Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)
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Parseval pairing on Schwartz space

Statement

Assume countable choice. For f,gS(Rn), f^(ξ)g^(ξ)dξ=f(x)g(x)dx. The pairing is complex-linear in the first variable. In particular f^2=f2.

Facts & Assumptions

[F1]

Schwartz inversion holds everywhere with an absolutely integrable transform (Fourier inversion on Schwartz space).

[F2]

Schwartz functions are integrable and bounded (Schwartz derivatives are integrable).

[F3]

Fubini applies to absolutely integrable complex product functions on sigma-finite spaces (Fubini's theorem for L^1 functions on a sigma-finite product).

Proof

technique · direct
1.1

By [F1], g(x)=g^(ξ)e2πixξdξ. The integrand after multiplying by f(x) is jointly measurable and has absolute double integral f1g^1<, by [F1], [F2] and product integration. Hence [F3] gives fg=g^(ξ)[f(x)e2πixξdx]dξ=f^g^. This also proves absolute integrability of the final product; the original product is integrable since g is bounded and f integrable.

F1F2F3given
2.1

Taking g=f gives equality of the nonnegative square integrals, finite by [F2] for the input and by step 1.1 for its transform. Taking nonnegative square roots proves the norm identity. The displayed pairing is linear in its first entry and conjugate-linear in its second directly from integration and conjugation.

step 1.1F2algebra

Depends on

Used by

Dependency tree · two levels

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Sources