JMATHJournal of Mathematics2314-47852314-4629Hindawi10.1155/2021/66350266635026Research ArticleStrong Convergence Analysis of Iterative Algorithms for Solving Variational Inclusions and Fixed-Point Problems of Pseudocontractive OperatorsYaoZhangsong1https://orcid.org/0000-0001-7788-8157WuYan-Kuen2https://orcid.org/0000-0001-8900-761XWenChing-Feng34YaoJen-Chih1School of Information EngineeringNanjing Xiaozhuang UniversityNanjing 211171Chinanjxzc.edu.cn2College of International Business and Shaoxing Key Laboratory of Intelligent Monitoring and Prevention of Smart CityZhejiang Yuexiu University of Foreign LanguagesShaoxinZhejiangChina3Center for Fundamental Science and Research Center for Nonlinear Analysis and OptimizationKaohsiung Medical UniversityKaohsiung 80708Taiwankmu.edu.tw4Department of Medical ResearchKaohsiung Medical University HospitalKaohsiung 80708Taiwankmuh.org.tw2021124202120211210202010112020103202112420212021Copyright © 2021 Zhangsong Yao et al.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Iterative methods for solving variational inclusions and fixed-point problems have been considered and investigated by many scholars. In this paper, we use the Halpern-type method for finding a common solution of variational inclusions and fixed-point problems of pseudocontractive operators. We show that the proposed algorithm has strong convergence under some mild conditions.

Ministry of Science and Technology of the People's Republic of China109-2115-M-037-001
1. Introduction

Let H be a real Hilbert space with inner product , and induced norm . Let C be a nonempty closed and convex subset of H. Let f:CH and g:H2H be two nonlinear operators. Recall that the variational inclusion () is to solve the following problem of finding x2H verifying(1)0f+gx.Here, use f+g10 to denote the set of solutions of (1).

Special Case 1.

Let δC:H0,+ be defined by(2)δC=0,xC,+,xC.

Setting g=δC, variational inclusion (1) reduces to find xC such that(3)fx,xx0,xC.

Problem (3) is the well-known variational inequality which has been studied, extended, and developed in a broad category of jobs (see, e.g., ).

Special Case 2.

Let φ:H+ be a proper lower semicontinuous convex function and φ be the subdifferential of φ. Setting g=φ, variational inclusion (1) reduces to find xH such that(4)fx,xx+φxφx0,xH.

Problem (4) is called the mixed quasi-variational inequality  which is a very significant extension of variational inequality (3) involving the nonlinear function φ. It is well known that a large number of practical problems arising in various branches of pure and applied sciences can be formulated as the model of mixed quasi-variational inequality (4).

Problem (1) plays a key role in minimization, convex feasibility problems, machine learning, and others. A popular algorithm for solving problem (1) is the forward-backward algorithm  generated by(5)xn+1=I+λg1Iλfxn,n1,where Iλf is a forward step and I+λg1 is a backward step with λ>0. This algorithm is a splitting algorithm which solves the difficulty of calculating of the resolvent of f+g.

Recently, there has been increasing interest for studying common solution problems relevant to (1) (see for example, ). Especially, Zhao, Sahu, and Wen  presented an iterative algorithm for solving a system of variational inclusions involving accretive operators. Ceng and Wen  introduced an implicit hybrid steepest-descent algorithm for solving generalized mixed equilibria with variational inclusions and variational inequalities. Li and Zhao  considered an iterate for finding a solution of quasi-variational inclusions and fixed points of nonexpansive mappings.

Motivated by the results in this direction, the main purpose of this paper is to research a common solution problem of variational inclusions and fixed point of pseudocontractions. We suggest a Halpern-type algorithm for solving such problem. We show that the proposed algorithm has strong convergence under some mild conditions.

2. Preliminaries

Let H be a real Hilbert space. Let g:H2H be an operator. Write domg=xH:gx. Recall that g is called monotone if x,ydomg, ugx and vgy, xy,uv0.

A monotone operator g is maximal monotone if and only if its graph is not strictly contained in the graph of any other monotone operator on H.

For a maximal monotone operator g on H,

Set g10=xH:0gx

Denote its resolvent by Jλg=I+λg1 which is single-valued from H into domg

It is known that g10=FixJλg,λ>0 and Jλg is firmly nonexpansive, i.e.,(6)JλgxJλgy2JλgxJλgy,xy,for all x,yC.

Let C be a nonempty closed convex subset of a real Hilbert space H. Recall that an operator T:CC is said to be

L-Lipschitz if there exists a positive constant L such that(7)TxTyLxy,x,yC.

If L=1, T is nonexpansive.

