Split hierarchical variational inequality problems and fixed point problems for nonexpansive mappings

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1 Ansari et al. Journal of Inequalities and Applications (015) 015:74 DOI /s R E S E A R C H Open Access Split hierarchical variational inequality problems and fixed point problems for nonexpansive mappings Qamrul Hasan Ansari 1, Aisha Rehan 1 and Ching-Feng Wen,3* * Correspondence: cfwen@kmu.edu.tw Center for Fundamental Science, Kaohsiung Medical University, Kaohsiung, 807, Taiwan, ROC 3 Center for Nonlinear Analysis and Optimization, Kaohsiung Medical University, Kaohsiung, 807, Taiwan, ROC Full list of author information is available at the end of the article Abstract The present paper deals with the common solution method for finding a fixed point of a nonexpansive mapping and a solution of split hierarchical Minty variational inequality problems. We discuss the weak convergence of the sequences generated by the proposed method to a common solution of a fixed point problem and a split hierarchical Minty variational inequality problem. An example is presented to illustrate the proposed algorithm and result. MSC: 49J40; 49J5; 47J0 Keywords: split hierarchical variational inequality problems; fixed point problems; iterative method; nonexpansive mappings; convergence analysis 1 Introduction Since its origin in 1994 by Censor and Elfving [1], the split feasibility problem (SFP) has been rapidly investigated and studied because of its applications in different areas such as signal processing, phase retrievals, image reconstruction, intensity-modulated radiation therapy, etc. (see, forexample, [ 8] and the references therein). Recently, Censor and Segal [9]introducedasplit common fixed point problem (SCFPP) which is to find a common element of a family of operators in one space such that its image under a linear transformation is a common fixed point of another family of operators in the image space. The SCFPP generalizes convex feasibility problem (CFP), split feasibility problem (SFP) and multiple sets split feasibility problem (MSSFP). They developed a parallel algorithm for solving SCFPP for the class of directed operators in the setting of finite-dimension spaces. Further, Cui et al. [10] proposed a damped projection method for SCFPP and studied its convergence result. Moudafi [11] further proposed and analyzed an iterative scheme for solving SCFPP for the class of demicontractive operators in the setting of Hilbert spaces. He studied the weak convergence of the sequence generated by the proposed algorithm to a solution of SCFPP. Subsequently, Cui and Wang [1] suggested a new algorithm that does not require any prior information of the operator norm to find a solution of SCFPP. They studied the weak convergence of the proposed algorithm. In 011, Moudafi [13]considered the relaxed algorithm for computing the approximate solution of SCFPP for quasinonexpansive operators and studied the weak convergence of the sequence generated by the suggested algorithm. Very recently, SCFPP was considered by Kraikaew and Saejung 015 Ansari et al. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page of 16 [14] for quasi-nonexpansive and strongly quasi-nonexpansive operators. They proposed an algorithm and showed that their algorithm converges strongly to a solution of SCFPP. They also considered split variational inequality problem [15], split common null point problem and Moudafi s split feasibility problem, and derived the algorithm for these problems from the main algorithm for SCFPP. Also, the strong convergence of these algorithms is derived from the main convergence result. Very recently, Ansari et al. [16]introduced the split hierarchical variational inequality problem (SHVIP). A variational inequality problem in which the underlying set is a set of fixed points of a nonlinear operator is called hierarchical variational inequality problem. For further details on hierarchical variational inequality problems, we refer to [17] and the references therein. More precisely, they considered the following split hierarchical Minty variational inequality problem (SHMVIP) which requires to find a solution of a hierarchical Minty variational inequality problem (HMVIP) such that its image under a nonlinear operator is a solution of another HMVIP. Let H 1 and H be real Hilbert spaces, f, T : H 1 H 1 be operators such that Fix(T), and h, S : H H be operators with Fix(S), wherefix(t) andfix(s) aredenoted by the set of fixed points of T and S, respectively.leta : H 1 H be an operator with R(A) Fix(S), where R(A) denotes the range of A. The split hierarchical variational inequality problem (SHVIP) is to find x Fix(T)suchthat f ( x ), x x 0 for all x Fix(T) (1) such that Ax Fix(S) and it satisfies h ( Ax ), y Ax 0 for all y Fix(S). () The solution set of the SHVIP is denoted by. Another problem which is closely related to (SHVIP) is the following split hierarchical Minty variational inequality problem (SHMVIP): Find x Fix(T)suchthat f (x), x x 0 for all x Fix(T), (3) and such that Ax Fix(S)satisfies h(y), y Ax 0 for all y Fix(S). (4) We denote by Ɣ the set of solutions of SHMVIP, that is, Ɣ = { x solves (3):Ax solves (4) }. ItcanbeeasilyseenbytheMintylemma[18], Lemma 1, that if Fix(T) andfix(s) are nonempty closed convex and f and h are monotone and continuous, then SHVIP (1)-() and SHMVIP (3)-(4)areequivalent. Ansari et al. [16] showed thatseveral problems, namelysplit convex minimization problem, split variational inequality problem over the solution set of monotone variational inclusion problem, andsplit variational inequality problem over the solution set of equilibrium problem, are particular cases of SHVIP. They proposed an iterative scheme for solving SHVIP and studied the weak convergence of the sequence generated by the proposed algorithm.

