Some Results for Split Equality Equilibrium Problems in Banach Spaces

In this paper, a new algorithm for finding a common element of a split equality fixed point problem for nonexpansive mappings and split equality equilibrium problem in three Banach spaces is introduced. Also, some strong and weak convergence theorems for the proposed algorithm are proved. Finally, the main results obtained in this paper are applied to solve the split equality convex minimization problem.


Introduction
Let C be a closed convex subset of real Banach space E with the dual space E * , and let f : C × C → be a bifunction, where is the set of real numbers.The equilibrium problem (for short, (EP)) is to find u * ∈ C such that: f (u * , u) ≥ 0 for all u ∈ C.
The solutions set of the problem (EP) is denoted by EP( f ), that is, It is well known that many problems in physics, optimization, economics and other applied sciences reduce to find a solution of the problem (EP).Equilibrium problems and variational inequality problems in Hilbert spaces or Banach spaces have been extensively studied by many authors (see, for example, [1][2][3][4][5][6][7][8] and the references therein).
In order to model inverse problems in phase retrievals and medical image reconstruction [9], Censor and Elfving [10] introduced the following split feasibility problem (shortly, (SFP)) in 1994: Find u * such that u * ∈ C and g(u * ) ∈ Q, where C and Q are nonempty closed convex subsets of Hilbert spaces H 1 and H 2 , respectively, g : H 1 → H 2 is a bounded linear operator.
As a matter of fact, many problems appeared in image restoration, computer tomograph and radiation therapy treatment planing can be formulated as the problem (SFP) [11][12][13].For approximating solutions of the problem (SFP), some methods have been proposed by some authors (see, for instance [9,[14][15][16][17]. Further, Moudafi [18] developed the problem (SFP) and proposed the split equality problem as follows: Let C, Q be two nonempty closed convex subsets of real Hilbert spaces H 1 and H 2 , respectively, H 3 be a real Hilbert space, g : H 1 → H 3 and h : H 2 → H 3 be two bounded linear operators.The split equality problem (shortly, (SEP)) is as follows: It is easy to see that the problem (SEP) may reduce to the problem (SFP) when H 2 = H 3 and h is the identity mapping I on H 2 .If C and Q are the sets of nonempty fixed points of the mappings T and S on H 1 and H 2 , respectively, then, the split equality problem is called the split equality fixed point problem (shortly, (SEFP)) [19]).The set of solutions of the problem (SEFP) on T and S is denoted as follows: Based on the idea of the split feasibility problem, in 2013, Kazmi and Rizvi [20] proposed the split equilibrium problems in Hilbert spaces.
Assume that f 1 : C × C → and f 2 : Q × Q → are nonlinear bifunctions, where C and Q are closed convex subsets of H 1 and H 2 , respectively, and g : H 1 → H 2 is a bounded linear operator.The split equilibrium problem (shortly, (SEQP)) is as follows: Here, based on the ideas of the problems (SEP) and (SEQP), we consider the following so-called split equality equilibrium problem in Banach spaces: two bifunctions and g : E 1 → E 3 , h : E 2 → E 3 be two bounded linear operators.The split equality equilibrium problem (shortly, (SEEP)) is as follows: The set of solutions of the problem (SEEP) is denoted by SEEP( The problems (SFP), (SEQP) and (SEP) in Hilbert spaces have attracted the attention of many authors.Some iteration algorithms have been proposed for finding a solution of these problems (see, for instance, ref. [20][21][22][23] and the references therein).Especially, the split equality mixed equilibrium problem was investigated in [24] and the convergence results on solutions were obtained in Hilbert spaces.
The recent research on the probelms (SFP), (SEP), the split common null point problem and the split common fixed point problem have been developed in Banach spaces by some authors (see, for example, ref. [25][26][27] and the references therein).But, according to the literature, we can not find out the results on the problems (SEP) and (SEQP) in Banach spaces.
In this paper, motivated and inspired by the recent works in [20,[23][24][25][26][27]), we construct a new algorithm to find a common element of the problem (SEFP) and the problem (SEEP) for nonexpansive mappings in three Banach spaces.Also, some strong and weak convergence theorems for the proposed algorithm are proved.Finally, our main results are applied to study the convergence of solutions of a split equality convex minimization problem.

