Strongly convergent approximations to fixed points of total asymptotically nonexpansive mappings

Strongly Convergent Approximations to Fixed Points of Total Asymptotically Nonexpansive Mappings Yakov Alber Department of Mathematics The Technion-Israel Institute of Technology 32000 Haifa, Israel. Email: [email protected] Rafa Esp´ınola∗and Pepa Lorenzo Departamento de An´alisis Matem´atico Universidad de Sevilla, P.O. Box 1160 41080-Seville, Spain. Emails: [email protected], [email protected] ∗Corresponding author Abstract In this work we prove a new strong convergence result of the regularized successive approximation method given by yn+1 =qnz0+ (1 −qn)Tnyn, n = 1,2, ..., where lim n→∞ qn= 0 and ∞ X n=1 qn=∞, for Ta total asymptotically nonexpansive mapping, i.e., Tis such that kTnx−Tnyk ≤ kx−yk+k(1) nφ(kx−yk) + k(2) n, where k1 nand k2 nare real null convergent sequences and φ:R+→R+is continuous and such that φ(0) = 0 and limt→∞ φ(t) t≤Cfor a certain constant C > 0. Among other features, our results essentially generalize existing results on strong convergence for Tnonexpansive and asymptotically nonexpansive. The convergence and stability analysis is given for both self- and nonself-mappings. Running title: Strong convergence to fixed points Key words: asymptotically nonexpansive mappings, best approximation, fixed point, duality map, iteration schemes. Mathematics Subject Classification 2000: 47A58, 47H10. 1 2 1 Introduction Iterative procedures for nonlinear operators have been largely studied by many authors in the last decades. One of the first results of this nature was obtained by Browder [5] for nonexpansive self-mappings defined on Hilbert spaces. Here Browder studied the iterative method: xω=ωz0+ (1 −ω)T xω.(1.1) for Ω a closed and convex subset of H,z0∈Ω an arbitrary point and T: Ω →Ω a nonexpansive mapping with nonempty fixed point set N(T) := {x∈Ω : T x =x}. In [5], Browder proved that lim ω→0xωexists and is a fixed point of T. This result was extended by Reich [17] to the case when Xis a uniformly smooth Banach space. Furthermore, he showed that the fixed point set of Tis a sunny nonexpansive retract of Ω. The recursive formula (explicit scheme) y1∈Ω, yn+1 =qnz0+ (1 −qn)Tyn, n = 1,2, ..., (1.2) was introduced by Halpern [12] who discussed its convergence in the framework of Hilbert spaces. Later it has been investigated in [12, 18, 19, 20] with different additional properties on the sequence {qn}, the operator Tand the space X. Browder’s and Halpern’s iterative procedures have motivated different schemes to find fixed points of asymptotically nonexpansive mappings (see Remark 1.2 for definition). In this way, T.C. Lim and H.K. Xu [15] studied the algorithm for Tasymptotically nonexpansive which generates the sequence (implicit scheme) xn=qnz0+ (1 −qn)Tnxn.