Download 1. Haar measure Let G be a locally compact group. Denote

1. Haar measure
Let G be a locally compact group. Denote by C00 (G) the algebra of continuous
+
functions on G with compact support. Endow it with the Sup-norm. Let C00
(G)
denote the cone of non-negative functions. Given f ∈ C00 (G) and g ∈ G, we
let Rg (f )(x) := f (xg) and Lg (f )(x) := f (g −1 x), x ∈ G. Then g 7→ Rg is a
homomorphism 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 ∈ C00
(G), φ 6= 0, there is a finite family of elements gj ∈ G
and reals cj ≥ 0 such that
X
f≤
cj Rgj (φ).
j
We let (f : φ) := inf
P
j cj
over all such families.
Lemma 2. (Properties of (f : φ))
(1)
(2)
(3)
(4)
(5)
(6)
(Rg (f ) : φ) = (f : φ)
(cf : φ) = c(f : φ), c ≥ 0
(f1 + f2 : φ) ≤ (f1 : φ) + (f2 : φ)
(f : ψ) ≤ (f : φ)(φ : ψ)
(f1 : φ) ≤ (f2 : φ) if f1 ≤ f2
(f : φ) ≥ kf k/kφk
+
Fix f0 ∈ C00
(G), f0 6= 0. We now let
Iφ (f ) =
(f : φ)
.
(f0 : φ)
Then from (4) we deduce
1
≤ Iφ (f ) ≤ (f : f0 ).
(f0 : f )
Lemma 3. (Properties of Iφ )
(1)
(2)
(3)
(4)
Iφ (Rg (f )) = Iφ (f )
Iφ (cf ) = cIφ (f )
Iφ (f1 + f2 ) ≤ Iφ (f1 ) + Iφ (f2 )
Iφ (f1 ) ≤ Iφ (f2 ) if f1 ≤ f2
This is almost what we are looking for. The functionals Iφ are not linear, but we
will show that if the support of φ is small enough, then Iφ is approximately linear.
+
Lemma 4. For f1 , f2 ∈ C00
(G) and > 0, there is a neighborhood U of e such
that if supp(φ) ⊂ U then
Iφ (f1 ) + Iφ (f2 ) ≤ Iφ (f1 + f2 ) + .
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 ∈ C00
(G) such that h0 = 1 on K. We let
h = f1 + f2 + δh0 for some δ > 0. We now set
f10 :=
f1
,
h
f20 :=
f2
,
h
assuming f10 (x) = f20 (x) = 0 if h(x) = 0.
+
Of course, f10 , f20 ∈ C00
(G) and f10 + f20 ≤ 1. Choose a symmetric neighborhood U
0
0
0
−1
of e such that |f1 (x) − f10 (y)|
P < and |f2 (x) − f2 (y)| < whenever xy ∈ U . Let
Supp(φ) ⊂ U and let h ≤ j cj Rgj (φ) for some cj ≥ 0 and gj ∈ G.
X
X
f1 (g) = f10 (g)h(g) ≤ f10 (g)
cj φ(ggj ) =
f10 (g)cj φ(ggj )
ggj ∈U
j
≤
X
(f10 (gj−1 ) + )cj φ(ggj ) =
ggj ∈U
Hence (f1 : φ) ≤
Therefore
X
(f10 (gj−1 ) + )cj φ(ggj )
j
0 −1
j (f1 (gj )
P
+ )cj and, similarly, (f2 : φ) ≤
X
(1 + 2)cj .
(f1 : φ) + (f2 : φ) ≤
0 −1
j (f2 (gj )
P
+ )cj
j
It follows that
(f1 : φ) + (f2 : φ) ≤ (1 + 2)(h : φ) ≤ (1 + 2)((f1 + f2 : φ) + δ(h0 : φ)).
Therefore
Iφ (f1 ) + Iφ (f2 ) ≤ (1 + 2)(Iφ (f1 + f2 ) + δIφ (h0 ))
We recall that Iφ (h0 ) ≤ (h0 : f0 ). Since δ is arbitrary, we conclude that
Iφ (f1 ) + Iφ (f2 ) ≤ (1 + 2)(Iφ (f1 + f2 )) ≤ Iφ (f1 + f2 ) + 2(f1 + g : f0 ).
Theorem 5. There is a non-trivial right invariant non-trivial integral I on C00 (G).
Proof. Let
X :=
Y
[
+
06=f ∈C00
(G)
1
, (f, f0 )].
(f0 : f )
It is a compact space. For each neighborhood U of e, we set
MU := {(Iφ (f ))06=f ∈C + (G) | Supp(φ) ∈ U }.
00
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 is
I = (I(f ))06=f ∈C + (G) ∈ X such that I belongs to the closure of each MU . Thus,
00
+
for each > 0 and each finite sequence f1 , . . . , fn ∈ C00
(G) \ {0} and each U , there
+
is φ ∈ C00 (G) with Supp(φ) ⊂ U such that
|I(fj ) − Iφ (fj )| < .
It now
(1)
(2)
(3)
(4)
follows from Lemma 3 that
I(Rg (f )) = I(f )
I(cf ) = cI(f )
I(f1 + f2 ) ≤ I(f1 ) + I(f2 )
I(f1 ) ≤ I(f2 ) if f1 ≤ f2 .
2
Fix f1 , f2 . Lemma 4 yields that
I(f1 ) + I(f2 ) ≤ I(f1 + f2 ) + for an arbitrary > 0. Therefore, jointly with (3) this gives
I(f1 ) + I(f2 ) ≤ I(f1 + f2 ).
Also, from the definition of X it follows that
1
≤ I(f ) ≤ (f : f0 ).
(f0 : f )
+
+
Hence I(f ) > 0 if f 6= 0, f ∈ C00
(G). Now extend I from the cone C00
(G) to the
entire C00 (G) in a usual way. In particular, I(0) = 0. Recall
Lemma 6. (Riesz Representation Theorem) Let X be a locally compact Hausdorff
space and let I be a positive linear functional on C00 (X). Then there is a unique
Radon measure µ (i.e. Ra 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 ∈ C00
(X), f ≤ 1U } and µ(K) = inf{I(f )|f ∈ C00
(X), f ≥ 1K }, for all
open U ⊂ X and compact K ⊂ X.
