Fuzzy nonlinear set-valued variational inclusions

(1)

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Fuzzy nonlinear set-valued variational inclusions

I

Byung Soo Lee

a,∗

, M. Firdosh Khan

b

, Salahuddin

c a_{Department of Mathematics, Kyungsung University, Busan 608-736, Republic of Korea} b_{S.S. School (Boys) Aligarh Muslim University, Aligarh-202002, India}

c_{Department of Mathematics, Aligarh Muslim University, Aligarh-202002, India}

a r t i c l e i n f o

Article history:

Received 4 March 2010

Received in revised form 7 July 2010 Accepted 7 July 2010

Keywords:

Fuzzy nonlinear set-valued variational inclusions

Fuzzy resolvent operator equation problem Nadler’s theorem

Hausdorff metric Closed fuzzy mapping

a b s t r a c t

The purpose of this paper is to study a new class of fuzzy nonlinear set-valued variational inclusions in real Banach spaces. By using the fuzzy resolvent operator techniques for m-accretive mappings, we establish the equivalence between fuzzy nonlinear set-valued variational inclusions and fuzzy resolvent operator equation problem. Applying this equivalence and Nadler’s theorem, we suggest some iterative algorithms for solving fuzzy nonlinear set-valued variational inclusions in real Banach spaces. By using the inequality of Petryshyn, the existence of solutions for these kinds of fuzzy nonlinear set-valued variational inclusions without compactness is proved and convergence criteria of iterative sequences generated by the algorithm are also discussed.

1. Introduction and preliminaries

Recently, there have been many research on the variational inequality problems and complementarity problems for fuzzy mappings [1–13]. In particular, Chang [6] introduced fuzzy quasivariational inclusions and established existence problem and iterative approximation problem of solutions for such inclusions by using the resolvent operator technique, Nadler’s fixed point theorem and analytic techniques in Banach spaces. In 2004, Liu et al. [12] introduced fuzzy mixed quasivariational inclusions and some mixed resolvent equations and established some existence theorems of solutions for the inclusions in Hilbert spaces.

In this paper, we introduce a class of fuzzy nonlinear set-valued variational inclusions in real Banach spaces. By using the fuzzy resolvent operator techniques form-accretive mappings, we establish the equivalence between fuzzy nonlinear set-valued variational inclusions and fuzzy resolvent operator equation problem. Applying this equivalence and Nadler’s theorem, we suggest some iterative algorithms for solving fuzzy nonlinear set-valued variational inclusions in real Banach spaces. By using the inequality of Petryshyn [14,15], the existence of solutions for the inclusions without compactness is proved and convergence criteria of iterative sequences generated by the algorithm are also discussed.

A mappingT from a setXto the collectionF

(

X

)

= {

B

:

X

→ [

0

,

1

]

a function

}

of fuzzy sets overXis called a fuzzy mapping, which means that for eachx

∈

Xa fuzzy setT

(

x

)

, denoted byTx, is a function fromXto

[

0

,

1

]

. For eachy

∈

X, Tx

(

y

)

denotes the membership grade ofyinTx.

A fuzzy mappingT

:

X

→

F

(

X

)

is said to be closed if for eachx

∈

X, the functiony

→

Tx

(

y

)

is upper semicontinuous,

that is, for any given net

{

y_α

} ⊂

X, satisfyingy_α

→

y0

∈

X, we have

lim sup

α Tx

(

yα

)

≤

Tx

(

y0

).

I_{This research was supported by Kyungsung University Research Grants in 2010.} ∗_{Corresponding author. Tel.: +82 51 66 4613; fax: +82 51 622 7510.}

(2)

ForB

∈

F

(

X

)

and

λ

∈ [

0

,

1

]

, the set

(

B

)

_λ

= {

x

∈

X

:

B

(

x

)

≥

λ

}

is called a

λ

-cut set ofB. LetT

:

X

→

F

(

X

)

be a closed fuzzy mapping satisfying the following condition:

Condition

(

∗

)

. If there exists a functiona

:

X

→ [

0

,

1

]

such that for eachx

∈

X, the set

(

Tx

)

a(x)

= {

y

∈

X

:

Tx

(

y

)

≥

a

(

x

)

}

is

a nonempty bounded subset ofX.

