1 Introduction

Let C and Q be nonempty closed convex subsets of real Hilbert spaces H 1 and H 2 , respectively. The split feasibility problem (SFP) is formulated as

to finding  x C and A x Q,
(1.1)

where A: H 1 H 2 is a bounded linear operator. In 1994, Censor and Elfving [1] first introduced the (SFP) in finite-dimensional Hilbert spaces for modeling inverse problems which arise from phase retrievals and in medical image reconstruction [2]. It has been found that the (SFP) can also be used in various disciplines such as image restoration, and computer tomograph and radiation therapy treatment planning [35]. The (SFP) in an infinite-dimensional real Hilbert space can be found in [2, 4, 610].

Assuming that the (SFP) is consistent, it is not hard to see that x C solves (SFP) if and only if it solves the fixed-point equation

x = P C ( I γ A ( I P Q ) A ) x ,

where P C and P Q are the metric projection from H 1 onto C and from H 2 onto Q, respectively, γ>0 is a positive constant and A is the adjoint of A.

A popular algorithm to be used to solves the SFP (1.1) is due to Byrne’s CQ-algorithm [2]:

x k + 1 = P C ( I γ A ( I P Q ) A ) x k ,k1,
(1.2)

where γ(0,2/λ) with λ being the spectral radius of the operator A A.

Recently, Moudafi [11, 12] introduced the following split equality feasibility problem (SEFP):

to find xC,yQ such that Ax=By,
(1.3)

where A: H 1 H 3 and B: H 2 H 3 are two bounded linear operators. Obviously, if B=I (identity mapping on H 2 ) and H 3 = H 2 , then (1.3) reduces to (1.1). The kind of split equality problems (1.3) allows asymmetric and partial relations between the variables x and y. The interest is to cover many situations, such as decomposition methods for PDEs, and applications in game theory and intensity-modulated radiation therapy.

In order to solve the split equality feasibility problem (1.3), Moudafi [11] introduced the following simultaneous iterative method:

{ x k + 1 = P C ( x k γ A ( A x k B y k ) ) , y k + 1 = P Q ( y k + β B ( A x k + 1 B y k ) ) ,
(1.4)

and under suitable conditions he proved the weak convergence of the sequence {( x n , y n )} to a solution of (1.3) in Hilbert spaces.

At the same time, he raised the following open question.

Moudafi’s Open Question 1.1 Is there any strong convergence theorem of an alternating algorithm for the split equality feasibility problem (1.3) in real Hilbert spaces?

More recently, Eslamian and Latif [13], Chen et al. [14], Chuang [15] and Chang and Wang [16] introduced and studied some kinds of general split feasibility problem, general split equality problem, and split variational inclusion problem in real Hilbert spaces. Under suitable conditions some strong convergence theorems are proved. Also a comprehensive survey and update bibliography on split feasibility problems are given in Ansari and Rehan [17].

Motivated by the above works and related literature, in this paper, we continue to consider the problem (1.3). We obtain some strongly convergent theorems to a solution of the problem (1.3) which provide an affirmative answer to Moudafi’s open question.

For the purpose we first introduce and consider the following more general problems.

(I) General split equality variational inclusion problem:

(GSEVIP) to find  x H 1  and  y H 2  such that 0 i = 1 U i ( x ) , 0 i = 1 K i ( y ) and A x = B y ,
(1.5)

where H 1 , H 2 and H 3 are three real Hilbert spaces, U i : H 1 H 1 and K i : H 2 H 2 , i=1,2, are two families of set-valued maximal monotone mappings, A: H 1 H 3 and B: H 2 H 3 are two linear and bounded operators.

(II) General split equality optimization problem:

(GSEOP) to find  x H 1  and  y H 2  such that for each  i 1 h i ( x ) = min x H 1 h i ( x ) , g i ( y ) = min y H 2 g i ( y ) and A x = B y ,
(1.6)

where H 1 , H 2 , and H 3 are three real Hilbert spaces, A: H 1 H 3 and B: H 2 H 3 are two linear and bounded operators, h i : H 1 R and g i : H 2 R are two countable families of proper, convex, and lower semicontinuous functions.

The following problems are special cases of Problem I and II.

(III) Split equality feasibility problems.

Let C H 1 and Q H 2 be two nonempty closed convex subsets and A: H 1 H 3 , B: H 2 H 3 be two bounded linear operators. As mentioned above the so-called ‘split equality feasibility problem’ (SEFP) is to find

(1.3**)

Let i C and i Q be the indicator functions of C and Q, respectively, i.e.,

i C (x)= { 0 , if  x C , + , if  x C ; i Q (y)= { 0 , if  y C , + , if  y Q .
(1.7)

Denote by N C (x) and N Q (y) the normal cones of C and Q at x and y, respectively:

N C ( x ) = { z H 1 : z , v x 0 , v C } , N Q ( y ) = { z H 2 : z , v y 0 , v Q } .

