# Strong Convergence Results of Split Equilibrium Problems and Fixed Point Problems

Authors: Li-Jun Zhu; Hsun-Chih Kuo; Ching-Feng Wen
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we investigate the split equilibrium problem and fixed point problem in Hilbert spaces. We propose an iterative scheme for solving such problem in which the involved equilibrium bifunctionsf and g are pseudomonotone and monotone, respectively, and the operators S and T are all pseudocontractive. We show that the suggested scheme converges strongly to a solution of the considered problem.


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## Body

## 1. Introduction

Letℋ1 and ℋ2 be two real Hilbert spaces. Let C and Q be two nonempty, closed, and convex subsets of ℋ1 and ℋ2, respectively. Let f:C×C⟶ℝ be a bifunction. Recall that the equilibrium problem is to find a point x∗∈C such that(1)fx∗,x≥0,∀x∈C.UseSEPC,f to denote the solution set of equilibrium problem (1).Equilibrium problems have been considered broadly in the literature (see e.g. [1–5]). Now, it is known that variational inequalities ([6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17]) and fixed point problems ([18, 19, 20, 21, 22, 23]) can be transformed in the form of (1). For every σ>0 and x∈H, there exists a unique point z∈C such that fz,y+1/σz−x,y−x≥0,∀y∈C (see [2]). Thus, for solving equilibrium problem (1), an important technique is to use the resolvent of bifunction f ([2]). Another important method for solving equilibrium problem (1) is to use linear search technique [4].LetS:C⟶C and T:Q⟶Q be two operators. Let FixS and FixT be the fixed point sets of S and T, respectively. Let g:Q×Q⟶ℝ be a bifunction. Let A:H1⟶H2 be a bounded linear operator. In this paper, we concern the following split problem of finding a point u˜∈C such that(2)u˜∈SEPC,f∩FixS,Au˜∈SEPQ,g∩FixT.Denote the solution set of (2) by Γ, i.e., Γ=x∗∈SEPC,f∩FixS,Ax∗∈SEPQ,g∩FixT.The split problem has received many concerns (see [13, 24–28]) due to its extensive applications in image recovery and signal processing, control theory, and so on. Note that the split problem (2) includes the following split problems as special cases:(i)
The split equilibrium problem studied in [29, 30] can be formulated to find an element u˜∈C such that(3)u˜∈SEPC,f,Au˜∈SEPQ,g.The solution set of (3) is denoted by Γ1.(ii)
The split fixed point problem considered in [31, 32, 33, 34] reduces to find a point u˜∈C such that(4)u˜∈FixS,Au˜∈FixT.The solution set of (4) is denoted by Γ2.Numerical iterative algorithms have been proposed for finding a split problem of the set of solutions of equilibrium problems and the set of fixed points of nonexpansive operators; see, for example, [35–39] and the references therein. Recently, Yao et al. [40] proposed an iterative scheme for solving the split problem (2) and they obtained the weak convergence of the suggested scheme.In this paper, we continuously study the split problem (2) in which the involved equilibrium bifunctions f and g are pseudomonotone and monotone, respectively, and the operators S and T are all pseudocontractive. We propose an iterative scheme for solving the split problem (2) and strong convergence results are obtained.


