H (·, ·)-η-cocoercive operators and variational-like inclusions in Banach spaces

Research Article

H (·, ·)-η-cocoercive operators and variational-like inclusions in Banach spaces

Rais Ahmad^a, Mohammad Dilshad^a,∗

aDepartment of Mathematics, Aligarh Muslim University, Aligarh-202002, India.

Dedicated to George A Anastassiou on the occasion of his sixtieth birthday Communicated by Professor G. Sadeghi

Abstract

In this paper, we defineH(·,·)-η-cocoercive operators inq-uniformly smooth Banach spaces and its resolvent operator. We prove the Lipschitz continuity of the resolvent operator associated with H(·,·)-η-cocoercive operator and estimate its Lipschitz constant. By using the techniques of resolvent operator, an iterative algorithm for solving a variational-like inclusion problem is constructed. The existence of solution for the variational-like inclusions and the convergence of iterative sequences generated by the algorithm is proved.

Some examples are given. c2012 NGA. All rights reserved.

Keywords: H(·,·)-η-cocoercive, Algorithm, Inclusion, Banach spaces, Lipschitz continuity.

2010 MSC: Primary 47H19; Secondary 49J40.

1. Introduction

It is well known that variational inequality theory has become very effective and powerful tool for studying a wide range of problems arising in differential equations, mechanics, contact problems in elasticity, optimization and control problems, operation research, etc.. A useful and important generalization of variational inequalities is a generalized mixed type variational inequality containing nonlinear term. Due to the appearance of this nonlinear term, the projection method can not be used to study the existence and algorithm of solutions for the generalized mixed type variational inequalities. These facts motivated Hassouni and Moudafi [8] to suggest the resolvent operator technique which does not depend on the projection. He studied mixed type of variational inequalities, called variational inclusions.

∗Corresponding author

Email addresses: [email protected]( Rais Ahmad),[email protected](Mohammad Dilshad) Received 2011-6-3

The concept of H-monotone, H-accretive, (H, η)-accretive, (H, η)-monotone, (A, η)-accretive, H(·,·)- accretive, (H(·,·)−η)-monotone and H(·,·)-cocoercive operators are introduced and applied by Fang and Huang [4], Fang and Huang [5], Fang, Cho and Kim [6], Fang, Huang and Thompson [7], Lan, Cho and verma [9], Zou and Huang [13], Xu and Wong [11], Ahmad et al.[1]. The concept ofη-cocoercivity,η-strong monotonicity and η-strong convexity of a mapping was introduced and studied by Ansari and Yao [2].

Impressed by the noble work mentioned above, in this paper, we introduceH(·,·)-η-cocoercive operators inq-uniformly smooth Banach spaces. The resolvent operator associated withH(·,·)-η-cocoercive operator is defined and its Lipschitz continuity is shown. With the help of resolvent operator an iterative algorithm is constructed for solving a variational-like inclusion problem in q-uniformly smooth Banach spaces. Some examples are constructed in support of definition of H(·,·)-η-cocoercive operators.

2. Preliminaries

LetX is a real Banach space with dual spaceX^?,h·,·ibe the duality coupling betweenX and X^?, and 2^X denotes the family of all the non-empty subsets ofX. The generalized duality mappingJq :X→2^X? is defined by

J_q(x) =

f^?∈X^?:hx, f^?i=kxk^q, kf^?k=kxk^q−1 , ∀x∈X,

whereq >1 is a constant. The modulus of smoothness of X is the functionρX : [0,∞)→[0,∞) defined by ρX(t) = sup 1

2(kx+yk+kx−yk)−1 :kxk ≤1,kyk ≤t . A Banach space X is called uniformly smooth if

limt→0

ρX(t) t = 0.

X is called q-uniformly smooth if there exists a constantC >0 such thatρ_X(t)≤Ct^q, q >1.

Note thatJ_q is single-valued ifX is uniformly smooth. In connection with the characteristic inequalities inq-uniformly smooth Banach spaces, Xu [12] proved the following nice result.

