University of Southern California Dissertations and Theses

Analysis of ergodic and mixing dynamical systems

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Analysis of ergodic and mixing dynamical systems

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Content Analysis of Ergodic and Mixing Dynamical Systems by Joseph Antonio Fantini A Thesis Presented to the FACULTY OF THE USC DORNSIFE COLLEGE OF LETTERS, ARTS AND SCIENCES UNIVERSITY OF SOUTHERN CALIFORNIA In Partial Fulfillment of the Requirements for the Degree MASTER OF SCIENCE (APPLIED MATHEMATICS) August 2023 Copyright 2023 Joseph Antonio Fantini Acknowledgements I would like to thank my advisor Professor Cymra Haskell for the countless hours we spent together dedicated to this project. Her unwavering encouragement and guidance throughout the past few months have been indispensable. Writing this thesis with her was the highlight of my short mathematics career. I would also like to thank my committee members Professor Nicolai Haydyn and Professor Steven Heilman for their support and contributions. ii Table of Contents Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v Chapter 1: Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Preliminaries and Notation . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Recurrence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Chapter 2: Functional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.1 Functional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.2 Applications to Dynamical Systems . . . . . . . . . . . . . . . . . . . 16 2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Chapter 3: Ergodic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 3.1 The Maximal Ergodic Theorem . . . . . . . . . . . . . . . . . . . . . 31 3.2 The Ergodic Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.3 Ergodicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 3.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 Chapter 4: Mixing Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 4.1 Weak Mixing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 4.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 4.3 Strong Mixing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 4.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 iii List of Figures 1.1 The Baker’s Transform . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.1 The PDFs of x ∗ and Tx ∗ . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.2 The PDFs of T 2 x ∗ and T 3 x ∗ . . . . . . . . . . . . . . . . . . . . . . . 23 2.3 The PDFs of T 4 x ∗ and T 10 x ∗ . . . . . . . . . . . . . . . . . . . . . . 24 2.4 The PDFs of x ∗ , Tx ∗ , and T 2 x ∗ . . . . . . . . . . . . . . . . . . . . . 24 2.5 The PDF of T 6 x ∗ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.6 Iterations of 1 n P n k=0 U k T |sin(2πx )| . . . . . . . . . . . . . . . . . . . . 26 2.7 Eigenvalues of Rational Rotation . . . . . . . . . . . . . . . . . . . . 27 2.8 Iterations of 1 n P n− 1 k=0 T k x . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.9 Iteration 100 ( 1 100 P 99 k=0 T k x) . . . . . . . . . . . . . . . . . . . . . . . 29 4.1 Ergodic Components of T × T . . . . . . . . . . . . . . . . . . . . . . 64 iv Abstract Ergodictheoryisthestudyofdynamicalsystemsandtheirlong-termaveragebehav- ior. Originally motivated by physicists’ concern with systems in which the time and space averages coincide, ergodic theory answers questions about the nature of these systems and their convergence properties. In this work, we establish the founda- tional theorems of ergodic theory, analyze the characteristics of ergodic and mixing systems, and provide examples of dynamical systems with different mixing proper- ties. This work focuses on ergodic and mixing properties that are characterized in terms of functional analysis. v Chapter 1 Dynamical Systems 1.1 Preliminaries and Notation In this thesis the objects of interest are dynamical systems (X,B,µ,T ). The set X is a measure space andB is aσ -algebra of measurable sets in X. The set functionµ is a measure onB such that µ (X)=1. The σ -algebraB contains all µ -null sets of X. In other words, (X,B,µ ) is a complete probability space. The transformation T : X → X is measurable and measure-preserving. This property of T admits an interesting structure and is fundamental to ergodic theory. Definition 1.1.1. Given a system (X,B,µ,T ), a measurable transformation T : X → X is called a measure-preserving transformation (m.p.t.) if µ (T − 1 E) = µ (E) for all E∈B. For the sake of simplicity, we will generally assume T is invertible, though we will discuss examples of systems with non-invertible T. If T is invertible then T is measure preserving if and only if µ (TE) = µ (E) for all E ∈ B. If T is non- invertible it is possible for µ (E)<µ (TE) for some E∈B. Notice that to establish the measure-preserving property of T it suffices to show that measure preservation holds for a collection of sets that generatesB. 1 Definition1.1.2. Giventwomeasurespaces(X,B)and(Y,C),ameasurabletrans- formation T : X → Y, and a measure µ :B → [0,+∞], the pushforward measure of µ is the map T ∗ µ :C →[0,+∞] defined by, T ∗ µ (C)=µ (T − 1 (C)). In a dynamical system (X,B,µ,T ), since T is measure preserving the pushfor- ward measure T ∗ µ =µ . Definition 1.1.3. Let (X,B,µ,T ) be a dynamical system. The orbit of a point x (under T), orO x , is the set{T n x:n∈Z} for invertible T and{T n x:n∈N∪{0}} for non-invertible T. In other words, O x is the set of images (and pre-images) of x under the action of T. We can view applying T to a point x as the location of x one time unit in the future. Likewise, T − 1 x is where x was one time unit in the past. From the temporal perspective, the orbit is the entire infinite past, present, and future of x. The primary motivation behind ergodic theory stems from physicists’ interest in the change of a system’s statistical properties over time. Analyzing the long-term behavior of systems reveals how quickly or slowly T spreads sets out over X. The ergodic hierarchy is a hierarchy of properties dynamical systems can pos- sess. The properties are hierarchical in the sense that "stronger" properties imply "weaker" ones but not vice-versa. Ergodic⇐ Weak Mixing⇐ Strong Mixing⇐ Kolmogorov⇐ Bernoulli This work will forego discussion of Kolmogorov and Bernoulli properties. 2 The ergodic hierarchy is often associated with the characterization of the "ran- domness" inherent in a system. However, it is crucial to emphasize that dynamical systems are entirely deterministic in nature. Given exact knowledge of the initial statex∈X and the action of the transformationT, the entire trajectory ofx can be predicted. However, we don’t usually have exact knowledge of the initial state of x, we may only know x∈E⊂ X. As n increases, if T n E is spread out over the whole of X (in a topological sense rather than measure-theoretic sense) then information about the state of the system is lost. 1.2 Recurrence Poincaré proved that, for all dynamical systems (X,B,µ,T ), almost every point is recurrent with respect to all sets in a B. This property can be thought of as the lowest level of the ergodic hierarchy. Definition 1.2.1. Let (X,B,µ,T ) be a dynamical system. For B ∈B, x∈ B is recurrent with respect to B if there exists n∈N where T n x∈B. In other words, a point is recurrent with respect to a set if it returns to said set at least once in its orbit. Theorem 1.2.2. Poincaré Recurrence Theorem. Given a dynamical system (X,B,µ,T ), for each B ∈B, a.e. x ∈ B is recurrent with respect to B. Proof. Let B ∈B be given and A ={x∈ B : T k x / ∈ B, for all k ∈N} be the set of points in B that are not recurrent with respect to B. Then A=B\∪ ∞ k=1 T − k B =B∩T − 1 (X\B)∩T − 2 (X\B)∩... 