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Convergence to common solutions of various problems for nonexpansive mappings in Hilbert spaces

Abstract

In this paper, motivated and inspired by Ceng and Yao (J. Comput. Appl. Math. 214(1):186-201, 2008), Iiduka and Takahashi (Nonlinear Anal. 61(3):341-350, 2005), Jaiboon and Kumam (Nonlinear Anal. 73(5):1180-1202, 2010), Kim (Nonlinear Anal. 73:3413-3419, 2010), Marino and Xu (J. Math. Anal. Appl. 318:43-52, 2006) and Saeidi (Nonlinear Anal. 70:4195-4208, 2009), we introduce a new iterative scheme for finding a common element of the set of solutions of a mixed equilibrium problem for an equilibrium bifunction, the set of fixed points of an infinite family of nonexpansive mappings, the set of solutions of some variational inequality problem, and the set of fixed points of a left amenable semigroup { T t :tS} of nonexpansive mappings with respect to W-mappings and a left regular sequence { μ n } of means defined on an appropriate space of bounded real-valued functions of the semigroup S. Furthermore, we prove that the iterative scheme converges strongly to a common element of the above four sets. Our results extend and improve the corresponding results of many others.

MSC:43A65, 47H05, 47H09, 47H10, 47J20, 47J25, 74G40.

1 Introduction

Let H be a real Hilbert space, let C be a nonempty closed convex subset of H, and let P C be the metric projection of H onto C. Let φ:CR be a real-valued function and θ:C×CR be an equilibrium bifunction with θ(u,u)=0 for each uC. We consider the mixed equilibrium problem (for short, MEP) is to find x C such that

MEP:θ ( x , y ) +φ(y)φ ( x ) 0,yC.

In particular, if φ0, this problem reduces to the equilibrium problem (for short, EP), which is to find x C such that

EP:θ ( x , y ) 0,yC.

Denote the set of solutions of MEP by Ω. The mixed equilibrium problems include fixed point problems, optimization problems, variational inequality problems, Nash equilibrium problems and the equilibrium problems as special cases.

A mapping T of C into itself is called nonexpansive if

TxTyxy,

for all x,yC. We denote by F(T) the set of fixed points of T. It is well known that F(T) is closed convex. Recall that a mapping f:CC is called contractive if there exists a constant α(0,1) such that

f ( x ) f ( y ) αxy,

for all x,yC.

In 2000, Moudafi [1] introduced the viscosity approximation method for nonexpansive mappings (see [2] for further developments in both Hilbert and Banach spaces).

Starting with an arbitrary initial x 0 H, define a sequence { x n } recursively by

x n + 1 =(1 α n )T x n + α n f( x n ),n0,
(1.1)

where α n is a sequence in (0,1). It is proved that under certain appropriate conditions imposed on { α n }, the sequence { x n } generated by (1.1) strongly converges to the unique solution x in F(T) of the variational inequality

( f I ) x , x x 0,xF(T)

(see [1, 2]).

Let A be a strongly positive bounded linear operator on H, that is, there exists a constant γ ¯ >0 such that

Ax,x γ ¯ x 2 ,

for all xH.

In 2006, Marino and Xu [3] considered the following iterative method:

x n + 1 =(I α n A)T x n + α n γf( x n ),n0,
(1.2)

where 0<γ< γ ¯ α , α is a contraction coefficient of f. They proved that if the sequence { α n } satisfies appropriate conditions, then the sequence { x n } generated by (1.2) converges strongly to the unique solution of the variational inequality

( A γ f ) x , x x 0,xF(T),

which is the optimality condition for the minimization problem

min x F ( T ) 1 2 Ax,xh(x),

where h is a potential function for γf (i.e., h (x)=γf(x), for xH).

A set-valued mapping T:H 2 H is called monotone if for all x,yH, fTx and gTy imply xy,fg0. A monotone mapping T:H 2 H is maximal if its graph G(T) is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping T is maximal if and only if for (x,f)H×H, xy,fg0 for every (y,g)G(T) implies fTx. Let A be a monotone mapping of C into H, and let N C v be the normal cone to C at vC, i.e.,

N C v= { w H : v u , w 0 , u C }

and define

Tv={ A v + N C v , if  v C , , if  v C .

Then T is maximal monotone, and 0Tv if and only if vVI(C,A); see [4].

In 2005, for finding an element of F(T)VI(C,A), Iiduka and Takahashi [5] proposed a new iterative sequence: x 1 =xC and

x n + 1 = α n x+(1 α n )T P C ( x n λ n A x n ),n1
(1.3)

and obtained a strong convergence theorem in a Hilbert space.

Let { T n } be a sequence of nonexpansive mappings of C into itself, and let { λ n } be a sequence of nonnegative numbers in [0,1]. For each n1, define a mapping W n of C into itself as follows:

U n , n + 1 = I , U n , n = λ n T n U n , n + 1 + ( 1 λ n ) I , U n , n 1 = λ n 1 T n 1 U n , n + ( 1 λ n 1 ) I , U n , k = λ k T k U n , k + 1 + ( 1 λ k ) I , U n , k 1 = λ k 1 T k 1 U n , k + ( 1 λ k 1 ) I , U n , 2 = λ 2 T 2 U n , 3 + ( 1 λ 2 ) I , W n = U n , 1 = λ 1 T 1 U n , 2 + ( 1 λ 1 ) I .
(1.4)

Such a mapping W n is called the W-mapping generated by T 1 , T 2 ,, T n and λ 1 , λ 2 ,, λ n . The concept of W-mapping was introduced in [6, 7] and [8].

In 2008, Ceng and Yao [9] introduced the hybrid iterative scheme

{ x 0 C arbitrary , θ ( y n , x ) + φ ( x ) φ ( y n ) + 1 r K ( y n ) K ( x n ) , η ( x , y n ) 0 , x C , x n + 1 = α n f ( W n x n ) + β n x n + γ n W n y n ,
(1.5)

where K (x) is the Fréchet derivative of K at x. They proved the sequences { x n } and { y n } generated by the hybrid iterative scheme (1.5) converge strongly to a common element of the set of solutions of MEP and the set of common fixed points of finitely many nonexpansive mappings.

Recall the mapping B is said to be relaxed (ξ,ν)-cocoercive, if there exist two constants ξ,ν>0 such that

BxBy,xyξ B x B y 2 +ν x y 2 ,x,yC.

This class of mappings has been considered by many authors; for example, [10, 11].

