Modified semi-implicit midpoint rule for nonexpansive mappings

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1 Yao et al. Fixed Point Theory and Applications 015) 015:166 DOI /s R E S E A R C H Open Access Modified semi-implicit midpoint rule for nonexpansive mappings Yonghong Yao 1, Naseer Shahzad * and Yeong-Cheng Liou 3,4 * Correspondence: nshahzad@kau.edu.sa Operator Theory and Applications Research Group, Department of Mathematics, King Abdulaziz University, P.O. Box 8003, Jeddah, 1589, Saudi Arabia Full list of author information is available at the end of the article Abstract The purpose of the paper is to construct iterative methods for finding the fixed points of nonexpansive mappings. We present a modified semi-implicit midpoint rule with the viscosity technique. We prove that the suggested method converges strongly to a special fixed point of nonexpansive mappings under some different control conditions. Some applications are also included. MSC: 47J5; 47N0; 34G0; 65J15 Keywords: modified semi-implicit midpoint rule; nonexpansive mapping; strong convergence 1 Introduction The implicit midpoint rule is one of the powerful numerical methods for solving ordinary differential equations and differential algebraic equations. For related works, please refer to [1 9]. For the ordinary differential equation x = f t), x0) = x 0, 1.1) the implicit midpoint rule generates a sequence {x n } by the recursion procedure ) x n+1 = x n + hf, n 0, 1.) where h > 0 is a stepsize. It is known that if f : R N R N is Lipschitz continuous and sufficiently smooth, then the sequence {x n } generated by 1.) converges to the exact solution of 1.1)ash 0uniformlyovert [0, t] for any fixed t >0. If we write the function f in the form f t)=gt) t, then differential equation 1.1) becomes x = gt) t. Then the equilibrium problem associated with the differential equation is the fixed point problem t = gt). Based on the above fact, Alghamdi et al. [10] presented the following semi-implicit midpoint rule for nonexpansive mappings: ) x n+1 =1 α n )x n + α n T, n 0, 1.3) 015 Yao 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 authors) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Yao et al. Fixed Point Theory and Applications 015) 015:166 Page of 15 where α n 0, 1) and T : H H is a nonexpansive mapping. They proved the weak convergence of 1.3) under some additional conditions on {α n }. Furthermore, in [11], Xu et al. used contractions to regularize the semi-implicit midpoint rule 1.3) and presented the following viscosity implicit midpoint rule for nonexpansive mappings: ) x n+1 = α n Qx n )+1 α n )T, n 0, 1.4) where α n 0, 1) and Q is a contraction. Xu et al. [11] showed the following strong convergence theorem. Theorem 1.1 Let H be a Hilbert space, C be a nonempty, closed, and convex subset of H, and T : C C be a nonexpansive mapping such that FixT). Let Q : C Cbea contraction with coefficient α [0, 1). Assume that the sequence {α n } satisfies the following three restrictions: C1): lim α n =0; C): n=0 α n = ; C3): either n=0 α α n+1 α n < or lim n+1 α n =1. Then the sequence {x n } generated by 1.4) converges in norm to a fixed point q of T, which is also the unique solution of the variational inequality I Q)q, x q 0, x FixT). 1.5) In other words, q is the unique fixed point of the contraction P FixT) Q, that is, P FixT) Qq)=q. Remark 1. The usefulness of 1.4) is that it can be used to find a periodic solution of the time-dependent nonlinear evolution equation see [11]) du + At)u = gt, u), t 0, dt where At) is a family of closed linear operators in a Hilbert space H and g maps R 1 H into H. Remark 1.3 Note that the proof of Theorem 1.1 in [11] is technical. However, Step 6 in the proof of Theorem 1.1 is also complicated. Fixed point method has attracted so much attention. Now we briefly recall some related historic approaches. Browder [1] introduced an implicit scheme as follows. Fix u C and, for each t 0, 1), let x t be the unique fixed point in C of the contraction T t which maps C into C: T t x = tu +1 t)tx, x C.Browderprovedthats-lim t 0 x t = P FixT) u. That is, the strong limit of {x t } as t 0 + is the fixed point of T which is nearest from FixT)tou. Halpern [13], on the other hand, introduced an explicit scheme. Again fix a u C.Then with a sequence {t n } in 0, 1) and an arbitrary initial guess x 0 C,wecandefineasequence {x n } through the recursive formula x n+1 = α n u +1 α n )Tx n, n )