Pseudocontractive if(8)TxTy,xyxy2,x,yC.

Inverse-strongly monotone if

(9)TxTy,xyαTxTy2,x,yC,where α>0 is a constant and T is also called α-ism.

Recall that the projection PC is an orthographic projection from H onto C, which is defined by xPCx=minyCxy. It is known that PC is nonexpansive.

Lemma 1 (see [<xref ref-type="bibr" rid="B23">23</xref>, <xref ref-type="bibr" rid="B31">31</xref>]).

Let C be a nonempty closed convex subset of a real Hilbert space H. Let T:CC be an L-Lipschitz pseudocontractive operator. Then,

T is demiclosed, i.e., xnp and TxnqTp=q

For 0<ζ<1/1+L2+1, xC and yFixT, we have(10)T1ζx+ζTxy2xy2+1ζxT1ζx+ζTx2.

Lemma 2 (see [<xref ref-type="bibr" rid="B16">16</xref>, <xref ref-type="bibr" rid="B32">32</xref>]).

Let H be a real Hilbert space and let g be a maximal monotone operator on H. Then, we have(11)JsgxJtgx2sttJsgxJtgx,Jsgxx,for all s,t>0 and xH.

Lemma 3 (see [<xref ref-type="bibr" rid="B33">33</xref>]).

Assume that a real number sequence an0, satisfies(12)an+11γnan+δnγn,where γn0,1 and δn,+ satisfy the following conditions:

n=1γn=

lim supnδn0 or n=1δnγn<

Then, limnan=0.

Lemma 4 (see [<xref ref-type="bibr" rid="B8">8</xref>]).

Let sn0, be a sequence. Assume that there exists at least a subsequence sni of sn verifying snisni+1 for all i0. Let τn be an integer sequence defined as τn=maxin:sni<sni+1. Then τn as n and(13)maxsτn,snsτn+1.

3. Main Results

Let C be a nonempty closed convex subset of a real Hilbert space H. Let the operator f:CH be an α-ism. Let g:H2H be a maximal monotone operator with domgC. Let T:CC be an LLipschitz pseudocontractive operator with L>1. Let αn0,1 and λn0, be two sequences. Let ν and ζ be two constants.

Next, we introduce a Halpern-type algorithm for finding a common solution of variational inclusion (1) and fixed point of pseudocontractive operator T.

Algorithm 1.

Let uC be a fixed point. Choose x0C. Set n=0.

Step 1. For given xn, compute yn by(14)yn=1νxn+νT1ζxn+ζTxn.

Step 2. Compute xn+1by(15)xn+1=αnu+1αnJλngIλnfyn.

Next, we prove the convergence of Algorithm 1.

Theorem 1.

Suppose that Γ:=FixTf+g10. Assume that the following conditions are satisfied:

limnαn=0 and n=1αn=

0<d1<λn<d2<2α and 0<ν<ζ<1/1+L2+1

Then, the sequence xn generated by Algorithm 1 converges strongly to PΓu.

Proof.

Let xFixTf+g10. Set un=JλngIλnfyn,n0. Since f is α-ism, we have(16)fynfx,ynxαfynfx2.

By the nonexpansivity of Jλng, we have(17)unx2=JλngIλnfynJλngIλnfx2ynxλnfynfx2=ynx22λnfynfx,ynx+λn2fynfx2ynx22λnαfynfx2+λn2fynfx2=ynx2λn2αλnfynfx2ynx2d12αd2fynfx2by condition r2ynx2.

Using Lemma 1, we get(18)T1ζI+ζTxnx2xnx2+1ζxnT1ζI+ζTxn2.

This together with (14) implies that(19)ynx2=1νxn+νT1ζI+ζTxnx2=1νxnx+νT1ζI+ζTxnx2=1νxnx2+νT1ζI+ζTxnx2ν1νT1ζI+ζTxnxn2xnx2νζνT1ζI+ζTxnxn2xnx2.

According to (15)-(19), we obtain(20)xn+1x=αnux+1αnunxαnux+1αnxnxmaxux,x0x.

Then, the sequence xn is bounded. The sequences un and yn are also bounded.

Again, by (15)-(19), we deduce(21)xn+1x2=αnux+1αnunx2αnux2+1αnunx2by the convexity of 2αnux2+unx2αnux2+xnx2d12αd2fynfx2νζνT1ζI+ζTxnxn2.

It follows that(22)d12αd2fynfx2+νζνT1ζI+ζTxnxn2αnux2+xnx2xn+1x2.

