Abstract

Let {𝑇𝑖}𝑁𝑖=1 be 𝑁 strictly pseudononspreading mappings defined on closed convex subset 𝐶 of a real Hilbert space 𝐻. Consider the problem of finding a common fixed point of these mappings and introduce cyclic algorithms based on general viscosity iteration method for solving this problem. We will prove the strong convergence of these cyclic algorithm. Moreover, the common fixed point is the solution of the variational inequality (𝛾𝑓𝜇𝐵)𝑥,𝑣𝑥0,𝑣𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖).

1. Introduction

Throughout this paper, we always assume that 𝐶 is a nonempty, closed, and convex subset of a real Hilbert space 𝐻. Let 𝐵𝐶𝐻 be a nonlinear mapping. Recall the following definitions.

Definition 1.1. 𝐵 is said to be(i)monotone if 𝐵𝑥𝐵𝑦,𝑥𝑦0,𝑥,𝑦𝐶,(1.1)(ii)strongly monotone if there exists a constant 𝛼>0 such that 𝐵𝑥𝐵𝑦,𝑥𝑦𝛼𝑥𝑦2,𝑥,𝑦𝐶,(1.2) for such a case, 𝐵 is said to be 𝛼-strongly-monotone, (iii)inverse-strongly monotone if there exists a constant 𝛼>0 such that 𝐵𝑥𝐵𝑦,𝑥𝑦𝛼𝐵𝑥𝐵𝑦2,𝑥,𝑦𝐶,(1.3) for such a case, 𝐵 is said to be 𝛼-inverse-strongly monotone, (iv)𝑘-Lipschitz continuous if there exists a constant 𝑘0 such that 𝐵𝑥𝐵𝑦𝑘𝑥𝑦𝑥,𝑦𝐶.(1.4)

Remark 1.2. Let 𝐹=𝜇𝐵𝛾𝑓, where 𝐵 is a 𝑘-Lipschitz and 𝜂-strongly monotone operator on 𝐻 with 𝑘>0 and 𝑓 is a Lipschitz mapping on 𝐻 with coefficient 𝐿>0, 0<𝛾𝜇𝜂/𝐿. It is a simple matter to see that the operator 𝐹 is (𝜇𝜂𝛾𝐿)-strongly monotone over  𝐻,  that is: 𝐹𝑥𝐹𝑦,𝑥𝑦(𝜇𝜂𝛾𝐿)𝑥𝑦2,(𝑥,𝑦)𝐻×𝐻.(1.5)
Following the terminology of Browder-Petryshyn [1], we say that a mapping 𝑇𝐷(𝑇)𝐻𝐻 is(1)𝑘-strict pseudocontraction if there exists 𝑘[0,1) such that 𝑇𝑥𝑇𝑦2𝑥𝑦2+𝑘𝑥𝑇𝑥(𝑦𝑇𝑦)2,𝑥,𝑦𝐷(𝑇),(1.6)(2)𝑘-strictly pseudononspreading if there exists 𝑘[0,1) such that 𝑇𝑥𝑇𝑦2𝑥𝑦2+𝑘𝑥𝑇𝑥(𝑦𝑇𝑦)2+2𝑥𝑇𝑥,𝑦𝑇𝑦,(1.7) for all 𝑥,𝑦𝐷(𝑇),(3)nonspreading in [2] if 𝑇𝑥𝑇𝑦2𝑇𝑥𝑦2+𝑇𝑦𝑥2,𝑥,𝑦𝐶.(1.8) It is shown in [3] that (1.8) is equivalent to 𝑇𝑥𝑇𝑦2𝑥𝑦2+2𝑥𝑇𝑥,𝑦𝑇𝑦,𝑥,𝑦𝐶.(1.9)

Clearly every nonspreading mapping is 0-strictly pseudononspreading. Iterative methods for strictly pseudononspreading mapping have been extensively investigated; see [2, 46].

Let 𝐶 be a closed convex subset of 𝐻, and let {𝑇𝑖}𝑁𝑖=1 be 𝑛  𝑘𝑖-strictly pseudocontractive mappings on 𝐶 such that 𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖). Let 𝑥1𝐶 and {𝛼𝑛}𝑛=1 be a sequence in (0,1). In [7], Acedo and Xu introduced an explicit iteration scheme called the followting cyclic algorithm for iterative approximation of common fixed points of {𝑇𝑖}𝑁𝑖=1 in Hilbert spaces. They define the sequence {𝑥𝑛} cyclically by 𝑥1=𝛼0𝑥0+1𝛼0𝑇0𝑥0;𝑥2=𝛼1𝑥1+1𝛼1𝑇1𝑥1;𝑥𝑁=𝛼𝑁1𝑥𝑁1+1𝛼𝑁1𝑇𝑁1𝑥𝑁1;𝑥𝑁+1=𝛼𝑁𝑥𝑁+1𝛼𝑁𝑇0𝑥𝑁;(1.10).(1.11) In a more compact form, they rewrite 𝑥𝑛+1 as 𝑥𝑁+1=𝛼𝑁𝑥𝑁+1𝛼𝑁𝑇𝑁𝑥𝑛,(1.12) where 𝑇𝑁=𝑇𝑖, with 𝑖=𝑛 (mod 𝑁), 0𝑖𝑁1. Using the cyclic algorithm (1.12), Acedo and Xu [7] show that this cyclic algorithm (1.12) is weakly convergent if the sequence {𝛼𝑛} of parameters is appropriately chosen.

