LemmaStatement: AI-generatedProof: AI-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

One flat reflection with simultaneous Lq and W1q bounds

Statement

Assume AC. Let $n \geq 1$ be an integer, $1 < q < \infty$ , and $K \in {R, C}$ . Write $x = (y, t) \in R^{n - 1} \times R$ and $H_{+} = R^{n - 1} \times (0, \infty)$ . Define one map $R : L^{q} (H_{+}; K) \to L^{q} (R^{n}; K)$ by $(R f) (y, t) = f (y, ∣ t ∣) (t \neq 0), (R f) (y, 0) = 0.$ It is independent of representatives, linear, restricts to $f$ almost everywhere on $H_{+}$ , and satisfies $∥ R f ∥_{L^{q} (R^{n})} = 2^{1 / q} ∥ f ∥_{L^{q} (H_{+})} .$ The same $R$ maps $W^{1, q} (H_{+}; K)$ into $W^{1, q} (R^{n}; K)$ and satisfies $∥ R f ∥_{W^{1, q} (R^{n})} = 2^{1 / q} ∥ f ∥_{W^{1, q} (H_{+})}, ∥ u ∥_{W^{1, q}}^{q} = ∥ u ∥_{L^{q}}^{q} + \sum_{j = 1}^{n} ∥ \partial_{x_{j}} u ∥_{L^{q}}^{q} .$

The compact-cylinder assertion is an almost-everywhere vanishing assertion: for any compact $K \subset R^{n - 1}$ and $a > 0$ , set $C_{+} = K \times [0, a]$ and $C = K \times [- a, a]$ . If $f = 0$ almost everywhere on $H_{+} ∖ C_{+}$ , then $R f = 0$ almost everywhere on $R^{n} ∖ C$ . For $n = 1$ , $R^{0}$ is the one-point space and the same conventions apply.

Facts & Assumptions

Given: The hypotheses and conventions of the Statement.

[F1]

Assume the Axiom of Choice The Axiom of Choice. Its countable instances supply the choice hypotheses of the Sobolev conventions and change-of-variables inputs below; the imported half-space theorem uses AC through its ACL, one-dimensional absolutely continuous and Fubini inputs. The reflection formula itself is explicit.

[F2]

Integer-order Sobolev spaces and their norms defines real and complex almost-everywhere Sobolev classes, with the component $ℓ^{q}$ -sum norm displayed in the Statement.

[F3]

Integer-order Sobolev extension from a half-space applies with $k = 1$ , $p = q$ and coefficient $a_{1} = 1$ . Its candidate derivatives and interface cancellation (Proof, steps 2.1 and 3.2) give the tangential even and normal odd weak derivatives of this reflection, without imposing zero boundary values on $f$ .

[F4]

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 implies that the linear involution $S (y, t) = (y, - t)$ preserves Lebesgue measurability and null sets, since $∣ \det S ∣ = 1$ . Its singular-map clause also makes ${t = 0}$ null: this hyperplane is the image of the singular linear map $(y, t) \mapsto (y, 0)$ .

[F5]

A C¹ diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions applies to the $C^{1}$ diffeomorphism $S : H_{-} \to H_{+}$ , where $H_{-} = R^{n - 1} \times (- \infty, 0)$ , with $∣ \det D S ∣ = 1$ . For any nonnegative Lebesgue measurable $g$ on $H_{+}$ it gives $\int_{H_{-}} g (y, - t) d y d t = \int_{H_{+}} g (y, t) d y d t$ .

Proof

Given: The hypotheses and data in the Statement.

Goal: Prove all stated conclusions for the single reflection map $R$ .

1.1F1F2F4F5given

By [F4], composition with $S$ on $H_{-}$ preserves measurability and carries the null exceptional set for equality of two representatives to a null set. The prescribed zero values on ${t = 0}$ introduce no ambiguity, since that hyperplane is null. Hence the formula defines a class independent of representatives and is linear; its restriction to $H_{+}$ is $f$ . Apply [F5] with $g = ∣ f ∣^{q}$ and split the integral over the two half-spaces, obtaining $\int_{R^{n}} ∣ R f ∣^{q} = 2 \int_{H_{+}} ∣ f ∣^{q}$ . This proves both membership in $L^{q}$ and the exact norm identity, before any differentiability is assumed.

2.1F4step 1.1given

Suppose $f = 0$ almost everywhere outside $C_{+}$ in $H_{+}$ , and let $N \subset H_{+} ∖ C_{+}$ be its null exceptional set. On the upper half-space outside $C$ the reflected function vanishes except on $N$ , and on the lower half-space outside $C$ it vanishes except on $S (N)$ . By [F4] the latter set is null as well, so the asserted almost-everywhere vanishing outside $C$ follows. This is invariant under changing representatives. If $K$ is contained in an open chart base $B$ and $a < b$ , then $C$ is a compact subset of the larger doubled cylinder $B \times (- b, b)$ , so lateral and top margins are retained.

3.1F1F2F3F5step 1.1∎

For $f \in W^{1, q} (H_{+}; K)$ , [F3] identifies the same formula with the $k = 1$ half-space extension. Its weak tangential derivatives are $(\partial_{y_{j}} f) (y, ∣ t ∣)$ for $1 \leq j \leq n - 1$ , and its weak normal derivative is $sgn (t) (\partial_{t} f) (y, ∣ t ∣)$ away from the null hyperplane. The supplier's interface cancellation proves these identities on all of $R^{n}$ ; the two zeroth-order traces match, so no surface term remains. Apply the integral calculation of step 1.1 to each derivative class, whose reflected modulus is unchanged by the normal sign. Each derivative's $q$ th power norm doubles. Summing these identities with the identity for $f$ using [F2] proves the exact $W^{1, q}$ norm identity for this same $R$ . When $n = 1$ the tangential list is empty.

Depends on

Used by · 1

Uniform endpoint localization in a smooth boundary chart

Dependency tree · 2 levels

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