# Strong Convergence to a Solution of a Variational Inequality Problem in Banach Spaces

Authors: Yasunori Kimura; Kazuhide Nakajo
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

We consider the variational inequality problem for a
family of operators of a nonempty closed convex subset of a 2-uniformly convex Banach space with a uniformly Gâteaux differentiable norm, into its dual space. We assume some properties for the operators and get strong convergence to a common solution to the variational inequality problem by the hybrid
method proposed by Haugazeau. Using these results, we obtain several results
for the variational inequality problem and the proximal point algorithm.


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## Body

## 1. Introduction

LetN and R be the set of all positive integers and the set of all real numbers, respectively. Throughout this paper, E is a real Banach space with norm ∥·∥ and E* is the dual of E. For x∈E and x*∈E*, let 〈x,x*〉 be the value of x* at x. Suppose that C is a nonempty closed convex subset of E and A is a monotone operator of C into E*; that is, 〈x-y,Ax-Ay〉≥0 holds for all x,y∈C. Then, we consider the variational inequality problem [1], that is, the problem of finding an element z∈C such that
(1)〈x-z,Az〉≥0∀x∈C.The set of all solutions to the variational inequality problem forA is denoted by VI⁡(C,A). For α>0, we say that A is α-inverse strongly monotone [2–5] if
(2)〈x-y,Ax-Ay〉≥α∥Ax-Ay∥2∀x,y∈C.Haugazeau [6] introduced a sequence {xn} generated by the hybrid method by the following way. Let {Tn} be a family of mappings of a real Hilbert space H into itself with ⋂n=0∞‍F(Tn)≠∅, where F(Tn) is the set of all fixed points of Tn. Let {xn} be a sequence generated by
(3)x1=x∈H,yn=Tnxn,Cn={z∈H:〈xn-yn,yn-z〉≥0},Qn={z∈H:〈xn-z,x-xn〉≥0},xn+1=PCn∩Qn(x)
for each n∈N, where PCn∩Qn is the metric projection of H onto Cn∩Qn. He proved a strong convergence theorem when Tn=P(nmod⁡m)+1 for every n∈N, where Pi is the metric projection of H onto a nonempty closed convex subset Ci of H for each i=1,2,…,m with ⋂i=1m‍Ci≠∅. Later, Solodov and Svaiter [7], Bauschke and Combettes [8], Nakajo and Takahashi [9], and many researchers studied the hybrid method in a real Hilbert space. In a real Banach space, Kamimura and Takahashi [10], Ohsawa and Takahashi [11], Kohsaka and Takahashi [12], Matsushita and Takahashi [13], Matsushita et al. [14], Nakajo et al. [15], and several researchers studied the hybrid method.In a real Hilbert spaceH, Iiduka et al. [16] considered a sequence {xn} generated by the following hybrid method:
(4)x1=x∈C,yn=PC(xn-λnAxn),Cn={z∈C:∥yn-z∥≤∥xn-z∥},Qn={z∈C:〈xn-z,x-xn〉≥0},xn+1=PCn∩Qnx
for each n∈N, where A is an α-inverse strongly monotone operator of C into H with VI⁡(C,A)≠∅, PC is the metric projection of H onto a nonempty closed convex subset C of H, and {λn}⊂[0,2α]. They proved that {xn} converges strongly to PVI⁡(C,A)x; see also [17, 18]. In a 2-uniformly convex and uniformly smooth Banach space E, Iiduka and Takahashi [19] proved the following.Theorem 1 (Iiduka and Takahashi [19]).
LetA be an α-inverse strongly monotone operator of E into E* with A-10≠∅ and {λn}⊂[a,c1α] for some a∈]0,c1α[, where c1 is a positive constant satisfying that ∥x+y∥2≥∥x∥2+2〈y,Jx〉+c1∥y∥2 for every x,y∈E. Let {xn} be a sequence generated by
(5)x1=x∈E,yn=J-1(Jxn-λnAxn),Cn={z∈E:ϕ(z,yn)≤ϕ(z,xn)},Qn={z∈E:〈xn-z,Jx-Jxn〉≥0},xn+1=ΠCn∩Qnx
for each n∈N, where ΠCn∩Qn is the generalized projection of E onto Cn∩Qn and ϕ(x,y)=∥x∥2-2〈x,Jy〉+∥y∥2 for x,y∈E. Then, {xn} converges strongly to ΠA-10x.Motivated by [19], we propose a new family of operators and prove strong convergence theorems of the sequence generated by these mappings. Using these results, we get several additional results for the problem of variational inequalities and the proximal point algorithm.


