CHAPTER 2

Generalized variational and variational-likeinclusions

2.1. Introduction

In 1979, Robinson [64] studied variational inclusion problem, that is for each

x ∈ Rn, find y ∈ Rm such that 0 ∈ g(x, y)+Q(x, y), where g : Rn×Rm → Rp

is a single-valued mapping and Q : Rn × Rm ( Rp is a set-valued mapping.

In the last decade, various classes of variational (-like) and quasi-variational

(-like) inclusions have been extensively studied and generalized in different

directions as they have potential and significant applications in optimization

theory, structural analysis and economics etc. (see for example [1, 5, 13, 26]).

Lin [51] studied generalized quasi-variational inclusion problems and applied

them to solve simultaneous equilibrium problems and optimization problems.

Huang and Fang [37] introduced generalized m-accretive mappings and

defined resolvent operator for generalized m-accretive mappings. Lan et al.

[44] introduced the concept of (A, η)-accretive mappings and their resolvent

operators.

The resolvent operator technique for solving variational inequalities and

variational inclusions is interesting and important. The resolvent operator

technique is used to establish an equivalence between variational inequalities

and resolvent equations. The resolvent equation technique is used to de-

velop powerful and efficient numerical techniques for solving various classes

of variational inequalities (inclusions) and related optimization problems.

The concept of fuzzy set theory was introduced by Zadeh [78] in 1965.

The applications of the fuzzy set theory can be found in many branches of

mathematical and engineering sciences including artificial intelligence, com-

puter science, control engineering, management science and operations re-

search [83].

Heilpern [34] introduced the concept of fuzzy mappings and proved a

fixed point theorem for fuzzy contraction mappings which is fuzzy analogue

of Nadler’s fixed point theorem for set-valued mappings. Ansari [8], Chang

and Zhu [16] first introduced the concept of variational inequalities for fuzzy

mappings, separately and extended the results of Lassonde [48], Shih and

Tan [65], Takahashi [69] and Yen [77] in the fuzzy setting.

Since then several classes of variational inequalities (inclusions) with

fuzzy mappings were considered and studied by Chang and Huang [15],

Huang [35], Park and Jeong [59, 60], Ding and Park [21], Ding [22], Wu and

Xu [75], Kumam and Petrol [41], Lee et al. [49], Lan [47], etc.. A well known

fact in mathematical programming is that variational inequality (inclusion)

problems have a close relation with optimization problems. Similarly, the

fuzzy variational inequality (inclusion) problems have a close relation with

fuzzy optimization problems.

In Section 2.2, we study a generalized quasi-variational-like inclusion

problem with (A, η)-accretive and relaxed cocoercive mappings, which is

more general problem than the problem considered by Peng [62], Lan [45]

and Bi et al. [10]. An iterative algorithm is suggested for finding the approx-

imate solutions of generalized quasi-variational-like inclusion problem with

(A, η)-accretive and relaxed cocoercive mappings. Convergence analysis is

also discussed. An example and some applications are given.

In Section 2.3, we introduce and study variational-like inclusion problem

and its resolvent equation involving infinite family of set-valued mappings.

The problem of this section is much more general problem than the problem

considered by Wang [74], Chang [12], Chang et al. [11] etc..

In Section 2.4, we consider a mixed variational inclusion problem in-

volving infinite family of fuzzy mappings. The mixed variational inclusion

problem involving infinite family of fuzzy mappings is a natural generaliza-

20

tion of problem considered by Siddiqi et al. [67], Wang [74] and includes

many other problems exist in literature. An equivalence of mixed variational

inclusions with a fixed point problem is shown. By using this equivalence an

iterative algorithm is defined for finding the solutions of mixed variational

inclusion problem involving infinite family of fuzzy mappings. An existence

and convergence result is proved for mixed variational inclusion problem in-

volving infinite family of fuzzy mappings.

The following definitions and results will be used to prove the results of

this chapter.

Definition 2.1.1. Let A : E → E, η : E × E → E be the single-valued

mappings. Then a set-valued mappingM : E → 2E is called (A, η)-accretive,

if M is m-relaxed η-accretive and (A+ ρM)(E) = E, for every ρ > 0.

Remark 2.1.1.

(i) If m = 0, then Definition 2.1.1 reduces to the definition of (H, η)-

accretive operators [23] which includes generalized m-accretive opera-

tors [37], H-accretive operators [25].

(ii) When m = 0 and E = X (Hilbert space), then Definition 2.1.1 reduces

to the definition of (H, η)-monotone operators [27, 29], which includes

classical maximal monotone operators [79].

Example 2.1.1. Let E = R, A(x) = x5, M(x) = x4, η(x, y) = (−2x4) −(−2y4), for all x, y ∈ R. ThenM is 2-relaxed η-accretive and (A+ρM)(R) =R, for all ρ > 0, that is, M is (A, η)-accretive mapping.

Definition 2.1.2[45]. Let A : E → E be a strictly η-accretive mapping and

M : E → 2E be an (A, η)-accretive mapping. Then the resolvent operator

Rρ,Aη,M : E → E is defined by

Rρ,Aη,M(u) = (A+ ρM)−1(u), for all u ∈ E.

Proposition 2.1.1[45]. Let E be a q-uniformly smooth Banach space and

η : E × E → E be τ -Lipschitz continuous, A : E → E be an r-strongly

21

η-accretive mapping and M : E → 2E be an (A, η)-accretive mapping. Then

the resolvent operator Rρ,Aη,M : E → E is τq−1

r−ρm-Lipschitz continuous, i.e.,

∥Rρ,Aη,M(u)−Rρ,A

η,M(v)∥ ≤ τ q−1

r − ρm∥u− v∥, for all u, v ∈ E,

where ρ ∈ (0, rm) is a constant.

Remark 2.1.2. If E be a real Banach space then the resolvent operator

Rρ,Aη,M is τ

r−ρm-Lipschitz continuous.

2.2. Generalized quasi-variational-like inclusions with

(A, η)-accretive and relaxed cocoercive mappings

Throughout this section, we take E to be q-uniformly smooth Banach space.

This section is devoted to study a generalized quasi-variational-like inclusion

problem involving (A, η)-accretive and relaxed cocoercive mappings. An it-

erative algorithm is constructed to approximate the solutions of generalized

quasi-variational-like inclusion problem. Finally, some applications are given.

Let N,W, η : E × E → E, g,m,A : E → E be the single-valued

mappings, B,C,D, F,G : E → 2E be the set-valued mappings. Let M :

E ×E → 2E be an (A, η)-accretive mapping in the first argument such that

g(u) − m(w) ∈ dom(M(·, u)), for all u,w ∈ E. We consider the following

generalized quasi-variational-like inclusion problem:

Find u ∈ E, x ∈ B(u), y ∈ C(u), z ∈ D(u), v ∈ F (u) and w ∈ G(u)

such that

0 ∈ N(x, y)−W (z, v) +m(w) +M(g(u)−m(w), u). (2.2.1)

Below are some special cases of problem (2.2.1).

Some special cases:

(i) Ifm = 0,M(g(u)−m(w), u) =M(g(u)) andW,D,F = 0, then problem

(2.2.1) reduces to the problem of finding u ∈ E, x ∈ B(u), y ∈ C(u)

such that

0 ∈ N(x, y) +M(g(u)). (2.2.2)

22

Problem (2.2.2) is considered by Chang [12].

(ii) If B and C are single-valued mappings, then problem (2.2.2) can be

replaced by finding u ∈ E such that

0 ∈ N(B(u), C(u)) +M(g(u)). (2.2.3)

A problem similar to problem (2.2.3) is considered by Lan [45].