Lemma 2.1. Let X be a real uniformly smooth Banach space. Then X isq-uniformly smooth if and only if there exists a constant C_q >0 such that, for all x, y∈X,

kx+yk^q ≤ kxk^q+qhy, J_q(x)i+Cqkyk^q. We need the following definitions for proving our main result.

Definition 2.2. Let A, B : X → X, η : X×X → X be the mappings and let Jq : X → 2^X? be the generalized duality mapping. ThenA is called

(i) η-cocoercive, if there exists a constant µ₁ >0 such that

hAx−Ay, Jq(η(x, y))i ≥µ1kAx−Ayk^q, ∀ x, y∈X;

(ii) η-accretive, if

hAx−Ay, J_q(η(x, y))i ≥0, ∀ x, y∈X;

(iii) η-strongly accretive, if there exists a constant β₁ >0 such that

hAx−Ay, J_q(η(x, y))i ≥β₁kx−yk^q, ∀x, y∈X;

ifη(x, y) =x−y, then it is called strongly accretive.

(iv) η-relaxed cocoercive, if there exists a constant γ₁>0 such that

hAx−Ay, J_q(η(x, y))i ≥(−γ₁)kAx−Ayk^q, ∀ x, y∈X;

(v) α-expansive, if there exists a constant α >0 such that

kAx−Ayk ≥αkx−yk, ∀ x, y∈X;

(vi) B is said to be β-Lipschitz continuous, if there exists a constantβ >0 such that kBx−Byk ≤βkx−yk, ∀x, y∈X;

(vii) η is said to Lipschitz continuous, if there exists a constantτ >0 such that kη(x, y)k ≤τkx−yk, ∀ x, y∈X.

Definition 2.3. LetA, B :X→X,H:X×X →X,η:X×X→X be three single-valued mappings and Jq:X →2^X? be the generalized duality mapping. Then

(i) H(A,·) is said to beη-cocoercive with respect toA, if there exists a constant µ >0 such that hH(Ax, u)−H(Ay, u), Jq(η(x, y))i ≥µkAx−Ayk^q, ∀ x, y∈X;

(ii) H(·, B) is said toη-relaxed cocoercive with respect to B, if there exists a constant γ >0 such that hH(u, Bx)−H(u, By), J_q(η(x, y))i ≥(−γ)kBx−Byk^q, ∀ x, y∈X;

(iii) H(A,·) is said to ber₁-Lipschitz continuous with respect to A, if there exists a constant r₁ >0 such that

kH(Ax, u)−H(Ay, u)k ≤r1kx−yk, ∀ x, y∈X;

(iv) H(·, B) is said to be r2-Lipschitz continuous with respect to B, if there exists a constantr2>0 such that

kH(u, Bx)−H(u, By)k ≤r2kx−yk, ∀x, y∈X.

Definition 2.4. A multi-valued mappingM :X→2^X is said to beη-cocoercive, if there exists a constant µ2>0 such that

hu−v, J_q(η(x, y))i ≥µ₂ku−vk^q, ∀ x, y∈X, u∈M x, v∈M y.

Definition 2.5. A multi-valued mapping T : X → CB(X) is said to be D-Lipschitz continuous, if there exists a constant λ_T >0 such that

D(T x, T y)≤λ_Tkx−yk, ∀x, y∈X, whereD(·,·) is the Hausd¨orff metric on CB(X).

Definition 2.6. Let η : X ×X → X be a mapping and let T, Q : X → CB(X) be two multi-valued mappings. A mappingN :X×X→X is said to beη-strongly accretive with respect toT and Q, if there exists a constant t >0 such that

hN(x₁, y₁)−N(x₂, y₂), J_q(η(u₁, u₂))i ≥tku₁−u₂k^q, ∀u₁, u₂ ∈X, x₁∈T(u₁), y₁ ∈Q(u₁), x₂ ∈T(u₂), y₂ ∈Q(u₂).

