of u−v is greater than 1/m or less than −1/m, for m≥1, shows that each part vanishes almost everywhere; hence u=v in L2(Ω).

3.1F1step 2.1given∎

Every subsequence of (uj) therefore has a further subsequence converging in L2(Ω) to u. If the whole sequence did not converge to u, some ε>0 would admit a subsequence staying at distance at least ε from u, contradicting the further-subsequence conclusion. Thus uj→u in L2(Ω). The Axiom of Choice is inherited through [F1].

RemarkRemark: Literature-sourcedProof: Not applicableOpen item page →

Rellich compactness is strictly subcritical

Remarks

The compactness statements of this page are strictly subcritical ; the companion page gives witnesses for the critical target exponents and for escape to infinity.

For the W1,p compactness theorems on extension domains, boundedness and the extension hypothesis are part of the stated setting. The W01,p compactness results need only bounded openness and zero extension, with no boundary regularity. On Rn, translations and dilations destroy compactness when the corresponding tail control is absent.

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

The level-set kernel measure estimate for the Slobodeckij kernel

Statement

Assume the Axiom of Countable Choice. Let d≥1, 0<θ<1, 1≤p<∞ with pθ<d, let x∈Rd and let E⊆Rd be Lebesgue measurable with 0<∣E∣<∞. Then ∫Rd∖E∣x−y∣−d−pθ dy ≥ c(d,p,θ) ∣E∣−pθ/d. One admissible constant is c=dpθ ω1+pθ/d, where ω:=∣B(0,1)∣ denotes the Lebesgue measure of the unit ball.

Facts & Assumptions

Given: the Axiom of Countable Choice, d≥1, 0<θ<1, 1≤p<∞ with pθ<d, a point x∈Rd, and a Lebesgue measurable set E⊆Rd with 0<∣E∣<∞. Write ω:=∣B(0,1)∣ for the unit-ball measure and ρ:=(∣E∣/ω)1/d>0.

[F2]

C1 change of variables for nonnegative Borel functions. If U,V⊆Rm are open and T:U→V is a C1 diffeomorphism, then every nonnegative Borel h:V→[0,∞] satisfies ∫Vh(y) dy=∫Uh(T(w))∣det⁡DT(w)∣ dw, with 0⋅∞=0 allowed. (Borel change of variables from the compact-support formula and Radon uniqueness)

[F3]

Polar coordinates. For every nonnegative Borel f:Rd→[0,∞], ∫Rdf(z) dz=∫0∞∫Sd−1f(rζ)rd−1 dσ(ζ) dr, where σ is the finite Borel surface measure on the unit sphere. (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, The polar surface set function on the unit sphere)

[F4]

Measures of set differences. If A⊆B are measurable with ∣A∣<∞, then ∣B∣=∣A∣+∣B∖A∣. (Measure of a set difference when the smaller set has finite measure)

[F5]

Additivity and monotonicity of the nonnegative integral. For measurable g,h:X→[0,∞]: ∫(g+h)=∫g+∫h, and g≤h implies ∫g≤∫h; moreover ∫cg=c∫g for real c>0; for c=0 the zero function has integral 0. (Additivity of the nonnegative Lebesgue integral, Monotonicity and nonnegative homogeneity of the nonnegative integral)

Proof

technique · Choose the radius $\rho$ carrying the mass of $E$, split the complement of $E$ into its part inside and outside $B(x,\rho)$, use the lower bound $|x-y|\ge\rho$ on the outer part and on $E\setminus B(x,\rho)$ to reach all of $B(x,\rho)^c$, and evaluate the resulting radial integral in polar coordinates
1.1F1F2F4given

The map T(w)=x+ρw is a C1 diffeomorphism of Rd with det⁡DT=ρd, so [F2] applied to the indicator of B(x,ρ) gives ∣B(x,ρ)∣=∫1B(x,ρ)(y) dy=ρd∫1B(0,1)(w) dw=ρdω=∣E∣. Since E∩B(x,ρ)⊆B(x,ρ) and E∩B(x,ρ)⊆E are measurable with ∣E∩B(x,ρ)∣≤∣E∣<∞, [F4] gives ∣(Rd∖E)∩B(x,ρ)∣=∣B(x,ρ)∣−∣E∩B(x,ρ)∣=∣E∣−∣E∩B(x,ρ)∣=∣E∖B(x,ρ)∣.

2.1F5step 1.1

Write C1:=(Rd∖E)∩B(x,ρ) and C2:=(Rd∖E)∩B(x,ρ)c; these are disjoint measurable sets with union Rd∖E. On C1 one has ∣x−y∣−d−pθ≥ρ−d−pθ, on C2 and on E∖B(x,ρ) one has ∣x−y∣≥ρ; hence [F5] gives ∫Rd∖E∣x−y∣−d−pθdy=∫C1+∫C2≥ρ−d−pθ∣C1∣+∫C2∣x−y∣−d−pθdy=ρ−d−pθ∣E∖B(x,ρ)∣+∫C2∣x−y∣−d−pθdy≥∫E∖B(x,ρ)∣x−y∣−d−pθdy+∫C2∣x−y∣−d−pθdy=∫B(x,ρ)c∣x−y∣−d−pθdy, where the last equality uses that E∖B(x,ρ) and C2 partition B(x,ρ)c.