## 2. Preliminaries

Letℋ1 be a real Hilbert space with its inner product ⋅,⋅ and norm ⋅. Let C be a nonempty, convex, and closed subset of ℋ1. Let PC:ℋ1⟶C be the metric projection defined by(5)PCx=argminy∈Cy−x.PC satisfies: for given x∈H1,(6)x−PCx,y−PCx≤0,∀y∈C.Letf:C×C⟶ℝ be a bifunction. Recall that f is said to be monotone if(7)fu†,v†+fv†,u†≤0,∀u†,v†∈C.f is said to be pseudomonotone if(8)fu†,v†≥0impliesfv†,u†≤0,∀u†,v†∈C.LetS:C⟶C be an operator. S is called pseudocontractive if(9)Sx−Sx†2≤x−x†2+I−Sx−I−Sx†2,∀x,x†∈C.S is called L-Lipschitz if there exists a constant L≥0 such that(10)Sx−Sx†≤Lx−x†,∀x,x†∈C.IfL=1, then S is said to be nonexpansive. If L<1, then S is said to be L-contraction.In the sequel, we use the following symbols. Letxk be a sequence in C:(i)
xk⇀x† means the weak convergence of xk to x† as k⟶∞(ii)
xk⟶x† means the strong convergence of xk to x† as k⟶∞(iii)
ωwxk=x†:∃xki⊂xksuch thatxki⇀x†i⟶∞Recall thatf is said to be jointly sequently weakly continuous on C×C, if for two sequences xk∈C and yk∈C satisfy xk⇀u† and yk⇀v†, then we have fxk,yk⟶fu†,v†.Letℋ2 be a real Hilbert space with its inner product ⋅,⋅ and norm ⋅. Let Q be a nonempty, convex, and closed subset of ℋ2. Let φ:Q⟶−∞,+∞ be a proper, lower semicontinuous, and convex function. Then, the subdifferential ∂φ of φ is defined by(11)∂φu≔v†∈H2:φu+v†,u†−u≤φu†,∀u†∈Q,for each u∈Q.It is well known that(12)u†=argminu∈Qφu⟺0∈∂φu†+NQu†,where NQu†=ω∈H2:ω,u−u†≤0,∀u∈Q.The following lemma can be found in [41]. For the completeness, we include the detail of proof.Lemma 1.
(see [41]). Let S:C⟶C be an L1-Lipschitz pseudocontractive operator. Then, for all u˜∈C and u†∈FixS, we have(13)u†−S1−ηu˜+ηSu˜2≤u˜−u†2+1−ηu˜−S1−ηu˜+ηSu˜2,where 0<η<1/1+L12+1.Proof.
Sinceu†∈Fiu˜S, we have from (9) that(14)S1−ηI+ηSu˜−u†2≤1−ηu˜−u†+ηSu˜−u†2+1−ηu˜+ηSu˜−S1−ηu˜+ηSu˜2,(15)Su˜−u†2≤u˜−u†2+Su˜−u˜2,for all u˜∈C.
SinceS is L1-Lipschitzian and u˜−1−ηu˜+ηSu˜=ηu˜−Su˜, we have(16)Su˜−S1−ηu˜+ηSu˜≤ηL1u˜−Su˜.
According to (15), we obtain(17)1−ηu˜−u†+ηSu˜−u†2=1−ηu˜−u†2+ηSu˜−u†2−η1−ηu˜−Su˜2≤1−ηu˜−u†2+ηu˜−u†2+Su˜−u˜2−η1−ηu˜−Su˜2=u˜−u†2+η2Su˜−u˜2.
Based on (16), we conclude(18)1−ηu˜+ηSu˜−S1−ηu˜+ηSu˜2=1−ηu˜−S1−ηu˜+ηSu˜+ηSu˜−S1−ηu˜+ηSu˜2=1−ηu˜−S1−ηu˜+ηSu˜2+ηSu˜−S1−ηu˜+ηSu˜2−η1−ηu˜−Su˜2≤1−ηu˜−S1−ηu˜+ηSu˜2−η1−η−η2L12u˜−Su˜2.
By (14), (17), and (18), we obtain(19)S1−ηI+ηSu˜−u†2≤u˜−u†2+η2u˜−Su˜2+1−ηu˜−S1−ηu˜+ηSu˜2−η1−η−η2L12u˜−Su˜2=u˜−u†2+1−ηu˜−S1−ηI+ηSu˜2−η1−2η−η2L12u˜−Su˜2.
Sinceη<1/1+L12+1, 1−2η−η2L12>0. Hence, we can deduce the desired result from (19).Lemma 2.
(see [42]). Let S:C⟶C be a continuous pseudocontractive operator. Then,(i)
FixS⊂C is closed and convex(ii)
S is demiclosedness, i.e., if xk⇀u˜ and Sxk⟶z† as k⟶∞, then Su˜=z†.
Here, we state some conditions onf and g which will be used in the sequel.
LetC and Q be two nonempty, closed, and convex subsets of real Hilbert spaces ℋ1 and ℋ2, respectively. Let f:C×C⟶ℝ and g:Q×Q⟶ℝ be two bifunctions. Assume that(i)
(A1):fz†,z†=0 for all z†∈C(ii)
(A2):f is pseudomonotone on SEPC,f(iii)
(A3):f is jointly sequently weakly continuous on C×C(iv)
(A4):fz†,⋅ is convex and subdifferentiable on C for all z†∈C(v)
(B1):gz†,z†=0 for all z†∈Q(vi)
(B2):g is monotone on Q(vii)
(B3):gu,⋅ is convex and lower semicontinuous on Q for each u∈Q(viii)
(B4): for allu,v,w∈Q, limsupλ↓0gλw+1−λu,v≤gu,vLemma 3.