(iii) If C = 0 and B, g = I, the identity mappings, then (2.2.3) reduces to

the problem of finding u ∈ E such that

0 ∈ N(u) +M(u), (2.2.4)

which is considered by Bi et al. [10].

Lemma 2.2.1. u ∈ E, x ∈ B(u), y ∈ C(u), z ∈ D(u), v ∈ F (u) and w ∈G(u) is the solution of problem (2.2.1) if and only if (u, x, y, z, v, w) satisfies

the relation:

g(u) = m(w) +Rρ,Aη,M(·,u)

[A(g(u)−m(w))− ρ(N(x, y)−W (z, v) +m(w))

],

(2.2.5)

where Rρ,Aη,M(·,u) = (A+ ρM(·, u))−1 and ρ ∈ (0, r

m) is a constant.

Proof. The proof follows directly from the Definition 2.1.2. �

Algorithm 2.2.1. For any given uo ∈ E, we choose xo ∈ B(uo), yo ∈C(uo), zo ∈ D(uo), vo ∈ F (uo), w0 ∈ G(u0) and compute {un}, {xn}, {yn},{zn}, {vn} and {wn} by the following iterative schemes:

g(un+1) = m(wn) +Rρ,Aη,M(·,un)

[A(g(un)−m(wn))

− ρ(N(xn, yn)−W (zn, vn) +m(wn))]; (2.2.6)

xn+1 ∈ B(un+1), ∥xn+1 − xn∥ ≤ H(B(un+1), B(un)); (2.2.7)

yn+1 ∈ C(un+1), ∥yn+1 − yn∥ ≤ H(C(un+1), C(un)); (2.2.8)

zn+1 ∈ D(un+1), ∥zn+1 − zn∥ ≤ H(D(un+1), D(un)); (2.2.9)

vn+1 ∈ F (un+1), ∥vn+1 − vn∥ ≤ H(F (un+1), F (un)); (2.2.10)

23

wn+1 ∈ G(un+1), ∥wn+1 − wn∥ ≤ H(G(un+1), G(un)); (2.2.11)

where n = 0, 1, 2, 3, . . ., ρ ∈ (0, rm) is a constant.

Now, we prove the following existence and convergence result for gener-

alized quasi-variational-like inclusion problem.

Theorem 2.2.1. Let E be a q-uniformly smooth Banach space and η :

E×E → E be Lipschitz continuous mapping with constant τ . Let A : E → E

be r-strongly η-accretive and Lipschitz continuous mapping with constant

λA, m : E → E be Lipschitz continuous mapping with constant λm and

M : E×E → 2E be (A, η)-accretive mapping in the first argument such that

g(u)−m(w) ∈ dom(M(·, u)), for all u,w ∈ E. Suppose N,W : E ×E → E

be Lipschitz continuous mappings in both arguments with constants λN1 ,

λN2 , λW1 and λW2 , respectively and B,C,D, F and G : E → CB(E) be H-

Lipschitz continuous mappings with constants α, β, γ, µ and δ, respectively.

Let g : E → E be (b, ξ)-relaxed cocoercive, Lipschitz continuous mapping

with constant λg and strongly accretive with constant l.

Suppose that there exist ρ ∈ (0, rm) and t > 0 such that the following

conditions hold:

∥Rρ,Aη,M(·,un)

(x)−Rρ,Aη,M(·,un−1)

(x)∥ ≤ t∥un−un−1∥, for all un, un−1 ∈ E, (2.2.12)

and

0 < λmδ(ρ+ λA) + λAq√(1− qξ + (qb+ Cq)λg

q

+ ρ q√

(λN1α + λN2β)q − (q − Cq)(λW1γ + λW2µ)

q

+ λA <[l−(λmδ+t)](r−ρm)

τq−1 , l > (λmδ + t),.

(2.2.13)

where Cq is the constant as in Lemma 1.2.1, then the iterative sequences

{un}, {xn}, {yn}, {zn},{vn} and {wn} generated by Algorithm 2.2.1 converge

strongly to u, x, y, z, v and w, respectively and (u, x, y, z, v, w) is a solution

of problem (2.2.1).

24

Proof. From Algorithm 2.2.1, Proposition 2.1.1 and (2.2.12), we have

∥g(un+1)− g(un)∥ = ∥m(wn) +Rρ,Aη,M(·,un)

[A(g(un)−m(wn))− ρ(N(xn, yn)

−W (zn, vn) +m(wn))]−{m(wn−1) +Rρ,A

η,M(·,un−1)

[A(g(un−1)−m(wn−1))

− ρ(N(xn−1, yn−1)−W (zn−1, vn−1) +m(wn−1))]}∥

≤ ∥m(wn)−m(wn−1)∥+ ∥Rρ,Aη,M(·,un)

[A(g(un)−m(wn))− ρ(N(xn, yn)

−W (zn, vn) +m(wn))]−Rρ,A

η,M(·,un)

[A(g(un−1)−m(wn−1))

− ρ(N(xn−1, yn−1)−W (zn−1, vn−1) +m(wn−1))]∥

+ ∥Rρ,Aη,M(·,un)

[A(g(un−1)−m(wn−1))− ρ(N(xn−1, yn−1)−W (zn−1, vn−1)

+m(wn−1))]−Rρ,A

η,M(·,un−1)

[A(g(un−1)−m(wn−1))− ρ(N(xn−1, yn−1)

−W (zn−1, vn−1) +m(wn−1))]∥

≤ ∥m(wn)−m(wn−1)∥+τ q−1

r − ρm

[∥A(g(un)−m(wn))− A(g(un−1)

−m(wn−1))− ρ{(N(xn, yn)−W (zn, vn) +m(wn)− (N(xn−1, yn−1)

−W (zn−1, vn−1)−m(wn−1))}∥]+ t∥un − un−1∥

≤ (1 +τ q−1ρ

r − ρm)∥m(wn)−m(wn−1)∥+

τ q−1

r − ρm∥A(g(un)−m(wn))

− A(g(un−1) +m(wn−1))∥+τ q−1

r − ρmρ∥N(xn, yn)

−N(xn−1, yn−1)− (W (zn, vn))−W (zn−1, vn−1))∥+ t∥un − un−1∥. (2.2.14)

Since A is λA-Lipschitz continuous, we have

∥g(un+1)− g(un)∥ ≤[1 +

τ q−1

r − ρm(ρ+ λA)

]∥m(wn)−m(wn−1)∥

+τ q−1

r − ρmλA∥un − un−1 − (g(un)− g(un−1))∥

+τ q−1

r − ρmρ∥N(xn, yn)−N(xn−1, yn−1)

− (W (zn, vn)−W (zn−1, vn−1))∥

+( τ q−1

r − ρmλA + t

)∥un − un−1∥. (2.2.15)

Since m is Lipschitz continuous with constant λm and G is H-Lipschitz con-

tinuous with constant δ, we have

25

∥m(wn)−m(wn−1)∥ ≤ λm∥wn − wn−1∥ ≤ λmH(G(un), G(un−1))

≤ λmδ∥un − un−1∥. (2.2.16)

Since g is (b, ξ)-relaxed cocoercive and λg-Lipschitz continuous, we have

∥un − un−1 − (g(un)− g(un−1))∥q

≤ ∥un − un−1∥q − q⟨g(un)− g(un−1),Jq(un − un−1)⟩+ Cq∥g(un)− g(un−1)∥q

≤ ∥un − un−1∥q + qb∥g(un)− g(un−1)∥q − qξ∥un − un−1∥q + Cqλqg∥un − un−1∥q

≤ ∥un − un−1∥q + qbλqg∥un − un−1∥q − qξ∥un − un−1∥q + Cqλqg∥un − un−1∥q

= (1− qξ + (qb+ Cq)λqg)∥un − un−1∥q.