Remark 1.1. LetXbe a normed vector space. IfTis a closed fuzzy mapping satisfying the condition

(

∗

)

, then for eachx

∈

X, the set

(

Tx

)

a(x)belongs to the collectionCB

(

X

)

of all nonempty closed and bounded subsets ofX. In fact, let

{

yα

} ⊂

(

Tx

)

a(x)

be a net andy_α

→

y0

∈

X, then

(

Tx

)(

yα

)

≥

a

(

x

)

for each

α

. SinceTis closed, we have Tx

(

y0

)

≥

lim sup

α Tx

(

yα

)

≥

a

(

x

),

which implies thaty0

∈

(

Tx

)

a(x)and so

(

Tx

)

a(x)

∈

CB

(

X

)

.

LetJ

:

X

→

2X∗be the normalized duality mapping defined by

J

(

x

)

= {

f

∈

X∗

: h

x

,

f

i = k

x

k k

f

k

,

k

f

k = k

x

k}

forx

∈

X

,

where

h·

,

·i

is the dual pairing betweenXand its dualX∗.

IfX∗is uniformly convex, thenJis single-valued and uniformly continuous on a bounded subset ofX. We denote a single-valued normalized duality mapping byj.

LetXbe a real Banach space. LetT

,

F

,

EandP

:

X

→

F

(

X

)

be closed fuzzy mappings satisfying the condition

(

∗

)

with functionsa

,

b

,

cande

:

X

→ [

0

,

1

]

, respectively, andG

,

g

:

X

→

Xbe surjective single-valued mappings. LetA

:

X

×

X

→

2X be anm-accretive mapping with respect to the first argument. For a given nonlinear mappingN

:

X

×

X

→

X, we consider a problem of findingx

,

u

, v, w,

q

∈

Xsuch that

Tx

(

u

)

≥

a

(

x

),

Fx

(v)

≥

b

(

x

),

Ex

(w)

≥

c

(

x

),

Px

(

q

)

≥

e

(

x

),

i.e.,u

∈

(

Tx

)

a(x)

, v

∈

(

Fx

)

b(x)

, w

∈

(

Ex

)

c(x)

,

q

∈

(

Px

)

e(x),

0

∈

G

(w)

−

N

(

u

, v)

+

A

(

g

(

x

),

q

).

(1.1)

Problem(1.1)is called a fuzzy nonlinear variational inclusion.

Problem(1.1)includes many types of variational inclusions, variational inequalities and complementarity problems for fuzzy mappings [1,5–7,12].

Puttinga

(

x

)

=

b

(

x

)

=

c

(

x

)

=

e

(

x

)

=

1 for allx

∈

X, problem(1.1)includes many kinds of variational inclusions, variational inequalities and complementarity problems [16–21].

Definition 1.1. LetA

:

D

(

A

)

⊂

X

→

2Xbe a set-valued mapping, whereD

(

A

)

is the domain ofA.

(i) Ais said to be accretive if for anyx

,

y

∈

D

(

A

),

u

∈

A

(

x

), v

∈

A

(

y

)

there exists aj

(

x

−

y

)

∈

J

(

x

−

y

)

such that

h

u

−

v,

j

(

x

−

y

)

i ≥

0

.

(ii)Ais said to bem-accretive ifAis accretive and

(

I

+

ρ

A

)(

D

(

A

))

=

Xfor every (equivalently, for some)

ρ >

0, whereIis the identity mapping.

Definition 1.2 ([22]).LetA

:

D

(

A

)

⊂

X

→

2Xbe anm-accretive mapping. For any given

η >

0, the mappingJA

:

X

→

D

(

A

)

associated withAdefined by

JA

(

x

)

=

(

I

+

η

A

)

−1

(

x

)

forx

∈

X

,

is called the resolvent operator ofA.

Remark 1.2. Barbu [22] pointed out that ifAis anm-accretive mapping, then for every

η >

0 the operator

(

I

+

η

A

)

−1is well-defined, single-valued and nonexpansive on the rangeR

(

I

+

η

A

)

, i.e.,

k

JA

(

x

)

−

JA

(

y

)

k ≤ k

x

−

y

k

forx,y

∈

R

(

I

+

η

A

).

FromRemark 1.2, we have the following result.

