# Iterative Solutions for Solving Variational Inequalities and Fixed-Point Problems

Authors: Li-Jun Zhu; Naseer Shahzad; Asim Asiri
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we are interested in variational inequalities and fixed-point problems in Hilbert spaces. We present an iterative algorithm for finding a solution of the studied variational inequalities and fixed-point problems. We show the strong convergence of the suggested algorithm.


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## Body

## 1. Introduction

LetH be a real Hilbert space with inner product ·,· and norm ∥·∥. Let C⊂H be a nonempty closed convex set. Let f:C⟶H, g:C⟶H, φ:C⟶C, and T:C⟶C be four nonlinear operators. Use FixT to denote the fixed-point set of T.In this paper, we will investigate the following variational inequalities and fixed-point problems of finding a pointu† such that(1)u†∈GVIC,f,φ,φu†∈VIC,g∩FixT,where GVIC,f,φ denotes the solution set of the generalized variational inequality (shortly, GVI) which is to find a point x†∈C such that(2)fx†,φx−φx†≥0,∀x∈C,and VIC,g means the solution set of the variational inequality (shortly, VI) which is to find a point x†∈C such that(3)gx†,x−x†≥0,∀x∈C.Throughout, we useΘ to denote the solution set of problem (1), that is,(4)Θ=GVIC,f,φ⋂φ−1VIC,g∩FixT.It is well known that variational inequalities play key roles and provide a useful mathematical framework, theory, and method for studying many valuable problems arising in water resources, finance, economics, medical images, and so on ([1–6]). A lot of work and a great deal of algorithms for solving GVI or VI have been introduced and investigated, see, e.g., [7–15]. Among them, a basic and important algorithm is the projected algorithm which generates a sequence xn with the form(5)xn+1=projCxn−κnfxn,n≥0,where κn is step-size and projC:H⟶C is the orthogonal projection.At the same time, we are also interested in the fixed-point problem of finding a pointu† such that Tu†=u†. Iterative solution for solving a fixed-point problem is an active research field, see, e.g., [16–24]. Recently, iterative algorithms for solving variational inequalities and fixed-point problems have been investigated extensively by many authors [25–33].Motivated by the work in this direction, in this paper, we devote to research variational inequalities and fixed-point problem (1). We introduce an iterative algorithm for finding a solution of problem (1). We show the strong convergence of the suggested algorithm.


## 2. Preliminaries

LetC be a nonempty closed convex subset of a real Hilbert space H. Recall that an operator f:C⟶H is said to be(i)
strongly monotone if(6)fu−fv,u−v≥λu−v2,∀u,v∈C(ii)
ϑ-inverse strongly φ-monotone if there exists a constant ϑ>0 such that(7)fu−fv,φu−φv≥ϑfu−fv2,∀u,v∈C(iii)
relaxedμ,ν-cocoercive [34, 35], if there exist two constants μ>0,ν>0 such that(8)fu−fv,u−v≥−μfu−fv2+νu−v2,∀u,v∈CAn operatorT:C⟶C is said to be(i)
pseudocontractive [36] if(9)∥Tu†−Tv†∥2≤∥u†−v†∥2+∥I−Tu†−I−Tv†∥2,∀u†,v†∈C(ii)
L-Lipschitz if(10)Tu†−Tv†≤Lu†−u†,∀u†,v†∈C,where L>0 is a constantIfL<1, then T is said to be L-contraction. If L=1, then T is said to be nonexpansive.An operatorA:H⟶2H is said to be monotone if x−y,u−v≥0 for all x,y∈domA, u∈Ax, and v∈Ay. A monotone operator A on H is said to be maximal if its graph is not strictly contained in the graph of any other monotone operator on H.For∀x†∈H, there exists a unique point in C, denoted by projCx† satisfying(11)x†−projCx†≤x−x†,∀x∈C.Moreover,projC is firmly nonexpansive, that is,(12)projCu†−projCv†2≤projCu†−projCv†,u†−v†,∀u†,v†∈H.Further,projC has the following property:(13)u†−projCu†,x†−projCu†≤0,∀u†∈H,x†∈C.Lemma 1 ([37]).
