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Pontryagin biduality: the evaluation map is a topological isomorphism

Statement

Assume the Axiom of Choice (The Axiom of Choice) and Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let G be a locally compact Hausdorff abelian group with dual G^ and bidual G^^. Then Φ:G→G^^,Φ(x)(γ):=γ(x), is an isomorphism of topological groups.

Facts & Assumptions

Given: A locally compact Hausdorff abelian group G with dual G^ and bidual G^^, and the evaluation map Φ.

[F1]

Φ is a continuous group homomorphism whose image is a subgroup of G^^; the sets Nx1(K,ϵ)={x:∣χ(x)−χ(x1)∣<ϵ for all χ∈K} over compact K⊆G^ and ϵ>0 form a neighbourhood basis at x1, and Φ is a homeomorphism onto its image. (Compact-open neighbourhoods on the dual give a neighbourhood basis on the group, The Pontryagin dual with the compact-open topology, Continuity of a map of topological spaces at a point and globally)

[F2]

Φ is injective: continuous characters separate points. (Continuous characters separate points of an LCA group)

[F5]

Bump on the dual of G^. Applied to the locally compact abelian group G^ with its Haar measure mG^: for ξ0∈G^^ and a compact neighbourhood K of ξ0 there is f∈L1(G^,mG^) with f^≥0 on G^^, f^(ξ0)>0 and f^=0 on G^^∖K, where f^(ξ)=∫G^f(γ)ξ(γ)‾ dmG^(γ). (Compactly supported nonnegative transform bumps on the dual, The Fourier transform on an LCA group, A compact identity neighbourhood in the dual)

[F6]

A finite regular complex Borel measure μ on G^ whose inverse transform x↦∫G^γ(x) dμ(γ) vanishes for every x∈G is zero. (Fourier-Stieltjes transforms determine finite Radon measures, Regular complex Borel measures)

[F7]

For f∈L1(G^) the measure μ=f mG^ is finite and regular: approximate f in L1 by hn∈Cc(G^), each hn mG^ is finite regular as follows. For bounded h supported in compact K, outer Haar approximations O1⊇E∩K and O2⊇K∖E give an open superset O1∪(G^∖K) of E and a compact subset K∖O2 of E, with weighted errors bounded by ∥h∥∞ times the arbitrarily small Haar errors (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones, A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact). Finally, ∥(f−hn)mG^∥TV=∥f−hn∥1→0, while total-variation limits of finite regular complex measures are finite regular (outer and inner regularity transfer from an approximant with error control). (A complex L^1 density defines a complex measure whose total variation is |h| dmu, Radon measure on an LCH space, C_c(X) is dense in L^p(mu) for a Radon measure, Compact support, Cc(X), and C0(X))

Proof

1.1F1F2

The evaluation map is a group homomorphism: Φ(x+y)(γ)=γ(x+y)=γ(x)γ(y)=(Φ(x)Φ(y))(γ) for all x,y∈G and γ∈G^; it is continuous and a homeomorphism onto its image by [F1], and injective by [F2]. Hence Φ(G) is a subgroup of G^^ isomorphic to G as a topological group.

2.1F1F3F4step 1.1

Since G is locally compact and Φ is a homeomorphism onto its image, the subgroup Φ(G) is locally compact in the subspace topology, so it is closed in the Hausdorff group G^^ by [F4].

3.1F3step 2.1

Suppose that Φ(G)≠G^^. Since Φ(G) is closed and G^^ is locally compact Hausdorff, pick ξ0∈G^^∖Φ(G) and a compact neighbourhood K of ξ0 contained in G^^∖Φ(G).

4.1F5step 3.1

Apply [F5] to G^: there is f∈L1(G^,mG^) with f^≥0, f^(ξ0)>0 and f^=0 on G^^∖K.

5.1F6F7step 2.1step 3.1step 4.1

Let μ:=f mG^, a finite regular complex measure on G^ by [F7]. For every x∈G the inverse transform of μ at x is ∫G^γ(x) dμ(γ)=∫G^f(γ)γ(x) dmG^(γ)=f^(Φ(−x))=0, because Φ(−x)∈Φ(G) lies outside K and f^ vanishes off K. Hence [F6] gives μ=0, so f=0 almost everywhere and therefore f^=0 everywhere, contradicting f^(ξ0)>0. Thus Φ(G)=G^^.

6.1step 1.1step 5.1∎

Consequently Φ is an injective, continuous, open map onto G^^, hence an isomorphism of topological groups; this is the statement.

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