# Extragradient Methods for Solving Split Feasibility Problem and General Equilibrium Problem and Resolvent Operators in Banach Spaces

Authors: Mostafa Ghadampour
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this paper, we introduce a new extragradient algorithm by using generalized metric projection. We prove a strong convergence theorem for finding a common element of the solution set of split feasibility problem and the set of fixed points of relatively nonexpansive mapping and a finite family of resolvent operator and the set of solutions of an equilibrium problem.


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## Body

## 1. Introduction

LetC be a nonempty closed convex subset of a real Banach space X with norm . and X∗ be the dual of X. We consider the following variational inequality problem (VI), which consists in finding a point x∈C such that
(1)x∗,y−x≥0∀y∈C,∀x∗∈Ax,where A:C⟶2X∗ is a mapping and .,. denotes the duality pairing. The solution set of the variational inequality problem is denoted by VIC,A.The operatorA:X⟶2X∗ is called
(i)
Monotone if(2)x−y,x∗−y∗≥0∀x,y∈X,∀x∗∈Ax,y∗∈Ay.(ii)
α-inverse strongly monotone if there exists a constant α>0 such that(3)x−y,x∗−y∗≥αx∗−y∗2,∀x,y∈X,∀x∗∈Ax,y∗∈Ay.(iii)
Demiclosed if for allxn⊂X with xn⇀x in X, and yn∈Axn with yn⟶y in X∗, we have x∈X and y∈AxA monotone mappingB is said to be maximal if its graph GB=x,Bx:x∈DB is not properly contained in the graph of any other monotone mapping. Obviously, the monotone mapping B is maximal if and only if for x,x∗∈X×X∗, x−y,x∗−y∗≥0 for all y,y∗∈GB, then it is implied that x∗∈Bx.Assume thatA:C⟶2X∗ is a nonlinear mapping and f:C×C⟶ℝ is a bifunction. The equilibrium problem (EP) is as follows: find x∈C such that
(4)fx,y+Ax,y−x≥0,∀y∈C.The solution set of (4) is denoted by EPf. The equilibrium problem is very general because it includes many well-known problems such as variational inequality problems and saddle point problems (see [1–4]). Several methods have been proposed to solve the equilibrium problem in Hilbert space (see [5]), and some authors obtained weak and strong convergence algorithms for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points of a nonexpansive mapping in a Hilbert space (see [6–9]). Then, the authors proved the strong convergence of the algorithms in a uniformly convex and uniformly smooth Banach space (see [10]).Suppose thatC and D are nonempty, closed, and convex subsets of real Banach spaces X1 and X2, respectively. The split feasibility problem (SFP) is to find a point
(5)x∈Csuchthatx∈A−1D,which A:X1⟶X2 is a bounded linear operator. The solution set of (5) is denoted by Ω.In 1994, the split feasibility problem was first studied by Censor and Elfving [11] in finite dimensional Hilbert spaces. In solving (SFP), Schöpfer et al. [12] proposed the next algorithm in p-uniformly convex real Banach spaces: x1∈X1 is chosen arbitrarily and for n≥1,
(6)xn+1=ΠCJX1∗JX1xn−tnA∗JX2Axn−PDAxn,where J is the duality mapping, ΠC denotes the Bregman projection, A is a bounded linear operator, and A∗ is the adjoint of A. Also, they have proven the generated sequence xn by algorithm (6) is weakly convergent under suitable conditions. The split feasibility problems were studied extensively by many authors [13, 14].In this paper, motivated by Schöpfer et al. [12], we present a new hybrid algorithm using the inverse strongly monotone operation and a finite family of resolvent operator. Then, we show that our generated sequence is strongly converges to a common point, the set of solution set of split feasibility problem, and the fixed point of relatively nonexpansive mapping and the fixed point of resolvent operator.