3 Yao et al. Fixed Point Theory and Applications 015) 015:166 Page 3 of 15 Lions [14] proved the strong convergence of 1.6) under conditions C1), C) and α n α n 1 C4): lim =0. αn It is now known that this sequence {x n } converges in norm to the same limit P FixT) u as Browder s implicit scheme if the sequence {α n } satisfies assumptions C1), C), and C3) above. Moudafi [15] presented an explicit viscosity method for nonexpansive mappings which generates a sequence {x n } through the iteration process x n+1 = α n Qx n )+1 α n )Tx n, n ) Moudafiprovedthestrong convergenceof 1.7) under conditions C1), C), and α n α n 1 C5): lim =0. α n α n 1 Refinements in Hilbert spaces and extensions to Banach spaces were obtained by Xu [16]. This technique uses strict) contractions to regularize a nonexpansive mapping for the purpose of selecting a particular fixed point of the nonexpansive mapping, for instance, the fixed point of minimal norm, or of a solution to another variational inequality. Motivated and inspired by the above work, in this paper we aim to construct a unified iterative algorithm for finding the fixed points of nonexpansive mappings. We present a modified semi-implicit midpoint rule with the viscosity technique for nonexpansive mappings. We prove that the suggested algorithm converges strongly to a special fixed point of nonexpansive mappings under some different conditions. Some applications are also included. Tools.1 Some notations Let H be a real Hilbert space with inner product, and norm,respectively.letc be a nonempty closed convex subset of H. Recall that a mapping T : C C is said to be nonexpansive if Tx Ty x y for all x, y C.WeuseFixT) to denote the set of fixed points of T. A mapping Q : C C is said to be contractive if there exists a constant α 0,1) such that Qx) Qy) α x y for all x, y C.Inthiscase,Q is called α-contraction. Let C be a nonempty closed convex subset of H. For every point x H, thereexistsa unique nearest point in C denoted by P C x such that x P C x x y, y C.

4 Yao et al. Fixed Point Theory and Applications 015) 015:166 Page 4 of 15 The mapping P C is called the metric projection of H onto C.ItiswellknownthatP C is a nonexpansive mapping and is characterized by the following property: x P C x, y P C x 0, x H, y C..1). Existing algorithm and convergence result Let C be a nonempty closed convex subset of a real Hilbert space H.LetQ : C C be an α-contraction and T : C C be a nonexpansive mapping with FixT). Algorithm.1 For given y 0 C arbitrarily, let the sequence {y n } be defined iteratively by the manner y n = α n Qy n )+1 α n )Ty n, n 0,.) where {α n } is a sequence in 0, 1). Theorem. [16]) The sequence {y n } generated by.) converges strongly to q = P FixT) Qq) provided lim α n =0. 3 Some lemmas The following demiclosedness principles for nonexpansive mappings are well known. Lemma 3.1 [17]) Let C be a nonempty closed convex subset of a Hilbert space H, and let T : C C be a nonexpansive mapping with FixT). Assume that {y n } is a sequence in Csuchthaty n x and I T)y n 0. Then x FixT). Lemma 3. [18]) Let {x n } and {y n } be bounded sequences in a Banach space E and {β n } be a sequence in [0, 1] with 0<lim inf β n lim sup β n <1.Suppose that x n+1 =1 β n )x n +β n z n for all n 0 and lim sup z n+1 z n x n+1 x n ) 0. Then lim z n x n =0. Lemma 3.3 [19]) Let {a n } n N be a sequence of nonnegative real numbers satisfying the following relation: a n+1 1 α n )a n + α n σ n + δ n, n 0, where i) {α n } n N [0, 1] and n=1 α n = ; ii) lim sup σ n 0; iii) n=1 δ n <. Then lim a n =0. 4 Main results In this section, we firstly present the following unified algorithm. Let C be a nonempty closed convex subset of a real Hilbert space H. LetT : C C be a nonexpansive mapping with FixT).LetQ : C C be an α-contraction.