## 2. Preliminaries

Throughout this paper, we writexn⇀x to indicate that a sequence {xn} converges weakly to x and xn→x will symbolize strong convergence. We denote by SE the unit sphere of a Banach space E; that is, SE={x∈E:∥x∥=1}.We define the modulusδE of convexity of E as follows: δE is a function of [0,2] into [0,1] such that
(6)δE(ϵ)=inf⁡{1-∥x+y∥2:x,y∈SE,∥x-y∥≥ϵ}
for every ϵ∈[0,2]. E is said to be uniformly convex if δE(ϵ)>0 for each ϵ>0. Let p>1. E is said to be p-uniformly convex if there exists a constant c>0 such that δE(ϵ)≥cϵp for every ϵ∈[0,2]. It is obvious that a p-uniformly convex Banach space is uniformly convex. E is said to be strictly convex if ∥x+y∥/2<1 for all x,y∈SE with x≠y. We know that a uniformly convex Banach space is strictly convex and reflexive. The duality mapping J:E→2E* of E is defined by
(7)J(x)={f∈E*:〈x,f〉=∥x∥2=∥f∥2}
for every x∈E. It is also known that if E is strictly convex and reflexive, then the duality mapping J of E is bijective and J-1:E*→2E is the duality mapping of E*. E is said to be smooth if the limit
(8)lim⁡t→0∥x+ty∥-∥x∥t
exists for every x,y∈SE. The norm of E is said to be uniformly Gâteaux differentiable if, for each y∈SE, the limit (8) is attained uniformly for x∈SE. E is said to be uniformly smooth if the limit (8) is attained uniformly for (x,y)∈SE×SE. We know that the duality mapping J of E is single-valued if and only if E is smooth. It is also known that if E is uniformly smooth, then the duality mapping J is uniformly continuous on bounded subsets of E and if the norm of E is uniformly Gâteaux differentiable, then J is norm-to-weak∗ uniformly continuous on bounded subsets of E; see [20, 21] for more details. The following is proved by Xu [22]; see also [23].Theorem 2 (Xu [22]).
LetE be a smooth Banach space. Then, the following are equivalent. (i)
Eis 2-uniformly convex.(ii)
There exists a constantc1>0 such that ∥x+y∥2≥∥x∥2+2〈y,Jx〉+c1∥y∥2 holds for each x,y∈E.Remark 3.
In the case whereE is a real Hilbert space, J is the identity mapping and we can choose c1=1.LetE be a smooth Banach space. The function ϕ:E×E→R is defined by
(9)ϕ(y,x)=∥y∥2-2〈y,Jx〉+∥x∥2
for every x,y∈E. It is obvious that (∥y∥-∥x∥)2≤ϕ(y,x)≤(∥y∥+∥x∥)2 for each x,y∈E and ϕ(z,x)+ϕ(x,y)=ϕ(z,y)+2〈x-z,Jx-Jy〉 for all x,y,z∈E. It is also known that if E is strictly convex and smooth, then, for x,y∈E, ϕ(y,x)=0 if and only if x=y; see also [13]. We have the following result from Theorem 2.Lemma 4.
LetE be a 2-uniformly convex and smooth Banach space. Then, for each x,y∈E, ϕ(x,y)≥c1∥x-y∥2 holds, where c1 is a constant in Theorem 2.Proof.
Letx,y∈E. By Theorem 2, we have
(10)ϕ(x,y)=∥x∥2-∥y∥2-2〈x-y,Jy〉≥c1∥x-y∥2,
which is the desired result.LetC be a nonempty closed convex subset of a strictly convex, reflexive, and smooth Banach space E and let x∈E. Then, there exists a unique element y0∈C such that
(11)ϕ(y0,x)=inf⁡y∈Cϕ(y,x).We denotey0 by ΠCx and call ΠC the generalized projection of E onto C; see [10, 24, 25]. We have the following well-known results [10, 24, 25] for the generalized projection.Lemma 5.
LetC be a nonempty convex subset of a smooth Banach space E, x∈E, and y0∈C. Then, ϕ(y0,x)=inf⁡y∈Cϕ(y,x) if and only if 〈y0-z,Jx-Jy0〉≥0 for all z∈C.LetC be a nonempty closed convex subset of a strictly convex and reflexive Banach space E and let x∈E. Then, there exists a unique element y0∈C such that ∥y0-x∥=inf⁡y∈C∥y-x∥. Putting y0=PCx, we call PC the metric projection of E onto C; see [26]. We have the following result for the metric projection; see [20] for more details.Lemma 6.
LetC be a nonempty closed convex subset of a strictly convex, reflexive, and smooth Banach space E, x∈E, and y0∈C. Then, y0=PCx if and only if 〈y0-z,J(x-y0)〉≥0 for all z∈C.An operatorT:E→2E* is said to be monotone if 〈x-y,x*-y*〉≥0 for every (x,x*),(y,y*)∈T. Notice that we often identify a set-valued operator with its graph; x*∈Tx if and only if (x,x*)∈T.A monotone operatorT⊂E×E* is said to be maximal if the graph of T is not properly contained in the graph of any other monotone operator. It is easy to see that a monotone operator T⊂E×E* is maximal if and only if, for (u,u*)∈E×E*, 〈x-u,x*-u*〉≥0 for every (x,x*)∈T implies that (u,u*)∈T. We know the following result.Theorem 7 (Rockafellar [27]; see also [28]).
LetE be a strictly convex, reflexive, and smooth Banach space and let T be a monotone operator of E into E*. Then, T is maximal if and only if R(J+rT)=E* for all r>0, where R(J+rT) is the range of J+rT.From this fact, we also know that ifE is a strictly convex, reflexive, and smooth Banach space and T is a maximal monotone operator of E into E*, then, for any x∈E and r>0, there exists a unique element xr∈D(T) such that J(xr-x)+rTxr∋0, where D(T) is the domain of T. We define Jr:E→E by Jrx=xr for every x∈E and r>0, and such Jr is called the resolvent of T; see [21, 29] for more details.