It is easy to know that i C and i Q both are proper convex and lower semicontinuous functions on H 1 and H 2 , respectively, and the sub-differentials i C and i Q both are maximal monotone operators. We define the resolvent operator J β i C of i C by

J β i C (x)= ( I + β i C ) 1 (x),β>0,x H 1 .

Here

i C ( x ) = { z H 1 : i C ( x ) + z , u x i C ( u ) , u H 1 } = { z H 1 : z , u x 0 , u C } = N C ( x ) , x C .

Hence we have

u = J β i C ( x ) x u β N C ( u ) x u , y u 0 , y C u = P C ( x ) .

This implies that J β i C = P C for any β>0. Similarly, we also have i Q (y)= N Q (y), and J β i Q = P Q for any β>0. Therefore the (SEFP) (1.3) is equivalent to the following split equality optimization problem, i.e., to find x H 1 , and y H 2 such that

i C ( x ) = min x H 1 i C ( x ) , i Q ( y ) = min y H 2 i Q ( y ) and A x = B y ; 0 i C ( x ) , 0 i Q ( y ) and A x = B y .

(IV) Split equality equilibrium problem.

Let D be a nonempty closed and convex subset of a real Hilbert space H. A bifunction g:D×D(,+) is said to be a equilibrium function, if it satisfies the following conditions:

  1. (A1)

    g(x,x)=0, for all xD;

  2. (A2)

    g is monotone, i.e., g(x,y)+g(y,x)0 for all x,yD;

  3. (A3)

    lim sup t 0 g(tz+(1t)x,y)g(x,y) for all x,y,zD;

  4. (A4)

    for each xD, yg(x,y) is convex and lower semicontinuous.

The so-called equilibrium problem with respect to the equilibrium function g is

to find  x D such that g ( x , y ) 0,yD.
(1.8)

Its solution set is denoted by EP(g).

For given λ>0 and xH, the resolvent of the equilibrium function g is the operator R λ , g :HD defined by

R λ , g (x):= { z D : g ( z , y ) + 1 λ y z , z x 0 , y D } .
(1.9)

Proposition 1.2 [18]

The resolvent operator R λ , g of the equilibrium function g has the following properties:

  1. (1)

    R λ , g is single-valued;

  2. (2)

    F( R λ , g )=EP(g) and EP(g) is a nonempty closed and convex subset of D;

  3. (3)

    R λ , g is a firmly nonexpansive mapping.

Let h,g:D×D(,+) be two equilibrium functions. For given λ>0, let R λ , h and R λ , g be the resolvent of h and g (defined by (1.9)), respectively.

The so-called split equality equilibrium problem with respective to h, g, and D (SEEP(h,g,D)) is to find x D, y D such that

h ( x , u ) 0,uD,g ( y , v ) 0,vDandA x =B y ,
(1.10)

where A,B:DD are two linear and bounded operators.

By Proposition 1.2, the (SEEP(h,g,D)) (1.10) is equivalent to find x D, y D such that for each λ>0

x E P ( h , D ) , y E P ( g , D ) and A x = B y x F ( R λ h ) , y F ( R λ g ) and A x = B y .

Letting C=F( R λ h ), Q=F( R λ g ), by Proposition 1.2, C and Q both are nonempty closed and convex subset of D. Hence the problem (1.10) is equivalent to the following split equality feasibility problem:

to find  x C, y Q such that A x =B y .
(1.11)

(V) Split optimization problem.

Let H 1 and H 2 be two real Hilbert spaces, A: H 1 H 2 be a linear and bounded operators, h: H 1 R and g: H 2 R be two proper convex and lower semicontinuous functions. The split optimization problem (SOP) is to find x H 1 , A x H 2 such that

h ( x ) = min x H 1 h i (x)andg ( A x ) = min z H 2 g(z).
(1.12)

Denote by U=h and K=g, then the (SOP) (1.12) is equivalent to the following split variational inclusion problem (SVIP): to find x H 1 such that

0U ( x ) ,0K ( A x ) .
(1.13)

For solving (GSEVIP) (1.5) and (GSEOP) (1.6), in Sections 3 and 4, we propose a new simultaneous type iterative algorithm. Under suitable conditions some strong convergence theorems for the sequences generated by the algorithm are proved in the setting of infinite-dimensional Hilbert spaces. As special cases, we shall utilize our results to study the split feasibility problem, split equality equilibrium problem and the split optimization problem. By the way, we obtain a strongly convergent iterative sequence to a solution of the problem (1.3), which provides an affirmative answer to the open question raised by Moudafi [11]. The results presented in the paper extend and improve the corresponding results announced by Moudafi et al. [11, 12, 19], Eslamian and Latif [13], Chen et al. [14], Censor et al. [1, 35, 20], Chuang [15], Naraghirad [21], Chang and Wang [16], Ansari and Rehan [17], and some others.

