𝐻((⋅,⋅),(⋅,⋅))-Mixed Cocoercive Operators with an Application for Solving Variational Inclusions in Hilbert Spaces

Abstract

We investigate a class of new 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operators in Hilbert spaces. We extend the concept of resolvent operators associated with 𝐻(⋅,⋅)-cocoercive operators to the 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operators and prove that the resolvent operator of 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator is single valued and Lipschitz continuous. Some examples are given to justify the definition of 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operators. Further, by using resolvent operator technique, we discuss the approximate solution and suggest an iterative algorithm for the generalized mixed variational inclusions involving 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operators in Hilbert spaces. We also discuss the convergence criteria for the iterative algorithm under some suitable conditions. Our results can be viewed as a generalization of some known results in the literature.

1. Introduction

In recent past, monotone mappings have a large number of applications, especially in differential equations, integral equations, mathematical economics, optimal control, and so forth. There are various kinds of generalized monotonicity such as pseudomonotone, quasimonotone, and paramonotone; see for example, [1–4]. The cocoercive mappings were studied by Tseng [5], Magnanti and Perakis [6], and Zhu and Marcotte [7] which are also the generalized forms of monotone mappings.

The resolvent operator techniques are important to study the existence of solutions and to design iterative schemes for different kinds of variational inequalities and their generalizations, which are providing mathematical models to some problems arising in optimization and control, economics, and engineering sciences. In order to study various variational inequalities and variational inclusions, Fang and Huang, Kazmi and Khan, and Lan et al. investigated many generalized operators such as 𝐻-monotone [8], 𝐻-accretive [9], (𝑃,𝜂)-proximal point [10], (𝑃,𝜂)-accretive [11], (𝐻,𝜂)-monotone [12], and (𝐴,𝜂)-accretive mappings [13]. Recently, Zou and Huang [14] introduced and studied 𝐻(⋅,⋅)-accretive operators; Kazmi et al. [15–17] introduced and studied generalized 𝐻(⋅,⋅)-accretive operators and 𝐻(⋅,⋅)-𝜂-proximal point mapping; Xu and Wang [18] introduced and studied (𝐻(⋅,⋅),𝜂)-monotone operators; Ahmad et al. [19] introduced and studied 𝐻(⋅,⋅)-cocoercive operators and Husain and Gupta [20, 21] introduced and studied 𝐻(⋅,⋅)-mixed operator and generalized 𝐻(⋅,⋅,⋅)-𝜂-cocoercive operators.

Motivated by the recent work going in this direction, we consider a class of 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operators, a natural generalization of monotone (accretive) operators in Hilbert (Banach) spaces. For details, we refer to see [8, 9, 12–14, 18–22]. We extend the concept of resolvent operators associated with 𝐻(⋅,⋅)-cocoercive operators to the 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operators and prove that the resolvent operator of 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator is single valued and Lipschitz continuous. Further, we consider the generalized mixed variational inclusion problem involving 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator in Hilbert spaces. Using new resolvent operator technique, we prove the existence of solutions and suggest an iterative algorithm for the generalized mixed variational inclusions. Furthermore, we discuss the convergence criteria of the iterative algorithm under some suitable conditions. Our results can be viewed as an extension and generalization of some known results [14–22]. For illustration of Definitions 4 and 7 and Theorem 20, Examples 5, 8, and 21 are given, respectively.

2. Preliminaries

Let 𝑋 be a real Hilbert space endowed with a norm ‖⋅‖ and an inner product ⟨⋅,⋅⟩, 𝑑 the metric induced by the norm ‖⋅‖, 2^𝑋 (resp., 𝐶𝐵(𝑋)) the family of all nonempty (resp., closed and bounded) subsets of 𝑋, and 𝒟(⋅,⋅) the Hausdorff metric on 𝐶𝐵(𝑋) defined by 𝒟(𝑃,𝑄)=maxsup_𝑥∈𝑃𝑑(𝑥,𝑄),sup_𝑦∈𝑄𝑑𝑃,𝑦, (1) where 𝑑(𝑥,𝑄)=inf_𝑦∈𝑄‖𝑥−𝑦‖ and 𝑑(𝑃,𝑦)=inf_𝑥∈𝑃‖𝑥−𝑦‖.

In the sequel, we recall important basic concepts and definitions, which will be used in this work.