(see [1, 2]). Assume that g satisfies conditions (B1)–(B4). For ς>0 and u∈H2, there exists w∈Q such that(20)gw,v+1ςv−w,w−u≥0,∀v∈Q.
Let the operatorJςg be defined by(21)Jςgu=w∈Q:gw,v+1ςv−w,w−u≥0,∀v∈Q.
We have the following conclusions:(i)
Jςg is single-valued and firmly nonexpansive, that is, for any u,v∈H2,(22)Jςgu−Jςgv2≤Jςgu−Jςgv,u−v.(ii)
SEPQ,g is closed and convex and SEPQ,g=FixJςg.(iii)
Forς1,ς2>0 and u,v∈H2, we have(23)Jς1gu−Jς2gv≤u−v+ς2−ς1ς2Jς2gv−v.Lemma 4.
(see [4]). Assume that f satisfies conditions (A1)-(A4). Let βk be a sequence satisfying βk∈β¯,β¯⊂0,1. For given vk∈C, let the sequence yk be generated by(24)yk=argminu†∈Cfvk,u†+12βkvk−u†2.
Then the boundedness ofvk implies that yk is bounded.Lemma 5.
(see [5]). Assume that f satisfies conditions (A1)–(A4). For given two points u¯,v¯∈C and two sequences ak⊂C and bk⊂C, if ak⇀u¯ and bk⇀v¯, respectively, then, for any ɛ>0, there exist ϑ>0 and Nɛ∈ℕ such that(25)∂2fbk,ak⊂∂2fv¯,u¯+ɛϑB,for every k≥Nɛ, where B:=b∈H1:b≤1.Lemma 6.
(see [43]). Let an⊂0,∞, bn⊂0,1, and cn be three real number sequences. If(26)an+1≤1−bnan+cn,for all n≥0 with ∑n=1∞bn=∞ and limsupn⟶∞cn/bn≤0 or ∑n=1∞cn<∞, then limn⟶∞an=0.


## 3. Main Results

In this section, in order to solve problem (2), we first present an iterative algorithm and consequently prove its strong convergence.Letℋ1 and ℋ2 be two real Hilbert spaces. Let C and Q be two nonempty, closed, and convex subsets of ℋ1 and ℋ2, respectively. Assume that(i)
h:C⟶C is a κ-contractive operator(ii)
S:C⟶C is an L1-Lipschitz pseudocontractive operator and T:Q⟶Q is an L2-Lipschitz pseudocontractive operator with L1>1 and L2>1(iii)
f and g are two bifunctions satisfying conditions (A1)–(A4) and conditions (B1)–(B4), respectively(iv)
A:ℋ1⟶ℋ2 is a bounded linear operator and A∗ is its adjointLetδk, ηk, βk, τk, ςk, ζk, λk, and μk be real number sequences and α, ϑ, and γ be constants. Next, we introduce our iterative algorithm.Algorithm 1.
Fix an initial pointx0∈C. Set k=0.Step 1: assume thatxk is known and compute(27)vk=1−δkxk+δkS1−ηkxk+ηkSxk.Step 2: compute(28)yk=argminy†∈Cfvk,y†+12βkvk−y†2.Ifyk=vk, then set uk=vk and go to Step 5. Otherwise, go to Step 3.Step 3: letmk=min1,2,…,k,… such that(29)fzk,mk,vk−fzk,mk,yk≥α2βkvk−yk2,where(30)zk,mk=1−ϑmkvk+ϑmkyk.Writeϑk=ϑmk and zk=zk,mk.Step 4: compute(31)uk=PCvk−τkιkνk,whereνk∈∂2fzk,vk and ιk=fzk,vk/νk2.Step 5:For anyv∈Q, find wk such that(32)gwk,v+1ςkv−wk,wk−Auk≥0.Compute(33)qk=1−ζkwk+ζkT1−λkwk+λkTwk.Step 6: compute(34)xk+1=μkhxk+1−μkPCuk+γA∗qk−Auk.Step 7: setk:=k+1 and return to Step 1.