Motivated and inspired by Acedo and Xu [7], we consider the following cyclic algorithm for finding a common element of the set of solutions of 𝑘𝑖-strictly pseudononspreading mappings {𝑇𝑖}𝑁𝑖=1. The sequence {𝑥𝑛}𝑖=1 generated from an arbitrary 𝑥1𝐻 as follows: 𝑥1=𝛼0𝑥𝛾𝑓0+𝐼𝜇𝛼0𝐵𝑇𝜔0𝑥0;𝑥2=𝛼1𝑥𝛾𝑓1+𝐼𝜇𝛼1𝐵𝑇𝜔1𝑥1;𝑥𝑁=𝛼𝑁1𝑥𝛾𝑓𝑁1+𝐼𝜇𝛼𝑁1𝐵𝑇𝜔𝑁1𝑥𝑁1;𝑥𝑁+1=𝛼𝑁𝑥𝛾𝑓𝑁+𝐼𝜇𝛼𝑁𝐵𝑇𝜔0𝑥𝑁;(1.13).(1.14) Indeed, the algorithm above can be rewritten as 𝑥𝑛+1=𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝜇𝛼𝑛𝐵𝑇𝜔[𝑛]𝑥𝑛,(1.15) where 𝑇𝜔[𝑛]=(𝐼𝜔[𝑛])𝐼+𝜔[𝑛]𝑇[𝑛], 𝑇[𝑛]=𝑇𝑛mod𝑁; namely, 𝑇[𝑁] is one of 𝑇1,𝑇2,,𝑇𝑁 circularly.

2. Preliminaries

Throughout this paper, we write 𝑥𝑛𝑥 to indicate that the sequence {𝑥𝑛} converges weakly to 𝑥. 𝑥𝑛𝑥 implies that {𝑥𝑛} converges strongly to 𝑥. The following definitions and lemmas are useful for main results.

Definition 2.1. A mapping 𝑇 is said to be demiclosed, if for any sequence {𝑥𝑛}which weakly converges to 𝑦, and if the sequence {𝑇𝑥𝑛} strongly converges to 𝑧, then 𝑇(𝑦)=𝑧.

Definition 2.2. 𝑇𝐻𝐻 is called demicontractive on 𝐻, if there exists a constant 𝛼<1 such that 𝑇𝑥𝑞2𝑥𝑞2+𝛼𝑥𝑇𝑥2,(𝑥,𝑞)𝐻×𝐹𝑖𝑥(𝑇).(2.1)

Remark 2.3. Every 𝑘-strictly pseudononspreading mapping with a nonempty fixed point set 𝐹𝑖𝑥(𝑇) is demicontractive (see [8, 9]).

Remark 2.4 (see [10]). Let 𝑇 be a 𝛼-demicontractive mapping on 𝐻 with 𝐹𝑖𝑥(𝑇) and 𝑇𝜔=(1𝜔)𝐼+𝜔𝑇 for 𝜔(0,): (A1)𝑇𝛼-demicontractive is equivalent to 1𝑥𝑇𝑥,𝑥𝑞2(1𝛼)𝑥𝑇𝑥2,(𝑥,𝑞)𝐻×𝐹𝑖𝑥(𝑇),(2.2)(A2)𝐹𝑖𝑥(𝑇)=𝐹𝑖𝑥(𝑇𝜔) if 𝜔0.

Remark 2.5. According to 𝐼𝑇𝜔=𝜔(𝐼𝑇) with 𝑇 being a 𝑘-strictly pseudononspreading mapping, we obtain 𝑥𝑇𝜔𝑥,𝑥𝑞𝜔(1𝑘)2𝑥𝑇𝑥2,(𝑥,𝑞)𝐻×𝐹𝑖𝑥(𝑇).(2.3)

Proposition 2.6 (see [2]). Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻, and let 𝑇𝐶𝐶 be a 𝑘-strictly pseudononspreading mapping. If 𝐹𝑖𝑥(𝑇), then it is closed and convex.

Proposition 2.7 (see [2]). Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻, and let 𝑇𝐶𝐶 be a 𝑘-strictly pseudononspreading mapping. Then (𝐼𝑇) is demiclosed at 0.

Lemma 2.8 (see [11]). Let {𝒯𝑛} be a sequence of real numbers that does not decrease at infinity, in the sense that there exists a subsequence {𝒯𝑛𝑗}𝑗0 of {𝒯𝑛} which satisfies 𝒯𝑛𝑗<𝒯𝑛𝑗+1 for all 𝑗0. Also consider the sequence of integers {𝛿(𝑛)}𝑛𝑛0 defined by 𝛿(𝑛)=max𝑘𝑛𝒯𝑘<𝒯𝑘+1.(2.4) Then {𝛿(𝑛)}𝑛𝑛0 is a nondecreasing sequence verifying lim𝑛𝛿(𝑛)=,𝑛𝑛0; it holds that 𝒯𝛿(𝑛)<𝒯𝛿(𝑛)+1 and we have 𝒯𝑛<𝒯𝛿(𝑛)+1.(2.5)

Lemma 2.9. Let 𝐻 be a real Hilbert space. The following expressions hold: (i)𝑡𝑥+(1𝑡)𝑦2=𝑡𝑥2+(1𝑡)𝑦2𝑡(1𝑡)𝑥𝑦2,𝑥,𝑦𝐻,𝑡[0,1], (ii)𝑥+𝑦2𝑥2+2𝑦,𝑥+𝑦,𝑥,𝑦𝐻.