In this paper, motivated and inspired by Ceng and Yao [9], Iiduka and Takahashi [5], Jaiboon and Kumam [12], Kim [13], Marino and Xu [3] and Saeidi [14], we introduce a new iterative scheme:

{ x 0 C arbitrary , θ ( z n , x ) + φ ( x ) φ ( z n ) + 1 r n K ( z n ) K ( x n ) , η ( x , z n ) 0 , y n = ( 1 γ n ) x n + γ n T μ n W n P C ( I δ n B ) z n , x n + 1 = α n γ f ( W n x n ) + β n x n + ( ( 1 β n ) I α n A ) T μ n W n y n ,
(1.6)

for all xC, n0, for finding a common element of the set of solutions of a mixed equilibrium problem for an equilibrium bifunction, the set of fixed points of an infinite family of nonexpansive mappings, the set of solutions of some variational inequality problem and the set of fixed points of a left amenable semigroup { T t :tS} of nonexpansive mappings with respect to W-mappings and a left regular sequence { μ n } of means defined on an appropriate space of bounded real-valued functions of the semigroup S. Furthermore, we prove that the proposed iterative scheme (1.6) converges strongly to a common element of the above four sets. Our result extends and improves the corresponding results of many others.

2 Preliminaries

Let S be a semigroup. We denote by B(S) the space of all bounded real-valued functions defined on S with supremum norm. For each sS, we define the left and right translation operators l s and r s on B(S) by

( l s f)(t)=f(st)and( r s f)(t)=f(ts)

for each tS and fB(S), respectively. Let X be a subspace of B(S) containing 1. An element μ in the dual space X of X is said to be a mean on X if μ=μ(1)=1. For sS, we can define a point evaluation δ s by δ s (f)=f(s) for each fX. It is well known that μ is a mean on X if and only if

inf s S f(s)μ(f) sup s S f(s)

for each fX.

Let X be a translation invariant subspace of B(S) (i.e., l s XX and r s XX for each sS) containing 1. Then a mean μ on X is said to be left invariant (resp. right invariant) if

μ( l s f)=μ(f) ( resp.  μ ( r s f ) = μ ( f ) )

for each sS and fX. A mean μ on X is said to be invariant if μ is both left and right invariant [1517]. X is said to be left (resp. right) amenable if X has a left (resp. right) invariant mean. X is amenable if X is left and right amenable. Moreover, B(S) is amenable when S is a commutative semigroup or a solvable group. However, the free group or semigroup of two generators is not left or right amenable. In this case, we say that the semigroup S is an amenable semigroup (see [18, 19]). A semigroup S is left reversible if S has the finite intersection property for right ideals. Every left reversible semigroup S, WAP(S) the space of weakly almost period functions on S has a left invariant mean. If S is both left and right reversible, then WAP(S) has an invariant mean. Each group or amenable semigroup is left and right reversible (see [20, 21]).

A net { μ α } of means on X is said to be asymptotically left (resp. right) invariant if

lim α ( μ α ( l s f ) μ α ( f ) ) =0 ( resp.  lim α ( μ α ( r s f ) μ α ( f ) ) = 0 )

for each fX and sS, and it is said to be left (resp. right) strongly asymptotically invariant (or strong regular) if

lim α l s μ α μ α =0 ( resp.  lim α r s μ α μ α = 0 )

for each sS, where l s and r s are the adjoint operators of l s and r s , respectively. Such nets were first studied by Day in [18] where they were called weak invariant and norm invariant, respectively.

It is easy to see that if a semigroup S is left (resp. right) amenable, then the semigroup S =S{e}, where e s = s e= s for all s S, is also left (resp. right) amenable and conversely.

Let S be a semigroup, and let C be a closed and convex subset of H. Let F(T) denote the fixed point set of T. Then ={ T s :sS} is called a representation of S as nonexpansive mappings on C if T s is nonexpansive with T e =I and T s t = T s T t for each s,tS (cf. [2230]). We denote by F() the set of common fixed points of { T s :sS}, i.e.,

F()= s S F( T s )= s S {xC: T s x=x}.

Lemma 2.1 ([19, 31])

Let S be a semigroup and C be a closed convex subset of a Hilbert space H. Let ={ T s :sS} be a nonexpansive semigroup on C such that { T t u:tS} is bounded for some uC, let X be a subspace of B(S) such that 1X and the mapping t T t x,y is an element of X for each xC and yH, and μ be a mean on X. If we write T μ x instead of T t xdμ(t), then the following hold:

  1. (i)

    T μ is a nonexpansive mapping from C into itself,

  2. (ii)

    T μ x=x for each xF(),

  3. (iii)

    T μ x co ¯ { T t x:tS} for each xC, where co ¯ A is the closed convex hull of A.

Let C be a nonempty subset of a Hilbert space H and T:CH be a mapping. Then T is said to be demiclosed at vH if, for any sequence { x n } in C, the following implication holds:

x n uCandT x n vimplyTu=v,

where → (resp. ) denotes strong (resp. weak) convergence.

Lemma 2.2 ([32])

Let C be a nonempty closed convex subset of a Hilbert space H and suppose that T:CH is nonexpansive. Then, the mapping IT is demiclosed at zero.

Let C be a nonempty subset of a normed space E, and let xE. An element y 0 C is said to be the best approximation to x if

x y 0 =d(x,C),

where d(x,C)= inf y C xy. The number d(x,C) is called the distance from x to C. Let H be a real Hilbert space with inner product , and norm . Let C be a nonempty closed convex subset of H. Then, for any xH, there exists a unique nearest point in C, denoted by P C x, such that

x P C xxy,yC.

The mapping P C is called the metric projection of H onto C. It is well known that P C is a nonexpansive mapping of H onto C and satisfies

xy, P C x P C y P C x P C y 2

for every x,yH. Moreover, P C x is characterized by the following properties: P C xC and for all xH, yC,

x P C x,y P C x0
(2.1)

and

x y 2 x P C x 2 + y P C x 2 .

It is easy to see that the following is true:

uVI(C,B)u= P C (uλBu),λ>0.
(2.2)

In this paper, for solving the mixed equilibrium problems for an equilibrium bifunction θ:C×CR, we assume that θ satisfies the following conditions:

(E1) θ(x,x)=0 for all xC;

(E2) θ is monotone, i.e., θ(x,y)+θ(y,x)0 for all x,yC;

(E3) for each x,y,zC,

lim t 0 θ ( t z + ( 1 t ) x , y ) θ(x,y);

(E4) for each xC, the function yθ(x,y) is convex and lower semicontinuous.

Definition 2.1 (1) Let F:CH and η:C×CH be two mappings. Then F is called:

  1. (i)

    η-monotone if

    F ( x ) F ( y ) , η ( x , y ) 0,x,yC,
  2. (ii)

    η-strongly monotone with constant α if there exists a constant α>0 such that

    F ( x ) F ( y ) , η ( x , y ) α x y 2 ,x,yC,
  3. (iii)

    Lipschitz continuous with constant β if there exists a constant β>0 such that

    F ( x ) F ( y ) βxy,x,yC.

If η(x,y)=xy, for all x,yC, then the definitions (i) and (ii) reduce to the definition of monotonicity and strong monotonicity, respectively.

