Differential equations: Dynamics and Chaos. Pierre Berger

Transcription

1 Differential equations: Dynamics and Chaos Pierre Berger

2 2

3 Contents 1 Introduction What is a dynamical system? Plan Phase spaces, regularity of the dynamics and basic tools Phase spaces Regularity of the maps of these spaces Orbits and dynamical system Some basic theorems Local theory in Dynamical system 13 3 A geometry of chaos: Fractals Motivations Compact subsets of R n Construction of mathematical Fractals Coding chaotic dynamics Symbolic dynamics Expanded compact subset Link between symbolic dynamics and expanded compact subsets Structural stability in chaotic dynamics Structural stability o north-south dynamics of the circle

4 4 CONTENTS 5.2 structural stability of the doubling angle map of the circle Quasi-periodic phenomena : homeomorphisms of the circle 37 7 Completeness of some spaces Definitions and basic properties Proof of the conmplete spaces of some spaces Proof of the completeness of R Completeness of the space of continuous, bounded, real functions of R n Completeness of the space of compact subsets Midterm at home Hausdorf dimension Hyperbolic Julia set Structural stability of hyperbolic compact subset Exam at home Examples Some preliminary implementations Expansivity of f on K Completeness of a space Definition of h f as a fixed point

5 Chapter 1 Introduction 1.1 What is a dynamical system? A dynamical system is a phase space which parametrizes the states of the system together with an evolution rule which predicts the short term evolution. theory is to understand the long term evolution. The aim of the dynamical system Example 1. If the system consist of an object falling down, the phase space is formed by 3 coordinates for the position p and 3 coordinates for the velocity v, that is R 3 R 3. The short time evolution is given by the Newton rule: { t p(t) = v(t) The orbits of the system are the trajectories. t v(t) = g 0 Example 2. In Canada, there are rabbits and Lynxes. Lynxes eat rabbits. Let x n be the number of Lynxes at the year n. Let c be the coefficient of reproduction of Lynxes. This coefficient is the half mean number of descendant of Lynxes minus the probability to death, such that x n+1 = c x n. The more Lynxes there are, the less rabbit there are, the less food there are, the weaker c is. Thus c is a decreasing function of x. Therefore we can approximate c as a decreasing affine maps of x: c = c(x) = ax + b, with a < 0 and b constant. 5

6 6 CHAPTER 1. INTRODUCTION Consequently: x n+1 = ax 2 n + bx n = f(x n ), with f(x) = ax 2 = bx. Here the phase space is the number of lynxes, that is R and the short times evolution is f. The orbit of the system are (f n (x)) n 0. Notations 3. Most of the times, f n denote the n th iterate of f, that is: f f f. }{{} n times Example 4. For many sciences such as biology, economy,..., all phenomena are assumed linear. Then the phase space is R n and the evolution rules is the linear map induced by some matrix A M n (R). Then the orbit of the point x R n is (A n x) n 0. Exercise 5. Compute A n when A is: diagonal, diagonalizable, trigonal, trigonalizable. Find a strategy, like a plan for a computer program, that return for any A M n (R), (a very good approximation of) A n in function of n Plan In the first chapter we will study the deterministic dynamical system, that is when every point is attracted by a point a, and we will show that when the dynamics is a C 1 -diffeomorphism f of R, the dynamics is identifiable to the one induced by the differential d a f of f at a. Beyond determinism, the most important and standard dynamics are related to two mains categories:

7 1.3. PHASE SPACES, REGULARITY OF THE DYNAMICS AND BASIC TOOLS 7 the quasi-periodic behaviors, whose paradigm is a rotation on a circle; the hyperbolic (or chaotic) behavior, whose paradigm is the doubling angle map on a circle. α R/Z 2α R/Z. For chaotic dynamics, Fractal geometry occurs. That is why in chapter 2 we will study them. In chapter 3, we will see that some chaotic dynamics are very well understood since they are typically identifiable to the dynamics on the space {0, 1} N of infinite words in the alphabet {0, 1}, and their dynamics are typically identifiable to the shift map σ: σ : (x i ) i 0 {0, 1} N (x i+1 ) i 0 In the 4 th chapter we will see that any perturbation of some chaotic dynamics are identifiable to the initial dynamics. In the 5 ve chapter, we will study the quasi-periodic dynamics by asking when does a map of the circle is identifiable to a rotation. 1.3 Phase spaces, regularity of the dynamics and basic tools Phase spaces Our frame work, as a phase space will be either: the euclidean space R n, for some n 1, the torus T n := R n /Z n. The euclidean space is canonically endowed with the euclidean metric: d(x, y) = n (x i y i ) 2, for any x, y R n, with x = (x i ) i and y = (y i ) i. Let us now describe the torus T n. We notice that the binary relation on R n defined by: (x i ) i (y i ) i ( x i y i ) Z, i {1,..., n} ) i=1

8 8 CHAPTER 1. INTRODUCTION is an equivalence relation. The equivalence class of x R n, that is the set of the point y R n equivalent to x, is equal to x + Z n := {(x i + n i ) i ; (n i ) i Z n }, with x =: (x i ) i. Let T n be the space of the equivalence classes. Such a space is endowed with the following canonical metric: d(x + Z n, y + Z n ) = min d(x + (m i), y + (n i ) i ) = min d(x, y + (n i) i ). (m i ) i, (n i ) i Z n (n i ) i Z n For instance, T 1 is a circle. Thus T 2 is the product of two circles, or a donut. And so, T n is the product of n-circles. Let π : x R n π(x) := x + Z n T n. We notice that for all p T n and x R n such that π(x) is equal to p, the subset V x := {x nr n ; d(x, x) < 1/2} is sent isometrically onto {p T n ; d(p, p) < 1/2} by π. Thus the spaces R n and T n are locally the same. Finally we notice that T n has a structure of group + since (Z n, +) is a subgroup of the commutative group (R n, +) Regularity of the maps of these spaces Let E be the space R n or T n. Let F be the space R k or T k. A continuous function f from E into F is a map such that for any x E, for any ɛ > 0 there exists δ > 0 such that for every y δ-distant to x the point f(y) is ɛ-distant to f(x). An homeomorphism of E is a bijective, continuous map of E whose inverse is continuous. A map f from R n into R k is differentiable if for any x R n, there exists a linear map d x f from R n into R k such that: f(y) = f(x) + d x f(y x) + o(d(x, y)), where o(d(x, y)) is a quantity which is small compare to d(x, y) when y approaches x. In other words: o(d(x, y)) d(x, y) 0, as y x.

