The Fourier Algebra of a Locally Compact Group · Given a locally compact group G, the Fourier algebra A(G) is a closed ideal inside the Fourier-Stieltjes algebra B(G) consisting

The Fourier Algebra of a Locally Compact

Group

John Sawatzky

August 2018

Abstract

Given a locally compact group G, the Fourier algebra A(G) is a

closed ideal inside the Fourier-Stieltjes algebra B(G) consisting of the

matrix coefficients associated with the left regular representation. The

Fourier algebra possesses a number of interesting properties, such as

the the fact that its spectrum is just G itself, that it has a bounded

approximate identity if and only if G is amenable, and that if H is a

closed subgroup of G then the map of restriction is surjective: A(G)|H =

A(H).

1

CONTENTS CONTENTS

Contents

1 Introduction 3

2 Preliminaries 4

3 The Fourier Algebra 9

3.1 The Construction of the Fourier-Stieltjes Algebra . . . . . . . 9

3.2 The Fourier Algebra . . . . . . . . . . . . . . . . . . . . . . . 11

4 The Spectrum of A(G) 18

5 Leptin’s Theorem 21

6 Restriction Map to Subgroups 27

2

1 INTRODUCTION

1 Introduction

Let G be a locally compact group. In this project we will construct the Fourier-

Stieltjes algebra B(G) and the Fourier algebra A(G), which are Banach alge-

bras that serve to generalize the notion of Pontraygin duality from the theory

of abelian groups. In particular, we will focus on A(G). Our approach will for

the most part be the same as that of Kaniuth and Lau [7], which is actually

the initial approach by Eymard [4].

The Fourier algebra is a fairly well-behaved algebra; an application of Fell’s

Absorption Principle yields that A(G) is actually an ideal inside of B(G). Fur-

thermore, a lemma of Eymard tells us that A(G) is a regular algebra. One

reason to think that A(G) is a “good choice” for the generalization of Pon-

tryagin duality is because the Gelfand spectrum of A(G) behaves very nicely.

Indeed, we can show that the spectrum can be identified with exactly G it-

self. Furthermore, we will see that A(G) behaves nicely with subgroups in

the expected way. That is, if H is a closed subgroup of G then not only is

A(G)|H ⊆ A(H), but actually A(G)|H = A(H). Our proof of that result will

follow that of Spronk [9], which is itself based on the approach of [2]

Much can be said about A(G) if we restrict our focus to specific classes

of groups. If G is compact then A(G) = B(G), and if G is abelian then

A(G) = L1(G), where G is the Pontryagin dual of G. The case where G

is amenable turns out to be particularly illuminating, because a theorem of

Leptin allows us to completely characterize when A(G) possesses a bounded

(indeed, contractive) approximate identity.

3

2 PRELIMINARIES

This document is written with the implicit assumption that the reader is

familiar with all fundamental results from functional analysis and abstract

measure theory. Knowledge of abstract harmonic analysis would be helpful,

however a brief introduction to the topic is done in the preliminaries section.

In particular, a newcomer to this subject should take note of the equivalent

definitions of amenability that we will be using.

2 Preliminaries

Definition 2.1. For a given group G and topology τ on G, we call (G, τ)

a topological group (usually just denoted G if the topology is clear) if group

multiplication and the taking of inverses is continuous with respect to τ .

We will list a collection of important and easy to prove properties of topo-

logical groups.

Proposition 2.2. Suppose that G is a topological group. Then the following

hold.

i. If U ⊆ G is open and A ⊆ G is any set, then the translated sets AU =

{ay : a ∈ A, y ∈ U} and UA = {ya : a ∈ A, y ∈ U} are open. In

particular, translating an open set by a single element results in another

open set.

ii. If U is an open neighborhood of e then there exists an open neighborhood

V of e such that V V ⊆ U and V is symmetric, that is, V = {y−1 : y ∈ V }.

iii. If C and K are compact subsets of G, then KC is compact.

iv. If C is compact in G and U is an open set such that C ⊆ U , then there

exists a symmetric neighborhood V of e such that CV 2 ∪ V 2C ⊆ U .

4

2 PRELIMINARIES

Following the example of most abstract harmonic analysts, from here on

out we will assume that G is always a locally compact group. The reason for

this restriction of focus is because of the following fundamental result.

Theorem 2.3 (Existence of Haar Measure). If G is a locally compact group,

then there exists a positive Radon measure m that satisfies the condition that

if x ∈ G and E is measurable, then m(xE) = m(E) (this property is called

left-invariance). Furthermore, m is unique up to scaling and if U is an open

set then m(U) > 0.

While technically this theorem gives us an infinite number of choices for our

measure because of the option of scaling, the actual explicit choice of a specific

measure is typically not important. Thus without ambiguity for a given locally

compact group G we can work with the measure m given by the above theo-

rem, which we will call Haar measure. Now from elementary measure theory

we know that we can in fact integrate against that measure. That is, if f is an

integrable function on G then we can write the integral∫Gf(x)dm(x), which

we will call the Haar integral. Because it is understood that we are always

integrating against Haar measure, we will typically omit the actual measure

m and just write∫Gf(x)dx. Following typical notational convention, if the ex-

act form of f is left arbitrary then we will allow ourselves to write∫Gf instead.

At this point we want to set some notation that will be used throughout

this project. Let f : G→ C and x ∈ G. Then let

f(x) = f(x−1)

f(x) = f(x−1)

Rxf(y) = f(yx)

5

2 PRELIMINARIES

f ∗ g(x) =

∫G

f(xy)g(y)dy

Remark 1. It turns out that if we consider the above defined ∗ operation

(also called convolution ) as algebra multiplication, we can actually consider

L1(G) as a Banach algebra

Note that if x ∈ G then the map E 7→ m(Ex) also defines a measure on

G, hence by uniqueness up to scaling we can find a constant ∆(x) such that

m(Ex) = ∆(x)m(E). This defines a homomorphism ∆ : G → R which we

will call the modular function. Due to the following proposition, this map is

actually quite useful.

