We introduce a new mixed equilibrium problem with a relaxed monotone mapping in a reflexive Banach space and prove the existence of solution of the equilibrium problem. Using Bregman distance, we introduce the concept of Bregman K-mapping for a finite family of Bregman quasiasymptotically nonexpansive mappings and show the fixed point set of the Bregman K-mapping is the set of common fixed points of {Ti}i=1N. Using the Bregman K-mapping, we introduce an iterative sequence for finding a common point in the set of a common fixed points of the finite family of Bregman quasiasymptotically nonexpansive mappings and the set of solutions of some mixed equilibrium problems. Strong convergence of the iterative sequence is proved. Our results generalise and improve many recent results in the literature.
1. Introduction
Let E be a real Banach space and let E∗ be the dual of E. Let C be a nonempty closed and convex subset of E. A mapping T:C→C is called nonexpansive if Tx-Ty≤x-y∀x,y∈C. A point x∈C is a fixed point of T if Tx=x. We denote by F(T) the fixed point set of T; that is, F(T)={x∈C:Tx=x}. Let g:C×C→R be a bifunction. The equilibrium problem with respect to g and C in the sense of Blum and Oettli [1] is to find z∈C such that(1)gz,y≥0∀y∈C.The set of solutions of equilibrium problem is denoted by EP(g); that is,(2)EPg=z∈C:gz,y≥0∀y∈C.In order to solve equilibrium problem (1), the bifunction g is usually assumed to satisfy the following conditions:
g(x,x)=0 for all x,y∈C;
g is monotone; that is, g(x,y)+g(y,x)≤0 for all x,y∈C;
for all x,y,z∈C, limsupt→0g(tz+(1-t)x,y)≤g(x,y);
for all x∈C,g(x,·) is convex and lower semicontinuous.
Fang and Huang [2] introduced the concept of relaxed η-α monotone mapping for solving mixed equilibrium problems.
A mapping A:C→E∗ is said to be relaxed η-α monotone (see also [3]) if there exist a mapping η:C×C→E and a function α:E→R with α(tz)=tpα(z) for all t>0 and z∈E, where p>1 such that(3)Ax-Ay,ηx,y≥αx-y∀x,y∈C.Particularly if η(x,y)=x-y∀x,y∈C and α(z)=kzp, where p>1 and k>1 are two constants, then A is called p monotone; see [4, 5].
Fang and Huang [2] proved that under some suitable assumptions, the following variational inequality is solvable: find z∈C such that(4)Az,ηy,z+ψy-ψz≥0∀y∈C,where ψ is a function from C to R∪{∞}. They also proved that the following inequality is equivalent to variational inequality (4): find z∈C such that(5)Az,ηy,z+ψy-ψz≥αy-z∀y∈C.The mixed equilibrium problem (see [6, 7]) is to find z∈C such that(6)gz,y+Az,ηy,z+ψy-ψz≥0∀y∈C.We denote the set of solutions of mixed equilibrium problem (6) by EP(g,A). It is easily seen that if g(z,y)=0∀z,y∈C, then mixed equilibrium problem (6) reduces to variational inequality (4). In the case of A=0 and ψ=0, then, EP(g,A) coincides with EP(g).
Equilibrium problems and mixed equilibrium problems have been used as tools for solving problems arising from linear and nonlinear programming, optimization problems, variational inequalities, fixed point problems, and also problems in physics, economics, engineering, and so forth (see, e.g., [1, 6, 8–15] and the references therein).
Let J:E→2E∗ be the normalised duality mapping defined by (7)Jx=f∗∈E∗:x,f∗=x2=f∗2,where 〈·,·〉 denotes the generalised duality pairing. It is well known that if E is smooth, strictly convex, and reflexive, then J is single-valued, one-to-one, and onto.
Let f:E→(-∞,+∞] be a convex function. We denote by domf the domain of f; that is, domf={x∈E:f(x)<+∞}. The function f is said to be coercive if limx→∞f(x)=+∞. f is said to be strongly coercive if limx→∞f(x)/x=+∞. The Fenchel conjugate of f is the function f∗:E∗→(-∞,+∞] defined by (8)f∗x∗=supx∗,x-fx:x∈E.The subdifferential of f is a mapping ∂f:E→E∗ defined by (9)∂fx=x∗∈E∗:fy≥fx+x∗,y-x∀y∈E∀x∈E.It is well known that (see [16]) x∗∈∂f(x) if and only if f(x)+f∗(x∗)=〈x∗,x〉 for all x∈E. It is also known that if f:E→(-∞,+∞] is a proper, convex, and lower semicontinuous function, then f∗:E∗→(-∞,+∞] is a proper, convex, and weak∗ lower semicontinuous function; see, for example, [17].
For any convex function f:E→(-∞,+∞], let x∈intdomf and y∈E. The right-hand derivative of f at x in the direction y is defined by(10)f∘x,y=limt→0fx+ty-fxt.The function f is said to be Gâteaux differentiable at x if limt→0(f(x+ty)-f(x))/t exists for any y∈E. In this case f∘(x,y) coincides with ∇f(x), the value of the gradient ∇f of f at x. f is said to be Gâteaux differentiable if it is Gâteaux differentiable at each x∈intdomf. If the limit in (10) is attained uniformly in y=1, then f is said to be Fréchet differentiable at x. f is said to be uniformly Fréchet differentiable on a subset C of E if the limit in (10) is attained uniformly for every x∈C and y=1. We know that if f is uniformly Fréchet differentiable on bounded subset of E, then f is uniformly continuous on bounded set of E (see, e.g., [18]).
The function f is said to be essentially smooth if ∂f is both bounded and single-valued on its domain. It is called essentially strictly convex if (∂f)-1 is locally bounded on its domain and f is strictly convex on every convex subset of dom∂f. f is said to be a Legendre function if it is both essentially smooth and essentially strictly convex. When the subdifferential of f is single-valued, it coincides with the gradient; that is, ∂f=∇f; see, for example, [19].
For a Legendre function f, the following properties are well known:
f is essentially smooth if and only if f∗ is essentially strictly convex; see [16];
(∂f)-1=∂f∗; see [20];
f is Legendre if and only if f∗ is Legendre function; see [16];
if f is Legendre function, then ∇f is bijection satisfying ∇f=(∇f∗)-1, ran∇f=dom∇f∗=intdomf∗, and ran∇f∗=dom∇f=intdomf; see [16].
If E is smooth and strictly convex, the function f(x)=1/pxp, 1<p<∞, is Legendre function; see, for example, [21]. In this case ∇f=Jp, 1<p<∞. In particular if E=H is a Hilbert space we have ∇f=I, the identity mapping.
Let f:E→R be a convex and Gâteaux differentiable function. The function Df:E×E→R defined by(11)Dfx,y=fx-fy-∇fy,x-y∀x,y∈Eis called Bregman distance corresponding to f; see [22, 23]. It follows from the strict convexity of f that Df(x,y)≥0∀x,y∈E and Df(x,y)=0 if and only if x=y; see [24].
Bregman projection with respect to f of x∈E onto the nonempty closed convex subset C of E is the unique vector PCf(x)∈C satisfying (12)DfPCfx,x=infDfy,x:y∈C.
Remark 1.
If E is smooth and strictly convex Banach space and f(x)=x2∀x∈E, then we have ∇f(x)=2J(x)∀x∈E and hence Df(x,y)=x2-2〈x,Jy〉+y2=ϕ(x,y)∀x,y∈E which is the Lyapunov function introduced by Alber [25] and the Bregman projection PCf(x) reduces to the generalised projection ΠC(x) which is defined by (13)ϕΠCx,x=infϕy,x:y∈C.If E=H, a Hilbert space, then the Bregman projection PCf(x) reduces to the metric projection PC(x) of H onto C.
