Markov-Kakutani Theorem and Haar Measure

Transcription

1 Markov-Kakutani Theorem and Haar Measure 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 = λ λa. 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 fundamental base of neighborhoods of zero formed with convex sets. Note that if p : E R + is a continuous seminorm and E is a TVS, then the set p 1 ([0, 1)) is an open, balanced, convex and absorbing neighborhood of 0. Viceversa, if A is convex, balanced and absorbing in a vector space E then the Minkowski functional p A (x) = inf{t > 0 x ta} is a seminorm and {p A < 1} A {p A 1}. It turns out that 1

2 Proposition 1 E is locally convex if and only if there exists a fundamental base of neighborhoods of origin formed with convex, balanced and absorbing sets. Definition 5 A family P of seminorms in the vector space E is said to be suficient if for any x E there exists p P so that p(x) 0. The locally convex topology generated by a sufficient familiy P of seminorms is the weakest locally convex topology that makes all of them continuous. A base of neighborhoods of 0 is given by the sets B = {x E p i (x) < ɛ i, p i F } where F P is an arbitrary finite set and ɛ i > 0 are arbitrary positive numbers. Definition 6 A TVS is said to be metrizable if the topology is given by a translation-invariant metric. Example 1 Let Ω R n be an open set, k N. The space C k (Ω) is the space of k continuosly differentiable functions in Ω endowed with the familiy of seminorms P K,m (f) = sup x K, α m α f(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 strong topology and the weak topology. The norm topology is given by the norm T = sup T x F x E 1 L(E, F ) is a Banach space with this topology. The strong topology is given by seminorms p x1,...,x n (T ) = max 1 i n T x i F 2

3 This topology is the topology of strong pointwise convergence. L(E, F ) is a LCS 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 weakstar topology on E is the strong topology defined above. Finally, the weak topology on L(E, F ) is given by the seminorms p x1,...,x n;l(t ) = max 1 i n L(T (x i) where x i E and L F. This is also a LCS topology, weaker than the strong 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 G in 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 E is invariant under π if π(g)a A holds for all g G. (Note that equality follows, 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 group in E. Assume that K is invariant under π. Then there exists x K, fixed by π. Proof (J. von Neumann) Let n 0 be an integer, and g G and let M n (g) := 1 ( I + π(g) + π(g 2 ) + + π(g n ) ). n + 1 Note that M n (g) is a continuous linear operator M n (g) : E E, but not necessarily invertible. Because K is invariant under π and convex, it follows that M n (g)(k) K for all n. Let S be the semigroup generated by M n (g) as n 0, g G. This is the set of finite words (products) of elements of the form described above. Let A = s S s(k) 3

4 Note that s(k) K for all s S, and s(k) is compact. We claim that A. Because of compactness, it is enough to prove that s 1 (K) s 2 (K) s n (K) for any n and s j S. We take s = s 1 s 2 s n. We claim s(k) n i=1s i (K). Indeed, s 1 (s 2 s n (K)) s 1 (K). Now we commute s 1 and s 2 to find s(k) s 2 (K) and we continue thus. This proves the claim and hence A. Let x A. It follows that for every g G and every n 0 there exists y K so that x = M n (g)y. (This just says that x M n (g)k). Now M n (g)(π(g)y) = 1 n+1 π(g k )y = M n (g)y + 1 ( π(g n+1 )y y ). n + 1 n + 1 Therefore, k=1 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) C K n + 1 where C K = sup z 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. We say that G acts continuously in X is there exist continuous maps (we denote φ g (x) = gx) such that φ g : X X g 1 (g 2 x) = (g 1 g 2 )x, and ex = x Let G act continuously in X. We define the representation T : G Aut(C(X)) 4

5 by and its dual by T g (f)(x) := f(g 1 x) T : G (C(K)) (T (g)(l))(f) := L((T (g 1 )f)). We say that a probability measure on X is G invariant if f(x)dµ = f(gx)dµ holds for any g G. X Theorem 2 Let G be an abelian group acting continuously in a compact metric 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 is compact with the weak-star topology by the Alaoglu-Bourbaki theorem, it is convex, and it is invariant under T. By the Markov-Kakutani theorem there exists µ K that is fixed by T, X T (g)µ = µ. This means µ(t (g 1 f) = µ(f) i.e., f(gx)dµ = X X f(x)dµ. Definition 12 Let π : G Aut(E) be a representation of the group G in the 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 a Banach space. Assume that π is a representation and that it is continuous 5

6 when Aut(E) has the strong topology. (Topology of strong convergence of the operators, i.e. pointwise convergence in norm.) Let x E and define by L G x = sup g G G x : E R + L(π(g)x) = sup π (g)(l)x g G Then: 1) G x is a seminorm on E. 2) π (g)l G x = L G x for all g G, L E. 3) G x : B E (0, 1) R + is weak-star continuous. 4) Lx 0 L G x 0. Proof. We address only point 3). The weak-star topology on E is defined by seminorms, p x1,...x k (L) = max 1 j k L(x j). Let x E be fixed. Then the orbit of the action O(x) = {π(g)x g G} is compact in E because of the compactness of G and strong continuity of the representation. Then if ɛ > 0 there exist points π(g 1 )x,... π(g k )x, so that for every g G there exists a j 1,... k such that π(g)x π(g j )x E ɛ 4. Then, if we consider the seminorm p = p π(g1 )x,...π(g k )x and take L 1, L 2 B E (0, 1) such that p(l 1 L 2 ) ɛ 2 then it follows that L 1 L 2 G x ɛ. Theorem 3 Let G be a compact group, let π : G Aut(E) be a strongly continuous representation in a Banach space E. Let A E be a normbounded, weak-star closed, convex set that is invariant under π. Then there exists L A that is fixed by π Proof. Let A = {B A B, convex, weak-star compact, invariant under π } By Zorn s lemma, there exists A A that is minimal under inclusion. We claim that A is a singleton, A = {L}. Indeed, if not, there exist L 1 L 2 L 1, L 2 A. There exists x E such that L 1 x L 2 x. Let d := sup{ T S G x T, S A} 6

7 d > 0 by assumption. Now A is weak-star compact and, for any r > 0, the balls {T T T 0 G x < r} = B G x (T 0, r) are weak-star open and cover A, as T 0 A. We take r = d and cover A with 2 finitely many such balls: ( A n i=1bx G T i, d ) 2 with T i A. Let now T = 1 n n T i. because A is convex, it follows that T A. Let T A be arbitrary. Because T T i G x d for at least one 1 i n, it follows that 2 T T G x 1 [ ] ( d + (n 1)d = 1 1 ) d n 2 2n and so Take A B G x i=1 ( ( T, 1 1 ) ) d 2n A 1 := T A {S A T S G x ( 1 1 ) } d 2n Because T A 1, we know that A 1. Moreover, A 1 is weak-star compact, convex and invariant under π. This is absurd, because its diameter is strictly less than d, so A 1 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 rightinvariant and is unique. Proof (J. von Neumann). Let π : G Aut(C(G)) 7

8 be defined by (π(g)f)(h) := f(g 1 h) for f C(G), g, h G. Then π is a 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, G f(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 probability measure (which exists by a similar argument). Then ( ) [ ] f(xy)dν(x) dµ(y) = f(x)dν(x) 1dµ(y) = f(x)dν(x) G G G G G and by changing order of integration, the same quantity equals G f(y)dµ(y), and so, ν = µ, which implies also the uniqueness. µ is the Haar measure on G. 8