Proposition 2.4. If f ∈ L1(G) then

∫G

1

∆(x)f(x)dx =

∫G

f(x)dx

With Haar measure in hand, we can construct the Lebesgue spaces Lp(G,m)

(which we will simply write as Lp(G) because Haar measure is understood).

We recall that in the case of p = 2 then L2(G) is a Hilbert space with inner

product 〈f, g〉 =∫Gf(x)g(x)dx.

Proposition 2.5. The following are equivalent:

i. If K is a compact subset of G and ε > 0, then there exists a compact set

C such that m(KC) < (1 + ε)m(C);

ii. If f ∈ C+C (G), then ‖f‖1 = ‖λ(f)‖.

The above equivalent conditions are equivalent to a notion of a group be-

ing amenable. There are many, many different equivalent conditions for being

amenable which there is not time to get into here, but are covered extensively

6

2 PRELIMINARIES

in [8].

For a Hilbert space H, let U(H) = {U ∈ B(H) : U∗U = I = UU∗} be

the unitary operators on H. A continuous unitary representation on G is

a homomorphism π : G → U(H(π)) such that for any ζ, η ∈ H(π) then

the function πζ,η(x) = 〈π(x)ζ, η〉 is continuous (we often call these functions

matrix coefficients). We will denote the collection of equivalence classes of uni-

tary representations on G by Σ(G), where two representations are equivalent

if they are equal up to a conjugation by a unitary.

Recall that if we have two Hilbert spaces H and K, we can form the ten-

sor product H⊗K with inner product

〈ζ ⊗ η, φ⊗ ψ〉H⊗K = 〈ζ, φ〉H〈η, ψ〉K where ζ, φ ∈ H and η, ψ ∈ K

Then given two unitary representations π and ρ, we can form another (con-

tinuous unitary) representation called the tensor product of π and ρ denoted

by π ⊗ ρ : G→ U(Hπ ⊗Hρ) defined as

π ⊗ ρ(x)(ζ ⊗ η) = π(x)ζ ⊗ ρ(x)η where ζ ∈ Hπ, η ∈ Hρ, x ∈ G

If we have a continuous unitary representation π on G, then we can determine a

non-degenerate ∗−representation on L1(G) by writing π(f) =∫Gf(x)π(x)dx.

This a bit of abuse of notation because the resulting integral is operator-valued,

so the method of actually working with this integral in practice is by letting

7

2 PRELIMINARIES

π(f) be defined by satisfying

〈π(f)ζ, η〉 =

∫G

f(x)〈π(x)ζ, η〉dx, where ζ, η ∈ H(π) and f ∈ L1(G)

A unitary representation of particular importance to this project is known as

the left regular representation and is defined by λ : G → U(L2(G)), where if

x, y ∈ G and f ∈ L2(G) then λ(y)f(x) = f(y−1x). Using the notion above

of representations defining operator valued integrals on L1(G), we can observe

the useful fact that λ(f)g = f ∗ g, where f ∈ L1(G) and g ∈ L2(G).

Assume that G is abelian. Let the group

G = {σ : G→ T : σ a continuous homomorphism}

be equipped with the topology of uniform convergence on compact sets. We

call G the dual group of G. For f ∈ L1(G), let the map f : G→ C, called the

Fourier transform of f, be defined by f(σ) =∫Gfσdm.

Theorem 2.6 (Plancherel’s Theorem). Let G be abelian and f ∈ L1(G) ∩

L2(G). Then ‖f‖L2(G) = ‖f‖L2(G). Furthermore, there is a unitary U :

L2(G)→ L2(G) such that Uf = f

Theorem 2.7 (Pontryagin duality). If G is abelian, then G ∼= ˆG.

Before we end this section, we will quickly mention some fundamental

theorems from the theory of von Neumann algebras.

Theorem 2.8 (Kaplansky’s Density Theorem). Let H be a Hilbert space. If

A is a self adjoint algebra of operators in B(H) then B1(ASOT) = B1(A)SOT,

8

3 THE FOURIER ALGEBRA

and in fact, if T ∈ B1(ASOT) is self adjoint then it is in the strong operator

closure of the self adjoint operators in B1(A).

Theorem 2.9 (von Neumann’s Double Commutant Theorem). Let H be a

Hilbert space and A ⊆ B(H) be a non-degenerate *-subalgebra. Denote the

commutant of A by A′ = {T ∈ B(H) : TS = ST for each S ∈ A}. Then

ASOT= AWOT

= A′′

3 The Fourier Algebra

3.1 The Construction of the Fourier-Stieltjes Algebra

Let f ∈ L1(G). Then we define the norm

‖f‖∗ = sup{‖π(f)‖ : π ∈ Σ(G)}

In general L1(G) is not complete under this new norm, so we let C∗(G) denote

the completion of L1(G) with respect to this ‖ · ‖∗ norm. The space C∗(G) is

called the group C∗-algebra of G. Now define a space of continuous functions

B(G) = {πζ,η : π ∈ Σ(G) and ζ, η ∈ Hπ}

Lemma 3.1. If u ∈ B(G), then sup{|∫Gf(x)u(x)dx| : f ∈ L1(G), ‖f‖∗ ≤

1} <∞

Proof. Let u(x) = πζ,η. If f ∈ L1(G), ‖f‖∗ ≤ 1 then we have that

|∫G

f(x)u(x)dx| = |〈π(f)ζ, η〉|

9

3.1 The Construction of the Fourier-Stieltjes Algebra3 THE FOURIER ALGEBRA

≤ ‖f‖∗‖ζ‖‖η‖

≤ ‖ζ‖‖η‖

This result allows us to equip B(G) with the following norm. If u ∈ B(G)

then set

‖u‖B(G) = sup{|∫G

f(x)u(x)dx : f ∈ L1(G), ‖f‖∗ ≤ 1}

We call the normed space (B(G), ‖ · ‖B(G)) the Fourier-Stieltjes Algebra of G.

