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Convergence theorems for continuous pseudocontractive mappings in Banach spaces

Abstract

In this paper, we prove some strong and weak convergence theorems for continuous pseudocontractive mapping and a weak convergence theorem for nonexpansive mapping in real uniformly convex Banach spaces. As an application of the strong convergence theorem, we give an interesting example.

1 Introduction and preliminaries

Throughout this paper, we assume that E is a real Banach space, E is the dual space of E and J:E 2 E is the normalized duality mapping defined by

J(x)= { f E : x , f = x f , f = x }

for all xE, where , denotes duality pairing between E and E . A single-valued normalized duality mapping is denoted by j.

Let C be a nonempty closed convex subset of a real Banach space E. A mapping T:CC is said to be pseudocontractive [1] if, for any x,yC, there exists j(xy)J(xy) such that

T x T y , j ( x y ) x y 2 .
(1.1)

It is well known [2] that (1.1) is equivalent to the following:

xy x y + s [ ( I T ) x ( I T ) y ]
(1.2)

for all s>0 and x,yC.

A mapping T:CC is said to be nonexpansive if

TxTyxy

for all x,yC.

Remark 1.1 It is easy to see that, if T is nonexpansive, then T is continuous pseudocontractive, but the converse is not true in general.

Example 1.1 Let E=(,) with the usual norm ||. Then E =E, x,f=xf for all xE and f E and the normalized duality mapping J:E 2 E is as follows:

J(x)= { f E : x , f = | x | | f | , | f | = | x | } ={x}

for all xE. Let C=E and define a mapping T:CC by Tx=2x for all xC. It is easy to prove that T is continuous pseudocontractive, but not nonexpansive.

In 1974, Deimling [3] proved the following fixed point theorem.

Theorem 1.1 Let E be a real Banach space, C be a nonempty closed convex subset of E, and T:CC be a continuous strongly pseudocontractive mapping. Then T has a unique fixed point in C.

Let E be a real Banach space, C be a nonempty closed convex subset of E and T:CC be a continuous pseudocontractive mapping. For all uC and t(0,1), define the mapping S t :CC by

S t x=tu+(1t)Tx

for all xC. It is easy to prove that S t is a continuous strongly pseudocontractive mapping. By Theorem 1.1, there exists a unique fixed point x t C of S t such that

x t =tu+(1t)T x t

for all t(0,1).

Let T:CC be a continuous pseudocontractive mapping and define an implicit iteration process { x n } by

{ x 0 C , x n = α n x n 1 + ( 1 α n ) T x n
(1.3)

for each n1, where { α n } is a sequence in (0,1) with some conditions.

In 2008, Zhou [4] studied the implicit iteration process (1.3) and proved a weak convergence theorem for the strict pseudocontraction in a real reflexive Banach space which satisfies Opial’s condition.

The purpose of this paper is to discuss the implicit iteration process (1.3) and to prove some strong and weak convergence theorems for a continuous pseudocontractive mapping and a weak convergence theorem for a nonexpansive mapping in real uniformly convex Banach spaces. As an application of the strong convergence theorem, we give an interesting example.

In order to prove the main results, we need the following:

A Banach space E is said to satisfy Opial’s condition [5] if, for any sequence { x n } of E, x n x weakly as n implies that

lim sup n x n x< lim sup n x n y

for all yE with yx. A Banach space E is said to have the Kadec-Klee property [6] if, for every sequence { x n } in E, x n x weakly and x n x, it follows that x n x strongly.

Lemma 1.1 ([7])

Let E be a uniformly convex Banach space with the modulus of uniform convexity δ E . Then δ E :[0,2][0,1] is continuous, increasing, δ E (0)=0, δ E (t)>0 for t>0 and, further,

c u + ( 1 c ) v 12min{c,1c} δ E ( u v )

whenever 0c1 and u,v1.