Lemma 2.10 (see [6]). Let 𝐶 be a closed convex subset of a Hilbert space 𝐻, and let 𝑇𝐻𝐻 be a 𝑘-strictly pseudononspreading mapping with a nonempty fixed point set. Let 𝑘𝜔<1 be fixed and define 𝑇𝜔𝐶𝐶 by 𝑇𝜔(𝑥)=(1𝜔)(𝑥)+𝜔𝑇(𝑥),𝑥𝐶.(2.6) Then 𝐹𝑖𝑥(𝑇𝜔)=𝐹𝑖𝑥(𝑇).

Lemma 2.11. Assume 𝐶 is a closed convex subset of a Hilbert space 𝐻. (a)Given an integer 𝑁1, assume, 𝑇𝑖𝐻𝐻 is a 𝑘𝑖-strictly pseudononspreading mapping for some 𝑘𝑖[0,1), (𝑖=1,2,,𝑁). Let {𝜆𝑖}𝑁𝑖=1 be a positive sequence such that 𝑁𝑖=1𝜆𝑖=1. Suppose that {𝑇𝑖}𝑁𝑖=1 has a common fixed point and 𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖). Then, 𝐹𝑖𝑥𝑁𝑖=1𝜆𝑖𝑇𝑖=𝑁𝑖=1𝐹𝑖𝑥𝑇𝑖.(2.7)(b)Assuming 𝑇𝑖𝐻𝐻 is a 𝑘𝑖-strictly pseudononspreading mapping for some 𝑘𝑖[0,1), (𝑖=1,2,,𝑁), let 𝑇𝜔𝑖=(1𝜔𝑖)𝐼+𝜔𝑖𝑇𝑖, 1𝑖𝑁. If 𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖), then 𝐹𝑖𝑥𝑇𝜔1𝑇𝜔2𝑇𝜔𝑁=𝑁𝑖=1𝐹𝑖𝑥𝑇𝜔𝑖.(2.8)

Proof. To prove (a), we can assume 𝑁=2. It suffices to prove that 𝐹𝑖𝑥(𝐹)𝐹𝑖𝑥(𝑇1)𝐹𝑖𝑥(𝑇2), where 𝐹=(1𝜆)𝑇1+𝜆𝑇2 with 0<𝜆<1. Let 𝑥𝐹𝑖𝑥(𝐹) and write 𝑉1=𝐼𝑇1 and 𝑉2=𝐼𝑇2.
From Lemma 2.9 and taking 𝑧𝐹𝑖𝑥(𝑇1)𝐹𝑖𝑥(𝑇2) to deduce that 𝑧𝑥2=(1𝜆)(𝑧𝑇1𝑥)+𝜆(𝑧𝑇2𝑥)2=(1𝜆)𝑧𝑇1𝑥2+𝜆𝑧𝑇2𝑥2𝑇𝜆(1𝜆)1𝑥𝑇2𝑥2(1𝜆)𝑧𝑥2+𝑘𝑥𝑇1𝑥2+𝜆𝑧𝑥2+𝑘𝑥𝑇2𝑥2𝑇𝜆(1𝜆)1𝑥𝑇2𝑥2=𝑧𝑥2𝑉+𝑘(1𝜆)1𝑥2𝑉+𝜆2𝑥2𝑉𝜆(1𝜆)1𝑥𝑉2𝑥2,(2.9) it follows that 𝑉𝜆(1𝜆)1𝑥𝑉2𝑥2𝑉𝑘(1𝜆)1𝑥2𝑉+𝜆2𝑥2.(2.10) Since (1𝜆)𝑉1𝑥+𝜆𝑉2𝑥=0, we obtain 𝑉(1𝜆)1𝑥2𝑉+𝜆2𝑥2𝑉=𝜆(1𝜆)1𝑥𝑉2𝑥2.(2.11)
This together with (2.10) implies that 𝑉(1𝑘)𝜆(1𝜆)1𝑥𝑉2𝑥20.(2.12) Since 0<𝜆<1 and 𝑘<1, we get 𝑉1𝑥𝑉2𝑥=0 which implies 𝑇1𝑥=𝑇2𝑥 which in turn implies that 𝑇1𝑥=𝑇2𝑥=𝑥 since (1𝜆)𝑇1𝑥+𝜆𝑇2𝑥=𝑥. Thus, 𝑥𝐹𝑖𝑥(𝑇1)𝐹𝑖𝑥(𝑇2).
By induction, we also claim that 𝐹𝑖𝑥(𝑁𝑖=1𝜆𝑖𝑇𝑖)=𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖) with {𝜆𝑖}𝑁𝑖=1 is a positive sequence such that 𝑁𝑖=1𝜆𝑖=1, (𝑖=1,2,,𝑁).
To prove (b), we can assume 𝑁=2. Set 𝑇𝜔1=(1𝜔1)𝐼+𝜔1𝑇1 and 𝑇𝜔2=(1𝜔2)𝐼+𝜔1𝑇2, 0<𝑘𝑖<𝜔𝑖<1/2, 𝑖=1,2. Obviously 𝐹𝑖𝑥𝑇𝜔1𝐹𝑖𝑥𝑇𝜔2𝐹𝑖𝑥𝑇𝜔1𝑇𝜔2.(2.13) Now we prove 𝐹𝑖𝑥𝑇𝜔1𝐹𝑖𝑥𝑇𝜔2𝐹𝑖𝑥𝑇𝜔1𝑇𝜔2,(2.14) for all 𝑞𝐹𝑖𝑥(𝑇𝜔1𝑇𝜔2) and 𝑇𝜔1𝑇𝜔2𝑞=𝑞. If 𝑇𝜔2𝑞=𝑞, then 𝑇𝜔1𝑞=𝑞; the conclusion holds. From Lemma 2.10, we can know that 𝐹𝑖𝑥(𝑇𝜔1)𝐹𝑖𝑥(𝑇𝜔2)=𝐹𝑖𝑥(𝑇1)𝐹𝑖𝑥(𝑇2). Taking 𝑝𝐹𝑖𝑥(𝑇𝜔1)𝐹𝑖𝑥(𝑇𝜔2), then 𝑝𝑞2=𝑝𝑇𝜔1𝑇𝜔2𝑞2=𝑝1𝜔1𝑇𝜔2𝑞+𝜔1𝑇1𝑇𝜔2𝑞2=1𝜔1𝑝𝑇𝜔2𝑞+𝜔1𝑝𝑇1𝑇𝜔2𝑞2=1𝜔1𝑝𝑇𝜔2𝑞2+𝜔1𝑝𝑇1𝑇𝜔2𝑞2𝜔11𝜔1𝑇𝜔2𝑞𝑇1𝑇𝜔2𝑞21𝜔1𝑝𝑇𝜔2𝑞2𝜔11𝜔1𝑇𝜔2𝑞𝑇1𝑇𝜔2𝑞2+𝜔1𝑝𝑇𝜔2𝑞2+𝑘1𝑇𝜔2𝑞𝑇1𝑇𝜔2𝑞2+2𝑝𝑇1𝑇𝜔2𝑝,𝑇𝜔2𝑞𝑇1𝑇𝜔2𝑞=1𝜔1𝑝𝑇𝜔2𝑞2𝜔11𝜔1𝑇𝜔2𝑞𝑇1𝑇𝜔2𝑞2+𝜔1𝑝𝑇𝜔2𝑞2+𝑘1𝑇𝜔2𝑞𝑇1𝑇𝜔2𝑞2𝑝𝑇𝜔2𝑞2𝜔11𝜔1𝑘1𝑇𝜔2𝑞𝑇1𝑇𝜔2𝑞2𝑝𝑞2𝜔11𝜔1𝑘1𝑇𝜔2𝑞𝑇1𝑇𝜔2𝑞2.(2.15) Since 0<𝑘1<𝜔1<1/2, we obtain 𝑇𝜔2𝑞𝑇1𝑇𝜔2𝑞20.(2.16) Namely, 𝑇𝜔2𝑞=𝑇1𝑇𝜔2𝑞, that is: 𝑇𝜔2𝑞𝐹𝑖𝑥𝑇1=𝐹𝑖𝑥𝑇𝜔1,𝑇𝜔2𝑞=𝑇𝜔1𝑇𝜔2𝑞.(2.17)
By induction, we also claim that the Lemma 2.11(b) holds.