(1.3) They showed that the sequence {xn}converges strongly to a fixed point of Tin the framework of a uniformly smooth Banach space, under suitable conditions on the coefficients. Very recently, in [6], the strong convergence of the explicit scheme given by y1∈Ωyn+1 =qnz0+ (1 −qn)Tnyn, n = 1,2, ..., (1.4) where z0∈Ω,and lim n→∞ qn= 0 and ∞ X n=1 qn=∞,(1.5) has been studied in uniformly smooth spaces. It is worthwhile to point out that the convergence of the implicit scheme given by (1.3) is an important tool in order to prove the strong convergence of explicit schemes as (1.2). In this paper we will consider the class of the total asymptotically nonexpansive mappings which have been introduced very recently in [2]. Definition 1.1 (cf. [2]) A mapping T: Ω →Ωis called total asymptotically nonexpansive if there exist nonnegative real sequences {k(1) n}and {k(2) n}with k(1) n, k(2) n→0as n→ ∞, and a continuous function φ:R+→R+with φ(0) = 0 such that kTnx−Tnyk ≤ kx−yk+k(1) nφ(kx−yk) + k(2) n.(1.6) 3 Remark 1.2 If φ(λ)≡0then (1.6) takes the form kTnx−Tnyk ≤ kx−yk+k(2) n. Hence, if Ωis a bounded set and TNis continuous for some integer N≥1the mapping T is of asymptotically nonexpansive type. If φ(λ) = λthen we can write kTnx−Tnyk ≤ (1 + k(1) n)kx−yk+k(2) n. In addition, if k(2) n= 0 for all n≥1then we obtain the definition of asymptotically nonexpansive mapping: kTnx−Tnyk ≤ knkx−yk, kn→1. If k(1) n= 0 and k(2) n= 0 for all n≥1then we obtain the class of nonexpansive mappings: kTx −Tyk ≤ kx−yk. If k(2) 1= 0 then it follows from (1.6) that Tis uniformly continuous, however, it can be uniformly continuous even if k(2) 16= 0. To construct the strong convergent approximations to solutions of the equation Tx =x(1.7) with a total asymptotically nonexpansive mapping T, we apply the iterative scheme given by (1.4). For nonexpansive operators T, the algorithm (1.4) is written down in the following form: y1∈Ω, yn+1 =qnz0+ (1 −qn)Tyn, n = 1,2, ..., (1.8) where lim n→∞ qn= 0. We show next that (1.8) is the regularized successive approximation method for (1.7). As it is known, equation (1.7) is equivalent to Ax = 0 (1.9) with the accretive operator A=I−T: Ω →Ω.That is, in this case hAx −Ay, J(x−y)i ≥ 0,(1.10) where Jstands for the normalized duality map. If x∗is a solution of (1.7) then Ax∗= 0. In the sequel, we assume that the fixed point set N(T) of Tis not empty. We emphasize that the problem (1.7) belongs to the class of ill-posed problems (for more on ill-posted problems see [4]). Strongly convergent approximations to x∗can be obtained only by using some regularization procedure. Let ωbe a parameter such that 0 < ω < 1 and ω→1.Obviously, if x∗is a solution of (1.9) then it is solution of the equation ωAx = 0 (1.11) 4 for any fixed ω > 0.Using the general theory (see for example [4], Section 2.7), construct for (1.11) the operator regularization method with regularization parameter α= 1−ω→0, namely, ωAx + (1 −ω)(x−z0) = 0,(1.12) where z0∈Ω.It is easy to see that (1.12) is equivalent to x= (1 −ω)z0+ωT x. (1.13) Denote Tωx= (1 −ω)z0+ωT x. Since Ω is convex and closed, we have that Tω: Ω →Ω,and (1.13) can be rewritten as x=Tωx. (1.14) Consequently, by Banach Contraction Principle, equation (1.12) has a unique solution xωand the successive approximation method x1∈Ωxn+1 = (1 −ω)z0+ωT xn converges strongly to xω.Let Xbe uniformly smooth and ωk→1 as k→ ∞.Consider now the regularized equation ωkAx + (1 −ωk)(x−z0) = 0 (1.15) with kfixed and denote by xkits unique solution. Then there exists ¯x∗∈ N(T) such that xk→¯x∗as k→ ∞.Moreover (see [4]), ¯x∗satisfies the inequality h¯x∗−z0, J(¯x∗−x∗)i ≥ 0∀x∗∈ N(T). It can be shown in the same way that (1.8) with Tnin place of Tis the regularized successive approximation