3 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page 3 of 16 In this paper, we give a common solution method for finding a fixed point of a nonexpansive operator and a solution of split hierarchical variational inequality problems. The weak convergence of such algorithm is studied. We also present an example to illustrate the proposed algorithm and the convergence result. Preliminaries Let H be a real Hilbert space whose inner product and norm are denoted by, and,respectively.letc be a nonempty closed convex subset of H.Wedenotebyx n x (respectively, x n x) the strong (respectively, weak) convergence of the sequence {x n } to x.lett : H H be an operator whose range is denoted by R(T). The set of all fixed points of T is denoted by Fix(T), that is, Fix(T)={x H : x = Tx}. Definition.1 An operator T : H H is said to be: (a) nonexpansive if Tx Ty x y for all x, y H; (b) strongly nonexpansive [19, 0] ift is nonexpansive and lim (xn y n ) (Tx n Ty n ) =0, whenever {x n } and {y n } are bounded sequences in H and lim ( x n y n Tx n Ty n )=0; (c) averaged nonexpansive if it can be written as T =(1 α)i + αs, where α (0, 1), I is the identity operator of H,andS : H H is a nonexpansive mapping; (d) firmly nonexpansive if Tx Ty x y, Tx Ty for all x, y H; (e) cutter [0]ifx Tx, z Tx 0 for all x H and z Fix(T); (f) monotone if x y, Tx Ty 0 for all x, y H; (g) α-inverse strongly monotone if there exists a constant α >0such that Tx Ty, x y α Tx Ty for all x, y H. Remark.1 Every strongly nonexpansive operator is nonexpansive, but a nonexpansive operator need not be strongly nonexpansive. Also, a nonexpansive cutter operator need not be strongly nonexpansive. Example.1 Let T :[ 1,1] R be defined by Tx = x for all x [ 1, 1]. Then T is nonexpansive but not strongly nonexpansive. Indeed, let x n =1andy n =0foralln.Then{x n } and {y n } are bounded sequences. Also, lim (xn y n ) (Tx n Ty n ) = lim 1+1 = 0. Thus, T is not strongly nonexpansive. Example. Let H 1 = H = R with inner product and norm be given by x, y = x 1 y 1 + x y and x = x 1 + x,respectively,wherex =(x 1, x )andy =(y 1, y ). Let C = {x R :

