Chapter 2

Common solutions of variational inequality, systemof variational inequalities, mixed equilibriumproblem and fixed point problem for apseudocontractive mapping

2.1 Introduction

In 2008, Ceng et al. [32] introduced the relaxed extragradient method for approximate

a common solution of system of variational inequality problems (SVIP): Find (x, y) ∈

C × C such that

〈µ1B1y + x− y, z − x〉 ≥ 0, ∀z ∈ C,

〈µ2B2x+ y − x, z − y〉 ≥ 0, ∀z ∈ C,(2.1.1)

where, for each i = 1, 2, µi > 0 and Bi : C → C is a nonlinear mapping, C is a

nonempty, closed and convex subset of Hilbert space H; and fixed point problem (FPP)

for a nonexpansive mapping. Further Yao et al. [175] extended the iterative method

given in [32] for SVIP(2.1.1) and FPP for a psudocontractive mapping. Recently, Ceng

et al. [30] extended the methods given in [32,175] to approximate a common solution of

SVIP(2.1.1); variational inequality problem (VIP): Find x ∈ C such that

〈Dx, y − x〉 ≥ 0, ∀y ∈ C, (2.1.2)

where D : C → H is a nonlinear mapping; and FPP for a psudocontractive mapping.

27

Next, we consider mixed equilibrium problem (MEP) [127]: Find x ∈ C such that

F (x, y) + 〈Ax, y − x〉 ≥ 0, ∀y ∈ C, (2.1.3)

where F : C × C → R is bifunction and A : C → H is a nonlinear mapping. If A = 0,

then MEP (2.1.3) is reduced to the equilibrium problem (EP): Find x ∈ C such that

F (x, y) ≥ 0, ∀y ∈ C, (2.1.4)

In 2008, Moudafi [123] extended Mann type iterative method to approximate a common

solution of MEP(2.1.3) and FPP for a nonexpansive mapping. He proved some weak con-

vergence theorems for the sequences generated by the proposed iterative method. Fur-

ther, Takahashi and Takahashi [162] extended the work of Moudafi [123] to an Ishikawa

type iterative method to approximate a common solution of MEP(2.1.3) and FPP for a

nonexpansive mapping. Recently, Yao et al. [174] extended the iterative methods given

in [123,162] to the viscosity approximation method.

Motivated by the work of Ceng et al. [30], Moudafi [123], Takahashi and Takahashi

[162], Yao et al. [174] and by the recent work going in this direction, we combine the

relaxed extragradient method with viscosity approximation method to introduced a

new iterative method for approximating a common solution of VIP(2.1.2), SVIP(2.1.1),

MEP(2.1.3) and FPP for a strictly pseudocontractive mapping in a real Hilbert space.

We establish a strong convergence theorem for the sequences generated by the proposed

iterative method. Further, we derive some consequences from the strong convergence

theorem. The results presented here extend and generalize the work given in [30, 32,

169,175].

2.2 Preliminaries

We recall some results related to SVIP(2.1.1) and EP(2.1.4) which are needed in the

sequel. First, we have the following technical lemma which is the fixed point formulation

of SVIP(2.1.1):

28

Lemma 2.2.1. [32] For any (x∗, y∗) ∈ C × C, (x∗, y∗) is a solution of SVIP(2.1.1) if

and only if x∗ is a fixed point of the mapping Q : C → C defined by

Q(x) = PC [PC(x− µ2B2x)− µ1B1PC(x− µ2B2x)], ∀x ∈ C, (2.2.1)

where y∗ = PC(x∗ − µ2B2x

∗), µi ∈ (0, 2βi) and Bi : C → H is a βi-inverse strongly

monotone mapping for each i = 1, 2.

Next, we have the following assumption:

Assumption 2.2.1. [18] Let F : C × C → R be a bifunction satisfying the following

assumptions:

(i) F (x, x) = 0, ∀x ∈ C;

(ii) F is monotone, i.e., F (x, y) + F (y, x) ≤ 0, ∀x ∈ C;

(iii) F is upper hemicontinuous, i.e., for each x, y, z ∈ C,

lim supt→0

F (tz + (1− t)x, y) ≤ F (x, y);

(iv) For each x ∈ C, y → F (x, y) is convex and lower semicontinuous.

We consider an auxiliary problem related to EP(2.1.4): Let r > 0 and x ∈ H, find z ∈ C

such that

F (z, y) +1

r〈y − z, z − x〉 ≥ 0, ∀y ∈ C.

The following lemma give the properties of solution set Sol(EP(2.1.4)) of EP(2.1.4).

Lemma 2.2.2. [53] Assume that F : C ×C → R satisfies Assumption 2.2.1. For r > 0

and for all x ∈ H, define a mapping Tr : H → C as follows:

Tr(x) = {z ∈ C : F (z, y) +1

r〈y − z, z − x〉 ≥ 0, ∀y ∈ C}.

Then the following hold:

29

(i) Tr(x) is nonempty for each x ∈ H;

(ii) Tr is single-valued;

(iii) Tr is firmly nonexpansive, i.e.,

‖Trx− Try‖2 ≤ 〈Trx− Try, x− y〉, ∀x, y ∈ H;

(iv) Fix(Tr) = Sol(EP(2.1.4));

(v) Sol(EP(2.1.4)) is closed and convex.

2.3 Iterative method

We prove a strong convergence theorem based on viscosity approximation and relaxed

extragradient method for computing an approximate common solution of VIP(2.1.2),

SVIP(2.1.1), MEP(2.1.3) and FPP for a strictly pseudocontractive mapping in a real

Hilbert space.