Proposition 1.1. Let A

:

D

(

A

)

⊂

X

×

X

→

2X be an m-accretive mapping with respect to the first argument. For a constant

η >

0we get

JA(·,z)

=

(

I

+

η

A

(

· ,

z

))

−1 for z

∈

X

.

Then for any given z

∈

X , the resolvent operator JA(·,z)

:

X

→

D

(

A

)(

⊂

X

×

X

)

is well-defined, single-valued and nonexpansive, that is,

(3)

Definition 1.3. LetT

:

X

→

F

(

X

)

be a closed fuzzy mapping satisfying the condition

(

∗

)

with a functiona

:

X

→ [

0

,

1

]

.T

is said to be

ξ

-Lipschitz continuous if for anyx

,

y

∈

X H

((

Tx

)

a(x)

, (

Ty

)

a(y)

)

≤

ξ

k

x

−

y

k

,

where

ξ >

0 is a constant andHis the Hausdorff metric onCB

(

X

)

.

Definition 1.4. LetXbe a real Banach space andg

:

X

→

Xa single-valued mapping; (i) gis said to be accretive if

h

g

(

x

)

−

g

(

y

),

j

(

x

−

y

)

i ≥

0 forj

(

x

−

y

)

∈

J

(

x

−

y

)

.

(ii) gis said to be

γ

-strongly accretive if there exists a constant

γ >

0 such that

h

g

(

x

)

−

g

(

y

),

j

(

x

−

y

)

i ≥

γ

k

x

−

y

k

2 forx

,

y

∈

X

,

j

(

x

−

y

)

∈

J

(

x

−

y

).

Related to the fuzzy nonlinear variational inclusion problem(1.1)we consider the following fuzzy resolvent operator equation problem:

Findx

,

z

,

u

, v, w,

q

∈

Xsuch that

(

Tx

)(

u

)

≥

a

(

x

),

(

Fx

)(v)

≥

b

(

x

),

(

Ex

)(w)

≥

c

(

x

),

(

Px

)(

q

)

≥

e

(

x

),

G

(w)

+

η

−1RA(·_,q)

(

z

)

=

N

(

u

, v),

(1.2)

where

η >

0 is a constant andRA(·,q)

=

(

I

−

JA(·,q)

)

.

The following known lemma plays an important role in proving our main results.

Lemma 1.1 ([14,15]).Let X be a real Banach space and J

:

X

→

2X∗be the normalized duality mapping. Then for any x

,

y

∈

X

k

x

+

y

k

2

≤ k

x

k

2

+

2

h

y

,

j

(

x

+

y

)

i

for j

(

x

+

y

)

∈

J

(

x

+

y

).

2. Main results

Now we establish the equivalence between fuzzy nonlinear set-valued variational inclusions and fuzzy resolvent operator equation problem.

Theorem 2.1. Let X be a Banach space. Assume that T

,

F

,

E

,

P

:

X

→

F

(

X

)

are fuzzy mappings satisfying the condition

(

∗

)

, with functions a

,

bc and e

:

X

→ [

0

,

1

]

, respectively. Let G, g

:

X

→

X and N

:

X

×

X

→

X be single-valued mappings. Then the followings are equivalent:

(i)

(

x

,

u

, v, w,

q

)

, where x

∈

X

, (

Tx

)(

u

)

≥

a

(

x

), (

Fx

)(v)

≥

b

(

x

), (

Ex

)(w)

≥

c

(

x

), (

Px

)(

q

)

≥

e

(

x

)

is a solution of problem(1.1),

(ii) x

∈

X is a fixed point of a mapping S

:

X

→

2X_{defined by}

S

(

x

)

=

G

(w)

−

N

(

u

, v)

+

A

(

g

(

x

),

q

)

+

x

,

(2.1)

(iii)

(

x

,

u

, v, w,

q

)

, where x

∈

X

, (

Tx

)(

u

)

≥

a

(

x

), (

Fx

)(v)

≥

b

(

x

), (

Ex

)(w)

≥

c

(

x

), (

Px

)(

q

)

≥

e

(

x

)

, is a solution of the following equation

g

(

x

)

=

JA(·,q)

[

g

(

x

)

−

η(

G

(w)

−

N

(

u

, v))

]

,

(2.2)

(iv)

(

x

,

z

,

u

, v, w,

q

)

, where x

,

z

∈

X

, (

Tx

)(

u

)

≥

a

(

x

), (

Fx

)(v)

≥

b

(

x

), (

Ex

)(w)

≥

c

(

x

), (

Px

)(

q

)

≥

e

(

x

)

, is a solution of problem (1.2), where

z

=

g

(

x

)

−

η(

G

(w)

−

N

(

u

, v)),

g

(

x

)

=

JA(·_,q)

(

z

).