LetC be a nonempty closed convex subset of a real Hilbert space H. Let T:C⟶C be an L-Lipschitz pseudocontractive operator. Then, ∀x†∈C and y†∈FixT, we have(14)1−ςx†+ςT1−λx†+λTx†−y†2≤ςς−λT1−λx†+λTx†−x†2+x†−y†2,where 0<ς<λ<1/1+L2+1.Lemma 2 ([24]).
LetC be a nonempty, convex, and closed subset of a Hilbert space H. Let T:C⟶C be a continuous pseudocontractive operator. Then,(i)
FixT⊂C is closed and convex(ii)
T is demiclosedness, i.e., un⇀z~ and Tun⟶z† imply that Tz~=z†Lemma 3 ([23]).
Letϖn⊂0,∞, ϑn⊂0,1, and ηn be real number sequences. Suppose the following conditions are satisfied:(i)
ϖn+1≤1−ϑnϖn+ηn,∀n≥1(ii)
∑n=1∞ϑn=∞(iii)
limsupn⟶∞ηn/ϑn≤0 or ∑n=1∞∣ηn∣<∞
Then,limn⟶∞ϖn=0.Lemma 4 ([38, 39]).
Letxn be a real number sequence. Assume there exists at least a subsequence xnk of xn such that(15)xnk≤xnk+1,for all k≥0. For every n≥N0, define an integer sequence μn as(16)μn=maxi≤n:xni<xni+1.Then,μn⟶∞ as n⟶∞ and for all n≥N0, maxxμn,xn≤xμn+1.


## 3. Main Results

In this section, we present our iterative algorithm and convergence theorem. LetC be a nonempty closed convex subset of a real Hilbert space H. Assume that(i)
ϕ:C⟶C is a ρ-contractive operator(ii)
φ:C⟶C is a weakly continuous and λ-strongly monotone operator such that its rang Rφ=C(iii)
f:C⟶H is a ϑ-inverse strongly φ-monotone operator(iv)
g:C⟶H is an L1-Lipschitz and relaxed μ,ν-cocoercive operator(v)
T:C⟶C is an L2-Lipschitz pseudocontractive operator with L2>1Letϑn, ςn, λn, and τn be four real number sequences in 0,1 and κn and γn be two real number sequences in 0,∞.Now, we present our algorithm for solving problem (1).Algorithm 5.
Letx0∈C be an initial value. Define the sequence xn by the following form:(17)un=ϑnϕxn+1−ϑnprojCφxn−κnfxn,yn=1−ςnun+ςnT1−λnun+λnTun,zn=projCyn−γngyn,φxn+1=1−τnφxn+τnzn,n≥0.Theorem 6.
Suppose thatΘ≠∅. Assume that the following conditions are satisfied:
C1:limn⟶∞ϑn=0 and ∑n=1∞ϑn=∞
C2:0<a1<ςn<c1<λn<b1<1/1+L22+1 for all n≥0
C3:ν>μL12 and 0<a2≤γn≤b2<2ν−μL12/L12 for all n≥0
C4:0<liminfn⟶∞τn≤limsupn⟶∞τn<1
C5:0<ρ<λ<2ϑ and 0<liminfn⟶∞κn≤limsupn⟶∞κn<2ϑ
Then, the sequencexn generated by (17) converges strongly to u†∈Θ verifying(18)ϕu†−φu†,φx†−φu†≤0,∀x†∈Θ.Proof.
Sinceφ is λ-strongly monotone, we can get from (6) that(19)φu−φv≥λu−v,∀u,v∈C.
Thus, VI (18) has a unique solution, denoted by u†. It follows that u†∈GVIC,f,φ and φu†∈FixT∩VIC,g. Using inequality (13), we can obtain that φu†=projCφu†−κnfu† for all n≥0.
Sincef is ϑ-inverse strongly φ-monotone, for any u∈C, we have(20)φu−κfu−φu†−κfu†2=φu−φu†2−2κfu−fu†,φu−φu†+κ2fu−fu†2≤φu−φu†2−2κϑfu−fu†2+κ2fu−fu†2≤φu−φu†2+κκ−2ϑfu−fu†2.
