c

with p+ . Hence its itinerary is

ι(p+ ) = (111111 . . . ).

The point ’p+ is carried into p+ under one application of Qc and then stays

there forever. Hence its itinerary is

ι(’p+ ) = (01111111 . . . ).

It follows from the very de¬nition that

ι(Qc (x)) = S(ι(x))

where S is our old friend, the shift map,

S : (s0 s1 s2 s3 . . . ) ’ (s1 s2 s3 s4 . . . )

applied to the space Σ. In other words,

ι —¦ Qc = S —¦ ι.

The map ι conjugates Qc , acting on Λ into the shift map, acting on Σ. To show

that this is a legitimate conjugacy, we must prove that ι is a homeomorphism.

That is, we must show that ι is one-to one, that it is onto, that it is continuous,

and that its inverse is continuous:

One-to one: Suppose that ι(x) = ι(y) for x, y ∈ Λ. This means that Qn (x) c

n

and Q (y) always lie in the same interval, I0 or I1 . Thus the interval [x, y] lies

entirely in either I0 or I1 and hence Qc maps it in one to one fashion onto an

interval contained in either I0 or I1 . Applying Qc once more, we conclude that

Q2 is one-to-one on [x, y]. Continuing, we conclude that Qn is one-to-one on

c c

the interval [x, y], and we also know that (3.9) implies that the length of [x, y]

is increased by a factor of »n . This is impossible unless the length of [x, y] is

zero, i.e. x = y.

Onto. We start with a point s = (s0 s1 s2 . . . ) ∈ Σ. We are looking for a point

x with ι() = s. Consider the set of y ∈ Λ such that

1

d(s, ι(y)) ¤ .

2n

83

3.7. SEQUENCE SPACE AND SYMBOLIC DYNAMICS.

This is the same as requiring that y belong to

Λ © Is0 s1 ...sn

where Is0 s1 ...sn is the interval

Is0 s1 ...sn = {y ∈ I| y ∈ Is0 , Qc (y) ∈ Is1 , . . . Q—¦n (y) ∈ Isn }.

c

So

= Is0 © Q’1 (Is1 ) © · · · © Q’n (Isn )

Is0 s1 ...sn c c

’1 ’(n’1)

= Is0 © Qc (Is1 © · · · © Qc (Isn ))

= Is0 © Q’1 (Is1 ...sn ) (3.13)

c

= Is0 s1 ...sn’1 © Q’n (Isn ) ‚ Is0 ...sn’1 . (3.14)

c

The inverse image of any interval, J under Qc consists of two intervals, one

lying in I0 and the other lying in I1 . For n = 0, Is0 is either I0 or I1 and hence

is an interval. By induction, it follows from (3.13) that Is0 s1 ...sn is an interval.

By (3.14), these intervals are nested. By construction these nested intervals

are closed. Since every sequence of closed nested intervals on the real line has a

non-empty intersection, there is a point x which belongs to all of these intervals.

Hence all the iterates of x lie in I, so x ∈ Λ and ι(x) = s.

Continuity. The above argument shows that the interiors of the intervals

Is0 s1 ...sn (intersected with Λ) form neighborhoods of x that map into small

neighborhoods of ι(x).

Continuity of ι’1 . Conversely, any small neighborhood of x in Λ will contain

one of the intervals Is0 ...sn and hence all of the points t whose ¬rst n coordinates

agree with s = ι(x) will be mapped by ι’1 into the given neighborhood of x.

To summarize: we have proved

Theorem 3.7.1 Suppose that c satis¬es (3.9). Let Λ ‚ [’p+ , p+ ] consist of

those points whose images under Qn lie in [’p, p+ ] for all n ≥ 0. Then Λ is a

c

closed, non-empty, disconnected set. The itinerary map ι is a homeomorphism

of Λ onto the sequence space, Σ, and conjugates Qc to the shift map, S.

Just as in the case of the space X in section 3.4, the periodic points for S

are precisely the periodic or “repeating” sequences. Thus we can conclude from

the theorem that there are exactly 2n points of period (at most) n for Qc . Also,

the same argument as in section 3.4 shows that the periodic points for S are

dense in Σ, and hence the periodic points for Qc are dense in Λ. Finally, the

same argument as in section 3.4 shows that S is transitive on Σ. Hence, the

restriction of Qc to Λ is chaotic.

84 CHAPTER 3. CONJUGACY

Chapter 4

Space and time averages

4.1 histograms and invariant densities

Let us consider a map, F : [0, 1] ’ [0, 1], pick an initial seed, x0 , and compute

its iterates, x0 , x1 , x2 , . . . , xm under F . We would like to see which parts of the

unit interval are visited by these iterates, and how often. For this purpose let

us divide the unit interval up into N subintervals of size 1/N given by

k’1 k N ’1

, k = 1, . . . , N ’ 1, IN =

Ik = , ,1 .

NN N

We count how many of the iterates x0 , x1 , . . . , xm lie in Ik . Call this number

nk . There are m + 1 iterates (starting with, and counting, x0 ) so the numbers

nk

pk =

m+1

add up to one:

p1 + · · · + pN = 1.

We would like to think of these numbers as “probabilities” - the number pk

representing the “probability” that an iterate belongs to Ik . Strictly speaking,

we should write pk (m). In fact, we should write pk (m, x0 ) since the procedure

depends on the initial seed, x0 . But the hope is that as m gets large the pk (m)

tend to a limiting value which we denote by pk , and that this limiting value

will be independent of x0 if x0 is chosen “generically”. We will continue in this

vague, intuitive vein a while longer before passing to a precise mathematical

formulation. If U is a union of some of the Ik , then we can write

p(U ) = pk

Ik ‚U

and think of p(U ) as representing the “probability” that an iterate of x0 be-

longs to U . If N is large, so the intervals Ik are small, every open set U can be

85

86 CHAPTER 4. SPACE AND TIME AVERAGES

closely approximated by a union of the Ik ™s, so we can imagine that the “prob-

abilities”, p(U ), are de¬ned for all open sets, U . If we buy all of this, then we

can write down an equation which has some chance of determining what these

“probabilities”, p(U ), actually are: A point y = F (x) belongs to U if and only if

x ∈ F ’1 (U ). Thus the number of points among the x1 , dots, xm+1 which belong

to U is the same as the number of points among the x0 , . . . , xm which belong

to F ’1 (U ). Since our limiting probability is una¬ected by this shift from 0 to

1 or from m to m + 1 we get the equation

p(U ) = p(F ’1 (U )). (4.1)

To understand this equation, let us put it in a more general context. Suppose

that we have a “measure”, µ, which assigns a size, µ(A), to every open set,

A. Let F be a continuous transformation. We then de¬ne the push forward

measure, F— µ by

(F— µ)(A) = µ(F ’1 (A)). (4.2)