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Halpern-Mann’s iterations for Bregman strongly nonexpansive mappings in reflexive Banach spaces with applications

Abstract

We investigate strong convergence for Bregman strongly nonexpansive mappings by modifying Halpern and Mann’s iterations in the framework of a reflexive Banach space. As applications, we apply our main result to problems of finding zeros of maximal monotone operators and equilibrium problems in reflexive Banach spaces.

MSC:47H05, 47H09, 47J25.

1 Introduction

Throughout this paper, we denote by and the sets of positive integers and real numbers, respectively. Let E be a real reflexive Banach space, and let C be a nonempty, closed and convex subset of E. Let T:CC be a nonlinear mapping. The fixed point set of T is denoted by F(T), that is, F(T)={xC:x=Tx}. A mapping T is said to be nonexpansive if

TxTyxy

for all x,yC.

Many problems in nonlinear analysis can be reformulated as a problem of finding a fixed point of a nonexpansive mapping. In 1953, Mann [1] introduced the following iterative sequence { x n } which is defined by

x n + 1 = α n x n +(1 α n )T x n ,

where the initial guess x 1 C is arbitrary and { α n } is a real sequence in [0,1]. It is known that under appropriate settings, the sequence { x n } converges weakly to a fixed point of T. However, even in a Hilbert space, Mann’s iteration may fail to converge strongly; for example, see [2].

Some attempts to construct an iteration method guaranteeing the strong convergence have been made. For example, Halpern [3] proposed the following so-called Halpern iteration:

x n + 1 = α n u+(1 α n )T x n ,

where u, x 1 C are arbitrary and { α n } is a real sequence in [0,1].

Because of a simple construction, Halpern’s iteration is widely used to approximate a solution of fixed points for nonexpansive mappings and other classes of nonlinear mappings by mathematicians in different styles [412].

The purpose of this work is to consider strong convergence results for Bregman strongly nonexpansive mappings in reflexive Banach spaces by modifying Halpern and Mann’s iterations. We note that there are many examples which are Bregman strongly nonexpansive such as the Bregman projection, the resolvents of maximal monotone operators, the resolvents of equilibrium problems, the resolvents of variational inequality problems and others (see, for example, [1316]). Finally, we give some applications concerning the problems of finding zeros of maximal monotone operators and equilibrium problems.

2 Preliminaries and lemmas

In the sequel, we begin by recalling some preliminaries and lemmas which will be used in the proof.

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

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 following Young-Fenchel inequality holds:

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

Furthermore, the equality holds if x f(x) (see also [[17], Theorem 23.5]).

The set lev f (r)={xE:f(x)r} for some rR is called a sublevel of f.

A function f on E is coercive [18] if the sublevel set of f is bounded; equivalently,

lim x + f(x)=+.

A function f on E is said to be strongly coercive [19] 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 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 [20, 21]). We will need the following result.

Lemma 2.1 [22]

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 2.2 [23]

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 2.3 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 [[23], Theorem 5.4]).

  2. (ii)
    ( f ) 1 = f

    (see [21]).

  3. (iii)

    f is Legendre if and only if f is Legendre (see [[23], 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 [[23], Theorem 5.10]).

Examples of Legendre functions were given in [23, 24]. 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

is called the Bregman distance with respect to f [25].

Recall that the Bregman projection [26] 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 2.4 [27]

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.
    (2.1)

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

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

The function f is called totally convex at x if ν 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 ν f (B,t)>0 for any nonempty bounded subset B of E and t>0, where the modulus of total convexity of the function f on the set B is the function ν f :intdomf×[0,+)[0,+] defined by

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

We know that f is totally convex on bounded sets if and only if f is uniformly convex on bounded sets (see [[27], Theorem 2.10]).

The next lemma turns out to be very useful in the proof of our main results.

Proposition 2.5 [28]

If xintdomf, 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 is called sequentially consistent [27] 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 2.6 [29]

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

Let C be a convex subset of intdomf, and let T be a self-mapping of C. A point pC is called an asymptotic fixed point of T (see [30, 31]) if C contains a sequence { x n } 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.

