Reminder:
Let (E, d) a metric space complet.D is a subset of ET
farm is an application of D DT said:
i) an isometry if d (Tx, Ty) = d (x, y) dxd.
ii) contracting if d (Tx, Ty)
iii) contractive if d (Tx, Ty)
Examples. I. Applicaion isometric
Applications that have no fixed point:
1) E = D is a Banach space and Tx = x + a (a in E-{0})
2) Let Banach space C (0) = {x =
(x (n)): lim x (n) = 0 when n -> infinity}) <=d(x,y) sur DxD.
equipped with the usual norm: II ( x (n)) II = sup Ix (n) I. < d(x,y), x different de y.
B denotes the closed unit ball of C (0).
T is an application of B in B defined by:
T ({x (n )})=( 1, x (0), x (1 )..., x (n )...)
It is obvious that T is an isometry and the only
fixed T is the vector
(1,1,1 ,..., 1 ,...) not owned
C (0), but it is al (infinite) space
of bounded sequences.
3) Example1-Edelstein (1964)
Exercise.
Let the Banach space C _ {0} with the usual standard.
T is a mapping defined on C:
Tx = y or {x = x_ {n}} and {y = y_ {n}} with
y_ {n} = [exp (2i \\ pi / n !)] x_ {n} - (-1) ^ {n} [exp (2i \\ pi / n !) -1]
n = 1,2, ...
i) Verify that y belongs to C
ii) Show that T is an isometry on C.
iii) For (0) = (0,0, ....), Show that {T ^ {k} (0)} is bounded in C
and that lim ^ {T} k! (0) = 0 ,
when k ---> infinity.
iv) Show that T has no fixed point in C.
4.Exemple2-Edelstein (1964)
Exercise.
Let the space of the suites ^ {2} with the usual standard:
IXI = I (x_ {0}, x_ {1}, ...) I = (sigma (k = 0 to infinity) ix_ {k} I) ^ {1 / 2}
known that (s ^ {2}, II) is a Hilbert space.
Let T be a mapping defined on ^ {2}:
Tx = y or {y = y_ {n}) with y_ {n} = [exp (2i \\ pi / n!)] (X_ {n } -1) +1
n = 1,2, ...
i) Show that the ^ {2} is stable T.
ii) Show that T is an isometry on ^ {2}
iii) Check that:
IT ^ {k} (0) I ^ {2} = 4 (Sigma (k = 1 to infinity ) sin ^ {2} (\\ pi k / n!)
Deduce that IT k! ^ {} (0) I ---> 0, when k ---> infinity, and if
Q_ {k} = (1 / 2) Sigma (j = 1 k) (2jk)! So
IT ^ {Q_ {k}} (0) I ---> infinity, as k ---> infinity
iii) Show that T has no fixed point in {^ 2}
NOTE.The Edelstein two examples above show that
the result of the same author (see 03/06/2007 post)
is not valid if T is an isometry.
5. The cons-instance of D. Alspach (important)
At the open question:
Let E be a Banach espcace and K is a subset of E
nonempty convex and weakly compact.T: K --- -> K a
application contractante.T she admits a fixed point?
Alspach replied negatively by the following example
Exercise.
Notation.I (f, [0,1]) = the integral of f on [0,1]
Let I = [0,1] and E = L ^ {1} (I) space integrable functions on I
, equipped with the usual standard.
We consider the set:
K = {f \\ in E: I (f, [0,1]) = 1, 0
T is an application defined on K by:
(Tm) ( t) = min {2f (2t), 2}, if 0
(Tf) (t) = max {2f (2t-1) -2.0}, if 1 / 2
Show that
i ) K is convex and weakly compact.
ii) K is stable by T.
iii) T is an isometry. <=f<= 2] }
iv) T has no fixed point in K.
evidente.Voir The proof is not, for example,
[Geobel-Kirk] (Message References 02/02/2007) <=t<=1/2]
< t<=1]
II.Applications Contracting.
i) Applications Contracting admit fixed points:
6) E = IR and T = sin.