Lemma 1.2 ([8])

Let X be a uniformly convex Banach space and C be a convex subset of X. Then there exists a strictly increasing continuous convex function γ:[0,)[0,) with γ(0)=0 such that, for each S:CC with Lipschitz constant L,

α S x + ( 1 α ) S y S [ α x + ( 1 α ) y ] L γ 1 ( x y 1 L S x S y )

for all x,yC and 0<α<1.

Lemma 1.3 ([8])

Let X be a uniformly convex Banach space such that its dual space X has the Kadec-Klee property. Suppose that { x n } is a bounded sequence and f 1 , f 2 W w ({ x n }) (where W w ({ x n }) denotes the set of all weak subsequential limits of a bounded sequence { x n } in X) such that

lim n α x n + ( 1 α ) f 1 f 2

exists for all α[0,1]. Then f 1 = f 2 .

Lemma 1.4 ([4])

Let E be a real uniformly convex Banach space, C be a nonempty closed convex subset of E and T:CC be a continuous pseudocontractive mapping. Then IT is semi-closed at zero, i.e., for each sequence { x n } in C, if { x n } converges weakly to qC and {(IT) x n } converges strongly to 0, then (IT)q=0.

2 Convergence for continuous pseudocontractive mappings

Lemma 2.1 Let E be a real uniformly convex Banach space, C be a nonempty closed convex subset of E and T:CC be a continuous pseudocontractive mapping with F(T)={xC:Tx=x}. Let { x n } be defined by (1.3), where α n (0,1) and lim sup n α n <1. Then

  1. (1)

    x n p x n 1 p for all n1 and all pF(T);

  2. (2)

    lim n x n p exists for all pF(T);

  3. (3)

    lim n d( x n ,F(T)) exists;

  4. (4)

    lim n x n T x n =0.

Proof For all n1 and pF(T), since T is pseudocontractive, it follows from (1.3) that

x n p 2 = x n p , j ( x n p ) = α n x n 1 p , j ( x n p ) + ( 1 α n ) T x n p , j ( x n p ) α n x n 1 p x n p + ( 1 α n ) x n p 2 ,

which implies that x n p x n 1 p and d( x n ,F(T))d( x n 1 ,F(T)). Therefore, lim n x n p and lim n d( x n ,F(T)) exist. Thus (1), (2), and (3) are proved.

Now, we prove (4). By using (1.2), (1.3), and Lemma 1.1, it follows that, for all pF(T),

x n p x n p + 1 α n 2 α n ( x n T x n ) = x n p + 1 α n 2 ( x n 1 T x n ) = x n p + x n 1 x n 2 = x n 1 + x n 2 p = x n 1 p 1 2 x n 1 p x n 1 p + 1 2 x n p x n 1 p x n 1 p [ 1 δ E ( x n x n 1 x n 1 p ) ] .

This implies that

x n 1 p δ E ( x n x n 1 x n 1 p ) x n 1 p x n p.

If x n p0, then we have x n x n 1 0 as n by the properties of δ E . It follows from x n 1 T x n = 1 1 α n x n x n 1 and lim sup n α n <1 that x n 1 T x n 0 as n and so

x n T x n = α n x n 1 T x n x n 1 T x n 0

as n.

If x n p0, then, since T is continuous, it follows that

lim n x n T x n =pTp=0.

This completes the proof. □

Theorem 2.1 Under the assumptions of Lemma  2.1, { x n } converges strongly to a fixed point of T if and only if lim inf n d( x n ,F(T))=0.

Proof The necessity is obvious. So, we will prove the sufficiency. Assume that

lim inf n d ( x n , F ( T ) ) =0.

By Lemma 2.1, limit lim n d( x n ,F(T)) exists and so lim n d( x n ,F(T))=0.

Now, we show that { x n } is a Cauchy sequence in C. In fact, it follows from Lemma 2.1 that x m p x n p for all positive integers m, n with m>n1 and pF(T). So,

x m x n x n p+ x m p2 x n p.