Definition 2.7 A mapping T with a nonempty asymptotic fixed point set is said to be:

  1. (i)

    Bregman strongly nonexpansive (see [15, 32]) with respect to a nonempty F ˆ (T) if

    D f (p,Tx) D f (p,x),xC,p F ˆ (T),
    (2.2)

and if whenever { x n }C is bounded, p F ˆ (T) and

lim n ( D f ( p , x n ) D f ( p , T x n ) ) =0,

it follows that

lim n D f ( x n ,T x n )=0.
  1. (ii)

    Bregman firmly nonexpansive [13, 16, 33] 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).

The existence and approximation of Bregman firmly nonexpansive mappings was studied in [16]. It is also known that if T is Bregman firmly nonexpansive and f is a Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E, then F(T)= F ˆ (T) and F(T) is closed and convex (see [16]). It also follows that every Bregman firmly nonexpansive mapping is Bregman strongly nonexpansive with respect to F(T)= F ˆ (T).

Let f:ER be a convex, Legendre and Gâteaux differentiable function. Following [34] and [25], we make use of the function V f :E× E [0,+) associated with f, which is defined by

V f ( x , x ) =f(x) x , x + f ( x )
(2.3)

for all xE, x E . Then V f is nonnegative and V f (x, x )= D f (x, f ( x )) for all xE and x E . We know the following lemma (see [35]).

Lemma 2.8 Let E be a reflexive Banach space, let f:ER be a convex, Legendre and Gâteaux differentiable function, and let V f be as in (2.3). Then

V f ( x , x ) + y , f ( x ) x V f ( x , x + y )
(2.4)

for all xE and x , y E .

Let E be a real reflexive Banach space, let f:E(,+] be a proper lower semi-continuous function, then f : E (,+] is a proper weak lower semi-continuous and convex function (see [36]). Hence V f is convex in the second variable. 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 ),
(2.5)

where { x i } i = 1 N E and { t i } i = 1 N (0,1) with i = 1 N t i =1.

The following results are of fundamental importance for the techniques of analysis used in this paper.

Lemma 2.9 [37]

Assume that { α n } is a sequence of nonnegative real numbers such that

α n + 1 (1 γ n ) α n + γ n δ n ,nN,

where { γ n } is a sequence in (0,1) and { δ n } is a sequence such that

  1. (a)
    lim n γ n =0

    , n = 1 γ n =;

  2. (b)
    lim sup n δ n 0

    .

Then lim n α n =0.

Lemma 2.10 [38]

Let { α n } be a sequence of real numbers such that there exists a subsequence { n i } of {n} such that α n i < α n i + 1 for all iN. Then there exists a nondecreasing sequence { m k }N such that m k , and the following properties are satisfied for all (sufficiently large) numbers kN:

α m k α m k + 1 and α k α m k + 1 .

In fact, m k =max{jk: α j < α j + 1 }.

3 Main results

In this section, we modify Halpern and Mann’s iterations for finding a fixed point of a Bregman strongly nonexpansive mapping in a real reflexive Banach space.

Lemma 3.1 [39]

Let C be a nonempty, closed and convex subset of a real reflexive Banach space E. Let f:ER be a Gâteaux differentiable and totally convex function, and let T:CC be a mapping such that F ˆ (T)=F(T) is nonempty, closed and convex. Suppose that uC and { x n } is a bounded sequence in C such that lim n x n T x n =0. Then

lim sup n f ( u ) f ( p ) , x n p 0,

where p= P F ( T ) f (u) and P F ( T ) f is the Bregman projection of C onto F(T).

Theorem 3.2 Let E be a real reflexive Banach space E, and let f:ER be a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E. Let T be a Bregman strongly nonexpansive mapping on E such that F(T)= F ˆ (T). Suppose that uE and define the sequence { x n } as follows: x 1 E and

x n + 1 = f ( α n f ( u ) + ( 1 α n ) ( β n f ( x n ) + ( 1 β n ) f ( T x n ) ) ) ,n1,
(3.1)

where { α n } and { β n } are sequences in (0,1) satisfying

(C1) lim n α n =0;

(C2) n = 1 α n =;

(C3) 0< lim inf n β n lim sup n β n <1.

Then { x n } converges strongly to P F ( T ) f (u), where P F ( T ) f is the Bregman projection of E onto F(T).