2 Preliminaries

We first recall some definitions, notations, and conclusions.

Throughout this paper, we assume that H is a real Hilbert space and C is a nonempty closed convex subset of H. In the sequel, we denote by F(T) the set of fixed points of a mapping T and by x n x and x n x , the strong convergence, and weak convergence of a sequence { x n } to a point x , respectively.

Recall that a mapping T:HH is said to be nonexpansive, if TxTyxy, x,yH. A typical example of nonexpansive mapping is the metric projection P C from H onto CH defined by x P C x= inf y C xy. The metric projection P C is firmly nonexpansive, if

P C x P C y 2 xy, P C x P C yx,yH,
(2.1)

and it can be characterized by the fact that

P C (x)Cand y P C ( x ) , x P C ( x ) 0,xH,yC.
(2.2)

A mapping T:HH is said to be quasinonexpansive, if F(T), and

Txpxp,for each xH and pF(T).

It is easy to see that if T is a quasi-nonexpansive mapping, then F(T) is a closed and convex subset of C. Besides, T is said to be a firmly nonexpansive, if

T x T y 2 x y , T x T y x , y C T x T y 2 x y 2 ( I T ) x ( I T ) y 2 x , y C .

Lemma 2.1 [22]

Let H be a real Hilbert space, and { x n } be a sequence in H. Then, for any given sequence { λ n } of positive numbers with i = 1 λ n =1 for any positive integers i, j with i<j the following holds:

i = 1 λ n x n 2 i = 1 λ n x n 2 λ i λ j x i x j 2 .

Lemma 2.2 [23]

Let H be a real Hilbert space. For any x,yH, the following inequality holds:

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

Lemma 2.3 [24]

Let { t n } be a sequence of real numbers. If there exists a subsequence { n i } of {n} such that t n i < t n i + 1 for all i1, then there exists a nondecreasing sequence {τ(n)} with τ(n) such that for all (sufficiently large) positive integer number n, the following holds:

t τ ( n ) t τ ( n ) + 1 , t n t τ ( n ) + 1 .

In fact,

τ(n)=max{kn: t k t k + 1 }.

Definition 2.4 (Demiclosedness principle)

Let C be a nonempty closed convex subset of a real Hilbert space H, and T:CC be a mapping with F(T). Then IT is said to be demiclosed at zero, if for any sequence { x n }C with x n x and x n T x n 0, x=Tx.

Remark 2.5 [25]

It is well known that if T:CC is a nonexpansive mapping, then IT is demiclosed at zero.

Lemma 2.6 Let { a n }, { b n } and { c n } be sequences of positive real numbers satisfying a n + 1 (1 b n ) a n + c n for all n1. If the following conditions are satisfied:

  1. (1)

    b n (0,1) and n = 1 b n =,

  2. (2)

    n = 1 c n <, or lim sup n c n b n 0,

then lim n a n =0.

Lemma 2.7 [15]

Let H be a real Hilbert space, B:H 2 H be a set-valued maximal monotone mapping, β>0, and let J β B be the resolvent mapping of B defined by J β B := ( I + β B ) 1 , then

  1. (i)

    for each β>0, J β B is a single-valued and firmly nonexpansive mapping;

  2. (ii)

    D( J β B )=H and F( J β B )= B 1 (0);

  3. (iii)

    (I J β B ) is a firmly nonexpansive mapping for each β>0;

  4. (iv)

    suppose that B 1 (0), then for each xH, each x B 1 (0) and each β>0

    x J β B x 2 + J β B x x x x 2 ;
  5. (v)

    suppose that B 1 (0). Then x J β B x, J β B xw0 for each xH, each w B 1 (0), and each β>0.

Lemma 2.8 Let H 1 , H 2 be two real Hilbert spaces, A: H 1 H 2 be a linear bounded operator and A be the adjoint of A. Let B: H 2 2 H 2 be a set-valued maximal monotone mapping, β>0, and let J β B be the resolvent mapping of B, then

  1. (i)

    ( I J β B ) A x ( I J β B ) A y 2 (I J β B )Ax(I J β B )Ay,AxAy;

  2. (ii)

    A ( I J β B ) A x A ( I J β B ) A y 2 A 2 (I J β B )Ax(I J β B )Ay,AxAy;

  3. (iii)

    if ρ(0, 2 A 2 ), then (Iρ A (I J β B )A) is a nonexpansive mapping.

