Markov-Kakutani Theorem and HaarMeasure

Analysis III

1 Markov-Kakutani

Definition 1 E is a TVS (topological vector space) if(a) E is a vector space (over R or C).(b) E is a Hausdorff topological space.(c) The operations + : E × E → E and · : C× E → E are continuous.

Definition 2 A ⊂ E is(a) balanced, if αA ⊂ A, for all |α| ≤ 1,(b) convex, if (1− t)A + tA ⊂ A for all 0 ≤ t ≤ 1,(c) absorbing, if E = ∪λλA.

Definition 3 Let E be a real or complex vector space.

p : E → R+

is a seminorm if(a) p(0) = 0,(b) p(x + y) ≤ p(x) + p(y), for all x, y ∈ E,(c) p(λx) = |λ|p(x), for all λ ∈ C, x ∈ E.

Definition 4 A TVS is locally convex space (LCS) if it has a fundamentalbase of neighborhoods of zero formed with convex sets.

Note that if p : E 7→ R+ is a continuous seminorm and E is a TVS, thenthe set p−1([0, 1)) is an open, balanced , convex and absorbing neighborhoodof 0. Viceversa, if A is convex, balanced and absorbing in a vector space Ethen the Minkowski functional

pA(x) = inft > 0 | x ∈ tA

is a seminorm and pA < 1 ⊂ A ⊂ pA ≤ 1. It turns out that

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Proposition 1 E is locally convex if and only if there exists a fundamentalbase of neighborhoods of origin formed with convex, balanced and absorbingsets.

Definition 5 A family P of seminorms in the vector space E is said to besuficient if for any x ∈ E there exists p ∈ P so that p(x) 6= 0.

The locally convex topology generated by a sufficient familiy P of seminormsis the weakest locally convex topology that makes all of them continuous. Abase of neighborhoods of 0 is given by the sets

B = x ∈ E | pi(x) < εi, pi ∈ F

where F ⊂ P is an arbitrary finite set and εi > 0 are arbitrary positivenumbers.

Definition 6 A TVS is said to be metrizable if the topology is given by atranslation-invariant metric.

Example 1 Let Ω ⊂ Rn be an open set, k ∈ N. The space Ck(Ω) is thespace of k continuosly differentiable functions in Ω endowed with the familiyof seminorms

PK,m(f) = supx∈K,|α|≤m

|∂αf(x)|

where K ⊂⊂ Ω.

This is the topology of uniform convergence with derivatives on compacts.It is a metrizable topology.

Definition 7 Let E, F be Banach spaces. Let

L(E, F ) = T : E → F | T linear, continuous

There are three natural topologies on L(E, F ): the norm topology, the strongtopology and the weak topology. The norm topology is given by the norm

‖T‖ = sup‖x‖E≤1

‖Tx‖F

L(E, F ) is a Banach space with this topology. The strong topology is givenby seminorms

px1,...,xn(T ) = max1≤i≤n

‖Txi‖F

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This topology is the topology of strong pointwise convergence. L(E, F ) is aLCS with this topology. This is a weaker topology than the norm topology.The dual E∗ is the special case L(E, F ) when F = C or F = R. The weak-star topology on E∗ is the strong topology defined above. Finally, the weaktopology on L(E, F ) is given by the seminorms

px1,...,xn;L(T ) = max1≤i≤n

|L(T (xi)|

where xi ∈ E and L ∈ F ∗. This is also a LCS topology, weaker than thestrong topology.

Definition 8 Let E be a LCS, then

Aut(E) = T : E → E |T, T−1 linear, continuous

Definition 9 Let G be a group. π : G → Aut(E) is a representation of Gin E if it is a group homomorphism.

Definition 10 x ∈ E is fixed by π if π(g)x = x holds for all g ∈ G. A ⊂ Eis invariant under π if π(g)A ⊂ A holds for all g ∈ G. (Note that equalityfollows, π(g)A = A).

Theorem 1 (Markov-Kakutani) Let E be a LCS and K ⊂ E be a convex ,compact subset. Let π : G → Aut(E) be a representation of an abelian groupin E. Assume that K is invariant under π. Then there exists x ∈ K, fixedby π.

