Davidson K.R., Donsig A.P. Real Analysis with Real Applications

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11.6 Topological Conjugacy 373

point y = (.y

. . . )

base 3

is the image τ(x

), where α

= y

/2. Thus τ is a

bijection.

Next we establish the continuity of τ. Let x

∈ X and ε > 0 be given. Choose

N so large that 3

−N

< ε. Now x

belongs to J

...α

. Note that J

...α

and

...α

are disjoint closed sets. Let δ be the positive distance between them.

Suppose that x ∈ X and |x − x

| < δ. Then x also belongs to J

...α

. Hence

τ(x) belongs to T

...α

. This is an interval of length 3

−N

containing τ(x

) as

well. Hence

|τ(x) − τ(x

)| ≤ 3

−N

< ε.

Finally, we use Theorem 11.6.3 to conclude that τ is a homeomorphism. Al-

ternatively, the continuity of τ

−1

can be proved in the same way as for τ . Note that

the map τ preserves the order on the 2

intervals in I

for every n. Hence it follows

easily that τ is monotone increasing.

Now we study those homeomorphisms between two spaces that carry a dynam-

ical system on one space to a different system on the other.

11.6.5. DEFINITION. Let S be a dynamical system on a set X and let T be

a dynamical system on a set Y . These two systems are said to be topologically

conjugate if there is a homeomorphism σ from X onto Y such that σS = T σ or,

equivalently, T = σSσ

−1

. The map σ is called a topological conjugacy between

S and T .

It is clear that if σ is a topological conjugacy between S and T , then

σS

= T

σ for all n ≥ 1.

Hence if x ∈ X is a periodic point for S with period n, then y = σ(x) will be

periodic for T of the same order. Moreover, the fact that σ is a homeomorphism

means that convergent sequences in X correspond exactly to convergent sequences

in Y under this map. Hence a periodic point y = σ(x) will be attracting or repelling

exactly as x is. Indeed, we have O

(σ(x)) = σ(O

(x)) for every point x ∈ X.

Topological conjugacy is an equivalence relation. First, if σ conjugates S onto

T , then σ

−1

conjugates T back onto S. If R is conjugate to S and S is conjugate

to T , then R and T are conjugate. Evidently, the identity map id : S → S is a

topological conjugacy from S to itself.

We will study topological conjugacy by examining a few examples.

11.6.6. PROPOSITION. The tent map of Exercise 11.3.E and the quadratic

map Q

x = 4x − 4x

on [0, 1] are topologically conjugate. Hence Q

is chaotic

on [0, 1].

PROOF. We will pull the appropriate homeomorphism out of the air. So the rest

of this proof will be easy to understand, but it won’t explain how to choose the

homeomorphism.

374 Discrete Dynamical Systems

Let σ(x) = sin

(

x). It is easily checked that σ is strictly increasing and

continuous on [0, 1] and that σ(0) = 0 and σ(1) = 1. Hence by Example 11.6.2,

it follows that σ is a homeomorphism of [0, 1]. Now using Q

x = 4x(1 − x),

compute

σ(x) = 4sin

(

x) cos

(

x) = sin

(πx).

Likewise,

σ(T x) =

(

sin

(2x)

= sin

(πx) if 0 ≤ x ≤

sin

(2 − 2x)

= sin

(π − πx) if

≤ x ≤ 1.

Thus σT = Q

σ; and so σ is a topological conjugacy intertwining T and Q

By Exercise 11.5.B, the tent map is chaotic. Hence Q

is also. ¥

The goal of the rest of this section is to continue our analysis of the quadratic

maps Q

for a > 2+

√

5. We will establish a topological equivalence with the shift

on the Cantor set. Hence these quadratic maps are all topologically equivalent to

each other, so that dynamically they all behave in exactly the same way.

11.6.7. THEOREM. For a > 2 +

√

5, the quadratic maps Q

on the set X

topologically conjugate to the shift S on the Cantor set C.

PROOF. Recall from Example 11.5.6 that X

is a Cantor set. We will construct a

homeomorphism along the lines of Example 11.6.4, except that the ordering will

be determined by the dynamics rather than by the usual order on the line. Recall

the notation from that example.