Proof By Lemma 2.7(iii), the mapping (I J β B ) is firmly nonexpansive, hence the conclusions (i) and (ii) are obvious.

Now we prove the conclusion (iii).

In fact, for any x,y H 1 , it follows from the conclusions (i) and (ii) that

( I ρ A ( I J β B ) A ) x ( I ρ A ( I J β B ) A ) y 2 = x y 2 2 ρ x y , A ( I J β B ) A x A ( I J β B ) A y + ρ 2 A ( I J β B ) A x A ( I J β B ) A y 2 x y 2 2 ρ A x A y , ( I J β B ) A x ( I J β B ) A y + ρ 2 A 2 ( I J β B ) A x ( I J β B ) A y 2 x y 2 ρ ( 2 ρ A 2 ) ( I J β B ) A x ( I J β B ) A y 2 x y 2 ( since  ρ ( 2 ρ A 2 ) 0 ) .

This completes the proof of Lemma 2.8. □

3 General split equality variational inclusion problem and strong convergence theorems

Throughout this section we assume that

  1. (1)

    H 1 , H 2 , H 3 are three real Hilbert spaces;

  2. (2)

    { U i } i = 1 : H 1 2 H 1 and { K i } i = 1 : H 2 2 H 2 are two families of set-valued maximal monotone mappings, β>0 and γ>0 are given positive numbers;

  3. (3)

    A: H 1 H 3 and B: H 2 H 3 are two bounded linear operators and A , B are the adjoint of A and B, respectively;

  4. (4)

    f= [ f 1 f 2 ] , where f i , i=1,2 is a k-contractive mapping on H i with k(0,1);

  5. (5)

    the set of solutions of (GSEVIP) (1.5) Ω,

    J μ i ( U i , K i ) :=[ J μ i U i J μ i K i ],G=[AB], G =[ A B ], G G=[ A A A B B A B B ],
  6. (6)

    for any given w 0 H 1 × H 2 , the iterative sequence { w n } H 1 × H 2 is generated by

    w n + 1 = α n w n + β n f( w n )+ i = 1 γ n , i ( J μ i ( U i , K i ) ( I λ n , i G G ) w n ) ,n0,
    (3.1)

or its equivalent form:

((3.1)′)

where { α n }, { β n }, { γ n , i } are the sequences of nonnegative numbers satisfying

α n + β n + i = 1 γ n , i =1,for each n0.

We are now in a position to give the following results.

Lemma 3.1 Let H 1 , H 2 , H 3 , A, B, A , B , { U i }, { K i }, J μ i ( U i , K i ) , G, G be the same as above. If Ω (the solution set of (GSEVIP) (1.5)), then w :=( x , y ) H 1 × H 2 is a solution of (GSEVIP) (1.5) if and only if for each i1, and for any given γ>0 and μ>0

w = J μ ( U i , K i ) ( I γ G G ) w .
(3.2)

Proof Indeed, if w =( x , y ) H 1 × H 2 is a solution of (GSEVIP) (1.5), then by Lemma 2.7(ii), for each i1, and for any γ>0 and μ>0 we have

x U i 1 ( 0 ) = F ( J μ U i ) , y K i 1 ( 0 ) = F ( J μ K i ) and A x = B y x = J μ U i x , y = J μ K i y and A x = B y .

Hence we have G( w )=A x B y =0, and so

J μ i ( U i , K i ) ( I γ G G ) ( w ) = J μ ( U i , K i ) ( w ) = ( J μ U i x , J μ K i y ) = w .

This implies that (3.2) is true.

Conversely, if w =( x , y ) H 1 × H 2 satisfies (3.2), then we have

{ x = J μ U i [ x γ A ( A x B y ) ] , y = J μ K i [ y + γ B ( A x B y ) ] .
(3.3)

We make the assumption that the solution set Ω of (GSEVIP) (1.5) is nonempty. Hence the sets U i 1 (0) and K i 1 (0) both are nonempty. By Lemma 2.7(v) and (3.3), we have

x ( x γ A ( A x B y ) ) , x x 0,x U i 1 (0),

and so

A x B y , A x A x 0,x U i 1 (0).
(3.4)

Similarly, by Lemma 2.7(v) and (3.3) again, one gets

A x B y , B y B y 0,y K i 1 (0).
(3.5)

Adding up (3.4) and (3.5), we have

A x B y , A x A x + B y B y 0,x U i 1 (0) and y K i 1 (0).

