Báo cáo hóa học: "Research Article Construction of Fixed Points by Some Iterative Schemes" ppt

El-Sayed Ahmed,ahsayed80@hotmail.com Received 23 October 2008; Revised 5 February 2009; Accepted 23 February 2009 Recommended by Massimo Furi We obtain strong convergence theorems of two

Volume 2009, Article ID 612491, 17 pages

doi:10.1155/2009/612491

Research Article

Construction of Fixed Points by

Some Iterative Schemes

1 Mathematics Department, Faculty of Science, Sohag University, Sohag 82524, Egypt

2 Mathematics Department, Faculty of Science, Taif University,

P.O Box 888 El-Hawiyah, El-Taif 5700, Saudi Arabia

3 Mathematics Department, The High Institute of Computer Science, Al-Kawser City, 82524 Sohag, Egypt

Correspondence should be addressed to A El-Sayed Ahmed,ahsayed80@hotmail.com

Received 23 October 2008; Revised 5 February 2009; Accepted 23 February 2009

Recommended by Massimo Furi

We obtain strong convergence theorems of two modifications of Mann iteration processes with errors in the doubly sequence setting Furthermore, we establish some weakly convergence theorems for doubly sequence Mann’s iteration scheme with errors in a uniformly convex Banach space by a Frech´et differentiable norm

Copyrightq 2009 A El-Sayed Ahmed and A Kamal This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

Let X be a real Banach space and let C be a nonempty closed convex subset of X A

x ∈ C is a fixed point of T provided Tx  x Denote by FixT the set of fixed points of T; that

is, FixT  {x ∈ C : Tx  x} It is assumed throughout this paper that T is a nonexpansive

important subject in the theory of nonexpansive mappings and its applications in a number

via the recursive sequence manner:

x n 1  α nxn 1− α n

Txn, n ≥ 0. 1.1

by 1.1 converges weakly to a fixed point of T However, this scheme has only weak

x0 x ∈ C,

y n  α n x n 1− α n

Tx n ,

Cnz ∈ C : y n − z ≤ x n − z,

Qnz ∈ C :xn − z, x0− x n

≥ 0,

x n 1  Pc n Qn

x0

,

1.2

at each iteration step, an additional projection is needed to calculate

x0 x ∈ X,

y n  α n x n 1− α n

Tx n ,

x n 1  β nu 1− β n

yn,

1.3

that{x n} defined by 1.3 converges to a fixed point of T see 7

x0 x ∈ C,

y n  J r n x n ,

x n 1  β nu 1− β n

y n ,

1.4

where for each r > 0, J r  I rA−1is the resolvent of A In7, it is proved, in a uniformly

{r n }, that {x n} defined by 1.4 converges strongly to a zero of A.

2 Preliminaries

dual space of X, is given by

Jx x∗∈ X∗:

x, x∗

 x2x∗2

, x ∈ X. 2.1 Now, we define Opial’s condition in the sense of doubly sequence

Definition 2.1 A Banach space X is said to satisfy Opial’s condition if for any sequence {x k,n}

lim

k,n → ∞supx k,n − x < lim

k,n → ∞supx k,n − y ∀y ∈ X with y/x, 2.2

We are going to work in uniformly smooth Banach spaces that can be characterized by

Lemma 2.2 see 8 A Banach space X is uniformly smooth if and only if the duality map J is

single-valued and norm-to-norm uniformly continuous on bounded sets of X.

Lemma 2.3 see 8 In a Banach space X, there holds the inequality

where jx y ∈ Jx y.

If C and D are nonempty subsets of a Banach space X such that C is a nonempty closed

Qx  Qx for all x ∈ C and t ≥ 0 whenever x tx − Qx ∈ C A sunny nonexpansive

retraction is a sunny retraction, which is also nonexpansive A sunny nonexpansive retraction plays an important role in our argument

and only if there holds the inequality

2.4

Lemma 2.4 see 9 Let X be a uniformly smooth Banach space and let T : C → C be a

nonexpansive mapping with a fixed point For each fixed u ∈ C and every t ∈ 0, 1, the unique fixed point xt

point of T Define Q : C → FixT by Qu  s−lim t → 0 x t Then, Q is the unique sunny nonexpansive retract from C onto FixT; that is, Q satisfies the property

2.5

Lemma 2.5 see 10,11 Let {a n}∞n0 be a sequence of nonnegative real numbers satisfying the property

a n 1≤1− γ n

a n γ n σ n, n ≥ 0, 2.6

where {γ n}∞

n0 ⊂ 0, 1 and {σ n}∞

n0 are such that

i limn → ∞ γ n  0, and ∞

n0 γ n  ∞,

ii either lim n → ∞ sup σ n ≤ 0 or∞n0 |γ nσn | < ∞.

