a-hm-ERG - DynSys MTG 6401 Prof. JLF King 19Oct2009 Home-A...

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Unformatted text preview: DynSys MTG 6401 Prof. JLF King 19Oct2009 Home-A for which the Mean Ergodic Thm conclusion fails in the L∞ -norm. ii Hello. Take-home is due by 3PM, Thursday, 15Oct2009, slid completely under my office door, LIT402. Here, we only consider bi-mpts on a probability space, e.g ( T : X, X , µ) or ( S : Y, Y , ν) . All mentioned subsets of a measure-space are assumed measurable. A1: Please prove this result. ∞ 1: First Ces`ro Lemma. Suppose that b := ( bn) n=1 is a dea creasing (non-increasing) sequence of real numbers. Then the following limit exists in [ ∞, ∞) and this equality holds: 1 N →∞ N ∗: N bk lim = inf bn . n k=1 I.e Ak (bk ) = inf k bk . ∞ ♦ Aside: You may use either Ak (bk ) or A∞ (b) to abbrev. ∞ LhS(∗). µ A2: Fix B > 0. For z ∈ X , define R = RB := {n ∈ Z | µ(B ∩ T n B ) > 0} ; ρ(z ) = ρB (z ) := {n ∈ Z | T n z ∈ B } . a Prove that R has bounded gaps. (The general term is “syndetic”. In an abelian topological group G, a subset R ⊂ G is syndetic if there exists a compact K ⊂ G st. R + K (the set of all sums) is the whole group.) b The Birkhoff thm implies that a.e z ∈ X hits B with a limiting frequency.♥1 Now assume that T is ergodic. Thus for a.e z : Den ρ(z ) equals µ(B ). µ Construct an ergodic T and set B > 0 st. for a.e z in X : The set ρ(z ) does not have bounded gaps. Z A3: Let S be the shift on Y := {0, 1} , equipped with 11 independent ( 2 , 2 ) -measure. (I.e, S is the Bernoulli 2-shift.) i Construct an f ∈ L∞ (ν ) st. f=0 Y |f | > 0 , and Y ♥1 A frequency that depends on z , or rather, on the ergodic component that z lies in. Can you construct a rank-1 trn T and L∞ -fnc f for where the same failure occurs? [Hint: For both parts, LARGE, colorful, pictures may be of use.] End of Home-A ...
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This note was uploaded on 01/26/2012 for the course MTG 6401 taught by Professor Staff during the Fall '09 term at University of Florida.

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