Simplifying it, we have

A x B y 2 A x B y , A x B y ,x U i 1 (0) and y K i 1 (0).
(3.6)

Since Ω, taking w ¯ =( x ¯ , y ¯ )Ω, for each i1, we have x ¯ U i 1 (0) and y ¯ K i 1 (0) and A x ¯ =B y ¯ . In (3.6), taking x= x ¯ and y= y ¯ , we have

A x B y =0,i.e.,A x =B y .
(3.7)

Hence from (3.3) and (3.7)

{ x = J μ U i ( x ) , y = J μ K i ( y ) , 0 U i ( x ) ,0 K i ( y ) ,i1.
(3.8)

It follows from (3.7) and (3.8) that w is a solution of (GSEVIP) (1.5).

This completes the proof of Lemma 3.1. □

Lemma 3.2 If λ(0, 2 L ), where L= G 2 , then (Iλ G G): H 1 × H 2 H 1 × H 2 is a nonexpansive mapping.

Proof In fact, for any w,u H 1 × H 2 , we have

( I λ G G ) u ( I λ G G ) w 2 = ( u w ) λ G G ( u w ) 2 = u w 2 + λ 2 G G ( u w ) 2 2 λ u w , G G ( u w ) u w 2 + λ 2 L G ( u w ) 2 2 λ G ( u w ) , G ( u w ) = u w 2 + λ 2 L G ( u w ) 2 2 λ G ( u w ) 2 = u w 2 λ ( 2 λ L ) G ( u w ) 2 u w 2 .

This completes the proof. □

Theorem 3.3 Let H 1 , H 2 , H 3 , A, B, A , B , { U i }, { K i }, J μ i ( U i , K i ) , G, G , f be the same as above. Let { w n } be the sequence defined by (3.1). If the solution set Ω of (GSEVIP) (1.5) is nonempty and the following conditions are satisfied:

  1. (i)

    α n + β n + i = 1 γ n , i =1, for each n0;

  2. (ii)

    lim n β n =0, and n = 0 β n =;

  3. (iii)

    lim inf n α n γ n , i >0 for each i1;

  4. (iv)

    { λ n , i }(0, 2 L ) for each i1, where L= G 2 ,

then the sequence { w n } converges strongly to w = P Ω f( w ), which is a solution of (GSEVIP) (1.5).

Proof (I) First we prove that the sequence { w n } is bounded.

In fact, for any given zΩ, it follows from Lemma 3.1, Lemma 3.2, and condition (iv) that

z= J μ ( U i , K i ) ( I λ n , i G G ) z,for each i1,

and (I λ n , i G G): H 1 × H 2 H 1 × H 2 is a nonexpansive mapping. Also by Lemma 2.7(i), for each i1, J μ i ( U i , K i ) is a firmly nonexpansive mapping. Hence we have

w n + 1 z = ( α n w n + β n f ( w n ) + i = 1 γ n , i J μ i ( U i , K i ) ( I λ n , i G G ) w n ) z α n w n z + β n f ( w n ) z + i = 1 γ n , i J μ i ( U i , K i ) ( I λ n , i G G ) w n z α n w n z + β n f ( w n ) z + i = 1 γ n , i ( I λ n , i G G ) w n z α n w n z + β n f ( w n ) z + i = 1 γ n , i ( I λ n , i G G ) w n ( I λ n , i G G ) z α n w n z + β n f ( w n ) z + i = 1 γ n , i w n z = ( 1 β n ) w n z + β n f ( w n ) z ( 1 β n ) w n z + β n f ( w n ) f ( z ) + β n f ( z ) z ( 1 β n ) w n z + k β n w n z + β n f ( z ) z = ( 1 ( 1 k ) β n ) w n z + ( 1 k ) β n 1 1 k f ( z ) z max { w n z , 1 1 k f ( z ) z } .

By induction, we can prove that

w n zmax { w 0 z , 1 1 k f ( z ) z } ,n0.

This shows that { w n } is bounded, and so is {f( w n )}.

(II) Now we prove that the following inequality holds:

α n γ n , i w n J μ i ( U i , K i ) ( I λ n , i G G ) w n 2 w n z 2 w n + 1 z 2 + β n f ( w n ) z 2 for each  i 1 .
(3.9)

Indeed, it follows from (3.1) and Lemma 2.1 that for each i1

w n + 1 z 2 = α n ( w n z ) + β n ( f ( w n ) z ) + j = 1 γ n , j ( J μ j ( U j , K j ) ( I λ n , j G G ) w n z ) 2 α n w n z 2 + β n f ( w n ) z 2 + j = 1 γ n , j J μ j ( U j , K j ) ( I λ n , j G G ) w n z 2 α n γ n , i w n J μ i ( U i , K i ) ( I λ n , i G G ) w n 2 α n w n z 2 + β n f ( w n ) z 2 + j = 1 γ n , j w n z 2 α n γ n , i w n J μ i ( U i , K i ) ( I λ n , i G G ) w n 2 = ( 1 β n ) w n z 2 + β n f ( w n ) z 2 α n γ n , i w n J μ i ( U i , K i ) ( I λ n , i G G ) w n 2 .

