# Algorithm for Solving a New System of Generalized Nonlinear Quasi-Variational-Like Inclusions in Hilbert Spaces

Authors: Shamshad Husain; Sanjeev Gupta; Huma Sahper
Journal: Chinese Journal of Mathematics
 (2014)

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## Abstract

We introduce and study a new system of generalized nonlinear quasi-variational-like inclusions withH(·,·)-cocoercive operator in Hilbert spaces. We suggest and analyze a class of iterative algorithms for solving the system of generalized nonlinear quasi-variational-like inclusions. An existence theorem of solutions for the system of generalized nonlinear quasi-variational-like inclusions is proved under suitable assumptions which show that the approximate solutions obtained by proposed algorithms converge to the exact solutions.


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## Body

## 1. Introduction

Variational inclusion problems are important generalization of classical variational inequalities and have wide applications to many fields including mechanics, physics, optimization and control, nonlinear programming, economics, and engineering sciences; see, for example, [1]. For these reasons, various variational inclusions have been intensively studied in recent years. Many efficient ways have been studied to find solutions for variational inclusions. Those methods include the projection method and its various forms, linear approximation, descent and Newton’s method, and the method based on auxiliary principle technique. The method based on the resolvent operator technique is a generalization of the projection method and has been widely used to solve variational inclusions. For details, we refer to see [2–19].Recently, Fang and Huang, Kazmi and Khan, and Lan et al. investigated several resolvent operators for generalized operators such asH-monotone [3, 17], H-accretive [4], A-maximal relaxed accretive [14], (H,η)-monotone [5], (A,η)-accretive [13], (P,η)-proximal point [8], and (P,η)-accretive [9] operators. Very recently, Zou and Huang [19] introduced and studied H(·,·)-accretive operators, Kazmi et al. [10–12] introduced and studied generalized H(·,·)-accretive operators and H(·,·)-η-proximal point mapping, and Xu and Wang [18] introduced and studied (H(·,·),η)-monotone operators. Ahmad et al. [2, 8] introduced and studied H(·,·)-cocoercive operators, showed some properties of the resolvent operator for the H(·,·)-cocoercive operators, and obtained an application for solving variational inclusions in Hilbert spaces. They also gave some examples to illustrate their results.Inspired and motivated by the researches going on in this area, we introduce and discuss a new system of generalized nonlinear quasi-variational-like inclusions involvingH(·,·)-cocoercive operators in Hilbert spaces. By using the resolvent operators associated with H(·,·)-cocoercive operators due to Ahmad et al. [2], we prove that the approximate solutions obtained by the iterative algorithms converge to the exact solutions of our system of generalized nonlinear quasi-variational-like inclusions. Our results can be viewed as an extension and generalization of some known results in the literature.


## 2. Preliminaries

Throughout this paper, we suppose thatX is a real Hilbert space endowed with a norm ∥·∥ and an inner product 〈·,·〉, d is the metric induced by the norm ∥·∥, 2X (resp., CB(X)) is the family of all the nonempty (resp., closed and bounded) subsets of X and 𝒟(·,·) is the Hausdorff metric on CB(X) defined by
(1)𝒟(P,Q)=max⁡{sup⁡x∈P⁡d(x,Q),sup⁡y∈P⁡d(P,y)},11111111111111111111111∀P,Q∈CB(X),
where d(x,Q)=inf⁡y∈Q∥x-y∥ and d(P,y)=inf⁡x∈P⁡∥x-y∥.In the sequel, let us recall some concepts.Definition 1 (see [20]).