3 is the set of points in B that are mapped to B c by T k for all k∈N. If x∈ A then T n x / ∈A for all n∈N. Therefore, A∩T − n A=∅ for all n∈N and T − k (A∩T − n A)=T − k A∩T − (n+k) A=∅ for all k,n ∈ N. Let m < k be given. Notice that T − m A∩ T − k A = T − m (A∩ T − (k− m) A)=∅. Thus the sets{T − k A} ∞ k=0 are pairwise disjoint so 1=µ (X)=≥ µ (∪ ∞ k=0 T − k A)= ∞ X k=0 µ (T − k A)= ∞ X k=0 µ (A). Therefore, µ (A)=0. This theorem states that a m.p.t. T almost surely will return any point x∈ B back to B in finite time. Measure preservation forces T to move points about the space X in such a way that most points will eventually return to where they have been. Recurrence precludes the existence of a source and, if T is invertible, a sink. 1.3 Examples Example 1.3.1. Rotations on a Circle. • X =[0,1) (the unit interval); • Tx=x+α mod 1 for x∈X and α ∈R; • µ =m (the Lebesgue Measure) . The transformation rotates a point on the circle by α . This map is also commonly represented multiplicatively with X ={z ∈C :|z| = 1} =K and Tz = ze 2πiα . A point x∈ [0,1) corresponds to z = e 2πix in K. It is easy to see T is a m.p.t. since 4 the Lebesgue measure is shift invariant. The case where α ∈R\Q is particularly interesting. Example 1.3.2. The Doubling Map. • X =[0,1) (the unit interval); • Tx=2x mod 1; • µ =m. Thedoublingmapdoublesapointthenremovesitsintegerpart. Thistransformation is not invertible since T maps two points to every x∈X, T x 2 =x T x+1 2 =x+1 mod 1=x. Recall that for non-invertible T it is possible for some E ∈ B to be such that µ (E)<µ (TE). For instance, the interval [0, 1 2 ) is such that µ ([0, 1 2 ))= 1 2 but µ (T([0, 1 2 )))=µ ([0,1))=1. ThetransformationT ismeasurepreservingsince, foranyintervalI =[a,a+b)∈X, T − 1 I = a 2 , a+b 2 ∪ a 2 + 1 2 , a+b 2 + 1 2 . 5 Notice the two intervals making up the preimage of I are disjoint, so µ (T − 1 I)= a+b 2 − a 2 + a+b 2 + 1 2 − a 2 − 1 2 , = b 2 + b 2 =b=µ (I). This map also can be represented as a map on a circle much like in Example 1.3.2. by taking z = e 2πix and letting Tz = z 2 . This map can be generalized by replacing 2 with any m∈N. Example 1.3.3. The Baker’s Transform. • X =[0,1)× [0,1) (the unit square); • T(x,y)=        (2x, y 2 ) for 0≤ x< 1 2 (2x− 1, y+1 2 ) for 1 2 ≤ x≤ 1 ; • µ =m (the Lebesgue measure). The baker’s transformation is named after the similarities between the action of T and kneading dough. Imagine T stretching the unit square about the x-axis, doubling its length horizontally. The unit square is deformed by T into the rectangle [0,2)× [0,1). This is an issue since we want T to be measure preserving and µ ([0,2)× [0,1))=2. To remedy this distortion, T then cuts the rectangle about x = 1 so that it is now two separate pieces [0,1)× [0,1) and [1,2)× [0,1). Next, T stacks the second piece on top of the first piece creating the rectangle [0,1)× [0,2). Finally, T squishes this rectangle down until it is the size of the unit square again. Bakers do not typically cut dough in half during the kneading process (the cutting and layering 6 A B T TA TB T T 2 A T 2 B T 2 A T 2 B Figure 1.1: The Baker’s Transform is more reminiscent of making croissants or baklava). There exist formulations of the baker’s transform that fold then squish rather than cut then squish but those transformations are not invertible. Since rectangles generateB, it is sufficient to show, for a,b,c,d∈ [0,1),a < b, c < d, the rectangle [a,b)× [c,d) has its measure preserved to show T is measure preserving. µ ([a,b)× [c,d))=(b− a)(d− c). There are three cases to consider when transforming such a set, T([a,b)× [c,d))=                [2a,2b)× [ c 2 , d 2 ) if a 0. There is also no particular reason σ should shift to the left, it is equally reasonable to define a right shift. Finally, we could have developed our system using A N∪{0} rather than A Z , however σ would be non-invertible in this case. 10 Chapter 2 Functional Analysis 2.1 Functional Analysis The study of functions on dynamical systems constitutes much of our focus. In particular, we are interested in analyzing L 1 (X,B,µ ) because it contains the class of distribution functions and L 2 (X,B,µ ) because of the nice structure it admits. Since X is a finite measure space, we have L 2 (X,B,T)⊂ L 1 (X,B,µ ) by Hölder’s inequality, ∥f∥ 1 ≤ µ (X) 1 2 ∥f∥ 2 =∥f∥ 2 . The following results from functional analysis help us understand long-term prop- erties of dynamical systems. Definition 2.1.1. The operator norm of a linear operator U :Y →Y on a normed vector space Y is ∥U∥ op =inf{c≥ 0:∥Uf∥ Y ≤ c∥f∥ Y for all f ∈Y}. 11 The operator norm of a transformation is the maximum factor by which the transformation scales vectors. Moving forward, all references to the norm of an operator will refer to the operator norm. Definition 2.1.2. A bounded linear operator U :Y →Y on a normed vector space Y is a linear contraction if∥U∥≤ 1. If U is a linear contraction then, for all f ∈Y, ∥Uf∥ Y ≤ ∥ f∥ Y . In other words, linear contractions either decrease the lengths of vectors or leave their lengths unchanged. Definition 2.1.3. Given a Hilbert space H and a bounded operator U : H → H, the operator U ∗ :H →H is adjoint to U if, for all f,g∈H, ⟨Uf,g⟩=⟨f,U ∗ g⟩. Theorem 2.1.4. Given a Hilbert space H with a linear contraction U : H → H, U ∗ is also a linear contraction. Proof. Given any f ∈H, ∥U ∗ f∥ 2 H =|⟨U ∗ f , U ∗ f⟩| =|⟨UU ∗ f , f⟩| ≤∥ UU ∗ f∥ H ∥f∥ H ≤∥ U ∗ f∥ H ∥f∥ H . So∥U ∗ f∥ H (∥U ∗ f∥ H −∥ f∥ H )≤ 0. Since∥U ∗ f∥ H ≥ 0, it follows that ∥U ∗ f∥ H ≤∥ f∥ H . 12 Theorem 2.1.5. Let H be a Hilbert space, U :H →H be a linear contraction, M be the set{f ∈H :Uf =f}, and N be the closed linear span of{g− Ug :g∈H}. Then N and M are orthogonal compliments. Proof. Note that M is closed because U is bounded. It suffices to show that N ⊥ = M, where N ⊥ is the orthogonal compliment of M. We first show that N ⊥ ⊂ M, then M ⊂ N ⊥ . Let h∈ N ⊥ be given. Then, by definition of orthogonal compliment, for all g∈ H, ⟨h, g− Ug⟩=0. So, 0=⟨h, g− Ug⟩=⟨h, g⟩−⟨ h, Ug⟩=⟨h, g⟩−⟨ U ∗ h, g⟩=⟨h− U ∗ h, g⟩. This holds for all g∈H, so h=U ∗ h. Now consider ∥Uh− h∥ 2 2 =⟨Uh− h, Uh− h⟩ =∥Uh∥ 2 2 −⟨ h, Uh⟩−⟨ Uh, h⟩+∥h∥ 2 2 =∥Uh∥ 2 2 −⟨ U ∗ h, h⟩−⟨ h, U ∗ h⟩+∥h∥ 2 2 . Since U is a contraction and h=U ∗ h, ∥Uh− h∥ 2 2 ≤∥ h∥ 2 2 −⟨ h, h⟩−⟨ h, h⟩+∥h∥ 2 2 =0. So Uh=h, therefore h∈M. Thus, N ⊥ ⊂ M. Now take h∈M. It follows, similar to the above argument, that U ∗ h=h, since, ∥U ∗ h− h∥ 2 2 =∥U ∗ h∥ 2 −⟨ Uh,h⟩−⟨ Uh,h⟩+∥h∥ 2 2 ≤∥ U ∗ h∥ 2 −∥ h∥ 2 2 −∥ h∥ 2 2 +∥h∥ 2 2 =0. 