3.1F1F2F3step 2.1∎

Substituting y=x+w by [F2] and evaluating the radial integrand by [F3], ∫B(x,ρ)c∣x−y∣−d−pθdy=∫∣w∣>ρ∣w∣−d−pθdw=σ(Sd−1)∫ρ∞r−1−pθdr=σ(Sd−1)pθρ−pθ. Applying [F3] to the indicator of B(0,1) gives ω=∫Sd−1 ⁣ ⁣∫01rd−1dr dσ=σ(Sd−1)/d, so σ(Sd−1)=dω and the lower bound is dωpθ(∣E∣/ω)−pθ/d=c ∣E∣−pθ/d with c=dpθω1+pθ/d>0 by [F1]; combined with step 2.1 this is the assertion.

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

A dyadic summability estimate for decreasing level-set sequences

Statement

Let d≥1, 0<θ<1, 1≤p<∞ with pθ<d and T>1. Let (ak)k∈Z be a bounded nonnegative nonincreasing sequence of real numbers with ak=0 for all sufficiently large k. Then ∑k∈Zak(d−pθ)/d Tk ≤ C(d,p,θ,T)∑k∈Z: ak≠0ak+1 ak−pθ/d Tk. The constant is explicit: C=Td/(d−pθ). The argument uses no choice principle; all sums are series of nonnegative terms.

Facts & Assumptions

Given: integers d≥1 and k∈Z indices, numbers 0<θ<1, 1≤p<∞ with pθ<d, a real T>1, and a bounded nonnegative nonincreasing sequence (ak)k∈Z with ak=0 for all sufficiently large k. Put t:=pθ/d∈(0,1), α:=1/t>1 and β:=1/(1−t)>1, so that 1/α+1/β=1.

[F1]

Hölder's inequality. On a measure space (X,A,μ), for conjugate exponents α,β∈(1,∞) and nonnegative measurable f,g with finite respective norms, ∫fg dμ≤(∫fαdμ)1/α(∫gβdμ)1/β, which is the finite-norm form used below. Step 1.1 establishes the required finite sums before the application. (Holder's inequality for integrals, including the endpoint cases)

Proof

technique · Shift the index, factor each term of the shifted sum into a product whose two factors have the two critical exponents, apply Hölder's inequality on the counting measure, and solve the resulting inequality for the unknown sum
1.1givenalgebra

Since ak=0 for all k≥N and (ak) is bounded by some M≥0, the sum A:=∑k∈Zak(d−pθ)/dTk satisfies A≤M(d−pθ)/d∑k<NTk<∞, and the sum B:=∑k: ak≠0ak+1ak−pθ/dTk satisfies 0≤B≤∑k: ak≠0ak(d−pθ)/dTk=A<∞ because ak+1≤ak with ak>0 implies ak+1ak−pθ/d≤ak1−pθ/d=ak(d−pθ)/d. Also ak=0 implies ak+1≤ak=0, so ak+1=0.

2.1F1step 1.1algebra

Shifting the index in step 1.1 and dropping exactly the vanishing terms, 1TA=∑k∈Zak+1(d−pθ)/dTk=∑k: ak≠0ak+1(d−pθ)/dTk. For each k with ak≠0 the factorization ak+11−tTk=(akt/βTk/α)(ak+11/βak−t/βTk/β) holds, because t/β−t/β=0, 1/β=1−t and 1/α+1/β=1. Applying [F1] with the counting measure on the set {k:ak≠0} to these two factors gives 1TA≤(∑kak1−tTk)t(∑k:ak≠0ak+1ak−tTk)1−t=AtB1−t, since raising the first factor-sum to the power α=1/t returns ∑kak1−tTk and raising the second to β=1/(1−t) returns B.

3.1step 1.1step 2.1algebra∎

If A=0 then every ak=0 and both sides of the asserted inequality are 0. Otherwise 0<A<∞ by step 1.1, so dividing step 2.1 by TAt gives (1/T)A1−t≤B1−t, hence B1−t≥(1/T)A1−t>0 and therefore A≤T1/(1−t)B. Substituting t=pθ/d gives 1/(1−t)=d/(d−pθ) and A≤Td/(d−pθ)B, which is the assertion with C=Td/(d−pθ). No choice principle is used: both series are sums of nonnegative real terms over a countable index set, evaluated as suprema of finite partial sums.

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

The Slobodeckij seminorm bounds the dyadic level-set sum

Statement

Assume the Axiom of Countable Choice. Let d≥1, 0<θ<1, 1≤p<∞ with pθ<d, let f∈L∞(Rd) have compact support, and put Ak:={∣f∣>2k}, ak:=∣Ak∣. Then [f]θ,pp ≥ c(d,p,θ)∑k∈Z: ak≠0ak+1 ak−pθ/d 2pk, where [⋅]θ,p is the Slobodeckij seminorm of The Gagliardo--Slobodeckij space on Euclidean space.

Facts & Assumptions

Given: the Axiom of Countable Choice, d≥1, 0<θ<1, 1≤p<∞ with pθ<d, a compactly supported f∈L∞(Rd), and the sets Ak={∣f∣>2k} with ak=∣Ak∣. Write α:=pθ/d∈(0,1), T:=2p>1, and Dk:=Ak∖Ak+1, dk:=∣Dk∣.

