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Strong convergence theorem for quasi-Bregman strictly pseudocontractive mappings and equilibrium problems in Banach spaces

Abstract

In this paper, we introduce a new iterative scheme by a hybrid method and prove a strong convergence theorem of a common element in the set of fixed points of a finite family of closed quasi-Bregman strictly pseudocontractive mappings and common solutions to a system of equilibrium problems in reflexive Banach space. Our results extend important recent results announced by many authors.

MSC:47H09, 47J25.

1 Introduction

Let E be a real Banach space and C a nonempty closed convex subset of E. The normalized duality map from E to 2 E ( E is the dual space of E) denoted by J is defined by

J(x)= { f E : x , f = x 2 = f 2 } .

Let T:CC be a map, a point xC is called a fixed point of T if Tx=x, and the set of all fixed points of T is denoted by F(T). The mapping T is called L-Lipschitzian or simply Lipschitz if there exists L>0, such that TxTyLxy, x,yC and if L=1, then the map T is called nonexpansive.

Let g:C×CR be a bifunction. The equilibrium problem with respect to g is to find

zC such that g(z,y)0,yC.

The set of solution of equilibrium problem is denoted by EP(g). Thus

EP(g):= { z C : g ( z , y ) 0 , y C } .

Numerous problems in physics, optimization and economics reduce to finding a solution of equilibrium problem. Some methods have been proposed to solve the equilibrium problem in Hilbert spaces; see for example Blum and Oettli [1], Combettes and Hirstoaga [2]. Recently, Tada and Takahashi [3, 4] and Takahashi and Takahashi [5] obtain weak and strong convergence theorems for finding a common element of the set of solutions of an equilibrium problem and set of fixed points of a nonexpansive mapping in Hilbert space. In particular, Takahashi and Zembayashi [4] established a strong convergence theorem for finding a common element of the two sets by using the hybrid method introduced in Nakajo and Takahashi [6]. They also proved such a strong convergence theorem in a uniformly convex and uniformly smooth Banach space.

Reich and Sabach [7] and Kassay et al. [8] proved some convergence theorems for the solution of some equilibrium and variational inequality problems in the setting of reflexive Banach spaces.

Let ϕ:E×E[0,) denote the Lyapunov functional defined by

ϕ(x,y)= x 2 2x,Jy+ y 2 ,x,yE.

A mapping T:CC is said to be quasi-ϕ strictly pseudocontractive, see [9], if F(T) and there exists a constant k(0,1] such that

ϕ(p,Tx)ϕ(p,x)+kϕ(x,Tx),xC and pF(T).

Let E be a real reflexive Banach space with norm and E the dual space of E. Throughout this paper, we shall assume f:E(,+] is a proper, lower semi-continuous and convex function. We denote by domf:={xE:f(x)<+} the domain of f.

Let xintdomf; the subdifferential of f at x is the convex set defined by

f(x)= { x E : f ( x ) + x , y x f ( y ) , y E } ,

where the Fenchel conjugate of f is the function f : E (,+] defined by

f ( x ) =sup { x , x f ( x ) : x E } .

We know that the Young-Fenchel inequality holds:

x , x f(x)+ f ( x ) ,xE, x E .

A function f on E is coercive [10] if the sublevel set of f is bounded; equivalently,

lim x + f(x)=+.

A function f on E is said be strongly coercive [11] if

lim x + f ( x ) x =+.

For any xintdomf and yE, the right-hand derivative of f at x in the direction y is defined by

f (x,y):= lim t 0 + f ( x + t y ) f ( x ) t .

The function f is said to be Gâteaux differentiable at x if lim t 0 + f ( x + t y ) f ( x ) t exists for any y. In this case, f (x,y) coincides with f(x), the value of the gradient f of f at x. The function f is said to be Gâteaux differentiable if it is Gâteaux differentiable for any xintdomf. The function f is said to be Fréchet differentiable at x if this limit is attained uniformly in y=1. Finally, f is said to be uniformly Fréchet differentiable on a subset C of E if the limit is attained uniformly for xC and y=1. It is well known that if f is Gâteaux differentiable (resp. Fréchet differentiable) on intdomf, then f is continuous and its Gâteaux derivative f is norm-to-weak continuous (resp. continuous) on intdomf (see also [12, 13]). We will need the following results.

