Some results on Rockafellar-type iterative algorithms for zeros of accretive operators

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1 Jung Journal of Inequalities and Applications 2013, 2013:255 R E S E A R C H Open Access Some results on Rockafellar-type iterative algorithms for zeros of accretive operators Jong Soo Jung * * Correspondence: jungjs@dau.ac.kr; jungjs@mail.donga.ac.kr Department of Mathematics, Dong-A University, Busan, , Korea Abstract We provide a new proof technique to obtain strong convergence of the sequences generated by viscosity iterative methods for a Rockafellar-type iterative algorithm and a Halpern-type iterative algorithm to a zero of an accretive operator in Banach spaces. By using a new method different from previous ones, the main results improve and develop the recent well-known results in this area. MSC: 47H06; 47H09; 47H10; 47J25; 49M05; 65J15 Keywords: Rockafellar proximal point algorithm; accretive operator; resolvent; zeros; nonexpansive mapping; contractive mapping; fixed points; variational inequalities; weakly continuous duality mapping 1 Introduction Let E be a real Banach space with the norm and the dual space E.Thevalueof x E at y E is denoted by y, x and the normalized duality mapping J from E into 2 E is defined by J x= { x E : x, x = x x, x = x }, x E. Recall that a possibly multivalued operator A : DA E 2 E with the domain DA and the range RAinE is accretive if, for each x i DAandy i Ax i i =1,2,thereexists a j J x 1 x 2 suchthaty 1 y 2, j 0. Here J is the normalized duality mapping. In a Hilbert space, an accretive operator is also called a monotone operator. The set of zeros of A is denoted by A 1 0, that is, A 1 0:= { z DA:0 Az }. If A 1 0, then the inclusion 0 Ax is solvable. Iterative methods have extensively been studied over the last forty years for constructions of zeros of accretive operators see, e.g., [1 7]. In particular, in order to find a zero of a monotone operator, Rockafellar [7] introduced a powerful and successful algorithm in a Hilbert space H, which is recognized as the Rockafellar proximal point algorithm: For any initial point x 0 H,asequence{x n } is generated by x n+1 = J rn x n + e n, n 0, 2013 Jung; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

2 JungJournal of Inequalities and Applications 2013, 2013:255 Page 2 of 19 where J r =I + ra 1 for all r > 0 is the resolvent of A and {e n } is an error sequence in H. Bruck [2]proposedthefollowinginaHilbertspaceH: For any fixed point u H, x n+1 = J rn u, n 0. In 1991, Güler [8] gave an example showing that Rockafellar s proximal algorithm does not converge strongly. Solodov and Svaitor [9] in 2000 proposed a modified proximal point algorithm which converges strongly to a solution of the equation 0 Ax by using the projection method. In 2000, Kamimura and Takahashi [10 12] introduced the following iterative algorithms of Halpern type [13] and Mann type [14] in Hilbert spaces and Banach spaces: For any initial point x 0, x n+1 = α n x 0 +1 α n J rn x n, x n+1 = α n x 0 +1 α n J rn x n + e n, n 0 and x n+1 = α n x n +1 α n J rn x n, x n+1 = α n x n +1 α n J rn x n + e n, n 0, where {α n } 0, 1, {r n } 0,, and {e n } is an error sequence, and obtained strong and weak convergence of sequences generated by these algorithms. Xu [15] in 2006 and Song and Yang [16] in 2009 obtained the strong convergence of the regularization method for Rockafellar s proximal point algorithm in a Hilbert space H:For any initial point x 0 H x n+1 = J rn αn u +1 α n x n + e n, n 0, where {α n } 0, 1, {e n } H and {r n } 0,. In 2012, as in [17], Zhang and Song [18] considered the following Rockafellar-type iterative algorithm 1.1 and Halpern-type iterative algorithm 1.2 for finding a zero of an accretive operator A in a uniformly convex Banach space E with a weakly continuous duality mapping J ϕ with gauge function ϕ or with a uniformly Gâteaux differentiable norm: For any initial point x 0 E and fixed point u E, x n+1 = β n x n +1 β n J rn αn u +1 α n x n, n and x n+1 = α n u + β n x n +1 α n β n J rn x n, n 0, 1.2 where the sequences {α n }, {β n } 0, 1 and {r n } 0, satisfy the conditions: i lim α n = 0; ii n=0 α n = ; iii lim sup β n <1;andivlim inf r n >0. In particular, in order to obtain strong convergence of the sequence generated by 1.2to a zero of an accretive operator A, they utilized the well-known inequality in uniformly convex Banach spaces see Xu [19]. The results of Zhang and Song [18] inabanach space with a uniformly Gâteaux differentiable norm and the corresponding results of