In order to demonstrate the convergence of Algorithm1, we need some additional assumptions on the iterative parameters. Suppose that the following conditions are satisfied:(C1):0<δ¯<δk<δ¯<ηk<η¯<1/1+L12+1∀k≥0 and α,ϑ∈0,1(C2):βk∈γ1,γ2⊂0,1; τk∈τ1,τ2⊂0,2 and 0<ς≤ςk<+∞(C3):0<ζ¯<ζk<ζ¯<λk<λ¯<1/1+L22+1∀k≥0 and γ∈0,1/A2(C4):limk⟶+∞μk=0 and ∑k=0+∞μk=+∞
We have the following remark which can be found in [4].Remark 1.
(1)
Ifyk=vk, then yn∈SEPC,f(2)
The linesearch rule (29) is well defined(3)
0∉∂2fzk,vk(4)
fzk,vk>0(5)
uk−p2≤vk−p2−τk2−τkιkνk2 for all p∈SEPC,fNext, we prove our main result.Theorem 1.
Suppose thatΓ≠∅. Then, the sequence xk generated by (34) converges strongly to q†=PΓhq†.Proof.
Letx∗∈Γ. We have x∗∈SEPC,f∩FixS and Ax∗∈SEPQ,g∩FixT. By (27) and Lemma 1, we get(35)vk−x∗2=1−δkxk−x∗+δkS1−ηkxk+ηkSxk−x∗2=1−δkxk−x∗2+δkS1−ηkxk+ηkSxk−x∗2−1−δkδkS1−ηkxk+ηkSxk−xk2≤1−δkxk−x∗2+δk1−ηkS1−ηkxk+ηkSxk−xk2+δkxk−x∗2−1−δkδkS1−ηkxk+ηkSxk−xk2=xk−x∗2−δkηk−δkS1−ηkxk+ηkSxk−xk2≤xk−x∗2.
From (31) and Remark 1, we have(36)uk−x∗2≤vk−x∗2−τk2−τkιkνk2≤vk−x∗2.
According to (32) and Lemma 3, we have wk=JςkgAuk and Ax∗∈FixJςkg. Since Jςkg is firmly nonexpansive, we deduce(37)wk−Ax∗2=JςkgAuk−JςkgAx∗2≤JςkgAuk−JςkgAx∗,Auk−Ax∗=wk−Ax∗,Auk−Ax∗=12wk−Ax∗2+Auk−Ax∗2−wk−Auk2.
It follows that(38)wk−Ax∗2≤Auk−Ax∗2−wk−Auk2.
By virtue of (33) and Lemma 1, we obtain(39)qk−Ax∗2=1−ζkwk−Ax∗+ζkT1−λkwk+λkTwk−Ax∗2=1−ζkwk−Ax∗2+ζkT1−λkwk+λkTwk−Ax∗2−1−ζkζkT1−λkwk+λkTwk−wk2≤wk−Ax∗2+ζk1−λkT1−λkwk+λkTwk−wk2−1−ζkζkT1−λkwk+λkTwk−wk2=wk−Ax∗2−λk−ζkζkT1−λkwk+λkTwk−wk2≤wk−Ax∗2.
Thanks to (38) and (39), we get(40)qk−Ax∗2≤Auk−Ax∗2−wk−Auk2.
Consequently,(41)uk−x∗,A∗qk−Auk=Auk−Ax∗,qk−Auk=qk−Ax∗,qk−Auk−qk−Auk2=12qk−Ax∗2+qk−Auk2−Auk−Ax∗2−qk−Auk2=12qk−Ax∗2−Auk−Ax∗2−12qk−Auk2≤−12wk−Auk2−12qk−Auk2.
Settk=PCuk+γA∗qk−Auk for all k≥0. In view of (35), (36), and (41), using the nonexpansivity of PC, we have(42)tk−x∗2=PCuk+γA∗qk−Auk−PCx∗2≤uk−x∗+γA∗qk−Auk2=uk−x∗2+γA∗qk−Auk2+2γA∗qk−Auk,uk−x∗≤uk−x∗2+γ2A2qk−Auk2−γwk−Auk2−γqk−Auk2=uk−x∗2−γ1−γA2qk−Auk2−γwk−Auk2≤xk−x∗2−τk2−τkιkνk2−δkηk−δkS1−ηkxk+ηkSxk−xk2−γ1−γA2qk−Auk2−γwk−Auk2≤xk−x∗2.