Lemma 1.1 [14]

If f:ER is uniformly Fréchet differentiable and bounded on bounded subsets of E, then f is uniformly continuous on bounded subsets of E from the strong topology of E to the strong topology of E .

Definition 1.2 [15]

The function f is said to be:

  1. (i)

    essentially smooth, if ∂f is both locally bounded and single-valued on its domain;

  2. (ii)

    essentially strictly convex, if ( f ) 1 is locally bounded on its domain and f is strictly convex on every convex subset of domf;

  3. (iii)

    Legendre, if it is both essentially smooth and essentially strictly convex.

Remark 1.3 Let E be a reflexive Banach space. Then we have:

  1. (i)

    f is essentially smooth if and only if f is essentially strictly convex (see [15], Theorem 5.4);

  2. (ii)

    ( f ) 1 = f (see [13]);

  3. (iii)

    f is Legendre if and only if f is Legendre (see [15], Corollary 5.5);

  4. (iv)

    if f is Legendre, then f is a bijection satisfying f= ( f ) 1 , ranf=dom f =intdom f and ran f =domf=intdomf (see [15], Theorem 5.10).

Examples of Legendre functions were given in [15, 16]. One important and interesting Legendre function is 1 p p (1<p<) when E is a smooth and strictly convex Banach space. In this case the gradient f of f is coincident with the generalized duality mapping of E, i.e., f= J p (1<p<). In particular, f=I the identity mapping in Hilbert spaces. In the rest of this paper, we always assume that f:E(,+] is Legendre.

Let f:E(,+] be a convex and Gâteaux differentiable function. The function D f :domf×intdomf[0,+), defined as follows:

D f (y,x):=f(y)f(x) f ( x ) , y x ,
(1.1)

is called the Bregman distance with respect to f (see [17]). It is obvious from the definition of D f that

D f (z,x)= D f (z,y)+ D f (y,x)+ f ( y ) f ( x ) , z y .
(1.2)

Recall that the Bregman projection [18] of xintdomf onto the nonempty, closed, and convex set Cdomf is the necessarily unique vector P C f (x)C satisfying

D f ( P C f ( x ) , x ) =inf { D f ( y , x ) : y C } .

Concerning the Bregman projection, the following are well known.

Lemma 1.4 [19]

Let C be a nonempty, closed, and convex subset of a reflexive Banach space E. Let f:ER be a Gâteaux differentiable and totally convex function and let xE. Then:

  1. (a)

    z= P C f (x) if and only if f(x)f(z),yz0, yC;

  2. (b)

    D f (y, P C f (x))+ D f ( P C f (x),x) D f (y,x), xE, yC.

Let f:E(,+] be a convex and Gâteaux differentiable function. The modulus of the total convexity of f at xintdomf is the function v f (x,):[0,+)[0,+] defined by

v f (x,t):=inf { D f ( y , x ) : y dom f , y x = t } .

The function f is called totally convex at x if v f (x,t)>0 whenever t>0. The function f is called totally convex if it is totally convex at any point xintdomf and is said to be totally convex on bounded sets if v f (B,t)>0 for any nonempty bounded subset B of E and t>0, where the modulus of the total convexity of the function f on the set B is the function v f :intdomf×[0,+)[0,+] defined by

v f (B,t):=inf { v f ( x , t ) : x B dom f } .

Lemma 1.5 [20]

If xdomf, then the following statements are equivalent:

  1. (i)

    the function f is totally convex at x;

  2. (ii)

    for any sequence { y n }domf,

    lim n + D f ( y n ,x)=0 lim n + y n x=0.

Recall that the function f called sequentially consistent [19] if for any two sequences { x n } and { y n } in E such that the first one is bounded

lim n + D f ( y n , x n )=0 lim n + y n x n =0.