T is a nonexpansive (check)
and 0 is a unique fixed point of T. (Justify
the existence of fixed point, taking D = [-1,1].
then shows that this fixed point is unique).
7) The space C (0) and the unit closed ball B
are defined in Example 2). Let T
application of B in B defined by:
T ({x (n )})=( x (0) ,1-Ix (0) I x (1 )..., x (n )...)
It is obvious that T is contractive (check).
As lim x (n) = 0, and if x = Tx, then x (1) = x (2 )=...= 0.
It follows that Ix (0) I = 1, and Fix (T) = {e (1),-e (1)
or e (1) = (1,0,0 ,..., 0 ,...)
Fix (T) discontinue.Remarquer that the space C (0), provided
from the norm, is not strictly convex.
(See message 02/23/2008)
ii) Applications Contracting admit no fixed points:
8) E = IR and D = [-2, -1] U [1,2 ]. T is an application of
D in D, defined by:
Tx = 1 if x belongs to [-2, -1]
=- 1 if x belongs to [1,2]
It is obvious that T is contractive and does not allow a point
fixe.On remark that the domain D is closed,
terminal and non-convex.
III) nonexpansive
Applications Contracting which do not admit fixed points:
9) E = IR, IR = D ^ {+} and Tx = \\ sqrt {x ^ {2} +1}
10) D = E = IR and Tx = log (1 + exp (x))
11) E = IR, IR = D ^ {+} and Tx = x +1 / (x +1)
12) E = C ([0,1] IR) is the space of continuous functions
[0,1] in IR with the norm u:
is easy to verify that
T is contractive, the only fixed point
T is the discontinuous function x (t) = 0 for 0
T has no fixed point in D.
13) Exercise
Let E be a countable set whose points are
Notes (n), n = 1,2 ,.... We equip E of the distance defined by:
d (a (p ), a (p)) = 0 and d (a (p), a (q)) = 1 + (1 / p) + (1 / q)
if p is different from q.
<= x(t)<= x(1)=1}.Alors D est
i) Check that d is a metric on E and (E, d)
is complete.
ii) Let T be the mapping of E into E defined by
T (a (p)) = a (p +1).
Show that T is contractive and Fix (T) is empty.
14) Exercise.
<= t<1 et x(1)=1.
Let B be the closed unit ball of the space of sequences C (0)
(see Example 2). Applicaction T is a C (0) in
C (0) such that T (e ( n)) = (1-2 ^ {- (n +1)} e (n +1), n> = 0, where {e (n)}
is the canonical basis of C (0). Let
U (x) = 1 / 2 (1 + IIxII) e (0) + T (x).
Show that U is contractive in B and B Fix (U) is empty.
IV) strictly contracting Applications .
15) Exercise.
Show that the cos function is strictly
Contracting of [-1,1] in [-1,1].
Calculate the fixed point.
16) Let the function Tx = x / 2] 0,1]] 0,1].
It is obvious that T is strictly contracting
but has no fixed point and the field n is not
farm.
v) Applications of the compact type (or alpha-compact).
17) is the function x (t) = t ^ {2} [0,1] [0,1].
Fix (x) = {0,1}
18) is the function x (t) = t ITI [-1,1] in [-1,1].
Fix (x) = {-1,0,1}
19) Let (E, II) an area designated Banach.B
the closed unit ball.
T is a mapping from B to B defined by:
T (x) = xixi.
Fix (T) = {0} U {x: Ixi = 1}.
Note.
The theorem explains the existence of Brouwer fixed points
in examples 15) and 16) by cons,
the same theorem explains the existence of fixed points,
in Example 17), where E is finite dimensional.
is found that, in these examples, the uniqueness of the fixed point
is not guaranteed.
20.Soient IR ^ (n) Euclidean space and S
sphere (= {x in IR ^ (n): IXI = 1}). Let the mapping T of S in
S defined by Tx =- x, then the fixed point can
T is 0 outside S.
(S is not convex).
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