Definition 2.7. Let T, Q :X → CB(X) be the multi-valued mappings. A mapping N :X×X → X is said to be

(i) Lipschitz continuous for the first argument with respect to T, if there exists a constantλ_N1 >0 such that

kN(x1,·)−N(x2,·)k ≤λ_N1kx₁−x2k, ∀u1, u2 ∈X, x1 ∈T(u1), x2∈T(u2);

(ii) Lipschitz continuous for the second argument with respect to Q, if there exists a constant λ_N2 > 0 such that

kN(·, y₁)−N(·, y₂)k ≤λ_N2ky₁−y₂k, ∀ u₁, u₂∈X, y₁ ∈Q(u₁), y₂∈Q(u₂).

Example 2.8. Let us consider the 2-uniformly smooth Banach space X = R². Let A, B : R² → R² are defined by

A(x1, x2) = (x1,3x2), B(y1, y2) = (−y₁,−y₁−y2), ∀(x1, x2),(y1, y2)∈R². SupposeH(A, B), η :R²×R²→R² are defined as

H(Ax, By) =Ax+By, η(x, y) =x−y, ∀ x, y∈R². Then

hH(Ax, u)−H(Ay, u), η(x, y)i=hAx+u−Ay−u, x−yi

=hAx−Ay, x−yi

=h((x₁,3x₂)−(y₁,3y₂)), (x₁−y₁, x₂−y₂)i

=h(x₁−y₁,3(x₂−y₂)), (x₁−y₁, x₂−y₂)i

= (x1−y1)²+ 3(x2−y2)² and

kAx−Ayk² =k(x₁−y1,3(x2−y2))k² = (x1−y1)²+ 9(x2−y2)²

≤3(x₁−y₁)²+ 9(x₂−y₂)²

= 3{(x₁−y1)²+ 3(x2−y2)²}

= 3{hH(Ax, u)−H(Ay, u), η(x, y)i}

i.e. hH(Ax, u)−H(Ay, u), η(x, y)i ≥ ¹₃kAx−Ayk²,which implies thatH is ¹₃-η-cocoercive with respect to A. Also

hH(u, Bx)−H(u, By), η(x, y)i=hBx−By, x−yi

=h((−x₁,−x₁−x₂)−(−y₁,−y₁−y₂)), (x₁−y₁, x₂−y₂)i

=h(−(x₁−y₁),−(x₁−y₁)−(x₂−y₂)), (x₁−y₁, x₂−y₂)i

=−(x₁−y1)²−(x1−y1)(x2−y2)−(x2−y2)²

=−{(x₁−y₁)²+ (x₁−y₁)(x₂−y₂) + (x₂−y₂)²} and

kBx−Byk² =k(−(x₁−y₁),−(x₁−y₁)−(x₂−y₂))k²

= (x₁−y₁)²+ ((x₁−y₁) + (x₂−y₂))²

= (x1−y1)²+ (x1−y1)²+ (x2−y2)²+ 2(x1−y1)(x2−y2)

≤2(x₁−y₁)²+ 2(x₂−y₂)²+ 2(x₁−y₁)(x₂−y₂)

= 2{(x₁−y1)²+ (x2−y2)²+ (x1−y1)(x2−y2)}

= 2{−hH(u, Bx)−H(u, By), η(x, y)i}

i.e. hH(u, Bx)−H(u, By), η(x, y)i ≥ −¹₂kBx−Byk²,which implies thatH is ¹₂-η-relaxed cocoercive with respect toB.

3. H(·,·)-η-cocoercive operator

In this section, we introduce H(·,·)-η-cocoercive operator and discuss some of its properties.

Definition 3.1. LetA, B:X→X,H :X×X→X,η:X×X→X be the single-valued mappings. Let M :X →2^X be a set-valued mapping. M is said to beH(·,·)-η-cocoercive with respect to the mappingA and B, ifM isη-cocoercive and (H(A, B) +λM)(X) =X, for everyλ >0.

