# Strong Convergence Analysis of Iterative Algorithms for Solving Variational Inclusions and Fixed-Point Problems of Pseudocontractive Operators

Authors: Zhangsong Yao; Yan-Kuen Wu; Ching-Feng Wen
Journal: Journal of Mathematics
 (2021)

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## Abstract

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.


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## Body

## 1. Introduction

LetH be a real Hilbert space with inner product ⋅,⋅ and induced norm ⋅. Let C be a nonempty closed and convex subset of H. Let f:C⟶H and g:H⟶2H be two nonlinear operators. Recall that the variational inclusion ([1]) is to solve the following problem of finding x‡∈2H verifying(1)0∈f+gx‡.Here, use f+g−10 to denote the set of solutions of (1).Special Case 1.
LetδC:H⟶0,+∞ be defined by(2)δC=0,x∈C,+∞,x∉C.
Settingg=∂δC, variational inclusion (1) reduces to find x‡∈C such that(3)fx‡,x−x‡≥0,∀x∈C.
Problem (3) is the well-known variational inequality which has been studied, extended, and developed in a broad category of jobs (see, e.g., [2–14]).Special Case 2.
Letφ:H⟶ℝ∪+∞ be a proper lower semicontinuous convex function and ∂φ be the subdifferential of φ. Setting g=∂φ, variational inclusion (1) reduces to find x‡∈H such that(4)fx‡,x−x‡+φx−φx‡≥0,∀x∈H.
Problem (4) is called the mixed quasi-variational inequality [15] 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 [16] generated by(5)xn+1=I+λg−1I−λfxn,n≥1,where I−λf is a forward step and I+λg−1 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, [17–27]). Especially, Zhao, Sahu, and Wen [28] presented an iterative algorithm for solving a system of variational inclusions involving accretive operators. Ceng and Wen [29] introduced an implicit hybrid steepest-descent algorithm for solving generalized mixed equilibria with variational inclusions and variational inequalities. Li and Zhao [30] 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

LetH be a real Hilbert space. Let g:H⟶2H be an operator. Write domg=x∈H:gx≠∅. Recall that g is called monotone if ∀x,y∈domg, u∈gx and v∈gy, x−y,u−v≥0.A monotone operatorg 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 operatorg on H,(i)
Setg−10=x∈H:0∈gx(ii)
Denote its resolvent byJλg=I+λg−1 which is single-valued from H into domgIt is known thatg−10=FixJλg,∀λ>0 and Jλg is firmly nonexpansive, i.e.,(6)Jλgx−Jλgy2≤Jλgx−Jλgy,x−y,for all x,y∈C.LetC be a nonempty closed convex subset of a real Hilbert space H. Recall that an operator T:C⟶C is said to be(i)
L-Lipschitz if there exists a positive constant L such that(7)Tx−Ty≤Lx−y,∀x,y∈C.IfL=1, T is nonexpansive.(ii)
Pseudocontractive if(8)Tx−Ty,x−y≤x−y2,∀x,y∈C.(iii)
Inverse-strongly monotone if(9)Tx−Ty,x−y≥αTx−Ty2,∀x,y∈C,where α>0 is a constant and T is also called α-ism.Recall that the projectionPC is an orthographic projection from H onto C, which is defined by x−PCx=miny∈Cx−y. It is known that PC is nonexpansive.Lemma 1 (see [23, 31]).
LetC be a nonempty closed convex subset of a real Hilbert space H. Let T:C⟶C be an L-Lipschitz pseudocontractive operator. Then,(i)
T is demiclosed, i.e., xn⇀p and Txn⟶q⇒Tp=q(ii)
For0<ζ<1/1+L2+1, ∀x∈C and y∈FixT, we have(10)T1−ζx+ζTx−y2≤x−y2+1−ζx−T1−ζx+ζTx2.Lemma 2 (see [16, 32]).
LetH be a real Hilbert space and let g be a maximal monotone operator on H. Then, we have(11)Jsgx−Jtgx2≤s−ttJsgx−Jtgx,Jsgx−x,for all s,t>0 and x∈H.Lemma 3 (see [33]).
Assume that a real number sequencean⊂0,∞ satisfies(12)an+1≤1−γnan+δnγn,where γn⊂0,1 and δn⊂−∞,+∞ satisfy the following conditions:(i)
∑n=1∞γn=∞(ii)
limsupn⟶∞δn≤0 or ∑n=1∞δnγn<∞
Then,limn⟶∞an=0.Lemma 4 (see [8]).
Letsn⊂0,∞ be a sequence. Assume that there exists at least a subsequence sni of sn verifying sni≤sni+1 for all i≥0. Let τn be an integer sequence defined as τn=maxi≤n:sni<sni+1. Then τn⟶∞ as n⟶∞ and(13)maxsτn,sn≤sτn+1.