## 3. Main Results

LetC be a nonempty closed convex subset of a strictly convex, reflexive, and smooth Banach space E and {An}n∈N a family of operators of C into E* satisfying the following:(i)
F=⋂n=1∞‍VI⁡(C,An)≠∅;(ii)
〈x-z,Anx-Anz〉≥0 for all n∈N, x∈C, and z∈F;(iii)
there exists a sequence{αn}n∈N in ]0,∞[ such that 0<liminf⁡n→∞αn≤limsup⁡n→∞αn<∞ and 〈x-z,Anx〉≥αn∥Anx-Anz∥2 for every n∈N, x∈C, and z∈F;(iv)
for allz∈F, sup⁡n∈N∥Anz∥<∞;(v)
for every bounded sequence{zn}⊂C, z∈F, and {rn}⊂]0,∞[ with inf⁡n∈Nrn>0, if lim⁡n→∞∥zn+1-zn∥=lim⁡n→∞∥zn-ΠCJ-1(Jzn-rnAnzn)∥=lim⁡n→∞∥Anzn-Anz∥=0, then there exists a subsequence {zni} of {zn} such that zni⇀z∈F.Let us observe some properties of the mappings and the subsets deduced from the assumptions above.First, we know that, for anyn∈N, the image of F by An is a singleton. Indeed, for z1,z2∈F, we have 〈z1-z2,Anz1〉≥αn∥Anz1-Anz2∥2 by the condition (iii). On the other hand, since z1∈VI⁡(C,An), it follows that 〈z2-z1,Anz1〉≥0. Thus, we get Anz1=Anz2 for all z1,z2∈F.Next, if we assume⋂n=1∞‍An-10≠∅, then we have ⋂n=1∞‍An-10=⋂n=1∞‍VI⁡(C,An). Indeed, the inclusion ⋂n=1∞‍An-10⊂⋂n=1∞‍VI⁡(C,An) is trivial. To show the opposite inclusion, let z∈⋂n=1∞‍VI⁡(C,An) and u∈⋂n=1∞‍An-10. By the condition (iii), we have 0=〈u-z,Anu〉≥αn∥Anu-Anz∥2=αn∥Anz∥2, which implies Anz=0; that is, z∈An-10 for all n∈N. Hence, we get ⋂n=1∞‍An-10⊃⋂n=1∞‍VI⁡(C,An).We also know thatF is closed and convex. Indeed, for z1,z2∈F and β∈]0,1[, let z=βz1+(1-β)z2. By the condition (iii), 〈z-z1,Anz〉≥αn∥Anz-Anz1∥2 and 〈z-z2,Anz〉≥αn∥Anz-Anz2∥2 hold for all n∈N. Thus we get
(12)0=〈z-(βz1+(1-β)z2),Anz〉=β〈z-z1,Anz〉+(1-β)〈z-z2,Anz〉≥αn(β∥Anz-Anz1∥2+(1-β)∥Anz-Anz2∥2)≥0,
which implies that Anz=Anz1=Anz2 for each n∈N. Since 〈x-z1,Anz〉=〈x-z1,Anz1〉≥0 and 〈x-z2,Anz〉=〈x-z2,Anz2〉≥0 for every n∈N and x∈C, we have
(13)〈x-z,Anz〉=β〈x-z1,Anz〉+(1-β)〈x-z2,Anz〉≥0
for all n∈N and x∈C; that is, z∈VI⁡(C,An) for each n∈N. Hence, F is convex.To seeF being closed, let {zm} be a sequence in F such that zm→z. Since we have 〈z-zm,Anz〉≥αn∥Anz-Anzm∥2 for every m,n∈N from the condition (iii), we get lim⁡m→∞∥Anz-Anzm∥=0 for all n∈N. Since 〈x-zm,Anzm〉≥0 for each m,n∈N and x∈C, we obtain 〈x-z,Anz〉≥0 for every n∈N and x∈C; that is, z∈VI⁡(C,An) for all n∈N. Therefore, F is closed.Now, we get the following result by the hybrid method using the generalized projections.Theorem 8.
LetC be a nonempty closed convex subset of a 2-uniformly convex Banach space E whose norm is uniformly Gâteaux differentiable, and let {An} be a sequence of operators of C into E* satisfying the conditions (i)–(v). Let {λn} be a sequence in ]0,∞[ such that inf⁡n∈Nλn>0 and inf⁡n∈N(2c1αn-λn)>0, where c1 is the constant in Theorem 2. Let x∈C and {xn} a sequence in C generated by
(14)x1=x,yn=ΠCJ-1(Jxn-λnAnxn),Cn={z∈C:ϕ(z,yn)≤ϕ(z,xn)+2λn〈xn-z,Anxn〉},Qn={z∈C:〈xn-z,Jx-Jxn〉≥0},xn+1=ΠCn∩Qnx
for eachn∈N. Then, {xn} converges strongly to ΠFx.Proof.