## 2. Preliminaries

LetX be a real smooth Banach space with norm . and let X∗ be the dual space of X. We denote the strong convergence and the weak convergence xn to x in X by xn⟶x and xn⇀x, respectively. A function δ:0,2⟶0,1 is said to be the modulus of convexity of X as follows:
(7)δε=inf1−x+y2:x≤1,y≤1,x−y≥ϵ,for every ε∈0,2. A Banach space X is said to be uniformly convex if and only if δε>0 for all ε>0. It is known that a uniformly convex Banach space has the Kadec-Klee property, that is, xn⇀u and xn⟶u imply that xn⟶u (see [15]). Let p be a fixed real number with p≥2. A Banach space X is called p-uniformly convex [16], if there exists a constant c>0 such that δ≥cϵp for all ϵ∈0,2. Let SE=x∈X:x=1. A Banach space X is said to be smooth if for all x∈SX, there exists a unique functional jx∈X∗ such that x,jx=x and jx=1 (see [17]).The norm ofX is said to be Ga^teaux differentiable if for all x,y∈SX, the limit
(8)limt⟶0x+ty−xt,exists. In this case, X is said to be smooth, and X is called uniformly smooth if the limit (8) is attained uniformly for all x,y∈SX[18]. If a Banach space X is uniformly convex, then X is reflexive and strictly convex, and X∗ is uniformly smooth [17]. The duality mapping JXp on X is defined by
(9)JXpx=f∈X∗:x,f=xp,f=xp−1,for every x∈X. If X is a p-uniformly convex and uniformly smooth, then JXp is single valued, one-to-one and satisfies JXp=JX∗−1=JXq−1, where JX∗=JXq is the duality mapping of X (see [19]). If p=2, then JX=J2=J is the normalized duality mapping. It is well known that if X is a reflexive, strictly convex, and smooth Banach space and JX∗:X∗⟶2X is the duality mapping on X∗, then JX−1=JX∗. If X is a uniformly smooth and uniformly convex Banach space, then JX is uniformly norm to norm continuous on bounded sets of X, and JX−1=JX∗ is also uniformly norm to norm continuous on bounded sets of X∗. Let X be a smooth Banach space and let JX be the duality mapping on X. The function ϕ:X×X⟶ℝ is defined by
(10)ϕx,y=x2−2x,JXy+y2,∀x,y∈X.Clearly, from (10), we can conclude that
(11)x−y2≤ϕx,y≤x+y2.IfX is a reflexive, strictly convex, and smooth Banach space, then for all x,y∈X(12)ϕx,y=0⇔x=y.Also, it is clear from the definition of the functionϕ that the following condition holds for all x,y∈X,
(13)ϕx,y=x,JXx−JXy+y−x,JXy≤xJXx−JXy+y−xy.Now, the functionV:X×X∗⟶ℝ is defined as follows:
(14)Vx,x∗=x2−2x,x∗+x∗2,for all x∈X and x∗∈X∗. Moreover, Vx,x∗=ϕx,JX−1x∗ for all x∈X and x∗∈X∗. If X is a reflexive strictly convex and smooth Banach space with X∗ as its dual, then
(15)Vx,x∗+2JX−1x∗−x,y∗≤Vx,x∗+y∗,for all x∈X and all x∗,y∗∈X∗[20].An operatorA:C⟶X∗ is hemicontinuous at x0∈C, if for any sequence xn converging to x0 along a line implies that the sequence Axn is weakly convergent to Ax0, i.e., Axn=Ax0+tnx⇀Ax0 as tn⟶0 for all x∈C.The generalized projectionΠC:X⟶C is a mapping that assigns to an arbitrary point x∈X, the minimum point of the functional ϕy,x; that is, ΠCx=x0, where x0 is the solution of the minimization problem
(16)ϕx0,x=miny∈Cϕy,x.The existence and uniqueness of the operatorΠC follows from the properties of the functional ϕx,y and strict monotonicity of the mapping J [21]. Suppose that C is a nonempty closed convex subset of X and T is a self mapping on C. We denote the set of fixed points of T by FT, that is FT=x∈C:x∈Tx. A point p∈C is called an asymptotically fixed point of T if C contains a sequence xn which converges weakly to p such that Txn−xn⟶0 [17]. The set of asymptotical fixed points of T will be denoted by F^T. A mapping T from C into itself is said to be relatively nonexpansive if F^T=FT and ϕp,Tx≤ϕp,x for all x∈C and p∈FT. The asymptotic behavior of a relatively nonexpansive mapping was studied in [22, 23].We need the following lemmas for proving our main results.Lemma 1. (see [24]).
LetX be a smooth and uniformly convex Banach space and let xn and yn be two sequences of X. If ϕxn,yn⟶0 and either xn or yn is bounded, then xn−yn⟶0.Lemma 2. (see [21]).
LetC be a nonempty closed convex subset of a smooth, strictly convex, and reflexive Banach space X and let y∈X. Then,
(17)ϕx,ΠCy+ϕΠCy,y≤ϕx,y,∀x∈C.Lemma 3. (see [21]).
LetC be a nonempty closed convex subset of a smooth, strictly convex, and reflexive Banach space X, let x∈X, and let z∈C. Then,
(18)z=ΠCx⇔y−z,JXx−JXz≤0,forally∈C.Lemma 4. (see [25]).
LetX be a 2-uniformly convex and smooth Banach space. Then, for all x, y∈X, we have that
(19)x−y≤2c2JXx−JXy,where 1/c0≤c≤1 is the 2-uniformly convex constant of X.Lemma 5. (see [25]).