Proof We note, by Reich and Sabach [16], that F(T) is closed and convex. Let p= P F ( T ) f (u)F(T)= F ˆ (T) and y n = f ( β n f( x n )+(1 β n )f(T x n )) for all nN. Then

x n + 1 = f ( α n f ( u ) + ( 1 α n ) f ( y n ) )
(3.2)

for all nN. By using (2.5) and (2.2), we have

D f ( p , y n ) = D f ( p , f ( β n f ( x n ) + ( 1 β n ) f ( T x n ) ) ) β n D f ( p , x n ) + ( 1 β n ) D f ( p , T x n ) β n D f ( p , x n ) + ( 1 β n ) D f ( p , x n ) = D f ( p , x n )
(3.3)

and

D f ( p , x n + 1 ) = D f ( p , f ( α n f ( u ) + ( 1 α n ) f ( y n ) ) ) α n D f ( p , u ) + ( 1 α n ) D f ( p , y n ) α n D f ( p , u ) + ( 1 α n ) D f ( p , x n ) max { D f ( p , u ) , D f ( p , x n ) } .

By induction, we have

D f (p, x n + 1 )max { D f ( p , u ) , D f ( p , x 1 ) }

for all nN. This implies that { D f (p, x n )} is bounded, and hence { D f (p, y n )} is bounded.

We next show that the sequence { x n } is also bounded. We follow the proof line as in [16]. Since { D f (p, x n )} is bounded, there exists M>0 such that

f(p) f ( x n ) , p + f ( f ( x n ) ) = V f ( p , f ( x n ) ) = D f (p, x n )M.

Hence {f( x n )} is contained in the sublevel set lev ψ (Mf(p)), where ψ= f ,p. Since f is lower semicontinuous, f is weak lower semicontinuous. Hence the function ψ is coercive by Moreau-Rockafellar theorem (see [[40], Theorem 7A]). This shows that {f( x n )} is bounded. Since f is strongly coercive, f is bounded on bounded sets (see [[23], Theorem 3.3]). Hence f is also bounded on bounded subsets of E (see [[29], Proposition 1.1.11]). Since f is a Legendre function, it follows that x n = f (f( x n )) is bounded for all nN. Therefore { x n } is bounded. So are { y n }, {T x n }, {f( y n )} and {f(T x n )}. Indeed, since f is a bounded function defined on bounded subsets of E, f is also bounded on bounded subsets of E (see [[29], Proposition 1.1.11]). Therefore {f(T x n )} is bounded.

We next show that if there exists a subsequence { x n k } of { x n } such that

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

then

lim k ( D f ( p , T x n k ) D f ( p , x n k ) ) =0.

In fact, since {f( y n k )} is bounded and α n k 0, from (3.2) we have

lim k f ( x n k + 1 ) f ( y n k ) = lim k α n k f ( u ) f ( y n k ) =0.
(3.4)

Since f is strongly coercive and uniformly convex on bounded subsets of E, f is uniformly Fréchet differentiable on bounded subsets of E (see [[19], Proposition 3.6.2]). Since f is Legendre, by Lemma 2.1, we have

lim k x n k + 1 y n k = lim k f ( f ( x n k + 1 ) ) f ( f ( y n k ) ) =0.
(3.5)

On the other hand, since f is uniformly Fréchet differentiable on bounded subsets of E, f is uniformly continuous on bounded subsets of E (see [[41], Theorem 1.8]). It follows that

lim k | f ( x n k + 1 ) f ( y n k ) | =0.
(3.6)

We now consider the following equality.

D f ( p , y n k ) D f ( p , x n k ) = f ( p ) f ( y n k ) f ( y n k ) , p y n k D f ( p , x n k ) = f ( p ) f ( x n k + 1 ) + f ( x n k + 1 ) f ( y n k ) f ( x n k + 1 ) , p x n k + 1 + f ( x n k + 1 ) , p x n k + 1 f ( y n k ) , p y n k D f ( p , x n k ) = D f ( p , x n k + 1 ) + ( f ( x n k + 1 ) f ( y n k ) ) + f ( x n k + 1 ) , p x n k + 1 f ( y n k ) , p y n k D f ( p , x n k ) = ( D f ( p , x n k + 1 ) D f ( p , x n k ) ) + ( f ( x n k + 1 ) f ( y n k ) ) + f ( x n k + 1 ) f ( y n k ) , p x n k + 1 f ( y n k ) , x n k + 1 y n k .
(3.7)