Observing (11), we have(14)Dfz,x-Dfz,y=Dfy,x+∇fy-∇fx,z-y∀x,y,z∈E,which is called the three-point identity.
Let f:E→(-∞,+∞] be a convex, Legendre, and Gâteaux differentiable function. Following [23, 25] we make use of the function Vf:E×E∗→[0,+∞) associated with f defined by(15)Vfx,x∗=fx-x,x∗+f∗x∗∀x∈E,x∗∈E∗.Then Vf is nonnegative and Vf(x,x∗)=Df(x,∇f∗(x∗))∀x∈E and x∗∈E∗. Also from definition (15), it is obvious that Df(x,y)=Vf(x,∇f(y)) and Vf is convex in the second variable. Therefore for t∈(0,1) and x,y∈E, we have(16)Dfz,∇f∗t∇fx+1-t∇fy≤tDfz,x+1-tDfz,y.Moreover by subdifferential inequality [26], we have(17)Vfx,x∗+y∗,∇f∗x∗-x≤Vfx,x∗+y∗∀x∈E,x∗∈E∗.Recall that a mapping T:C→C is said to be ϕ-quasi nonexpansive if F(T)≠∅ and ϕ(p,Tx)≤ϕ(p,x)∀x∈C and p∈F(T). T is ϕ-quasiasymptotically nonexpansive if F(T)≠∅ and there exists a real sequence {vn}⊂[0,∞) such that vn→0 as n→∞ and ϕ(p,Tnx)≤(1+vn)ϕ(p,x)∀x∈C and p∈F(T). T is called Bregman quasi nonexpansive if F(T)≠∅ and Df(p,Tx)≤Df(p,x)∀x∈C and p∈F(T). T is Bregman quasiasymptotically nonexpansive if F(T)≠∅ and there exists a real sequence {vn}⊂[0,∞) such that vn→0 as n→∞ and Df(p,Tnx)≤(1+vn)Df(p,x)∀x∈C and p∈F(T). T is said to be closed if for any sequence {xn}⊂C with xn→x and Txn→y, Tx=y.
It is worth mentioning that several iterative schemes have been constructed and proposed for finding points which solve fixed point problems and mixed equilibrium problems with relaxed monotone mappings in various settings. In 2010 Wang et al. [7] introduced the following iterative scheme for finding a common element of the set of solutions of a mixed equilibrium problem with relaxed monotone mapping and the set of fixed points of nonexpansive mappings in Hilbert spaces:(18)x1∈Cchosenarbitrarily,Φun,y+Tun,ηy,un+1λny-un,un-xn≥0∀y∈C,yn=αnxn+1-αnβnSxn+1-αn1-βnun,Cn=z∈C:yn-z≤xn-z,Dn=⋂j=1nCj,xn+1=PDnx1,n≥1,where T is a relaxed η-α monotone mapping and S:C→C is a nonexpansive mapping. Under some mild conditions on the three control sequences {αn}, {βn}, and {λn}, they obtained strong convergence of scheme (18) to common solution of mixed equilibrium problems and fixed point of nonexpansive mapping.
Recently, Chen et al. [27] introduced a new mixed equilibrium problem with the relaxed monotone mapping in uniformly convex and uniformly smooth Banach spaces and proved the existence of solutions of the mixed equilibrium problem. They also proposed the following iterative scheme to find the common element of the set of solutions of the mixed equilibrium problem and the set of fixed points of a quasi-ϕ-nonexpansive mapping:(19)x1=x∈Cchosenarbitrarily,yn=J-1αnJxn+1-αnJSxn,un∈Csuchthatθun,y+Aun,ηy,un+fy-fun+1rny-un,Jun-Jyn≥0∀y∈C,Cn=z∈C:ϕz,un≤ϕz,xn,Dn=⋂j=1nCj,xn+1=ΠDnx,n≥1,where S is a quasi-ϕ-nonexpansive mapping from C into itself and J is the normalised duality mapping. Under some assumptions on the parameter sequences {αn} and {rn}, they obtained strong convergence of scheme (19) to common solution of mixed equilibrium problems and fixed point of nonexpansive mapping.
Motivated and inspired by the above results, in this paper we introduce and prove the existence of solutions of the mixed equilibrium problem with relaxed monotone mapping in reflexive Banach spaces. Using Bregman distance, we introduce the concept of Bregman K-mapping of a finite family of Bregman quasiasymptotically nonexpansive mappings and propose an iterative sequence for finding a common element of the set of fixed points of a finite family of Bregman quasiasymptotically nonexpansive mappings and the set of solutions of mixed equilibrium problem.
2. Preliminaries
Let f:E→(-∞,+∞] be a convex and Gâteaux differentiable function. The modulus of total convexity of f at x∈intdomf is the function vf(x,·):[0,+∞)→[0,+∞) defined by (20)vfx,t=infDfy,x:y∈domf,y-x=t.The function f is totally convex at x if vf(x,t)>0 for all t>0 and f is totally convex if it is totally convex at each point x∈domf. Let B be a bounded subset of E. For t>0, define a functional on B, vf:intdomf×[0,+∞)→[0,+∞) defined by (21)vfB,t=infvfx,t:x∈B∪domf.f is totally convex on bounded set B if vf(B,t)>0 for any bounded subset D of E and t>0, where vf(·,t) is the total convexity of the function f on the set B.
Let Br={z∈E:z≤r}, for all r>0 and SE={x∈E:x=1}. The function f is bounded if f(Br) is bounded for all r>0 and f is uniformly convex on bounded subsets of E [28] if the function ρr:[0,+∞)→[0,+∞) defined by (22)ρrt=infx,y∈Br,x-y=t,α∈0,1αfx+1-αfy-fαx+1-αyα1-αsatisfies (23)ρrt>0∀r,t>0,where ρr is called the gauge of uniform convexity of f.
The gauge of uniform smoothness of f is the function σr:[0,+∞)→[0,+∞) defined by (24)σrt=supx∈Br,y∈SE,α∈0,1αfx+1-αty+1-αfx-tαy-fxα1-α.The function f is said to be uniformly smooth on E if σr(t)>0∀r,t>0. We know that from [28, 29] f is totally convex on bounded sets if and only if f is uniformly convex on bounded sets.
Definition 2 (see [2]).
Let E be a Banach space with the dual E∗ and C be a nonempty closed convex subset of E. Let A:C→E∗ and η:C×C→E be two mappings. Then A:C→E∗ is said to be η-hemicontinuous if for any fixed x,y∈C, the function h:[0,1]→R defined by h(t)=〈A((1-t)x+ty),η(x,y)〉 is continuous at 0+.
Definition 3 (see [30]).
Let E be a Banach space with the dual E∗ and C be a nonempty subset of E. A mapping H:C→2E is called a KKM mapping if for any finite set B⊂C one has coB⊂⋃x∈BH(x).
Remark 4.
Observe that from Definition 3, if H is a KKM mapping and W:C→2E such that H(x)⊂W(x) for all x∈C, then W is a KKM mapping.
In the sequel we will need the following lemmas.
Lemma 5 (see [31]).
Let M be a nonempty subset of a Hausdorff topological vector space X and let H:M→2X be a KKM mapping. If H(x) is closed in X for all x∈M and compact for some x∈M, then ⋂x∈MH(x)≠∅.
Lemma 6 (see [32]).
If x∈domf, then the following statements are equivalent:
the function f is totally convex at x;
for any sequence {yn}⊂domf(25)limn→∞Dfyn,x=0implies limn→∞yn-x=0.