A mappingg:X→X is said to be(i)
monotone if(2)〈g(x)-g(y),x-y〉≥0,∀x,y∈X;(ii)
ξ-strongly monotone if there exists a constant ξ>0 such that
(3)〈g(x)-g(y),x-y〉≥ξ∥x-y∥2,∀x,y∈X;(iii)
μ-cocoercive if there exists a constant μ>0 such that
(4)〈g(x)-g(y),x-y〉≥μ∥g(x)-g(y)∥2,∀x,y∈X;(iv)
γ-relaxed cocoercive if there exists a constant γ>0 such that
(5)〈g(x)-g(y),x-y〉≥(-γ)∥g(x)-g(y)∥2,∀x,y∈X;(v)
λg-Lipschitz continuous if there exists a constant λg>0 such that
(6)∥g(x)-g(y)∥≤λg∥x-y∥,∀x,y∈X;(vi)
α-expansive if there exists a constant α>0 such that
(7)∥g(x)-g(y)∥≥α∥x-y∥,∀x,y∈X;
if α=1, then it is expansive.Definition 2.
LetM:X→2X be set-valued mapping. Then, M is said to be μ′-cocoercive if there exists a constant μ′>0 such that
(8)111〈u-v,η(x,y)〉≥μ′∥u-v∥2,∀x,y∈X,u∈M(x),v∈M(y).Definition 3 (see [2]).
LetH:X×X→X and A,B:X→X be the single-valued mappings. Then,(i)
H(A,·) is said to be μ-cocoercive with respect to A if there exists a constant μ>0 such that
(9)〈H(Ax,u)-H(Ay,u),x-y〉≥μ∥Ax-Ay∥2,111111111111111111111111111111,1∀x,y∈X;(ii)
H(·,B) is said to be γ-relaxed cocoercive with respect to B if there exists a constant γ>0 such that
(10)〈H(u,Bx)-H(u,By),x-y〉≥(-γ)∥Bx-By∥2,111111111111111111111111111111111∀x,y∈X;(iii)
H(A,·) is said to be r1-Lipschitz continuous with respect to A if there exists a constant r1>0 such that
(11)∥H(Ax,·)-H(Ay,·)∥≤r1∥x-y∥,∀x,y∈X;(iv)
H(·,B) is said to be r2-Lipschitz continuous with respect to B if there exists a constant r2>0 such that
(12)∥H(·,Bx)-H(·,By)∥≤r2∥x-y∥,∀x,y∈X.Definition 4 (see [2]).
LetH:X×X→X and A,B:X→X be the single-valued mappings. Then, the set-valued mapping M:X→2X is said to be H(·,·)-cocoercive with respect to A and B (or simply H(·,·)-cocoercive in the sequel), if(i)
M is cocoercive;(ii)
(H(A,B)+λM)(X)=X, for every λ>0.Example 5.
LetX=ℝ2 with the usual inner product. Let A,B:ℝ2→ℝ2 be defined by
(13)Ax=(2x1-2x2,-2x1+4x2),By=(-y1+y2,-y2),∀x,y∈ℝ2.
Suppose thatH(A,B):ℝ2×ℝ2→ℝ2 is defined by
(14)H(Ax,By)=Ax+By,∀x,y∈ℝ2.
Then, it is easy to check thatH(A,B) is 1/6-cocoercive with respect to A and 1/2-relaxed cocoercive with respect to B.
LetM=I, where I is the identity mapping. Then, M is H(·,·)-cocoercive mapping with respect to A and B.Example 6.
LetX=𝕊2, where 𝕊2 denotes the space of all 2×2 real symmetric matrices. Let H(Ax,By)=x2-y, for all x,y∈𝕊2 and M=I. Then, for λ=1, we have
(15)(H(A,B)+M)(x)=x2-x+x=x2,
but
(16)[000-1]∉(H(A,B)+M)(𝕊2),
because [000-1] is not the square of any 2×2 real symmetric matrix. Hence, M is not H(·,·)-cocoercive with respect to A and B.Proposition 7 (see [2]).
LetH(A,B) be μ-cocoercive with respect to A and γ-relaxed cocoercive with respect to B; A is α-expansive, B is β-Lipschitz continuous, and μ>γ, α>β. Let M:X→2X be an H(·,·)-cocoercive operator. If the inequality
(17)〈u-v,x-y〉≥0
holds for all (y,v)∈Graph(M), then u∈Mx, where
(18)Graph(M)={(u,x)∈X×X:x∈M(u)}.Theorem 8 (see [2]).