## 3. Main Results

LetC be a nonempty closed convex subset of a real Hilbert space H. Let the operator f:C⟶H be an α-ism. Let g:H⟶2H be a maximal monotone operator with domg⊂C. Let T:C⟶C be an L−Lipschitz pseudocontractive operator with L>1. Let αn⊂0,1 and λn⊂0,∞ 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.
Letu∈C be a fixed point. Choose x0∈C. Set n=0.Step 1. For given xn, compute yn by(14)yn=1−νxn+νT1−ζxn+ζTxn.Step 2. Compute xn+1by(15)xn+1=αnu+1−αnJλngI−λnfyn.Step 3. Set n:=n+1 and return to Step 1.
Next, we prove the convergence of Algorithm1.Theorem 1.
Suppose thatΓ:=FixT∩f+g−10≠∅. Assume that the following conditions are satisfied:limn⟶∞αn=0 and ∑n=1∞αn=∞0<d1<λn<d2<2α and 0<ν<ζ<1/1+L2+1
Then, the sequencexn generated by Algorithm 1 converges strongly to PΓu.Proof.
Letx∗∈FixT∩f+g−10. Set un=JλngI−λnfyn,∀n≥0. Since f is α-ism, we have(16)fyn−fx∗,yn−x∗≥αfyn−fx∗2.
By the nonexpansivity ofJλng, we have(17)un−x∗2=JλngI−λnfyn−JλngI−λnfx∗2≤yn−x∗−λnfyn−fx∗2=yn−x∗2−2λnfyn−fx∗,yn−x∗+λn2fyn−fx∗2≤yn−x∗2−2λnαfyn−fx∗2+λn2fyn−fx∗2=yn−x∗2−λn2α−λnfyn−fx∗2≤yn−x∗2−d12α−d2fyn−fx∗2by conditionr2≤yn−x∗2.
Using Lemma1, we get(18)T1−ζI+ζTxn−x∗2≤xn−x∗2+1−ζxn−T1−ζI+ζTxn2.
This together with (14) implies that(19)yn−x∗2=1−νxn+νT1−ζI+ζTxn−x∗2=1−νxn−x∗+νT1−ζI+ζTxn−x∗2=1−νxn−x∗2+νT1−ζI+ζTxn−x∗2−ν1−νT1−ζI+ζTxn−xn2≤xn−x∗2−νζ−νT1−ζI+ζTxn−xn2≤xn−x∗2.
According to (15)-(19), we obtain(20)xn+1−x∗=αnu−x∗+1−αnun−x∗≤αnu−x∗+1−αnxn−x∗≤⋯≤maxu−x∗,x0−x∗.
Then, the sequencexn is bounded. The sequences un and yn are also bounded.
Again, by (15)-(19), we deduce(21)xn+1−x∗2=αnu−x∗+1−αnun−x∗2≤αnu−x∗2+1−αnun−x∗2by the convexity of⋅2≤αnu−x∗2+un−x∗2≤αnu−x∗2+xn−x∗2−d12α−d2fyn−fx∗2−νζ−νT1−ζI+ζTxn−xn2.
It follows that(22)d12α−d2fyn−fx∗2+νζ−νT1−ζI+ζTxn−xn2≤αnu−x∗2+xn−x∗2−xn+1−x∗2.
SinceJλng is firmly nonexpansive, using (6), we have(23)un−x∗2=JλngI−λnfyn−JλngI−λnfx∗2≤I−λnfyn−I−λnfx∗,un−x∗=yn−x∗,un−x∗−λnun−x∗,fyn−fx∗=12yn−x∗2+un−x∗2−yn−un2−λnyn−x∗,fyn−fx∗−λnun−yn,fyn−fx∗≤12yn−x∗2+un−x∗2−yn−un2+λnun−ynfyn−fx∗,which leads to(24)un−x∗2≤yn−x∗2−yn−un2+2λnun−ynfyn−fx∗≤xn−x∗2−yn−un2+2λnun−ynfyn−fx∗.
Combining (21) with (24), we obtain(25)xn+1−x∗2≤αnu−x∗2+un−x∗2≤αnu−x∗2+xn−x∗2−yn−un2+2λnun−ynfyn−fx∗.which results in that(26)yn−un2≤αnu−x∗2+xn−x∗2−xn+1−x∗2+2λnun−ynfyn−fx∗.
Next, we analyze two cases. (i)∃n0∈ℕsuch thatxn+1−x∗≤xn−x∗,∀n≥n0. (ii) For any n0∈ℕ, ∃m≥n0 such that xm−x∗≤xm+1−x∗.
In case of (i),limn⟶∞xn−x∗ exists. From (22), we deduce(27)limn⟶∞fyn−fx∗=0and(28)limn⟶∞T1−ζI+ζTxn−xn=0.
It follows from (14) that(29)limn⟶∞yn−xn=limn⟶∞νT1−ζI+ζTxn−xn=0.