Lemma 2.12. Let 𝐾 be a closed convex subset of a real Hilbert space 𝐻, given 𝑥𝐻 and 𝑦𝐾. Then 𝑦=𝑃𝐾𝑥 if and only if there holds the inequality 𝑥𝑦,𝑦𝑧0,𝑧𝐾.(2.18)

3. Cyclic Algorithm

In this section, we are concerned with the problem of finding a point 𝑝 such that 𝑝𝑁𝑖=1𝐹𝑖𝑥𝑇𝜔𝑖=𝑁𝑖=1𝐹𝑖𝑥𝑇𝑖,𝑁1,(3.1) where 𝑇𝜔𝑖=(1𝜔𝑖)𝐼+𝜔𝑖𝑇𝑖, {𝜔𝑖}𝑁𝑖=1(0,1/2] and {𝑇𝑖}𝑁𝑖=1 are 𝑘𝑖-strictly pseudononspreading mappings with 𝑘𝑖[0,𝜔𝑖), (𝑖=1,2,,𝑁), defined on a closed convex subset 𝐶 in Hilbert space 𝐻. Here 𝐹𝑖𝑥(𝑇𝜔𝑖)={𝑞𝐶𝑇𝜔𝑖𝑞=𝑞} is the set of fixed points of 𝑇𝑖, 1𝑖𝑁.

Let 𝐻 be a real Hilbert space, and let 𝐵𝐻𝐻 be 𝜂-strongly monotone and 𝜌-Lipschitzian on 𝐻 with 𝜌>0, 𝜂>0. Let 0<𝜇<2𝜂/𝜌2, 0<𝛾<𝜇(𝜂(𝜇𝜌2/2))/𝐿=𝜏/𝐿. Let 𝑁 be a positive integer, and let 𝑇𝑖𝐻𝐻 be a 𝑘𝑖-strictly pseudononspreading mapping for some 𝑘𝑖[0,1), (𝑖=1,2,,𝑁), such that 𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖). We consider the problem of finding 𝑝𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖) such that (𝛾𝑓𝜇𝐵)𝑝,𝑣𝑝0,𝑣𝑁𝑖=1𝐹𝑖𝑥𝑇𝑖.(3.2)

Since 𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖) is a nonempty closed convex subset of 𝐻, VI (3.2) has a unique solution. The variational inequality has been extensively studied in literature; see, for example, [1216].

Remark 3.1. Let 𝐻 be a real Hilbert space. Let 𝐵 be a 𝜌-Lipschitzian and 𝜂-strongly monotone operator on 𝐻 with 𝜌>0,𝜂>0. Leting 0<𝜇<2𝜂/𝜌2 and leting 𝑆=(𝐼𝑡𝜇𝐵) and 𝜇(𝜂(𝜇𝜌2/2))=𝜏, then for 𝑥,𝑦𝐻 and 𝑡(0,min{1,1/𝜏}), 𝑆 is a contraction.

