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LCA Fourier inversion is not an everywhere statement for arbitrary L^1 functions

Statement

Let G be a nondiscrete locally compact Hausdorff abelian group with Haar measure mG, and fix a Haar measure mG^ on its dual. If f∈L1(G,mG) and f^∈L1(G^,mG^), the inverse integral Ff(x):=∫G^f^(γ)γ(x) dmG^(γ) is defined at every x∈G and depends only on the L1 class of f. For every x0∈G, that class has a measurable representative whose value at x0 differs from Ff(x0) and which agrees with any given representative away from x0. Thus the inverse integral cannot recover an arbitrary representative pointwise.

Facts & Assumptions

Given: A nondiscrete locally compact Hausdorff abelian group G, Haar measure mG, a fixed Haar measure mG^ on its dual, a class f∈L1(G,mG) with f^∈L1(G^,mG^), a measurable representative u of f, and x0∈G.

[F2]

The L1 space is the quotient by null functions, so measurable representatives agreeing almost everywhere determine the same class (The space Lp(μ) as the quotient by null functions, Measure-null sets and almost-everywhere statements relative to a measure).

[F3]

The Fourier transform is defined on the L1 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 f^∈L1(G^,mG^).

Proof technique: direct.

Counterexample

1.1F1algebra

(Singletons are Haar-null.) Put a:=mG({0}), which is finite because {0} is compact. If a>0, translation invariance gives mG({x})=a for every x∈G. The open neighborhood U in [F1] is infinite: if it were finite, then U∖{0} would be closed in the Hausdorff space G, making {0}=U∩(G∖(U∖{0})) open and G discrete. For every positive integer N, choose N distinct points of U; their disjoint singletons lie in K, so finite additivity gives Na≤mG(K). Since this holds for all N and mG(K)<∞, a=0. Translation invariance then gives mG({x})=0 for every x∈G.

2.1F2step 1.1

(A point change preserves the class.) Define u~ to agree with u off {x0} and choose its value at x0 to be any complex number different from u(x0) and from Ff(x0). Such a value exists because C is infinite. By step 1.1, u and u~ agree almost everywhere, so [F2] gives [u]=[u~]=f.

3.1F3step 2.1

(The inverse integral cannot distinguish the representatives.) By [F3], u^=u~^=f^ on G^. Therefore both representatives give the same absolutely convergent inverse integral Ff(x0), while u~(x0)≠Ff(x0) by construction. This is an explicit failure of pointwise recovery for an arbitrary representative.

4.1step 1.1step 3.1∎

The modification leaves the L1 class and its Fourier transform unchanged but changes the value at x0; hence no inverse formula determined by the transform can hold everywhere for every representative.

Depends on

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