Thus, we have

∥un−un−1−(g(un)−g(un−1))∥ ≤ q

√1− qξ + (qb+ Cq)λ

qg∥un−un−1∥. (2.2.17)

Also

∥N(xn, yn)−N(xn−1, yn−1)− (W (zn, vn)−W (zn−1, vn−1))∥q

≤ ∥N(xn, yn)−N(xn−1, yn−1)∥q

− (q − Cq)∥W (zn, vn)−W (zn−1, vn−1)∥q. (2.2.18)

By using Lipschitz continuity of N with constant λN1 for the first argument

and λN2 for the second argument and H-Lipschitz continuity of B and C

with constants α and β, respectively, we have

∥N(xn, yn)−N(xn−1, yn−1)∥

= ∥N(xn, yn)−N(xn, yn−1) +N(xn, yn−1)−N(xn−1, yn−1)∥

≤ ∥N(xn, yn)−N(xn, yn−1)∥+ ∥N(xn, yn−1)−N(xn−1, yn−1)∥

≤ λN2∥yn − yn−1∥+ λN1∥xn − xn−1∥

≤ λN2β∥un − un−1∥+ λN1α∥un − un−1∥

= (λN1α + λN2β)∥un − un−1∥. (2.2.19)

Thus

∥N(xn, yn)−N(xn−1, yn−1)∥q ≤ (λN1α+ λN2β)q∥un − un−1∥q. (2.2.20)

26

Using the similar arguments as for (2.2.19), we have

∥W (zn, vn)−W (zn−1, vn−1)∥q ≤ (λW1γ + λW2µ)q∥un − un−1∥q. (2.2.21)

Using (2.2.20) and (2.2.21), (2.2.18) becomes

∥N(xn, yn)−N(xn−1, yn−1)− [W (zn, vn)−W (zn−1, vn−1)]∥q

=[(λN1α + λN2β)

q − (q − Cq)(λW1γ + λW2µ)q]∥un − un−1∥q. (2.2.22)

It follows that

∥N(xn, yn)−N(xn−1, yn−1)− (W (zn, vn)−W (zn−1, vn−1))∥

≤ q

√(λN1α + λN2β)

q − (q − Cq)(λW1γ + λW2µ)q∥un − un−1∥. (2.2.23)

Combining (2.2.16), (2.2.17), (2.2.23) with (2.2.15), we obtain

∥g(un+1)− g(un)∥ ≤ (λmδ +τ q−1λmδ

r − ρm(ρ+ λA))∥un − un−1∥

+τ q−1ρ

r − ρmλA

q

√(1− qξ + (qb+ Cq)λg

q∥un − un−1∥

+τ q−1ρ

r − ρmq

√(λN1α + λN2β)

q − (q − Cq)(λW1γ + λW2µ)q∥un − un−1∥

+ (τ q−1λAr − ρm

+ t)∥un − un−1∥

=

[λmδ +

τ q−1λmδ

r − ρm(ρ+ λA) +

τ q−1ρ

r − ρmλA

q

√(1− qξ + (qb+ Cq)λg

q

+τ q−1

r − ρmρ q

√(λN1α + λN2β)

q − (q − Cq)(λW1γ + λW2µ)q

+τ q−1

r − ρmλA + t

]∥un − un−1∥. (2.2.24)

By the strong accretivity of g with constant l, we have

∥g(un+1)− g(un)∥.∥un+1 − un∥q−1 ≥ ⟨g(un+1)− g(un),Jq(un+1 − un)⟩

≥ l∥un+1 − un∥q,

which implies that

∥un+1 − un∥ ≤ 1

l∥g(un+1)− g(un)∥. (2.2.25)

27

Combining (2.2.24) and (2.2.25), we have

∥un+1 − un∥ ≤ θ∥un − un−1∥, (2.2.26)

where

θ =[λmδ +

τ q−1λmδ

r − ρm(ρ+ λA) +

τ q−1ρ

r − ρmλA

q

√(1− qξ + (qb+ Cq)λg

q

+τ q−1

r − ρmρ q

√(λN1α + λN2β)

q − (q − Cq)(λW1γ + λW2µ)q +

τ q−1

r − ρmλA + t

]/l.

By (2.2.13), we know that θ < 1 and so (2.2.26) implies that {un} is a Cauchysequence. Thus, there exists u ∈ E such that un → u as n → ∞. By the

H-Lipschitz continuity of set-valued mappings B,C,D, F and G and (2.2.6)-

(2.2.11) of Algorithm 2.2.1, it follows that xn → x, yn → y, zn → z, vn → v

and wn → w.

As A, η,M,N,W,B,C,D, F,G,m, g and Rρ,Aη,M are all continuous and by

Algorithm 2.2.1, it follows that u, x, y, z, v, w satisfy the following relation:

g(u) = m(w) +Rρ,Aη,M(·,u)

[A(g(u)−m(w))− ρ(N(x, y)−W (z, v) +m(w))

].

It follows that (u, x, y, z, v, w) is a solution of generalized quasi-variational-

like inclusion problem (2.2.1). This completes the proof. �

Applications:

(1) We show that generalized quasi-variational-like inclusion problem (2.2.1)

includes optimization problem.

If E = X, a Hilbert space, B,C,D, F,G,W,m = 0, g = I, identity

mapping, N : X → 2X is a set-valued mapping andM(u, u) =M(u) = ∂φ(·),where φ : X → R ∪ {+∞} is a proper function and ∂φ denotes the η-

subdifferential of φ. If, in addition ∂φ = δK , the indicator function of a

nonempty closed convex set K ⊂ X, then generalized quasi-variational-like

inclusion problem (2.2.1) reduces to the problem finding u ∈ K, ξ ∈ N(u)

such that

⟨ξ, η(a, u)⟩ ≥ 0, for all a ∈ K. (2.2.27)

28

Let ψ : K → R be a given function, then the optimization problem is to

find u ∈ K such that

ψ(a)− ψ(u) ≥ 0, for all a ∈ K. (2.2.28)

By using the definition of Clarke generalized subdifferential of ψ and invexity,

the equivalence of (2.2.27) and (2.2.28) can be shown easily. See for example,

Ansari and Yao [6] and references therein.

(2) Here we show that generalized quasi-variational-like inclusion problem

(2.2.1) is equivalent to an A-resolvent equation problem. We consider the

following problem:

Find s, u ∈ E, x ∈ B(u), y ∈ C(u), z ∈ D(u), v ∈ F (u) and w ∈ G(u)

such that

N(x, y)−W (z, u) +m(w) + ρ−1Jρ,Aη,M(·,u)(s) = 0, (2.2.29)

where Jρ,Aη,M(·,u) = I − A(Rρ,A

η,M(·,u)), A[Rρ,A

η,M(·,u)(s)]=

[A(Rρ,A

η,M(·,u))](s) and I

is the identity operator, Jρ,Aη,M(·,u) is the resolvent operator and ρ ∈ (0, r

m) is

a constant. We call (2.2.29) an A-resolvent equation problem, which is new

and different from those resolvent equations given in the literature.