Lemma 1.6 [21]

The function f is totally convex on bounded sets if and only if the function f is sequentially consistent.

Lemma 1.7 [22]

Let f:ER be a Gâteaux differentiable and totally convex function. If x 0 E and the sequence { D f ( x n , x 0 )} is bounded, then the sequence { x n } is bounded too.

Lemma 1.8 [22]

Let f:ER be a Gâteaux differentiable and totally convex function, x 0 E and let C be a nonempty, closed, and convex subset of E. Suppose that the sequence { x n } is bounded and any weak subsequential limit of { x n } belongs to C. If D f ( x n , x 0 ) D f ( P C f ( x 0 ), x 0 ) for any nR, then { x n } converges strongly to P C f ( x 0 ).

A mapping T is said to be Bregman firmly nonexpansive [23], if for all x,yC,

f ( T x ) f ( T y ) , T x T y f ( x ) f ( y ) , T x T y

or, equivalently,

D f (Tx,Ty)+ D f (Ty,Tx)+ D f (Tx,x)+ D f (Ty,y) D f (Tx,y)+ D f (Ty,x).

A point pC is said to be asymptotic fixed point of a map T, if there exists a sequence { x n } in C which converges weakly to p such that lim n x n T x n =0. We denote by F ˆ (T) the set of asymptotic fixed points of T. A point pC is said to be strong asymptotic fixed point of a map T, if there exists a sequence { x n } in C which converges strongly to p such that lim n x n T x n =0. We denote by F ˜ (T) the set of strong asymptotic fixed points of T. Let f:ER, a mapping T:CC is said to be Bregman relatively nonexpansive [24] if F(T), F ˆ (T)=F(T) and D f (p,T(x)) D f (p,x) for all xC and pF(T). The map T:CC is said to be Bregman weak relatively nonexpansive if F(T), F ˜ (T)=F(T) and D f (p,T(x)) D f (p,x) for all xC and pF(T). T is said to be quasi-Bregman relatively nonexpansive if F(T), and D f (p,T(x)) D f (p,x) for all xC and pF(T). In [22] quasi-Bregman relatively nonexpansive is called left quasi-Bregman relatively nonexpansive. A map T:CC is called right quasi-Bregman relatively nonexpansive [25] if F(T), and D f (T(x),p) D f (x,p) for all xC and pF(T). T is said to be quasi-Bregman strictly pseudocontractive if there exist a constant k[0,1) and F(T) such that D f (p,Tx) D f (p,x)+k D f (x,Tx) for all xC and pF(T). In particular, T is said to be quasi-Bregman relatively nonexpansive if k=0 and T is said to be quasi-Bregman pseudocontractive if k=1.

Very recently, Zhou and Gao [9] introduced this definition of a quasi-strict pseudocontraction related to the function ϕ and proved the convergence of a hybrid projection algorithm to a fixed point of a closed and quasi-strict pseudocontraction in a smooth and uniformly convex Banach space. They studied the strong convergence of the following scheme:

{ x 0 E , C 1 = C , x 1 = C 1 ( x 0 ) , C n + 1 = { z C n : ϕ ( x n , T x n ) 2 1 k x n z , J x n J T x n } , x n + 1 = C n + 1 ( x 0 ) ,

where C n + 1 is the generalized projection from E onto C n + 1 . They proved that the sequence { x n } converges strongly to F ( T ) ( x 0 ).

Recently, Zegeye and Shahzad [26] proved a strong convergence theorem for the common fixed point of a finite family of right Bregman strongly nonexpansive mappings in a reflexive Banach space. Alghamdi et al. [27] proved a strong convergence theorem for the common fixed point of a finite family of quasi-Bregman nonexpansive mappings. Pang et al. [28] proved weak convergence theorems for Bregman relatively nonexpansive mappings. Shahzad and Zegeye [29] proved a strong convergence theorem for multivalued Bregman relatively nonexpansive mappings, while Zegeye and Shahzad [30] proved a strong convergence theorem for a finite family of Bregman weak relatively nonexpansive mappings.

