# Hybrid Extragradient Methods for Finding Zeros of Accretive Operators and Solving Variational Inequality and Fixed Point Problems in Banach Spaces

Authors: Lu-Chuan Ceng; Ching-Feng Wen
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

We introduce and analyze hybrid implicit and explicit extragradient methods for finding a zero of an accretive operator and solving a general system of variational inequalities and a fixed point problem of an infinite family of nonexpansive self-mappings in a uniformly convex Banach spaceX which has a uniformly Gateaux differentiable norm. We establish some strong convergence theorems for hybrid implicit and explicit extra-gradient algorithms under suitable assumptions. Furthermore, we derive the strong convergence of hybrid implicit and explicit extragradient algorithms for finding a common element of the set of zeros of an accretive operator and the common fixed point set of an infinite family of nonexpansive self-mappings and a self-mapping whose complement is strictly pseudocontractive and strongly accretive in X. The results presented in this paper improve, extend, supplement, and develop the corresponding results announced in the earlier and very recent literature.


---

## Body

## 1. Introduction

LetX be a real Banach space whose dual space is denoted by X*. Let U={x∈X:∥x∥=1} denote the unite sphere of X. A Banach space X is said to be uniformly convex if, for each ϵ∈(0,2], there exists δ>0 such that for all x,y∈U,
(1)∥x-y∥≥ϵ⟹∥x+y∥2≤1-δ.
It is known that a uniformly convex Banach space is reflexive and strict convex. The normalized duality mapping J:X→2X* is defined by
(2)J(x)={x*∈X*:〈x,x*〉=∥x∥2=∥x*∥2},∀x∈X,
where 〈·,·〉 denotes the generalized duality pairing. It is an immediate consequence of the Hahn-Banach theorem that J(x) is nonempty for each x∈X. Moreover, it is known that J is single-valued if and only if X is smooth, whereas if X is uniformly smooth, then J is norm-to-norm uniformly continuous on bounded subsets of X. If X has a uniformly Gateaux differentiable norm, then the duality mapping J is norm-to-weak* uniformly continuous on bounded subsets of X; see for example [1].LetC be a nonempty closed convex subset of a real Banach space X. A mapping T:C→C is called nonexpansive if
(3)∥Tx-Ty∥≤∥x-y∥,∀x,y∈C.
The set of fixed points of T is denoted by Fix(T). We use the notation ⇀to indicate the weak convergence and → to indicate the strong convergence.Recall that (possibly multivalued) operatorA⊂E×E with domain D(A) and range R(A) in E is accretive if, for each xi∈D(A) and yi∈Axi (i=1,2), there exists a j(x1-x2)∈J(x1-x2) such that 〈y1-y2,(x1-x2)〉≥0 (here J is the duality mapping). An accretive operator A is said to satisfy the range condition if D(A)¯⊂R(I+rA) for all r>0. An accretive operator A is m-accretive if R(I+rA)=E for each r>0. If A is an accretive operator which satisfies the range condition, then we can define, for each r>0, a mapping Jr:R(I+rA)→D(A) defined by Jr=(I+rA)-1, which is called the resolvent of A. We know that Jr is nonexpansive and Fix(Jr)=A-10 for all r>0. Hence
(4)Fix(Jr)=A-10={z∈D(A):0∈Az}.
If A-10≠∅, then the inclusion 0∈Az is solvable.The following resolvent identity is well known to us; see [2], where more details on accretive operators can be found.Proposition 1 (resolvent identity).
Forλ>0,μ>0, and x∈X,
(5)Jλx=Jμ(μλx+(1-μλ)Jλx).Recently, Aoyama et al. [3] studied the following iterative scheme in a uniformly convex Banach space having a uniformly Gareaux differentiable norm: for resolvents Jrn of an accretive operator A such that A-10≠∅ and D(A)¯⊂C⊂⋂r>0‍R(I+rA) and {αn}⊂(0,1),
(6)x0=x∈C,xn+1=αnx+(1-αn)Jrnxn.
They proved that the sequence {xn} generated by (6) converges strongly to a zero of A under appropriate assumptions on {αn} and {rn}. Subsequently, Ceng et al. [4] introduced and analyzed the following composite iterative scheme in either a uniformly smooth Banach space or a reflexive Banach space having a weakly sequentially continuous duality mapping,
(7)x0=x∈E,yn=αnu+(1-αn)Jrnxn,xn+1=(1-βn)yn+βnJrnyn,
where u∈D(A)¯ is an arbitrary (but fixed) element, under the following control conditions:(H1)
limn→∞αn=0;(H2)
∑n=0∞αn=∞, or, equivalently, ∏n=0∞(1-αn)=0;(H3)
∑n=1∞|αn-αn-1|<∞;(H4)
rn≥ε, for all n≥0, for some ε>0 and ∑n=1∞‍|rn-rn-1|<∞;(H5)
βn∈[0,a) for some a∈(0,1) and ∑n=1∞‍|βn-βn-1|<∞.Furthermore, as the viscosity approximation method, Jung [5] purposed and analyzed the following composite iterative scheme for finding a zero of an accretive operator A: for resolvent Jrn of an accretive operator A such that A-10≠∅ and D(A)¯⊂C⊂⋂r>0‍R(I+rA), f∈ΞC (ΞC denotes the set of all contractions on C) and {αn},{βn}⊂(0,1),
(8)x0=x∈C,yn=αnf(xn)+(1-αn)Jrnxn,xn+1=(1-βn)yn+βnJrnyn.
He established the strong convergence of the sequence {xn} generated by (8) to a zero of A under certain appropriate conditions.Theorem 2 (see [5, Theorem 3.1]).
LetX be a strictly convex and reflexive Banach space having a uniformly Gateaux differentiable norm. Let C be a nonempty closed convex subset of X and A⊂X×X an accretive operator in X such that A-10≠∅ and D(A)¯⊂C⊂⋂r>0‍R(I+rA). Let {αn} and {βn} be sequences in (0,1) which satisfy the following conditions:(C1)
limn→∞αn=0 and ∑n=0∞‍αn=∞;(B1)
βn∈[0,a) for some 0<a<1 for all n≥0. Let f∈ΞC and x0∈C be chosen arbitrarily. Let {xn} be a sequence generated by (8) for rn>0. If {xn} is asymptotically regular, that is, limn→∞∥xn+1-xn∥=0, then {xn} converges strongly to q∈A-10, which is the unique solution of the variational inequality problem (VIP)
(9)〈(I-f)q,J(q-p)〉≤0,∀f∈ΞC,p∈A-10.On the other hand, we first recall the following concepts.Definition 3.
LetC be a nonempty closed convex subset of a real Banach space X and let A:C→X be a mapping of C into X. Then A is said to be:
(i) accretive if, for eachx,y∈C, there exists j(x-y)∈J(x-y) such that
(10)〈Ax-Ay,j(x-y)〉≥0,
where J is the normalized duality mapping;
(ii)α-strongly accretive if, for each x,y∈C, there exists j(x-y)∈J(x-y) such that
(11)〈Ax-Ay,j(x-y)〉≥α∥x-y∥2,
for some α∈(0,1);
(iii)β-inverse-strongly-accretive if, for each x,y∈C, there exists j(x-y)∈J(x-y) such that
(12)〈Ax-Ay,j(x-y)〉≥β∥Ax-Ay∥2,
for some β>0;
(iv)λ-strictly pseudocontractive [6] if, for each x,y∈C, there exists j(x-y)∈J(x-y) such that
(13)〈Ax-Ay,j(x-y)〉≤∥x-y∥2-λ∥x-y-(Ax-Ay)∥2
for some λ∈(0,1).It is worth emphasizing that the definition of the inverse strongly accretive mapping is based on that of the inverse strongly monotone mapping, which was studied by so many authors; see, for example, [7–9].Very recently, Cai and Bu [10] considered the following general system of variational inequalities (GSVI) in a real smooth Banach space X, which involves finding (x*,y*)∈C×C such that
(14)〈μ1B1y*+x*-y*,J(x-x*)〉≥0,∀x∈C,〈μ2B2x*+y*-x*,J(x-y*)〉≥0,∀x∈C,where C is a nonempty, closed, and convex subset of X, B1,B2:C→X are two nonlinear mappings, and μ1 and μ2 are two positive constants. Here the set of solutions of GSVI (14) is denoted by GSVI(C,B1,B2). In particular, if X=H, a real Hilbert space, then GSVI (14) reduces to the following GSVI of finding (x*,y*)∈C×C such that
(15)〈μ1B1y*+x*-y*,x-x*〉≥0,∀x∈C,〈μ2B2x*+y*-x*,x-y*〉≥0,∀x∈C,
where μ1 and μ2 are two positive constants. The set of solutions of problem (15) is still denoted by GSVI(C,B1,B2). In particular, if B1=B2=A, then problem (15) reduces to the new system of variational inequalities (NSVI), introduced and studied by Verma [11]. Furthermore, if x*=y* additionally, then the NSVI reduces to the classical variational inequality problem (VIP) of finding x*∈C such that
(16)〈Ax*,x-x*〉≥0,∀x∈C.
The solution set of the VIP (16) is denoted by VI(C,A). Variational inequality theory has been studied quite extensively and has emerged as an important tool in the study of a wide class of obstacle, unilateral, free, moving, equilibrium problems. It is now well known that the variational inequalities are equivalent to the fixed point problems, the origin of which can be traced back to Lions and Stampacchia [12]. This alternative formulation has been used to suggest and analyze projection iterative method for solving variational inequalities under the conditions that the involved operator must be strongly monotone and Lipschitz continuous.Recently, Ceng et al. [13] transformed problem (15) into a fixed point problem in the following way.Lemma 4 (see [13]).
For a givenx-,y-∈C,(x-,y-) is a solution of problem (15) if and only if x- is a fixed point of the mapping G:C→C defined by
(17)G(x)=PC[PC(x-μ2B2x)-μ1B1PC(x-μ2B2x)],∀x∈C,
where y-=PC(x--μ2B2x-) and PC is the projection of H onto C.
In particular, if the mappingBi:C→H is βi-inverse strongly monotone for i=1,2, then the mapping G is nonexpansive provided μi∈(0,2βi) for i=1,2.In 1976, Korpelevič [14] proposed an iterative algorithm for solving the VIP (16) in Euclidean space Rnas follows:
(18)yn=PC(xn-τAxn),xn+1=PC(xn-τAyn),n≥0,with τ>0 a given number, which is known as the extragradient method (see also [15]). The literature on the VIP is vast and Korpelevich's extragradient method has received great attention given by many authors, who improved it in various ways; see, for example, [10, 13, 16–23] the and references therein, to name but a few.In particular, wheneverX is still a real smooth Banach space, B1=B2=A, and x*=y*, then GSVI (17) reduces to the variational inequality problem (VIP) of finding x*∈C such that
(19)〈Ax*,J(x-x*)〉≥0,∀x∈C.
which was considered by Aoyama et al. [24]. Note that VIP (19) is connected with the fixed point problem for nonlinear mapping (see, e.g., [15, 25]), the problem of finding a zero point of a nonlinear operator (see, e.g., [1, 26]), and so on. It is clear that VIP (19) extends VIP (16) from Hilbert spaces to Banach spaces.In order to find a solution of VIP (19), Aoyama et al. [24] introduced the following iterative scheme for an accretive operator A:
(20)xn+1=αnxn+(1-αn)ΠC(xn-λnAxn),∀n≥1,
where ΠC is a sunny nonexpansive retraction from X onto C. Then they proved a weak convergence theorem.Beyond doubt, it is an interesting and valuable problem of constructing some algorithms with strong convergence for solving GSVI (14) which contains VIP (19) as a special case. Very recently, Cai and Bu [10] constructed an iterative algorithm for solving GSVI (14) and a common fixed point problem of an infinite family of nonexpansive mappings in a uniformly convex and 2-uniformly smooth Banach space. They proved the strong convergence of the proposed algorithm by virtue of the following inequality in a 2-uniformly smooth Banach space X.Lemma 5 (see [27]).