Proof. Consider 𝑆𝑥𝑆𝑦2=(𝐼𝑡𝜇𝐵)𝑥(𝐼𝑡𝜇𝐵)𝑦2=(𝐼𝑡𝜇𝐵)𝑥(𝐼𝑡𝜇𝐵)𝑦,(𝐼𝑡𝜇𝐵)𝑥(𝐼𝑡𝜇𝐵)𝑦=𝑥𝑦2+𝑡2𝜇2𝐵𝑥𝐵𝑦22𝑡𝜇𝑥𝑦,𝐵𝑥𝐵𝑦𝑥𝑦2+𝑡2𝜇2𝜌2𝑥𝑦22𝑡𝜇𝜂𝑥𝑦212𝑡𝜇𝜂𝜇𝜌22𝑥𝑦2=(12𝑡𝜏)𝑥𝑦2(1𝑡𝜏)2𝑥𝑦2.(3.3) It follows that 𝑆𝑥𝑆𝑦(1𝑡𝜏)𝑥𝑦.(3.4) So 𝑆 is a contraction.

Next, we consider the cyclic algorithm (1.15), respectively, for solving the variational inequality over the set of the common fixed points of finite strictly pseudononspreading mappings.

Lemma 3.2. Assume that {𝑥𝑛} is defined by (1.15); if 𝑝 is solution of (3.2) with 𝑇𝐻𝐻 being strictly pseudononspreading mapping and demiclosed and {𝑦𝑛}𝐻 is a bounded sequence such that 𝑇𝑦𝑛𝑦𝑛0, then liminf𝑛(𝛾𝑓𝜇𝐵)𝑝,𝑦𝑛𝑝0.(3.5)

Proof. By 𝑇𝑦𝑛𝑦𝑛0 and 𝑇𝐻𝐻 demi-closed, we know that any weak cluster point of {𝑦𝑛} belongs to 𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖). Furthermore, we can also obtain that there exists 𝑦 and a subsequence {𝑦𝑛𝑗} of {𝑦𝑛} such that 𝑦𝑛𝑗𝑦 as 𝑗 (hence 𝑦𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖)) and liminf𝑛(𝛾𝑓𝜇𝐵)𝑝,𝑦𝑛𝑝=lim𝑗(𝛾𝑓𝜇𝐵)𝑝,𝑦𝑛𝑗.𝑝(3.6) From (3.2), we can derive that liminf𝑛(𝛾𝑓𝜇𝐵)𝑝,𝑦𝑛𝑝=(𝛾𝑓𝜇𝐵)𝑝,𝑦𝑝0.(3.7) It is the desired result. In addition, the variational inequality (3.7) can be written as (𝐼𝜇𝐵+𝛾𝑓)𝑝𝑝,𝑦𝑝0,𝑦𝑁𝑖=1𝐹𝑖𝑥𝑇𝑖.(3.8) So, by Lemma 2.12, it is equivalent to the fixed point equation 𝑃𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖)(𝐼𝜇𝐵+𝛾𝑓)𝑝=𝑝.(3.9)

Theorem 3.3. Let 𝐶 be a nonempty closed convex subset of 𝐻 and for 1𝑖𝑁. Let 𝑇𝑖𝐻𝐻 be 𝑘𝑖-strictly pseudononspreading mappings for some 𝑘𝑖[0,𝜔𝑖), 𝜔𝑖(0,1/2), (𝑖=1,2,,𝑁), and 𝑘=max{𝑘𝑖1𝑖𝑁}. Let 𝑓 be 𝐿-Lipschitz mapping on 𝐻 with coefficient 𝐿>0, and let 𝐵𝐻𝐻 be 𝜂-strongly monotone and 𝜌-Lipschitzian on 𝐻 with 𝜌>0, 𝜂>0. Let {𝛼𝑛} being a sequence in (0,min{1,1/𝜏}) satisfying the following conditions: (c1)lim𝑛𝛼𝑛=0, (c2)𝑛=0𝛼𝑛=.
Given 𝑥0𝐶, let {𝑥𝑛}𝑛=1 be the sequence generated by the cyclic algorithm (1.15). Then {𝑥𝑛} converges strongly to the unique element 𝑝 in 𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖) verifying 𝑝=𝑃𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖)(𝐼𝜇𝐵+𝛾𝑓)𝑝,(3.10) which equivalently solves the variational inequality problem (3.2).