Let µ correspond to I. Then µ is called a right-invariant Haar measure. In a
similar way a left invariant Haar measure is defined. If J is the inversion in G, a
measure µ is a right invariant Haar measure if and only if µ ◦ J is left-invariant
Haar measure.
Corollary 7. It follows that the Haar measure of each compact set is finite and
the Haar measure of each open set is strictly positive.
Theorem 8. (Uniqueness of Haar measure). Given two non-trivial right-invariant
integrals I and I 0 , there is c > 0 such that I = cI 0 .
R
Proof. Let I(f ) = f dµ
+
Let I 0 be another right invariant integral on C00
(G) and let µ0 stand for the
+
corresponding measure. Since I 0 is nontrivial, there is 0 6= f0 ∈ C
00 (G) with
P
+
I 0 (f0 ) > 0. For each 0 6= f ∈ C00
(G), we can find gj and cj with f0 ≤ j cj Rg (f ).
P
0
0
Hence I (f0 ) ≤ j cj I (f ). It follows that I 0 (f0 ) ≤ (f0 : f )I 0 (f ). In particular,
I 0 (f ) > 0.
+
+
Take f1 , f2 ∈ C00
(G). Fix a neighborhood V of e. Select ψ ∈ C00
(G) such that
ψ = 1 on the union of Supp(f1 )·V ∪U ·Supp(f1 ) and Supp(f2 )·V ∪U ·Supp(f2 ). For
each > 0, there is a symmetric neighborhood U ⊂ V of e with |fj (g) − fj (g 0 )| < +
whenever g 0 g −1 ∈ U . Select 0 6= φ ∈ C00
such that Supp(φ) ⊂ U .
Then we have
Z
Z
0
0
I (φ)I(f1 ) = φ(g1 )f1 (g2 ) dµ (g1 )dµ(g2 ) = φ(g1 )f1 (g2 g1 ) dµ0 (g1 )dµ(g2 ).
On the other hand
Z
Z
0
0
I(φ)I (f1 ) = φ(g1 g2 )f1 (g2 ) dµ(g1 )dµ (g2 ) = φ(y)f1 (g1−1 y) dµ(g1 )dµ0 (y).
3
Hence
0
0
Z
φ(g1 )|f1 (g2 g1 ) − f1 (g1−1 g2 )| dµ0 (g1 )dµ(g2 )
Z
φ(g1 )|f1 (g2 g1 ) − f1 (g1−1 g2 )|ψ(g2 ) dµ0 (g1 )dµ(g2 )
|I (φ)I(f1 ) − I(φ)I (f1 )| ≤
=
≤ I 0 (φ)I(ψ)
Therefore
I(f1 )
I(ψ)
I(φ) I 0 (f1 ) − I 0 (φ) ≤ I 0 (f1 )
In a similar way,
0
I(f2 )
I(ψ)
I
(φ)
I 0 (f2 ) − I 0 (φ) ≤ I 0 (f2 )
Therefore
I(f1 )
I(ψ)
I(f2 ) I(ψ)
I 0 (f1 ) − I 0 (f2 ) ≤ I 0 (f1 ) + I 0 (f2 ) .
Hence
I(f2 )
I(f1 )
= 0
,
0
I (f1 )
I (f2 )
i.e. I = cI 0 . Theorem 9. Let G be a locally compact group G and λ be a left Haar measure on
it. 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 not
compact, fix U , a neighborhood ofSe with compact closure. There is an infinite
sequence (gn )∞
n=1 such that gn 6∈
j<n gj U . Take a symmetric neighborhood V
of
⊂ U . Then gj V , j = 1, 2, . . . are pairwise disjoint. Hence λ(G) ≥
P e with V V P
λ(g
V
)
=
j
j
j λ(V ) = ∞. Modular homomorphism. Let I be a right Haar integral. Then I ◦ Lg is also a
right 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 verify
that ∆ : g 7→ ∆(g) is a group homomorphism from G to R∗+ .
Theorem 10. ∆ is continuous.
+
Proof. Take 0 6= φ ∈ C00
(G). Select a symmetric precompact neighborhood U of e
+
Let ψ ∈ C00
(G) be such that ψ = 1 on U · Supp(φ).
Fix > 0. Select a symmetric precompact neighborhood V ⊂ U of e such that
kLg (φ) − φk < . Then
|I(Lg (φ)) − I(φ)| ≤ I(|Lg (φ) − φ|) = I(|Lg (φ) − φ|ψ) ≤ I(ψ).
Thus |∆(g) − 1| ≤ I(ψ)
I(φ) . G is called unimodular if ∆ is trivial.
4
Proposition 11. Each locally compact Abelian group is unimodular. Each compact
group is unimodular.
Exercise 12. Let I be a right Haar integral. (1)Show that I(f ) = I(J(f )·∆). (2)As
a corollary: G is unimodular iff I ◦ J = I. Hint: a) Show that the right-hand-side is
also a right Haar integral. b) Prove that the corresponding proportional coefficient
is 1.
Examples. 1) G is a finite group.
2) G is infinite discrete group.
3) G = Rn .
4) G = Tn
5) Suppose now that G is an open subset in Rn . Suppose that the multiplication
is given by the formula
(x ∗ y)j = hA(j) x, yi + dj for all x = (xj )j , y = (yj )j ,
for some matrices A(j) and reals dj . We let r(g) := | det Rg0 | and l(g) := | det L0g−1 |.
Then
Z
Z
f (x)
f (x)
dx and Il (f ) :=
dx
Ir (f ) :=
G l(x)
G r(x)
are right and left Haar integrals. Indeed,
Z
Ir (Rg (f )) =
G
f (x ∗ g)
dx =
r(x)
Z
G
f (y)
1
dy =
−1
r(y ∗ g ) r(g)
Z
G
f (y)
dy = Ir (f ).
r(y)
Moreover,
Ir (Lg f ) =
Hence ∆(g) =
l(g)
Ir (f ).
r(g)
l(g)
r(g) .
R (x)
5a) G = R∗ . Ir (f ) = f|x|
dx.
x y
5b) G =
| x 6= 0 . Then
0 1
Z
Ir (f ) =
f (x, y)
dxdy,
|x|
Z
Il (f ) =
f (x, y)
dxdy,
x2
∆(x, y) =
5c) G = GL2 (R). Then r(A) = l(A) = (det A)2 .
6) Haar measures for product of l.c.g.
Literature.