Proposition 2.2.1. The generalized quasi-variational-like inclusion problem

(2.2.1) has a solution (u, x, y, z, v, w) with u ∈ E, x ∈ B(u), y ∈ C(u), z ∈D(u), v ∈ F (u) and w ∈ G(u), if and only if A-resolvent equation prob-

lem (2.2.29) has a solution (s, u, x, y, z, v, w) with s, u ∈ E, x ∈ B(u), y ∈C(u), z ∈ D(u), v ∈ F (u) and w ∈ G(u) where

g(u) = m(w) +Rρ,Aη,M(·,u)(s), (2.2.30)

and

s = A(g(u)−m(w))− ρ{N(x, y)−W (z, v) +m(w)}, (2.2.31)

where ρ ∈ (0, rm) is a constant.

29

Proof. Let u, x, y, z, v and w is a solution of problem (2.2.1). Then by

Lemma 2.2.1, it is the solution of following equation

g(u) = m(w)+Rρ,Aη,M(·,u)

[A(g(u)−m(w))−ρ{N(x, y)−W (z, v)+m(w)}

].(2.2.32)

Let s = A(g(u)−m(w))−ρ{N(x, y)−W (z, v)+m(w)}, then above equation

(2.2.32), becomes

g(u) = m(w) +Rρ,Aη,M(·,u)(s).

Using the fact that Jρ,Aη,M(.,u) = I − A(Rρ,A

η,M(·,u)), where A[Rρ,A

η,M(·,u)(s)]=[

A(Rρ,Aη,M(·,u))

](s), we obtain

s = A(m(w) +Rρ,Aη,M(·,u)(s)−m(w))− ρ{N(x, y)−W (z, v) +m(w)}

⇔ s− A(Rρ,Aη,M(·,u)(s)) = −ρ{N(x, y)−W (z, v) +m(w)}

⇔[I − A(Rρ,A

η,M(·,u))](s) = −ρ{N(x, y)−W (z, v) +m(w)}

⇔ Jρ,Aη,M(·,u)(s) = −ρ{N(x, y)−W (z, v) +m(w)}.

Hence N(x, y)−W (z, v)+m(w)+ρ−1Jρ,Aη,M(·,u)(s) = 0. �

2.3. Variational-like inclusions involving infinite family

of set-valued mappings

In this section, we takeE to be a real Banach space and consider a variational-

like inclusion problem involving infinite family of set-valued mappings. An

equivalence of variational-like inclusion problem involving infinite family of

set-valued mappings is shown with a fixed point problem as well as with

a resolvent equation problem. A new algorithm is constructed to prove an

existence and convergence result.

Let Ti : E → CB(E), i = 1, 2, . . . be an infinite family of set-valued

mappings and N : E∞ = E × E × E . . . → E be a nonlinear mapping. Let

W, η : E×E → E, A, g,m : E → E be single-valued mappings and B,C,D :

E → CB(E) be set-valued mappings. Suppose that M(·, ·) : E × E → 2E

be (A, η)-accretive mapping in the first argument. We consider the following

problem:

30

Find u ∈ E,wi ∈ Ti(u), i = 1, 2, . . . , a ∈ B(u), x ∈ C(u) and y ∈ D(u)

such that

0 ∈ N(w1, w2, · · · )−W (x, y) +m(a) +M(g(u)−m(a), u). (2.3.1)

The problem (2.3.1) is called variational-like inclusion problem involving in-

finite family of set-valued mappings.

Some special cases:

(i) IfW = 0,m = 0, then problem (2.3.1) reduces to the problem of finding

u ∈ E,wi ∈ Ti(u), i = 1, 2, . . ., such that

0 ∈ N(w1, w2, · · · ) +M(g(u), u). (2.3.2)

Problem (2.3.2) is introduced and studied by Y. H. Wang [74].

(ii) If W = 0, m = 0, N(·, · · · ) = N(·, ·), and M(·, ·) =M(·) then problem

(2.3.1) reduces to a problem considered by Chang [12] and Chang et

al. [11]. Find u ∈ X,w1 ∈ T1(u), w2 ∈ T2(u) such that

0 ∈ N(w1, w2) +M(g(u), u). (2.3.3)

It is now clear that for a suitable choices of mappings involved in the

formulation of problem (2.3.1), we can drive many known variational (-like)

inclusions considered and studied in the literature.

In connection with problem (2.3.1), we consider the following resolvent

equation problem. Find z, u ∈ E,wi ∈ Ti(u), i = 1, 2, . . . , a ∈ B(u), x ∈C(u), y ∈ D(u) such that

N(w1, w2, · · · )−W (x, y) +m(a) + ρ−1Jρ,Aη,M(·,u)(z) = 0, (2.3.4)

where ρ is a constant and Jρ,Aη,M(·,u) = I−A(Rρ,A

η,M(·,u)), where A[Rρ,Aη,M(·,u)(z)] =

[A(Rρ,Aη,M(·,u))](z) and I is the identity mapping. Equation (2.3.4) is called the

resolvent equation problem.

31

We mention the following equivalence between the problem (2.3.1) and

a fixed point problem, which can be easily proved by using the definition of

resolvent operator.

Lemma 2.3.1. Let (u, a, x, y, (w1, w2, · · · )) where u ∈ E,wi ∈ Ti(u), i =

1, 2, . . . , a ∈ B(u), x ∈ C(u) and y ∈ D(u) is a solution of (2.3.1) if and only

if it is a solution of the following equation

g(u) = m(a)+Rρ,Aη,M(·,u)[A(g(u)−m(a))−ρ{N(w1, w2, · · · )−W (x, y)+m(a)}].

Now we show that the problem (2.3.1) is equivalent to a resolvent equa-

tion problem.

Lemma 2.3.2. Let u ∈ E,wi ∈ Ti(u), i = 1, 2, . . . , a ∈ B(u), x ∈ C(u), y ∈D(u), then the following are equivalent:

(i) (u, a, x, y, (w1, w2, · · · )) is a solution of problem (2.3.1),

(ii) (z, u, a, x, y, (w1, w2, · · · )) is a solution of the problem (2.3.4),

where

z = A(g(u)−m(a))− ρ{N(w1, w2, · · · )−W (x, y) +m(a)} (2.3.5)

and

g(u) = m(a)+Rρ,Aη,M(·,u)[A(g(u)−m(a))−ρ{N(w1, w2, · · · )−W (x, y)+m(a)}].

Proof. Let (u, a, x, y, (w1, w2, · · · )) is a solution of the problem (2.3.1) then

by Lemma 2.3.1, it is a solution of the problem

g(u) = m(a)+Rρ,Aη,M(·,u)[A(g(u)−m(a))−ρ{N(w1, w2, · · · )−W (x, y)+m(a)}].

Using the fact that Jρ,Aη,M(·,u) = I − A(Rρ,A

η,M(·,u)), we have

Jρ,Aη,M(·,u)(z) = Jρ,A

η,M(·,u)[A(g(u)−m(a))− ρ{N(w1, w2, · · · )−W (x, y) +m(a)}

]= I − A(Rρ,A

η,M(·,u))[(A(g(u)−m(a))− ρ{N(w1, w2, · · · )

−W (x, y) +m(a)}]

32

= A(g(u)−m(a))− ρ{N(w1, w2, · · · )−W (x, y) +m(a)}

− A(Rρ,A

η,M(·,u)[A(g(u)−m(a))− ρ{N(w1, w2, · · · )

−W (x, y) +m(a)}])

= A(g(u)−m(a))− ρ{N(w1, w2, · · · )−W (x, y) +m(a)}

− A(g(u)−m(a)),

which implies that

N(w1, w2, · · · )−W (x, y) +m(a) + ρ−1Jρ,Aη,M(·,u)(z) = 0,

with

z = A(g(u)−m(a))− ρ{N(w1, w2, · · · )−W (x, y) +m(a)},

i.e., (z, u, a, x, y, (w1, w2, · · · )) is a solution of problem (2.3.4).