Based on (20), we deduce(21)φxn−κnfxn−φu†−κnfu†2≤φxn−φu†2+κnκn−2ϑfxn−fu†2≤φxn−φu†2,(22)φxn+1−κn+1fxn+1−φxn−κn+1fxn2≤φxn+1−φxn2+κn+1κn+1−2ϑfxn+1−fxn2.
By (17), (19), and (21), we derive(23)un−φu†=ϑnϕxn+1−ϑnprojCφxn−κnfxn−projCφu†−κnfu†≤ϑnϕxn−φu†+κnfu†+1−ϑnφxn−κnfxn−φu†−κnfu†≤ϑnϕxn−ϕu†+ϑnϕu†−φu†+κnfu†+1−ϑnφxn−κnfxn−φu†−κnfu†≤ϑnρxn−u†+ϑnϕu†−φu†+κnfu†+1−ϑnφxn−φu†≤ϑnρλφxn−φu†+ϑnϕu†−φu†+κnfu†+1−ϑnφxn−φu†=1−1−ρλϑnφxn−φu†+ϑnϕu†−φu†+κnfu†≤1−1−ρλϑnφxn−φu†+ϑnϕu†−φu†+2ϑfu†.
According to (21) and (23), we obtain(24)∥un−φu†∥2≤∥ϑnϕxn−φu†+κnfu†+1−ϑnφxn−κnfxn−φu†−κnfu†∥2≤ϑn∥ϕxn−φu†+κnfu†∥2+1−ϑn∥φxn−κnfxn−φu†−κnfu†∥2≤ϑn∥ϕxn−φu†+κnfu†∥2+1−ϑn∥φxn−φu†∥2+κnκn−2ϑ∥fxn−fu†∥2.
Applying Lemma1 to (17), we have(25)∥yn−φu†∥2=∥1−ςnun+ςnT1−λnun+λnTun−φu†∥2≤∥un−φu†∥2+ςnςn−λn∥T1−λnun+λnTun−un∥2≤∥un−φu†∥2.
Sinceg is relaxed μ,ν-cocoercive and L1-Lipschitz, for all u,v∈C, we have(26)∥I−γngu−I−γngv∥2=∥u−v∥2−2γngu−gv,u−v+γn2∥gu−gv∥2≤∥u−v∥2−2γn−μ∥gu−gv∥2+ν∥u−v∥2+γn2∥gu−gv∥2≤∥u−v∥2+2γnμL12∥u−v∥2−2γnν∥u−v∥2+γn2L12∥u−v∥2=1+2γnμL12−2γnν+γn2L12∥u−v∥2.
Since0<γn<2ν−μL12/L12, 1+2γnμL12−2γnν+γn2L12≤1. Thus, from (26), we obtain(27)∥I−γngu−I−γngv∥≤∥u−v∥,∀u,v∈C.
Hence,(28)∥zn−φu†∥=∥projCI−γngyn−projCI−γngφu†∥≤∥I−γngyn−I−γngφu†∥≤∥yn−φu†∥.
Combining (17), (23), (25), and (28), we obtain(29)∥φxn+1−φu†∥≤1−τn∥φxn−φu†∥+τn∥zn−φu†∥≤1−τn∥φxn−φu†∥+τn∥un−φu†∥≤1−τn∥φxn−φu†∥+τn1−1−ρλϑn∥φxn−φu†∥+τnϑn∥ϕu†−φu†∥+2ϑ∥fu†∥=1−1−ρλτnϑn∥φxn−φu†∥+1−ρλτnϑn∥ϕu†−φu†∥+2ϑ∥fu†∥1−ρ/λ.
By induction, we have(30)∥φxn−φu†∥≤max∥φx0−φu†∥,∥ϕu†−φu†∥+2ϑ∥fu†∥1−ρ/λ.
It follows that(31)∥xn−u†∥≤1λ∥φxn−φu†∥≤1λmax∥φx0−φu†∥,∥ϕu†−φu†∥+2ϑ∥fu†∥1−ρ/λ.
So,φxn, xn, yn, zn, and un are bounded.
From (17), we have(32)φxn+1−φxn=τnzn−φxn,n≥0.