It follows from (3.4)-(3.6) that

lim k ( D f ( p , y n k ) D f ( p , x n k ) ) =0
(3.8)

and

D f ( p , y n k ) D f ( p , x n k ) β n k D f ( p , x n k ) + ( 1 β n k ) D f ( p , T x n k ) D f ( p , x n k ) = ( 1 β n k ) ( D f ( p , T x n k ) D f ( p , x n k ) ) .

By virtue of condition (C3), (2.2) and (3.8), we have

lim k ( D f ( p , T x n k ) D f ( p , x n k ) ) =0.
(3.9)

We next consider the following two cases.

Case 1. D f (p, x n + 1 ) D f (p, x n ) for all sufficiently large n. Hence the sequence { D f (p, x n )} is bounded and nonincreasing. So, the limit lim n D f (p, x n ) exists. This shows that lim n ( D f (p, x n + 1 ) D f (p, x n ))=0, and hence

lim n ( D f ( p , T x n ) D f ( p , x n ) ) =0.

Since T is Bregman strongly nonexpansive, we have

lim n D f ( x n ,T x n )=0.

Since f is totally convex on bounded subsets of E, by Lemma 2.6, we have

lim n x n T x n =0.
(3.10)

From (2.5) we have

D f ( T x n , y n ) = D f ( T x n , f ( β n f ( x n ) + ( 1 β n ) f ( T x n ) ) ) β n D f ( T x n , x n ) + ( 1 β n ) D f ( T x n , T x n ) = β n D f ( T x n , x n ) 0
(3.11)

and

D f ( y n , x n + 1 ) α n D f ( y n ,u)+(1 α n ) D f ( y n , y n )= α n D f ( y n ,u)0.
(3.12)

From (3.11), (3.12) and Lemma 2.6, we get

lim n T x n y n =0and lim n y n x n + 1 =0.
(3.13)

From (3.10), (3.13) and invoking Lemma 3.1, we have

lim n sup f ( u ) f ( p ) , x n + 1 p = lim n sup f ( u ) f ( p ) , x n p 0.
(3.14)

Finally, we show that x n p. In fact, by using (2.4), we obtain that

D f ( p , x n + 1 ) = D f ( p , f ( α n f ( u ) + ( 1 α n ) f ( y n ) ) ) = V f ( p , α n f ( u ) + ( 1 α n ) f ( y n ) ) V f ( p , α n f ( u ) + ( 1 α n ) f ( y n ) α n ( f ( u ) f ( p ) ) ) + α n ( f ( u ) f ( p ) ) , f ( α n f ( u ) + ( 1 α n ) f ( y n ) ) p = V f ( p , α n f ( p ) + ( 1 α n ) f ( y n ) ) + α n f ( u ) f ( p ) , x n + 1 p = D f ( p , f ( α n f ( p ) + ( 1 α n ) f ( y n ) ) ) + α n f ( u ) f ( p ) , x n + 1 p α n D f ( p , p ) + ( 1 α n ) D f ( p , y n ) + α n f ( u ) f ( p ) , x n + 1 p ( 1 α n ) D f ( p , x n ) + α n f ( u ) f ( p ) , x n + 1 p .
(3.15)

By Lemma 2.9, we can conclude that lim n D f (p, x n )=0. Therefore, by Lemma 2.6, x n p since f is totally convex on bounded subsets of E.

Case 2. Suppose that there exists a subsequence { D f (p, n i )} of { D f (p, x n )} such that

D f (p, x n i )< D f (p, x n i + 1 )

for all iN. Then, by Lemma 2.10, there exists a nondecreasing sequence { m k }N such that m k