Preliminaries
In this paper, we denote the strong convergence and weak convergence of a sequence {x n } to a point x ∈ E by x n → x and x n x, respectively.Let E be a real normed linear space and C be a nonempty closed convex subset of E .A mapping T : C → C is said to be nonexpansive if If C is a bounded closed convex subset of a uniformly convex Banach space E and T : C −→ C is the nonexpansive, then the fixed point set Fix(T ) is nonempty (see [28] for more details).
Let E be a real Banach space with the dual space E * .The normalized duality mapping J from E to 2 E * is defined by where •, • denotes the generalized duality pairing between E and E * .Note that, by the Hahn-Banach theorem, J (x) is nonempty (see [28] for more details) and, if E := H is a Hilbert space, then J is the identity mapping on E .Proposition 1 ([28,29]).Assume that E is a Banach space and J is the normalized duality mapping from E into E * .Then 1.If E is reflexive, strictly convex and smooth Banach space, then J is single-valued, one-to-one and surjective, and J −1 : E * → E is the inverse of J . 2. If E is a uniformly smooth Banach spaces, then J is uniformly norm-to-norm continuous on each bounded subset of E .
The normalized duality mapping J is said to be weakly sequentially continuous if the weak convergence of a sequence {x n } to x ∈ E implies the weak * convergence of {J (x n )} to J (x) in E * .Definition 2. Let C be a nonempty closed convex subset of a Banach space E .The mapping A : C → E is said to be:

accretive if
Ax − Ay, J (x − y) ≥ 0 for all x, y ∈ C; 2. strongly accretive if there exists a constant c > 0 such that Ax − Ay, J (x − y) ≥ c x − y 2 for all x, y ∈ C; 3. α-inverse strongly accretive if there exists a constant α > 0 such that Ax − Ay, J (x − y) ≥ α Ax − Ay 2 for all x, y ∈ C.
For solving the equilibrium problem, we assume that the bifunction f : C × C → satisfies the following conditions: (C4) for all x ∈ C, the function y −→ f (x, y) is convex and lower semi-continuous.

Lemma 1 ([2]
).Let C be a closed convex subset of a smooth, strictly convex and reflexive Banach space E and f : C × C → be a bifunction satisfying (C1)-(C4).For any r > 0 and x ∈ E , there exists z ∈ C such that f (z, y) + 1 r y − z, J z − J x ≥ 0 for all y ∈ C.

Lemma 2 ([30]
).Let C be a closed convex subset of a smooth, strictly convex and reflexive Banach space E and f : C × C → be a bifunction satisfying (C1)-(C4).For any r > 0 and x ∈ E , define a mapping T F r : C → C as follows: Then the following hold: Lemma 3 ([31]).For any number r > 0, a real Banach space E is uniformly convex if and only if there exists a continuous strictly increasing function g for all u, v ∈ E with u ≤ r and v ≤ r and t ∈ [0, 1].
Let T : C → C be a mapping with Fix(T ) = ∅.T is said to be demi-closed at zero if, for any {x n } ⊂ C with x n x and x n − T x n → 0, then x = T x.A mapping T : C → C is said to be semi-compact if for any bounded sequence {x n } in C such that x n − T x n → 0, (n → ∞), there exists a subsequence {x n j } of {x n } such that {x n j } converges strongly to x * ∈ C.
A Banach space E is said to satisfy Opial's property if, for any sequence {x n } in E with x n x, for any y ∈ E with y = x, we have lim inf

Lemma 4 ([31]
).Let E be a 2-uniformly smooth Banach space with the best smoothness constants K > 0. Then the following inequality holds: x + y 2 ≤ x 2 + 2 y, J x + 2 Ky 2 for all x, y ∈ E .Lemma 5 ([32]).Let C be a nonempty closed subset of a real uniformly convex Banach space E and T : C → C be a nonexpansive mapping.Then T is demi-closed at zero.

Main Results
Throughout the rest of this paper, we always assume the following conditions are satisfied: (A) E 1 , E 2 are real uniformly convex and 2-uniformly smooth Banach spaces satisfying Opial's condition and with the best smoothness constant k satisfying 0 < k ≤ 1 √ 2 ; (B) E 3 is a smooth, reflexive and strictly convex Banach space; (C) two bounded linear operators with adjoints g * , h * , respectively.
T , S, g and h be the same as above.Let {(x n , y n )} be the iteration scheme in E 1 × E 2 defined as follows: for any (x 1 , where r ∈ (0, ∞), ( then we have the following: Proof.Since E 1 , E 2 are real uniformly convex and 2-uniformly smooth Banach spaces, E 3 is a smooth, reflexive and strictly convex Banach space, by the properties of the the normalized duality mapping J , we know that the iteration scheme ( 1) is well defined.

