1. Haar measure

Let G be a locally compact group. Denote by C00(G) the algebra of continuousfunctions on G with compact support. Endow it with the Sup-norm. Let C+

00(G)denote the cone of non-negative functions. Given f ∈ C00(G) and g ∈ G, welet Rg(f)(x) := f(xg) and Lg(f)(x) := f(g−1x), x ∈ G. Then g 7→ Rg is ahomomorphism from G to the group of invertible positive isometries of C00(G).

Lemma 0. (On uniform continuity) For each f ∈ C00(G), the maps G 3 g 7→Rg(f) ∈ C00(G) and G 3 g 7→ Lg(f) ∈ C00(G) are continuous.

Lemma 1. Given f, φ ∈ C+00(G), φ 6= 0, there is a finite family of elements gj ∈ G

and reals cj ≥ 0 such that

f ≤∑j

cjRgj (φ).

We let (f : φ) := inf∑j cj over all such families.

Lemma 2. (Properties of (f : φ))

(1) (Rg(f) : φ) = (f : φ)(2) (cf : φ) = c(f : φ), c ≥ 0(3) (f1 + f2 : φ) ≤ (f1 : φ) + (f2 : φ)(4) (f : ψ) ≤ (f : φ)(φ : ψ)(5) (f1 : φ) ≤ (f2 : φ) if f1 ≤ f2

(6) (f : φ) ≥ ‖f‖/‖φ‖

Fix f0 ∈ C+00(G), f0 6= 0. We now let

Iφ(f) =(f : φ)

(f0 : φ).

Then from (4) we deduce

1

(f0 : f)≤ Iφ(f) ≤ (f : f0).

Lemma 3. (Properties of Iφ)

(1) Iφ(Rg(f)) = Iφ(f)(2) Iφ(cf) = cIφ(f)(3) Iφ(f1 + f2) ≤ Iφ(f1) + Iφ(f2)(4) Iφ(f1) ≤ Iφ(f2) if f1 ≤ f2

This is almost what we are looking for. The functionals Iφ are not linear, but wewill show that if the support of φ is small enough, then Iφ is approximately linear.

Lemma 4. For f1, f2 ∈ C+00(G) and ε > 0, there is a neighborhood U of e such

that if supp(φ) ⊂ U then

Iφ(f1) + Iφ(f2) ≤ Iφ(f1 + f2) + ε.1

Proof. Let K be a compact subset of G such that Supp(f1) ⊂ K and Supp(f2) ⊂K. By Urysohn lemma, there is h0 ∈ C+

00(G) such that h0 = 1 on K. We leth = f1 + f2 + δh0 for some δ > 0. We now set

f ′1 :=f1

h, f ′2 :=

f2

h, assuming f ′1(x) = f ′2(x) = 0 if h(x) = 0.

Of course, f ′1, f′2 ∈ C+

00(G) and f ′1 + f ′2 ≤ 1. Choose a symmetric neighborhood Uof e such that |f ′1(x)− f ′1(y)| < ε and |f ′2(x)− f ′2(y)| < ε whenever xy−1 ∈ U . LetSupp(φ) ⊂ U and let h ≤

∑j cjRgj (φ) for some cj ≥ 0 and gj ∈ G.

f1(g) = f ′1(g)h(g) ≤ f ′1(g)∑j

cjφ(ggj) =∑ggj∈U

f ′1(g)cjφ(ggj)

≤∑ggj∈U

(f ′1(g−1j ) + ε)cjφ(ggj) =

∑j

(f ′1(g−1j ) + ε)cjφ(ggj)

Hence (f1 : φ) ≤∑j(f′1(g−1

j ) + ε)cj and, similarly, (f2 : φ) ≤∑j(f′2(g−1

j ) + ε)cjTherefore

(f1 : φ) + (f2 : φ) ≤∑j

(1 + 2ε)cj .

It follows that

(f1 : φ) + (f2 : φ) ≤ (1 + 2ε)(h : φ) ≤ (1 + 2ε)((f1 + f2 : φ) + δ(h0 : φ)).

ThereforeIφ(f1) + Iφ(f2) ≤ (1 + 2ε)(Iφ(f1 + f2) + δIφ(h0))

We recall that Iφ(h0) ≤ (h0 : f0). Since δ is arbitrary, we conclude that

Iφ(f1) + Iφ(f2) ≤ (1 + 2ε)(Iφ(f1 + f2)) ≤ Iφ(f1 + f2) + 2ε(f1 + g : f0).

�

Theorem 5. There is a non-trivial right invariant non-trivial integral I on C00(G).

Proof. Let

X :=∏

06=f∈C+00(G)

[1

(f0 : f), (f, f0)].

It is a compact space. For each neighborhood U of e, we set

MU := {(Iφ(f))06=f∈C+00(G) | Supp(φ) ∈ U}.

Then MU 6= ∅, MU ⊂ X and MU1∩ MU2

= MU1∩U2. Thus the family {MU |

U is a neighborhood of e} is a base of a filter in X. Since X is compact, there isI = (I(f))06=f∈C+

00(G) ∈ X such that I belongs to the closure of each MU . Thus,

for each ε > 0 and each finite sequence f1, . . . , fn ∈ C+00(G) \ {0} and each U , there

is φ ∈ C+00(G) with Supp(φ) ⊂ U such that

|I(fj)− Iφ(fj)| < ε.

It now follows from Lemma 3 that

(1) I(Rg(f)) = I(f)(2) I(cf) = cI(f)(3) I(f1 + f2) ≤ I(f1) + I(f2)(4) I(f1) ≤ I(f2) if f1 ≤ f2.

2

Fix f1, f2. Lemma 4 yields that

I(f1) + I(f2) ≤ I(f1 + f2) + ε

for an arbitrary ε > 0. Therefore, jointly with (3) this gives

I(f1) + I(f2) ≤ I(f1 + f2).

Also, from the definition of X it follows that

1

(f0 : f)≤ I(f) ≤ (f : f0).

Hence I(f) > 0 if f 6= 0, f ∈ C+00(G). Now extend I from the cone C+

00(G) to theentire C00(G) in a usual way. In particular, I(0) = 0. �

Recall

Lemma 6. (Riesz Representation Theorem) Let X be a locally compact Hausdorffspace and let I be a positive linear functional on C00(X). Then there is a uniqueRadon measure µ (i.e. a Borel measure which is locally finite and inner regular)on X such that I(f) =

∫f dµ, for all f ∈ C00(X). Moreover, µ satisfies: µ(U) =

sup{I(f)|f ∈ C+00(X), f ≤ 1U} and µ(K) = inf{I(f)|f ∈ C+

00(X), f ≥ 1K}, for allopen U ⊂ X and compact K ⊂ X.

Let µ correspond to I. Then µ is called a right-invariant Haar measure. In asimilar way a left invariant Haar measure is defined. If J is the inversion in G, ameasure µ is a right invariant Haar measure if and only if µ ◦ J is left-invariantHaar measure.

Corollary 7. It follows that the Haar measure of each compact set is finite andthe Haar measure of each open set is strictly positive.

Theorem 8. (Uniqueness of Haar measure). Given two non-trivial right-invariantintegrals I and I ′, there is c > 0 such that I = cI ′.

Proof. Let I(f) =∫f dµ

Let I ′ be another right invariant integral on C+00(G) and let µ′ stand for the

corresponding measure. Since I ′ is nontrivial, there is 0 6= f0 ∈ C+00(G) with

I ′(f0) > 0. For each 0 6= f ∈ C+00(G), we can find gj and cj with f0 ≤

∑j cjRg(f).