4 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page 4 of 16 x } be a nonempty closed subspace of R.LetT : C C be defined by ( T(x, y)= 3 x y, 1 3 x + ) 3 y for all (x, y) C. Then T is a nonexpansive cutter operator, but T is notstrongly nonexpansive. Indeed, let {x n } = (1, 0) and {y n } = (0, 0) for all n. Then{x n } and {y n } are two bounded sequences of C,andTx n =( 3, 1 3 )andty n = (0, 0). Note that ( lim xn y n Tx n Ty n ) ( ( (1, = lim 0) 3, 1 )) = lim (1 1) = 0. 3 But lim (x n y n ) (Tx n Ty n ) ( = lim (1, 0) 3 3), 1 = lim 1 3, 1 3 = 3 0. Thus, T is not strongly nonexpansive. InordertoshowthatT is a nonexpansive cutter operator, we first prove that it is nonexpansive. Let x =(x 1, x ), y =(y 1, y ) C.Then ( Tx Ty = 3 x x, 1 3 x 1 + ) ( 3 x 3 y y, 1 3 y 1 + ) 3 y = 3 (x 1 y 1 )+ 1 3 (x y ), 1 3 (x 1 y 1 )+ 3 (x y ) = 3 (x 1 y 1 )+ 1 3 (x y ) (x 1 y 1 )+ 3 (x y ) 3 x 1 y x y x 1 y x y = x 1 y 1 + x y = (x1 y 1 ), (x, y ) = (x1, x ) (y 1, y ) = x y. Thus, T is nonexpansive. We next show that T is cutter. Note that Fix(T)={(x, y) C : x = y}. Letx =(x, y) C and (p, p) Fix(T), we have ( x Tx, p Tx = (x, y) 3 x y, 1 3 x + ) ( 3 y,(p, p) 3 x y, 1 3 x + ) 3 y ( 1 = 3 (x y), 1 ) (y x), (p 13 3 (x + y), p 13 ) (x +y) = 1 ( 3 (x y) p 1 ) 3 (x + y) + 1 ( 3 (y x) p 1 ) 3 (x +y) = 1 9 (x y){ 3p (x + y) 3p +(x +y) } = 1 9 (x y)(y x)= 1 9 (x y) 0. This shows that T is cutter.

5 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page 5 of 16 The following lemma provides some fundamental properties of Hilbert spaces. These properties will be used in the sequel. Lemma.1 Let H be a real Hilbert space. Then, for all x, y H, we have (a) x y = x y x y, y ; (b) x y x +y, y x ; (c) λx +(1 λ)y = λ x +(1 λ) y λ(1 λ) x y for all λ [0, 1]. Lemma. ([1], Lemma, Demiclosedness principle) Let C be a nonempty closed convexsubsetofarealhilbertspacehandt: C C be a nonexpansive operator with Fix(T). If the sequence {x n } C converges weakly to x and the sequence {(I T)x n } converges strongly to y, then (I T)x = y; in particular, if y =0,then x Fix(T). Definition. Let T : H H be a set-valued operator with domain D(T) ={x H : T(x) }, ranger(t)= x D(T) T(x) andtheinverseoft is T 1 (y) ={x H : y T(x)}. T is said to be: (a) monotone if x y, f g 0, whenever f Tx, g Ty; (b) maximal monotone if it is monotone and the graph G(T)= { (x, f ) H H : f Tx } of T is not properly contained in the graph of any other monotone operator; (c) α-inverse strongly monotone if there exists a constant α >0such that Tx Ty, x y α Tx Ty, whenever x, y D(T). It is well known that when T is maximal monotone, then for each x H and λ >0,there is a unique z H such that x (I + λt)z. In this case, the operator Jλ T := (I + λt) 1 is called resolvent of T with parameter λ. ItisknownthatJλ T is a single-valued and firmly nonexpansive mapping. The following lemma will be used in our main result. Lemma.3 ([]) Let {a n } n=1 and {b n} n=1 be sequences of nonnegative real numbers such that a n+1 a n + b n for all n 1. If n=1 b n <, then the limit lim a n exists. 3 Algorithms and convergence results Let K : H 1 H 1 be a nonexpansive operator with Fix(K) Ɣ. We propose the following algorithm to compute a common element of the set of fixed points of K and the set of solutions of SHMVIP.