The ﬁrst step in the construction of the Cantor set X

is the set

= J

∪ J

= {x ∈ [0, 1] : Q

x ∈ [0, 1]}.

For each point x in X

, Q

n−1

x belongs to X

, and thus to either J

or J

. Deﬁne

the itinerary of x to be the sequence Γx = γ

. . . of 0s and 1s deﬁned by the

condition that

n−1

x ∈ J

for all n ≥ 1.

The interval J

...α

n−1

is mapped bijectively by Q

n−1

onto the whole unit interval.

And X

∩ J

...α

n−1

is mapped into X

, and in particular into I

= J

∪ J

. This

dichotomy determines the sets J

...α

n−1

and J

...α

n−1

, as one is mapped onto J

by Q

n−1

and the other is mapped onto J

. The order we need to keep track of is

this itinerary ordering, not the usual order on R. Notice that this discussion shows

that if x and y both belong to X

∩ J

...α

, then the itineraries Γx and Γy agree

for the ﬁrst n terms.

Deﬁne a map σ from X

to C by

σ(x) = y

Γx

k≥1

2γ

−k

Let us verify that σ is a homeomorphism.

For any x ∈ X

and ε > 0, choose N so that 3

−N

< ε. Then x belongs to

one of the Nth level intervals J

...α

. Let δ be the positive distance between this

11.6 Topological Conjugacy 375

interval and the remaining I

\ J

...α

. Then any y ∈ X

with |x − y| < δ also

belongs to J

...α

. Hence the itinerary of y agrees with x for the ﬁrst N terms.

This means that σ(x) and σ(y) belong to the same N th-level interval for C. Hence

|σ(x) − σ(y)| ≤ 3

−N

< ε.

So σ is continuous.

To see that σ is a bijection, consider any point y ∈ C. As usual, we write

y = .y

. . .

base 3

in ternary using a sequence of 0s and 2s. This is the image of

all points x ∈ X

with itinerary Γ = γ

. . . given by γ

= y

/2. However, Γ

determines another unique sequence a = α

. . . by the relation

n−1

...α

= J

for all n ≥ 1.

So the points x with itinerary Γ are the points in

n≥1

...α

= {x

As we have noted before, this intersection consists of exactly one point. So σ is

onto because this set is nonempty for each Γ; and it is one-to-one because the set is

always a singleton.

Now we may apply Theorem 11.6.3 to see that σ is a homeomorphism.

We must show that σ intertwines Q

and S. Suppose that x ∈ X

has itinerary

Γ = γ

. . . . Then the itinerary of Q

x is evidently γ

. . . because

n−1

x = Q

x ∈ J

n+1

for all n ≥ 1.

This is just saying that

σ(Q

x) = Sσ(x).

Therefore, Q

is topologically conjugate to the shift. ¥

11.6.8. COROLLARY. The quadratic maps Q

for a > 2 +

√

5 have a dense

set of transitive points in X

PROOF. This follows from our discussion of the shift in Example 11.5.16. The

shift has a dense set of transitive points. So any map topologically conjugate to the

shift must have such a set as well. ¥

11.6.9. REMARK. We have been studying the quadratic logistic maps in detail

throughout this chapter. Our early arguments depended on speciﬁc calculations for

these functions. However, all of the arguments for chaos depend only on a few

fairly general properties.

Suppose that f is a function on [0, 1] with f(0) = f(1) = 0 that is unimodal,

meaning that f increases to a maximum at a point (x

, y

) and then decreases back

down to (1, 0). In order for our arguments to work, all we need is that y

> 1 and

(x)| ≥ c > 1 for all x in

= {x ∈ [0, 1] : f(x) ∈ [0, 1]}.

376 Discrete Dynamical Systems

Then the argument of Example 11.5.6 would apply to show that the set

X = {x ∈ [0, 1] : f

(x) ∈ [0, 1] for all n ≥ 1}

is a Cantor set.