method for the equation (1.7) with total asymptotically nonexpansive mapping. Indeed, if total asymptotically nonexpansive mappings are considered in place of nonexpansive mappings, then hAnx−Any, J(x−y)i ≥ −k(1) nφ(kx−yk)kx−yk − k(2) nkx−yk,(1.16) where An=I−Tn.It is clear that the analysis of strong convergence is more difficult in this situation, moreover, very little is known about the structure of the solution set. In particular the same holds for asymptotically nonexpansive mappings for which (1.16) is hAnx−Any, J(x−y)i ≥ −k(1) nkx−yk2.(1.17) The main result of this paper, Theorem 3.1 in Section 3, states a strong convergence result for the iterative scheme (1.4) in reflexive Banach spaces with a weakly continuous duality map on unbounded domains. Notice that it is an open question wether a reflexive Banach space admitting a weakly sequentially continuous duality mapping is uniformly smooth. An implicit scheme convergence result is also proved. This result is used to guarantee the existence of sunny nonexpansive retractions. In Section 4 we study the same iterative scheme for total asymptotically nonexpansive nonself-mapping. Finally, in Section 5, our last section we investigate the stability problem for iterative schemes with respect to perturbations of constraint sets for nonexpansive nonself-mappings. 5 2 Preliminaries Let Xbe a real Banach space with norm k·k,let X∗be its dual space with the norm k · k∗ and, as usual, denote the duality pairing of Xand X∗by hϕ, xi,where x∈Xand ϕ∈X∗ (in other words, hϕ, xiis the value of ϕat x). It is said that Xis uniformly smooth if for any given ε > 0,there exists δ > 0 such that for all x, y ∈Xwith kxk= 1 and kyk ≤ δ, the inequality 2−1(kx+yk+kx−yk)−1≤εkyk holds. The function ρX(τ) = sup{2−1(kx+yk+kx−yk)−1 : kxk= 1,kyk=τ} is called the modulus of smoothness of the space X. This function is increasing and approaches to zero as τ→0.Denote hX(τ) = ρX(τ) τ. Observe that the space Xis uniformly smooth if and only if lim τ→0hX(τ) = 0. Let ψ: [0,∞)→[0,∞) be a continuous strictly increasing function such that ψ(t)→ ∞ as t→ ∞ and ψ(0) = 0. The generalized duality mapping Jψ:X→2X∗associated to a gauge function ψis defined as Jψ(x) = {x∗∈X∗:hx, x∗i=ψ(kxk)kxk,kx∗k=ψ(kxk)}, x ∈X. In the case that ψ(t) = tthen Jψ=Jwhich is the normalized duality map. We say that a Banach space Xhas a weakly continuous duality map ([5]) if there exists a gauge function ψfor which the generalized duality map Jψis single-valued and weak-to- weak* sequentially continuous. It is well-known that Jψis the subdifferential, in the sense of convex analysis, of the convex function Φ(t) = Zt 0 ψ(τ)dτ, for τ≥0, and that Jψis single-valued if and only if Xis smooth. We will need the following subdifferential inequality which is known to hold in smooth spaces: Φ(kx+yk)≤Φ(kxk) + hy, Jψ(x+y)i for any x, y ∈X. Next we introduce some definitions and auxiliary results that will be needed in the sequel. Definition 2.1 Let Xbe a Banach space and Ca nonempty closed convex subset of X. An operator T:C→Xis demiclosed (at y) if T(x) = ywhenever {xn} ⊆ Cis a sequence weakly convergent to xand T(xn)→yas n→ ∞. 