Motivated and inspired by the above works, in this paper, we prove a new strong convergence theorem for a finite family of closed quasi-Bregman strictly pseudocontractive mapping and a system of equilibrium problems in a real reflexive Banach space. These results generalize and improve several recent results. We showed by an example that the class of quasi-Bregman strictly pseudocontractive mappings is a proper generalization of the class of quasi-ϕ-Bregman strictly pseudocontractive mappings.

2 Preliminaries

The next lemma will be useful in the proof of our main results.

Lemma 2.1 Let f:ER be a Legendre function which is uniformly Fréchet differentiable and bounded on subsets of E, let C be a nonempty, closed, and convex subset of E and let T:CC be a quasi-Bregman strictly pseudocontractive mapping with respect to f. Then, for any xC, pF(T) and k[0,1) the following hold:

D f (x,Tx) 1 1 k f ( x ) f ( T x ) , x p .
(2.1)

Proof Let xC, pF(T) and k[0,1), by definition of T, we have

D f (p,Tx) D f (p,x)+k D f (x,Tx)

and, from (1.2), we obtain

D f (p,x)+ D f (x,Tx)+ f ( x ) f ( T x ) , p x D f (p,x)+k D f (x,Tx),

which implies

D f (x,Tx) 1 1 k f ( x ) f ( T x ) , x p .

This completes the proof. □

Lemma 2.2 [31]

Let E be a real reflexive Banach space, f:E(,+] be a proper lower semi-continuous function, then f : E (,+] is a proper weak lower semi-continuous and convex function. Thus, for all zE, we have

D f ( z , f ( i = 1 N t i f ( x i ) ) ) i = 1 N t i D f (z, x i ).
(2.2)

In order to solve the equilibrium problem, let us assume that a bifunction g:C×CR satisfies the following conditions [1]:

(A1) g(x,x)=0, xC;

(A2) g is monotone, i.e., g(x,y)+g(y,x)0, x,yC;

(A3) lim sup t 0 g(x+t(zx),y)g(x,y), x,z,yC;

(A4) the function yg(x,y) is convex and lower semi-continuous.

The resolvent of a bifunction g [2] is the operator Res g f :E 2 C defined by

Res g f (x)= { z C : g ( z , y ) + f ( z ) f ( x ) , y z 0 , y C } .
(2.3)

From Lemma 1, in [32], if f:(,+]R is a strongly coercive and Gâteaux differentiable function, and g satisfies conditions (A1)-(A4), then dom( Res g f )=E. The following lemma gives some characterization of the resolvent Res g f .

Lemma 2.3 [32]

Let E be a real reflexive Banach space and C be a nonempty closed convex subset of E. Let f:E(,+] be a Legendre function. If the bifunction g:C×CR satisfies the conditions (A1)-(A4), then the following hold:

  1. (i)

    Res g f is single-valued;

  2. (ii)

    Res g f is a Bregman firmly nonexpansive operator;

  3. (iii)

    F( Res g f )=EP(g);

  4. (iv)

    EP(g) is closed and convex subset of C;

  5. (v)

    for all xE and for all qF( Res g f ), we have

    D f ( q , Res g f ( x ) ) + D f ( Res g f ( x ) , x ) D f (q,x).
    (2.4)

3 Main result

Lemma 3.1 Let f:ER be a Legendre function which is uniformly Fréchet differentiable on bounded subsets of E, let C be a nonempty, closed, and convex subset of E and let T:CC be a quasi-Bregman strictly pseudocontractive mapping with respect to f. Then F(T) is closed and convex.

Proof Let F(T) be nonempty set. First we show that F(T) is closed. Let { x n } be a sequence in F(T) such that x n z as n, we need to show that zF(T). From Lemma 2.1, we obtain

D f (z,Tz) 1 1 k f ( z ) f ( T z ) , z x n .
(3.1)

From (3.1), we have D f (z,Tz)0, and from [15], Lemma 7.3, it follows that Tz=z. Therefore F(T) is closed.