The preceding proof showing that Q

is topologically conjugate to the shift

only relied on the fact that Q

J = [0, 1] for every component interval of the nth-

level set I

for each n ≥ 1. It is easy to see that this property also holds in the

generality of the unimodal function f . Thus, in particular, f is chaotic on X. This

enables us to recognize chaos in many situations.

A simple example of this which is very similar to the quadratic family is the

function f(x) = 3x − 3x

. However, it is more instructive to look at the quadratic

maps again.

Consider the graph of Q

3.75

in Figure 11.10. Notice that Q

3.75

has a ﬁxed point

at 1 − 1/3.75 =

; and Q

3.75

as well. On the interval J = [

], Q

3.75

decreases from (

) to a local minimum at (

225

1024

), and increases again to the

point (

). Since

225

1024

≈ .22 < .267 ≈

, this graph “escapes” the square

J ×J. The qualitative behaviour of this part of the graph is just like that of Q

for

a > 2 +

√

FIGURE 11.10. The graph of Q

3.75

, with J × J and K × K marked.

To verify that our proof applies, we need to see that the absolute value of the

derivative is greater than 1 on

= {x ∈ J : Q

3.75

(x) ∈ J}.

It is notationally easier to use the generic parameter a and substitute 3.75 for a later.

What exactly is J

? To compute it, we ﬁrst solve the quadratic Q

(x) =

. This

has solutions

√

− 4

11.6 Topological Conjugacy 377

These are roughly .077 and .923 for a = 3.75. The points that Q

maps to the

endpoints of J

are seen from the graph to be solutions of

√

− 4

= Q

x = ax − ax

This quadratic has solutions

− 2a − 2

√

− 4

For a = 3.75, we obtain J

= [.2667, .4377] ∪ [.5623, .7333]. Now the derivative

of Q

is monotone increasing on J

, changing sign at x = .5, and is symmetric

about the midpoint. So the minimal slope is obtained at the two interior endpoints

)

) = Q

) Q

)

= a

(1 − 2Q

)(1 − 2x

)

= a

√

− 4

− 2a − 2

√

− 4

− 4)(a

− 2a − 2

− 4) ≈ 1.482.

Thus our earlier arguments apply to show that there is a Cantor set X contained

in J on which Q

3.75

acts chaotically, and in fact is topologically conjugate to the

shift map.

Now Q

3.75

maps J

onto the interval K = [.733, .923]. The restriction of Q

3.75

to K × K behave in exactly the same way, and there is another Cantor set Y on

which Q

3.75

behaves chaotically. Moreover, Q

3.75

maps the Cantor set X into Y

and vice versa. From this, it is not difﬁcult to see that Q

3.75

acts chaotically on the

Cantor set X ∪ Y . See the Exercises.

Just as Q

is actually chaotic for a > 4 with a more delicate proof, the same

is true for this analysis of Q

. It can be shown that the preceding argument works

whenever the graph in the interval J ×J escapes in the middle. This occurs at about

a = 3.6786. So once a > 3.6786, the quadratic map Q

is chaotic on a Cantor set.

Exercises for Section 11.6

A. Use the sequential characterization of continuity Theorem 5.3.1 (2) to provide another

proof of Theorem 11.6.3. HINT: Let (x

)

∞

n=1

be a convergent sequence in Y with

lim

n→∞

= a. Use Exercise 4.4.F on (f(x

))

∞

n=1

B. Consider the map of [0, 2π) onto the circle T by wrapping around exactly once. Show

that this map is continuous, one-to-one, and onto. Show that this is not a homeomor-

phism. Explain why this does not contradict Theorem 11.6.3.

C. Show that the circle T is not homeomorphic to the interval [0, 1]. HINT: If σ maps

T onto [0, 1], show that there are at least two points mapping to each 0 < y < 1.

D. Consider the map of the circle to itself given by T θ ≡ θ +

2π

+ ε sin(2πnx), where

0 < ε < 1/2πn.

378 Discrete Dynamical Systems

(a) Compute T

(θ) and deduce that T is a homeomorphism.

(b) Show that 0 and

are periodic points.

x, O(

)) is strictly decreasing.

(d) Show that O(0) is a repelling orbit and O(

) is attracting, and that ω(x) = O(

)

except for x ∈ O(0).