Theorem 2.3.1. Let C be a nonempty, closed and convex subset of a real Hilbert space

H. For each i = 1, 2, let A,D,Bi : C → H be θ, α, βi-inverse strongly monotone

mappings, respectively. Let F : C × C → R be a bifunction satisfying the Assumption

2.2.1 and T : C → C be a k-strict pseudocontractive mapping such that Θ := Fix(T ) ∩

Sol(SVIP(2.1.1)) ∩ Sol(MEP(2.1.3)) ∩ Sol(VIP(2.1.2)) 6= ∅. Let f be a ρ-contraction

mapping with ρ ∈ [0, 12). For a given x0 ∈ C arbitrarily, let the iterative sequences {un},

{xn}, {yn} and {zn} be generated by

F (un, y) + 〈Axn, y − un〉+1

rn〈y − un, un − xn〉 ≥ 0, ∀y ∈ C,

zn = PC(un − λnDun),

yn = αnf(xn) + (1− αn)PC [PC(zn − µ2B2zn)− µ1B1PC(zn − µ2B2zn)],

xn+1 = βnxn + γnyn + δnTyn,

(2.3.1)

where µi ∈ (0, 2βi), for each i = 1, 2, {rn} ⊂ (0, 2θ), {λn} ⊂ (0, 2α), and {αn}, {βn},

{γn} and {δn} are the sequences in (0, 1) satisfying the following conditions:

30

(i) βn + γn + δn = 1 and (γn + δn)k ≤ γn, for all n ≥ 0;

(ii) limn→∞

αn = 0 and∞∑

n=0

αn =∞;

(iii) 0 < lim infn→∞

βn ≤ lim supn→∞

βn < 1 and lim infn→∞

δn > 0;

(iv) lim infn→∞

rn > 0,∞∑

n=1

|rn+1 − rn| <∞;

(v) limn→∞

( γn+1

1− βn+1

−γn

1− βn

)

= 0;

(vi) 0 < lim infn→∞

λn ≤ lim supn→∞

λn < 2α and limn→∞

| λn+1 − λn |= 0.

Then the sequence {xn} converges strongly to z ∈ Θ where z = PΘf(z).

Proof. First, we show that the mapping (I − rnA) is nonexpansive. For any x, y ∈ C,

‖(I − rnA)x− (I − rnA)y‖2 = ‖(x− y)− rn(Ax− Ay)‖2

= ‖x− y‖2 − 2rn〈x− y, Ax− Ay〉+ r2n‖Ax− Ay‖2

≤ ‖x− y‖2 − rn(2θ − rn)‖Ax− Ay‖2

≤ ‖x− y‖2. (2.3.2)

Similarly, we can show that the mappings (I − λnD) and (I − µiBi) are nonexpansive

for each i = 1, 2. It follows from Lemma 2.2.2 that, un = Trn(xn − rnAxn). Let x∗ ∈ Θ,

we have x∗ = Trn(x∗ − rnAx

∗). Now, we estimate

‖un − x∗‖2 = ‖Trn(xn − rnAxn)− Trn(x∗ − rnAx

∗)‖2

≤ ‖(xn − rnAxn)− (x∗ − rnAx∗)‖2

= ‖(xn − x∗)− rn(Axn − Ax∗)‖2

≤ ‖xn − x∗‖2 + r2n‖Axn − Ax∗‖2 − 2rn〈xn − x∗, Axn − Ax∗〉

≤ ‖xn − x∗‖2 + r2n‖Axn − Ax∗‖2 − 2rnθ‖Axn − Ax∗‖

≤ ‖xn − x∗‖2 − rn(2θ − rn)‖Axn − Ax∗‖2

≤ ‖xn − x∗‖2. (2.3.3)

31

Since x∗ ∈ Θ, we have

x∗ = PC [PC(x∗ − µ2B2x

∗)− µ1B1PC(x∗ − µ2B2x

∗)].

Putting

y∗ = PC(x∗ − µ2B2x

∗),

we see that

x∗ = PC(y∗ − µ1B1y

∗). (2.3.4)

Since the mapping D : C → H is α-inverse strongly monotone, we have

‖zn − x∗‖2 = ‖PC(un − λnDun)− PC(x∗ − λnDx∗)‖2

≤ ‖(un − λnDun)− (x∗ − λnDx∗)‖2

≤ ‖(un − x∗)− λn(Dun −Dx∗)‖2

≤ ‖un − x∗‖2 − λn(2α− λn)‖Dun −Dx∗‖2

≤ ‖un − x∗‖2 ≤ ‖xn − x∗‖2. (2.3.5)

Setting tn := PC [PC(zn − µ2B2zn)− µ1B1PC(zn − µ2B2zn)] and vn := PC(zn − µ2B2zn).

It follows that

‖vn − y∗‖2 = ‖PC(zn − µ2B2zn)− PC(x∗ − µ2B2x

∗)‖2

≤ ‖(zn − µ2B2zn)− (x∗ − µ2B2x∗)‖2

≤ ‖zn − x∗‖2 − µ2(2β2 − µ2)‖B2zn − B2x∗‖2

≤ ‖zn − x∗‖2 ≤ ‖xn − x∗‖2. (2.3.6)

Further, we have

‖tn − x∗‖2 = ‖PC [PC(zn − µ2B2zn)− µ1B1PC(zn − µ2B2zn)]

−PC [PC(x∗ − µ2B2x

∗)− µ1B1PC(x∗ − µ2B2x

∗)]‖2

32

≤ ‖[PC(zn − µ2B2zn)− µ1B1PC(zn − µ2B2zn)]

−[PC(x∗ − µ2B2x

∗)− µ1B1PC(x∗ − µ2B2x

∗)]‖2

≤ ‖[PC(zn − µ2B2zn)− PC(x∗ − µ2B2x

∗)]

−µ1[B1PC(zn − µ2B2zn)− B1PC(x∗ − µ2B2x

∗)]‖2

≤ ‖PC(zn − µ2B2zn)− PC(x∗ − µ2B2x

∗)‖2

−µ1(2β1 − µ1)‖B1PC(zn − µ2B2zn)− B1PC(x∗ − µ2B2x

∗)‖2

≤ ‖(zn − µ2B2zn)− (x∗ − µ2B2x∗)‖2

−µ1(2β1 − µ1)‖B1vn − B1y∗‖2

≤ ‖zn − x∗‖2 − µ2(2β2 − µ2)‖B2zn − B2x∗‖2

−µ1(2β1 − µ1)‖B1vn − B1y∗‖2 (2.3.7)

≤ ‖zn − x∗‖2 ≤ ‖xn − x∗‖2. (2.3.8)

Next, we estimate

‖yn − x∗‖ = ‖αn(f(xn)− x∗) + (1− αn)(tn − x∗)‖

≤ αn‖f(xn)− x∗‖+ (1− αn)‖tn − x∗‖

≤ αn(ρ‖xn − x∗‖+ ‖f(x∗)− x∗‖) + (1− αn)‖xn − x∗‖

= [1− (1− ρ)αn]‖xn − x∗‖+ (1− ρ)αn

‖f(x∗)− x∗‖

1− ρ

≤ max{

‖xn − x∗‖,‖f(x∗)− x∗‖

1− ρ

}

. (2.3.9)

Since (γn + δn)k ≤ γn for all n ≥ 0, utilizing Lemma 1.2.2, we have

‖xn+1 − x∗‖ = ‖βn(xn − x∗) + γn(yn − x∗) + δn(Tyn − x∗)‖

≤ βn‖xn − x∗‖+ ‖γn(yn − x∗) + δn(Tyn − x∗)‖

≤ βn‖xn − x∗‖+ (γn + δn)‖yn − x∗‖

≤ βn‖xn − x∗‖+ (γn + δn)max{

‖xn − x∗‖,‖f(x∗)− x∗‖

1− ρ

}

≤ max{

‖xn − x∗‖,‖f(x∗)− x∗‖

1− ρ

}

. (2.3.10)

33

By induction on n, we obtain ‖xn − x∗‖ ≤ max{

‖x0 − x∗‖,‖f(x∗)− x∗‖

1− ρ

}

, for every

n ≥ 0 and x0 ∈ C.

