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LOCAL COMPACTNESS OF THE DUAL GROUP USING ASCOLI
b be its dual group, which is also abelian.
Let G be a locally compact abelian group and G
b
We will explain how to make G into a locally compact group using the compact-open
topology.
b is a topological group in the
Theorem 1. If G is a locally compact abelian group, then G
compact-open topology.
b the basic open sets around χ in the compact-open topology on G
b are of
Proof. For χ ∈ G,
b
the form Nχ (K, ε) = {ψ ∈ G : |ψ(x) − χ(x)| < ε for all x ∈ K}, where K is a nonempty
b×G
b → G
b and
compact subset of G and ε > 0. We need to show multiplication m : G
b
b
b
inversion G → G are continuous for the compact-open topology on G.
b a nonempty compact
To show multiplication is continuous, pick characters χ and χ0 in G,
subset K in G, and ε > 0. Then Nχ (K, ε/2) × Nχ0 (K, ε/2) is an open set around (χ, χ0 )
b×G
b that m maps into Nχχ0 (K, ε) by the triangle inequality: if ψ ∈ Nχ (K, ε/2) and
in G
0
ψ ∈ Nχ0 (K, ε/2), then for all x ∈ K we have
|ψ(x)ψ 0 (x) − χ(x)χ0 (x)| =
≤
<
=
|(ψ(x) − χ(x))ψ 0 (x) + (ψ 0 (x) − χ0 (x))χ(x)|
|ψ(x) − χ(x)| + |ψ 0 (x) − χ0 (x)|
ε/2 + ε/2
ε,
so ψψ 0 ∈ Nχχ0 (K, ε).
b a nonempty compact subset K of G, and
To show inversion is continuous, pick χ ∈ G,
ε > 0. Then Nχ (K, ε) is an open set containing χ and inversion maps it into Nχ−1 (K, ε):
χ(x) − ψ(x) −1
−1
= |χ(x) − ψ(x)| < ε,
ψ ∈ Nχ (K, ε), x ∈ K =⇒ |ψ (x) − χ (x)| = ψ(x)χ(x) so ψ −1 ∈ Nχ−1 (K, ε).
b is locally compact in the compact-open topology, the usual proof proceeds
To show G
through Banach algebras, Alaoglu’s theorem, L1 -L∞ duality, and a comparison between
b We will give
the compact-open topology and the topology of pointwise convergence on G.
b
a proof in Theorem 5 that G is locally compact in the compact-open topology using no
additional topologies, no Banach algebras, etc. Our main tool will be the standard theorem
describing when sets of functions are compact: Ascoli’s theorem.
Lemma 2. Fix f ∈ L1 (G). For y ∈ G, let Ly f : G → C by (Ly f )(x) = f (yx). The map
y 7→ Ly f from G to L1 (G) is continuous.
The lemma is proved by checking it directly on the dense subset Cc (G) ⊂ L1 (G) and then
extending it to all of L1 (G) by an approximation argument. Details are left to the reader.
Next we prove a result about the decay of Fourier transforms. For f ∈ L1 (G), its Fourier
R
transform is fb(χ) = G f (x)χ(x) dx, where dx is some choice of left Haar measure on G.
b the function fb: G
b → C is (uniformly) continuous.
Using the compact-open topology on G,
1
2
LOCAL COMPACTNESS OF THE DUAL GROUP USING ASCOLI
b → C “vanishes at ∞”: for any ε > 0 there is a
Theorem 3. If f ∈ L1 (G) then fb: G
b such that |fb(χ)| < ε for all χ 6∈ C.
compact set C ⊂ G
b → C is continuous when G
b has the compact-open topology, our task
Proof. Since fb: G
would follow from showing for any ε > 0 that the (closed) set
b : |fb(χ)| ≥ ε}
C := {χ ∈ G
b using the compact-open topology.
is compact in G
b
Since G is a closed subset of the space C(G, S 1 ) of continuous functions from G to S 1 ,
what we need to do is show the above set is compact in C(G, S 1 ). For this, Ascoli’s theorem
tells us exactly what has to be checked: equicontinuity of the characters in C at each point
of G. Since we’re dealing with characters and the compact-open topology, it is enough to
check equicontinuity of the characters in C at the identity e of G. So for each δ > 0 we
want to find an open neighborhood U = Uδ of e such that
y ∈ U, |fb(χ)| ≥ ε =⇒ |χ(y) − 1| < δ.