LetX be a uniformly convex Banach space and r>0. Then, there exists a continuous strictly increasing convex function g:0,2r⟶0,∞ such that g0=0 and
(20)tx+1−ty2≤tx2+1−ty2−t1−tgx−y,for all x,y∈Br0=z∈X:z≤r and t∈0,1.Lemma 6. (see [24]).
LetX be a uniformly convex Banach space and r>0. Then, there exists a continuous strictly increasing convex function g:0,2r⟶0,∞ such that g0=0 and
(21)gx−y≤ϕx,y,for all x,y∈Br0=z∈X:z≤r.Lemma 7. (see [25]).
Letx,y∈X. If X is p-uniformly smooth, then there is a c>0 so that
(22)x−yp≤xp−py,JXpx+cyp.Throughout this paper, we assume thatf:C×C⟶ℝ is a bifunction satisfying the following conditions(A1)fx,x=0 for all x∈C(A2)f is monotone, i.e., fx,y+fy,x≤0, for all x,y∈C(A3)limt↓0ftz+1−tx,y≤fx,y, for all x,y,z∈C(A4) For eachx∈C,y↦fx,y is convex and lower semicontinuous.Lemma 8. (see [26]).
LetC be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space X. Let A:C⟶X∗ be an α-inverse-strongly monotone operator and f be a bifunction from C×C to ℝ satisfying A1−A4. Then, for all r>0 the following hold
(i)
Forx∈X, there exists u∈C such that(23)fu,x+Au,y−u+1ry−u,JXu−JXx≥0,∀y∈C,(ii)
IfX is additionally uniformly smooth and Kr:X⟶C is defined as(24)Krx=u∈C:fu,y+Au,y−u+1ry−u,JXu−JXx≥0,∀y∈C,then, the following conditions hold:
Kr is single-valued
Kr is firmly nonexpansive, i.e., for all x,y∈X,
(25)Krx−Kry,JXKrx−JXKry≤Krx−Kry,JXx−JXy,FKr=F^Kr=EPf.
EP is a closed convex subset of C.
(26)ϕp,Krx+ϕKrx,x≤ϕp,x,∀p∈FKr.Definition 9.
LetX be a real smooth and uniformly convex Banach space and let M:X⟶2X∗ be a maximal monotone operator. For all ι>0, define the operator QιM:X⟶X by QιM=JX+ιM−1JXx for all x∈X.Lemma 10. (see [18]).
LetX be a real smooth and uniformly convex Banach space, and let M:X⟶2X∗ be a maximal monotone operator. Then, M−10 is a closed and convex subset of X, and the graph GM of M is demiclosed.Lemma 11.
LetX be a real reflexive, strictly convex, and let smooth Banach space and M:X⟶2X∗ be a maximal monotone operator with M−10≠∅. Then, for all x∈X,y∈M−10 and ι>0, then ϕy,QιMx+ϕQιMx,x≤ϕy,x.


## 3. Main Results

In this section, we introduce our new extragradient algorithm.Theorem 12.
LetX1 and X2 are real 2-uniformly convex and uniformly smooth Banach spaces. Suppose that C and D are nonempty closed and convex subsets of X1 and X2, respectively. Suppose that g is a bifunction from C×C to ℝ which satisfies the conditions A1-A4, A:X1⟶X2 is a bounded linear operator and A∗:X2∗⟶X1∗ is the adjoint of A. Let Mi:X1⟶2X1∗ be a maximal monotone operator with Mi−10≠∅ for all i=1,2,⋯,k. Assume that B:C⟶X∗ is an α-inverse strongly monotone operator, and f is a relatively nonexpansive mappings from C into itself and Γ=Ω∩Ff∩∩i=1kFQιMi∩EPg≠∅. Let xn is a sequence generated by v1∈C and
(27)xn∈Cs.tgxn,y+Bxn,y−xn+1rny−xn,JX1xn−JX1vn≥0,un=ΠCJX1−1snJX1xn+1−snJX1QιM1QιM2⋯QιMkxn,zn=ΠCJX1−1JX1un−τA∗JX2Aun−PDAun,yn=ΠCJX1−1JX1zn−τA∗JX2Azn−PDAzn,wn=ΠCJX1−1βnJX1QιM1QιM2⋯QιMkzn+1−βnJX1QιM1QιM2⋯QιMkyn,vn+1=ΠCJX1−1αn,1JX1fxn+αn,2JX1un+αn,3JX1wn,where rn∈a,∞ for some a>0, sn and βn are real sequences in a,b⊂0,1, and τ and αn,ii=13 satisfy the following conditions:
(i)
αn,ii=13⊂0,1, ∑i=13αn,i=1, liminfn⟶∞αn,1αn,2>0, and liminfn⟶∞αn,3>0(ii)
τ is real number such that 0<τ<2/cA2, where c depends on 2-uniformly smoothness of X1∗
Then,xn converges strongly to q=ΠΩ∩∩i=1kFQιMi∩EPg∘fq.Proof.