[F1]

Level sets and annuli. Ak+1⊆Ak, so ak+1≤ak and dk=ak−ak+1; the Dk are pairwise disjoint, Ak=⋃ℓ≥kDℓ up to a null set, ak=∑ℓ≥kdℓ, and ak=0 for all large k because f is bounded with compact support. With Z:={f=0}, for every i we have Ai−1c=Z∪⋃j≤i−2Dj up to a null set. All these sets are measurable. (Lebesgue measurable sets, the family L(Rn), and the restricted set function λn, Finite and countable subadditivity of measures, Measure of a set difference when the smaller set has finite measure)

[F2]

Kernel estimate. If E is measurable with 0<∣E∣<∞ and x∈Rd, then ∫Rd∖E∣x−y∣−d−pθ dy≥c1∣E∣−pθ/d with c1=c1(d,p,θ)>0 independent of x and E. (The level-set kernel measure estimate for the Slobodeckij kernel)

[F3]

Slobodeckij seminorm on disjoint blocks. For measurable B⊆Rd×Rd, ∬B∣f(x)−f(y)∣p∣x−y∣−d−pθ dx dy≤[f]θ,pp; sums over pairwise disjoint such blocks of a nonnegative integrand are bounded by the total integral. (The Gagliardo--Slobodeckij space on Euclidean space, Tonelli and Fubini for the completed product, with only almost-everywhere section measurability)

Proof

technique · Pair the annulus $D_i$ with the complement of $A_{i-1}$, where the level gap is at least $2^{i-1}$; sum the resulting block estimates against the geometric weights, control the overlap of the nested tails by a geometric series, and relabel
1.1F1given

By [F1] the annuli Dk are pairwise disjoint measurable sets with dk=ak−ak+1 and ak=∑ℓ≥kdℓ, and ak=0 for all k large. Let Z:={f=0}. If x∈Di and y∈Dj with j≤i−2, then ∣f(x)∣>2i and ∣f(y)∣≤2j+1≤2i−1, so ∣f(x)−f(y)∣≥2i−1. The same bound holds for y∈Z, since ∣f(y)∣=0 and ∣f(x)∣>2i. Thus the low positive bands together with Z cover Ai−1c up to a null set.

2.1F2F3step 1.1algebra

If [f]θ,p=∞, the conclusion is immediate. Otherwise the disjoint-block sum I below is finite by [F3]. The sums X,S,Y are finite: aiai−1−α≤ai−11−α, the ai are bounded and eventually zero, and T>1, so the negative tail is geometric. Fix i with ai−1≠0. By [F2] applied to E=Ai−1 (which has 0<ai−1<∞), for every x∈Di the integral of ∣x−y∣−d−pθ over Ai−1c is at least c1ai−1−α. Since Ai−1c=Z∪⋃j≤i−2Dj up to a null set, the lower bound from step 1.1 gives Ii:=∑j≤i−2∬Di×Dj∣f(x)−f(y)∣p∣x−y∣−d−pθdxdy+∬Di×Z∣f(x)−f(y)∣p∣x−y∣−d−pθdxdy≥c02piai−1−αdi, where c0:=2−pc1. Writing di=ai−∑ℓ≥i+1dℓ and summing over i with ai−1≠0, I:=∑iIi≥c0X−c0Y where X:=∑i:ai−1≠02piai−1−αai and Y:=∑i:ai−1≠0∑ℓ≥i+12piai−1−αdℓ. Swapping the order of summation in Y and using ai−1≥aℓ−1 whenever i≤ℓ gives Y≤∑ℓ:aℓ−1≠0dℓaℓ−1−α∑i≤ℓ−12pi=T−11−T−1∑ℓ:aℓ−1≠02pℓaℓ−1−αdℓ=1T−1S, where S:=∑ℓ:aℓ−1≠02pℓaℓ−1−αdℓ. On the other hand, the block bound itself gives I≥c0S, so S≤I/c0 and therefore I≥c0X−c0T−1S≥c0X−1T−1I, that is I≥c0(T−1)TX.

3.1F3step 2.1algebra∎

The blocks Di×Dj with j≤i−2 and Di×Z are pairwise disjoint: the Di are disjoint in the first coordinate, and for each i the second-coordinate bands and Z are disjoint. Hence [F3] gives I≤[f]θ,pp. Relabelling k=i−1 in X=∑i:ai−1≠02piai−1−αai yields X=2p∑k:ak≠02pkak−αak+1, so [f]θ,pp≥c0(T−1)2pT∑k:ak≠0ak+1ak−pθ/d2pk, which is the assertion with c=c0(T−1)2pT>0.

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The critical fractional Sobolev inequality on Rd

Statement

Assume the Axiom of Countable Choice. Let d≥1, 0<θ<1, 1≤p<∞ with pθ<d and p⋆:=dpd−pθ. There is C=C(d,p,θ)>0 such that every measurable, compactly supported f:Rd→R satisfies ∥f∥Lp⋆(Rd)p ≤ C [f]θ,pp. Consequently, for every bounded open U⊆Rd and every 1≤q≤p⋆ there is C′=C′(d,p,θ,U,q) with ∥f∥Lq(U)≤C′(∥f∥Lp(Rd)+[f]θ,p) for every compactly supported measurable f.

Facts & Assumptions

Given: Countable Choice, d≥1, 0<θ<1, 1≤p<∞ with pθ<d, and p⋆=dp/(d−pθ); write α:=pθ/d, so that p/p⋆=(d−pθ)/d=1−α and p⋆/p=1/(1−α).