Hence I ′(f0) ≤∑j cjI

′(f). It follows that I ′(f0) ≤ (f0 : f)I ′(f). In particular,

I ′(f) > 0.Take f1, f2 ∈ C+

00(G). Fix a neighborhood V of e. Select ψ ∈ C+00(G) such that

ψ = 1 on the union of Supp(f1)·V ∪U ·Supp(f1) and Supp(f2)·V ∪U ·Supp(f2). Foreach ε > 0, there is a symmetric neighborhood U ⊂ V of e with |fj(g)− fj(g′)| < εwhenever g′g−1 ∈ U . Select 0 6= φ ∈ C+

00 such that Supp(φ) ⊂ U .Then we have

I ′(φ)I(f1) =

∫φ(g1)f1(g2) dµ′(g1)dµ(g2) =

∫φ(g1)f1(g2g1) dµ′(g1)dµ(g2).

On the other hand

I(φ)I ′(f1) =

∫φ(g1g2)f1(g2) dµ(g1)dµ′(g2) =

∫φ(y)f1(g−1

1 y) dµ(g1)dµ′(y).

3

Hence

|I ′(φ)I(f1)− I(φ)I ′(f1)| ≤∫φ(g1)|f1(g2g1)− f1(g−1

1 g2)| dµ′(g1)dµ(g2)

=

∫φ(g1)|f1(g2g1)− f1(g−1

1 g2)|ψ(g2) dµ′(g1)dµ(g2)

≤ εI ′(φ)I(ψ)

Therefore ∣∣∣∣ I(f1)

I ′(f1)− I(φ)

I ′(φ)

∣∣∣∣ ≤ ε I(ψ)

I ′(f1)

In a similar way, ∣∣∣∣ I(f2)

I ′(f2)− I ′(φ)

I ′(φ)

∣∣∣∣ ≤ ε I(ψ)

I ′(f2)

Therefore ∣∣∣∣ I(f1)

I ′(f1)− I(f2)

I ′(f2)

∣∣∣∣ ≤ ε( I(ψ)

I ′(f1)+

I(ψ)

I ′(f2)

).

HenceI(f1)

I ′(f1)=

I(f2)

I ′(f2),

i.e. I = cI ′. �

Theorem 9. Let G be a locally compact group G and λ be a left Haar measure onit. Then G is compact if and only if λ(G) <∞.

Proof. If G is compact then the assertion of the theorem is trivial. If G is notcompact, fix U , a neighborhood of e with compact closure. There is an infinitesequence (gn)∞n=1 such that gn 6∈

⋃j<n gjU . Take a symmetric neighborhood V

of e with V V ⊂ U . Then gjV , j = 1, 2, . . . are pairwise disjoint. Hence λ(G) ≥∑j λ(gjV ) =

∑j λ(V ) =∞. �

Modular homomorphism. Let I be a right Haar integral. Then I ◦Lg is also aright Haar integral. Hence there is a number ∆(g) > 0 such that I ◦ Lg = ∆(g)I.It is easy to see that ∆(g) does not depend on the choose of I. It is easy to verifythat ∆ : g 7→ ∆(g) is a group homomorphism from G to R∗+.

Theorem 10. ∆ is continuous.

Proof. Take 0 6= φ ∈ C+00(G). Select a symmetric precompact neighborhood U of e

Let ψ ∈ C+00(G) be such that ψ = 1 on U · Supp(φ).

Fix ε > 0. Select a symmetric precompact neighborhood V ⊂ U of e such that‖Lg(φ)− φ‖ < ε. Then

|I(Lg(φ))− I(φ)| ≤ I(|Lg(φ)− φ|) = I(|Lg(φ)− φ|ψ) ≤ εI(ψ).

Thus |∆(g)− 1| ≤ ε I(ψ)I(φ) . �

G is called unimodular if ∆ is trivial.4

Proposition 11. Each locally compact Abelian group is unimodular. Each compactgroup is unimodular.

Exercise 12. Let I be a right Haar integral. (1)Show that I(f) = I(J(f) ·∆). (2)Asa corollary: G is unimodular iff I ◦J = I. Hint: a) Show that the right-hand-side isalso a right Haar integral. b) Prove that the corresponding proportional coefficientis 1.

Examples. 1) G is a finite group.2) G is infinite discrete group.3) G = Rn.4) G = Tn5) Suppose now that G is an open subset in Rn. Suppose that the multiplication

is given by the formula

(x ∗ y)j = 〈A(j)x, y〉+ dj for all x = (xj)j , y = (yj)j ,

for some matrices A(j) and reals dj . We let r(g) := |detR′g| and l(g) := |detL′g−1 |.Then

Ir(f) :=

∫G

f(x)

r(x)dx and Il(f) :=

∫G

f(x)

l(x)dx

are right and left Haar integrals. Indeed,

Ir(Rg(f)) =

∫G

f(x ∗ g)

r(x)dx =

∫G

f(y)

r(y ∗ g−1)

1

r(g)dy =

∫G

f(y)

r(y)dy = Ir(f).

Moreover,

Ir(Lgf) =l(g)

r(g)Ir(f).

Hence ∆(g) = l(g)r(g) .

5a) G = R∗. Ir(f) =∫ f(x)|x| dx.

5b) G =

{(x y0 1

)| x 6= 0

}. Then

Ir(f) =

∫f(x, y)

|x|dxdy, Il(f) =

∫f(x, y)

x2dxdy, ∆(x, y) =

1

|x|.

5c) G = GL2(R). Then r(A) = l(A) = (detA)2.6) Haar measures for product of l.c.g.

Literature.1. A. Weil, Integration in topological groups and its applications2. Hewitt, Ross, Abstract harmonic analysis, vol 1.3. Naimark, Normed rings

5

2. Amenability for discrete groups

Invariant means on discrete groups. Let G be a group endowed with discretetopology. Denote by BC(G) the algebra of bounded continuous (real) functions onG.

Definition 1. A left invariant mean (LIM) on G is a functional M : BC(G)→ Rsuch that

(1) inf f ≤M(f) ≤ sup f for each f ∈ BC(G) and

M ◦ Lg = M for all g ∈ G. In this case G is called left amenable. In a similar waywe can define a right invariant mean, 2-side invariant mean, right amenability and(two-sided) amenability.

(1) is equivalent to M(f) ≥ 0 if f ≥ 0 plus M(1) = 1.

Proposition 2. G is left amenable iff G is right amenable iff G is amenable.

Proof. Indeed, if M is LIM then M ◦ J is RIM. Moreover, given f ∈ BC(B), letf ′(g) := M(Rg−1(f)). Then f ′ ∈ BC(G) and f 7→ M(J(f ′)) is a 2-sided invariantmean. �

Of course, every finite group is amenable.

Theorem 3 (Dixmier criterium). Let Sl := {∑j(Lgj (fj) − fj) | f1, . . . , fn ∈

BC(X), g1, . . . , gn ∈ G} and Sl,r := {f1 + f2 ◦ J | f1, f2 ∈ Sl}. Then(1) A left invariant mean on BC(G) exists iff sup f ≥ 0 for each f ∈ Sl.(2) A 2-side invariant mean on BC(G) exists iff sup f ≥ 0 for each f ∈ Sl,r.

Proof. (1) Let p(f) := sup f . Then p is a sub linear function on BC(X). Suppose(1) holds. We put M(f) := 0 on Sl. Then p ≥ M on Sl. By Hahn-Banachtheorem, M extends as a linear functional to BC(X) with p ≥M . For f ∈ BC(G),inf f = − sup(−f) ≤ −M(−f) = M(f). And since M(f − Lgf) = 0, M is leftinvariant. �

Definition 4. Let every finite subset of G generate a finite subgroup. Then G iscalled locally finite.

Theorem 5. Every locally finite group is amenable.

Proof. Take f ∈ Sl. Then f =∑j(Lgj (fj) − fj) for some gj , fj . Then there is a

finite subgroup H ⊂ G such that gj ∈ H. Let f ′j := fj � H, f ′ := f � H. Thenf ′ =

∑j(Lgj (f

′j)−f ′j). By Theorem 2(i), sup(f � H) ≥ 0. It follows that sup f ≥ 0.