To justify the use of the word algebra here, we just note that if u = πζ,η and

v = ρφ,ψ where ζ, η ∈ Hπ and φ, ψ ∈ Hρ then we have that

uv = 〈πζ, η〉〈ρφ, ψ〉 = 〈π ⊗ ρ(ζ ⊗ φ), η ⊗ ψ〉 ∈ B(G)

For π ∈ Σ(G), define the subspace

Aπ(G) = span{πf,g : f, g ∈ L2(G)}B(G)

Let us also define a subalgebra of B(Hπ) by

V Nπ(G) = π(L1(G))WOT

= π(L1(G))w*

This definition is permissible because the weak* and weak operator topologies

agree on bounded sets in B(Hπ). Now with these two spaces in hand, we can

look at the following useful theorem:

10

3.2 The Fourier Algebra 3 THE FOURIER ALGEBRA

Theorem 3.2. If π ∈ Σ(G), then

i. Aπ(G)∗ ∼= V Nπ(G) via the dual pairing 〈πζ,η, T 〉 = 〈Tζ, η〉

ii. Aπ = 〈u ∈ B(G) : u =∑∞

i=1 πζi,ηi where∑∞

i=1 ‖ζi‖ · ‖ηi‖ <∞〉

The proof of this result is outside the scope of this project. The curious

reader might enjoy a reference that actually focuses on the Fourier-Stieltjes

algebra specifically, such as the one by Anderson-Sackaney [1].

3.2 The Fourier Algebra

While B(G) is a perfectly interesting algebra in its own right, for the purposes

of this project we want to focus on a particular subalgebra of B(G). Recall

that λ : G → U(L2(G)), called the left regular representation, is the map

such that λ(x)f(y) = f(x−1y) where x, y ∈ G and f ∈ L2(G). Using the

definitions from the last section, we will let A(G) = Aλ(G) be called the

Fourier Algebra of G and V N(G) = V Nλ(G) the von Neumann algebra of G.

We will have to justify the use of the word “algebra” here, because if π ∈ Σ(G)

then it is by no means necessarily true that Aπ is any more than just a closed

subspace, which gives us an indication of how special a representation λ really

is. It turns out that λ possesses a remarkable “absorbing” property that allows

A(G) to not only be a subalgbra of B(G), but actually an ideal. In order to

show this, we will need a theorem by Fell.

Theorem 3.3 (Fell’s Absorption Principle). Let π ∈ Σ(G) and {ei}i∈I be an

orthonormal basis for Hπ. Then we have that π⊗ λ ∼= I · λ, where I · λ : G→

U(Hπ)I is the unitary representation defined by I · λ(x)(ζi)i∈I = (λ(x)ζi)i∈I

11


Proof. We begin by fixing x ∈ G and defining the three following unitaries:

V : L2(G)I → L2(G)⊗2 Hπ, V (fi)i∈I =∑i∈I

fi ⊗ ei

U : L2(G)⊗2 Hπ → L2(G,Hπ), U(∑i∈I

fi ⊗ ei) =∑i∈I

fi(·)ei

W : L2(G,Hπ)→ L2(G,Hπ),W (H) = π(x)H

It is not to hard to see why each of these maps are unitaries. It follows easily

that V is an isometry and has dense range, but because isometries must al-

ways have closed range then it follows that V is a surjective isometry, hence

a unitary. As for U , it is easy to see that U−1H =∑

i∈I〈H(·), ei〉 ⊗ ei and by

elementary computation U−1 = U∗.

I claim that the unitary L = V ∗U∗W ∗U satisfies L(λ ⊗ π)L∗ = I · λ, which

would show that λ ⊗ π and I · λ are unitarily equivalent. This calculation

requires multiple steps, so for notational convenience after each successive ap-

plication of one of the above unitaries and its adjoint I will denote the result

as a function ∆i. Now consider the following calculations:

U(λ⊗ π(x))U∗(H) = Uλ⊗ π(x)(∑i∈I

〈H(·), ei〉 ⊗ ei)

= U(∑i∈I

〈H(x−1·), ei〉 ⊗ π(x)ei)

= π(x)H(x−1y)

= ∆1(x)H(y)

W ∗∆1(x)WH(y) = W ∗∆1(x)π(x)H(y)

12


= W ∗π(x)π(x−1y)H(x−1y)

= H(x−1y)

= ∆2(x)H(y)

U∗∆2(x)U(∑j∈I

fj ⊗ ej)(y) = U∗∆2(x)∑j∈I

fj(y)ej

= U∗∑j∈I

〈fj(x−1y)ej|ei〉 ⊗ ei

=∑i∈I

fi(x−1y)⊗ ei

= λ(x)⊗ I(∑j∈I

fj ⊗ ej)(y)

= ∆3(x)(∑j∈I

fj ⊗ ej)(y)

V ∗∆3(x)V (fj)j∈I = V ∗∆3(x)∑j∈I

fj ⊗ ej

= V ∗∑j∈I

λ(x)fj ⊗ ej

= (λ(x)fj)j∈I

= I · λ(x)(fj)j∈I

Corollary 3.4. A(G) is an ideal of B(G)

Proof. Let πζ,η ∈ B(G) and u ∈ A(G). By Theorem 3.2, we know that u

is of the form u =∑∞

j=1 λfj ,gj ∈ A(G). An application of Fell’s Absorption

13


Principle yields that that

πζ,ηu(x) =∞∑j=1

〈λ⊗ π(x)fj ⊗ ζ, jj ⊗ η〉

=∞∑j=1

〈L∗I · λ(x)Lfj ⊗ ζ, gj ⊗ η〉

=∞∑j=1

〈(λ(x)fj,i)i∈I , (gi,j)i∈I〉

=∞∑j=1

∑i∈I

〈λ(x)fj,i, gj,i〉 ∈ A(G)

It turns out that A(G) possesses a regularity property that will prove to

be of fundamental importance.