Recall that a function f:E→(-∞,+∞] is called sequentially consistent [29] if for any two sequences {xn} and {yn} in E such that {xn} is bounded, (26)limn→∞Dfyn,xn=0implieslimn→∞yn-xn=0.
Lemma 7 (see [24]).
The function f:E→(-∞,+∞] is totally convex on bounded sets if and only if f is sequentially consistent.
Lemma 8 (see [26]).
Let f:E→(-∞,+∞] be a Legendre function such that ∇f∗ is bounded on bounded subsets of E∗. Let x∈E. If the sequence {Df(x,xn)} is bounded, then the sequence {xn} is bounded.
Lemma 9 (see [33]).
Let r>0 be a constant and let f:E→R be a uniformly convex function on bounded subsets of E. Then for any x,y∈Br and α∈(0,1),(27)fαx+1-αy≤αfx+1-αfy-α1-αρrx-y,where ρr is the gauge of the uniform convexity of f.
Lemma 10 (see [34]).
Let f:E→(-∞,+∞] be a uniformly Fréchet differentiable function and bounded on bounded subsets of E. Then ∇f is uniformly continuous on bounded subsets of E from the strong topology of E to strong topology of E∗.
Lemma 11 (see [28]).
Let f:E→(-∞,+∞] be a strongly coercive function. If ∇f is uniformly continuous on bounded subsets of E, then f∗:E∗→(-∞,+∞] is uniformly convex on bounded subsets of E∗.
Lemma 12 (see [28]).
Let f:E→(-∞,+∞] be a convex function which is bounded on bounded subsets of E. Then the following assertions are equivalent:
f is strongly coercive and uniformly convex on bounded subsets of E;
f∗ is Fréchet differentiable and ∇f∗ is uniformly norm-to-norm continuous on bounded subsets of domf∗=E∗.
Lemma 13 (see [29]).
Let C be a nonempty closed and convex subset of E. Let f:E→R be a Gâteaux differentiable and totally convex function. Let x∈E. Then
z=PCf(x) if and only if 〈∇f(x)-∇f(y),y-z〉≤0 for all y∈C;
Df(y,PCf(x))+Df(PCf(x),x)≤Df(y,x) for all y∈C.
Lemma 14 (see [35]).
Let E be a reflexive Banach space and C be a nonempty closed and convex subset of E. Let f:E→(-∞,+∞] be a Legendre function which is bounded, uniformly Fréchet differentiable, and totally convex which is bounded on bounded subsets of E. Let T:C→C be a closed and Bregman quasiasymptotically nonexpansive mapping. If F(T)≠∅, then it is closed and convex.
Lemma 15.
Let E be a reflexive Banach space and C be a nonempty closed and convex subset of E. Let f:E→(-∞,+∞] be a Legendre, uniformly Fréchet differentiable, strongly coercive, and totally convex function on bounded subsets of E. Then the Bregman projection PCf:E→C is continuous.
Proof.
Let {xn} be a sequence in E such that xn→x as n→∞. Let xn′=PCfxn and x′=PCfx. By Lemma 13(ii), we have(28)Dfy,xn′+Dfxn′,xn≤Dfy,xn,∀y∈C.From inequality (28), we have (29)Dfy,xn′≤Dfy,xn,∀y∈C.Since {Df(y,xn)} converges, it is bounded and using the above inequality it follows that Df(y,xn′) is bounded. The function f is strongly coercive and totally convex which is bounded on bounded subsets of E; therefore in view of Lemma 12∇f∗ is uniformly norm-to-norm continuous on bounded subsets of domf∗=E∗ and consequently ∇f∗ is bounded. Hence by Lemma 8 we obtain that {xn′} is bounded.
Since x′=PCfx we have Df(x′,x)≤Df(xn′,x). Therefore (30)Dfxn′,xn-Dfx′,x≥Dfxn′,xn-Dfxn′,x=fx-fxn+∇fxn-∇fx,xn-xn′+∇fx,xn-x.Since f is uniformly Fréchet differentiable on bounded subsets of E, it follows that f is uniformly continuous on bounded subsets of E (see, e.g., [18]). Thus, taking liminf as n→∞ of both sides of the above inequality, we obtain (31)liminfn→∞Dfxn′,xn-Dfx′,x≥0.Now let ϵ>0. Using (28), we have (32)Dfx′,xn′≤Dfx′,xn-Dfxn′,xn≤fx-fxn+∇fx-∇fxn,x′-x+∇fxn,xn-x+ϵ∀n≥N0,where N0 is some natural number. As n→∞ and ϵ→0, we obtain (33)Dfx′,xn′⟶0.By total convexity of f, we get xn′→x′ as n→∞. This completes the proof.
3. Main ResultsLemma 16.
Let C be a nonempty, closed, and convex subset of a reflexive Banach space E with the dual E∗. Let f:E→(-∞,+∞] be a convex and Gâteaux differentiable function. Let A:C→E∗ be η-hemicontinuous and relaxed η-α monotone mapping and g:C×C→R be a bifunction satisfying (C1) and (C4). Let ψ:C→R be proper, convex, and lower semicontinuous. For r>0 and x∈E, suppose the following conditions hold:
η(z,z)=0∀z∈C;
〈Av,η(·,u)〉 is convex for fixed u,v∈C.
Then problems (34) and (35) are equivalent.
Find z∈C such that(34)gz,y+Az,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z≥0∀y∈C.Find z∈C such that(35)gz,y+Ay,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z≥αy-z∀y∈C.
Proof.
Suppose (34) holds. Let z∈C be a solution of (34); then (36)gz,y+Az,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z≥0∀y∈C.Since A is relaxed η-α monotone, we obtain (37)gz,y+Ay,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z≥gz,y+Az,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z+αy-z∀y∈C≥αy-z∀y∈C.This shows that z∈C is a solution of (35).
Conversely, suppose (35) holds; that is, z∈C is a solution of (35). Let y∈C such that ψ(y)<∞; then ψ(z)<∞. Let yt=(1-t)z+ty, t∈(0,1). Since z∈C is a solution of (35), we have (38)gz,yt+Ayt,ηyt,z+ψyt-ψz+1r∇fz-∇fx,yt-z≥αyt-z=αty-z≥tpαy-z.Hence by (C1), (C4), (i), and (ii), we obtain (39)tpαy-z≤gz,1-tz+ty+Ayt,η1-tz+ty,z+ψ1-tz+ty-ψz+1r∇fz-∇fx,1-tz+ty-z≤1-tgz,z+tgz,y+1-tAyt,ηz,z+tAyt,ηy,z+tψy-tψz+t1r∇fz-∇fx,y-z.Thus, (40)tp-1αy-z≤gz,y+Ayt,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z=gz,y+A1-tz+ty,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z.Since A is η-hemicontinuous and p>1, by allowing t→0+ we obtain (41)0≤gz,y+Az,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z.This shows that z∈C is a solution of (34)
Lemma 17.
Let E be a reflexive Banach space with the dual E∗ and let C be a nonempty closed, convex, and bounded subset of E. Let f:E→(-∞,+∞] be a Legendre function and uniformly Fréchet differentiable on bounded subsets of E. Let A:C→E∗ be η-hemicontinuous and relaxed η-α monotone mapping and g:C×C→R be a bifunction satisfying (C1) and (C4). Let ψ:C→R be proper, convex lower semicontinuous. For r>0 and x∈E, suppose
η(x,x)=0 for all x∈C,
η(z,y)+η(y,z)=0∀z,y∈C,
〈Au,η(·,v)〉 is convex and lower semicontinuous for fixed u,v∈C,
α:E→R is weakly lower semicontinuous.