Taking the infimum over all pF(T), we have

x m x n 2d ( x n , F ( T ) ) .

It follows from lim n d( x n ,F(T))=0 that { x n } is a Cauchy sequence. C is a closed subset of E and so { x n } converges strongly to some qC. Further, by the continuity of T, it is easy to prove that F(T) is closed and it follows from lim n d( x n ,F(T))=0 that qF(T). This completes the proof. □

Corollary 2.1 Under the assumptions of Lemma  2.1, { x n } converges strongly to a fixed point p of T if and only if there exists a subsequence { x n k } of { x n } such that { x n k } converges strongly to p.

Proof Since

lim inf n d ( x n , F ( T ) ) lim inf k d ( x n k , F ( T ) ) lim k x n k p,

it follows from Theorem 2.1 that Corollary 2.1 holds. This completes the proof. □

Theorem 2.2 Under the assumptions of Lemma  2.1, if there exists a nondecreasing function f:[0,)[0,) with f(0)=0 and f(r)>0 for all r(0,) such that

f ( d ( x , F ( T ) ) ) xTx

for all xC, then { x n } converges strongly to a fixed point of T.

Proof Since lim n x n T x n =0 by Lemma 2.1 and so lim n f(d( x n ,F(T)))=0. Further, by using Lemma 2.1 lim n d( x n ,F(T)) exists, and we assume lim n d( x n ,F(T))=r.

If r>0, there exists a positive integer N such that d( x n ,F(T))> r 2 for all n>N. Thus we have

lim n f ( d ( x n , F ( T ) ) ) f ( r 2 ) >0,

which is a contradiction. Therefore, r=0. It follows from Theorem 2.1 that Theorem 2.2 holds. This completes the proof. □

Definition 2.1 ([9])

Let E be a real normed linear space, C be a nonempty subset of E, and T:CE be a mapping. The pair (T,C) is said to satisfy the condition (A) if, for any bounded closed subset G of C, {z=xTx:xG} is a closed subset of E.

Now, we prove strong convergence and weak convergence theorems for a continuous pseudocontractive mapping in real uniformly convex Banach spaces.

Theorem 2.3 Under the assumptions of Lemma  2.1, if the pair (T,C) satisfies the condition (A), then { x n } converges strongly to a fixed point of T.

Proof The sequence { x n } is bounded in C by Lemma 2.1. Letting G= { x n } ¯ , where A ¯ denotes the closure of A, G is a bounded closed subset of C and so M={z=xTx:xG} is a closed subset of E since the pair (T,C) satisfies the condition (A). It follows from { x n T x n }M and x n T x n 0 as n by Lemma 2.1(3) that the zero vector 0M and so there exists qG such that q=Tq. This shows that q is a fixed point of T and so there exists a positive integer n 0 such that x n 0 =q or there exists a subsequence { x n k } of { x n } such that x n k q as n.

If x n 0 =q, then it follows from Lemma 2.1(1) that x n =q for all n n 0 and so x n q as n.

If x n k q, then, since lim n x n q exists by Lemma 2.1(2), x n q as n. This completes the proof. □

As an application of Theorem 2.1, we give the following.

Example 2.1 Let E=(,) with the usual norm ||. Then E =E, x,f=xf for all xE and f E and J(x)={x} for all xE. Let C=[0,). Define a mapping T:CC by

Tx={ 1 4 , 0 x 1 2 , 1 2 x , 1 2 < x < .