Lemma 3.5 (Eymard’s Trick). Let U ⊆ G be an open set, and let K ⊆ U be

a compact set. Then there exists u ∈ A(G) such that u|K = 1, uUc = 0.

Proof. Let V be a relatively compact neighborhood of e such that KV 2 ⊆ U ,

the existence of which is guaranteed by Proposition 2.2. Let 1V and 1KV

denote the characteristic functions for V and KV respectively. Then consider

the function

u(x) =1

m(V )λ1V ,1KV

It is immediate that u ∈ A(G), and a simple calculation yields that

u(x) =1

m(V )

∫G

1V (x−1y)1KV (y)dy

=1

m(V )

∫G

1xV ∩KV (y)dy

=m(xV ∩KV )

m(V )

14


If x ∈ K by the left-invariance of Haar measure we have that u(x) = m(xV )m(V )

= 1.

Conversely, if x /∈ U then xV ∩KV = ∅ so u(x) = m(∅)m(V )

= 0.

Remark 2. Note that if we let U = G, an important consequence of this

lemma is that for any compact set we can always find a function in A(G) that

is constantly one on that set. This is actually quite useful, because in general

A(G) will not be a unital algebra. After all, A(G) is an ideal in B(G) so if

1 ∈ A(G) then A(G) = B(G). As will be addressed later in this section, it

turns out that this is only possible if G is a compact group.

Lemma 3.6. Let u = λf,g ∈ A(G). Then ‖u‖A(G) ≤ ‖f‖2‖g‖2

Proof. Observe that if u = λf,g then

‖u‖A(G) = sup{|∫G

h(x)u(x)dx| : h ∈ L1(G), ‖h‖∗ ≤ 1}

= sup{〈λ(h)f, g〉 : h ∈ L1(G), ‖h‖∗ ≤ 1}

≤ ‖f‖2‖g‖2

Theorem 3.7. span{λf,g : f, g ∈ CC(G)}‖·‖B(G)

= B(G) ∩ CC(G)‖·‖B(G)

= A(G)

Proof. Take v ∈ B(G) ∩ CC(G). Then supp(v) is compact, so by Eymard’s

Trick we can find u ∈ A(G) such that u|supp(v) = 1. Then because A(G) is

an ideal of B(G) it follows that uv = v ∈ A(G), which means that B(G) ∩

CC(G) ⊆ A(G).

Because clearly span{λf,g : f, g ∈ CC(G)} ⊆ B(G) ∩ CC(G), then it suffices

to show that span{λf,g : f, g ∈ CC(G)}‖·‖B(G)

= A(G). Let λf,g ∈ A(G) with

f, g ∈ L2(G), and ε > 0 be given. Then by the density of CC(G) in L2(G), we

15


can find h, k ∈ CC(G) such that ‖f − h‖2, ‖g − k‖2 < ε. Then by Lemma 3.6

we have that

‖λf,g − λh,k‖B(G) = ‖λf−h,g − λh,g−k‖B(G)

≤ ‖f − h‖2 · ‖g‖2 + ‖h‖2 · ‖g − k‖2

≤ ε · ‖g‖2 + ε(ε+ ‖f‖2)

Thus by letting ε get small our result follows.

Remark 3. Because A(G) ⊆ B(G) and B(G)∩CC(G) ⊆ A(G), then it follows

that B(G) ∩ CC(G) = A(G) ∩ CC(G) so our above result is really telling us

that A(G) ∩ CC(G)‖·‖A(G)

= A(G)

Corollary 3.8. A(G) is uniformly dense in C0(G)

Proof. First note that if u ∈ A(G) then

‖u‖A(G) = sup{∫G

u(s)f(s)ds|f ∈ L1(G), ‖f‖∗ ≤ 1}

≥ sup{∫G

u(s)f(s)ds|f ∈ L1(G), ‖f‖1 ≤ 1}

= ‖u‖∞

Thus because A(G) ∩ CC(G) is dense in A(G) it follows that A(G) ⊆ C0(G).

Now it is clear that A(G) is self-adjoint and by Eymard’s Trick we have that

A(G) is point separating, hence by the Stone-Weierstrass Theorem we have

that A(G) is dense in C0(G) under the uniform norm.

We will end this section by seeing what we can say about A(G) if we restrict

our focus to specific classes of groups.

Proposition 3.9. G is compact if and only if A(G) = B(G)

16


Proof. Suppose that G is compact. We have that by Theorem 3.7 that B(G)∩

CC(G) ⊆ A(G), but because CC(G) = C(G) then we in fact have that

B(G) ⊆ A(G), which can only be possible if A(G) = B(G).

Conversely, suppose that A(G) = B(G). Now Corollary 3.8 tells us that

A(G) ⊆ C0(G), so because A(G) = B(G) then it follows that 1 ∈ C0(G). This

of course is only possible if G is compact.

Proposition 3.10. If G is abelian, then A(G) ∼= L1(G)

Proof. By Theorem 3.2 we know that A(G)∗ ∼= V N(G), and of course it is

well-known that L1(G)∗ ∼= L∞(G). Because the predual of a von Neumann

algebra is unique, it suffices to show that V N(G) ∼= L∞(G).

Consider the representation λ : G→ U(L2(G)) defined by λ(x)f(σ) = σ(x)f(σ).

Let U be the unitary given from Plancherel’s Theorem and choose any x ∈

G, f ∈ L1 ∩ L2(G). Then

Uλ(x)U∗f(σ) =

∫G

U∗f(x−1y)σ(y)dm(y)

=

∫G

U∗f(y)σ(xy)dm(y)

= σ(x)

∫G

U∗f(y)σ(y)dm(y)

= σ(x)U(U∗f)σ

= σ(x)f(σ)

Because L1 ∩ L2(G) is dense in L1(G), then it follows that λ and λ are uni-

tarily equivalent. An application of the Hahn-Banach theorem yields that

spanλ(G)w* ∼= L∞(G). Because λ and λ are unitarily equivalent, then we have

17

4 THE SPECTRUM OF A(G)

that V N(G) ∼= L∞(G), as desired.