LetH(A,B) be μ-cocoercive with respect to A and γ-relaxed cocoercive with respect to B; A is α-expansive, B is β-Lipschitz continuous, and  μ>γ and α>β. Let M be an H(·,·)-cocoercive operator with respect to A and B. Then, the operator (H(A,B)+λM)-1 is single-valued.Definition 9 (see [2]).
LetH(A,B) be μ-cocoercive with respect to A and γ-relaxed cocoercive with respect to B; A is α-expansive, B is β-Lipschitz continuous, and μ>γ, α>β. Let M be an H(·,·)-cocoercive operator with respect to A and B. The resolvent operator Rλ,MH(·,·):X→X is defined by
(19)Rλ,MH(·,·)(u)=(H(A,B)+λM)-1(u),∀u∈X.Theorem 10 (see [2]).
LetH(A,B) be μ-cocoercive with respect to A and γ-relaxed cocoercive with respect to B, A is α-expansive, B is β-Lipschitz continuous, and μ>γ, α>β with r=μα2-γβ2. Let M be an H(·,·)-cocoercive operator with respect to A and B. Then, the resolvent operator Rλ,MH(·,·):X→X is 1/(μα2-γβ2)-Lipschitz continuous; that is,
(20)∥Rλ,MH(·,·)(u)-Rλ,MH(·,·)(v)∥≤1μα2-γβ2∥u-v∥,∀u,v∈X.


## 3. The System of Generalized Nonlinear Quasi-Variational-Like Inclusions and Iterative Algorithm

LetX1 and X2 be real Hilbert spaces. Let P:X1×X2→X1, Q:X1×X2→X2, g,A, B:X1→X1, h,C, D:X2→X2, G:X1×X1→X1, and H:X2×X2→X2 be the single-valued mappings. Let S:X1→CB(X1) and T:X2→CB(X2) be the set-valued mappings, let M:X1×X1→2X1 be a set-valued mapping such that, for each a∈X1, M(·,a) is a G(·,·)-cocoercive operator with respect to A and B, and let  N:X2×X2→2X2 be a set-valued mapping such that, for each b∈X2, N(·,b) is an H(·,·)-cocoercive operator with respect to C and D. Assume that g(X1)∩dom⁡(M(·,a))≠∅ for each a∈X1, andh(X2)∩dom⁡(N(·,b))≠∅ for eachb∈X2. Then, we consider the following system of generalized nonlinear quasi-variational-like inclusions.Find(a,b)∈X1×X2 with p∈S(a), q∈T(b) such that
(21)0∈P(p,q)+M(g(a),a),0∈Q(p,q)+N(h(b),b).Next are some special cases of problem (21).(1)
LetS, T be identity mappings, for each (a,b)∈X1×X2, M(g(a),a)=M(x) and N(h(b),b)=N(x); then problem (21) reduces to the following problem considered in [15]:
(22)0∈P(p,q)+M(a),0∈Q(p,q)+N(b).(2)
X1=X2,  a=b, S=T is an identity mapping, g=h, P(p,q)=Q(p,q)=P(·), for each (a,b)∈X1×X2, M(g(a),a)=N(h(b),b)=M(g(x)); then problem (21) reduces to the following problem considered in [16]:
(23)0∈P(x)+M(g(a)).For a suitable choice of the mappingsg, h, P, Q, G, H, A, B, C, D, S, T and the space X1=X2, a number of known systems of quasi-variational inequalities, systems of variational inequalities, systems of quasi-variational inclusions, and variational inclusions can be obtained as special cases of the generalized nonlinear quasi-variational inclusion problem (21). We would like to mention that the problem of finding zero of the sum of two maximal monotone operators is also a special case of problem (21). Furthermore, these types of variational inclusion enable us to study many important problems arising in mathematical, physical, and engineering science in a general and unified framework.Lemma 11.
