Open Access

Some results on zero points of m-accretive operators in reflexive Banach spaces

Fixed Point Theory and Applications20142014:118

https://doi.org/10.1186/1687-1812-2014-118

Received: 16 January 2014

Accepted: 30 April 2014

Published: 14 May 2014

Abstract

A modified proximal point algorithm is proposed for treating common zero points of a finite family of m-accretive operators. A strong convergence theorem is established in a reflexive, strictly convex Banach space with the uniformly Gâteaux differentiable norm.

Keywords

accretive operator nonexpansive mapping resolvent fixed point zero point

1 Introduction and preliminaries

Let E be a Banach space and let E be the dual of E. Let , denote the pairing between E and E . The normalized duality mapping J : E 2 E is defined by
J ( x ) = { f E : x , f = x 2 = f 2 } , x E .

A Banach space E is said to strictly convex if and only if x = y = ( 1 λ ) x + λ y for x , y E and 0 < λ < 1 implies that x = y . Let U E = { x E : x = 1 } . The norm of E is said to be Gâteaux differentiable if the limit lim t 0 x + t y x t exists for each x , y U E . In this case, E is said to be smooth. The norm of E is said to be uniformly Gâteaux differentiable if for each y U E , the limit is attained uniformly for all x U E . The norm of E is said to be Fréchet differentiable if for each x U E , the limit is attained uniformly for all y U E . The norm of E is said to be uniformly Fréchet differentiable if the limit is attained uniformly for all x , y U E . It is well known that (uniform) Fréchet differentiability of the norm of E implies (uniform) Gâteaux differentiability of the norm of E.

Let ρ E : [ 0 , ) [ 0 , ) be the modulus of smoothness of E by
ρ E ( t ) = sup { x + y x y 2 1 : x U E , y t } .

A Banach space E is said to be uniformly smooth if ρ E ( t ) t 0 as t 0 . It is well known that if the norm of E is uniformly Gâteaux differentiable, then the duality mapping J is single valued and uniformly norm to weak continuous on each bounded subset of E.

Recall that a closed convex subset C of a Banach space E is said to have a normal structure if for each bounded closed convex subset K of C which contains at least two points, there exists an element x of K which is not a diametral point of K, i.e., sup { x y : y K } < d ( K ) , where d ( K ) is the diameter of K.

Let D be a nonempty subset of a set C. Let Proj D : C D . Q is said to be
  1. (1)

    sunny if for each x C and t ( 0 , 1 ) , we have Proj D ( t x + ( 1 t ) Proj D x ) = Proj D x ;

     
  2. (2)

    a contraction if Proj D 2 = Proj D ;

     
  3. (3)

    a sunny nonexpansive retraction if Proj D is sunny, nonexpansive, and a contraction.

     

D is said to be a nonexpansive retract of C if there exists a nonexpansive retraction from C onto D. The following result, which was established in [13], describes a characterization of sunny nonexpansive retractions on a smooth Banach space.

Let E be a smooth Banach space and let C be a nonempty subset of E. Let Proj C : E C be a retraction and J φ be the duality mapping on E. Then the following are equivalent:
  1. (1)

    Proj C is sunny and nonexpansive;

     
  2. (2)

    x Proj C x , J φ ( y Proj C x ) 0 , x E , y C ;

     
  3. (3)

    Proj C x Proj C y 2 x y , J φ ( Proj C x Proj C y ) , x , y E .

     

It is well known that if E is a Hilbert space, then a sunny nonexpansive retraction Proj C is coincident with the metric projection from E onto C. Let C be a nonempty closed convex subset of a smooth Banach space E, let x E , and let x 0 C . Then we have from the above that x 0 = Proj C x if and only if x x 0 , J φ ( y x 0 ) 0 for all y C , where Proj C is a sunny nonexpansive retraction from E onto C. For more additional information on nonexpansive retracts, see [4] and the references therein.