9 1.3. PHASE SPACES, REGULARITY OF THE DYNAMICS AND BASIC TOOLS 9 A map f from T n into R k is differentiable if for any p T n, there exists a linear map d p f from R n into R k such that: with x π 1 (p) and with π Vx f(p ) = f(p) + d p f π 1 V x (p p) + o(d(p, p ) ), the restriction of π to the neighborhood V x defined above. A map f from T n into T k is differentiable if for any p T n, there exists a linear map d p f from R n into R k such that: with x π 1 (p). f(p ) = f(p) + π d p f π 1 V x (p p) + o(d(p, p ) ), A map f from R n into T k is differentiable if for any p R n, there exists a linear map d p f from R n into R k such that: f(p ) = f(p) + π d p f(p p) + o(d(p, p ) ). One can show easily that d x f is then unique. Also, the differentiability of a map implies the continuity of the same map. A map f from E into F is continuously differentiable or of class C 1, if it is differentiable and its differential: is continuous. df : x R n d x f A C 1 -diffeomorphism of R n (resp. T n ) is a bijective, C 1 -map of R n (resp. T n ) whose inverse is of class C 1. Exercise 6. Show that a function f is continuous (resp. C 1 ) from T n into R iff there exists a continuous (resp. C 1 ) map g from R n into R such that: g = f π. In other words the continuous or C 1 -functions on T n are the Z n -periodic functions on R n. Exercise 7. Is the following map from T k2 into R differentiable? What is the differential of x R k x 2? (x, y) T k2 d(x, y) 2

10 10 CHAPTER 1. INTRODUCTION Orbits and dynamical system As we saw we restrict our study to the phase space M that are Euclidean space or tori. We will restrict our study to: the homeomorphisms of M, the continuous endomorphisms of a M (that a continuous map from the M into itself), the diffeomorphisms of M, the C 1 -endomorphisms of M. When the dynamics f is non-invertible, the orbit of a point x M is the sequence (f n (x)) n 0. When the dynamics f is invertible, the (resp. forward, back ward) orbit of a point x M is the sequence (f n (x)) n Z (resp. (f n (x)) n 0, resp. (f n (x)) n 0 ) Some basic theorems In this section, we just remind some basic differential theorems of the euclidean space (see... for a proof). As a torus is locally an euclidean space, and as these result are local, the same statement hold for tori. Proposition 8 (Chain Rule). Let E, F, G be euclidean spaces or tori. Let f be a differentiable map from E to F and g be a differentiable map from F to G. Then we have for every x E: d x (g f) = d f(x) g d x f. Exercise 9. Does the following map is differentiable : (x, y) R k2 d(x, y)? Theorem 10 (Means values inequality). Let f be a C 1 map from R k to R n. Let x and y in R k. Then we have: d(f(x), f(y)) max z [x,y] d zf d(x, y), where d z f is the norm of d z f subordinated to the Euclidean norms.

11 1.3. PHASE SPACES, REGULARITY OF THE DYNAMICS AND BASIC TOOLS 11 Theorem 11 (Local inversion theorem). Let E and F be a torus or/and an euclidean space. Let f be a C 1 map from E to F. If for any x E such that d x f is invertible, then there exists r > 0, r 1 such that f is a diffeomorphism from {x R n : d(x, x ) < r} onto its image that contains {y R n : d(f(x), y) < r }. Theorem 12 (Global inversion theorem). Let f be a bijective, C 1 map from R k onto itself. If for any x E such that d x f is invertible, then f is a diffeomorphism of R k.

12 12 CHAPTER 1. INTRODUCTION

13 Chapter 2 Local theory in Dynamical system A basic idea to understand a dynamical system, is to consider its fixed points, that is a point sends into itself by the dynamics. This can be generalized to the the periodic points, that is a point sends into itself by some iterate of the dynamics. These particular points allow us to understand some times the dynamics around them by using the tools explain below. Let f be a C 1 -function of R n into R n which fixes the point 0: f(0) = 0. We want to compare the dynamics of the differential d 0 f of f at 0 with the dynamics of f. Suppose that there exists a homeomorphism h of R n (that is a continuous, bijective map h of R n whose inverse is continuous) such that: f h = h d 0 f. Then the orbits of d 0 f are sent by h to the orbits of f. As d 0 f is a linear map, we understand very well its dynamics. Since such maps f are very rare, we are more interested in the following concept: Definition 13. Let f be a map of R n into itself which fixes a point a R n. f is linearizable at a if there exists a homeomorphism h of R such that, for any x small enough: f h(x) = h d a f(x). Remarks : We notice that the maps h sends the (forward/backward) orbits of d a f which lands close to a to the orbits of f. 13

14 14 CHAPTER 2. LOCAL THEORY IN DYNAMICAL SYSTEM Not all the maps fixing a are linearizable. For instance f : x x x 3 is not linearizable at 0. Its derivative at 0 is the identity, the orbits of the identity are single points, but the forward orbit of any point close to zero is a sequence close to zero and converging to zero: f(x) = x 1 x 2 < x, if x 0. Nevertheless we have the following theorem: Theorem 14. Let f be a C 1 -diffeomorphism which fixes a R and satisfies that d a f is contracting: there exists λ < 1 such that for any u R: d a f(u) λ u. Then f is linearizable at a. Remark: The contraction of d a f is equivalent to the existence λ < 1 such that d a f(u) d a f(v) λ u v, u, v R n. Example 15. Let f : x x 3 + 1/3 x such a map is linearizable at 0 Example 16. Let f : (x, y) R 2 R x 2 +y 2(x/2, y/2), where R α is the rotation centered at 0 and with angle α. Such a map is linearizable at 0. In other words, there exists a homeomorphism h of R 2 such that for x close to 0: f h(x) = h H 1/2 (x), with H 1/2 (x, y) = (x/2, y/2). Figures... and... show the orbits of the maps f and H 1/2. fig The homeomorphism h sends the orbits of H 1/2 starting at points close to zero to orbits of f. Actually the above theorem is a special case of the following admitted theorem:

15 15 Theorem 17. Let f be a C 1 -diffeomorphism which fixes a R and satisfies that d a f has no eigenvalue of modulus 1, then f is linearizable at a. Proof of theorem 14 when n = 1, a = 0 and f (0) > 0. Let λ < 1 be greater than the derivative f (0). f(x) < λ x. By the mean value theorem, for r > 0 small enough, any point x [ r, r] satisfies Thus, f sends [ r, r] into ( r, r). Also, for all x [ r, r] and n 0, the norm f n (x) is less than λ n x and so (f n (x)) n converges to 0. Thus n 0 f n( [ r, r] \ f([ r, r]) ) is equal to [ r, r] punctured of 0. Let h 0 be a homeomorphism from [ r, r] \ f(( r, r)) onto [ r, r] \ ( f (0) r, f (0) r), such that h 0 (r) = r and h 0 ( r) = r. For instance one can choose the map: r f (0) r r f(r) (x r) + r if x > 0 x (x + r) r if x < 0 r f (0) r r+f( r) Let h be the map of R defined by : x if x > r h(x) = 0 if x = 0 d 0 f n h 0 f n (x) if x f n( [ r, r] \ f([ r, r]) ) We notice that h is a homeomorphism of R. Moreover, h satisfies for any x f n( [ r, r]\f([ r, r]) ) with n > 1: d 0 f h(x) = d 0 f n+1 (0) h 0 f n (x) = d 0 f n+1 h 0 f n 1 f(x) = h f(x) as f(0) = h(0) = 0, for any x f([ r, r]), d 0 f(0) h(x) = h f(x). Proof of theorem 14 when n 1. First of all we may assume that a = 0 by considering the map x f(x a) instead of f. Then if the map h is convenient for f(x a), the map h(x + a) is convenient for f.