1. A. Weil, Integration in topological groups and its applications
2. Hewitt, Ross, Abstract harmonic analysis, vol 1.
3. Naimark, Normed rings
5
1
.
|x|
2. Amenability for discrete groups
Invariant means on discrete groups. Let G be a group endowed with discrete
topology. Denote by BC(G) the algebra of bounded continuous (real) functions on
G.
Definition 1. A left invariant mean (LIM) on G is a functional M : BC(G) → R
such 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 way
we 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), let
f 0 (g) := M (Rg−1 (f )). Then f 0 ∈ BC(G) and f 7→ M (J(f 0 )) is a 2-sided invariant
mean. Of course, every finite group is amenable.
P
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-Banach
theorem, 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 − Lg f ) = 0, M is left
invariant. Definition 4. Let every finite subset of G generate a finite subgroup. Then G is
called locally finite.
Theorem 5. Every locally finite group is amenable.
P
Proof. Take f ∈ Sl . Then f = j (Lgj (fj ) − fj ) for some gj , fj . Then there is a
finiteP
subgroup H ⊂ G such that gj ∈ H. Let fj0 := fj H, f 0 := f H. Then
f 0 = j (Lgj (fj0 )−fj0 ). 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 limα Gα and each Gα is amenable then G is amenable. In
particular, G is amenable if and only if every finitely generated subgroup of it is
amenable.
In particular, a group G is amenable off each finitely generated subgroup of G is
amenable. This means that the amenability is a “local” concept.
6
Markov-Kakutani theorem. Let V be a locally convex vector space. Let K be a
compact convex subset in V . Let (Tx )x∈X be a family of pairwise commuting affine
transformations of K. Then there is a common fixed point for (Tx )x∈X .
Proof. Let FTx := {k ∈ K | Tx k = k}. Let us show that FTx 6= ∅. Set
n
Tx,n
1X j
:=
T .
n j=1 x
Of course, Tx,n : K → K. Fix k ∈ K and put kn := Tx,n k. Let k0 be a limit
point of (kn )∞
n=1 . Then k0 ∈ K. It is easy to check that k0 ∈ FTx . Next, we note
that FTx is a compact convex set. Moreover,
Ty FTx ⊂ FTx for each y ∈ G. By
T
induction,Tfor 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)0 | F (f ) ≥ 0 for each f ≥ 0, F (1) = 1}
is compact (follows from the Banach-Alaoglu theorem) and convex.
c) Tg F := F ◦ Lg is affine.
It remains to apply Markov-Kakutani. Remark 8. Let G = Z. Put Fn := {−n, . . . , n} and
Mn (f )(k) :=
X
1
f (k + j).
2n + 1
j∈Fn
Then Mn ∈ M. Hence there is a subsequence Mnm that ∗-weakly converges to
a LIM M . It is interesting to note that if f is a converging sequence of reals
then f ∈ BC(Z) and limn→∞ f (n) = M (f ). However if f is bounded but it does
not 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
S
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 0 (g) := f (h). Then
f 0 ∈ BC(G). We now set M 0 (f ) := M (f 0 ). Of course, M 0 (f ) ≥ 0 if f ≥ 0 and
M 0 (1) = 1. (Lh (f ))0 = Lh (f 0 ). Hence M 0 is LIM on H. Theorem 10. Let H be a normal subgroup in G. Then G is amenable if and only
if H and G/H are both amenable.
Proof. (⇒) By Theorem 9, H is amenable. Let π : G → G/H be the natural
projection and let π ∗ denote the corresponding map f → f ◦ π from BC(H) to
BC(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 → H
be a map with Hs(x) = x and Hs(G/H) = G. Given f ∈ BC(G), we let
fx0 (h) := f (hs(x)),
x ∈ G/H, h ∈ H.
Let f 00 (x) := MH (fx0 ). Finally we let M (f ) := MG/H (f 00 ). Then M is RIM on G.
Let g = hs(x), g0 = h0 s(x0 ). Then
gg0 = hs(x)h0 s(x)−1 s(x)s(x0 ) = hh1 h2 s(xx0 ).
0
0
(Rg0 f )0x (h) = (Rg f )(hs(x)) = f (hh1 h2 s(xx0 )) = fxx
(hh1 h2 ) = (Rh1 h2 fxx
)(h).
0
0
0
(Rg0 f )00 (x) = MH (fxx
) = f 00 (xx0 ) = (Rx0 (f 00 ))(x).
0
Hence M is right invariant. Finitely additive invariant measures. Let M be LIM on G. For each subset
A ⊂ 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 on
the 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 that
fn takes finitely many values for each n and kf − fn k → 0.
(b) Of course, one can define naturally M (fn ). Now define M (f ) := limn M (fn ).
It is well defined.
(c) M is LIM on CB(G). Example 12. Let G = F2 (free group with 2 generators). Then G F
is not amenable.
Let a, b be the free generators of F2 . Consider a partition F2 = n∈Z Bn , where
w ∈ Bn if w = an bj · · · , j 6= 0 unless w = an . Suppose that F2 is amenable. Let µ be
the left invariant finitely invariant normalized
measure on F2 . Since aBn = Bn+1 ,
F
µ(Bn ) = µ(Bn+1 ). For each n, 1 ≥ µ( 1≤j≤n Bj ) = nµ(B0 ). Hence µ(B0 ) = 0.
On the other hand, let B = F2 \ B0 . Then bB ⊂ B0 .
1 = µ(B t B0 ) = 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 not
true. A counterexample was constructed by Olshansky. However the conjecture is
true within the class of linear groups due to Tits alternative.
Tits alternative. (without proof ) Let V be a finite dimensional vector space over
a field F of characteristic 0. Let G be a subgroup of GL(V ). Then G is amenable
if and only if it does not contain F2 as a a subgroup. If G is amenable then G
contains 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 elementary
group is amenable. The converse is true within the class of linear groups (follows
from Tits alternative). However, in general there are non-elementary amenable
groups (Grigorchuk).
Example 15.
group P SL2 (R) is non-amenable. We first note that each
The α β
matrix A :=
∈ SL2 (R) generates a linear fractional transformation QA
γ δ
of C by
αz + β
.
QA (z) :=
γz + δ
The map A 7→ QA is a homomorphism from SL2 (R) and its kernel is {I, −I}. Hence
this map identifies P SL2 (R) with the group of linear fractional transformation of
C preserving the upper half plane.