[F1]

Dyadic summability. For a bounded nonnegative nonincreasing sequence (ak) vanishing for all large k and T=2p>1, ∑kak1−αTk≤C1∑k:ak≠0ak+1ak−αTk. (A dyadic summability estimate for decreasing level-set sequences)

[F2]

Level-set bound. For f∈L∞ compactly supported with ak=∣{∣f∣>2k}∣, [f]θ,pp≥c∑k:ak≠0ak+1ak−α2pk. (The Slobodeckij seminorm bounds the dyadic level-set sum)

[F3]

Fatou and dominated convergence. Fatou's lemma bounds the integral of a pointwise limit below by the lower limit of the integrals; dominated convergence applies under an integrable dominating function. (Fatou's lemma, Dominated convergence)

[F4]

Interpolation and H"older. For p<q<p⋆, let σ∈(0,1) be defined by 1/q=(1−σ)/p+σ/p⋆. Lyapunov interpolation, with its parameter 1−σ, gives ∥g∥Lq≤∥g∥Lp1−σ∥g∥Lp⋆σ. For q≤p on a set U of finite measure, ∥g∥Lq(U)≤∣U∣1/q−1/p∥g∥Lp(U). (Lyapunov interpolation inequality for Lp norms, Holder's inequality for integrals, including the endpoint cases)

[F5]

Seminorm and classes. [⋅]θ,p is the Slobodeckij seminorm, finite on Wθ,p. The scalar truncation TN(t)=max⁡(−N,min⁡(N,t)) is 1-Lipschitz, hence ∣TN(f(x))−TN(f(y))∣≤∣f(x)−f(y)∣ and [TNf]θ,p≤[f]θ,p. (The Gagliardo--Slobodeckij space on Euclidean space, The space Lp(μ) as the quotient by null functions)

Proof

technique · prove the inequality for bounded compactly supported $f$ by the layer-cake expansion and the two dyadic estimates, then pass to general $f$ by truncation and Fatou. If the seminorm is infinite, the inequality is immediate
1.1F1F2F5algebra

Let first f∈L∞ have compact support, put Ak={∣f∣>2k} and ak=∣Ak∣. On Dk=Ak∖Ak+1 one has ∣f∣≤2k+1, and the Dk partition {f≠0}, and f vanishes on the remaining set, so ∥f∥p⋆p⋆=∫∣f∣p⋆≤∑k2(k+1)p⋆∣Dk∣≤2p⋆∑k2kp⋆ak; raising to the power p/p⋆<1 and using the concavity bound (∑kbk1/(1−α))1−α≤∑kbk for bk:=ak1−α2pk, whose 1/(1−α)-th powers are ak2kp⋆, gives ∥f∥p⋆p≤2p∑kak1−α2pk. Since the sequence ak is bounded, nonincreasing and eventually 0, [F1] followed by [F2] bounds the last sum by a constant times [f]θ,pp, proving the inequality for this f.

2.1F3F5step 1.1

For general compactly supported measurable f with [f]θ,p<∞ put fN:=max⁡{−N,min⁡{N,f}}. Then fN→f pointwise with ∣fN∣≤∣f∣, so [fN]θ,p≤[f]θ,p by the pointwise contraction in [F5]; the bounded case of step 1.1 gives ∥fN∥p⋆p≤C[fN]θ,pp≤C[f]θ,pp, and Fatou's lemma [F3] passes to the limit: ∥f∥p⋆p≤lim inf⁡N∥fN∥p⋆p≤C[f]θ,pp, which is the asserted inequality.

3.1F1F2F3F4step 1.1step 2.1algebra∎

Let U be bounded open and 1≤q≤p⋆. If U=∅ or ∥f∥Lp(Rd)+[f]θ,p=∞, the conclusion is immediate. For q≤p, Holder [F4] on the finite-measure set U gives ∥f∥Lq(U)≤∣U∣1/q−1/p∥f∥Lp(U)≤∣U∣1/q−1/p∥f∥Lp(Rd). If p<q<p⋆, choose σ∈(0,1) so that 1/q=(1−σ)/p+σ/p⋆. Lyapunov [F4], the embedding of the restricted Lp and Lp⋆ norms below their global norms, and step 2.1 give ∥f∥Lq(U)≤∥f∥Lp(Rd)1−σ(C1/p[f]θ,p)σ. Put a=∥f∥Lp(Rd) and b=C1/p[f]θ,p. Weighted AM--GM yields a1−σbσ≤(1−σ)a+σb≤max⁡{1,C1/p}(a+[f]θ,p). At q=p⋆, step 2.1 directly gives ∥f∥Lp⋆(U)≤C1/p[f]θ,p. These estimates, and the q≤p Holder bound, give the claimed consequence with a constant depending on d,p,θ,U,q. Countable Choice is inherited through [F1], [F2] and [F3].

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Mollification rates for compactly supported Slobodeckij functions

Statement

Assume the Axiom of Countable Choice. Let d≥1, 0<θ<1, 1≤p<∞, let g∈Wθ,p(Rd) have compact support, and let gδ:=g∗ηδ be the mollification of g by a radial mollifier η with ∫η=1. Then gδ∈Cc∞(Rd) is supported in the δ-neighbourhood of supp⁡g; There are constants C1=C1(d,p,θ) and C2=C2(d,p,θ,η) such that (i) ∥g−gδ∥Lp(Rd)≤C1(d,p,θ) δθ[g]θ,p;(ii) ∥gδ∥W1,p(Rd)≤C2(d,p,θ,η)(∥g∥Lp(Rd)+δθ−1[g]θ,p).

Facts & Assumptions

Given: the Axiom of Countable Choice, d≥1, 0<θ<1, 1≤p<∞, a compactly supported g∈Wθ,p(Rd), a radial mollifier η with ∫η=1, and gδ=g∗ηδ. Write ω(t):=sup⁡∣z∣≤t∥τzg−g∥Lp(Rd) for t≥0.