16 16 CHAPTER 2. LOCAL THEORY IN DYNAMICAL SYSTEM Let us denote by B r the ball centered at 0 and with radius r: B r := {x R n ; x < r}. Let us denote by S r the sphere centered at 0 and with radius r: S r := {x R n ; x = r}. h 0 The idea of the proof is the same as in the case n = 1 unless that the homeomorphism : B r \ f(b r ) B r \ d 0 f(b r ) for r > 0 small is harder to construct since: 1. the existence of such a homeomorphism is not obvious, 2. h 0 has to glue well with d 0 f h 0 f 1 at f(s r ): h 0 (x) = d 0 f h 0 f 1 (x), x f(s r ). As f is contracting, there exists λ such that for r sufficiently small, f sends B r into B λ r. Let ρ be a C 1 -function equal to 1 at [0, λ] and equal to 0 on [1, ). Let h 0 (x) = x + ρ( x 2 /r 2 ) (d 0 f f 1 (x) x), for r > 0 small and with f 1 the inverse function of f. We will show below that, for r > 0 small enough, h 0 is a homeomorphism from B r \ f(b r ) onto B r \ d 0 f(b r ) such that condition 2 above is satisfied. Then, we can proceed as in the proof of the theorem for n = 1 to conclude. To show that h 0 is a diffeomorphism, we will prove that h 0 is equal to the identity id of R n minus a 1/2-contracting maps 1 ɛ of R n, for r > 0 small enough. Therefore the differential of h 0 is invertible for every x R n. To use the global inversion theorem (th. 12), we shall prove that h 0 is a bijection onto R n. Let x, y R n sent by h 0 to a same point. We have: h 0 (x) = h 0 (y) x + ɛ(x) = y + ɛ(y) x y = ɛ(y) ɛ(x) Thus by contraction of ɛ, the point x and y are the same. Consequently, h 0 is injective. 1 for all x, y, d(ɛ(x) ɛ(y)) 1/2d(x, y)

17 17 As h 0 is equal to id on R n \B r, the image of h 0 contains R n \B r. As h 0 is equal to d 0 f f 1 (x) on f(b r ), h 0 sends f(b r ) onto d 0 f(b r ). Consequently, by injectivity, h 0 sends B r \ f(b r ) into B r \d 0 f(b r ). It remains to prove that h 0 sends B r \f(b r ) onto B r \d 0 f(b r ). Let y B r \d 0 f(b r ). The line (0, y) intersects S r B r \d 0 f(b r ) into to segments. Let [y 1, y 2 ] be the one that contains y, with y 1 in S r and y 2 in d 0 f(s r ). Let t be the supremum of the t [0, 1] such that [y 1, y 1 +t(y 2 y 1 )] is includes in the image of h 0. If t = 1, then y is in the image of h 0 and so we have shown the surjectivity of h 0. Let us suppose that t < 1. Let us show that y 0 := y 1 +t(y 2 y 1 ) belongs to the image of h 0. By definition of t, there exists (x n ) n 0 h 1 ([y 1, y 1 + t(y 2 y 1 )]) B r \ d 0 f(b r ) such that (h 0 (x n )) n converges to y 0. As B r \ d 0 f(b r ) is bounded, we can find a subsequence (x ni ) i 0 which converges in R n (see section 3.2 for more details). Let x its limit. We notice that x belongs to S r B r \ f(b r ). By continuity, h 0 sends x to y 0. By the local inversion theorem, there exists a r > 0, such that {y R n : d(y 0, y) < r} is contained in the image of h 0. This is a contraction with the fact that t is the supremum. Let us show that h 0 is equal to id ɛ with ɛ 1/2-contracting and fixing 0 when r is small enough. Let ɛ(x) = x h 0 (x) = ρ( x 2 /r 2 ) (d 0 f f 1 (x) x). We notice that ɛ fixes 0 since h 0 and id fix 0. Moreover < 2x, > d x ɛ = d x 2 /r2ρ d0f f 1 (x) x ρ( x 2 /r 2 ) (d 0 f d x f 1 id). r r As f is of class C 1, the differential d x f is close to d 0 f for x close to 0. Thus d 0 f d x f 1 id is close to 0 when x is close to 0. Thus, for r small enough, for any x B r and u R n : (d 0 f d x f 1 id)(u) 1/4 u ρ( x 2 /r 2 ) (d 0 f d x f 1 id)(u) 1/4 u By definition of the derivative, f(x) d 0 f(x) / x is small when x is small. Thus for r small and any x B r, d 0f f 1 (x) x r is small. Consequently, for r small enough, for any x B r and u R n : d < 2x, u > d 0 f f 1 (x) x x 2 /r2ρ r r u 4. Therefore, d x ɛ(u) u /2, for any r small enough, x B r and u R n. By the mean values theorem ɛ is well 1/2-contracting.

18 18 CHAPTER 2. LOCAL THEORY IN DYNAMICAL SYSTEM Exercise 18. Find a diffeomorphism of R 2 fixing 0, whose differential at 0 is rotation with irrational angle but which is not linearizable at 0. Exercise 19. linearization of z z 2, regularity in dimension 1.

