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Pointwise restriction is not defined on Lp equivalence classes

Statement refuted

Assume Countable Choice and n≥2. Statement refuted: for 1<p<2 the pointwise restriction f^∣Sn−1 is well defined by the ambient Lp′ class of the Hausdorff-Young transform. Data: let f be a nonzero Schwartz function on Rn and let F=Fpf∈Lp′(Rn) be its transform class, with representative f^; put G:=f^+1Sn−1 pointwise. Since λn(Sn−1)=0, G represents the same Lp′ class, but G∣Sn−1=f^∣Sn−1+1 differs from f^∣Sn−1 everywhere on the sphere. Hence restriction cannot be read off an ambient Lp′ representative; it begins on Schwartz functions and, when a restriction estimate holds at the chosen exponent, extends by density as in Fourier restriction and adjoint extension operators.

Facts & Assumptions

[F1]

Hausdorff–Young: for 1≤p≤2 the transform extends to a bounded map Fp:Lp→Lp′ that agrees almost everywhere with the integral transform on L1∩Lp; in particular f^ is a representative of Fpf when f is Schwartz. (Hausdorff–Young for the Euclidean Fourier transform)

[F2]

Complex Lp classes are quotients of measurable functions by almost-everywhere equality, with representative-independent norm; S(Rn) consists of actual smooth functions and 1Sn−1 is Borel measurable. (Complex Lp classes and Euclidean test-function conventions, Schwartz space and its seminorms)

[F3]

The unit sphere is Lebesgue null: λn(Sn−1)=0, and a nonnegative measurable function with finite integral is finite almost everywhere; adding an indicator of a null set changes a function only on a null set. (The unit sphere is Lebesgue null, A nonnegative measurable function with finite integral is finite almost everywhere)

[F4]

For 1<p<∞, a restriction bound on Schwartz data yields the unique bounded extension R:Lp→L2(σ); its existence is conditional on that bound. (Restriction and extension estimates are dual)

Counterexample

Given: Countable Choice, n≥2, 1<p<2, a nonzero Schwartz function f∈S(Rn), its integral transform f^, the Hausdorff–Young class F=Fpf∈Lp′(Rn) with representative f^, and G:=f^+1Sn−1.

1.1F1F2F3

The two representatives coincide almost everywhere. The function 1Sn−1 is Borel measurable by [F2], and G=f^+1Sn−1 is measurable. Since Sn−1 is Lebesgue null by [F3], G=f^ almost everywhere. Hence ∫∣G∣p′=∫∣f^∣p′<∞ by the Hausdorff–Young membership in [F1], and G belongs to Lp′(Rn) and represents the same class as f^, namely F=Fpf: two almost-everywhere equal integrable functions define the same quotient class by [F2].

1.2givenalgebra

The restrictions differ at every point of the sphere. By construction G(ω)=f^(ω)+1 for every ω∈Sn−1, so the pointwise restrictions satisfy G∣Sn−1=f^∣Sn−1+1Sn−1. The difference is 1 at every point of the nonempty sphere Sn−1, so the two restrictions are different functions on Sn−1.

2.1step 1.1step 1.2

The refutation. Suppose that the pointwise restriction were well defined by the ambient Lp′ class, that is, that two representatives of one class always have equal restrictions to Sn−1. Steps 1.1 and 1.2 exhibit two representatives f^ and G of the same class F=Fpf whose restrictions differ everywhere on Sn−1; this contradicts the supposition. Therefore pointwise restriction is not well defined on Lp′ classes.

3.1F2F4given∎

The correct convention. The restriction operator R0 of Fourier restriction and adjoint extension operators is defined on the actual functions f^ with f∈S(Rn), where the pointwise values exist. If a bound ∥R0f∥L2(σ)≤C∥f∥Lp holds on Schwartz data, density gives its unique bounded extension R:Lp(Rn)→L2(σ), as in Restriction and extension estimates are dual; no pointwise restriction of a general ambient class is asserted.

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