From (34), we get(43)xk+1−x∗=μkhxk−x∗+1−μktk−x∗≤μkhxk−hx∗+μkhx∗−x∗+1−μktk−x∗≤μkκxk−x∗+μkhx∗−x∗+1−μkxk−x∗=1−1−κμkxk−x∗+μkhx∗−x∗≤maxxk−x∗,hx∗−x∗1−κ.
By induction, we can obtain thatxk−x∗≤maxx0−x∗,hx∗−x∗/1−κ. Thus, the sequences xk, uk, and vk are all bounded.
Based on (34), we have(44)xk+1−x∗2=μkhxk−x∗+1−μktk−x∗2≤1−μk2tk−x∗2+2μkhxk−x∗,xk+1−x∗≤1−μk2tk−x∗2+2μkκxk−x∗xk+1−x∗+2μkhx∗−x∗,xk+1−x∗≤1−μk2tk−x∗2+μkκxk−x∗2+μkκxk+1−x∗2+2μkhx∗−x∗,xk+1−x∗.
It follows that(45)xk+1−x∗2≤1−μk21−κμktk−x∗2+κμk1−μkκxk−x∗2+2μk1−κμkhx∗−x∗,xk+1−x∗≤1−μk21−κμkxk−x∗2−τk2−τkιkνk2−γ1−γA2qk−Auk2−δkηk−δkS1−ηkxk+ηkSxk−xk2−γwk−Auk2+κμk1−μkκxk−x∗2+2μk1−κμkhx∗−x∗,xk+1−x∗=1−21−κ−μk1−κμkμkxk−x∗2+1−μk2μk1−κμk−τk2−τkιkνk2μk−δkηk−δkS1−ηkxk+ηkSxk−xk2μk−γ1−γA2qk−Auk2μk−γwk−Auk2μk+21−μk2hx∗−x∗,xk+1−x∗.
Setak=xk−x∗2, bk=21−κ−μk/1−κμkμk, and(46)ck=1−μk221−κ−μk−τk2−τkιkνk2μk−γ1−γA2qk−Auk2μk−γwk−Auk2μk+21−μk2hx∗−x∗,xk+1−x∗−δkηk−δkS1−ηkxk+ηkSxk−xk2μk.for all k≥0.
Sinceμk⟶0 as k⟶+∞, without loss of generality, we assume that μk≤1−κ for all k≥0. From (46), we have(47)ck≤221−κ−μkhx∗−x∗,xk+1−x∗≤21−κhx∗−x∗xk+1−x∗.
So,limsupk⟶+∞ck<+∞. Next, we show that limsupk⟶+∞ck≥−1. Assume that limsupk⟶+∞ck<−1. Then, there exists a positive integer number K0 such that ck<−1 when k≥K0. We can rewrite (45) as ak+1≤1−bkak+bkck. Thus, for all k≥K0, from (45), we deduce(48)ak+1≤1−bkak+bkck≤ak−bk,which leads to ak+1≤aK0−∑i=K0kbk. Therefore,(49)limsupk⟶+∞ak+1≤aK0−limsupk⟶+∞∑i=K0kbk.
Note thatbk=21−κ−μk/1−κμkμk≥1−κμk. This together with the last inequality implies that limsupk⟶+∞ak+1≤−∞. It is impossible. Hence, −1≤limsupk⟶+∞ck<+∞. As a result, we can select a subsequence ki of k such that xki⇀p† and(50)limsupk⟶+∞ck=limi⟶+∞cki=limi⟶+∞1−μki221−κ−μki−τki2−τkiιkiνki2μki−γwki−Auki2μki−γ1−γA2qki−Auki2μki+21−μki2hx∗−x∗,xki+1−x∗−δkiηki−δkiS1−ηkixki+ηkiSxki−xki2μki.
Since the sequencexki+1 is bounded, without loss of generality, we assume that limi⟶+∞hx∗−x∗,xki+1−x∗ exists. Consequently, from (50), we obtain(51)limi⟶+∞τki2−τkiιkiνki2μki+δkiηki−δkiS1−ηkixki+ηkiSxki−xki2μki+γ1−γA2qki−Auki2μki+γwki−Auki2μki,exists.