Then, {a n}∞n0 converges to zero.

Lemma 2.6 see 8 Assume that X has a weakly continuous duality map J ϕ with gauge ϕ Then,

A is demiclosed in the sense that A is closed in the product space X w × X, where X is equipped with

the norm topology and Xw with the weak topology That is, if x n, yn  ∈ A, x n  x, yn → y, then

x, y ∈ A.

Lemma 2.7 see 12 Let X be a Banach space and γ ≥ 2 Then,

i X is uniformly convex if and only if, for any positive number r, there is a strictly increasing

continuous function gr :0, ∞ → 0, ∞, g r 0  0, such that

where t ∈ 0, 1, x, y ∈ Br : {u ∈ X : u ≤ r}, the closed ball of X centered at the origin

with radius r, and W γ t  t γ 1 − t t1 − t γ

ii X is γ-uniformly convex if and only if there holds the inequality

where cγ > 0 is a constant.

Lemma 2.8 see 4 Let C be a closed convex subset of a uniformly convex Banach space with

a Fr´echet differentiable norm, and let Tn  be a sequence of nonexpansive self mapping of C with a

nonempty common fixed point set F If x1 ∈ C and x n 1  T n x n for n ≥ 1, then lim n → ∞ n, Jf1−

weak limit points of {x n }.

Lemma 2.9 the demiclosedness principle of nonexpansive mappings 13 Let T be a

nonexpansive selfmapping of a closed convex subset of E of a uniformly convex Banach space Suppose that T has a fixed point Then I − T is demiclosed This means that

Theorem A Let C be a closed convex subset of a uniformly smooth Banach space X, and let T : C →

C be a nonexpansive mapping such that FixT / ∅ Given a point u ∈ C and given sequences {αn}∞

n0 and {β n}∞n0 in 0, 1, the following conditions are satisfied.

ii∞n0 αn  ∞, ∞n0 βn  ∞,

iii∞

n0 |α n 1 − α n | < ∞, ∞

n0 |β n 1 − β n | < ∞.

Define a sequence {x n}∞

n0 in C by

yn  α nxn 1− α n

Txn, n ≥ 0,

x n 1  β nu 1− β n

y n, n ≥ 0.

2.10

Then {x n}∞

n0 is strongly converges to a fixed point of T.

Recently, the study of fixed points by doubly Mann iteration process began by Moore

with errors, then we obtained some fixed point theorems for some different classes of mappings In this paper, we will continue our study in the doubly sequence setting We propose two modifications of the doubly Mann iteration process with errors in uniformly smooth Banach spaces: one for nonexpansive mappings and the other for the resolvent of accretive operators The two modified doubly Mann iterations are proved to have strong convergence Also, we append this paper by obtaining weak convergence theorems for Mann’s doubly sequence iteration with errors in a uniformly convex Banach space by a Fr´echet differentiable norm Our results in this paper extend, generalize, and improve a lot of

of doubly sequence settings as well as by adding the error part in the iteration processes

3 A Fixed Point of Nonexpansive Mappings

In this section, we propose a modification of doubly Mann’s iteration method with errors to have strong convergence Modified doubly Mann’s iteration process is a convex combination

of a fixed point in C, and doubly Mann’s iteration process with errors can be defined as

yk,n  α nxk,n 1− α n

Txk,n α nwk,n, k, n ≥ 0,

x k,n 1  β nu 1− β n

yk,n β nvk,n, k, n ≥ 0.

3.1

The advantage of this modification is that not only strong convergence is guaranteed, but also computations of iteration processes are not substantially increased

Theorem 3.1 Let C be a closed convex subset of a uniformly smooth Banach space X and let T :

C → C be a nonexpansive mapping such that FixT / ∅ Given a point u ∈ C and given sequences

{α n}∞

n0 and {β n}∞

n0 in 0, 1, the following conditions are satisfied.

i α n → 0, β n → 0,

ii∞n0 αn  ∞, ∞n0 βn  ∞.

Define a sequence {x k,n}∞k,n0 in C by 3.1 Then, {x k,n}∞k,n0 converges strongly to a fixed point of T Proof First, we observe that {x k,n}∞

k,n0 is bounded Indeed, if we take a fixed point p of T

noting that

y k,n − p  α nxk,n 1− α n

Txk,n α nwk,n − p

≤ α n x k,n − p 1 − α n Tx k,n − p α n w k,n

we obtain

x k,n 1 − p  β nu 1− β n

y k,n β n v k,n − p

≤ β n u − p 1− β n y k,n − p β n v k,n

≤ β n u − p 1− β n x k,n − p α n w k,n  β n v k,n

≤ maxx k,n − p,u − p β n v k,n  1 − β n

α n w k,n .