It is obvious thatQn is closed and convex for every n∈N. Since ϕ(z,yn)≤ϕ(z,xn)+2λn〈xn-z,Anxn〉 if and only if 2〈z,Jxn-Jyn〉+∥yn∥2-∥xn∥2-2λn〈xn-z,Anxn〉≤0, we have that Cn is closed and convex for all n∈N. Next, we show that, for n∈N, xn∈C implies that F⊂Cn. Let z∈F. We have
(15)(ϕ(z,yn)+2λn〈yn-z,Anz〉)-(ϕ(z,xn)+2λn〈xn-z,Anz〉)=(ϕ(z,yn)-ϕ(z,xn))+2λn〈yn-xn,Anz〉=-ϕ(yn,xn)+2〈yn-z,Jyn-Jxn〉+2λn〈yn-xn,Anz〉
for every n∈N. Since yn=ΠCJ-1(Jxn-λnAnxn) and by Lemma 5, we get 〈yn-z,Jxn-λnAnxn-Jyn〉≥0. Further, by Lemma 4, we obtain
(16)(ϕ(z,yn)+2λn〈yn-z,Anz〉)-(ϕ(z,xn)+2λn〈xn-z,Anz〉)≤-c1∥yn-xn∥2-2λn〈yn-z,Anxn〉+2λn〈yn-xn,Anz〉=-c1∥yn-xn∥2-2λn(〈yn-xn,Anxn〉+〈xn-z,Anxn〉)+2λn〈yn-xn,Anz〉.
Usingxn∈C and the condition (iii), we have 〈xn-z,Anxn〉≥αn∥Anxn-Anz∥2 and thus
(17)(ϕ(z,yn)+2λn〈yn-z,Anz〉)-(ϕ(z,xn)+2λn〈xn-z,Anz〉)≤-c1∥yn-xn∥2-2λn(〈yn-xn,Anxn〉+αn∥Anxn-Anz∥2)+2λn〈yn-xn,Anz〉=-c1∥yn-xn∥2-2λnαn∥Anxn-Anz∥2-2λn〈yn-xn,Anxn-Anz〉≤-c1∥yn-xn∥2-2λnαn∥Anxn-Anz∥2+2λn∥yn-xn∥∥Anxn-Anz∥≤-c1∥yn-xn∥2-2λnαn∥Anxn-Anz∥2+2λn(β2∥yn-xn∥2+12β∥Anxn-Anz∥2)=(βλn-c1)∥yn-xn∥2+λn(1β-2αn)∥Anxn-Anz∥2
for any β>0. Since a sequence {λn} satisfies 0<λn<2c1αn for all n∈N, lim⁡⁡sup⁡n→∞λn<∞, and inf⁡n∈N(c1/λn-1/(2αn))>0, we can choose a positive sequence {βn} such that
(18)βnλn-c1<0,λn(1βn-2αn)<0∀n∈N,lim⁡sup⁡n→∞(βnλn-c1)<0,lim⁡sup⁡n→∞λn(1βn-2αn)<0.
So, we obtain(19)(ϕ(z,yn)+2λn〈yn-z,Anz〉)-(ϕ(z,xn)+2λn〈xn-z,Anz〉)≤0.
Sincez∈VI⁡(C,An), we have 〈yn-z,Anz〉≥0. Using xn∈C and the condition (ii), we have 〈xn-z,Anz〉≤〈xn-z,Anxn〉. Thus, we get
(20)ϕ(z,yn)≤ϕ(z,xn)+2λn〈xn-z,Anxn〉;
that is, F⊂Cn. From this fact, we get that F⊂Cn∩Qn for every n∈N and {xn} is well defined. Indeed, x1=x∈C is given and since Q1=C, we have F⊂C1∩Q1. Assume that xk is well defined and F⊂Ck∩Qk for some k∈N. There exists a unique element xk+1=ΠCk∩Qkx and we get 〈xk+1-z,Jx-Jxk+1〉≥0 for all z∈Ck∩Qk by Lemma 5. Since F⊂Ck∩Qk, we have 〈xk+1-z,Jx-Jxk+1〉≥0 for every z∈F; that is, F⊂Qk+1. Since xk+1∈Ck∩Qk⊂C, we have F⊂Ck+1. Hence, we obtain F⊂Ck+1∩Qk+1. By mathematical induction, we get F⊂Cn∩Qn for every n∈N and {xn} is well defined. Since xn+1=ΠCn∩Qnx and F⊂Cn∩Qn, we have ϕ(xn+1,x)≤ϕ(ΠFx,x) for every n∈N, which implies that {xn} is bounded. Further, since xn+1∈Qn, we have
(21)ϕ(xn+1,xn)+ϕ(xn,x)=ϕ(xn+1,x)+2〈xn-xn+1,Jxn-Jx〉≤ϕ(xn+1,x)
for all n∈N. Thus, there exists lim⁡n→∞ϕ(xn,x) and
(22)lim⁡n→∞ϕ(xn+1,xn)=0.
By Lemma4, we get
(23)lim⁡n→∞∥xn+1-xn∥=0.