Let C be a nonempty closed convex subset of E. Let T : C C be a mapping. In this paper, we use F ( T ) to denote the set of fixed points of T. Recall that T is said to be an α-contractive mapping iff there exists a constant α [ 0 , 1 ) such that T x T y α x y , x , y C . The Picard iterative process is an efficient method to study fixed points of α-contractive mappings. It is well known that α-contractive mappings have a unique fixed point. T is said to be nonexpansive iff T x T y x y , x , y C . It is well known that nonexpansive mappings have fixed points if the set C is closed and convex, and the space E is uniformly convex. The Krasnoselski-Mann iterative process is an efficient method for studying fixed points of nonexpansive mappings. The Krasnoselski-Mann iterative process generates a sequence { x n } in the following manner:
x 1 C , x n + 1 = α n T x n + ( 1 α n ) x n , n 1 .
It is well known that the Krasnoselski-Mann iterative process only has weak convergence for nonexpansive mappings in infinite-dimensional Hilbert spaces; see [57] for more details and the references therein. In many disciplines, including economics, image recovery, quantum physics, and control theory, problems arise in infinite-dimensional spaces. In such problems, strong convergence (norm convergence) is often much more desirable than weak convergence, for it translates the physically tangible property that the energy x n x of the error between the iterate x n and the solution x eventually becomes arbitrarily small. To improve the weak convergence of a Krasnoselski-Mann iterative process, so-called hybrid projections have been considered; see [822] for more details and the references therein. The Halpern iterative process was initially introduced in [23]; see [23] for more details and the references therein. The Halpern iterative process generates a sequence { x n } in the following manner:
x 1 C , x n + 1 = α n u + ( 1 α n ) T x n , n 1 ,

where x 1 is an initial and u is a fixed element in C. Strong convergence of Halpern iterative process does not depend on metric projections. The Halpern iterative process has recently been extensively studied for treating accretive operators; see [2431] and the references therein.

Let I denote the identity operator on E. An operator A E × E with domain D ( A ) = { z E : A z } and range R ( A ) = { A z : z D ( A ) } is said to be accretive if for each x i D ( A ) and y i A x i , i = 1 , 2 , there exists j ( x 1 x 2 ) J ( x 1 x 2 ) such that y 1 y 2 , j ( x 1 x 2 ) 0 . An accretive operator A is said to be m-accretive if R ( I + r A ) = E for all r > 0 . In this paper, we use A 1 ( 0 ) to denote the set of zero points of A. For an accretive operator A, we can define a nonexpansive single valued mapping J r : R ( I + r A ) D ( A ) by J r = ( I + r A ) 1 for each r > 0 , which is called the resolvent of A.

Now, we are in a position to give the lemmas to prove main results.

Lemma 1.1 [32]

Let { a n } , { b n } , { c n } , and { d n } be four nonnegative real sequences satisfying a n + 1 ( 1 b n ) a n + b n c n + d n , n n 0 , where n 0 is some positive integer, { b n } is a number sequence in ( 0 , 1 ) such that n = n 0 b n = , { c n } is a number sequence such that lim sup n c n 0 , and { d n } is a positive number sequence such that n = n 0 d n < . Then lim n a n = 0 .

Lemma 1.2 [33]

Let C be a closed convex subset of a strictly convex Banach space E. Let N 1 be some positive integer and let T i : C C be a nonexpansive mapping for each i { 1 , 2 , , N } . Let { δ i } be a real number sequence in ( 0 , 1 ) with i = 1 N δ i = 1 . Suppose that i = 1 N F ( T i ) is nonempty. Then the mapping i = 1 N T i is defined to be nonexpansive with F ( i = 1 N T i ) = i = 1 N F ( T i ) .

Lemma 1.3 [34]

Let { x n } and { y n } be bounded sequences in a Banach space E and let β n be a sequence in [ 0 , 1 ] with 0 < lim inf n β n lim sup n β n < 1 . Suppose that x n + 1 = ( 1 β n ) y n + β n x n for all n 0 and
lim sup n ( y n + 1 y n x n + 1 x n ) 0 .

Then lim n y n x n = 0 .

Lemma 1.4 [35]

Let E be a real reflexive Banach space with the uniformly Gâteaux differentiable norm and the normal structure, and let C be a nonempty closed convex subset of E. Let f : C C be α-contractive mapping and let T : C C be a nonexpansive mapping with a fixed point. Let { x t } be a sequence generated by the following: x t = t f ( x t ) + ( 1 t ) T x t , where t ( 0 , 1 ) . Then { x t } converges strongly as t 0 to a fixed point x of T, which is the unique solution in F ( T ) to the following variational inequality: f ( x ) x , j ( x p ) 0 , p F ( T ) .