By the assumptions, we haveliminfi⟶+∞τki2−τki>0 and liminfi⟶+∞δkiηki−δki>0, Therefore, we deduce(52)limi⟶+∞ιkiνki=0,(53)limi⟶+∞S1−ηkixki+ηkiSxki−xki=0,(54)limi⟶+∞qki−Auki=0,(55)limi⟶+∞wki−Auki=0.
By (54) and (55), we get(56)limi⟶+∞qki−wki=0.
In addition, from (31), we have(57)uk−vk=PCvk−τkιkνk−PCvk≤τkιkνk.
So, we get from (52) that(58)limi⟶+∞uki−vki=0.
Observe that(59)Sxk−xk≤Sxk−S1−ηkxk+ηkSxk+S1−ηkxk+ηkSxk−xk≤LηkSxk−xk+S1−ηkxk+ηkSxk−xk.
It follows that(60)Sxk−xk≤11−LηkS1−ηkxk+ηkSxk−xk.
This together with (53) implies that(61)limi⟶+∞Sxki−xki=0.
In addition, by (27) and (53), we have(62)vki−xki≤δkiS1−ηkixki+ηkiSxki−xki⟶0.
Sincevk is bounded, by Lemma 4, yk is bounded. Consequently, the sequence zk is bounded. Applying Lemma 5, we deduce that νk is bounded. According to (52), we derive(63)limi⟶+∞fzki,vki=limi⟶+∞ιkiνkiνki=0.
Sincefzki,⋅ is convex, we have(64)0=fzki,zki=fzki,1−ϑkivki+ϑkiyki≤1−ϑkifzki,vki+ϑkifzki,yki.
So, we get from (29) that(65)fzki,vki≥ϑkifzki,vki−fzki,yki≥α2βkiϑkivki−yki2.
Combining the above inequality with (63), we have(66)limi⟶+∞ϑkivki−yki2=0.
Note thatxki⇀p†∈C. Then, it follows from (55), (58), and (62) that uki⇀p†, vki⇀p†, Auki⇀Ap†, Avki⇀Ap†, and wki⇀Ap†∈Q.
There are two possible cases.Case 1.
limsupk⟶+∞ϑki>0. Then, there exist ϑ¯>0 and a subsequence of ϑki, still denoted by ϑki such that for some I0>0, ϑki>ϑ¯ for all i≥I0. Consequently, by (66), we deduce(67)limi⟶+∞vki−yki=0.
Noting thatvki⇀p†, thus yki⇀p†. According to (28), we obtain(68)0∈∂2fvki,yki+1βkiyki−vki+NCyki,so, there exists ν^ki∈∂2fvki,yki such that(69)ν^ki,y−yki+1βkiyki−vki,y−yki≥0,∀y∈C.
By the subdifferential inequality, we have(70)fvki,y−fvki,yki≥ν^ki,y−yki,∀y∈C.
Therefore,(71)fvki,y−fvki,yki+1βkiyki−vki,y−yki≥0,∀y∈C.
Since(72)yki−vki,y−yki≤yki−vkiy−yki,from (71), we get(73)fvki,y−fvki,yki+1βkiyki−vkiy−yki≥0.
Lettingi⟶+∞ in (73), from (A1), (A3), and (67), we obtain(74)fp†,y≥fp†,p†=0,∀y∈C,hence p†∈SEPC,f.Case 2.
limi⟶+∞ϑki=0. Since the sequence yki is bounded, without loss of generality, we may assume that yki⇀y¯ as i⟶+∞. Replacing y by vki in (71), we get(75)fvki,yki≤−1βkiyki−vki2.
According to (29), for mki−1, we have(76)fzki,mki−1,vki−fzki,mki−1,yki<α2βkiyki−vki2.
From (75) and (76), we obtain(77)fvki,yki≤2αfzki,mki−1,yki−fzki,mki−1,vki.
Lettingi⟶+∞ in (77) and noting that vki⇀p†, yki⇀y¯ and zki,mki−1⇀p† as i⟶+∞, we obtain(78)fp†,y¯≤2αfp†,y¯.
Therefore,fp†,y¯=0 and limi⟶+∞yki−vki=0. Consequently, by the similar argument as that in Case 1, we get p†∈SEPC,f.