On the basic of (26) and (27), we have(30)limn⟶∞yn−un=limn⟶∞yn−JλngI−λnfyn=0.
Note thatxn+1−xn≤αnu−xn+1−αnun−xn. Thanks to (29) and (30), we derive that(31)limn⟶∞xn+1−xn=0.
However,(32)xn−Txn≤T1−ζI+ζTxn−xn+Txn−T1−ζI+ζTxn≤T1−ζI+ζTxn−xn+ζLxn−Txn,
We have(33)xn−Txn≤11−ζLT1−ζI+ζTxn−xn.
This together with (28) implies that(34)limn⟶∞xn−Txn=0.
Setp=PΓu. Next, we prove that(35)limsupn⟶∞u−p,xn+1−p≤0.
Sincexn+1 is bounded, there exists a subsequence xni+1 of xn+1 satisfying(1)
xni+1⇀x˜ (hence, xni⇀x˜ by (31))(2)
limsupn⟶∞u−p,xn+1−p=limi⟶∞u−p,xni+1−p
From (34) and Lemma 1, we obtain x˜∈FixT.
Owing to (29) and (30), we have that yni⇀x˜ and(36)limi⟶∞JλnigI−λnifyni−yni=0.
Sinceλn∈d1,d2, without loss of generality, we assume that λni⟶λ†>0i⟶∞. Observe that(37)JλnigI−λnifyni−Jλ†gI−λ†fyni≤JλnigI−λnifyni−Jλ†gI−λnifyni+Jλ†gI−λnifyni−Jλ†gI−λ†fyni≤JλnigI−λnifyni−Jλ†gI−λnifyni+λni−λ†fyni.
Applying Lemma2, we obtain(38)JλnigI−λnifyni−Jλ†gI−λnifyni2≤λni−λ†λ†JλnigI−λnifyni−Jλ†gI−λnifyni,JλnigI−λnifyni−I−λnifyni≤λni−λ†λ†JλnigI−λnifyni−Jλ†gI−λnifyniJλnigI−λnifyni−I−λnifyni.
It follows that(39)JλnigI−λnifyni−Jλ†gI−λnifyni≤λni−λ†λ†JλnigI−λnifyni−I−λnifyni.
Thanks to (37) and (39), we get(40)JλnigI−λnifyni−Jλ†gI−λ†fyni≤λni−λ†fyni+λni−λ†λ†JλnigI−λnifyni−I−λnifyni.
Noting thatλni⟶λ†i⟶∞, from (36) and (40), we get(41)limi⟶∞yni−Jλ†gI−λ†fyni=0.
By Lemma1, we deduce that x˜∈FixJλ†gI−λ†f=f+g−10. Therefore, x˜∈Γ and(42)limsupn⟶∞u−p,xn+1−p=limi⟶∞u−p,xni+1−p=u−p,x˜−p≤0.
From (15), we have(43)xn+1−p2=αnu−p+1−αnun−p2≤1−αnun−p2+2αnu−p,xn+1−p≤1−αnxn−p2+2αnu−p,xn+1−p.
Applying Lemma3 to (43) to deduce xn⟶p.
In case of (ii), letsn=xn−x∗. So, we have sn0≤sn0+1. Define an integer sequence τn,∀n≥n0, by τn=maxi∈ℕ|n0≤i≤n,si≤si+1. It is obvious that limn⟶∞τn=∞ and sτn≤sτn+1 for all n≥n0. Similarly, we can prove that limn⟶∞xτn−Txτn=0 and limn⟶∞JλτngI−λτnfxτn=0. Therefore, all weak cluster points ωwxτn⊂Γ. Consequently,(44)limsupn⟶∞u−p,xτn−p≤0.
Note thatsτn≤sτn+1. From (43), we deduce(45)sτn2≤sτn+12≤1−ατnsτn2+2ατnu−p,xτn+1−p.
It follows that(46)sτn2≤2u−p,xτn+1−p.
Combining (44) and (46), we have limsupn⟶∞sτn≤0 and hence(47)limk⟶∞sτk=0.
From (45), we deduce that limsupn⟶∞sτn+12≤limsupn⟶∞sτn2. This together with (47) implies that limn⟶∞sτn+1=0. According to Lemma 4, we get 0≤sn≤maxsτn,sτn+1. Therefore, sn⟶0 and xn⟶p. This completes the proof.Remark 1.
Since the pseudocontractive operator is nonexpansive, Theorem1 still holds if T is nonexpansive.Remark 2.
Assumptionr1 imposed on parameter αn is essential and we do not add any other assumptions.



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