Conversely, let (z, u, a, x, y, (w1, w2, · · · )) is a solution of problem (2.3.4)

then

ρ{N(w1, w2, · · · )−W (x, y) +m(a)} = −Jρ,Aη,M(·,u)(z)

ρ{N(w1, w2, · · · )−W (x, y) +m(a)} = A[Rρ,Aη,M(·,u)(z)]− z. (2.3.6)

From equation (2.3.5) and (2.3.6), we have

ρ{N(w1, w2, · · · )−W (x, y) +m(a)} = A[Rρ,A

η,M(·,u)(A(g(u)−m(a))

− ρ{N(w1, w2, · · · )−W (x, y) +m(a)})]

−[A(g(u)−m(a))− ρ{N(w1, w2, · · · )

−W (x, y) +m(a)}],

which implies that

g(u) = m(a)+Rρ,Aη,M(·,u)[A(g(u)−m(a))−ρ{N(w1, w2, · · · )−W (x, y)+m(a)}].

that is, (u, a, x, y, (w1, w2, · · · )) is a solution of (2.3.1). �

We now invoke Lemma 2.3.1 and Lemma 2.3.2 to suggest the following

iterative algorithm for solving resolvent equation problem (2.3.4).

33

Algorithm 2.3.1. For a given z0, u0 ∈ E,w0i ∈ Ti(u0), i = 1, 2, . . . , a0 ∈

B(u0), x0 ∈ C(u0) and y0 ∈ D(u0). Let

z1 = A(g(u0)−m(a0))− ρ{N(w01, w

02, · · · )−W (x0, y0) +m(a0)}.

Take z1, u1 ∈ E such that

g(u1) = m(a1) +Rρ,Aη,M(·,u1)

(z1).

Since for each i, w0i ∈ Ti(u0), a0 ∈ B(u0), x0 ∈ C(u0) and y0 ∈ D(u0) by

Nadler’s theorem [56] there exist w1i ∈ Ti(u1), a1 ∈ B(u1), x1 ∈ C(u1) and

y1 ∈ D(u1) such that

∥w0i − w1

i ∥ ≤ H(Ti(u0), Ti(u1));

∥a0 − a1∥ ≤ H(B(u0), B(u1));

∥x0 − x1∥ ≤ H(C(u0), C(u1));

∥y0 − y1∥ ≤ H(D(u0), D(u1)).

Let z2 = A(g(u1)−m(a1))− ρ{N(w11, w

12, · · · )−W (x1, y1) +m(a1)},

and take any u2 ∈ E such that

g(u2) = m(a2) +Rρ,Aη,M(·,u2)

(z2).

Continuing the above process inductively, we obtain the following:

For any z0, u0 ∈ E, w0i ∈ Ti(u0), i = 1, 2, . . ., a0 ∈ B(u0), x0 ∈ C(u0)

and y0 ∈ D(u0). Compute the sequences {zn}, {un}, {wni } i = 1, 2, . . .,

{a0}, {x0}, {y0} by following iterative schemes:

(i) g(un) = m(an)+Rρ,Aη,M(.,un)

(zn); (2.3.7)

(ii) an ∈ B(un), ∥an − an+1∥ ≤ H(B(un), B(un+1)); (2.3.8)

(iii) xn ∈ C(un), ∥xn−xn+1∥ ≤ H(C(un), C(un+1)); (2.3.9)

(iv) yn ∈ D(un), ∥yn−yn+1∥ ≤ H(D(un), D(un+1)); (2.3.10)

34

(v) for each i, wni ∈ Ti(un), ∥wn

i −wn+1i ∥ ≤ H(Ti(un), Ti(un+1)); (2.3.11)

(vi) zn+1 = A(g(un)−m(an))−ρ{N(wn1 , w

n2 , · · · )−W (xn, yn)+m(an)}; (2.3.12)

where ρ > 0 is a constant and n = 0, 1, 2, . . . .

Theorem 2.3.1. Let E be a real Banach space. Let Ti, B, C,D : E →CB(E) be H-Lipschitz continuous mappings with constants δi, i = 1, 2, 3 . . .,

α, t, γ, respectively. Let N = E∞ = E × E × E . . . → E be Lipschitz con-

tinuous with constant βi and A, g,m : E → E be Lipschitz continuous with

constants λA, λg, λm, respectively and A is r-strongly η-accretive mapping.

Suppose W, η : E ×E → E be mappings such that η is Lipschitz continuous

with constant τ and W is Lipschitz continuous in both the arguments with

constants λW1 and λW2 respectively. LetM : E×E → 2E be (A, η)-accretive

mapping in first argument such that the following holds for µ > 0,

∥Rρ,ηM(·,un)

(zn)−Rρ,ηM(·,un−1)

(zn)∥ ≤ µ∥un − un−1∥. (2.3.13)

Suppose there exists a ρ > 0 such that

λAλg + λmα(λA + ρ) + ρ∞∑i=1

βiδi + ρ(λW1t+ λW2γ)

< (r−ρm)2τ

√1− (2λ2mα

2 + 4µ2 − 2k),

m < rρ, 2λ2mα

2 < 1 + 2k − 4µ2.

(2.3.14)

Then there exist z, u,∈ E, a ∈ B(E), x ∈ C(E), y ∈ D(E), wi ∈ Ti(u)

satisfying resolvent equation problem (2.3.4) and the iterative sequences {zn},{un}, {an}, {xn}, {yn} and {wn

i }, i = 1, 2, . . ., n = 0, 1, . . . generated by

Algorithm 2.3.1 converge strongly to z, u, a, x, y, wi, respectively.

Proof. From Algorithm 2.3.1, we have

∥zn+1 − zn∥ = ∥A(g(un)−m(an))− ρ{N(wn1 , w

n2 , · · · )−W (xn, yn) +m(an)}

− [A(g(un−1)−m(an−1))− ρ{N(wn−11 , wn−1

2 , · · · )

−W (xn−1, yn−1) +m(an−1)}]∥

35

≤ ∥A(g(un)−m(an))− (A(g(un−1)−m(an−1))∥

+ ρ∥N(wn1 , w

n2 , · · · )−N(wn−1

1 , wn−12 , · · · )∥

+ ρ∥W (xn, yn)−W (xn−1, yn−1)∥+ ρ∥m(an)−m(an−1)∥.(2.3.15)

By using the Lipschitz continuity of A, g and m with constants λA, λg and

λm, respectively and by Algorithm 2.3.1, we have

∥A(g(un)−m(an))− (A(g(un−1)−m(an−1)))∥

≤ λA∥g(un)− g(un−1)∥+ λA∥m(an)−m(an−1)∥

≤ λAλg∥un − un−1∥+ λAλm∥an − an−1∥

≤ λAλg∥un − un−1∥+ λAλmH(B(un), B(un−1))

≤ λAλg∥un − un−1∥+ λAλmα∥un − un−1∥

= (λAλg + λAλmα)∥un − un−1∥. (2.3.16)

SinceN is Lipschitz continuous in all the arguments with constants βi, respec-

tively and using H-Lipschitz continuity of Ti with constants δi, i = 1, 2, . . .,

we have

∥N(wn1 , w

n2 , · · · )−N(wn−1

1 , wn−12 , · · · )∥

= ∥N(wn1 , w

n2 , · · · )−N(wn−1

1 , wn2 , · · · ) +N(wn−1

1 , wn2 , · · · ) + . . . ∥

≤ ∥N(wn1 , w

n2 , · · · )−N(wn−1

1 , wn2 , · · · )∥

+ ∥N(wn−11 , wn

2 , · · · )−N(wn−11 , wn−1

2 , · · · )∥+ . . .

