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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Critical bubbles converge weakly but not strongly

Example

Assume Countable Choice. Let n≥2, 1≤p<n, p∗=npn−p, and choose a nonzero real φ∈Cc∞(B(0,1)) with ∥φ∥Lp∗(Rn)=1 (for instance a normalised smooth bump). On Ω=B(0,1) put uj(x)=(j+1)(n−p)/pφ((j+1)x) for j≥0. Then uj∈W01,p(Ω), ∥uj∥Lp∗(Ω)=1, ∥Duj∥Lp(Ω)=∥Dφ∥Lp(Rn) and ∥uj∥Lp(Ω)=(j+1)−1∥φ∥Lp(Rn). Moreover uj⇀0 in Lp∗(Ω) and uj→0 almost everywhere, but no subsequence converges strongly in Lp∗(Ω): the unit mass concentrates at the origin, while the weak limit is the zero class and the norms remain one.

Facts & Assumptions

Given: the Axiom of Countable Choice, n≥2, 1≤p<n, p∗=npn−p, a nonzero real φ∈Cc∞(B(0,1)) with ∥φ∥p∗=1, and uj(x)=(j+1)(n−p)/pφ((j+1)x) on Ω=B(0,1).

[F1]

Scaling. For measurable nonnegative h and j≥0, ∫h((j+1)x)(j+1)n dx=∫h(y) dy, and supp⁡(uj)⊆B(0,1/(j+1)). (A linear map T of Rn sends Lebesgue measurable sets to Lebesgue measurable sets, with λn(T[E])=∣det⁡T∣ λn(E) when T is invertible and T[E] Lebesgue null when it is not, Integer-order Sobolev spaces and their norms)

[F2]

H"older's inequality. ∫∣fg∣≤∥f∥p∗∥g∥(p∗)′ for conjugate exponents. (Holder's inequality for integrals, including the endpoint cases)

[F3]

Duality of Lp∗. Since 1<p∗<∞ and Ω has finite measure, every bounded linear functional on Lp∗(Ω) is integration against some g∈L(p∗)′(Ω). (For 1<p<∞, the same representation theorem holds on arbitrary measure spaces)

[F4]

Absolute continuity of the integral. If ∣g∣(p∗)′ is integrable, then ∥g1B(0,1/(j+1))∥(p∗)′→0 as j→∞. (Dominated convergence)

[F5]

Weak convergence. uj⇀0 in Lp∗ means ∫ujg→0 for every g∈L(p∗)′; strong convergence implies weak convergence. (Weak convergence of nets and sequences)

[F6]

Membership in W01,p. The function uj is smooth and compactly supported in Ω, hence belongs to W01,p(Ω) and its classical derivatives represent Duj. (Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms)

Verification

technique · direct
1.1F1F6given

By [F1], ∥uj∥p∗p∗=(j+1)(n−p)p∗/p(j+1)−n∥φ∥p∗p∗=1, ∥uj∥p=(j+1)(n−p)/p(j+1)−n/p∥φ∥p=(j+1)−1∥φ∥p, and ∥Duj∥p=(j+1)(n−p)/p(j+1)1−n/p∥Dφ∥p=∥Dφ∥p; [F6] gives uj∈W01,p(Ω), and uj(x)→0 for every x≠0 because φ((j+1)x)=0 once (j+1)∣x∣>1.

2.1F2F3F4F5step 1.1

Let g∈L(p∗)′(Ω) and extend it by zero to Rn. By [F2] and step 1.1, ∣∫Ωujg dx∣≤∥uj∥p∗∥g1B(0,1/(j+1))∥(p∗)′=∥g1B(0,1/(j+1))∥(p∗)′, which tends to 0 by [F4]; by [F3] and [F5] this says uj⇀0 in Lp∗(Ω).

3.1F2F5step 1.1step 2.1∎

If a subsequence converged strongly in Lp∗(Ω) to some v, then ∥v∥p∗=1 by continuity of the norm and step 1.1, while [F5] would give v=0 because strong convergence and step 2.1 imply ∫Ωvg=0 for every g∈L(p∗)′. Taking g=sgn⁡(v)∣v∣p∗−1 gives ∫∣v∣p∗=0; this contradiction shows that no subsequence converges strongly in Lp∗(Ω). The example therefore exhibits the failure of compactness at the critical exponent while for 1≤q<p∗ the scaling exponent (n−p)/p−n/q is negative, so the subcritical Lq norms tend to zero.

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