6 Definition 2.2 A Banach space Xsatisfies the Opial’s condition if for each sequence {xn} in X, the relation xn* x implies that lim sup n→∞ kxn−xk<lim sup n→∞ kxn−yk for all y∈Xwith x6=y. Definition 2.3 A Banach space Xsatisfies the Generalized Gossez-Lami Dozo property (GGLD-property) if lim inf n→∞ kxnk<lim sup m→∞ lim sup n→∞ kxm−xnk whenever {xn}is a weak null sequence which is not norm convergent. The following demiclosedness principle can be found in [11]. Theorem 2.4 Let Xbe a Banach space with GGLD-property and Opial’s condition. Let Cbe a weakly compact convex subset of Xand T:C→Ca uniformly continuous mapping of asymptotically nonexpansive type. Then I−Tis demiclosed at zero. The following result is well-known (see [13] and [14]). Proposition 2.5 If in a reflexive Banach space Xthe duality mapping Jis weakly continuous then Xsatisfies GGLD-property and Opial’s condition. The next corollary follows as a consequence of this proposition and a careful reading of the original proof of Theorem 2.4. Corollary 2.6 Let Xbe a reflexive Banach space with a weakly continuous duality mapping J. Let Cbe a closed convex subset of Xand T:C→Ca uniformly continuous mapping and total asymptotically nonexpansive with bounded orbits. Then I−Tis demiclosed at zero. We will also use the concept of a sunny nonexpansive retraction [10] and, in particular, its characterization by means of the duality map in a smooth Banach space. Definition 2.7 Let Cbe a non-empty subset of a Banach space Xand Da subset of C. A mapping Q:C→Dis said to be (i) a retraction onto Dif Q2=Q; (ii) a nonexpansive retraction if it also satisfies the inequality kQx −Qyk ≤ kx−yk,for all x, y ∈C; (iii) a sunny retraction if for all x∈Cand for all 0≤t < ∞, Q(Qx +t(x−Qx)) = Qx, whenever Qx +t(x−Qx)∈C. 7 Proposition 2.8 Assume that Cis a non-empty closed convex subset of a smooth Banach space Xand Dis a subset of C. Then a nonexpansive mapping Q:C→Dis a sunny retraction if and only if for all x∈Cand for all ξ∈D, hx−Qx, J(ξ−Qx)i ≤ 0. In particular, there is at most one sunny nonexpansive retraction on D. Remark 2.9 Proposition 2.5 and Corollary 2.6 remain still valid if the normalized duality mapping Jis replaced by the duality mapping Jψwith the gauge function ψ(t).Moreover, we can use Jψto characterize sunny nonexpansive retractions in a smooth Banach space given by Proposition 2.8. Let G1and G2be nonempty closed subsets of X. The Hausdorff distance between G1 and G2is defined by the following formula: H(G1, G2) = max{sup z1∈G1 inf z2∈G2 kz1−z2k,sup z1∈G2 inf z2∈G1 kz1−z2k}. We need the following lemma [3] in order to prove the main result of Section 5. Lemma 2.10 If Xis a uniformly smooth Banach space, Ω1and Ω2are closed convex subsets of Xsuch that the Hausdorff distance H(Ω1,Ω2)≤σand QΩ1and QΩ2are the (unique) sunny nonexpansive retractions onto the subsets Ω1and Ω2,respectively, then kQΩ1x−QΩ2xk2≤16R(2r+d)hX(16LR−1σ),(2.1) where r=kxk, d = max{d1, d2}, R = 2(2r+d) + σand 1<L<1.7is the Figiel constant [1, 2, 9]. Here di=dist (θ, Ωi), i = 1,2,and θis the origin of the space X. We will often apply the following lemma on numerical recurrent inequalities. Lemma 2.11 Let {λn}and {γn}be nonnegative, {αn}be positive real numbers such that λn+1 ≤λn−αnλn+γn,∀n≥1. Let for all n > 1γn αn ≤c1and αn≤α. (2.2) Then λn≤max{λ1, K∗},where K∗= (1 + α)c1. In addition, if ∞ X 1 αn=∞and γn αn →0 then λn→0as n→ ∞. 