# Tseng-Type Algorithms for the Split Variational Inclusion

Authors: Li-Jun Zhu; Tzu-Chien Yin
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we study iterative methods for solving the split variational inclusion in Hilbert spaces. We consider a Tseng-type algorithm with self-adaptive step sizes for finding a solution of the split variational inclusion. We show the weak convergence of the proposed sequence under some mild assumptions.


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## Body

## 1. Introduction

LetH be a real Hilbert space with inner product ⋅,⋅ and induced norm ⋅. Let ϕ:H⇉2H and f:H⟶H be two operators. Recall that the variational inclusion is to find p∈H such that(1)0∈fp+ϕp.Throughout, we usef+ϕ−10 to denote the solution set of the variational inclusion (1).LetF,Φ:H⟶R∪+∞ be two proper, lower semicontinuous and convex functions. It is known that the variational inclusion (1) is closely related to the following minimization problem:(2)minp∈HFp+Φp.Assume thatF is differentiable and Φ is subdifferentiable. If choose f=∇F and ϕ=∂Φ, then solving (1) is equivalent to solving (2) [1].Variational inclusion is at the core of the modeling of many problems, such as variational inequalities [2–4], optimization problems [5], split problems [6–9], equilibrium problems [10], and fixed point problems [11]. Variational inclusion (1) has been extended and studied in different ways, see [12–18]. An efficient way for solving (1) is the forward-backward iterate [19–21] defined by(3)xk+1=ResϖϕI−ϖfxk,k≥0,where Resϖϕ is the resolvent of ϕ, and ϖ is a positive constant.Iff is (inverse) strongly monotone and ϕ is maximal monotone, then the sequence xk defined by (3) converges weakly to some point in f+ϕ−10. In order relaxed the conditions imposed on f, Tseng [22] proposed the following algorithm:(4)yk=Resϖkϕxk−ϖkfxk,xk+1=yk+ϖkfxk−fyk,where ϖk is a sequence in 0,+∞. Note that the Lipschitz constant of f is usually unknown or difficult to estimate in many problems. Tseng’s method has been applied extensively, and some self-adaptive techniques are used, see [14, 23–26]. Especially, Cholamjiak, Hieu, and Cho [23] suggested an iterative method described as follows:(5)yk=Resϖkϕxk−ϖkfxk,xk+1=1−αkxk+αkyk+ϖkfxk−fyk,where(6)ϖk+1=minϖk,ςxk−ykfxk−fyk,ifxk≠yk,ϖk,else.LetH1 and H2 be two real Hilbert spaces. Let h:H1⟶H1 and φ:H2⟶H2 be two single-valued operators. Let g:H1⇉2H1 and ψ:H2⇉2H2 be two multivalued operators. Let A:H1⟶H2 be a bounded linear operator. In this paper, we investigate the split variational inclusion of finding a point q^∈H1 such that(7)q^∈g+h−10andAq^∈φ+ψ−10.Observe that the prototype of the split variational inclusion is the following split feasibility problem:(8)Findx∈Csuch thatAx∈Q,where C⊂H1 and Q∈H2 are two nonempty closed convex sets. The split feasibility problem in finite-dimensional Hilbert spaces was first introduced by Censor and Elfving [27] for modeling inverse problems which arise from phase retrievals and in medical image reconstruction. Now, the split feasibility problem has been extended in many different directions by many scholars, see [2, 6, 7, 10, 28]. The split variational inclusion (7) was first studied in [29]. Note that the following equivalent relation holds(9)x∗∈g+h−10⇔x∗=Resϖgx∗−ϖhx∗,∀ϖ>0.By applying (9), Moudafi [29] suggested the following iterate for solving the split variational inclusion (7):(10)xk+1=Uxk+γA∗T−IAxk,k≥0,where U=ResϖgI−ϖh and T=ResϖψI−ϖφ.It should be pointed out that the operatorsh and φ in (10) were assumed to be inverse strongly monotone. The split variational inclusion (7) has been studied extensively, see [24, 28, 30–36].The main purpose of this paper is to further study the split variational inclusion (7) in which the operators h and φ are plain monotone. We suggest a self-adaptive Tseng-type algorithm for finding a solutions of the split variational inclusion (7). Weak convergence theorem is shown under some additional conditions.


