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My title - Departamento de Matemática da Universidade do Minho

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11<br />

RECORRÊNCIAS 76<br />

Poincaré recurrence theorem (topologic version). Let f : X → X be a continuous<br />

transformation of a separable metrizable topological space X. The support of any invariant Borel<br />

probability measure µ is contained in the closure of the set of recurrent points, namely<br />

supp (µ) ⊂ Rec f .<br />

If, in particular, f admits an invariant measure µ which is diffuse (i.e. gives positive measure to<br />

any nonempty open set) then the set of recurrent points is <strong>de</strong>nse in X, namely<br />

Rec f = X.<br />

Observe that if f is a homeomorphism, then the same applies to Rec f −1, and the support of<br />

any invariant Borel probability measure is contained in the closure of Rec f ∩Rec f −1.<br />

If you <strong>do</strong>n’t like the above proof, here is another, perhaps more elemenary, of the last statement.<br />

Alternative proof. Assume that the continuous map f : X → X preserves a diffuse Borel<br />

probability measure µ. For each n ≥ 1, let<br />

R n = { x ∈ X s.t. ∃ k ≥ 1 s.t. d(f k (x), x) < 1/n }<br />

be the set of ”1/n-recurrent” points. Of course, Rec f = ∩ ∞ n=1R n . The sets R n are clearly open.<br />

To show that Rec f is <strong>de</strong>nse we must show that each R n is, since then the Baire theorem implies<br />

that also their countable intersection is <strong>de</strong>nse. So, take any nonempty ball B with diameter<br />

< 1/n. Its inverse images f −1 (B), f −2 (B), f −3 (B),... have all the same measure µ(B) > 0. Since<br />

µ(X) = 1, they cannot be disjoint. There follows that there exist k > 0 and n ≥ 0 such that<br />

f −(n+k) (B) ∩ f −n (B) ≠ 0, and this implies that B contains a 1/n-recurrent point (for a point x in<br />

the intersection has both images f n (x) and f n+k (x) = f k (f n (x)) in B, hence at distance < 1/n).<br />

Since B was arbitrary, this proves that each R n is <strong>de</strong>nse, and Baire theorem implies that Rec f is<br />

<strong>de</strong>nse too.<br />

Example. Rotations of the circle preserve the Lebesgue probability measure. Hence, almost any<br />

point x + Z is recurrent for the rotation R α : x + Z ↦→ x + α + Z. If α is rational this is trivial, since<br />

all points are periodic. When α is irrational, this says that for any ε > 0 there exist an infinity of<br />

times q ∈ N that<br />

d(x + Z, x + qα + Z) < ε<br />

hence for any ε > 0, there exist an infinity of rationals p/q such that<br />

|qα − p| < ε i.e.<br />

∣ α − p q ∣ < ε q<br />

We know from continued factions that the estimate is in<strong>de</strong>ed better, since ε/q may be substituited<br />

by 1/q 2 .<br />

11.4 Conjunto não-errante<br />

Por volta <strong>do</strong>s anos trinta, George D. Birkhoff teve a i<strong>de</strong>ia <strong>de</strong> dividir o espaço <strong>do</strong>s esta<strong>do</strong>s <strong>de</strong> um<br />

sistema dinâmico em duas classes <strong>de</strong> pontos com dinâmicas qualitativamente distintas.<br />

O ponto x é errante 26 se admite uma vizinhança disjunta <strong>de</strong> to<strong>da</strong>s as suas itera<strong>da</strong>s, i.e. se<br />

existe um aberto U que contém x tal que U ∩ f n (U) = ∅ para to<strong>do</strong> tempo n ≥ 1. O ponto x não<br />

é errante se para to<strong>da</strong> vizinhança U <strong>de</strong> x existe um tempo n ≥ 1 tal que f n (U) ∩ U ≠ ∅.<br />

O conjunto não-errante NW f (<strong>do</strong> inglês “non-wan<strong>de</strong>ring set”) é o conjunto <strong>do</strong>s pontos x que não<br />

são errantes. A i<strong>de</strong>ia informal é que conjunto não-errante é on<strong>de</strong> acontece a dinâmica interessante,<br />

enquanto o conjunto errante é o conjunto <strong>do</strong>s pontos que a dinâmica esquece.<br />

26 A palavra grega por “errante”, ou seja, “vagabun<strong>do</strong>”, “que vagueia ao acaso”, era πλανητης, ou seja, planeta.

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