(2.3)

Proof. (i)

⇒

(ii) Addingxto both sides of(1.1), we have 0

∈

G

(w)

−

N

(

u

, v)

+

A

(

g

(

x

),

q

)

⇒

x

∈

G

(w)

−

N

(

u

, v)

+

A

(

g

(

x

),

q

)

+

x

=

S

(

x

).

(ii)

⇒

(iii) Letxbe a fixed point ofS, then

x

∈

G

(w)

−

N

(

u

, v)

+

A

(

g

(

x

),

q

)

+

x

⇒

0

∈

G

(w)

−

N

(

u

, v)

+

A

(

g

(

x

),

q

)

⇒

0

∈

η(

G

(w)

−

N

(

u

, v))

+

η

A

(

g

(

x

),

q

)

⇒

0

∈ −

(

g

(

x

)

−

η(

G

(w)

−

N

(

u

, v)))

+

g

(

x

)

+

η

A

(

g

(

x

),

q

)

⇒

0

∈ −

(

g

(

x

)

−

η(

G

(w)

−

N

(

u

, v)))

+

(

I

+

η

A

(

· ,

q

))(

g

(

x

)).

Henceg

(

x

)

=

JA(·,q)

[

g

(

x

)

−

η(

G

(w)

−

N

(

u

, v))

]

.

(iii)

⇒

(iv) Takingz

=

g

(

x

)

−

η(

G

(w)

−

N

(

u

, v))

, from(2.2)we haveg

(

x

)

=

JA(·,q)

(

z

)

so, z

=

JA(·,q)

(

z

)

−

η(

G

(w)

−

N

(

u

, v)),

(4)

which implies that

(

I

−

JA(·,q)

)(

z

)

= −

η(

G

(w)

−

N

(

u

, v))

η

−1

₍

_I

₋

_J A(·,q)

)(

z

)

= −

G

(w)

+

N

(

u

, v)

⇒

G

(w)

+

η

−1

(

I

−

JA(·_,q)

)(

z

)

=

N

(

u

, v)

⇒

G

(w)

+

η

−1RA(·,q)

(

z

)

=

N

(

u

, v).

Consequently,

(

x

,

z

,

u

, v, w,

q

)

is a solution of the fuzzy resolvent operator equation problem(1.2). (iv)

⇒

(i), From(2.3)we have

g

(

x

)

=

JA(·,q)

(

z

)

=

JA(·,q)

[

g

(

x

)

−

η(

G

(w)

−

N

(

u

, v))

]

,

i.e., g

(

x

)

=

(

I

+

η

A

(

· ,

q

))

−1

[

g

(

x

)

−

η(

G

(w)

−

N

(

u

, v))

]

so

[

g

(

x

)

−

η(

G

(w)

−

N

(

u

, v))

] ∈ [

I

+

η

A

(

· ,

q

)

]

(

g

(

x

)),

which implies 0

∈

G

(w)

−

N

(

u

, v)

+

A

(

g

(

x

),

q

).

Therefore

(

x

,

u

, v, w,

q

)

, wherex

∈

X

, (

Tx

)(

u

)

≥

a

(

x

), (

Fx

)(v)

≥

b

(

x

), (

Ex

)(w)

≥

c

(

x

), (

Px

)(

q

)

≥

e

(

x

)

, is a solution

of(1.1).

We now invokeLemma 1.1and(2.3)to suggest the following algorithm for solving problem(1.1)in Banach spaces. Algorithm 2.1.We assume thatgis surjective. For any givenx0

,

z0

∈

X

,

u0

∈

(

Tx0

)

a(x0)

, v

0

∈

(

Fx0

)

b(x0)

, w

0

∈

(

Ex0

)

c(x0)

,

q0

∈

(

Px0

)

e(x0), let

z1

=

g

(

x0

)

−

η(

G

(w

0

)

−

N

(

u0

, v

0

)).