LetX1 and X2 be real Hilbert spaces. Let P:X1×X2→X1, Q:X1×X2→X2,g,A,B:X1→X1,h,C,D:X2→X2 be the single-valued mappings. Let G:X1×X1→X1be a single-valued mapping such that G(A,B) is μ1-cocoercive with respect to A and γ1-relaxed cocoercive with respect to B; A is α1-expansive, B is β1-Lipschitz continuous, and μ1>γ1, α1>β1. Let  H:X2×X2→X2 be a single-valued mapping such that H(C,D) is μ2-cocoercive with respect to C and γ2-relaxed cocoercive with respect to D; C is α2-expansive, D is β2-Lipschitz continuous, and μ2>γ2, α2>β2. Let S:X1→CB(X1) and T:X2→CB(X2) be the set-valued mappings, let M:X1×X1→2X1be a set-valued mapping such that, for each a∈X1, M(·,a) is a G(·,·)-cocoercive operator, and let  N:X2×X2→2X2 be a set-valued mapping such that, for each b∈X2, N(·,b) is an H(·,·)-cocoercive operator. Assume that g(X1)∩dom⁡(M(·,a))≠∅ for each a∈X1, andh(X2)∩dom⁡(N(·,b))≠∅ for eachb∈X2. Then, for any (a,b)∈X1×X2 with p∈S(a), q∈T(b) is a solution of the problem (21), if and only if
(24)g(a)=Rλ1,M(·,a)G(·,·)[G(A(g(a)),B(g(a)))-λ1P(p,q)],h(b)=Rλ2,N(·,b)H(·,·)[H(C(h(b)),D(h(b)))-λ2Q(p,q)],
where Rλ1,M(·,a)G(·,·)=(G(·,·)+λ1M(·,a))-1, Rλ2,N(·,b)H(·,·)=(H(·,·)+λ2N(·,b))-1, and λ1,λ2>0 are constants.Proof.
By using the definitions of the resolvent operatorsRλ1,M(·,a)G(·,·) and Rλ2,N(·,b)H(·,·), the conclusion follows directly.The preceding lemma allows us to suggest the following iterative algorithm for problem (21).Algorithm 12.
For(a0,b0)∈X1×X2 with p∈S(a), q∈T(b), compute the sequences {an}, {bn}, {pn}, {qn} as follows:
(25)g(an+1)=Rλ1,M(·,a)G(·,·)[G(A(g(an)),B(g(an)))11111111111111111-λ1P(pn,qn)[G(A(g(an)),B(g(an)))],λ1>0,h(bn+1)=Rλ2,N(·,b)H(·,·)[H(C(h(bn)),D(h(bn)))11111111111111111-λ2Q(pn,qn)[H(C(h(bn)),D(h(bn)))],λ2>0,pn∈S(an),∥pn-pn+1∥≤(1+1n+1)pn∈S(an),∥pn-pn+1∥≤×𝔇(S(an),S(an+1)),qn∈T(bn),∥qn-qn+1∥≤(1+1n+1)pn∈S(an),∥pn-pn+1∥≤×𝔇(T(bn),T(bn+1)),
for all n=0, 1, 2,…, and λ1 and λ2 are constants.Definition 13.
LetS:X1→CB(X1) and T:X2→CB(X2) be two set-valued mappings. A single-valued mapping P:X1×X2→X1 is said to be(i)
ϵ1-Lipschitz continuous in the first argument with respect to S, if there exists a constant ϵ1>0 such that
(26)11∥P(p1,·)-P(p2,·)∥≤ϵ1∥p1-p2∥,∀p1∈S(a1),p2∈S(a2),a1,a2∈X1;(ii)
ϵ2-Lipschitz continuous in the second argument with respect to T, if there exists a constant ϵ2>0 such that
(27)11,∥P(·,q1)-P(·,q2)∥≤ϵ2∥q1-q2∥,∀q1∈T(b1),q2∈T(b2),b1,b2∈X2.Definition 14.