Proof. Take a 𝑝𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖). Let 𝑇𝜔𝑥=(1𝜔)𝑥+𝜔𝑇𝑥 and 0<𝑘<𝜔<1/2. Then 𝑥,𝑦𝐶, we have 𝑇𝜔𝑥𝑇𝜔𝑦2=𝜔𝑥𝑦2+1𝜔𝑇𝑥𝑇𝑦2𝜔1𝜔𝑥𝑇𝑥(𝑦𝑇𝑦)2𝜔𝑥𝑦2+1𝜔𝑥𝑦2+𝑘𝑥𝑇𝑥(𝑦𝑇𝑦)2+2𝑥𝑇𝑥,𝑦𝑇𝑦𝜔1𝜔𝑥𝑇𝑥(𝑦𝑇𝑦)2=𝑥𝑦2+21𝜔𝑥𝑇𝑥,𝑦𝑇𝑦1𝜔𝜔𝑘𝑥𝑇𝑥(𝑦𝑇𝑦)2𝑥𝑦2+21𝜔𝑥𝑇𝑥,𝑦𝑇𝑦=𝑥𝑦2+21𝜔𝜔2𝑥𝑇𝜔𝑥,𝑦𝑇𝜔𝑦.(3.11)
From 𝑝𝐹𝑖𝑥(𝑇) and (3.11), we also have 𝑇𝜔𝑥𝑛𝑥𝑝𝑛.𝑝(3.12) Using (1.15) and (3.12), we obtain 𝑥𝑛+1=𝛼𝑝𝑛𝛾𝑓𝑥𝑛𝑓(𝑝)+𝛼𝑛(𝛾𝑓(𝑝)𝑝)+𝐼𝜇𝛼𝑛𝐵𝑇𝜔[𝑛]𝑥𝑛𝑝𝛼𝑛𝛾𝑓𝑥𝑛𝑓(𝑝)+𝛼𝑛𝛾𝑓(𝑝)𝑝+1𝛼𝑛𝜏𝑥𝑛,𝑝(3.13) which combined with (3.12) and 𝑓(𝑥𝑛)𝑓(𝑝)𝐿𝑥𝑛𝑝 amounts to 𝑥𝑛+1𝑝1𝛼𝑛𝑥(𝜏𝛾𝐿)𝑛𝑝+𝛼𝑛𝛾𝑓(𝑝)𝑝.(3.14) Putting 𝑀1=max{𝑥0𝑝,𝛾𝑓(𝑝)𝑝}, we clearly obtain 𝑥𝑛𝑝𝑀1. Hence {𝑥𝑛} is bounded. We can also prove that the sequences {𝑓(𝑥𝑛)} and {𝑇𝜔[𝑛]𝑥𝑛} are all bounded.
From (1.15) we obtain that 𝑥𝑛+1𝑥𝑛+𝛼𝑛𝜇𝐵𝑥𝑛𝑥𝛾𝑓𝑛=𝐼𝜇𝛼𝑛𝐵𝑇𝜔[𝑛]𝑥𝑛𝑥𝑛,(3.15) hence 𝑥𝑛+1𝑥𝑛+𝛼𝑛𝜇𝐵𝑥𝑛𝑥𝛾𝑓𝑛,𝑥𝑛=𝑝1𝛼𝑛𝑇𝜔[𝑛]𝑥𝑛𝑥𝑛,𝑥𝑛𝑝+𝛼𝑛𝑇(𝐼𝜇𝐵)𝜔[𝑛]𝑥𝑛𝑥𝑛,𝑥𝑛=𝑝1𝛼𝑛𝑇𝜔[𝑛]𝑥𝑛𝑥𝑛,𝑥𝑛𝑝+𝛼𝑛𝑇(𝐼𝜇𝐵)𝜔[𝑛]𝑥𝑛𝑥𝑛𝑥𝑛=𝑝1𝛼𝑛𝑇𝜔[𝑛]𝑥𝑛𝑥𝑛,𝑥𝑛𝑝+𝛼𝑛𝑇(1𝜏)𝜔[𝑛]𝑥𝑛𝑥𝑛𝑥𝑛=𝑝1𝛼𝑛𝑇𝜔[𝑛]𝑥𝑛𝑥𝑛,𝑥𝑛𝑝+𝜔[𝑛]𝛼𝑛𝑇(1𝜏)[𝑛]𝑥𝑛𝑥𝑛𝑥𝑛.𝑝(3.16) Moreover, by 𝑝𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖) and using Remark 2.5, we obtain 𝑥𝑛𝑇𝜔[𝑛]𝑥𝑛,𝑥𝑛1𝑝2𝜔[𝑛]1𝑘[𝑛]𝑥𝑛𝑇[𝑛]𝑥𝑛2,(3.17) which combined with (3.16) entails 𝑥𝑛+1𝑥𝑛+𝛼𝑛(𝜇𝐵𝛾𝑓)𝑥𝑛,𝑥𝑛1𝑝2𝜔[𝑛]1𝑘[𝑛]1𝛼𝑛𝑥𝑛𝑇[𝑛]𝑥𝑛2+𝜔[𝑛]𝛼𝑛(𝑇1𝜏)[𝑛]𝑥𝑛𝑥𝑛𝑥𝑛,𝑝(3.18) or equivalently 𝑥𝑛𝑥𝑛+1,𝑥𝑛𝑝𝛼𝑛(𝜇𝐵𝛾𝑓)𝑥𝑛,𝑥𝑛1𝑝2𝜔[𝑛]1𝑘[𝑛]1𝛼𝑛𝑥𝑛𝑇[𝑛]𝑥𝑛2+𝜔[𝑛]𝛼𝑛𝑇(1𝜏)[𝑛]𝑥𝑛𝑥𝑛𝑥𝑛.𝑝(3.19) Furthermore, using the following classical equality 1𝑢,𝑣=2𝑢212𝑢𝑣2+12𝑣2,𝑢,𝑣𝐶,(3.20) and setting 𝒯𝑛=1/2𝑥𝑛𝑝2, we have 𝑥𝑛𝑥𝑛+1,𝑥𝑛𝑝=𝒯𝑛𝒯𝑛+1+12𝑥𝑛𝑥𝑛+12.(3.21) So (3.19) can be equivalently rewritten as 𝒯𝑛+1𝒯𝑛12𝑥𝑛𝑥𝑛+12𝛼𝑛(𝜇𝐵𝛾𝑓)𝑥𝑛,𝑥𝑛1𝑝2𝜔[𝑛]1𝑘[𝑛]1𝛼𝑛𝑥𝑛𝑇[𝑛]𝑥𝑛2+𝜔[𝑛]𝛼𝑛𝑇(1𝜏)[𝑛]𝑥𝑛𝑥𝑛𝑥𝑛.𝑝(3.22) Now using (3.15) again, we have 𝑥𝑛+1𝑥𝑛2=𝛼𝑛(𝛾𝑓(𝑥𝑛)𝜇𝐵𝑥𝑛)+(𝐼𝜇𝛼𝑛𝐵)(𝑇𝜔[𝑛]𝑥𝑛𝑥𝑛)2.(3.23) Since 𝐵𝐻𝐻 is 𝜂-strongly monotone and 𝑘-Lipschitzian on 𝐻, hence it is a classical matter to see that 𝑥𝑛+1𝑥𝑛22𝛼2𝑛𝛾𝑓(𝑥𝑛)𝜇𝐵𝑥𝑛2+21𝛼𝑛𝜏2𝑇𝜔[𝑛]𝑥𝑛𝑥𝑛2,(3.24) which by 𝑇𝜔[𝑛]𝑥𝑛𝑥𝑛=𝜔[𝑛]𝑥𝑛𝑇[𝑛]𝑥𝑛 and (1𝛼𝑛𝜏)2(1𝛼𝑛𝜏) yields 12𝑥𝑛+1𝑥𝑛2𝛼2𝑛𝛾𝑓(𝑥𝑛)𝜇𝐵𝑥𝑛2+1𝛼𝑛𝜏𝜔2[𝑛]𝑥𝑛𝑇[𝑛]𝑥𝑛2.(3.25) Then from (3.22) and (3.25), we have 𝒯𝑛+1𝒯𝑛+𝜔[𝑛]121𝑘[𝑛]1𝛼𝑛𝜔[𝑛]1𝛼𝑛𝜏𝑥𝑛𝑇[𝑛]𝑥𝑛2𝛼𝑛𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐵𝑥𝑛2(𝜇𝐵𝛾𝑓)𝑥𝑛,𝑥𝑛𝑝+𝜔[𝑛]𝑇(1𝜏)[𝑛]𝑥𝑛𝑥𝑛𝑥𝑛.𝑝(3.26) The rest of the proof will be divided into two parts.