Sincegis surjective, there existsx1

∈

Xsuch that

g

(

x1

)

=

JA(·,q0)

(

z1

).

On the other hand, by Nadler [23], there existu1

∈

(

Tx1

)

a(x1)

, v

1

∈

(

Fx1

)

b(x1)

, w

1

∈

(

Ex1

)

c(x1)andq1

∈

(

Px1)e(x1)such that

k

u0

−

u1

k ≤

(

1

+

1

)

H

(

Tx0)a(x0)

, (

Tx1

)

a(x1)

,

k

v

₀

−

v

₁

k ≤

(

1

+

1

)

H

(

Fx0

)

b(x0)

, (

Fx1

)

b(x1)

,

k

w

₀

−

w

₁

k ≤

(

1

+

1

)

H

(

Ex0)c(x0)

, (

Ex1

)

c(x1)

,

k

q0

−

q1

k ≤

(

1

+

1

)

H

(

Px0

)

e(x0)

, (

Px1

)

e(x1)

.

Let z2

=

g

(

x1

)

−

η(

G

(w

1

)

−

N

(

u1

, v

1

)).

Again by the surjectivity ofg, there existsx2

∈

Xsuch that

g

(

x2

)

=

JA(·,q1)

(

z2

).

Similarly, there existu2

∈

(

Tx2

)

a(x2)

, v

2

∈

(

Fx2

)

b(x2)

, w

2

∈

(

Ex2

)

c(x2)andq2

∈

(

Px2

)

e(x2)such that

k

u1

−

u2

k ≤

1

+

1 2

H

(

Tx1

)

a(x1)

, (

Tx2

)

a(x2)

,

k

v

₁

−

v

₂

k ≤

1

+

1 2

H

(

Fx1

)

b(x1)

, (

Fx2

)

b(x2)

,

k

w

₁

−

w

₂

k ≤

1

+

1 2

H

(

Ex1

)

c(x1)

, (

Ex2

)

c(x2)

,

k

q₁

−

q₂

k ≤

1

+

1 2

H

(

P_x₁

)

_e₍_x₁₎

, (

P_x₂

)

_e₍_x₂₎

.

(5)

(i) un

∈

(

Txn

)

a(xn)

k

un

−

un−1

k ≤

1

+

1 n

H

(

Txn

)

a(xn)

, (

Txn−1)a(xn−1)

,

(ii)

v

n

∈

(

Fxn

)

b(xn)

k

v

_n

−

v

_n−1

k ≤

1

+

1 n

H

(

Fxn

)

b(xn)

, (

Fxn−1)b(xn−1)

,

(iii)

w

n

∈

(

Exn

)

c(xn)

k

w

_n

−

w

_n−1

k ≤

1

+

1 n

H

(

Exn

)

c(xn)

, (

Exn−1

)

c(xn−1)

,

(iv) qn

∈

(

Pxn

)

e(xn)

k

qn

−

qn−1

k ≤

1

+

1 n

H

(

Pxn

)

e(xn)

, (

Pxn−1

)

e(xn−1)

,

(v) zn+1

=

g

(

xn

)

−

η(

G

(w

n

)

−

N

(

un

, v

n

)),

(2.4) (vi) g

(

xn+1

)

=

JA(·,qn)

(

zn+1

).

(2.5)

Now we consider our main result.

Theorem 2.2. Let X be a real Banach space. Let T , F , E and P

:

X

→

F

(

X

)

be closed fuzzy mappings satisfying the condition

(

∗

)

with functions a

,

b

,

c and e

:

X

→ [

0

,

1

]

, respectively. Let N

:

X

×

X

→

X be continuous, G

,

g

:

X

→

X surjective and A

:

X

×

X

→

2X_{m-accretive with respect to the first argument.}

Suppose that the following conditions hold;

(i) g is

β

-Lipschitz continuous and

γ

-strongly accretive,0

< β <

1

, γ

∈

(

0

,

1

)

\ {

1

2

}

,

(ii) T

,

F

,

E and P are Lipschitz continuous with their Lipschitz constants

ξ, α, ζ

and

σ

, respectively,

(iii) G is

-Lipschitz continuous,

(iv) for any given y

∈

X , a mapping x

→

N

(

x

,

y

)

is

λ

-Lipschitz continuous,

(v) for any given x

∈

X , a mapping y

→

N

(

x

,

y

)

is

µ

-Lipschitz continuous. If the following conditions

(

a

)

k

JA(·_,x)