D f (p, x m k ) D f (p, x m k + 1 )and D f (p, x k ) D f (p, x m k + 1 )

for all kN. So, we have

0 lim k ( D f ( p , x m k + 1 ) D f ( p , x m k ) ) lim sup n ( D f ( p , x n + 1 ) D f ( p , x n ) ) lim sup n ( α n D f ( p , u ) + ( 1 α n ) D f ( p , y n ) D f ( p , x n ) ) lim sup n ( α n D f ( p , u ) + ( 1 α n ) ( β n D f ( p , x n ) + ( 1 β n ) D f ( p , T x n ) ) D f ( p , x n ) ) = lim sup n ( α n D f ( p , u ) + ( 1 α n ) ( 1 β n ) ( D f ( p , T x n ) ) D f ( p , x n ) ) α n ( D f ( p , x n ) ) lim sup n α n ( D f ( p , u ) D f ( p , x n ) ) = 0 .
(3.16)

This implies

lim k ( D f ( p , x m k + 1 ) D f ( p , x m k ) ) =0.
(3.17)

Following the proof line in Case 1, we can verify

lim k f ( u ) f ( p ) , x m k + 1 p 0
(3.18)

and

D f (p, x m k + 1 )(1 α m k ) D f (p, x m k )+ α m k f ( u ) f ( p ) , x m k + 1 p .

Since D f (p, x m k ) D f (p, x m k + 1 ), we have

α m k D f ( p , x m k ) D f ( p , x m k ) D f ( p , x m k + 1 ) + α m k f ( u ) f ( p ) , x m k + 1 p α m k f ( u ) f ( p ) , x m k + 1 p .

In particular, since α m k >0, we get

D f (p, x m k ) f ( u ) f ( p ) , x m k + 1 p .

Hence it follows from (3.18) that lim k D f (p, x m k )=0. Using this and (3.17) together, we conclude that

lim k sup D f (p, x k ) lim k D f (p, x m k + 1 )=0.

This completes the proof. □

Letting β n β gives the following result.

Corollary 3.3 Let E be a real reflexive Banach space E, and let f:ER be a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E. Let T be a Bregman strongly nonexpansive mapping on E such that F(T)= F ˆ (T). Suppose that uE and define the sequence { x n } as follows: x 1 E and

x n + 1 = f ( α n f ( u ) + ( 1 α n ) ( β f ( x n ) + ( 1 β ) f ( T x n ) ) )
(3.19)

for all nN, where { α n } is a sequence in (0,1) satisfying conditions (C1) and (C2), and β(0,1). Then { x n } converges strongly to P F ( T ) f (u).

4 Application to a zero point problem of maximal monotone mappings and equilibrium problems

Let E be a real reflexive Banach space. Let A:E 2 E be a set-valued mapping. The domain of A is denoted by domA={xE:Ax}, and also the graph of A is denoted by G(A)={(x, x )E× E : x Ax}. A is said to be monotone if xy, x y 0 for each (x, x ),(y, y )G(A). It is said to be maximal monotone if its graph is not contained in the graph of any other monotone operator on E. It is known that if A is maximal monotone, then the set A 1 ( 0 )={zE: 0 Az} is closed and convex. Now, we apply Theorem 3.2 to the problem of finding xE such that 0 Ax, which strongly relates to the convex minimization problems in optimization, economics and applied sciences.

The resolvent of A, denoted by Res A f :E 2 E , is defined as follows [13]:

Res A f (x)= ( f + A ) 1 f(x).

It is known that F( Res A f )= A 1 ( 0 ), and Res A f is single-valued and Bregman firmly nonexpansive (see [13]). If f is a Legendre function which is bounded, uniformly Fréchet differentiable on bounded subsets of E, then F ˆ ( Res A f )=F( Res A f ) (see [16]). The Yosida approximation A λ :EE, λ>0, is also defined by

A λ (x)= 1 λ ( f ( x ) f ( Res λ A f ( x ) ) )

for all xE. From Proposition 2.7 in [42], we know that ( Res λ A f (x), A λ (x))G(A) and 0 Ax if and only if 0 A λ x for all xE and λ>0. Using these facts, we obtain the following result by replacing T= Res λ A f , λ>0 in Theorem 3.2.

Theorem 4.1 Let C be a nonempty, closed and convex subset of a real reflexive Banach space E, and let f:ER be a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E. Let A:E 2 E be a maximal monotone operator such that A 1 ( 0 ). Suppose that uE and define the sequence { x n } as follows: x 1 E and

x n + 1 = f ( α n f ( u ) + ( 1 α n ) ( β n f ( x n ) + ( 1 β n ) f ( Res λ A f x n ) ) ) ,n1,

where λ>0 and { α n } and { β n } are sequences in (0,1) satisfying

(C1) lim n α n =0;

(C2) n = 1 α n =;

(C3) 0< lim inf n β n lim sup n β n <1.