19 Chapter 3 A geometry of chaos: Fractals 3.1 Motivations This field of dynamical system was initiated by Benoit Mandelbrot in the 80 s to model many natural geometry such as: the geometry of geography phenomena like the design of a coast(fig..), a mountains (fig..), rivers (fig...., the geometry of many organism like a cauliflower (fig...., a tree (fig..), a lung (fig..), the geometry of a many object in chaotic dynamics that we will see further (fig...). One can define informally the fractal by satisfying some of these properties: It has a fine structure at arbitrarily small scales. It is too irregular to be easily described in traditional Euclidean geometric language. It is self-similar (at least approximately or stochastically). It has a a length, surface, volume, equal to infinity or 0 ; more precisely its (Hausdorff) dimension is not an integer. We propose to give a dynamical construction of most of these objects. For this we need some basic elements of topology. 19

20 20 CHAPTER 3. A GEOMETRY OF CHAOS: FRACTALS 3.2 Compact subsets of R n Let us fix n 1. We remind that R n is canonically equipped with the following norm: u = (u i ) 1 i n R n u := n This is a norm define the distance: d : (u, v) (R n ) 2 u v R + i=1 u 2 i To say that a function d is a distance means that it satisfies: Symmetry property: u, v R n, d(u, v) = d(v, u). Definition property: u, v R n, d(u, v) = 0 u = v. Triangular inequality: u, v, w R n, d(u, w) d(u, v) + d(v, u). We remind that a ball centered at x R n with radius r 0 is the subset: B(x, r) := {y R n ; d(x, y) < r}. A subset O of R n is said to be open if its is empty or satisfy that for any x O there exists r > 0 such that B(x, r) is included in O. Example 20. R is open, a ball is open, a union of ball is open, R n 1 (0, ) is open. Property 21. Any (possibly infinite) union of open subsets of R n is open. Any finite intersection of open subsets is open. Proof. Let (O i ) i be a possibly infinite family of open subsets of R n. Let x i O i. Thus x belongs to the subset O j, for some j. As O j is open there exists r > 0 such that B(x, r) is contains into O j, and so in i O i. Let (O i ) 1 i k be a finite family of open subsets of R n. Let x i O i. Thus x belongs to each subset O i. As O i is open there exists r i > 0 such that B(x, r i ) is contains into O i. Thus B(x, r) is contains in i O i, with r := min i r i > 0. A subset C of R n is said to be closed if its complement F c := {x R : x / F } is open.

21 3.2. COMPACT SUBSETS OF R N 21 Property 22. Any (possibly infinite) intersection of closed subsets of R n is closed. Any finite union of closed subsets is closed. Proof. Exercise. Remark: A subset can be neither closed, neither open. For instance: [0, 1) R. Exercise 23. Say for each of these subsets if they are closed, open or nothing in R 2 : 1. B(x, r) \ {0}, 2. {y R 2 : 0 < y 1}, 3. {y R 2 : 0 < y 1}, 4. (0, 1] {0} {0} R. Property 24. Let F be a a subset of R n. (x n ) n F N converging to some x R n has its limit x in F. Then F is closed if and only if for any sequence Proof. ) We suppose F closed. Let (x n ) n be a sequence of F N converging to some x R n. Suppose for the sake of a contraction that x belongs to R n \ F. As R n \ F is open, there exists r > 0 s.t. B(x, r) is included in R n \ F. As (x n ) n converges to x, for n sufficiently large, x n belongs to B(x, r) and so to R n \ F. Contradiction. ) for any sequence (x n ) n F N converging to some x R n has its limit x in F. Let us suppose by the sake of a contraction that F is not closed that is R n \ F is not open. Therefor there exists x R n \ F s.t. for any r > 0 the ball B(x, r) is not included in R N \ F. Thus there exists x n B(x, 1/n) F for any n 0. Such a sequence (x n ) n F N converges to x R n \ F. Contradiction The diameter of a subset E of R n is the positive number: diam E := sup d(x, y). x,y E 2 for instance the diameter of [ 1, 1] is 2, the diameter of [ 1, 1] 2 is 2, the diameter of [ 1, 1] n is n, the diameter of B(x, r) is 2r. A subset B of R n is bounded if its has finite diameter. A subset K of R n is compact if it is bounded and closed.

22 22 CHAPTER 3. A GEOMETRY OF CHAOS: FRACTALS Exercise 25. Prove that the subset K of [0, 1] formed by the points [0, 1] without 4, 5 and 6 in their decimal expression (for instance 0, but not 0, 0, ) is a compact subset of R. Felix Hausdorff ( ) has defined a very useful distance on the set of nonempty compact subset of R n. The Hausdorff distance d HD between two compact subsets K, K is defined by : d HD (K, K ) := sup d(x, K ) + sup d(x, K) x K x K with d(x, K ) := inf x K d(x, x ) and d(x, K) := inf x K d(x, x ). d HD is a distance on the set K n of nonempty compact subsets of R n since it satisfies the triangular inequality, the symmetry and definition properties explain above. Proposition 26. Let (K p ) p 0 be sequence of compact subsets of R n. If N p=0 K p is not empty for any N 0 then p 0 K p is empty. Proof. For any N 0, let x N be in N p=0 K p. The idea of the proof is to find an increasing sequence (N i ) i N N such that (x Ni ) i converges to some x R. As each K p is closed and (x Ni ) i p lands in K p, the point x belongs to K p and so to p K p. Consequently this intersection is not empty. Let us construct (n i ) i N N. As K 1 is compact it is include in [ M, M] n for some M 0. Thus N p=0 K p is included in [ M, M] n for any N 0. Let N i 0 and (a ik ) 1 k n { 2 i, 2 i + 1,, 0, 2 i } such that n k=1 [ M a ik/2 i, M (a ik + 1)/2 i ] contains x Ni and infinitely many elements of (x N ) N. As we can share n k=1 [ M a ik/2 i, M (a ik + 1)/2 i ] in finitely many (2 n ) pieces of the form n k=1 [ M b ik/2 i+1, M (b ik + 1)/2 i+1 ] with b ik {2a ik, 2a ik + 1}, there exists a i+1 k {2a ik, 2a ik + 1} such that n k=1 [ M a i+1 k/2 i+1, M (a i+1 k + 1)/2 i+1 ] contains infinitely many (x n ) n. Let n i+1 > n i be in n k=1 [ M a i+1 k/2 i+1, M (a i+1 k + 1)/2 i+1 ]. We notice that by induction (N i ) i is an increasing sequence. Moreover, the sequence (x Ni ) i converges since its i first digits of the binary expression of each coordinate of x Ni the one of x Nj for j i. Exercise 27. Prove that d HD is a distance, that is for any K, K, K K n we have: is the same as d(k, K ) = 0 K = K, d(k, K ) = d(k, K), d(k, K ) d(k, K ) + d(k, K ).