## 2. Preliminaries

In this section, we assume thatH is a real Hilbert space. The following equality is well known:(11)ζx+1−ζx†2=ζx2+1−ζx†2−ζ1−ζx−x†2,for all x,x†∈H and for all ζ∈R.Letf:H⟶H be a single-valued operator. Recall that f is said to be(i)
κ-Lipschitz if there is a constant κ>0 fulfilling(12)fx−fx†≤κx−x†‖,∀x,x†∈H.(ii)
monotone if(13)fx−fx†,x−x†≥0,∀x,x†∈H.LetF:H⇉2H be a multivalued operator. F is said to be monotone if(14)x−x^,u−u^≥0,∀x,x^∈H,where u∈Fx and u^∈Fx^.A multivalued monotone operatorF is said to be maximal monotone if and only if its graph GF is not strictly contained in the graph of any other monotone operator.LetF:H⇉2H be a maximal monotone operator. Let γ>0 be a constant. Define an operator ResγF:H⟶H by(15)ResγFx:=I+γF−1x,∀x∈H,which is called the resolvent of F.It is well-known that the resolventResγF is a single-valued firmly nonexpansive operator [1] and F−10=x∈H:ResγFx=x.In what follows, we use the following expressions:”⇀ indicates weak convergence“⟶” indicates strong convergenceωwxk indicates the set of all weak cluster points of the sequence xk, that is,(16)ωwxk:=p∈H:there is a subsequencexkiofxkfulfillingxki⇀pasi⟶+∞.Lemma 1 (see [37]).
LetΘ⊂H be a nonempty closed convex set. Let xk⊂H be a sequence. Assume that(i)
∀x∗∈Θ, limk⟶∞xk−x∗exists(ii)
ωwxk⊂Θa2+b2
Then,xk⇀b^∈H.