2 Main results

Theorem 2.1 Let E be a real reflexive, strictly convex Banach space with the uniformly Gâteaux differentiable norm. Let N 1 be some positive integer. Let A m be an m-accretive operator in E for each m { 1 , 2 , , N } . Assume that C : = m = 1 N D ( A m ) ¯ is convex and has the normal structure. Let f : C C be an α-contractive mapping. Let { α n } , { β n } , and { γ n } be real number sequences in ( 0 , 1 ) with the restriction α n + β n + γ n = 1 . Let { δ n , i } be a real number sequence in ( 0 , 1 ) with the restriction δ n , 1 + δ n , 2 + + δ n , N = 1 . Let { r m } be a positive real numbers sequence and { e n , i } a sequence in E for each i { 1 , 2 , , N } . Assume that i = 1 N A i 1 ( 0 ) is not empty. Let { x n } be a sequence generated in the following manner:
x 1 C , x n + 1 = α n f ( x n ) + β n x n + γ n i = 1 N δ n , i J r i ( x n + e n , i ) , n 1 ,
where J r i = ( I + r i A i ) 1 . Assume that the control sequences { α n } , { β n } , { γ n } , and { δ n , i } satisfy the following restrictions:
  1. (a)

    lim n α n = 0 , n = 1 α n = ;

     
  2. (b)

    0 < lim inf n β n lim sup n β n < 1 ;

     
  3. (c)

    n = 1 e n , m < ;

     
  4. (d)

    lim n δ n , i = δ i ( 0 , 1 ) .

     

Then the sequence { x n } converges strongly to x ¯ , which is the unique solution to the following variational inequality: f ( x ¯ ) x ¯ , J ( p x ¯ ) 0 , p i = 1 N A i 1 ( 0 ) .