Next, we show that F(T) is convex. Let z 1 , z 2 F(T), for any t(0,1); putting z=t z 1 +(1t) z 2 , we need to show that zF(T). From Lemma 2.1, we obtain, respectively,

D f (z,Tz) 1 1 k f ( z ) f ( T z ) , z z 1
(3.2)

and

D f (z,Tz) 1 1 k f ( z ) f ( T z ) , z z 2 .
(3.3)

Multiplying (3.2) by t and (3.3) by (1t) and adding the results, we obtain

D f (z,Tz) 1 1 k f ( z ) f ( T z ) , z z ,
(3.4)

which implies D f (z,Tz)0, and from [15], Lemma 7.3, it follows that Tz=z. Therefore F(T) is also convex. This completes the proof. □

We now prove the following theorem.

Theorem 3.2 Let C be a nonempty, closed, and convex subset of a real reflexive Banach space E and f:ER a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subset of E. For each k=1,2,,m, let g k be a bifunction from C×C to satisfying (A1)-(A4) and let { T i = 1 N } be a finite family of L i -Lipschitzian, i=1,2,3,,N, closed and quasi-Bregman strictly pseudocontractive self mappings of C such that F:=( k = 1 m EP( g k ))( i = 1 N F( T i )). Let { x n } n = 1 be a sequence generated by x 1 =xC, C 1 =C and

{ x 1 C , y n = f ( α n f ( x n ) + ( 1 α n ) f ( T n x n ) ) , u j , n = Res g j f y n , j = 1 , 2 , 3 , , m , w n = f ( j = 1 m β j , n f ( u j , n ) ) , C n + 1 = { w C n : D f ( x n , w n ) 1 1 k f ( x n ) C n + 1 = f ( T n x n ) , x n w + f ( T n x n ) f ( w n ) , x n w } , x n + 1 = P C n + 1 f ( x ) , n N ,
(3.5)

where T n = T n ( mod N ) , and k[0,1), for each i=1,2,,N, T i is uniformly continuous; suppose { α n } n = 1 and { β j , n } n = 1 , j=1,2,,m are sequences in (0,1) such that (i) lim inf n (1 α n )>0, (ii) j = 1 m β j , n =1, n1. Then { x n } n = 1 converges strongly to P F f (x), where P F f is the Bregman projection of E onto F.

Proof The proof is divided into six steps.

Step I. Show that F=( j = 1 m EP( g j ))( i = 1 N F( T i )) is closed and convex. From Lemma 3.1, i = 1 N F( T i ) is closed and convex and from (iv) of Lemma 2.3, j = 1 m EP( g j ) is closed and convex. So, F=( j = 1 m EP( g j ))( i = 1 N F( T i )) is closed and convex.

Step II. Show that C n is closed and convex for all n1. For n=1, C 1 =C is closed and convex. Assume that C h is closed and convex for some h>1. For w C h + 1 , one obtains

D f ( x h , w h ) 1 1 k f ( x h ) f ( T h x h ) , x h w + f ( T h x h ) f ( w h ) , x h w ;

using the fact that f( x h )f( T h x h ), and f( T h x h )f( w h ), are continuous and linear in E, for h1, C h + 1 is closed and convex.

Step III. Show that F C n for every n1. Note that F C 1 =C. Suppose F C h , for h1, then for all wF C h , since u j , h = Res g j f ( y h ) for each j=1,2,,m, from (2.2) and Lemma 2.3, we have

D f ( w , w h ) = D f ( w , f ( j = 1 m β j , n f ( u j , n ) ) ) j = 1 m β j h D f ( w , u j h ) j = 1 m β j h D f ( w , y h ) = D f ( w , y h ) ;
(3.6)

also from (2.2) and (2.1), we obtain

D f ( w , y h ) = D f ( w , f ( α h f ( x h ) + ( 1 α h ) f ( T h x h ) ) ) α h D f ( w , x h ) + ( 1 α h ) D f ( w , T h x h ) α h D f ( w , x h ) + ( 1 α h ) [ D f ( w , x h ) + k D f ( x h , T h x h ) ] D f ( w , x h ) + k D f ( x h , T h x h ) D f ( w , x h ) + k 1 k f ( x h ) f ( T h x h ) , x h w .
(3.7)