This implies that for each i1

α n γ n , i w n J μ i ( U i , K i ) ( I λ n , i G G ) w n 2 w n z 2 w n + 1 z 2 + β n f ( w n ) z 2 .

Inequality (3.3) is proved.

It is easy to see that the solution set Ω of (GSEVIP) (1.5) is a closed and convex subset in H 1 × H 2 . By the assumption that Ω is nonempty, so it is a nonempty closed and convex subset in H 1 × H 2 . Hence the metric projection P Ω is well defined. In addition, since P Ω f: H 1 × H 2 Ω is a contractive mapping, there exists a unique w Ω such that

w = P Ω f ( w ) .
(3.10)

(III) Now we prove that { w n } converges strongly to w .

For the purpose, we consider two cases.

Case I. Suppose that the sequence { w n w } is monotone. Since { w n w } is bounded, { w n w } is convergent. Since w Ω, in (3.9) taking z= w and letting n, in view of conditions (ii) and (iii), we have

lim n w n J μ i ( U i , K i ) ( I λ n , i G G ) w n =0,for each i1.
(3.11)

On the other hand, by Lemma 2.2 and (3.1), we have

w n + 1 w 2 = ( α n w n + β n f ( w n ) + i = 1 γ n , i J μ i ( U i , K i ) ( I λ n , i G G ) w n ) w 2 = α n ( w n w ) + β n ( f ( w n ) w ) + i = 1 γ n , i ( J μ i ( U i , K i ) ( I λ n , i G G ) w n w ) 2 α n ( w n w ) + i = 1 γ n , i ( J μ i ( U i , K i ) ( I λ n , i G G ) w n w ) 2 + 2 β n f ( w n ) w , w n + 1 w ( by Lemma  2.2 ) { α n w n w + i = 1 γ n , i w n w } 2 + 2 β n f ( w n ) f ( w ) , w n + 1 w + 2 β n f ( w ) w , w n + 1 w = ( 1 β n ) 2 w n w 2 + 2 β n k w n w w n + 1 w + 2 β n f ( w ) w , w n + 1 w ( 1 β n ) 2 w n w 2 + β n k { w n w 2 + w n + 1 w 2 } + 2 β n f ( w ) w , w n + 1 w .

Simplifying it we have

w n + 1 w 2 ( 1 β n ) 2 + β n k 1 β n k w n w 2 + 2 β n 1 β n k f ( w ) w , w n + 1 w = 1 2 β n + β n k 1 β n k w n w 2 + β n 2 1 β n k w n w 2 + 2 β n 1 β n k f ( w ) w , w n + 1 w = ( 1 2 ( 1 k ) β n 1 β n k ) w n w 2 + 2 ( 1 k ) β n 1 β n k { β n M 2 ( 1 k ) + 1 1 k f ( w ) w , w n + 1 w } = ( 1 η n ) w n w 2 + η n δ n ,
(3.12)

where

η n = 2 ( 1 k ) β n 1 β n k , δ n = β n M 2 ( 1 k ) + 1 1 k f ( w ) w , w n + 1 w , M = sup n 0 w n w 2 .

By condition (ii), lim n β n =0 and n = 1 β n =, and so is n = 1 η n =.

Next we prove that

lim sup n δ n 0.
(3.13)

In fact, since { w n } is bounded in H 1 × H 2 , there exists a subsequence { w n k }{ w n } with w n k v (some point in C×Q), and λ n k , i λ i (0, 2 L ) such that

lim n f ( w ) w , w n k w = lim sup n f ( w ) w , w n w .

Since

w n k J μ i ( U i , K i ) ( I λ n k , i G G ) w n k 0,for each i1

and J μ i ( U i , K i ) (I λ n k , i G G) is a nonexpansive mapping, by Remark 2.5, I J μ i ( B i , K i ) (I λ n , i G G) is demiclosed at zero, hence we have

v = J μ i ( U i , K i ) ( I λ n , i G G ) v ,i1.
(3.14)

By Lemma 3.1, this implies that v Ω. In addition, since w = P Ω f( w ), we have

lim sup n f ( w ) w , w n w = lim n f ( w ) w , w n k w = f ( w ) w , v w 0 .

This shows that (3.13) is true. Taking a n = w n w 2 , b n = η n , and c n = δ n η n in Lemma 2.6, therefore all conditions in Lemma 2.6 are satisfied. We have w n w .