3 JungJournal of Inequalities and Applications 2013, 2013:255 Page 3 of 19 Song [17] are mutuallycomplementarysince Zhang and Song [18] assumed uniform convexity on the space instead of reflexivity on the space in Song [17] and relaxed the conditions 0 < lim inf β n lim sup β n <1andlim inf r n >0,lim n r n+1 r =1on sequences {β n } and {r n } in Song [17]. Yu [20] filled the gaps in the result of Zhang and Song [18] for the Halpern-type iterative algorithm 1.2 by utilizing the result on the sequence of real numbers in [21], which is of fundamental importance for the techniques of analysis. Also, Zhang and Song [18] studied the Rockafellar-type iterative algorithm 1.1 in a uniformly convex Banach space with a weakly continuous normalized duality mapping J or with a uniformly Gâteaux differentiable norm. In this paper, motivated by the above mentioned results, we consider viscosity iterative methods for the Rockafellar-type iterative algorithm 1.1 and Halpern-type iterative algorithm 1.2. By using a new method different from ones in [18, 20] which recover the gaps in [18] asin[20], we establish results on strong convergence of the sequences generated by the proposed iterative methods to a zero of an accretive operator A,whichsolvesacertain variational inequality in a uniformly convex Banach space having a weakly continuous duality mapping J ϕ with gauge function ϕ or having a uniformly Gâteaux differentiable norm. Our results improve, develop and complement the corresponding results of Song [17], Zhang and Song [18], Yu [20] and Song et al. [22]aswellasmanyexistingones. 2 Preliminaries and lemmas Let E be a real Banach space with the norm and let E be its dual. When {x n } is a sequence in E,thenx n x resp., x n x, x n x will denote strong resp., weak, weak convergence of the sequence {x n } to x. Recall that a mapping f : E E is said to be contractive mapping on E if there exists a constant k 0, 1 such that f x fy k x y, x, y E. An accretive operator A is said to satisfy the range condition if DA RI + ra for all r >0,whereI is an identity operator of E and DA denotes the closure of the domain DA ofa. An accretive operator A is called m-accretive if RI + ra=efor each r >0.IfAis an accretive operator which satisfies the range condition, then we can define, for each r > 0, a mapping J r : RI +ra DA defined by J r =I + ra 1, which is called the resolvent of A.WeknowthatJ r is nonexpansive i.e., J r x J r y x y, x, y RI + ra and A 1 0=FJ r ={x DJ r :J r x = x} for all r >0.Moreover,forr >0,t >0andx E, t J r x = J t r x + 1 t J r x, 2.1 r which is referred to as the resolvent identity see [23, 24] where more details on accretive operators can be found. The norm of E is said to be Gâteaux differentiable if x + ty x lim t 0 t exists for each x, y in its unit sphere S = {x E : x =1}. SuchanE is called a smooth Banach space. The norm is said to be uniformly Gâteaux differentiable if for y S, the limit is attained uniformly for x S.

4 JungJournal of Inequalities and Applications 2013, 2013:255 Page 4 of 19 A Banach space E is said to be uniformly convex if for all ε [0, 2], there exists δ ε >0 such that x = y = 1 implies x + y 2 <1 δ ε whenever x y ε. Let l >1andM > 0 be two fixed real numbers. Then a Banach space is uniformly convex if and only if there exists a continuous strictly increasing convex function g :[0, [0, with g0 = 0 such that λx +1 λy l λ x l +1 λ y l ω l λg x y 2.2 for all x, y B M 0 = {x E : x M},whereω l λ=λ l 1 λ+λ1 λ l. For more detail, see Xu [19]. By a gauge function we mean a continuous strictly increasing function ϕ defined on R + := [0, suchthatϕ0 = 0 and lim r ϕr=. The mapping J ϕ : E 2 E defined by J ϕ x= { f E : x, f = x f, f = ϕ x }, x E is called the duality mapping with gauge function ϕ. In particular, the duality mapping with gauge function ϕt=t denoted by J, is referred to as the normalized duality mapping. The following property of duality mapping is well known [23]: ϕ λ x J ϕ λx=sign λ J x, x E \ 0, λ R, 2.3 x where R is the set of all real numbers; in particular, J x= J x, x E.Itiswellknown that E is smooth if and only if the normalized duality mapping J is single-valued, and that in a Hilbert space H, the normalized duality mapping J is the identity. We say that a Banach space E has a weakly continuous duality mapping if there exists a gauge function ϕ such that the duality mapping J ϕ is single-valued and continuous from the weak topology to the weak topology, that is, for any {x n } E with x n x, J ϕ x n J ϕ x. For example, every l p space 1 < p < has a weakly continuous duality mapping with gauge function ϕt=t p 1 [23]. Let LIM be a continuous linear functional on l and a 1, a 2,... l.wewritelim n a n instead of LIMa 1, a 2,... LIM is said to be a Banach limit if LIM satisfies LIM = LIM n 1 = 1 and LIM n a n+1 =LIM n a n for all a 1, a 2,... l.iflim is a Banach limit, thefollowingarewellknown[25]: i for all n 1, a n c n implies LIM n a n LIM n c n, ii LIM n a n+n =LIM n a n for any fixed positive integer N, iii lim inf a n LIM n a n lim sup a n for all a 1, a 2,... l. We need the following lemmas for the proofs of our main results. Lemma 2.1 [23, 25] Let E be a real Banach space and ϕ be a continuous strictly increasing function on R + such that ϕ0 = 0 and lim r ϕr=. Define t= t 0 ϕτ dτ, t R +.