[F1]

Slobodeckij seminorm as a translation integral. Because the diagonal is null and Tonelli's theorem together with the substitution z=y−x applies, [g]θ,pp=∫Rd∥τzg−g∥Lp(Rd)p∣z∣−d−pθ dz. (The Gagliardo--Slobodeckij space on Euclidean space, Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, Translation of a function on Rn)

[F3]

The translation modulus is subadditive. ω is nondecreasing, and ω(s+t)≤ω(s)+ω(t) for all s,t≥0, because τzg−g=τz2(τz1g−g)+(τz2g−g) when z=z1+z2 and translations are isometries of Lp. (Translation of a function on Rn, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, The space Lp(μ) as the quotient by null functions)

[F4]

Minkowski's integral inequality. For measurable F on a product of sigma-finite measure spaces with ∫Y∥F(⋅,y)∥p dν(y)<∞, ∥∫YF(⋅,y) dν(y)∥p≤∫Y∥F(⋅,y)∥p dν(y). (Minkowski's integral inequality)

[F5]

Young's convolution inequality. ∥f∗h∥Lp≤∥f∥L1∥h∥Lp. (Young's convolution inequality under Countable Choice)

[F6]

Mollifying a locally integrable function. For g∈Lloc1 the convolution g∗ηδ is smooth with ∂α(g∗ηδ)=g∗(∂αηδ); an Lp function is locally integrable, and a compactly supported Lp function lies in L1. (Convolution with a mollifier is smooth, and derivatives pass under the integral sign, Holder's inequality for integrals, including the endpoint cases)

[F7]

Support of a convolution. For Borel representatives of f,h∈L1, supp⁡(f∗h)⊆supp⁡f+supp⁡h‾. (The support of a convolution lies in the closure of the support sumset)

[F8]

The radial mollifier. ηδ=δ−dη(⋅/δ) satisfies ∫ηδ=1, supp⁡ηδ⊆B(0,δ), and ∫Rd∂jηδ=0 for every j, since η is radial and its gradient is odd in each coordinate. (A radial mollifier family in Rn)

Proof

technique · Use subadditivity of the translation modulus to dominate its supremum by an average over a ball, compare that average to the Slobodeckij translation integral, and then estimate the approximation error and gradient by Minkowski's integral inequality
1.1F1F2F3

Fix t>0 and ∣h∣≤t. By [F3], for every z∈B(0,t) we have ∥τhg−g∥p≤∥τzg−g∥p+∥τh−zg−g∥p. Raising to the p-th power and averaging over z∈B(0,t) gives ∥τhg−g∥pp≤2p−1∣B(0,t)∣(∫B(0,t)∥τzg−g∥ppdz+∫B(0,t)∥τh−zg−g∥ppdz)≤2pvdtd∫∣w∣≤2t∥τwg−g∥ppdw, because both B(0,t) and h−B(0,t) lie in B(0,2t) and [F2] gives ∣B(0,t)∣=vdtd. Taking the supremum over ∣h∣≤t, then using [F1] and ∣w∣−d−pθ≥(2t)−d−pθ for 0<∣w∣≤2t, yields ω(t)p≤2pvdtd(2t)d+pθ[g]θ,pp. Thus ω(t)≤C1tθ[g]θ,p for a constant C1=C1(d,p,θ), which is the translation-modulus estimate needed below.

2.1F4F8step 1.1

Since ∫ηδ=1 and ηδ≥0 is supported in B(0,δ), g(x)−gδ(x)=∫ηδ(y)(g(x)−g(x−y))dy; taking Lp-norms and applying [F4] with Y=B(0,δ) gives ∥g−gδ∥p≤∫ηδ(y)∥g−τyg∥p dy≤ω(δ)≤C1δθ[g]θ,p, which is (i).

3.1F5F6F7F8step 1.1∎

By [F6], gδ is smooth and ∇gδ=g∗∇ηδ; [F6] also gives g∈L1 because g has compact support and lies in Lp. By [F8], ∫∂jηδ=0, so ∂jgδ(x)=∫(g(x−y)−g(x))∂jηδ(y) dy, and [F4] gives ∥∂jgδ∥p≤ω(δ)∫∣∂jηδ∣≤C2δθ−1[g]θ,p because ∫∣∇ηδ∣=δ−1∫∣∇η∣. Moreover ∥gδ∥p≤∥g∥p by [F5] with ∥ηδ∥1=1, so ∥gδ∥W1,p≤∥g∥p+∑j∥∂jgδ∥p, which is (ii) after enlarging the constant. Finally, gδ∈C∞ by [F6], and [F7] applied to the Borel representative of g and to ηδ gives supp⁡gδ⊆supp⁡g+B(0,δ)‾, the δ-neighbourhood of supp⁡g; this is compact because supp⁡g is compact, so gδ∈Cc∞(Rd).

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Subcritical compactness for compactly supported Slobodeckij functions

Statement

Assume the Axiom of Choice. Let d≥1, 0<θ<1, 1≤p<∞ with pθ<d and p⋆=dpd−pθ. Let F⊆Wθ,p(Rd) be a family of functions all supported in one fixed bounded set. In the displayed nonnegative supremum, take the value 0 if F=∅. Assume it satisfies sup⁡g∈F(∥g∥Lp(Rd)+[g]θ,p)<∞. Then F is relatively compact in Lq(Rd) for every 1≤q<p⋆: every sequence in F has a subsequence converging in Lq(Rd).

Facts & Assumptions

Given: the Axiom of Choice, d≥1, 0<θ<1, 1≤p<∞ with pθ<d, p⋆=dp/(d−pθ), a family F⊆Wθ,p(Rd) supported in one fixed bounded set and bounded in the norm ∥⋅∥p+[⋅]θ,p by M<∞, and 1≤q<p⋆.