It follows that(33)φxn+1−φxn,φxn−φu†=τnzn−φxn,φxn−φu†.
Thanks to (33), we deduce(34)∥φxn+1−φu†∥2−∥φxn−φu†∥2−∥φxn+1−φxn∥2=τn∥zn−φu†∥2−∥φxn−φu†∥2−∥zn−φxn∥2.
Combining (32) and (34), we obtain(35)∥φxn+1−φu†∥2−∥φxn−φu†∥2=τn∥zn−φu†∥2−∥φxn−φu†∥2−∥zn−φxn∥2+τn2∥zn−φxn∥2=τn∥zn−φu†∥2−∥φxn−φu†∥2−τn1−τn∥zn−φxn∥2≤τn∥un−φu†∥2−∥φxn−φu†∥2−τn1−τn∥zn−φxn∥2.
By virtue of (23), we get(36)∥un−φu†∥2≤1−1−ρλϑn∥φxn−φu†∥2+1−ρλϑn∥ϕu†−φu†∥+2ϑ∥fu†∥1−ρ/λ2.
Now, we consider two cases.
Case 1. There exists some integer N0>0 such that ∥φxn−φu†∥ is decreasing when n≥N0. Then, limn⟶∞∥φxn−φu†∥ exists. According to (35), (36), and C1, we have(37)τn1−τn∥zn−φxn∥2≤∥φxn−φu†∥2−∥φxn+1−φu†∥2+τn∥un−φu†∥2−∥φxn−φu†∥2≤∥φxn−φu†∥2−∥φxn+1−φu†∥2+1−ρλϑn∥ϕu†−φu†∥+2ϑ∥fu†∥1−ρ/λ2⟶0.
This together withC4 implies that(38)limn⟶∞∥zn−φxn∥=0.
Therefore, by (32), we have(39)limn⟶∞∥φxn+1−φxn∥=0.
By (24), we have(40)∥φxn+1−φu†∥2=∥1−τnφxn−φu†+τnzn−φu†∥2≤1−τn∥φxn−φu†∥2+τn∥zn−φu†∥2≤1−τn∥φxn−φu†∥2+τn∥un−φu†∥2≤1−τn∥φxn−φu†∥2+τnϑn∥ϕxn−φu†+κnfu†∥2+τn1−ϑnκnκn−2ϑ∥fxn−fu†∥2+τn1−ϑn∥φxn−φu†∥2≤∥φxn−φu†∥2+τnϑn∥ϕxn−φu†+κnfu†∥2+τn1−ϑnκnκn−2ϑ∥fxn−fu†∥2.
It results in that(41)τn1−ϑnκn2ϑ−κn∥fxn−fu†∥2≤∥φxn−φu†∥2−∥φxn+1−φu†∥2+τnϑn∥ϕxn−φu†+κnfu†∥2≤∥φxn−φu†∥+∥φxn+1−φu†∥∥φxn+1−φxn∥+τnϑn∥ϕxn−φu†+κnfu†∥2⟶0.
Hence,(42)limn⟶∞∥fxn−fu†∥=0.
Setvn=φxn−κnfxn−φu†−κnfu† for all n≥0. Using (13) and (21), we have(43)∥projCφxn−κnfxn−φu†∥2=∥projCφxn−κnfxn−projCφu†−κnfu†∥2≤vn,projCφxn−κnfxn−φu†=12∥vn∥2+∥projCφxn−κnfxn−φu†∥2−∥φxn−projCφxn−κnfxn−κnfxn−fu†∥2≤12∥φxn−φu†∥2+∥projCφxn−κnfxn−φu†∥2−∥φxn−projCφxn−κnfxn∥2−κn2∥fxn−fu†∥+2κnφxn−projCφxn−κnfxn,fxn−fu†.
It yields(44)∥projCφxn−κnfxn−φu†∥2≤∥φxn−φu†∥2−κn2∥fxn−fu†∥−∥φxn−projCφxn−κnfxn∥2+2κnφxn−projCφxn−κnfxn,fxn−fu†.