13 Thus, for any g∈H, ⟨h,g− Ug⟩=⟨h,g⟩−⟨ h,Ug⟩=⟨h,g⟩−⟨ U ∗ h,g⟩=⟨h− U ∗ h,g⟩=⟨0,g⟩=0, so h∈N ⊥ . Therefore M ⊂ N ⊥ . Theorem 2.1.6. The Mean Ergodic Theorem in a Hilbert Space. Let H be a Hilbert space, U : H → H be a linear contraction, M be the set {f ∈ H : Uf = f}, and P : H → M be the orthogonal projection of H onto M. Then, for all f ∈H, lim n→∞ 1 n n− 1 X k=0 U k f− Pf 2 =0. Proof. Let N be the closed linear span of {g− Ug : g ∈ H}. By Theorem 2.1.5. N = M ⊥ . Suppose first f = g− Ug for some g ∈ H. Then f ∈ N, so Pf = 0. Moreover, 1 n n− 1 X k=0 U k f 2 = 1 n (g− Ug+Ug− U 2 g+···− U n− 1 g+U n− 1 g− U n g) 2 = 1 n (g− U n g) 2 ≤ 1 n (∥g∥ 2 +∥g∥ 2 )→0. Therefore, lim n→∞ 1 n P n− 1 k=0 U k f 2 =Pf. Now suppose f ∈ N and choose a sequence {g i } in H such that f i = g i − Ug i 14 converges to f. For arbitrary ϵ> 0, choose i such that∥f− f i ∥ 2 < ϵ 2 , and choose n such that 1 n P n− 1 k=0 U k f i < ϵ 2 . Then, 1 n n− 1 X k=0 U k f 2 ≤ 1 n n− 1 X k=0 U k (f− f i ) 2 + 1 n n− 1 X k=0 U k f i 2 ≤ 1 n n− 1 X k=0 ∥U k (f− f i )∥ 2 + 1 n n− 1 X k=0 U k f i 2 = 1 n n− 1 X k=0 ∥(f− f i )∥ 2 + 1 n n− 1 X k=0 U k f i 2 =∥(f− f i )∥ 2 + 1 n n− 1 X k=0 U k f i 2 < ϵ 2 + ϵ 2 =ϵ. Thus, lim n→∞ 1 n P n− 1 k=0 U k f =0=Pf. Now consider arbitraryf ∈H. Recall thatH =N⊕ M, so we can writef =f 0 +Pf where f 0 ∈N. Then, since Pf ∈M, UPf =Pf, so, 1 n n− 1 X k=0 U k f− Pf 2 = 1 n n− 1 X k=0 U k (f 0 +Pf)− Pf 2 = 1 n n− 1 X k=0 U k f 0 + 1 n n− 1 X k=0 U k Pf− Pf 2 = 1 n n− 1 X k=0 U k f 0 2 →0. Definition 2.1.7. A linear operator U :H →H on a Hilbert space H is unitary if, for all f,g∈H, ⟨Uf,Ug⟩=⟨f,g⟩. 15 Theorem 2.1.8. Let H be a Hilbert space and U :H →H be a unitary operator. Then, for all eigenvalues λ ∈ σ (U) ={λ ∈C : U− λI has no continuous inverse}, λ ∈K. Proof. Letf ∈H be an eigenfunction with eigenvalue λ . Sincef is an eigenfunction ⟨Uf,Uf⟩=⟨f,f⟩. Therefore, ⟨f,f⟩=⟨Uf,Uf⟩=λ ¯λ ⟨f,f⟩=|λ | 2 ⟨f,f⟩, So|λ 2 |=1, thus λ ∈K. 2.2 Applications to Dynamical Systems The functional operator induced by T, referred to as U T , is of particular importance in the ergodic theorems. This operator is defined as U T f(x) = f(Tx) for functions in L 2 (X,B,µ ) or L 1 (X,B,µ ). The operator U T is also referred to as the pullback of a function. Lemma 2.2.1. Let (X,B,µ,T ) be a dynamical system and U T : L p (X,B,µ ) → L p (X,B,µ ), p≥ 1, be the operator induced by T. Then ∥f∥ p =∥U T f∥ p for all f ∈L p (X,B,µ ). Proof. By the change of variables formula, Z TX |f| p dT ∗ µ = Z X U T (|f| p )dµ. 16 Since T is a m.p.t. we have TX =X and T ∗ µ =µ , therefore, Z X |f| p dµ = Z X U T (|f| p )dµ. Since U T (|f| p )=|U T f| p . It follows that Z X |f| p dµ = Z X |U T f| p dµ. Corollary 2.2.2. Let (X,B,µ,T ) be a dynamical system and U T :L p (X,B,µ )→ L p (X,B,µ ), p ≥ 1, be the operator induced by T. Then U T is a bounded linear operator where∥U T ∥=1. Proof. This follows immediately from Lemma 2.2.1. Proposition2.2.3. Let(X,B,µ,T )beadynamicalsystemandU T :L 2 (X,B,µ )→ L 2 (X,B,µ ) be the operator induced by T. Then U T is unitary and if T is invertible U ∗ T =U T − 1. Proof. Let f,g∈L 2 (X,B,µ ) be given, then, ⟨Uf,Ug⟩= Z X f(Tx)g(Tx)dµ = Z X f(Tx)g(Tx)dT ∗ µ = Z X f(y)g(y)dµ =⟨f,g⟩. 17 Now suppose T is invertible. Since U T is unitary we have ⟨U T − 1f,g⟩=⟨U T U T − 1f,U T g⟩ =⟨f,U T g⟩. The following theorem shows how applications of U T and U T − 1 model how our uncertainty in the initial state of the system evolves over time. Theorem 2.2.4. Suppose (X,B,µ,T ) is a dynamical system, T is invertible, U T is the operator induced by T on L 1 (X,B,µ ), and f ∈ L 1 (X,B,µ ) is the probability distribution of the state of the system at time 0. Then U T f is the probability distribution of the state of the system at time -1 and U T − 1f is the probability distribution of the state of the system at time 1. In other words, P(T − 1 x∈B)= Z B U T f(x)dµ P(Tx∈B)= Z B U T − 1f(x)dµ. Proof. Letf ∈L 1 (X,B,µ ) be the probability distribution of the state of the system at time 0. Then, P(T − 1 x∈B)=P(x∈TB) = Z TB f(x)dµ (x) = Z TB f(x)dT ∗ µ (x) = Z B f(Ty)dµ (y) = Z B U T f(y)dµ (y). 18 By replacing T with T − 1 in the argument above, we have P(Tx∈B)= Z B U T − 1f(y)dµ (y). Theorem 2.2.5. The set eigenvalues of U T , σ (U T ), includes 1. If there exists an eigenfunction f that is non-zero a.e., then σ (U T ) forms a semigroup ofK. Proof. The set of eigenvalues must include 1 since, for any constant function f ∈ L 2 (X,B,µ ), Uf =f. By theorem 2.1.8. we have σ (U T )⊂ K. Suppose f and g are eigenfunctions with eigenvalues λ and ζ respectively and f(x) is non-zero a.e. Then U T g f = ζ λ g f so g f isaneigenfunctionwitheigenvalue ζ λ . SotheeigenvaluesofU T formasemigroup ofK. 2.3 Examples Example 2.3.1. U T on the Doubling Map. Consider Example 1.3.2., the doubling map. Recall that T is non-invertible. For f ∈L 1 (X,B,µ ), U T f(x)=        f(2x) if x≤ 1 2 f(2x− 1) if x> 1 2 . 19 Let f,g ∈ L 1 (X,B,µ ) and restrict U T to L 2 (X,B,µ ). Then by the substitutions 2x=y and 2x− 1=u, ⟨U T f,g⟩= Z 1 0 f(2x mod 1)g(x)dµ (x) = Z 1 2 0 f(2x)g(x)dµ (x)+ Z 1 1 2 f(2x− 1)g(x)dµ (x) = Z 1 0 f(y) g( y 2 ) 2 dµ (y)+ Z 1 0 f(u) g( u+1 2 ) 2 dµ (u) = Z 1 0 f(y) g( y 2 )+g( y+1 2 ) 2 dµ (y). Therefore,U ∗ T g = g( y 2 )+g( y+1 2 ) 2 . The probability distribution of the state of the system at time 1 is given by U ∗ T since, by the substitution 2x=y, P(Tx∈(a,b))=P(x∈( a 2 , b 2 )∪( a+1 2 , b+1 2 )) = Z b 2 a 2 f(x)dµ (x)+ Z b+1 2 a+1 2 f(x)dµ (x) = Z b a f( y 2 ) 1 2 dµ (y)+ Z b a f( y 2 )+f( y+1 2 ) 2 dµ (y) = Z b a U ∗ T f(y)dµ (y). Since T − 1 x={ x 2 , x+1 2 }, when considering probability distribution of the state of the system at time -1 we will assume all past states are equally likely, i.e. P(T − 1 x∈(a,b))= 1 2 P( x 2 ∈(a,b))+ 1 2 P( x+1 2 ∈(a,b)). 20 To observe that U T f is the probability distribution of the state of the system at time -1, we will consider the three followings cases. When 0≤ a 0 and J = (z,z+ϵ ). Notice that µ (J) = ϵ . Since T is an invertible m.p.t., µ (T n J)=ϵ for all n∈N. Let M > 1 ϵ , then, if for each m,n∈N, T n J∩T m J are disjoint, the P M k=0 µ (T k J) > 1. This is impossible since P M k=0 (T k J) ⊂ K and µ (K) = 1, so there exists m,n ∈ N,m > n, such that T n J ∩T m J is not empty. In fact, T m z ∈ T n J ∩T m J since T m J = (T m z,T m (z +ϵ )). Since T is invertible, T m− n z ∈ J∩T m− n J. Notice the interval I = [z,T m− n z] ⊂ J, therefore µ (I) < ϵ . The infinite union ∪ ∞ i=1 T i(m− n) I coversK with a collection of intervals that all have measure less than ϵ . Therefore, for any x ∈ K, there exists a y ∈ O z such that x and y both belong to an interval that has measure less than ϵ , i.e. |x− y|<ϵ . From the proposition it is believable that the only invariant functions are con- stants. Proof of this fact is in Chapter 3. We illustrate the case where α = √ 2. Consider the function f(x) = x. As n increase, 1 n P n− 1 k=0 U k T f(x) seems to converge to the constant function f(x)= 1 2 . The eigenvalues of U T are {e 2πiαn } for n ∈ Z. All the multiples of a rotation by α are eigenvalues because the function f(z) = z n is invariant under U T for a.e. z∈S 1 and n∈Z U T f(z)=f(Tz)=(Tz) n =(ze 2πiα ) n =ze 2πiαn . Clearly the spectrum of U T forms a semigroup ofK since, for any m,n∈Z, multi- plying any two eigenvalues e 2πiαm · e 2πiαn =e 2πiα (m+n) gives another multiple of the rotation. 