E. Show that f(x) = 1 − 2|x| and g(x) − 1 − 2x

as dynamical systems on [−1, 1] are

topologically conjugate as follows:

(a) If ϕ is a homeomorphism of [−1, 1] such that ϕ(f (x)) = g(ϕ(x)), show that ϕ is

an odd function such that ϕ(−1) = −1 and ϕ(0) = 0.

(b) Use ﬁxed points to show that ϕ(1/3) = 1/2. Deduce that ϕ(2/3) =

√

3/2.

F. Let f be a homeomorphism of [0, 1] with no ﬁxed points in (0, 1).

(a) Show that f is strictly monotone increasing, and either f(x) < x for all x in (0, 1)

or f(x) > x for all x in (0, 1).

(b) If f(x) > x, prove that the orbit of x under f converges to 1 and orbit under f

−1

converges to 0.

(.5), f

k+1

(.5)) for k ∈ Z.

(d) Let f and g be two homeomorphisms of [0, 1] with no ﬁxed points in (0, 1). Prove

that they are topologically conjugate.

HINT: Assume ﬁrst that f(x) > x and g(x) > x for all x in (0, 1). Deﬁne ϕ from

[.5, f(.5)] onto [.5, g(.5)]. Extend this to the whole interval to obtain a conjugacy.

G. (a) Show that every quadratic function p(x) = ax

+ bx + c on R is topologically

conjugate to some q(x) = x

+ d. HINT: Use a linear map τ (x) = mx + e.

Compute p(τ(x)) and τ (q(x)) and equate coefﬁcients to solve for m, e, and d.

(b) For which values of d is q(x) = x

+d topologically conjugate to one of the logistic

maps Q

for a > 0? What are the dynamics of q when d is outside this range?

H. Suppose that T : I → I is given, and T

maps an inﬁnite compact subset X into itself

and is chaotic on X. Show that T is chaotic on X ∪ T X.

11.7. Iterated Function Systems

An iterated function system, or IFS, is a multivariable discrete dynamical

system. Under reasonable hypotheses, these systems have a unique compact in-

variant set. This invariant set exhibits certain self-similarity properties. Such sets

have become known as fractals.

We begin with a ﬁnite set T = {T

, . . . , T

} of contractions on a closed subset

X of R

. This family of maps determines a multivariable dynamical system. The

orbit of a point x will consist of the set of all points obtained by repeated application

of the maps T

in any order with arbitrary repetition. That is, for each ﬁnite word

. . . i

in the alphabet {1, . . . , r}, the point T

. . . T

x is in the orbit O(x).

We wish to ﬁnd a compact set A with the property that

A = T

A ∪ T

A ∪ ··· ∪ T

Surprisingly this set turns out to be unique!

11.7 Iterated Function Systems 379

A similitude is a map T that is a scalar multiple of an isometry. That is, there

is a constant r > 0 so that kTx −T yk = rkx −yk for all x, y ∈ X. Such maps are

obtained as rotations, translations, and scalings (see Exercise 11.7.F). In particular,

these maps are a similarity in the geometric sense that they map sets to similar sets

and so preserve shape, up to a scaling factor.

In the event that each T

is a similitude, the fact that A = T

A∪T

A∪···∪T

means that each T

A is similar to A. This will be especially evident in examples

in which these r sets are disjoint. The process repeats and T

A decomposes as

A = T

A∪T

A∪···∪T

A. After k steps, A is decomposed into r

similar

pieces. This symmetry property is called self-similarity and is characteristic of

fractals arising from iterated function systems.

11.7.1. EXAMPLES.

(1) Let X = R

, and consider three afﬁne maps T

x =

x, T

x =

x + (2, 0),

and T

x =

x + (1,

√

3). Notice that each T

is a similitude with scaling factor

. It is easy to verify that the ﬁxed points of these three maps are v

= (0, 0),

= (4, 0), and v

= (2, 2

√

3), respectively.

Let ∆ be the solid equilateral triangle with these three vertices. A computa-

tion shows that T

∆ for i = 1, 2, 3 are the three equilateral triangles with half the

dimensions of the original that lie inside ∆ and share the vertex v

with ∆. So

∆

= T

∆ ∪ T

∆ equals ∆ with the middle triangle removed.

