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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.

Memory Access C0000005

Demarcation and periodicity of solutions of differential equations (based Massera)

Exercise, below, presents the first result

in the famous article Massera (1950), which highlights

relief, the link between the existence of bounded solutions and the existence

periodic solutions.

Exercise.

Consider the following scalar differential equation:

x '= f (t, x) (*)

where f is a continuous function of IR XIR ^ {+} in IR, and

p-periodic report at (p> 0).

(ie d, for x in IR, f (t + p, x) = f (t, x) <1.Sans utiliser le principe de

for all t in IR ^{+}).

Assume that equation (*), satisfies the conditions

the uniqueness of solutions with respect to conditions

cause will.

Let y be a solution of equation (*), bounded on IR ^{+}.

i) Show that if y (0) = y (p), then y is p-periodic.

ii) Assume that y (0) is different from y (p), for example

y (0)

{y_ {n} (t) = y (t + pn)} is strictly increasing on IR .

iii) In the space of continuous functions, C ([0, t], IR),

with the norm sup, demonstrate that the sequence {y_ {n}}

is uniformly convergent to a solution

p-periodic Equation (*).

(Using the Ascoli theorem and Dini's theorem)

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modular formulation of the local version of the set of fixed point theorem of Banach.

In modular spaces, the following result offers a generalization

the local version of the principle set

contraction in Banach spaces:

Execice (provisional text).

Let X be any set and (E, rho) a complete modular space

, where the modular rho satisfies the property of Fatou < y(p).Montrer que,pour tout t dans [0,p], la suite

Let the ball B = B (z, r) = {x: rho (XZ)

T is an application deXxB in E satisfying:

There are two constants, k in] 0,1 [and p> 1, such as:

rho (p (T (x, y)-T (x, y ')))

for all x in X and all (y, y ') in BxB.

Assume that for any x in X, rho (q (T (x, z)-z))

the conjugate of p, ie d (1 / p) + (1 / q) = 1

demonstrate that there is a unique application of X f in B such that:

f (x) = T (x, f (x)) on X. Assume further that X is a modular space

any (or topological space satisfying the first axiom of countability

) T is continuous with respect to its first argument and rho

satisfies condition delta2.

Demonstrate that f is continuous on X.
< r}.

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Local version of parameterized fixed point theorem Banach

Exercise

Let X be any set and E a metric space

complet.B is an open ball in E with center z and radius <= k rho(y-y')

rT is an application of XXB in E satisfying: There

k in] 0,1 [such that d (T (x, y), T (x, y ')) \\ leq kd (y, y') < r(1-k), où q est

for all x in X and all (y, y ') in BxB. Assume that for all x

in X, d (T (x, z), z)

Prove that there exists a unique f X

B such that f (x) = T (x, f (x)) on X. If moreover X is a topological space

and T is continuous with respect to its first argument,

show that f is continuous on X.

Proof: From

after the local version (see results below), for

each x in X, the mapping T (x,.) B in E has a fixed point

single Note f (x), ie d T (x, f (x)) = f (x), which

can define a map f from X to B. We prove that f

is unique; Indeed: Let g

another application of X in B such that T (x, g (x)) = g (x)

then for all x in X:
< r(1-k).
d (f (x), g (x)) = d (T (x, f (x)), T (x, g (x))) \\ leq kd (f (x), g (x))

This entails that f = g on X.

for the continuity of f, x, x 'in X, then we have:

d (f (x), f (x')) = d (T (x, f (x)), T (x, f (x ')))

\\ leq d (T (x, f (x)), T (x, f (x')))+ d (T (x, f (x') ), T (x, f (x ')))

\\ leq kd (f (x), f (x')) + d (T (x, f (x ')), T (x', f (x ')))

From: d (f (x), f (x')) \\ leq (1/1-k) d (T (x, f (x ')), T (x '(x')) or

On the continuity of f at x '

Note.

For a direct demonstration in Banach spaces,

of this result, see [Dieudonne]. Note that this result is utlise

for a direct proof of the theorem of implicit functions.

see [Dieudonne] - [Deimling].