Orthogonal convexity and generalized nonexpansive mappingsgrupo.us.es/gfqm127/vworkshop/charlas/e_Moreno.pdf · Orthogonal convexity and generalized nonexpansive mappings 1 Condition

Orthogonal convexity and generalized nonexpansive mappings

Orthogonal convexity and generalized

nonexpansive mappings

Elena Moreno Gálvez

Universidad Católica de Valencia

V Workshop on Metric Fixed Point Theory

November 16, 2012


1 Condition (L)

Classes of mappings contained in class (L)

Properties of type (L) mappings

2 Results

Karlovitz lemma for type (L) mappings

Maurey lemma for type (L) mappings

Fixed point theorem in OC Banach spaces


Condition (L)

C will be a nonempty, closed, convex and bounded set of a Banach

space (X , ‖ · ‖).An almost fixed point sequence for T : C → C (s.p.c.f. from nowon) is a sequence (xn) on C such that

limn→∞

‖xn − Txn‖ = 0.

Definition (S. Dhompongsa, N. Nanan (2011))

Given a mapping T : C → X we will say that T satisfies condition(A) on C whenever there exists an a.f.p.s. for T in each nonempty,

closed, convex and T -invariant subset D of C , that is, if

.inf{‖x − Tx‖ : x ∈ D} = 0.

Each nonexpansive mapping satisfies condition (A) on any nonempty,

closed, bounded and convex subset of its domain.


Condition (L)

Definition

Let T : C → C be a mapping such that

T satisfies condition (A) on C .

For each a.f.p.s. (xn) for T on C and each x ∈ C

lim supn→∞

‖xn − Tx‖ ≤ lim supn→∞

‖xn − x‖.

We say that T satisfies condition (L) or is an (L)-type mapping.


Condition (L)



Nonexpansive mappings:

‖Tx − Ty‖ ≤ ‖x − y‖, for every x , y ∈ C .

Generalized nonexpansive mappings (Goebel, Kirk y Shimi, 1973;

Wong 1976):

∃ a, b, c ≥ 0 with b 6= 12 satisfying a + 2b + 2c ≤ 1 such that

‖Tx−Ty‖ ≤ a‖x−y‖+b(‖x−Tx‖+‖y−Ty‖)+c(‖x−Ty‖+‖y−Tx‖)

for every x , y ∈ C .Type (C ) mappings (Suzuki, 2008):

∀x , y ∈ C/ 12‖x − Tx‖ ≤ ‖x − y‖ ⇒ ‖Tx − Ty‖ ≤ ‖x − y‖.

Type (E ) mappings that satisfy condition (A) (Garćıa-Falset,

Llorens-Fuster y Suzuki, 2010):

‖x − Ty‖ ≤ µ‖x − Tx‖+ ‖x − y‖ for any x , y ∈ C .


Condition (L)

Properties of type (L) mappings

Let T : C → C be a mapping satisfying condition (L), then

T is continuous on Fix(T ) := {x ∈ C : Tx = x}.

If an a.f.p.s. converges strongly, the limit is a fixed point.

T is invariant under translations and dilations.

FPP in reflexive Banach spaces with normal structure

There are nonreflexive Banach spaces with the FPP

In spaces with Opial condition, condition (L) together with

asymptotically regularity implies convergence of the Picard

iterates to a fixed point whenever the domain is weakly

compact and convex.


Results


Lemma

Let K be a nonempty, weakly compact and convex subset of a Banach

space X . Let T : K → K be a mapping satisfying condition (L). Let Cbe a minimal subset for T on K . Hence, there exists λ > 0 such that for

any a.f.p.s. for T on C (xn) and any x ∈ C ,

lim supn→∞

‖xn − x‖ = λ.

Sketch of the proof:

The function x 7→ lim supn→∞

‖xn − x‖ defined on C is constant.


Results


Lemma




lim supn→∞

‖xn − x‖ = λ.



‖xn − x‖ defined on C is constant.Else, the function takes two values α1 < α2. Hence,

C ′ = {x ∈ C : lim supn→∞

‖xn − x‖ ≤α1 + α2

2}

satisfies that C ′ $ C , C ′ 6= ∅ and it is closed, bounded and convex. By condition

(L) for T , it is T -invariant, which contradicts the minimality of C .


Results


Lemma




lim supn→∞

‖xn − x‖ = λ.




Let (xn), (yn) a.f.p.s., x0, y0 such that xn ⇀ x0 and yn ⇀ y0. Then,

λ(xn) := lim supn→∞

‖xn−x0 + y0

2‖ ≤ lim sup

n→∞(

1

2‖xn−x0‖+

1

2λ(yn)) ≤

1

2λ(xn)+

1

2λ(yn),


Results


Lemma




lim supn→∞

‖xn − x‖ = λ.