6 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page 6 of 16 Algorithm 3.1 Initialization: Choose {α n } n=1, {β n} n=1, {λ n} n=1 (0, 1). Take arbitrary x 0 H 1. Iterative step: For a given current x n H 1,compute z n = x n γ A ( I S(I β n h) ) Ax n, y n = T(I α n f )z n, (5) x n+1 = λ n x n +(1 λ n )Ky n, where γ (0, ). A Last step: Update n := n +1. When K is the identity operator, Algorithm 3.1 reduces to the following algorithm. Algorithm 3. Initialization: Choose {α n } n=1, {β n} n=1, {λ n} n=1 (0, 1). Take arbitrary x 0 H 1. Iterative step: For a given current x n H 1,compute z n = x n γ A ( I S(I β n h) ) Ax n, y n = T(I α n f )z n, (6) x n+1 = λ n x n +(1 λ n )y n, where γ (0, ). A Last step: Update n := n +1. Next we prove the weak convergence of the sequences generated by Algorithm 3.1. Theorem 3.1 Let f : H 1 H 1 be a monotone continuous mapping, T : H 1 H 1 be a nonexpansive cutter operator such that Fix(T), h : H H be a monotone continuous mapping and S : H H be a strongly nonexpansive cutter operator such that Fix(S). Let A : H 1 H be a bounded linear operator with R(A) Fix(S) and let K : H 1 H 1 be a nonexpansive operator with Fix(K) Ɣ. Let {x n } and {y n } be the sequences generated by Algorithm 3.1 such that the following conditions hold: (i) There exists a natural number n such that Ɣ n=n { z H1 : h(ax n ), S(I β n h)ax n Az 0 } ; (ii) {f (z n )} n=1 is a bounded sequence; (iii) n=0 α n < ; (iv) lim β n =0; (v) x n+1 x n = o(α n ) and α n = o(βn ); (vi) {h(ax n )} n=1 is a bounded sequence. Then the sequences {x n } and {y n } converge weakly to an element x Fix(K) Ɣ.

7 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page 7 of 16 Proof Let p Fix(K) Ɣ. ThenT(p) =p, K(p) =p and S(Ap) =Ap. Consider z n p = x n γ A ( I S(I β n h) ) Ax n p = x n p + γ A ( I S(I β n h) ) Ax n γ x n p, A ( I S(I β n h) ) Ax n x n p + γ A ( I S(I β n h) ) Ax n γ x n p, A ( I S(I β n h) ) Ax n for all n 1. (7) Since S is a cutter operator, we have xn p, A ( S(I β n h) I ) Ax n = Ax n Ap, ( S(I β n h) I ) Ax n = S(I β n h)ax n Ap + Ax n S(I β n h)ax n, ( S(I β n h) I ) Ax n = S(I β n h)ax n Ap, ( S(I β n h) I ) ( Ax n S(I βn h) I ) Ax n = S(I β n h)ax n Ap, ( S(I β n h) I ) Ax n + β n hax n β n hax n ( S(I β n h) I ) Ax n = S(I β n h)ax n Ap, S(I β n h)ax n (I β n h)ax n β n S(I βn h)(ax n ) Ap, h(ax n ) ( S(I β n h) I ) Ax n ( S(I β n h) I ) Ax n βn S(I βn h)(ax n ) Ap, h(ax n ). Since p Ɣ, by condition (i), we have S(I βn h)(ax n ) Ap, h(ax n ) 0. Since β n (0, 1) for all n N,wefurtherhave β n S(I βn h)(ax n ) Ap, h(ax n ) 0. Therefore, xn p, A ( S(I β n h) I ) Ax n ( S(I βn h) I ) Ax n. Thus, (7)becomes z n p x n p + γ A ( I S(I β n h) ) Ax n γ ( S(I β n h) I ) Ax n = x n p γ ( γ A ) ( S(I β n h) I ) Ax n for all n 1. (8) Since γ (0, A ), we observe that γ ( γ A )>0,andhence z n p x n p for all n 1. (9)

8 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page 8 of 16 Let M := sup{ f (z n ) : n 1}.Then,foralln 1, we have y n p = T ( zn α n f (z n ) ) T(p) z n p + α n f (z n ) z n p + α n M x n p + α n M, (10) x n+1 p = λn x n +(1 λ n )Ky n p = λ n (x n p)+(1 λ n )(Ky n p) λ n x n p +(1 λ n ) y n p λ n x n p +(1 λ n ) x n p +(1 λ n )α n M x n p +(1 λ n )α n M. Since α n < and 0 < (1 λ n )<1,wehave n=1 (1 λ n)α n <. Thus, by Lemma.3, the limit lim x n p exists. Also, from (9)-(10), the limits lim z n p and lim y n p exist. This implies that {x n }, {y n } and {z n } are bounded sequences. Now, consider y n p = T(I αn f )(z n ) T(p) (zn p) α n f (z n ) z n p + αn f (zn ). (11) From (9), (11) and by Lemma.1(c), we have x n+1 p = λn x n +(1 λ n )Ky n p which is equivalent to = λ n (x n p)+(1 λ n )(Ky n p) = λ n x n p +(1 λ n ) y n p λ n (1 λ n ) Ky n x n λ n x n p +(1 λ n ) { z n p + αn f (z n ) } λ n (1 λ n ) Ky n x n λ n x n p +(1 λ n ) x n p (1 λ n )γ ( γ A ) ( S(I β n h) I ) Ax n +(1 λ n )αn f (zn ) λ n (1 λ n ) Ky n x n = x n p (1 λ n )γ ( γ A ) ( S(I β n h) I ) Ax n +(1 λ n )αn f (zn ) λ n (1 λ n ) Ky n x n, (1) (1 λ n )γ ( γ A ) ( S(I β n h) I ) Ax n + λn (1 λ n ) Ky n x n x n p x n+1 p +(1 λ n )αn f (zn ). (13)

