TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30

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.

Hausdorff–Young for the Euclidean Fourier transform

Statement

Assume Countable Choice, let $n \geq 1$ , and let $\hat{f} (ξ) = \int_{R^{n}} f (x) e^{- 2 π i x \cdot ξ} d x$ be the integral Fourier transform on $L^{1} (R^{n}; C)$ . For $1 \leq p \leq 2$ with conjugate exponent $p^{'}$ , the transform extends compatibly to a complex-linear bounded map $F_{p} : L^{p} (R^{n}; C) ⟶ L^{p^{'}} (R^{n}; C), ∥ F_{p} f ∥_{p^{'}} \leq ∥ f ∥_{p} .$ At $p = 1$ this map is the integral transform of The L1 transform is bounded and uniformly continuous, at $p = 2$ it is the unitary Plancherel transform of Plancherel theorem, and on $L^{1} \cap L^{2}$ the two interpretations agree almost everywhere (Agreement of the integral and L2 transforms). For $1 < p < 2$ the extension agrees almost everywhere with the integral transform on its intersection with $L^{1}$ and with the Plancherel transform on its intersection with $L^{2}$ . No statement for $p > 2$ and no pointwise representative identity is claimed.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , an exponent $1 \leq p \leq 2$ , and, where required, $f \in L^{p} (R^{n}; C)$ .

[A1]

Countable Choice is carried by the Plancherel and interpolation interfaces cited below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $f \in L^{1} (R^{n}; C)$ the integral $\hat{f} (ξ)$ converges absolutely for every $ξ$ , is unchanged by null-set modifications, and satisfies $∣ \hat{f} (ξ) ∣ \leq ∥ f ∥_{1}$ (The integral transform is representative independent).

[F2]

The integral transform is a complex-linear map $L^{1} (R^{n}; C) \to B U C (R^{n}; C)$ with $\sup_{ξ} ∣ \hat{f} (ξ) ∣ \leq ∥ f ∥_{1}$ , so its classes are bounded measurable classes on the sigma-finite Lebesgue space (The L1 transform is bounded and uniformly continuous).

[F3]

Plancherel extends the Schwartz transform to a surjective complex-linear isometry $F_{2} : L^{2} \to L^{2}$ (Plancherel theorem).

[F4]

If $f \in L^{1} \cap L^{2}$ , the bounded continuous integral transform $\hat{f}$ represents $F_{2} f$ almost everywhere (Agreement of the integral and L2 transforms).

[F5]

On sigma-finite measure spaces a complex-linear finite-simple-core operator with $∥ T g ∥_{\infty} \leq A ∥ g ∥_{1}$ and $∥ T g ∥_{2} \leq B ∥ g ∥_{2}$ satisfies, for $1 < p < 2$ , $∥ T g ∥_{p^{'}} \leq A^{2 / p - 1} B^{2 - 2 / p} ∥ g ∥_{p}$ , retains the endpoint estimates at $p = 1, 2$ , and has unique compatible bounded extensions to the full $L^{p}$ spaces under countable choice (Interpolate L1 to Linfinity and L2 to L2 bounds).

[F6]

Every two extensions of the same finite-simple core operator agree as measurable almost-everywhere classes on their domain intersection (Compatible extensions from the finite simple core).

[F7]

Complex finite simple functions with finite-measure nonzero sets are dense in $L^{q} (R^{n}; C)$ for $1 \leq q < \infty$ (Complex finite-simple and smooth compact-support density for finite p).

[F8]

Complex $L^{p}$ classes, their norms and almost-everywhere equality are those of Complex Lp classes and Euclidean test-function conventions.

Proof

technique · interpolate the L1 and L2 endpoint bounds on the finite-simple core and identify the extensions with the two known transforms

1.1F1F2F8given

Let $T$ send the almost-everywhere class of a complex finite simple function $s$ with finite-measure nonzero set on $R^{n}$ to the class of its integral transform $\hat{s}$ . The class is well defined and $T$ is complex-linear by [F1], [F2] and [F8].

2.1F1F2step 1.1

For such an $s$ one has $∥ T s ∥_{\infty} = \sup_{ξ} ∣ \hat{s} (ξ) ∣ \leq ∥ s ∥_{1}$ , the L^1-to-L-infinity endpoint bound with $A = 1$ .

2.2F3F4step 1.1

Such an $s$ lies in $L^{1} \cap L^{2}$ , so [F4] identifies $\hat{s}$ with $F_{2} s$ almost everywhere; [F3] then gives $∥ T s ∥_{2} = ∥ F_{2} s ∥_{2} = ∥ s ∥_{2}$ , the L^2-to-L^2 endpoint bound with $B = 1$ .

3.1F5step 2.1step 2.2

Applying [F5] to $T$ with the endpoints $(p_{0}, q_{0}) = (1, \infty)$ , $A = 1$ and $(p_{1}, q_{1}) = (2, 2)$ , $B = 1$ on the sigma-finite Lebesgue space $R^{n}$ gives, for every $1 < p < 2$ , a unique compatible bounded extension $F_{p} : L^{p} \to L^{p^{'}}$ with $∥ F_{p} f ∥_{p^{'}} \leq ∥ f ∥_{p}$ , while the endpoint estimates of steps 2.1 and 2.2 hold at $p = 1$ and $p = 2$ .

3.2F2F3F7step 2.2

The $L^{1}$ -extension of $T$ is the integral transform: by [F1] and [F2] the transform is a bounded linear map $L^{1} \to L^{\infty}$ agreeing with $T$ on the core, and the core is dense in $L^{1}$ by [F7]. Likewise the $L^{2}$ -extension of $T$ is $F_{2}$ , since by step 2.2 $F_{2}$ agrees with the core map on that dense core.

4.1F6step 3.1step 3.2

Fix $1 < p < 2$ and $f \in L^{p}$ . By [F6] the extension $F_{p} f$ agrees almost everywhere with the $L^{1}$ -extension on $L^{p} \cap L^{1}$ and with the $L^{2}$ -extension on $L^{p} \cap L^{2}$ ; by step 3.2 these are the integral transform and the Plancherel transform respectively.

5.1A1F3F4F5step 3.1step 4.1∎

The claims at $p = 1$ and $p = 2$ are steps 2.1 and 2.2 together with step 3.2, while for $1 < p < 2$ steps 3.1 and 4.1 give the bounded compatible extension and its agreement with the integral and Plancherel transforms on the respective intersections; the case $f \in L^{1} \cap L^{2}$ is [F4]. Countable Choice is used only through [F3] and [F5].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

59 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Loukas Grafakos, Classical Fourier Analysis, 3rd ed. (standard reference, not scraped)
Richard S. Laugesen, Harmonic Analysis Lecture Notes (standard reference, not scraped)