My title - Departamento de Matemática da Universidade do Minho

11
RECORRÊNCIAS 75
Exercícios.
• Defina uma ordem parcial em X da seguinte maneira: x ≺ x ′ se para todas vizinhanças U
de x e V de x ′ existe um tempo n ≥ 1 tal que f n (U) ∩ V ≠ ∅. Prove que x é recorrente sse
x ≺ x.
• Mostre que Per f ⊂ Rec f .
• Dê exemplos que mostram que Rec f e Rec f −1 podem ser vazios.
Pseudo-trajetórias e “chain-recurrent points”. Dado ε > 0, a sequência de pontos
{x k } k=0,1,...,n
com x 0 = x é dita ε-pseudo-trajetória de x se
d (x k+1 , f (x k )) < ε
para todo k = 0, 1, 2, ..., n−1. O ponto x ∈ X é dito chain-recurrent se, para todo ε > 0, existe uma
ε-trajetória {x k } k=0,1,...,n
com x 0 = x n = x. Rec ε f
denota o conjunto dos pontos chain-recurrent
de f.
so what
11.3 Invariant measures and recurrent points: Poincaré theorem
If f satisfies a condition (natural in physics) like ”preserving a probability measure”, then there
are a lot of recurrent points, actually almost any point is recurrent. If, moreover, the probability
measure is diffuse, i.e. any non-empty open set has positive measure, then the set of recurrent
points is also dense. These results, discovered by Jules Henri Poincaré around 1890, motivated the
modern theory of dynamical systems. They show how weak informations on the transformation
f may yeld significative qualitative information about ”almost all” orbits of the system. Here
follow the precise statements, together with all the necessary technical details. If you don’t know
the meaning of some words, like ”measurable” or ”Borel set”, don’t worry, just try to understand
what’s going on. Poincaré himself didn’t know, yet!
Poincaré recurrence theorem (probabilistic version). Let f : X → X be an endomorphism
of a probability space (X, E, µ), and let A ∈ E. Then the set
A rec = {x ∈ A t.q. f n (x) ∈ i.o. A}
of those points of A whose orbit passes through A infinitely often has total probability, namely
µ (A rec ) = µ (A).
proof. For k ≥ 1, let B k = {x ∈ A s.t. f n (x) /∈ A ∀ n ≥ k} be the set of those points of A
which never return in A after n ≥ k iterates. Observe that B k = A ∩ (∩ n≥k f −n (X\A)), and
that A rec = A\ (∪ k≥1 B k ), and this shows in particular that A rec is measurable. One sees that
f −nk (B k ) ∩ B k = ∅ for any n ≥ 1 (for a point x in the intersection would be a point of B k with
f kn (x) ∈ A for some kn ≥ k), and this implies that f −nk (B k ) ∩ f −mk (B k ) = ∅ for any n > m ≥ 0.
The sets f −nk (B k ) all have the same measure µ (B k ), because µ is invariant, and are pairwise
disjoint. This implies that µ (B k ) = 0, because
∑
µ ( f −nk (B k ) ) = µ ( ∪ n≥1 f −nk (B k ) ) ≤ µ (X) = 1
n≥1
µ (B k )) = ∑ n≥1
and so µ (A rec ) = µ (A). □
Now, let f : X → X be a continuous transformation of a metrizable topological space X, and
let µ be an invariant Borel probability measure. If X admits a countable basis (U i ) i∈N , we can
apply the above theorem to every open set U i . This easily implies that the set of recurrent points
has full measure, i.e.
µ (Rec f ) = 1
In particular, since any set of full measure is dense in the support of a Borel measure, we get the
following general result.