[F1]

Fractional Sobolev inequality. For real compactly supported g, ∥g∥Lp⋆p≤C1[g]θ,pp; hence ∥g−h∥p⋆≤C11/p([g]+[h]) for compactly supported g,h. For complex g, apply the real inequality to its real and imaginary parts, whose seminorms are at most [g]θ,p, and use the Lp⋆ triangle inequality, enlarging the constant by at most 2. (The critical fractional Sobolev inequality on Rd)

[F2]

Mollification rates. With the radial mollifier at scale δ, ∥g−gδ∥Lp≤C2δθ[g]θ,p and ∥gδ∥W1,p≤C2(∥g∥p+δθ−1[g]θ,p); the mollified functions are supported in the δ-neighbourhood of the fixed support set, and [gδ]θ,p≤[g]θ,p because ∣gδ(x)−gδ(y)∣≤∫ηδ(z)∣g(x−z)−g(y−z)∣ dz and Minkowski's inequality applies in the weighted Lp-space of the seminorm. (Mollification rates for compactly supported Slobodeckij functions, The Gagliardo--Slobodeckij space on Euclidean space)

[F3]

First-order compactness on a ball. For fixed δ>0, the mollified family is bounded in W1,p on a smooth ball containing all its supports. Its closure in Lp is compact by the first-order Rellich theorem, including dimension one. (Compactness of W1,p(Ω)↪Lp(Ω) on bounded extension domains)

Proof

technique · Obtain finite $L^p$ nets from uniform mollification error and first-order Rellich, then interpolate pairwise differences against the fractional critical bound
1.1F2F3F4given

The empty family is immediate. Otherwise fix a ball S containing the common bounded support and its distance-one neighbourhood, and use only 0<δ≤1. By [F2], Gδ is supported in S and bounded in W1,p; [F3] makes it totally bounded in Lp(Rd), since restriction to S and zero extension preserve distances on this family. Also sup⁡g∥g−gδ∥p≤C2δθM→0. Given ε>0, choose δ making this error less than ε/4 and a finite ε/4-net for Gδ. Its centres cover F with radius ε/2; choosing one point of F in every nonempty such ball moves the centres into F and gives an ε-net. Thus F is totally bounded in Lp.

2.1F1F4step 1.1

Fix 1≤q<p⋆. If q=p, step 1.1 applies. If q<p, all members and their differences vanish off S, so ∥g−h∥q≤∣S∣1/q−1/p∥g−h∥p by [F4]. If p<q<p⋆, [F1] gives ∥g−h∥p⋆≤2C11/pM for g,h∈F, and [F4] gives ∥g−h∥q≤∥g−h∥pλ(2C11/pM)1−λ with 0<λ<1 and 1/q=λ/p+(1−λ)/p⋆. Hence a sufficiently fine finite Lp net with centres in F is an Lq net in each case. If M=0, the family contains only the zero class.

3.1F4step 2.1∎

By [F4] the totally bounded closure in the complete space Lq(Rd) is compact and sequentially compact, giving the asserted subsequence for every sequence in F. The assumed Axiom of Choice supplies the first-order Rellich interface and Countable and Dependent Choice in [F4].

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Subcritical compactness of the Sobolev trace

Statement

Assume the Axiom of Choice. Let n≥2, let Ω⊂Rn be a bounded C1 domain, let T:W1,p(Ω)→Lp(∂Ω) be the trace of The Lp trace operator on a bounded C1 domain, and let 1<p<n with p∗:=(n−1)pn−p. Then T is compact as a map into Lq(∂Ω) for every 1≤q<p∗: every sequence bounded in W1,p(Ω) has a subsequence whose traces converge in Lq(∂Ω). If n<p<∞, then the traces of a suitable subsequence converge in C0,β(∂Ω) for every 0≤β<1−np, hence also in every Lq(∂Ω), 1≤q<∞; the endpoint case p=n is not claimed.

Facts & Assumptions

Given: the Axiom of Choice, a bounded C1 domain Ω⊂Rn, n≥2, and 1<p<∞, p≠n, with a sequence (uj) bounded in W1,p(Ω).

[F1]

Sharp trace boundedness. For 1<p<∞ and θ=1−1p, the trace satisfies ∥Tu∥Wθ,p(∂Ω)≤C∥u∥W1,p(Ω), where the boundary norm is the finite sum over a finite atlas of Euclidean Wθ,p-norms of compactly supported chart representations, and it is independent of the atlas up to equivalence. (The sharp trace theorem: boundedness and range in the fractional space, The fractional Sobolev space on a compact C1 boundary, Chart independence of the fractional boundary norm)

[F2]

Fractional compactness in dimension n−1. For 1<p<n and θ=1−1p one has (n−1)−pθ=n−p>0 and the critical exponent of Wθ,p(Rn−1) is (n−1)pn−p=p∗; a family of functions supported in one fixed bounded set and bounded in Wθ,p(Rn−1) is therefore relatively compact in Lq(Rn−1) for every 1≤q<p∗. (Subcritical compactness for compactly supported Slobodeckij functions, The fractional Sobolev space on a compact C1 boundary)

[F3]

Trace and chart cutoffs. The trace commutes with multiplication by smooth ambient cutoffs and with the chart parametrisations; on a compact boundary patch the surface-measure density of the parametrisation is continuous and positive, so Lq convergence of the finitely many chart representations gives Lq(∂Ω) convergence of their sum. (The trace commutes with smooth cutoffs and is chart local, Surface integration on compact C1 hypersurfaces, Finite ambient partitions near compact sets, Bounded C1 domains and their outward normals)

[F4]

The Morrey branch. For n<p<∞, the extension theorem at k=1 makes the bounded C1 domain Ω a W1,p-extension domain; hence a bounded sequence in W1,p(Ω) has a subsequence whose representatives converge in C0,β(Ω‾) for every 0≤β<1−np, and the trace of such a class is its classical boundary restriction. (Bounded C^k domains admit integer-order Sobolev extension, Morrey--Rellich compactness for p>n, The trace agrees with classical restriction for continuous Sobolev functions)

Proof

technique · bound the traces in the boundary fractional space, apply fractional compactness chart by chart, and take the Morrey branch for $p>n$
1.1F1given

Assume 1<p<n. By [F1] the traces satisfy sup⁡j∥Tuj∥Wθ,p(∂Ω)≤Csup⁡j∥uj∥W1,p(Ω)<∞ with θ=1−1p; by the definition of the boundary norm this means that each of the finitely many compactly supported chart representations of the traces is bounded in Wθ,p(Rn−1).