We show that P SL2 (R) contains F2 . Consider 4 circles in C: C1 := {z | |z + 2| =
1}, C10 := {z | |z − 2| = 1}, C2 := {z | |z + 5| = 1}, C20 := {z | |z − 5| = 1}. Consider
two 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 P SL2 (R) in a natural way: if
w = aj1 bj2 aj3 · · · ∈ F2 is a nontrivial reduced word, i.e. each jk 6= 0, then Tw :=
T1j1 T2j2 T1j3 · · · . In a similar way we define Tw for a word w starting with b. Our
purpose to prove that the kernel of T is trivial, i.e. if w is a nontrivial word then
Tw 6= I.
It is easy to see that T1 (C1 ) = C10 and T2 (C2 ) = C20 . Moreover,
T1 (e(C1 )) = i(C10 ) and T2 (e(C2 )) = i(C20 ).
Let z0 = 4i ∈ C. Then Tw z0 is inside the interior of the 4 circles. Hence Tw z0 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 applications
3. 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 Dg f (x, y) = f (xgy).
Definition 1. A function f ∈ BC(G) is called almost periodic (AP) if of of the
following is satisfied:
(1)
(2)
(3)
(4)
{Lg (f ) | g ∈ G} is relatively compact in BC(G).
{Rg (f ) | g ∈ G} is relatively compact in BC(G).
{Rh Lg (f ) | h, g ∈ G} is relatively compact in BC(G).
{Dg f | 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 totally
bounded.
(1) ⇒ (3) Given > 0, there are a1 , . . . , an ∈ G such that La1 (f ), . . . , Lan (f ) is
an -net in {Lg (f ) | g ∈ G}. Since Laj (f ) is AF, there is an -net Rbj,k , k = 1, . . . Kj ,
in {Rh (Laj (f )) | h ∈ G} by (ii). Then Rbj,k (Laj ) is 2-net in {Rh Lg (f ) | h, g ∈ G}.
Thus (1), (2), (3) are equivalent. Of course, (4) implies (1) and (2).
Given , we find a1 , . . . , an ∈ G such that Ra1 (f ), . . . , Ran (f ) is -netSin {Rg (f ) |
g ∈ G}. Next, let Aj := {g ∈ G | kRg (f ) − Raj (f )k < }. Then j Aj = G.
T
Consider the family of sets j Alj a−1
when l1 , . . . , ln runs 1, . . . , n. Enumerate
j
Sm
them (only those which are nonempty) as B1 , . . . , Bm . Then k=1 Bm = 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 (xbk0 y)| ≤ |f (xcy) − f (xcaj0 )|
+ |f (xcaj0 ) − f (xbk0 aj0 )| + |f (xbk0 aj0 ) − f (xbk0 y)|
≤ kRcy f − Rcaj0 f k + kRcaj0 f − Rbk0 aj0 f k + kRbk0 aj0 f − Rbk0 y f k
≤ kRy f − Raj0 f k + kRcaj0 f − Rbk0 aj0 f k + kRaj0 f − Ry f k < 4
because caj0 and bk0 aj0 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. It
is invariant under Lg and Rg , g ∈ G. If f ∈ AP (G) then Re(f ), Im(f ) and f are
in AP (G).
Marriage Lemma. Let X, Y be two nonempty sets. Let F be a map from X to
the set of all non-empty finite subsets of Y . Suppose that for each X1 ⊂ X,
#(
[
F (x)) ≥ #(X1 )
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 },
[
F (x)) > #(X1 )
#(
x∈X1
then we put F 0 (x) := F (x) \ {y0 } and apply the inductive assumption: there is a
1-to-1 map s : X \ {x0 } → Y \ {y0 } with s(x) ∈ F 0 (x), which we extend to x0 as
s(x0 ) := y0 .
S
If
there
is
X
1 ⊂ X, #(X1 ) < n with #( x∈X1 F (x)) = #(X1 ), we put Y1 :=
S
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 lemmaSare satisfied for
X2 , Y2 , F2 . If not, there is a non-empty subset X3 ⊂ X2 with #( x∈X3 F2 (x)) <
#(X3 ). However then
!
[
[
F (x) < #(X3 ) + #(X1 ), i.e.
F2 (x) t
#
x∈X1
x∈X3
!
#
[
F (x)
< #(X3 t X1 ).
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.
Q
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 ∈ Z
such 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
\
ZS 6= ∅.
S is a finite subset of X
Lemma 4. Let X be a mertic space and let x1 , . . . , xn be an -net of smallest
possible cardinality (for this ). Let Y be an -net in X. Then there is a 1-to-1 map
s : {x1 , . . . , xn } → Y with xj and s(xj ) being inside the same -ball for each j.
Proof. Given j ≤ n, let F (xj ) := {y
S ∈ Y | xj and y are inside the same -ball}.
All that we need to prove is that #( j∈I F (xj )) ≥ #(I) for each nonempty subset
S
I ⊂ {1, . . . , n}. Indeed, assume that #( j∈I F (xj )) < #(I) for some I. We claim
S
that the set A := i∈I F (xi ) ∪ {xj | j 6∈ I} is a -net. Take z ∈ X. If z is -far
from each xj , j 6∈ I, then there is j0 ∈ I such that z and xSj0 are -close. On the
other hand, there is y ∈ Y which is -close to z. Hence y ∈ j∈I F (xj ). Thus A is
-net and hence #(A) ≥ n. Corollary 5. Let f ∈ AP (G), > 0 and Da1 f, . . . , Dan f be an -net in {Dg f |
g ∈ G} of minimal cardinality among the -nets. Then
X
n
n
X
1
1
f (aj ) −
Daj f ≤ 2.
n
n j=1
j=1
11
Proof. Let Db1 , . . . , Dbn be another -net in {Dg f | g ∈ G}. Then there is a
bijection σ of {1, . . . , n} such that kDaj f − Dbσ(j) f k < 2. Therefore
(1)
X
n
n
n
X
1
1 X
1
Daj f − Db f < 2.
Daj f −
Dbj f ≤
σ(j)
n
n j=1
n j=1
j=1
Let u, v ∈ G. Then Dua1 v , . . . , Duan v is an -net in {Dg f | g ∈ G}. Indeed, given
g ∈ G, we have kDu−1 gv−1 f − Daj f k < for some j. Hence kDg f − Duaj v f k < .
