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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
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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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