Then T is continuous pseudocontractive with F(T)={ 1 4 } and the pair (T,C) satisfies the condition (A). In fact, for all x[0, 1 2 ] and y( 1 2 ,), we have 4x<2y+1 and so

0< 1 2 y 1 4 <yx.
(2.1)

For all x( 1 2 ,) and y[0, 1 2 ], we obtain 4y<2x+1 and so

0< 1 2 x 1 4 <xy.
(2.2)

Thus, for all x,yC, taking j(xy)=xyJ(xy), it follows from (2.1) and (2.2) that

T x T y , j ( x y ) = { 1 4 1 4 , x y = 0 | x y | 2 , x , y [ 0 , 1 2 ] , 1 4 1 2 y , x y = ( 1 2 y 1 4 ) ( y x ) < | x y | 2 , x [ 0 , 1 2 ] , y ( 1 2 , ) , 1 2 x 1 4 , x y = ( 1 2 x 1 4 ) ( x y ) < | x y | 2 , x ( 1 2 , ) , y [ 0 , 1 2 ] , 1 2 x 1 2 y , x y = 1 2 ( x y ) 2 | x y | 2 , x , y ( 1 2 , ) .

This shows that T is continuous pseudocontractive.

Now, we prove that the pair (T,C) satisfies the condition (A). For any bounded closed subset G of C, we denote M={z=xTx:xG}. Then M is closed. Indeed, for any z n M with z n z, there exists x n G such that z n = x n T x n . We consider the following cases.

Case 1. There exists a positive integer n 0 such that x n [0, 1 2 ] for all n n 0 .

Case 2. There exists a subsequence { x n k } of { x n } such that x n k ( 1 2 ,) for all k1.

If Case 1 holds, then z n = x n T x n = x n 1 4 for all n n 0 and so x n = z n + 1 4 z+ 1 4 [0, 1 2 ] as n. Since G is closed, it follows that z+ 1 4 G and so z=(z+ 1 4 )T(z+ 1 4 )M.

If Case 2 holds, then z n k = x n k T x n k = 1 2 x n k for all k1 and so x n k =2 z n k 2zG. If 2z= 1 2 , we have z= 1 4 = 1 2 T 1 2 =2zT(2z)M. Otherwise, we have 2z( 1 2 ,) and so z=2zT(2z)M.

By Theorem 1.1, it is easy to prove that, for any x 0 C, there exists a unique x n C such that

x n = α n x n 1 +(1 α n )T x n

for all n1, where α n = 2 n 3 n + 1 (0,1) for all n1 and

lim sup n α n = lim n α n = 2 3 <1.

Thus it follows from Theorem 2.3 that the sequence { x n } converges to 1 4 .

Theorem 2.4 Under the assumptions of Lemma  2.1, if E satisfies Opial’s condition, then { x n } converges weakly to a fixed point of T.

Proof Since { x n } is bounded by Lemma 2.1 and E is reflexive, there exists a subsequence { x n k } of { x n } which converges weakly to a point pC. By Lemma 2.1, we have lim k x n k T x n k =0. It follows from Lemma 1.4 that pF(T).

Now, we prove that { x n } converges weakly to p. Suppose that there exists a subsequence { x m j } of { x n } such that { x m j } converges weakly to a point p C. Then p= p . In fact, if p p , then it follows from Opial’s condition that

lim n x n p = lim sup k x n k p < lim sup k x n k p = lim n x n p = lim sup j x n j p < lim sup j x n j p = lim n x n p ,

which is a contradiction. So p= p . Therefore, { x n } converges weakly to a fixed point of T. This completes the proof. □

Remark 2.1 By Remark 1.1, clearly, Theorems 2.1, 2.2, 2.3 and 2.4 still hold for nonexpansive mappings.

3 Weak convergence for nonexpansive mappings

In this section, we prove a weak convergence theorem for a nonexpansive mapping in real uniformly convex Banach spaces.

Lemma 3.1 Let E be a real uniformly convex Banach space and C be a nonempty closed convex subset of E. Let T:CC be a nonexpansive mapping with F(T) and { x n } be the sequence defined by (1.3), where α n (0,1) and lim sup n α n <1. Then, for all p 1 , p 2 F(T), the limit lim n t x n +(1t) p 1 p 2 exists for all t[0,1].