  1. (2)

    A mapping η:C×CH is called Lipschitz continuous with constant λ if there exists a constant λ>0 such that

    η ( x , y ) λxy,x,yC.
  2. (3)

    A differentiable function K:CR on a convex set C is called:

  3. (i)

    η-convex [33] if

    K(y)K(x) K ( x ) , η ( y , x ) ,x,yC,

where K (x) is the Fréchet derivative of K at x,

  1. (ii)

    η-strongly convex with constant σ [34] if there exists a constant σ>0 such that

    K(y)K(x) K ( x ) , η ( y , x ) σ 2 x y 2 ,x,yC.
  1. (4)

    A mapping F:CR is called sequentially continuous at x 0 [35], if F( x n )F( x 0 ) for each sequence { x n } satisfying x n x 0 . F is called sequentially continuous on C if it is sequentially continuous at each point of C.

Lemma 2.3 ([9])

Let K:CR be differentiable η-strongly convex with a constant σ>0, and let η:C×CH be a mapping such that η(x,y)+η(y,x)=0 for all x,yC. Then K :CH is η-strongly monotone with constant σ>0.

Lemma 2.4 ([36])

A Hilbert space H is said to satisfy Opial’s condition if for each sequence { x n } in H, the condition x n x implies that

lim inf n x n x< lim inf n x n y

for all yH with yx.

Lemma 2.5 Let H be a real Hilbert space. Then

x + y 2 x 2 +2y,x+y,

for all x,yH.

Let C be a nonempty closed convex subset of a real Hilbert space H, φ:CR be a real-valued function and θ:C×CR be an equilibrium bifunction. Let r be a positive parameter. For a given point xC, consider the auxiliary problem for the mixed equilibrium problem (for short, MEP(x,r)) which consists of finding yC such that

θ(y,z)+φ(z)φ(y)+ 1 r K ( y ) K ( x ) , η ( z , y ) 0,zC,

where η:C×CH and K (x) is the Fréchet derivative of a functional K:CR at x. Let S r :CC be the mapping such that for each xC, S r (x) is the solution set of MEP(x,r), i.e.,

S r ( x ) = { y C : θ ( y , z ) + φ ( z ) φ ( y ) + 1 r K ( y ) K ( x ) , η ( z , y ) 0 , z C }
(2.3)

for all xC.

We first need the following important and interesting result.

Lemma 2.6 ([9])

Let C be a nonempty closed convex subset of a real Hilbert space H, and let φ:CR be a lower semicontinuous and convex functional. Let θ:C×CR be an equilibrium bifunction satisfying conditions (E1)-(E4). Assume that

  1. (i)

    η:C×CH is Lipschitz continuous with constant λ>0 such that

  2. (a)

    η(x,y)+η(y,x)=0, x,yC,

  3. (b)

    for each fixed yC, xη(y,x) is sequentially continuous from the weak topology to the weak topology,

  4. (ii)

    K:CR is η-strongly convex with constant σ>0 and its derivative K is sequentially continuous from the weak topology to the strong topology,

  5. (iii)

    for each xC, there exist a bounded subset D x C and z x C such that for any yC D x ,

    θ(y, z x )+φ( z x )φ(y)+ 1 r K ( y ) K ( x ) , η ( z x , y ) <0.

Then the following hold:

  1. (1)

    S r is single-valued;

  2. (2)

    S r is a firmly nonexpansive-type mapping, i.e., for all x,yH,

    S r x S r y 2 S r x S r y,xy;
  3. (3)

    (j) S r is nonexpansive if K is Lipschitz continuous with constant ν>0 such that σλν;

(jj) K ( x 1 ) K ( x 2 ),η( u 1 , u 2 ) K ( u 1 ) K ( u 2 ),η( u 1 , u 2 ), ( x 1 , x 2 )C×C, where u i = S r ( x i ), i=1,2;

  1. (4)

    F( S r )=Ω;

  2. (5)

    Ω is a closed and convex subset of C.

Remark 2.1 In particular, from Lemma 2.6, whenever K(x)= x 2 2 and η(x,y)=xy for each (x,y)C×C, then S r is firmly nonexpansive, i.e.,

x 1 x 2 , S r ( x 1 ) S r ( x 2 ) S r ( x 1 ) S r ( x 2 ) 2 ,( x 1 , x 2 )C×C.

We need the following results concerning the W-mapping W n .

Lemma 2.7 ([37])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let T 1 , T 2 , be nonexpansive mappings of C into H such that i = 1 F( T i ) is nonempty, and let λ 1 , λ 2 , be real numbers such that 0< λ i b<1 for any iN. Then, for every xC and kN, the limit lim n U n , k x exists.

Using Lemma 2.7, one can define a mapping W of C into H as

Wx= lim n W n x= lim n U n , 1 x,

for every xC.

Remark 2.2 ([37])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let T 1 , T 2 , be nonexpansive mappings of C into H such that i = 1 F( T i ) is nonempty, and let λ 1 , λ 2 , be real numbers such that 0< λ i b<1 for any iN. Then F(W)= i = 1 F( T i ).

Remark 2.3 ([38])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let T 1 , T 2 , be nonexpansive mappings of C into H such that i = 1 F( T i ) is nonempty. If { x n } is an arbitrary bounded sequence in C, then we have

lim n W x n W n x n =0.

Lemma 2.8 ([39])

Let { x n } and { z n } be bounded sequences in a Hilbert space H and let { β n } be a sequence in [0,1] with 0< lim inf n β n and lim sup n β n <1. Suppose

x n + 1 = β n x n +(1 β n ) z n

for all integers n0 and

lim sup n ( z n + 1 z n x n + 1 x n ) 0.

Then lim n z n x n =0.

Lemma 2.9 ([3])

Assume A is a strongly positive linear bounded operator on a Hilbert space H with a coefficient γ ¯ >0 and 0<ρ A 1 . Then IρA1ρ γ ¯ .

Lemma 2.10 ([2])

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

a n + 1 (1 b n ) a n + b n c n ,

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

  1. (1)

    n = 1 b n =,

  2. (2)

    lim sup n c n 0 or n = 1 | b n c n |<.

Then

lim n a n =0.

3 Main result: strong convergence theorems

In this section, we deal with the strong convergence of hybrid viscosity approximation scheme (1.6) for finding a common element of the set of solutions of a mixed equilibrium problem, the set of fixed points of an infinite family of nonexpansive mappings, the set of fixed points of a left amenable semigroup of nonexpansive mappings and the set of solutions of variational inequality in a Hilbert space.