5 JungJournal of Inequalities and Applications 2013, 2013:255 Page 5 of 19 Then the following inequalities hold: kt k t, 0 < k <1, x + y x + y, j ϕ x + y, x, y E, where j ϕ x + y J ϕ x + y. In particular, if E is smooth, then one has x + y 2 x 2 +2 y, J x + y, x, y E. Lemma 2.2 [26] Let a R be a real number and let a sequence {a n } l satisfy the condition LIM n a n a for all Banach limit LIM. If lim sup a n+1 a n 0, then lim sup a n a. Lemma 2.3 [27] Let {s n } be a sequence of non-negative real numbers satisfying s n+1 1 λ n s n + λ n δ n, n 0, where {λ n } and {δ n } satisfy the following conditions: i {λ n } [0, 1] and n=0 λ n = ; ii lim sup δ n 0 or n=0 λ nδ n <. Then lim s n =0. Also, we will use the next lemma which is of fundamental importance for our proof. Lemma 2.4 [21] Let {s n } be a sequence of real numbers that does not decrease at infinity, in the sense that there exists a subsequence {s ni } of {s n } such that s ni < s ni +1 for all i 0. For every n n 0, define the sequence of integers {τn} by τn:=max{k n : s k < s k+1 }. Then {τn} n n0 is a nondecreasing sequence verifying lim τn= and, for all n n 0, thefollowingtwoestimateshold: s τn s τn+1, s n s τn+1. 3 Main results In this section, we study the convergence of the following two iterative algorithms: For an initial value x 0 C, x n+1 = β n x n +1 β n J rn αn f x n +1 α n x n, n and x n+1 = α n f x n +β n x n +1 α n β n J rn x n, n

6 JungJournal of Inequalities and Applications 2013, 2013:255 Page 6 of 19 Throughout this section, it is assumed that A : DA E 2 E is an accretive operator satisfying the range condition with A 1 0 ; C is a nonempty closed convex subset of E such that DA C r>0 RI + ra; f : C C is a contractive mapping with a constant k 0, 1; and {α n }, {β n } 0, 1 and {r n } 0, are sequences satisfying the conditions: C1 lim α n =0; C2 n=0 α n = ; C3 0 β n a <1for some a; C4 lim inf r n >0. We need the following result for the existence of solutions of a certain variational inequality. Theorem J [28, 29] Let E be a reflexive Banach space with a weakly continuous duality mapping J ϕ with gauge function ϕ. Let C be a nonempty closed convex subset of E, let T : C C be a nonexpansive mapping with FT and f : C C be a contractive mapping with a constant k 0, 1. For t 0, 1, let {x t } be the unique solution in C to the equation x t = tf x t +1 ttx t. Then {x t } converges as t 0 + strongly to a point q in FT, which solves the variational inequality I f q, Jϕ q p 0, p FT. Using Theorem J, we have thefollowingresult. Theorem 3.1 Let E be a reflexive Banach space having a weakly continuous duality mapping J ϕ with gauge function ϕ. Let {x n } be a sequence generated by 3.1 and y n = α n f x n +1 α n x n for all n 0. Let LIM be a Banach limit. If lim y n J rn y n =0, then LIM n I f q, Jϕ q x n 0, where q := lim t 0 + x t with x t being defined by x t = tf x t +1 tj r x t for each r >0. Proof Let x t be defined by x t = tf x t +1 tj r x t for 0 < t <1andr > 0. Then, since A 1 0, {x t } is bounded. In fact, for p A 1 0=FJ r forr >0,wehave x t p t f xt p +1 t Jr x t J rn p t f x t p +1 t x t p. This gives that x t p f x t p f x t f p + f p P k x t p + f p p. Thus x t p 1 f p p, t 0, 1, 1 k