5 Yao et al. Fixed Point Theory and Applications 015) 015:166 Page 5 of 15 Algorithm 4.1 For given x 0 C arbitrarily, let the sequence {x n } be generated by the manner ) x n+1 = α n Qx n )+β n x n + γ n T, n 0, 4.1) where {α n } 0, 1), {β n } [0, 1), and {γ n } 0, 1) are three sequences satisfying α n + β n + γ n =1foralln 0. Remark 4. Equation 4.1) is well defined.as a matter of fact,for fixed u C, we can define a mapping ) u + x x T u x := αqu)+βu + γ T. Then we have ) ) T u x T u y = γ u + x u + y T T γ x y. This means T u is a contraction with coefficient γ 0, 1 ). Hence, Algorithm 4.1 is well defined. Now, we show the boundedness of the sequence {x n }. Conclusion 4.3 The sequence {x n } generated by 4.1) is bounded. Proof Pick any x FixT). From 4.1), we have xn+1 x = α n Qxn ) Q x )) + α n Q x ) x ) + β n xn x ) It follows that ) ) + γ n T x α Qxn n ) Q x ) + αn Q x ) x + βn xn x ) + γ n T x α n α x n x + α n Q x ) x + β n x n x + γ n xn x γ n + xn+1 x. xn+1 x 1+β n +α 1)α n xn x α n + Q x ) x 1+β n + α n 1+β n + α n [ = 1 1 α)α ] n xn x 1 α)α n 1 + Q x ) x 1+β n + α n 1+β n + α n 1 α max{ x n x 1, Q x ) x }. 1 α

6 Yao et al. Fixed Point Theory and Applications 015) 015:166 Page 6 of 15 By induction, we can deduce x n x max{ x 0 x, 1 Q x ) x }. 1 α Hence, {x n } is bounded. This completes the proof. Now we give the following result. Theorem 4.4 The sequence {x n } generated by 4.1) converges strongly to q = P FixT) Qq) provided {α n } satisfies C1)-C3) as stated in Theorem 1.1) and {β n } satisfies β n C6): lim =0. α n Proof Set y n+1 = α n Qy n )+1 α n )T y n+y n+1 ) for all n.thenwehave x n+1 y n+1 = α n Qxn ) Qy n ) ) )) yn + y n+1 + β n x n T ) )) yn + y n+1 + γ n T T ) yn + y n+1 αα n x n y n + β n x n T + γ n x n y n + γ n x n+1 y n+1. 4.) It is obvious that {x n } and {y n } are bounded by Conclusion 4.3. Hence,wededucefrom 4.)that [ x n+1 y n α)α ] n x n y n + β n M 1 γ n =1 σ n ) x n y n + σ n β n σ n M 1, 4.3) where σ n = 1 α)α n γ n and M 1 is a constant such that sup n { x n T y n+y n+1 that lim sup β n σ n ) } M 1.Note 0. Apply Lemma 3.3 to 4.3) toconcludethat x n+1 y n+1 0. Consequently, x n q = P FixT) Qq) according to Theorem 1.1. This completes the proof. Remark 4.5 The proof of Theorem 4.4 is very simple. Remark 4.6 In 4.1), if we choose β n 0 for all n,then4.1) is reduced to 1.4). Thus, our Algorithm 4.1 includes Algorithm 1.4 as a special case, and Theorem 1.1 is also a special case of our Theorem 4.4. Next, we can define the following algorithm. Algorithm 4.7 For given y 0 C arbitrarily, let the sequence {y n } be defined iteratively by the manner y n = α n Qy n )+β n y n + γ n Ty n, n 0, 4.4) where {α n }, {β n },and{γ n } are the same sequences as stated in Algorithm 4.1.