(

z

)

−

JA(·_,y)

(

z

)

k ≤

κ

k

x

−

y

k

,

for x

,

y

,

z

∈

X ,

κ >

0

,

(2.6)

(

b

)

"

1

−

1

+

2

γ

c

+

p

c2

₊

₄

β

2 2

+

2

κσ

2

!#

1₂

<

1 (2.7) for t

=

2

ζ

2

+

λ

2

ξ

2

+

µ

2

α

2

+

2

λξµα >

0and c

=

2

√

2

η

t

are satisfied, then there exist x

,

z

∈

X

,

u

∈

(

Tx

)

a(x)

, v

∈

(

Fx

)

b(x)

, w

∈

(

Ex

)

c(x)

,

q

∈

(

Px

)

e(x)satisfying the fuzzy resolvent operator equation(1.2)and so

(

x

,

u

, v, w,

q

)

is a solutions of the fuzzy nonlinear variational inclusions(1.1)and the iterative sequences

{

xn

}

,

{

un

}

,

{

v

n

}

,

{

w

n

}

and

{

qn

}

generated byAlgorithm2.1converge strongly x

,

u

, v, w

and q in X , respectively.

Proof. By the

λ

-Lipschitz continuity ofNwith respect to the first argument and the

µ

-Lipschitz continuity with respect to the second, the

ξ

-Lipschitz continuity ofT, the

α

-Lipschitz continuity ofF, the

ζ

-Lipschitz continuity ofEand the

σ

-Lipschitz continuity ofP, we have

k

N

(

un

, v

n

)

−

N

(

un−1

, v

n−1

)

k ≤

λ

k

un

−

un−1

k +

µ

k

v

n

−

v

n−1

k

≤

λ

1

+

1 n

H

(

T_x_n

)

_a₍_x_n₎

, (

T_x_n₋₁

)

_a₍_x_n₋₁₎

+

µ

1

+

1 n

H

(

F_x_n

)

_b₍_x_n₎

, (

F_x_n₋₁

)

_b₍_x_n₋₁₎

≤

λξ

1

+

1 n

k

xn

−

xn−1

k +

µα

1

+

1 n

k

xn

−

xn−1

k

≤

(λξ

+

µα)

1

+

1 n

k

xn

−

xn−1

k

,

(2.8)

k

w

_n

−

w

_n−1

k ≤

1

+

1 n

H

(

Exn

)

c(xn)

, (

Exn−1

)

c(xn−1)

≤

ζ

1

+

1 n

k

xn

−

xn−1

k

,

(2.9)

(6)

k

qn

−

qn−1

k ≤

1

+

1 n

H

(

Pxn

)

e(xn)

, (

Pxn−1)e(xn−1)

≤

σ

1

+

1 n

k

xn

−

xn−1

k

.

(2.10)

From(2.4)byLemma 1.1, for anyj

(

zn+1

−

zn

)

∈

J

(

zn+1

−

zn

)

we have

k

zn+1

−

zn

k

2

= k

g

(

xn

)

−

η(

G

(w

n

)

−

N

(

un

, v

n

))

−

g

(

xn−1

)

+

η(

G

(w

n−1

)

−

N

(

un−1

, v

n−1

))

k

2

= k

g

(

xn

)

−

g

(

xn−1

)

−

η(

G

(w

n

)

−

G

(w

n−1

)

−

N

(

un

, v

n

)

+

N

(

un−1

, v

n−1

))

k

2

≤ k

g

(

xn

)

−

g

(

xn−1

)

k

2

−

2

η

h

G

(w

n

)

−

G

(w

n−1

)

−

(

N

(

un

, v

n

)

−

N

(

un−1

, v

n−1

)),

j

(

zn+1

−

zn

)

i

≤

β

2

k

xn

−

xn−1

k

2

+

2

η

k

G

(w

n

)

−

G

(w

n−1

)

−

(

N

(

un

, v

n

)

−

N

(

un−1

, v

n−1

))

kk

zn+1

−

zn

k

.