Hence G is amenable by Theorem 2. �

In a similar way one can prove the following

Theorem 6. If G = inj limαGα and each Gα is amenable then G is amenable. Inparticular, G is amenable if and only if every finitely generated subgroup of it isamenable.

In particular, a group G is amenable off each finitely generated subgroup of G isamenable. This means that the amenability is a “local” concept.

6

Markov-Kakutani theorem. Let V be a locally convex vector space. Let K be acompact convex subset in V . Let (Tx)x∈X be a family of pairwise commuting affinetransformations of K. Then there is a common fixed point for (Tx)x∈X .

Proof. Let FTx := {k ∈ K | Txk = k}. Let us show that FTx 6= ∅. Set

Tx,n :=1

n

n∑j=1

T jx .

Of course, Tx,n : K → K. Fix k ∈ K and put kn := Tx,nk. Let k0 be a limitpoint of (kn)∞n=1. Then k0 ∈ K. It is easy to check that k0 ∈ FTx . Next, we notethat FTx is a compact convex set. Moreover, TyFTx ⊂ FTx for each y ∈ G. Byinduction, for each finite subset S ⊂ G,

⋂x∈S FTx 6= ∅. Therefore, because K is

compact,⋂x∈G FTx 6= ∅. �

Theorem 7. If G is Abelian then it is amenable.

Proof. a) BC(G) is a Banach space.b) Endow BC(G) with the *-weak topology. Then the subset

M := {F ∈ BC(G)′ | F (f) ≥ 0 for each f ≥ 0, F (1) = 1}

is compact (follows from the Banach-Alaoglu theorem) and convex.c) TgF := F ◦ Lg is affine.It remains to apply Markov-Kakutani. �

Remark 8. Let G = Z. Put Fn := {−n, . . . , n} and

Mn(f)(k) :=1

2n+ 1

∑j∈Fn

f(k + j).

Then Mn ∈ M. Hence there is a subsequence Mnm that ∗-weakly converges toa LIM M . It is interesting to note that if f is a converging sequence of realsthen f ∈ BC(Z) and limn→∞ f(n) = M(f). However if f is bounded but it doesnot converge then M(f) is still well defined. Thus M(f) can be considered as a“generalized” (so-called Banach) limit of f .

Theorem 9. If H is a subgroup in G and G is amenable then H is amenable.

Proof. Let M be LIM on G. Fix a subset X ⊂ G such that Hx ∩ Hy = ∅ and⋃x∈X Hx = G. then every element g ∈ G can be represented uniquely as g = hx

for some h ∈ H and h ∈ H. Given f ∈ BC(H), we let f ′(g) := f(h). Thenf ′ ∈ BC(G). We now set M ′(f) := M(f ′). Of course, M ′(f) ≥ 0 if f ≥ 0 andM ′(1) = 1. (Lh(f))′ = Lh(f ′). Hence M ′ is LIM on H. �

Theorem 10. Let H be a normal subgroup in G. Then G is amenable if and onlyif H and G/H are both amenable.

Proof. (⇒) By Theorem 9, H is amenable. Let π : G → G/H be the naturalprojection and let π∗ denote the corresponding map f → f ◦ π from BC(H) toBC(G). Let M be LIM on G. Then M ◦ π∗ is LIM on BC(G).

7

(⇐) Let MH and MG/H be RIM on H and G/H respectively. Let s : G/H → Hbe a map with Hs(x) = x and Hs(G/H) = G. Given f ∈ BC(G), we let

f ′x(h) := f(hs(x)), x ∈ G/H, h ∈ H.

Let f ′′(x) := MH(f ′x). Finally we let M(f) := MG/H(f ′′). Then M is RIM on G.Let g = hs(x), g0 = h0s(x0). Then

gg0 = hs(x)h0s(x)−1s(x)s(x0) = hh1h2s(xx0).

(Rg0f)′x(h) = (Rgf)(hs(x)) = f(hh1h2s(xx0)) = f ′xx0(hh1h2) = (Rh1h2

f ′xx0)(h).

(Rg0f)′′(x) = MH(f ′xx0) = f ′′(xx0) = (Rx0(f ′′))(x).

Hence M is right invariant. �

Finitely additive invariant measures. Let M be LIM on G. For each subsetA ⊂ G, we set µ(A) := M(1A). Then

(1) 0 ≤ µ(A) ≤ 1,(2) µ(A) + µ(B) = µ(A ∪B) if A ∩B = ∅.(3) µ(gA) = µ(A)(4) µ(∅) = 0, µ(X) = 1.

Conversely,

Proposition 11. If µ is a finitely additive left invariant normalized measure onthe set of all subsets of G then G is amenable.

Proof. (a) Given f ∈ BC(G), there is a sequence of functions fn on G such thatfn takes finitely many values for each n and ‖f − fn‖ → 0.

(b) Of course, one can define naturally M(fn). Now define M(f) := limnM(fn).It is well defined.

(c) M is LIM on CB(G). �

Example 12. Let G = F2 (free group with 2 generators). Then G is not amenable.Let a, b be the free generators of F2. Consider a partition F2 =

⊔n∈ZBn, where

w ∈ Bn if w = anbj · · · , j 6= 0 unless w = an. Suppose that F2 is amenable. Let µ bethe left invariant finitely invariant normalized measure on F2. Since aBn = Bn+1,µ(Bn) = µ(Bn+1). For each n, 1 ≥ µ(

⊔1≤j≤nBj) = nµ(B0). Hence µ(B0) = 0.

On the other hand, let B = F2 \B0. Then bB ⊂ B0.

1 = µ(B tB0) = µ(B) + µ(B0) = µ(B) = µ(bB) ≤ µ(B0) = 0,

a contradiction.

Corollary 13. If G contains F2 as a subgroup then G is not amenable.

There was a conjecture that every non-amenable group contains F2. It is nottrue. A counterexample was constructed by Olshansky. However the conjecture istrue within the class of linear groups due to Tits alternative.

Tits alternative. (without proof) Let V be a finite dimensional vector space overa field F of characteristic 0. Let G be a subgroup of GL(V ). Then G is amenableif and only if it does not contain F2 as a a subgroup. If G is amenable then Gcontains a normal solvable subgroup of finite index.

8

Definition 14. Let EG be the smallest class of groups such that

(1) EG contains the abelian groups and the finite groups and(2) EG is closed under operation of taking subgroups, quotient groups, group

extensions and inductive limits.

In particular all nilpotent, all solvable groups are elementary. Every elementarygroup is amenable. The converse is true within the class of linear groups (followsfrom Tits alternative). However, in general there are non-elementary amenablegroups (Grigorchuk).

Example 15. The group PSL2(R) is non-amenable. We first note that each

matrix A :=

(α βγ δ

)∈ SL2(R) generates a linear fractional transformation QA

of C by

QA(z) :=αz + β

γz + δ.

The map A 7→ QA is a homomorphism from SL2(R) and its kernel is {I,−I}. Hencethis map identifies PSL2(R) with the group of linear fractional transformation ofC preserving the upper half plane.

We show that PSL2(R) contains F2. Consider 4 circles in C: C1 := {z | |z+2| =1}, C ′1 := {z | |z−2| = 1}, C2 := {z | |z+ 5| = 1}, C ′2 := {z | |z−5| = 1}. Considertwo linear fractional transformations of C:

T1(z) :=2z + 3

z + 2, T2(z) :=

5z + 24

z + 5.

Extend them to a homomorphism T of F2 into PSL2(R) in a natural way: ifw = aj1bj2aj3 · · · ∈ F2 is a nontrivial reduced word, i.e. each jk 6= 0, then Tw :=T j11 T j22 T j31 · · · . In a similar way we define Tw for a word w starting with b. Ourpurpose to prove that the kernel of T is trivial, i.e. if w is a nontrivial word thenTw 6= I.

It is easy to see that T1(C1) = C ′1 and T2(C2) = C ′2. Moreover,

T1(e(C1)) = i(C ′1) and T2(e(C2)) = i(C ′2).