Hence {xn} is bounded and consequently, we deduce that {un}, {yn}, {zn}, {vn} and

{tn} are bounded. On the other hand, from the nonexpansivity of the mapping (I−λnD),

we have

‖zn+1 − zn‖ = ‖PC(un+1 − λn+1Dun+1)− PC(un − λnDun)‖

≤ ‖(un+1 − λn+1Dun+1)− (un − λnDun)‖

= ‖(un+1 − un)− λn+1(Dun+1 −Dun) + (λn+1 − λn)Dun‖

≤ ‖(un+1 − un)− λn+1(Dun+1 −Dun)‖+ |λn+1 − λn|‖Dun‖

≤ ‖un+1 − un‖+ |λn+1 − λn|‖Dun‖. (2.3.11)

We next estimate

‖tn+1 − tn‖2 = ‖PC [PC(zn+1 − µ2B2zn+1)− µ1B1PC(zn+1 − µ2B2zn+1)]

−PC [PC(zn − µ2B2zn)− µ1B1PC(zn − µ2B2zn)]‖2

≤ ‖[PC(zn+1 − µ2B2zn+1)− µ1B1PC(zn+1 − µ2B2zn+1)]

−[PC(zn − µ2B2zn)− µ1B1PC(zn − µ2B2zn)]‖2

≤ ‖[PC(zn+1 − µ2B2zn+1)− PC(zn − µ2B2zn)]

−µ1[B1PC(zn+1 − µ2B2zn+1)− B1PC(zn − µ2B2zn)]‖2

≤ ‖[PC(zn+1 − µ2B2zn+1)− PC(zn − µ2B2zn)]‖2

−µ1(2β1 − µ1)‖B1PC(zn+1 − µ2B2zn+1)− B1PC(zn − µ2B2zn)‖2

≤ ‖PC(zn+1 − µ2B2zn+1)− PC(zn − µ2B2zn)‖2

≤ ‖(zn+1 − zn)− µ2(B2zn+1 −B2zn)‖2

≤ ‖zn+1 − zn‖2 − µ2(2β2 − µ2)‖B2zn+1 − B2zn‖

2

≤ ‖zn+1 − zn‖2. (2.3.12)

From (2.3.11) and (2.3.12), we have

34

‖tn+1 − tn‖ ≤ ‖un+1 − un‖+ |λn+1 − λn|‖Dun‖. (2.3.13)

We observe that

‖yn+1 − yn‖2 = ‖tn+1 − αn+1[f(xn+1)− tn+1]− tn − αn[f(xn)− tn]‖

≤ ‖tn+1 − tn‖+ αn+1‖f(xn+1)− tn+1‖+ αn‖f(xn)− tn‖

≤ ‖un+1 − un‖+ |λn+1 − λn|‖Dun‖+ αn+1‖f(xn+1)− tn+1‖

+αn‖f(xn)− tn‖. (2.3.14)

On the other hand un = Trn(xn− rnAxn) and un+1 = Trn+1(xn+1− rn+1Axn+1), we have

F (un, y) + 〈Axn, y − un〉+1

rn〈y − un, un − xn〉 ≥ 0, ∀y ∈ C (2.3.15)

and

F (un+1, y) + 〈Axn+1, y − un+1〉+1

rn+1

〈y − un+1, un+1 − xn+1〉 ≥ 0, ∀y ∈ C. (2.3.16)

Take y = un+1 in (2.3.15) and y = un in (2.3.16), we have

F (un, un+1) + 〈Axn, un+1 − un〉+1

rn〈un+1 − un, un − xn〉 ≥ 0 (2.3.17)

and

F (un+1, un) + 〈Axn+1, un − un+1〉+1

rn+1

〈un − un+1, un+1 − xn+1〉 ≥ 0. (2.3.18)

Adding inequalities (2.3.17) and (2.3.18), we have the resultant inequality, after using

the monotonicity of F ,

〈Axn+1 − Axn, un − un+1〉+

⟨

un − un+1,un+1 − xn+1

rn+1

−un − xn

rn

⟩

≥ 0,

35

which implies that

0 ≤ 〈un − un+1, rn(Axn+1 − Axn) +rnrn+1

(un+1 − xn+1)− (un − xn)〉

≤ 〈un+1 − un, un − un+1 +

(

1−rnrn+1

)

un+1 + (xn+1 − rnAxn+1)

−(xn − rnAxn)− xn+1 +rnrn+1

xn+1〉

= 〈un+1 − un, un − un+1 +

(

1−rnrn+1

)

(un+1 − xn+1)

+(xn+1 − rnAxn+1)− (xn − rnAxn)〉

‖un+1 − un‖2 ≤ ‖un+1 − un‖

{

‖xn+1 − xn‖+

∣

∣

∣

∣

1−rnrn+1

∣

∣

∣

∣

‖un+1 − xn+1‖

}

and hence

‖un+1 − un‖ ≤ ‖xn+1 − xn‖+

∣

∣

∣

∣

1−rnrn+1

∣

∣

∣

∣

‖un+1 − xn+1‖

≤ ‖xn+1 − xn‖+1

rn+1

|rn+1 − rn| ‖un+1 − xn+1‖.

Without loss of generality, let us assume that there exists a real number c such that

rn > c > 0, for all positive integers n. Then the preceding inequality implies

‖un+1 − un‖ ≤ ‖xn+1 − xn‖+1

c|rn+1 − rn|M1, (2.3.19)

where M1 = sup {‖un − xn‖ : n ∈ N}. From (2.3.13) and (2.3.19), we have

‖tn+1 − tn‖ ≤ ‖xn+1 − xn‖+M1

c|rn+1 − rn|+ |λn+1 − λn|‖Dun‖. (2.3.20)

Using (2.3.14) and (2.3.19), we have

‖yn+1 − yn‖ ≤ ‖xn+1 − xn‖+M1

c|rn+1 − rn|+ |λn+1 − λn|‖Dun‖

+αn+1‖f(xn+1)− tn+1‖+ αn‖f(xn)− tn‖. (2.3.21)

36

Setting xn+1 = βnxn + (1− βn)en, which implies from (2.3.1) that

en =xn+1 − βnxn

1− βn

=γnyn + δnTyn

1− βn

.