(2.11)

On the other hand, since

k

x

+

y

k

2

≤

2

(

k

x

k

2

+ k

y

k

2

)

forx

,

y

∈

X

,

by(2.8),(2.9)and the

-Lipschitz continuity ofG,

k

G

(w

n

)

−

G

(w

n−1

)

+

N

(

un−1

, v

n−1

)

−

N

(

un

, v

n

)

k

2

≤

2

k

G

(w

n

)

−

G

(w

n−1

)

k

2

+

2

k

N

(

un

, v

n

)

−

N

(

un−1

, v

n−1

)

k

2

≤

2

ζ

2

1

+

1 n

2

k

xn

−

xn−1

k

2

+

2

λ

2

ξ

2

+

4

λξµα

+

2

µ

2

α

2

1

+

1 n

2

k

xn

−

xn−1

k

2

=

2

1

+

1 n

2

_ζ

2

₊

_λ

2

_ξ

2

₊

_µ

2

_α

2

₊

₂

_λξµα

2

_k

_x n

−

xn−1

k

2

.

(2.12)

From(2.11)and(2.12), we get

k

zn+1

−

zn

k

2

≤

β

2

k

xn

−

xn−1

k

2

+

2

√

2

η

1

+

1 n

2

_ζ

2

₊

_λ

2

_ξ

2

₊

_µ

2

_α

2

₊

₂

_λξµα

_k

_x n

−

xn−1

k k

zn+1

−

zn

k

.

Hence

k

zn+1

−

zn

k ≤

bn

+

p

b2 n

+

4

β

2 2

k

xn

−

xn−1

k

,

(2.13) where bn

=

2

√

2

η

1

+

1 n

t

(

n

∈

_N

)

andt

=

2

ζ

2

+

λ

2

ξ

2

+

µ

2

α

2

+

2

λξµα.

On the other hand, sincegis

γ

-strongly accretive, byLemma 1.1, from(2.5)

for anyj

(

xn

−

xn−1

)

∈

J

(

xn

−

xn−1

),

k

xn

−

xn−1

k

2

= k

JA(·,qn−1)

(

zn

)

−

JA(·,qn−2)

(

zn−1

)

− [

g

(

xn

)

−

xn

−

(

g

(

xn−1

)

−

xn−1

)

]k

2

≤ k

JA(·,qn−1)

(

zn

)

−

JA(·,qn−2)

(

zn−1

)

k

2

₋

₂

_h

_g

₍

_x n

)

−

xn

−

(

g

(

xn−1

)

−

xn−1

),

j

(

xn

−

xn−1

)

i

= k

JA(·,qn−1)

(

zn

)

−

JA(·,qn−2)

(

zn−1

)

k

2

₋

₂

_h

_g

₍

_x n

)

−

g

(

xn−1

),

j

(

xn

−

xn−1

)

i +

2

h

xn

−

xn−1

,

j

(

xn

−

xn−1

)

i

≤ k

JA(·,qn−1)

(

zn

)

−

JA(·,qn−2)

(

zn−1

)

k

2

₊

₂

₍

₁

₋

_{γ )}

_k

_x n

−

xn−1

k

2

.

(2.14)

On the other hand, from(2.6)and(2.10)byProposition 1.1, we obtain

k

JA(·,qn−1)

(

zn

)

−

JA(·,qn−2)

(

zn−1

)

k

2

_{≤ k}

_J A(·,qn−1)

(

zn

)

−

JA(·,qn−1)

(

zn−1

)

+

JA(·,qn−1)

(

zn−1

)

−

JA(·,qn−2)

(

zn−1

)

k

2

≤

2

k

JA(·_,qn−1)

(

zn

)

−

JA(·,qn−1)

(

zn−1

)

k

2

₊

₂

_k

_J A(·_,qn−1)

(

zn−1

)

−

JA(·,qn−2)

(

zn−1

)

k

2

≤

2

k

zn

−

zn−1

k

2

+

2

κ

k

qn−1

−

qn−2

k

2

≤

2

k

zn

−

zn−1

k

2

+

2

κσ

2

1

+

1 n

−

1

2

k

xn−1

−

xn−2

k

2

.

(2.15)

From(2.14)and(2.15), we obtain

k

xn

−

xn−1

k

2

≤

2

k

zn

−

zn−1

k

2

+

2

(

1

−

γ )

k

xn

−

xn−1

k

2

+

2

κσ

2

1

+

1 n

−

1

2

k

xn−1

−

xn−2

k

2

.