Let z0 = 4i ∈ C. Then Twz0 is inside the interior of the 4 circles. Hence Twz0 6= z0.Therefore Tw 6= I.

Corollary 16. SL2(R) is not amenable.

Literature.1. Hewitt, Ross, Abstract harmonic analysis, vol 1.2. Greenleaf, Invariant means on topological groups and their applications3. Paterson, Amenability.

9

3. Invariant mean on almost periodic functions

As before, G is a group with discrete topology. Given f ∈ BC(G) and g ∈ G,we define Dg : G×G→ C by Dgf(x, y) = f(xgy).

Definition 1. A function f ∈ BC(G) is called almost periodic (AP) if of of thefollowing is satisfied:

(1) {Lg(f) | g ∈ G} is relatively compact in BC(G).(2) {Rg(f) | g ∈ G} is relatively compact in BC(G).(3) {RhLg(f) | h, g ∈ G} is relatively compact in BC(G).(4) {Dgf | g ∈ G} is relative compact in BC(G2).

Proposition 2. The conditions (1)–(4) are equivalent.

Proof. BC(G) is a Banach space. A subset of it is relatively compact off it is totallybounded.

(1)⇒ (3) Given ε > 0, there are a1, . . . , an ∈ G such that La1(f), . . . , Lan(f) isan ε-net in {Lg(f) | g ∈ G}. Since Laj (f) is AF, there is an ε-netRbj,k , k = 1, . . .Kj ,in {Rh(Laj (f)) | h ∈ G} by (ii). Then Rbj,k(Laj ) is 2ε-net in {RhLg(f) | h, g ∈ G}.Thus (1), (2), (3) are equivalent. Of course, (4) implies (1) and (2).

Given ε, we find a1, . . . , an ∈ G such that Ra1(f), . . . , Ran(f) is ε-net in {Rg(f) |g ∈ G}. Next, let Aj := {g ∈ G | ‖Rg(f) − Raj (f)‖ < ε}. Then

⋃j Aj = G.

Consider the family of sets⋂j Alja

−1j when l1, . . . , ln runs 1, . . . , n. Enumerate

them (only those which are nonempty) as B1, . . . , Bm. Then⋃mk=1Bm = G. Select

bk ∈ Bk for each k = 1, . . . ,m. Now given c ∈ G, take k0 such that c ∈ Bk0 . Let(x, y) ∈ G2. Take j0 such that y ∈ Aj0 . Then

|f(xcy)− f(xbk0y)| ≤ |f(xcy)− f(xcaj0)|+ |f(xcaj0)− f(xbk0aj0)|+ |f(xbk0aj0)− f(xbk0y)|≤ ‖Rcyf −Rcaj0 f‖+ ‖Rcaj0 f −Rbk0aj0 f‖+ ‖Rbk0aj0 f −Rbk0yf‖≤ ‖Ryf −Raj0 f‖+ ‖Rcaj0 f −Rbk0aj0 f‖+ ‖Raj0 f −Ryf‖ < 4ε

because caj0 and bk0aj0 are in the same set Alj0 . �

Denote by AP (G) the set of all AP-functions on G.

Proposition 3. AP (G) is a closed subalgebra of BC(G). It contains constants. Itis invariant under Lg and Rg, g ∈ G. If f ∈ AP (G) then Re(f), Im(f) and f arein AP (G).

Marriage Lemma. Let X,Y be two nonempty sets. Let F be a map from X tothe set of all non-empty finite subsets of Y . Suppose that for each X1 ⊂ X,

#(⋃x∈X1

F (x)) ≥ #(X1)

then there is a 1-to-1 map s : X → Y such that s(x) ∈ F (x).

Proof. Suppose that X is finite. Apply an induction argument.— If #(X) = 1 then the claim is trivial.

10

— Let #(X) = n. Fix x0 ∈ X. Choose y0 ∈ F (x0). If for each X1 ⊂ X \ {x0},

#(⋃x∈X1

F (x)) > #(X1)

then we put F ′(x) := F (x) \ {y0} and apply the inductive assumption: there is a1-to-1 map s : X \ {x0} → Y \ {y0} with s(x) ∈ F ′(x), which we extend to x0 ass(x0) := y0.

If there is X1 ⊂ X, #(X1) < n with #(⋃x∈X1

F (x)) = #(X1), we put Y1 :=⋃x∈X1

F (x). Then by inductive assumption, there is a 1-to-one map s : X1 → Y1

with s(x) ∈ F (x) ⊂ Y1 for each x ∈ X1. We now set X2 := X \X1, Y2 := Y \ Y1,F2(x) := F (x) \ Y1. Let us verify that the conditions of the lemma are satisfied forX2, Y2, F2. If not, there is a non-empty subset X3 ⊂ X2 with #(

⋃x∈X3

F2(x)) <#(X3). However then

#

( ⋃x∈X3

F2(x) t⋃x∈X1

F (x)

)< #(X3) + #(X1), i.e.

#

( ⋃x∈X3tX1

F (x)

)< #(X3 tX1).

Therefore by inductive assumption, there is a 1-to-1 map r : X2 → Y2 with r(x) ∈F2(x). It remains to concatenate s and r.

Consider now the case of infinite X. The infinite product Z :=∏x∈X F (x) is

a compact space. For each finite subset S ⊂ X, let ZS be the subset of all z ∈ Zsuch that z � S is one-to-one. By the above argument, ZS 6= ∅ and it is closed.Moreover, ZS1

∩ ZS2∩ · · · ∩ ZSn ⊃ ZS1∩···Sn . Hence these finite intersections are

non-empty. Therefore ⋂S is a finite subset of X

ZS 6= ∅.

�

Lemma 4. Let X be a mertic space and let x1, . . . , xn be an ε-net of smallestpossible cardinality (for this ε). Let Y be an ε-net in X. Then there is a 1-to-1 maps : {x1, . . . , xn} → Y with xj and s(xj) being inside the same ε-ball for each j.

Proof. Given j ≤ n, let F (xj) := {y ∈ Y | xj and y are inside the same ε-ball}.All that we need to prove is that #(

⋃j∈I F (xj)) ≥ #(I) for each nonempty subset

I ⊂ {1, . . . , n}. Indeed, assume that #(⋃j∈I F (xj)) < #(I) for some I. We claim

that the set A :=⋃i∈I F (xi) ∪ {xj | j 6∈ I} is a ε-net. Take z ∈ X. If z is ε-far

from each xj , j 6∈ I, then there is j0 ∈ I such that z and xj0 are ε-close. On theother hand, there is y ∈ Y which is ε-close to z. Hence y ∈

⋃j∈I F (xj). Thus A is

ε-net and hence #(A) ≥ n. �

Corollary 5. Let f ∈ AP (G), ε > 0 and Da1f, . . . ,Danf be an ε-net in {Dgf |g ∈ G} of minimal cardinality among the ε-nets. Then∥∥∥∥∥∥ 1

n

n∑j=1

f(aj)−1

n

n∑j=1

Dajf

∥∥∥∥∥∥ ≤ 2ε.

11

Proof. Let Db1 , . . . , Dbn be another ε-net in {Dgf | g ∈ G}. Then there is abijection σ of {1, . . . , n} such that ‖Dajf −Dbσ(j)f‖ < 2ε. Therefore

(1)

∥∥∥∥∥∥ 1

n

n∑j=1

Dajf −1

n

n∑j=1

Dbjf

∥∥∥∥∥∥ ≤ 1

n

n∑j=1

∥∥Dajf −Dbσ(j)f∥∥ < 2ε.