It’s not evident how to turn a lower bound on the Fourier transform at χ into an upper
bound on χ(y) − 1. The trick is to get a bound on |χ(y) − 1| where y doesn’t show up in
χ(y) anymore.
b such that |fb(χ)| ≥ ε and any y ∈ G, we have
For any χ ∈ G
ε|χ(y) − 1| ≤ |(χ(y) − 1)fb(χ)|
Z
= (χ(y) − 1)
f (x)χ(x) dx
G
Z
Z
f (x)χ(x) dx
= f (x)χ(xy) dx −
G
ZG
Z
−1
f (x)χ(x) dx
= f (xy )χ(x) dx −
G
ZG
−1
= (f (xy ) − f (x))χ(x) dx
ZG
f (xy −1 ) − f (x) dx
≤
G
= |Ly−1 f − f |1 ,
so
1
|χ(y) − 1| ≤ |Ly−1 f − f |1 .
ε
From continuity of y 7→ Ly f and continuity of inversion on G, |Ly−1 f − f |1 → 0 as y → e
in G. Therefore |χ(y) − 1| < δ for all y near e, and that level of nearness to e gives us the
desired set U .
Remark 4. Theorem 3 is a generalization of the Riemann–Lebesgue lemma, which is the
special case G = S 1 = R/2πZ: if f : R → C is 2π-periodic and integrable then its Fourier
R 2π
coefficients fb(n) = 0 f (x)e−2πinx dx tend to 0 as |n| → ∞.
b is locally compact in
Theorem 5. When G is a locally compact abelian group, the group G
the compact-open topology.
LOCAL COMPACTNESS OF THE DUAL GROUP USING ASCOLI
3
b is a topological group, it suffices to show there is a neighborhood basis of the
Proof. Since G
trivial character 1 consisting of open subsets with compact closures. Every neighborhood
b contains some N1 (K, ε) = {χ ∈ G
b : |χ(x) − 1| < ε for all x ∈ K}, where K is a
of 1 in G
nonempty compact subset of G and ε > 0.
When 0 < ε < 1 and K has positive Haar measure, we will show N1 (K, ε) has compact
b by copying the argument from [1, p. 362]. Let f = ξK be the characteristic
closure in G
function of K, so f ∈ L1 (G) because compact subsets of G have finite Haar measure. Let
b
µ(K) be the Haar measure of K. For any χ ∈ G,
Z
Z
Z
ξK · χ dx.
ξK · (1 − χ) dx +
ξK dx =
µ(K) =
G
G
G
Taking absolute values, if χ ∈ N1 (K, ε) then µ(K) ≤ µ(K)ε + |ξbK (χ)|, so
|ξbK (χ)| ≥ (1 − ε)µ(K) > 0.
By Theorem 3, the set of χ fitting this inequality is a compact set, so N1 (K, ε) has compact
b
closure in G.
What if ε ≥ 1 or K has measure 0? We want to reduce these cases to the previous case
by passing to a smaller neighborhood where both ε < 1 and K has positive Haar measure.
Making ε smaller makes N1 (K, ε) smaller, so by passing to a suitable smaller neighborhood
of 1 we can assume ε < 1. Making K larger makes N1 (K, ε) smaller, so we can reduce to
the case that K has positive measure if we show any compact subset of G with measure 0 is
contained in a compact subset of G with positive measure. Since G itself has positive Haar
measure, by inner regularity of Haar measure there’s a compact subset L of G such that L
has positive measure. Then K ∪ L is compact with positive measure and contains K. References
[1] E. Hewitt and K. A. Ross, “Abstract Harmonic Analysis, I”, Springer-Verlag, New York, 1963.