4 The Spectrum of A(G)

As mentioned in the introduction, we want to think of A(G) as representing

some sort of generalization of Pontrayagin duality. One way of testing if A(G)

performs an adequate job at this task is if we can recover any sort of group

structure from A(G), which it turns out that we can. But first, we need a

technical lemma which tells us that singletons are sets of spectral synthesis for

A(G).

Lemma 4.1. Let a ∈ G, f ∈ A(G) such that f(a) = 0. Then given ε > 0,

there exists h ∈ A(G) ∩ CC(G) vanishing on a neighborhood of a such that

‖h− f‖A(G) < ε

Proof. Without loss let ε < 1. Because A(G) ∩ CC(G) is dense in A(G) by

Theorem 3.7, we can find f ′ ∈ A(G) ∩ CC(G) such that ||f − f ′||A(G) < ε. As

noted in Corollary 3.8, we have that || · ||A(G) ≥ || · ||∞, so ||f − f ′||A(G) < ε.

In particular, this means we can find a relatively compact open neighborhood

V ′ of e such that sup{|f ′(ay)| : y ∈ V ′} < ε. Define the set

W = {y ∈ G : ‖f ′ −Ryf′‖A(G) < ε}.

We have that W is a neighborhood of e by the continuity of translation, so if

we set V = V ′ ∩W then sup{|f ′(ay)| : y ∈ V } < ε. Then by the regularity of

Haar measure, there exists a compact neighborhood K of e with K ⊆ V and

18


m(K) ≥ (1− ε)m(V ). Now define the following functions:

u =1

m(K)1K

g = 1aV f′

h(x) = (f ′ − g) ∗ u =1

m(K)

∫K

f ′(xy)[1− 1aV (xy)]dy ∈ A(G)

All of the above functions have compact support because f ′ has compact sup-

port and V has compact closure. Now pick an open neighborhood O of a such

that OK ⊆ aV . It follows that h = 0 on a neighborhood of a. Now observe

the following calculations:

‖u‖2 = (

∫K

(1

m(K))2)

12 =

1

m(K)12

≤ 1

m(V )12

(1

1− ε)12

‖g‖2 = (

∫aV

|f ′(y)|2dy)12 ≤ εm(V )

12

‖f ′ − f ′ ∗ u‖A(G) = ‖f ′ − 1

m(K)

∫K

Ryf′dy‖A(G)

= ‖ 1

m(K)

∫K

f ′(y)−Ryf′dy‖A(G)

≤ supy∈K‖f ′ −Ryf

′‖A(G)

< ε

Then with all of these calculations in hand, an application of Lemma 3.6 yields

that

‖f − h‖A(G) ≤ ‖f − f ′‖A(G) + ‖f ′ − h‖A(G)

19


≤ ε+ ‖f ′ − f ′ ∗ u‖A(G) + ‖g ∗ u‖A(G)

< 2ε+ ‖g‖2‖u‖2

< 2ε+ ε(1

1− ε)12

Definition 4.2. If A is a commutative algebra, let σ(A) = {φ : A → C :

φ a nonzero algebra homomorphism} denote the spectrum of A.

Theorem 4.3. G can be identified with σ(A(G)) via the mapping x 7→ φx,

where φx is point evaluation at x ∈ G.

Proof. First, we note that the above map is clearly injective because A(G)

separates points by Eymard’s Trick. Now let us suppose that the map is not

surjective, that is, there exists φ ∈ σ(A(G)) such that φ is not point evaluation.

Then for any x ∈ G, φ 6= φx. We can find ux ∈ A(G) such that φ(ux) 6= 0

and φx(ux) = 0. By scaling we can assume that in fact φ(ux) = 1. From

the above lemma we can approximate ux with a sequence from A(G)∩CC(G)

each vanishing on on a neighborhood of x, so again without loss assume that

ux vanishes on an open neighborhood Vx of x. Then by Theorem 3.7, we can

find u0 ∈ A(G)∩CC(G) such that φ(u0) = 1 (once again scaling if necessary).

Now {Vx}x∈G is an open cover of supp(u0), so by compactness we can find

x1, ..., xn ∈ supp(u0) such that supp(u0) ⊆ ∪nj=1Vxj . Set the function

u = u0

n∏j=1

uxj ∈ A(G)

For x ∈ G if x ∈ supp(u0) then we can find xi such that x ∈ Vxi , so uxi(x) = 0.

Meanwhile, by definition we have that u0|supp(u) = 0, so it follows that u = 0.

20

5 LEPTIN’S THEOREM

However, note that

φ(u) = φ(u0)n∏j=1

φ(uxj) = 1

which creates a contradiction because u = 0. Thus φ = φx for some x, as

desired.

5 Leptin’s Theorem

Recall from our discussion in Section 3.2 that knowing certain facts about the

structure of G, such as whether it is abelian or compact, allows us to know

more about the structure of A(G). In this section we explore the more general

case where G is amenable.

Definition 5.1. Let A be a commutative normed algebra. Then an

approximate identity for A is a net (eα)α∈I ⊆ A such that limα ‖eαa− a‖ = 0

for all a ∈ A. We call (eα)α∈I a bounded approximate identity if in addition

it is bounded as a net, that is, there exists c ∈ R such that ‖eα‖ ≤ c for all

α ∈ I.

Remark 4. Let {Uα}α be the neighborhood basis of e consisting of relatively

compact sets. Then it turns out that the net { 1m(Uα)

1Uα}α is a bounded ap-

proximate identity for L1(G)

Remark 5. It is not the case that all Banach algebras possess approxi-

mate identities that are bounded. For example, consider `1(N). Let a =∑∞i=1 a(i)δi, b =

∑∞i=1 b(i)δi ∈ `1(N) and define multiplication by

a · b = (∞∑n=1

a(n)δn)(∞∑n=1

b(n)δn) =∞∑n=1

a(n)b(n)δn

21


Then it is easy to see that if

en =n∑i=1

δn

then (en)n∈N is an approximate identity for `1(N) such that ||en||1 = n, so

(en)n∈N is an unbounded approximate identity. Observe that if (eα)α∈I is any

approximate identity for `1(N) then if n ∈ N it follows that

lim supα∈I

‖eα‖1 ≥ limα

n∑i=1

|eα(i)| = n.