Let u, v ∈ G. Then Dua1v, . . . , Duanv is an ε-net in {Dgf | g ∈ G}. Indeed, giveng ∈ G, we have ‖Du−1gv−1f −Dajf‖ < ε for some j. Hence ‖Dgf −Duajvf‖ < ε.It now follows from (1) with bj := uajv that

2ε >

∣∣∣∣∣∣ 1nn∑j=1

Dajf(e, e)− 1

n

n∑j=1

Dbjf(e, e)

∣∣∣∣∣∣ =

∣∣∣∣∣∣ 1nn∑j=1

f(aj)−1

n

n∑j=1

f(uajv)

∣∣∣∣∣∣ .It remains to take supremum over u, v ∈ G. �

Theorem 6 (Existence of 2-sided means on the AP-functions). There isa 2-sided invariant mean M on AP (G). Moreover, M is strictly positive, i.e. iff ≥ 0 and f 6= 0 then M(f) > 0.

Proof. Take f ∈ AP (G) and ε > 0. We put

Eε :=

{z ∈ R |

∥∥∥∥∥z − 1

#A

∑a∈A

Daf

∥∥∥∥∥ < ε for a finite A ⊂ G

}.

By Corollary 5, Eε 6= ∅. Let us verify that the diameter of Eε is less than 2ε.Indeed, if z1, z2 ∈ Eε, then

|z1 −1

#A

∑a∈A

f(xay)| < ε,(2)

|z2 −1

#B

∑b∈B

f(xby)| < ε(3)

for all x, y ∈ G and some finite subsets A,B in G. From (2) and (3) we deduce

|z1 −1

#A#B

∑a∈A,b∈B

f(ab)| < ε and

|z2 −1

#A#B

∑a∈A,b∈B

f(ab)| < ε

respectively. Hence

(4) |z1 − z2| ≤ 2ε.

Moreover, Eε is bounded. Indeed, |z| < ‖f‖+ ε for each z ∈ Eε. Since Eε1 ∩ · · · ∩Eεk ⊃ Emin(ε1,...,εk), it follows that

⋂ε>0Eε 6= ∅. From (4) we deduce that

⋂ε>0Eε

12

is a singleton. Denote it by M(f). Thus M(f) is the only number such that foreach ε > 0, there is a finite A ⊂ G with∥∥∥∥∥M(f)− 1

#A

∑a∈A

Daf

∥∥∥∥∥ < ε.

It it easy to see that f 7→M(f) is a 2-sided invariant functional, M(αf) = M(f) foreach number α, M is non-negative and M(1) = 1. To show additivity, let ‖M(f)−

1#A

∑a∈ADaf‖ < ε and ‖M(f1) − 1

#B

∑b∈B Dbf1‖ < ε for some f, f,∈ AP (G)

and A,B ⊂ G. Then we obtain easily that ‖M(f) − 1#A#B

∑a∈A,b∈B Dabf‖ < ε

and ‖M(f1)− 1#B#A

∑b∈B,a∈ADabf1‖ < ε and hence

‖M(f) +M(f1)− 1

#B#A

∑b∈B,a∈A

Dab(f + f1)‖ < ε.

Since ε is arbitrary, M(f + f1) = M(f) + M(f1). It remains to prove that M isstrictly positive. Let f ∈ AP (G), f ≥ 0 and f(g0) > 0 for some g0 ∈ G. Let A ⊂ Gbe an f(g0)/2-net in {Dgf | g ∈ G}. Then for x, y, g ∈ G, we have

∑a∈A

f(xay) ≥ maxa∈A

Daf(x, y) > Dgf(x, y)− f(g0)

2= f(xgy)− f(g0)

2.

Therefore ∑a∈A

f(xa) >f(g0)

2

for each x ∈ G. Therefore∑a∈ARa(f) > f(g0)/2 and hence #(A)M(f) ≥

f(g0)/2. �

Theorem 7 (Uniqueness of RIM on AP (G)). Let M ′ be a RIM on AP (G).Then M ′ = M .

Proof. Given f ∈ AP (G) and ε > 0, we have

−ε < M(f)− 1

n

∑a∈A

f(xay) < ε

for all x, y ∈ G and a finite subset A ⊂ G. Hence −ε < M(f)− 1n

∑a∈ARa(f) < ε.

Therefore −ε ≤M(f)−M ′(f) ≤ ε. �

Remark 8. If G is locally compact (or Polish) then every bounded function f withrelatively compact {Lg(f) | g ∈ G} is (uniformly) continuous. Indeed, for eachε > 0, there are subsets Aj ⊂ G and elements aj ∈ G with ‖Lgf −Lajf‖ ≤ ε for allg ∈ Aj . It follows from our assumption on G that there is j0 such that Aj0 has the

Baire property. Then ‖Lb(f)− f‖ ≤ 2ε for all b ∈ A−1j0Aj0 . By the Pettis theorem,

there is an open neighborhood U of e with U ⊂ A−1j0Aj0 .

Literature.1. Hewitt, Ross, Abstract harmonic analysis, vol 1.

13

4. Amenability for locally compact groups

Let G be a locally compact group. Fix a left Haar measure λ on G. Denote byL∞(X) the space of essentially bounded functions. Recall that L∞(G) = L1(G)′.It is a Banach space when endowed with the norm

‖f‖∞ := vraisup|f | = inf

{sup

g∈G\N|f(g)| | N is a λ-locally 0 subset of G

}.

We consider the following subspaces of L∞(G):

— BC(X),— BUCl(G), the left uniformly continuous bounded functions on G, i.e.

{f ∈ BC(G) | the map G 3 g 7→ Lg(f) is continuous}.

— BUCr(G), the right uniformly continuous bounded functions on G,— BUC(G), the 2-sided uniformly continuous bounded functions on G, i.e.

BUCl(G) ∩BUCr(G).

Lemma 1. BUCl(G), BUCr(G), BUC(G) are 2-sided invariant closed subspacesof BC(G).

In each of these spaces (and also in L∞(G)) one can define concepts of LIM,RIM and 2-sided invariant mean.

Definition 2. A locally compact group G is called amenable if there is a LIM (or,equivalently, RIM or, equivalently, 2-sided mean) on L∞(G).

This extend the concept of amenability for the discrete groups. Indeed, anydiscrete group G is locally compact and L∞(G) = B(G). Each compact group isamenable with LIM equals the normalized Haar integral.

Let C0(G) denote the space of continuous functions f on G with limg→∞ f(g) =0. By M(G) denote the dual space C0(G)′ of bounded regular measures on G. Itis a Banach space with the norm equal to the full variation. The convolution iscontinuous on M(G):

〈µ ∗ ν, f〉 :=

∫G×G

f(g1g2) dµ(g1)dν(g2).

We note that L1(G) is embedded isometrically into M(G) via the one-to-one mapφ 7→ µφ, dµφ(g) := φ(g)dµ(g).

Exercise 3. L1(G) is an 2-sided ideal in the algebra (M(G), ∗). More precisely, if

µ ∗ φ(g) :=

∫G

φ(x−1g) dµ(x),

φ ∗ µ(g) :=

∫G

φ(gx−1)∆(x−1) dµ(x))

then µ ∗ µφ = µµ∗φ and µφ ∗ µ = µφ∗µ. Moreover, ‖µ ∗ φ‖1 ≤ ‖µ‖ · ‖φ‖1 and‖φ ∗ µ‖1 ≤ ‖µ‖ · ‖φ‖1. In particular, µφ ∗ µψ = µφ∗ψ, where

φ ∗ ψ(g) =

∫G

φ(x)ψ(x−1g) dλ(g).

Since L∞(G) = L1(G,µ)′, we use this duality to define the following convolution.14

Definition 4. Given φ ∈ L1(G) and f ∈ L∞(G) we let

φ ∗ f(g) :=

∫G

φ(y)f(yg)dλ(y), f ∗ φ(g) :=

∫g

f(gy)φ(y)dλ(y).

We see that ‖φ ∗ f‖∞ ≤ ‖φ‖1 · ‖f‖∞ and ‖f ∗ φ‖∞ ≤ ‖φ‖1 · ‖f‖∞. Moreover,

〈φ ∗ ψ, f〉 = 〈ψ, φ ∗ f〉 = 〈φ, f ∗ ψ〉.