7 JungJournal of Inequalities and Applications 2013, 2013:255 Page 7 of 19 and hence {x t } is bounded. Also, by Theorem J, {x t } converges as t 0 + strongly to a point in FJ r =A 1 0, which is denoted by q := lim t 0 + x t. First we show that {x n } and {y n } are bounded. Since A 1 0, wetakep A 1 0=FJ r for all r >0.From3.1 and the nonexpansivity of J rn for all n,wehave x n+1 p β n x n p +1 β n J rn y n p β n x n p +1 β n αn f x n +1 α n x n p β n x n p +1 β n [ α n f xn fp ] +1 αn x n p + α n f p p β n x n p +1 β n [ α n k x n p +1 α n x n p + α n f p p ] = f p p 1 1 β n 1 kα n xn p +1 β n 1 kα n 1 k { } f p p max x n p, 1 k { max x 0 p, } f p p. 1 k Hence {x n } is bounded. Also, for p A 1 0, we get y n p α n f xn f p +1 αn x n p + α n f p p α n k x n p +1 α n x n p + α n f p p = f p p 1 1 kα n xn p +1 kα n 1 k { } f p p max x n p, 1 k and so {y n } is bounded. Moreover, since J rn y n p y n p, it follows that {J rn y n } is bounded. Also, {f x n } is bounded. As a consequence, with the control condition C1, we get y n x n = α n f x n x n α n f x n + x n 0 n. 3.3 Since lim y n J rn y n =0,by3.1and3.3, we obtain lim x n J rn y n lim xn y n + y n J rn y n =0 3.4 and lim x n+1 J rn y n = lim β n x n J rn y n lim a x n J rn y n = Now, we show that LIM n I f q, J ϕ q x n 0, where q = lim t 0 + x t with x t being defined by x t = tf x t +1 tj r x t for each r > 0. Indeed, it follows that x t x n+1 =1 tj r x t x n+1 +t f x t x n+1.

8 JungJournal of Inequalities and Applications 2013, 2013:255 Page 8 of 19 Applying Lemma 2.1,wehave x t x n+1 1 t J r x t x n+1 + t f x t x n+1, J ϕ x t x n Along with using the resolvent identity 2.1, noting J r y n J rn y n = r J r y n + 1 r J rn y n J r y n r n r n 1 r r y n J rn y n n 1 r yn x n + x n J rn y n, we observe also that r n J r x t x n+1 J r x t J r x n + J r x n J r y n + J r y n J rn y n + J rn y n x n+1 x t x n + x n y n + 1 r yn x n + x n J rn y n + J rn y n x n+1 r n = x t x n + ε n, where ε n =1+ 1 r r n x n y n + 1 r r n x n J rn y n + x n+1 J rn y n 0asn by 3.3, 3.4and3.5, and f xt x n+1, J ϕ x t x n+1 = f x t x t, J ϕ x t x n+1 + x t x n+1 ϕ x t x n+1. Thus it follows from 3.6that x t x n+1 1 t x t x n +1 t ε n ϕ x t x n+1 + t f x t x t, J ϕ x t x n+1 + x t x n+1 ϕ x t x n Applying the Banach limit LIM to 3.7, we have LIM n xt x n+1 LIM n 1 t xt x n +1 t LIM n εn ϕ x t x n+1 + t LIM n f xt x t, J ϕ x t x n+1 + t LIM n xt x n+1 ϕ x t x n Hence, noting lim ε n = 0 and applying the property LIM n a n+1 =LIM n a n ofthe Banachlimit LIM to 3.8, we obtain LIM n xt f x t, J ϕ x t x n 1 t LIM n 1 t xt x n x t x n + LIM n xt x n ϕ x t x n

9 JungJournal of Inequalities and Applications 2013, 2013:255 Page 9 of 19 = 1 t LIM n xt x n ϕτ dτ + LIM n xt x n ϕ x t x n 1 t x t x n = LIM n xt x n ϕ x t x n ϕθ n 3.9 for some τ n satisfying 1 t x t x n θ n x t x n.sinceϕ is uniformly continuous on compact intervals on R + and x t x n θ n t x t x n 2 t f p p + x 0 p 0 t 0, 1 k we conclude from 3.9andq = lim t 0 + x t that LIM n I f q, Jϕ q x n lim sup LIM n xt f x t, J ϕ x t x n t 0 lim sup LIM n xt x n ϕ x t x n ϕθ n t 0 0. This completes the proof. By using Theorem 3.1, we establish the strong convergence of the Rockafellar-type iterative algorithm 3.1. Theorem 3.2 Let E be a uniformly convex Banach space having a weakly continuous duality mapping J ϕ with gauge function ϕ. Then the sequence {x n } generated by 3.1 converges strongly to q A 1 0, where q is the unique solution of the variational inequality I f q, Jϕ q p 0, p A Proof First, we note that by Theorem J,thereexistsasolutionq of a variational inequality I f q, Jϕ q p 0, p A 1 0, where q = lim t 0 + x t A 1 0withx t being defined by x t = tf x t +1 tj r x t for each r >0 and 0 < t <1.Fromnow,weputy n = α n f x n +1 α n x n for all n 0. 1 We know that x n p max{ x 0 p, 1 k f p p } for all n 0andallp A 1 0and {x n }, {y n }, {f x n } and {J rn y n } are bounded by the proof of Theorem 3.1. First, by using arguments similar to those of [18]withu = f x n and the inequality 2.2 l =2,λ = 1, we have 2 x n+1 q 2 x n q 2 + α n f xn q 2 1 β n 1 4 g y n J rn y n, where g :[0, [0, is a continuous strictly increasing convex function in 2.2. From the condition C3, it follows that 1 β n 1 a > 0 for all n 0and 1 a 1 4 g y n J rn y n α n f xn q 2 x n q 2 x n+1 q