Case II. If the sequence { w n w } is not monotone, by Lemma 2.3, there exists a sequence of positive integers: {τ(n)}, n n 0 (where n 0 large enough) such that

τ(n)=max { k n : w k w w k + 1 w } .
(3.15)

Clearly {τ(n)} is a nondecreasing, τ(n) as n, and for all n n 0

w τ ( n ) w w τ ( n ) + 1 w ; w n w w τ ( n ) + 1 w .
(3.16)

Therefore { w τ ( n ) w } is a nondecreasing sequence. According to Case I, lim n w τ ( n ) w =0 and lim n w τ ( n ) + 1 w =0. Hence we have

0 w n w max { w n w , w τ ( n ) w } w τ ( n ) + 1 w 0,as n.

This implies that w n w and w = P Ω f( w ) is a solution of (GSEVIP) (1.5).

This completes the proof of Theorem 3.3. □

Remark 3.4 Theorem 3.3 extends and improves the main results in Moudafi et al. [11, 12, 19], Eslamian and Latif [13], Chen et al. [14], Chuang [15], Naraghirad [21] and Ansari and Rehan [17].

4 General split equality optimization problem and strong convergence theorems

Let H 1 , H 2 , and H 3 be three real Hilbert spaces. Let A: H 1 H 3 and B: H 2 H 3 be two linear and bounded operators. The so-called general split equality optimization problem (GSEOP) is to find x H 1 , and y H 2 such that for each i1

h i ( x ) = min x H 1 h i (x), g i ( y ) = min z H 2 g i (z)andA x =B y ,
(4.1)

where h i : H 1 R and g i : H 2 R are two families of proper, lower semicontinuous, and convex functions.

For each i1 denote by h i = U i and g i = K i . Then the mappings U i : H 1 2 H 1 and K i : H 2 2 H 2 , i=1,2, both are set-valued maximal monotone mappings, and

h i ( x ) = min x H 1 h i ( x ) 0 h i ( x ) = U i ( x ) , g i ( y ) = min z H 2 g i ( z ) 0 g i ( y ) = K i ( y ) .

Therefore (GSEOP) (4.1) is equivalent to the following general split equality variational inclusion problem (GSEVIP): to find x H 1 and y H 2 such that

0 i = 1 U i ( x ) ,0 i = 1 K i ( y ) andA x =B y .
(4.2)

Therefore, the following theorem can be obtained from Theorem 3.3 immediately.

Theorem 4.1 Let H 1 , H 2 , H 3 , A, B, A , B , { U i }, { K i } be the same as above. Let J μ i ( U i , K i ) , G, G , f be the same as in Theorem  3.3. Let { w n } be the sequence defined by (3.1). If the solution set Ω 1 of (GSEVIP) (4.1) is nonempty and the following conditions are satisfied:

  1. (i)

    α n + β n + i = 1 γ n , i =1, for each n0;

  2. (ii)

    lim n β n =0, and n = 0 β n =;

  3. (iii)

    lim inf n α n γ n , i >0 for each i1;

  4. (iv)

    { λ n , i }(0, 2 L ) for each i1, where L= G 2 ,

then the sequence { w n } converges strongly to w = P Ω 1 f( w ), which is a solution of (GSEOP) (4.1).

By using Theorem 3.3 and Theorem 4.1, now we give some corollaries for the split equality feasibility problem, the split equality equilibrium problem, and the split optimization problem.

Let H 1 , H 2 , H 3 , C, Q, A, B be the same as in the split equality feasibility problem (1.3). Let i C and i Q be the indicator function of C and Q, respectively, defined by (1.7). In Theorem 4.1, take {U}={ i C }, {K}={ i Q }, and J μ ( U , K ) = P C × Q := [ P C P Q ] , therefore we have the following.

Corollary 4.2 Let H 1 , H 2 , H 3 , A, B, A , B , P C × Q be the same as above. Let G, G , f be the same as in Theorem  4.1. Let { w n } be the sequence generated by w 0 H 1 × H 2

w n + 1 = α n w n + β n f( w n )+ γ n ( P C × Q ( I λ n G G ) w n ) ,n0,
(4.3)

or its equivalent form

{ x n + 1 = α n x n + β n f 1 ( x n ) + γ n ( P C ( x n λ n ( A ( A x n B y n ) ) ) ) , y n + 1 = α n y n + β n f 2 ( y n ) + γ n ( P Q ( y n + λ n ( B ( A x n B y n ) ) ) ) .
(4.4)

If the solution set Γ 1 of (SEFP) (1.3) is nonempty and the following conditions are satisfied:

  1. (i)

    α n + β n + γ n =1, for each n0;

  2. (ii)

    lim n β n =0, and n = 0 β n =;

  3. (iii)

    lim inf n α n γ n >0;

  4. (iv)

    { λ n }(0, 2 L ) for each i1, where L= G 2 ,

then the sequence { w n } converges strongly to w = P Γ 1 f( w ), which is a solution of (SEFP) (1.3).