9 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page 9 of 16 From the existence of the limit lim x n p and the facts that α n 0, f (z n ) is bounded, 0 < λ n <1andγ (0, ), it follows that A lim ( S(I β n h) I ) Ax n =0, (14) and lim Ky n x n = 0. (15) From (5), we have lim x n+1 x n lim Ky n x n =0. (16) Since T is a cutter operator, we have p y n, z n y n = y n p, y n z n = T(I α n f )z n p, T(I α n f )z n (I α n f )z n +(I α n f )z n z n = T(I α n f )z n p, T(I α n f )z n (I α n f )z n + T(I α n f )z n p, α n fz n T(I α n f )z n p, T(I α n f )z n (I α n f )z n + α n T(I αn f )z n p fzn, and T(I αn f )z n p, T(I α n f )z n (I α n f )z n 0. This implies that p y n, z n y n α n T(I αn f )z n p fzn. (17) From (9), (17) and by Lemma.1(a), we have y n p = z n p z n y n y n p, z n y n x n p x n γ A ( I S(I β n h) ) Ax n y n y n p, z n y n = x n p x n y n γ A ( I S(I β n h) ) Ax n +γ x n y n, A ( ) I S(I β n h)ax n +p yn, z n y n x n p x n y n γ A ( I S(I β n h) ) Ax n +α n T(I α n f )z n p fz n +γ x n y n A ( S(I β n h) I ) Ax n. (18)

10 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page 10 of 16 Thus, from (1)and(18), we have x n+1 p = λ n x n p +(1 λ n ) y n p λ n (1 λ n ) Ky n x n λ n x n p +(1 λ n ) x n p (1 λ n )γ A ( I S(I β n h) ) Ax n (1 λ n ) x n y n λ n (1 λ n ) Ky n x n +(1 λ n )α n T(I α n f )z n p f (z n ) +γ (1 λ n ) x n y n A ( S(I β n h) I ) Ax n x n p (1 λ n ) x n y n (1 λ n )γ A ( I S(I β n h) ) Ax n +(1 λ n )α n T(I αn f )z n p f (zn ) λn (1 λ n ) Ky n x n +γ (1 λ n ) x n y n A ( S(I β n h) I ) Ax n, which is equivalent to (1 λ n ) x n y n x n p x n+1 p λ n (1 λ n ) Ky n x n +(1 λ n )α n T(I αn f )z n p f (zn ) +γ (1 λ n ) x n y n A ( S(I β n h) I ) Ax n. Taking limit as n, and taking into account α n 0, 0 < λ n <1,γ (0, )andfrom A equations (14), (15)wehave x n y n 0 asn, (19) y n Ky n = y n x n + x n Ky n y n x n + x n Ky n. From (15)and(19), we obtain y n Ky n 0 asn. (0) Since {x n } is a bounded sequence, there exists a convergent subsequence {x ni } of {x n } that convergesweaklytosomex H 1.Since x n y n 0, it is known that y ni x H 1.By the demiclosed principle, y ni x and y ni Ky ni 0, we have Kx = x. From (5), we obtain z n x n = γ A ( I S(I β n h) ) Ax n. By (14), we have z n x n 0 asn, (1)