Further, it follows that

en+1 − en =γn+1yn+1 + δn+1Tyn+1

1− βn+1

−γnyn + δnTyn

1− βn

=γn+1yn+1 + δn+1Tyn+1

1− βn+1

−γn+1yn + δn+1Tyn

1− βn+1

+γn+1yn + δn+1Tyn

1− βn+1

−γnyn + δnTyn

1− βn

=γn+1(yn+1 − yn) + δn+1(Tyn+1 − Tyn)

1− βn+1

r

+( γn+1

1− βn+1

−γn

1− βn

)

yn +( δn+1

1− βn+1

−δn

1− βn

)

Tyn.

Since (γn + δn)k ≤ γn, for all n ≥ 0, from Lemma 1.2.2, we get

‖γn+1(yn+1 − yn) + δn+1(Tyn+1 − Tyn)‖ ≤ (γn+1 + δn+1)‖yn+1 − yn‖. (2.3.22)

Hence, we obtain

‖en+1 − en‖ ≤

∥

∥

∥

∥

γn+1(yn+1 − yn) + δn+1(Tyn+1 − Tyn)

1− βn+1

∥

∥

∥

∥

+

∣

∣

∣

∣

γn+1

1− βn+1

−γn

1− βn

∣

∣

∣

∣

‖yn‖+

∣

∣

∣

∣

δn+1

1− βn+1

−δn

1− βn

∣

∣

∣

∣

‖Tyn‖

≤γn+1 + δn+1

1− βn+1

‖yn+1 − yn‖+∣

∣

∣

γn+1

1− βn+1

−γn

1− βn

∣

∣

∣

(

‖yn‖+ ‖Tyn‖)

= ‖yn+1 − yn‖+∣

∣

∣

γn+1

1− βn+1

−γn

1− βn

∣

∣

∣

(

‖yn‖+ ‖Tyn‖)

.

From (2.3.21), we obtain

‖en+1 − en‖ ≤ ‖xn+1 − xn‖+M1

c|rn+1 − rn|+ |λn+1 − λn|‖Dun‖

+αn+1‖f(xn+1)− tn+1‖+ αn‖f(xn)− tn‖

+∣

∣

∣

γn+1

1− βn+1

−γn

1− βn

∣

∣

∣

(

‖yn‖+ ‖Tyn‖)

, (2.3.23)

37

which implies that

‖en+1 − en‖ − ‖xn+1 − xn‖ ≤M1

c|rn+1 − rn|+ |λn+1 − λn|‖Dun‖

+αn+1‖f(xn+1)− tn+1‖+ αn‖f(xn)− tn‖

+∣

∣

∣

γn+1

1− βn+1

−γn

1− βn

∣

∣

∣

(

‖yn‖+ ‖Tyn‖)

.

Hence, it follows by conditions (i)-(vi) that

lim supn→∞

[

‖en+1 − en‖ − ‖xn+1 − xn‖]

≤ 0. (2.3.24)

From Lemma 1.2.8 and (2.3.24), we get limn→∞

‖en − xn‖ = 0 and

limn→∞

‖xn+1 − xn‖ = limn→∞

(1− βn)‖en − xn‖ = 0. (2.3.25)

Next, we show that ‖xn− un‖ → 0 as n→∞. Since x∗ ∈ Θ, by using Lemma 1.2.2 and

(2.3.7), we obtain

‖xn+1 − x∗‖2 = ‖βn(xn − x∗) + γn(yn − x∗) + δn(Tyn − x∗)‖2

≤ βn‖xn − x∗‖2

+(γn + δn)∥

∥

∥

1

γn + δn

(

γn(yn − x∗) + δn(Tyn − x∗))∥

∥

∥

2

≤ βn‖xn − x∗‖2 + (γn + δn)‖yn − x∗‖2

≤ βn‖xn − x∗‖2 + (γn + δn)[

αn‖f(xn)− x∗‖2 + (1− αn)‖tn − x∗‖2]

≤ βn‖xn − x∗‖2 + αn‖f(xn)− x∗‖2 + (γn + δn)‖tn − x∗‖2

≤ βn‖xn − x∗‖2 + αn‖f(xn)− x∗‖2 + (γn + δn)

×[

‖zn − x∗‖2 − µ2(2β2 − µ2)‖B2zn − B2x∗‖2

−µ1(2β1 − µ1)‖B1vn − B1y∗‖2

]

≤ βn‖xn − x∗‖2 + αn‖f(xn)− x∗‖2 + (γn + δn)[

‖un − x∗‖2

−λn(2α− λn)‖Dun −Dx∗‖2 − µ2(2β2 − µ2)‖B2zn − B2x∗‖2

−µ1(2β1 − µ1)‖B1vn − B1y∗‖2

]

38

≤ βn‖xn − x∗‖2 + αn‖f(xn)− x∗‖2 + (γn + δn)

×[

‖xn − x∗‖2 − rn(2θ − rn)‖Axn − Ax∗‖2

−λn(2α− λn)‖Dun −Dx∗‖2 − µ2(2β2 − µ2)‖B2zn − B2x∗‖2

−µ1(2β1 − µ1)‖B1vn − B1y∗‖2

]

≤ ‖xn − x∗‖2 + αn‖f(xn)− x∗‖2

−(γn + δn)[

rn(2θ − rn)‖Axn − Ax∗‖2

+λn(2α− λn)‖Dun −Dx∗‖2 + µ2(2β2 − µ2)‖B2zn − B2x∗‖2

+µ1(2β1 − µ1)‖B1vn − B1y∗‖2

]

. (2.3.26)

Hence from (2.3.25) and (2.3.26), we have

(γn + δn)[

rn(2θ − rn)‖Axn − Ax∗‖2 + λn(2α− λn)‖Dun −Dx∗‖2

+µ2(2β2 − µ2)‖B2zn − B2x∗‖2 + µ1(2β1 − µ1)‖B1vn − B1y

∗‖2]

≤ ‖xn − x∗‖2 − ‖xn+1 − x∗‖2 + αn‖f(xn)− x∗‖2

≤ ‖xn − xn+1‖(

‖xn − x∗‖+ ‖xn+1 − x∗‖)

+ αn‖f(xn)− x∗‖2. (2.3.27)

Since 0 < lim infn→∞

λn ≤ lim supn→∞

λn < 2α, ‖xn+1 − xn‖ → 0, αn → 0 and lim infn→∞

(γn +

δn) > 0, we obtain limn→∞

‖Axn−Ax∗‖ = 0, lim

n→∞‖Dun−Dx∗‖ = 0, lim

n→∞‖B2zn−B2x

∗‖ = 0

and limn→∞

‖B1vn − B1y∗‖ = 0.