Letu^∈Γ. By (10), Lemma 2 and the convexity of .2, we have
(28)ϕu^,un≤ϕu^,JX1−1snJX1xn+1−snJX1QιM1QιM2⋯QιMkxn=u∧2−2u^,snJX1xn+1−snJX1QιM1QιM2⋯QιMkxn+snJX1xn+1−snJX1QιM1QιM2⋯QιMkxn2≤u∧2−2snu^,JX1xn−21−snu^,JX1QιM1QιM2⋯QιMkxn+snxn2+1−snQιM1QιM2⋯QιMkxn2=snϕu^,xn+1−snϕu^,QιM1QιM2⋯QιMkxn.Now, it follows from Lemma11 and the above that
(29)ϕu^,un≤snϕu^,xn+1−snϕu^,QιM2⋯QιMkxn≤snϕu^,xn+1−snϕu^,QιM3⋯QιMkxn,(30)⋮(31)≤snϕu^,xn+1−snϕu^,xn=ϕu^,xn.Letkn=Aun−PDAun. From (10) and Lemmas 2 and 7, we have that
(32)ϕu^,zn≤ϕu^,JX1−1JX1un−τA∗JX2kn=u∧2−2u^,JX1un−τA∗JX2kn+JX1un−τA∗JX2kn2=u∧2−2u^,JX1un+2τu^,A∗JX2kn+JX1un−τA∗JX2kn2≤u∧2−2u^,JX1un+2τu^,A∗JX2kn+JX1un2−2τA∗JX2kn,JX1∗JX1un+cτ2A∗JX2kn2=ϕu^,un+2τAu^,JX2kn−2τJX2kn,Aun+cτ2A2JX2kn2=ϕu^,un+2τAu^−Aun,JX2kn+cτ2A2kn2.SinceJX2x−PDx,y−PDx≤0 for each y∈D and for each x∈X2, we have that
(33)JX2kn,Aun−Au^=JX2kn,PDAun−Au^+JX2kn,Aun−PDAun=JX2kn,PDAun−Au^+PDAun−Aun2≥PDAun−Aun2.From (32), our assumptions, and the above, we conclude that
(34)ϕu^,zn≤ϕu^,un−2τPDAun−Aun2+cτ2A2kn2=ϕu^,un−τ2−cτA2PDAun−Aun2≤ϕu^,un.In a similar way as above, we obtain that(35)ϕu^,yn≤ϕu^,zn−τ2−cτA2PDAzn−Azn2≤ϕu^,zn.It follows from (10), (29), (34), (35), Lemma 11, and the convexity of .2 that
(36)ϕu^,wn≤ϕu^,JX1−1βnJX1QιM1QιM2⋯QιMkzn+1−βnJX1QιM1QιM2⋯QιMkyn=u∧2−2u^,βnJX1QιM1QιM2⋯QιMkzn+1−βnJX1QιM1QιM2⋯QιMkyn+βnJX1QιM1QιM2⋯QιMkzn+1−βnJX1QιM1QιM2⋯QιMkyn2≤u∧2−2βnu^,JX1QιM1QιM2⋯QιMkzn−21−βnu^,JX1QιM1QιM2⋯QιMkyn+βnQιM1QιM2⋯QιMkzn2+1−βnQιM1QιM2⋯QιMkyn2=βnϕu^,QιM1QιM2⋯QιMkzn+1−βnϕu^,QιM1QιM2⋯QιMkyn,(37)≤βnϕu^,QιM2QιM3⋯QιMkzn+1−βnϕu^,QιM2QιM3⋯QιMkyn,(38)⋮(39)≤βnϕu^,zn+1−βnϕu^,yn≤ϕu^,xn.By (10), (29), (37), Lemmas 2, 8, the condition (i), the convexity of .2, and the relatively nonexpansiveness of f, we have that
(40)ϕu^,xn+1=ϕu^,Krnvn+1≤ϕu^,vn+1≤ϕu^,JX1−1αn,1JX1fxn+αn,2JX1un+αn,3JX1wn=u∧2−2u^,αn,1JX1fxn+αn,2JX1un+αn,3JX1wn+αn,1JX1fxn+αn,2JX1un+αn,3JX1wn2≤u∧2−2αn,1u^,JX1fxn−2αn,2u^,JX1un−2αn,3u^,JX1wn+αn,1fxn2+αn,2un2+αn,3wn2=αn,1ϕu^,fxn+αn,2ϕu^,un+αn,3ϕu^,wn≤αn,1ϕu^,xn+αn,2ϕu^,un+αn,3ϕu^,wn≤ϕu^,xn.Therefore,ϕu^,xn is bounded, and limn⟶∞ϕu^,xn exists. Now, by (11), we conclude that xn is bounded. It follows from (29), (34), (35), (37), (40), and relatively nonexpansiveness of f that the sequences un,zn,yn, wn, vn, and fxn are bounded.Next, by (28), (37), (40), and Lemma 11, we conclude that