Thus we can see that supα∈I ||eα||1 ≥ ∞, so (eα)α∈I also must be unbounded.

Hence `1(N) is a Banach algebra with only unbounded approximate identities.

Lemma 5.2. If f ∈ L(G) only attains nonnegative real values, then ‖f ∗n‖1 =

‖f‖n1 , where f ∗n denotes the convolution of f with itself n times.

Proof. It suffices to prove the result for n = 2, and let the rest follow by

induction. In order to achieve this, we will confirm that the functional f 7→∫Gf is multiplicative with respect to convolution. By Fubini’s theorem and

the fact that the Haar integral is left-invariant, we have the following:

∫G

f ∗ g =

∫G

∫G

f(x)g(x−1y)dxdy

=

∫G

f(x)

∫G

g(x−1y)dydx

=

∫G

f(x)

∫G

g(y)dydx

=

∫G

f

∫G

g

Because f, g ≥ 0 a.e., it follows that

‖f ∗ g‖1 =

∫G

f ∗ g =

∫g

f

∫G

g = ‖f‖1‖g‖1.

22


Remark 6. Observe that if we allow ourselves to consider a general f ∈ L1(G)

then above proof easily extends to show that ‖f ∗n‖1 ≤ ‖f‖n1 , which is why

L1(G) is a Banach algebra with convolution as multiplication.

Theorem 5.3. (Leptin’s Theorem) A locally compact group G is amenable if

and only if A(G) has a bounded approximate identity.

Proof. Suppose that G is amenable. Let K ⊆ G be compact and ε > 0.

Then by Proposition 2.5 we can find a compact set C such that m(KC) <

(1 + ε)m(C). Now let us consider the following collection of functions:

uK,ε(x) =1

(1 + ε)m(C)λ1C ,1KC (x) =

1

(1 + ε)m(C)

∫C

1KC(xy)dy

Clearly by construction each uK,ε ∈ A(G) ∩ CC(G). If we let K(G) denote

the set of all compact subsets of G, then we can define a preorder on the set

K(G)×R+ by letting (K, ε) < (K ′, ε′) if K ⊆ K ′ and ε < ε. This allows us to

consider (uK,ε)K,ε as a net. By the following calculation, we have that

‖uK,ε‖A(G) ≤‖1C‖2‖1KC‖2

(1 + ε)m(C)m(C) =

m(KC)12

(1 + ε)m(C)12

≤ 1

Thus we have that (uK,ε)K,ε is in fact a bounded net. Now let v ∈ A(G)∩CC(G)

and ε > 0. SetK = supp(v). Observe that because uK,ε|K = 11+ε

then it follows

that uK,εv(x) = v(x)1+ε

. We then have that

‖uK,εv − v‖A(G) =ε

1 + ε‖v‖A(G)

Thus we have that lim(K,ε) ‖uK,εv − v‖A(G) = 0 for all v ∈ A(G) ∩ CC(G).

This is not quite yet enough to satisfy the condition of being an approximate

23


identity because this condition needs to hold for all elements of A(G), not

just a subset. Fortunately, it turns out the density of A(G) ∩ CC(G) in A(G)

carries us the rest of the way. To see that this is true, let w ∈ A(G) and ε > 0.

Then by density there exists v ∈ A(G) ∩ CC(G) such that ‖w − v‖A(G) < ε.

Set K = supp(v). Then we have that

‖uK,εw − w‖A(G) = ‖uK,εw − uK,εv + uK,εv − v + v − w‖A(G)

≤ ‖uK,εw − uK,εv‖A(G) + ‖uK,εv − v‖A(G) + ‖v − w‖A(G)

≤ ‖uK,ε‖A(G)‖w − v‖A(G) + ‖uK,εv − v‖A(G) + ‖v − w‖A(G)

≤ ‖uK,εv − v‖A(G) + 2ε

From our work above we know that we can allow the above quantity to get

arbitrarily small, hence we have our desired result that (uK,ε)K,ε is a bounded

approximate identity for A(G).

Conversely, suppose that (uα)α is a bounded approximate identity for A(G)

with bound c. We wish to show that Proposition 2.5(ii) is satisfied. Let

f ∈ C+C (G) and set K = supp(f). By Eymard’s Trick we can find u ∈ A(G)

such that u|K = 1. We have that limα ‖uαu − u‖A(G) = 0, so because

‖ · ‖∞ ≤ ‖ · ‖A(G) then it follows that limα ‖uαu− u‖∞ = 0. Because u|K = 1,

then we have that uα converges uniformly to 1 on K. Thus if ε > 0 we can

find αε such that Re(uαε) ≥ 1− ε for all x ∈ K. It then follows that

Re〈uαε , f〉 =

∫G

Re(uαε(x)f(x)dx ≥ (1− ε)‖f‖1

24


However, also observe that

|〈uαε , f〉| ≤ ‖λ(f)‖ · ‖uαε‖A(G) ≤ c‖λ(f)‖

Because ε is allowed to vary, then it must hold that ‖f‖1 ≤ c‖λ(f)‖. Then by

Lemma 5.2, we have that

‖f‖n1 = ‖f ∗n‖1 ≤ c‖λ(f ∗n)‖ ≤ c‖λ(f)‖n

By taking limits on n we can see that

‖f‖1 ≤ ‖λ(f)‖ limn→∞

c1n = ‖λ(f)‖ ≤ ‖f‖1

Thus by Proposition 2.5 we have that G is amenable.