LetX be a 2-uniformly smooth Banach space. Then
(21)∥x+y∥2≤∥x∥2+2〈y,J(x)〉+2∥κy∥2,∀x,y∈X,
where κ is the 2-uniformly smooth constant of X and J is the normalized duality mapping from X into X*.Define the mappingG:C→C as follows:
(22)G(x)∶=ΠC(I-μ1B1)ΠC(I-μ2B2)x,∀x∈C.
The fixed point set of G is denoted by Ω. Then their strong convergence theorem on the proposed method is stated as follows.Theorem 6 (see [10, Theorem 3.1]).
LetC be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X. Let ΠC be a sunny nonexpansive retraction from X onto C. Let the mapping Bi:C→X be βi-inverse strongly accretive with 0<μi<βi/κ2 for i=1,2. Let f be a contraction of C into itself with coefficient δ∈(0,1). Let {Sn}n=1∞ be an infinite family of nonexpansive mappings of C into itself such that F=⋂i=1∞‍Fix(Si)∩Ω≠∅, where Ω is the fixed point set of the mapping G defined by (22). For arbitrarily given x1∈C, let {xn} be the sequence generated by
(23)xn+1=βnxn+(1-βn)Snyn,yn=αnf(xn)+(1-αn)zn,zn=ΠC(un-μ1B1un),un=ΠC(xn-μ2B2xn),∀n≥1.Suppose that {αn} and {βn} are two sequences in (0,1) satisfying the following conditions:(i)
limn→∞αn=0 and ∑n=1∞‍αn=∞;(ii)
0<liminfn→∞βn≤limsupn→∞βn<1.
Assume that∑n=1∞‍supx∈D∥Sn+1x-Snx∥<∞ for any bounded subset D of C and let S be a mapping of C into X defined by Sx=limn→∞Snx for all x∈C and suppose that Fix(S)=⋂n=1∞‍Fix(Sn). Then {xn} converges strongly to q∈F, which solves the following VIP:
(24)〈q-f(q),J(q-p)〉≤0,∀p∈F.Corollary 7 (see [10, Corollary 3.2]).
LetC be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X. Let ΠC be a sunny nonexpansive retraction from X onto C. Let the mapping Bi:C→X be βi-inverse strongly accretive with 0<μi<βi/κ2 for i=1,2. Let f be a contraction of C into itself with coefficient δ∈(0,1). Let S be a nonexpansive mapping of C into itself such that F=Fix(S)∩Ω≠∅, where Ω is the fixed point set of the mapping G defined by (22). For arbitrarily given x1∈C, let {xn} be the sequence generated by
(25)xn+1=βnxn+(1-βn)Syn,yn=αnf(xn)+(1-αn)zn,zn=ΠC(un-μ1B1un),un=ΠC(xn-μ2B2xn),∀n≥1.Suppose that {αn} and {βn} are two sequences in (0,1) satisfying the following conditions:(i)
limn→∞αn=0 and ∑n=1∞‍αn=∞;(ii)
0<liminfn→∞βn≤limsupn→∞βn<1.Then {xn} converges strongly to q∈F, which solves the following VIP:
(26)〈q-f(q),J(q-p)〉≤0,∀p∈F.We remark that in Theorem6, the Banach space X is both uniformly convex and 2-uniformly smooth. According to Lemma 5, the 2-uniform smoothness of X guarantees the nonexpansivity of the mapping I-μiBi for αi-inverse-strongly accretive mapping Bi:C→X with 0≤μi≤αi/κ2 for i=1,2, and hence the composite mapping G:C→C is nonexpansive where G=ΠC(I-μ1B1)ΠC(I-μ2B2). In the meantime, for the convenience of implementing the argument techniques in [13], they have applied the following inequality in a real smooth and uniform convex Banach space X.Proposition 8 (see [28]).
LetX be a real smooth and uniform convex Banach space and let r>0. Then there exists a strictly increasing, continuous, and convex function g:[0,2r]→R, g(0)=0 such that
(27)g(∥x-y∥)≤∥x∥2-2〈x,J(y)〉+∥y∥2,∀x,y∈Br,
where Br={x∈X:∥x∥≤r}.LetC be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gateaux differentiable norm. Let ΠC be a sunny nonexpansive retraction from X onto C. Motivated and inspired by the research going on this area, we introduce and analyze hybrid implicit and explicit extragradient methods for finding a zero of an accretive operator A⊂X×X such that D(A)¯⊂C⊂⋂r>0‍R(I+rA) and solving GSVI (14) and a fixed point problem of an infinite family of nonexpansive self-mappings on C. We establish some strong convergence theorems for hybrid implicit and explicit extragradient algorithms under suitable assumptions. Furthermore, we derive the strong convergence of hybrid implicit and explicit extragradient algorithms for finding a common element of the set of zeros of an accretive operator and the common fixed point set of an infinite family of nonexpansive self-mappings on C and a self-mapping whose complement is strictly pseudocontractive and strongly accretive on C. The results presented in this paper improve, extend, supplement, and develop the corresponding results announced in the earlier and very recent literature; see, for example, [5, 10, 13, 16].


## 2. Preliminaries

LetX be a real Banach space. X is said to be smooth if the limit
(28)limt→0∥x+ty∥-∥x∥t,
exists for all x,y∈U; in this case, X is also said to have a Gateaux differentiable norm. X is said to have a uniformly Gateaux differentiable norm if for each y∈U, the limit is attained uniformly for x∈U. Moreover, it is said to be uniformly smooth if this limit is attained uniformly for x,y∈U. The norm of X is said to be the Frechet differential if, for each x∈U, this limit is attained uniformly for y∈U. In addition, we define a function ρ:[0,∞)→[0,∞) called the modulus of smoothness of X as follows:
(29)ρ(τ)=sup{12(∥x+y∥+∥x-y∥)-1:x,y∈X,∥x∥=1,∥y∥=τ12}.
It is known that X is uniformly smooth if and only if limτ→0ρ(τ)/τ=0. Let q be a fixed real number with 1<q≤2. Then a Banach space X is said to be q-uniformly smooth if there exists a constant c>0 such that ρ(τ)≤cτq for all τ>0. As pointed out in [29], no Banach space is q-uniformly smooth for q>2.We list some lemmas that will be used in the sequel. Lemma9 can be found in [30]. Lemma 10 is an immediate consequence of the subdifferential inequality of the function (1/2)∥·∥2.Lemma 9.
Let{sn} be a sequence of nonnegative real numbers satisfying
(30)sn+1≤(1-αn)sn+αnβn+γn,∀n≥0,
where {αn}, {βn}, and {γn} satisfy the following conditions:(i)
{αn}⊂[0,1] and ∑n=0∞αn=∞;(ii)
limsupn→∞βn≤0;(iii)
γn≥0, for all n≥0, and ∑n=0∞γn<∞.Then limsupn→∞sn=0.Lemma 10.
In a smooth Banach spaceX, there holds the inequality
(31)∥x+y∥2≤∥x∥2+2〈y,J(x+y)〉,∀x,y∈X.Lemma 11 (see [6]).
LetC be a nonempty closed convex subset of a real smooth Banach space X and let A:C→X be a λ-strictly pseudocontractive mapping. Then A is Lipschitz continuous with constant 1+1/λ.Proof.
SinceA:C→X is a λ-strictly pseudocontractive mapping, we have for all x,y∈C(32)〈Ax-Ay,J(x-y)〉≤∥x-y∥2-λ∥x-y-(Ax-Ay)∥2,
and hence
(33)λ∥(I-A)x-(I-A)y∥2≤〈(I-A)x-(I-A)y,J(x-y)〉.
This yields
(34)∥(I-A)x-(I-A)y∥≤1λ∥x-y∥,∀x,y∈C.
Therefore,
(35)∥Ax-Ay∥≤∥x-y∥+∥(I-A)x-(I-A)y∥≤(1+1λ)∥x-y∥,∀x,y∈C.LetD be a subset of C and let Π be a mapping of C into D. Then Π is said to be sunny if
(36)Π[Π(x)+t(x-Π(x))]=Π(x),
whenever Π(x)+t(x-Π(x))∈C for x∈C and t≥0. A mapping Π of C into itself is called a retraction if Π2=Π. If a mapping Π of C into itself is a retraction, then Π(z)=z for every z∈R(Π) where R(Π) is the range of Π. A subset D of C is called a sunny nonexpansive retract of C if there exists a sunny nonexpansive retraction from C onto D. The following lemma concerns the sunny nonexpansive retraction.Lemma 12 (see [31]).
LetC be a nonempty closed convex subset of a real smooth Banach space X. Let D be a nonempty subset of C. Let Π be a retraction of C onto D. Then the following are equivalent:(i)
Π is sunny and nonexpansive;(ii)
∥Π(x)-Π(y)∥2≤〈x-y,J(Π(x)-Π(y))〉, for all x,y∈C;(iii)
〈x-Π(x),J(y-Π(x))〉≤0, for all x∈C,y∈D.It is well known that ifX=H a Hilbert space, then a sunny nonexpansive retraction ΠC is coincident with the metric projection from X onto C; that is, ΠC=PC. If C is a nonempty closed convex subset of a strictly convex and uniformly smooth Banach space X and if T:C→C is a nonexpansive mapping with the fixed point set Fix(T)≠∅, then the set Fix(T) is a sunny nonexpansive retract of C.Lemma 13 (see [32]).
LetX be a uniformly convex Banach space and Br={x∈X:∥x∥≤r},r>0. Then there exists a continuous, strictly increasing, and convex function g:[0,∞]→[0,∞],g(0)=0 such that
(37)∥αx+βy+γz∥2≤α∥x∥2+β∥y∥2+γ∥z∥2-αβg(∥x-y∥)for all x,y,z∈Br and all α,β,γ∈[0,1] with α+β+γ=1.Lemma 14 (see [33]).
LetC be a nonempty closed convex subset of a Banach space X. Let S0,S1,… be a sequence of mappings of C into itself. Suppose that ∑n=1∞‍sup{∥Snx-Sn-1x∥:x∈C}<∞. Then for each y∈C, {Sny} converges strongly to some point of C. Moreover, let S be a mapping of C into itself defined by Sy=limn→∞Sny for all y∈C. Then limn→∞sup{∥Sx-Snx∥:x∈C}=0.LetC be a nonempty closed convex subset of a Banach space X and let T:C→C be a nonexpansive mapping with Fix(T)≠∅. As previously mentioned, let ΞC be the set of all contractions on C. For t∈(0,1) and f∈ΞC, let xt∈C be the unique fixed point of the contraction x↦tf(x)+(1-t)Tx on C; that is,
(38)xt=tf(xt)+(1-t)Txt.Lemma 15 (see [15]).
LetX be a reflexive and strictly convex Banach space with a uniformly Gateaux differentiable norm. Let C a nonempty closed convex subset of X,T:C→C a nonexpansive mapping with Fix(T)≠∅, and f∈ΞC. Then the net {xt} defined by xt=tf(xt)+(1-t)Txt converges strongly to a point in Fix(T). If we define a mapping Q:ΞC→Fix(T) by Q(f):=s-limt→0xt, for all f∈ΞC, then Q(f) solves the following VIP:
(39)〈(I-f)Q(f),J(Q(f)-p)〉≤0,∀f∈ΞC,p∈Fix(T).Lemma 16 (see [34]).
LetC be a nonempty closed convex subset of a strictly convex Banach space X. Let {Tn}n=0∞ be a sequence of nonexpansive mappings on C. Suppose ⋂n=0∞‍Fix(Tn) is nonempty. Let {λn} be a sequence of positive numbers with ∑n=0∞‍λn=1. Then a mapping S on C defined by Sx=∑n=0∞λnTnx for x∈C is well defined, nonexpansive and Fix(S)=⋂n=0∞‍Fix(Tn) holds.Lemma 17 (see [27]).
Given a numberr>0. A real Banach space X is uniformly convex if and only if there exists a continuous strictly increasing function g:[0,∞)→[0,∞), g(0)=0, such that
(40)∥λx+(1-λ)y∥2≤λ∥x∥2+(1-λ)∥y∥2-λ(1-λ)g(∥x-y∥)for all λ∈[0,1] and x,y∈X such that ∥x∥≤r and ∥y∥≤r.Lemma 18 (see [24]).
