6.2. AFFINE EXAMPLES

and so on. We then must take the Hausdor¬ limit of this increasing colletion

of sets. To describe the limiting set c from this point of view, it is useful to use

triadic expansions of points in [0, 1]. Thus

0 = .0000000 · · ·

2/3 = .2000000 · · ·

2/9 = .0200000 · · ·

8/9 = .2200000 · · ·

and so on. Thus the set Bn will consist of points whose triadic exapnsion has

only zeros or twos in the ¬rst n positions followed by a string of all zeros. Thus

a point will lie in C (be the limit of such points) if and only if it has a triadic

expansion consisting entirely of zeros or twos. This includes the possibility of

an in¬nite string of all twos at the tail of the expansion. for example, the point

1 which belongs to the Cantor set has a triadic expansion 1 = .222222 · · · .

2 2

Simialrly the point 3 has the triadic expansion 3 = .0222222 · · · and so is in

the limit of the sets Bn . But a point such as .101 · · · is not in the limit of the

Bn and hence not in C. This description of C is also due to Cantor. Notice

that for any point a with triadic expansion a = .a1 a2 a2 · · ·

T1 a = .0a1 a2 a3 · · · , T2 a = .2a1 a2 a3 · · · .

while

Thus if all the entries in the expansion of a are either zero or two, this will also

be true for T1 and T2 a. This shows that the C (given by this second Cantor

description) satis¬es T C ‚ C. On the other hand,

T1 (.a2 a3 · · · ) = .0a2 a3 · · · , T2 (.a2 a3 · · · ) = .2a2 a3 · · ·

which shows that .a1 a2 a3 · · · is in the image of T1 if a1 = 0 or in the image of

T2 if a1 = 2. This shows that T C = C. Since C (according to Cantor™s second

description) is closed, the uniqueness part of the ¬xed point theorem guarantees

that the second description coincides iwth the ¬rst.

The statement that T C = C implies that C is “self-similar”.

6.2.2 The Sierpinski Gasket

Consider the three a¬ne transformations of the plane:

1 1

x x x x 1

’ ’

T1 : , T2 : + ,

y y y y 0

2 2

1 1

x x 0

’

T3 : + .

y y 1

2 2

The ¬xed point of the Hutchinson operator for this choice of T1 , T2 , T3 is called

the Sierpinski gasket, S. If we take our initial set A0 to be the right triangle

with vertices at

0 1 0

, , and

0 0 1

122 CHAPTER 6. HUTCHINSON™S THEOREM AND FRACTAL IMAGES.

then each of the Ti A0 is a similar right triangle whose linear dimensions are one-

half as large, and which shares one common vertex with the original triangle.

In other words,

A1 = T A 0

is obtained from our original triangle be deleting the interior of the (reversed)

right triangle whose vertices are the midpoints of our origninal triangle. Just

as in the case of the Cantor set, successive applications of T to this choice of

original set amounts to sussive deletions of the “middle” and the Hausdor¬ limit

is the intersection of all them: S = Ai .

We can also start with the one element set

0

B0

0

Using a binary expansion for the x and y coordinates, application of T to B0

gives the three element set

0 .1 0

, , .

0 0 .1

The set B2 = T B1 will contain nine points, whose binary expansion is obtained

from the above three by shifting the x and y exapnsions one unit to the right

and either inserting a 0 before both expansions (the e¬ect of T1 ), insert a 1

before the expansion of x and a zero before teh y or vice versa. Proceding

in this fashion, we see that Bn consists of 3n points which have all 0 in the

binary expansion of the x and y coordinates, past the n-th position, and which

are further constrained by the condition that at no earler point do we have

both xi = 1 and y1 = 1. Passing to the limit shows that S consists of all

points for which we can ¬nd (possible ini¬nite) binary expansions of the x and

y coordinates so that xi = 1 = yi never occurs. (For example x = 2 , y = 1

1

2

belongs to S because we can write x = .10000 · · · , y = .011111 . . . ). Again,

from this (second) description of S in terms of binary expansions it is clear that

T S = S.

Chapter 7

Hyperbolicity.

C 0 linearization near a hyperbolic point

7.1

Let E be a Banach space. A linear map

A:E’E

is called hyperbolic if we can ¬nd closed subspaces S and U of E which are

invariant under A such that we have the direct sum decomposition

E =S•U (7.1)

and a positive constant a < 1 so that the estimates

As ¤ a < 1, As = A|S (7.2)

and

A’1 ¤ a < 1, Au = A|U (7.3)

u

hold. (Here, as part of hypothesis (7.3), it is assumed that the restriction of A

to U is an isomorphism so that A’1 is de¬ned.)

u

If p is a ¬xed point of a di¬eomorphism f , then it is called a hyperbolic ¬xed

point if the linear transformation dfp is hyperbolic.

The main purpose of this section is prove that any di¬eomorphism, f is

conjugate via a local homeomorphism to its derivative, dfp near a hyperbolic

¬xed point. A more detailed statement will be given below. We discussed the

one dimensional version of this in Chapter 3.

Proposition 7.1.1 Let A be a hyperbolic isomorphism (so that A’1 is bounded)

and let

1’a

< . (7.4)

A’1

If φ and ψ are bounded Lipschitz maps of E into itself with

Lip[φ] < , Lip[ψ] <

123

124 CHAPTER 7. HYPERBOLICITY.

then there is a unique solution to the equation

(id + u) —¦ (A + φ) = (A + ψ) —¦ (id + u) (7.5)

in the space, X of bounded continuous maps of E into itself. If φ(0) = ψ(0) = 0

then u(0) = 0.

Proof. If we expand out both sides of (7.5) we get the equation

Au ’ u(A + φ) = φ ’ ψ(id + u).

Let us de¬ne the linear operator, L, on the space X by

L(u) = Au ’ u —¦ (id + φ).

So we wish to solve the equation

L(u) = φ ’ ψ(A + u).

We shall show that L is invertible with

A’1

’1