Further, we observe that

‖un − x∗‖2 = ‖Trn(xn − rnAxn)− Trn(x∗ − rnAx

∗)‖2

≤ 〈un − x∗, (xn − rnAxn)− (x∗ − rnAx∗)〉

=1

2

{

‖un − x∗‖2 + ‖(xn − rnAxn)− (x∗ − rnAx∗)‖2

−∥

∥(un − x∗)− [(xn − rnAxn)− (x∗ − rnAx∗)]‖2

}

.

Hence,

‖un − x∗‖2 ≤ ‖(xn − rnAxn)− (x∗ − rnAx∗)‖2 − ‖(un − xn) + rn(Axn − Ax∗)‖2

39

≤ ‖xn − x∗‖2 − ‖(un − xn) + rn(Axn − Ax∗)‖2

≤ ‖xn − x∗‖2 − ‖un − xn‖2 + 2rn‖un − xn‖‖Axn − Ax∗‖. (2.3.28)

It follows that

‖xn+1 − x∗‖2 = ‖βn(xn − x∗) + γn(yn − x∗) + δn(Tyn − x∗)‖2

≤ βn‖xn − x∗‖2 + (γn + δn)∥

∥

∥

1

γn + δn

(

γn(yn − x∗) + δn(Tyn − x∗))∥

∥

∥

2

≤ βn‖xn − x∗‖2 + (γn + δn)‖yn − x∗‖2

≤ βn‖xn − x∗‖2 + (γn + δn)[

αn‖f(xn)− x∗‖2 + (1− αn)‖tn − x∗‖2]

≤ βn‖xn − x∗‖2 + αn‖f(xn)− x∗‖2 + (γn + δn)‖tn − x∗‖2

≤ βn‖xn − x∗‖2 + αn‖f(xn)− x∗‖2 + (γn + δn)‖un − x∗‖2

≤ βn‖xn − x∗‖2 + αn‖f(xn)− x∗‖2 + (1− βn)

×[

‖xn − x∗‖2 − ‖un − xn‖2 + 2rn‖un − xn‖‖Axn − Ax∗‖

]

= ‖xn − x∗‖2 + αn‖f(xn)− x∗‖2 − (1− βn)‖un − xn‖2

+(1− βn)2rn‖un − xn‖‖Axn − Ax∗‖. (2.3.29)

Therefore,

(1− βn)‖un − xn‖2 ≤ ‖xn − x∗‖2 − ‖xn+1 − x∗‖2 + αn‖f(xn)− x∗‖2

+(1− βn)2rn‖un − xn‖‖Axn − Ax∗‖

≤ (‖xn − x∗‖+ ‖xn+1 − x∗‖)‖xn − xn+1‖+ αn‖f(xn)− x∗‖2

+(1− βn)2rn‖un − xn‖‖Axn − Ax∗‖.

Since αn → 0, 0 < lim infn→∞

βn ≤ lim supn→∞

βn < 1, limn→∞

‖Axn − Ax∗‖ = 0 and ‖xn+1 −

xn‖ → 0, we obtain

limn→∞

‖un − xn‖ = 0. (2.3.30)

Furthermore, we observe that

‖zn − x∗‖2 = ‖PC(un − λnDun)− PC(x∗ − λnDx∗)‖2

40

≤ 〈(un − λnDun)− (x∗ − λnDx∗), zn − x∗〉

≤1

2

{

‖un − x∗ − λn(Dun −Dx∗)‖2 + ‖zn − x∗‖2

−‖un − x∗ − λn(Dun −Dx∗)− (zn − x∗)‖2}

≤1

2

{

‖un − x∗‖2 + ‖zn − x∗‖2

−‖un − zn − λn(Dun −Dx∗)‖2}

≤1

2

{

‖xn − x∗‖2 + ‖zn − x∗‖2 − ‖un − zn‖2

+2λn‖un − zn‖‖Dun −Dx∗‖2}

.

Hence,

‖zn − x∗‖2 ≤ ‖xn − x∗‖2 − ‖un − zn‖2 + 2λn‖un − zn‖‖Dun −Dx∗‖2. (2.3.31)

Next, we estimate

‖vn − y∗‖2 = ‖PC(zn − µ2B2zn)− PC(x∗ − µ2B2x

∗)‖2

≤ 〈(zn − µ2B2zn)− (x∗ − µ2B2x∗), vn − y∗〉

≤1

2

{

‖zn − x∗ − µ2(B2zn − B2x∗)‖2 + ‖vn − y∗‖2

−‖zn − x∗ − µ2(B2zn − B2x∗)− (vn − y∗)‖2

}

≤1

2

{

‖zn − x∗‖2 + ‖vn − y∗‖2 − ‖zn − vn − (x∗ − y∗)‖2 =

+2µ2〈zn − vn − (x∗ − y∗), B2zn − B2x∗〉 − µ2

2‖B2zn − B2x∗‖2

}

≤1

2

{

‖zn − x∗‖2 + ‖vn − y∗‖2 − ‖zn − vn − (x∗ − y∗)‖2

+2µ2‖zn − vn − (x∗ − y∗)‖‖B2zn − B2x∗‖}

.

Hence,

‖vn − y∗‖2 ≤ ‖zn − x∗‖2 − ‖zn − vn − (x∗ − y∗)‖2

+2µ2‖zn − vn − (x∗ − y∗)‖‖B2zn − B2x∗‖

≤ ‖xn − x∗‖2 − ‖un − zn‖2 + 2λn‖un − zn‖‖Dun −Dx∗‖2

41

−‖zn − vn − (x∗ − y∗)‖2

+2µ2‖zn − vn − (x∗ − y∗)‖‖B2zn − B2x∗‖. (2.3.32)

Similarly, we also estimate

‖tn − x∗‖2 = ‖PC(vn − µ1B1vn)− PC(y∗ − µ1B1y

∗)‖2

≤ 〈(vn − µ1B1vn)− (y∗ − µ1B1y∗), tn − x∗〉

≤1

2

{

‖vn − y∗ − µ1(B1vn − B1y∗)‖2 + ‖tn − x∗‖2

−‖vn − y∗ − µ1(B1vn − B1y∗)− (tn − x∗)‖2

}

≤1

2

{

‖vn − y∗‖2 + ‖tn − x∗‖2 − ‖vn − tn + (x∗ − y∗)‖2

+2µ1〈vn − tn + (x∗ − y∗), B1vn − B2y∗〉 − µ2

1‖B1vn − B1y∗‖2

}

≤1

2

{

‖vn − y∗‖2 + ‖tn − x∗‖2 − ‖vn − tn + (x∗ − y∗)‖2

+2µ1‖vn − tn + (x∗ − y∗)‖‖B1vn − B1y∗‖}

.