Usingxn+1∈Cn, we have
(24)ϕ(xn+1,yn)≤ϕ(xn+1,xn)+2λn〈xn-xn+1,Anxn〉≤ϕ(xn+1,xn)+2λn∥xn-xn+1∥∥Anxn∥
for all n∈N. From the condition (iii), we have
(25)∥xn-z∥∥Anxn∥≥αn(∥Anxn∥-∥Anz∥)2≥αn(∥Anxn∥2-2∥Anxn∥∥Anz∥),
which implies that
(26)∥xn-z∥≥αn∥Anxn∥-2αn∥Anz∥
for every n∈N and z∈F. Since 0<lim⁡inf⁡n→∞αn≤lim⁡sup⁡n→∞αn<∞, using the condition (iv) and the boundedness of {xn}, we get that {Anxn} is bounded. Since (22)–(24) hold and {λn} is bounded, we have lim⁡n→∞ϕ(xn+1,yn)=0, which implies that
(27)lim⁡n→∞∥xn+1-yn∥=0
by Lemma4. From (23) and (27), we also have lim⁡n→∞∥xn-yn∥=0. Using the facts that
(28)(ϕ(z,yn)+2λn〈yn-z,Anz〉)-(ϕ(z,xn)+2λn〈xn-z,Anz〉)=2〈z,Jxn-Jyn〉+(∥yn∥2-∥xn∥2)+2λn〈yn-xn,Anz〉
for all n∈N, lim⁡n→∞∥xn-yn∥=0, the condition (iv) holds, {xn}, {yn}, and {λn} are bounded, and the duality mapping J is norm-to-weak∗ uniformly continuous on bounded subset of E, we obtain
(29)lim⁡n→∞((ϕ(z,yn)+2λn〈yn-z,Anz〉)vvvvvv-(ϕ(z,xn)+2λn〈xn-z,Anz〉)(ϕ(z,yn)+2λn〈yn-z,Anz〉))=0.
Since (17) holds for every n∈N, it follows from (29) and lim⁡n→∞∥xn-yn∥=0 that
(30)lim⁡n→∞∥Anxn-Anz∥=0.
From the condition (v), there exists a subsequence{xni} of {xn} such that xni⇀z∈F. Since the norm of E is weakly lower semicontinuous, we get
(31)ϕ(z,x)=∥z∥2-2〈z,Jx〉+∥x∥2≤lim⁡inf⁡i→∞(∥xni∥2-2〈xni,Jx〉+∥x∥2)=lim⁡inf⁡i→∞ϕ(xni,x)=lim⁡n→∞ϕ(xn,x)≤ϕ(ΠFx,x),
which implies z=ΠFx and
(32)lim⁡n→∞ϕ(xn,x)=ϕ(ΠFx,x).
Usingxn+1=ΠCn∩Qnx, F⊂Cn∩Qn, and Lemma 5, we have
(33)0≥〈xn+1-ΠFx,Jxn+1-Jx〉=12(ϕ(ΠFx,xn+1)+ϕ(xn+1,x)-ϕ(ΠFx,x))
which implies that
(34)ϕ(ΠFx,x)≥ϕ(ΠFx,xn+1)+ϕ(xn+1,x)
for all n∈N. From (32), we get lim⁡n→∞ϕ(ΠFx,xn+1)=0 and by Lemma 4, we obtain xn→ΠFx, which is the desired result.Next, we have the following result by the hybrid method using the metric projections.Theorem 9.
Assume thatE, C, {An}, {λn}, and c1 are the same as in Theorem 8. Let x∈C and {xn} a sequence in C generated by
(35)x1=x,yn=ΠCJ-1(Jxn-λnAnxn),Cn={z∈C:ϕ(z,yn)≤ϕ(z,xn)+2λn〈xn-z,Anxn〉},Qn={z∈C:〈xn-z,J(x-xn)〉≥0},xn+1=PCn∩Qnx
for each n∈N. Then, {xn} converges strongly to PFx.Proof.
Q
n is closed convex for every n∈N. As in the proof of Theorem 8, we have that Cn is closed and convex. We also get that, for n∈N, xn∈C implies that F⊂Cn. By this fact, we obtain F⊂Cn∩Qn for every n∈N and {xn} is well defined. Indeed, x1=x∈C is given and F⊂C1∩Q1 since Q1=C. Assume that xk is well defined and F⊂Ck∩Qk for some k∈N. There exists a unique element xk+1=PCk∩Qkx and we get 〈xk+1-z,J(x-xk+1)〉≥0 for all z∈Ck∩Qk by Lemma 6. Since F⊂Ck∩Qk, we have 〈xk+1-z,J(x-xk+1)〉≥0 for every z∈F; that is, F⊂Qk+1. Since xk+1∈Ck∩Qk⊂C, we also have F⊂Ck+1. Thus, we obtain F⊂Ck+1∩Qk+1. By mathematical induction, we get F⊂Cn∩Qn for every n∈N and {xn} is well defined.
Sincexn+1=PCn∩Qnx and F⊂Cn∩Qn, we have
(36)∥xn+1-x∥≤∥x-PFx∥
for every n∈N and, hence, {xn} is bounded. Using xn+1∈Qn and Theorem 2, we have
(37)∥xn+1-x∥2-∥xn-x∥2≥2〈xn+1-xn,J(xn-x)〉+c1∥xn+1-xn∥2≥c1∥xn+1-xn∥2
for each n∈N, which implies that there exists lim⁡n→∞∥xn-x∥ and
(38)lim⁡n→∞∥xn+1-xn∥=0.
Sinceyn=ΠCJ-1(Jxn-λnAnxn), by Lemma 5, we have
(39)〈yn-xn,Jxn-Jyn-λnAnxn〉≥0,
which implies that
(40)-c1∥xn-yn∥2≥12(-ϕ(yn,xn)-ϕ(xn,yn))=〈yn-xn,Jxn-Jyn〉≥λn〈yn-xn,Anxn〉≥-λn∥yn-xn∥∥Anxn∥
from Lemma 4. As in the proof of Theorem 8, {Anxn} is bounded. Thus, we get that {∥yn-xn∥} is also bounded by the boundedness of {λn} and so is {yn}. Since xn+1∈Cn, we have ϕ(xn+1,yn)≤ϕ(xn+1,xn)+2λn〈xn-xn+1,Anxn〉; that is,
(41)ϕ(xn,yn)≤2〈xn-xn+1,Jxn-Jyn〉+2λn〈xn-xn+1,Anxn〉
for all n∈N. By the boundedness of {xn}, {yn}, {λn}, and {Anxn} with (38) and Lemma 4, we have
(42)lim⁡n→∞∥xn-yn∥=0.