23 3.3. CONSTRUCTION OF MATHEMATICAL FRACTALS 23 Definition 28 (Property). Let E be a subsets of R n. The closure of E is the smallest closed subset of R n containing E, that is the intersection of all closed subset of R n containing E and consists of the limits of all converging sequences of E. Proof. Let C 1 be the intersection of all closed subset of R n containing E. By property..., the subset C 1 is closed. As E is included in C 1 by property..., C 1 contains the subset C 2 consisting of the limits of all converging sequences of E. To show that C 1 and C 2 are equal, it is sufficient to show that C 2 contains C 1 and so that C 2 is closed. Let (x k ) k be a sequence of points in C 2 converging to some point x. Let us show that x belongs to C 2. For any p 1, there exists k p such that d(x kp, x) is less than 1/p. As x p belongs to C 2, there exists y p E 1/p-close to x p. Thus y p E and x are 2/p-close. Therefore (y p ) p E N converge to x. Exercise 29. Find the closure of the subsets of exercise Construction of mathematical Fractals Let (f i ) 1 i k be a family of contracting homeomorphism of R n. That is for any i {1,..., k}, there exists λ i < 1 such that for all (x, y) R n : d(f i (x), f i (y)) λ i d(x, y). Theorem 30. For K K n, let F (K) be the subset k i=1 f i(k) of R n. Then F (K) is compact and the map: F : K K n F (K) K n, has a unique fixed point. Example snow flake, fractal cube, trees + function Exercise 32. Find for each of these following fractals some functions (f i ) making them. Proof of theorem F (K) is compact. Let K K n. We will prove that each subset f i (K) is a nonempty compact. Then as a finite union of bounded subset is bounded, F (K) is bounded. And as a finite union of closed subset is closed (propr..), the F (K) is closed and so compact.

24 24 CHAPTER 3. A GEOMETRY OF CHAOS: FRACTALS Let us prove that f i (K) is compact. First of all, as f i is contracting, the diameter of f i (K) is less than the diameter of K. Thus f i (K) is bounded. To show that f i (K) is closed, we have to prove that its complementary f i (K) c is open. As f i is bijective, f i (K) c is equal to the image f i (K c )of the complementary of K by f i. Let y = f i (x) be in f i (K c ). As K is close, its complementary K c is open. Thus their exists r > 0 such that the ball B(x, r) is included in K c. As f 1 i points close to x. Thus, for δ > 0 small enough B(y, δ) is sent by f 1 i is continuous, the points close to y are sent by fi 1 to into B(x, r). Consequently B(y, δ) is included in fi 1 1 (B(x, r)) = f i (B(x, r)). As B(x, r) is contained into K c, the ball B(y, δ) is included in f i (K c ) = f i (K) c. As this true for any y f(k i ) c, f(k i ) c is open and f(k i ) is closed. 2. F is contracting for d HD. We want to show that for any K, K K n, we have: with λ := max(λ i, i {1,..., k}). we remind that d HD (F (K), F (K )) = d HD (F (K), F (K )) λ d HD (K, K ), sup y F (K) Let us now evaluate sup y F (K) d(y, F (K )): d(y, F (K )) + sup y F (K ) d(y, F (K)). sup y F (K) d(y, F (K )) = max i max i sup y f i (K) sup y f i (K) d(y, F (K )) = max i d(y, f i (K )) = max i min j sup y f i (K) sup d(f i (x), f i (K )) x K d(y, f j (K )) Thus And similarly: max λ i sup d(x, K ) = λ sup d(x, K ) i x K x K sup d(y, F (K )) λ sup d(x, K ) y F (K) x K sup d(y, F (K)) λ sup d(x, K) y F (K ) x K

25 3.3. CONSTRUCTION OF MATHEMATICAL FRACTALS 25 Consequently: d HD (F (K), F (K )) λ sup d(x, K ) + λ sup d(x, K) λ d HD (K, K ). x K x K 3. F has a unique fixed point in K n We will prove in..., that any contracting maps of K n has a unique fix point.

26 26 CHAPTER 3. A GEOMETRY OF CHAOS: FRACTALS Exercise 33. Draw the shape of the fractal obtained by the following functions: R n = C, f 1 := z (1/2)(z i), f 2 := z e iπ/6 2 (z + i), f 3 := z eiπ/6 2 (z + i), R n = C, f 1 := z eiπ/6 2 (z 2), f 1 := z eiπ/6 2 (z + 2) Picture obtained by computer programming are even better! Exercise 34. show that the maps f : x [ 2, 2] x 2 2, g : x R/2πZ 2x and h : x R/2πZ cos(x) satisfy: f h = h g conclude that the set of points of [ 2, 2] whose orbit does not intersect an open neighborhood of {0} is (homeomorphic) to a fractal.

27 Chapter 4 Coding chaotic dynamics 4.1 Symbolic dynamics Giving an integer n 0, we denote by Σ n the set {1,..., n} N of sequences (a i ) i 0 of integers in {1,..., n}. On Σ n, we define the shift σ dynamics: Σ n Σ n (a i ) i 0 (a i+1 ) i Such a dynamical system (σ, Σ n ) is quite simple and thus would be a good model to describe the expanding maps. Let us improve this family of models by considering some restrictions of these maps. Let M n ({0, 1}) be the set of matrices with entries equal to 0 or 1 and of dimension n n. For A M n ({0, 1}), let Σ A be the subset of Σ n formed by the sequences (a i ) i such that for every i, the entry in the a th i column and the a th i+1 line is positive. For n 0 and A A n, we endow Σ A with the following distance: d((a i ) i, (a i) i ) = i 0 a i a i 2 i We notice that for this distance, the balls are of the form: {(a i) i Σ A : a i = a i, i N}, where (a i ) i is some element in Σ A and N is an integer. We notice that these balls (which generate the topology) are open and closed. 27

28 28 CHAPTER 4. CODING CHAOTIC DYNAMICS Proposition 35. The space Σ A is totally disconnected: for any two distinct points in a, a Σ A, there exist two open neighborhoods V a and V a of respectively a and a, that are disjoint but cover Σ A. Proof. Let k the first integer such that a k a k. Let V a be the ball centered at a and with radius 1/2 k. Let V a be the complement of a. As V a is open and closed, V a is well open, moreover V a contains a. We remark that the shift map σ is continuous for this distance. Proposition 36. The space Σ A is compact. Proof. Let (a k ) k be a sequence in Σ A. We want to construct a sequence of infinite subsets (J l ) j of N such that for every k, k J l the l first coordinates of a k and a k are equal. The construction of the (J l ) l is done by induction as in the proof of the completeness of R. (for instance as the first digit of a k belongs to the finite set {1,..., n} and as there infinitely (a k ) k, on of these digits j 0 {1,..., n} is taken infinitely many times. Let J 0 be the infinite subset of N formed by the integer k such that a 1 is equal to j 0 ). Let us now define an increasing sequence of integers (n i ) i by induction: n 0 is the minimum of J 0, n i+1 is the minimum of J n+1 [n n + 1,..., [. We notice that the sequence (a n i ) i converges. A matrix A of M n ({0, 1}) is said to be non nilpotent if, for every k 0, the product A k is nonzero. We denote by A n the subset of non nilpotent matrices. Proposition 37. For A M n ({0, 1}), the set Σ A is non-empty if and only if the A is non nilpotent. Proof. If Σ A is non empty there exists (a i ) i Σ A. We notice that if A =: [α kl ] and A i := [α (i) We notice that α a1,a 0 = α (1) a 1,a 0 is non zero for some i 0. Also, we have: kl ]. is nonzero, by definition of Σ A. Suppose by induction that α (i) a i a 0 α (i+1) kl = n p=1 α kp α (i) pl. (4.1.1)