Example 3.2. Let X = R² and A, B, H(A, B) and η are defined as in Example 2.8. Suppose that M :X→2^X is defined by

M(x1, x2) = (x1,0), ∀(x1, x2)∈R². Then it can be easily verify thatM is η-cocoercive and

(H(A, B) +λM)(R²) =R², ∀λ >0, which shows thatM is H(·,·)-η-cocoercive with respect toA andB.

Theorem 3.3. Let H(A, B) be η-cocoercive with respect to A with constant µ > oand η-relaxed cocoercive with respect to B with constantγ >0, A isα-expansive and B isβ-Lipschitz continuous and µ > γ, α > β.

Let M :X →2^X be H(·,·)-η-cocoercive operator with respect to A and B. If the following inequality hx−y, Jq(η(u, v))i ≥0, holds for all (v, y)∈ Graph(M),

thenx∈M u, where Graph(M) ={(u, x)∈X×X:x∈M u}.

Proof. Suppose that there exists some (u₀, x₀) such that

hx₀−y, Jq(η(u0, v))i ≥0, ∀ (v, y)∈ Graph(M). (3.1) SinceM is H(·,·)-η-cocoercive operator with respect toA and B, we know that (H(A, B) +λM)(X) =X, holds for everyλ >0 and so there exists (u₁, x₁)∈ Graph(M) such that

H(Au1, Bu1) +λx1 =H(Au0, Bu0) +λu0 ∈X. (3.2) It follows from (3.1) and (3.2) that

0≤λhx₀−x1, Jq(η(u0, u1))i=−hH(Au₀, Bu0)−H(Au1, Bu1), Jq(η(u0, u1))i

=−hH(Au₀, Bu0)−H(Au1, Bu0), Jq(η(u0, u1))i

− hH(Au₁, Bu₀)−H(Au₁, Bu₁), J_q(η(u₀, u₁))i

≤ −µkAu₀−Au₁k^q+γkBu₀−Bu₁k^q

≤ −µα^qku₀−u₁k^q+γβ^qku₀−u₁k^q

=−(µα^q−γβ^q)ku₀−u1k^q≤0,

which gives u₁ = u₀, since µ > γ and α > β. By (3.2), we have x₁ = x₀. Hence (u₀, x₀) = (u₁, x₁) ∈ Graph(M) and so x₀ ∈M u₀.

Theorem 3.4. Let H(A, B) be η-cocoercive with respect to A with constant µ >0 and η-relaxed cocoercive with respect toB with constantγ >0,A isα-expansive and B is β-Lipschitz continuous, µ > γ and α > β.

Let M be H(·,·)-η-cocoercive operator with respect to A and B. Then the operator (H(A, B) +λM)⁻¹ is single-valued.

Proof. For anyu∈X, letx, y∈(H(A, B) +λM)⁻¹(u). It follows that

−H(Ax, Bx) +u∈λM x and

−H(Ay, By) +u∈λM y asM isη-cocoercive (thusη-accretive), we have

0≤ h−H(Ax, Bx) +u−(−H(Ay, By) +u), J_q(η(x, y))i

=−hH(Ax, Bx)−H(Ay, By), J_q(η(x, y))i

=−hH(Ax, Bx)−H(Ay, Bx) +H(Ay, Bx)−H(Ay, By), Jq(η(x, y))i

=−hH(Ax, Bx)−H(Ay, Bx), Jq(η(x, y))i

− hH(Ay, Bx)−H(Ay, By), J_q(η(x, y))i. (3.3) Since H is η-cocoercive with respect to A with constant µ and η-relaxed cocoercive with respect to B with constantγ,A is α-expansive and B is β-Lipschitz continuous, thus (3.3) becomes

0≤ −µα^qkx−yk^q+γβ^qkx−yk^q=−(µα^q−γβ^q)kx−yk^q≤0, (3.4) sinceµ > γ and α > β. Thus, we havex=y and so (H(A, B) +λM)⁻¹ is single-valued.

Definition 3.5. LetH(A, B) beη-cocoercive with respect toAwith constantµ >0 andη-relaxed cocoercive with respect toB with constantγ >0,Aisα-expansive andB isβ-Lipschitz continuous,µ > γ andα > β.