11 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page 11 of 16 and z n y n x n y n + γ A ( I S(I β n h) ) Ax n. Equations (14)and(19)yield that z n y n 0 asn. () From the definition of y n,wehave y n Tz n = T ( z n α n f (z n ) ) Tz n z n α n f (z n ) z n α n f (z n ). This implies that lim y n Tz n =0, (3) y n Ty n = y n Tz n + Tz n Ty n y n Tz n + z n y n. From ()and(3), we get lim y n Ty n 0. Since y ni x and y ni Ty ni 0, by the demiclosed principle, we obtain Tx = x. Let v n := Ax n β n h(ax n ) for all n 1. We observe that 0 v n Ap Sv n SAp = Ax n β n h(ax n ) Ap Sv n SAp = Axn Sv n + Sv n β n h(ax n ) Ap Svn SAp Ax n Sv n + Sv n SAp + β n h(ax n ) Sv n SAp = Axn S ( ( Ax n β n h(axn ) )) + βnh(axn ) = ( S(I β n h) I ) Ax n + β n h(ax n ). From condition (iv) and (14), we have ( lim vn Ap Sv n SAp ) =0. The boundedness of v n and strong nonexpansiveness of S imply that lim Sv n v n =0. (4)

12 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page 1 of 16 From the definition of v n and condition (iv), we get and lim v n Ax n =0 (5) v n SAx n v n Sv n + Sv n S(Ax n ) From (4)and(5), we have = v n Sv n + v n Ax n. and thus lim v n SAx n =0, lim Ax n SAx n =0. (6) Since x ni x H 1,wehaveAx ni Ax H.From(6) and by the demiclosed principle, we obtain SAx = Ax. Let q n := z n α n f (z n ). By Lemma.1(b) and inequality (9), we have y n p = T(I αn f )z n Tp zn p α n f (z n ) z n p + α n f (z n ), α n f (z n ) z n + p = z n p + α n f (z n ), p z n +α n f (z n ) x n p +α n f (zn ), p z n +α n f (z n ). (7) From the definition of x n+1,(7) and using the monotonicity of f,wehave x n+1 p λ n x n p +(1 λ n ) y n p λ n x n p +(1 λ n ) { x n p +α n f (zn ), p z n +αn f (z n ) } x n p +(1 λ n )α n f (zn ), p z n +(1 λn )αn f (zn ) = x n p +(1 λ n )α n f (zn ) f(p)+f(p), p z n +(1 λ n )αn f (z n ) = x n p +(1 λ n )α n f (zn ) f(p), p z n +(1 λ n )α n f (p), p zn +(1 λn )αn f (z n ) = x n p (1 λ n )α n f (p) f(zn ), p z n

13 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page 13 of 16 +(1 λ n )α n f (p), p zn +(1 λn )αn f (zn ) = x n p +(1 λ n )α n f (p), p zn +(1 λn )αn f (zn ), (8) which is equivalent to (1 λ n ) f (p), z n p ( xn p x n+1 p ) +(1 λ n )α n f (z n ) α n { } ( xn p + x n+1 p )( x n p x n+1 p ) +(1 λ n )α n f (z n ) α n M 1 ( xn p x n+1 p α n M 1 ( xn x n+1 α n ) +(1 λ n )α n f (zn ) ) +(1 λ n )α n f (zn ), where M 1 = sup{ x n p + x n+1 p, n 1} <. Taking limit of both sides and taking into account that 0 < (1 λ n )<1,α n 0, x n x n+1 = o(α n )andz ni x,wehave f (p), x p 0, and thus f (p), x p 0 for all p Fix(T), that is, x Fix(T) solves(3). Since α n = o(βn ), we may assume that α n βn for all n 1. From (13), for all n 1, we have (1 λ n )γ ( γ A ) ( S(I β n h) I ) Ax n x n p x n+1 p + αn f (zn ) {( x n p + x n+1 p )( x n p x n+1 p )} + αn f (zn ) M x n x n+1 + αn f (zn ), where M = sup{ x n p + x n+1 p : n 1} <. Therefore, for all n 1, we have (1 λ n )γ ( γ A ) Ax n Sv n β n x n x n+1 β n M + α n f (zn ) β n x n x n+1 α n M + α n f (z n ). Subsequently, since α n 0, x n+1 x n = o(α n ), γ ( γ A )>0and0<(1 λ n )<1,we have lim Ax n Sv n β n =0. (9)