Remark 4.3 Since the simultaneous iterative sequence {( x n , y n )} (4.4) converges strongly to a solution of (SEFP) (1.3). Therefore it provides an affirmative answer to Moudafi’s open question 1.1 [11].

Let h,g:D×D(,+) be two equilibrium functions. For given λ>0, let R λ , h and R λ , g be the resolvents of h and g (defined by (1.9)), respectively.

The so-called split equality equilibrium problem with respective to h, g, and D (SEEP(h,g,D)) is to find x D, y D such that

h ( x , u ) 0,uD,g ( y , v ) 0,vDandA x =B y ,
(4.5)

where A,B:DD are two linear and bounded operators.

By Proposition 1.2, the (SEEP(h,g,D)) (4.5) is equivalent to find x D, y D such that for each λ>0

x E P ( h , D ) , y E P ( g , D ) and A x = B y x F ( R λ h ) , y F ( R λ g ) and A x = B y .
(4.6)

Letting C=F( R λ h ), Q=F( R λ g ), by Proposition 1.2, C and Q both are nonempty closed and convex subset of D. Hence the problem (4.5) (and so the problem (4.6)) is equivalent to the following split equality feasibility problem:

to find  x C, y Q such that A x =B y .
(4.7)

In Corollary 4.2 taking H 1 = H 2 = H 3 =D, from Corollary 4.2 we have the following.

Corollary 4.4 Let D, C, Q be the same as above. Let A, B, A , B , P C × Q , G, G , f be the same as in Corollary  4.2. For any given w 0 D×D, let { w n } be the sequence generated by

w n + 1 = α n w n + β n f( w n )+ γ n ( P C × Q ( I λ n G G ) w n ) ,n0.
(4.8)

If the solution set Γ 2 of (SEEP(h,g,D)) (4.5) is nonempty and the following conditions are satisfied:

  1. (i)

    α n + β n + γ n =1, for each n0;

  2. (ii)

    lim n β n =0, and n = 0 β n =;

  3. (iii)

    lim inf n α n γ n >0;

  4. (iv)

    { λ n }(0, 2 L ) for each i1, where L= G 2 ,

then the sequence { w n } converges strongly to w = P Γ 2 f( w ), which is a solution of (SEEP(h,g,D)) (4.5).

Let H 1 and H 2 be two real Hilbert spaces, A: H 1 H 2 be a linear and bounded operators, h: H 1 R and g: H 2 R be two proper convex and lower semicontinuous functions. The split optimization problem (SOP) is to find x H 1 , A x H 2 such that

h ( x ) = min x H 1 h i (x)andg ( A x ) = min z H 2 g(z).
(4.9)

Denote U=h and K=g, then the (SOP) (4.9) is equivalent to the following split variational inclusion problem (SVIP): to find x H 1 such that

0U ( x ) ,0K ( A x ) .
(4.10)

In Theorem 4.1 taking H 3 = H 2 , B=I (the identity mapping on H 2 ) and

G ˜ =[AI], G ˜ =[ A I ], G ˜ G ˜ =[ A A A A I ],

then from Theorem 4.1 we have the following.

Corollary 4.5 Let H 1 , H 2 , A, I, G ˜ , G ˜ , U, K, be the same as above. Let J μ ( U , K ) , f be the same as in Theorem  4.1. For any given w 0 =( x 0 , y 0 ) H 1 × H 2 , let { w n =( x n , y n )} be the sequence defined by

{ x n + 1 = α n x n + β n f 1 ( x n ) + γ n J μ U ( x n λ n A ( A x n y n ) ) , y n + 1 = α n y n + β n f 2 ( y n ) + γ n J μ K ( y n + λ n ( A x n y n ) ) ,
(4.11)

or its equivalent form:

w n + 1 = α n w n + β n f( w n )+ γ n ( J μ ( U , K ) ( I λ n G ˜ G ˜ ) w n ) ,n0,
(4.12)

If Γ 3 :={ x U 1 (0) A 1 K 1 (0)}, the solution set of (SOP) (4.9) is nonempty, and the following conditions are satisfied:

  1. (i)

    α n + β n + γ n =1, for each n0;

  2. (ii)

    lim n β n =0, and n = 0 β n =;

  3. (iii)

    lim inf n α n γ n >0;

  4. (iv)

    { λ n }(0, 2 L ), where L= G ˜ 2 ,

then the sequence { w n } converges strongly to w = P Γ 3 f( w ), which is a solution of (SOP) (4.9).