2.1F2F3step 1.1given

Choose exponents qℓ↑p∗ with 1≤qℓ<p∗. For each ℓ, [F2] applied successively on the finitely many charts supplies a common subsequence converging in every chart in Lqℓ. Dependent Choice selects nested subsequences for ℓ=1,2,…; their diagonal converges in each chart for each qℓ. For any 1≤q<p∗, choose ℓ with q<qℓ and use finite-measure inclusion on the common bounded chart supports. The chart Jacobian is bounded on each compact support, so [F3] transfers convergence of the finitely many chart pieces to convergence of their sum in Lq(∂Ω). Thus the same subsequence works throughout the stated range.

3.1F1F2F3F4step 2.1∎

If n<p<∞, [F4] first verifies the extension-domain hypothesis and then provides a subsequence of the uj whose representatives converge in C0,β(Ω‾) for every 0≤β<1−np, and their traces, being the classical boundary restrictions, converge in C0,β(∂Ω) and hence in every Lq(∂Ω), 1≤q<∞. The endpoint p=n would require the limiting fractional embedding at θ=1−1/n=d/p in dimension d=n−1 and is deliberately not claimed. The Axiom of Choice is inherited through the published trace theorem [F1] and the Morrey branch [F4].

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Strong convergence of subcritical powers

Statement

Assume Countable Choice. Let Ω⊂Rn have finite Lebesgue measure, let 1≤q<r<∞ and m≥1, and let (uj) be a sequence in Lq(Ω;R) with uj→u in Lq(Ω;R) and sup⁡j∥uj∥Lr(Ω)<∞. Then u∈Lr(Ω), and for every 1≤s<r/m the nonlinear maps converge: ∣uj∣m−1uj⟶∣u∣m−1uin Ls(Ω). The range is nonempty only when m<r; the endpoint s=r/m is not asserted.

Facts & Assumptions

Given: Countable Choice, a finite-measure set Ω⊆Rn, exponents 1≤q<r<∞, a real number m≥1, and real-valued measurable classes uj,u on Ω with uj→u in Lq(Ω) and M:=sup⁡j∥uj∥Lr(Ω)<∞.

[F1]

Hölder inclusion on a finite-measure space. If 1≤a<b<∞ and g is measurable on the finite-measure space Ω, then g∈La whenever g∈Lb, with ∥g∥La≤∣Ω∣1/a−1/b∥g∥Lb; this is Hölder applied to ∣g∣a and the constant function 1. (Holder's inequality for integrals, including the endpoint cases, The space Lp(μ) as the quotient by null functions)

[F2]

Lyapunov interpolation. If 1≤p0<p<p1<∞, θ∈(0,1) and 1/p=θ/p0+(1−θ)/p1, then every f∈Lp0∩Lp1 lies in Lp with ∥f∥Lp≤∥f∥Lp0θ∥f∥Lp11−θ. (Lyapunov interpolation inequality for Lp norms)

[F3]

Almost-everywhere subsequences. Every sequence converging in Lq, 1≤q≤∞, has a subsequence whose representatives converge almost everywhere to a representative of the limit. (Assuming Countable Choice, Lp-convergent sequences have almost-everywhere convergent subsequences)

[F4]

Fatou's lemma. For nonnegative measurable functions fk, ∫lim inf⁡kfk≤lim inf⁡k∫fk. (Fatou's lemma)

[F5]

Mean value theorem. If f:[a,b]→R is continuous on [a,b] and differentiable on (a,b), then f(b)−f(a)=f′(c)(b−a) for some c∈(a,b). (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∈(a,b) with f(b)−f(a)=f′(c)(b−a))

[F6]

Hölder's inequality for products. For conjugate exponents 1≤p,q≤∞ and measurable f,g, ∫∣fg∣≤∥f∥Lp∥g∥Lq. (Holder's inequality for integrals, including the endpoint cases)

Proof

Proof technique: Extract an almost-everywhere subsequence to obtain the Lr-bound of the limit, transfer convergence to the exponent ms<r by interpolation, and apply the pointwise mean value bound followed by Hölder.

1.1F3F4given

By [F3] fix a subsequence (ujk) convergent almost everywhere to u. Then lim inf⁡k∣ujk∣r=∣u∣r pointwise almost everywhere, so [F4] gives ∥u∥Lrr≤lim inf⁡k∥ujk∥Lrr≤Mr<∞, hence u∈Lr(Ω) with ∥u∥Lr≤M. Consequently ∥uj−u∥Lr≤∥uj∥Lr+∥u∥Lr≤2M for every j.