Then { x n } converges strongly to P A 1 ( 0 ) f (u), where P A 1 ( 0 ) f is the Bregman projection of E onto A 1 ( 0 ).

Let C be a nonempty, closed and convex subset of E, and let Θ:C×CR be a bifunction. Now, we apply Theorem 3.2 to the problem of finding xC such that Θ(x,y)0 for all yC. Such a problem is called an equilibrium problem and the solutions set is denoted by EP(Θ). Numerous problems in economics, physics and applied sciences can be reduced to finding solutions of equilibrium problems.

In order to solve the equilibrium problem, let us assume that a bifunction Θ:C×CR satisfies the following conditions [43]:

( A 1 ) Θ(x,x)=0, xC.

( A 2 ) Θ is monotone, i.e., Θ(x,y)+Θ(y,x)0, x,yC.

( A 3 ) lim sup t 0 Θ(x+t(zx),y)Θ(x,y), x,z,yC.

( A 4 ) The function yΘ(x,y) is convex and lower semi-continuous.

The resolvent of a bifunction Θ [44] is the operator Res Θ f :E 2 C defined by

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

From Lemma 1 in [15], if f:E(,+] is a strongly coercive and Gâteaux differentiable function, and Θ satisfies conditions ( A 1 )-( A 4 ), then dom( Res Θ f )=E.

The following lemma gives us some characterizations of the resolvent Res Θ f .

Lemma 4.2 [15]

Let E be a real reflexive Banach space, and let C be a nonempty closed convex subset of E. Let f:E(,+] be a Legendre function. If the bifunction Θ:C×CR satisfies the conditions ( A 1 )-( A 4 ). Then the following hold:

  1. (i)
    Res Θ f

    is single-valued;

  2. (ii)
    Res Θ f

    is a Bregman firmly nonexpansive operator;

  3. (iii)
    F( Res Θ f )=EP(Θ)

    ;

  4. (iv)
    EP(Θ)

    is a closed and convex subset of C;

  5. (v)

    for all xE and for all pF( Res Θ f ), we have

    D f ( p , Res Θ f ( x ) ) + D f ( Res Θ f ( x ) , x ) D f (p,x).

In addition, by Reich and Sabach [16], if f is uniformly Fréchet differentiable and bounded on bounded subsets of E, then we have from Lemma 4.2 that F( Res Θ f )= F ˆ ( Res Θ f )=EP(Θ) is closed and convex. Also, by replacing T= Res Θ f in Theorem 3.2, we obtain the following result.

Theorem 4.3 Let C be a nonempty, closed and convex subset of a real reflexive Banach space E, and let f:ER be a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E. Let Θ:C×CR be a bifunction which satisfies the conditions ( A 1 )-( A 4 ) such that EP(Θ). Suppose that uE and define the sequence { x n } as follows: x 1 E and

x n + 1 = f ( α n f ( u ) + ( 1 α n ) ( β n f ( x n ) + ( 1 β n ) f ( Res Θ f x n ) ) ) ,n1,

where { α n } and { β n } are sequences in (0,1) satisfying

(C1) lim n α n =0;

(C2) n = 1 α n =;

(C3) 0< lim inf n β n lim sup n β n <1.

Then { x n } converges strongly to P EP ( Θ ) f (u), where P EP ( Θ ) f is the Bregman projection of E onto EP(Θ).

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Acknowledgements

The authors would like to express their thanks to the referees for their helpful comments and suggestions. This work was supported by the Scientific Research Fund of Sichuan Provincial Education Department (12ZA295), the Scientific Research Fund of Science Technology Department of Sichuan Province (2011JYZ011) and the Natural Science foundation of Yibin university (2011Z08).

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Correspondence to Shih-sen Chang.

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Keywords

  • Bregman strongly nonexpansive mapping
  • Bregman projection
  • Legendre function
  • totally convex function
  • Halpern’s iteration
  • Mann’s iteration