Proof Letting a n (t)=t x n +(1t) p 1 p 2 for all t[0,1], lim n a n (0)= p 1 p 2 and lim n a n (1)= lim n x n p 2 exists by Remark 1.1 and Lemma 2.1. Thus it remains to prove Lemma 3.1 for any t(0,1). For all n1 and xC, we define a mapping A x , n 1 :CC by

A x , n 1 y= α n x+(1 α n )Ty

for all yC. Then we have

A x , n 1 u A x , n 1 v=(1 α n )TuTv(1 α n )uv

for all u,vC. It follows from 0<1 α n <1 that A x , n 1 is contractive and so it has a unique fixed point in C, which is denoted by G n 1 x. Define a mapping G n :CC by

G n = α n + 1 I+(1 α n + 1 )T G n ,
(3.1)

where I is a identity mapping. It follows from (3.1) that

G n x G n y α n + 1 xy+(1 α n + 1 ) G n x G n y

for all x,yC and so

G n x G n yxy.
(3.2)

Using (3.1) and (1.3), we obtain

G n x n x n + 1 (1 α n + 1 ) G n x n x n + 1
(3.3)

and

G n pp(1 α n + 1 ) G n pp
(3.4)

for all pF(T). It follows from (3.3), (3.4) and 0<1 α n + 1 <1 that G n x n = x n + 1 and G n p=p. For each m1, let

S n , m = G n + m 1 G n + m 2 G n

and

b n , m = S n , m ( t x n + ( 1 t ) p 1 ) ( t S n , m x n + ( 1 t ) S n , m p 1 ) .
(3.5)

By (3.2), we have

S n , m x S n , m yxy

for all x,yC. This shows that S n , m is nonexpansive, S n , m x n = x n + m and S n , m p=p for all pF(T). By Lemma 1.2, we obtain

b n , m γ 1 ( x n p 1 x n + m p 1 ) .
(3.6)

It is easy to prove that

a n + m ( t ) t x n + ( 1 t ) p 1 p 2 + b n , m a n ( t ) + b n , m a n ( t ) + γ 1 ( x n p 1 x n + m p 1 ) .

For fixed n and letting m, we have

lim sup m a m (t) a n (t)+ γ 1 ( x n p 1 lim m x m p 1 ) .

Again, letting n, we obtain

lim sup n a n (t) lim inf n a n (t)+ γ 1 (0)= lim inf n a n (t).

This shows that

lim n t x n + ( 1 t ) p 1 p 2

exists for all t(0,1). This completes the proof. □

Theorem 3.1 Under the assumptions of Lemma  3.1, if the dual space E of E has the Kadec-Klee property, then { x n } converges weakly to a fixed point of T.

Proof Using the same method as in the proof of Theorem 2.4, we can prove that there exists a subsequence { x n k } of { x n }, which converges weakly to a point pF(T).

Now, we prove that { x n } converges weakly to p. Suppose that there exists a subsequence { x m j } of { x n } such that { x m j } converges weakly to a point p C. Then p= p . In fact, it follows from Lemma 3.1 that the limit

lim n t x n + ( 1 t ) p p

exists for all t[0,1]. Again, since p, p W w ({ x n }), we have p =p by Lemma 1.3. This shows that { x n } converges weakly to p. This completes the proof. □

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Acknowledgements

The project was supported by the National Natural Science Foundation of China (Grant Number: 11271282) and the third author was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant Number: 2012-0008170).

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Correspondence to Yeol Je Cho.

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All authors contributed equally to this paper and they read and approved the final manuscript.

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Guo, W., Choi, M.S. & Cho, Y.J. Convergence theorems for continuous pseudocontractive mappings in Banach spaces. J Inequal Appl 2014, 384 (2014). https://doi.org/10.1186/1029-242X-2014-384

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Keywords

  • uniformly convex Banach space
  • pseudocontractive mapping
  • nonexpansive mapping
  • implicit iteration process
  • convergence theorem