2.1F1F2step 1.1

Fix 1≤s<r/m and put b:=ms<r. If b=q then ∥uj−u∥Lb=∥uj−u∥Lq→0; if b<q, then [F1] gives ∥uj−u∥Lb≤∣Ω∣1/b−1/q∥uj−u∥Lq→0. If q<b<r, choose θ∈(0,1) with 1/b=θ/q+(1−θ)/r; [F2] applied to the classes uj−u∈Lq∩Lr gives ∥uj−u∥Lb≤∥uj−u∥Lqθ∥uj−u∥Lr1−θ≤∥uj−u∥Lqθ(2M)1−θ→0 by step 1.1. In both cases ∥uj−u∥Lms→0, and sup⁡j∥uj∥Lms≤∣Ω∣1/(ms)−1/rsup⁡j∥uj∥Lr<∞ by [F1], while ∥u∥Lms≤∣Ω∣1/(ms)−1/rM by step 1.1.

3.1F5F6step 2.1given∎

If m=1 then ∣uj∣m−1uj=uj and ∣u∣m−1u=u, so the claim is step 2.1 itself with b=s<r. If m>1, consider N(t):=∣t∣m−1t on R; N is differentiable with N′(t)=m∣t∣m−1, and for real a≠b every point t of the closed interval between them satisfies ∣t∣m−1≤∣a∣m−1+∣b∣m−1. By [F5] applied to N on that interval, ∣N(a)−N(b)∣≤m(∣a∣m−1+∣b∣m−1)∣a−b∣. Writing a=uj(x), b=u(x) and applying [F6] with exponents m/(m−1) and m to the product (∣uj∣m−1+∣u∣m−1)∣uj−u∣ gives ∥N(uj)−N(u)∥Ls≤m(∥uj∥Lmsm−1+∥u∥Lmsm−1)∥uj−u∥Lms, and the right-hand side tends to 0 by the bounds and convergence of step 2.1. Hence ∣uj∣m−1uj→∣u∣m−1u in Ls(Ω).

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

A bounded map into H01 yields a compact L2 operator

Statement

Assume the Axiom of Choice. Let Ω⊂Rn be a bounded open set and T:L2(Ω)→H01(Ω) a bounded linear operator. Then ιT:L2(Ω)→L2(Ω) is compact, where ι:H01(Ω)↪L2(Ω) is the inclusion. No boundary regularity of Ω is needed.

Facts & Assumptions

Given: the Axiom of Choice, a bounded open set Ω⊆Rn, and a bounded linear operator T:L2(Ω)→H01(Ω), with ι:H01(Ω)→L2(Ω) the inclusion.

[F1]

Zero-boundary Rellich theorem. W01,2(Ω)=H01(Ω) is compactly embedded in L2(Ω): every sequence bounded in H01(Ω) has a subsequence converging in L2(Ω). (Compactness of W01,p(Ω)↪Lp(Ω) on bounded open sets, Compactly embedded normed spaces, The notation Hk and the reserved zero-boundary symbol)

[F2]

Bounded operators map bounded sequences to bounded sequences. If (fj) satisfies ∥fj∥L2≤M, then ∥Tfj∥H01≤∥T∥M for the operator norm of A bounded linear operator between normed spaces. (A bounded linear operator between normed spaces)

[F3]

Compactness is the sequential extraction criterion. A bounded operator S is compact exactly when the image of every bounded sequence has a convergent subsequence. (Compact linear operator, Compactly embedded normed spaces)

Proof

technique · direct
1.1F1F2given

Let (fj) be bounded in L2(Ω) with ∥fj∥L2≤M. By [F2], (Tfj) is bounded in H01(Ω), so [F1] supplies a subsequence with Tfjk→g in L2(Ω), that is, ιTfjk→g.

2.1F1F3step 1.1∎

Since every bounded sequence in L2(Ω) has an image under ιT with a convergent subsequence, [F3] makes ιT a compact operator. The inclusion ι is bounded because ∥u∥L2≤∥u∥H01, and the Axiom of Choice is inherited through the Rellich theorem [F1].

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

Closed target constraints survive compact extraction

Statement

Assume Countable and Dependent Choice. Let X be a normed space with compact continuous inclusion J:X↪Lp(Ω), 1≤p<∞, and let C⊆Lp(Ω) be closed in norm. Every bounded sequence uj∈X with Juj∈C admits a subsequence Jujk→v in Lp(Ω) with v∈C. In particular this applies to any of this page's Rellich inclusions, with their stated domain, exponent and choice hypotheses. This statement does not assert that v belongs to X or that C is weakly closed.

Facts & Assumptions

Given: Countable and Dependent Choice, a normed space X with compact continuous inclusion J:X↪Lp(Ω), a norm-closed set C⊆Lp(Ω), and a bounded sequence (uj) in X with Juj∈C for all j.

[F1]

The sequential form of a compact embedding. Under Countable and Dependent Choice, a compact continuous inclusion J sends every bounded sequence in X to a sequence with a subsequence converging in Lp(Ω). (Compactly embedded normed spaces)

[F2]

Closed sets contain sequential limits. A closed subset of a metric space contains the limit of every convergent sequence of its points: otherwise the open complement contains a ball about the limit, contradicting eventual membership of the sequence in that ball. (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, The closure of a nonempty A is {x:d(x,A)=0}, equals A together with its limit points, and is the smallest closed superset)

Proof

technique · extract a convergent subsequence by compactness and use closedness of the target constraint
1.1F1given

By [F1] the bounded sequence (Juj) has a subsequence (Jujk) converging in Lp(Ω) to some v.

2.1F2step 1.1given∎

Since Jujk∈C for every k and C is closed, [F2] gives v∈C; the statement makes no claim that v lies in the image of J. Countable and Dependent Choice are used exactly through the compact-embedding interface [F1].

5 · Examples, counterexamples and false statements

None yet.

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