Since ∆

⊂ ∆, it followsfairly easily (see Corollary 11.7.5) that when we iterate

the procedure by setting

∆

k+1

= T

∆

∪ T

∆

∪ T

∆

a decreasing sequence of compact sets is obtained. The intersection ∆

∞

of these

sets is the Sierpinski snowﬂake of Exercise4.4.J (see Figure 4.4). Ithas the property

that we are looking for,

∆

∞

= T

∆

∞

∪ T

∆

∞

∪ T

∆

∞

(2) Not all fractals are solid ﬁgures. The von Koch curve is obtained from an IFS

using the following four similitudes:

, T

−

√

−

√

, T

Let B

= {(x, 0) : x ∈ [0, 1]} and deﬁne B

k+1

= T

∪T

Graphing these ﬁgures shows an increasingly complex curve emerging in which the

previous curve is scaled by 1/3 and used to replace the four line segments of B

See Figure 11.11.

380 Discrete Dynamical Systems

FIGURE 11.11. The sets B

through B

As we will see, this construction works in great generality, and manysets can be

obtained as the invariant sets for iterated function systems. To establish these facts,

we need a framework, in this case, a metric space. Let K(X) denote the collection

of all nonempty compact subsets of X. This is a metric space with respect to the

Hausdorff metric of Example 9.1.2(5),

(A, B) = max

sup

a∈A

dist(a, B), sup

b∈B

dist(b, A)

Our ﬁrst result is the completeness of K(X) in the Hausdorff metric.

11.7.2. THEOREM. If X is a closed subset of R

, the metric space K(X) of

all compact subsets of X with the Hausdorff metric is complete.

PROOF. Let A

be a Cauchy sequence of compact sets in K(X). Deﬁne

A =

k≥1

[

i≥k

Observe that for any ε > 0, there is an integer N so that d

, A

) < ε for all

i, j ≥ N. In particular, it follows from the deﬁnition of the Hausdorff metric that

⊂ (A

)

for all i ≥ N. Consequently, A ⊂ (A

)

. Now, (A

)

is a closed

and bounded subset of R

and so, by the Heine–Borel Theorem, is compact. It

follows that A is the decreasing intersection of nonempty compact sets; by Cantor’s

Intersection Theorem, A is a nonempty compact set.

Having shown that A ∈ K(X), it remains only to prove that (A

) converges to

A. We also have A ⊂ (A

)

⊂ ((A

)

= (A

)

2ε

for all i ≥ N . Conversely, ﬁx

∈ A

with i ≥ N. For each j > i, A

⊂ (A

)

and thus there is a point a

∈ A

11.7 Iterated Function Systems 381

with ka

−a

k < ε. Each a

lies in the bounded set (A

)

. By compactness, there

is a convergent subsequence lim

l→∞

= a. Clearly, ka

− ak ≤ ε.

We must show that a belongs to A. However, all but the ﬁrst few terms of (a

)

lie in

j≥k

, and so a belongs to

j≥k

. Thus a also lies in the intersection

of these sets, A. We deduce that dist(a

, A) ≤ ε for each point in A

and therefore

⊂ A

. Combining the two estimates, d

, A) ≤ 2ε for all i ≥ N. Therefore,

converges to A. ¥

Next we deﬁne a map from K(X) into itself by T A = T

A ∪T

A ∪···∪T

We need to verify that T A is in K(X) (i.e., that T A is compact). Each T

continuous, and the continuous image of a compact set is compact (Theorem 5.4.3).

Also, the ﬁnite union of compact sets is compact, and therefore T A is a compact

set. Our goal is to show that this is a contraction. First we need an easy lemma

whose proof is left as Exercise 11.7.A.