## 3. Main Results

LetH1 and H2 be two real Hilbert spaces. Let g:H1⇉2H1 and ψ:H2⇉2H2 be two multivalued maximal monotone operators. Let h:H1⟶H1 be a κ1-Lipschitz monotone operator and φ:H2⟶H2 be a κ2-Lipschitz monotone operator. Let A:H1⟶H2 be a nonzero bounded linear operator and A∗ be the adjoint of A. Throughout, suppose that Θ:=u†|u†∈g+h−10,Au†∈φ+ψ−10≠∅.Letτk and αk be two sequences in 0,1. Let ς1, ς2, and γ be three constants. Suppose that the following restrictions are fulfilled:(i)
τk∈[τ¯,τ¯]⊂0,1 and αk∈α¯,α¯⊂0,1 for all k≥0(ii)
ς1∈0,1, ς2∈0,1, and γ∈0,1/A2Next, we present an iterative sequence for solving (7).Algorithm 1.
Letx0 be a fixed point in H1. Let ϑ0 and ϖ0 be two initial points in 0,+∞. Let k=0.Step 1. Let the iterate xk be known. Calculate(17)vk=Resϑkgxk−ϑkhxk,(18)uk=1−τkxk+τkvk−ϑkhvk−hxk.Step 2. Compute(19)zk=ResϖkψAuk−ϖkφAuk,(20)yk=1−αkAuk+αkzk−ϖkφzk−φAuk.Step 3. Compute(21)xk+1=uk+γA∗yk−Auk.Step 4. Set(22)ϑk+1=minϑk,ς1xk−vkhxk−hvk,ifxk≠vk,ϑk,else.(23)ϖk+1=minϖk,ς2zk−Aukφzk−φAuk,ifzk≠Auk,ϖk,else.
Letk:=k+1 and return to Step 1.Remark 1.
According to (22), we have ϑk+1≤ϑk for all k≥0. Meanwhile, owing to κ1-Lipschitz continuity of h, we obtain hxk−hvk≤κ1xk−vk. It follows that ϑk≥minϑ0,ς1/κ1 for all k≥0. Therefore, limk⟶+∞ϑk exists. Similarly, from (23), we deduce that minϖ0,ς2/κ2≤ϖk+1≤ϖk for all k≥0, and hence, limk⟶+∞ϖk exists.Next, we state our main convergence theorem.Theorem 1.
The sequencexk generated by Algorithm 1 converges weakly to some point in Θ.Next, we will divide the proof details into several propositions. In the sequel, assume thatx∗∈Θ. Then, x∗∈g+h−10 and Ax∗∈ψ+φ−10.Proposition 1.
limk⟶+∞xk−x∗exists.Proof.
Observe that(24)vk−x∗+ϑkhxk−hvk2=vk−x∗2+ϑk2hxk−hvk2+2ϑkhxk−hvk,vk−x∗,(25)vk−x∗2=xk−x∗2−vk−xk2+2vk−xk,vk−x∗.
which together with (24) implies that(26)vk−x∗+ϑkhxk−hvk2=xk−x∗2+2ϑkhxk−hvk,vk−x∗−vk−xk2+2vk−xk,vk−x∗+ϑk2hxk−hvk2=xk−x∗2+ϑk2hxk−hvk2+2vk−xk+ϑkhxk−hvk,vk−x∗−vk−xk2.
Based on (17), we get(27)xk−ϑkhxk∈I+ϑkgvk.
It follows that(28)xk−vk−ϑkhxk−hvk∈ϑkg+hvk.
As a result of0∈ϑkg+hx∗ and the monotonicity of ϑkg+h, by (28), we receive(29)vk−xk+ϑkhxk−hvk,vk−x∗≤0.
Using (22), we have hvk−hxk≤ς1/ϑk+1vk−xk. With the help of (26) and (29), we have(30)vk−x∗+ϑkhxk−hvk2≤xk−x∗2−1−ς12ϑk2ϑk+12vk−xk2.(31)uk−x∗2=1−τkxk−x∗2+τkvk−x∗+ϑkhxk−hvk2−1−τkτkvk−xk+ϑkhxk−hvk2.
Utilizing (11), from (18), we obtain
Combining (30) and (31), we have(32)uk−x∗2≤xk−x∗2−τk1−ς12ϑk2ϑk+12vk−xk2−1−τkτkvk−xk+ϑkhxk−hvk2≤xk−x∗2.
The following two equalities hold:(33)zk−Ax∗+ϖkφAuk−φzk2=zk−Ax∗2+ϖk2φAuk−φzk2+2ϖkφAuk−φzk,zk−Ax∗,(34)2zk−Auk,zk−Ax∗=zk−Auk2+zk−Ax∗2−Auk−Ax∗2.
Then,(35)zk−Ax∗+ϖkφAuk−φzk2=Auk−Ax∗2+2ϖkφAuk−φzk,zk−Ax∗−zk−Auk2+2zk−Auk,zk−Ax∗+ϖk2φAuk−φzk2=Auk−Ax∗2+ϖk2φAuk−φzk2+2zk−Auk+ϖkφAuk−φzk,zk−Ax∗−zk−Auk2.
Thanks to (19), we get(36)Auk−zk−ϖkφAuk−φzk∈ϖkψ+φzk.
Applying the monotonicity ofϖkψ+φ and 0∈ϖkψ+φAx∗, by (36), we gain(37)zk−Auk+ϖkφAuk−φzk,zk−Ax∗≤0.
Owing to (23), we get φzk−φAuk≤ς2/ϖk+1zk−Auk. In the light of (35) and (37), we acquire(38)zk−Ax∗+ϖkφAuk−φzk2≤Auk−Ax∗2−1−ς22ϖk2ϖk+12zk−Auk2.
Applying (11) to (20), we attain(39)yk−Ax∗2=1−αkAuk−Ax∗2+αkzk−Ax∗+ϖkφAuk−φzk2−1−αkαkzk−Auk+ϖkφAuk−φzk2.
In view of (38) and (39), we deduce(40)yk−Ax∗2≤Auk−Ax∗2−αk1−ς12ϖk2ϖk+12zk−Auk2−1−αkαkzk−Auk+ϖkφAuk−φzk2≤Auk−Ax∗2.