7 Yao et al. Fixed Point Theory and Applications 015) 015:166 Page 7 of 15 Proposition 4.8 The sequence {y n } generated by 4.4) converges strongly to q = P FixT) Qq) provided lim α n =0. In fact, we can rewrite 4.4) asy n = α n 1 β n Qy n )+1 α n 1 β n )Ty n for all n. Thus,Proposition 4.8 can be deduced from Theorem.. Next we use Proposition 4.8 to show the convergence analysis of Algorithm 4.1 under other control conditions. Let the sequences {x n } and {y n } be generated by 4.1) and4.4), respectively. Note that the sequences {x n } and {y n } are all bounded. First, we have the following estimation: x n+1 y n = α n Qxn ) Qy n ) ) ) ) + β n x n y n )+γ n T Ty n It follows that α n α x n y n + β n x n y n + γ n x n y n [ x n+1 y n 1 1 α)α n 1+α n + β n [ 1 1 α)α n 1+α n + β n ] x n y n ] x n y n 1 + y n y n 1. x n+1 y n + γ n. It is easily seen that if n=0 α n = and lim y n y n 1 α n =0,thenwegetlim x n+1 y n = 0 by Lemma 3.3.Consequently,x n q = P FixT) Qq) provided lim α n =0. Next, we estimate y n y n 1.From4.4), we have y n y n 1 = α n Qyn ) Qy n 1 ) ) +α n α n 1 )Qy n 1 )+β n y n y n 1 ) +β n β n 1 )y n 1 + γ n Ty n Ty n 1 )+γ n γ n 1 )Ty n 1 αα n + β n + γ n ) y n y n 1 + α n α n 1 Qy n 1 ) + β n β n 1 y n 1 + γ n γ n 1 Ty n 1. Hence, y n y n 1 α n α n 1 Qy α n 1 α)αn n 1 ) + Ty n 1 ) + β n β n 1 yn 1 + Ty 1 α)αn n 1 ). = 0. So, we ob- α If lim n α n 1 αn tain immediately the following theorem. = lim β n β n 1 α n =0,wederivethatlim y n y n 1 α n Theorem 4.9 Assume that {α n } satisfies C1), C), and C4) and {β n } satisfies β n β n 1 C7): lim =0. αn Then the sequence {x n } generated by 4.1) converges strongly to q = P FixT) Qq).

8 Yao et al. Fixed Point Theory and Applications 015) 015:166 Page 8 of 15 Remark 4.10 Note that conditions C1), C), and C4) were presented by Lions in [14]. At the same time, C7) is different from C6). In fact, we can choose β n = β 0, 1) in C7). Next, we will give another control condition instead of C4) and C7). Theorem 4.11 Assume that {α n } satisfies C1) and C) and {β n } satisfies and C8): 0 < lim inf β n lim sup β n <1 C9): lim β n+1 β n )=0. Then the sequence {x n } generated by 4.1) converges strongly to q = P FixT) Qq). Proof From Conclusion 4.3, we can choose a constant M such that { 3 sup Qx n ) ) ) )} + n 1 β T + n x n T M. Set y n = x n+1 β n x n 1 β n for all n 0. Thus, we have y n+1 y n = x n+ β n+1 x n+1 1 β n+1 x n+1 β n x n 1 β n = α n+1qx n+1 )+1 α n+1 β n+1 )T x n+1+x n+ ) 1 β n+1 α nqx n )+1 α n β n )T x n+x n+1 ) 1 β n = α n+1 Qxn+1 ) Qx n ) ) 1 β n α [ ) )] n+1 β n+1 xn+1 + x n+ T T 1 β n+1 αn+1 + α ) )) n Qx n ) T. 1 β n+1 1 β n It follows that αn+1 y n+1 y n + α ) ) n Qx n ) T 1 β n+1 1 β n + 1 α [ n+1 β n+1 xn+1 x n + x ] n+ x n+1 1 β n+1 + αα n+1 1 β n+1 x n+1 x n. 4.5) From 4.1), we have x n+ x n+1 = α n+1 Qxn+1 ) Qx n ) ) +α n+1 α n )Qx n ) )) + β n+1 x n+1 x n )+β n+1 β n ) x n T