(41)ϕu^,xn+1≤αn,1ϕu^,xn+αn,2ϕu^,un+αn,3ϕu^,wn≤1−αn,2ϕu^,xn+αn,2ϕu^,un≤1−αn,2ϕu^,xn+αn,2snϕu^,xn+1−snϕu^,QιM1QιM2⋯QιMkxn≤1−αn,2ϕu^,xn+αn,2snϕu^,xn+1−snϕu^,QιM2⋯QιMkxn−ϕQιM1QιM2⋯QιMkxn,QιM2⋯QιMkxn≤1−αn,2ϕu^,xn+αn,2snϕu^,xn+1−snϕu^,QιM3⋯QιMkxn−ϕQιM2⋯QιMkxn,QιM3⋯QιMkxn−ϕQιM1QιM2⋯QιMkxn,QιM2⋯QιMkxn,(42)⋮(43)≤1−αn,2ϕu^,xn+αn,2snϕu^,xn+1−snϕu^,xn−ϕQιMkxn,xn−⋯−ϕQιM1QιM2⋯QιMkxn,QιM2⋯QιMkxn=ϕu^,xn−αn,21−snϕQιMkxn,xn−⋯−ϕQιM1QιM2⋯QιMkxn,QιM2⋯QιMkxn.Now, from (11) and (41), we have the following inequalities:
(44)ϕu^,xn+1≤ϕu^,xn−αn,21−snϕQιMkxn,xn,ϕu^,xn+1≤ϕu^,xn−αn,21−snϕQιMk−1QιMkxn,QιMkxn,⋮ϕu^,xn+1≤ϕu^,xn−αn,21−snϕQιM1QιM2⋯QιMkxn,QιM2⋯QιMkxn.Now, sinceϕu^,xn is convergent, it follows from (44), the conditions (i), and our assumptions that
(45)limn⟶∞ϕQιMkxn,xn=0,limn⟶∞ϕQιMk−1QιMkxn,QιMkxn=0,⋮limn⟶∞ϕQιM1QιM2⋯QιMkxn,QιM2⋯QιMkxn=0.Therefore, from Lemma1, we have that
(46)limn⟶∞QιMkxn−xn=0,limn⟶∞QιMk−1QιMkxn−QιMkxn=0,⋮limn⟶∞QιM1QιM2⋯QιMkxn−QιM2⋯QιMkxn=0.Then,(47)limn⟶∞QιM1QιM2⋯QιMkxn−xn=0.From (13), (47), the boundedness of the sequences xn, QιM1QιM2⋯QιMkxn, and using uniformly norm-to-norm continuity of J on bounded sets, it is clear that
(48)limn⟶∞ϕxn,QιM1QιM2⋯QιMkxn=0.By (29), (34), (35), (36), (40), and Lemma 11, we conclude that
(49)ϕu^,xn+1≤αn,1ϕu^,xn+αn,2ϕu^,un+αn,3ϕu^,wn≤1−αn,3ϕu^,xn+αn,3ϕu^,wn≤1−αn,3ϕu^,xn+αn,3βnϕu^,QιM1QιM2⋯QιMkzn+1−βnϕu^,QιM1QιM2⋯QιMkyn≤1−αn,3ϕu^,xn+αn,3βnϕu^,QιM2⋯QιMkzn−ϕQιM1QιM2⋯QιMkzn,QιM2⋯QιMkzn+1−βnϕu^,QιM2⋯QιMkyn−ϕQιM1QιM2⋯QιMkyn,QιM2⋯QιMkyn,⋮≤1−αn,3ϕu^,xn+αn,3βnϕu^,zn−ϕQιMkzn,zn−⋯−ϕQιM1QιM2⋯QιMkzn,QιM2⋯QιMkzn+1−βnϕu^,yn−ϕQιMkyn,yn−⋯−ϕQιM1QιM2⋯QιMkyn,QιM2⋯QιMkyn≤ϕu^,xn−αn,3βnϕQιMkzn,zn+⋯+ϕQιM1QιM2⋯QιMkzn,QιM2⋯QιMkzn−αn,31−βnϕQιMkyn,yn+⋯+ϕQιM1QιM2⋯QιMkyn,QιM2⋯QιMkyn.Hence, from (11), the above, and our assumptions, we obtain the following results
(50)ϕu^,xn+1≤ϕu^,xn−αn,3βnϕQιMkzn,zn,ϕu^,xn+1≤ϕu^,xn−αn,3βnϕQιMk−1QιMkzn,QιMkzn,⋮ϕu^,xn+1≤ϕu^,xn−αn,3βnϕQιM1QιM2⋯QιMkzn,QιM2⋯QιMkzn,(51)ϕu^,xn+1≤ϕu^,xn−αn,31−βnϕQιMkyn,yn,ϕu^,xn+1≤ϕu^,xn−αn,31−βnϕQιMk−1QιMkyn,QιMkyn,⋮ϕu^,xn+1≤ϕu^,xn−αn,31−βnϕQιM1QιM2⋯QιMkyn,QιM2⋯QιMkyn.Sinceϕu^,xn is convergent, we conclude from (i), (50), (51), and our assumptions that