It follows that(41)uk−x∗,A∗yk−Auk=Auk−Ax∗,yk−Auk=12yk−Ax∗2−Auk−Ax∗2−12yk−Auk2≤−12αk1−ς22ϖk2ϖk+12zk−Auk2−12yk−Auk2.
On account of (21), (32), and (41), we achieve(42)xk+1−x∗2=uk−x∗+γA∗yk−Auk2=uk−x∗2+γA∗yk−Auk2+2γA∗yk−Auk,uk−x∗≤uk−x∗2+γ2A2yk−Auk2−γαk1−ς22ϖk2ϖk+12zk−Auk2−γyk−Auk2=uk−x∗2−γ1−γA2yk−Auk2−γαk1−ς22ϖk2ϖk+12zk−Auk2≤xk−x∗2−γ1−γA2yk−Auk2−γαk1−ς22ϖk2ϖk+12zk−Auk2≤xk−x∗2,which implies that limk⟶+∞xk−x∗ exists.Proposition 2.
(43)limk⟶+∞vk−xk=0,limk⟶+∞uk−xk=0,andlimk⟶+∞zk−Auk=0,limk⟶+∞yk−Auk=0.Proof.
In virtue of (32) and (42), we receive(44)xk+1−x∗≤uk−x∗≤xk−x∗,which results in that(45)limk⟶+∞uk−x∗=limk⟶+∞xk−x∗.
Then, the sequencesxk, yk, vk, uk, and zk are all bounded.
By (42), we have(46)γ1−γA2yk−Auk2+γαk1−ς22ϖk2ϖk+12zk−Auk2≤xk−x∗2−xk+1−x∗2⟶0.
This together with conditions (i) and (ii) and Remark1 implies that(47)limk⟶+∞yk−Auk=0,(48)limk⟶+∞zk−Auk=0.
Due to (32), we obtain(49)τk1−ς22ϑk2ϑk+12vk−xk2≤xk−x∗2−uk−x∗2⟶0,which results in that(50)limk⟶+∞vk−xk=0.
From (18), we get(51)uk−xk=τkvk−xk−ϑkhvk−hxk≤τkvk−xk+τkϑkhvk−hxk.
Based on the Lipschitz continuity ofh and (50), we derive(52)limk⟶+∞uk−xk=0.
By (21), (47), and (52), we deduce(53)limk⟶+∞xk+1−xk=0.By Proposition1, the sequence xk is bounded. Let b^ be any point in ωwxk and xki be a subsequence of xk satisfying xki⇀b^ as i⟶∞.Proposition 3.
b^∈g+h−10.Proof.
Letp^,q^∈Gg+h. Then, q^−hp^∈gp^. From (17), we have vki=Resϑkigxki−ϑkihxki=I+ϑkig−1xki−ϑkihxkiA which yields that(54)xki−vkiϑki−hxki∈gvki.
By the monotonicity ofg and (54), we attain q^−hp^−xki−vki/ϑki−hxki,p^−vki≥0. Thus,(55)q^,p^−vki≥hp^−hxki+xki−vkiϑki,p^−vki=hp^−hvki,p^−vki+hvki−hxki,p^−vki+1ϑkixki−vki,p^−vki.
Note thathp^−hvki,p^−vki≥0 because of the monotonicity of h. It follows from (55) that(56)q^,p^−vki≥hvki−hxki,p^−vki+1ϑkixki−vki,p^−vki.
Sincexki⇀b^, by Proposition 3, we conclude that uki⇀b^, vki⇀b^, yki⇀Ab^, and zki⇀Ab^. With the help of (50) and (56), we obtain q^,p^−b^≥0 for all p^,q^∈Gg+h which leads to that b^∈g+h−10.Proposition 4.
Ab^∈ψ+φ−10.Proof.
Letp†,q†∈Gφ+ψ. Thus, q†−φp†∈ψp†. From (19), we get zki=ResϖkiψAuki−ϖkiφAuki=I+ϖkiψ−1Auki−ϖkiφAuki which results in that(57)Auki−zkiϖki−φAuki∈ψzki.
Applying the monotonicity ofψ to (57), we derive q†−φp†−Auki−zki/ϖki−φAuki,p†−zki≥0. It follows that(58)q†,p†−zki≥φp†−φAuki+Auki−zkiϖki,p†−zki=φp†−φzki,p†−zki+φzki−φAuki,p†−zki+1ϖkiAuki−zki,p†−zki.
Notice thatφp†−φzki,p†−zki≥0. This together with (58) implies that(59)q†,p†−zki≥φzki−φAuki,p†−zki+1ϖkiAuki−zki,p†−zki.
Owing tozki⇀Ab^ and Auki−zki⟶0, from (59), we conclude that q†,p†−Ab^≥0 for all p†,q†∈Gφ+ψ which implies that Ab^∈φ+ψ−10.According to Propositions1–4, we can prove Theorem 1.Proof.
In fact, we have obtained the following results:(i)
∀x∗∈Θ, limk⟶∞‖xk−x∗‖ exists by Propositions 1(ii)
ωwxk⊂Θ due to Propositions 3 and 4.
Utilizing Lemma1, we can conclude that xk converges weakly to some point in Θ. This completes the proof.Remark 2.
In this paper, we study the split variational inclusion problem (7). We construct an iterative algorithm (Algorithm 1) for finding a solution of (7). It should be pointed out that the involved two operators h and φ are all plain monotone. Our result improves and extends some corresponding results in [2, 23, 29].



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