8 3 Convergence Analysis of Successive Approximation Method The goal of this section is to prove strong convergence of the regularized successive approximation method (1.4). Let us consider the explicit scheme (1.4) given by y0∈Ω, yn+1 =qnz0+ (1 −qn)Tnyn, n = 1,2, ..., with lim n→∞ qn= 0 and ∞ X n=1 qn=∞.(3.1) Theorem 3.1 Let Ωbe a nonempty closed and convex subset of a smooth reflexive Banach space Xwith a weakly sequentially continuous duality map Jψ,T: Ω →Ωa uniformly continuous mapping which is total asymptotically nonexpansive with nonempty fixed point set N(T).Let N(T)be such that there exists a sunny nonexpansive retraction Q: Ω → N (T). Let z0∈Ωand {qn} ⊂ (0,1] a sequence satisfying (3.1). Let the sequence {yn}be generated by (1.4). Assume that lim n→∞ k(1) n+k(2) n qn = 0,(3.2) and that there exist positive constants M0and M1such that φ(λ)≤M0λfor λ≥M1. Suppose that lim nkyn−Tynk= 0, then {yn}converges strongly to the fixed point ¯x∗=Qz0 of T. Proof. Firstly we observe that the sequence {yn} ⊂ Ω because Ω is convex. Take x∗∈ N(T).It follows from (1.4) that kyn+1 −x∗k ≤ qnkz0−x∗k+ (1 −qn)kTnyn−Tnx∗k ≤qnkz0−x∗k+ (1 −qn)kyn−x∗k+k(1) nφ(kyn−x∗k) + k(2) n. Denoting λn=kyn−x∗kwe have λn+1 ≤(1 −qn)λn+qnkz0−x∗k+ (1 −qn)k(1) nφ(λn) + k(2) n.(3.3) Since φis continuous it attains its maximum Mon [0, M1]. Then it is easy to verify that for all λ∈[0,∞) φ(λ)≤M+M0λ. The inequality (3.3) is rewritten as λn+1 ≤λn−qn−(1 −qn)k(1) nM0λn+γn, where γn= (1 −qn)k(1) nM+k(2) n+qnkz0−x∗k. 9 Without loss of generality, in view of (3.2), we assume that there exist constants α∈(0,1) and M2>0 such that for all n≥1 k(1) n qn ≤M0(1 −α) 1−qn ,(3.4) and γn qn ≤αM2. Then λn+1 ≤λn−αqnλn+γn. By Lemma 2.11, we conclude that λn≤max{λ1,(1 + α)M2}. Thus, the sequence {yn−x∗}is bounded, which, clearly, implies that {yn}is a bounded sequence. Applying the subdifferential inequality to yn+1 −Qz0= (1 −qn)(Tnyn−Qz0) + qn(z0−Qz0) we deduce that Φ(kyn+1 −Qz0k)≤Φ((1 −qn)kTnyn−Qz0k) + qnhz0−Qz0, Jψ(yn+1 −Qz0)i. Since Tis total asymptotically nonexpansive and Qz0∈ N(T), we have kTnyn−Qz0k ≤ kyn−Qz0k+νn, where νn=Mk(1) n+M0k(1) nkyn−Qz0k+k(2) nis bounded and vanishes as n→ ∞. Now, since Φ is a convex and nondecreasing, for nlarge enough we have Φ(kTnyn−Qz0k)≤(1 −νn)Φ(kyn−Qz0k) + νnΦ(kyn−Qz0k+ 1) ≤Φ(kyn−Qz0k) + νnM3 for M3a suitable constant. Consequently Φ(kyn+1 −Qz0k)≤(1 −qn)Φ(kyn−Qz0k) + (1 −qn)νnM3+ (3.5) +qnhz0−Qz0, Jψ(yn+1 −Qz0)i. We claim that lim sup n→∞ hz0−Qz0, Jψ(yn−Qz0)i ≤ 0. Indeed, since the sequence {yn}is bounded and the space Xreflexive there exists a subsequence {ynk}which is weakly convergent in Ω. Let ¯ybe its weak limit. We can fix this subsequence so that lim sup n→∞ hz0−Qz0, Jψ(yn−Qz0)i= lim k→∞hz0−Qz0, Jψ(ynk−Qz0)i. 16 Proof. First of all, we note that N(QΩT) is non-empty because N(T)⊆ N (QΩT). We show next that {yn} ⊆ Ω is bounded. Take x∗∈ N(T). In view of the inequalities kyn+1 −x∗k ≤ qnkz0−x∗k+ (1 −qn)k(QΩT)nyn−(QΩT)nx∗k ≤qnkz0−x∗k+ (1 −qn)kQΩT(QΩT)n−1yn−QΩT(QΩT)n−1x∗k ≤qnkz0−x∗k+ (1 −qn)kT(QΩT)n−1yn−T(QΩT)n−1x∗k ≤qnkz0−x∗k+ (1 −qn)kyn−x∗k+k(1) nφ(kyn−x∗k) + k(2) n, we conclude, by analogy with Theorem 3.1, that there exist positive constants C, C1and ¯ Msuch that kynk ≤ C, kyn−x∗k ≤ C1and φ(kyn−x∗k)≤¯ M. Since QΩis a nonexpansive mapping, it is not difficult to verify that there exists a constant ¯ C > 0 such that k(QΩT)nynk ≤ ¯ C. It follows from (4.1) that lim n→∞ yn+1 −(QΩT)nyn= 0.