Proof (J. von Neumann) Let n ≥ 0 be an integer, and g ∈ G and let

Mn(g) :=1

n + 1

(I + π(g) + π(g2) + · · ·+ π(gn)

).

Note that Mn(g) is a continuous linear operator Mn(g) : E → E, but notnecessarily invertible. Because K is invariant under π and convex, it followsthat

Mn(g)(K) ⊂ K

for all n. Let S be the semigroup generated by Mn(g) as n ≥ 0, g ∈ G. Thisis the set of finite words (products) of elements of the form described above.Let

A = ∩s∈Ss(K)

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Note that s(K) ⊂ K for all s ∈ S, and s(K) is compact. We claim thatA 6= ∅. Because of compactness, it is enough to prove that

s1(K) ∩ s2(K) ∩ · · · ∩ sn(K) 6= ∅

for any n and sj ∈ S. We take s = s1 · s2 · · · · · · sn. We claim

s(K) ⊂ ∩ni=1si(K).

Indeed, s1(s2 · · · sn(K)) ⊂ s1(K). Now we commute s1 and s2 to find s(K) ⊂s2(K) and we continue thus. This proves the claim and hence A 6= ∅. Letx ∈ A. It follows that for every g ∈ G and every n ≥ 0 there exists y ∈ K sothat x = Mn(g)y. (This just says that x ∈ Mn(g)K). Now

Mn(g)(π(g)y) =1

n + 1

n+1∑k=1

π(gk)y = Mn(g)y +1

n + 1

(π(gn+1)y − y

).

Therefore,

π(g)x− x =1

n + 1(π(gn+1)y − y).

We take any continuous seminorm p on E. It follows that

p(π(g)x− x) ≤ CK

n + 1

where CK = supz∈K−K p(z). Letting n →∞ it follows that p(π(g)x−x) = 0.Therefore π(g)x = x and the theorem is proved.

Definition 11 Let G be an abelian group and let X be a metric space. Wesay that G acts continuously in X is there exist continuous maps

φg : X → X

(we denote φg(x) = gx) such that

g1(g2x) = (g1g2)x, and ex = x

Let G act continuously in X. We define the representation

T : G → Aut(C(X))

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byTg(f)(x) := f(g−1x)

and its dualT ∗ : G → (C(K))∗

by(T ∗(g)(L))(f) := L((T (g−1)f)).

We say that a probability measure on X is G invariant if∫X

f(x)dµ =

∫X

f(gx)dµ

holds for any g ∈ G.

Theorem 2 Let G be an abelian group acting continuously in a compactmetric space. Then there exists a G-invariant probability measure on X.

Proof. Let K = P(X) = µ ∈ C(X)∗ | µ ≥ 0, µ(1) = 1. Then K iscompact with the weak-star topology by the Alaoglu-Bourbaki theorem, itis convex, and it is invariant under T ∗. By the Markov-Kakutani theoremthere exists µ ∈ K that is fixed by T ∗,

T ∗(g)µ = µ.

This means µ(T (g−1f) = µ(f) i.e.,∫X

f(gx)dµ =

∫X

f(x)dµ.

Definition 12 Let π : G → Aut(E) be a representation of the group G inthe automorphism group of the Banach space E. The dual representation

π∗ : G → Aut(E∗)

is defined by(π∗(g)L)(x) = L(π(g−1)x)

Lemma 1 Let G be a compact group and let π : G → Aut(E) where E is aBanach space. Assume that π is a representation and that it is continuous

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when Aut(E) has the strong topology. (Topology of strong convergence of theoperators, i.e. pointwise convergence in norm.) Let x ∈ E and define

‖ · ‖Gx : E∗ → R+

by‖L‖G

x = supg∈G

|L(π(g)x)| = supg∈G

|π∗(g)(L)x|

Then:1) ‖ · ‖G

x is a seminorm on E∗.2) ‖π∗(g)L‖G

x = ‖L‖Gx for all g ∈ G, L ∈ E∗.

3) ‖ · ‖Gx : BE∗(0, 1) → R+ is weak-star continuous.