As in the proof of Theorem8, using (17) and (29), we have
(43)lim⁡n→∞∥Anxn-Anz∥=0.
From the condition (v), there exists a subsequence{xni} of {xn} such that xni⇀z∈F. Since the norm of E is weakly lower semicontinuous and (36) holds, we get
(44)∥z-x∥≤lim⁡⁡inf⁡i→∞∥xni-x∥=lim⁡n→∞∥xn-x∥≤∥PFx-x∥
which implies that z=PFx and
(45)lim⁡n→∞∥xn-x∥=∥PFx-x∥.
Using the facts thatxn+1=PCn∩Qnx and F⊂Cn∩Qn and by Theorem 2, we obtain
(46)∥x-PFx∥2≥∥x-xn+1∥2+2〈xn+1-PFx,J(x-xn+1)〉+c1∥xn+1-PFx∥2≥∥x-xn+1∥2+c1∥xn+1-PFx∥2
for all n∈N. By (45), we have
(47)lim⁡n→∞∥xn+1-PFx∥=0,
which is the desired result.Remark 10.
Even though we replace the definition ofCn in Theorems 8 and 9 with
(48)Cn={{z∈C:ϕ(z,yn)≤ϕ(z,xn)}(Anz=0∀z∈F),{z∈C:ϕ(z,yn)≤ϕ(z,xn)+2λn〈xn-z,Anxn〉}(Anz≠0forsomez∈F),
the theorems are still valid.


## 4. The Variational Inequality Problem for Monotone Operators

LetI be a countable set and i:N→I a mapping. Nakajo et al. [30] propose the condition (NST) as follows: i satisfies the condition (NST) if there exists a subsequence {nk} of N such that, for any j∈I, there is Mj∈N with j∈{i(nk),i(nk+1),…,i(nk+Mj-1)} for all sufficiently large k∈N. Using the condition (NST), we get the following result for the variational inequality problem by Theorem 8.Theorem 11.
LetC be a nonempty closed convex subset of a 2-uniformly convex and uniformly smooth Banach space E, I a countable set, and {Bi}i∈I a family of operators of C into E* such that (i)
F=⋂i∈I‍VI(C,Bi)≠∅;(ii)
Bi is an inverse strongly monotone operator for each i∈I; that is, there exists {βi:i∈I}⊂]0,∞[ such that for every i∈I and x,y∈C, the inequality 〈x-y,Bix-Biy〉≥βi∥Bix-Biy∥2 holds;(iii)
for allz∈F, sup⁡i∈I∥Biz∥<∞.
Suppose that the index mappingi:N→I satisfies the condition (NST) and 0<lim⁡inf⁡n→∞βi(n)≤lim⁡⁡sup⁡n→∞βi(n)<∞. Let {λn} be a sequence in ]0,∞[ such that inf⁡n∈Nλn>0 and inf⁡n∈N(2c1βi(n)-λn)>0, where c1 is the constant in Theorem 2. Let x∈C and {xn} a sequence in C generated by
(49)x1=x,yn=ΠCJ-1(Jxn-λnBi(n)xn),Cn={z∈C:ϕ(z,yn)≤ϕ(z,xn)vvvvv+2λn〈xn-z,Bi(n)xn〉},Qn={z∈C:〈xn-z,Jx-Jxn〉≥0},xn+1=ΠCn∩Qnx
for each n∈N. Then, {xn} converges strongly to ΠFx.Proof.
We apply Theorem8 with An=Bi(n) for all n∈N. Then, the conditions (i)–(iv) are satisfied, and we will verify the condition (v). Let {zn} be a bounded sequence in C with
(50)lim⁡n→∞∥zn+1-zn∥=lim⁡n→∞∥zn-ΠCJ-1(Jzn-rnAnzn)∥=0,{rn}⊂]0,∞[ with lim⁡inf⁡n→∞rn>0, and z∈F such that lim⁡n→∞∥Anzn-Anz∥=0. By the condition (NST), there exists a weakly convergent subsequence {znk} of {zn} such that, for any i∈I, there is Mi∈N with i∈{i(nk),i(nk+1),…,i(nk+Mi-1)} for all sufficiently large k∈N. Let znk⇀u and fix i∈I. There exists jk∈{0,1,…,Mi-1} such that i(nk+jk)=i for every sufficiently large k∈N. We consider a subsequence of {nk+jk} for all k∈{k∈N:nk+jk<nk+1+jk+1} and denote it by {nk+jk} again. We have
(51)∥znk+jk-znk∥≤∑l=nknk+Mi-1‍∥zl+1-zl∥
for all k∈N, which implies that znk+jk⇀u. Let yn=ΠCJ-1(Jzn-rnAnzn). By Lemma 5, we have
(52)∥ynk+jk-y∥∥Jznk+jk-Jynk+jk∥≥〈ynk+jk-y,Jznk+jk-Jynk+jk〉≥rnk+jk〈ynk+jk-y,Ank+jkznk+jk〉
for every sufficiently large k∈N and y∈C. Since Ank+jk=Bi(nk+jk)=Bi for each sufficiently large k∈N, ynk+jk⇀u, ∥znk+jk-ynk+jk∥→0, lim⁡inf⁡n→∞rn>0, lim⁡k→∞∥Biznk+jk-Biz∥=0, and the duality mapping J is uniformly continuous on bounded subset of E, we have
(53)〈y-u,Biz〉≥0
for all y∈C. Since Bi is inverse strongly monotone, we have
(54)〈znk+jk-u,Biznk+jk-Biu〉≥βi∥Biznk+jk-Biu∥2
for every k∈N, which implies that
(55)0=〈u-u,Biz-Biu〉≥βi∥Biz-Biu∥2;
that is, Biz=Biu. From (53),
(56)〈y-u,Biu〉≥0
for all y∈C. Therefore, u∈VI⁡(C,Bi) for every i∈I; that is, u∈F. Hence, the condition (v) is satisfied. Consequently, we obtain xn→ΠFx by Theorem 8.As in the proof of Theorem11, we get the following result for the variational inequality problem by Theorem 9.Theorem 12.