Theorem 3.1 Let S be a semigroup, ={ T t :tS} be a nonexpansive semigroup on H such that F(), X be a left invariant subspace of B(S) such that 1X, and the function t T t x,y is an element of X for each x,yH. Let { μ n } be a left strong regular sequence of means on X such that lim n μ n + 1 μ n =0. Let C be a nonempty closed convex subset of a real Hilbert space H and { T i } be an infinite family of nonexpansive mappings from C into itself such that T i (F()Ω)F() for each iN. Let φ:CR be a lower semicontinuous and convex functional. Let θ:C×CR be an equilibrium bifunction satisfying conditions (E1)-(E4), and let T 1 , T 2 , be an infinite family of nonexpansive mappings of C into H. Let r>0, γ>0 be two constants. Let f be a contraction of C into itself with a coefficient α(0,1), and let A be a strongly positive bounded linear operator with a coefficient γ ¯ >0 such that 0<αγ< γ ¯ <αγ+1. Let B:CH be an L-Lipschitzian and relaxed (ξ,ν)-cocoercive mapping. Suppose that F= n = 1 F( T n )F()ΩVI(C,B). Let { α n }, { β n } and { γ n } be sequences in [0,1] such that α n + β n 1, and let the sequence { δ n }(0,). Assume that:

(C1) η:C×CH is Lipschitz continuous with constant λ>0 such that

  1. (a)

    η(x,y)+η(y,x)=0, x,yC,

  2. (b)

    for each fixed yC, xη(y,x) is sequentially continuous from the weak topology to the weak topology,

(C2) K:CR is η-strongly convex with constant σ>0 and its derivative K is not only sequentially continuous from the weak topology to the strong topology, but also Lipschitz continuous with constant ν>0 such that σλν,

(C3) for each xC, there exist a bounded subset D x C and z x C such that for any yC D x ,

θ(y, z x )+φ( z x )φ(y)+ 1 r K ( y ) K ( x ) , η ( z x , y ) <0,

(C4)

  1. (i)

    lim n α n =0, n = 0 α n =,

  2. (ii)

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

0< lim inf n γ n lim sup n γ n <1,

  1. (iii)

    lim n | δ n + 1 δ n |=0, a δ n b for some a, b with 0ab 2 ( ν ξ L 2 ) L 2 ,

  2. (iv)

    lim n | γ n + 1 γ n |=0,

(C5) lim inf n r n >0 and lim n | r n + 1 r n |=0.

Given x 0 C is arbitrary, then the sequences { x n }, { y n } and { z n } generated iteratively by (1.6) converge strongly to x F, where x = P F (γf+(IA)) x , which solves the following variational inequality:

( γ f A ) x , x x 0,xF.

Lemma 3.1 (1 β n )I α n A1 β n α n γ ¯ .

Proof Since lim n α n =0, we may assume, without loss of generality, that α n (1 β n ) A 1 . Since A is a linear bounded self-adjoint operator on H, we have

A=sup { | A x , x | : x H , x = 1 } .

Observe that

( ( 1 β n ) I α n A ) x , x = 1 β n α n A x , x 1 β n α n A 0 ,

which shows that (1 β n )I α n A is positive. By Lemma 2.9, we have

( 1 β n ) I α n A 1 β n α n γ ¯ .

 □

Lemma 3.2 Let B be an L-Lipschitzian and relaxed (ξ,ν)-cocoercive mapping and δ n 2 ( ν ξ L 2 ) L 2 , then

( I δ n B ) x ( I δ n B ) y xy,

for all x,yC.

Proof Since B is an L-Lipschitzian and relaxed (ξ,ν)-cocoercive mapping and δ n 2 ( ν ξ L 2 ) L 2 , we have

( I δ n B ) x ( I δ n B ) y 2 = x y 2 2 δ n B x B y , x y + δ n 2 B x B y 2 x y 2 2 δ n ( ξ B x B y 2 + ν x y 2 ) + δ n 2 L 2 x y 2 x y 2 + 2 δ n ξ L 2 x y 2 2 δ n ν x y 2 + δ n 2 L 2 x y 2 = ( 1 + 2 δ n ξ L 2 2 δ n ν + δ n 2 L 2 ) x y 2 x y 2 ,

for all x,yC. Thus,

( I δ n B ) x ( I δ n B ) y xy,

for all x,yC. □

Lemma 3.3 z n p x n p, pF.

Proof From (2.3), we note that z n = S r n x n . From Lemma 2.6, we get

z n p S r n x n S r n p x n p

for all pF. □

Lemma 3.4 { x n }, { y n }, { z n }, { W n y n }, { W n x n } and {f( W n x n )} are all bounded.

Proof Let pF. Since pVI(C,B), from (2.2), we get p= P C (I δ n B)p. From Lemma 2.1, Lemma 3.3 and W n , P C being nonexpansive, we have

y n p = ( 1 γ n ) x n + γ n T μ n W n P C ( I δ n B ) z n p ( 1 γ n ) x n p + γ n T μ n W n P C ( I δ n B ) z n p ( 1 γ n ) x n p + γ n ( I δ n B ) z n ( I δ n B ) p ( 1 γ n ) x n p + γ n z n p x n p .

From (1.6) and Lemma 3.1, we obtain

(3.1)

for all n0. It follows by mathematical induction that

x n + 1 pmax { x 0 p , γ f ( p ) A p γ ¯ γ α } ,n0.

Therefore, { x n } is bounded. We also deduce that { y n }, { z n }, { W n y n }, { W n x n } and {f( W n x n )} are all bounded. □

Lemma 3.5 Let the mapping W n be generated iteratively by (1.5). If { ω n } is a bounded sequence in H, then

  1. (1)

    lim n W n + 1 ω n W n ω n =0.

  2. (2)

    lim n T μ n + 1 ω n T μ n ω n =0.

Proof (1) We shall use M to denote the possible different constants appearing in the following argument. From (1.5), since T i and U n , i are nonexpansive, we have

W n + 1 ω n W n ω n = λ 1 T 1 U n + 1 , 2 ω n + ( 1 λ 1 ) ω n λ 1 T 1 U n , 2 ω n ( 1 λ 1 ) ω n λ 1 U n + 1 , 2 ω n U n , 2 ω n = λ 1 λ 2 T 2 U n + 1 , 3 ω n + ( 1 λ 2 ) ω n λ 2 T 2 U n , 3 ω n ( 1 λ 2 ) ω n λ 1 λ 2 U n + 1 , 3 ω n U n , 3 ω n λ 1 λ 2 λ n U n + 1 , n + 1 ω n U n , n + 1 ω n M i = 1 n λ i ,

which implies that

lim n W n + 1 ω n W n ω n =0.
  1. (2)

    Let qF(). Then T s ω n q ω n q. Also, we have

    T s ω n ω n q+q

for all sS and n0. Since { ω n } is bounded and lim n μ n + 1 μ n =0, we get

T μ n + 1 ω n T μ n ω n = sup { | T μ n + 1 ω n T μ n ω n , z | : z = 1 } = sup { | μ n + 1 ( s ) T s ω n , z μ n ( s ) T s ω n , z | : z = 1 } μ n + 1 μ n sup s S T s ω n μ n + 1 μ n ( ω n q + q ) 0 .

 □

Lemma 3.6 lim n x n + 1 x n = lim n z n + 1 z n =0.