It now follows from (1) with bj := uaj v that
X
X
n
n
n
n
X
X
1
1
1
1
2 > Daj f (e, e) −
Dbj f (e, e) = f (aj ) −
f (uaj v) .
n j=1
n j=1
n j=1
n j=1
It remains to take supremum over u, v ∈ G. Theorem 6 (Existence of 2-sided means on the AP-functions). There is
a 2-sided invariant mean M on AP (G). Moreover, M is strictly positive, i.e. if
f ≥ 0 and f 6= 0 then M (f ) > 0.
Proof. Take f ∈ AP (G) and > 0. We put
(
E :=
)
X
1
z ∈ R | z −
Da f < for a finite A ⊂ G .
#A
a∈A
By Corollary 5, E 6= ∅. Let us verify that the diameter of E is less than 2.
Indeed, if z1 , z2 ∈ E , then
1 X
f (xay)| < ,
#A
a∈A
1 X
|z2 −
f (xby)| < #B
|z1 −
(2)
(3)
b∈B
for all x, y ∈ G and some finite subsets A, B in G. From (2) and (3) we deduce
|z1 −
|z2 −
1
#A#B
1
#A#B
X
f (ab)| < and
a∈A,b∈B
X
f (ab)| < a∈A,b∈B
respectively. Hence
(4)
|z1 − z2 | ≤ 2.
Moreover, E is bounded. Indeed, |z|
T < kf k + for each z ∈ E . Since E1T∩ · · · ∩
Ek ⊃ Emin(1 ,...,k ) , it follows that >0 E 6= ∅. From (4) we deduce that >0 E
12
is a singleton. Denote it by M (f ). Thus M (f ) is the only number such that for
each > 0, there is a finite A ⊂ G with
X
1
M
(f
)
−
D
f
a < .
#A
a∈A
It it easy to see that f 7→ M (f ) is a 2-sided invariant functional, M (αf ) = M (f ) for
eachP
number α, M is non-negative and MP
(1) = 1. To show additivity, let kM (f ) −
1
1
a∈A Da f k < and kM (f1 ) − #B
b∈B Db f1 k < for some f, f, ∈ AP (G)
#A
P
1
and A, B ⊂ G. Then we obtain easily that kM (f ) − #A#B
a∈A,b∈B Dab f k < P
1
and kM (f1 ) − #B#A b∈B,a∈A Dab f1 k < and hence
1
#B#A
kM (f ) + M (f1 ) −
X
Dab (f + f1 )k < .
b∈B,a∈A
Since is arbitrary, M (f + f1 ) = M (f ) + M (f1 ). It remains to prove that M is
strictly positive. Let f ∈ AP (G), f ≥ 0 and f (g0 ) > 0 for some g0 ∈ G. Let A ⊂ G
be an f (g0 )/2-net in {Dg f | g ∈ G}. Then for x, y, g ∈ G, we have
X
f (xay) ≥ max Da f (x, y) > Dg f (x, y) −
a∈A
a∈A
Therefore
X
f (xa) >
a∈A
for each x ∈ G.
f (g0 )/2. Therefore
P
a∈A
f (g0 )
f (g0 )
= f (xgy) −
.
2
2
f (g0 )
2
Ra (f ) > f (g0 )/2 and hence #(A)M (f ) ≥
Theorem 7 (Uniqueness of RIM on AP (G)). Let M 0 be a RIM on AP (G).
Then M 0 = M .
Proof. Given f ∈ AP (G) and > 0, we have
− < M (f ) −
1 X
f (xay) < n
a∈A
for all x, y ∈ G and a finite subset A ⊂ G. Hence − < M (f ) − n1
Therefore − ≤ M (f ) − M 0 (f ) ≤ . P
a∈A
Ra (f ) < .
Remark 8. If G is locally compact (or Polish) then every bounded function f with
relatively compact {Lg (f ) | g ∈ G} is (uniformly) continuous. Indeed, for each
> 0, there are subsets Aj ⊂ G and elements aj ∈ G with kLg f − Laj f k ≤ for all
g ∈ Aj . It follows from our assumption on G that there is j0 such that Aj0 has the
Baire property. Then kLb (f ) − f k ≤ 2 for all b ∈ A−1
j0 Aj0 . By the Pettis theorem,
−1
there is an open neighborhood U of e with U ⊂ Aj0 Aj0 .
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 by
L (X) the space of essentially bounded functions. Recall that L∞ (G) = L1 (G)0 .
It is a Banach space when endowed with the norm
(
)
∞
kf k∞ := vraisup|f | = inf
sup |f (g)| | N is a λ-locally 0 subset of G .
g∈G\N
We consider the following subspaces of L∞ (G):
— BC(X),
— BU Cl (G), the left uniformly continuous bounded functions on G, i.e.
{f ∈ BC(G) | the map G 3 g 7→ Lg (f ) is continuous}.
— BU Cr (G), the right uniformly continuous bounded functions on G,
— BU C(G), the 2-sided uniformly continuous bounded functions on G, i.e.
BU Cl (G) ∩ BU Cr (G).
Lemma 1. BU Cl (G), BU Cr (G), BU C(G) are 2-sided invariant closed subspaces
of 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, any
discrete group G is locally compact and L∞ (G) = B(G). Each compact group is
amenable 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)0 of bounded regular measures on G. It
is a Banach space with the norm equal to the full variation. The convolution is
continuous on M (G):
Z
hµ ∗ ν, f i :=
f (g1 g2 ) dµ(g1 )dν(g2 ).
G×G
We note that L1 (G) is embedded isometrically into M (G) via the one-to-one map
φ 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
Z
µ ∗ φ(g) :=
φ(x−1 g) dµ(x),
ZG
φ ∗ µ(g) :=
φ(gx−1 )∆(x−1 ) dµ(x))
G
then µ ∗ µφ = µµ∗φ and µφ ∗ µ = µφ∗µ . Moreover, kµ ∗ φk1 ≤ kµk · kφk1 and
kφ ∗ µk1 ≤ kµk · kφk1 . In particular, µφ ∗ µψ = µφ∗ψ , where
Z
φ ∗ ψ(g) =
φ(x)ψ(x−1 g) dλ(g).
G
∞
1
0
Since L (G) = L (G, µ) , we use this duality to define the following convolution.