28 Figure 2.8: Iterations of 1 n P n− 1 k=0 T k x Figure 2.9: Iteration 100 ( 1 100 P 99 k=0 T k x) 29 Chapter 3 Ergodic Systems Suppose (X,B,µ,T ) and f ∈ L 1 (X,B,µ ). In this section we will explore the behavior of the time averages of x f n (x)= 1 n n− 1 X k=0 f(T k x)= 1 n n− 1 X k=0 U k T f(x). Physicists conjectured that, for a system in equilibrium, it should be the case that f n (x) converges to R X f(x)dµ , i.e. the time average off(x) equals the space average. The mean ergodic theorem tell us that if f ∈ L 2 (X.B,µ ) then f n converges to Pf where Pf is the projection of f onto the space of invariant functions. Suppose we only know that f ∈ L 1 (X,B,µ ). What can we say about the limit of f n (x) and in what sense does it converge? Under what conditions is the limit the same for all x ∈ X? This chapter addresses these questions and starts with the study of the maximal time average of f. 30 3.1 The Maximal Ergodic Theorem Definition 3.1.1. The maximal time average of f for a given x∈X is f ∗ (x)=sup n≥ 1 1 n n− 1 X k=0 f(T k x). Theorem 3.1.2. The Maximal Ergodic Theorem. Suppose (X,B,µ,T ) is a dynamical and f ∈L 1 (X,B,µ ). For each α ∈R, Z {f ∗ >α } fdµ ≥ αµ {f ∗ >α }. In other words, if we integrate a function over the set of points where its maxi- mum average is greater than someα , then that integral is at least α times the size of the set where the integral is greater than α . To prove the maximal ergodic theorem we will use the following lemma. Lemma 3.1.3. Let (X,B,µ,T ) be a dynamical system. For f ∈ L 1 (X,B,µ ), f :X →R, Z {f ∗ ≥ 0} fdµ ≥ 0. Notice this lemma at first appears to be a special case of the maximal ergodic theorem attained by takingα =0 but in fact it is actually equivalent to the maximal ergodic theorem as shown in the following proof. 31 Proof of Theorem 3.1.2. Let g =f− α and apply the maximal ergodic theorem. 0≤ Z {g ∗ >0} gdµ = Z {f ∗ >α } (f− α )dµ = Z {f ∗ >α } fdµ − Z {f ∗ >α } αdµ = Z {f ∗ >α } fdµ − αµ {f ∗ >α }. Proof of Lemma 3.1.2. We will partition the set {f ∗ > 0} into a disjoint union of sets B i for all i ∈ N where it is easy to show R ∪ ∞ i=1 B i fdµ ≥ 0. Define the B i ’s as follows B 1 ={x:f(x)>0} B 2 ={x:f(x)≤ 0,f(x)+f(Tx)>0} . . . B n ={x:f(x)≤ 0,...,f(x)+...+f(T n− 2 x)≤ 0,f(x)+...+f(T n− 1 x)>0} . . . In other words B i for i>2 is the set of x where for all j≤ i− 2, P j k=0 f(T k x)≤ 0, i− 1 X k=0 f(T k x)>0. The set B i is the set of points where the time average of x after i− 2 units of time is non-positive and then is positive for the first time after i− 1 units of time. Note that though B technically isn’t considering the time average since there is no division by n, these statements about the sign of the sums are the same for the averages. 32 We claim that, for k∈{1,2,...,n− 1}, T k B n ⊂ B 1 ∪...∪B n− k . This is because if x∈B n then f(x)+f(Tx)+...+f(T k− 1 x)≤ 0, for all k <n and f(x)+f(Tx)+...+f(T k− 1 x)+f(T k x)+...+f(T n− 1 x)>0. Therefore, f(T k x)+...+f(T n− 1 x)>0. Let x ′ =T k x∈T k B n , then the above inequality is equivalent to the following, f(x ′ )+f(Tx ′ )+...+f(T n− k− 1 x ′ )>0. The fact that this sum is positive means x ′ ∈ B 1 ∪...∪B n− k , thus T k B n ⊂ B 1 ∪ ...∪B n− k . We claim that the sets B n ,TB n ,...,T n− 1 B n are pairwise disjoint. For i,j∈N,j > i, let B ⊂ T i B n ∩ T j B n . Then T − i B ⊂ B n ∩ T j− i B n . By the previous result, T j− i B n ⊂ B 1 ∪...∪B n− j+i , soB n ∩T j− i B n =∅ since theB k ’s are pairwise disjoint. 33 Therefore T − i B⊂ B n ∩T j− i B n =∅, so B =∅. Finally, for fixed n∈N, define the following sets, B ′ n =B n , C n =B n ∪TB n ∪...∪T n− 1 B n , B ′ n− 1 =B n− 1 \C n , C n− 1 =B ′ n− 1 ∪TB ′ n− 1 ∪...∪T n− 2 B ′ n− 1 , . . . . . . B ′ 1 =B 1 \(C 2 ∪...∪C n ), C 1 =B ′ 1 . The C n are pairwise disjoint by construction and the B ′ k ,TB ′ k ,...,T k− 1 B ′ k are also pairwise disjoint. Thus, for all n∈N, B 1 ∪...∪B n =C 1 ∪...∪C n . Z B 1 ∪...∪Bn fdµ = n X k=1 Z C k fdµ = n X k=1 Z B ′ k ∪...∪T k− 1 B ′ k fdµ = n X k=1 Z B ′ k (f +fT +...+fT k− 1 )dµ ≥ 0, since f +...+fT k− 1 > 0 on the set B ′ k ⊂ B k . Notice that∪ ∞ n=1 B n ={f ∗ > 0}, so as n increases, 1 B 1 ∪...∪Bn f converges to 1 {f ∗ >0} f. By the Dominated Convergence Theorem, 0≤ lim n→∞ Z B 1 ∪...∪Bn fdµ = Z {f ∗ >0} fdµ. 34 The maximal ergodic theorem tells us that if we integrate some function over the set of points where the maximum of its time averages is non-negative, the integral is non-negative as well. 3.2 The Ergodic Theorem AttritubedtoG.D.Birkhoff, theergodictheoremorpointwiseergodictheoremisthe most fundamental theorem of ergodic theory. Whereas the mean ergodic theorem describedconvergenceinL 2 toaprojection,thistheoremshowsthatthisconvergence isalsosensibleinL 1 andinthealmostsuresensewherethenotionofsuchprojections are not well defined. Theorem 3.2.1. The Ergodic Theorem. For (X,B,µ,T ) where T is a m.p.t. and f ∈L 1 (X,B,µ ), the following are true; (1) lim n→∞ 1 n P n− 1 k=0 f(T k x)= ¯ f(x) exists a.e.; (2) ¯ f(Tx)= ¯ f(x) a.e., i.e. U ¯ f = ¯ f; (3) ¯ f ∈L 1 and∥ ¯ f∥ 1 ≤∥ f∥ 1 ; (4) 1 n P n− 1 k=0 fT k → ¯ f in L 1 . Proof. (1): To prove this limit exists a.e. we will show the set on which the limsup and liminf of the sum disagree by a certain margin has measure zero. Define the set E α,β ={x∈X :liminf n→∞ 1 n n− 1 X k=0 f(T k x)<α<β < limsup n→∞ 1 n n− 1 X k=0 f(T k x)} 35 for α,β ∈R where α<β . If x∈{f ∗ >β } then Tx∈{f ∗ >β } as well since limsup n→∞ 1 n n− 1 X k=0 f(T k+1 x)=limsup n→∞ 1 n n− 1 X k=0 f(T k x) since the time averages only differ by one term which does not affect the limsup. So E α,β is an invariant set and a subset of{f ∗ >β }. If we restrict the domain of T to just x∈E α,β then the maximal ergodic theorem gives Z {E α,β |f ∗ >β } fdµ = Z E α,β fdµ ≥ βµ (E α,β ). For x∈E α,β there is an n such that 1 n n− 1 X k=0 f(T k x)<α so E α,β ⊂{ (− f) ∗ >− α }. Then, by the maximal ergodic theorem, Z E α,β − fdµ ≥− αµ (E α,β ) So βµ (E α,β )≤ Z E α,β fdµ ≤ αµ (E α,β ) Sinceα<β,µ (E α,β )=0.LetE ={x∈X :lim n→∞ 1 n P n− 1 k=0 f(T k+1 x) does not exist}. Notice that E = ∪ {α,β ∈Q,α<β } E α,β . Notice this is a countable union of measure 0 36 sets, therefore the limit exists a.e. (2): We will show the difference between the two limits is 0. ¯ f(x)− ¯ f(Tx)= lim n→∞ 1 n n− 1 X k=0 f(T k x)− lim n→∞ 1 n n− 1 X k=0 f(T k+1 x) = lim n→∞ 1 n (f(x)− f(T n x)) = lim n→∞ 1 n f(T n x). Note that lim n→∞ 1 n f(T n ) = 0 implies if and only if for all ϵ > 0, 1 n f(T n x)≥ ϵ for at most a finite number of n’s. Let ϵ > 0 be given. Define for all n ∈ N the set B n,ϵ = {x ∈ X : |f(x)| > nϵ }. Notice B 1,ϵ ⊃ B 2,ϵ ⊃ B 3,ϵ ⊃ .... Moreover, for all n∈N, ∥f∥ 1 = Z X |f(x)|dµ ≥ Z B 1,ϵ \B 2,ϵ |f(x)|dµ + Z B 2,ϵ \B 3,ϵ |f(x)|dµ +... Z B n− 1,ϵ \Bn,ϵ |f(x)|dµ ≥ ϵ [µ (B 1,ϵ )− µ (B 2,ϵ )]+2ϵ [µ (B 2,ϵ )− µ (B 3,ϵ )]+...+(n− 1)ϵ [µ (B n− 1,ϵ )− µ (B n,ϵ )] =ϵ [µ (B 1,ϵ +...µ (B n− 1,ϵ )] thus P ∞ n=1 µ (B n,ϵ )≤ ∥f∥ 1 ϵ . For alln∈N letC n,ϵ =T − n B n,ϵ . The setC n,ϵ is the set of x’s such that 1 n f(T n x)≥ ϵ . Since T is a m.p.t., P ∞ n=1 µ (C n,ϵ )= P ∞ n=1 µ (B n,ϵ )<∞. Thus, by the Borel-Cantelli lemma µ ({x∈X :x∈C n,ϵ i.o.})=0. 37 Therefore, µ ({x∈X :x∈C n,ϵ i.o.} c )=µ (x∈X :x∈C n,ϵ for a finite number of n’s})=1. Let A={x∈X : lim n→∞ 1 n f(T n x)=0} ={x∈X : for all ϵ> 0, 1 n f(T n x)≥ ϵ for at most a finite number of n’s} =∩ ∞ m=1 A m , where, for all m∈N, Am={x∈X : 1 n f(T n x)≥ 1 m for at most a finite number of m’s}. Notice A 1 ⊃ A 2 ⊃ ... and µ (A m )=1 for all m∈N by the above result. So µ (A)=µ (∩ ∞ n=1 A m )= lim n→∞ µ (A m )=1. Therefore, lim n→∞ 1 n f(T n x)=0 a.e. (3): Let ¯ f abs be the maximal average of the absolute value of f ¯ f abs = lim n→∞ 1 n n− 1 X k=0 |f|(T k x) = lim n→∞ 1 n n− 1 X k=0 |f(T k x)|. 