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Dilations force the Sobolev conjugate
Example
Assume Countable Choice (The Axiom of Countable Choice ()). Let , , and let and ; for put . Then If the inequality is to hold for all with a constant independent of , then the two powers of must balance: which is equivalent to . The exponent is therefore forced by scaling alone.
Facts & Assumptions
Given: Countable Choice; ; ; ; a fixed nonzero ; and .
The Sobolev conjugate is and satisfies (The Sobolev conjugate exponent and the scaling identity).
An invertible linear map scales Lebesgue measure by : substituting gives ; an class is determined up to null sets (A linear map of sends Lebesgue measurable sets to Lebesgue measurable sets, with when is invertible and Lebesgue null when it is not, The space as the quotient by null functions).
The classical chain rule computes for the linear map : (The chain rule for total derivatives: ).
The fundamental theorem on smooth line segments (If is differentiable with integrable then ; and a bounded derivative makes Lipschitz) implies that a smooth function with zero gradient is constant.
Verification
The norm scalings. For finite , by [F2] with , , so . For , the superlevel sets scale in measure by , so the essential supremum is unchanged; take . By the chain rule [F3], , so and , that is (if this norm were zero, continuity would give everywhere; [F4] would make constant, and compact support would force ).
The balance condition. Dividing the two identities of step 1.1, the proposed inequality reads for every ; after multiplying by this is . Since and are fixed positive numbers, a finite satisfying this for all exists exactly when the exponent vanishes, i.e. ; solving, , so 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 the conjugate exponent is selected by requiring the two sides to carry the same power of ; the exponent arises from the balance . 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.
Depends on
- The Sobolev conjugate exponent and the scaling identity
- The space $L^p(\mu)$ as the quotient by null functions
- A linear map $T$ of $\mathbb{R}^n$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $\lambda_n(T[E])=|\det T|\,\lambda_n(E)$ when $T$ is invertible and $T[E]$ Lebesgue null when it is not
- The chain rule for total derivatives: $D(g\circ f)(a)=Dg(f(a))\circ Df(a)$
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
- If $f : [a,b] \to \mathbb{R}^m$ is differentiable with integrable $f'$ then $\int_a^b f' = f(b)-f(a)$; and a bounded derivative makes $f$ Lipschitz
Used by
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Dependency tree · two levels
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Sources
- Juha Kinnunen, Sobolev Spaces (Aalto University, 2026, complete graduate lecture notes) (standard reference, not scraped)
- John K. Hunter, Notes on Partial Differential Equations (UC Davis, revised 18 June 2014, complete 242-page two-quarter notes) (standard reference, not scraped)