14
Definition 4. Given φ ∈ L1 (G) and f ∈ L∞ (G) we let
Z
Z
φ ∗ f (g) :=
φ(y)f (yg)dλ(y),
f ∗ φ(g) := f (gy)φ(y)dλ(y).
G
g
We see that kφ ∗ f k∞ ≤ kφk1 · kf k∞ and kf ∗ φk∞ ≤ kφk1 · kf k∞ . Moreover,
hφ ∗ ψ, f i = hψ, φ ∗ f i = hφ, f ∗ ψi.
Definition 5. A linear functional m on L∞ (G) is called a topological LIM on
L∞ (G) if it is positive, normalized and m(φ ∗ f ) = m(f ) for all
φ ∈ P(G) := {φ ∈ L1 (G) | φ ≥ 0, kφk1 = 1}.
A linear functional m on L∞ (G) is called a topological RIM on L∞ (G) if it is
positive, normalized and m(f ∗ φ) = m(f ) for all φ ∈ P(G).
In a similar way one can define topological LIM and RIM on BC(G), BU Cr (G), BU C(G).
Lemma 6 (On regularization). Let f ∈ L∞ (G) and φ ∈ P(G). Then φ ∗ f ∈
BU Cl (G), f ∗ φ ∈ BU Cr (G). If f1 ∈ BU Cr (G) then φ ∗ f1 ∈ BU C(G). If
f2 ∈ BU Cl (G) then f ∗ φ ∈ BU C(G).
Proof. We check only the first claim.
Z
Z
|φ ∗ f (x) − φ ∗ f (yx)| = | φ(t)f (tx)dλ(t) − φ(t)f (tyx)dλ(t)|
Z
≤ |φ(tx−1 ) − φ(tx−1 y −1 )∆(y)||f (t)|∆(x)dλ(t)
Z
≤ kf k∞ |φ(t) − φ(ty −1 )∆(y)|dλ(t)
= kf k∞ kφ − Ry−1 (φ)k1
Recall that Lg and Rg denote the following isometries in L1 (G). Lg f (x) :=
f (g −1 x), Rg f (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) are
continuous.
(2) For each > 0, there is a neighborhood U of e suchR that for each nonnegative ψ ∈ L1 (G) vanishing out of U and such that ψ(g)dλ(g) = 1,
kψ ∗ f − f k1 < and kf ∗ ψ − f k1 < .
Proof. (1) Approximate f with a function from C00 (G) in k.k1 norm.
(2)
Z
Z
kψ ∗ f − f k1 ≤ ψ(y) · |f (yx) − f (x)|dλ(x)dλ(y)
Z
= ψ(y)kLy−1 (f ) − f k1 dλ(y)
15
It remains to apply (1). In a similar way,
Z
Z
kf ∗ ψ − f k1 ≤ ψ(y) |f (xy) − f (x)|dλ(x)dλ(y)
Z
Z
= ψ(y) |Ry f (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 will
additionally assume that eα ∈ P and eα is compactly supported.
We need an analogue of Lemma 7(2) for BU C(G).
Lemma 9. Let f ∈ BU C(G). Given > 0, there is a neighborhood U of e with
compact
support such that for each ψ ∈ L1+ (G) vanishing out of U and such that
R
ψ dλ = 1, kψ ∗ f − f k∞ < .
Proof.
Z
|ψ ∗ f (x) − f (x)| ≤
ψ(y) · |f (yx) − f (x)|dλ(y)
Z
=
ψ(y)kLy−1 (f ) − f k∞ dλ(y)
≤ sup kLy−1 (f ) − f k∞
y∈U
Theorem 10.
converse is not
(1) there is
(2) there is
(3) there is
(4) there is
(5) there is
If there is a LIM on B(G) then there is a LIM on BC(G). The
true. The following are equivalent:
a topological LIM on L∞ (G),
a LIM on L∞ (G),
a LIM on BC(G),
a LIM on BU Cl (G),
a LIM on BU C(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),
Z
(φ ∗ Lg f )(x) = φ(t)f (gtx) dλ(t)
Z
= φ(g −1 t)f (tx) dλ(u) = Lg (φ) ∗ f.
Hence
m(Lg f ) = m(φ ∗ Lg f ) = m(Lg (φ) ∗ f ) = m(f ).
(2)⇒(3)⇒(4)⇒(5) are obvious.
16
(5)⇒(1) Let m be a LIM on BU C(G). First of all we show that m is a topological
LIM on BU C(G). Take f ∈ BU C(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 ).
Z
J(Lg (φ)) ∗ f (x) =
φ(g −1 t−1 )f (tx)∆G (t)dλ(t)
ZG
−1
=
φ(t−1
x)∆G (g −1 t1 )∆G (g)dλ(t1 )
1 )f (t1 g
ZG
=
J(φ)(t1 )f (t1 g −1 x)dλ(t)
G
= J(φ) ∗ f (g −1 x).
Then J(Lg (φ)) ∗ f and J(φ) ∗ f are in BU C(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 on
L1 (G). Hence (uniqueness of Haar measure) there is a constant cf such that
Z
m(J(φ) ∗ f ) = cf φ dλ.
We note that J(φ) ∈ P(G) if φ ∈ P(G) and φ 7→ J(φ) is an affine isomorphism of
P(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 k.k∞ 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 ∈ BU C(G) for each
f ∈ L∞ (G). Hence
m0 (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) = m0 (φ ∗ f ) = m0 (f ) = m(ψ ∗ f ∗ ψ).
It will follow from the more general assertion (in view of Lemma 6):
m(τ1 ∗ f 0 ) = m(τ2 ∗ f 0 )
for all τ1 , τ2 ∈ P(G) and f 0 ∈ BU Cr (G). To prove the latter, we note that τj ∗eα →
τj in L1 (G) and hence kτj ∗ eα ∗ f 0 − τj ∗ f 0 k∞ ≤ kτj ∗ eα − τj k1 kf 0 k∞ → 0.
m(τj ∗ f 0 ) = lim m(τj ∗ eα ∗ f 0 ) = lim m(eα ∗ f 0 ).
α
α
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 continuous
onto 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 is
amenable.
(3) Let N be an amenable closed normal subgroup of G and let the quotient
group G/N is also amenable. Then G is amenable.