38 By the triangle inequality, | ¯ f|= lim n→∞ 1 n n− 1 X k=0 f(T k x) ≤ lim n→∞ 1 n n− 1 X k=0 |f(T k x)|= ¯ f abs . By Fatou’s lemma, Z | ¯ f|dµ ≤ Z ¯ f abs dµ ≤ liminf n→∞ Z 1 n n− 1 X k=0 |f(T k x)|dµ = Z |f|dµ< ∞. (4): We will first show convergence for bounded functions in L 1 , then approximate general L 1 functions with bounded functions. Suppose first that f ∈L 1 (X,B,µ ) is bounded. Then there exists M ∈N such that f(x) < M for all x∈ X. So for all n ∈ N 1 n P n− 1 k=0 f(T k x) < M. Moreover, by (1), 1 n P n− 1 k=0 f(T k x) converges to ¯ f(x) a.e. Then by the bounded convergence theorem lim n→∞ ∥ 1 n n− 1 X k=0 f(T k x)− ¯ f(x)∥ 1 =∥lim n→∞ 1 n n− 1 X k=0 f(T k x)− ¯ f(x)∥ 1 =0. 39 Now let f ∈ L 1 (X,B,µ ) and assume f ≥ 0. Fix ϵ > 0. Choose a bounded non- negative g≤ f where∥g− f∥ 1 < ϵ 3 . Then, ∥ 1 n n− 1 X k=0 f(T k x)− ¯ f(x)∥ 1 ≤∥ 1 n n− 1 X k=0 (f(T k x)− g(T k x))∥ 1 +∥ 1 n n− 1 X k=0 g(T k x)− ¯g(x)∥ 1 +∥¯g(x)− ¯ f(x)∥ 1 . By (3), ∥¯g(x)− ¯ f(x)∥ 1 ≤∥ g(x)− f(x)∥ 1 < ϵ 3 . Since∥f∥ 1 =∥U n f∥ 1 for all n∈N, ∥ 1 n n− 1 X k=0 f(T k x)− g(T k x)∥ 1 =∥ 1 n (f(x)+f(Tx)+...+f(T n− 1 x)− g(x)− g(Tx)− ...− g(T n− 1 x))∥ 1 ≤ 1 n (∥f− g∥ 1 +∥f(Tx)− g(Tx)∥ 1 +...+∥f(T n− 1 x)− g(T n− 1 x)∥ 1 =∥f− g∥ 1 < ϵ 3 . Since g is bounded lim n→∞ 1 n n− 1 X k=0 g(T k x)− ¯g(x) 1 =0. 40 so there exists an N ∈ N such that ∥ 1 n P n− 1 k=0 g(T k x)− ¯g(x)∥ 1 < ϵ 3 for all n > N. Choose n>N, then 1 n n− 1 X k=0 f(T k x)− ¯ f(x) 1 < ϵ 3 + ϵ 3 + ϵ 3 =ϵ so lim n→∞ ∥ 1 n P n− 1 k=0 f(T k x)− ¯ f(x)∥ 1 = 0. If f is not non-negative then write f = f + − f − and apply the result to the positive and negative parts of f. Corollary 3.2.2. If A∈B is such that T − 1 A=A, then, Z A fdµ = Z A ¯ fdµ. The most useful result of this corollary is that R X fdµ = R X ¯ fdµ . Proof. Z A fdµ − Z A ¯ fdµ = Z A 1 n n− 1 X k=0 f(T k )− ¯ f ! dµ ≤ Z A 1 n n− 1 X k=0 f(T k )− ¯ f dµ = 1 n n− 1 X k=0 f(T k x)− ¯ f L 1 (A) . By (5) the last term converges to 0. 3.3 Ergodicity Ergodic systems (X,B,µ,T ) can be thought of as dynamical systems where O x visits "all" of X for almost every x ∈ X. In other words, T moves almost every 41 x ∈ X all over X in some sense to be defined. This section formulates equivalent notions of ergodicity. Definition 3.3.1. A set B∈B is called an invariant set if µ (T − 1 B∆ B)=0. Definition3.3.2. Asystem(X,B,µ,T )isergodicifeveryinvariantsethasmeasure 0 or 1. We use T is ergodic and (X,B,µ,T ) is ergodic interchangeably. Example 3.3.3. Let X = [0,1] and µ = m. We will examine three different transformations on the space (X,B,µ ). (1): Let Tx = x for all x ∈ X. Then every set is invariant under T, including sets that are not measure 0 or 1, so T is not ergodic. Moving sets around X is an essential quality of (non-trivial) ergodic transformations. (2): Let us manipulate T to introduce movement. Let Tx = x + 1 2 mod 1, a rational rotation on a circle (Example 1.2.1). Despite introducing movement, this transformation is not ergodic since for any a,b∈ [0, 1 2 ], a < b, since the set [a,b]∪ [ a+1 2 , b+1 2 ] is invariant and not of measure 0 or 1. Simply translating points back and forth is insufficient for ergodicity. (3): For α ∈ R\Q, let Tx = x + α mod 1, an irrational rotation on a circle (Example 1.2.1). It turns out this transformation is ergodic but its difficult to show this directly simply because characterizing invariant sets is hard. Clearly TX = X is invariant so all sets equivalent toX mod 0 are measure1 and invariant. Likewise, for all x∈ [0,1], O x is invariant by definition. Each O x is a countably infinite set of points so, for all x∈ [0,1], µ (O x ) = 0. Likewise, all countably infinite unions of orbits are invariant and have measure0. It is not immediately clear if other invariant sets exist. Checking all the invariant sets is often too difficult a task for non-trivial systems. There are multiple equivalent conditions to ergodicity that are much easier to check in practice. 42 Theorem 3.3.4. For (X,B,µ,T ) the following are equivalent: (1) T is ergodic. (2) All invariant measurable functions on X are constant a.e. (3) The eigenvalues of the transformation U T induced on L 2 (X,B,µ ) are all sim- ple, i.e. the eigenspace of the eigenvalues has dimension 1. (4) For all f ∈ L 1 (X,B,µ ) and a.e. x ∈ X, ¯ f(x) := lim n→∞ 1 n P n− 1 k=0 f(T k x) = R X fdµ. (5) For each f,g∈L 2 (X,B,µ ) we have lim n→∞ 1 n P n− 1 k=0 ⟨U k T f,g⟩=⟨f,1⟩⟨1,g⟩. Proof. (1) ⇐⇒ (2): First, assume all invariant measurable functions are constant almost everywhere. Let E∈B be invariant. Then 1 E (x) is invariant so it is constant a.e. and therefore is either 1 or 0 almost everywhere, implying µ (E)=0 or 1. Now assume (X,B,µ,T ) is ergodic and f is an invariant measurable function. For all a ∈ R, define the set E a = {x ∈ X : f(x) > a}. Since f is invariant for all a∈R each E a is invariant as well, so µ (E a ) = 0 or 1. If f is not constant almost everywhere there exists a∈R where 0<µ (E a )<1. This contradicts the ergodicity of the system, therefore f must be constant almost everywhere. (2) ⇐⇒ (3): Supposeallinvariantmeasurablefunctionsareconstanta.e.,thismeanstheeigenspace of the eigenvalue 1 has dimension 1. Thus 1 is simple. Recall by Theorem 2.2.5. that U T ’s eigenvalues include 1 and form a semigroup of K if a nonzero a.e. f exists. If f is an eigenfunction with eigenvalue λ , then U|f|=|Uf|=|λf |=|f|. 43 Since invariant functions are constant a.e. |f| is a nonzero constant a.e. therefore f ̸=0 a.e. All the eigenvalues are simple since if some eigenfunctions f,g both have eigenvalue λ then U g f = Ug Uf = λ λ g f = g f is an invariant function and thus is constant a.e. Now suppose the eigenvalues of the transformation U T induced on L 2 (X,B,µ ) by T are all simple. Then the eigenvalue 1 is simple so U T f =f is constant a.e. (2) =⇒ (4): Suppose all invariant measurable functions are constant a.e. Let f ∈ L 1 (X,B,µ ) be given. Since ¯ f is invariant it is constant a.e. By Corollary 3.2.2., Z X fdµ = Z X ¯ fdµ = ¯ fµ (X)= ¯ f(x) for a.e. x. (4) =⇒ (1): Suppose for all f ∈L 1 (X,B,µ ), ¯ f is constant a.e. Suppose A⊂ B is invariant and consider 1 A (x). For all n∈N and a.e. x∈X, ¯ 1 A (x)= 1 n n− 1 X k=0 1 A (T k x)= 1 n n− 1 X k=0 1 A (x)= 1 A (x). But ¯ 1 A (x) is constant a.e. so 1 A is constant a.e. so µ (A) = 0 or 1. Therefore T is ergodic. 44 (4) ⇐⇒ (5): Suppose for all f ∈L 1 (X,B,µ ) and a.e. x∈X, ¯ f(x)= R X fdµ. Then lim n→∞ 1 n n− 1 X k=0 ⟨U k f,g⟩=⟨ ¯ f,g⟩ = Z X fdµ,g =⟨⟨f,1⟩,g⟩ =⟨f,1⟩⟨1,g⟩ Now suppose for all f,g∈ L 2 (X,B,µ )⊂ L 1 (X,B,µ ) we have⟨f,g⟩ =⟨f,1⟩⟨1,g⟩. Then let f be given and let g =1 a.e. Then we have ⟨f,1⟩=⟨f,1⟩⟨1,1⟩=⟨f,1⟩= Z X fdµ a.e. Notice that (4) in Theorem 3.3.4. shows that the physicists’ hypothesis about the interchangeability of time and space averages holds when T is ergodic. Corollary 3.3.5. If T is ergodic then, for all A,B∈B, lim n→∞ 1 n n− 1 X k=0 µ (T − k A∩B)=µ (A)µ (B). 45 Proof. Suppose T is ergodic. Let A,B ∈ B be given and consider f = 1 A and g = 1 B . Then, lim n→∞ 1 n n− 1 X k=0 ⟨U k T 1 A (x), 1 B (x)⟩= lim n→∞ 1 n n− 1 X k=0 ⟨1 A (T k x), 1 B (x)⟩ = lim n→∞ 1 n n− 1 X k=0 ⟨1 T − k A (x), 1 B (x)⟩ = lim n→∞ 1 n n− 1 X k=0 µ (T − k A∩B) On the other hand, by (5) in theorem 3.3.4., lim n→∞ 1 n n− 1 X k=0 ⟨U k T 1 A (x), 1 B (x)⟩=⟨1 A (x),1⟩⟨1, 1 B (x)⟩ = Z X 1 A dµ Z X 1 B dµ =µ (A)µ (B). Therefore, lim n→∞ 1 n n− 1 X k=0 µ (T − k A∩B)=µ (A)µ (B). 