9 Yao et al. Fixed Point Theory and Applications 015) 015:166 Page 9 of 15 ) )) xn+1 + x n+ +1 α n+1 β n+1 ) T T ) +α n α n+1 )T αα n+1 x n+1 x n +α n+1 + α n ) Qx n ) + β n+1 x n+1 x n xn+1 x n +1 α n+1 β n+1 ) + x ) n+ x n+1 ) ) +α n + α n+1 ) T + β n+1 β n x n T. It follows that [ x n+ x n α)α ] n+1 x n+1 x n 1+α n+1 + β n+1 + α n+1 + α n ) Qxn ) ) ) + T 1+α n+1 + β n+1 + β ) n+1 β n 1+α n+1 + β x n T n+1 x n+1 x n + M α n + α n+1 + β n+1 β n ). 4.6) Substitute 4.6)into4.5)toget [ y n+1 y n 1 1 α)α ] n+1 x n+1 x n +M α n+1 + α n + β n+1 β n ). 1 β n+1 Hence, lim sup yn+1 y n x n+1 x n ) 0. This together with Lemma 3. implies that lim y n x n =0. Note that So, y n x n = x n+1 x n 1 β n. lim x n+1 x n =0. 4.7) Again, from 4.1), we have x n Tx n x n x n+1 + x n+1 Tx n x n x n+1 + α n Qx n ) Tx n + β n x n Tx n +1 α n β n ) 1 x n x n+1.

10 Yao et al. Fixed Point Theory and Applications 015) 015:166 Page 10 of 15 It follows that x n Tx n α n 1 β n Qxn ) Tx n + 3 α n β n 1 β n ) x n+1 x n. This together with C1) and 4.7)impliesthat lim x n Tx n =0. 4.8) Next, we prove that lim sup q Qq), q xn 0, 4.9) where q FixT) is the unique fixed point of the contraction P FixT) Q,thatis,q = P FixT) Qq). Since {x n } is bounded, there exists a subsequence {x ni } of {x n } such that {x ni } converges weakly to a point x and lim sup PFixT) Qq) Qq), P FixT) Qq) x n = lim i PFixT) Qq) Qq), P FixT) Qq) x ni. 4.10) By Lemma 3.1 and 4.8), we deduce x FixT). This together with.1)impliesthat lim sup PFixT) Qq) Qq), P FixT) Qq) x n = lim PFixT) Qq) Qq), P FixT) Qq) x ni i = P FixT) Qq) Qq), P FixT) Qq) x 0. Finally, we prove that x n q.from4.1), we have x n+1 q = α n Qxn ) Qq), x n+1 q + α n Qq) q, xn+1 q ) +1 α n β n ) T q, x n+1 q + β n x n q, x n+1 q α n α x n q x n+1 q + α n Qq) q, xn+1 q +1 α n β n ) 1 xn q + x n+1 q ) x n+1 q + β n x n q x n+1 q 1+β n +α 1)α n 4 + α n Qq) q, xn+1 q. x n q + 3 β n 3 α)α n x n+1 q 4