For 1, we divide the proof of the Conclusion 1 into four steps as follows:
Step 1. Show that the limit of the sequence { x n+1 − x 2 + y n+1 − y } 2 exists for any (x, y) ∈ Γ.In fact, taking (x, y) ∈ Γ, from Lemma 2, we know that x = T f 1 r x and y = T f 2 r y.Furthermore, we have and Because of the nonexpansiveness of S and T , using (2), (3), Lemma 3 and Lemma 4, we have where Since (x, y) ∈ Γ, we know that g(x) = h(y) and so, by ( 4) and ( 5), Let Γ n (x, y) := x n − x 2 + y n − y 2 .Then, by ( 6), we have and so, from (7), it follows that Γ n (x, y) = x n − x 2 + y n − y 2 is decreasing.So, lim n→∞ Γ n (x, y) exists.Further, it is easy to see that {x n } and {y n } are bounded.
Step 2. Show that In fact, it follows from ( 7) that Since ( for some a, b ∈ (0, 1), by ( 8), we have lim and lim Applying the properties of g 1 , g 2 , (9) and Lemma 3, we have In addition, since by (11), we have lim Similarly, we obtain lim Again, since 12) and ( 13), it follows that Since (x, y) ∈ Γ, we have x = T f 1 r x and y = T f 2 r y.In addition, it follows from Lemma 2 that T r is firmly nonexpansive.Further, we have and So, it follows from ( 15) and ( 16) that Also, it follows from (1) and Lemma 4 that and Adding the inequalities ( 18), ( 19) and taking into account the fact that g(x) = h(y), we obtain and so Since lim n→∞ Γ n (x, y) exists, by (20), we have Step 3. Show that lim n→∞ x n − T x n = 0 and lim n→∞ y n − Sy n = 0.In fact, using the nonexpansiveness of T and S, we have and, further, By ( 14), ( 21) and ( 22), we have lim Similarly, we have lim Since it follows from ( 21) and ( 23) that lim In addition, we have and so it follows from ( 21) and ( 24) that Step 4. Show that {(x n , y n )} has the unique weak cluster points (x * , y * ) ∈ Γ.In fact, since E 1 and E 2 are reflexive, {x n } and {y n } are bounded, we may assume that {(x n , y n )} has a weak cluster points (x * , y * ).Since S and T are nonespansive, T and S are demiclosed and so, from Lemma 5, ( 25) and (26), it follows that x * ∈ F(T ) and y * ∈ F(S ).Now, we show that x * ∈ EP( f 1 ) and y * ∈ EP( f 2 ).Without loss of generality, we may suppose that the subsequence {(x n i , y n i )} of {(x n , y n )} converges weakly to (x * , y * ).Also, by (21), we know that {(u n , v n )} converges weakly to (x * , y * ).Using the uniformly norm-to-norm continuity of J 1 , it follows from (21) that lim n→∞ x n , we have From the condition (C2), we obtain for all u ∈ E 1 .Taking the limit as n → ∞ in the inequality above , it follows from the condition (C4) x * for all t ∈ (0, 1] and u ∈ E 1 .Thus we have z t ∈ E 1 and f 1 (z t , x * ) ≤ 0. Applying the conditions (C1) and (C4), it follows that that is, f 1 (z t , u) ≥ 0. As t → 0, from the condition (C3), it follows that f 1 (x * , u) ≥ 0 for all u ∈ E 1 .This means that x * ∈ EP( f 1 ).Following the same argument above, we also have y * ∈ EP( f 2 ).Since g and h are bounded linear operators, the point g(x * ) − h(y * ) is a weak cluster point of {g(u n ) − h(v n )}.Again, applying the weakly lower semi-continuous property of the norm and (10), we obtain g and so g(x * ) = h(y * ).Therefore, we have (x * , y * ) ∈ Γ.Now, we show that (x * , y * ) is the unique weak cluster point of {(x n , y n )}.Suppose that there exists another subsequence {(x n k , y n k )} of {(x n , y n )} such that {(x n k , y n k )} converges weakly to a point (p, q) with (p, q) = (x * , y * ).It is easy to see that (p, q) ∈ Γ.By Opial's properties of E 1 and E 2 , we obtain which are contradictions and so (p, q) = (x * , y * ).This completes the proof of the Conclusion 1.
Let φ : E 1 → be a proper lower semi-continuous and convex functions, ψ : E 1 → E * 1 be a continuous and β-inverse strongly accretive mapping.Define We can see that H(ξ, y) also satisfies the conditions (C1)-(C4) if f satisfies the conditions (C1)-(C4).So, the problem (EP) reduces to the problem: Find which is also called the generalized mixed equilibrium problem (shortly, (GMEP)).
The set of solutions of the problem (GMEP) is denoted by GMEP( f , ψ, φ).
In Theorem 1, putting B = I, E 2 = E 3 and J 2 = J 3 , then, by the similar proof in Theorem 1, the following result is obtained.