(7)

Thus, from(2.13)

k

xn

−

xn−1

k

2

≤

2

−

1

+

2

γ

k

zn

−

zn−1

k

2

₊

2

κσ

2 ₁

₊

1 n−1

2

−

1

+

2

γ

k

xn−1

−

xn−2

k

2

≤

2

−

1

+

2

γ

cn

+

p

c2 n

+

4

β

2 2

!

2

k

xn−1

−

xn−2

k

2

+

2

κσ

2 ₁

₊

1 n−1

2

−

1

+

2

γ

k

xn−1

−

xn−2

k

2

=

1

−

1

+

2

γ







cn

+

p

c2 n

+

4

β

2

2 2

+

2

κσ

2

1

+

1 n

−

1

2







k

xn−1

−

xn−2

k

2 wherecn

=

2

√

2

η(

1

+

1 n−1

)

2_t

₍

_n

_∈

N

)

. Finally, we have

k

xn

−

xn−1

k ≤

θ

n

k

xn−1

−

xn−2

k

,

(2.16) where

θ

n

=







1

−

1

+

2

γ







cn

+

p

c2 n

+

4

β

2

2 2

+

2

κσ

2

1

+

1 n

−

1

2













1 2

.

Let

θ

=







1

−

1

+

2

γ







c

+

p

c2

₊

₄

β

2

2 2

+

2

κσ

2













1 2

,

wherec

=

2

√

2

η

t.

Obviously

θ

n

→

θ

asn

→ ∞

. It is easy to prove that condition(2.7)implies that 0

< θ <

1 and so 0

< θ

n

<

1

for sufficiently largen. It follows from(2.16)that

{

xn

}

is a Cauchy sequence and then from(2.13)that

{

zn

}

is also a Cauchy

sequence. Letxn

→

xandzn

→

z. By condition (ii), it follows from(2.8)–(2.10)that

{

un

}

,

{

v

n

}

,

{

w

n

}

and

{

qn

}

are also Cauchy

sequences. Letun

→

u

, v

n

→

v, w

n

→

w,

qn

→

q(asn

→ ∞

), then by(2.4)and(2.5)we have zn+1

=

g

(

xn

)

−

η(

G

(w

n

)

−

N

(

un

, v

n

))

=

JA(·,qn−1)

(

zn

)

−

η(

G

(w

n

)

−

N

(

un

, v

n

)).

By the continuities ofgandGfrom(2.6), lettingn

→ ∞

in the above expression to obtain

z

=

g

(

x

)

−

η(

G

(w)

−

N

(

u

, v))

=

JA(·,q)

(

z

)

−

η(

G

(w)

−

N

(

u

, v)).

Finally, we prove thatu

∈

(

Tx

)

a(x)

, v

∈

(

Fx

)

b(x)

, w

∈

(

Ex

)

c(x)

,

q

∈

(

Px

)

e(x). Sinceun

∈

(

Txn

)

a(xn), we have

dist

(

u

, (

Tx

)

a(x)

)

≤ k

u

−

un

k +

dist

(

un

, (

Tx

)

a(x)

)

≤ k

u

−

un

k +

dist

(

un

, (

Txn

)

a(xn)

)

+

H

((

Txn

)

a(xn)

, (

Tx

)

a(x)

)

≤ k

u

−

un

k +

0

+

ξ

k

xn

−

x

k →

0 asn

→ ∞

,

which shows thatu

∈

(

Tx

)

a(x)due to the closedness of

(

Tx

)

a(x).

In similar ways, we also can prove that

v

∈

(

Fx

)

b(x),

w

∈

(

Ex

)

c(x)andq

∈

(

Px

)

e(x), which imply that

(

x

,

z

,

u

, v, w,

q

)

is a solution of fuzzy resolvent operator equation problem (1.2). Hence by Theorem 2.1

(

x

,

u

, v, w,

q

)

is a solution of fuzzy nonlinear variational inclusion problem(1.1), and the iterative sequences

{

xn

}

,

{

un

}

,

{

v

n

}

,

{

w

n

}

and

{

qn

}

generated

byAlgorithm 2.1, converge stronglyx

,

u

, v, w

andqinX, respectively. Remark 2.1. Our result improves and generalizes many results in [1,5–7,12].

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(8)

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