But, from (1.2),

D f (w, w h )= D f (w, x h )+ D f ( x h , w h )+ f ( x h ) f ( w h ) , w x h .
(3.8)

From (3.6), (3.7), and (3.8), we obtain

D f ( x h , w h ) k 1 k f ( x h ) f ( T h x h ) , x h w + f ( x h ) f ( w h ) , x h w = k 1 k f ( x h ) f ( T h x h ) , x h w + f ( x h ) f ( T h x h ) , x h w + f ( T h x h ) f ( w h ) , x h w = ( k 1 k + 1 ) f ( x h ) f ( T h x h ) , x h w + f ( T h x h ) f ( w h ) , x h w = 1 1 k f ( x h ) f ( T h x h ) , x h w + f ( T h x h ) f ( w h ) , x h w .
(3.9)

This shows that w C h + 1 , which implies F C n for every n1.

Step IV. Show that lim n D f ( x n ,x) exists. From (3.5), x n = P C n f x, which from (a) of Lemma 1.4 implies

f ( x ) f ( x n ) , y x n 0,y C n .

Since F C n , we have

f ( x ) f ( x n ) , w x n 0,wF.
(3.10)

From (b) of Lemma 1.4 we have

D f ( x n , x ) = D f ( P C n f x , x ) D f ( w , x ) D f ( w , P C n f x ) D f ( w , x ) , n 1 , w F .
(3.11)

This implies that { D f ( x n ,x)} is bounded, from Lemma 1.7, { x n } is bounded. By the construction of C n , we have x m C m C n , and x n = P C n f x, for any positive integer mn. Then we obtain

D f ( x m , x n ) = D f ( x m , P C n f x ) D f ( x m , x ) D f ( P C n f x , x ) = D f ( x m , x ) D f ( x n , x ) .
(3.12)

In particular,

D f ( x n + 1 , x n ) D f ( x n + 1 ,x) D f ( x n ,x).

Since x n = P C n f x and x n + 1 = P C n + 1 f x C n + 1 C n , we obtain D f ( x n ,x) D f ( x n + 1 ,x), n1. This shows that { D f ( x n ,x)} is nondecreasing and hence the limit lim n D f ( x n ,x) exists. Thus from (3.12), taking the limit as m,n, we obtain lim n D f ( x m , x n )=0. Since f is totally convex on bounded subsets of E, f is sequentially consistent (see [17]). It follows that x m x n 0 as m,n. Hence { x n } is Cauchy sequence in C. As { x n } is Cauchy in a complete space E, there exists pE such that x n p as n. Clearly pC.

Since D f ( x m , x n )0, as m,n, we have in particular

lim n D f ( x n + 1 , x n )=0,
(3.13)

and this further implies that

lim n x n + 1 x n =0.
(3.14)

Step V. Next we show that x n pF.

Since x n + 1 = P C n + 1 f x C n + 1 , we have from (3.5)

D f ( x n , w n ) 1 1 k f ( x n ) f ( T n x n ) , x n x n + 1
(3.15)
+ f ( T n x n ) f ( w n ) , x n x n + 1 ,
(3.16)

which implies that lim n D f ( x n , w n )=0. Since f is totally convex on bounded subsets of E, f is sequentially consistent (see [17]). It follows that

lim n x n w n =0.
(3.17)

From (3.14) and (3.17), we have

lim n x n + 1 w n =0.
(3.18)

Since f is uniformly Fréchet differentiable, it follows from Lemma 1.1 that f is uniformly continuous and f is uniformly continuous on bounded subsets of E (see [33], Theorem 1.8). Hence

lim n f ( x n + 1 ) f ( w n ) =0
(3.19)

and

lim n |f( x n + 1 )f( w n )|=0.
(3.20)

Since x n + 1 C n + 1 , it follows from (3.6), (3.7) that

f ( x n + 1 ) f ( w n ) f ( w n ) , x n + 1 w n = D f ( x n + 1 , w n ) D f ( x n + 1 , y n ) D f ( x n + 1 , x n ) + k 1 k f ( x n ) f ( T n x n ) , x n x n + 1 ,

which implies from (3.20), (3.18), (3.13), and (3.14) that

lim n D f ( x n + 1 , y n )=0.