(4.3) On the other hand, we can show that lim n→∞ yn+1 −(QΩT)nyn+1= 0.(4.4) Indeed, kyn+1 −(QΩT)nyn+1k≤kyn+1 −(QΩT)nynk+k(QΩT)nyn−(QΩT)nyn+1k. Due to the total asymptotical nonexpansiveness of T, we obtain k(QΩT)nyn−(QΩT)nyn+1k ≤ kT(QΩT)n−1yn−T(QΩT)n−1yn+1k ≤ kyn−yn+1k+k(1) nφ(kyn−yn+1k) + k(2) n. Now the boundedness of {(QΩT)nyn}, the fact that kyn−yn+1k → 0 and (4.3) prove (4.4). Further, we have kyn−QΩTynk ≤ kyn−yn+1k+kyn+1 −(QΩT)nynk+k(QΩT)nyn−QΩT ynk. From the uniform continuity of T, we estimate the last term in the form: kQΩT(QΩT)n−1yn−QΩTynk ≤ kT(QΩT)n−1yn−Tynk ≤ ω(k(QΩT)n−1yn−ynk). By (4.3) and (4.4), one gets lim(yn−QΩTyn) = 0. 17 Notice that QΩTis a mapping as required in Theorem 3.1 and then we can apply Theorem 3.1 to the sequence {yn}. Indeed, we write yn+1 −Qz0= (1 −qn)((QΩT)nyn−Qz0) + qn(z0−Qz0). we use the subdifferential inequality as in Theorem 3.1, and the rest of the proof follows the pattern with the only difference that we obtain that {yn}strongly converges to Qz0= ¯x∗ which is a fixed point of QΩT. 5 Stability Analysis for Nonexpansive Nonself-mappings Next we study the stability problem for iterative processes with respect to perturbations of constraint sets. Specifically, we consider a process which involves nonexpansive nonselfmappings in the following form: y1∈Ω1, yn+1 =qnz0+ (1 −qn)QΩn+1 Tyn, n = 1,2, ..., (5.1) where QΩn:X→Ωnis a sunny nonexpansive retraction, and the proximity between the original set Ω and Ωnwith n= 1,2, ... is given by the Hausdorff distance: H(Ωn,Ω) ≤σn.(5.2) Let G=∩nΩnand ¯ G= Ω ∩G6=∅. Theorem 5.1 Let Xbe a uniformly smooth Banach space which has a weakly sequentially continuous duality map Jψ. Assume that D⊂Xis a closed convex set, Ω⊂Dand Ωn⊂D, n = 1,2, ... are closed convex subsets of Xwith property (5.2). Let T:D→Xbe a nonexpansive mapping with fixed point set N(T)such that N(T)∩Ω6=∅.Take z0some point in ¯ G, x∗∈ N(T)∩Ω,{qn}a sequence in (0,1) satisfying (3.1) and a non-increasing sequence σn≤σ, such that σn→0as n→ ∞.Let the sequence {yn}be generated by (5.1). We suppose that the sequence {T yn}is bounded, lim n→∞ |qn−qn−1| qn = 0 (5.3) and lim n→∞ phX(σn) qn = 0.(5.4) Then {yn}converges strongly to the fixed point ¯x∗=Qz0of QΩT, where Q: Ω → N (QΩT) is the unique sunny nonexpansive retraction onto N(QΩT). Proof. We show first that {yn}is bounded. It is not difficult to see that kyn+1 −x∗k=kqnz0+ (1 −qn)QΩTyn−QΩTx∗+ (1 −qn)(QΩn+1 Tyn−QΩTyn)k ≤qnkz0−x∗k+ (1 −qn)kQΩTyn−QΩTx∗k+ (1 −qn)kQΩn+1 Tyn−QΩTynk. 18 Since {Tyn}is bounded and due to Lemma 2.10, there exist positive constants M5and M6 such that kQΩn+1 Tyn−QΩTynk ≤ M5qhX(M6σn). This implies kyn+1 −x∗k ≤ qnkz0−x∗k+ (1 −qn)kyn−x∗k+M5qhX(M6σn). Denoting λn=kyn−x∗kwe obtain λn+1 ≤(1 −qn)λn+qnkz0−x∗k+M5qhX(M6σn). From Lemma 2.11 it follows that {yn−x∗}is bounded. Let kyn−x∗k ≤ C1for all n. Next we evaluate the following difference: yn+1 −yn= (1 −qn)(QΩn+1 T yn−QΩnT yn−1)+(qn−qn−1)(z0−x∗) +(qn−1−qn)(QΩnTyn−1−QΩTx∗) = (1 −qn)(QΩn+1 T yn−QΩn+1 T yn−1) + (1 −qn)(QΩn+1 T yn−1−QΩnT yn−1) +(qn−qn−1)(z0−x∗)+(qn−1−qn)(QΩnTyn−1−QΩnTx∗) +(qn−1−qn)(QΩnTx∗−QΩT x∗). Using the following estimates kQΩn+1 Tyn−QΩn+1 T yn−1k ≤ kyn−yn−1k, kQΩnTyn−1−QΩnT x∗k ≤ kyn−1−x∗k, kQΩnTx∗−QΩT x∗k ≤ M7qhX(M8σn), and kQΩn+1 Tyn−1−QΩnT yn−1k ≤ M9qhX(M10σn), for suitable constants M7, ..., M10 (see Lemma 2.10), we obtain kyn+1 −ynk ≤ (1 −qn)kyn−yn−1k+ (1 −qn)M9qhX(M10σn) +|qn−qn−1|kz0−x∗k+C1) + M7qhX(M8σn). Denoting λn=kyn−yn−1kone has λn+1 ≤(1 −qn)λn+γn, where γn=|qn−qn−1|(kz0−x∗k+C1) + |qn−qn−1|M7qhX(M8σn) + (1 −qn)M9qhX(M10σn). 