≤ β1∥wn1 − wn−1

1 ∥+ β2∥wn2 − wn−1

2 ∥+ . . .

≤∞∑i=1

βi∥wni − wn−1

i ∥

≤∞∑i=1

βiH(Ti(un), Ti(un−1))

≤∞∑i=1

βiδi∥un − un−1∥, n = 0, 1, 2, . . . . (2.3.17)

SinceW is Lipschitz continuous in both the arguments with constants λW1 , λW2 ,

respectively and C and D are H-Lipschitz continuous with constants t and

γ, respectively, we have

36

∥W (xn, yn)−W (xn−1, yn−1)∥ ≤ λW2∥yn − yn−1∥+ λW1∥xn − xn−1∥

≤ λW2γ∥un − un−1∥+ λW1t∥un − un−1∥

= (λW1t+ λW2γ)∥un − un−1∥. (2.3.18)

Combining (2.3.16), (2.3.17) and (2.3.18) with (2.3.15), we have

∥zn+1 − zn∥ ≤ (λAλg + λAλmα)∥un − un−1∥+ ρ

∞∑i=1

βiδi∥un − un−1∥

+ ρ(λW1t+ λW2γ)∥un − un−1∥+ ρλmα∥un − un−1∥

=[(λAλg + λmα(λA + ρ)) + ρ

∞∑i=1

βiδi

+ ρ(λW1t+ λW2γ)]∥un − un−1∥. (2.3.19)

Since ∥x + y∥2 ≤ 2(∥x∥2 + ∥y∥2) and using Proposition 1.2.1 and k-strong

accretivity of g, we have

∥un − un−1∥2 = ∥m(an) +Rρ,ηM(·,un)

(zn)−m(an−1)

−Rρ,ηM(·,un−1)

(zn−1)− [g(un)− un − (g(un−1)− un−1)]∥2

≤ ∥m(an)−m(an−1) +Rρ,ηM(·,un)

(zn)−Rρ,ηM(·,un−1)

(zn−1)∥2

− 2⟨g(un)− un − (g(un−1)− un−1), un − un−1⟩

≤ 2∥m(an)−m(an−1)∥2 + 2∥Rρ,ηM(·,un)

(zn)−Rρ,ηM(·,un−1)

(zn−1)∥2

− 2⟨g(un)− un − (g(un−1)− un−1), j(un − un−1)

⟩≤ 2∥m(an)−m(an−1)∥2 + 2∥Rρ,η

M(·,un)(zn)−Rρ,η

M(·,un−1)(zn)

+Rρ,ηM(·,un−1)

(zn)−Rρ,ηM(·,un−1)

(zn−1)∥2

− 2⟨g(un)− un − (g(un−1)− un−1), j(un − un−1)

⟩≤ 2λ2mα

2∥un − un−1∥2 + 4∥Rρ,ηM(·,un)

(zn)−Rρ,ηM(·,un−1)

(zn)∥2

+ 4∥Rρ,ηM(·,un−1)

(zn)−Rρ,ηM(·,un−1)

(zn−1)∥2

− 2⟨g(un)− un − (g(un−1)− un−1), j(un − un−1)

⟩≤ 2λ2mα

2∥un − un−1∥2 + 4µ2∥un − un−1∥2

+ 4(τ

r − ρm)2∥zn − zn−1∥2 − 2k∥un − un−1∥2.

37

It follows that

∥un − un−1∥2 ≤ (2λ2mα2 + 4µ2 − 2k)∥un − un−1∥2 + 4(

τ

r − ρm)2∥zn − zn−1∥2

∥un − un−1∥2 ≤4τ 2

(r − ρm)2(1− (2λ2mα2 + 4µ2 − 2k))

∥zn − zn−1∥2.

Thus,

∥un − un−1∥ ≤ 2τ

(r − ρm)√1− (2λ2mα

2 + 4µ2 − 2k)∥zn − zn−1∥. (2.3.20)

Using (2.3.20), (2.3.19) becomes

∥zn+1−zn∥ ≤

[λAλg + λmα(λA + ρ) + ρ

∞∑i=1

βiδi + ρ(λW1t+ λW2γ)]2τ

(r − ρm)√

1− (2λ2mα2 + 4µ2 − 2k)

∥zn−zn−1∥,

or

∥zn+1 − zn∥ ≤ θ∥zn − zn−1∥, (2.3.21)

where

θ =

[λAλg + λmα(λA + ρ) + ρ

∞∑i=1

βiδi + ρ(λW1t+ λW2γ)]2τ

(r − ρm)√

1− (2λ2mα2 + 4µ2 − 2k)

.

From (2.3.14), we have θ < 1, and consequently {zn} is a Cauchy sequence

in E. Since E is a Banach space, there exists z ∈ E such that zn → z. From

(2.3.20), we know that {un} is also a Cauchy sequence in E. Therefore,

there exists u ∈ E such that un → u. Since the mappings Ti, i = 1, 2, 3 . . .,

B,C and D are H-Lipschitz continuous, it follows from (2.3.8)-(2.3.11) of

Algorithm 2.3.1 that {an}, {xn}, {yn} and {wni } are also Cauchy sequences.

We can assume that wni → wi, an → a, xn → x and yn → y.

Now, we prove that wi ∈ Ti(u). Infact, since wni ∈ Ti(un) and

d(wi, Ti(u)) ≤ ∥wi − wni ∥+ d(wn

i , Ti(u))

≤ ∥wi − wni ∥+H(Ti(un), Ti(u))

≤ ∥wi − wni ∥+ δi∥un − u∥ → 0, (n→ ∞),

38

which implies that d(wi, Ti(u)) = 0. As Ti(u) ∈ CB(E), we have wi ∈Ti(u), i = 1, 2, . . . .

Finally, by the continuity of A, g,m,N and W and by Algorithm 2.3.1,

it follows that

zn+1 = A(g(un)−m(an))− ρ{N(wn1 , w

n2 , · · · )−W (xn, yn) +m(an)}

→ z = A(g(u)−m(a))−ρ{N(w1, w2, · · · )−W (x, y)+m(a)}, (n→ ∞).(2.3.22)

Jρ,ηM(·,un)

(zn) = g(un)−m(an) → g(u)−m(a) = Rρ,ηM(·,u)(z), (n→ ∞). (2.3.23)

From (2.3.22), (2.3.23) and Lemma 2.3.1, it follows that

N(w1, w2, · · · )−W (x, y) +m(a) + ρ−1[z − A(Rρ,ηM(·,u)(z))] = 0.

N(w1, w2, · · · )−W (x, y) +m(a) + ρ−1Jρ,ηM(·,u)(z) = 0,

that is, (z, u, a, x, y, (w1, w2, · · · )) is a solution of resolvent equation problem

(2.3.4). �

2.4. Mixed variational inclusions involving infinite

family of fuzzy mappings

In this section, we consider mixed variational inclusion problem involving

infinite family of fuzzy mappings. An iterative algorithm is suggested to find

the approximate solutions of mixed variational inclusion problem involving

infinite family of fuzzy mappings. The convergence criteria is also discussed.

Throughout this section we take E to be uniformly smooth Banach space.

Let F(E) be the collection of all fuzzy sets over E. A mapping F : E →F(E) is called a fuzzy mapping. For each u ∈ E,F (u) ( denote it by Fu, in

the sequel ) is a fuzzy set on E and Fu(y) is the membership function of y

in Fu.