Proof Define a sequence { u n } by

x n + 1 = β n x n +(1 β n ) u n

for all n0. Observe that from the definition of u n , we get

u n + 1 u n = x n + 2 β n + 1 x n + 1 1 β n + 1 x n + 1 β n x n 1 β n = α n + 1 γ f ( W n + 1 x n + 1 ) + ( ( 1 β n + 1 ) I α n + 1 A ) T μ n + 1 W n + 1 y n + 1 1 β n + 1 α n γ f ( W n x n ) + ( ( 1 β n ) I α n A ) T μ n W n y n 1 β n = α n + 1 1 β n + 1 γ f ( W n + 1 x n + 1 ) α n 1 β n γ f ( W n x n ) + T μ n + 1 W n + 1 y n + 1 T μ n W n y n + α n 1 β n A T μ n W n y n α n + 1 1 β n A T μ n + 1 W n + 1 y n + 1 = α n + 1 1 β n + 1 ( γ f ( W n + 1 x n + 1 ) A T μ n + 1 W n + 1 y n + 1 ) + α n 1 β n ( A T μ n W n y n γ f ( W n x n ) ) + T μ n + 1 W n + 1 y n + 1 T μ n + 1 W n + 1 y n + T μ n + 1 W n + 1 y n T μ n W n y n .
(3.2)

From (2.3), we note that z n = S r n x n and z n + 1 = S r n + 1 x n + 1 , we have

(3.3)
(3.4)

for all xC. Putting x= z n + 1 in (3.3) and x= z n in (3.4), we have

(3.5)
(3.6)

After multiplying (3.5) and (3.6) by r n and adding them together, we obtain

η ( z n + 1 , z n ) , K ( z n ) K ( x n ) r n r n + 1 ( K ( z n + 1 ) K ( x n + 1 ) ) 0.

Hence,

η ( z n + 1 , z n ) , K ( z n ) K ( z n + 1 ) + K ( x n + 1 ) K ( x n ) + ( 1 r n r n + 1 ) ( K ( z n + 1 ) K ( x n + 1 ) ) 0 .

Then, by Lemma 2.3, we have

η ( z n + 1 , z n ) , K ( x n + 1 ) K ( x n ) + ( 1 r n r n + 1 ) ( K ( z n + 1 ) K ( x n + 1 ) ) η ( z n , z n + 1 ) , K ( z n ) K ( z n + 1 ) σ z n z n + 1 2 ,

and hence

σ z n + 1 z n 2 η ( z n + 1 , z n ) ( K ( x n + 1 ) K ( x n ) + ( 1 r n r n + 1 ) K ( z n + 1 ) K ( x n + 1 ) ) λ z n + 1 z n ( ν x n + 1 x n + ( 1 r n r n + 1 ) ν z n + 1 x n + 1 ) .

Without loss of generality, we assume that there exists a real number k such that r n >k>0 for all n0, we have

z n + 1 z n λ ν σ x n + 1 x n + ν σ 1 k | r n + 1 r n | z n + 1 x n + 1 x n + 1 x n + ν σ k | r n + 1 r n | M ,
(3.7)

where M =sup{ z n x n :n0}.

Setting v n = P C (I δ n B) z n for all n0, from Lemma 3.2, we have

v n + 1 v n P C ( I δ n + 1 B ) z n + 1 P C ( I δ n B ) z n ( I δ n + 1 B ) z n + 1 ( I δ n B ) z n ( I δ n + 1 B ) z n + 1 ( I δ n + 1 B ) z n + ( δ n + 1 δ n ) B z n z n + 1 z n + | δ n + 1 δ n | B z n .

From (3.7), we get

(3.8)

Also, we have

(3.9)

From (3.2), we obtain

(3.10)

Combining (3.8), (3.9) and (3.10), we obtain

(3.11)

Thus, it follows from (3.11), Lemma 3.4, Lemma 3.5 and condition (C4) that

lim sup n ( u n + 1 u n x n + 1 x n ) 0.

By Lemma 2.8, we get

lim n u n x n =0.

Consequently, we have

lim n x n + 1 x n = lim n (1 β n ) u n x n =0.
(3.12)

From (3.7), we get

lim n z n + 1 z n =0.

 □

Lemma 3.7 lim n x n T t x n =0 for all tS.

Proof Let pF and put

M 0 =max { x 0 p , γ f ( p ) A p γ ¯ γ α } .

Set D={yH:yp M 0 }. We remark that D is a bounded closed convex set, { x n },{ y n },{ z n }D and it is invariant under and W n for all nN. We will show that

lim sup n sup y D T μ n y T t T μ n y=0,
(3.13)

for all tS. Let ε>0, by [40] (Theorem 1.2), there exists δ>0 such that

co ¯ F δ ( T t ;D)+ B δ F ε ( T t ;D)
(3.14)

for all tS. By [40] (Corollary 1.1), there exists a natural number N such that

1 N + 1 i = 0 N T t i s y T t ( 1 N + 1 i = 0 N T t i s y ) δ,
(3.15)

for all t,sS and yD. Since { μ n } is left strong regular, there exists n 0 N such that μ n l t i μ n δ M 0 + p for n n 0 and i=1,2,,N. Then we have

(3.16)

for all n n 0 . By Lemma 2.1, we have

1 N + 1 i = 0 N T t i s yd μ n (s) co ¯ { 1 N + 1 i = 0 N T t i ( T s y ) : s S } .
(3.17)

It follows from (3.14)-(3.17) that

T μ n y = 1 N + 1 i = 0 N T t i s y d μ n ( s ) + ( T μ n y 1 N + 1 i = 0 N T t i s y d μ n ( s ) ) co ¯ { 1 N + 1 i = 0 N T t i s y : s S } + B δ co ¯ F δ ( T t ; D ) + B δ T ε ( T t ; D ) ,

for all yD and n n 0 . Therefore,

lim sup n sup y D T t T μ n y T μ n yε.

Since ε>0 is arbitrary, we obtain (3.13). Now, let tS and ε>0. Then there exists δ>0, which satisfies (3.14). Take L 0 =(γα+A) M 0 +γf(p)Ap. From lim n α n =0, (3.12) and (3.13), there exists kN such that α n < δ ( 1 β n ) 2 L 0 , T μ n y F δ ( T t ;D) for all yD and x n x n + 1 < δ ( 1 β n ) 2 β n for all nk. We note that

α n ( γ f ( W n x n ) A T μ n W n y n ) α n ( γ f ( W n x n ) f ( p ) + γ f ( p ) A p + A T μ n W n y n A p ) α n ( γ α W n x n p + γ f ( p ) A p + A T μ n W n y n p ) α n ( γ α x n p + γ f ( p ) A p + A y n p ) α n ( ( γ α + A ) x n p + γ f ( p ) A p ) α n ( ( γ α + A ) M 0 + γ f ( p ) A p ) δ ( 1 β n ) 2 ,

for all nk. Since

x n + 1 β n x n + 1 = ( 1 β n ) T μ n W n y n + β n x n + α n ( γ f ( W n x n ) A T μ n W n y n ) β n x n + 1 ,

we get

x n + 1 = T μ n W n y n + β n 1 β n ( x n x n + 1 ) + α n 1 β n ( γ f ( W n x n ) A T μ n W n y n ) F δ ( T t ; D ) + B δ 2 + B δ 2 F δ ( T t ; D ) + B δ F ε ( T t ; D ) ,

for all nk. This shows that

lim sup n x n T t x n ε.