Assume thatE, C, I, {Bi}, F, c1, i, {βi}, and {λn} are the same as Theorem 11. Let x∈C and {xn} a sequence in C generated by
(57)x1=x,yn=ΠCJ-1(Jxn-λnBi(n)xn),Cn={z∈C:ϕ(z,yn)≤ϕ(z,xn)vvvvvvvvvvvv+2λn〈xn-z,Bi(n)xn〉},Qn={z∈C:〈xn-z,J(x-xn)〉≥0},xn+1=PCn∩Qnx
for each n∈N. Then, {xn} converges strongly to PFx.Remark 13.
In Theorems11 and 12, under the assumption that ⋂i∈I‍Bi-10≠∅, we have VI⁡(C,Bi)=Bi-10 for all i∈I. Indeed, Bi-10⊂VI⁡(C,Bi) is trivial. Let i∈I, u∈VI⁡(C,Bi), and z∈Bi-10. From the condition (ii), we have 〈z-u,Biz-Biu〉≥βi∥Biz-Biu∥2 which implies that -〈z-u,Biu〉≥βi∥Biu∥2. On the other hand, 〈z-u,Biu〉≥0 from u∈VI⁡(C,Bi). So, we obtain u∈Bi-10; that is, VI⁡(C,Bi)⊂Bi-10. Therefore, Bi-10=VI⁡(C,Bi) for all i∈I. Now suppose that F=⋂i∈I‍Bi-10≠∅ instead of the condition (i) and Cn={z∈C:ϕ(z,yn)≤ϕ(z,xn)}. By the argument mentioned above and Remark 10, Theorems 11 and 12 hold under the conditions (i) and (ii) and we get the result of [19].Remark 14.
We know that, for a continuously Fréchet differentiable and convex functionalf on a Banach space E, if ∇f is Lipschitz continuous with constant 1/α, then ∇f is α-inverse strongly monotone operator; see [2, 19]. So, we can apply Theorems 11 and 12 and Remark 13 to such a functional; see [19].


## 5. The Proximal Point Algorithm

LetE be a strictly convex, reflexive, and smooth Banach space, T⊂E×E* a maximal monotone operator with T-10≠∅, r>0, and Ax=J(x-Jrx) for all x∈E, where Jr is the resolvent of T. Then, A is well defined as a mapping of E into E* for all r>0. We also have
(58)VI⁡(E,A)=T-10,〈x-u,Ax〉≥∥Ax∥2∀x∈E,u∈T-10.In fact,VI⁡(E,A)=T-10 since u∈T-10 is equivalent to Jru=u and Au=0. Let x∈E and u∈T-10. Since (1/r)J(x-Jrx)∈TJrx and 0∈Tu, we have 〈Jrx-u,J(x-Jrx)〉≥0 which implies that 〈x-u,Ax〉≥∥Ax∥2. By Theorem 8 and Remark 10, we get the following result using the index mapping which satisfies the condition (NST).Theorem 15.
LetI be a countable set, E a 2-uniformly convex Banach space whose norm is uniformly Gâteaux differentiable, and {Ti}i∈I a family of maximal monotone operators of E into E* such that F=⋂i∈I‍Ti-10≠∅. Let {rn} be a sequence in ]0,∞[ with lim⁡inf⁡n→∞rn>0 and {λn}n∈N a sequence in ]0,∞[ such that inf⁡n∈Nλn>0 and inf⁡n∈N(2c1-λn)>0, where c1 is the constant in Theorem 2. Let x∈E and {xn}n∈N a sequence in E generated by
(59)x1=x,yn=J-1(Jxn-λnJ(xn-JrnTi(n)xn)),Cn={z∈E:ϕ(z,yn)≤ϕ(z,xn)},Qn={z∈E:〈xn-z,Jx-Jxn〉≥0},xn+1=ΠCn∩Qnx
for each n∈N, where the index mapping i:N→I satisfies the condition (NST) and JrnTi(n) is the resolvent of Ti(n). Then, {xn} converges strongly to ΠFx.Proof.