Hence,

‖tn − x∗‖2

≤ ‖vn − y∗‖2 − ‖vn − tn + (x∗ − y∗)‖2

+2µ1‖vn − tn + (x∗ − y∗)‖‖B1vn − B1y∗‖

≤ ‖xn − x∗‖2 − ‖un − zn‖2 + 2λn‖un − zn‖‖Dun −Dx∗‖2

−‖zn − vn − (x∗ − y∗)‖2 + 2µ2‖zn − vn − (x∗ − y∗)‖‖B2zn − B2x∗‖

−‖vn − tn + (x∗ − y∗)‖2 + 2µ1‖vn − tn + (x∗ − y∗)‖‖B1vn − B1y∗‖.

(2.3.33)

From Lemma 1.2.2, (2.3.1) and (2.3.33), we have

‖xn+1 − x∗‖2 = ‖βn(xn − x∗) + γn(yn − x∗) + δn(Tyn − x∗)‖2

≤ βn‖xn − x∗‖2 + (γn + δn)∥

∥

∥

1

γn + δn

(

γn(yn − x∗) + δn(Tyn − x∗))∥

∥

∥

2

42

≤ βn‖xn − x∗‖2 + (γn + δn)‖yn − x∗‖2r

≤ βn‖xn − x∗‖2 + (γn + δn)[

αn‖f(xn)− x∗‖2 + (1− αn)‖tn − x∗‖2]

≤ βn‖xn − x∗‖2 + αn‖f(xn)− x∗‖2 + (γn + δn)‖tn − x∗‖2

≤ βn‖xn − x∗‖2 + αn‖f(xn)− x∗‖2 + (1− βn)

×[

‖xn − x∗‖2 − ‖un − zn‖2 + 2λn‖un − zn‖‖Dun −Dx∗‖2

−‖zn − vn − (x∗ − y∗)‖2 + 2µ2‖zn − vn − (x∗ − y∗)‖‖B2zn − B2x∗‖

−‖vn − tn + (x∗ − y∗)‖2 + 2µ1‖vn − tn + (x∗ − y∗)‖‖B1vn − B1y∗‖]

= ‖xn − x∗‖2 + αn‖f(xn)− x∗‖2

+(1− βn)[

2λn‖un − zn‖‖Dun −Dx∗‖2

+2µ2‖zn − vn − (x∗ − y∗)‖‖B2zn − B2x∗‖

+2µ1‖vn − tn + (x∗ − y∗)‖‖B1vn − B1y∗‖]

− (1− βn)[

‖un − zn‖2

+‖zn − vn − (x∗ − y∗)‖2 + ‖vn − tn + (x∗ − y∗)‖2]

which yields

(1 − βn)[

‖un − zn‖2 + ‖zn − vn − (x∗ − y∗)‖2 + ‖vn − tn + (x∗ − y∗)‖2

≤ ‖xn − x∗‖2 − ‖xn+1 − x∗‖2 + αn‖f(xn)− x∗‖2 + (1− βn)

×[

2λn‖un − zn‖‖Dun −Dx∗‖2 + 2µ2‖zn − vn − (x∗ − y∗)‖‖B2zn − B2x∗‖

+2µ1‖vn − tn + (x∗ − y∗)‖‖B1vn − B1y∗‖]

≤ (‖xn − x∗‖+ ‖xn+1 − x∗‖)‖xn − xn+1‖+ αn‖f(xn)− x∗‖2 + (1− βn)

×[

2λn‖un − zn‖‖Dun −Dx∗‖2 + 2µ2‖zn − vn − (x∗ − y∗)‖‖B2zn − B2x∗‖

+2µ1‖vn − tn + (x∗ − y∗)‖‖B1vn − B1y∗‖]

. (2.3.34)

Since 0 < lim supn→∞

βn ≤ 1, 0 < λn ≤ 2α, ‖xn+1 − xn‖ → 0, αn → 0 and limn→∞

‖Axn −

Ax∗‖ = 0, limn→∞

‖Dun−Dx∗‖ = 0, limn→∞

‖B2zn−B2x∗‖ = 0 and lim

n→∞‖B1vn−B1y

∗‖ = 0,

it follows from the boundedness of {xn}, {un}, {zn} {tn} and {vn} that

43

limn→∞

‖un − zn‖ = 0; limn→∞

‖zn − vn − (x∗ − y∗)‖ = 0 and limn→∞

‖vn − tn + (x∗ − y∗)‖ = 0.

(2.3.35)

Consequently, it follows that

‖zn − tn‖ ≤ ‖zn − vn − (x∗ − y∗)‖+ ‖vn − tn + (x∗ − y∗)‖ → 0 as n→∞. (2.3.36)

Also limn→∞

‖un − tn‖ = 0 and limn→∞

‖yn − tn‖ ≤ αn‖f(xn)− tn‖ → 0 as n→∞

‖un − yn‖ ≤ ‖un − tn‖+ ‖tn − yn‖ → 0 as n→∞, (2.3.37)

and

‖yn − xn‖ ≤ ‖yn − un‖+ ‖un − xn‖ → 0 as n→∞. (2.3.38)

Since

‖δn(Tyn − xn)‖ ≤ ‖xn+1 − xn‖+ γn‖yn − xn‖,

it follows that

limn→∞

‖Tyn − xn‖ = 0, limn→∞

‖Tyn − yn‖ = 0.

Also, since H is reflexive and {yn} is bounded, without loss of generality we can assume

that yn ⇀ w for some w ∈ C. It follows from Lemma 1.2.3 (ii) that w ∈ Fix(T ).