Since ε>0 is arbitrary, we get lim n x n T t x n =0. □

Lemma 3.8 lim n x n T μ n W n y n =0.

Proof Since x n + 1 = α n γf( W n x n )+ β n x n +((1 β n )I α n A) T μ n W n y n , we have

x n T μ n W n y n x n x n + 1 + x n + 1 T μ n W n y n = x n x n + 1 + α n γ f ( W n x n ) A T μ n W n y n + β n x n T μ n W n y n ,

that is,

x n T μ n W n y n 1 1 β n x n x n + 1 + α n 1 β n γ f ( W n x n ) A T μ n W n y n .

It follows from condition (C4) and Lemma 3.6 that

lim n x n T μ n W n y n =0.

 □

Lemma 3.9 lim n x n z n =0.

Proof For pF, since S r is firmly nonexpansive, we have

z n p 2 = S r n x n S r n p 2 S r n x n S r n p , x n p = z n p , x n p = 1 2 ( z n p 2 + x n p 2 z n x n 2 )

and hence

z n p 2 x n p 2 z n x n 2 .
(3.18)

Note that the following equality holds:

t x + ( 1 t ) y 2 =t x 2 +(1t) y 2 t(1t) x y 2

for all t[0,1] and x,yH. So, from (1.6) and (3.18), we get

y n p 2 = ( 1 γ n ) ( x n p ) + γ n ( T μ n W n P C ( I δ n B ) z n p ) 2 γ n T μ n W n P C ( I δ n B ) z n p 2 + ( 1 γ n ) x n p 2 γ n ( 1 γ n ) T μ n W n P C ( I δ n B ) z n x n 2 = γ n z n p 2 + ( 1 γ n ) x n p 2 ( 1 γ n ) y n x n 2 γ n ( x n p 2 z n x n 2 ) + ( 1 γ n ) x n p 2 = x n p 2 γ n z n x n 2 .
(3.19)

Therefore, from Lemma 2.5, Lemma 3.1 and (3.19), we have

x n + 1 p 2 = α n ( γ f ( W n x n ) A p ) + β n ( x n T μ n W n y n ) + ( I α n A ) ( T μ n W n y n p ) 2 β n ( x n T μ n W n y n ) + ( I α n A ) ( T μ n W n y n p ) 2 + 2 α n γ f ( W n x n ) A p , x n + 1 p ( I α n A y n p + β n x n T μ n W n y n ) 2 + 2 α n γ f ( W n x n ) A p x n + 1 p ( ( 1 α n γ ¯ ) y n p + β n x n T μ n W n y n ) 2 + 2 α n γ f ( W n x n ) A p x n + 1 p ( 1 α n γ ¯ ) 2 y n p 2 + β n 2 x n T μ n W n y n 2 + 2 ( 1 α n γ ¯ ) β n y n p x n T μ n W n y n + 2 α n γ f ( W n x n ) A p x n + 1 p ( 1 α n γ ¯ ) 2 ( x n p 2 γ z n x n 2 ) + β n 2 x n T μ n W n y n 2 + 2 ( 1 α n γ ¯ ) β n y n p x n T μ n W n y n + 2 α n γ f ( W n x n ) A p x n + 1 p .

Then we derive

(3.20)

So, from (C4), Lemma 3.6, Lemma 3.8 and (3.20), we obtain

lim n x n z n =0.

 □

Lemma 3.10 lim n y n v n = lim n z n v n =0, where v n = P C (I δ n B) z n for all n0.

Proof Let pF. Setting v n = P C (I δ n B) z n for all n0, since pVI(C,B), we have p= P C (I δ n B)p. From the L-Lipschitzian and relaxed (ξ,ν)-cocoercive mapping on B and Lemma 3.3, we have

(3.21)

From (1.6) and (3.21), we get

y n p 2 = ( 1 γ n ) x n + γ n T μ n W n v n p 2 = ( 1 γ n ) ( x n p ) + γ n ( T μ n W n v n p ) 2 ( 1 γ n ) x n p 2 + γ n v n p 2 x n p 2 + γ n ( 2 δ n ξ + δ n 2 2 δ n ν L 2 ) B z n B p 2 .
(3.22)

From (1.6), Lemma 3.1 and (3.22), we have

(3.23)

It follows that

0 ( 1 α n γ ¯ ) 2 γ n ( 2 δ n ξ + δ n 2 2 δ n ν L 2 ) B z n B p 2 α n M + x n p 2 x n + 1 p 2 α n M + ( x n p + x n + 1 p ) x n + 1 x n 0 ,

which implies that

lim n B z n Bp=0.
(3.24)

On the other hand, since P C is firmly nonexpansive, by Lemma 3.2, we have

v n p 2 = P C ( I δ n B ) z n P C ( I δ n B ) p 2 ( I δ n B ) z n ( I δ n B ) p , v n p = 1 2 { ( I δ n B ) z n ( I δ n B ) p 2 + v n p 2 z n v n γ n ( B z n B p ) 2 } 1 2 { z n p 2 + v n p 2 z n v n 2 + 2 γ n z n v n B z n B p γ n 2 B z n B p 2 } ,

which yields that

v n p 2 z n p 2 z n v n 2 + 2 γ n z n v n B z n B p x n p 2 z n v n 2 + 2 γ n z n v n B z n B p .
(3.25)

Combining (3.22) and (3.25), we obtain

y n p 2 ( 1 γ n ) x n p 2 + γ n v n p 2 ( 1 γ n ) x n p 2 + γ n ( x n p 2 z n v n 2 + 2 γ n z n v n B z n B p ) = x n p 2 γ n z n v n 2 + 2 γ n z n v n B z n B p .
(3.26)

Therefore, from (3.23) and (3.26), we get

x n + 1 p 2 α n M + β n ( 1 α n γ ¯ ) x n p 2 + ( 1 α n γ ¯ ) ( 1 β n α n γ ¯ ) y n p 2 α n M + β n ( 1 α n γ ¯ ) x n p 2 + ( 1 α n γ ¯ ) ( 1 β n α n γ ¯ ) × ( x n p 2 γ z n v n 2 + 2 γ n 2 z n v n B z n B p ) α n M + ( 1 α n γ ¯ ) 2 x n p 2 ( 1 α n γ ¯ ) ( 1 β n α n γ ¯ ) γ n z n v n 2 + 2 ( 1 α n γ ¯ ) ( 1 β n α n γ ¯ ) γ n 2 z n v n B z n B p .