14 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page 14 of 16 For all n 1, by Lemma.1(b) and the monotonicity of h,wecompute Sv n SAp This gives v n Ap Axn β n h(ax n ) Ap Ax n Ap + β n h(ax n ), β n h(ax n ) Ax n + Ap Ax n Ap +β n h(axn ), Ap Ax n +β h(axn n ) Ax n Ap β n h(ap) h(axn ), Ap Ax n +βn h(axn ) +β n h(ap), Ap Axn Ax n Ap +βn h(axn ) +β n h(ap), Ap Axn. (30) h(ap), Ax n Ap { Axn Ap Sv n SAp } +β n h(ax n ) β n { } ( Axn Ap Sv n SAp )( Ax n Ap + Sv n SAp ) +β h(axn n ) β n β n ( ) Axn Sv n M 4 +β h(axn n ), where M 4 := sup{ Ax n Ap + Sv n SAp : n 1} <. From(9), condition (iv) and Ax n Ap,weobtain h(ap), Ax Ap 0 for all Ap Fix(S), that is, Ax solves (4). Finally, it remains to show that x n x.notethat,bytheboundedness of {x n }, it suffices to show that there is no subsequence {x ni } of {x n } such that x ni y H 1 and y x. Indeed, if this is not true, then the well-known Opial theorem would imply lim x n y = lim x nj x < lim x nj y j j = lim xn y = lim xni y i < lim x ni x = lim x n y, i which leads to a contradiction. Therefore, the sequence {x n } n=1 converges weakly to a solution x Ɣ. Thus, x n y n 0and x n z n 0implythaty n x and z n x, respectively. Remark 3.1 By taking K I, the identity operator in Theorem 3.1, we can derive a weak convergence result for Algorithm 3. in which the operator T is nonexpansive cutter, but not strongly nonexpansive as in [16].

15 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page 15 of 16 Now, we illustrate Algorithm 3.1 and Theorem 3.1 by the followingexample. Example 3.1 Let H 1, H, C and T be the same as in Example..LetS : C C be defined as ( 1 S(x, y)= 10 x y, 1 10 x + 1 ) 10 y for all (x, y) C. Then S is strongly nonexpansive cutter and firmly nonexpansive, and it has a fixed point (0, 0). Thus being firmly nonexpansive, S is strongly nonexpansive. Also, every firmly nonexpansive operator with a fixed point is cutter (see [16, 0]). Thus, S is a strongly nonexpansive cutter operator. Let f, h : C C be operators defined by and ( 1 f (x, y)= 3 x 1 3 y, 1 3 x + 1 ) 3 y ( 1 h(x, y)= x 1 y, 1 x + 1 ) y for all (x, y) C, for all (x, y) C. Then f and h are monotone. Let K : C C be defined by ( 1 K(x, y)= 5 x y, 1 5 x + 1 ) 5 y for all (x, y) C. Then K is nonexpansive. Let A : C C be defined by ( 1 A(x, y)= x, 1 ) y for all (x, y) C. Then A is a bounded linear operator and A = 1 4. Table 1 Convergence table of Example 3.1 No. of iterations (n) y n x n 1 ( 0., 0.) (1, 1) (1.0e 004) ( 0.158, 0.158) (1.0e 004) (0.5000, ) 3 (1.0e 008) ( , ) (1.0e 008) (0.5000, ) 4 (1.0e 01) ( 0.447, 0.447) (1.0e 01) (0.7500, ) 5 (1.0e 016) ( , ) (1.0e 015) (0.1500, ) 6 (1.0e 019) ( 0.138, 0.138) (1.0e 019) (0.3750, ) 7 (1.0e 03) ( 0.37, 0.37) (1.0e 0) (0.115, 0.115) 8 (1.0e 06) ( , ) (1.0e 06) (0.3938, ) 9 (1.0e 030) ( 0.53, 0.53) (1.0e 09) (0.1575, ) 10 (1.0e 033) ( 0.35, 0.35) (1.0e 033) (0.7088, ) 11 (1.0e 036) ( , ) (1.0e 036) (0.5544, ) 1 (0, 0) (0, 0) 13 (0, 0) (0, 0) 14 (0, 0) (0, 0) 15 (0, 0) (0, 0) 16 (0, 0) (0, 0)