Definition 5. A linear functional m on L∞(G) is called a topological LIM onL∞(G) if it is positive, normalized and m(φ ∗ f) = m(f) for all

φ ∈ P(G) := {φ ∈ L1(G) | φ ≥ 0, ‖φ‖1 = 1}.

A linear functional m on L∞(G) is called a topological RIM on L∞(G) if it ispositive, normalized and m(f ∗ φ) = m(f) for all φ ∈ P(G).

In a similar way one can define topological LIM and RIM onBC(G), BUCr(G), BUC(G).

Lemma 6 (On regularization). Let f ∈ L∞(G) and φ ∈ P(G). Then φ ∗ f ∈BUCl(G), f ∗ φ ∈ BUCr(G). If f1 ∈ BUCr(G) then φ ∗ f1 ∈ BUC(G). Iff2 ∈ BUCl(G) then f ∗ φ ∈ BUC(G).

Proof. We check only the first claim.

|φ ∗ f(x)− φ ∗ f(yx)| = |∫φ(t)f(tx)dλ(t)−

∫φ(t)f(tyx)dλ(t)|

≤∫|φ(tx−1)− φ(tx−1y−1)∆(y)||f(t)|∆(x)dλ(t)

≤ ‖f‖∞∫|φ(t)− φ(ty−1)∆(y)|dλ(t)

= ‖f‖∞‖φ−Ry−1(φ)‖1

�

Recall that Lg and Rg denote the following isometries in L1(G). Lgf(x) :=f(g−1x), Rgf(x) := f(xg)∆(g−1).

Lemma 7 (On approximation in L1(G)). Let f ∈ L1(G).

(1) The maps G 3 g 7→ Lg(f) ∈ L1(G) and G 3 g 7→ Rg(f) ∈ L1(G) arecontinuous.

(2) For each ε > 0, there is a neighborhood U of e such that for each non-negative ψ ∈ L1(G) vanishing out of U and such that

∫ψ(g)dλ(g) = 1,

‖ψ ∗ f − f‖1 < ε and ‖f ∗ ψ − f‖1 < ε.

Proof. (1) Approximate f with a function from C00(G) in ‖.‖1 norm.(2)

‖ψ ∗ f − f‖1 ≤∫ψ(y) ·

∫|f(yx)− f(x)|dλ(x)dλ(y)

=

∫ψ(y)‖Ly−1(f)− f‖1dλ(y)

15

It remains to apply (1). In a similar way,

‖f ∗ ψ − f‖1 ≤∫ψ(y)

∫|f(xy)− f(x)|dλ(x)dλ(y)

=

∫ψ(y)

∫|Ryf(x)∆(y)− f(x)|dλ(x)dλ(y)

�

Corollary 8. There is an approximate unit (eα)α∈A in L1(G), i.e. eα ∈ L1(G)and limα eα ∗ f = limα f ∗ eα = f , where A is a directed set. Moreover, we willadditionally assume that eα ∈ P and eα is compactly supported.

We need an analogue of Lemma 7(2) for BUC(G).

Lemma 9. Let f ∈ BUC(G). Given ε > 0, there is a neighborhood U of e withcompact support such that for each ψ ∈ L1

+(G) vanishing out of U and such that∫ψ dλ = 1, ‖ψ ∗ f − f‖∞ < ε.

Proof.

|ψ ∗ f(x)− f(x)| ≤∫ψ(y) · |f(yx)− f(x)|dλ(y)

=

∫ψ(y)‖Ly−1(f)− f‖∞ dλ(y)

≤ supy∈U‖Ly−1(f)− f‖∞

�

Theorem 10. If there is a LIM on B(G) then there is a LIM on BC(G). Theconverse is not true. The following are equivalent:

(1) there is a topological LIM on L∞(G),(2) there is a LIM on L∞(G),(3) there is a LIM on BC(G),(4) there is a LIM on BUCl(G),(5) there is a LIM on BUC(G).

Proof. The first claim is obvious.To show the second one, consider O3(R). It is amenable as a compact group but

it is non-amenable as a discrete group since it contains F2.(1)⇒(2) Let m be a topological LIM on L∞(G). Let us show that m is a LIM

on L∞(G). Take g ∈ G and φ ∈ P(G). For each f ∈ L∞(G),

(φ ∗ Lgf)(x) =

∫φ(t)f(gtx) dλ(t)

=

∫φ(g−1t)f(tx) dλ(u) = Lg(φ) ∗ f.

Hencem(Lgf) = m(φ ∗ Lgf) = m(Lg(φ) ∗ f) = m(f).

(2)⇒(3)⇒(4)⇒(5) are obvious.16

(5)⇒(1) Let m be a LIM on BUC(G). First of all we show that m is a topologicalLIM on BUC(G). Take f ∈ BUC(G), φ ∈ L1(G) and g ∈ G. We set J(φ)(g) :=φ(g−1)∆G(g). Of course, J(φ) ∈ L1(G). We claim that J(Lg(φ))∗f = Lg(J(φ)∗f).

J(Lg(φ)) ∗ f(x) =

∫G

φ(g−1t−1)f(tx)∆G(t)dλ(t)

=

∫G

φ(t−11 )f(t1g

−1x)∆G(g−1t1)∆G(g)dλ(t1)

=

∫G

J(φ)(t1)f(t1g−1x)dλ(t)

= J(φ) ∗ f(g−1x).

Then J(Lg(φ)) ∗ f and J(φ) ∗ f are in BUC(G) by Lemma 6 and

m(J(Lg(φ)) ∗ f) = m(J(φ) ∗ f).

Thus the map φ 7→ m(J(φ) ∗ f) is a bounded linear right-invariant functional onL1(G). Hence (uniqueness of Haar measure) there is a constant cf such that

m(J(φ) ∗ f) = cf

∫φdλ.

We note that J(φ) ∈ P(G) if φ ∈ P(G) and φ 7→ J(φ) is an affine isomorphism ofP(G). Hence

m(φ ∗ f) = cf

for each φ ∈ P(G). Take an approximate unit (eα)α∈A in L1(G). Take φ ∈ P(G).We have eα ∗ f → f in ‖.‖∞ by Lemma 9 and

m(φ ∗ f) = cf = m(eα ∗ f)→ m(f),

as desired.Take a symmetric neighborhood U of e with compact closure. Let ψ := 1U/λ(U).

Then ψ ∈ P(G) and J(ψ) = ψ. By Lemma 6, ψ ∗ f ∗ ψ ∈ BUC(G) for eachf ∈ L∞(G). Hence

m′(f) := m(ψ ∗ f ∗ ψ)

is a well defined mean on L∞(G), i.e. it is positive, normalized linear functional.We want to have for each φ ∈ P(G) and f ∈ L∞(G),

m(ψ ∗ φ ∗ f ∗ ψ) = m′(φ ∗ f) = m′(f) = m(ψ ∗ f ∗ ψ).

It will follow from the more general assertion (in view of Lemma 6):

m(τ1 ∗ f ′) = m(τ2 ∗ f ′)

for all τ1, τ2 ∈ P(G) and f ′ ∈ BUCr(G). To prove the latter, we note that τj ∗eα →τj in L1(G) and hence ‖τj ∗ eα ∗ f ′ − τj ∗ f ′‖∞ ≤ ‖τj ∗ eα − τj‖1‖f ′‖∞ → 0.

m(τj ∗ f ′) = limαm(τj ∗ eα ∗ f ′) = lim

αm(eα ∗ f ′).

�

17

Theorem 11. Let G be a locally compact group and N a closed subgroup of G.

(1) Let G and H be a locally compact groups and let π : G→ H be a continuousonto homomorphism. If G is amenable then H is amenable.

(2) If H is a closed subgroup of a locally compact amenable group G then H isamenable.

(3) Let N be an amenable closed normal subgroup of G and let the quotientgroup G/N is also amenable. Then G is amenable.

(4) Let G = inj limα∈AGα for a directed set A, every Gα is an amenable closedsubgroup of G. Then G is also amenable.

