L . (7.6)

(1 ’ a)

Assume, for the moment that we have proved (7.6). We are then looking for a

solution of

u = K(u)

where

K(u) = L’1 [φ ’ ψ(id + u)].

But

L’1 [φ ’ ψ(id + u1 ) ’ φ + ψ(id + u2 )]

K(u1 ) ’ K(u2 ) =

L’1 [ψ(id + u2 ) ’ ψ(id + u1 )]

=

L’1 Lip[ψ] u2 ’ u1

¤

< c u 2 ’ u1 , c<1

if we combine (7.6) with (7.4). Thus K is a contraction and we may apply the

contraction ¬xed point theorem to conclude the existence and uniqueness of the

solution to (7.5). So we turn our attention to the proof that L is invertible and

of the estimate (7.6). Let us write

Lu = A(M u)

where

M u = u ’ A’1 u —¦ (A + φ).

Composition with A is an invertible operator and the norm of its inverse is

A’1 . So we are reduced to proving that M is invertible and that we have the

estimate

1

M ’1 ¤ . (7.7)

1’a

7.1. C 0 LINEARIZATION NEAR A HYPERBOLIC POINT 125

Let us write

u = f • g, f : E ’ S, g : E ’ U

in accordance with the decomposition (7.1). So if we let Y denote the space of

bounded continuous maps from E to S, and let Z denote the space of bounded

continuous maps from E to U , we have

X =Y •Z

and the operator M sends each of the spaces Y and Z into themselves since A’1

preserves S and U . We let Ms denote the restriction of M to Y , and let Mu

denote the restriction of M to Z. It will be enough for us to prove that each of

the operators Ms and Mu is invertible with a bounds (7.7) with M replaced by

Ms and by Mu . For f ∈ Y let us write

N f = A’1 f —¦ (A + φ).

Ms f = f ’ N f,

We will prove

Lemma 7.1.1 The map N is invertible and we have

N ’1 ¤ a.

Proof.We claim that he map A + φ a homeomorphism with Lipschitz inverse.

Indeed

1

Ax ≥ x

A’1

so

1

Ax + φ(x) ’ Ay ’ φ(y) ≥ ’ Lip[φ] x’y

A’1

a

≥ x’y

A’1

by (7.4). This shows that A + φ is one to one. Furthermore, to solve

Ax + φ(x) = y

for x, we apply the contraction ¬xed point theorem to the map

x ’ A’1 (y ’ φ(x)).

The estimate (7.4) shows that this map is a contraction. Hence A + φ is also

surjective.

Thus the map N is invertible, with

N ’1 f = As f —¦ (A + φ)’1 .

Since As ¤ a, we have

N ’1 f ¤ a f .

126 CHAPTER 7. HYPERBOLICITY.

(This is in terms of the sup norm on Y .) In other words, in terms of operator

norms,

N ’1 ¤ a.

’1

We can now ¬nd Ms by the geometric series

’1

= (I ’ N )’1

Ms

= [(’N )(I ’ N ’1 )]’1

= (’N )’1 [I + N ’1 + N ’2 + N ’3 + · · · ]

and so on Y we have the estimate

a

’1

¤

Ms .

1’a

The restriction, Mu , of M to Z is

Mu g = g ’ Qg

with

Qg ¤ a g

so we have the simpler series

Mu = I + Q + Q 2 + · · ·

’1

giving the estimate

1

Mu ¤ .

1’a

Since

a 1

<

1’a 1’a

the two pieces together give the desired estimate

1

M¤ ,

1’a

completing the proof of the ¬rst part of the proposition. Since evaluation at zero

is a continuous function on X, to prove the last statement of the proposition

it is enough to observe that if we start with an initial approximation satisfying

u(0) = 0 (for example u ≡ 0) Ku will also satisfy this condition and hence so

will K n u and therefor so will the unique ¬xed point.

Now let f be a di¬erentiable, hyperbolic transformation de¬ned in some

neighborhood of 0 with f (0) = 0 and df0 = A. We may write

f =A+φ

where

φ(0) = 0, dφ0 = 0.

We wish to prove

7.1. C 0 LINEARIZATION NEAR A HYPERBOLIC POINT 127

Theorem 7.1.1 There exists neighborhoods U and V of 0 and a homeomor-

phism h : U ’ V such that

h —¦ A = f —¦ h. (7.8)

We prove this theorem by modifying φ outside a su¬ciently small neighborhood

of 0 in such a way that the new φ is globally de¬ned and has Lipschitz constant

less than where satis¬es condition (7.4). We can then apply the proposition

to ¬nd a global h which conjugates the modi¬ed f to A, and h(0) = 0. But since

we will not have modi¬ed f near the origin, this will prove the local assertion

of the theorem. For this purpose, choose some function ρ : R ’ R with