In the light of (17) and (44), we have(45)∥un−φu†∥2≤ϑn∥ϕxn−φu†∥2+1−ϑn∥projCφxn−κnfxn−φu†∥2≤ϑn∥ϕxn−φu†∥2+1−ϑn∥φxn−φu†∥2+2κn∥φxn−projCφxn−κnfxn∥∥fxn−fu†∥−1−ϑn∥φxn−projCφxn−κnfxn∥2.
Based on (40) and (45), we obtain(46)∥φxn+1−φu†∥2≤1−τn∥φxn−φu†∥2+τn∥un−φu†∥2≤∥φxn−φu†∥2+τnϑn∥ϕxn−φu†∥2−τn1−ϑn∥φxn−projCφxn−κnfxn∥2+2τnκn∥φxn−projCφxn−κnfxn∥∥fxn−fu†∥.
Then,(47)τn1−ϑn∥φxn−projCφxn−κnfxn∥2≤∥φxn−φu†∥+∥φxn+1−φu†∥∥φxn+1−φxn∥+τnϑn∥ϕxn−φu†∥2+2τnκn∥φxn−projCφxn−κnfxn∥∥fxn−fu†∥.
According toC1, C4, (39), (42), and (47), we deduce(48)limn⟶∞∥φxn−projCφxn−κnfxn∥=0.
Sinceun−ϕxn=1−ϑnprojCφxn−κnfxn−ϕxn, from (38), (39), and (48), we have(49)limn⟶∞∥φxn−un∥=limn⟶∞∥φxn+1−un∥=limn⟶∞∥zn−un∥=0.
From (26) and (28), we get(50)∥zn−φu†∥2≤∥yn−φu†∥2−2γn−μ∥gyn−gφu†∥2+ν∥yn−φu†∥2+γn2∥gyn−gφu†∥2≤∥yn−φu†∥2+2γnμ+γn2−2γnνL12∥gyn−gφu†∥2.
It follows that(51)∥φxn+1−φu†∥2≤1−τn∥φxn−φu†∥2+τn∥un−φu†∥2+τn2γnμ+γn2−2γnνL12∥gyn−gφu†∥2,which together with (49) implies that(52)−τn2γnμ+γn2−2γnνL12∥gyn−gφu†∥2≤1−τn∥φxn−φu†∥2−∥φxn+1−φu†∥2+τn∥un−φu†∥2−∥φxn+1−φu†∥2⟶0.
Therefore,(53)limn⟶∞∥gyn−gφu†∥=0.
SinceprojC is firmly nonexpansive, from (12) and (28), we have(54)∥zn−φu†∥2=∥projCI−γngyn−projCI−γngφu†∥2≤I−γngyn−I−γngφu†,zn−φu†=12∥I−γngyn−I−γngφu†∥2+∥zn−φu†∥2−∥I−γngyn−I−γngφu†−zn−φu†∥2≤12∥yn−φu†∥2+∥zn−φu†∥2−∥yn−zn−γngyn−gφu†∥2≤12∥un−φu†∥2+∥zn−φu†∥2−∥yn−zn∥2−γn2∥gyn−gφu†∥2+2γngyn−gφu†,yn−zn,which yields(55)∥zn−φu†∥2≤∥un−φu†∥2−∥yn−zn∥2+2γn∥gyn−gφu†∥∥yn−zn∥.
This together with (40) implies that(56)∥φxn+1−φu†∥2≤1−τn∥φxn−φu†∥2+τn∥un−φu†∥2−τn∥yn−zn∥2+2τnγn∥gyn−gφu†∥∥yn−zn∥.
It follows that(57)τn∥yn−zn∥2≤1−τn∥φxn−φu†∥2+τn∥un−φu†∥2−∥φxn+1−φu†∥2+2τnγn∥gyn−gφu†∥∥yn−zn∥⟶0by49and53.
So,(58)limn⟶∞∥yn−zn∥=0.
By (49) and (58), we have(59)limn⟶∞∥yn−un∥=0.
In view of (25) and (59), we get(60)ςnλn−ςn∥T1−λnun+λnTun−un∥2≤∥un−φu†∥2−∥yn−φu†∥2≤∥un−yn∥∥un−φu†∥+∥yn−φu†∥⟶0.