Hence, we have

( 1 α n γ ¯ ) ( 1 β n α n γ ¯ ) γ n z n v n 2 α n M + ( x n p + x n + 1 p ) x n + 1 x n + 2 ( 1 α n γ ¯ ) ( 1 β n α n γ ¯ ) γ n 2 z n v n B z n B p ,

which implies that

lim n z n v n =0.
(3.27)

Observe that

y n v n y n x n + x n z n + z n v n γ n T μ n W n v n x n + x n z n + z n v n γ n ( T μ n W n v n T μ n W n y n + T μ n W n y n x n ) + x n z n + z n v n γ n ( v n y n + T μ n W n y n x n ) + x n z n + z n v n ,

and hence

(1 γ n ) y n v n γ n T μ n W n y n x n + x n z n + z n v n .

Thus, from Lemma 3.8, Lemma 3.9, (3.27) and (C4), we derive

lim n y n v n =0.

 □

Lemma 3.11 P F (γf+(IA)) is a contraction of H into itself.

Proof From Lemma 3.1, we have

P F ( γ f + ( I A ) ) x P F ( γ f + ( I A ) ) y γ f ( x ) + ( I A ) x γ f ( y ) ( I A ) y γ f ( x ) f ( y ) + ( I A ) ( x y ) γ α x y + ( 1 γ ¯ ) x y = ( 1 ( γ ¯ γ α ) ) x y ,

for all x,yH. From the condition γ ¯ , 0<αγ< γ ¯ <αγ+1, we obtain 1( γ ¯ γα)(0,1). Therefore, P F (γf+(IA)) is a contraction. □

Now, we prove Theorem 3.1.

Proof of Theorem 3.1 From Lemma 3.11 and the Banach contraction principle, P F (γf+(IA)) has a unique fixed point, say x H. That is, x = P F (γf+(IA)) x . Then, using (2.1), x is the unique solution of the variational inequality

( γ f A ) x , x x 0
(3.28)

for all xF. Now, we show that

lim sup n ( γ f A ) x , x n x 0.
(3.29)

To show this, we can choose a subsequence { x n i } of { x n } such that

lim sup n ( γ f A ) x , x n x = lim i ( γ f A ) x , x n i x .
(3.30)

Since { x n i } is bounded, there exists a subsequence { x n i j } of { x n i } which converges weakly to z. Without loss of generality, we can assume that x n i z. We need to show that zF=F()Ω( n = 1 F( T n ))VI(C,B).

  1. (I)

    Since x n i z, by Lemma 2.2 and Lemma 3.7, we get

    T t z=z

for all tS. Therefore, zF().

  1. (II)

    Now, we show that zΩ. Since z n = S r n x n , we derive

    θ( z n ,x)+φ(x)φ( z n )+ 1 r n K ( z n ) K ( x n ) , η ( x , z n ) 0,

for all xC. From the monotonicity of θ, we have

θ(x, z n )θ( z n ,x)φ(x)φ( z n )+ 1 r n K ( z n ) K ( x n ) , η ( x , z n ) ,

and hence

θ(x, z n i )φ(x)φ( z n i )+ K ( z n i ) K ( x n i ) r n i , η ( x , z n i ) .

Since K ( z n i ) K ( x n i ) r n i 0 and z n i z, from the lower semicontinuity of φ and (E4), we have

θ(x,z)+φ(z)φ(x)0,

for all xC. For t with 0<t1 and xC, let x t =tx+(1t)z. Since xC and zC, we have x t C and

θ( x t ,z)+φ(z)φ( x t )0.

From (E1), (E4) and the convexity of φ, we get

0 = θ ( x t , x t ) + φ ( x t ) φ ( x t ) t θ ( x t , x ) + ( 1 t ) θ ( x t , z ) + t φ ( x ) + ( 1 t ) φ ( z ) φ ( x t ) t ( θ ( x t , x ) + φ ( x ) φ ( x t ) ) .

Hence,

θ( x t ,x)+φ(x)φ( x t )0,

for all xC. From (E3) and the lower semicontinuity of φ, we have

θ(z,x)+φ(x)φ(z)0,

for all xC. Therefore, zΩ.

  1. (III)

    We show that zF(W)= i = 1 F( T i ). Assume that zF(W), then zWz. Since zF()Ω, by our assumption, we have T i zF(), iN and then W n zF(). From Lemma 2.1, we get

    T μ n W n z= W n z,
    (3.31)

for all nN. Since

y n x n y n v n + v n z n + z n x n ,

from Lemma 3.9 and Lemma 3.10, we get

lim n y n x n =0.
(3.32)

By Lemma 2.4, Lemma 3.8, (3.31) and (3.32), we obtain

lim inf i x n i z < lim inf i x n i W z lim inf i ( x n i T μ n i W n i y n i + T μ n i W n i y n i T μ n i W n i x n i + T μ n i W n i x n i T μ n i W n i z + T μ n i W n i z W z ) lim inf i ( y n i x n i + x n i z ) lim inf i x n i z .

This is a contraction. Therefore, zF(W)= i = 1 F( T i ).

  1. (IV)

    We show that zVI(C,B). Let U:H 2 H be a set-valued mapping defined by

    Ux={ B x + N C x , if  x C , , if  x C ,

where N C x is the normal cone to C at xC. By assumption of B, we have

B x B y , x y ξ B x B y 2 + ν x y 2 ( ν ξ L 2 ) x y 2 0 ,
(3.33)

which implies that B is monotone. Thus, U is a maximal monotone. Let (u,v)G(U). Since vBu N C u and v n = P C (I δ n B) z n C, we have

u v n ,vBu0.
(3.34)

On the other hand, from (2.1), we have

u v n , v n ( I δ n B ) z n 0

and hence

u v n , v n z n δ n + B z n 0.

It follows by (3.33) and (3.34) that

u v n i , v u v n i , B u u v n i , B u u v n i , v n i z n i δ n i + B z n i = u v n i , B u B z n i v n i z n i δ n i = u v n i , B u B v n i + u v n i , B v n i B z n i u v n i , v n i z n i δ n i u v n i , B v n i B z n i u v n i , v n i z n i δ n i .