10 JungJournal of Inequalities and Applications 2013, 2013:255 Page 10 of 19 In order to prove that lim x n q = 0, we consider two possible cases as in the proof of Yu [20]. Case 1. Assume that { x n q } is a monotone sequence. In other words, for n 0 large enough, { x n q } is either nondecreasing or nonincreasing. Hence { x n q } converges since { x n q } is bounded. Thus, by 3.11weobtain lim g y n J rn y n =0. Thus, from the property of the function g in 2.2, it follows that lim y n J rn y n =0. Now, we proceed with the following steps. Step 1. We know from 3.5thatlim x n+1 x n =0. Step 2. We show that lim sup I f q, J ϕ q x n 0. To this end, put a n := I f q, J ϕ q x n, n 0. Then Theorem 3.1 implies that LIM n a n 0 for any Banach limit LIM. Since{x n } is bounded, there exists a subsequence {x nj } of {x n } such that lim supa n+1 a n = lim a nj +1 a nj j and x nj z E.Thisimpliesthatx nj +1 z since {x n } is weakly asymptotically regular by Step 1. From the weak continuity of a duality mapping J ϕ,wehave and so w- lim j J ϕ q x nj +1=w- lim j J ϕ q x nj =J ϕ q z lim supa n+1 a n = lim I f q, Jϕ q x nj j +1 J ϕ q x nj =0. Then Lemma 2.2 implies that lim sup a n 0, that is, lim sup I f q, Jϕ q x n 0. Step 3. We show that lim sup I f q, J ϕ q y n 0. In fact, let {y ni } be a subsequence of {y n } such that y ni v E and lim sup I f q, Jϕ q y n = lim I f q, J q yni. i Since lim x n y n =0by3.3 in the proof of Theorem 3.1, wehavealsox ni v. From the weak continuity of J ϕ, it follows that w- lim i J ϕ q y ni =w- lim i J ϕ q x ni =J ϕ q v.

11 JungJournal of Inequalities and Applications 2013, 2013:255 Page 11 of 19 Hence, by Step 2, we have lim sup I f q, Jϕ q y n = lim I f q, Jϕ q y ni J ϕ q x ni + lim I f q, Jϕ q x ni i i = lim I f q, Jϕ q x ni i lim sup I f q, Jϕ q x n 0. Step 4. We show that lim x n q = 0. Indeed, by using 3.1, we obtain x n+1 q ϕ x n+1 q = β n x n q+1 β n J rn y n q, J ϕ x n+1 q β n x n q ϕ x n+1 q +1 β n y n q ϕ x n+1 q and so x n+1 q β n x n q +1 β n y n q Since y n q = α n f xn f q + α n f q q +1 αn x n q, by Lemma 2.1,wealsoget y n q α n f xn fq +1 αn x n q + α n f q q, Jϕ y n q α n k x n q +1 α n x n q + α n f q q, Jϕ y n q 1 1 kα n xn q + α n f q q, Jϕ y n q As a consequence, since in Lemma 2.1 is an increasing convex function with 0 = 0, by 3.12and3.13, we have x n+1 q β n x n q +1 β n y n q β n x n q +1 β n y n q β n x n q +1 β n 1 1 kα n xn q +1 β n α n f q q, Jϕ y n q = 1 1 β n 1 kα n xn q +1 β n α n f q q, Jϕ y n q Put λ n =1 β n 1 kα n and δ n = 1 I f q, Jϕ q y n. 1 k

12 JungJournal of Inequalities and Applications 2013, 2013:255 Page 12 of 19 From the conditions C1-C3 and Step 3, it follows that λ n 0, n=0 λ n = and lim sup δ n 0. Since 3.14reducesto x n+1 q 1 λ n x n q + λ n δ n, from Lemma 2.3,weconcludethatlim x n q =0andlim x n q =0. Case 2. Assume that { x n q } is not a monotone sequence. Then, we can define a sequence of integers {τn} for all n n 0 for some n 0 large enough by τn:=max { k N : k n, x k q < x k+1 q }. Clearly, {τn} is a nondecreasing sequence such that τn as n and x τn q x τn+1 q for all n n 0.Inthiscase,wederivefrom3.11that lim g y τn J rτn y τn =0. So, by the property of the function g in 2.2, we have y τn J rτn y τn =0. From 3.3, 3.4and3.5, we also have and lim x τn y τn =0, lim x τn J rτn y τn =0 lim x τn+1 J rτn y τn =0. By using the same argument as in Theorem 3.1 with {x τn }, {y τn } and {J rτn y τn },weobtain LIM n I f q, Jϕ q x τn 0. Moreover, by using the same argument as in Step 1-Step 4 of Case 1 with {x τn }, {y τn } and {J rτn y τn }, we obtain the following: Step 1 lim x τn+1 x τn =0; Step 2 lim sup I f q, J ϕ q x τn 0; Step 3 lim sup I f q, J ϕ q y τn 0; Step 4 lim x τn q =0and lim x τn+1 q =0.Hence lim x τn q =0 and lim x τn+1 q =0.