LetC be a nonempty closed convex subset of a smooth Banach space X. Let ΠC be a sunny nonexpansive retraction from X onto C and let A be an accretive operator of C into X. Then, for all λ>0,
(41)VI(C,A)=Fix(ΠC(I-λA)).Lemma 19.
LetC be a nonempty closed convex subset of a smooth Banach space X and let the mapping Bi:C→X be ζi strictly pseudocontractive and θi strongly accretive with θi+ζi≥1 for i=1,2. Then, for μi∈(0,1], we have
(42)∥(I-μiBi)x-(I-μiBi)y∥≤{1-θiζi+(1-μi)(1+1ζi)}×∥x-y∥,∀x,y∈C,
for i=1,2. In particular, if 1-(ζi/(1+ζi))(1-(1-θi)/ζi)≤μi≤1, then I-μiBi is nonexpansive for i=1,2.Proof.
Taking into account theζi-strict pseudocontractivity of Bi, by Lemma 11 we derive for every x,y∈C(43)∥Bix-Biy∥≤(1+1ζi)∥x-y∥.
Utilizing the θi-strong accretivity and ζi-strict pseudocontractivity of Bi, we get
(44)ζi∥(I-Bi)x-(I-Bi)y∥2≤∥x-y∥2-〈Bix-Biy,J(x-y)〉≤(1-θi)∥x-y∥2.
So, we have
(45)∥(I-Bi)x-(I-Bi)y∥≤1-θiζi∥x-y∥.
Therefore, for μi∈(0,1], we have
(46)∥(I-μiBi)x-(I-μiBi)y∥≤∥(I-Bi)x-(I-Bi)y∥+(1-μi)∥Bix-Biy∥≤1-θiζi∥x-y∥+(1-μi)(1+1ζi)∥x-y∥={1-θiζi+(1-μi)(1+1ζi)}∥x-y∥.
Since 1-(ζi/(1+ζi))(1-(1-θi)/ζi)≤μi≤1, it follows immediately that
(47)1-θiζi+(1-μi)(1+1ζi)≤1.
This implies that I-μiBi is nonexpansive for i=1,2.Lemma 20.
LetC be a nonempty closed convex subset of a smooth Banach space X. Let ΠC be a sunny nonexpansive retraction from X onto C and let the mapping Bi:C→X be ζi strictly pseudocontractive and θi strongly accretive with θi+ζi≥1 for i=1,2. Let G:C→C be the mapping defined by
(48)G(x)=ΠC[ΠC(x-μ2B2x)-μ1B1ΠC(x-μ2B2x)],∀x∈C.
If 1-(ζi/(1+ζi))(1-(1-θi)/ζi)≤μi≤1, then G:C→C is nonexpansive.Proof.
According to Lemma19, we know that I-μiBi is nonexpansive for i=1,2. Hence, for all x,y∈C, we have
(49)∥G(x)-G(y)∥=∥ΠC[ΠC(x-μ2B2x)-μ1B1ΠC(x-μ2B2x)]-ΠC[ΠC(y-μ2B2y)-μ1B1ΠC(y-μ2B2y)]∥=∥ΠC(I-μ1B1)ΠC(I-μ2B2)x-ΠC(I-μ1B1)ΠC(I-μ2B2)y∥≤∥(I-μ1B1)ΠC(I-μ2B2)x-(I-μ1B1)ΠC(I-μ2B2)y∥≤∥ΠC(I-μ2B2)x-ΠC(I-μ2B2)y∥≤∥(I-μ2B2)x-(I-μ2B2)y∥≤∥x-y∥.
This shows that G:C→C is nonexpansive. This completes the proof.Lemma 21.
LetC be a nonempty closed convex subset of a smooth Banach space X. Let ΠC be a sunny nonexpansive retraction from X onto C and let B1,B2:C→X be two nonlinear mappings. For a given x*,y*∈C,(x*,y*) is a solution of GSVI (14) if and only if x*=ΠC(y*-μ1B1y*) where y*=ΠC(x*-μ2B2x*).Proof.
We can rewrite GSVI (14) as
(50)〈x*-(y*-μ1B1y*),J(x-x*)〉≥0,∀x∈C,〈y*-(x*-μ2B2x*),J(x-y*)〉≥0,∀x∈C,
which is obviously equivalent to
(51)x*=ΠC(y*-μ1B1y*),y*=ΠC(x*-μ2B2x*),
because of Lemma 12. This completes the proof.Remark 22.
By Lemma21, we observe that
(52)x*=ΠC[ΠC(x*-μ2B2x*)-μ1B1ΠC(x*-μ2B2x*)],
which implies that x* is a fixed point of the mapping G=ΠC(I-μ1B1)ΠC(I-μ2B2). Throughout this paper, the set of fixed points of the mapping G is denoted by Ω.


## 3. Hybrid Implicit Extragradient Algorithm

In this section, letC be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gateaux differentiable norm. We suggest and analyze a hybrid implicit extragradient algorithm for finding a zero of A⊂X×X an accretive operator in X with D(A)¯⊂C⊂⋂r>0‍R(I+rA) and solving a general system of variational inequalities and a common fixed point problem of an infinite family of nonexpansive self-mappings in X.Theorem 23.
LetC be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gateaux differentiable norm. Let ΠC be a sunny nonexpansive retraction from X onto C. Let A⊂X×X be an accretive operator in X such that D(A)¯⊂C⊂⋂r>0‍R(I+rA). Let Bi:C→X be ζi strictly pseudocontractive and θi strongly accretive with θi+ζi≥1 for each i=1,2. Define the mapping G:C→C by G=ΠC(I-μ1B1)ΠC(I-μ2B2) where 1-(ζi/(1+ζi))(1-(1-θi)/ζi)≤μi≤1 for i=1,2. Let f:C→C be a contraction with coefficient ρ∈(0,1). Let {Si}i=0∞ be an infinite family of nonexpansive mappings of C into itself such that F=⋂i=0∞‍Fix(Si)∩Ω∩A-10≠∅ where Ω=Fix(G). For arbitrarily given x0∈C, let {xn} be the sequence generated by
(53)yn=αnf(xn)+βnxn+γnJrnxn+δnSnyn,xn+1=σnyn+(1-σn)ΠC(I-μ1B1)×ΠC(I-μ2B2)yn,∀n≥0.
Suppose that {rn}⊂(0,∞), {σn},{αn},{βn},{γn},{δn}⊂[0,1], αn+βn+γn+δn=1 and the following conditions hold:(i)
limn→∞αn=0 and ∑n=0∞‍αn=∞;(ii)
{γn},{δn}⊂[c,d] for some c,d∈(0,1);(iii)
∑n=1∞‍(|σn-σn-1|+|αn-αn-1|+|βn-βn-1|+|γn-γn-1|+|δn-δn-1|)<∞;(iv)
∑n=1∞‍|rn-rn-1|<∞ and rn≥ε>0, for all n≥0 for some ε>0;(v)
0<liminfn→∞βn≤limsupn→∞βn<1 and 0<liminfn→∞σn≤limsupn→∞σn<1.Assume that ∑n=0∞‍supx∈D∥Sn+1x-Snx∥<∞ for any bounded subset D of C and let S be a mapping of C into itself defined by Sx=limn→∞Snx for all x∈C and suppose that Fix(S)=⋂i=0∞‍Fix(Si). Then {xn} converges strongly to q∈F, which solves the following VIP:
(54)〈q-f(q),J(q-p)〉≤0,∀p∈F.Proof.
It is easy to see that (53) can be rewritten as follows:
(55)yn=αnf(xn)+βnJrnxn+γnBxn+δnSnyn,xn+1=σnyn+(1-σn)Gyn,∀n≥0,
where G=ΠC(I-μ1B1)ΠC(I-μ2B2). By Lemma 20 we know that G is a nonexpansive self-mapping on C.
Now, let us show that the sequence{xn} is bounded. Indeed, take a fixed p∈F arbitrarily. Then from (55), we have
(56)∥yn-p∥≤αn∥f(xn)-p∥+βn∥xn-p∥+γn∥Jrnxn-p∥+δn∥Snyn-p∥≤αn(∥f(xn)-f(p)∥+∥f(p)-p∥)+βn∥xn-p∥+γn∥xn-p∥+δn∥yn-p∥≤αn(ρ∥xn-p∥+∥f(p)-p∥)+βn∥xn-p∥+γn∥xn-p∥+δn∥yn-p∥=(1-δn-(1-ρ)αn)∥xn-p∥+αn∥f(p)-p∥+δn∥yn-p∥,
which hence implies that
(57)∥yn-p∥≤(1-(1-ρ)αn1-δn)∥xn-p∥+αn1-δn∥f(p)-p∥.
So, we have
(58)∥xn+1-p∥≤σn∥yn-p∥+(1-σn)∥Gyn-p∥≤σn∥yn-p∥+(1-σn)∥yn-p∥=∥yn-p∥≤(1-(1-ρ)αn1-δn)∥xn-p∥+αn1-δn∥f(p)-p∥=(1-(1-ρ)αn1-δn)∥xn-p∥+(1-ρ)αn1-δn∥f(p)-p∥1-ρ≤max{∥xn-p∥,∥f(p)-p∥1-ρ}.
By induction, we obtain
(59)∥xn-p∥≤max{∥x0-p∥,∥f(p)-p∥1-ρ},∀n≥0.
Hence {xn} is bounded and so are {yn}, {Gyn}, {f(xn)}, {Jrnxn}, and {Snyn}.
Let us show that(60)limn→∞∥xn+1-xn∥=0.
As a matter of fact, observe that yn can be rewritten as follows:
(61)yn=βnxn+(1-βn)zn,
where zn=(αnf(xn)+γnJrnxn+δnSnyn)/(1-βn). Observe that
(62)∥zn-zn-1∥=∥αnf(xn)+γnJrnxn+δnSnyn1-βn-αn-1f(xn-1)+γn-1Jrn-1xn-1+δn-1Sn-1yn-11-βn-1∥=∥yn-βnxn1-βn-yn-1-βn-1xn-11-βn-1∥=∥yn-βnxn1-βn-yn-1-βn-1xn-11-βn+yn-1-βn-1xn-11-βn-yn-1-βn-1xn-11-βn-1∥≤∥yn-βnxn1-βn-yn-1-βn-1xn-11-βn∥+∥yn-1-βn-1xn-11-βn-yn-1-βn-1xn-11-βn-1∥=11-βn∥yn-βnxn-(yn-1-βn-1xn-1)∥+|11-βn-11-βn-1|∥yn-1-βn-1xn-1∥=11-βn∥yn-βnxn-(yn-1-βn-1xn-1)∥+|βn-βn-1|(1-βn-1)(1-βn)∥yn-1-βn-1xn-1∥=11-βn∥αnf(xn)+γnJrnxn+δnSnyn-αn-1f(xn-1)-γn-1Jrn-1xn-1-δn-1Sn-1yn-1∥+|βn-βn-1|(1-βn-1)(1-βn)∥yn-1-βn-1xn-1∥≤11-βn[αn∥f(xn)-f(xn-1)∥+γn∥Jrnxn-Jrn-1xn-1∥+δn∥Snyn-Sn-1yn-1∥+|αn-αn-1|∥f(xn-1)∥+|γn-γn-1|∥Jrn-1xn-1∥+|δn-δn-1|∥Sn-1yn-1∥]+|βn-βn-1|(1-βn-1)(1-βn)×∥yn-1-βn-1xn-1∥.
On the other hand, if rn-1≤rn, using the resolvent identity in Proposition 1,
(63)Jrnxn=Jrn-1(rn-1rnxn+(1-rn-1rn)Jrnxn),
we get
(64)∥Jrnxn-Jrn-1xn-1∥=∥Jrn-1(rn-1rnxn+(1-rn-1rn)Jrnxn)-Jrn-1xn-1∥≤rn-1rn∥xn-xn-1∥+(1-rn-1rn)∥Jrnxn-xn-1∥≤∥xn-xn-1∥+rn-rn-1rn∥Jrnxn-xn-1∥≤∥xn-xn-1∥+1ε|rn-rn-1|∥Jrnxn-xn-1∥.