3.4 Examples Example 3.4.1. Irrational Rotations on a Circle are Ergodic. In Example 2.3.3. it appeared that 1 n P n− 1 k=0 U k T f(x) converged to the invariant mea- surable function f(x) = 1 2 . This is because irrational rotations on a circle are ergodic, therefore, by Theorem 3.3.4., all invariant measurable functions on X are constant a.e. To prove this we will consider the additive representation of rotation 46 on a circle. Assume f is invariant on [0,1]. Write f(z)= P n∈Z a n e 2πizn as a Fourier series. Since f is invariant, f(Tz)=f(z), i.e. X n∈Z a n e 2πi (z+α )n = X n∈Z a n e 2πizn . So a n e 2πiαn = a n for all n∈Z. But α is irrational so e 2πiαn = 1 only when n = 0. Therefore a n =0 for all n∈Z\{0}. So f(z)=f(Tz)=a 0 is constant a.e. Example 3.4.2. Rational Rotations on a Circle are Not Ergodic. This follows immediately from (2) in Theorem 3.3.4. Example 2.3.2. describes how to construct non-constant invariant functions on K when α ∈Q. 47 Chapter 4 Mixing Systems Suppose you have a cup of black coffee mixed with a proportionate amount of milk. As you pour the milk into the coffee, gravity causes the two fluids to homogenize. This process results in a change from the coffee’s original deep black color to a more subdued brown hue, owing to the dispersion of the white milk throughout the coffee. These fluids are mixing. While ergodic suggests every orbit visits all parts of X, mixing suggests the orbit also mixes up X. 4.1 Weak Mixing The definition of weak mixing differs subtly from that of ergodicity. Definition 4.1.1. A dynamical system (X,B,µ,T ) is weak mixing if lim n→∞ 1 n n− 1 X k=0 |µ (T − k A∩B)− µ (A)µ (B)|=0 An equivalent formulation is lim n→∞ 1 n n− 1 X k=0 | µ (T − k A∩B) µ (B) − µ (A)|=0 48 for all A,B ∈B and µ (B) > 0. Inline with the coffee cup analogy, weak mixing systems guarantee that, on average, the cup’s contents exhibit uniformity. Proposition 4.1.2. Let (X,B,µ,T ) be a dynamical system. Then T is weak mixing implies T is ergodic. Proof. Suppose (X,B,µ,T ) is weak mixing. Let A∈B be an invariant set. Then, µ (A) 2 = lim n→∞ 1 n n− 1 X k=0 µ (T − k A∩A) = lim n→∞ 1 n n− 1 X k=0 µ (A) =µ (A), so µ (A)=0 or 1. If T is ergodic then by Corollary 3.3.5. we have for all A,B∈B, lim n→∞ 1 n n− 1 X k=0 (µ (T − k A∩B)− µ (A)µ (B))=0. Whereas if T is weak mixing, we have, lim n→∞ 1 n n− 1 X k=0 |µ (T − k A∩B)− µ (A)µ (B)|=0. The key to understanding the difference between ergodic and weak mixing trans- formations is understanding the difference between the nature of sequences {a n } ∞ n=0 where 1 2 P ∞ n= a n =0 and sequences{b n } ∞ n=0 where 1 n P ∞ n=0 |b n |=0. In the first case, the a n need not ever be close to zero because cancellation is possible. For example, the sequence{a n } ∞ n=0 where, for all n∈N, a n =(− 1) n has this property. 49 Example 4.1.3. Consider the dynamical system of three atoms X ={a,b,c} where µ (a)=µ (b)=µ (c)= 1 3 . Let Ta=b,Tb=c, and Tc=a. Let A=B ={a}. Then, 1 n n− 1 X k=0 (µ (T − k {a}∩{a})− µ ({a}) 2 )= 1 n n− 1 X k=0 µ (T − k {a}∩{a})− 1 9 = 1 n 2 9 − 1 9 − 1 9 + 2 9 − 1 9 − 1 9 +... , since whenever n=3m for m∈N, µ (T − n {a}∩{a})= 1 3 , otherwise this intersection is 0. Thus, as n increases this sum tends to 0. What kinds of sequences{b n } ∞ n=0 have 1 n P n− 1 k=0 |b n |=0. Theb n need not converge pointwise to zero. For example, the sequence{1,0,1,0,0,1,0,0,0,1,0,...} has this property. The following definition helps us characterize such sequences. Definition 4.1.4. A set E⊂ N has density zero if lim n→∞ 1 n n− 1 X k=0 1 E (k)=0. A familiar example of such a set is the set of all primes. It can be shown that the set of primes has density zero using the Prime Number Theorem. The Fibonacci sequence {1,1,2,3,5,8,13,...} has density zero as well. These sets become incredibly sparse in N. A set that doesn’t have density zero has entries that occur relatively frequently. One such set is {1,2,3,4,5,11,12,13,14,15,21,22,23,24,25,...} 50 where every other 5 natural numbers are excluded. However, if the gap between the n th and (n+1) th block of 5 numbers increases by 2 n− 1 · 5 {1,2,3,4,5,11,12,13,14,15,26,27,28,29,30,51,...} the set has density zero. A set can have dense pockets and still be density zero so long as the gaps between said pockets tend to increase as n increases. Lemma 4.1.5. Let f :N→N be non-negative and bounded. Then lim n→∞ 1 n n− 1 X k=0 f(k)=0 if and only if there exists a subset E⊂ N of density zero such that lim n→∞ 1 E c(n)f(n)=0. Proof. First we will suppose such an E exists. We can rewrite the average value of f by splitting the sum into the indices in and out of E, 1 n n− 1 X k=0 f(k)= 1 n X k≤ n− 1 1 E (k)f(k)+ 1 n X k≤ n− 1 1 E c(k)f(k). Since f is bounded there exists M ∈N such that f(k) < M for all k∈N. The set E has density zero so, for any ϵ > 0 there exists N 1 ∈N such that, for all n > N 1 , we have, 1 n n− 1 X k=0 1 E (k)< ϵ 2M . 51 Likewise, by our hypothesis, lim n→∞ 1 E c(n) 1 n f(n) = 0. Therefore, for any ϵ > 0, there exists a N 2 ∈N such that, for all n>N 2 , we have, 1 n n− 1 X k=0 1 E c(k)f(k)< ϵ 2 Let N =max{N 1 ,N 2 }, and consider n>N. Notice 1 n n− 1 X k=0 f(k)= 1 n n− 1 X k=0 1 E (k)f(k)+ 1 n n− 1 X k=0 1 E c(k)f(k) ≤ M 1 n n− 1 X k=0 1 E (k)+ 1 n n− 1 X k=0 1 E c(k)f(k) <M ϵ 2M + ϵ 2 =ϵ. Now suppose the average of the function converges to 0. Begin by defining the set E m = {k ∈ N|f(k) > 1 m } for all m ∈ N. Then each E m has density zero since 1 m 1 Em <f and E 1 ⊂ E 2 ⊂ ... so, lim n→∞ 1 n n− 1 X k=0 1 Em (k)<m· lim n→∞ 1 n n− 1 X k=0 f(k)=m· 0=0. For each m∈N we can choose an i m >0 such that i 1 1 m so lim n→∞ 1 E c(n)f(n)=0. E has density zero since, for i m− 1 <n≤ i m , 1 n n− 1 X k=0 1 E (k)= 1 n i m− 1 − 1 X k=0 1 E (k)+ 1 n n− 1 X k=i m− 1 1 E (k) ≤ 1 i m− 1 i m− 1 − 1 X k=0 1 E m− 1 (k)+ 1 n n− 1 X k=0 1 Em(k) < 1 m− 1 + 1 m . Corollary 4.1.6. Let (X,B,µ,T ) be a dynamical system. The transformation T is weak mixing if and only if, given A,B ∈B, there exists a set J ⊂ Z + of density 0 such that lim n→∞,n/ ∈J µ (T n A∩B)=µ (A)µ (B). Proof. We will use Lemma 4.1.5. For given A,B∈B the function f(n)=|µ (T n A∩B)− µ (A)µ (B)| is non-negative and bounded by 2. If T is weak mixing then, lim n→∞ 1 n n− 1 X k=0 |µ (T n A∩B)− µ (A)µ (B)|=0, 53 therefore lim n→∞ P n− 1 k=0 f(k) = 0. By Lemma 4.1.5., there exists E ⊂ N of density zero such that lim n→∞ 1 E c(n)f(n)=0. Likewise, if said E of density zero exists, then Lemma 4.1.5. implies, 0= lim n→∞ n− 1 X k=0 f(k) = lim n→∞ 1 n n− 1 X k=0 |µ (T k A∩B)− µ (A)µ (B)|. Theorem 4.1.7. Let(X,B,µ,T ) be a dynamical system. Then the following state- ments are all equivalent: (1) T is weakly mixing. (2) T × T is weakly mixing. (3) T × S is ergodic on X× Y for each ergodic (Y,C,ν,S ). (4) T × T is ergodic. Proof. (1) =⇒ (2): Let A,B,C,D ∈B be given. By Corollary 4.1.5., there exists sets of density zero J 1 ,J 2 ∈N such that lim n→∞,n/ ∈J 1 |µ (T n A∩C)− µ (A)µ (C)|=0 lim n→∞,n/ ∈J 2 |µ (T n B∩D)− µ (B)µ (D)|=0 54 Then J =J 1 ∪J 2 is of density zero and lim n→∞,n/ ∈J |µ × µ ((T × T) n ((A× B)∩(C× D)))− µ × µ (A× B) µ × µ (C× D))| = lim n→∞,n/ ∈J |µ (T n A∩C)µ (T n B∩D)− µ (A)µ (B)µ (C)µ (D)| ≤ lim n→∞,n/ ∈J [µ (T n A∩C)|µ (T n B∩D)− µ (B)µ (D)| +µ (B)µ (D)|µ (T n A∩C)− µ (A)µ (C)|] ≤ lim n→∞,n/ ∈J [|µ (T n B∩D)− µ (B)µ (D)|+|µ (T n A∩C)− µ (A)µ (C)|] =0. Therefore, there exists a set J of density zero where lim n→∞,n/ ∈J µ (T n A∩C)µ (T n B∩D)=µ (A)µ (B)ν (C)ν (D), so the hypothesis of Corollary 4.1.5. is also true for T× T, therefore T× T is weak mixing. (2) =⇒ (3): If T × T is weak mixing then so is T since, given A,B∈B, 0= lim n→∞ 1 n n− 1 X k=0 |µ × µ (T × T) k (A× X)∩(B× X)− µ (A)µ (B)µ (X) 2 | = lim n→∞ 1 n n− 1 X k=0 |µ (T k A∩B)µ (T k X∩X)− µ (A)µ (B)| = lim n→∞ 1 n n− 1 X k=0 |µ (T k A∩B)− µ (A)µ (B)|. 