(52)limn⟶∞ϕQιMkzn,zn=0,limn⟶∞ϕQιMk−1QιMkzn,QιMkzn=0,⋮limn⟶∞ϕQιM1QιM2⋯QιMkzn,QιM2⋯QιMkzn=0.(53)limn⟶∞ϕQιMkyn,yn=0,limn⟶∞ϕQιMk−1QιMkyn,QιMkyn=0,⋮limn⟶∞ϕQιM1QιM2⋯QιMkyn,QιM2⋯QιMkyn=0.Now, from (29), (34), (35), (37), and (40), we have
(54)ϕu^,xn+1≤αn,1ϕu^,xn+αn,2ϕu^,un+αn,3ϕu^,wn≤1−αn,3ϕu^,xn+αn,3ϕu^,wn≤1−αn,3ϕu^,xn+αn,3βnϕu^,zn+1−βϕu^,yn≤1−αn,3ϕu^,xn+αn,3ϕu^,zn≤1−αn,3ϕu^,xn+αn,3ϕu^,un−τ2−cτA2PDAun−Aun2≤ϕu^,xn−αn,3τ2−cτA2PDAun−Aun2.Hence, it follows from (54) that
(55)αn,3τ2−cτA2PDAun−Aun2≤ϕu^,xn−ϕu^,xn+1.Then, it follows from (i) and our assumptions that(56)limn⟶∞PDAun−Aun2=0.From (29), (34), (35), (37), and (40), we have
(57)ϕu^,xn+1≤αn,1ϕu^,xn+αn,2ϕu^,un+αn,3ϕu^,wn≤1−αn,3ϕu^,xn+αn,3ϕu^,wn≤1−αn,3ϕu^,xn+αn,3βnϕu^,zn+1−βnϕu^,yn≤1−αn,3ϕu^,xn+αn,3βnϕu^,zn+1−βnϕu^,zn−τ2−cτA2PDAzn−Azn2≤1−αn,3ϕu^,xn+αn,3ϕu^,zn−αn,31−βnτ2−cτA2PDAzn−Azn2≤ϕu^,xn−αn,31−βnτ2−cτA2PDAzn−Azn2.So,(58)αn,31−βnτ2−cτA2PDAzn−Azn2≤ϕu^,xn−ϕu^,xn+1.Therefore, it follows from (i) and our assumptions that(59)limn⟶∞PDAzn−Azn2=0.Suppose thatr1=supnfxn,un. Therefore, from Lemma 5, there exists a continuous strictly increasing convex function g1:0,2r1⟶0,∞ such that g10=0 and using (29), (37), Lemmas 2 and 8, the convexity of .2, and the condition relatively nonexpansiveness of f, we have that
(60)ϕu^,xn+1=ϕu^,Krnvn+1≤ϕu^,vn+1≤ϕu^,JX1−1αn,1JX1fxn+αn,2JX1un+αn,3JX1wn=u∧2−2u^,αn,1JX1fxn+αn,2JX1un+αn,3JX1wn+αn,1JX1fxn+αn,2JX1un+αn,3JX1wn2≤u∧2−2αn,1u^,JX1fxn−2αn,2u^,JX1un−2αn,3u^,JX1wn+αn,1fxn2+αn,2un2+αn,3wn2−αn,1αn,2g1JX1fxn−JX1un≤αn,1ϕu^,fxn+αn,2ϕu^,un+αn,3ϕu^,wn−αn,1αn,2g1JX1fxn−JX1un≤αn,1ϕu^,xn+αn,2ϕu^,un+αn,3ϕu^,wn−αn,1αn,2g1JX1fxn−JX1un≤ϕu^,xn−αn,1αn,2g1JX1fxn−JX1un.So,(61)αn,1αn,2g1JX1fxn−JX1un≤ϕu^,xn−ϕu^,xn+1.Sincelimn⟶∞ϕu^,xn exists. Therefore, it follows from the condition (i) that
(62)limn⟶∞g1JX1fxn−JX1un=0.Becauseg1 is continues function, we conclude that
(63)g1limn⟶∞JX1fxn−JX1un=limn⟶∞g1JX1fxn−JX1un=0=g10.Therefore,(64)limn⟶∞JX1fxn−JX1un=0.SinceJX1−1 is uniformly norm-to-norm continuous on bounded sets, it imply that
(65)limn⟶∞fxn−un=0.Using (13), (65), the uniformly norm-to-norm continuity of JX1 on bounded sets, and the boundedness of the sequences fxn and un, we conclude that
(66)limn⟶∞ϕun,fxn=0.By (48) and using our assumptions, we obtain that
(67)ϕxn,un≤ϕxn,JX1−1snJX1xn+1−snJX1QιM1QιM2⋯QιMkxn=xn2−2xn,snJX1xn+1−snJX1QιM1QιM2⋯QιMkxn+snJX1xn+1−snJX1QιM1QιM2⋯QιMkxn2≤xn2−2snxn,JX1xn−21−snxn,JX1QιM1QιM2⋯QιMkxn+snxn2+1−snQιM1QιM2⋯QιMkxn2=snϕxn,xn+1−snϕxn,QιM1QιM2⋯QιMkxn=1−snϕxn,QιM1QιM2⋯QιMkxn⟶0asn⟶∞.Then, it follows from Lemma1 that
