So the map K : X ’ X

K(u) = ’v —¦ (id + u)

114 CHAPTER 5. THE CONTRACTION FIXED POINT THEOREM

is a contraction. Hence there is a unique ¬xed point. This proves the proposi-

tion.

Now let f be a homeomorphism from an open subset, U of a Banach space,

E1 to an open subset, V of a Banach space, E2 whose inverse is a Lipschitz

map. Suppose that h : U ’ E2 is a Lipschitz map satisfying

Lip[h]Lip[f ’1 ] < 1. (5.5)

Let

g = f + h. (5.6)

We claim that g is open. That is, we claim that if y = g(x), then the image

of a neighborhood of x under g contains a neighborhood of g(x). Since f is a

homeomorphism, it su¬ces to establish this for g—¦f ’1 = id+h—¦f ’1. Composing

by translations if necessary, we may apply the proposition. QED

We now want to conclude that g is a homeomorphism = continuous with

continuous inverse, and, in fact, is Lipschitz:

Proposition 5.5.2 Let f and g be two continuous maps from a metric space

X to a Banach space E. Suppose that f is injective and f ’1 is injective with

Lipschitz constant Lip[f ’1 ]. Suppose that g satis¬es

1

Lip[g ’ f ] < .

Lip[f ’1 ]

Then g is injective and

Lip[f ’1 ]

1

Lip[g ’1 ] ¤ = . (5.7)

1 1 ’ Lip[g ’ f ]Lip[f ’1 ]

’ Lip[g ’ f ]

Lip[f ’1 ]

Proof. By de¬nition,

d(x, y) ¤ Lip[f ’1 ] f (x) ’ f (y)

where d denotes the distance in X and denotes the norm in E. We can write

this as

d(x, y)

f (x) ’ f (y) ≥ .

Lip[f ’1 ]

So

g(x) ’ g(y) ≥ f (x) ’ f (y) ’ (g ’ f )(x) ’ (g ’ f )(y)

1

≥ ’ Lip[g ’ f ] d(x, y).

Lip[f ’1 ]

Dividing by the expression in parenthesis gives the proposition. QED

We can now be a little more precise as to the range covered by g:

115

5.5. THE LIPSCHITZ IMPLICIT FUNCTION THEOREM

Proposition 5.5.3 Let U be an open subset of a Banach space E1 , and g be a

homeomorphism of U onto an open subset of a Banach space, E2 . Let x ∈ U

and suppose that

Br (x) ‚ U.

If g ’1 is Lipschitz with

Lip[g ’1 ] < c

then

B r (g(x)) ‚ g Br (x) .

c

Proof. By composing with translations, we may assume that x = 0, g(x) = 0.

Let

v ∈ B r (0).

c

Let

T = T (v) = sup{t|[0, t]v ‚ g Br (0) }.

We wish to show that T = 1. Since g(Br (0)) contains a neighborhood of 0, we

know that T (v) > 0, and, by de¬nition,

(0, T )v ‚ g Br (0) .

By the Lipschitz estimate for g ’1 we have

g ’1 (tv) ’ g ’1 (sv) ¤ c|t ’ s| v .

This implies that the limit limt’T g ’1 (tv) exists and

lim g ’1 (tv) ∈ Br (0).

t’T

So

T v ∈ (g Br (0) .

If T = T (v) < 1 we would have

g ’1 (T v) g ’1 (T v) ’ g ’1 (0)

¤

¤ c Tv

= cT v

< cv

¤ r.

This says that

T v ∈ g (Br (0)) .

But since g is open, we could ¬nd an > 0 such that [T, T + ]v ‚ g (Br (0))

contradicting the de¬nition of T . QED

The above three propositions, taken together, constitute what we might call

the inverse function theorem for Lipschitz maps. But the contraction ¬xed point

116 CHAPTER 5. THE CONTRACTION FIXED POINT THEOREM

theorem allows for continuous dependence on parameters, and gives the ¬xed

point as a continuous function of the parameters. So this then yields the implicit

function theorem.

The di¬erentiability of the solution, in the case that the implicit function is

assumed to be continuously di¬erentiable follows as in Chapter 1.

Chapter 6

Hutchinson™s theorem and

fractal images.

6.1 The Hausdor¬ metric and Hutchinson™s the-

orem.

Let X be a complete metric space. Let H(X) denote the space of non-empty

compact subsets of X. For any A ∈ H(X) and any positive number , let

A = {x ∈ X|d(x, y) ¤ , for some y ∈ A}.

We call A the -collar of A. Recall that we de¬ned

d(x, A) = inf d(x, y)

y∈A

to be the distance from any x ∈ X to A, then we can write the de¬nition of the

-collar as

A = {x|d(x, A) ¤ }.

Notice that the in¬mum in the de¬nition of d(x, A) is actually achieved, that

is, there is some point y ∈ A such that

d(x, A) = d(x, y).

This is because A is compact. For a pair of non-empty compact sets, A and B,

de¬ne

d(A, B) = max d(x, B).

x∈A

So

d(A, B) ¤ if f A ‚ B .

Notice that this condition is not symmetric in A and B. So Hausdor¬ introduced

h(A, B) = max{d(A, B), d(B, A)} (6.1)

= inf{ | A ‚ B and B ‚ A }. (6.2)