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Dilations force the Sobolev conjugate

Example

Assume Countable Choice (The Axiom of Countable Choice (ACω)). Let n≥2, 1≤p<n, and let 1≤q≤∞ and u∈Cc∞(Rn)∖{0}; for λ>0 put uλ(x):=u(λx). Then ∥uλ∥Lq=λ−n/q∥u∥Lq,∥Duλ∥Lp=λ 1−n/p∥Du∥Lp. If the inequality ∥uλ∥Lq≤C∥Duλ∥Lp is to hold for all λ>0 with a constant C independent of λ, then the two powers of λ must balance: −nq−(1−np)=0,that is1−np+nq=0, which is equivalent to q=npn−p=p∗. The exponent p∗ is therefore forced by scaling alone.

Facts & Assumptions

Given: Countable Choice; 1≤q≤∞; n≥2; 1≤p<n; a fixed nonzero u∈Cc∞(Rn); and λ>0.

[F1]

The Sobolev conjugate is p∗=np/(n−p) and satisfies 1p∗=1p−1n (The Sobolev conjugate exponent and the scaling identity).

[F2]

An invertible linear map scales Lebesgue measure by ∣det⁡∣: substituting y=λx gives ∫Rng(λx) dx=λ−n∫Rng(y) dy; an Lq class is determined up to null sets (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, The space Lp(μ) as the quotient by null functions).

[F3]

The classical chain rule computes D(u∘T) for the linear map T(x)=λx: D(uλ)(x)=λ (Du)(λx) (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

[F4]

The fundamental theorem on smooth line segments (If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz) implies that a smooth function with zero gradient is constant.

Verification

technique · direct
1.1F2F3F4givenalgebra

The norm scalings. For finite q, by [F2] with y=λx, ∥uλ∥Lqq=∫Rn∣u(λx)∣q dx=λ−n∫Rn∣u(y)∣q dy, so ∥uλ∥Lq=λ−n/q∥u∥Lq>0. For q=∞, the superlevel sets scale in measure by λ−n, so the essential supremum is unchanged; take 1/q=0. By the chain rule [F3], D(uλ)(x)=λ (Du)(λx), so ∣Duλ(x)∣=λ∣Du(λx)∣ and ∥Duλ∥Lpp=λpλ−n∫Rn∣Du(y)∣p dy=λp−n∥Du∥Lpp, that is ∥Duλ∥Lp=λ1−n/p∥Du∥Lp>0 (if this norm were zero, continuity would give Du=0 everywhere; [F4] would make u constant, and compact support would force u=0).

2.1F1step 1.1givenalgebra∎

The balance condition. Dividing the two identities of step 1.1, the proposed inequality reads λ−n/q∥u∥Lq≤Cλ1−n/p∥Du∥Lp for every λ>0; after multiplying by λn/q this is ∥u∥Lq≤Cλ 1−n/p+n/q∥Du∥Lp. Since ∥u∥Lq and ∥Du∥Lp are fixed positive numbers, a finite C satisfying this for all λ>0 exists exactly when the exponent vanishes, i.e. 1−np+nq=0; solving, nq=np−1=n−pp, so q=npn−p=p∗ by [F1].

Source notes

The computation is the dilation argument that opens Kinnunen's Chapter 3, printed p. 61, and the same scaling discussion precedes Hunter's Theorem 3.28. For uλ(x)=u(λx) the conjugate exponent is selected by requiring the two sides to carry the same power of λ; the exponent p∗ arises from the balance 1−np+nq=0. No optimality of constants beyond this power balance is claimed, and the example does not construct a counterexample for other exponents, which is done on the companion page by the opposite dilation convention.

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