LetM beH(·,·)-η-cocoercive operator with respect toA andB. The resolvent operator R^{H(·,·)−η}_λ,M :X→X is defined by

R^{H(·,·)−η}_λ,M (u) = (H(A, B) +λM)⁻¹(u), ∀ u∈X. (3.5) The following theorem shows that the resolvent operator is Lipschitz continuous.

Theorem 3.6. Let H(A, B) be η-cocoercive with respect to A with constant µ >0 and η-relaxed cocoercive with respect to B with constant γ >0, A is α-expansive, B is β-Lipschitz continuous and η is τ-Lipschitz continuous andµ > γ,α > β. LetM be an H(·,·)-η-cocoercive operator with respect to A and B. Then the resolvent operator R^{H(·,·)−η}_λ,M :X→X is τ^q−1

µα^q−γβ^q-Lipschitz continuous, that is kR^{H(·,·)−η}_λ,M (u)−R^{H(·,·)−η}_λ,M (v)k ≤ τ^q−1

µα^q−γβ^qku−vk, ∀ u, v∈X.

Proof. Letu and v be any given points inX. It follows from (3.5) that R^H_λ,M^(·,·)−η(u) = (H(A, B) +λM)⁻¹(u), and

R^H_λ,M^(·,·)−η(v) = (H(A, B) +λM)⁻¹(v).

This implies that 1

λ(u−H(A(R^H_λ,M^(·,·)−η(u)), B(R^{H(·,·)−η}_λ,M (u))))∈M(R^{H(·,·)−η}_λ,M (u)),

and 1

λ(v−H(A(R^H_λ,M^(·,·)−η(v)), B(R^{H(·,·)−η}_λ,M (v))))∈M(R^{H(·,·)−η}_λ,M (v)).

For the sake of clarity, we take

P u=R_λ,M^{H(·,·)−η}(u) and P v=R^{H(·,·)−η}_λ,M (v).

Since M isη-cocoercive (thusη-accretive), we have 1

λhu−H(A(P u), B(P u))−(v−H(A(P v), B(P v))), Jq(η(P u, P v))i

= 1

λhu−v−(H(A(P u), B(P u))−H(A(P v), B(P v))), Jq(η(P u, P v))i ≥0.

Therefore we have

hu−v, J_q(η(P u, P v))i ≥ hH(A(P u), B(P u))−H(A(P v), B(P v)), J_q(η(P u, P v))i.

It follows that

ku−vkkη(P u, P v)k^q−1 ≥ hu−v, J_q(η(P u, P v))i

≥ hH(A(P u), B(P u))−H(A(P v), B(P u)), Jq(η(P u, P v))i +hH(A(P v), B(P u))−H(A(P v), B(P v)), Jq(η(P u, P v))i

≥µkA(P u)−A(P v)k^q−γkB(P u)−B(P v)k^q

≥µα^qkP u−P vk^q−γβ^qkP u−P vk^q and so

ku−vkkη(P u, P v)k^q−1 ≥(µα^q−γβ^q)kP u−P vk^q or

(µα^q−γβ^q)kP u−P vk^q≤ ku−vkkη(P u, P v)k^q−1

≤ ku−vkτ^q−1kP u−P vk^q−1 kP u−P vk ≤ τ^q−1

µα^q−γβ^qku−vk, ∀ u, v∈X, or

kR^{H(·,·)−η}_λ,M (u)−R^{H(·,·)−η}_λ,M (v)k ≤ τ^q−1

µα^q−γβ^qku−vk, ∀u, v∈X.

This completes the proof.

4. An application for solving variational-like inclusions

In this section, we shall show that under suitable assumption, H(·,·)-η-cocoercive operator plays an important role for solving a variational-like inclusion problem.

Let η, N, W :X×X→X,g:X →X,H :X×X→ X,A, B :X →X be the single-valued mappings and T, Q, R, S:X →CB(X),M :X→2^X be the set-valued mappings such thatM is H(·,·)-η-cocoercive with respect toA and B. Then we consider the following problem.