11 Yao et al. Fixed Point Theory and Applications 015) 015:166 Page 11 of 15 It follows that [ ] x n+1 q 41 α)α n 1 x n q 1+β n +3 α)α n 4α n + Qq) q, xn+1 q. 4.11) 1+β n +3 α)α n Apply Lemma 3.3 and 4.9)to4.11)todeducethatx n q. This completes the proof. Remark 4.1 Note that condition C8) has been used in a large number of references. Theorems 4.4, 4.9, and4.11 demonstrate the strong convergence of Algorithm 4.1 under different control conditions on parameters {α n } and {β n }. Our algorithm and results provide a unified framework for the class problem of algorithmic approach to the fixed point of nonlinear operators. 5 Applications 5.1 Application to variational inequalities Let C be a nonempty closed convex subset of a real Hilbert space H. LetA : H H be a single-valued monotone operator such that C doma). Now we consider the following variational inequality: Ax, x x 0, x C. 5.1) It is known that 5.1) is equivalent to the fixed point problem, for any λ >0, P C I λa)x = x. 5.) If A is Lipschitzian and α-inverse-strongly monotone, then P C I λa) isnonexpansive provided 0 < λ < α. Thus, we can get the following theorem. Theorem 5.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let A : H H be a Lipschitzian and α-inverse-strongly monotone operator. Let Q : C Cbe a contraction. Assume 5.1) is solvable. Let {x n } be a sequence generated by the manner ) x n+1 = α n Qx n )+β n x n + γ n P C I λa), n 0, 5.3) where λ 0, α) and {α n } and {β n } satisfy one of the following conditions: C1), C), C4), and C7) or C1), C), C3), and C6) or C1), C), C8), and C9). Then the sequence {x n } converges strongly to a solution x of 5.1) which is also a solution to the variational inequality I Q)x, x x 0, x A 1 0). 5. Application to hierarchical minimization Consider the following hierarchical minimization problem: min ψ 1 x), 5.4) x S 0

12 Yao et al. Fixed Point Theory and Applications 015) 015:166 Page 1 of 15 where S 0 := arg min x H ψ 0 x)andψ 0, ψ 1 are two lower semi-continuous convex functions from H to R. Assume that S 0.SetS = arg min x S0 ψ 1 x) and assume S. Assume that ψ 0 and ψ 1 are differentiable and their gradients satisfy the Lipschitz continuity conditions ψ0 x) ψ 0 y) L0 x y, ψ1 x) ψ 1 y) L1 x y 5.5) for all x, y H. Note that the Lipschitz continuity 5.5) impliesthat ψ i is 1 L i -inversestrongly monotone. Consequently, I γ i ψ i ) is nonexpansive provided 0 < γ i </L i and S 0 = FixI γ 1 ψ 0 )). The optimality condition for x S 0 to be a solution of the hierarchical minimization 5.4) is the variational inequality x S 0, ψ1 x ), x x 0, x S ) Hence, we have the following theorem. Theorem 5. Assume that the hierarchical minimization problem 5.4) is solvable. Assume 5.5) and 0<γ i </L i. Let Q : C C be a contraction. Define a sequence {x n } by the manner ) x n+1 = α n Qx n )+β n x n + γ n P S0 I λ ψ 1 ), n 0, 5.7) where {α n } and {β n } satisfy one of the following conditions: C1), C), C4), and C7) or C1), C), C3), and C6) or C1), C), C8), and C9). Then the sequence {x n } converges strongly to a solution x of 5.6) which is also a solution to the variational inequality I Q)x, x x 0, x S. 5.3 Periodic solution of a nonlinear evolution equation Consider the time-dependent nonlinear equation of evolution in H given by du + At)u = f t, u), dt t 0, 5.8) where At) is a family of closed linear operators in a Hilbert space H and f maps R 1 H into H. We assume that At)andf t, u) are periodic in t with a common period ξ >0. An interesting result on the existence of periodic solutions of equation 5.8) isdueto Browder [0]. Theorem 5.3 Suppose that At) and f t, u) are periodic in T of period ξ >0and satisfy the following assumptions: i) For each t and each pair u, v H, Re f t, u) f t, v), u v 0.