From the sequential consistency of f, we have

lim n x n + 1 y n =0;
(3.21)

from (3.14) and (3.21), we obtain

lim n x n y n =0,
(3.22)

which implies that y n pC, since x n pC. From the uniform continuity of f, we have

lim n f ( x n ) f ( y n ) =0.
(3.23)

From (3.5), we have

f ( T n x n ) f ( x n ) = 1 1 α n f ( x n ) f ( y n ) ,

which implies from (3.23) that

lim n f ( T n x n ) f ( x n ) =0.
(3.24)

Since f is strongly coercive and uniformly convex on bounded subsets of E, f is uniformly Fréchet differentiable on bounded sets. Moreover, f is bounded on bounded sets, and from (3.24) we obtain

lim n T n x n x n =0.
(3.25)

On the other hand, we see that

x n T n + l x n x n x n + l + x n + l T n + l x n + l + T n + l x n + l T n + l x n ( 1 + L ) x n x n + l + x n + l T n + l x n + l

for all l{1,2,,N}, where L:= sup 1 i N L i . It follows from (3.14) and (3.25) that

lim n x n T n + l x n =0

for all l{1,2,,N}. Thus

lim n x n T l x n =0
(3.26)

for all l{1,2,,N}. Since x n p as n, by the closedness of T l for each l{1,2,,N}, we obtain p l = 1 N F( T l ).

Also, since y n p as n, we have from Lemma 2.3, for each j=1,2,,m,

0 D f (p, u j n )= D f ( p , Res g j f y n ) D f (p, y n )0as n.

Then we have from Lemma 1.5 that lim n p u j n =0, for each j=1,2,,m. Consequently, we have

u j n y n u j n p+p y n 0as n.
(3.27)

From the uniform continuity of f, for each j=1,2,,m we have

lim n f ( u j n ) f ( y n ) =0.
(3.28)

From (2.3), we have, for j=1,2,,m,

g j ( u j n ,y)+ f ( u j n ) f ( y n ) , y u j n 0,yC.

Furthermore, using (A2) in the last inequality, we obtain

f ( u j n ) f ( y n ) , y u j n g j (y, u j n ),yC.

By (A4), (3.28), and u j n p as n, we have

g j (y,p)0,yC.
(3.29)

Let z t :=ty+(1t)p for t(0,1] and yC. This implies that z t C. This yields g j ( z t ,p)0. It follows from (A1) and (A4) that

0 = g j ( z t , z t ) t g j ( z t , y ) + ( 1 t ) g j ( z t , p ) t g j ( z t , y ) ,

and hence

0 g j ( z t ,y).

From condition (A3), we obtain

g j (p,y)0,yC and j{1,2,3,,m}.

This implies that pEP( g j ), for each j=1,2,,m. Thus, p j = 1 m EP( g j ). Hence we have pF=( i = 1 N F( T i ))( j = 1 m EP( g j )).

Step VI. Finally, we show that p= P F f x. Setting n in (3.10), we obtain

f ( x ) f ( p ) , w p 0,wF.

By (a) of Lemma 1.4, we have p= P F f x. □

Here we give an example of a quasi-Bregman strictly peudocontractive mapping which is not quasi-ϕ strictly pseudocontractive mapping; this shows that the former class is a generalization of the latter.

Example 3.3 Let E=R, C=[1,0] and define T,f:[1,0]R by f(x)=x and Tx=2x, for all x[1,0]. We want to show that T is a quasi-Bregman strictly pseudocontractive but not quasi-ϕ strictly pseudocontractive.