(68)limn⟶∞xn−un=0.Now, by (15), (56), and Lemmas 2 and 4, we conclude that
(69)ϕun,zn≤ϕun,JX1−1JX1un−τA∗JX2kn=Vun,JX1un−τA∗JX2kn≤Vun,JX1un−2JX1−1JX1un−τA∗JX2kn−un,τA∗JX2kn=ϕun,un−2JX1−1JX1un−τA∗JX2kn−JX1−1JX1un,τA∗JX2kn≤2JX1−1JX1un−τA∗JX2kn−JX1−1JX1unτA∗JX2kn≤4τ2c2A∗JX2kn2≤4τ2c2A2Aun−PDAun2⟶0asn⟶∞.Then, using Lemma1, we obtain
(70)limn⟶∞un−zn=0.Also, from (15), (59), and Lemma 2, and the same way used for proving (70), we can conclude that
(71)ϕzn,yn≤ϕzn,JX1−1JX1zn−τA∗JX2Azn−PDAzn≤4τ2c2A2Azn−PDAzn2⟶0asn⟶∞.Then, using Lemma1, we get
(72)limn⟶∞zn−yn=0.Now, it follows from (13), (52), and Lemma 1 that
(73)zn−QιM1QιM2⋯QιMkzn≤zn−QιMkzn+QιMkzn−QιMk−1QιMkzn+⋯+QιM2⋯QιMkzn−QιM1QιM2⋯QιMkzn⟶0asn⟶∞.Similarly, from (13), (53), and Lemma 1, we have
(74)limn⟶∞yn−QιM1QιM2⋯QιMkyn⟶0,then, by (72), we obtain that
(75)limn⟶∞zn−QιM1QιM2⋯QιMkyn=0.Now, by (13), (73), (75), and using uniformly norm-to-norm continuity of JX1 on bounded sets, it is implied that
(76)limn⟶∞ϕzn,QιM1QιM2⋯QιMkzn=0,limn⟶∞ϕzn,QιM1QιM2⋯QιMkyn=0.It follows from (10), (76), Lemma 2, and the convexity of .2 that
(77)ϕzn,wn≤ϕzn,JX1−1βnJX1QιM1QιM2⋯QιMkzn+1−βnJX1QιM1QιM2⋯QιMkyn=zn2−2zn,βnJX1QιM1QιM2⋯QιMkzn+1−βnJX1QιM1QιM2⋯QιMkyn+βnJX1QιM1QιM2⋯QιMkzn+1−βnJX1QιM1QιM2⋯QιMkyn2≤zn2−2βnzn,JX1QιM1QιM2⋯QιMkzn−21−βnzn,JX1QιM1QιM2⋯QιMkyn+βnQιM1QιM2⋯QιMkzn2+1−βnQιM1QιM2⋯QιMkyn2=βnϕzn,QιM1QιM2⋯QιMkzn+1−βnϕzn,QιM1QιM2⋯QιMkyn⟶0asn⟶∞.Now, by Lemma1, we have limn⟶∞zn−wn=0. Therefore, we obtain from (70) that limn⟶∞un−wn=0, then by (13), we conclude that
(78)limn⟶∞ϕun,wn=0.From (66), (78), Lemma 2, and our assumptions, it implied that
(79)ϕun,vn+1≤ϕun,JX1−1αn,1JX1fxn+αn,2JX1un+αn,3JX1wn=un2−2un,αn,1JX1fxn+αn,2JX1un+αn,3JX1wn+αn,1JX1fxn+αn,2JX1un+αn,3JX1wn2≤un2−2αn,1un,JX1fxn−2αn,2un,JX1un−2αn,3un,JX1wn+αn,1fxn2+αn,2un2+αn,3wn2=αn,1ϕun,fxn+αn,2ϕun,un+αn,3ϕun,wn⟶0asn⟶∞.Therefore, by Lemma1, we have
(80)limn⟶∞vn+1−un=0.Letr2=supnvn,xn. Therefore, by Lemma 6, there exists a continuous, convex, and strictly increasing function g2:0,2r2⟶0,∞ such that g20=0 and
(81)g2xn−vn≤ϕxn,vn.It follows from (40), (81), Lemma 8, and the fact that xn=Krnvn, we conclude that