It follows fromC2 and (60) that(61)limn⟶∞∥T1−λnun+λnTun−un∥=0.
SinceT is L2-Lipschitz, we have(62)∥Tun−un∥≤∥Tun−T1−λnun+λnTun∥+∥T1−λnun+λnTun−un∥≤λnL2∥Tun−un∥+∥T1−λnun+λnTun−un∥.
Hence,(63)∥Tun−un∥≤11−λnL2∥T1−λnun+λnTun−un∥.
Owing toC2, (61) and (63), we deduce(64)limn⟶∞∥Tun−un∥=0.
Next, we show thatlimsupn⟶∞ϕu†−φu†,un−φu†≤0. Let uni be a subsequence of un such that(65)limsupn⟶∞ϕu†−φu†,un−φu†=limi⟶∞ϕu†−φu†,uni−φu†.
Sincexni is bounded, there exists a subsequence xnij of xni which converges weakly to some point z∈C. Without loss of generality, we may assume that xni⇀z. This implies that φxni⇀φz due to the weak continuity of φ. Thus, uni⇀φz, yni⇀φz, and zni⇀φz. Applying Lemma 2 to (64) to deduce φz∈FixT.
Now, we show thatφz∈VIC,g. Let(66)S1v=gv+NCv,v∈C,∅,v∉C.
Sinceg is relaxed μ,ν-cocoercive, for all x,y∈C, we have(67)gx−gy,x−y≥−μ∥gx−gy∥2+ν∥x−y∥2≥ν−μL2∥x−y∥2≥0,which implies that g is monotone and so S1 is maximal monotone. Let v,u∈GS1. Owing to u−gv∈NCv and zn∈C, we get(68)v−zn,u−gv≥0.
According tozn=projCI−γngyn, we obtain(69)v−zn,zn−I−γngyn≥0.
Then,(70)v−zn,zn−ynγn+gyn≥0.
It follows that(71)v−zni,u≥v−zni,gv−gzni+v−zni,gzni−gyni−v−zni,zni−yniγni≥v−zni,gzni−gyni−v−zni,zni−yniγni.
Sincezni⇀φz, ∥zni−yni∥⟶0, it follows from (71) that v−φz,u≥0. Therefore, φz∈S1−10 and φz∈VIC,g.
Next, we provez∈GVIC,f,φ. Let(72)S2v=fv+NCv,v∈C,∅,v∈C.
It is known thatS2 is maximal φ-monotone. Let v,w∈GS2. Since w−fv∈NCv and xn∈C, we have φv−φxn,w−fv≥0. Set wn=projCφxn−κnfxn. Then,(73)φv−wn,wn−φxn−κnfxn≥0.
It follows that(74)φv−wn,wn−φxnκn+fxn≥0.
Thus,(75)φv−φxni,w≥φv−φxni,fv−fxni+φv−φxni,fxni−φv−wni,wni−φxniκni−φv−wni,fxni≥−φv−wni,wni−φxniκni−φxni−wni,fxni.
Since∥φxni−wni∥⟶0 and φxni⇀φz, we deduce that φv−φz,w≥0 by taking i⟶∞ in (75). Thus, z∈S2−10 by the maximal φ-monotonicity of S2. Hence, z∈GVIC,f,φ. Therefore, z∈φ−1FixT∩VIC,g⋂GVIC,f,φ=Θ.
From (49) and (65), we obtain(76)limsupn⟶∞ϕu†−φu†,un−φu†=limi⟶∞ϕu†−φu†,φxni−φu†=ϕu†−φu†,φz−φu†≤0.
By (17), we have(77)∥un−φu†∥2=∥ϑnϕxn−φu†+1−ϑnprojCφxn−κnfxn−φu†∥2≤1−ϑn2∥projCφxn−κnfxn−φu†∥2+2ϑnϕxn−φu†,un−φu†≤1−ϑn2∥φxn−φu†∥2+2ϑnϕxn−ϕu†,un−φu†+2ϑnϕu†−φu†,un−φu†≤1−ϑn2∥φxn−φu†∥2+2ϑnρλ∥φxn−φu†∥∥un−φu†∥+2ϑnϕu†−φu†,un−φu†≤1−ϑn2∥φxn−φu†∥2+ϑnρλ∥φxn−φu†∥2+ϑnρλ∥un−φu†∥2+2ϑnϕu†−φu†,un−φu†.
