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LCA Fourier inversion is not an everywhere statement for arbitrary L^1 functions
Statement
Let be a nondiscrete locally compact Hausdorff abelian group with Haar measure , and fix a Haar measure on its dual. If and , the inverse integral is defined at every and depends only on the class of . For every , that class has a measurable representative whose value at differs from and which agrees with any given representative away from . Thus the inverse integral cannot recover an arbitrary representative pointwise.
Facts & Assumptions
Given: A nondiscrete locally compact Hausdorff abelian group , Haar measure , a fixed Haar measure on its dual, a class with , a measurable representative of , and .
There is a compact neighborhood of the identity and an open neighborhood with (Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space, Topological group: multiplication and inversion are continuous, In a locally compact Hausdorff space every open set containing a point contains an open set containing it whose closure is compact and still inside; such a space is regular). Haar measure is finite on compact sets, so (Haar measure is positive on nonempty open sets and finite on compact sets).
The space is the quotient by null functions, so measurable representatives agreeing almost everywhere determine the same class (The space as the quotient by null functions, Measure-null sets and almost-everywhere statements relative to a measure).
The Fourier transform is defined on the class; equal representatives have equal transforms (The Fourier transform on an LCA group, The Pontryagin dual with the compact-open topology). Since every character has modulus one, the inverse integral is absolutely convergent at each point when .
Proof technique: direct.
Counterexample
(Singletons are Haar-null.) Put , which is finite because is compact. If , translation invariance gives for every . The open neighborhood in [F1] is infinite: if it were finite, then would be closed in the Hausdorff space , making open and discrete. For every positive integer , choose distinct points of ; their disjoint singletons lie in , so finite additivity gives . Since this holds for all and , . Translation invariance then gives for every .
(A point change preserves the class.) Define to agree with off and choose its value at to be any complex number different from and from . Such a value exists because is infinite. By step 1.1, and agree almost everywhere, so [F2] gives .
(The inverse integral cannot distinguish the representatives.) By [F3], on . Therefore both representatives give the same absolutely convergent inverse integral , while by construction. This is an explicit failure of pointwise recovery for an arbitrary representative.
The modification leaves the class and its Fourier transform unchanged but changes the value at ; hence no inverse formula determined by the transform can hold everywhere for every representative.
Depends on
- Measure-null sets and almost-everywhere statements relative to a measure
- The Fourier transform on an LCA group
- The space $L^p(\mu)$ as the quotient by null functions
- Left Haar integral and left Haar measure
- Radon measure on an LCH space
- Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space
- Topological group: multiplication and inversion are continuous
- Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not
- In a locally compact Hausdorff space every open set containing a point contains an open set containing it whose closure is compact and still inside; such a space is regular
- Haar measure is positive on nonempty open sets and finite on compact sets
- The Pontryagin dual with the compact-open topology
Used by
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Sources
- Manfred Einsiedler and Thomas Ward, Ergodic Theory with a View Towards Number Theory, Appendix C.2-C.3 (course-hosted full text) (standard reference, not scraped)