13 JungJournal of Inequalities and Applications 2013, 2013:255 Page 13 of 19 From Lemma 2.4,wehave x n q x τn+1 q. Therefore, lim x n q = 0. This completes the proof. By taking β n =0inTheorem3.2, we obtain the following result, which is an extension of Corollary3.4 of Zhang and Song [18] to the viscosity iteration method. Corollary 3.1 Let E be a uniformly convex Banach space having a weakly continuous duality mapping J ϕ with gauge function ϕ. Let {x n } be a sequence generated by x n+1 = J rn αn f x n +1 α n x n, n 0. Then {x n } converges strongly to q A 1 0, where q is the unique solution of the variational inequality Theorem 3.3 Let E be a uniformly convex Banach space having a uniformly Gâteaux differentiable norm. Then the sequence {x n } generated by 3.1 converges strongly to q A 1 0, where q is the unique solution of the variational inequality I f q, J q p 0, p A Proof We also note that by [30], there exists a solution q of the variational inequality I f q, J q p 0, p A 1 0, where q = lim t 0 + x t A 1 0withx t being defined by x t = tf x t +1 tj r x t for each r >0 and 0 < t <1.Fromnow,weputy n = α n f x n +1 α n x n for n 0. We also know that {x n }, {y n }, {J rn x n } and {f x n } are bounded by the proof of Theorem 3.1. As in the proof of Theorem 3.2, we divide the proof into several steps. We only include the differences. Step 1. By considering two cases as in the proof of Theorem 3.2, we have that lim y n J rn y n =0andlim y τn J rτn y τn =0,whereτn isasincase2 in the proof of Theorem 3.2. Step 2. 1 In the case when lim y n J rn y n = 0, we show that lim sup I f q, J q yn 0. To prove this, let a subsequence {y nj } of {y n } be such that lim sup I f q, J q yn = lim I f q, J q ynj j and y nj z for some z E.Since x t y n =1 tj r x t y n +t f x t y n,

14 JungJournal of Inequalities and Applications 2013, 2013:255 Page 14 of 19 by Lemma 2.1,wehave x t y n 2 1 t 2 J r x t y n 2 +2t f x t y n, J x t y n. Along with using the resolvent identity 2.1, noting J r y n J rn y n = r J r y n + 1 r J rn y n J r y n r n r n 1 r y n J rn y n, we observe also that r n J r x t y n J r x t J r y n + J r y n J rn y n + J rn y n y n x t y n r y n J rn y n = x t y n + ε n, r n where ε n =1+ 1 r r n y n J rn y n 0asn by Step 1 and condition C4. Putting a j t=1 t 2 ε nj 2 xt y nj + ε nj 0 j and using Lemma 2.1,weobtain x t y nj 2 1 t 2 J r x t y nj 2 +2t f x t y nj, J x t y nj 1 t 2 J r x t J r y nj + J r y nj y nj 2 +2t f x t x t, J x t y nj +2t x t y nj 2 1 t 2 x t y nj 2 + a j t +2t f x t x t, J x t y nj +2t x t y nj 2. The last inequality implies xt f x t, J x t y nj t 2 x t y nj t a jt. It follows that lim xt f x t, J x t y nj t M, 3.16 j 2 where M > 0 is a constant such that M x t y n 2 for all n 0andt 0, 1. Taking the lim sup as t 0 in3.16 and noticing the fact that the two limits are interchangeable due to the fact that J is uniformly continuous on bounded subsets of E from the strong topology of E to the weak topology of E,wehave lim sup I f q, J q yn = lim I f q, J q ynj 0. j