Proof. (1) Just note that BC(H) is embedded into BC(G).(3) Let mN be a LIM on BC(N) and let mG/N be a LIM on BC(G/N). Given

f ∈ BC(G), we let f ′(g) := mN (Lg(f)). Since g 7→ Lg(f) is a continuous mapfrom G to BC(G), f ′ ∈ C(G). Of course, f ′ is bounded. Since Ln(f ′) = f ′ forall n ∈ N , we conclude that f ′ is a continuous function on G/N . We now setm(f) := mG/N (f ′). Then m is a LIM on G.

(4) Let mα be a LIM on BC(Gα). Then the mα can be viewed as a map definedon BC(G). It is a mean and it is Gα-invariant. Denote by Mα the set of allGα-invariant LIMs on G. Then Mα is *-weakly compact subset of the unit ball inBC(G)′. It is non-empty. Given α, β ∈ A, there is γ ∈ A with γ � α and γ � β.Then ∅ 6= Mγ ⊂Mα ∩Mβ . Hence

⋂γ∈AMγ 6= ∅.

(2) Suppose that G is Polish. Then there is a Borel map s : G/H → G suchthat Hs(Hg) = Hg. Every g uniquely decomposes into g = hs(Hg), where h =gs(Hg)−1. Given f ∈ CB(H), let f ′(g) := f(h). Then f ′ ∈ B(G). Moreover, it isBorel. Hence f ′ ∈ L∞(G). Moreover, the map f 7→ f ′ is linear and ‖f ′‖∞ ≤ ‖f‖∞.It remains to put mH(f) := m(f ′). If m is LIM on L∞(G) then mH is a LIM onBC(H). �

Literature.1. Greenleaf, Invariant means on topological groups and their applications

18

5. Følner condition and its applications

Let M(G) ⊂ L∞(G)′ denote the space of means on L∞(G). Then P(G) ⊂M(G).

Lemma 12. P(G) is *-weakly dense in M(G).

Proof. We first note that M(G) is a *-weakly compact subset of L∞(G)′. Let Kbe the closure of P(G) in L∞(G)′. Then K is compact. If there is k0 ∈M(G) \Kthen by Hahn-Banach separation theorem, there is f ∈ L∞(G) and δ > 0 suchthat k0(f) > 〈φ, f〉+ δ for all φ ∈ P(G). There is a compact set C ⊂ G such thatλ(C) > 0 and f(g) ≥ ‖f‖∞− δ/2 for all g ∈ C. Then φ0 := λ(C)−11C ∈ P(G) andk0(f) >

∫Cf dλ/λ(C) + δ ≥ ‖f‖∞ + δ/2, a contradiction. �

Theorem 13.

(1) G is amenable if and only if there is a net φα ∈ P(G), α ∈ A, such that foreach g ∈ G, Lg(φα)− φα → 0 *-weakly in L∞(G)′.

(2) G is amenable if and only if there is a net φα ∈ P(G), α ∈ A, such that foreach φ ∈ P(G), φ ∗ φα − φα → 0 *-weakly in L∞(G)′.

Proof. We prove only (2).(⇐) There is a subnet φα converging to some m ∈M(G). Then

m(φ ∗ f)−m(f) = limα

(〈φα, φ ∗ f)〉 − 〈φα, f〉) = limα〈φ ∗ φα − φα, f〉 = 0.

(⇒) Take a topological LIM m on L∞(G). By Lemma 12, there is a net φα ∈P(G) with limα φα = m. Then

〈φ ∗ φα − φα, f〉 = 〈φα, φ ∗ f〉 − 〈φα, f〉 → m(φ ∗ f)−m(f) = 0.

�

Theorem 14.

(1) If there is a net φα ∈ P(G), α ∈ A such that for each g ∈ G, Lg(φα)−φα →0 *-weakly in L∞(G)′ then there is a net ψβ ∈ P(G), β ∈ B, such that foreach g ∈ G, ‖Lg(φβ)− φβ‖1 → 0.

(2) If there is a net φα ∈ P(G), α ∈ A such that for each φ ∈ P(G), φ ∗ φα −φα → 0 *-weakly in L∞(G)′ then there is a net ψβ ∈ P(G), β ∈ B, suchthat for each φ ∈ P(G), ‖φ ∗ ψβ − ψβ‖1 → 0.

Proof. We prove only (2).Let E =

∏φ∈P(G) L

1(G) with the product topology, each L1 endowed with

‖.‖1 norm. Then E is a locally convex space and E′ =⊕

φ∈P(G) L∞(G). Let

T : L1(G) → E be defined by Tψ(φ) = φ ∗ ψ − ψ. Then T is a linear map andhence T (P(G)) is a convex subset of E. For each l ∈ E′, finite subset A ⊂ P(G)and lφ ∈ L∞(G), φ ∈ A,

〈l, Tφα〉 =∑φ∈A

〈lφ, φ ∗ φα − φα〉 → 0.

Thus Tφα → 0 ∗-weakly. Hence 0 belongs to the ∗-weal closure of T (P(G)). SinceT (P(G)) is convex, the weak closure equals the strong closure. Hence there is a netψβ ∈ P(G) such that Tψβ → 0 strongly in E. �

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Definition 15 (Reiter condition). A locally compact group G satisfies Reitercondition if for each ε > 0 and compact subset K ⊂ G, there is φ ∈ P(G) such that

(R) supg∈K‖Lg(φ)− φ‖1 < ε.

Theorem 16. G satisfies the Reiter condition if and only if G is amenable.

Proof. (⇒) Let J = {(ε,K)} and select φj , j ∈ J , satisfying (R). Order J in ausual way. By Theorem 13(1), G is amenable.

(⇐) By Theorems 13(2) + 14(2), there is a net (φα)α∈A in P(G) such that‖φ∗φα−φα‖1 → 0 for each φ ∈ P(G). Fix ε > 0, a compact K in G and β ∈ P(G).Take a small neighborhood U of e with compact closure such that

‖φU ∗ β − β‖1 < ε, and supg∈U‖Lg(β)− β‖1 < ε,

where φU = 1U/λ(U) ∈ P(G). There are g1, . . . , gN ∈ G with⋃Nj=1 gjU ⊃ K. We

may assume that g1 = e. Select φα such that

max1≤j≤N

‖Lgj (φU ) ∗ φα − φα‖1 < ε, ‖β ∗ φα − φα‖1 < ε.

Let us show that φ := β ∗ φα satisfies (R). Indeed,

‖φU ∗ φ− Lg(φ)‖1 ≤ ‖φU ∗ φ− φ‖1 + ‖φ− Lg(φ)‖1= ‖(φU ∗ β − β) ∗ φα‖1 + ‖(β − Lg(β)) ∗ φα‖1 < 2ε

for each g ∈ U . Hence

2ε > ‖Lgj (φU ∗ φ)− Lgj (Lg(φ))‖1 = ‖Lgj (φU ) ∗ φ− Lgjg(φ)‖1.

Now

‖Lgjg(φ)− φ‖1 ≤ 2ε+ ‖Lgj (φU ) ∗ φ− φ‖1 = 2ε+ ‖Lgj (φU ) ∗ β ∗ φα − β ∗ φα‖1≤ 4ε+ ‖Lgj (φU ) ∗ φα − φα‖1 < 5ε.

�

Theorem 17. A locally compact group G is amenable if and only if for each ε > 0and a compact set K ⊂ G, there is a Borel set F such that 0 ≤ λ(F ) <∞ and

λ(gF4F )

λ(F )< ε.

Proof. (⇐) If φ := 1F /λ(F ) then φ ∈ P(G) and ‖Lg(φ)− φ‖1 = λ(gF4F )λ(F ) .

(⇒) We first prove a weaker assertion.(F) Given ε > 0 and δ > 0 and compact set K ⊂ G, there are Borel subsets

F ⊂ G and N ⊂ F such that 0 < λ(F ) < ∞, λ(N) < δ and λ(gF4F )λ(F ) < ε for all

g ∈ K \N .20

From the Reiter condition, there is φ ∈ P(G) with ‖Lg(φ)− φ‖1 < εδ/λ(K) forall k ∈ K. Without loss of generality we may assume that φ is a simple function.

