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thoereme Banach's fixed point and invariant field

Theorem.

Let (E, II) a space is open in Banach.U E.

T is a mapping from U into E, k-strictly contracting.

Let f (x) = xT (x). Then:

a) f is a open, in particular, f (U) is open.

b) f, U f (U) is a homeomeophisme


Proof

a) To show that f is an open application, just check

that for any x in U, if B (x, r) (= the open ball of center x and radius r

) is in U, then B [f (x), r (1-k)] is in f (B (x, r) ).

Then, a z in the ball B [f (x), (1-k) r] and we define an application

S B (x, r) in E by S (y) = z + T (y) then S is strictly k-

Contracting and:


IS (x) = Iz-xI + T (x)-Iz-xI = f (x) I

So, the local version of the theorem of Banach (see message

20 / 2 / 07), there exists an x 'in B (x, r) unique such that x' = z + Tx ', ie,

f (x) = z; the

B [f (x) ( 1-k) r] is in f [B (x, r)]. Finally f is a

open, in particular, f (U) is open.

b) We observe that if u, v in U, then:

If (u)-f (v) I> = Iu-vI-IT (u)-T (v) I> = (1 - k) Iu-vI

this entails that f is injective, so f, U f (U) is a continuous mapping

, open and bijective, ie, f is a homeomorphism.


As consequence we have the following result: Corollary



E is a Banach space and T is a mapping of E into E, < (1-k)r

strictly contractante.Alors f = IT is a homeomorphism of E onto E.

(Hint: use the Banach contraction principle to prove

that f (E) = E)


NOTE.The corollary above (generalized) matches in a

the Banach algebra L (E) (= the space of continuous linear

E in E) with the usual standard:

Exercice.Soit A in L (E) such that IAI

contraction Banach demonstrate that IA is a topological isomorphism

E E.