15 JungJournal of Inequalities and Applications 2013, 2013:255 Page 15 of 19 2 In the case when lim y τn J rτn y τn =0,byusingthesameargumentwith {y τn } and {J rτn y τn },wealsohavelim sup I f q, J q y τn 0. Step 3. 1 In the case when lim y n J rn y n =0,weconcludelim x n q =0. Indeed, by using 3.1 and applying Lemma 2.1,weobtain and x n+1 q 2 = β n x n q+1 β n J rn y n q, J x n+1 q β n x n q x n+1 q +1 β n y n q x n+1 q β n x n q 2 + x n+1 q β n y n q 2 + x n+1 q 2 2 y n q 2 = α n f x n +1 α n x n p, J y n q = α n f xn fq +1 α n x n q+α n f q q, J yn q α n k x n q +1 α n x n q y n q + α n f q q, J yn q = x n q 2 + y n q kα n + α n f q q, J yn q. 2 Thus x n+1 q 2 β n x n q 2 +1 β n y n q and y n q kα n x n q kα n Combining 3.17and3.18yields Put + 2α n 1+1 kα n f q q, J yn q x n+1 q 2 β n x n q 2 +1 β n 1 21 kα n x n q kα n = + 21 β nα n f q q, J yn q 1+1 kα n 1 1 β n 21 kα n x n q kα n 21 kα n +1 β n f q q, J yn q kα n 1 k λ n =1 β n 21 kα n and δ n = 1 I f q, J q yn. 1+1 kα n 1 k

16 JungJournal of Inequalities and Applications 2013, 2013:255 Page 16 of 19 From the conditions C1-C3 and 1 of Step 2, it follows that λ n 0, n=0 λ n = and lim sup δ n 0. Since 3.19reducesto x n+1 q 2 1 λ n x n q 2 + λ n δ n, from Lemma 2.3,weconcludethatlim x n q =0. 2 In the case when lim y τn J rτn y τn =0,byusingthesameargumentwith {x τn }, {y τn } and {J rτn y τn } and 2 of Step 2, we can obtain lim x τn q =0 and lim x τn+1 q =0. From Lemma 2.4,wehave x n q x τn+1 q. Therefore, lim x n q = 0. This completes the proof. By taking β n = 0, we also have the following. Corollary 3.2 Let E be a uniformly convex Banach space having a uniformly Gâteaux differentiable norm. Let {x n } be a sequence generated by x n+1 = J rn αn f x n +1 α n x n, n 0. Then {x n } converges strongly to q A 1 0, where q is the unique solution of the variational inequality Corollary 3.3 Let H be a Hilbert space. Assume that A : DA H 2 H is a monotone operator satisfying the range condition with A 1 0 and that C is a nonempty closed convex subset of H such that DA C r>0 RI + ra. Let {x n} be a sequence generated by 3.1. Then {x n } converges strongly to q A 1 0, whereqistheuniquesolutionofthe variational inequality I f q, q p 0, p A By taking β n =0inCorollary3.3, we also have the following. Corollary 3.4 Let H be a Hilbert space. Assume that A : DA H 2 H is a maximal monotone operator with A 1 0. Let {x n } be a sequence generated by x n+1 = J rn αn f x n +1 α n x n, n 0. Then {x n } converges strongly to q A 1 0, where q is the unique solution of the variational inequality Proof Since A is maximal monotone, A is monotone and satisfies the range condition DA H = RI +ra for all r > 0. Putting C = H in Corollary 3.3, we can obtain the desired result.

17 JungJournal of Inequalities and Applications 2013, 2013:255 Page 17 of 19 By using arguments similar to those in the proofs of Theorems 3.1, 3.2 and 3.3 and [20], we can obtain the following theorems for the Halpern-type iterative algorithm 3.2. Theorem 3.4 Let E be a reflexive Banach space having a weakly continuous duality mapping J ϕ with gauge function ϕ. Let {x n } be a sequence generated by 3.2 and LIM be a Banach limit. If lim x n J rn x n =0,then LIM n I f q, Jϕ q x n 0, where q := lim t 0 + x t with x t being defined by x t = tf x t +1 tj r x t for each r >0. Theorem 3.5 Let E be a uniformly convex Banach space having a weakly continuous duality mapping J ϕ with gauge function ϕ. Then the sequence {x n } generated by 3.2 converges strongly to q A 1 0, where q is the unique solution of the variational inequality Theorem 3.6 Let E be a uniformly convex Banach space having a uniformly Gâteaux differentiable norm. Then the sequence {x n } generated by 3.2 converges strongly to q A 1 0, where q is the unique solution of the variational inequality Corollary 3.5 Let H be a Hilbert space. Assume that A : DA H 2 H is a maximal monotone operator with A 1 0. Let {x n } be a sequence generated by 3.2. Then {x n } converges strongly to q A 1 0, where q is the unique solution of the variational inequality Remark Theorem 3.2 improves and develops Theorem 3.7 of Zhang and Song [18]inthe following aspects. a The following gaps, which authors in [18] overlooked, are corrected: there exist two subsequences {z ni } and {z nj } of {z n } satisfying and β n i g z ni J rn z ni α ni u p 2, i β n j g z nj J rnj z nj > α nj u p 2, j 0, where x n+1 = β n x n +1 β n J rn z n, z n = α n u +1 α n x n and p A 1 0. b The case of an iterative scheme z n = α n u +1 α n x n in [11,Theorem3.7]is extended to the case of a viscosity iterative scheme y n = α n f x n +1 α n x n, where f : C C is a contractive mapping with a constant k 0, 1. c We utilize the weakly continuous duality mapping J ϕ with gauge function ϕ instead of the weakly continuous normalized duality mapping J in [11, Theorem 3.7]. 2 Theorem 3.3 extends Theorem 3.8 of Zhang and Song [18]totheviscosityiterative method together with our proof, which corrects the gap in the proof of [18].