LetS:X1→CB(X1) and T:X2→CB(X2) be two set-valued mappings. A single-valued mapping Q:X1×X2→X2 is said to be(i)
δ1-Lipschitz continuous in the first argument with respect to S, if there exists a constant δ1>0 such that
(28)1,∥Q(p1,·)-Q(p2,·)∥≤δ1∥p1-p2∥,∀p1∈S(a1),p2∈S(a2),a1,a2∈X1;(ii)
δ2-Lipschitz continuous in the second argument with respect to T, if there exists a constant δ2>0 such that
(29)1,∥Q(·,q1)-Q(·,q2)∥≤δ2∥q1-q2∥,∀q1∈T(b1),q2∈T(b2),b1,b2∈X2.Definition 15.
A set-valued mappingA:X→CB(X) is said to be 𝔇-Lipschitz continuous, if there exists a constant ρ>0 such that
(30)𝔇(A(x),A(y))≤ρ∥x-y∥,∀x,y∈X.Theorem 16.
LetX1 and X2 be real Hilbert spaces. Let P:X1×X2→X1, Q:X1×X2→X2,g,A,B:X1→X1, h,C,D:X2→X2, G:X1×X1→X1, and H:X2×X2→X2 be the single-valued mappings. Let S:X1→CB(X1) and T:X2→CB(X2) be the set-valued mappings, let M:X1×X1→2X1be a set-valued mapping such that, for each a∈X1, M(·,a) is a G(·,·)-cocoercive operator with respect to A and B, and let N:X2×X2→2X2 be a set-valued mapping such that for each b∈X2, N(·,b) is an H(·,·)-cocoercive operator with respect to C and D. Assume that g(X1)∩dom⁡(M(·,a))≠∅ for each a∈X1, andh(X2)∩dom⁡(N(·,b))≠∅ for eachb∈X2. Assume that (i)
S, T are ρ, τ-Lipschitz continuous in the Hausdorff metric 𝔇(·,·), respectively;(ii)
G(A,B) is μ1-cocoercive with respect to A and γ1-relaxed cocoercive with respect to B;(iii)
H(C,D) is μ2-cocoercive with respect to C and γ2-relaxed cocoercive with respect to D;(iv)
A, C is α1, α2-expansive, respectively;(v)
B, D is β1, β2-Lipschitz continuous, respectively;(vi)
g is λg-Lipschitz continuous and ξ1-strongly monotone;(vii)
h is λh-Lipschitz continuous and ξ2-strongly monotone;(viii)
G(A,B) is r1-Lipschitz continuous with respect to A and r2-Lipschitz continuous with respect to B;(ix)
H(C,D) is s1-Lipschitz continuous with respect to C and s2-Lipschitz continuous with respect to D;(x)
P(·,·) is ϵ1-Lipschitz continuous in the first argument and ϵ2-Lipschitz continuous in the second argument;(xi)
Q(·,·)is δ1-Lipschitz continuous in the first argument and δ2-Lipschitz continuous in the second argument;(xii)
(31)0<(r1+r2)λg+λ1ϵ1ρξ1(μ1α12-γ1β12)+λ2δ1(1+1/n)ρξ2(μ2α22-γ2β22)<1,0<(s1+s2)λh+λ2δ2τξ2(μ2α22-γ2β22)+λ1ϵ2(1+1/n)τξ1(μ1α12-γ1β12)<1.
Then, the iterative sequences{an}, {bn}, {pn}, and {qn}, generated by Algorithm 12, converge strongly to a, b, p, and q,respectively, and (a, b, p, q) is a solution of problem (21).Proof.
SinceS, T are Lipschitz continuous with constants ρ, τ, respectively, it follows from Algorithm 12 that
(32)∥pn-pn+1∥≤(1+1n)𝔇(S(an),S(an+1))∥pn-pn+1∥≤(1+1n)ρ∥an-an+1∥,∥qn-qn+1∥≤(1+1n)𝔇(T(bn),T(bn+1))∥qn-qn+1∥≤(1+1n)τ∥bn-bn+1∥,
for all n=0,1,2,….
Using theξ-strong monotonicity of g, we have
(33)∥g(an+1)-g(an)∥∥an+1-an∥≥〈g(an+1)-g(an),an+1-an〉≥ξ1∥an+1-an∥2,
which implies that
(34)∥an+1-an∥≤1ξ1∥g(an+1)-g(an)∥.