Following [21], a fuzzy mapping F : E → F(E) is said to be closed if

for each x ∈ E, the function y → Fx(y) is upper semi-continuous, that is, for

any given net {yα} ⊂ E, satisfying yα → y0 ∈ E, we have

limα

supFx(yα) ≤ Fx(y0),

39

where α ∈ Γ (index set).

For B ∈ F(E) and λ ∈ [0, 1], such that (B)λ = {u ∈ E : Bu ≥ λ} is

called a λ-cut set of B.

A closed fuzzy mapping F : E → F(E) is said to satisfy the following

condition (∗): If there exists a function a : E → [0, 1] such that for each

u ∈ E, (Fu)a(u) is a non-empty bounded subset of E. It is clear that if F is

a closed fuzzy mapping satisfying condition (∗), then for each u ∈ E, the set

(Fu)a(u) ∈ CB(E).

In fact, let {yα}α∈Γ ⊂ (Fu)a(u) be a net and yα → y0 ∈ E. Then (Fu)yα ≥a(u) for each α ∈ Γ. Since F is closed, we have

(Fu)(y0) ≥ lim supα∈Γ

Fu(yα) ≥ a(u),

which implies that y0 ∈ (Fu)a(u) and so (Fu)a(u) ∈ CB(E).

Let Ti : E → F(E), i = 1, 2, . . . be an infinite family of closed fuzzy map-

pings satisfying condition (∗). Then there exist functions ai : E → [0, 1], i =

1, 2, . . . such that for each u ∈ E, we have (T1u)a1(u), (T2u)a2(u), . . . ∈ CB(E).

Therefore, we can define set-valued mappings T̃i : E → CB(E), i = 1, 2, . . .

by T̃1(u) = (T1u)a1(u), T̃2(u) = (T2u)a2(u), . . ., for all u ∈ E. In the sequel,

T̃1, T̃2, T̃3, . . ., are called the set-valued mappings induced by the fuzzy map-

pings T1 , T2, T3, . . ., respectively.

Let N : E∞ = E × E × E . . . → E, g : E → E and η : E × E → E

be three single-valued mappings such that g ̸≡ 0. Let Ti : E → F(E) be

the fuzzy mappings and ai : E → [0, 1] be given functions, i = 1, 2, . . . . Let

M : E×E → 2E be a set-valued mapping such that for each u ∈ E,M(·, u) isan (A, η)-accretive mapping in the first argument with g(u) ∈ dom M(·, u).We consider the following problem:

Find u, x1, x2, ..... ∈ E such that

(T1u)(x1) ≥ a1(u), (T2u)(x2) ≥ a2(u), . . . and

0 ∈ N(x1, x2, · · · ) +M(g(u), u).(2.4.1)

40

Problem (2.4.1) is called mixed variational inclusion problem involving infi-

nite family of fuzzy mappings.

Some special cases:

(i) If N(·, · · · ) = N(·, ·), T1 = T , T2 = F , T3 = T4 = · · · = 0, a1, a2 :

E → [0, 1], a3 = a4 = · · · = 0, η(u, v) = u − v for all u, v ∈ E

and M(·, ·) = M(·) is an m-accretive mapping. Then problem (2.4.1)

reduces to the problem of finding u,w, q ∈ E such that

Tu(w) ≥ a1(u), Fu(q) ≥ a2(u) and

0 ∈ N(w, q) +M(g(u)).(2.4.2)

Problem (2.4.2) is introduced and studied by Siddiqi et al. [67].

(ii) If Ti : E → CB(E), i = 1, 2, . . ., are classical set-valued mappings, we

can define fuzzy mappings Ti : E → F(E) by

u→ χTi(u), i = 1, 2, . . .

where χTi(u) are characteristic functions of Ti(u), i = 1, 2, . . . .

Taking ai(u) = 1, i = 1, 2, . . . for all u ∈ E, η(u, v) = u − v and

M : E × E → 2E be an m-accretive mapping then problem (2.4.1) is

equivalent to following problem considered by Y. H. Wang [74]. Find

u ∈ E, xi ∈ Ti(u), i = 1, 2, . . . such that

0 ∈ N(x1, x2, · · · ) +M(g(u), u). (2.4.3)

For suitable choices of mappings involved in the formulation of problem

(2.4.1), a number of known classes of variational inequalities (inclusions)

studied in [22, 35], can be obtained from problem (2.4.1).

Lemma 2.4.1. The (u, x1, x2, · · · ) is a solution of problem (2.4.1) if and

only if (u, x1, x2, · · · ) satisfies the following equation:

g(u) = Rρ,Aη,M(·,u)[A(g(u))− ρN(x1, x2, · · · )]. (2.4.4)

41

Proof. It can be proved easily by using the definition of resolvent operator.�

Using Lemma 2.4.1 and Nadler’s theorem [56], we define an iterative

algorithm for finding the approximate solution of problem (2.4.1).

Iterative Algorithm 2.4.1. Let Ti : E → F(E) i = 1, 2, . . . be an infinite

family of closed fuzzy mappings satisfying the condition (∗) and let T̃i : E →CB(E), i = 1, 2, . . . be an infinite family of set-valued mappings induced by

the fuzzy mappings Ti, i = 1, 2, . . ., respectively. Let N = E∞ = E × E ×E . . . → E, η : E × E → E, g, A : E → E be the single-valued mappings

such that g ̸≡ 0 and M : E × E → 2E be a set-valued mapping such that

for each u ∈ E,M(·, u) is an (A, η)-accretive mapping in the first argument

satisfying g(u) ∈ domM(·, u).

Let Q : E → 2E be a set-valued mapping such that for each u ∈E,Q(u) ⊆ g(E), where Q is defined by

Q(u) =∪

x1∈T̃1(u)

∪x2∈T̃2(u)

∪x3∈T̃3(u)

· · ·∪

· · · [A(g(u))− ρN(x1, x2, · · · )]. (2.4.5)

For given u0 ∈ E, x01 ∈ T̃1(u0), x02 ∈ T̃2(u0), . . . and let

t0 = Rρ,Aη,M(·,u0)

[A(g(u0))− ρN(x01, x02, · · · )] ⊆ Q(u0) ⊆ g(E).

Hence, there exists u1 ∈ E such that t0 = g(u1). Further, since T̃i(u0) ∈CB(E), i = 1, 2, . . . by Nadler’s [56], there exist x11 ∈ T̃1(u1), x

12 ∈ T̃2(u1), . . .

such that

∥x01 − x11∥ ≤ H(T̃1(u0), T̃1(u1));

∥x02 − x12∥ ≤ H(T̃2(u0), T̃2(u1));

.............................................

............................................. .

Let t1 = Rρ,Aη,M(·,u1)

[A(g(u1))− ρN(x11, x12, · · · )] ⊆ Q(u1) ⊆ g(E). Hence there

exists u2 ∈ E such that t1 = g(u2). Continuing the above process inductively,

we can define iterative sequences {un}, {xni }, i = 1, 2, . . . satisfying

g(un+1) = Rρ,Aη,M(·,un)

[A(g(un))− ρN(xn1 , xn2 , · · · )], (2.4.6)

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xni ∈ T̃i(un), ∥xni − xn−1i ∥ ≤ H(T̃ (un), T̃ (un−1)),

for all i = 1, 2, . . . and n = 0, 1, 2, . . . .