13 Yao et al. Fixed Point Theory and Applications 015) 015:166 Page 13 of 15 ii) For each t and each u DAt)), ReAt)u, u 0. iii) There exists a mild solution u of equation 5.1) on R + foreachinitialvaluev H. iv) There exists some R >0such that Re f t, u), u <0 for u = R and all t [0, ξ]. Then there exists an element v of H with v < R such that the mild solution of equation 5.8) with the initial condition u0)=v isofperiodξ. Define a mapping T : H H by Tv = uξ), v H, 5.9) where u solves 5.8)withu0) = v. Then each fixed point of T corresponds to a periodic solution of equation 5.8) with period ξ.since 1 d { ut) } ) = Re At)ut), ut) + Re f t, ut), ut) dt Re f t, ut) ), ut), we see that for any value of t in [0, ξ]forwhich ut) = R,wehave d dt { ut) } <0.Hence, uξ) R,andT maps the closed ball B := {v H : v R} into itself. At the same time, we note that T is nonexpansive. As a matter of fact, if v and v 1 are two elements of B, ut)andu 1 t) are the corresponding mild solutions, we have 1 d { ut) u 1 t) } = Re At) ut) u1 t) ), ut) u 1 t) dt + Re f t, ut) ) f t, u 1 t) ), ut) u 1 t) 0. Hence, uξ) u 1 ξ) u0) u 1 0), i.e., Tv Tv 1 v v 1. Consequently, T has a fixed point which we denote by v, and the corresponding solution u of 5.8) with the initial condition u0) = v is a desired periodic solution of 5.8) with period ξ. In other words, to find a periodic solution u of 5.8) is equivalent to finding a fixed point of T. Thus, our method is applicable to 5.8). It turns out that under one of the following conditions: C1), C), C4), and C7) or C1), C), C3), and C6) or C1), C), C8), and C9), the sequence {v n } generated by the manner ) vn + v n+1 v n+1 = α n Qv n )+β n v n + γ n T, n 0 converges strongly to a fixed point v of T, and the mild solution of 5.8) with the initial value u0) = ξ is a periodic solution of 5.8).

14 Yao et al. Fixed Point Theory and Applications 015) 015:166 Page 14 of Fredholm integral equation Consider a Fredholm integral equation of the form xt)=gt)+ t 0 F t, s, xs) ) ds, t [0, 1], 5.10) where g is a continuous function on [0, 1] and F :[0,1] [0, 1] R R is continuous. Note that if F satisfies the Lipschitz continuity condition Ft, s, x) Ft, s, y) x y, t, s [0, 1], x, y R, then equation 5.10) has at least one solution in L [0, 1] see [14]). Define a mapping T : L [0, 1] L [0, 1] by Tx)t)=gt)+ t 0 F t, s, xs) ) ds, t [0, 1]. 5.11) It is easily seen that T is nonexpansive. In fact, we have, for x, y L [0, 1], Tx Ty = = 1 Txt) Tyt) dt F t, s, xs) ) F t, s, ys) )) ds dt xs) ys) ds dt 0 xs) ys) ds = x y. This means that to find the solution of integral equation 5.10) is reduced to finding a fixed point of the nonexpansive mapping T in the Hilbert space L [0, 1]. Initiating with any function y 0 L [0, 1], define a sequence of functions {y n } in L [0, 1] by ) yn + y n+1 y n+1 = α n Qy n )+β n y n + γ n T, n 0. Then the sequence {y n } converges strongly in L [0, 1] to the solution of integral equation 5.10) under one of the following conditions: C1), C), C4), and C7) or C1), C), C3), and C6) or C1), C), C8), and C9). Competing interests The authors declare that they have no competing interests. Authors contributions All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

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