Find u∈X,x∈T(u), y∈Q(u),z∈R(u),v∈S(u) such that

0∈N(x, y)−W(z, v) +M(g(u)). (4.1)

Problem (4.1) is called variational-like inclusion problem.

Below are some special cases of problem (4.1).

(i) If W, R, S = 0, then problem (4.1) reduces to the problem of findingu ∈X,x∈T(u),y∈Q(u) such that

0∈N(x, y) +M(g(u)). (4.2)

Problem (4.2) is introduced and studied by Peng [10].

(ii) IfN(x, y) =N(x),T is single-valued andg=I, the identity mapping, then problem (4.2) reduces to the problem considered by Bi et al.[3]. That is to find u∈X such that

0∈N(u) +M(u). (4.3)

Lemma 4.1. (u, x, y, z, v), where u∈X, x∈T(u), y∈Q(u), z∈R(u), v∈S(u) is a solution of problem (4.1) if and only if (u, x, y, z, v) is the solution of the following equation.

g(u) =R^{H(·,·)−η}_λ,M [H(A(gu), B(gu))−λ{N(x, y)−W(z, v)}], (4.4) where λ >0 is a constant.

Proof. Proof is direct consequence of definition of resolvent operator.

Based on (4.4), we have the following iterative algorithm.

Algorithm 4.2. For any given u₀ ∈ X, x₀ ∈ T(u₀), y₀ ∈ Q(u₀), z₀ ∈ R(u₀), v₀ ∈ S(u₀), compute the sequences {u_n}, {x_n}, {y_n}, {z_n} and {v_n} by the following iterative procedure.

g(u_n+1) =R^{H(·,·)−η}_λ,M [H(A(gu_n), B(g(u_n))−λ{N(x_n, y_n)−W(z_n, u_n)}], (4.5) kx_n+1−xnk ≤ D(T(u_n+1), T(un)); (4.6) ky_n+1−y_nk ≤ D(Q(u_n+1), Q(u_n)); (4.7) kz_n+1−znk ≤ D(R(u_n+1), R(un)); (4.8) kv_n+1−v_nk ≤ D(S(u_n+1), S(u_n)); (4.9) where n is the iteration number,λ >0 is a constant and D is the Hausd¨orff metric on CB(X).

Theorem 4.3. LetXbe aq-uniformly smooth Banach space. LetA, B, g:X→X,H, N, η, W :X×X→X be the single-valued mappings. LetT, Q, R, S :X→CB(X) andM :X→2^X be the multi-valued mappings such thatM is H(·,·)-η-cocoercive mapping with respect to A and B. Suppose that

(i) g isδ-strongly accretive and λ_g-Lipschitz continuous,

(ii) N is Lipschitz continuous for the first argument with constant λN1 and λN2 for the second argument, η-strongly accretive with respect to T and Q with constant t,

(iii) W is Lipschitz continuous for the first argument with constantλW1 and λW2 for the second argument, (iv) η is τ-Lipschitz continuous,A is α-expansive andB is β-Lipschitz continuous,

(v) H(A, B) is η-cocoercive with respect to A with constant µ and η-relaxed-cocoercive with respect to B with constantγ,r1-Lipschitz continuous with respect toA and r2-Lipschitz continuous with respect to B,

(vi) T, Q, R, S are D-Lipschitz continuous mappings with constantsλT, λQ, λR and λS, respectively.

Suppose that the following condition is satisfied:

hqq

(r1+r2)^qλ^qg−qλt+qλ(λN1λT +λN2λQ)[(r1+r2)^q−1λ^q−1g +τ^q−1] i

+λ^qCq(λN1λT +λN2λQ)^q

< δ

τ^q(µα^q−γβ^q)−λ(λ_W1λ_R+λ_W2λ_S) , δ

τ^q(µα^q−γβ^q)> λ(λ_W1λ_R+λ_W2λ_S), µ > γ, α > β.