It follows that(78)∥un−φu†∥2≤1−21−ρ/λϑn1−ϑnρ/λ∥φxn−φu†∥2+ϑn21−ϑnρ/λ∥φxn−φu†∥2+2ϑn1−ϑnρ/λϕu†−φu†,un−φu†.
SetM=supn∥φxn−φu†∥2. Therefore,(79)∥φxn+1−φu†∥2≤1−τn∥φxn−φu†∥2+τn∥un−φu†∥2≤1−21−ρ/λϑnτn1−ϑnρ/λ∥φxn−φu†∥2+ϑn2τn1−ϑnρ/λ∥φxn−φu†∥2+2ϑnτn1−ϑnρ/λϕu†−φu†,un−φu†=1−21−ρ/λϑnτn1−ϑnρ/λ∥φxn−φu†∥2+21−ρ/λϑnτn1−ϑnρ/λ×ϑn21−ρ/λM+11−ρ/λϕu†−φu†,un−φu†.
By Lemma3 and (79), we conclude that φxn⟶φu† and xn⟶u†.
Case 2. For any N, there exists an integer n0>N such that ∥φxn0−φu†∥≤∥φxn0+1−φu†∥. Let ψn=∥φxn−φu†∥2. Then, we have ψn0≤ψn0+1. Let μn be an integer sequence defined by, for all n≥n0,(80)μn=maxl∈ℕ∣n0≤l≤n,ψl≤ψl+1.
Note thatμn is nondecreasing and satisfies limn⟶∞μn=∞ and ψμn≤ψμn+1,∀n≥n0.
Similarly, we can deduce(81)limsupn⟶∞ϕu†−φu†,uμn−φu†≤0(82)ψμn+1≤1−21−ρ/λϑμnτμn1−ϑμnρ/λψμn+21−ρ/λϑμnτμn1−ϑμnρ/λ×ϑμn21−ρ/λM+11−ρ/λϕu†−φu†,uμn−φu†.
Note thatψμn≤ψμn+1. By (82), we have(83)ψμn≤ϑμn21−ρ/λM+11−ρ/λϕu†−φu†,uμn−φu†.
Based on (81) and (83), we derive(84)limsupn⟶∞ψμn≤0,and thus,(85)limn⟶∞ψμn=0.
From (81) and (82), we can deduce(86)limsupn⟶∞ψμn+1≤limsupn⟶∞ψμn.
This together with (85) implies that(87)limn⟶∞ψμn+1=0.
By Lemma4, we obtain(88)0≤ψn≤maxψμn,ψμn+1.
Therefore,ψn⟶0. That is, φxn⟶φu† and thus xn⟶u†. This completes the proof.In Algorithm5, choose φ=I, identity operator, and f:C⟶H is a ϑ-inverse strongly monotone operator. Then, we have the following algorithm and corollary.Algorithm 7.
Letx0∈C be an initial value. Define the sequence xn by the following form:(89)un=ϑnϕxn+1−ϑnprojCxn−κnfxn,yn=1−ςnun+ςnT1−λnun+λnTun,zn=projCyn−γngyn,xn+1=1−τnxn+τnzn,n≥0.Corollary 8.
Suppose thatΘ1≔VIC,f∩VIC,g∩FixT≠∅. Assume that conditions C1-C5 are satisfied. Then, the sequence xn generated by (89) converges strongly to v†=projΘ1ϕv†.Algorithm 9.
Letx0∈C be an initial value. Define the sequence xn by the following form:(90)un=ϑnϕxn+1−ϑnprojCφxn−κnfxn,zn=projCun−γngun,φxn+1=1−τnφxn+τnzn,n≥0.Corollary 10.
Suppose thatΘ2≔GVIC,f,φ∩φ−1VIC,g≠∅. Assume that conditions C1 and C3-C5 are satisfied. Then, the sequence xn generated by (90) converges strongly to u†∈Θ2 verifying(91)ϕu†−φu†,φx†−φu†≤0,∀x†∈Θ2.



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