19 Now, by (5.3), (5.4) and Lemma 2.11, we conclude that kyn−yn−1k → 0. Next we show that lim n→∞ kyn−QΩTynk= 0.(5.5) Indeed, kyn−QΩTynk≤kyn−QΩnT yn−1k+kQΩnTyn−1−QΩT yn−1k +kQΩTyn−1−QΩT ynk ≤ kyn−QΩnTyn−1k+M7qhX(M8σn) + kyn−1−ynk, which states our claim since kyn−yn−1k → 0 and phX(M8σn)→0 as n→ ∞, and kyn−QΩnTyn−1k=qn(z0−QΩnT yn−1) which, from the boundedness of {QΩnT yn−1}, also tends to 0. Now we write yn+1 −Qz0= (1 −qn)(QΩn+1 T yn−Qz0) + qn(z0−Qz0), and apply the subdifferential inequality to Jψas in Theorem 3.1 to deduce that φ(kyn+1 −Qz0k)≤φ((1 −qn)kQΩn+1 Tyn−Qz0k) + qnhz0−Qz0, Jψ(yn+1 −Qz0)i. Since Qz0∈ N(QΩT)⊆Ω, kQΩn+1 Tyn−Qz0k ≤ kQΩn+1 T yn−QΩn+1 T Qz0k+kQΩn+1 T Qz0−QΩT Qz0k ≤ kyn−Qz0k+νn, where νn=M7phX(M8σn+1) is bounded and vanishes as n→ ∞. From (5.4) we can follow the same reasoning as in Theorem 3.1 to obtain (for nlarge enough) φ(kyn+1 −Qz0k)≤(1 −qn)φ(kyn−Qz0k) + (1 −qn)νnM11+ (5.6) +qnhz0−Qz0, Jψ(yn+1 −Qz0)i. for M11 a suitable constant. We claim that lim sup n→∞ hz0−Qz0, Jψ(yn−Qz0)i ≤ 0.(5.7) Since {yn}is bounded there exists a subsequence {ynk}, ynk∈Ωnkfor each k, which weakly converges to some point ¯yand lim sup n→∞ hz0−Qz0, Jψ(yn−Qz0)i= lim k→∞hz0−Qz0, Jψ(ynk−Qz0)i. Now we apply the following result: 20 Lemma 5.2 [16] If the set Ωis convex and closed set in reflexive Banach space Xand the sequence of sets Ωn⊆Xsatisfy the limit relation H(Ωn,Ω) →0as n→ ∞ then every weak limit point uof any sequence {un}, un∈Ωn,belongs to the subset Ω. To deduce that ¯y∈Ω.The rest of the proof follows the pattern of Theorem 3.1 once we prove that ¯yis a fixed point of QΩT. To prove this we make use of Opial’s condition used in the following form: lim inf l→∞ kynl−¯yk<lim inf l→∞ kynl−QΩT¯yk ≤lim inf l→∞ (kynl−QΩTynlk+kQΩT ynl−QΩT¯yk) ≤lim inf l→∞ kQΩTynl−QΩT¯yk ≤ lim inf l→∞ kynl−¯yk. Remark 5.3 Notice that in this theorem we do not obtain convergence to a fixed point of T. Convergence to a fixed point of Tcan be obtained if we impose certain boundary conditions on Tas, for instance, that T(∂Ω) ⊆Ω. It is not hard to see that in this case N(QΩT) = N(T). Remark 5.4 The sunny nonexpansive retraction onto N(QΩT)always exists in this case since QΩTis a nonexpansive self-mapping [8]. Remark 5.5 If the condition ¯ G= Ω∩G6=∅does not hold we prove the previous theorem in the following way. Instead of Ωnwe consider the collection of sets Ω0 n=co(Ω ∪Ωn), where co(A)stands for the closed convex closure of a set A. It is easy to see that H(Ω,Ω0 n)→0 as n→ ∞. Now it suffices to follow the same proof. Remark 5.6 The sequence {T yn}is bounded if, for instance, {yn}is bounded. This obviously holds if, for instance, Ωis bounded. References [1] Alber, Ya.I.: Generalized projection operators in Banach spaces: properties and applications. Functional Differential Equations 1, 1-21 (1994). 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