If rn≤rn-1, we derive in a similar way
(65)∥Jrnxn-Jrn-1xn-1∥≤∥xn-1-xn∥+1ε|rn-1-rn|∥Jrn-1xn-1-xn∥.
Thus, combining the above cases we obtain
(66)∥Jrnxn-Jrn-1xn-1∥≤∥xn-1-xn∥+M0|rn-1-rn|,∀n≥1,
where supn≥1{(1/ε)(∥Jrnxn-xn-1∥+∥Jrn-1xn-1-xn∥)}≤M0 for some M0>0. Substituting (66) into (62), we have
(67)∥zn-zn-1∥≤11-βn[αn∥f(xn)-f(xn-1)∥+γn(∥xn-1-xn∥+M0|rn-1-rn|)+δn∥Snyn-Sn-1yn-1∥+|αn-αn-1|∥f(xn-1)∥+|γn-γn-1|∥Jrn-1xn-1∥+|δn-δn-1|∥Sn-1yn-1∥]+|βn-βn-1|(1-βn-1)(1-βn)∥yn-1-βn-1xn-1∥≤11-βn[αnρ∥xn-xn-1∥+γn(∥xn-1-xn∥+M0|rn-1-rn|)+δn∥Snyn-Snyn-1∥+|αn-αn-1|∥f(xn-1)∥+|γn-γn-1|∥Jrn-1xn-1∥+|δn-δn-1|∥Sn-1yn-1∥+δn∥Snyn-1-Sn-1yn-1∥]+|βn-βn-1|(1-βn-1)(1-βn)×∥αn-1f(xn-1)+γn-1Jrn-1xn-1+δn-1Sn-1xn-1∥≤11-βn[(αnρ+γn)∥xn-xn-1∥+δn∥yn-yn-1∥+M(|αn-αn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+δn∥Snyn-1-Sn-1yn-1∥]+1(1-βn-1)(1-βn)|βn-βn-1|M,
where supn≥0{M0+∥f(xn)∥+∥Jrnxn∥+∥Snyn∥}≤M for some M>0. In the meantime, simple calculations show that
(68)yn-yn-1=βn(xn-xn-1)+(1-βn)(zn-zn-1)+(βn-βn-1)(xn-1-zn-1).
Taking into account condition (iv), we may assume, without loss of generality, that {βn}⊂[a,b] for some a,b∈(0,1). Hence, from (67) and (68) we deduce that
(69)∥yn-yn-1∥≤βn∥xn-xn-1∥+(1-βn)∥zn-zn-1∥+|βn-βn-1|∥xn-1-zn-1∥≤βn∥xn-xn-1∥+(1-βn)×{11-βn[(αnρ+γn)∥xn-xn-1∥+δn∥yn-yn-1∥+M(|αn-αn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+δn∥Snyn-1-Sn-1yn-1∥]+1(1-βn-1)(1-βn)|βn-βn-1|M}+|βn-βn-1|∥xn-1-zn-1∥=(αnρ+βn+γn)∥xn-xn-1∥+δn∥yn-yn-1∥+M(|αn-αn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+δn∥Snyn-1-Sn-1yn-1∥+11-βn-1|βn-βn-1|M+|βn-βn-1|∥xn-1-zn-1∥=(1-δn-αn(1-ρ))∥xn-xn-1∥+δn∥yn-yn-1∥+M(|αn-αn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+δn∥Snyn-1-Sn-1yn-1∥+11-βn-1|βn-βn-1|M+|βn-βn-1|∥xn-1-zn-1∥≤(1-δn-αn(1-ρ))∥xn-xn-1∥+δn∥yn-yn-1∥+M~0(|αn-αn-1|+|βn-βn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+δn∥Snyn-1-Sn-1yn-1∥,
where supn≥0{(M/(1-b))+∥xn-zn∥}≤M~0 for some M~0>0. Thus, from condition (ii) we immediately get
(70)∥yn-yn-1∥≤(1-1-ρ1-δnαn)∥xn-xn-1∥+M~01-δn(|αn-αn-1|+|βn-βn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+δn1-δn∥Snyn-1-Sn-1yn-1∥≤(1-1-ρ1-δnαn)∥xn-xn-1∥+M~01-d(|αn-αn-1|+|βn-βn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+d1-d∥Snyn-1-Sn-1yn-1∥.
Also, from (55) we have
(71)xn+1-xn=σn(yn-yn-1)+(σn-σn-1)(yn-1-Gyn-1)+(1-σn)(Gyn-Gyn-1).
Taking into account condition (v), we may assume, without loss of generality, that {βn}⊂[a,b] for some a,b∈(0,1). This together with (70) implies that
(72)∥xn+1-xn∥≤σn∥yn-yn-1∥+|σn-σn-1|∥yn-1-Gyn-1∥+(1-σn)∥Gyn-Gyn-1∥≤σn∥yn-yn-1∥+|σn-σn-1|∥yn-1-Gyn-1∥+(1-σn)∥yn-yn-1∥=∥yn-yn-1∥+|σn-σn-1|∥yn-1-Gyn-1∥≤(1-1-ρ1-δnαn)∥xn-xn-1∥+M~01-d(|αn-αn-1|+|βn-βn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+d1-d∥Snyn-1-Sn-1yn-1∥+|σn-σn-1|∥yn-1-Gyn-1∥≤(1-1-ρ1-δnαn)∥xn-xn-1∥+M1(|σn-σn-1|+|αn-αn-1|+|βn-βn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+d1-d∥Snyn-1-Sn-1yn-1∥,
where supn≥0{(M~0/(1-d))+∥yn-Gyn∥}≤M1 for some M1>0. Since ∑n=0∞‍αn=∞ and ((1-ρ)/(1-δn))αn≥(1-ρ)αn, we know that ∑n=0∞‍((1-ρ)/(1-δn))αn=∞. So, applying Lemma 9 to (72), we obtain from conditions (iii), (iv), and the assumption on {Sn} that
(73)limn→∞∥xn+1-xn∥=0.
Next we show that∥xn-Gxn∥→0 as n→∞.
Indeed, according to Lemma10 we have from (55)
(74)∥yn-p∥2=∥αn(f(xn)-f(p))+βn(xn-p)+γn(Jrnxn-p)+δn(Snyn-p)+αn(f(p)-p)∥2≤∥αn(f(xn)-f(p))+βn(xn-p)+γn(Jrnxn-p)+δn(Snyn-p)∥2+2αn〈f(p)-p,J(yn-p)〉≤αn∥f(xn)-f(p)∥2+βn∥xn-p∥2+γn∥Jrnxn-p∥2+δn∥Snyn-p∥2+2αn〈f(p)-p,J(yn-p)〉≤αnρ∥xn-p∥2+βn∥xn-p∥2+γn∥xn-p∥2+δn∥yn-p∥2+2αn〈f(p)-p,J(yn-p)〉=(αnρ+βn+γn)∥xn-p∥2+δn∥yn-p∥2+2αn〈f(p)-p,J(yn-p)〉≤(1-δn-αn(1-ρ))∥xn-p∥2+δn∥yn-p∥2+2αn〈f(p)-p,J(yn-p)〉,
which hence implies that
(75)∥yn-p∥2≤(1-αn(1-ρ)1-δn)∥xn-p∥2+αn1-δn2〈f(p)-p,J(yn-p)〉≤(1-αn(1-ρ)1-δn)∥xn-p∥2+αn1-δn2∥f(p)-p∥∥yn-p∥.
Utilizing Lemma 17, we get from (55) and (75)
(76)∥xn+1-p∥2=∥σn(yn-p)+(1-σn)(Gyn-p)∥2≤σn∥yn-p∥2+(1-σn)∥Gyn-p∥2-σn(1-σn)g1(∥yn-Gyn∥)≤σn∥yn-p∥2+(1-σn)∥yn-p∥2-σn(1-σn)g1(∥yn-Gyn∥)=∥yn-p∥2-σn(1-σn)g1(∥yn-Gyn∥)≤(1-αn(1-ρ)1-δn)∥xn-p∥2+αn1-δn2∥f(p)-p∥∥yn-p∥-σn(1-σn)g1(∥yn-Gyn∥)≤∥xn-p∥2+αn1-δn2∥f(p)-p∥∥yn-p∥-σn(1-σn)g1(∥yn-Gyn∥),
which hence yields
(77)σn(1-σn)g1(∥yn-Gyn∥)≤∥xn-p∥2-∥xn+1-p∥2+αn1-δn2∥f(p)-p∥∥yn-p∥≤(∥xn-p∥+∥xn+1-p∥)×∥xn-xn+1∥+αn1-d2∥f(p)-p∥∥yn-p∥.
Since αn→0 and ∥xn+1-xn∥→0, from condition (iv) and the boundedness of {xn} and {yn}, it follows that
(78)limn→∞g1(∥yn-Gyn∥)=0.
Utilizing the properties of g1, we have
(79)limn→∞∥yn-Gyn∥=0.
Observe that
(80)∥xn-yn∥≤∥xn-xn+1∥+∥xn+1-yn∥=∥xn-xn+1∥+(1-σn)∥Gyn-yn∥⟶0,asn⟶∞,
and hence
(81)∥xn-Gxn∥≤∥xn-yn∥+∥yn-Gyn∥+∥Gyn-Gxn∥≤2∥xn-yn∥+∥yn-Gyn∥⟶0,asn⟶∞.
That is,(82)limn→∞∥xn-yn∥=0,limn→∞∥xn-Gxn∥=0.
Next, let us show that
(83)limn→∞∥Jrnxn-xn∥=0,limn→∞∥Snxn-xn∥=0.
Indeed, observe that yn can be rewritten as follows:
(84)yn=αnf(xn)+βnxn+γnJrnxn+δnSnyn=αnf(xn)+βnxn+(γn+δn)γnJrnxn+δnSnynγn+δn=αnf(xn)+βnxn+enz^n,
where en=γn+δn and z^n=(γnJrnxn+δnSnyn)/(γn+δn). Utilizing Lemma 13 and (84), we have
(85)∥yn-p∥2=∥αn(f(xn)-p)+βn(xn-p)+en(z^n-p)∥2≤αn∥f(xn)-p∥2+βn∥xn-p∥2+en∥z^n-p∥2-βneng3(∥z^n-xn∥)=αn∥f(xn)-p∥2+βn∥xn-p∥2-βneng2(∥z^n-xn∥)+en∥γnJrnxn+δnSnynγn+δn-p∥2=αn∥f(xn)-p∥2+βn∥xn-p∥2-βneng2(∥z^n-xn∥)+en∥γnγn+δn(Jrnxn-p)+δnγn+δn(Snyn-p)∥2≤αn∥f(xn)-p∥2+βn∥xn-p∥2-βneng2(∥z^n-xn∥)+en[γnγn+δn∥Jrnxn-p∥2+δnγn+δn∥Snyn-p∥2]≤αn∥f(xn)-p∥2+βn∥xn-p∥2-βneng2(∥z^n-xn∥)+en[γnγn+δn∥xn-p∥+δnγn+δn∥yn-p∥2]=αn∥f(xn)-p∥2+(βn+γn)∥xn-p∥2+δn∥yn-p∥2-βneng2(∥z^n-xn∥)=αn∥f(xn)-p∥2+(1-αn-δn)∥xn-p∥2+δn∥yn-p∥2-βneng2(∥z^n-xn∥)≤αn∥f(xn)-p∥2+(1-δn)∥xn-p∥2+δn∥yn-p∥2-βneng2(∥z^n-xn∥),
which hence implies that
(86)βneng2(∥z^n-xn∥)≤αn∥f(xn)-p∥2+(1-δn)(∥xn-p∥2-∥yn-p∥2)≤αn∥f(xn)-p∥2+(1-δn)(∥xn-p∥+∥yn-p∥)∥xn-yn∥≤αn∥f(xn)-p∥2+(∥xn-p∥+∥yn-p∥)∥xn-yn∥.