Next, we show that w ∈ Sol(MEP(2.1.3)). Since un = Trn(xn − rnAxn), for any y ∈ C,

we have

F (un, y) + 〈Axn, y − un〉+1

rn〈y − un, un − xn〉 ≥ 0, ∀y ∈ C. (2.3.39)

It follows from monotonicity of F that

〈Axn, y − un〉+1

rn〈y − un, un − xn〉 ≥ F (y, un). (2.3.40)

44

Replacing n by ni, we get

〈Axni, y − uni

〉+

⟨

y − uni,uni− xni

rni

⟩

≥ F (y, uni). (2.3.41)

Since ‖un − yn‖ → 0 and yn ⇀ w, it is easy to observe that uni⇀ w. Further, for any t

with 0 < t ≤ 1 and y ∈ C, let yt = ty + (1− t)w. Since w ∈ C, y ∈ C, we have yt ∈ C.

So from (2.3.41), we have

〈yt − uni, Ayt〉 ≥ 〈yt − uni

, Ayt〉 − 〈Axni, yt − uni

〉

−

⟨

yt − uni,uni− xni

rni

⟩

+ F (yt, uni)

= 〈yt − uni, Ayt − Axni

〉+ 〈yt − uni, Auni

− Axni〉

−

⟨

yt − uni,uni− xni

rni

⟩

+ F (yt, uni). (2.3.42)

From condition (iv), Lipschitz continuity of A and limn→∞

‖un − xn‖ = 0, we obtain

‖Auni− Axni

‖ = 0, anduni− xni

rni

→ 0 as i → ∞. Further, since A is monotone and

uni⇀ w, it follows that from (2.3.42) that

〈yt − w,Ayt〉 ≥ F (yt, w). (2.3.43)

Hence, from Assumption 2.2.1 and (2.3.43), we have

0 = F (yt, yt) ≤ tF (yt, y) + (1− t)F (yt, w)

≤ tF (yt, y) + (1− t)〈yt − w,Ayt〉

≤ tF (yt, y) + (1− t)t〈y − w,Ayt〉 (2.3.44)

which implies that F (yt, y) + (1− t)〈y − w,Ayt〉 ≥ 0. Letting t→ 0+ we have

F (w, y) + 〈y − w,Aw〉 ≥ 0, ∀y ∈ C

which implies that w ∈ Sol(MEP(2.1.3)).

Further, we show that w ∈ Sol(SVIP(2.1.1)). Take any x, y ∈ C. Using (2.2.1), we

45

estimate

‖Q(x)−Q(y)‖2 = ‖PC [PC(x− µ2B2x)− µ1B1PC(x− µ2B2x)]

−PC [PC(y − µ2B2y)− µ1B1PC(y − µ2B2y)]‖2

≤ ‖[PC(x− µ2B2x)− PC(y − µ2B2y)]

−µ1[B1PC(x− µ2B2x)− B1PC(y − µ2B2y)]‖2

≤ ‖PC(x− µ2B2x)− PC(y − µ2B2y)‖2

−µ1(2β1 − µ1)‖PC(x− µ2B2x)− PC(y − µ2B2y)‖2

≤ ‖(x− µ2B2x)− (y − µ2B2y)‖2

≤ ‖x− y‖2 − µ2(2β2 − µ2)‖B2x− B2y‖2

≤ ‖x− y‖2.

This implies that Q : C → C is nonexpansive.

Now, we have

‖yn −Q(yn)‖ = αn‖f(xn)−Q(yn)‖+ (1− αn)‖PC [PC(zn − µ2B2zn)

−µ1B1PC(zn − µ2B2zn)]−Q(yn)‖

= αn‖f(xn)−Q(yn)‖+ (1− αn)‖Q(zn)−Q(tn)‖

≤ αn‖f(xn)−Q(yn)‖+ (1− αn)‖zn − tn‖. (2.3.45)

Since αn → 0 and ‖zn − tn‖ → 0 as n→∞, (2.3.45) implies lim infn→∞

‖yn −Q(yn)‖ = 0

and hence by Lemma 1.2.3 (ii), it follows that w ∈ Q(w). Further, it follows from Lemma

2.2.1 that w ∈ Sol(SVIP(2.1.1)).

Furthermore, we show that w ∈ Sol(VIP(2.1.2)). Indeed, let

T v =

Dv +NCv if v ∈ C,

∅ if v /∈ C,(2.3.46)

where NCv := {w ∈ H : 〈v − u, w〉 ≥ 0, ∀u ∈ C} is the normal cone to C at v ∈ C.

46

Then the multi-valued mapping T is maximal monotone and 0 ∈ T v if and only if

v ∈ Sol(VIP(2.1.2)), see [128]. Let G(T ) denote the graph of T and let (v, u) ∈ G(T ).

Then we have u ∈ T v = Dv + NCv and hence u − Dv ∈ NCv. Therefore, we have

〈v − t, u−Dv〉 ≥ 0, ∀t ∈ C. Since zn ∈ C, ∀n, so we have

〈v − zn, u−Dv〉 ≥ 0. (2.3.47)

On the other hand, it follows from zn = PC(un − λnDun) and v ∈ C that

〈v − zn, zn − (I − λnD)un〉 ≥ 0

and hence⟨

v − zn,zn − un

λn

+Dun

⟩

≥ 0.

Further, from (2.3.47), inverse strongly monotonicity of D and replacing n by ni, we

have

〈v − zni, u〉 ≥ 〈v − zni

, Dv〉

≥ 〈v − zni, Dv〉 −

⟨

v − zni,zni− uni

λni

+Duni

⟩

= 〈v − zni, Dv −Dzni

〉+ 〈v − zni, Dzni

− duni〉 −

⟨

v − zni,zni− uni

λni

⟩

≥ 〈v − zni, Dzni

−Duni〉 −

⟨

v − zni,zni− uni

λni

⟩

. (2.3.48)

Since zni⇀ w and ‖uni

− zni‖ → 0 as i → ∞, hence we obtain 〈v − w, u〉 ≥ 0. Since

T is maximal monotone, we have w ∈ T −10 and hence w ∈ Sol(VIP(2.1.2)). Thus we

have w ∈ Θ.

Next, we claim that lim supn→∞

〈f(z)− z, xn − z〉 ≤ 0, where z = PΘf(z).

Since {xn} is bounded, ‖yn − xn‖ → 0 and yn ⇀ w then there exists a subsequence

{xni} of {xn} such that xni

⇀ w and

lim supn→∞

〈f(z)− z, xn − z〉 = lim supi→∞

〈f(z)− z, xni− z〉 = 〈f(z)− z, w − z〉 ≤ 0.

47

Further, from (2.2.1) and definition of tn, we have

‖tn − z‖ = ‖Q(zn)−Q(z)‖

≤ ‖zn − z‖ ≤ ‖xn − z‖.