Proof From the definition it is clear that f is proper, lower semi-continuous, and convex, and also F(T)={0}. By the definition of quasi-Bregman strict pseudocontractivity, we find k[0,1) such that D f (p,Tx) D f (p,x)+k D f (x,Tx) for all xC and pF(T). Now,

D ( 0 , T x ) = f ( 0 ) f ( T x ) f ( T x ) , 0 T x D ( 0 , T x ) = 0 2 x f ( 2 x ) , 0 2 x D ( 0 , T x ) = 2 x 2 , 2 x D ( 0 , T x ) = 2 x + 4 x = 2 x ,
(3.30)
D ( 0 , x ) = f ( 0 ) f ( x ) f ( x ) , 0 x D ( 0 , x ) = 0 x 1 , x D ( 0 , x ) = x + x = 0
(3.31)

and

D ( x , T x ) = f ( x ) f ( T x ) f ( T x ) , x T x = x 2 x 2 , x 2 x = x + 2 x = x .
(3.32)

From (3.30), (3.31), and (3.32), we obtain

D ( 0 , T x ) = 2 x x 0 + k x , x [ 1 , 0 ] , k [ 0 , 1 ) D ( 0 , x ) + k D ( x , T x ) , x [ 1 , 0 ] , k [ 0 , 1 ) .

Therefore

D(0,Tx)D(0,x)+kD(x,Tx),x[1,0],k[0,1).

Hence, T is a quasi-Bregman strictly pseudocontractive map.

Further,

ϕ ( 0 , T x ) = | 0 | 2 2 0 , J ( T x ) + | T x | 2 ϕ ( 0 , T x ) = 0 2 0 , J ( 2 x ) + | 2 x | 2 ϕ ( 0 , T x ) = 4 | x | 2 ,
(3.33)
ϕ ( 0 , x ) = | 0 | 2 2 0 , J ( x ) + | x | 2 ϕ ( 0 , x ) = 0 2 0 , J ( x ) + | x | 2 ϕ ( 0 , x ) = | x | 2
(3.34)

and

ϕ ( x , T x ) = | x | 2 2 x , J ( T x ) + | T x | 2 = | x | 2 2 x , J ( 2 x ) + 4 | x | 2 = | x | 2 4 x , J ( x ) + 4 | x | 2 = | x | 2 4 | x | 2 + 4 | x | 2 = | x | 2 .
(3.35)

Since 4 | x | 2 > | x | 2 +k | x | 2 , for all k[0,1) and for all x[1,0],

ϕ(0,Tx)ϕ(0,x)+kϕ(x,Tx),x[1,0]

cannot hold for any k[0,1). Hence, T is not a quasi-ϕ strictly pseudocontractive map. □

4 Numerical example

In this section we discuss the direct application of Theorem 3.2 on a typical example on a real line. Consider the following:

E = R , C = [ 1 , 1 ] , g ( z , y ) = y 2 + y z 2 z 2 , f ( x ) = 2 3 x 2 , f ( x ) = 4 3 x , T x = 2 x , f ( x ) = sup { x , x f ( x ) : x E } , f ( z ) = 3 8 z 2 , f ( z ) = 3 4 z , α n = n + 1 4 n , α n f ( x n ) + ( 1 α n ) f ( T x n ) = ( 5 n 3 ) 3 n x n , k = 1 / 2 , x 1 = 1 / 2 C ,

then the scheme can be simplified as

y n = f ( α n f ( x n ) + ( 1 α n ) f ( T x n ) ) , y n = ( 5 n 3 ) 4 n x n , u n = Res g f y n = 4 13 y n , w n = f ( f ( u n ) ) = u n , C n + 1 = { w C n : w x n ( 1 k ) ( x n w n ) 2 2 [ ( 1 + 2 k ) x n ( 1 k ) w n ] } , x n + 1 = P C n + 1 f ( x 1 ) = x n ( 1 k ) ( x n w n ) 2 2 [ ( 1 + 2 k ) x n ( 1 k ) w n ] .

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Keywords

  • Bregman distance
  • quasi-Bregman strictly pseudocontractive map
  • fixed point