29 4.1. SYMBOLIC DYNAMICS 29 We notice that α ai+1,a i is non zero. Thus by the above formula α a (i+1) i+1,a 0 is nonzero since α a (i) i a 0 is nonzero (the terms of the sum are non negative). Thus by induction, A i is nonzero for every i. in other words, A is non nilpotent. Suppose that A is non nilpotent. Let us show by induction on i 1, if α (i) kl there exists a family (a i j )i k=0 such that: α a i j+1,a i j is non zero, is nonzero, then a i i is equal to k. This is obvious when i = 1. Let i 0. By equation (4.1.1), if α (i+1) kl is nonzero, then α kp and α (i) pl are non zero for some p. By induction hypothesis, there exists a family (a i j )i j=0 such that: α a i j+1,a i j is non zero, a i i is equal to p. We notice that a i+1 j := a i j if j i and ai+1 i+1 = k satisfies the requested properties. Let a i j := 1 for every j > i. Let ai := (a j ) j 0. Now we can extract extract as in the proof of the compactness of Σ A a subsequence (a n i ) i that converges to some a. We easy show that a belongs to Σ A. Definition 38. A matrix A M n ({0, 1}), is said to be irreducible if there exists n 0 such that A n has only positive entries. Remark: If A is irreducible, then A is non nilpotent. Exercise 39. Show that if A is irreducible then σ is transitive: it has a dense orbits. Definition 40. Let f be a map from a metric space (E, d) into itself. The map f is said to be ɛ-expansive, if the following property holds: For every points x, y E, if f n (x), f n (y) are at most ɛ distant for every n 0, then x and y are equal. The map f is said to be expansive if f is ɛ-expansive for some ɛ > 0. Proposition 41. for all n 0 and A A n, the shift σ is expansive on Σ A.

30 30 CHAPTER 4. CODING CHAOTIC DYNAMICS Proof. Let us show that σ is 1/2-expansive. Let x = (a i ) i, y = (a i ) i Σ A, s.t. f n (x), f n (y) are at most 1/2 distant for every n 0. Thus, for every i 0, a i a i d(σi (x), σ i (y)) 1/2. As a i and a i are integer, they are equal, for every i. Thus, x and y are equal. Parler de la decomposition en composante irreductible. Devaney Expanded compact subset In all that follow n is a positive integer and f is a C 1 map of R n (or T n ). Definition 42. A compact subset K of R n, f-invariant (f 1 (K) = K) is expanded if there exists λ < 1 such that for any x K, d x f is invertible and with inverse λ-contracting. Example 43. Let f expanded. : (x, y) C z 2 C 2. Then the unit circle of the complex plan R 2 is Property 44 (Admitted). If f expands a compact subset K, then there exists ɛ > 0 such that for any points x K, the restriction f B(x,ɛ) of f to the ball B(x, ɛ) is a diffeomorphism onto its image and its image contains B(f(x), ɛ). Moreover the inverse of f B(x,ɛ) is λ-contracting, with λ < 1 which does not depends on x. Proposition 45. If f expands a compact subset K, then f is expansive on K. 4.3 Link between symbolic dynamics and expanded compact subsets Actually there is a strong links between the symbolic (shift) dynamics and the dynamics over an expanded compact subset given by the following theorem: Theorem 46. There exists p 1 and A M p ({0, 1}) such that the dynamics induced by f on K is semi-conjugated to the shift σ of Σ A := {(x i ) i 0 {1,..., p} N : A xi+1 x i = 1}. In other word, there exists a continuous maps π : Σ A K continue onto K and such that: f π = π σ.

31 4.3. LINK BETWEEN SYMBOLIC DYNAMICS AND EXPANDED COMPACT SUBSETS31 Remark: Actually the map π is injective on (an intersection of dense open subset and so) a dense subset of Σ A. Let us illustrate this theorem by an example. Example 47. Let K be the circle (R/Z). Let f be the map x K 10 x K. Let π be the following map from the symbolic space Σ 10 = {1,..., 10} N which sends (x i ) i to i 0 x i/10 i. The map π is continuous and well defined. Also for any x S 1 if x = 0, x 1 x 2 x i we have π((x i ) i ) equals to x. Thus π surjective. Finally, we notice that f π((x i ) i ) = 10 0, x 1 x 2 x i [1] = x 1, x 2 x i [1] = 0, x 2 x i [1] = π σ((x i ) i ) The map π is mostly injective but is not injective: the point (1, 9, 9, 9, 9, 9, 9, ) and (2, 0, 0, 0, ) are sends to the same point. Example 48. example de l endo du tore avec une partition infy (fractal de Rauzy?). The proof of this theorem follows from the existence of a Markov Partition. Definition 49. A family R = {R 1,..., R N } of disjoint subsets of R n is a Markov partitionof K if: 1. N i=1 R i contains K, 2. for every i, j we have: f(r i ) R j f(r i ) R j. Let us show that the following proposition (showed further) is sufficient the prove theorem 46. Proposition 50. For any ɛ > 0, there exists a Markov partition for any expanded compact subset, whose elements have diameter less than ɛ. Proof of theorem 46. Let ɛ > 0 be less than the expansiveness constant of K and the constant of property 44. Let (R i ) N i=1 be a Markov partition of K whose elements have diameter less than ɛ. Let A M n ({0, 1}) such that with A := [a ij ], a ij = 1 iff f(r j ) R i. Let (x i ) i Σ A.