Suppose thatAnx=J(x-JrnTi(n)x) for every n∈N and x∈E in Theorem 8. Then, we have that An is a mapping of E into E* with ⋂n∈N‍VI⁡(E,An)=F≠∅, the condition (iii) is satisfied with αn=1 for all n∈N, and the conditions (ii) and (iv) hold by Anz=0 for all n∈N and all z∈F. Let {zn} be a bounded sequence in E, z∈F, and {rn}⊂]0,∞[ with lim⁡inf⁡n→∞rn>0. Assume that
(60)lim⁡n→∞∥zn+1-zn∥=lim⁡n→∞∥zn-J-1(Jzn-rnAnzn)∥=lim⁡n→∞∥Anzn-Anz∥=0.
By the condition (NST), there exists a subsequence{znk} of {zn} such that, for any i∈I, there is Mi∈N with i∈{i(nk),i(nk+1),…,i(nk+Mi-1)} for all sufficiently large k∈N. Let znk⇀u and i∈I. As in the proof of Theorem 11, there exists jk∈{0,1,…,Mi-1} such that i(nk+jk)=i for every sufficiently large k∈N and we get znk+jk⇀u. Let (v,v*)∈Ti. Since
(61)〈Jrnk+jkTiznk+jk-v,1rnk+jkJ(znk+jk-Jrnk+jkTiznk+jk)-v*〉≥0,lim⁡k→∞∥znk+jk-Jrnk+jkTiznk+jk∥=0,
we obtain
(62)〈u-v,-v*〉≥0
for each (v,v*)∈Ti. As Ti is a maximal monotone operator, u∈Ti-10 for every i∈I. So, we get u∈F. Therefore, the condition (v) holds. So, we get conclusion by Theorem 8 and Remark 10.As in the proof of Theorem15, we get the following result from Theorem 9 and Remark 10.Theorem 16.
Assume thatE, I, {Ti}, F, {rn}, {λn}, c1, i, and JrnTi(n) are the same as Theorem 15. Let x∈E and let {xn} be a sequence in E generated by(63)x1=x,yn=J-1(Jxn-λnJ(xn-JrnTi(n)xn)),Cn={z∈E:ϕ(z,yn)≤ϕ(z,xn)},Qn={z∈E:〈xn-z,J(x-xn)〉≥0},xn+1=PCn∩Qnx
for each n∈N. Then, {xn} converges strongly to PFx.Letf:E→]-∞,∞] be a proper, lower semicontinuous, and convex function. Then, it is known that the subdifferential ∂f of f defined by
(64)∂f(x)={x*∈E*:f(y)mmmmmmvvvvvv≥f(x)+〈y-x,x*〉∀y∈E}
for all x∈E is a maximal monotone operator [31, 32]. Moreover, when E is strictly convex, reflexive, and smooth, we know that, for the resolvent of ∂f,
(65)Jr∂fx=argminy∈E⁡(f(y)+12r∥y-x∥2)
for every r>0 and x∈E and ∂f-10=argminy∈Ef(y); see [21] for more details. Now, we have the following results from Theorems 15 and 16.Theorem 17.
LetI be a countable set, E a 2-uniformly convex Banach space whose norm is uniformly Gâteaux differentiable, and {fi}i∈I a family of proper, lower semicontinuous, and convex functions of E into ]-∞,∞] such that F=⋂i∈I‍argminy∈Efi(y)≠∅. Let {rn} be a sequence in ]0,∞[ with lim⁡inf⁡n→∞rn>0 and {λn}n∈N a sequence in ]0,∞[ such that inf⁡n∈Nλn>0 and inf⁡n∈N(2c1-λn)>0, where c1 is the constant in Theorem 2. Let x∈E and let {xn}n∈N be a sequence in E generated by
(66)x1=x,un=argminy∈E⁡(fi(n)(y)+12rn∥y-xn∥2),yn=J-1(Jxn-λnJ(xn-un)),Cn={z∈E:ϕ(z,yn)≤ϕ(z,xn)},Qn={z∈E:〈xn-z,Jx-Jxn〉≥0},xn+1=ΠCn∩Qnx
for eachn∈N, where the index mapping i:N→I satisfies the condition (NST). Then, {xn} converges strongly to ΠFx.Theorem 18.
Assume thatE, I, {fi}, F, {rn}, {λn}, c1, and i are the same as Theorem 17. Let x∈E and let {xn} be a sequence in E generated by
(67)x1=x,un=argminy∈E(fi(n)(y)+12rn∥y-xn∥2)yn=J-1(Jxn-λnJ(xn-un)),Cn={z∈E:ϕ(z,yn)≤ϕ(z,xn)},Qn={z∈E:〈xn-z,J(x-xn)〉≥0},xn+1=PCn∩Qnx
for each n∈N. Then, {xn} converges strongly to PFx.At the end of this section, we make a remark about a result of the problem of image recovery due to [33, 34]. Let iC be the indicator function of a nonempty closed convex subset C of E. We know that iC:E→]-∞,∞[ is proper lower semicontinuous and convex and, for a nonempty closed convex subset C of a strictly convex, reflexive, and smooth Banach space E, (∂iC)-10=C and Jr∂iCx=PC(x) for every r>0 and x∈E; see [15]. So, when fj=iCj for nonempty closed convex subset Cj of E for every j∈I with ⋂j∈I‍Cj≠∅ in Theorems 17 and 18, we get the strong convergence to a common point of {Cj}j∈I.



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