Now we estimate∥g(an+1)-g(an)∥by using Algorithm 12 and the Lipschitz continuity of Rλ1,M(·,a)G(·,·) as follows:
(35)∥g(an+1)-g(an)∥=∥Rλ1,M(·,a)G(·,·)[G(A(g(an)),B(g(an)))-λ1P(pn,qn)]11111-[Rλ1,M(·,a)G(·,·){G(A(g(an-1)),B(g(an-1)))11111111111111111-λ1P(pn-1,qn-1)(A(g(an-1)),B(g(an-1)))}[Rλ1,M(·,a)G(·,·){G(A(g(an-1)),B(g(an-1)))]∥≤1μ1α12-γ1β12∥G(A(g(an)),B(g(an)))11111111111111111-G(A(g(an-1)),B(g(an-1)))∥+λ1μ1α12-γ1β12∥P(pn,qn)-P(pn-1,qn-1)∥≤1μ1α12-γ1β12∥G(A(g(an)),B(g(an)))11111111111111111-G(A(g(an-1)),B(g(an)))∥+1μ1α12-γ1β12∥G(A(g(an-1)),B(g(an)))111111111111111111-G(A(g(an-1)),B(g(an-1)))∥+λ1μ1α12-γ1β12∥P(pn,qn)-P(pn-1,qn)∥+λ1μ1α12-γ1β12∥P(pn-1,qn)-P(pn-1,qn-1)∥.
SinceG(A,B) is r1-Lipschitz continuous with respect to A and r2-Lipschitz continuous with respect to B, P is ϵ1-Lipschitz continuous in the first argument and ϵ2-Lipschitz continuous in the second argument. g is λg-Lipschitz continuous and using (32), (34), and (35) it becomes
(36)∥g(an+1)-g(an)∥≤(r1+r2)λgμ1α12-γ1β12∥an-an-1∥∥g(an+1)-g(an)∥=+λ1ϵ1μ1α12-γ1β12∥pn-pn-1∥∥g(an+1)-g(an)∥=+λ1ϵ2μ1α12-γ1β12∥qn-qn-1∥∥g(an+1)-g(an)∥≤(r1+r2)λgμ1α12-γ1β12∥an-an-1∥∥g(an+1)-g(an)∥=+λ1ϵ1μ1α12-γ1β12(1+1n)ρ∥an-an-1∥∥g(an+1)-g(an)∥=+λ1ϵ2μ1α12-γ1β12(1+1n)τ∥bn-bn-1∥,∥an+1-an∥≤(r1+r2)λg+λ1ϵ1(1+1/n)ρξ1(μ1α12-γ1β12)∥an-an-1∥∥an+1-an∥=+λ1ϵ2(1+1/n)τξ1(μ1α12-γ1β12)∥bn-bn-1∥.
Let(37)∥an+1-an∥≤θ1n∥an-an-1∥+θ2n∥bn-bn-1∥,
where
(38)θ1n=(r1+r2)λg+λ1ϵ1(1+1/n)ρξ1(μ1α12-γ1β12),θ2n=λ1ϵ2(1+1/n)τξ1(μ1α12-γ1β12).
Now we estimate∥h(bn)-h(bn-1)∥ by using Algorithm 12 and the Lipschitz continuity of Rλ2,N(·,b)H(·,·) as follows:
(39)∥h(bn+1)-h(bn)∥=∥Rλ2,N(·,b)H(·,·)[H(C(h(bn)),D(h(bn)))-λ2Q(pn,qn)]111111-Rλ2,N(·,b)H(·,·)[H(C(h(bn-1)),D(h(bn-1)))1111111111111111-λ2Q(pn-1,qn-1)(C(h(bn-1)),D(h(bn-1)))]-Rλ2,N(·,b)H(·,·)[H(C(h(bn)),D(h(bn)))-λ2Q(pn,qn)]∥≤1μ2α22-γ2β22∥H(C(h(bn)),D(h(bn)))11111111111111111-H(C(h(bn-1)),D(h(bn-1)))∥+λ2μ2α22-γ2β22∥Q(pn,qn)-Q(pn-1,qn-1)∥≤1μ2α22-γ2β22∥H(C(h(bn)),D(h(bn)))11111111111111111-H(C(h(bn-1)),D(h(bn)))∥+1μ2α22-γ2β22∥H(C(h(bn-1)),D(h(bn)))111111111111111111-H(C(h(bn-1)),D(h(bn-1)))∥+λ2μ2α22-γ2β22∥Q(pn,qn)-Q(pn-1,qn)∥+λ2μ2α22-γ2β22∥Q(pn-1,qn)-Q(pn-1,qn-1)∥.