Then there exists z∈C such that (42)gz,y+Az,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z≥0∀y∈C.
Proof.
Define two set-valued mappings H,W:C→2E as follows: (43)Hy=z∈C:gz,y+Az,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z≥0∀y∈C,Wy=z∈C:gz,y+Ay,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z≥αy-z∀y∈C.We claim H is a KKM mapping. By contradiction suppose then there do not exist {y1,y2,…,yn}⊂C and ti>0(i=1,2,…,n) such that ∑i=1nti=1 and y=∑i=1ntiyi∉⋃i=1nH(yi). This implies (44)gy,yi+Ay,ηyi,y+ψyi-ψy+1r∇fy-∇fx,y-yi<0∀i=1,2,…,n.It follows that (45)0=gy,y+Ay,ηy,y=gy,∑i=1ntiyi+Ay,η∑i=1ntiyi,y≤∑i=1ntigy,yi+∑i=1ntiAy,ηyi,y<∑i=1ntiψy-ψyi+1r∇fy-∇fx,yi-y=ψy-∑i=1ntiψyi≤ψy-ψ∑i=1ntiyi=0,which is a contradiction. Thus H is a KKM mapping.
Next we show that ∀y∈CH(y)⊂W(y).
Let z∈H(y). Then (46)gz,y+Az,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z≥0.
Since A is relaxed η-α monotone, we have (47)gz,y+Ay,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z≥gz,y+Az,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z+αy-z≥αy-z.Showing that z∈W(y) for all y∈C. By Remark 4 it follows that W is a KKM mapping.
We claim also that W(y) is closed in the weak topology of E. Let y∈C and z¯ be the weak closure point of W(y). Since E is reflexive, there exists a sequence {zn}⊂W(y) such that zn⇀z¯∈C as n→∞. Observe that (48)gzn,y+Ay,ηy,zn+ψy-ψzn+1r∇fzn-∇fx,y-zn≥αy-znis equivalent to(49)1r∇fzn-∇fx,y-zn≥gy,zn+Ay,ηzn,y+ψzn-ψy+αy-zn.By (iii) and (iv) and taking liminf as n→∞ of both sides of (49) we obtain (50)1r∇fz¯-∇fx,y-z¯≥gy,z¯+Ay,ηz¯,y+ψz¯-ψy+αy-z¯.That is, z¯∈W(y)∀y∈C. This implies W(y) is weakly closed for all y∈C. Since C is weakly compact, then W(y) is weakly compact in C for all y∈C.
It is clear that the solution sets of problem (34) and (35) are ⋂y∈CH(y) and ⋂y∈CW(y). Using Lemmas 16 and 5 we obtain (51)⋂y∈CHy=⋂y∈CWy≠∅.Hence there exists z∈C such that (52)gz,y+Az,ηy,z+ψy-ψz+1r∇fz-∇fx,y-z≥0∀y∈C.This completes the proof.
Lemma 18.
Let E be a reflexive Banach space with the dual E∗ and let C be a nonempty closed, convex, and bounded subset of E. Let f:E→(-∞,+∞] be Legendre and Gâteaux differentiable function. Let A:C→E∗ be η-hemicontinuous and relaxed η-α monotone mapping and g:C×C→R be a bifunction satisfying (C1), (C2), and (C4). Let ψ:C→R be proper, convex, and lower semicontinuous. For r>0 and x∈E, define a map Tr:E→2C by (53)Trx=z∈C:gz,y+Ay,ηyz+ψy-ψz+1r∇fz-∇fx,y-z≥0∀y∈C.Assume that
η(z,y)+η(y,z)=0∀z,y∈C;
〈Au,η(·,v)〉 is convex and lower semicontinuous for fixed u,v∈C;
α:E→R is weakly lower semicontinuous;
α(x-y)+α(y-z)≥0∀x,y∈C.
Then
Tr is single-valued;
Tr is a Bregman firmly nonexpansive type mapping; that is,(54)∇fTrx-∇fTry,Trx-Try≤∇fx-∇fy,Trx-Try∀x,y∈C;
F(Tr)=EP(g,A);
Tr is a Bregman quasi nonexpansive satisfying (55)Dfu,Trx+DfTrx,x≤Dfu,x;
EP(g,A) is closed and convex.
Proof.
First we show that Tr is single-valued. Let z1,z2∈Trx; then(56)gz1,z2+Az1,ηz2,z1+ψz2-ψz1+1r∇fz1-∇fx,z2-z1≥0,gz2,z1+Az2,ηz1,z2+ψz1-ψz2+1r∇fz2-∇fx,z1-z2≥0.By using (C2), adding (56) yields (57)Az1,ηz2,z1+Az2,ηz1,z2+1r∇fz1-∇fz2,z2-z1≥0.By (i) we have (58)Az1-Az2,ηz2,z1+1r∇fz1-∇fz2,z2-z1≥0.Since A is relaxed η-α monotone, we obtain (59)1r∇fz1-∇fz2,z2-z1≥Az2-Az1,ηz2,z1≥αz2-z1.Thus(60)∇fz1-∇fz2,z2-z1≥rαz2-z1.Interchanging z1 and z2 in (60), we have(61)∇fz2-∇fz1,z1-z2≥rαz1-z2.Adding (60) and (61), we have (62)∇fz1-∇fz2,z2-z1+∇fz2-∇fz1,z1-z2≥rαz2-z1+αz1-z2.Hence (63)2∇fz1-∇fz2,z2-z1≥rαz2-z1+αz1-z2.By (iv), we have (64)∇fz1-∇fz2,z2-z1≥0.Thus,(65)∇fz2-∇fz1,z2-z1≤0.Since f is convex and Gâteaux differentiable we have(66)∇fz2-∇fz1,z2-z1≥0.By (65) and (66) we obtain (67)∇fz2-∇fz1,z2-z1=0.Since f is Legendre function, then z1=z2.
Next we show that Tr is Bregman firmly nonexpansive type. Let x,y∈C; then(68)gTrx,Try+ATrx,ηTry,Trx+ψTry-ψTrx+1r∇fTrx-∇fx,Try-Trx≥0,gTry,Trx+ATry,ηTrx,Try+ψTrx-ψTry+1r∇fTry-∇fy,Trx-Try≥0.Adding (68), using (i) and (C2), we obtain (69)ATrx-ATry,ηTry,Trx+1r∇fTrx-∇fTry-∇fx+∇fy,Try-Trx≥0,so that (70)1r∇fTrx-∇fTry-∇fx+∇fy,Try-Trx≥ATry-ATrx,ηTry,Trx.Since A is relaxed η-α monotone and r>0, we have(71)∇fTrx-∇fTry-∇fx+∇fy,Try-Trx≥rαTry-Trx.Also interchanging the roles of x and y in (71) and applying (iv), we have (72)2∇fTrx-∇fTry-∇fx+∇fy,Try-Trx≥0.Hence,(73)∇fTrx-∇fTry,Trx-Try≤∇fx-∇fy,Trx-Try,showing that Tr is Bregman firmly nonexpansive type.
We now show that F(Tr)=EP(g,A). Indeed (74)u∈FTr⟺u∈Tru⟺gu,y+Au,ηy,u+ψy-ψu+1r∇fu-∇fu,y-u≥0∀y∈C⟺gu,y+Au,ηy,u+ψy-ψu≥0∀y∈C⟺u∈EPg,A.Next we prove that Tr is Bregman quasi nonexpansive mapping.