Moreover, we can represent φ as φ =∑Nj=1 λj1Aj/λ(Aj) with A1 ⊂ A2 ⊂ · · · ⊂

AN and λj > 0. Hence∑j λj = 1. Since Lg(φ) − φ =

∑j λj1gAj\Aj/λ(Aj) −∑

j λj1Aj\gAj/λ(Aj) and (gAj \Aj) ∩ (Ai \ gAi) = ∅ then

‖Lg(φ)− φ‖1 =

N∑j=1

λjλ(gAj4Aj)λ(Aj)

< εδ/λ(K).

HenceN∑j=1

λj

∫K

λ(gAj4Aj)λ(Aj)

dλ(g) < εδ.

Hence there is j with ∫K

λ(gAj4Aj)λ(Aj)

dλ(g) < εδ.

Henceλ(gAj4Aj)λ(Aj)

> ε on a subset N whose measure λ(N) < δ. Thus (F) is proved.

Now we note that if λ(gF4F )λ(F ) < ε for each g ∈ K \N then λ(gF4F )

λ(F ) < 2ε for each

g ∈ (K \N)(K \N)−1. Indeed,

λ(g1g−12 F4F )

λ(F )=λ(g−1

2 F4g−11 F )

λ(F )≤ λ(g−1

1 F4F )

λ(F )+λ(g−1

2 F4F )

λ(F )

=λ(F4g1F )

λ(F )+λ(F4g2F )

λ(F ).

It remains to show

Lemma 18. If K is a compact in G then for each Borel subset N ⊂ K ∪KK, ifλ(N) < λ(K)/2 then (K \N)(K \N)−1 ⊃ K.

Proof. Let K1 := K ∪KK then

kK ⊂ kK1 ∩K1 ⊂ (k(K1 \N) ∩ (K1 \N)) ∪ kN ∪N.

Henceλ(K) ≤ λ(kK1 ∩K1) ≤ λ(k(K1 \N) ∩ (K1 \N)) + 2λ(N).

It follows that λ(k(K1 \N)∩(K1 \N)) > 0. In particular, k(K1 \N)∩(K1 \N) 6= ∅,i.e. k ∈ (K \N)(K \N)−1. �

Thus to prove the theorem, first given K, consider K1 and put δ := λ(K)/2.Apply (F). Then apply Lemma 18. �

Definition 19. A net of Borel subsets (Fα)α∈A of finite measure in a locallycompact group is called a (left) Følner net if

limα∈A

λ(gFα4Fα)

λ(Fα)= 0

for each g ∈ G.

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Corollary 20. Let G be a locally compact group. Then G is amenable if and onlyif there is a Følner net in it. If G is σ-compact then it is amenable if and only iftree is a Følner sequence it.

Proof. Let A be the directed set {(K, ε) | K is a compact in G, ε > 0}. For each

α ∈ A, let Fα be the corresponding set in G such that λ(gFα4Fα)λ(Fα) < ε for each g ∈ K

(by Theorem 17). Then (Fα)α∈A is a Følner net in G. Conversely, if (Fα)α∈A isa Følner net in G then ‖Lg(1Fα/λ(Fα)) − 1Fα/λ(Fα)‖1 → 0. It remains to applyTheorem 13(1) because the norm convergence implies the *-weak convergence. �

It follows from Theorem 17 that if G is amenable and σ-compact then there isa Følner sequence in it (the converse is also true). For example, if G = Z thenFn := {1, . . . , n} is a Følner sequence. If G = R then Fn := [0, n] is a Følnersequence.

Corollary 21 (The invariant mean on the almost periodic functions onamenable groups). Let (Fα)α be a Følner net in G. Let M be the (unique!)invariant mean on AP (G). Then

M(f) = limα

1

λ(Fα)

∫Fα

f(gx) dλ(g).

Proof. Consider a net 1Fα/λ(Fα) ∈ P(G). Take a limit point m of this net. Then mia a LIM m on L∞(G). We note that m is a LIM on AP (G) ⊂ BUC(G) ⊂ L∞(G).Since the LIM on AP (G) is unique, we conclude that m � AP (G) = M . Thus,though there can be many limit points m, there restrictions to the AP (G) is unique.Therefore there is limα〈1Fn/λ(Fn), f〉 = M(f) if f ∈ AP (G). �

Let V be a locally convex space and K a compact convex subset of V .

Definition 22. A continuous map T : G × K 3 (g, v) 7→ Tgv ∈ K is called anaffine action of G on K if Tg is an affine map of K and TgTh = Tgh for all g, h ∈ G.

Theorem 23 (Markov-Kakutani fixed point theorem). Let G be a locallycompact group. It is amenable if and only if each affine action of G has a fixedpoint.

Proof. (⇐) LetM be the set of means on BUCl(G). ThenM is ∗-compact convexsubset of BUCl(G)′. We let 〈Tgm, f〉 := 〈m,Lg−1(f)〉. Then (Tg)g∈G is an affineaction of G on M. If gj → g in G then ‖Lg(f) − Lgj (f)‖∞ → 0 because f ∈BUCl(G). It is easy to check that T is continuous in 2 variables. Hence there is afixed point for T , which is a LIM on BUC(G).

(⇒) Let T be an affine action of G on a compact convex set K. Since Gis amenable, there is a Følner net (Fα)α∈A in it. Take v ∈ K and set vα :=λ(Fα)−1

∫FαTgv dλ(g). Then vα ∈ K. Take a limit point w ∈ K of (vα)α. Take a

seminorm p on V . Then

p(Thxα − xα) ≤ 1

λ(Fα)

∫Fα4hFα

p(Tgv) dλ(g) ≤ λ(Fα4hFα) maxK p(k)

λ(Fα).

Thus p(Thw − w) = 0 for each h ∈ G and each seminorm p defining the topologyon V . �

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Corollary 24 (Generalized Bogolubov-Krylov theorem). Let G be a locallycompact group. It is amenable if and only if there is an invariant probability Radonmeasure for each continuous action of G on a compact (Hausdorff) space.

Proof. Let G be amenable. Let T be a continuous action of G on a compactspace K. Define an action T ∗of G on the space C(K)′ (endowed with the ∗-weaktopology) by setting 〈T ∗g F, f〉 = 〈F, f ◦ Tg〉. It is continuous. Restrict T ∗ to thesubset P ⊂ C(K)′ of probability measures. We note that P is ∗-weakly compactand convex. Hence there is a fixed point in it.

Conversely, consider the space M ⊂ L∞(G)′ of all mean on L∞(G). It is ∗-weakly compact and convex. Of course, (g,m) 7→ m ◦ Lg is a continuous action ofG on M. Hence there is a fixed point, which is an LIM on L∞(G). �

Theorem 25 (Von Neumann mean ergodic theorem). Let G be an amenablegroup and (Fα))α ∈ A a Føliner net in it. Let G 3 g 7→ Ug ∈ U(H) be a stronglycontinuous unitary representation of G in a Hilbert space H. Then

limα

1

λ(Fα)

∫Fα

Ug dλ(g) = P

strongly, where P is the orthogonal projection to the space of (Ug)g∈G-invariantvectors.

Proof. Let W be the linear span of the space {Uhv − v | h ∈ G, v ∈ H}. Let I bethe subspace of U -invariant vectors. Put Aα := 1

λ(Fα)

∫FαUgdλ(g).

— Aαw → 0 for each w ∈ W. Hence Aαw → 0 for each w ∈ W.— Aαv = v for each v ∈ I.— W ⊥ I.— W⊥ ⊂ I.Hence H = W ⊕ I. Therefore verify the von Neumann theorem separately on

vectors from W and from I. �

Literature.1. Greenleaf, Invariant means on topological groups and their applications

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