16 Ansari et al. Journal of Inequalities and Applications (015) 015:74 Page 16 of 16 Let α n = (1/n, 1/n ), β n =(1/n,1/n), γ (0, 8) and λ n (0, 1). Then the sequences x n and y n generated by Algorithm 3.1 with initial guess x 1 = (1, 1) converge to (0, 0) (see Table 1)whichisafixedpointofT and K,whereasA(0, 0) = (0, 0) which is the fixed point of S,wherex 1 =(x 1 1, x1 ). Thus, (0, 0) is the required solution. Competing interests The authors declare that they have no competing interests. Authors contributions All authors have equal contribution. Author details 1 Department of Mathematics, Aligarh Muslim University, Aligarh, India. Center for Fundamental Science, Kaohsiung Medical University, Kaohsiung, 807, Taiwan, ROC. 3 Center for Nonlinear Analysis and Optimization, Kaohsiung Medical University, Kaohsiung, 807, Taiwan, ROC. Acknowledgements In this research, the last author was partially supported by the Grant MOST M Received: 7 March 015 Accepted: 3 August 015 References 1. Censor, Y, Elfving, T: A multiprojection algorithm using Bregman projections in a product space. Numer. Algorithms 8, 1-39 (1994). Ansari, QH, Rehan, A: Split feasibility and fixed point problems. In: Ansari, QH (ed.) Nonlinear Analysis: Approximation Theory, Optimization and Applications, pp Springer, New York (014) 3. Byrne, C: Iterative oblique projection onto convex subsets and the split feasibility problem. Inverse Probl. 18, (00) 4. Byrne, C: A unified treatment of some iterative algorithms in signal processing and image reconstruction. Inverse Probl. 0, (004) 5. Censor, Y, Bortfeld, T, Martin, B, Trofimov, A: A unified approach for inversion problems in intensity-modulated radiation therapy. Phys. Med. Biol. 51, (006) 6. Ceng, L-C, Ansari, QH, Yao, J-C: An extragradient method for solving split feasibility and fixed point problems. Comput. Math. Appl. 64, (01) 7. Ceng, L-C, Ansari, QH, Yao, J-C: Mann type iterative methods for finding a common solution of split feasibility and fixed point problems. Positivity 16, (01) 8. Ceng, L-C, Wong, M-M, Yao, J-C: A hybrid extragradient-like approximation method with regularization for solving split feasibility and fixed point problems. J. Nonlinear Convex Anal. 14, (013) 9. Censor, Y, Segal, A: The split common fixed point problems for directed operators. J. Convex Anal. 16, (009) 10. Cui, H, Su, M, Wang, F: Damped projection method for split common fixed point problems. Fixed Point Theory Appl. 013, Article ID 13 (013) 11. Moudafi, A: The split common fixed-point problem for demicontractive mappings. Inverse Probl. 6, 1-6 (010) 1. Cui, H, Wang, F: Iterative method for the split common fixed point problem in Hilbert spaces. Fixed Point Theory Appl. 014, Article ID 78 (014) 13. Moudafi, A: A note on the split common fixed-point problem for quasi nonexpansive operators. Nonlinear Anal. 74, (011) 14. Kraikaew, R, Saejung, S: On split common fixed-point problems. J. Math. Anal. Appl. 415, (014) 15. Censor, Y, Gibali, A, Reich, S: Algorithms for the split variational inequality problem. Numer. Algorithms 59, (01) 16. Ansari, QH, Nimana, N, Petrot, N: Split hierarchical variational inequality problems and related problems. Fixed Point Theory Appl. 014, ArticleID08 (014) 17. Ansari, QH, Ceng, L-C, Gupta, H: Triple hierarchical variational inequalities. In: Ansari, QH (ed.) Nonlinear Analysis: Approximation Theory, Optimization and Applications, pp Springer, New York (014) 18. Minty, GJ: On the generalization of a direct method of the calculus of variations. Bull. Am. Math. Soc. 73, (1967) 19. Bruck, RE, Reich, S: Nonexpansive projections and resolvent of assertive operators in Banach spaces. Houst. J. Math. 3, (1977) 0. Cegeilski, A: Iterative Methods for Fixed Point Problems in Hilbert Spaces. Springer, New York (01) 1. Opial, Z: Weak convergence of the sequence of successive approximations for nonexpansive mappings. Bull. Am. Math. Soc. 73, (1976). Polyak, BT: Introduction to Optimization. Optimization Software Inc., New York (1987)