Since(75)DfTrx,Try+DfTry,Trx≤DfTrx,y+DfTry,x-DfTrx,x-DfTry,y,we have(76)DfTrx,Try+DfTry,Trx≤DfTrx,y+DfTry,x.Let u=y∈F(Tr); then from (76) we obtain (77)Dfu,Trx≤Dfu,x.This shows that Tr is Bregman quasi nonexpansive mapping which is Bregman quasiasymptotically nonexpansive. Also from (75), we have(78)DfTrx,Try+DfTry,Trx+DfTrx,x+DfTry,y≤DfTrx,y+DfTry,x.As u=y∈F(Tr), we obtain (79)Dfu,Trx+DfTrx,x≤Dfu,x.Lastly, using (3), (4), and Lemma 14 we obtain that EP(g,A) is closed and convex.
Definition 19.
Let C be a nonempty, closed, and convex subset of a real Banach space E. Let {Ti}i=1N be a finite family of Bregman quasiasymptotically nonexpansive mappings. For any n∈N, define a mapping Kn:C→C as follows:(80)Sn,0x=xSn,1x=PCf∇f∗αn,1∇fT1nx+1-αn,1∇fxSn,2x=PCf∇f∗αn,2∇fT2nSn,1x+1-αn,2∇fSn,1xSn,3x=PCf∇f∗αn,3∇fT3nSn,2x+1-αn,3∇fSn,2x⋮Sn,N-1x=PCf∇f∗αn,N-1∇fTN-1nSn,N-2x+1-αn,N-1∇fSn,N-2xKnx=Sn,Nx=PCf∇f∗αn,N∇fTNnSn,N-1x+1-αn,N∇fSn,N-1x.Such a mapping Kn is called the Bregman K-mapping generated by T1,T2,T3,…,TN and αn,i∈(0,1), i=1,2,3,…,N.
Using the above definition, we have the following Lemma.
Lemma 20.
Let E be a reflexive Banach space with the dual E∗ and let C be a nonempty, closed, convex, and bounded subset of E. Let f:E→(-∞,+∞] be strongly coercive, Legendre, uniformly Fréchet differentiable, and totally convex function which is bounded on bounded subsets of E. Let {Ti}i=1N be a finite family of continuous Bregman quasiasymptotically nonexpansive mappings of C into itself such that ⋂i=1NF(Ti)≠∅. Let {αn,i} be a real sequence in (0,1) such that liminfn→∞αn,i>0∀i∈{1,2,3,…,N}. Let Kn be Bregman K-mapping generated by T1,T2,T3,…,TN in (80). Then
Df(x∗,Knx)≤(1+tn)Df(x∗,x) for all x∗∈F(Kn) and x∈C, where tn→0 as n→∞;
F(Kn)=⋂i=1NF(Ti);
Kn is a closed mapping.
Proof.
Let x∗∈F(Kn) and x∈C. Then by Lemma 13(ii) and inequality (16) we have (81)Dfx∗,Knx=Dfx∗,PCf∇f∗αn,N∇fTNnSn,N-1x+1-αn,N∇fSn,N-1x≤Dfx∗,∇f∗αn,N∇fTNnSn,N-1x+1-αn,N∇fSn,N-1x≤αn,NDfx∗,TNnSn,N-1x+1-αn,NDfx∗,Sn,N-1x≤αn,N1+vn,NDfx∗,Sn,N-1x+1-αn,NDfx∗,Sn,N-1x=1+αn,Nvn,NDfx∗,Sn,N-1x=1+αn,Nvn,NDfx∗,∇f∗αn,N-1∇fTN-1nSn,N-2x+1-αn,N-1∇fSn,N-2x≤1+αn,Nvn,Nαn,N-1Dfx∗,TN-1nSn,N-2x+1-αn,N-1Dfx∗,Sn,N-2x≤1+αn,Nvn,Nαn,N-11+vn,N-1Dfx∗,Sn,N-2x+1-αn,N-1Dfx∗,Sn,N-2x=1+αn,Nvn,N1+αn,N-1vn,N-1Dfx∗,Sn,N-2x≤1+αn,Nvn,N1+αn,N-1vn,N-11+αn,N-2vn,N-2Dfx∗,Sn,N-3x⋮≤1+αn,Nvn,N1+αn,N-1vn,N-1⋯1+αn,2vn,2Dfx∗,Sn,1x.Hence, (82)Dfx∗,Knx≤1+αn,Nvn,N1+αn,N-1vn,N-1⋯1+αn,2vn,2Dfx∗,∇f∗αn,1∇fT1nx+1-αn,1∇fx≤1+αn,Nvn,N1+αn,N-1vn,N-1⋯1+αn,1vn,1Dfx∗,x=∏i=1N1+αn,ivn,iDfx∗,x.Observe that αn,ivn,i→0 as n→∞ for all i∈{1,2,3,…,N}. Let ∏i=1N(1+αn,ivn,i)=(1+tn); then tn→0 as n→∞ and (83)Dfx∗,Knx≤1+tnDfx∗,x.Next we show that F(Kn)=⋂i=1NF(Ti). It is obvious that ⋂i=1NF(Ti)⊂F(Kn). Now let z∈F(Kn) and x∗ be any point in ⋂i=1NF(Ti).
Let r1=sup∇fz,∇fSn,iz,∇fTinSn,iz for 1≤i≤N. Then (84)Dfx∗,z=Dfx∗,Knz=Dfx∗,PCf∇f∗αn,N∇fTNnSn,N-1z+1-αn,N∇fSn,N-1z≤Dfx∗,∇f∗αn,N∇fTNnSn,N-1z+1-αn,N∇fSn,N-1z=Vfx∗,αn,N∇fTNnSn,N-1z+1-αn,N∇fSn,N-1z=fx∗-x∗,αn,N∇fTNnSn,N-1z+1-αn,N∇fSn,N-1z+f∗αn,N∇fTNnSn,N-1z+1-αn,N∇fSn,N-1z.Since f is uniformly Fréchet differentiable function which is bounded on bounded subsets of E, then by Lemma 10∇f is uniformly continuous on bounded subsets and consequently from Lemma 11f∗ is uniformly convex. Therefore in view of Lemma 9 we have (85)Dfx∗,z≤fx∗-x∗,αn,N∇fTNnSn,N-1z+1-αn,N∇fSn,N-1z+αn,Nf∗∇fTNnSn,N-1z+1-αn,Nf∗∇fSn,N-1z-αn,N1-αn,Nρr1∗∇fTNnSn,N-1z-∇fSn,N-1z=αn,Nfx∗-x∗,∇fTNnSn,N-1z+f∗∇fTNnSn,N-1z+1-αn,Nfx∗-x∗,∇fSn,N-1z+f∗∇fSn,N-1z-αn,N1-αn,Nρr1∗∇fTNnSn,N-1z-∇fSn,N-1z=αn,NVfx∗,∇fTNnSn,N-1z+1-αn,NVfx∗,∇fSn,N-1z-αn,N1-αn,Nρr1∗∇fTNnSn,N-1z-∇fSn,N-1z=αn,NDfx∗,TNnSn,N-1z+1-αn,NDfx∗,Sn,N-1z-αn,N1-αn,Nρr1∗∇fTNnSn,N-1z-∇fSn,N-1z≤αn,N1+vn,NDfx∗,Sn,N-1z+1-αn,NDfx∗,Sn,N-1z-αn,N1-αn,Nρr1∗∇fTNnSn,N-1z-∇fSn,N-1z=1+αn,Nvn,NDfx∗,Sn,N-1z-αn,N1-αn,Nρr1∗∇fTNnSn,N-1z-∇fSn,N-1z.Continuing in this fashion we obtain (86)Dfx∗,z≤1+αn,Nvn,N1+αn,N-1vn,N-1⋯1+αn,1vn,1Dfx∗,z-1+αn,Nvn,N1+αn,N-1vn,N-1⋯1+αn,2vn,2αn,11-αn,1ρr1∗∇fT1nz-∇fz-1+αn,Nvn,N1+αn,N-1vn,N-1⋯1+αn,3vn,3αn,21-αn,2ρr1∗∇fT2nSn,1z-∇fSn,1z-1+αn,Nvn,N1+αn,N-1vn,N-1⋯1+αn,4vn,4αn,31-αn,3ρr1∗∇fT3nSn,2z-∇fSn,2z⋮-1+αn,Nvn,Nαn,N-11-αn,N-1ρr1∗∇fTN-1nSn,N-2z-∇fSn,N-2z-αn,N1-αn,Nρr1∗∇fTNnSn,N-1z-∇fSn,N-1z.From (86), we have (87)1+αn,Nvn,N1+αn,N-1vn,N-1⋯1+αn,2vn,2αn,11-αn,1ρr1∗∇fT1nz-∇fz≤1+αn,Nvn,N1+αn,N-1vn,N-1⋯1+αn,1vn,1Dfx∗,z-Dfx∗,z.Since liminfαn,1(1-αn,1)>0, we have (88)limn→∞ρr1∗∇fT1nz-∇fz=0.By the property of ρr1∗, we obtain(89)limn→∞∇fT1nz-∇fz=0.In similar fashion and assuming liminfαn,i(1-αn,i)>0∀i∈{2,3,…,N} we have(90)limn→∞∇fT2nSn,1z-∇fSn,1z=limn→∞∇fT3nSn,2z-∇fSn,2z=limn→∞∇fT4nSn,3z-∇fSn,3z⋮=limn→∞∇fTNnSn,N-1z-∇fSn,N-1z=0.Since by Lemma 12∇f∗ is uniformly continuous, we obtain from (89) that T1nz→z as n→∞. By our assumption, T1 is continuous and so T1z=z, that is, z∈F(T1). Also from (90), we obtain(91)T2nSn,1z=Sn,1zT3nSn,2z=Sn,2z⋮TNnSn,N-1z=Sn,N-1z.On the other hand (92)Dfz,Sn,1z≤Dfz,∇f∗αn,1∇fT1nx+1-αn,1∇fx≤αn,1Dfz,T1nz+1-αn,1Dfz,z.Since T1nz=z we get(93)Dfz,Sn,1z=0and so Sn,1z=z.From (91) and (93), we have z∈F(Sn,1) and z∈F(T2). Applying the same argument we can conclude that z∈F(Ti) for i=3,4,…,N. Thus, it follows that z∈⋂i=1NF(Ti).
Next we show that Kn is closed.
From (80)(94)Sn,1x=PCf∇f∗αn,1∇fT1nx+1-αn,1∇fx.Let {xm} be a sequence in C such that xm→x¯ and Sn,1xm→y as m→∞. Since Ti is continuous for each i=1,2,3,…,N, ∇f,∇f∗ are uniformly continuous and applying Lemma 15 we have(95)Sn,1xm⟶Sn,1x¯as m⟶∞.By uniqueness of limit, we get Sn,1x¯=y showing that Sn,1 is closed. Using (95), we have that Sn,2 is closed. Continuing in this way we obtain that Kn is closed.
Now we prove the strong convergence theorems.
Theorem 21.
Let E be a reflexive Banach space with the dual E∗ and let C be a nonempty, closed, convex, and bounded subset of E. Let f:E→(-∞,+∞] be a strongly coercive, Legendre, uniformly Fréchet differentiable, and totally convex function which is bounded on bounded subsets of E. For each j=1,2,3,…,m, let Aj:C→E∗ be η-hemicontinuous and relaxed η-α monotone mappings and gj:C×C→R be bifunctions satisfying (C1)–(C4). Let ψj:C→R be proper, convex, and lower semicontinuous mappings. Let {Ti}i=1N be a finite family of continuous Bregman quasiasymptotically nonexpansive mappings of C into itself such that F=[⋂i=1NF(Ti)]∩[EP(gj,Aj)]≠∅. Let Kn be the Bregman K-mapping generated by T1,T2,T3,…,TN in (80). Assume that the conditions of Lemma 18 and the following condition are satisfied:
For all x,y,z,w∈C one has (96)limsupt→0+Az,ηx,ty+1-tw≤Az,ηx,w.
Let {xn} be iteratively defined as follows:(97)x0=x∈Cchosen arbitrarily,C1,j=C=C0yn=∇f∗βn∇fxn+1-βn∇fKnxn,un,j∈Csuch thatgjun,j,y+Ajun,j,ηy,un,j+ψjy-ψjun,j+1rn∇fun,j-∇fyn,y-un,j≥0∀y∈C,Cn+1,j=z∈Cn:Dfz,un,j≤Dfz,xn+θn,Cn+1=⋂j=1mCn+1,j,xn+1=PCn+1fx0,n≥0,where {βn} is a real sequence in (0,1) satisfying liminfn→∞βn(1-βn)>0, {rn}⊂[a,∞) for some a>0 and θn=(1-βn)tnsupp∈FDf(p,xn). Then {xn} converges strongly to u=PFfx0.
Proof.
It follows from Lemmas 20(ii) and 14 that F(Kn) is closed and convex. On the other hand by Lemma 18(5) EP(gj,Aj) is closed and convex for each j∈{1,2,3,…,m}; consequently F is closed and convex.
Next we prove Cn is closed and convex. The proof is by induction.
For any fixed j, C1,j=C is closed and convex which implies C1 is closed and convex.
Now suppose Ck,j is closed and convex for some k∈N. Observe that (98)Dfz,un,j≤Dfz,xn+θn⟺∇fxn-∇fun,j,z≤fun,j-fxn+∇fun,j-∇fxn,un,j+∇fxn,xn-un,j+θn.It follows that Ck+1,j is closed and convex and so Ck+1 is closed and convex. Hence Cn is closed and convex for all n≥0. This implies that the iterative sequence is well defined.
Next we show that F⊂Cn∀n≥0. Obviously F⊂C1,j=C. Assume F⊂Ck,j for some k∈N and ∀j=1,2,…,m. Let x∗∈F; then x∗∈Ck,j∀j=1,2,…,m and so x∗∈Ck. Observe from scheme (97) that un,j=Trnjyn. Therefore by Lemma 18(4) we have(99)Dfx∗,un,j=Dfx∗,Trnjyn≤Dfx∗,yn.But (100)Dfx∗,yn=Dfx∗,∇f∗βn∇fxn+1-βn∇fKnxn=Vfx∗,βn∇fxn+1-βn∇fKnxn=fx∗-x∗,βn∇fxn+1-βn∇fKnxn+f∗βn∇fxn+1-βn∇fKnxn≤fx∗-βnx∗,∇fxn-1-βnx∗,∇fKnxn+βnf∗∇fxn+1-βnf∗∇fKnxn=βnVfx∗,∇fxn+1-βnVfx∗,∇fKnxn=βnDfx∗,xn+1-βnDfx∗,Knxn≤βnDfx∗,xn+1-βn1+tnDfx∗,xn=Dfx∗,xn+1-βntnDfx∗,xn≤Dfx∗,xn+1-βntnsupp∈FDfp,xn=Dfx∗,xn+θn.Thus, (101)Dfx∗,un,j≤Dfx∗,xn+θn.This shows that x∗∈Ck+1,j which implies F⊂Cn,j∀n≥0, ∀j∈{1,2,…,m}. Therefore we obtain F⊂Cn.
Next we show that limn→∞Df(xn,x0) exists.