Let (xn), (yn) a.f.p.s., x0, y0 such that xn ⇀ x0 and yn ⇀ y0. Then,

λ(yn) := lim supn→∞

‖yn−x0 + y0

2‖ ≤ lim sup

n→∞(

1

2‖yn−x0‖+

1

2λ(xn)) ≤

1

2λ(yn)+

1

2λ(xn).


Results


Lemma

Let K be a closed bounded and convex subset of a Banach space.

Let T : K → K satisfies condition (L) and in addition Tε : K → Kdefined by

Tεx := (1− ε)Tx + εx0

for any x0 ∈ K has a fixed point in any closed bounded convex andTε-invariant. Hence, if (xn) and (yn) are a.f.p.s. for T in K, there

exists another a.f.p.s. for T in K , say (zn), such that

limn→+∞

‖xn − zn‖ = limn→+∞

‖yn − zn‖ =1

2lim

n→+∞‖xn − yn‖


Results


Proof

Let λ = limn→+∞ ‖xn − yn‖. Fix n ∈ N and choose ε = ε(n) and δsuch that 0 < δ < 2ε2,

‖xn − yn‖ ≤ λ+ δ,‖xn − Tx‖ ≤ ‖xn − x‖+ δ2 and‖yn − Tx‖ ≤ ‖yn − x‖+ δ2

for any x ∈ K .Notice that, since lim sup

n→∞‖xn−Tx‖ ≤ lim sup

n→∞‖xn−x‖ and lim sup

n→∞‖yn−

Tx‖ ≤ lim supn→∞

‖yn − x‖, ε(n) can be chosen so that it tends to 0when n goes to infinity.

Consider

Kn = {z ∈ K : ‖xn − z‖ ≤λ

2+ ε, ‖yn − z‖ ≤

λ

2+ ε}.

It can be easily seen that wn =xn+yn

2 ∈ Kn.


Results


Let us define

Tε : K → Kx 7→ (1− ε)Tx + εwn.

Take z ∈ Kn, we claim that Tεz ∈ Kn

‖xn − Tεz‖ = ‖xn − ((1− ε)Tz + εwn)‖≤ (1− ε)‖xn − Tz‖+ ε‖xn − wn‖≤ (1− ε)(‖xn − z‖+ δ2) + ε

12‖xn − yn‖

≤ (1− ε)(λ2 + ε+δ2) + ε(

λ2 +

δ2)

= λ2 +δ2 + (1− ε)ε

< λ2 + ε2 + ε− ε2 = λ2 + ε

Analogously, it can be proven that ‖yn−Tεz‖ ≤ λ2+ε. Consequently,Tεz ∈ Kn and Kn is a closed bounded convex and invariant subsetfor Tε.


Results


By hypothesis, Tε has a fixed point in Kn, say zn. Then

zn = Tεzn = (1− ε)Tzn + εwn.

Taking into account this last equality,

‖Tzn − zn‖ = ‖Tzn − (1− ε)Tzn + εwn‖= ε‖wn − Tzn‖≤ ε2(‖xn − zn‖+

δ2 + ‖yn − zn‖+

δ2)

≤ ε2(λ2 + ε+

δ2 +

λ2 + ε+

δ2)

= ε(λ2 +δ2 + ε) ≤ ε(

λ2 + ε

2 + ε)

which tends to 0 when n tends to infinity. By the definition of Kn,

limn→+∞


‖yn − zn‖ =1

2lim

n→+∞‖xn − yn‖

�


Results


Definition (Orhogonal convexity)

Let (X , ‖ · ‖)) be a Banach space. Consider for a pair of points x ,y ∈ X the subset defined, for any λ > 0, by

Mλ(x , y) = {z ∈ X : max{‖z − x‖, ‖z − y‖} ≤1

2(1 + λ)‖x − y‖}.

If A is a bounded subset of X , let |A| = sup{‖x‖ : x ∈ A}, and fora bounded sequence (xn) in X and λ > 0, let

D[(xn)] = lim supn→∞

lim supm→∞

‖xn − xm‖

Aλ[(xn)] = lim supn→∞

lim supm→∞

|Mλ(xn, xm)|.

A Banach space is said to be orthogonally convex (OC) if for each

sequence (xn) in X which converges weakly to 0 and for which

D[(xn)] > 0, there exists λ > 0 such that Aλ[(xn)] < D[(xn)].