(4) Let G = inj limα∈A Gα for a directed set A, every Gα is an amenable closed
subgroup 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 0 (g) := mN (Lg (f )). Since g 7→ Lg (f ) is a continuous map
from G to BC(G), f 0 ∈ C(G). Of course, f 0 is bounded. Since Ln (f 0 ) = f 0 for
all n ∈ N , we conclude that f 0 is a continuous function on G/N . We now set
m(f ) := mG/N (f 0 ). 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 defined
on BC(G). It is a mean and it is Gα -invariant. Denote by Mα the set of all
Gα -invariant LIMs on G. Then Mα is *-weakly compact subset of the unit ball in
BC(G)0 . It is non-empty. Given α,
Tβ ∈ A, there is γ ∈ A with γ α and γ β.
Then ∅ =
6 Mγ ⊂ Mα ∩ Mβ . Hence γ∈A Mγ 6= ∅.
(2) Suppose that G is Polish. Then there is a Borel map s : G/H → G such
that Hs(Hg) = Hg. Every g uniquely decomposes into g = hs(Hg), where h =
gs(Hg)−1 . Given f ∈ CB(H), let f 0 (g) := f (h). Then f 0 ∈ B(G). Moreover, it is
Borel. Hence f 0 ∈ L∞ (G). Moreover, the map f 7→ f 0 is linear and kf 0 k∞ ≤ kf k∞ .
It remains to put mH (f ) := m(f 0 ). If m is LIM on L∞ (G) then mH is a LIM on
BC(H). Literature.
1. Greenleaf, Invariant means on topological groups and their applications
18
5. Følner condition and its applications
Let M (G) ⊂ L∞ (G)0 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)0 . Let K
be the closure of P(G) in L∞ (G)0 . Then K is compact. If there is k0 ∈ M (G) \ K
then by Hahn-Banach separation theorem, there is f ∈ L∞ (G) and δ > 0 such
that k0 (f ) > hφ, f i + δ for all φ ∈ P(G). There is a compact set C ⊂ G such that
λ(C) > 0R and f (g) ≥ kf k∞ − δ/2 for all g ∈ C. Then φ0 := λ(C)−1 1C ∈ P(G) and
k0 (f ) > C f dλ/λ(C) + δ ≥ kf k∞ + 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 for
each g ∈ G, Lg (φα ) − φα → 0 *-weakly in L∞ (G)0 .
(2) G is amenable if and only 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)0 .
Proof. We prove only (2).
(⇐) There is a subnet φα converging to some m ∈ M (G). Then
m(φ ∗ f ) − m(f ) = lim(hφα , φ ∗ f )i − hφα , f i) = limhφ ∗ φα − φα , f i = 0.
α
α
(⇒) Take a topological LIM m on L∞ (G). By Lemma 12, there is a net φα ∈
P(G) with limα φα = m. Then
hφ ∗ φα − φα , f i = hφα , φ ∗ f i − hφα , f i → 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)0 then there is a net ψβ ∈ P(G), β ∈ B, such that for
each g ∈ G, kLg (φβ ) − φβ k1 → 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)0 then there is a net ψβ ∈ P(G), β ∈ B, such
that for each φ ∈ P(G), kφ ∗ ψβ − ψβ k1 → 0.
Proof. We prove
Q only (2).
Let E = φ∈P(G) L1 (G) with the product topology, each L1 endowed with
L
∞
k.k1 norm. Then E is a locally convex space and E 0 =
φ∈P(G) L (G). Let
T : L1 (G) → E be defined by T ψ(φ) = φ ∗ ψ − ψ. Then T is a linear map and
hence T (P(G)) is a convex subset of E. For each l ∈ E 0 , finite subset A ⊂ P(G)
and lφ ∈ L∞ (G), φ ∈ A,
X
hl, T φα i =
hlφ , φ ∗ φα − φα i → 0.
φ∈A
Thus T φα → 0 ∗-weakly. Hence 0 belongs to the ∗-weal closure of T (P(G)). Since
T (P(G)) is convex, the weak closure equals the strong closure. Hence there is a net
ψβ ∈ P(G) such that T ψβ → 0 strongly in E. 19
Definition 15 (Reiter condition). A locally compact group G satisfies Reiter
condition if for each > 0 and compact subset K ⊂ G, there is φ ∈ P(G) such that
sup kLg (φ) − φk1 < .
(R)
g∈K
Theorem 16. G satisfies the Reiter condition if and only if G is amenable.
Proof. (⇒) Let J = {(, K)} and select φj , j ∈ J, satisfying (R). Order J in a
usual 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
kφ ∗ φα − φα k1 → 0 for each φ ∈ P(G). Fix > 0, a compact K in G and β ∈ P(G).
Take a small neighborhood U of e with compact closure such that
kφU ∗ β − βk1 < ,
and
sup kLg (β) − βk1 < ,
g∈U
where φU = 1U /λ(U ) ∈ P(G). There are g1 , . . . , gN ∈ G with
may assume that g1 = e. Select φα such that
max kLgj (φU ) ∗ φα − φα k1 < ,
1≤j≤N
SN
j=1 gj U
⊃ K. We
kβ ∗ φα − φα k1 < .
Let us show that φ := β ∗ φα satisfies (R). Indeed,
kφU ∗ φ − Lg (φ)k1 ≤ kφU ∗ φ − φk1 + kφ − Lg (φ)k1
= k(φU ∗ β − β) ∗ φα k1 + k(β − Lg (β)) ∗ φα k1 < 2
for each g ∈ U . Hence
2 > kLgj (φU ∗ φ) − Lgj (Lg (φ))k1 = kLgj (φU ) ∗ φ − Lgj g (φ)k1 .
Now
kLgj g (φ) − φk1 ≤ 2 + kLgj (φU ) ∗ φ − φk1 = 2 + kLgj (φU ) ∗ β ∗ φα − β ∗ φα k1
≤ 4 + kLgj (φU ) ∗ φα − φα k1 < 5.
Theorem 17. A locally compact group G is amenable if and only if for each > 0
and a compact set K ⊂ G, there is a Borel set F such that 0 ≤ λ(F ) < ∞ and
λ(gF 4F )
< .
λ(F )
4F )
Proof. (⇐) If φ := 1F /λ(F ) then φ ∈ P(G) and kLg (φ) − φk1 = λ(gF
λ(F ) .
(⇒) We first prove a weaker assertion.
(F) Given > 0 and δ > 0 and compact set K ⊂ G, there are Borel subsets
4F )
F ⊂ G and N ⊂ F such that 0 < λ(F ) < ∞, λ(N ) < δ and λ(gF
< for all
λ(F )
g ∈ K \ N.