32 32 CHAPTER 4. CODING CHAOTIC DYNAMICS We notice that n 1 f n (cl(r xn )) is a decreasing intersection of non empty compact subset. By expensiveness, this intersection consists of a unique point π((x i ) i ). We notice that we have well: f π = π σ. On the other hand, for y K, let x n {1,..., N} s.t. f n (y) is in R xn. We notice that x = (x n ) n is in Σ A and that π(x) is equal to y, by uniqueness. It remains to prove that π is continuous onto K. By property 44, for ɛ small enough the diameter of k n=1 f n (cl(r xn )) is less than λ k ɛ, for any (x i ) i Σ A. When (y i ) i is close to (x i ) i, the coordinate x i and y i are equal for every i less than a large k, thus the distance between π((x i ) i ) and π((y i ) i ) is less than λ k ɛ and so is small (since λ < 1). Consequently the map π is continuous. Proof of proposition 50. We can suppose ɛ less than the constant of property 44. We suppose f λ < 1-expansive on K. As K is bounded there exists a finite family of open subsets (U i ) N i=1 with diameter less than ɛ(1 λ)/2, such that K is contained in N i=1 U i. For any j, k 0 and x f k (U j ) K, let: ( ( Uxj k := f 1 B(x,ɛ) f 1 (f 1 B(f(x),ɛ) B(f k (x),ɛ) (U j) ) )) K. We notice that the diameter of U j xi is less than λj ɛ(1 λ)/2. For i {1,..., N}, let us construct U (n) i Let U (0) i intersecting U (k) i (U (k) i by induction. be equal to U i. For k 0, let U (k+1) i, for x K and j {1,..., N}. be the union of U (k) i with the subsets Uxj k+1 We notice that by induction the diameter of U (k) i is less than k ) k is an increasing sequence of open subsets of K. Thus the open subset U ( ) to k 0 U (k) i has diameter less than ɛ = p=0 λp ɛ(1 λ). p=0 λp ɛ(1 λ) and that i of K equal Let (R k ) k be the nonempty subsets of the form i I U ( ) i \ i I cu ( ) i for I {1,..., N}. We notice that (R k ) k is a partition. Let k, k such that f(r k ) intersects R k, with R k := i I U ( ) i \ i I cu ( ) i and R k := i I U ( ) i \ i I cu ( ) i. Let us show that (R k ) k is a markov partition: for every j, k, if f(r k ) intersects R j then f(r k ) contains R j,. For we pfirst prove the following lemma:

33 4.3. LINK BETWEEN SYMBOLIC DYNAMICS AND EXPANDED COMPACT SUBSETS33 Lemma 51. For every l, the φ(u ( ) i section of f (that is of this form f 1 B(x,ɛ) ) such that φ(u i ) intersects U l cover U ( ) l, where φ is a ) whose domain contains U ( ) i. Proof. Let x U ( ) l, then there exists a path (U ik ) j k=0 such that i 0 is l and f j (x) belongs to U ij and f(u ik+1 ) intersects U ik for every k. Then we notice that i = i 1 is convenient. This lemma implies that each times that some φ(r i ) intersects some U ( ) l, then φ(r i ) is contained in U ( ) l. To prove this implication we see that if φ(r i ) intersects some U ( ) l, then there exists U j such that φ(u j ) intersects U l and such that φ(u ( ) j statement implies that φ(r i ) is contained in φ(u ( ) j ) intersects φ(r i ). The last ). Also as φ(u j ) intersects U l, the set φ(u ( ) is contained in φ(u ( ) l ). Thus φ(r i ) is well contained in U ( ) l whenever φ(r i ) intersects U ( ) l. Consequently φ(r i ) is contained in R l whenever φ(r i ) intersects R l. This implies the markov partition property since, for any m, i if f(r m ) intersects R i then R m intersects φ(r i ) (for some good section φ) and so φ(r i ) is contained in R m, and R i is well contained in R m. Lien avec les fractals? j )

34 34 CHAPTER 4. CODING CHAOTIC DYNAMICS

35 Chapter 5 Structural stability in chaotic dynamics 5.1 Structural stability o north-south dynamics of the circle 5.2 structural stability of the doubling angle map of the circle 35

36 36 CHAPTER 5. STRUCTURAL STABILITY IN CHAOTIC DYNAMICS

37 Chapter 6 Quasi-periodic phenomena : homeomorphisms of the circle Let f be a homeomorphism of the circle S 1 which preserves the orientation. We wonder about the existence of a semi-conjugation between f and a rotation of the circle. That is the existence of a continuous map h from S 1 onto S 1 such that: h f = R α h, for some rotation R α of the circle with angle α. For in this section we explain the rotation number α f S 1 of f, which satisfies the following properties: Theorem 52 (Poincaré). For all homeomorphisms f and f, the rotation numbers satisfy the following properties: 1. if f and f are semi-conjugated then α f = α f, 2. if α f is irrational, then f is semi-conjugated to the rotation R αf, 3. if α f is rational then f has a periodic point. To define the rotation number, let us admit some properties about the liftings. For let us denote by π : R S 1 the canonical map. 37

38 38CHAPTER 6. QUASI-PERIODIC PHENOMENA : HOMEOMORPHISMS OF THE CIRCLE Proposition 53. For every homeomorphism f of the circle, there exists a homeomorphism F of the real line R such that: 1. the following diagram commutes: R F R π π, S 1 f S 1 2. F (x + n) = F (x) + n, for all x R and n Z, 3. if F is another map satisfying the first above property, then F F is a constant map equal to an integer, 4. if f preserves the orientation, then F is increasing. The map F is called a lifting of f. Let us fix such a lifting F of f. let u n : x F n (x) x, where the map F n is the n th composition of F. The rotation number is defined as: ( α f := π lim n ) u n (x). n ( 1 ) We notice that α f = π lim n 1 n n k=0 F k+1 (x) F k (x). The existence of this limit and these non-dependences in F nor in x are to be proved. Let us begin by showing this non-dependence. The non-dependence in F follows obviously from For we notice that, by proposition 53.2, for every x, y R we have: u n (x) u n (y) 2 (6.0.1) This prove the independence of α f in the point x. Also this shows that: u n+m (x) u n (x) + u m (x), (6.0.2) by taking y := F m (x). Now the existence of the limit follows only from this lemma:

39 39 Lemma 54. For every ɛ > 0, there exists N 0 and such that for every n N, for m sufficiently large we have: u m (x) m u n(x) n Proof. Let m = k n + r, with 0 r n. Then we have: u m (x) m = u k n+r(x) m u k n(x) + u r (x) + 2 m + ɛ. ku n(x) + 2k kn + u r(x) + 2 m u n(x) n For N sufficiently large, 2/n is less than ɛ/2 and for m sufficiently large ur(x)+2 m ɛ/ n + u r(x) + 2 m is less than As moreover the symmetric inequality is done by the same way, we have that (u m (x)/m) m 0 converges well to a real α f. Let us show now that if f and f are conjugated, then their rotation number is the same: Let h be the conjugacy such that: h f = f h. Then for two liftings F and H of respectively f and h, we notice that the following map is a lifting of f : F = H F H 1. Also we have the existence of M 0 s.t. x M H(x) x + M Thus F n H(x) F n (x) M. Thus α f and α f are equal. Proof of Poincaré theorem. Let us show first that f has a periodic point if and only if α f is a rational number. If f has a periodic point p of period l, then there exist ˆp π 1 ({p}), k Z and a lifting F of f such that From this we get that: F l (ˆp) = ˆp + k lim F n (x)/n = k/l n Thus the angle α l is rational. Suppose now that α f is rational. We may suppose that α f = 0 (by considering f n instead of f which satisfies α f n = nα f ). We suppose for the sake of contradiction