Utilizing (82), conditions (i), (ii), and (v), and the boundedness of {xn}, {yn}, and {f(xn)}, we get
(87)limn→∞g2(∥z^n-xn∥)=0.
From the properties of g2, we have
(88)limn→∞∥z^n-xn∥=0.
Utilizing Lemma 17 and the definition of z^n, we have
(89)∥z^n-p∥2=∥γnJrnxn+δnSnynγn+δn-p∥2=∥γnγn+δn(Jrnxn-p)+δnγn+δn(Snyn-p)∥2≤γnγn+δn∥Jrnxn-p∥2+δnγn+δn∥Snyn-p∥2-γnδn(γn+δn)2g3(∥Jrnxn-Snyn∥)≤γnγn+δn∥xn-p∥2+δnγn+δn∥yn-p∥2-γnδn(γn+δn)2g3(∥Jrnxn-Snyn∥),
which leads to
(90)γnδn(γn+δn)2g3(∥Jrnxn-Snxn∥)≤γnγn+δn(∥xn-p∥2-∥z^n-p∥2)+δnγn+δn(∥yn-p∥2-∥z^n-p∥2)≤γnγn+δn(∥xn-p∥+∥z^n-p∥)∥xn-z^n∥+δnγn+δn(∥yn-p∥+∥z^n-p∥)∥yn-z^n∥≤γnγn+δn(∥xn-p∥+∥z^n-p∥)∥xn-z^n∥+δnγn+δn(∥yn-p∥+∥z^n-p∥)×(∥yn-xn∥+∥xn-z^n∥).
Since {xn}, {yn}, and {z^n} are bounded, ∥xn-yn∥→0, and ∥z^n-xn∥→0 as n→∞, we deduce from condition (ii) that
(91)limn→∞g3(∥Jrnxn-Snyn∥)=0.
From the properties of g3, we have
(92)limn→∞∥Jrnxn-Snyn∥=0.
On the other hand, yn can also be rewritten as follows:
(93)yn=αnf(xn)+βnxn+γnJrnxn+δnSnyn=βnxn+γnJrnxn+(αn+δn)αnf(xn)+δnSnynαn+δn=βnxn+γnJrnxn+dnz~n,
where dn=αn+δn and z~n=(αnf(xn)+δnSnyn)/(αn+δn). Utilizing Lemma 13 and the convexity of ∥·∥2, we have
(94)∥yn-p∥2=∥βn(xn-p)+γn(Jrnxn-p)+dn(z~n-p)∥2≤βn∥xn-p∥2+γn∥Jrnxn-p∥2+dn∥z~n-p∥2-βnγng4(∥xn-Jrnxn∥)=βn∥xn-p∥2+γn∥Jrnxn-p∥2+dn∥αnf(xn)+δnSnynαn+δn-p∥2-βnγng4(∥xn-Jrnxn∥)=βn∥xn-p∥2+γn∥Jrnxn-p∥2+dn∥αnαn+δn(f(xn)-p)+δnαn+δn(Snyn-p)∥2-βnγng4(∥xn-Jrnxn∥)≤βn∥xn-p∥2+γn∥xn-p∥2+dn[αnαn+δn∥f(xn)-p∥2+δnαn+δn∥Snyn-p∥2]-βnγng4(∥xn-Jrnxn∥)≤αn∥f(xn)-p∥2+(βn+γn)∥xn-p∥2+δn∥yn-p∥2-βnγng4(∥xn-Jrnxn∥)=αn∥f(xn)-p∥2+(1-αn-δn)∥xn-p∥2+δn∥yn-p∥2-βnγng4(∥xn-Jrnxn∥)≤αn∥f(xn)-p∥2+(1-δn)∥xn-p∥2+δn∥yn-p∥2-βnγng4(∥xn-Jrnxn∥),
which hence implies that
(95)βnγng4(∥xn-Jrnxn∥)≤αn∥f(xn)-p∥2+(1-δn)(∥xn-p∥2-∥yn-p∥2)≤αn∥f(xn)-p∥2+(1-δn)(∥xn-p∥+∥yn-p∥)×∥xn-yn∥≤αn∥f(xn)-p∥2+(∥xn-p∥+∥yn-p∥)∥xn-yn∥.
From (82), conditions (i), (ii), and (v), and the boundedness of {xn}, {yn}, and {f(xn)}, we have
(96)limn→∞g4(∥xn-Jrnxn∥)=0.
Utilizing the properties of g4, we have
(97)limn→∞∥xn-Jrnxn∥=0.
Note that
(98)∥xn-Snxn∥≤∥xn-Jrnxn∥+∥Jrnxn-Snyn∥+∥Snyn-Snxn∥≤∥xn-Jrnxn∥+∥Jrnxn-Snyn∥+∥yn-xn∥.
From (82), (92), and (97), we get
(99)limn→∞∥xn-Snxn∥=0.
In terms of (99) and Lemma 14, we have
(100)∥xn-Sxn∥≤∥xn-Snxn∥+∥Snxn-Sxn∥⟶0,asn⟶∞.
That is,
(101)limn→∞∥xn-Sxn∥=0.
Furthermore, we claim that limn→∞∥xn-Jrxn∥=0 for a fixed number r such that ε>r>0. In fact, from the resolvent identity in Proposition 1, we have
(102)∥Jrnxn-Jrxn∥=∥Jr(rrnxn+(1-rrn)Jrnxn)-Jrxn∥≤(1-rrn)∥xn-Jrnxn∥≤∥xn-Jrnxn∥.
So, we get from (102)
(103)∥xn-Jrxn∥≤∥xn-Jrnxn∥+∥Jrnxn-Jrxn∥≤∥xn-Jrnxn∥+∥xn-Jrnxn∥=2∥xn-Jrnxn∥.
Thus, from (97) it follows that
(104)limn→∞∥xn-Jrxn∥=0.
Define a mapping Wx=(1-θ1-θ2)Jrx+θ1Sx+θ2Gx, where θ1,θ2∈(0,1) are two constants with θ1+θ2<1. Then, by Lemma 16, we have that Fix(W)=Fix(Jr)∩Fix(S)∩Fix(G)=F. We observe that
(105)∥xn-Wxn∥=∥(1-θ1-θ2)(xn-Jrxn)+θ1(xn-Sxn)+θ2(xn-Gxn)∥≤(1-θ1-θ2)∥xn-Jrxn∥+θ1∥xn-Sxn∥+θ2∥xn-Gxn∥.
From (82), (101), and (104), we obtain
(106)limn→∞∥xn-Wxn∥=0.
Now, we claim that(107)limsupn→∞〈f(q)-q,J(xn-q)〉≤0,
where q=s-limt→0xt with xt being the fixed point of the contraction
(108)x↦tf(x)+(1-t)Wx.
Then xt solves the fixed point equation xt=tf(xt)+(1-t)Wxt. Thus we have
(109)xt-xn=(1-t)(Wxt-xn)+t(f(xt)-xn).
By Lemma 10, we conclude that
(110)∥xt-xn∥2=∥(1-t)(Wxt-xn)+t(f(xt)-xn)∥2≤(1-t)2∥Wxt-xn∥2+2t〈f(xt)-xn,J(xt-xn)〉≤(1-t)2(∥Wxt-Wxn∥+∥Wxn-xn∥)2+2t〈f(xt)-xn,J(xt-xn)〉≤(1-t)2(∥xt-xn∥+∥Wxn-xn∥)2+2t〈f(xt)-xn,J(xt-xn)〉=(1-t)2[∥xt-xn∥2+2∥xt-xn∥∥Wxn-xn∥+∥Wxn-xn∥2]+2t〈f(xt)-xt,J(xt-xn)〉+2t〈xt-xn,J(xt-xn)〉=(1-2t+t2)∥xt-xn∥2+fn(t)+2t〈f(xt)-xt,J(xt-xn)〉+2t∥xt-xn∥2,
where
(111)fn(t)=(1-t)2(2∥xt-xn∥+∥xn-Wxn∥)×∥xn-Wxn∥⟶0,asn⟶∞.
It follows from (110) that
(112)〈xt-f(xt),J(xt-xn)〉≤t2∥xt-xn∥2+12tfn(t).
Letting n→∞ in (112) and noticing (111), we derive
(113)limsupn→∞〈xt-f(xt),J(xt-xn)〉≤t2M2,
where M2>0 is a constant such that ∥xt-xn∥2≤M2 for all t∈(0,1) and n≥0. Taking t→0 in (113), we have
(114)limsupt→0limsupn→∞〈xt-f(xt),J(xt-xn)〉≤0.
On the other hand, we have
(115)〈f(q)-q,J(xn-q)〉=〈f(q)-q,J(xn-q)〉-〈f(q)-q,J(xn-xt)〉+〈f(q)-q,J(xn-xt)〉-〈f(q)-xt,J(xn-xt)〉+〈f(q)-xt,J(xn-xt)〉-〈f(xt)-xt,J(xn-xt)〉+〈f(xt)-xt,J(xn-xt)〉=〈f(q)-q,J(xn-q)-J(xn-xt)〉+〈xt-q,J(xn-xt)〉+〈f(q)-f(xt),J(xn-xt)〉+〈f(xt)-xt,J(xn-xt)〉.
It follows that
(116)limsupn→∞〈f(q)-q,J(xn-q)〉≤limsupn→∞〈f(q)-q,J(xn-q)-J(xn-xt)〉+∥xt-q∥limsupn→∞∥xn-xt∥+ρ∥q-xt∥limsupn→∞∥xn-xt∥+limsupn→∞〈f(xt)-xt,J(xn-xt)〉.
Taking into account that xt→q as t→0, we have
(117)limsupn→∞〈f(q)-q,J(xn-q)〉=limsupt→0limsupn→∞〈f(q)-q,J(xn-q)〉≤limsupt→0limsupn→∞〈f(q)-q,J(xn-q)-J(xn-xt)〉.
Since X has a uniformly Gateaux differentiable norm, the duality mapping J is norm-to-weak* uniformly continuous on bounded subsets of X. Consequently, the two limits are interchangeable and hence (107) holds. From (82), we get (yn-q)-(xn-q)→0. Noticing the norm-to-weak* uniform continuity of J on bounded subsets of X, we deduce from (107) that
(118)limsupn→∞〈f(q)-q,J(yn-q)〉=limsupn→∞(〈f(q)-q,J(xn-q)〉+〈f(q)-q,J(yn-q)-J(xn-q)〉)=limsupn→∞〈f(q)-q,J(xn-q)〉≤0.
Finally, let us show thatxn→q as n→∞. Indeed, observe that
(119)∥yn-q∥2=∥αn(f(xn)-f(q))+βn(xn-q)+γn(Jrnxn-q)+δn(Snyn-q)+αn(f(q)-q)∥2≤∥αn(f(xn)-f(q))+βn(xn-q)+γn(Jrnxn-q)+δn(Snyn-q)∥2+2αn〈f(q)-q,J(yn-q)〉≤αn∥f(xn)-f(q)∥2+βn∥xn-q∥2+γn∥Jrnxn-q∥2+δn∥Snyn-q∥2+2αn〈f(q)-q,J(yn-q)〉≤αnρ∥xn-q∥2+βn∥xn-q∥2+γn∥xn-q∥+δn∥yn-q∥2+2αn〈f(q)-q,J(yn-q)〉=(1-δn-αn(1-ρ))∥xn-q∥2+δn∥yn-q∥2+2αn〈f(q)-q,J(yn-q)〉,
which implies that
(120)∥yn-q∥2≤(1-αn(1-ρ)1-δn)∥xn-q∥2+2αn1-δn〈f(q)-q,J(yn-q)〉.
Thus, we have
(121)∥xn+1-q∥2≤σn∥yn-q∥2+(1-σn)∥Gyn-q∥2≤∥yn-q∥2≤(1-αn(1-ρ)1-δn)∥xn-q∥2+2αn1-δn〈f(q)-q,J(yn-q)〉≤(1-αn(1-ρ)1-δn)∥xn-q∥2+αn(1-ρ)1-δn2〈f(q)-q,J(yn-q)〉1-ρ.
