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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-30
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Weak lower semicontinuity of the Sobolev norm

Statement

Assume the Axiom of Choice. Let Ω⊆Rn be open, n≥1, k∈N0, 1<p<∞, and K∈{R,C}. If a sequence (uj) in Wk,p(Ω;K) converges weakly to u∈Wk,p(Ω;K) with respect to the displayed norm of Integer-order Sobolev spaces and their norms, then ∥u∥Wk,p(Ω)≤lim inf⁡j→∞∥uj∥Wk,p(Ω).

The statement uses the exact norm convention of the Sobolev definition: the finite ℓp sum of derivative norms for 1≤p<∞ and the maximum for p=∞. Neither reflexivity nor completeness of the space is used; the exponent hypothesis 1<p<∞ is retained from the intended use, and the argument would apply to every 1≤p≤∞. If Ω=∅, then the space is the zero space and both sides are 0.

Facts & Assumptions

Given: AC, open Ω⊆Rn, k∈N0, 1<p<∞, K∈{R,C}, and a sequence uj⇀u in Wk,p(Ω;K).

[F1]

HB is the real dominated-extension principle: every real linear functional on a subspace dominated by a sublinear functional extends to the whole space with the same domination (The real dominated-extension principle as an additional hypothesis over ZF).

[F2]

Under the Axiom of Choice the dominated-extension theorem holds for every real vector space, sublinear functional and dominated linear functional, which is precisely the assertion HB (Hahn-Banach dominated extension theorem for real vector spaces).

[F3]

Under HB, if a net xi converges weakly to x in a real or complex normed space X, then ∥x∥≤lim inf⁡i∥xi∥, with no boundedness or completeness hypothesis (Weak convergence implies lower semicontinuity of the norm).

[F4]

Weak convergence of a sequence in a normed space means convergence against every bounded linear functional (Weak convergence of nets and sequences).

[F5]

Assuming Countable Choice, for either scalar field and all k∈N0, 1≤p≤∞, the displayed Wk,p(Ω) formula is finite, vanishes exactly on the zero class, and makes Wk,p(Ω;K) a normed space with the same derivative classes as Integer-order Sobolev spaces and their norms (The Sobolev norm descends to equivalence classes, Integer-order Sobolev spaces and their norms).

[F6]

The Axiom of Choice is the assertion that every family of nonempty sets has a choice function (The Axiom of Choice). In particular, applying it to families indexed by N gives Countable Choice.

Proof

technique · obtain HB from AC, recognise the Sobolev space as a normed space under the displayed norm, and invoke the published dual-norming lower-semicontinuity corollary
1.1F1F2F6given

By [F6], the Axiom of Choice is available, so [F2] gives the real dominated-extension theorem; comparing its conclusion with the definition [F1] shows that HB holds.

1.2F4F5F6given

By [F6], AC supplies the Countable Choice hypothesis of [F5]. By [F5] the set Wk,p(Ω;K) with the displayed formula is a normed space over K, for both K=R and K=C, and uj,u are points of it. The weak convergence in the statement is convergence against every bounded linear functional of this normed space, as recalled in [F4], so (uj) is a net in the sense required by [F3].

2.1F3step 1.1step 1.2

Apply [F3] with X=Wk,p(Ω;K), xj=uj and x=u. Its hypothesis is exactly the weak convergence of step 1.2, and its conclusion is ∥u∥≤lim inf⁡j∥uj∥, which is the displayed Sobolev-norm inequality. No reflexivity, uniform convexity, boundedness of the sequence, or completeness of X is used in this application.

3.1F1F2F3F5F6step 1.1step 1.2step 2.1

The choice accounting and boundary cases are explicit. The Axiom of Choice supplies HB in step 1.1 through [F2] and Countable Choice in step 1.2 through [F6] for the Sobolev definition and norm theorem [F5]. The abstract lower-semicontinuity corollary [F3] needs HB, and the Sobolev normed-space interface retains its Countable Choice hypothesis; neither use is asserted to be choice-free. The exponent hypothesis 1<p<∞ and the differentiability order k enter only through the normed-space structure of Wk,p, so the same argument covers p=1 and p=∞; they are recorded as a hypothesis rather than used. If Ω=∅, then by [F5] the space is the zero space, its norm is zero, and the inequality reads 0≤lim inf⁡j0=0. If the sequence is the zero sequence or a constant sequence, both sides equal the common norm. □

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