20
From the Reiter condition, there is φ ∈ P(G) with kLg (φ) − φk1 < δ/λ(K) for
all k ∈ K. Without loss of generality we may assume that φ is a simple function.
PN
Moreover, we can represent φ as φ = j=1 λj 1Aj /λ(Aj ) with A1 ⊂ A2 ⊂ · · · ⊂
P
P
AN and λj > 0. Hence j λj = 1. Since Lg (φ) − φ = j λj 1gAj \Aj /λ(Aj ) −
P
j λj 1Aj \gAj /λ(Aj ) and (gAj \ Aj ) ∩ (Ai \ gAi ) = ∅ then
kLg (φ) − φk1 =
N
X
j=1
Hence
N
X
Z
λj
K
j=1
λj
λ(gAj 4Aj )
< δ/λ(K).
λ(Aj )
λ(gAj 4Aj )
dλ(g) < δ.
λ(Aj )
Hence there is j with
Z
K
Hence
λ(gAj 4Aj )
λ(Aj )
λ(gAj 4Aj )
dλ(g) < δ.
λ(Aj )
> on a subset N whose measure λ(N ) < δ. Thus (F) is proved.
4F )
Now we note that if λ(gF
< for each g ∈ K \ N then
λ(F )
−1
g ∈ (K \ N )(K \ N ) . Indeed,
λ(gF 4F )
λ(F )
< 2 for each
λ(g1 g2−1 F 4F )
λ(g2−1 F 4g1−1 F )
λ(g1−1 F 4F ) λ(g2−1 F 4F )
=
≤
+
λ(F )
λ(F )
λ(F )
λ(F )
λ(F 4g1 F ) λ(F 4g2 F )
+
.
=
λ(F )
λ(F )
It remains to show
Lemma 18. If K is a compact in G then for each Borel subset N ⊂ K ∪ KK, if
λ(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.
Hence
λ(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 locally
compact group is called a (left) Følner net if
λ(gFα 4Fα )
=0
α∈A
λ(Fα )
lim
for each g ∈ G.
21
Corollary 20. Let G be a locally compact group. Then G is amenable if and only
if there is a Følner net in it. If G is σ-compact then it is amenable if and only if
tree is a Følner sequence it.
Proof. Let A be the directed set {(K, ) | K is a compact in G, > 0}. For each
α 4Fα )
α ∈ A, let Fα be the corresponding set in G such that λ(gF
< for each g ∈ K
λ(Fα )
(by Theorem 17). Then (Fα )α∈A is a Følner net in G. Conversely, if (Fα )α∈A is
a Følner net in G then kLg (1Fα /λ(Fα )) − 1Fα /λ(Fα )k1 → 0. It remains to apply
Theorem 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 is
a Følner sequence in it (the converse is also true). For example, if G = Z then
Fn := {1, . . . , n} is a Følner sequence. If G = R then Fn := [0, n] is a Følner
sequence.
Corollary 21 (The invariant mean on the almost periodic functions on
amenable groups). Let (Fα )α be a Følner net in G. Let M be the (unique!)
invariant mean on AP (G). Then
Z
1
f (gx) dλ(g).
M (f ) = lim
α λ(Fα ) F
α
Proof. Consider a net 1Fα /λ(Fα ) ∈ P(G). Take a limit point m of this net. Then m
ia a LIM m on L∞ (G). We note that m is a LIM on AP (G) ⊂ BU C(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 h1Fn /λ(Fn ), f i = 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→ Tg v ∈ K is called an
affine action of G on K if Tg is an affine map of K and Tg Th = Tgh for all g, h ∈ G.
Theorem 23 (Markov-Kakutani fixed point theorem). Let G be a locally
compact group. It is amenable if and only if each affine action of G has a fixed
point.
Proof. (⇐) Let M be the set of means on BU Cl (G). Then M is ∗-compact convex
subset of BU Cl (G)0 . We let hTg m, f i := hm, Lg−1 (f )i. Then (Tg )g∈G is an affine
action of G on M. If gj → g in G then kLg (f ) − Lgj (f )k∞ → 0 because f ∈
BU Cl (G). It is easy to check that T is continuous in 2 variables. Hence there is a
fixed point for T , which is a LIM on BU C(G).
(⇒) Let T be an affine action of G on a compact convex set K. Since G
is amenable,
there is a Følner net (Fα )α∈A in it. Take v ∈ K and set vα :=
R
λ(Fα )−1 Fα Tg v dλ(g). Then vα ∈ K. Take a limit point w ∈ K of (vα )α . Take a
seminorm p on V . Then
Z
λ(Fα 4hFα ) maxK p(k)
1
p(Tg v) dλ(g) ≤
.
p(Th xα − xα ) ≤
λ(Fα ) Fα 4hFα
λ(Fα )
Thus p(Th w − w) = 0 for each h ∈ G and each seminorm p defining the topology
on V . 22
Corollary 24 (Generalized Bogolubov-Krylov theorem). Let G be a locally
compact group. It is amenable if and only if there is an invariant probability Radon
measure 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 compact
space K. Define an action T ∗ of G on the space C(K)0 (endowed with the ∗-weak
topology) by setting hTg∗ F, f i = hF, f ◦ Tg i. It is continuous. Restrict T ∗ to the
subset P ⊂ C(K)0 of probability measures. We note that P is ∗-weakly compact
and convex. Hence there is a fixed point in it.
Conversely, consider the space M ⊂ L∞ (G)0 of all mean on L∞ (G). It is ∗weakly compact and convex. Of course, (g, m) 7→ m ◦ Lg is a continuous action of
G 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 amenable
group and (Fα ))α ∈ A a Føliner net in it. Let G 3 g 7→ Ug ∈ U(H) be a strongly
continuous unitary representation of G in a Hilbert space H. Then
Z
1
Ug dλ(g) = P
lim
α λ(Fα ) F
α
strongly, where P is the orthogonal projection to the space of (Ug )g∈G -invariant
vectors.
Proof. Let W be the linear span of the space {Uh v − Rv | h ∈ G, v ∈ H}. Let I be
the subspace of U -invariant vectors. Put Aα := λ(F1 α ) Fα Ug dλ(g).
— Aα w → 0 for each w ∈ W. Hence Aα w → 0 for each w ∈ W.
— Aα v = 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
23