Denise Milani Trying Bras

Generalized contraction principle (based Krasnoselskii and authors).

Among the generalizations Secures Breakthrough the principle of contraction
Banach, mention the following result
(Krasnoselskii and authors)



Theorem Let (E, d) a metric space T a complet.Soit

generalizee contraction of E in E, c- ie,

for all a, b in IR ^{+}-{ 0], there exists a constant 0



d (Tx, Ty)

Then T admits a unique fixed point z, and for all x in

E, the sequence {T} x ^ {n} converges to z. Two

demostration are proposees.L a constructive
< k= k(a,b)<1 telle que:
is similar to the original demonstration of the principle of contraction
<= k d(x,y) , a<=d(x,y)<=b
Banach (see [Krasnoselskii-Burd-Kolesov]
.
The other is based on reasoning by contradiction and the principle of

nested. (Cantor's Theorem)

(See Krasnoselskii-Zabreiko].


As a consequence, we find the following results:

corollary.

Let (E, d) a metric space is an application complet.T

E in E. Then T has a fixed point only if one of

following conditions is true:

i) T is strictly contracting

Banach (1922)

ii) E is compact and d (Tx, Ty)

with x different from y


Edelstein (1962)

iii) There exists a function f IR ^ {+} in

IR ^ {+}-upper semi-continuous, f (r) < d(x,y) sur ExE

on IR ^ {+} \\ { 0} and such that


d (Tx, Ty)

Boyd-Wong (1969)

< r
iv) There exists a function f IR ^ {+} in

IR ^ {+} growing, continues to right, f (r)

on IR ^ {+} \\ {0} and such that <= f(d(x,y))) sur ExE.


d (Tx, Ty)

Browder (1968)

v) There exists a function f IR ^ {+} in < r

IR ^ {+} growing, check for any r in

IR ^ {+} f ^ {n} (r ) -> o as n -> infinity and such that
<=f(d(x,y)) sur ExE.


d (Tx, Ty)

Matkowski (1975)

Proof: The proof is

facile.Pour i) is obvious.

show by example ii).
<= f(d(x,y)) sur ExE.
Let a, b in IR ^{+}-{ 0}. Since K is compact and

T is continuous, there exists (x_ {0}, y_ {0}) in kxk as:

sup {d (Tx, Ty) / d (x, y): a

= d (T (x_ {0}, {0} Ty_) / d (x_ {0}, y_ {0})

us prove v). Let a, b in IR ^ {+} - {0}. By recurrence, we have:

d (T ^ {n} x T ^ {n} y)

Since f ^ {n} (b) -> 0 as n ---> infinity, we can choose an n
<=d(x,y)<= b}
such that f ^ {n} (b) 0
<1.


IN, f ^ {n} is increasing, we have:
<=f^{n}(d(x,y)) sur ExE.
d (T ^ {n} x T ^ {n} y)


= [f ^ n} {(d (x, y)) / d (x, y)] d (x, y) < a ; il suit que, pour

< a <=d(x,y)<= b,et en tenant compte que ,pour tout n dans

By the above theorem, T ^ {n} admits a point single fixed z,

and hence z is also a unique fixed point of T. <=f^{n}(d(x,y))

Exercice.Démontrer direct corollary of the results.

(For bibliograhie, see references message 02/02/2007).
<= f^{n}(b)/a]d(x,y)