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The character topology on L^1 of an LCA group is the compact-open topology

Statement

Assume Dependent Choice. Let G be a locally compact Hausdorff abelian group with Haar measure mG and A=L1(G,mG). Under the bijection γ↦hγ of Nonzero multiplicative functionals on L^1 of an LCA group are Fourier evaluations, the compact-open topology of G^ is the topology of pointwise evaluation on all of A: a net (hi) converges to h0 in the topology of pointwise convergence on A if and only if the corresponding characters converge to γ0 uniformly on every compact subset of G. Consequently the algebraically defined character space of A carries exactly the compact-open topology of G^, and the resulting identification is a homeomorphism.

Facts & Assumptions

Given: Dependent Choice, a locally compact Hausdorff abelian group G with Haar measure mG, A=L1(G,mG), the bijection γ↦hγ(f)=f^(γ) onto the nonzero multiplicative linear functionals (Nonzero multiplicative functionals on L^1 of an LCA group are Fourier evaluations), a net (γi)i∈I in G^ with compact-open limit γ0, and a net (hi) evaluating pointwise to h0 on A (Directed preorders and nets, The topology of pointwise convergence on YX, which is the product topology, and its restriction to C(X,Y), The compact-open topology on C(X,Y) for arbitrary topological spaces).

[F1]

hγ(Txf)=γ(x)‾ hγ(f) for all x∈G, f∈A: substituting y↦y−x (a translation, so mG-preserving) gives hγ(Txf)=∫Gf(y−x)γ(y)‾ dmG(y)=∫Gf(z)γ(z+x)‾ dmG(z)=γ(x)‾∫Gf(z)γ(z)‾ dmG(z) (The Pontryagin dual with the compact-open topology, The multiplicative unit circle is a compact metrizable topological abelian group).

[F4]

For an arbitrary abelian topological group, the sets UK(γ,r):={η:∣η(x)−γ(x)∣<r for every x∈K}, with K compact and r>0, form a neighbourhood basis in its compact-open dual. Thus compact-open convergence of characters is exactly uniform convergence on every compact set; no metrizability of G or choice axiom is required (The compact-open character group is a Hausdorff topological abelian group, Statement and proof 1.2 and 2.1).

Proof

technique · direct
1.1F2F3F4

(Compact-open convergence gives evaluation convergence.) Assume γi→γ0 in the compact-open topology; by [F4] this is uniform convergence on compacta. Fix f∈A, ε>0. By [F2] choose f0∈Cc(G) with ∥f−f0∥1<ε/3 and put K:=supp⁡f0. Then for every i, using [F2] and the definition of hγ, ∣hγi(f)−hγ0(f)∣≤2∥f−f0∥1+∫K∣f0(y)∣ ∣γi(y)−γ0(y)∣ dmG(y)≤2ε3+∥f0∥∞ mG(K)sup⁡y∈K∣γi(y)−γ0(y)∣, and the supremum tends to 0 along the net; hence hγi(f)→hγ0(f) for every f∈A.

1.2F2F3

(Evaluation convergence gives compact-open convergence.) Conversely, assume hi→h0 pointwise on A, say hi=hγi and h0=hγ0. Choose f∈A with h0(f)≠0; then ∣hi(f)∣≥∣h0(f)∣/2 for all sufficiently large i. Let K⊆G be compact and δ>0. By [F2] the set {Txf:x∈K} is a continuous image of K, hence compact, so finitely many of its points Tx1f,…,Txmf cover it by δ-balls; put ηi:=max⁡j∣hi(Txjf)−h0(Txjf)∣, which tends to 0 by pointwise convergence. For every x∈K, choosing j with ∥Txf−Txjf∥1<δ and using [F2] gives ∣hi(Txf)−h0(Txf)∣≤2δ+ηi, so sup⁡x∈K∣hi(Txf)−h0(Txf)∣≤2δ+ηi.

2.1F1F4step 1.2

(Uniform convergence of the characters.) By [F1], hi(Txf)−h0(Txf)=γi(x)‾hi(f)−γ0(x)‾h0(f) for every x. For large i the denominator hi(f) satisfies ∣hi(f)∣≥∣h0(f)∣/2>0, and γi(x)‾−γ0(x)‾=hi(Txf)−h0(Txf)+γ0(x)‾(h0(f)−hi(f))hi(f). Taking suprema over x∈K and using step 1.2 together with hi(f)→h0(f) gives sup⁡x∈K∣γi(x)−γ0(x)∣≤2(2δ+ηi+∥h0(f)−hi(f)∥)/∣h0(f)∣ eventually, which tends to 4δ/∣h0(f)∣; since K and δ>0 are arbitrary, γi→γ0 uniformly on every compact subset of G, hence in the compact-open topology by [F4].

3.1step 1.1step 2.1

(The homeomorphism.) Steps 1.1 and 2.1 show that under the bijection γ↦hγ the compact-open topology of G^ corresponds exactly to the topology of pointwise convergence on A; thus the algebraically defined character space of A carries the compact-open topology of G^ and the identification is a homeomorphism.

4.1step 1.1step 2.1step 3.1∎

Together with the bijection of Nonzero multiplicative functionals on L^1 of an LCA group are Fourier evaluations, steps 1.1 and 2.1 prove both implications of the stated equivalence, and step 3.1 records the homeomorphism.

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