4) Lx 6= 0 ⇒ ‖L‖Gx 6= 0.

Proof. We address only point 3). The weak-star topology on E∗ is definedby seminorms,

px1,...xk(L) = max

1≤j≤k|L(xj)|.

Let x ∈ E be fixed. Then the orbit of the action O(x) = π(g)x | g ∈ G iscompact in E because of the compactness of G and strong continuity of therepresentation. Then if ε > 0 there exist points π(g1)x, . . . π(gk)x, so that forevery g ∈ G there exists a j ∈ 1, . . . k such that ‖π(g)x−π(gj)x‖E ≤ ε

4. Then,

if we consider the seminorm p = pπ(g1)x,...π(gk)x and take L1, L2 ∈ BE∗(0, 1)such that

p(L1 − L2) ≤ε

2then it follows that

‖L1 − L2‖Gx ≤ ε.

Theorem 3 Let G be a compact group, let π : G → Aut(E) be a stronglycontinuous representation in a Banach space E. Let A ⊂ E∗ be a norm-bounded, weak-star closed, convex set that is invariant under π∗. Then thereexists L ∈ A that is fixed by π∗

Proof. Let

A = B ⊂ A | B 6= ∅, convex, weak-star compact, invariant under π∗

By Zorn’s lemma, there exists A ∈ A that is minimal under inclusion. Weclaim that A is a singleton, A = L. Indeed, if not, there exist L1 6= L2

L1, L2 ∈ A. There exists x ∈ E such that L1x 6= L2x. Let

d := sup‖T − S‖Gx | T, S ∈ A

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d > 0 by assumption. Now A is weak-star compact and, for any r > 0, theballs

T | ‖T − T0‖Gx < r = BG

x (T0, r)

are weak-star open and cover A, as T0 ∈ A. We take r = d2

and cover A withfinitely many such balls:

A ⊂ ∪ni=1B

Gx

(Ti,

d

2

)with Ti ∈ A. Let now

T =1

n

n∑i=1

Ti.

because A is convex, it follows that T ∈ A. Let T ∈ A be arbitrary. Because‖T − Ti‖G

x ≤ d2

for at least one 1 ≤ i ≤ n, it follows that

‖T − T‖Gx ≤

1

n

[d

2+ (n− 1)d

]=

(1− 1

2n

)d

and so

A ⊂ BGx

(T ,

(1− 1

2n

)d

)Take

A1 := ∩T∈A

S ∈ A | ‖T − S‖G

x ≤(

1− 1

2n

)d

Because T ∈ A1, we know that A1 6= ∅. Moreover, A1 is weak-star compact,convex and invariant under π∗. This is absurd, because its diameter is strictlyless than d, so A1 6= A, but A was minimal under inclusion.

Theorem 4 Let G be a compact group, and let

P(G) = µ ∈ C(G)∗ | µ ≥ 0, µ(1) = 1.

Then there exists µ ∈ P(G) that is left invariant. Moreover, µ is also right-invariant and is unique.

Proof (J. von Neumann). Let

π : G → Aut(C(G))

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be defined by (π(g)f)(h) := f(g−1h) for f ∈ C(G), g, h ∈ G. Then π isa strongly continuous representation, P(G) is weak-star compact, convex,invariant under π∗, and hence, by the previous theorem, there exists µ ∈P(G) that is fixed by π∗, π∗(g)µ = µ. This means that, for any f ∈ C(G)and any g ∈ G, ∫

Gf(x)dµ(x) =

∫G

fd(π∗(g)(µ)(x))=

∫G(π(g−1)f)(x)dµ(x) =

∫G

f(gx)dµ(x)

that is, µ is left invariant. Now, let ν be any right invariant probabilitymeasure (which exists by a similar argument). Then∫

G

(∫G

f(xy)dν(x)

)dµ(y) =

[∫G

f(x)dν(x)

] ∫G

1dµ(y) =

∫G

f(x)dν(x)

and by changing order of integration, the same quantity equals∫

Gf(y)dµ(y),

and so, ν = µ, which implies also the uniqueness. µ is the Haar measure onG.

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