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L1 Fourier inversion with an integrable transform
Statement
Assume countable choice. If and , then is bounded and continuous, equals f almost everywhere, and equals the specified value at every Lebesgue point of f.
Facts & Assumptions
Given: , and The Axiom of Countable Choice ().
Gaussian Fourier means recover each Lebesgue value (Gaussian Fourier summability at Lebesgue points).
An transform is bounded and uniformly continuous (The L1 transform is bounded and uniformly continuous).
Dominated convergence permits passage under an integral with a single integrable majorant (Dominated convergence).
Almost every point of a locally integrable function is a Lebesgue point under countable choice (Almost every point is a Lebesgue point of a locally integrable function).
Proof
Since , F2 applied to it shows that is bounded and uniformly continuous. For fixed and the explicit sequence , , the damped integrands converge pointwise to and have majorant . F3 therefore gives .
At any Lebesgue point with value a, F1 gives the same sequence limit a, so . Since is locally integrable, F4 gives such points with a=f(x) outside a null set (for complex inputs apply the real conclusion to both components and bound the complex oscillation by their sum). Hence almost everywhere. In particular g represents the original class. Countable choice is precisely inherited from F1 and F4; no statement about all values of an arbitrary representative follows.
Depends on
- Gaussian Fourier summability at Lebesgue points
- The L1 transform is bounded and uniformly continuous
- Translation, modulation, linear dilation and reflection laws
- Dominated convergence
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
- Almost every point is a Lebesgue point of a locally integrable function
Used by
- Uniqueness of the L1 Fourier transform Corollary
- Null-set modifications defeat everywhere representative recovery Counterexample
- Poisson kernel transform and Abel summability on the line Example
- Fourier inversion on Schwartz space Theorem
- Fourier transform of a product with one integrable transform Theorem
Dependency tree · two levels
40 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
- Gerald Teschl, Topics in Real and Functional Analysis (2017) (standard reference, not scraped)