Applying Lemma 9 to (121), we conclude from condition (i) and (118) that xn→q as n→∞. This completes the proof.The following results can be obtained from Theorem23. We, therefore, omit the proof.Corollary 24.
LetC be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gateaux differentiable norm. Let ΠC be a sunny nonexpansive retraction from X onto C. Let A⊂X×X be an accretive operator in X such that D(A)¯⊂C⊂⋂r>0‍R(I+rA). Let Bi:C→X be ζi strictly pseudocontractive and θi strongly accretive with θi+ζi≥1 for each i=1,2. Define the mapping G:C→C by G=ΠC(I-μ1B1)ΠC(I-μ2B2) where 1-(ζi/(1+ζi))(1-(1-θi)/ζi)≤μi≤1 for i=1,2. Let f:C→C be a contraction with coefficient ρ∈(0,1). Let S:C→C be a nonexpansive mapping such that F=Fix(S)∩Ω∩A-10≠∅ where Ω=Fix(G). For arbitrarily given x0∈C, let {xn} be the sequence generated by
(122)yn=αnf(xn)+βnxn+γnJrnxn+δnSyn,xn+1=σnyn+(1-σn)ΠC(I-μ1B1)×ΠC(I-μ2B2)yn,∀n≥0.
Suppose that {rn}⊂(0,∞), {σn},{αn},{βn},{γn},{δn}⊂[0,1], αn+βn+γn+δn=1, and the following conditions hold:(i)
limn→∞αn=0 and ∑n=0∞‍αn=∞;(ii)
{γn},{δn}⊂[c,d] for some c,d∈(0,1);(iii)
∑n=1∞‍(|σn-σn-1|+|αn-αn-1|+|βn-βn-1|+|γn-γn-1|+|δn-δn-1|)<∞;(iv)
∑n=1∞‍|rn-rn-1|<∞ and rn≥ε>0, for all n≥0 for some ε>0;(v)
0<liminfn→∞βn≤limsupn→∞βn<1 and 0<liminfn→∞σn≤limsupn→∞σn<1.
Then{xn} converges strongly to q∈F, which solves the following VIP:
(123)〈q-f(q),J(q-p)〉≤0,∀p∈F.Corollary 25.
LetC be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gateaux differentiable norm. Let ΠC be a sunny nonexpansive retraction from X onto C. Let A⊂X×X be an accretive operator in X such that D(A)¯⊂C⊂⋂r>0‍R(I+rA). Let f:C→C be a contraction with coefficient ρ∈(0,1). Let V:C→C be a self-mapping such that I-V:C→X is ζ strictly pseudocontractive and θ strongly accretive with θ+ζ≥1. Let {Si}i=0∞ be an infinite family of nonexpansive mappings of C into itself such that F=⋂i=0∞‍Fix(Si)∩Fix(V)∩A-10≠∅. For arbitrarily given x0∈C, let {xn} be the sequence generated by
(124)yn=αnf(xn)+βnxn+γnJrnxn+δnSnyn,xn+1=σnyn+(1-σn)×((1-α)I+αV)yn,∀n≥0,
where 1-(ζ/(1+ζ))(1-(1-θ)/ζ)≤α≤1. Suppose that {rn}⊂(0,∞), {σn},{αn},{βn},{γn},{δn}⊂[0,1], αn+βn+γn+δn=1, and the following conditions hold:(i)
limn→∞αn=0 and ∑n=0∞‍αn=∞;(ii)
{γn},{δn}⊂[c,d] for some c,d∈(0,1);(iii)
∑n=1∞‍(|σn-σn-1|+|αn-αn-1|+|βn-βn-1|+|γn-γn-1|+|δn-δn-1|)<∞;(iv)
∑n=1∞‍|rn-rn-1|<∞ and rn≥ε>0, for all n≥0 for some ε>0;(v)
0<liminfn→∞βn≤limsupn→∞βn<1 and 0<liminfn→∞σn≤limsupn→∞σn<1.
Assume that∑n=0∞‍supx∈D∥Sn+1x-Snx∥<∞ for any bounded subset D of C and let S be a mapping of C into itself defined by Sx=limn→∞Snx for all x∈C and suppose that Fix(S)=⋂i=0∞‍Fix(Si). Then {xn} converges strongly to q∈F, which solves the following VIP:
(125)〈q-f(q),J(q-p)〉≤0,∀p∈F.Proof.
In Theorem23, we put B1=I-V,B2=0, and μ1=α where 1-(ζ/(1+ζ))(1-(1-θ)/ζ)≤α≤1. Then GSVI (14) is equivalent to the VIP of finding x*∈C such that
(126)〈B1x*,J(x-x*)〉≥0,∀x∈C.
In this case, B1:C→X is ζ strictly pseudocontractive and θ strongly accretive. It is not hard to see that Fix(V)=VI(C,B1). As a matter of fact, we have, for α>0,
(127)u∈VI(C,B1)⟺〈B1u,J(y-u)〉≥0,∀y∈C⟺〈u-αB1u-u,J(u-y)〉≥0,∀y∈C⟺u=ΠC(u-αB1u)⟺u=ΠC(u-αu+αVu)⟺〈u-αu+αVu-u,J(u-y)〉≥0,∀y∈C⟺〈u-Vu,J(u-y)〉≤0,∀y∈C⟺u=Vu⟺u∈Fix(V).
Accordingly, we know that F=⋂i=1∞‍Fix(Si)∩Ω∩A-10=⋂i=0∞‍Fix(Si)∩Fix(V)∩A-10, and
(128)ΠC(I-μ1B1)ΠC(I-μ2B2)yn=ΠC(I-μ1B1)yn=ΠC((1-α)yn+αVyn)=(1-α)yn+αVyn.
So, the scheme (53) reduces to (124). Therefore, the desired result follows from Theorem 23.Corollary 26.
LetC be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gateaux differentiable norm. Let ΠC be a sunny nonexpansive retraction from X onto C. Let A⊂X×X be an accretive operator in X such that D(A)¯⊂C⊂⋂r>0‍R(I+rA). Let f:C→C be a contraction with coefficient ρ∈(0,1). Let V:C→C be a self-mapping such that I-V:C→X is ζ strictly pseudocontractive and θ strongly accretive with θ+ζ≥1. Let S:C→C be a nonexpansive mapping such that F=Fix(S)∩Fix(V)∩A-10≠∅. For arbitrarily given x0∈C, let {xn} be the sequence generated by
(129)yn=αnf(xn)+βnxn+γnJrnxn+δnSyn,xn+1=σnyn+(1-σn)((1-α)I+αV)yn,∀n≥0,
where 1-(ζ/(1+ζ))(1-(1-θ)/ζ)≤α≤1. Suppose that {rn}⊂(0,∞), {σn},{αn},{βn},{γn},{δn}⊂[0,1], αn+βn+γn+δn=1, and the following conditions hold:(i)
limn→∞αn=0 and ∑n=0∞‍αn=∞;(ii)
{γn},{δn}⊂[c,d] for some c,d∈(0,1);(iii)
∑n=1∞‍(|σn-σn-1|+|αn-αn-1|+|βn-βn-1|+|γn-γn-1|+|δn-δn-1|)<∞;(iv)
∑n=1∞‍|rn-rn-1|<∞ and rn≥ε>0, for all n≥0 for some ε>0;(v)
0<liminfn→∞βn≤limsupn→∞βn<1 and 0<liminfn→∞σn≤limsupn→∞σn<1.
Then{xn} converges strongly to q∈F, which solves the following VIP:
(130)〈q-f(q),J(q-p)〉≤0,∀p∈F.Remark 27.
Our Theorem23 improves, extends, supplements, and develops Cai and Bu [10, Theorem 3.1 and Corollary 3.2] and Jung [5, Theorems 3.1] in the following aspects.
(i) The problem of finding a pointq∈⋂n‍Fix(Sn)∩Ω∩A-10 in our Theorem 23 is more general and more subtle than every one of both the problem of finding a point q∈⋂n‍Fix(Sn)∩Ω in Cai and Bu [10, Theorem 3.1] and the problem of finding a point q∈A-10 in Jung [5, Theorem 3.1].
(ii) Our Theorem23 drops the assumption of the asymptotical regularity of {xn} in [5, Theorems 3.1] (i.e., limn→∞∥xn-xn+1∥=0).
(iii) Cai and Bu's proof in [10, Theorem 3.1] depends on the argument techniques in [13], inequality (21) in 2-uniformly smooth Banach spaces, and inequality (27) in smooth and uniform convex Banach spaces. Jung's proof in [5, Theorem 3.1] depends on the resolvent identity in Proposition 1. It is worth emphasizing that the proof of our Theorem 23 does not depend on the argument techniques in [13], inequality (21) in 2-uniformly smooth Banach spaces, and inequality (27) in smooth and uniform convex Banach spaces. However, it depends on the resolvent identity in Proposition 1 and the inequalities in uniform convex Banach spaces; see Lemmas 13 and 17 in Section 2 of this paper.
(iv) The assumption of the uniformly convex and2-uniformly smooth Banach space X in [10, Theorem 3.1] is weakened to the one of the uniformly convex Banach space X having a uniformly Gateaux differentiable norm in our Theorem 23.
(v) The iterative scheme in our Theorem23 is very different from every one in both [10, Theorem 3.1] and [5, Theorem 3.1] because the first iteration step in our iterative scheme is implicit.
(vi) The problem of finding a pointq∈Fix(S)∩Ω in [10, Corollary 3.2] is extended to develop the problem of finding a point q∈Fix(S)∩Ω∩A-10 in our Corollary 24.


## 4. Hybrid Explicit Extragradient Algorithm

In this section, letC be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gateaux differentiable norm. We suggest and analyze a hybrid explicit extragradient algorithm for finding a zero of A⊂X×X an accretive operator in X with D(A)¯⊂C⊂⋂r>0‍R(I+rA) and solving a general system of variational inequalities and a common fixed point problem of an infinite family of nonexpansive self-mappings in X.Theorem 28.
LetC be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gateaux differentiable norm. Let ΠC be a sunny nonexpansive retraction from X onto C. Let A⊂X×X be an accretive operator in X such that D(A)¯⊂C⊂⋂r>0‍R(I+rA). Let Bi:C→X be ζi strictly pseudocontractive and θi strongly accretive with θi+ζi≥1 for each i=1,2. Define the mapping G:C→C by G=ΠC(I-μ1B1)ΠC(I-μ2B2) where 1-(ζi/(1+ζi))(1-(1-θi)/ζi)≤μi≤1 for i=1,2. Let f:C→C be a contraction with coefficient ρ∈(0,1). Let {Si}i=0∞ be an infinite family of nonexpansive mappings of C into itself such that F=⋂i=0∞‍Fix(Si)∩Ω∩A-10≠∅ where Ω=Fix(G). For arbitrarily given x0∈C, let {xn} be the sequence generated by
(131)yn=αnf(xn)+βnxn+γnJrnxn+δnSnxn,xn+1=σnyn+(1-σn)ΠC(I-μ1B1)×ΠC(I-μ2B2)yn,∀n≥0.
Suppose that {rn}⊂(0,∞), {σn},{αn},{βn},{γn},{δn}⊂[0,1], αn+βn+γn+δn=1, and the following conditions hold:(i)
limn→∞αn=0 and ∑n=0∞‍αn=∞;(ii)
{γn},{δn}⊂[c,d] for some c,d∈(0,1);(iii)
∑n=1∞‍(|σn-σn-1|+|αn-αn-1|+|βn-βn-1|+|γn-γn-1|+|δn-δn-1|)<∞;(iv)
∑n=1∞‍|rn-rn-1|<∞ and rn≥ε>0, for all n≥0 for some ε>0;(v)
0<liminfn→∞βn≤limsupn→∞βn<1 and 0<liminfn→∞σn≤limsupn→∞σn<1. Assume that ∑n=0∞‍supx∈D∥Sn+1x-Snx∥<∞ for any bounded subset D of C and let S be a mapping of C into itself defined by Sx=limn→∞Snx for all x∈C and suppose that Fix(S)=⋂i=0∞‍Fix(Si). Then {xn} converges strongly to q∈F, which solves the following VIP:
(132)〈q-f(q),J(q-p)〉≤0,∀p∈F.Proof.