Results


Theorem

Let X be an orthogonally convex Banach space. Let K be a weakly

compact closed convex subset of X and let T satisfy condition (L)

and in addition Tε : K → K defined by

Tεx := (1− ε)Tx + εx0

for any x0 ∈ K has a fixed point in any closed bounded convex andTε-invariant.

Then T has a fixed point.


Results


Proof

Let us suppose that T is fixed point free. There is no loss of gen-

erality in assuming that K is minimal w.r.t. the conditions stated.

Since T has no fixed points, K is not a singleton and we can assume

that diam (K ) = 1 and that 0X ∈ K .Since T satisfies condition (L), there is an a.f.p.s (xn) for T in K ,

that is,

xn − Txn → 0X .

We can assume again that xn ⇀ 0X . By the lemma above, for any

x ∈ K ,lim

n→+∞‖x − xn‖ = λ ≤ diam (K ) = 1.

Hence, D[(xn)] = lim supn→∞

(lim supm→∞

‖xn − xm‖) ≤ 1.


Results


Since X is orthogonally convex, there exists some λ > 0 such that

Aλ[(xn)] < D[(xn)] ≤ 1

and we can choose b < 1 such that

Aλ[(xn)] = lim supn→∞

(lim supm→∞

|Mλ(xn, xm)|) < b

and hence there exists some positive integer n0 such that for any

n ≥ n0,lim supm→∞

|Mλ(xn, xm)| < b.

Fixing n ≥ n0 we can obtain another positive integer mn such that

|Mλ(xn, xmn)| < b.

Moreover, such mn can be chosen so that mn < mn+1 for all n ≥ n0.We can assume, by passing to subsequences if necessary, that (‖xn−xmn‖)n is convergent to, say, L.


Results


By Lemma 2applied to the sequences (xn) and (xmn), there is an

a.f.p.s. (zn) for T in K such that

limn→+∞


‖xmn − zn‖ =1

2L.

Hence, there is some n1 ≥ n0 such that for any n ≥ n1

max{‖xn − zn‖, ‖xmn − zn‖} ≤1 + λ

2‖xn − xmn‖,

which implies that for n ≥ n1,

zn ∈ Mλ(xn, xmn).

Consequently, for n ≥ n1,

‖zn‖ ≤ |Mλ(xn, xmn)| < b < 1.

On the other hand, since 0X ∈ K , by Lemma 1,

‖zn‖ → 1,

which is a contradiction. �


Results


QUESTION

Do (L)-type mappings satisfy that Tε : K → K defined by

Tεx := (1− ε)Tx + εx0

where x0 ∈ K has a fixed point in any closed bounded convex andTε-invariant.


Results


Example (NEGATIVE ANSWER)

Let T : [0, 1]→ [0, 1] given by

Tx =

{1− x x ∈ [0, 13)x+13 x ∈ [

13 , 1].

T is a mapping that satisfies condition (L), but the mapping T1/2,

T1/2x =

{1−x2 x ∈ [0,

13)

x+16 x ∈ [

13 , 1].

does not have a fixed point.


Results


POSITIVE ANSWER

We get a positive answer in the case of

Nonexpansive mappings

(gne) mappings

But...

we already knew this results.


Results


QUESTION

And what about type (C ) mappings of Suzuki?

In order to get a counterexample we need to find a type C

such that a convex combination with a point of its domain is

fixed point free.

It would suffice to find a fixed point free mapping T whose

domain contains the point 0 ∈ X such that 2T satisfiescondition (C ).


Results


J. Garćıa Falset, E. Llorens Fuster, T. Suzuki, Fixed point

theory for a class of generalized nonexpansive mappings,

Journal of Mathematical Analysis and Applications, Appl. 375

(2011), 185–195.

A. Jiménez Melado, E. Llorens Fuster, A sufficient condition

for the fixed point property, Nonlinear Anal. 20 (1993), no. 7,

849â853.

E. Llorens-Fuster, E. Moreno-Gálvez The fixed point theory for

some generalized nonexpansive mappings., Abstract and

Applied Analysis, Volume 2011 (2011), Article ID 435686, 15

pages.

M. Smyth, The fixed point problem for generalised

nonexpansive maps, Bull. Austral. Math. Soc., 55 (1997),

45–61.

Condition (L)Classes of mappings contained in class (L)Properties of type (L) mappings

ResultsKarlovitz lemma for type (L) mappingsMaurey lemma for type (L) mappingsFixed point theorem in OC Banach spaces

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Orthogonal convexity and generalized nonexpansive mappingsgrupo.us.es/gfqm127/vworkshop/charlas/e_Moreno.pdf · Orthogonal convexity and generalized nonexpansive mappings 1 Condition