55 Let (Y,C,ν,S ) be an ergodic system. To show T × S is ergodic on X× Y we will show that for A,B∈B and C,D∈C we have lim n→∞ 1 n n− 1 X k=0 µ × ν [(T × S) k (A× C)∩(B× D)]=µ (A)µ (B)µ (C)µ (D). The limit is equivalent to lim n→∞ 1 n n− 1 X k=0 µ (T k A∩B) ν (S k C∩D) = 1 n n− 1 X k=0 (µ (A)µ (B) ν (S k C∩D)+[µ (T k A∩B)− µ (A)µ (B)]ν (S k C∩D)). By the ergodicity of S, the first term converges to µ (A)µ (B)ν (C)ν (D). The second term is bounded by 1 n n− 1 X k=0 |µ (T k A∩B)− µ (A)µ (B)| which converges to zero since T is weakly mixing. Therefore, lim n→∞ 1 n n− 1 X k=0 µ (T k A∩B)ν (S k C∩D)=µ (A)µ (B)ν (C)ν (D). (3) =⇒ (4): Consider a system where Y = {1}, ν ({1}) = 1, and S(1) = (1). This system is 56 ergodic because its only invariant set has measure 1. So by (3) T × S is ergodic. Then for A,B∈X, 0= lim n→∞ 1 n n− 1 X k=0 µ (T − k A∩B)ν (T − k {1}∩{1})− µ (A)µ (B)ν ({1}) 2 = lim n→∞ 1 n n− 1 X k=0 µ (T − k A∩B)− µ (A)µ (B). Since T is ergodic, T × T is ergodic by (3). (4) =⇒ (1): Let A,B∈B be given, then, 1 n n− 1 X k=0 [µ (T k A∩B)− µ (A)µ (B)] 2 = 1 n n− 1 X k=0 µ (T k A∩B) 2 − 2µ (A)µ (B) 1 n n− 1 X k=0 µ (T k A∩B)+[µ (A)µ (B)] 2 = 1 n n− 1 X k=0 µ × µ [(T × T) k (A× A)∩(B× B)] − 2µ (A)µ (B) 1 n n− 1 X k=0 µ × µ [(T × T) k (A× X)∩(B× X)]+[µ (A)µ (B)] 2 . By the ergodicity of T × T, the limit of the above quantity is µ × µ (A× A)µ × µ (B× B)− 2µ (A)µ (B)µ × µ (A× X)µ × µ (B× X)+[µ (A)µ (B)] 2 =µ (A) 2 µ (B) 2 − 2µ (A) 2 µ (B) 2 +µ (A) 2 µ (B) 2 =0. By Lemma 4.1.5., there exists a set J of density zero where lim n→∞,n/ ∈J |µ (T n A∩B)− µ (A)µ (B)| 2 =0. 57 Theorem 4.1.8. Let(X,B,µ,T ) be a dynamical system. Then the following state- ments are all equivalent: (1) T is weak mixing. (2) lim n→∞ 1 n P n− 1 k=0 |⟨U k T f,g⟩−⟨ f,1⟩⟨1,g⟩|=0 for all f,g∈L 2 (X,B,µ ). (3) All of U T ’s measurable eigenfunctions are constant. In other words, U only has the single eigenvalue 1 and it is a simple eigenvalue. Proof. (1) ⇐⇒ (2) : Suppose T is weak mixing. We will first show that (2) is true for characteristic functions of measurable sets then use linear combinations of these functions to approximate general functions. Let A,B ∈B and consider the indicator functions 1 A and 1 B . Notice, lim n→∞ 1 n n− 1 X k=0 |µ (T k A∩B)− µ (A)µ (B)| = lim n→∞ 1 n n− 1 X k=0 |⟨U k T 1 A , 1 B ⟩−⟨ 1 A ,1⟩⟨1, 1 B ⟩| = lim n→∞ 1 n n− 1 X k=0 |⟨U k T 1 A , 1 B ⟩−⟨ U k T 1 A ,1⟩⟨1, 1 B ⟩| = lim n→∞ 1 n n− 1 X k=0 Z 1 T − k A 1 B dµ − Z 1 T − k A dµ Z 1 B dµ =0. 58 Now fix f,g ∈ L 2 (X,B,µ ). The result follows from approximating f and g with linear combinations of characteristic functions. Now suppose (2). Then for any measurable sets A,B∈B we have, 0= lim n→∞ 1 n n− 1 X k=0 |⟨U k T 1 A , 1 B ⟩−⟨ 1 A ,1⟩⟨1, 1 B ⟩ = lim n→∞ 1 n n− 1 X k=0 |⟨1 T − k A , 1 B ⟩−⟨ 1 A ,1⟩⟨1, 1 B ⟩| = lim n→∞ 1 n n− 1 X k=0 |µ (T − k A∩B)− µ (A)µ (B)|. Proof. (2)⇒(3): Let f ∈L 2 be an eigenfunction with eigenvalue λ . For (x,y)∈X× X, let g(x,y)= f(x) ¯ f(y). Then, U T× T g(x,y)=g(Tx,Ty)=f(Tx) ¯ f(Ty)=λ ¯λg (x,y)=g(x,y) a.e. Therefore g is an invariant function. Since T× T is ergodic g must be constant a.e. So for some c∈R and all (x,y)∈ X× X, g(x,y) = c = f(x) ¯ f(y). Let x 1 ,x 2 ∈ X. Then c=g(x 1 ,y)=g(x 2 ,y) =f(x 1 ) ¯ f(y)=f(x 2 ) ¯ f(y) therefore f(x 1 )=f(x 2 ) for arbitrary x 1 ,x 2 ∈X, so f is constant a.e. Before we show how (3) implies (2) recall the spectral theorem. 59 Theorem 4.1.9. The Spectral Theorem. Let U be a self-adjoint operator on a separable Hilbert space H. Then, there exists a operator-valued measureE onC called the spectral measure that has the following properties: (1) E(C)= identity. (2) For measurable, disjoint A i ⊂ C and f ∈H, E(∪ ∞ i=1 A i )(f)= P ∞ i=1 (EA i )(f). (3) The support of E is the spectrum of U, σ (U). (4) U k = R σ (T) λ k dE λ ; (5) For all f,g∈H,⟨U k f,g⟩= R σ (T) λ k dµ f,g where µ f,g (λ )=⟨E(λ )f,g⟩. We write U = Z C λdE λ . Proof. (3)⇒(2): Let M be the closed linear span in L 2 (X) of the eigenfunctions of U. We begin by showing that for f ∈M ⊥ and g∈L 2 (X) we have lim n→∞ 1 n n− 1 X k=0 |⟨U k T f,g⟩| 2 =0. Consider for a fixed f ∈M ⊥ and g∈L 2 (X) µ f,g . For any λ ∗ ∈C we have UE({λ ∗ })f = Z σ (U) λ 1 λ ∗ dE λ f =λ ∗ Z σ (U) 1 λ ∗ dE λ f =λ ∗ E(λ ∗ )f 60 therefore E({λ ∗ })f is an eigenfunction of U or zero. Since f ∈M ⊥ 0=⟨E({λ ∗ })f,f⟩ =⟨E({λ ∗ }) 2 f,f⟩ =⟨E({λ ∗ })f,E({λ ∗ }) ∗ f⟩ =⟨E({λ ∗ })f,E({λ ∗ })f⟩, therefore E({λ ∗ })f =0, so µ ({λ })=0 for all λ ∈C. Now 1 n n− 1 X k=0 |⟨U k f,g⟩| 2 = 1 n n− 1 X k=0 | Z σ (U)) λ k d(⟨E(λ )f,g⟩)| 2 = 1 n n− 1 X k=0 | Z σ (U)) λ k dµ (λ )| 2 = 1 n n− 1 X k=0 ( Z σ (U) λ k dµ (λ ))( Z σ (U) ¯ζ k d¯µ (ζ )) = 1 n n− 1 X k=0 Z Z σ (U)× σ (U) λ k ¯ζ k dµ (λ )d¯µ (ζ ) = Z Z σ (U)× σ (U) ( 1 n n− 1 X k=0 λ k ¯ζ k )dµ (λ )d¯µ (ζ ) = Z Z σ (U)× σ (U) 1 n 1− (λ ¯ζ ) n 1− λ ¯ζ dµ (λ )d¯µ (ζ ). This integral is undefined if the set of points where λ = ζ and hence λ ¯ζ = 1 has positive measure. However, µ assigns 0 measure to all singletons so the set of points where λ =ζ has measure 0. The integrand is bounded by 1 n 1− (λ ¯ζ ) n 1− λ ¯ζ 61 which goes to 0 a.e. By the bounded convergence theorem we have lim n→∞ 1 n n− 1 X k=0 |⟨U k f,g⟩| 2 =0. By Lemma 4.1.5. lim n→∞ 1 n n− 1 X k=0 |⟨U k f,g⟩|=0 for f ∈M ⊥ and g∈L 2 (X). Considerf,g∈L 2 (X). Since the only measurable eigenfunctions ofU are constants, we have f−⟨ f,1⟩∈M ⊥ . Therefore, 0= lim n→∞ 1 n n− 1 X k=0 |⟨U k f−⟨ f,1⟩,g⟩| = lim n→∞ 1 n n− 1 X k=0 |⟨U k f,g⟩−⟨ f,1⟩⟨1,g⟩|. The fact that all of U T ’s eigenfunctions are constant is an incredibly deep re- sult that reveals much of the underlying structure of weak mixing transformations. Unlike ergodic transformations whose eigenvalues can be any point on the unit circle, the only eigenvalue a weak mixing transformation can have is 1 since all measurable eigenfunctions of weak mixing systems are constant. Whereas the spec- trum of ergodic and not weak transformations form a semigroup ofK and can span L 2 (X,B,µ ), the spectrum of a weak mixing transformation necessarily cannot span L 2 (X,B,µ ). 62 4.2 Examples Example 4.2.1. It is not unreasonable to suspect T is ergodic implies T × T is also ergodic but Theorem 4.1.7. suggests T needs to be weak mixing for this to be the case. Consider the system of three atoms such as in Example 4.1.3. Then T is ergodic since the only invariant set is X which has measure 1. However, T × T is not ergodic. Consider what happens when we transform the set a× a. (T × T)(a× a)=b× b, (T × T)(b× b)=c× c, (T × T)(c× c)=a× a. Then, for all k∈N, (T × T) 3k+1 (a× a)=b× b, (T × T) 3k+2 (a× a)=c× c, (T × T) 3k (a× a)=a× a. Now consider the set S =(a× a)∪(b× b)∪(c× c). S is an invariant set since (x,y)∈S if and only if (x,y) belongs to one of its respec- tivesubsets, i.e. (x,y)∈(a× a),(b× b), or(c× c). Sinceµ × µ [(a× a)∪(b× b)∪(c× c)]= 1 3 ̸=0 or 1, T × T is not ergodic. 