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Rellich compactness fails on Rn by translations

Statement refuted

Refuted claim. For 1≤p<∞ the inclusion W1,p(Rn)↪Lp(Rn) is compact: every sequence bounded in W1,p(Rn) has a subsequence converging in Lp(Rn).

The witness is the sequence of translates of one fixed nonzero compactly supported test function. Boundedness survives translation, but a pair of translates at large separation has two disjoint copies of the same mass, so the sequence is not even Cauchy in Lp.

Facts & Assumptions

Given: Countable Choice; n≥1, 1≤p<∞, and ψ∈Cc∞(Rn) nonzero; for k≥0 put uk(x):=ψ(x−ke1). Write K:=supp⁡ψ. A concrete choice is the bump of A Euclidean bump for a compact set inside an open set applied to the compact set {0} inside the open unit ball: it is smooth, supported in B(0,1), equal to 1 at the origin and hence nonzero.

[F1]

Classical derivatives of smooth compactly supported functions are weak derivatives. (Classical derivatives agree with weak derivatives)

[F2]

Translation is an Lp-isometry and commutes with classical differentiation. ∥τhg∥Lp=∥g∥Lp for every g∈Lp and every h, since Lebesgue measure is translation invariant, and ∂j(τhg)=τh(∂jg) for smooth g. (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, Translation of a function on Rn, The space Lp(μ) as the quotient by null functions)

[F3]

Supports of separated translates are disjoint. supp⁡uk=K+ke1; if ∣k−l∣>2R where K⊆B(0,R), then (K+ke1)∩(K+le1)=∅. A compact set is bounded, so such an R exists. (The support of a function on Rn and its compactly supported Riemann integral, A compact subset of a metric space is closed and bounded, Translation of a function on Rn)

[F4]

The Sobolev norm. ∥g∥W1,p(Rn)=(∥g∥Lpp+∑j=1n∥∂jg∥Lpp)1/p. (Integer-order Sobolev spaces and their norms)

Counterexample

technique · direct
1.1F1F2F4given

By [F2] each translate satisfies ∥uk∥Lp=∥ψ∥Lp and ∂juk=τke1(∂jψ) with ∥∂juk∥Lp=∥∂jψ∥Lp; [F1] identifies these classical derivatives with the weak derivatives, so by [F4] ∥uk∥W1,p(Rn)=∥ψ∥W1,p(Rn) for every k. Hence sup⁡k∥uk∥W1,p<∞.

2.1F2F3step 1.1

By [F3] fix R with K⊆B(0,R); if ∣k−l∣>2R then supp⁡uk and supp⁡ul are disjoint, so ∥uk−ul∥Lpp=∫supp⁡uk∣uk∣p+∫supp⁡ul∣ul∣p=2∥ψ∥Lpp>0.

3.1step 2.1given∎

Let (ukj) be any subsequence. Its indices tend to infinity, so for every tail there are two indices in it whose difference exceeds 2R. Step 2.1 makes their distance the fixed positive value 21/p∥ψ∥p, so this subsequence is not Cauchy and cannot converge. Hence no subsequence converges in Lp, and the refuted claim is false. Countable Choice is inherited through the smooth weak-derivative and Sobolev interfaces.

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