Case  1. Suppose that there exists 𝑛0 such that {𝒯𝑛}𝑛𝑛0 is nonincreasing. In this situation, {𝒯𝑛} is then convergent because it is also nonnegative (hence it is bounded from below), so that lim𝑛(𝒯𝑛+1𝒯𝑛)=0; hence, in light of (3.26) together with lim𝑛𝛼𝑛=0 and the boundedness of {𝑥𝑛}, we obtain lim𝑛𝑥𝑛𝑇[𝑛]𝑥𝑛=0.(3.27) It also follows from (3.26) that 𝒯𝑛𝒯𝑛+1𝛼𝑛𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐵𝑥𝑛2+(𝜇𝐵𝛾𝑓)𝑥𝑛,𝑥𝑛𝑝+𝜔[𝑛]𝑇(1𝜏)[𝑛]𝑥𝑛𝑥𝑛𝑥𝑛.𝑝(3.28) Then, by 𝑛=0𝛼𝑛=, we obviously deduce that liminf𝑛𝛼𝑛𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐵𝑥𝑛2+(𝜇𝐵𝛾𝑓)𝑥𝑛,𝑥𝑛𝑝+𝜔[𝑛]𝑇(1𝜏)[𝑛]𝑥𝑛𝑥𝑛𝑥𝑛𝑝0,(3.29) or equivalently (as 𝛼𝑛𝛾𝑓(𝑥𝑛)𝜇𝐵𝑥𝑛20) liminf𝑛(𝜇𝐵𝛾𝑓)𝑥𝑛,𝑥𝑛𝑝0.(3.30) Moreover, by Remark 1.2, we have 2(𝜇𝜂𝛾𝐿)𝒯𝑛+(𝜇𝐵𝛾𝑓)𝑝,𝑥𝑛𝑝(𝜇𝐵𝛾𝑓)𝑥𝑛,𝑥𝑛𝑝,(3.31) which by (3.30) entails liminf𝑛2(𝜇𝜂𝛾𝐿)𝒯𝑛+(𝜇𝐵𝛾𝑓)𝑝,𝑥𝑛𝑝0,(3.32) hence, recalling that lim𝑛𝒯𝑛 exists, we equivalently obtain 2(𝜇𝜂𝛾𝐿)lim𝑛𝒯𝑛+liminf𝑛(𝜇𝐵𝛾𝑓)𝑝,𝑥𝑛𝑝0,(3.33) namely 2(𝜇𝜂𝛾𝐿)lim𝑛𝒯𝑛liminf𝑛(𝜇𝐵𝛾𝑓)𝑝,𝑥𝑛𝑝.(3.34) From (3.27) and Lemma 3.2, we obtain liminf𝑛(𝜇𝐵𝛾𝑓)𝑝,𝑥𝑛𝑝0,(3.35) which yields lim𝑛𝒯𝑛=0, so that {𝑥𝑛} converges strongly to 𝑝.
Case  2. Suppose there exists a subsequence {𝒯𝑛𝑘}𝑘0 of {𝒯𝑛}𝑛0 such that 𝒯𝑛𝑘𝒯𝑛𝑘+1 for all 𝑘0. In this situation, we consider the sequence of indices {𝛿(𝑛)} as defined in Lemma 2.8. It follows that 𝒯𝛿(𝑛+1)𝒯𝛿(𝑛)>0, which by (3.26) amounts to 𝜔[𝑛]121𝑘[𝑛]1𝛼𝛿(𝑛)𝜔[𝑛]1𝛼𝛿(𝑛)𝜏𝑥𝛿(𝑛)𝑇[𝑛]𝑥𝛿(𝑛)2<𝛼𝛿(𝑛)𝛼𝛿(𝑛)𝑥𝛾𝑓𝛿(𝑛)𝜇𝐵𝑥𝛿(𝑛)2(𝜇𝐵𝛾𝑓)𝑥𝛿(𝑛),𝑥𝛿(𝑛)𝑝+𝜔[𝑛](𝑇1𝜏)[𝑛]𝑥𝛿(𝑛)𝑥𝛿(𝑛)𝑥𝛿(𝑛).𝑝(3.36) By the boundedness of {𝑥𝑛} and lim𝑛𝛼𝑛=0, we immediately obtain lim𝑛𝑥𝛿(𝑛)𝑇[𝑛]𝑥𝛿(𝑛)=0.(3.37) Using (1.15), we have 𝑥𝛿(𝑛)+1𝑥𝛿(𝑛)𝛼𝛿(𝑛)𝑥𝛾𝑓𝛿(𝑛)𝜇𝐵𝑥𝛿(𝑛)+1𝛼𝛿(𝑛)𝜏𝑇𝜔[𝑛]𝑥𝛿(𝑛)𝑥𝛿(𝑛)𝛼𝛿(𝑛)𝑥𝛾𝑓𝛿(𝑛)𝜇𝐵𝑥𝛿(𝑛)+𝜔[𝑛]1𝛼𝛿(𝑛)𝜏𝑇[𝑛]𝑥𝛿(𝑛)𝑥𝛿(𝑛),(3.38) which together with (3.37) and lim𝑛𝛼𝑛=0 yields lim𝑛𝑥𝛿(𝑛)+1𝑥𝛿(𝑛)=0.(3.39)
Now by (3.36) we clearly have 𝛼𝛿(𝑛)𝛾𝑓(𝑥𝛿(𝑛))𝜇𝐵𝑥𝛿(𝑛)2+𝜔[𝑛]𝑇(1𝜏)[𝑛]𝑥𝛿(𝑛)𝑥𝛿(𝑛)𝑥𝛿(𝑛)𝑝(𝜇𝐵𝛾𝑓)𝑥𝛿(𝑛),𝑥𝛿(𝑛),𝑝(3.40) which in the light of (3.31) yields 2(𝜇𝜂𝛾𝐿)𝒯𝛿(𝑛)+(𝜇𝐵𝛾𝑓)𝑝,𝑥𝛿(𝑛)𝑝𝛼𝛿(𝑛)𝛾𝑓(𝑥𝛿(𝑛))𝜇𝐵𝑥𝛿(𝑛)2+𝜔[𝑛]𝑇(1𝜏)[𝑛]𝑥𝛿(𝑛)𝑥[𝑛]𝑥𝛿(𝑛),𝑝(3.41) hence (as lim𝑛𝛼𝛿(𝑛)𝛾𝑓(𝑥𝛿(𝑛))𝜇𝐵𝑥𝛿(𝑛)2=0 and (3.37)) it follows that 2(𝜇𝜂𝛾𝐿)limsup𝑛𝒯𝛿(𝑛)liminf𝑛(𝜇𝐵𝛾𝑓)𝑝,𝑥𝛿(𝑛).𝑝(3.42) From (3.37) and Lemma 3.2, we obtain lim𝑛(𝜇𝐵𝛾𝑓)𝑝,𝑥𝛿(𝑛)𝑝0,(3.43) which by (3.42) yields limsup𝑛𝒯𝛿(𝑛)=0, so that lim𝑛𝒯𝛿(𝑛)=0. Combining (3.39), we have lim𝑛𝒯𝛿(𝑛)+1=0. Then, recalling that 𝒯𝑛<𝒯𝛿(𝑛)+1 (by Lemma 2.8), we get lim𝑛𝒯𝑛=0, so that 𝑥𝑛𝑝 strongly.