At the same time, from (61), xki⇀p† and Lemma 2, we deduce that p†∈FixS. Therefore, p†∈FixS∩SEPC,f.
Next, we show thatp†∈FixT∩SEPQ,g. First, by (39), we have(79)λk−ζkζkT1−λkwk+λkTwk−wk2≤wk−Ax∗2−qk−Ax∗2≤wk−qkwk−Ax∗+qk−Ax∗.
Sinceliminfk⟶+∞λk−ζkζk>0 and wk and qk are bounded, from (56) and (79), we deduce that(80)limk⟶+∞T1−λkwk+λkTwk−wk=0.
Observe that(81)Twk−wk≤Twk−T1−λkwk+λkTwk+T1−λkwk+λkTwk−wk≤L2λkTwk−wk+T1−λkwk+λkTwk−wk.
It follows that(82)Twk−wk≤11−L2λkT1−λkwk+λkTwk−wk.
This together with (80) implies that limk⟶+∞Twk−wk=0. Combining this with wki⇀Ap† and the fact that I−T is demiclosed at zero (Lemma 2), it is immediate that Ap†∈FixT.
By Lemma3, we have(83)JςkgAuk−JςgAuk≤ςk−ςςkJςkgAuk−Auk.
Hence,(84)JςgAuk−Auk≤JςkgAuk−Auk+JςkgAuk−JςgAuk≤2JςkgAuk−Auk.
It follows from (55) that limk⟶∞JςgAuk−Auk=0. Since Jςg is nonexpansive and Auki⇀Ap†, we deduce that Ap†∈FixJςg=SEPQ,g by Lemma 3. So, p†∈Γ and ωwxk⊂Γ.
Replacingx∗=PChq† in (45), we have(85)xk+1−PChq†2≤1−21−κ−μk1−κμkμkxk−PChq†2+21−κ−μk1−κμkμk×221−κ−μkhPChq†−PChq†,xk+1−PChq†.
Noting thatlimsupk⟶+∞hPChq†−PChq†,xk+1−PChq†≤0, applying Lemma 6 to the last inequality, we deduce that xk⟶PChq†. This completes the proof. □
Next, we can apply Algorithm1 and Theorem 1 for solving the split equilibrium problem (3). Setting S=I and T=I in Algorithm 1, we deduce that vk=xk and qk=wk. Consequently, we have the following algorithm and corollary.Algorithm 2.
Fix an initial pointx0∈C. Set k=0.Step 1: assume thatxk is known and compute(86)yk=argminy†∈Cfxk,y†+12βkxk−y†2.Ifyk=xk, then set uk=xk and go to Step 4. Otherwise, go to Step 2.Step 2: letmk=min1,2,…,k,… such that(87)fzk,mk,xk−fzk,mk,yk≥α2βkxk−yk2,where(88)zk,mk=1−ϑmkxk+ϑmkyk.Writeϑk=ϑmk and zk=zk,mk.Step 3: compute(89)uk=PCvk−τkιkνk,whereνk∈∂2fzk,xk and ιk=fzk,xk/νk2.Step 4: for anyv∈Q, find wk such that(90)gwk,v+1ςkv−wk,wk−Auk≥0.Step 5: compute(91)xk+1=μkhxk+1−μkPCuk+γA∗wk−Auk.Step 6: setk:=k+1 and return to Step 1.Corollary 1.
Assume thatΓ1≠∅. Then, the sequence xk generated by (91) strongly converges to a solution q1=PΓ1hq1.
Next, we can apply Algorithm1 and Theorem 1 for solving the split fixed point problem (4). Setting f=0 and g=0 in Algorithm 1, we deduce that yk=xk and wk=vk. Consequently, we have the following algorithm and corollary.Algorithm 3.
Fix an initial pointx0∈C. Define the sequence xk iteratively by(92)vk=1−δkxk+δkS1−ηkxk+ηkSxk,qk=1−ζkvk+ζkT1−λkvk+λkTvk,xk+1=μkhxk+1−μkPCxk+γA∗qk−Axk,k≥0.Corollary 2.
Assume thatΓ2≠∅. Then, the sequence xk generated by (92) strongly converges to a solution q2=PΓ2hq2.



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*Source: 6624321-2021-01-29.xml*