40 40CHAPTER 6. QUASI-PERIODIC PHENOMENA : HOMEOMORPHISMS OF THE CIRCLE that f has no fixed point. Then any lifting F of f cannot have a fixed point. Let us assume for instance that: F (x) > x, x R Then either F n (0) < 1 for every n 0, either there exists k 0 such that F k (0) 1. In the later case, then F mk (0) m. So F n (0) lim 1 n n k This is a contradiction. In the first case, (F n (0)) n is an increasing sequence bounded by 1. Such a sequence must converges to a fixed point of F. Contradiction. Let us now prove the second statement of the Poincaré theorem. We suppose that α f is irrational. We want to show that f is semi conjugated to the rotation of angle α f. For let us fix a lifting F of f and put α F = lim F n (0)/n. The proof of this theorem consists of constructing a map between G := {F n (0) + m; n, m Z} and G := {n α F + m; n, m Z} Since α F is irrational, the subgroup G of R is dense. We will show at the end the Lemma 55. For every n, m, n, m Z, if F n (0)+m < F n (0)+m then n α F +m < n α F +m. Then we can define: H : x R sup (nα F + m). {n,m Z:F n (0)+m x} The map H is well define, continuous and onto R by the above lemma and the fact that G is dense in R. Also by definition of H, we have for every x R and m Z: H(x + m) = H(x) + m (6.0.3) H F (x) = H(x) + α f (6.0.4) By (6.0.3), there exists a continuous map h of the circle such that: f π = π H

41 41 As H is surjective, h must be surjective. Also by (6.0.4), we have: h f = R αf h This implies the second statement of the Poincaré Theorem. Proof of lemma 55. For let us consider the function: φ : x F n (x) + m F n (x) m. We notice that φ is positive at zero, continuous, and cannot vanish (since else f as a periodic point). So φ is positive everywhere. In other words, for every x R, we have: F n (x) + m < F n (x) m. Also the maps f 1 : x F n (x) + m and f 2 : x F n (x) + m are increasing since f preserves the orientation. Let us show by induction on k > 0 that: f k 1 (0) < f k 2 (0) For k = 1 this is the hypothesis of the lemma. Let k > 0 and suppose that f1 k(0) < f 2 k (0). Since f 1 is increasing, we have: Since the map φ is positive, we have f k+1 1 (0) < f 1 f k 2 (0). f 1 f k 2 (0) < f k+1 2 (0). Thus, we have well f k+1 1 (0) < f k+1 2 (0). This induction implies that f2 k nα F + m = lim k k lim f1 k k k = n α F + m. Exercise 56. P Devaney 1.14.

42 42CHAPTER 6. QUASI-PERIODIC PHENOMENA : HOMEOMORPHISMS OF THE CIRCLE

43 Chapter 7 Completeness of some spaces 7.1 Definitions and basic properties Throughout this section, we denote by (E, d) a metric space. A sequence (x n ) n 0 E N is Cauchy if for every ɛ > 0, there exists N > 0 such that for p, q N, we have: d(x q, x p ) ɛ. The space (E, d) is complete if every Cauchy sequence of (E, d) converges. Property 57. Every Cauchy sequence is bounded. Proof. Let (x n ) n E N be a Cauchy sequence. Then there exists N 0 s.t. for every n 0, d(x n, x N ) 1. Thus the diameter of {x n } n is less than: max 2 d(x k, x N ) + 2 < k N Example The subset (0, 1] endowed with the euclidean distance is not complete, since the sequence ( 1 n+1 ) n 0 is Cauchy but does not converge in (0, 1]. 2. The set of rational numbers Q endowed with the euclidean distance is not complete, since the sequence (x n ) n defined by: x n := [10n 2] 10 n, with [ ] : R N equal to the entire part, 43

44 44 CHAPTER 7. COMPLETENESS OF SOME SPACES is a Cauchy sequence that does not converge in Q. 3. (non-trivial) The space of real numbers R endowed with the euclidean norm is complete. 4. (non-trivial) The space of continuous bounded functions of R endowed with the following norm is complete: f C 0 = sup f(x). x R 5. (non-trivial) The space of C k -functions of [0, 1] endowed with the following norm is complete: f C k = k sup f (j) (x). j=0 x R 6. (non-trivial) The space of non empty compact subsets of R n, endowed with the Hausdorff distance, is complete. All the interest of complete spaces follows from the very useful following theorem: Theorem 59. Every contracting map of a complete space has an unique fixed point. Proof. Let (E, d) be a metric space and let f : E E be a contracting map. To say that f is contracting means that there exists λ < 1 such that x, y E, d(f(x), f(y)) λ d(x, y). Let x E, and let (x n ) n be the sequence defined by: Existence Let ɛ > 0. There exists N 0 such that Then, for every p q N, we have: x n := f n (x), n 0. λ N 1 λ d(x 1, x 0 ) ɛ d(x q, x p ) d(x q, x q+1 ) + d(x q+1, x q+2 ) + + d(x p 1, x p ).

45 7.2. PROOF OF THE COMPLETE SPACES OF SOME SPACES 45 By induction we can show that for any n 0: Thus we have : d(x n, x n+1 ) λ n d(x 0, x 1 ). d(x q, x p ) (λ q + λ q λ p 1 )d(x 0, x1) = λ λ N k=0 λ k d(x 0, x1) = p q q( k=0 λn 1 λ d(x 1, x 0 ) ɛ. )λ k) d(x 0, x 1 ). Consequently, (x n ) n is a Cauchy sequence and so converges to some x. As f(x n ) = x n+1, when n approaches infinity, by continuity of f, we have f(x) = x. Uniqueness Let x, x be two fixed points of f. Then we have: Thus x and x are equal. d(x, x ) = d(f(x), f(x )) λd(x, x ). 7.2 Proof of the complete spaces of some spaces Proof of the completeness of R They are several ways to construct R. In my opinion, the most canonical way is to see the set of real numbers is as the set of number with possibly infinite decimal expression: { } R := x = x 0, x 1 x 2 x n ; x 0 Z, x n {0, 1,..., 9} n 1 s.t. N 0 k N x k 9 The last condition N 0 k N x k 9 is to avoid the fact that 0, is equal to 1. From the expression of R, one can easily define the addition, the subtraction, the absolute value and order relation. So we can define the following Euclidean distance: The following injection of Q respect : d (x, x ) R x x R +. q Q (q n ) n 0, with q 0 := [q] and q n+1 = [10 n+1 q] 10 q n