3.3

Now, an induction yields

x k,n 1 − y k,n   β nu − βn y k,n β n v k,n

We next show that

It suffices to show that

x k,n 1 − x k,n  −→ 0. 3.7 Indeed, if3.7 holds, in view of 3.5, we obtain

x k,n − Tx k,n  ≤ x k,n − x k,n 1  x k,n 1 − y k,n  y k,n − Tx k,n

≤x k,n − x k,n 1  x k,n 1 − y k,n  α n x k,n − Tx k,n  α n w k,n  −→ 0.

3.8

Hence,3.6 holds In order to prove 3.7, we calculate

x k,n 1 − x k,nβn − β n−1

u − Txn−1 1− β n

αn

xk,n − x k,n−1 αn − α n−1

1− β n

−βn − β n−1

α n−1

x k,n−1 − Tx k,n−1 1− α n

1− β n

Txk,n − Tx k,n−1

1− β n

αnwk,n

β nvk,n−1− β n

α n−1 w k,n−1 − β n−1 v k,n−1

3.9

It follows that

x k,n 1 − x k,n  ≤ 1 − α n

1− β n Tx k,n − Tx k,n−1  1 − β n

αn x k,n − x k,n−1

1− β n

−βn − β n−1

α n−1 k,n−1 − Tx k,n−1

n − β n−1 k,n−1  1 − β n

αn w k,n  β n v k,n

−1− β n

α n−1 w k,n−1  − β n−1 v k,n−1 .

3.10

Next, we claim that

lim

k,n → ∞sup

u − q, Jx k,n − q ≤ 0, 3.11

tTz In order to prove 3.11, we need some more information on q, which is obtained from that of z tcf 18 Indeed, z tsolves the fixed point equation

Thus we have

zt − x k,n  1 − tTzt − x k,n

tu − xk,n tv. 3.13

z t − x k,n2≤ 1 − t2Tz t − x k,n2 2tu v − xk,n , J

z t − x k,n

≤1− 2t t2z t − x k,n  a n t 2tu v − zt, Jzt − x k,n 2tz t − x k,n2,

3.14

a n t 2zt − x k,n  x k,n − Tx k,n x k,n − Tx k,n  −→ 0 as n −→ ∞. 3.15

It follows that



zt − u, Jzt − x k,n

2z t − x k,n2 1

Letting n → ∞ in 3.16 and noting 3.15 yield

lim

n → ∞sup

zt − u, Jzt − x k,n

where M > 0 is a constant such that M ≥ z t − x k,n2for all t ∈ 0, 1 and n ≥ 1 Since the set

x k,n 1 − q21− β n

y k,n β nu βn v k,n − q2

1− β n

y k,n − q β n u − q β n v k,n2

≤1− β n2y k,n − q2 2β n

u vk,n − q, Jx k,n 1 − q

≤1− β n2x k,n − q α n w k,n2

2β n

u vn − q, Jx k,n 1 − q .

3.18

We support our results by giving the following examples

Example 3.2 Let T : 0, 1 × 0, 1 → 0, 1 × 0, 1 be given by Tx  x Then, the modified

Picard and Mann iteration processes converge to the same point too

Proof. I Doubly Picards iteration converges

For every point in0, 1 × 0, 1 is a fixed point of T Let b 0,0be a point in0, 1 × 0, 1,

then

b k 1,k 1  Tb k,k  T n b 0,0  b 0,0 3.19 Hence,

lim

k → ∞ b k,k  b 0,0 3.20

Letx, y − a, b  |x − a|, |y − b|, for all x, y, a, b ∈ 0, 1 × 0, 1 Take p 0,0  0, 0 and

δk,k  p k 1,k 1 − Tp k,k

 1

kk 1 ,

1

kk 1



II Doubly Mann’s iteration converges

Let e 0,0be a point in0, 1 × 0, 1, then

e k 1,k 11− α k

ek,k α kek,k  e k,k  · · ·  e 0,0 3.22

Since doubly Mann’s iteration is defined by

e k 1,k 11− α k

Take u 0,0  e 0,0 , uk,k  1/k 1, 1/k 1 to obtain

ε k,k  u k 1,k 1−1− α k

u k,k α k Tu k,k





1

k 1k 2 ,

1

k 1k 2



III Modified doubly Mann’s iteration process with errors converges because the

y k,k  α k e k,k 1− α k

e k,k α k w k,k

In3.1, we suppose that u  e k,k,

e k,k 1  β ku 1− β k

ek,k α kwk,k

β kνk,k

 e k,k 1− β k

α k w k,k β k ν k,k ,

e k,k 1 − e k,k1− β k

α k w k,k β k ν k,k

3.26

Example 3.3 Let T : 0, ∞ × 0, ∞ → 0, ∞ × 0, ∞ be given by Tx  x/4 Then the doubly

iteration process with errors does not converge

Proof I Doubly Mann’s iteration converges because the sequence e k,k → 0, 0 as we can

see,

e k 1,k 11− α k

ek,k α k e k,k

4





1−3α k 4



e k,k

m1



1−3α m 4



e 0,0

≤ exp



4

n



k1 αk



−→ 0, 0.