From Lemma 3.9 and Lemma 3.10, we obtain uz,v0 as i. Since U is maximal monotone, we have z U 1 (0). Therefore, zVI(C,B). By (I)-(IV), zF=F()Ω( n = 1 F( T n ))VI(C,B). Since x = P F (γf+(IA)) x , from (3.28), we have

lim sup n ( γ f A ) x , x n x = lim sup i ( γ f A ) x , x n i x = lim sup i ( γ f A ) x , z n i x = ( γ f A ) x , z x 0 .
  1. (V)

    Finally, we prove that { x n }, { y n } and { z n } converge strongly to x . From (1.6), we obtain

    x n + 1 x 2 = α n ( γ f ( W n x n ) A x ) + β n ( x n x ) + ( ( 1 β n ) I α n A ) ( T μ n W n y n x ) 2 β n ( x n x ) + ( ( 1 β n ) I α n A ) ( T μ n W n y n x ) 2 + 2 α n γ f ( W n x n ) A x , x n + 1 x ( β n x n x + ( ( 1 β n ) I α n A ) ( T μ n W n y n x ) ) 2 + 2 α n γ f ( W n x n ) f ( x ) , x n + 1 x + 2 α n γ f ( x ) A x , x n + 1 x ( β n x n x + ( 1 β n α n γ ¯ ) y n x ) 2 + 2 α n γ α x n x x n + 1 x + 2 α n γ f ( x ) A x , x n + 1 x ( 1 α n γ ¯ ) 2 x n x 2 + α n γ α ( x n x 2 + x n + 1 x 2 ) + 2 α n γ f ( x ) A x , x n + 1 x ,

which implies that

x n + 1 x 2 1 2 α n γ ¯ + ( α n γ ¯ ) 2 + α n γ α 1 α n γ α x n x 2 + 2 α n 1 α n γ α γ f ( x ) A x , x n + 1 x = ( 1 2 ( γ ¯ γ α ) α n 1 α n γ α ) x n x 2 + ( α n γ ¯ ) 2 1 α n γ α x n x 2 + 2 α n 1 α n γ α γ f ( x ) A x , x n + 1 x = ( 1 2 ( γ ¯ γ α ) α n 1 α n γ α ) x n x 2 + 2 ( γ ¯ γ α ) α n 1 α n γ α ( α n γ ¯ 2 2 ( γ ¯ γ α ) x n x 2 + 1 γ ¯ γ α γ f ( x ) A x , x n + 1 x ) .

It follows that

x n + 1 x 2 (1 b n ) x n x + b n c n ,
(3.35)

where

b n = 2 ( γ ¯ γ α ) α n 1 α n γ α , c n = α n γ ¯ 2 2 ( γ ¯ γ α ) x n x 2 + 1 γ ¯ γ α γ f ( x ) A x , x n + 1 x .

From (C4)-(i), we have n = 0 b n =, and by (3.29), we get lim sup n c n 0. Consequently, applying Lemma 2.10 to (3.35), we get x n x 0. Therefore,

lim n x n = x .

From Lemma 3.9 and (3.32), we obtain

lim n y n = x and lim n z n = x .

This completes the proof of Theorem 3.1. □

Corollary 3.1 Let H, C, S, , X, { μ n }, φ, θ, T i , f, A, { α n }, { β n } and { γ n } be as in Theorem  3.1. Suppose that F= n = 1 F( T n )F()Ω. Assume that

(C1) η:C×CH is Lipschitz continuous with constant λ>0 such that

  1. (a)

    η(x,y)+η(y,x)=0, x,yC,

  2. (b)

    for each fixed yC, xη(y,x) is sequentially continuous from the weak topology to the weak topology;

(C2) K:CR is η-strongly convex with constant σ>0 and its derivative K is not only sequentially continuous from the weak topology to the strong topology, but also Lipschitz continuous with constant ν>0 such that σλν;

(C3) for each xC, there exist a bounded subset D x C and z x C such that for any yC D x ,

θ(y, z x )+φ( z x )φ(y)+ 1 r K ( y ) K ( x ) , η ( z x , y ) <0;

(C4)

  1. (i)

    lim n α n =0, n = 0 α n =,

  2. (ii)

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

0< lim inf n γ n lim sup n γ n <1,

  1. (iii)

    lim n | γ n + 1 γ n |=0;

(C5) lim inf n r n >0 and lim n | r n + 1 r n |=0.

Given x 0 C is arbitrary, let the sequences { x n }, { y n } and { z n } be generated by

{ θ ( z n , x ) + φ ( x ) φ ( z n ) + 1 r n K ( z n ) K ( x n ) , η ( x , z n ) 0 , y n = ( 1 γ n ) x n + γ n T μ n W n P C z n , x n + 1 = α n γ f ( W n x n ) + β n x n + ( ( 1 β n ) I α n A ) T μ n W n y n , n 0 .

Then { x n }, { y n } and { z n } converge strongly to x F, where x = P F (γf+(IA)) x , which solves the following variational inequality:

( γ f A ) x , x x 0,xF.

Proof Setting B=0 in Theorem 3.1, we obtain the required result. □

Corollary 3.2 Let H, C, S, , X, { μ n }, T i , f, A, { α n }, { β n } and { γ n } be as in Theorem  3.1. Suppose that F= n = 1 F( T n )F(). Assume that

(C1)

  1. (i)

    lim n α n =0, n = 0 α n =,

  2. (ii)

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

0< lim inf n γ n lim sup n γ n <1,

  1. (iii)

    lim n | γ n + 1 γ n |=0;

(C2) lim inf n r n >0 and lim n | r n + 1 r n |=0.

Given x 0 C is arbitrary, let the sequences { x n }, { y n } and { z n } be generated by

{ y n = ( 1 γ n ) x n + γ n T μ n W n P C x n , x n + 1 = α n γ f ( W n x n ) + β n x n + ( ( 1 β n ) I α n A ) T μ n W n y n , n 0 .

Then { x n }, { y n } and { z n } converge strongly to x F, where x = P F (γf+(IA)) x , which solves the following variational inequality:

( γ f A ) x , x x 0,xF.

Proof Set B=0, θ(x,y)=0 for all x,yC, φ=0 and r n =1 for all n1. Take K(x)= x 2 2 and η(x,y)=xy for all x,yC. From (1.6), we have

{ y n = ( 1 γ n ) x n + γ n T μ n W n P C x n , x n + 1 = α n γ f ( W n x n ) + β n x n + ( ( 1 β n ) I α n A ) T μ n W n y n , n 0 .

Then the conclusion immediately follows from Theorem 3.1. □

Corollary 3.3 Let H, C, S, , X, { μ n }, φ, θ, f, A, { α n }, { β n } and { γ n } be as in Theorem  3.1. Suppose that Ω. Assume that

(C1) η:C×CH is Lipschitz continuous with constant λ>0 such that

  1. (a)

    η(x,y)+η(y,x)=0, x,yC,

  2. (b)

    for each fixed yC, xη(y,x) is sequentially continuous from the weak topology to the weak topology;

(C2) K:CR is η-strongly convex with constant σ>0, and its derivative K is not only sequentially continuous from the weak topology to the strong topology, but also Lipschitz continuous with constant ν>0 such that σλν;

(C3) for each xC, there exist a bounded subset D x C and z x C such that for any yC D x ,

θ(y, z x )+φ( z x )φ(y)+ 1 r K ( y ) K ( x ) , η ( z x , y ) <0;

(C4)

  1. (i)

    lim n α n =0, n = 0 α n =,

  2. (ii)

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

0< lim inf n γ n lim sup n γ n <1,

  1. (iii)

    lim n | γ n + 1 γ n |=0;

(C5) lim inf n r n >0 and lim n |