11.7.3. LEMMA. Let A

, . . . , A

and B

, . . . , B

be compact subsets of R

Then d

∪ ··· ∪ A

, B

∪ ··· ∪ B

) ≤ max

, B

), . . . , d

, B

)

11.7.4. THEOREM. Let X be a closed subset of R

and let T

, . . . , T

be con-

tractions of X into itself. Let s

be the Lipschitz constants for each T

and set

s = max{s

, . . . , s

}. Then T is a contraction of K(X) into itself with Lipschitz

constant s. Hence there is a unique compact subset A of X such that

A = T

A ∪ T

A ∪ ··· ∪ T

Moreover, if B is any compact set, we have the estimates

B, A) ≤ s

(B, A) ≤

1 − s

(B, T B).

PROOF. Let A and B be any two compact subsets of X. Observe that

A, T

B) ≤ s

(A, B).

Indeed, if a ∈ A, then there is a b ∈ B with ka − bk ≤ d

(A, B). Hence

a − T

bk ≤ s

(A, B). So sup

a∈A

dist(T

a, T

B) ≤ s

(A, B). Reversing

the roles of A and B, we arrive at the desired estimate. By Lemma 11.7.3, it follows

that d

(T A, T B) ≤ sd

(A, B).

Theorem 11.7.2 shows that (K(X), d

) is a complete metric space. The proof

of the Contraction Principle goes through verbatim in the metric space case. Alter-

natively, the proof of Theorem 9.5.3 shows that any complete metric space Y can

be embedded as a complete (hence closed) subset of the complete normed linear

space C

(Y ). Thus we may apply the Banach Contraction Principle to T . It follows

that there is a unique ﬁxed point A = T A. By deﬁnition of T , this is the unique

compact set such that A = T

A ∪ T

A ∪ ··· ∪T

A. It is also an immediate conse-

quence of the Contraction Principle that A is obtained as the limit of iterates of T

applied to any initial set B, and the estimates follow directly from the estimates in

the Contraction Principle. ¥

382 Discrete Dynamical Systems

11.7.5. COROLLARY. Suppose that T is a contraction of K(X) into itself with

Lipschitz constant s. If B is a compact set such that T B ⊂ B, then the ﬁxed point

is given by A =

k≥0

PROOF. We show by induction that T

k+1

B ⊂ T

B for k ≥ 0. If k = 0, this is

true by hypothesis. Assuming that T

B ⊂ T

k−1

B, we have

k+1

B = T (T

B) ⊂ T (T

k−1

B) = T

By Theorem 11.7.4, the ﬁxed point A is the limit of the sequence T

B. From the

proof of Theorem 11.7.2, this limit is given by

A =

k≥1

[

i≥k

B =

k≥1

An excellent choice for the compact set B to use in computing the limit set is

modeled by the Sierpinski snowﬂake, Example 11.7.1(1). Another good choice for

an initial compact set B is a single point {x} that happens to belong to A. Such

points are easy to ﬁnd. Each T

is a contraction on X and thus has a unique ﬁxed

point x

that may be found by iteration of T

applied to any initial point. We give a

signiﬁcant strengthening of this fact, which hints at the dynamical properties of the

iterated function system.

11.7.6. THEOREM. For each word w = i

. . . i

in the alphabet {1, . . . , r},

there is a unique ﬁxed point a

of T

= T

. . . T

. Each a

belongs to the

ﬁxed set A, and the set of all of these ﬁxed points is dense in A.

If a is any point in A, the orbit O(a) =

a : w is a word in {1, . . . , r}

dense in A.

PROOF. First observe that the composition of contractions is a contraction (see

Exercise 11.1.E). Therefore, each T

is a contraction and hence has a unique ﬁxed

point a

. Moreover, starting with any point x, the iterates T

x converge to a

Take x to be any point in A. Since each T

maps A into itself, T

x is in A for all

k ≥ 0. As A is closed, the limit a

belongs to A.

Next we prove that the set of these ﬁxed points

:w is a word in {1, . . . , r}

is dense in A. Now A = T A = T

A ∪ ··· ∪ T

A. So

A = T

A ∪ T

A ∪ . . . ∪ T

A ∪

A ∪ T

A ∪ . . . ∪ T

A ∪

. . . ∪

A ∪ T

A ∪ . . . ∪ T

Repeating this N times, we obtain

A = T

A =

[

words w of length N