Since Cn+1⊂Cn∀n≥0, we have xn+1=PCn+1x0∈Cn+1⊂Cn∀n≥0. Therefore(102)Dfxn,x0=DfPCnfx0,x0≤Dfxn+1,x0,showing that {Df(xn,x0)} is nondecreasing sequence of real numbers. On the other hand we have from Lemma 13(ii) that(103)Dfxn,x0≤Dfx∗,x0-Dfx∗,xn≤Dfx∗,x0.Using (102) and (103) we have that limn→∞Df(xn,x0) exists. Now(104)Dfxn+1,xn=Dfxn+1,PCnfx0≤Dfxn+1,x0-Dfxn,x0.Since limn→∞Df(xn,x0) exists, we obtain(105)limn→∞Dfxn+1,xn=0.Since f is totally convex on bounded set, by Lemma 7f is sequentially consistent and so we have(106)limn→∞xn+1-xn=0.Let m>n where n,m∈N. Using (104) we have (107)Dfxm,xn=Dfxm,PCnfx0≤Dfxm,x0-Dfxn,x0.Hence (108)Dfxm,xn⟶0asn,m⟶∞.It follows that(109)xm-xn⟶0asn,m⟶∞.This shows that {xn} is Cauchy sequence in E. Since E is reflexive and C is weakly closed, then there exists u∈C such that(110)limn→∞xn-u=0.Observe that xn+1∈Cn+1=⋂j=1mCn+1,j, ∀j∈{1,2,…,m}. Hence we obtain (111)Dfxn+1,un,j≤Dfxn+1,xn+θn.By (105) and the fact that θn→0 as n→∞, we get (112)limn→∞Dfxn+1,un,j=0∀j∈1,2,…,m,and so(113)limn→∞xn+1-un,j=0,∀j∈1,2,…,m.Now (114)un,j-xn≤un,j-xn+1+xn+1-xn.Therefore using (106) and (113) we obtain(115)limn→∞un,j-xn=0,∀j∈1,2,…,m.Also by (110) and (115), it follows that(116)limn→∞un,j-u=0,∀j∈1,2,…,m.Let r2=sup{∇f(xn),∇f(Knxn)}. In view of Lemma 9, we have (117)Dfx∗,yn=Dfx∗,∇f∗βn∇fxn+1-βn∇fKnxn=Vfx∗,βn∇fxn+1-βn∇fKnxn=fx∗-x∗,βn∇fxn+1-βn∇fKnxn+f∗βn∇fxn+1-βn∇fKnxn≤βnVfx∗,∇fxn+1-βnVfx∗,∇fKnxn-βn1-βnρr2∗∇fxn-∇fKnxn=βnDfx∗,xn+1-βnDfx∗,Knxn-βn1-βnρr2∗∇fxn-∇fKnxn.From (99), we have (118)Dfx∗,un,j≤βnDfx∗,xn+1-βnDfx∗,Knxn-βn1-βnρr2∗∇fxn-∇fKnxn.Therefore by Lemma 20(i), we have(119)Dfx∗,un,j≤βnDfx∗,xn+1-βn1+tnDfx∗,xn-βn1-βnρr2∗∇fxn-∇fKnxn=Dfx∗,xn+1-βntnDfx∗,xn-βn1-βnρr2∗∇fxn-∇fKnxn≤Dfx∗,xn+1-βntnsupp∈FDfp,xn-βn1-βnρr2∗∇fxn-∇fKnxn=Dfx∗,xn+θn-βn1-βnρr2∗∇fxn-∇fKnxn.Observe that (120)Dfx∗,xn-Dfx∗,un,j=fun,j-fxn+∇fun,j,x∗-un,j-∇fxn,x∗-xn=fun,j-fxn+∇fun,j,x∗-xn+∇fun,j,xn-un,j-∇fxn,x∗-xn=fun,j-fxn+∇fun,j-∇fxn,x∗-xn+∇fun,j,xn-un,j.Therefore (121)Dfx∗,xn-Dfx∗,un,j≤fun,j-fxn+∇fun,j-∇fxn,x∗-xn+∇fun,j,xn-un,j≤fun,j-fxn+∇fun,j-∇fxnx∗-xn+fun,jxn-un,j.From (115), we obtain(122)Dfx∗,xn-Dfx∗,un,j⟶0asn⟶∞∀j∈1,2,…,m.By (122) and liminfn→∞βn(1-βn)>0, it follows from (119) that (123)limn→∞ρr2∗∇fxn-∇fKnxn=0.From the property of ρr2∗, we deduce limn→∞∇f(xn)-∇f(Knxn)=0. By uniform continuity of ∇f∗ on bounded subsets of E, we obtain(124)limn→∞xn-Knxn=0.On the other hand (125)Knxn-u≤Knxn-xn+xn-u.From (110) and (124), we have(126)limn→∞Knxn-u=0.Now using (110), (126), and Lemma 20(iii), we obtain Knu=u∀n∈N. This implies u∈F(Kn)=⋂i=1NF(Ti).
Next we show u∈EP(gj,Aj).
From scheme (97) (127)∇fxn-∇fyn=1-βn∇fxn-∇fKnxn.Therefore by (124) and uniform continuity of ∇f, we have (128)limn→∞∇fxn-∇fyn=0.As ∇f∗ is uniformly continuous, we obtain(129)limn→∞xn-yn=0.From (115) and (129), we have(130)limn→∞un,j-yn=0∀j=1,2,…,m.Again since ∇f is uniformly continuous and rn>a, we get(131)limn→∞∇fun,j-∇fynrn=0∀j=1,2,…,m.From scheme (97) (132)gjun,j,y+Ajun,j,ηy,un,j+ψjy-ψjun,j+1rn∇fun,j-∇fyn,y-un,j≥0∀y∈C.Using (C2) and Lemma 17(ii), it follows that (133)1rn∇fun,j-∇fynun,j-y≥Ajun,j,ηun,j,y+ψjun,j-ψjy-gjun,j,y∀y∈C≥Ajun,j,ηun,j,y+ψjun,j-ψjy+gjy,un,j∀y∈C.Using (116) and (131) and taking liminf as n→∞ of the above inequality, we get (134)0≥Aju,ηu,y+ψju-ψjy+gjy,u∀y∈C∀j=1,2,…,m.Now for any t∈(0,1) and y∈C, let yt=ty+(1-t)u. Then yt∈C and so(135)0≥Aju,ηu,yt+ψju-ψjyt+gjyt,u∀j=1,2,…,m.Therefore by (C1), (C4), Lemma 18(i), (ii), and (135), we have (136)0=gjyt,yt+Aju,ηyt,yt+ψjyt-ψjyt=gjyt,ty+1-tu+Aju,ηty+1-tu,yt+ψjty+1-tu-ψjyt≤tgjyt,y+Aju,ηy,yt+ψjy-ψjyt+1-tgjyt,u+Aju,ηu,yt+ψju-ψjyt≤tgjyt,y+Aju,ηy,yt+ψjy-ψjyt.That is, (137)gjyt,y+Aju,ηy,yt+ψjy-ψjyt≥0.Since yt=ty+(1-t)u, we have (138)gjty+1-tu,y+Aju,ηy,ty+1-tu+ψjy-ψjty+1-tu≥0.From (C3), (v), and lower semicontinuity of ψ, we have by allowing t→0+(139)gju,y+Aju,ηy,u+ψjy-ψju≥0∀y∈C.This shows that u∈EP(gj,Aj).
Lastly we show u=PFfx0.
From xn=PCnfx0 and Lemma 13(i) we have (140)∇fx0-∇fxn,xn-z≥0∀z∈Cn.Since F⊂Cn, this implies that(141)∇fx0-∇fxn,xn-h≥0∀h∈F.Letting n→∞ in (141), we obtain (142)∇fx0-∇fxn,xn-h≥0∀h∈F.Again by Lemma 13(i) we have u=PFfx0. This completes the proof.