(82)g2xn−vn≤ϕxn,vn≤ϕu^,vn−ϕu^,xn≤ϕu^,xn−1−ϕu^,xn⟶0asn⟶∞.Therefore,(83)limn⟶∞xn−vn=0,because g2 is a continuous strictly increasing convex function. Now, by (80) and (83), we have
(84)limn⟶∞un−xn+1=0.From (68) and (84), we obtain that
(85)limn⟶∞xn+1−xn=0.This shows thatxn is a Cauchy sequence, so xn converges strongly to a point q∈C. Therefore, by (68), (70), and (72), we imply that un, yn, and zn converge strongly to q.Next, we prove thatq∈∩i=1kFQιMi. It follows from (46) and uniformly continuity of JX1 on bounded subset of X1 that JX1QιM1QιM2⋯QιMkxn−JX1QιM2⋯QιMkxn⟶0 as n⟶∞. Get ηn=QιM1QιM2⋯QιMkxn; hence, by Definition 9, we have JX1ηn+ιM1ηn=JX1QιM2QιM3⋯QιMkxn. Therefore, there exists hn∈M1ηn such that
(86)hn=JX1QιM2QιM3⋯QιMkxn−JX1ηnι.So, by the above observation,hn⟶0 as n⟶∞. On the other hand, since xn⇀q, we can conclude from (47) that ηn⇀q. Then, from Lemma 10, 0∈M1q, i.e., q∈M1−10=FQιM1. Similar to the above, by using (46), we can also prove that q∈Mi−10=FQιMi for all i=2,3,⋯k. Hence, q∈∩i=1kFQιMi.Next, we show thatq∈Ff. From (65), (68), and the triangle inequality, we conclude that
(87)limn⟶∞fxn−xn=0.Hence,q is an asymptotic fixed point of f. Then, F^f=Ff because f is a relatively nonexpansive mapping. Hence, q∈Ff.Now, we prove thatq∈EPg. Since JX1 is uniformly norm-to-norm continuous on bounded sets, it follows from (83) that
(88)limn⟶∞JX1xn−JX1vn=0.Byxn=Krnvn, we conclude that gxn,y+Bxn,y−xn+1/rny−xn,JX1xn−JX1vn≥0 for all y∈C. Moreover, by the condition A2, gy,xn≤−gxn,y for all y∈C. Therefore,
(89)gy,xn≤Bxn,y−xn+1rny−xn,JX1xn−JX1vn,for all y∈C. Using (88), the condition A4, and letting n⟶∞, we have that
(90)gy,q≤Bq,y−q,for ally∈C. Let yλ=λy+1−λq for all y∈C and λ∈0,1. It follows from (90), the conditions A1, A4, and the monotonicity of B that
(91)0=gyλ,yλ+Byλ,yλ−yλ≤λgyλ,y+1−λgyλ,q+Byλ,λy+1−λq−yλ=λgyλ,y+1−λgyλ,q+λByλ,y−yλ+1−λByλ,q−yλ=λgyλ,y+1−λgyλ,q+λByλ,y−yλ+1−λByλ−Bq,q−yλ+1−λBq,q−yλ≤λgyλ,y+λByλ,y−yλ,for all y∈C. Therefore, 0≤gyλ,y+Byλ,y−yλ. Using the condition A3 and letting λ⟶0, we obtain that 0≤gq,y+Bq,y−q for all y∈C. Then, q∈EPg.Finally, we prove thatq∈Ω. From (56), we have that PDAq−Aq=limn⟶∞PDAun−Aun=0. Therefore, Aq∈D, i.e., q∈Ω. Hence, q=ΠΩ∩∩i=1kFQιMi∩EPg∘fq, and this completed the proof.



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*Source: 4035695-2022-07-18.xml*