3.27

k1 α k  ∞.

II The origin is the unique fixed point of T.

III Note that, modified doubly Mann’s iteration process with errors does not converge

yk,k  α kek,k 1− α k e k,k

4 α kwk,k 



1 3α k

4



ek,k α kwk,k. 3.28

Putting u  e k,k,

e k,k 1  β k e k,k 1− β k 1 3α k

4



e k,k α k w k,k

4 Convergence to a Zero of Accretive Operator

In this section, we prove a convergence theorem for m-accretive operator in Banach spaces.

DA and range RA in X is accretive if, for each xi ∈ DA and y i ∈ Ax i i  1, 2, there exists a j ∈ Jx2− x1 such that



y2− y1, j

An accretive operator A is m-accretive if RI rA  X for each r > 0 Throughout this section,

we always assume that A is m-accretive and has a zero The set of zeros of A is denoted by F.

Hence,

F  {z ∈ DA : 0 ∈ Az}  A−10. 4.2

Lemma 4.1 7 the resolvent identity For λ > 0, μ > 0 and x ∈ X,

J λx  Jμ μ

λ x



1−μ λ



J λ x



Theorem 4.2 Assume that X is a uniformly smooth Banach space, and A is an m-accretive operator

in X such that A−10 / ∅ Let {x k,n } be defined by

x 0,0  x ∈ X,

y k,n  J r n x k,n ,

x k,n 1  α nu 1− α n

yk,n α nwk,n.

4.4

Suppose {α n } and {r n } satisfy the conditions,

i limn → ∞ α n  0 and ∞

n0 α n  ∞,

ii∞

n0 |α n 1 − α n | < ∞,

iii r n ≥ ε for some ε > 0 and for all n ≥ 1 Also assume that

∞



n1

r n−1

Then, {x k,n } converges strongly to a zero of A.

Proof First of all we show that {x k,n } is bounded Take p ∈ F  A−10 It follows that

x k,n 1 − p  α nu 1− α n

Jr n xk,n α nwk,n − p

By induction, we get that

x k,n 1 − J r n xk,n  −→ 0. 4.8

A simple calculation shows that

x k,n 1 − x k,nα n − α n−1u − y k,n−1 1− α n

y k,n − y k,n−1 α n w k,n − α n−1 w k,n−1

4.9

y k,n  J r n−1

r

n−1

r n x k,n



1−r n−1 r

n



J r n x k,n



which in turn implies that

y k,n − y k,n−1 J r n−1

r

n−1

r n xk,n



1−r n−1 r

n



Jr n xk,n



− J r n−1 x k,n−1

,



J r n−1

r

n−1

rn xk,n − x k,n−1





1−r n−1 rn



Jr n xk,n





J r n−1

r

n−1

r n x k,n − x k,n x k,n − x k,n−1





1−r n−1 r

n



J r n x k,n





J r n−1

r

n−1

rn − 1



xk,n xk,n − x k,n−1



1−r n−1 rn



Jr n xk,n





1−r n−1

rn



J r n−1 − x k,n

J r n−1



x k,n − x k,n−1

n J r n xk,n − J r n−1 xk,n  J r n−1 xk,n − J r n−1 x k,n−1

rn J r n x k,n − J r n−1 x k,n  x k,n − x k,n−1 .

4.11

x k,n 1 − x k,n  ≤ 1 − α n x k,n − x k,n−1  M n − α n−1 r n−1

rn



|α n − α n−1 | |1 − r n−1 /r n | < ∞ Hence,Lemma 2.5is applicable to4.12, and we conclude thatx k,n 1 − x k,n  → 0.

find

J r n xk,n − J r xk,n J r

rn xk,n



1−rn r



Jr n xk,n



− J r xk,n



≤



1− r r

n

x

k,n − J r n x k,n

≤x k,n − x k,n 1  x k,n 1 − J r n x k,n  −→ 0.

4.13

It follows that

x k,n 1 − J r x k,n 1  ≤ x k,n 1 − J r n x k,n  J r n x k,n − J r x k,n  J r x k,n − J r x k,n 1

≤x k,n 1 − J r xk,n  J r xk,n − J r xk,n  x k,n − x k,n 1 . 4.14

Thus... converge to the same point too

Proof. I Doubly Picards iteration converges

For every point in0, 1 × 0, 1 is a fixed point of T Let b 0,0be a point... section,

we always assume that A is m-accretive and has a zero The set of zeros of A is denoted by F.

Hence,

F  {z ∈ DA : ∈ Az}  A−10.