Taking 𝑘𝑖=0, we know that 𝑘𝑖-strictly pseudononspreading mapping is nonspreading mapping and 𝑖=𝑛 (mod 𝑁), 0𝑖𝑁1. According to the proof Theorem 3.3, we obtain the following corollary.

Corollary 3.4. Let 𝐶 be a nonempty closed convex subset of 𝐻. Let 𝑇𝑖𝐶𝐶 be nonspreading mappings and 𝜔𝑖(0,1/2), (𝑖=1,2,,𝑁). Let 𝑓 be 𝐿-Lipschitz mapping on 𝐻 with coefficient 𝐿>0 and let 𝐵𝐻𝐻 be 𝜂-strongly monotone and 𝜌-Lipschitzian on 𝐻 with 𝜌>0, 𝜂>0. Let {𝛼𝑛} be a sequence in (0,min{1,1/𝜏}) satisfying the following conditions: (c1)lim𝑛𝛼𝑛=0, (c2)𝑛=0𝛼𝑛=.
Given 𝑥0𝐶, let {𝑥𝑛}𝑛=1 be the sequence generated by the cyclic algorithm (1.15). Then {𝑥𝑛} converges strongly to the unique element 𝑝 in 𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖) verifying 𝑝=𝑃𝑁𝑖=1𝐹𝑖𝑥(𝑇𝑖)(𝐼𝜇𝐵+𝛾𝑓)𝑝,(3.44) which equivalently solves the variational inequality problem (3.2).

Acknowledgment

This work is supported in part by China Postdoctoral Science Foundation (Grant no. 20100470783).