Next, we have

〈f(xn)− z, yn − z〉 = 〈f(xn)− z, xn − z〉+ 〈xn − z, yn − z〉

= 〈f(xn)− f(z), xn − z〉+ 〈f(z)− z, xn − z〉

+〈f(xn)− z, yn − xn〉

≤ ρ‖xn − z‖2 + 〈f(z)− z, xn − z〉

+‖f(xn)− z‖‖yn − xn‖. (2.3.49)

Finally, we show that xn → z.

‖xn+1 − z‖2 = ‖β(xn − z) + γn(yn − z) + δn(Tyn − z)‖2

= βn‖xn − z‖2 + (γn + δn)∥

∥

∥

1

(γn + δn)

(

γn(yn − z) + δn(Tyn − z))∥

∥

∥

2

≤ βn‖xn − z‖2 + (γn + δn)‖yn − z‖2

≤ βn‖xn − z‖2 + (γn + δn)[

(1− αn)‖tn − z‖2 + 2〈f(xn)− z, yn − z〉]

≤ βn‖xn − z‖2 + (γn + δn)[

(1− αn)‖xn − z‖2 + 2〈f(xn)− z, yn − z〉]

≤ [1− (γn + δn)αn]‖xn − z‖2 + (γn + δn)2αn〈f(xn)− z, yn − z〉

≤ [1− (γn + δn)αn]‖xn − z‖2 + (γn + δn)2αn

{

ρ‖xn − z‖2

+〈f(z)− z, xn − z〉+ ‖f(xn)− z‖‖yn − xn‖}

≤ [1− (1− 2ρ)(γn + δn)αn]‖xn − z‖2

+(γn + δn)2αn

{

〈f(z)− z, xn − z〉+ ‖f(xn)− z‖‖yn − xn‖}

≤ [1− (1− 2ρ)(γn + δn)αn]‖xn − z‖2

+(1− 2ρ)(γn + δn)αn

2{

〈f(z)− z, xn − z〉+ ‖f(xn)− z‖‖yn − xn‖}

1− ρ.

(2.3.50)

48

Since lim infn→∞

(1 − 2ρ)(γn + δn) > 0. It follows that∞∑

n=0

(1 − 2ρ)(γn + δn)αn = ∞ and

hence we notice that

lim supn→∞

2{

〈f(z)− z, xn − z〉+ ‖f(xn)− z‖‖yn − xn‖}

1− ρ≤ 0, (2.3.51)

since lim supn→∞

〈f(z)− z, xn − z〉 ≤ 0 and limn→∞

‖xn − yn‖ = 0. Thus all the conditions of

Lemma 1.2.11 are satisfied. Hence we deduce that xn → z. This completes the proof.

�

2.4 Consequences

We derive some consequences from Theorem 2.3.1.

Corollary 2.4.1. For each i = 1, 2, let A,Bi : C → H be θ, βi-inverse strongly mono-

tone mappings, respectively. Let F : C × C → R be a bifunction satisfying the As-

sumption 2.2.1 and let T : C → C be a k-strict pseudocontractive mapping such that

Θ1 := Fix(T )∩ Sol(SVIP(2.1.1))∩ Sol(MEP(2.1.3)) 6= ∅. Let f be a ρ-contraction map-

ping with ρ ∈ [0, 12). For a given x0 ∈ C arbitrarily, let the iterative sequences {un},

{xn} and {yn} be generated by

F (un, y) + 〈Axn, y − un〉+1

rn〈y − un, un − xn〉 ≥ 0, ∀y ∈ C,

yn = αnf(xn) + (1− αn)PC [PC(un − µ2B2un)− µ1B1PC(un − µ2B2un)],

xn+1 = βnxn + γnyn + δnTyn,

where µi ∈ (0, 2βi) for each i = 1, 2, {rn} ⊂ (0, 2θ) and {αn}, {βn}, {γn} and {δn} are

the sequences in (0, 1) satisfying conditions (i)-(v) of Theorem 2.3.1. Then the sequence

{xn} converges strongly to z ∈ Θ1, where z = PΘ1f(z).

Proof. Putting D = 0 in Theorem 2.3.1, then conclusion of Corollary 2.4.1 is obtained.

Corollary 2.4.2. [30] For each i = 1, 2, let D,Bi : C → H be α, βi-inverse strongly

monotone mappings, respectively. Let T : C → C be a k-strict pseudocontractive map-

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ping such that Θ2 := Fix(T ) ∩ Sol(SVIP(2.1.1)) ∩ Sol(VIP(2.1.2)) 6= ∅. Let f be a

ρ-contraction mapping with ρ ∈ [0, 12). For a given x0 ∈ C arbitrarily, let the iterative

sequences {un}, {xn} and {yn} be generated by

un = PC(xn − λnDxn),

yn = αnf(xn) + (1− αn)PC [PC(un − µ2B2un)− µ1B1PC(un − µ2B2un)],

xn+1 = βnxn + γnyn + δnTyn,

where µi ∈ (0, 2βi) for each i = 1, 2, {λn} ⊂ (0, 2α) and {αn}, {βn}, {γn} and {δn} are

the sequences in (0, 1) satisfying conditions (i)-(iii),(v), (vi) of Theorem 2.3.1. Then the

sequence {xn} converges strongly to z ∈ Θ2, where z = PΘ2f(z).

Proof. Putting F = A = 0 in Theorem 2.3.1, then conclusion of Corollary 2.4.2 is

obtained.

Corollary 2.4.3. [32] For each i = 1, 2, let Bi : C → H be βi-inverse strongly monotone

mappings, respectively. Let T : C → C be a nonexpansive mapping such that Θ3 :=

Fix(T )∩Sol(SVIP(2.1.1)) 6= ∅. For a given x0 ∈ C arbitrarily, let the iterative sequences

{yn} and {xn} be generated by

yn = PC [PC(xn − µ2B2xn)− µ1B1PC(xn − µ2B2xn)],

xn+1 = βnxn + γnyn + δnTyn,

where µi ∈ (0, 2βi) for each i = 1, 2 and {βn}, {γn} and {δn} are the sequences in (0, 1)

satisfying conditions (i)-(iii),(v) of Theorem 2.3.1. Then the sequence {xn} converges

strongly to z ∈ Θ3, where z = PΘ3f(z).

Proof. Putting F = A = D = 0, αn = 0, and taking T a nonexpansive mapping in

Theorem 2.3.1, then conclusion of Corollary 2.4.3 is obtained.

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