Proof Put y n = i = 1 N δ n , i J r i ( x n + e n , i ) . Fixing p i = 1 N A i 1 ( 0 ) , we have
y n p i = 1 N δ n , i J r i ( x n + e n , i ) p i = 1 N δ n , i ( x n + e n , i ) p x n p + i = 1 N e n , i .
Hence, we have
x n + 1 p α n f ( x n ) p + β n x n p + γ n y n p α n α x n p + α n f ( p ) p + β n x n p + γ n x n p + γ n i = 1 N e n , i ( 1 α n ( 1 α ) ) x n p + α n ( 1 α ) f ( p ) p 1 α + i = 1 N e n , i max { x n p , f ( p ) p } + i = 1 N e n , i max { x 1 p , f ( p ) p } + j = 1 i = 1 N e j , i .
This proves that the sequence { x n } is bounded, and so is { y n } . Since
y n y n 1 = i = 1 N δ n , i ( J r m ( x n + e n , i ) J r i ( x n 1 + e n 1 , i ) ) + i = 1 N ( δ n , i δ n 1 , i ) J r i ( x n 1 + e n 1 , i ) ,
we have
y n y n 1 i = 1 N δ n , i J r i ( x n + e n , i ) J r i ( x n 1 + e n 1 , i ) + i = 1 N | δ n , i δ n 1 , i | J r i ( x n 1 + e n 1 , i ) x n x n 1 + i = 1 N e n , i + i = 1 N e n 1 , i + i = 1 N | δ n , i δ n 1 , i | J r i ( x n 1 + e n 1 , i ) x n x n 1 + i = 1 N e n , i + i = 1 N e n 1 , i + M 1 i = 1 N | δ n , i δ n 1 , i | ,
where M 1 is an appropriate constant such that
M 1 = max { sup n 1 J r 1 ( x n + e n , 1 ) , sup n 1 J r 2 ( x n + e n , 2 ) , , sup n 1 J r N ( x n + e n , N ) } .
Define a sequence { z n } by z n : = x n + 1 β n x n 1 β n , that is, x n + 1 = β n x n + ( 1 β n ) z n . It follows that
y z n z n 1 α n 1 β n f ( x n ) y n + α n 1 1 β n 1 f ( x n 1 ) y n 1 + y n y n 1 α n 1 β n f ( x n ) y n + α n 1 1 β n 1 f ( x n 1 ) y n 1 + x n x n 1 + i = 1 N | δ n , i δ n 1 , i | J r i x n 1 α n 1 β n f ( x n ) y n + α n 1 1 β n 1 f ( x n 1 ) y n 1 + x n x n 1 + M 2 ( i = 1 N | δ n , i δ i | + i = 1 N | δ i δ n 1 , i | ) ,
where M 2 is an appropriate constant such that
M 2 = max { sup n 1 J r 1 x n , sup n 1 J r 2 x n , , sup n 1 J r N x n } .
This implies that
z n z n 1 x n x n 1 α n 1 β n f ( x n ) y n + α n 1 1 β n 1 f ( x n 1 ) y n 1 + M 2 ( i = 1 N | δ n , i δ i | + i = 1 N | δ i δ n 1 , i | ) .
From the restrictions (a), (b), (c), and (d), we find that
lim sup n ( z n z n 1 x n x n 1 ) 0 .
Using Lemma 1.4, we find that lim n z n x n = 0 . This further shows that lim sup n x n + 1 x n = 0 . Put T = i = 1 N δ i J r i . It follows from Lemma 1.3 that T is nonexpansive with F ( T ) = i = 1 N F ( J r i ) = i = 1 N A i 1 ( 0 ) . Note that
x n T x n x n x n + 1 + x n + 1 T x n x n x n + 1 + α n f ( x n ) T x n + β n x n T x n + γ n y n T x n x n x n + 1 + α n f ( x n ) T x n + β n x n T x n + M 2 i = 1 N | δ n , i δ i | .
This implies that
( 1 β n ) x n T x n x n x n + 1 + α n f ( x n ) T x n + M 2 i = 1 N | δ n , i δ i | .
It follows from the restrictions (a), (b), and (d) that
lim n T x n x n = 0 .
Now, we are in a position to prove that lim sup n f ( x ¯ ) x ¯ , J ( x n x ¯ ) 0 , where x ¯ = lim t 0 x t , and x t solves the fixed point equation
x t = t f ( x t ) + ( 1 t ) T x t , t ( 0 , 1 ) .
It follows that
x t x n 2 = t f ( x t ) x n , J ( x t x n ) + ( 1 t ) T x t x n , j ( x t x n ) = t f ( x t ) x t , J ( x t x n ) + t x t x n , J ( x t x n ) + ( 1 t ) T x t T x n , J ( x t x n ) + ( 1 t ) T x n x n , J ( x t x n ) t f ( x t ) x t , J ( x t x n ) + x t x n 2 + T x n x n x t x n , t ( 0 , 1 ) .
This implies that
x t f ( x t ) , J ( x t x n ) 1 t T x n x n x t x n , t ( 0 , 1 ) .
Since lim n T x n x n = 0 , we find that lim sup n x t f ( x t ) , J ( x t x n ) 0 . Since J is strong to weak uniformly continuous on bounded subsets of E, we find that
| f ( x ¯ ) x ¯ , J ( x n x ¯ ) x t f ( x t ) , J ( x t x n ) | | f ( x ¯ ) x ¯ , J ( x n x ¯ ) f ( x ¯ ) x ¯ , J ( x n x t ) | + | f ( x ¯ ) x ¯ , J ( x n x t ) x t f ( x t ) , J ( x t x n ) | | f ( x ¯ ) x ¯ , J ( x n x ¯ ) J ( x n x t ) | + | f ( x ¯ ) x ¯ + x t f ( x t ) , J ( x n x t ) | f ( x t ) x ¯ J ( x n x ¯ ) J ( x n x t ) + ( 1 + α ) x ¯ x t x n x t .
Since x t x ¯ , as t 0 , we have
lim t 0 | f ( x ¯ ) x ¯ , J ( x n x ¯ ) f ( x t ) x t , J ( x n x t ) | = 0 .
For ϵ > 0 , there exists δ > 0 such that t ( 0 , δ ) , we have
f ( x ¯ ) x ¯ , J ( x n x ¯ ) f ( x t ) x t , J ( x n x t ) + ϵ .