18 JungJournal of Inequalities and Applications 2013, 2013:255 Page 18 of 19 3 Theorem3.2 and Theorem 3.3 improve Theorem 3.3 and Theorem 3.4 of Yu [20], which were given without proofs, to the case of the viscosity iterative method together with our proofs. Theorem 3.3 also develops and complements Theorem 4.2 of Song [17]. In particular, the limit point q A 1 0 of the sequence {x n } in Theorem 3.3 is the unique solution of the variational inequality 3.15in comparison with [10, Theorem 4.2]. 4 Theorem 3.5 and Theorem 3.6 extend Theorems 3.1 and 3.2 of Zhang and Song [18] and Theorem3.1 and Theorem3.2 of Yu [20] to the viscosity iterative method. 5 Corollaries 3.1 and 3.2 improve the corresponding results of Zhang and Song [18] and Song et al. [22]. Corollary 3.4 also develops the corresponding results of Xu [15] and Song and Yang [16]. 6 As in [22, 31, 32], we can replace the contractive mapping f in our algorithms by the weakly contractive mapping g recall that a mapping g : C C is said to be weakly contractive [33]if gx gy x y ψ x y, x, y C,where ψ :[0,+ [0,+ is a continuous and strictly increasing function such that ψ is positive on 0, and ψ0 = 0. Competing interests The author declares that they have no competing interests. Acknowledgements The author would like to thank the anonymous referees for their valuable comments and suggestions, which improved the presentation of this manuscript. This research was supported by the Basic Science Research Program through the National Research Foundation of Korea NRF funded by the Ministry of Education, Science and Technology Received: 15 November 2012 Accepted: 3 May 2013 Published: 20 May 2013 References 1. Benavides, TD, Acedo, GL, Xu, HK: Iterative solutions for zeros of accretive operators. Math. Nachr , Bruck, RE: A strongly convergent iterative method for the solution of 0 Ux for a maximal monotone operator U in Hilbert space. J. Math. Anal. Appl. 48, Bréziz, H, Lions, PL: Produits infinis de resolvantes. Isr. J. Math. 29, Jung, JS, Takahashi, W: Dual convergence theorems for the infinite products of resolvents in Banach spaces. Kodai Math. J. 14, Jung, JS, Takahashi, W: On the asymptotic behavior of infinite products of resolvents in Banach spaces. Nonlinear Anal. 20, Reich, S: On infinite products of resolvents. Atti Accad. Naz. Lincei, Rend. Cl. Sci. Fis. Mat. Nat. 63, Rockafellar, RT: Monotone operators and the proximal point algorithm. SIAM J. Control Optim. 14, Güler, O: On the convergence of the proximal point algorithm for convex minimization. SIAM J. Control Optim. 29, Solodov, MV, Svaiter, BF: Forcing strong convergence of proximal point iterations in a Hilbert space. Math. Program., Ser. A 87, Kamimura, S, Takahashi, W: Approximating solutions of maximal monotone operators in Hilbert spaces. J. Approx. Theory 106, Kamimura, S, Takahashi, W: Iterative schemes for approximating solutions of accretive operators in Banach spaces. Sci. Math. 3, Kamimura, S, Takahashi, W: Weak and strong convergence of solutions of accretive operator inclusion and applications. Set-Valued Anal. 8, Halpern, B: Fixed points of nonexpansive maps. Bull. Am. Math. Soc. 73, Mann, WR: Mean value methods in iteration. Proc. Am. Math. Soc. 4, Xu, HK: A regularization method for the proximal point algorithm. J. Glob. Optim. 36, Song, Y, Yang, C: A note on a paper A regularization method for the proximal point algorithm. J. Glob. Optim. 43, Song, Y: New iterative algorithms for zeros of accretive operators. J. Korean Math. Soc. 46, Zhang, Q, Song, Y: Halpern type proximal point algorithm of accretive operators. Nonlinear Anal. 75, Xu, HK: Inequality in Banach spaces with applications. Nonlinear Anal. 16, Yu, Y: Convergence analysis of a Halpern type algorithm for accretive operators. Nonlinear Anal. 75, Maingé, P-E: Strong convergence of projected subgradient methods for nonsmooth and nonstrictly convex minimization. Set-Valued Anal. 16,

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