If N=1 and m=1, in Theorem 21 then we have the following corollary.
Corollary 22.
Let E be a reflexive Banach space with the dual E∗ and let C be a nonempty, closed, convex, and bounded subset of E. Let f:E→(-∞,+∞] be a strongly coercive, Legendre, uniformly Fréchet differentiable, and totally convex function which is bounded on bounded subsets of E. Let A:C→E∗ be η-hemicontinuous and relaxed η-α monotone mapping and g:C×C→R be bifunctions satisfying (C1)–(C4). Let ψ:C→R be a proper, convex, and lower semicontinuous mapping. Let T be a continuous Bregman quasiasymptotically nonexpansive mapping of C into itself such that F=F(T)∩EP(g,A)≠∅. Assume that the conditions of Lemma 18 and the following condition are satisfied:
For all x,y,z,w∈C one has (143)limsupt→0+Az,ηx,ty+1-tw≤Az,ηx,w.
Let {xn} be iteratively defined as follows:(144)x0=x∈Cchosen arbitrarily,yn=∇f∗βn∇fxn+1-βn∇fTnxn,un∈Csuch thatgun,y+Aun,ηy,un+ψy-ψun+1rn∇fun-∇fyn,y-un≥0∀y∈C,Cn+1=z∈Cn:Dfz,un≤Dfz,xn+θn,xn+1=PCn+1fx0,n≥0,where {βn} is a real sequence in (0,1) satisfying liminfn→∞βn(1-βn)>0, {rn}⊂[a,∞) for some a>0 and θn=(1-βn)tnsupp∈FDf(p,xn). Then {xn} converges strongly to u=PFfx0.
Setting Aj=0∀j=1,2,…,m in Theorem 21 then we obtain the following result.
Corollary 23.
Let E be a reflexive Banach space with the dual E∗ and let C be a nonempty, closed, convex, and bounded subset of E. Let f:E→(-∞,+∞] be a strongly coercive, Legendre, uniformly Fréchet differentiable, and totally convex function which is bounded on bounded subsets of E. For each j=1,2,3,…,m, let gj:C×C→R be bifunctions satisfying (C1)–(C4) and ψj:C→R be proper, convex, and lower semicontinuous mappings. Let {Ti}i=1N be a finite family of continuous Bregman quasiasymptotically nonexpansive mapping of C into itself such that F=[⋂i=1NF(Ti)]∩[EP(gj,ψj)]≠∅. Let Kn be the Bregman K-mapping generated by T1,T2,T3,…,TN in (80). Let {xn} be iteratively defined as follows:(145)x0=x∈Cchosen arbitrarily,C1,j=C=C0yn=∇f∗βn∇fxn+1-βn∇fKnxn,un,j∈Csuch thatgjun,j,y+ψjy-ψjun,j+1rn∇fun,j-∇fyn,y-un,j≥0∀y∈C,Cn+1,j=z∈Cn:Dfz,un,j≤Dfz,xn+θn,Cn+1=⋂j=1mCn+1,j,xn+1=PCn+1fx0,n≥0,where {βn} is a real sequence in (0,1) satisfying liminfn→∞βn(1-βn)>0, {rn}⊂[a,∞) for some a>0 and θn=(1-βn)tnsupp∈FDf(p,xn). Then {xn} converges strongly to u=PFfx0.
If Aj=0 and ψj=0∀j=1,2,…,m in Theorem 21 then we obtain the following result.
Corollary 24.
Let E be a reflexive Banach space with the dual E∗ and let C be a nonempty, closed, convex, and bounded subset of E. Let f:E→(-∞,+∞] be a strongly coercive, Legendre, uniformly Fréchet differentiable, and totally convex function which is bounded on bounded subsets of E. For each j=1,2,3,…,m, let gj:C×C→R be bifunctions satisfying (C1)–(C4). Let {Ti}i=1N be a finite family of continuous Bregman quasiasymptotically nonexpansive mappings of C into itself such that F=[⋂i=1NF(Ti)]∩[EP(gj)]≠∅. Let Kn be the Bregman K-mapping generated by T1,T2,T3,…,TN in (80). Let {xn} be iteratively defined as follows:(146)x0=x∈Cchosen arbitrarily,C1,j=C=C0yn=∇f∗βn∇fxn+1-βn∇fKnxn,un,j∈Csuch thatgjun,j,y+1rn∇fun,j-∇fyn,y-un,j≥0∀y∈C,Cn+1,j=z∈Cn:Dfz,un,j≤Dfz,xn+θn,Cn+1=⋂j=1mCn+1,j,xn+1=PCn+1fx0,n≥0,where {βn} is a real sequence in (0,1) satisfying liminfn→∞βn(1-βn)>0, {rn}⊂[a,∞) for some a>0 and θn=(1-βn)tnsupp∈FDf(p,xn). Then {xn} converges strongly to u=PFfx0.
As a direct consequence of Theorem 21 and Remark 1, we obtain the convergence result for system of the mixed equilibrium problems and finite family of quasi-ϕ-asymptotically nonexpansive mappings in uniformly convex and uniformly smooth Banach spaces.
Corollary 25.
Let E be a uniformly convex and uniformly smooth Banach space with the dual E∗ and let C be a nonempty, closed, convex, and bounded subset of E. Let J:E→E∗ be a normalised duality mapping. For each j=1,2,3,…,m, let Aj:C→E∗ be η-hemicontinuous and relaxed η-α monotone mappings and gj:C×C→R be bifunctions satisfying (C1)–(C4). Let ψj:C→R be proper, convex, and lower semicontinuous mappings. Let {Ti}i=1N be a finite family of continuous Bregman quasiasymptotically nonexpansive mappings of C into itself such that F=[⋂i=1NF(Ti)]∩[EP(gj,Aj)]≠∅. Let Kn be the Bregman K-mapping generated by T1,T2,T3,…,TN in (80). Assume that the conditions of Lemma 18 and the following condition are satisfied:
For all x,y,z,w∈C one has (147)limsupt→0+Az,ηx,ty+1-tw≤Az,ηx,w.
Let {xn} be iteratively defined as follows:(148)x0=x∈Cchosen arbitrarily,C1,j=C=C0yn=J-1βnJxn+1-βnJKnxn,un,j∈Csuch thatgjun,j,y+Ajun,j,ηy,un,j+ψjy-ψjun,j+1rnJun,j-Jyn,y-un,j≥0∀y∈C,Cn+1,j=z∈Cn:ϕz,un,j≤ϕz,xn+θn,Cn+1=⋂j=1mCn+1,j,xn+1=ΠCn+1x0,n≥0,where {βn} is a real sequence in (0,1) satisfying liminfn→∞βn(1-βn)>0, {rn}⊂[a,∞) for some a>0 and θn=(1-βn)tnsupp∈FDf(p,xn). Then {xn} converges strongly to u=PFfx0.
Remark 26.
Our theorems and corollaries generalise the main theorem of Chen et al. [27] in the following senses:
For the structure of Banach spaces, we extend the duality mapping to more general case: that is, a Legendre, strongly coercive, uniformly Fréchet differentiable, and totally convex function.
For the mapping, we consider Bregman quasiasymptotically nonexpansive mappings which contain Bregman quasi nonexpansive mappings as a special case which itself is generalisation of quasi-ϕ-nonexpansive mappings.
In Chen et al. [27] the authors considered mixed equilibrium problems while in this paper a system of equilibrium problems is considered.
Competing Interests
The authors declare that they have no competing interests.
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