This implies that lim sup n f ( x ¯ ) x ¯ , J ( x n x ¯ ) 0 .

Finally, we show that x n x ¯ as n . Since 2 is convex, we see that
y n x ¯ 2 = i = 1 N δ n , i J r i ( x n + e n , i ) x ¯ 2 i = 1 N δ n , i J r i ( x n + e n , i ) x ¯ 2 x n x ¯ 2 + i = 1 N e n , i ( e n , i + 2 x n x ¯ ) .
It follows that
x n + 1 x ¯ 2 = α n f ( x n ) x ¯ , J ( x n + 1 x ¯ ) + β n x n x ¯ , J ( x n + 1 x ¯ ) + γ n y n x ¯ , J ( x n + 1 x ¯ ) α n α x n x ¯ x n + 1 x ¯ + α n f ( x ¯ ) x ¯ , J ( x n + 1 x ¯ ) + β n x n x ¯ x n + 1 x ¯ + γ n y n x ¯ x n + 1 x ¯ α n α 2 ( x n x ¯ 2 + x n + 1 x ¯ 2 ) + α n f ( x ¯ ) x ¯ , J ( x n + 1 x ¯ ) + β n 2 ( x n x ¯ 2 + x n + 1 x ¯ 2 ) + γ n 2 x n x ¯ 2 + i = 1 N e n , i ( e n , i + 2 x n x ¯ ) + γ n 2 x n + 1 x ¯ 2 .
Hence, we have
x n + 1 x ¯ 2 ( 1 α n ( 1 α ) ) x n x ¯ 2 + 2 α n f ( x ¯ ) x ¯ , J ( x n + 1 x ¯ ) + i = 1 N e n , i ( e n , i + 2 x n x ¯ ) .

Using Lemma 1.1, we find x n x ¯ as n . This completes the proof. □

Remark 2.2 There are many spaces satisfying the restriction in Theorem 2.1, for example L p , where p > 1 .

Corollary 2.3 Let E be a Hilbert space and let N 1 be some positive integer. Let A m be a maximal monotone operator in E for each m { 1 , 2 , , N } . Assume that C : = m = 1 N D ( A m ) ¯ is convex and has the normal structure. Let f : C C be an α-contractive mapping. Let { α n } , { β n } , and { γ n } be real number sequences in ( 0 , 1 ) with the restriction α n + β n + γ n = 1 . Let { δ n , i } be a real number sequence in ( 0 , 1 ) with the restriction δ n , 1 + δ n , 2 + + δ n , N = 1 . Let { r m } be a positive real numbers sequence and { e n , i } a sequence in E for each i { 1 , 2 , , N } . Assume that i = 1 N A i 1 ( 0 ) is not empty. Let { x n } be a sequence generated in the following manner:
x 1 C , x n + 1 = α n f ( x n ) + β n x n + γ n i = 1 N δ n , i J r i ( x n + e n , i ) , n 1 ,
where J r i = ( I + r i A i ) 1 . Assume that the control sequences { α n } , { β n } , { γ n } , and { δ n , i } satisfy the following restrictions:
  1. (a)

    lim n α n = 0 , n = 1 α n = ;

     
  2. (b)

    0 < lim inf n β n lim sup n β n < 1 ;

     
  3. (c)

    n = 1 e n , m < ;

     
  4. (d)

    lim n δ n , i = δ i ( 0 , 1 ) .

     

Then the sequence { x n } converges strongly to x ¯ , which is the unique solution to the following variational inequality: f ( x ¯ ) x ¯ , p x ¯ 0 , p i = 1 N A i 1 ( 0 ) .

3 Applications

In this section, we consider a variational inequality problem. Let A : C E be a single valued monotone operator which is hemicontinuous; that is, continuous along each line segment in C with respect to the weak topology of E . Consider the following variational inequality:
find  x C  such that  y x , A x 0 , y C .
The solution set of the variational inequality is denoted by VI ( C , A ) . Recall that the normal cone N C ( x ) for C at a point x C is defined by
N C ( x ) = { x E : y x , x 0 , y C } .

