d(xn+1 , xn ) ¤ C n d(x1 , x0 ).

Thus for any m > n we have

m’1

d(x1 , x0 )

d(xi+1 , xi ) ¤ C n + C n+1 + · · · + C m’1 d(x1 , x0 ) ¤ C n

d(xm , xn ) ¤ .

1’C

n

This says that the sequence {xn } is Cauchy. Since X is complete, it must

converge to a limit x, and Kx = lim Kxn = lim xn+1 = x so x is a ¬xed point.

We already know that this ¬xed point is unique. QED

We often encounter mappings which are contractions only near a particular

point p. If K does not move p too much we can still conclude the existence of

a ¬xed point, as in the following:

111

5.4. DEPENDENCE ON A PARAMETER.

Proposition 5.3.1 Let D be a closed ball of radius r centered at a point p in

a complete metric space X, and suppose K : D ’ X is a contraction with

Lipschitz constant C < 1. Suppose that

d(p, Kp) ¤ (1 ’ C)r.

Then K has a unique ¬xed point in D.

Proof. We simply check that K : D ’ D and then apply the preceding theorem

with X replaced by D: For any x ∈ D, we have

d(Kx, p) ¤ d(Kx, Kp)+d(Kp, p) ¤ Cd(x, p)+(1’C)r ¤ Cr+(1’C)r = r QED.

Proposition 5.3.2 Let B be an open ball or radius r centered at p in a complete

metric space X and let K : B ’ X be a contraction with Lipschitz constant

C < 1. Suppose that

d(p, Kp) < (1 ’ C)r.

Then K has a unique ¬xed point in B.

Proof. Restrict K to any slightly smaller closed ball centered at p and apply

Prop. 5.3.1. QED

Proposition 5.3.3 Let K : X ’ X be a contraction with Lipschitz constant

C of a complete metric space. Let x be its (unique) ¬xed point. Then for any

y ∈ X we have

d(y, Ky)

d(y, x) ¤ .

1’C

Proof. We may take x0 = y and follow the proof of Theorem 5.3.1. Alterna-

tively, we may apply Prop. 5.3.1 to the closed ball of radius d(y, Ky)/(1 ’ C)

centered at y. Prop. 5.3.1 implies that the ¬xed point lies in the ball of radius

r centered at y. QED

Prop. 5.3.3 will be of use to us in proving continuous dependence on a

parameter in the next section. In the section on iterative function systems for

the construction of fractal images, Prop. 5.3.3 becomes the “collage theorem”.

We might call Prop. 5.3.3 the “abstract collage theorem”.

5.4 Dependence on a parameter.

Suppose that the contraction “depends on a parameter s”. More precisely,

suppose that S is some other metric space and that

K :S—X ’X

with

dX (K(s, x1 ), K(s, x2 )) ¤ CdX (x1 , x2 ), 0 ¤ C < 1, ∀s ∈ S, x1 , x2 ∈ X. (5.1)

112 CHAPTER 5. THE CONTRACTION FIXED POINT THEOREM

(We are assuming that the C in this inequality does not depend on s.) If we

hold s ∈ S ¬xed, we get a contraction

Ks : X ’ X, Ks (x) := K(s, x).

This contraction has a unique ¬xed point, call it ps . We thus obtain a map

S ’ X, s ’ ps

sending each s ∈ S into the ¬xed point of Ks .

Proposition 5.4.1 Suppose that for each ¬xed x ∈ X, the map

s ’ K(s, x)

of S ’ X is continuous. Then the map

s ’ ps

is continuous.

Proof. Fix a t ∈ S and an > 0. We must ¬nd a δ > 0 such that dX (ps , pt ) <

if dS (s, t) < δ. Our continuity assumption says that we can ¬nd a δ > 0 such

that

dX (K(s, pt ), pt ) = dX (K(s, pt ), K(t, pt ) ¤ (1 ’ C)

if dS (s, t) < δ. This says that Ks moves pt a distance at most (1 ’ C) . But

then the ˜abstract collage theorem”, Prop. 5.3.3, says that

dX (pt , ps ) ¤ . QED

It is useful to combine Proposition 5.3.1 and 5.4.1 into a theorem:

Theorem 5.4.1 Let B be an open ball of radius r centered at a point q in a

complete metric space. Suppose that K : S — B ’ X (where S is some other

metric space) is continuous, satis¬es (5.1) and

dX (K(s, q), q) < (1 ’ C)r, ∀ s ∈ S.

Then for each s ∈ S there is a unique ps ∈ B such that K(s, ps ) = ps , and the

map s ’ ps is continuous.

.

5.5 The Lipschitz implicit function theorem

In this section we follow the treatment in [?]. We begin with the inverse function

theorem which contains the guts of the argument. We will consider a map

F : Br (0) ’ E where Br (0) is the open ball of radius r about the origin in a

Banach space, E, and where F (0) = 0. We wish to conclude the existence of

an inverse to F , de¬ned on a possible smaller ball by means of the contraction

¬xed point theorem.

113

5.5. THE LIPSCHITZ IMPLICIT FUNCTION THEOREM

Proposition 5.5.1 Let F : Br (0) ’ E satisfy F (0) = 0 and

Lip[F ’ id] = » < 1. (5.2)

Then the ball Bs (0) is contained in the image of F where

s = (1 ’ »)r (5.3)

and F has an inverse, G de¬ned on Bs (0) with

»

Lip[G ’ id] ¤ . (5.4)

1’»

Proof. Let us set F = id + v so

id + v : Br (0) ’ E, v(0) = 0, Lip[v] < » < 1.

We want to ¬nd a w : Bs (0) ’ E with

w(0) = 0

and

(id + v) —¦ (id + w) = id.

This equation is the same as

w = ’v —¦ (id + w).

Let X be the space of continuous maps of Bs (0) ’ E satisfying

u(0) = 0

and

»

Lip[u] ¤ .

1’»

Then X is a complete metric space relative to the sup norm, and, for x ∈ Bs (0)

and u ∈ X we have

»

u(x) = u(x) ’ u(0) ¤ x ¤ r.

1’»

Thus, if u ∈ X then

u : Bs ’ B r .

If w1 , w2 ∈ X,