It is easy to see that (131) can be rewritten as follows
(133)yn=αnf(xn)+βnxn+γnJrnxn+δnSnxn,xn+1=σnyn+(1-σn)Gyn,∀n≥0,
where G=ΠC(I-μ1B1)ΠC(I-μ2B2). By Lemma 20 we know that G is a nonexpansive self-mapping on C.
Now, let us show that the sequence{xn} is bounded. Indeed, take a fixed p∈F arbitrarily. Then from (133), we have
(134)∥yn-p∥≤αn∥f(xn)-p∥+βn∥xn-p∥+γn∥Jrnxn-p∥+δn∥Snxn-p∥≤αn(∥f(xn)-f(p)∥+∥f(p)-p∥)+βn∥xn-p∥+γn∥xn-p∥+δn∥xn-p∥≤αn(ρ∥xn-p∥+∥f(p)-p∥)+(1-αn)∥xn-p∥=(1-(1-ρ)αn)∥xn-p∥+αn∥f(p)-p∥=(1-(1-ρ)αn)∥xn-p∥+(1-ρ)αn∥f(p)-p∥1-ρ,
and hence
(135)∥xn+1-p∥≤σn∥yn-p∥+(1-σn)∥Gyn-p∥≤∥yn-p∥≤(1-(1-ρ)αn)∥xn-p∥+(1-ρ)αn∥f(p)-p∥1-ρ≤max{∥xn-p∥,∥f(p)-p∥1-ρ}.
By induction, we obtain
(136)∥xn-p∥≤max{∥x0-p∥,∥f(p)-p∥1-ρ},∀n≥0.
It immediately follows that {xn} is bounded and so are {yn}, {Gyn}, {f(xn)}, {Jrnxn}, and {Snxn}.
Let us show that(137)limn→∞∥xn+1-xn∥=0.
As a matter of fact, observe that yn can be rewritten as follows:
(138)yn=βnxn+(1-βn)zn,
where zn=(αnf(xn)+γnJrnxn+δnSnxn)/(1-βn). Observe that
(139)∥zn-zn-1∥=∥αnf(xn)+γnJrnxn+δnSnxn1-βn-αn-1f(xn-1)+γn-1Jrn-1xn-1+δn-1Sn-1xn-11-βn-1∥=∥yn-βnxn1-βn-yn-1-βn-1xn-11-βn-1∥=∥yn-βnxn1-βn-yn-1-βn-1xn-11-βn+yn-1-βn-1xn-11-βn-yn-1-βn-1xn-11-βn-1∥≤∥yn-βnxn1-βn-yn-1-βn-1xn-11-βn∥+∥yn-1-βn-1xn-11-βn-yn-1-βn-1xn-11-βn-1∥=11-βn∥yn-βnxn-(yn-1-βn-1xn-1)∥+|11-βn-11-βn-1|∥yn-1-βn-1xn-1∥=11-βn∥yn-βnxn-(yn-1-βn-1xn-1)∥+|βn-βn-1|(1-βn-1)(1-βn)∥yn-1-βn-1xn-1∥=11-βn∥αnf(xn)+γnJrnxn+δnSnxn-αn-1f(xn-1)-γn-1Jrn-1xn-1-δn-1Sn-1xn-1∥+|βn-βn-1|(1-βn-1)(1-βn)∥yn-1-βn-1xn-1∥≤11-βn[αn∥f(xn)-f(xn-1)∥+γn∥Jrnxn-Jrn-1xn-1∥+δn∥Snxn-Sn-1xn-1∥+|αn-αn-1|∥f(xn-1)∥+|γn-γn-1|∥Jrn-1xn-1∥+|δn-δn-1|∥Sn-1xn-1∥]+|βn-βn-1|(1-βn-1)(1-βn)∥yn-1-βn-1xn-1∥.
On the other hand, repeating the same arguments as those of (66) in the proof of Theorem 23, we can derive
(140)∥Jrnxn-Jrn-1xn-1∥≤∥xn-1-xn∥+M0|rn-1-rn|,∀n≥1,
where supn≥1{(1/ε)(∥Jrnxn-xn-1∥+∥Jrn-1xn-1-xn∥)}≤M0 for some M0>0. Substituting (140) into (139), we have
(141)∥zn-zn-1∥≤11-βn[αn∥f(xn)-f(xn-1)∥+γn(∥xn-1-xn∥+M0|rn-1-rn|)+δn∥Snxn-Sn-1xn-1∥+|αn-αn-1|∥f(xn-1)∥+|γn-γn-1|∥Jrn-1xn-1∥+|δn-δn-1|∥Sn-1xn-1∥]+|βn-βn-1|(1-βn-1)(1-βn)∥yn-1-βn-1xn-1∥≤11-βn[αnρ∥xn-xn-1∥+γn(∥xn-1-xn∥+M0|rn-1-rn|)+δn∥Snxn-Snxn-1∥+|αn-αn-1|∥f(xn-1)∥+|γn-γn-1|∥Jrn-1xn-1∥+|δn-δn-1|∥Sn-1xn-1∥+δn∥Snxn-1-Sn-1xn-1∥]+|βn-βn-1|(1-βn-1)(1-βn)×∥αn-1f(xn-1)+γn-1Jrn-1xn-1+δn-1Sn-1xn-1∥≤11-βn[(αnρ+γn)∥xn-xn-1∥+δn∥xn-xn-1∥+M(|αn-αn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+δn∥Snxn-1-Sn-1xn-1∥]+1(1-βn-1)(1-βn)|βn-βn-1|M=11-βn[(1-αn-βn+αnρ)∥xn-xn-1∥+M(|αn-αn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+δn∥Snxn-1-Sn-1xn-1∥]+1(1-βn-1)(1-βn)|βn-βn-1|M=(1-1-ρ1-βnαn)∥xn-xn-1∥+M1-βn(|αn-αn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+δn1-βn∥Snxn-1-Sn-1xn-1∥+1(1-βn-1)(1-βn)|βn-βn-1|M,
where supn≥0{M0+∥f(xn)∥+∥Jrnxn∥+∥Snxn∥}≤M for some M>0. Also, from (133) we have
(142)xn+1-xn=σn(yn-yn-1)+(σn-σn-1)(yn-1-Gyn-1)+(1-σn)(Gyn-Gyn-1).
Taking into account condition (v), we may assume, without loss of generality, that {βn}⊂[a,b] for some a,b∈(0,1). This together with (141) implies that
(143)∥xn+1-xn∥≤σn∥yn-yn-1∥+|σn-σn-1|∥yn-1-Gyn-1∥+(1-σn)∥Gyn-Gyn-1∥≤σn∥yn-yn-1∥+|σn-σn-1|∥yn-1-Gyn-1∥+(1-σn)∥yn-yn-1∥=∥yn-yn-1∥+|σn-σn-1|∥yn-1-Gyn-1∥≤(1-1-ρ1-βnαn)∥xn-xn-1∥+M1-βn(|αn-αn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+δn1-βn∥Snxn-1-Sn-1xn-1∥+1(1-βn-1)(1-βn)|βn-βn-1|M+|σn-σn-1|∥yn-1-Gyn-1∥≤(1-1-ρ1-βnαn)∥xn-xn-1∥+M1(|αn-αn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+∥Snxn-1-Sn-1xn-1∥+|βn-βn-1|M1+|σn-σn-1|M1=(1-1-ρ1-βnαn)∥xn-xn-1∥+M1(|σn-σn-1|+|αn-αn-1|+|βn-βn-1|+|γn-γn-1|+|δn-δn-1|+|rn-1-rn|)+∥Snxn-1-Sn-1xn-1∥,
where supn≥0{(M/(1-b)2)+∥yn-Gyn∥}≤M1 for some M1>0. Since ∑n=0∞‍αn=∞ and ((1-ρ)/(1-βn))αn≥(1-ρ)αn, we know that ∑n=0∞‍((1-ρ)/(1-δn))αn=∞. So, applying Lemma 9 to (143), we obtain from conditions (iii) and (iv) and the assumption on {Sn} that
(144)limn→∞∥xn+1-xn∥=0.
Next we show that∥xn-Gxn∥→0 as n→∞.
Indeed, according to Lemma10 we have from (133)
(145)∥yn-p∥2=∥αn(f(xn)-f(p))+βn(xn-p)+γn(Jrnxn-p)+δn(Snxn-p)+αn(f(p)-p)∥2≤∥αn(f(xn)-f(p))+βn(xn-p)+γn(Jrnxn-p)+δn(Snxn-p)∥2+2αn〈f(p)-p,J(yn-p)〉≤αn∥f(xn)-f(p)∥2+βn∥xn-p∥2+γn∥Jrnxn-p∥2+δn∥Snxn-p∥2+2αn〈f(p)-p,J(yn-p)〉≤αnρ∥xn-p∥2+βn∥xn-p∥2+γn∥xn-p∥2+δn∥xn-p∥2+2αn〈f(p)-p,J(yn-p)〉=(αnρ+βn+γn+δn)∥xn-p∥2+2αn〈f(p)-p,J(yn-p)〉≤(1-αn(1-ρ))∥xn-p∥2+2αn〈f(p)-p,J(yn-p)〉.
Utilizing Lemma 17 we get from (133) and (145)
(146)∥xn+1-p∥2=∥σn(yn-p)+(1-σn)(Gyn-p)∥2≤σn∥yn-p∥2+(1-σn)∥Gyn-p∥2-σn(1-σn)g1(∥yn-Gyn∥)≤σn∥yn-p∥2+(1-σn)∥yn-p∥2-σn(1-σn)g1(∥yn-Gyn∥)=∥yn-p∥2-σn(1-σn)g1(∥yn-Gyn∥)≤(1-αn(1-ρ))∥xn-p∥2+2αn〈f(p)-p,J(yn-p)〉-σn(1-σn)g1(∥yn-Gyn∥)≤∥xn-p∥2+2αn∥f(p)-p∥∥yn-p∥-σn(1-σn)g1(∥yn-Gyn∥),
which hence yields
(147)σn(1-σn)g1(∥yn-Gyn∥)≤∥xn-p∥2-∥xn+1-p∥2+2αn∥f(p)-p∥∥yn-p∥≤(∥xn-p∥+∥xn+1-p∥)∥xn-xn+1∥+2αn∥f(p)-p∥∥yn-p∥.
Since αn→0 and ∥xn+1-xn∥→0, from condition (v) and the boundedness of {xn} and {yn}, it follows that
(148)limn→∞g1(∥yn-Gyn∥)=0.
Utilizing the properties of g1, we have
(149)limn→∞∥yn-Gyn∥=0.
Observe that
(150)∥xn-yn∥≤∥xn-xn+1∥+∥xn+1-yn∥=∥xn-xn+1∥+(1-σn)∥Gyn-yn∥⟶0,asn⟶∞,
and hence
(151)∥xn-Gxn∥≤∥xn-yn∥+∥yn-Gyn∥+∥Gyn-Gxn∥≤2∥xn-yn∥+∥yn-Gyn∥⟶0,asn⟶∞.
That is,
(152)limn→∞∥xn-yn∥=0,limn→∞∥xn-Gxn∥=0.
Next, let us show that
(153)limn→∞∥Jrnxn-xn∥=0,limn→∞∥Snxn-xn∥=0.