Remark 7. It is of interest to note that that the above proof actually implies

that if A(G) has a bounded approximate identity, then it has an approximate

identity that is bounded by 1 (we call such nets contractive). This is most

certainly not true for all Banach algebras. Let S be a finite semigroup, and

consider `1(S). If a =∑

s∈S a(s)δs, b =∑

s∈S b(s)δs ∈ `1(S) then we define

multiplication between a and b by

a ∗ b =∑s∈S

∑t∈S

a(s)b(t)δst =∑u∈S

(∑st=u

a(s)b(s))δu.

Notice that in particular that δs ∗ δt = δst. Furthermore, we can see that

‖a∗b‖1 =∑u∈S

|∑st=u

a(s)b(s)| ≤∑u∈S

(∑st=u

|a(s)||b(s)|) =∑s∈S

|a(s)|∑t∈S

|b(t)| = ‖a‖1‖b‖1

Hence it follows that `1(S) is a Banach algebra. Now let us consider the abelian

25


semigroup Sn = {s1 . . . , sn, o} where semigroup multiplication is defined as

sjsk =

sj if j = k

o otherwise,

osj = o = sjo = oo

Consider the element

e =n∑j=1

δsj − (n− 1)δo

We can see that if 1 ≤ j ≤ n then

δsje = (e(sj))δsj = δsj

and

δoe = (n∑i=1

e(sj) + e)so))δo = (n− (n− 1))δo = δo.

Hence it follows that e is an identity for `1(Sn). An identity is trivially an

approximate identity, however note that ||e||1 = 2n − 1 so e is a bounded

approximate identity that is not contractive.

Now let (eα)α∈I be a bounded approximate identity for `1(Sn). Because

Br(`1(Sn)) is compact, where r = supα ‖eα‖, then we can find a cluster point

e′ for (eα)α∈I . By passing to an appropriate subnet we can see that

‖e′‖1 ≤ r and ‖e′ − e‖1 = ‖ee′ − e‖1 = limβ‖eeα(β) − e‖1 = 0

It follows that e′ = e, so e′ is unique. Thus we have that r ≥ ||e||1 = 2n − 1.

Therefore we have that (eα)α∈I is not contractive, so we have shown that `1(Sn)

is a Banach algebra with bounded approximate identities but no contractive

26

6 RESTRICTION MAP TO SUBGROUPS

approximate identities.

For another example that is more directly related to the material in this

project, if H is a closed subgroup of G that is not open, then according to [6]

the ideal {u ∈ A(G) : u(h) = 0 for all h ∈ H} contains an approximate iden-

tity bounded by 2, and in fact it turns out that no number strictly less than

2 can act as a suitable bound for any approximate identities on this algebra.

6 Restriction Map to Subgroups

Recall that A(G) is the predual of V N(G) through the dual pairing 〈λf,g, T 〉 =

〈Tf, g〉. If u ∈ A(G) and T ∈ V N(G), we can define an action T · u by

〈T · u, S〉 = 〈u, ST 〉 where S ∈ V N(G)

Let u =∑∞

i=1 λfi,gi ∈ A(G). Then it follows that

u(t) =∞∑i=1

〈λ(t)fi, gi〉

=∞∑i=1

〈λfi,gi , λ(t)〉

= 〈u, λ(t)〉

We require the following useful lemma

Lemma 6.1. Let u =∑∞

i=1 λfi,gi ∈ A(G).

(i) If T ∈ V N(G), then T · u =∑∞

i=1 λTfi,gi

(ii) If f ∈ L1(G), then λ(f) · u = (f ∗ u)

27


Proof. (i) Let S ∈ V N(G). Then we calculate

〈T · u, S〉 = 〈u, ST 〉

= 〈∞∑i=1

λfi,gi , ST 〉

=∞∑i=1

〈λfi,gi , ST 〉

=∞∑i=1

〈STfi, gi〉

=∞∑i=1

〈λTfi,gi , S〉

= 〈∞∑i=1

λTfi,gi , S〉

Hence it follows that T · u =∑∞

i=1 λTfi,gi .

(ii)Let s ∈ G. Then by the above remark, we have that

λ(f) · u(s) = 〈λ(f) · u, λ(s)〉

= 〈u, λ(f)λ(s)〉

= 〈u, λ(λ(s)f)〉

=

∫G

f(s−1t)u(t)dt

=

∫G

f(t)u(st)d

=

∫G

f(t)u(t−1s−1)dt

= (f ∗ u)(s−1)

28


Thus we have that λ(f) · u = (f ∗ u)

Lemma 6.2. Let h ∈ CC(G). Then the operator Ch : L2(G)→ L2(G) defined

by f 7→ f ∗ h for f ∈ L2(G) commutes with every other operator in V N(G).

Proof. First we will show that Ch is bounded. Define the representation σ :

G→ U(L2(G)) by σ(s)f(t) = 1

∆(s)12f(ts), which we will call the right regular representation.

Observe that for f, g, k ∈ L2(G) then by Fubini’s Theorem and properties of

Haar measure we have that

〈σ(f)g, k〉 =

∫G

f(s)

∫G

1

δ(s)12

g(ts)k(t)dtds

=

∫G

∫G

f(s)1

∆(s)12

g(ts)k(t)dsdt

=

∫G

∫G

f(t−1s)1

∆(t−1s)12

g(s)k(t)dsdt

=

∫G

∫G

g ∗ (∆12 f)(t)k(t)dt

= 〈g ∗ (∆12 f), k〉

Because this holds for all k ∈ L2(G), it follows that σ(f)g = g ∗ (∆12 f). Now

clearly ∆12 f ∈ CC(G) ⊆ L1(G), hence we have that ‖Ch‖ = ‖σ(∆

12 f)‖ ≤

‖∆ 12 f‖1, so it follows that C is bounded operator.