Since Jλng is firmly nonexpansive, using (6), we have(23)unx2=JλngIλnfynJλngIλnfx2IλnfynIλnfx,unx=ynx,unxλnunx,fynfx=12ynx2+unx2ynun2λnynx,fynfxλnunyn,fynfx12ynx2+unx2ynun2+λnunynfynfx,which leads to(24)unx2ynx2ynun2+2λnunynfynfxxnx2ynun2+2λnunynfynfx.

Combining (21) with (24), we obtain(25)xn+1x2αnux2+unx2αnux2+xnx2ynun2+2λnunynfynfx.which results in that(26)ynun2αnux2+xnx2xn+1x2+2λnunynfynfx.

Next, we analyze two cases. (i) n0 such that xn+1xxnx,nn0. (ii) For any n0, mn0 such that xmxxm+1x.

In case of (i), limnxnx exists. From (22), we deduce(27)limnfynfx=0and(28)limnT1ζI+ζTxnxn=0.

It follows from (14) that(29)limnynxn=limnνT1ζI+ζTxnxn=0.

On the basic of (26) and (27), we have(30)limnynun=limnynJλngIλnfyn=0.

Note that xn+1xnαnuxn+1αnunxn. Thanks to (29) and (30), we derive that(31)limnxn+1xn=0.

However,(32)xnTxnT1ζI+ζTxnxn+TxnT1ζI+ζTxnT1ζI+ζTxnxn+ζLxnTxn,

We have(33)xnTxn11ζLT1ζI+ζTxnxn.

This together with (28) implies that(34)limnxnTxn=0.

Set p=PΓu. Next, we prove that(35)limsupnup,xn+1p0.

Since xn+1 is bounded, there exists a subsequence xni+1 of xn+1 satisfying

xni+1x˜ (hence, xnix˜ by (31))

limsupnup,xn+1p=limiup,xni+1p

From (34) and Lemma 1, we obtain x˜FixT.

Owing to (29) and (30), we have that ynix˜ and(36)limiJλnigIλnifyniyni=0.

Since λnd1,d2, without loss of generality, we assume that λniλ>0i. Observe that(37)JλnigIλnifyniJλgIλfyniJλnigIλnifyniJλgIλnifyni+JλgIλnifyniJλgIλfyniJλnigIλnifyniJλgIλnifyni+λniλfyni.

Applying Lemma 2, we obtain(38)JλnigIλnifyniJλgIλnifyni2λniλλJλnigIλnifyniJλgIλnifyni,JλnigIλnifyniIλnifyniλniλλJλnigIλnifyniJλgIλnifyniJλnigIλnifyniIλnifyni.

It follows that(39)JλnigIλnifyniJλgIλnifyniλniλλJλnigIλnifyniIλnifyni.

Thanks to (37) and (39), we get(40)JλnigIλnifyniJλgIλfyniλniλfyni+λniλλJλnigIλnifyniIλnifyni.

Noting that λniλi, from (36) and (40), we get(41)limiyniJλgIλfyni=0.

By Lemma 1, we deduce that x˜FixJλgIλf=f+g10. Therefore, x˜Γ and(42)limsupnup,xn+1p=limiup,xni+1p=up,x˜p0.

From (15), we have(43)xn+1p2=αnup+1αnunp21αnunp2+2αnup,xn+1p1αnxnp2+2αnup,xn+1p.

Applying Lemma 3 to (43) to deduce xnp.

In case of (ii), let sn=xnx. So, we have sn0sn0+1. Define an integer sequence τn,nn0, by τn=maxi|n0in,sisi+1. It is obvious that limnτn= and sτnsτn+1 for all nn0. Similarly, we can prove that limnxτnTxτn=0 and limnJλτngIλτnfxτn=0. Therefore, all weak cluster points ωwxτnΓ. Consequently,(44)limsupnup,xτnp0.

Note that sτnsτn+1. From (43), we deduce(45)sτn2sτn+121ατnsτn2+2ατnup,xτn+1p.

It follows that(46)sτn22up,xτn+1p.

Combining (44) and (46), we have limsupnsτn0 and hence(47)limksτk=0.

From (45), we deduce that limsupnsτn+12limsupnsτn2. This together with (47) implies that limnsτn+1=0. According to Lemma 4, we get 0snmaxsτn,sτn+1. Therefore, sn0 and xnp. This completes the proof.

Remark 1.

Since the pseudocontractive operator is nonexpansive, Theorem 1 still holds if T is nonexpansive.

Remark 2.

Assumption r1 imposed on parameter αn is essential and we do not add any other assumptions.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

Zhangsong Yao was partially supported by the Grant 19KJD100003. Ching-Feng Wen was partially supported by the Grant of MOST 109-2115-M-037-001.

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