Theorem 2.4.1. Let E be a uniformly smooth Banach space with modulus

of smoothness ρE(t) ≤ Ct2, for C > 0. Let Ti : E → F(E), i = 1, 2, . . . be

an infinite family of closed fuzzy mappings satisfying condition (∗) and let

T̃i : E → CB(E) be the set-valued mappings induced by the fuzzy map-

pings Ti, i = 1, 2, . . . such that T̃i are H-Lipschitz continuous mappings

with constants λHi, i = 1, 2, . . . . Let N : E∞ = E × E × E . . . → E

be a Lipschitz continuous mapping in all the arguments with constants

βi, i = 1, 2, . . . and η : E × E → E be τ -Lipschitz continuous mapping.

Let g : E → E is α-relaxed Lipschitz and λg-Lipschitz continuous with

g ̸≡ 0 and let (g− I) is δ-strongly accretive, where I is the identity mapping.

Suppose A : E → E is r-strongly η-accretive and λA-Lipschitz continuous

mapping. Let M : E × E → 2E be a set-valued mapping such that for

each u ∈ E,M(·, u) is an (A, η)-accretive mapping in the first argument with

g(u) ∈ dom M(·, u) and let for each u ∈ E, Q(u) ⊆ g(E) where Q is de-

fined by (2.4.5). Suppose that there exist ρ > 0 and µ > 0 such that for each

x, y, z ∈ E,

∥Rρ,Aη,M(·,x)(z)−Rρ,A

η,M(·,y)(z)∥ ≤ µ∥x− y∥, (2.4.7)

and

λA(1+√

1− 2α + 64Cλg2)+ρ∞∑i=1

βiλHi<

(√1 + 2δ − µ)(r − ρm)

τ, (2.4.8)

where√1 + 2δ > µ and r

ρ> m.

Then the iterative sequences {un}, {xn1}, {xn2}, . . . generated by Algo-

rithm 2.4.1 converge strongly to u, x1, x2, . . ., respectively and (u, x1, x2, · · · )is a solution of problem (2.4.1).

Proof. Since (g − I) is δ-strongly accretive mapping, we have

∥un+1 − un∥2 = ∥g(un+1)− g(un) + un+1 − un − (g(un+1)− g(un))∥2

≤ ∥g(un+1)− g(un)∥2 − 2⟨(g − I)(un+1)− (g − I)(un), j(un+1 − un)⟩

≤ ∥g(un+1)− g(un)∥2 − 2δ∥un+1 − un∥2,

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which implies that

∥un+1 − un∥ ≤ 1√1 + 2δ

∥g(un+1)− g(un)∥. (2.4.9)

By Proposition 2.1.1, Algorithm 2.4.1 and (2.4.6), we have

∥g(un+1)− g(un)∥ = ∥Rρ,Aη,M(.,un)

[A(g(un))− ρN(xn1 , xn2 , · · · )]

−Rρ,Aη,M(·,un−1)

[A(g(un−1))− ρN(xn−11 , xn−1

2 , · · · )]∥

= ∥Rρ,Aη,M(·,un)

[A(g(un))− ρN(xn1 , xn2 , · · · )]

−Rρ,Aη,M(·,un)

[A(g(un−1))− ρN(xn−11 , xn−1

2 , · · · )]

+Rρ,Aη,M(·,un)

[A(g(un−1))− ρN(xn−11 , xn−1

2 , · · · )]

−Rρ,Aη,M(·,un−1)

[A(g(un−1))− ρN(xn−11 , xn−1

2 , · · · )]∥

≤ τ

r − ρm∥A(g(un))− A(g(un−1))− ρ[N(xn1 , x

n2 , · · · )

−N(xn−11 , xn−1

2 , · · · )]∥+ µ∥un − un−1∥

≤ τλAr − ρm

∥g(un)− g(un−1)∥+τρ

r − ρm∥N(xn1 , x

n2 , · · · )

−N(xn−11 , xn−1

2 , · · · )∥+ µ∥un − un−1∥

≤ τλAr − ρm

∥un − un−1 + (g(un)− g(un−1))∥+( τλAr − ρm

+ µ)∥un − un−1∥

+τρ

r − ρm∥N(xn1 , x

n2 , · · · )−N(xn−1

1 , xn−12 , · · · )∥. (2.4.10)

Since g is α-relaxed Lipschitz continuous and using Proposition 1.2.1, we

have

∥un − un−1 + (g(un)− g(un−1))∥2

≤ ∥un − un−1∥2 + 2⟨g(un)− g(un−1), j(un − un−1 + (g(un)− g(un−1)))⟩

= ∥un − un−1∥2 + 2⟨g(un)− g(un−1), j(un − un−1)⟩

+ 2⟨g(un)− g(un−1), j(un − un−1 + g(un)− g(un−1))− j(un − un−1)⟩

≤ ∥un − un−1∥2 − 2α∥un − un−1∥2 + 4D2ρE(4∥g(un)− g(un−1)∥/D)

≤ ∥un − un−1∥2 − 2α∥un − un−1∥2 + 64C∥g(un)− g(un−1)∥2

≤ (1− 2α + 64Cλ2g)∥un − un−1∥2. (2.4.11)

Since N is Lipschitz continuous in all the arguments with constant βi, i =

1, 2, . . ., respectively and using H-Lipschitz continuity of T̃i with constants

44

λHi, i = 1, 2, . . ., respectively, we have

∥N(xn1 , xn2 , · · · )−N(xn−1

1 , xn−12 , · · · )∥ ≤

∞∑i=1

βiλHi∥un − un−1∥. (2.4.12)

Combining (2.4.11) and (2.4.12), (2.4.10) becomes

∥g(un)− g(un−1)∥ ≤[ τλAr − ρm

√1− 2α + 64Cλ2g + (

τλAr − ρm

+ µ)

+τρ

r − ρm

∞∑i=1

βiλHi

]∥un − un−1∥. (2.4.13)

Using (2.4.3), (2.4.9) becomes

∥un+1 − un∥ ≤ θ ∥un − un−1∥, (2.4.14)

where

θ =1√

1 + 2δ

[ τ

r − ρm(λA

√1− 2α+ 64Cλ2g + λA + ρ

∞∑i=1

βiλHi) + µ

].

Since θ < 1 by condition (2.4.8). Therefore, (2.4.14) implies that {un} is a

Cauchy sequence in E, and hence there exists u ∈ E such that un → u, as

n→ ∞. By H-Lipschitz continuity of T̃i and Algorithm 2.4.1, we have

∥x01 − x11∥ ≤ H(T̃1(u0), T̃1(u1)) ≤ λH1∥uo − u1∥;

∥x02 − x12∥ ≤ H(T̃2(u0), T̃2(u1)) ≤ λH2∥uo − u1∥;

...........................................................................

........................................................................... .

It follows that {xn1}, {xn2}, . . . all are Cauchy sequences in E. Hence there

exist x1, x2, . . . in E such that xn1 → x1, xn2 → x2, . . . as n→ ∞. Furthermore,

since xni ∈ T̃i(un), we have

d(xi, T̃i(u)) ≤ ∥xi − xni ∥+ d(xni , T̃i(u))

≤ ∥xi − xni ∥+H(T̃i(un), T̃i(u))

≤ ∥xi − xni ∥+ λHi∥un − u∥ → 0, as n→ ∞.

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Hence d(xi, T̃i(u)) = 0. Since T̃i(u) ∈ CB(E), it follows that xi ∈ T̃i(u),

for i = 1, 2, . . . . Hence, we have (T1u)(x1) ≥ a1(u), (T2u)(x2) ≥ a2(u), . . . .

Thus from Algorithm 2.4.1, we have

g(u) = Rρ,AM(·,u)[A(g(u))− ρN(x1, x2, · · · )].

By Lemma 2.4.1, it follows that (u, x1, x2, · · · ) is a solution of problem (2.4.1).

�

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