63 The issue with T is that it has a rotational component. Having a rotational compo- nent is what lead to difficulties with T × T "getting stuck" in one part of X× X. The transformation T × T can be thought of as having 3 ergodic components, (a× a)∪(b× b)∪(c× c) (a× b)∪(b× c)∪(c× a) (a× c)∪(b× a)∪(c× b), each isomorphic to (X,B,µ,T ). a× a b× b c× c a× b b× c c× a a× c b× a c× b Figure 4.1: Ergodic Components of T × T Example 4.2.2. Irrational Rotations on a Circle are not Weak Mixing. Let (X,B,µ,T ) be such as in Example 1.3.2. with α ∈ R\Q. Intuitively, this system should not be weak mixing because rotations obviously have a rotational element. Rotating a set only translates it, it does not "mix" the contents of the set up. We will show this result using Theorem 4.1.7. to show T × T is not ergodic 64 implies T is not weak-mixing. Consider the system (K× K,B× B,µ × µ,T × T). For (x 1 ,x 2 )∈K× K, (T × T)(x 1 ,x 2 )=(x 1 e 2πiα ,(x 2 e 2πiα )). The following function is invariant and non-constant a.e.: f(x 1 ,x 2 )= x 1 x 2 (T × T)f(x 1 ,x 2 )= x 1 e 2πiα x 2 e 2πiα = x 1 x 2 , therefore T × T is not ergodic so T is not weak mixing. 4.3 Strong Mixing Definition4.3.1. Let(X,B,µ,T )beadynamicalsystem. Thesystemisstrong mixing if lim n→∞ |µ (T − n A∩B)− µ (A)µ (B)|=0. Clearly strong mixing implies weak mixing. Equivalently, a system is strong mixing if lim k→∞ µ (T − k A∩B) µ (B) =µ (A) for all A,B ∈ B where µ (B) > 0. Imagine you have a cup of coffee with milk, and you are very particular. You cannot enjoy your drink unless it is perfectly homogeneous; no pockets of milk or coffee. To achieve this, you use a spoon to carefully stir the mixture continuous in hopes that it becomes entirely uniform. Unlike the result of simply pouring milk into coffee, this method guarantees in the 65 long run that no pockets of milk remain, ensuring each sip contains an equal amount of milk and coffee, regardless of the sip size. You could even take a straw and sip from any region of the cup, after enough time your drink will be just how you like it. When we say T is strong mixing, it signifies that, over time, the transformation T will evenly distribute the points in set A throughout the space. The proportion of points that map from A to B will be approximately the same as the overall proportion of points in A. Theorem 4.3.2. Let (X,B,µ,T ) be a dynamical system. Then T is strong mixing if and only if, for each A∈B, lim n→∞ µ (T − n A∩A)=µ (A) 2 . Proof. IfT isstrongmixingthenchoseAandB tobethesameset; theresultsfollows immediately. Now suppose that for each A∈B,lim n→∞ µ (T − n A∩A)=µ (A) 2 . Fix A ∈ B. Define M to be the closed linear subspace of L 2 (X,B,µ ) of constant functions along with{U k 1 A :k∈Z}. Then the following two identities hold; lim n→∞ ⟨U n 1 A ,1⟩= lim n→∞ µ (A)=⟨1 A ,1⟩⟨1,1⟩. lim n→∞ ⟨U n 1 A ,U k 1 A ⟩= lim n→∞ ⟨U n− k 1 A , 1 A ⟩= lim n→∞ µ (T − n+k A∩A) =µ (A) 2 =⟨1 A ,1⟩⟨U k 1 A ,1⟩. 66 Every f ∈ M takes the form f = c+ P k∈Z a k U k 1 A where c,a k ∈C. So combining the above identities gives lim n→∞ ⟨U n 1 A ,f⟩= lim n→∞ ⟨U n 1 A ,c+ X k∈Z a k U k 1 A ⟩ = lim n→∞ ⟨U n 1 A ,c⟩+ X k∈Z ¯a k ⟨U n 1 A ,U k 1 A ⟩ =⟨1 A ,1⟩⟨1,c⟩+ X k∈Z ¯a k ⟨1 A ,1⟩⟨1,U k 1 A ⟩ =⟨1 A ,1⟩(⟨1,c⟩+ X k∈Z ¯a k ⟨1,U k 1 A ⟩) =⟨1 A ,1⟩⟨1,c+ X k∈Z a k U k 1 A ⟩ =⟨1 A ,1⟩⟨1,f⟩ for all f ∈M. Any arbitrary f ∈L 2 (X,B,µ ) can be written as f =f M +f M ⊥ where f M ∈M and f M ⊥ ∈M ⊥ . Then lim n→∞ ⟨U n 1 A ,f⟩= lim n→∞ ⟨U n 1 A ,f M ⟩+ lim n→∞ ⟨U n 1 A ,f M ⊥⟩ =⟨1 A ,1⟩⟨1,f M ⟩ =⟨1 A ,1⟩⟨1,f⟩ Recall that ⟨U k 1 A ,f M ⊥⟩ = 0 and ⟨1,f M ⊥⟩ = 0 by definition of f M ⊥ ∈ M ⊥ . Then, for any B∈B and f = 1 B , lim n→∞ µ (T − n A∩B)= lim n→∞ ⟨U n 1 A , 1 B ⟩=⟨1 A ,1⟩⟨1, 1 B ⟩=µ (A)µ (B). 67 Example 4.3.3. Irrational Rotations on a Circle are not Strong Mixing. Let (X,B,µ,T ) be such as in Example 1.3.2. and let α ∈R\Q. Since we already showed in Example 4.2.2. that this system is not weak mixing we know it cannot be strongly mixing. A direct proof that these rotations are not strongly mixing is still useful to understand what properties these systems have that are not conducive with strong mixing. Let A = [0, 1 8 ] and B = [ 1 2 , 3 4 ]. Then µ (A) = 1 8 and µ (B) = 1 4 . Since O A is dense inK, there exists a sequence of integers{a n } such that T − an A⊂ B, so, µ (T − an A∩B)= 1 8 . Therefore, lim n→∞ µ (T − an A∩B)= 1 8 . Then one of the following must be true: (1): lim n→∞ µ (T − n A∩B) does not exist; (2): lim n→∞ µ (T − n A∩B)= 1 8 . If the limit does not exist T is not strong-mixing. Likewise, should the limit be 1 8 , T is not strong-mixing since µ (A)µ (B)= 1 32 ̸= 1 8 . It is not trivial to find a system that is weak but not strong mixing. Despite this, with respect to the weak topology, the property of being weak mixing is generic. 4.4 Examples Example 4.4.1. Bernoulli Shifts are Strong Mixing. Consider the Bernoulli shift as defined in Example 1.3.4. We can show T is strong mixing by first showing the mixing property is satisfied for all cylinder sets A,B∈ A Z . SupposeA andB are cylinder sets. By definition the values {H,T} are specified in the sequences in only a finite number of entries. Therefore, there exists a k,N ∈ 68 N,k > N, such that the specified entries in the sequences in T k A and B do not overlap anywhere. Thus, for all k >N, we have µ (T k A)µ (B)=µ (A)µ (B) therefore lim k→∞ µ (T k A)µ (B)=µ (A)µ (B). Since cylinder sets generateB, for any ϵ> 0 we can approximate any non-cylinder sets up to measure ϵ by unions of cylinder sets. Example 4.4.2. The Baker’s Transform and Doubling Map are Strong Mixing The content of this example is outside the scope of this work but is interesting nevertheless. Both the baker’s transform and the doubling map are measure theo- retically isomorphic to the Bernoulli shift defined in Example 1.3.4., therefore both the baker’s transform and the doubling map are strong mixing. All three of these maps are Kolmogorov and Bernoulli as well. 69 References [1] Patrick Billingsley. Ergodic theory and information. Vol. 1. Wiley New York, 1965. [2] Leonid A Bunimovich, Shrikrishna Gopalrao Dani, Roland Lvovich Dobrushin, Michael V Jakobson, Isaak Pavlovič Kornfeld, Nina Borisovna Maslova, Ya B Pesin, J Smillie, Yu M Sukhov, and AM Vershik. Dynamical systems, ergodic theory and applications. Vol. 100. Springer Science & Business Media, 2000. [3] Andrés Del Junco. “A simple measure-preserving transformation with trivial centralizer”. In: Pacific Journal of Mathematics 79.2 (1978), pp. 357–362. [4] Matthew Nicol and Karl Petersen. “Ergodic theory: basic examples and constructions”. In: Mathematics of Complexity and Dynamical Systems. 2009. [5] Karl E Petersen. Ergodic theory. Vol. 2. Cambridge university press, 1989. [6] Peter Walters. An introduction to ergodic theory. Vol. 79. Springer Science & Business Media, 2000. 70

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Creator Fantini, Joseph Antonio (author)

Core Title Analysis of ergodic and mixing dynamical systems

School College of Letters, Arts and Sciences

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Degree Conferral Date 2023-08

Publication Date 08/10/2023

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