Because V N(G) = spanλ(L1(G))w∗

, by continuity it suffices to show that

Ch commutes with λ(f) for all f ∈ L1(G). Then for g ∈ L2(G), we observe

that

Chλ(f)g = Ch(f ∗ g) = f ∗ g ∗ h = λ(f)(g ∗ h) = λ(f)Chg

29


Theorem 6.3. Let H be a closed subgroup of G. Then A(G)|H = A(H)

Proof. First we must confirm that A(G)|H ⊆ A(H). Let σ : H → G denote the

inclusion map. Then λG ◦ σ is a continuous unitary representation of H, and

using the characterization of AλG◦σ from Theorem 3.2, then we we can easily

see that A(G)|H = AλG◦σ, hence A(G)|H is a closed subspace in B(H). Fur-

thermore, if u ∈ A(G) ∩ CC(G) then certainly u|H ∈ B(H) ∩ CC(H) ⊆ A(H)

by Theorem 3.7, and furthermore the density which is guaranteed by that the-

orem implies that A(G)|H ⊆ A(H).

We claim that A(G)|H is invariant under the action of V N(H) on A(H).

Let s, t ∈ H and u ∈ A(G). Then

λ(s) · u(t) = 〈λ(s) · s, λ(t)〉 = 〈u, λ(t)λ(s)〉 = 〈u, λ(ts)〉 = u(ts)

Thus we have that λ(s) · u ∈ A(G)|H , so it follows that T · u ∈ A(G)|H for all

T ∈ spanλ(H). Because V N(H) = span(V N(H))sa by von Neumann’s Double

Commutant Theorem, we only need to show that the result holds for T ∈

B1(V N(H))sa. For such a T , by Kaplansky’s Density Theorem we have that

B1(V N(H))sa = B1(spanλ(H))sa

SOTbecause V N(H) = spanλ(H)

SOT, hence

we can find a net {Tα} ⊆ B1(spanλ(H)) convering to T in the strong operator

topology. Let u =∑∞

i=1 λfi,gi ∈ A(H) and ε > 0. Because∑∞

i=1 ‖fi‖2‖gi‖2 <

∞ we can find an nε ∈ N such that∑∞

i=nε+1 ‖fi‖2‖gi‖2 <ε4. Also, by strong

operator convergence we know that for 1 ≤ i ≤ n we can find αi such that for

all α ≥ αi, then ‖Tfi − Tαfi‖ <ε

2 sup{‖gj‖2 : 1 ≤ j ≤ n}(note that we can

always pick a large enough n such that this supremum is nonzero; otherwise

the result is trivial). Set αε such that αε ≥ αi for all 1 ≤ i ≤ n. Let α ≥ αε

30


and observe that

‖T · u− Tα · u‖A(H) = ‖∞∑i=1

λTfi,gi −∞∑i=1

λTαfi,gi‖A(H)

= ‖∞∑i=1

λ(T−Tα)fi,gi‖A(H)

≤n∑i=1

‖Tfi − Tαfi‖2‖gi‖2 + 2∞∑

i=n+1

‖fi‖2‖gi‖2

<ε

2+ε

2

= ε

Thus we have that T · u = limα Tα · u, but because A(G)|H is closed in A(H)

it follows that T · u ∈ A(G)|H , as desired.

Now suppose that A(G)|H 6= A(H). Then the Hahn-Banach Theorem yields

that there must exist some T ∈ V N(H) − {0} such that 〈u, T 〉 = 0 for all

u ∈ A(G)|H .

By the density guaranteed from Theorem 3.7, we can find some h ∈ span{λf,g :

f, g ∈ CC(G)} such that Th 6= 0 because T 6= 0. By Eymard’s Trick we find a

function u ∈ A(G)∩CC(G) such that u|H = 1, thus it follows that the product

uTh 6= 0. Then if v ∈ A(G)|H we have by Lemma 6.1 and Lemma 6.2 that

〈λ(h) · uv, T 〉 = 〈(h ∗ uv), T 〉

= T (h ∗ uv)(e)

= (Th) ∗ uv(e)

=

∫G

u(s)Th(s)v(s)ds

31


In particular, because A(G)|H is invariant under the action of T then it follows

that∫Gu(s)Th(s)v(s)ds = 0. The Stone-Weierstrass Theorem yields that

A(G)|H‖·‖∞

= C0(H) hence it follows that∫Gu(s)Th(s)w(s)ds = 0 for all

w ∈ C0(H). Thus uTh = 0, which is a contradiction of our choice of T .

Theorefore it must be true that A(G)|H = A(H), as desired.

32

REFERENCES REFERENCES

References

[1] B. Anderson-Sackaney, The Fourier-Stieltjes Algebra and Host’s Idempo-

tent Theorem, MMath Project, University of Waterloo, 2018.

[2] G. Arsac, Sur l’espace de Banach engendre par les coefficients d’une

representation unitaire, Publ. Dep. Math. (Lyon), Vol. 13, 1–101, 1976.

[3] J. Conway, A Course in Functional Analysis, Springer, 1990.

[4] P. Eymard, L’algebre de Fourier d’un groupe localement compact, Bull. Soc.

Math. France, Vol. 92, 1964.

[5] G. Folland, A Course in Abstract Harmonic Analysis, Ed. 2, CRC Press,

2015.

[6] B. Forrest, E. Kaniuth, A.T.-M. Lau and N. Spronk, Ideals with bounded

approximate identities in Fourier algebras , J. Funct. Anal. 203 (2003), 288-

306.

[7] E. Kaniuth and A.T.-M. Lau, Fourier and Fourier– Stieltjes Algebras on

Locally Compact Groups, Preliminary Version of Book in Press, 2018.

[8] J.P. Pier, Amenable locally compact groups. Wiley, 1984.

[9] N. Spronk, Banach Spaces of Coefficient Functions of Continuous Unitary

Representations of a locally Compact Group, Unpublished Seminar Notes,

2007.

33

Documents

The Fourier Algebra of a Locally Compact Group · Given a locally compact group G, the Fourier algebra A(G) is a closed ideal inside the Fourier-Stieltjes algebra B(G) consisting