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Continuous characters separate points of an LCA group

Facts & Assumptions

Given: A locally compact Hausdorff abelian group G and a point x∈G with x≠0.

[F2]

If K⊆U with K compact and U open in a locally compact Hausdorff space X, then under Dependent Choice there is f∈Cc(X) with 1K≤f≤1U. (LCH Urysohn cutoff)

[F3]

For g∈Cc(G) put g~(x):=g(−x)‾. Then g∗g~∈Cc(G) is continuous with compact support, is positive definite, and (g∗g~)(0)=∫G∣g∣2 dmG≥0; the convolution is (f∗g)(x)=∫Gf(y)g(x−y) dmG(y) and Cc(G)⊆L1(G)∩L2(G). A left Haar measure is strictly positive on nonzero nonnegative compactly supported functions, so ∫G∣g∣2>0 whenever g≢0. (Positive convolution squares form a dense inversion core, L^1 of an LCA group is a commutative Banach star algebra under convolution, Compact support, Cc(X), and C0(X), Haar measure is positive on nonempty open sets and finite on compact sets, Left Haar integral and left Haar measure)

[F4]

Bochner's theorem: a continuous function ϕ:G→C is positive definite if and only if there is a unique finite positive Radon measure μ on G^ with ϕ(x)=∫G^γ(x) dμ(γ) for all x, and then μ(G^)=ϕ(0). (Bochner's theorem for LCA groups, Positive definite functions on an abelian group, Fourier-Stieltjes transforms of positive measures are continuous positive definite)

[F5]

For g∈Cc(G) and y∈G one has g(y−x)‾=g~(x−y), and x∉S−S is equivalent to S∩(x+S)=∅. (Topological group: multiplication and inversion are continuous, Left and right translations and inversion in a topological group are homeomorphisms)

Proof

1.1F1

Because x≠0 and G is Hausdorff, addition is continuous at (0,0) and G∖{x} is open, so there are open neighbourhoods W1,W2 of 0 with W1+W2⊆G∖{x}; replacing them by their intersections with their negatives and with each other, we obtain a symmetric open neighbourhood W of 0 with (W+W)∩{x}=∅.

2.1F1F5step 1.1

Choose a symmetric compact neighbourhood S of 0 with S⊆W: a neighbourhood basis of open sets with compact closure at 0 supplies an open V with 0∈V⊆clG(V)⊆W, and S:=clG(V)∩(−clG(V)) is compact and symmetric with 0∈S⊆W. Then S−S⊆W+W, so x∉S−S; hence S∩(x+S)=∅, since s=x+s′ would give x=s−s′∈S−S.

3.1F2F3step 2.1

Apply the cutoff of [F2] with K={0} and U=intG(S) to obtain g∈Cc(G) with g≥0, g(0)=1 and supp⁡g⊆S. Then h:=g∗g~ is continuous with compact support, positive definite, and h(0)=∫G∣g∣2 dmG>0 because g≢0 and g≥0.

4.1F3F5step 2.1step 3.1

For this h one has h(x)=0: by the convolution formula and g~(x−y)=g(y−x)‾ from [F5], h(x)=∫Gg(y)g(y−x)‾ dmG(y), and the integrand vanishes identically because g(y)≠0 forces y∈S while g(y−x)≠0 forces y∈x+S, and S∩(x+S)=∅.

5.1F4step 3.1step 4.1

By Bochner's theorem [F4] there is a unique finite positive Radon measure μ on G^ with h(x)=∫G^γ(x) dμ(γ) for all x∈G and μ(G^)=h(0)>0. If γ(x)=1 for every γ∈G^, then h(x)=∫G^1 dμ(γ)=μ(G^)=h(0)>0, contradicting h(x)=0 from step 4.1. Hence some γ∈G^ satisfies γ(x)≠1.

6.1step 5.1∎

Since x≠0 was arbitrary, continuous characters separate points of G. Equivalently the evaluation map is injective: if Φ(x)=Φ(x′) then γ(x−x′)=γ(x)γ(−x′)=γ(x)γ(x′)‾=1 for every γ∈G^, the separation result forces x−x′=0, that is x=x′. Conversely, if Φ is injective and x≠0, then Φ(x)≠Φ(0), so some character has γ(x)≠1.

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Sources