(4.10)

Then there existu∈X, x∈T(u), y∈Q(u), z∈R(u) and v∈S(u) satisfying the variational-like inclusion problem (4.1) and the iterative sequences {u_n}, {x_n}, {y_n}, {z_n} and {v_n} generated by Algorithm 4.1 converge strongly to u, x,y, z, and v, respectively.

Proof. Since g isδ-strongly accretive, we have

kg(u_n+1)−g(un)kku_n+1−unk^q−1 =kg(u_n+1)−g(un)kkJ_q(un+1−un)k

≥ hg(u_n+1)−g(un), Jq(un+1−un)i

≥δku_n+1−u_nk^q. (4.11)

From (4.11), we get

ku_n+1−unk ≤ 1

δkg(u_n+1)−g(un)k. (4.12)

By Algorithm 4.2 and Theorem 3.6, we have

kg(u_n+1)−g(u_n)k=kR^{H(·,·)−η}_λ,M [H(A(gu_n), B(gu_n))−λ{N(x_n, y_n)−W(z_n, v_n)}]

−R^{H(·,·)−η}_λ,M [H(A(gun−1), B(gun−1))

−λ{N(xn−1, yn−1)−W(zn−1, vn−1)}]k

≤ τ^q−1

µα^q−γβ^qkH(A(gu_n), B(gu_n))−H(A(gun−1), B(gun−1))

−λ{N(xn, yn)−N(xn−1, yn−1)}

−λ{W(zn, vn)−W(zn−1, vn−1)}k

≤ τ^q−1

µα^q−γβ^qkH(A(gu_n), B(gu_n))−H(A(gun−1), B(gun−1))

−λ{N(x_n, y_n)−N(xn−1, yn−1)}k + τ^q−1λ

µα^q−γβ^qkW(zn, vn)−W(zn−1, vn−1)k. (4.13) Using Lipschitz continuity of N with constantλN1 for the first argument andλN2 for the second argument and D-Lipschitz continuity ofT and Qwith constants λT and λQ, respectively, we have

kN(xn, yn)−N(xn−1, yn−1)k=kN(xn, yn)−N(xn−1, yn)

+N(xn−1, yn)−N(xn−1, yn−1)k

≤ kN(x_n, y_n)−N(xn−1, y_n)k

+kN(xn−1, y_n)−N(xn−1, yn−1)k

≤λ_N1kx_n−xn−1k+λ_N2ky_n−yn−1k

≤λ_N1D(T(u_n), T(un−1)) +λ_N2D(Q(u_n), Q(un−1))

≤λN1λTku_n−un−1k+λN2λQku_n−un−1k

= (λN1λT +λN2λQ)ku_n−un−1k. (4.14) Also asH(A, B) isr1-Lipschitz continuous with respect toA andr2-Lipschitz continuous with respect to B and gis λ_g-Lipschitz continuous, we have

kH(A(gu_n), B(gun))−H(A(gun−1), B(gun−1))k ≤(r1+r2)λgku_n−un−1k. (4.15) By using Lemma 2.1, (4.14), (4.15) andη-strongly accretiveness ofN with respect toT andQwith constant tand τ-Lipschitz continuity ofη, we have

kH(A(gu_n), B(gun))−H(A(gun−1), B(gun−1))−λ{N(xn, yn)−N(xn−1, yn−1)}k^q

≤ kH(A(gu_n), B(gu_n))−H(A(gun−1), B(g(un−1))k^q

−qλhN(x_n, y_n)−N(xn−1, yn−1), J_q(η(u_n, un−1))i

−qλhN(x_n, y_n)−N(xn−1, yn−1), J_q[H(A(gu_n), B(gu_n))−H(A(gun−1), B(gun−1))]

−Jq(η(un, un−1))i+λ^qCqkN(xn, yn)−N(xn−1, yn−1)k^q

≤(r1+r2)^qλ^q_gku_n−un−1k^q−qλtku_n−un−1k^q+qλkN(xn, yn)−N(xn−1, yn−1)k×

kH(A(gu_n)−, B(gu_n))−H(A(gun−1), B(gun−1))k^q−1+kη(u_n, un−1)k^q−1 +λ^qC_q(λ_N1λ_T +λ_N2)^qku_n−un−1k^q