Indeed, observe that yn can be rewritten as follows:
(154)yn=αnf(xn)+βnxn+γnJrnxn+δnSnxn=αnf(xn)+βnxn+(γn+δn)γnJrnxn+δnSnxnγn+δn=αnf(xn)+βnxn+enz^n,
where en=γn+δn and z^n=(γnJrnxn+δnSnxn)/(γn+δn). Utilizing Lemma 13 and (154), we have
(155)∥yn-p∥2=∥αn(f(xn)-p)+βn(xn-p)+en(z^n-p)∥2≤αn∥f(xn)-p∥2+βn∥xn-p∥2+en∥z^n-p∥2-βneng3(∥z^n-xn∥)=αn∥f(xn)-p∥2+βn∥xn-p∥2-βneng2(∥z^n-xn∥)+en∥γnJrnxn+δnSnxnγn+δn-p∥2=αn∥f(xn)-p∥2+βn∥xn-p∥2-βneng2(∥z^n-xn∥)+en∥γnγn+δn(Jrnxn-p)+δnγn+δn(Snxn-p)∥2≤αn∥f(xn)-p∥2+βn∥xn-p∥2-βneng2(∥z^n-xn∥)+en[γnγn+δn∥Jrnxn-p∥2+δnγn+δn∥Snxn-p∥2]≤αn∥f(xn)-p∥2+βn∥xn-p∥2-βneng2(∥z^n-xn∥)+en[γnγn+δn∥xn-p∥+δnγn+δn∥xn-p∥2]=αn∥f(xn)-p∥2+(βn+γn+δn)∥xn-p∥2-βneng2(∥z^n-xn∥)≤αn∥f(xn)-p∥2+∥xn-p∥2-βneng2(∥z^n-xn∥),
which hence implies that
(156)βneng2(∥z^n-xn∥)≤αn∥f(xn)-p∥2+∥xn-p∥2-∥yn-p∥2≤αn∥f(xn)-p∥2+(∥xn-p∥+∥yn-p∥)∥xn-yn∥.
Utilizing (152), conditions (i), (ii), and (v) and the boundedness of {xn}, {yn}, and {f(xn)}, we get
(157)limn→∞g2(∥z^n-xn∥)=0.
From the properties of g2, we have
(158)limn→∞∥z^n-xn∥=0.
Utilizing Lemma 17 and the definition of z^n, we have
(159)∥z^n-p∥2=∥γnJrnxn+δnSnxnγn+δn-p∥2=∥γnγn+δn(Jrnxn-p)+δnγn+δn(Snxn-p)∥2≤γnγn+δn∥Jrnxn-p∥2+δnγn+δn∥Snxn-p∥2-γnδn(γn+δn)2g3(∥Jrnxn-Snxn∥)≤γnγn+δn∥xn-p∥2+δnγn+δn∥xn-p∥2-γnδn(γn+δn)2g3(∥Jrnxn-Snxn∥)=∥xn-p∥2-γnδn(γn+δn)2g3(∥Jrnxn-Snxn∥),
which leads to
(160)γnδn(γn+δn)2g3(∥Jrnxn-Snxn∥)≤∥xn-p∥2-∥z^n-p∥2≤(∥xn-p∥+∥z^n-p∥)∥xn-z^n∥.
Since {xn} and {z^n} are bounded and ∥z^n-xn∥→0 as n→∞, we deduce from condition (ii) that
(161)limn→∞g3(∥Jrnxn-Snxn∥)=0.
From the properties of g3, we have
(162)limn→∞∥Jrnxn-Snxn∥=0.
On the other hand, yn can also be rewritten as follows:
(163)yn=αnf(xn)+βnxn+γnJrnxn+δnSnxn=βnxn+γnJrnxn+(αn+δn)αnf(xn)+δnSnxnαn+δn=βnxn+γnJrnxn+dnz~n,
where dn=αn+δn and z~n=(αnf(xn)+δnSnxn)/(αn+δn). Utilizing Lemma 13 and the convexity of ∥·∥2, we have
(164)∥yn-p∥2=∥βn(xn-p)+γn(Jrnxn-p)+dn(z~n-p)∥2≤βn∥xn-p∥2+γn∥Jrnxn-p∥2+dn∥z~n-p∥2-βnγng4(∥xn-Jrnxn∥)=βn∥xn-p∥2+γn∥Jrnxn-p∥2+dn∥αnf(xn)+δnSnxnαn+δn-p∥2-βnγng4(∥xn-Jrnxn∥)=βn∥xn-p∥2+γn∥Jrnxn-p∥2+dn∥αnαn+δn(f(xn)-p)+δnαn+δn(Snxn-p)∥2-βnγng4(∥xn-Jrnxn∥)≤βn∥xn-p∥2+γn∥xn-p∥2+dn[αnαn+δn∥f(xn)-p∥2+δnαn+δn∥Snxn-p∥2]-βnγng4(∥xn-Jrnxn∥)≤αn∥f(xn)-p∥2+(βn+γn+δn)∥xn-p∥2-βnγng4(∥xn-Jrnxn∥)≤αn∥f(xn)-p∥2+∥xn-p∥2-βnγng4(∥xn-Jrnxn∥),
which hence implies that
(165)βnγng4(∥xn-Jrnxn∥)≤αn∥f(xn)-p∥2+∥xn-p∥2-∥yn-p∥2≤αn∥f(xn)-p∥2+(∥xn-p∥+∥yn-p∥)∥xn-yn∥.
From (152), conditions (i), (ii), and (v), and the boundedness of {xn}, {yn}, and {f(xn)}, we have
(166)limn→∞g4(∥xn-Jrnxn∥)=0.
Utilizing the properties of g4, we have
(167)limn→∞∥xn-Jrnxn∥=0.
Note that
(168)∥xn-Snxn∥≤∥xn-Jrnxn∥+∥Jrnxn-Snxn∥.
From (162) and (167), we get
(169)limn→∞∥xn-Snxn∥=0.
In terms of (169) and Lemma 14, we have
(170)∥xn-Sxn∥≤∥xn-Snxn∥+∥Snxn-Sxn∥⟶asn⟶∞.
That is,
(171)limn→∞∥xn-Sxn∥=0.
Furthermore, repeating the same arguments as those of (104) in the proof of Theorem 23, we can conclude that
(172)limn→∞∥xn-Jrxn∥=0
for a fixed number r such that ε>r>0.
Define a mappingWx=(1-θ1-θ2)Jrx+θ1Sx+θ2Gx, where θ1,θ2∈(0,1) are two constants with θ1+θ2<1. Then by Lemma 16, we have that Fix(W)=Fix(Jr)∩Fix(S)∩Fix(G)=F. We observe that
(173)∥xn-Wxn∥=∥(1-θ1-θ2)(xn-Jrxn)+θ1(xn-Sxn)+θ2(xn-Gxn)∥≤(1-θ1-θ2)∥xn-Jrxn∥+θ1∥xn-Sxn∥+θ2∥xn-Gxn∥.
From (152), (171), and (172), we obtain
(174)limn→∞∥xn-Wxn∥=0.
Repeating the same arguments as those of (107) in the proof of Theorem 23, we can deduce that
(175)limsupn→∞〈f(q)-q,J(xn-q)〉≤0.
Since J is norm-to-weak* uniformly continuous over bounded subsets of X, we obtain from ∥xn-yn∥→0 that
(176)limsupn→∞〈f(q)-q,J(yn-q)〉≤0.
Finally, let us show that xn→q as n→∞. Indeed, observe that
(177)∥yn-q∥2=∥αn(f(xn)-f(q))+βn(xn-q)+γn(Jrnxn-q)+δn(Snxn-q)+αn(f(q)-q)∥2≤∥αn(f(xn)-f(q))+βn(xn-q)+γn(Jrnxn-q)+δn(Snxn-q)∥2+2αn〈f(q)-q,J(yn-q)〉≤αn∥f(xn)-f(q)∥2+βn∥xn-q∥2+γn∥Jrnxn-q∥2+δn∥Snxn-q∥2+2αn〈f(q)-q,J(yn-q)〉≤αnρ∥xn-q∥2+βn∥xn-q∥2+γn∥xn-q∥+δn∥xn-q∥2+2αn〈f(q)-q,J(yn-q)〉=(αnρ+βn+γn+δn)∥xn-q∥2+2αn〈f(q)-q,J(yn-q)〉=(1-αn(1-ρ))∥xn-q∥2+2αn〈f(q)-q,J(yn-q)〉,
and hence
(178)∥xn+1-q∥2≤σn∥yn-q∥2+(1-σn)∥Gyn-q∥2≤∥yn-q∥2≤(1-αn(1-ρ))∥xn-q∥2+2αn〈f(q)-q,J(yn-q)〉.
Applying Lemma 9 to (178), we conclude from condition (i) and (178) that xn→q as n→∞. This completes the proof.The following results can be obtained from Theorem28. We, therefore, omit the proof.Corollary 29.
Under the same conditions as those in Corollary24, let {xn} be the sequence generated from any given x0∈C by
(179)yn=αnf(xn)+βnxn+γnJrnxn+δnSxn,xn+1=σnyn+(1-σn)ΠC(I-μ1B1)×ΠC(I-μ2B2)yn,∀n≥0.
Then {xn} converges strongly to q∈F=Fix(S)∩Ω∩A-10, which solves the following VIP:
(180)〈q-f(q),J(q-p)〉≤0,∀p∈F.Corollary 30.
Under the same conditions as those in Corollary25, let {xn} be the sequence generated from any given x0∈C by
(181)yn=αnf(xn)+βnxn+γnJrnxn+δnSnxn,xn+1=σnyn+(1-σn)((1-α)I+αV)yn,∀n≥0.
Then {xn} converges strongly to q∈F=⋂i=0∞‍Fix(Si)∩Fix(V)∩A-10   , which solves the following VIP:
(182)〈q-f(q),J(q-p)〉≤0,∀p∈F.Corollary 31.
Under the same conditions as those in Corollary26, let {xn} be the sequence generated from any given x0∈C by
(183)yn=αnf(xn)+βnxn+γnJrnxn+δnSxn,xn+1=σnyn+(1-σn)((1-α)I+αV)yn,∀n≥0.
Then {xn} converges strongly to q∈F=Fix(S)∩Fix(V)∩A-10, which solves the following VIP:
(184)〈q-f(q),J(q-p)〉≤0,∀p∈F.Remark 32.
Our Theorem28 improves, extends, supplements, and develops Cai and Bu [10, Theorem 3.1 and Corollary 3.2] and Jung [5, Theorems 3.1] in the following aspects.
(i) The problem of finding a pointq∈⋂n‍Fix(Sn)∩Ω∩A-10 in our Theorem 28 is more general and more subtle than every one of both the problem of finding a point q∈⋂n‍Fix(Sn)∩Ω in Cai and Bu [10, Theorem 3.1] and the problem of finding a point q∈A-10 in Jung [5, Theorem 3.1].
(ii) Our Theorem28 drops the assumption of the asymptotical regularity of {xn} in [5, Theorems 3.1] (i.e., limn→∞∥xn-xn+1∥=0).
(iii) Cai and Bu's proof in [10, Theorem 3.1] depends on the argument techniques in [13], inequality (21) in 2-uniformly smooth Banach spaces, and inequality (27) in smooth and uniform convex Banach spaces. Jung's proof in [5, Theorem 3.1] depends on the resolvent identity in Proposition 1. It is worth emphasizing that the proof of our Theorem 28 does not depend on the argument techniques in [13], inequality (21) in 2-uniformly smooth Banach spaces, and inequality (27) in smooth and uniform convex Banach spaces. However, it depends on the resolvent identity in Proposition 1 and the inequalities in uniform convex Banach spaces; see Lemmas 13 and 17 in Section 2 of this paper.
(iv) The assumption of the uniformly convex and2-uniformly smooth Banach space X in [10, Theorem 3.1] is weakened to the one of the uniformly convex Banach space X having a uniformly Gateaux differentiable norm in our Theorem 28.
(v) The iterative scheme in our Theorem28 is very different from every one in both [10, Theorem 3.1] and [5, Theorem 3.1] because the first iteration step in the iterative scheme of [10, Theorem 3.1] is given by yn=αnf(xn)+(1-αn)ΠC(I-μ1B1)ΠC(I-μ2B2)xn and the first iteration step in the iterative scheme of [5, Theorem 3.1] is given by yn=αnf(xn)+(1-αn)Jrnxn. In the meantime, it is clear that the second iteration steps for xn+1 in three iterative schemes are completely different.
(vi) The problem of finding a pointq∈Fix(S)∩Ω in [10, Corollary 3.2] is extended to develop the problem of finding a point q∈Fix(S)∩Ω∩A-10 in our Corollary 29.



---

*Source: 894926-2013-09-24.xml*