Now, we are in a position to give the convergence theorem.

Theorem 3.1 Let E be a real reflexive, strictly convex Banach space with the uniformly Gâteaux differentiable norm. Let N 1 be some positive integer and let C be nonempty closed and convex subset of E. Let A i : C E a single valued, monotone and hemicontinuous operator. Assume that i = 1 N VI ( C , A i ) is not empty and C has the normal structure. Let f : C C be an α-contractive mapping. Let { α n } , { β n } , and { γ n } be real number sequences in ( 0 , 1 ) with the restriction α n + β n + γ n = 1 . Let { δ n , i } be a real number sequence in ( 0 , 1 ) with the restriction δ n , 1 + δ n , 2 + + δ n , N = 1 . Let { r m } be a positive real numbers sequence and { e n , i } a sequence in E for each i { 1 , 2 , , N } . Let { x n } be a sequence generated in the following manner:
x 1 C , x n + 1 = α n f ( x n ) + β n x n + γ n i = 1 N δ n , i VI ( C , A i + 1 r i ( I x n ) ) , n 1 .
Assume that the control sequences { α n } , { β n } , { γ n } , and { δ n , i } satisfy the following restrictions:
  1. (a)

    lim n α n = 0 , n = 1 α n = ;

     
  2. (b)

    0 < lim inf n β n lim sup n β n < 1 ;

     
  3. (c)

    n = 1 e n , m < ;

     
  4. (d)

    lim n δ n , i = δ i ( 0 , 1 ) .

     

Then the sequence { x n } converges strongly to x ¯ , which is the unique solution to the following variational inequality: f ( x ¯ ) x ¯ , J ( p x ¯ ) 0 , p i = 1 N VI ( C , A i ) .

Proof Define a mapping T i E × E by
T i x : = { A i x + N C x , x C , , x C .
From Rockafellar [36], we find that T i is maximal monotone with T i 1 ( 0 ) = VI ( C , A i ) . For each r i > 0 , and x n E , we see that there exists a unique x r i D ( T i ) such that x n x r i + r i T i ( x r i ) , where x r i = ( I + r i T i ) 1 x n . Notice that
y n , i = VI ( C , A i + 1 r i ( I x n ) ) ,
which is equivalent to
y y n , i , A i y n , i + 1 r i ( y n , i x n ) 0 , y C ,

that is, A i y n , i + 1 r i ( x n y n , i ) N C ( y n , i ) . This implies that y n , i = ( I + r i T i ) 1 x n . Using Theorem 2.1, we find the desired conclusion immediately. □

From Theorem 3.1, the following result is not hard to derive.

Corollary 3.2 Let E be a real reflexive, strictly convex Banach space with the uniformly Gâteaux differentiable norm. Let C be nonempty closed and convex subset of E. Let A : C E a single valued, monotone and hemicontinuous operator with VI ( C , A ) . Assume that C has the normal structure. Let f : C C be an α-contractive mapping. Let { α n } , { β n } , and { γ n } be real number sequences in ( 0 , 1 ) with the restriction α n + β n + γ n = 1 . Let { x n } be a sequence generated in the following manner:
x 1 C , x n + 1 = α n f ( x n ) + β n x n + γ n VI ( C , A + 1 r ( I x n ) ) , n 1 .
Assume that the control sequences { α n } , { β n } , and { γ n } satisfy the following restrictions:
  1. (a)

    lim n α n = 0 , n = 1 α n = ;

     
  2. (b)

    0 < lim inf n β n lim sup n β n < 1 .

     

Then the sequence { x n } converges strongly to x ¯ , which is the unique solution to the following variational inequality: f ( x ¯ ) x ¯ , J ( p x ¯ ) 0 , p VI ( C , A i ) .

Declarations

Acknowledgements

The authors are grateful to the editor and the reviewers for useful suggestions which improved the contents of the article.

Authors’ Affiliations

(1)
School of Business and Administration, Henan University
(2)
School of Mathematics and Information Science, Shangqiu Normal University
(3)
Vietnam National University

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© Wu et al.; licensee Springer. 2014

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