≤(r₁+r₂)^qλ^q_gku_n−un−1k^q−qλtku_n−un−1k^q+qλ(λ_N1λ_T +λ_N2λ_Q)ku_n−un−1k×

(r1+r2)^q−1λ^q−1_g ku_n−un−1k^q−1+τ^q−1ku_n−un−1k^q−1 +λ^qCq(λN1λT +λN2λQ)^qku_n−un−1k^q

=

(r₁+r₂)^qλ^q_g−qλt+qλ(λ_N1λ_T +λ_N2λ_Q)[(r₁+r₂)^q−1λ^q−1_g +τ^q−1] +λ^qCq(λN1λT +λN2λQ)^q

ku_n−un−1k^q. (4.16)

Using Lipschitz continuity ofW with constantλ_W1 for the first argument andλ_W2 for the second argument and D-Lipschitz continuity ofR and S with constants λR and λS, respectively, we obtain

kW(zn, vn)−W(zn−1, vn−1)k ≤(λW1λR+λW2λS)ku_n−un−1k. (4.17) In view of (4.16) and (4.17), (4.13) becomes

kg(u_n)−g(un−1)k ≤h τ^q−1 µα^q−γβ^q

qq

(r1+r2)^qλ^qg−qλt+qλ(λN1λT +λN2λQ) [(r1+r2)^q−1λ^q−1g +τ^q−1] +λ^qCq(λ_N1λ_T +λ_N2λ_Q)^q

+ τ^q−1λ

µα^q−γβ^q(λ_W1λ_R+λ_W2λ_S)i

ku_n−un−1k. (4.18)

Using (4.18), (4.12) becomes

ku_n+1−u_nk ≤θku_n−un−1k. (4.19)

where

θ= 1 δ

h τ^q−1 µα^q−γβ^q

qq

(r₁+r₂)^qλ^qg−qλt+qλ(λ_N1λ_T +λ_N2λ_Q) [(r₁+r₂)^q−1λ^q−1_g +τ^q−1] +λ^qC_q(λ_N1λ_T +λ_N2λ_Q)^q

+ τ^q−1λ

µα^q−γβ^q(λW1λR+λW2λS) i

ku_n−un−1k.

By condition (4.10), 0 < θ <1 and hence {u_n} is Cauchy sequence in X, so there exists u∈ X such that un→u asn→ ∞. By theD-Lipschitz continuity of T, Q, Rand S, we have

kx_n+1−xnk ≤λ_Tku_n+1−unk;

ky_n+1−ynk ≤λ_Qku_n+1−unk;

kz_n+1−znk ≤λRku_n+1−unk;

kv_n+1−n_nk ≤λ_Sku_n+1−u_nk;

which shows that the sequences{x_n},{y_n},{z_n}and {v_n} are all Cauchy sequences inX, so there existx, y, z, andv ∈X such thatxn→x,yn→y,zn→z and vn→ v, as n→ ∞. By the Lipschitz continuity of the operations H,A,B,N,W,R^H_λ,M^(·,·)−η and Algorithm 4.2, it follows that

g(u) =R^{H(·,·)−η}_λ,M [H(A(gu), B(gu))−λ{N(x, y)−W(z, v)}].

It remain to show thatx∈T(u). In fact, since xn∈T(un). we have d(x, T(u))≤ kx−x_nk+d(x_n, T(u))

≤ kx−xnk+D(T(un), T(u))

≤ kx−xnk+λTku_n−uk →0, as n→ ∞,

which implies thatd(x, T(u)) = 0, since T(u) ∈ CB(X), it follows that x ∈ T(u). Similarly, we can show thaty∈Q(u),z∈R(u) and v∈S(u). This completes the proof.

Acknowledgements:

This work is supported by Department of Science and Technology, Government of India under grant no.

SR/S4/MS: 577/09.

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