SinceH(C,D) is s1-Lipschitz continuous with respect to C and s2-Lipschitz continuous with respect to D, Q is δ1-Lipschitz continuous in the first argument and δ2-Lipschitz continuous in the second argument. h is λh-Lipschitz continuous and using (32), (34), and (39) it becomes
(40)∥h(bn+1)-h(bn)∥≤(s1+s2)λhμ2α22-γ2β22∥bn-bn-1∥∥h(bn+1)-h(bn)∥=+λ2δ1μ2α22-γ2β22(1+1n)ρ∥an-an-1∥∥h(bn+1)-h(bn)∥=+λ2δ2μ2α22-γ2β22(1+1n)τ∥bn-bn-1∥.
In the light of (34), we have
(41)∥bn+1-bn∥≤(s1+s2)λh+λ2δ2(1+1/n)τξ2(μ2α22-γ2β22)∥bn-bn-1∥∥bn+1-bn∥=+λ2δ1ξ2(μ2α22-γ2β22)(1+1n)ρ∥an-an-1∥.
Let(42)∥bn+1-bn∥≤θ3n∥an-an-1∥+θ4n∥bn-bn-1∥,
where
(43)θ3n=λ2δ1(1+1/n)ρξ2(μ2α22-γ2β22),θ4n=(s1+s2)λh+λ2δ2(1+1/n)τξ2(μ2α22-γ2β22).
From adding (37) and (42), we get
(44)∥an+1-an∥+∥bn+1-bn∥≤(θ1n+θ3n)∥an-an-1∥∥an+1-an∥+∥bn+1-bn∥=+(θ2n+θ4n)∥bn-bn-1∥∥an+1-an∥+∥bn+1-bn∥≤Θn{∥an-an-1∥+∥bn-bn-1∥},
where Θn=max⁡{(θ1n+θ3n),(θ2n+θ4n)}.
Lettingn→∞, we obtain Θn→θ, where
(45)θ=max⁡{(θ1+θ3),(θ2+θ4)},θ1=(r1+r2)λg+λ1ϵ1ρξ1(μ1α12-γ1β12),θ2=λ1ϵ2τξ1(μ1α12-γ1β12),θ3=λ2δ1ρξ2(μ2α22-γ2β22),θ4=(s1+s2)λh+λ2δ2τξ2(μ2α22-γ2β22).
By (31), θ∈(0,1), and (44) {an} and {bn} both are Cauchy sequences. Thus, there exists (a,b)∈X1×X2 such that an→a, bn→b, as n→∞. From the Lipschitz continuity of S and T and (32), {pn},{qn} are also Cauchy sequences, and thus there exists (p,q)∈X1×X2 such that pn→p, qn→q, as n→∞.
Now, we prove thatp∈S(a) and q∈T(b). In fact, since pn∈S(an) and qn∈T(bn), we have
(46)d(p,S(a))≤∥p-pn∥+d(pn,S(a)),d(p,S(a))≤∥p-pn∥+𝔇(S(an),S(a)),d(p,S(a))≤∥p-pn∥+ρ∥an-a∥⟶0,asn⟶∞,
which implies that d(p,S(a))=0. Since S(a)∈CB(X1), it follows that p∈S(a). Similarly, we have qϵT(b). By Lemma 11, it follows that a, b, p, q is a solution of problem (21), and this completes the proof.



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*Source: 957482-2014-02-05.xml*