Definition 1. A mapping 𝑔∶𝑋→𝑋 is said to be(i)monotone if 𝑔(𝑥)−𝑔𝑦,𝑥−𝑦≥0,∀𝑥,𝑦∈𝑋; (2)(ii)𝜉-strongly monotone if there exists a constant 𝜉>0 such that 𝑔(𝑥)−𝑔𝑦,𝑥−𝑦≥𝜉𝑥−𝑦²,∀𝑥,𝑦∈𝑋; (3)(iii)𝜇-cocoercive if there exists a constant 𝜇>0 such that 𝑔(𝑥)−𝑔𝑦,𝑥−𝑦≥𝜇𝑔(𝑥)−𝑔𝑦²,∀𝑥,𝑦∈𝑋; (4)(iv)𝛾-relaxed cocoercive if there exists a constant 𝛾>0 such that 𝑔(𝑥)−𝑔𝑦,𝑥−𝑦≥−𝛾𝑔(𝑥)−𝑔𝑦²,∀𝑥,𝑦∈𝑋; (5)(v)𝜁_𝑔-Lipschitz continuous if there exists a constant 𝜁_𝑔>0 such that 𝑔(𝑥)−𝑔𝑦≤𝜁_𝑔𝑥−𝑦,∀𝑥,𝑦∈𝑋; (6)(vi)𝛼-expansive if there exists a constant 𝛼>0 such that 𝑔(𝑥)−𝑔𝑦≥𝛼𝑥−𝑦,∀𝑥,𝑦∈𝑋; (7) if 𝛼=1, then it is expansive.

Definition 2. A set-valued mapping 𝑀∶𝑋→2^𝑋 is said to be 𝑚-relaxed monotone if there exists a constant 𝑚>0 such that 𝑢−𝑣,𝑥−𝑦≥−𝑚𝑥−𝑦²(𝑥)𝑦.,∀𝑥,𝑦∈𝑋,𝑢∈𝑀,𝑣∈𝑀 (8)

Definition 3 (see [14, 19]). Let 𝐻∶𝑋×𝑋→𝑋 and 𝐴,𝐵∶𝑋→𝑋 be the single-valued mappings. Then(i)𝐻(𝐴,⋅) is said to be 𝛼-strongly monotone with respect to 𝐴 if there exists a constant 𝛼>0 such that 𝐻(𝐴𝑥,𝑢)−𝐻𝐴𝑦,𝑢,𝑥−𝑦≥𝛼𝑥−𝑦²,∀𝑥,𝑦,𝑢∈𝑋; (9)(ii)𝐻(⋅,𝐵) is said to be 𝛽-relaxed monotone with respect to 𝐵 if there exists a constant 𝛽>0 such that 𝐻(𝑢,𝐵𝑥)−𝐻𝑢,𝐵𝑦,𝑥−𝑦≥−𝛽𝑥−𝑦²,∀𝑥,𝑦,𝑢∈𝑋; (10)(iii)𝐻(𝐴,⋅) is said to be 𝜇-cocoercive with respect to 𝐴 if there exists a constant 𝜇>0 such that 𝐻(𝐴𝑥,𝑢)−𝐻𝐴𝑦,𝑢,𝑥−𝑦≥𝜇𝐴𝑥−𝐴𝑦²,∀𝑥,𝑦,𝑢∈𝑋; (11)(iv)𝐻(⋅,𝐵) is said to be 𝛾-relaxed cocoercive with respect to 𝐵 if there exists a constant 𝛾>0 such that 𝐻(𝑢,𝐵𝑥)−𝐻𝑢,𝐵𝑦,𝑥−𝑦≥−𝛾𝐵𝑥−𝐵𝑦²,∀𝑥,𝑦,𝑢∈𝑋; (12)(v)𝐻(𝐴,⋅) is said to be 𝜏₁-Lipschitz continuous with respect to 𝐴 if there exists a constant 𝜏₁>0 such that 𝐻(𝐴𝑥,⋅)−𝐻𝐴𝑦,⋅≤𝜏₁𝑥−𝑦,∀𝑥,𝑦∈𝑋, (13)(vi)𝐻(⋅,𝐵) is said to be 𝜏₂-Lipschitz continuous with respect to 𝐵 if there exists a constant 𝜏₂>0 such that 𝐻(⋅,𝐵𝑥)−𝐻⋅,𝐵𝑦≤𝜏₂𝑥−𝑦,∀𝑥,𝑦∈𝑋. (14)

Definition 4. Let 𝐻∶(𝑋×𝑋)×(𝑋×𝑋)→𝑋 and 𝐴,𝐵,𝐶,𝐷∶𝑋→𝑋 be the single-valued mappings. Then(i)𝐻((𝐴,⋅),(𝐶,⋅)) is said to be (𝜇₁,𝛾₁)-mixed cocoercive with respect to (𝐴,𝐶) if there exist constants 𝜇₁,𝛾₁>0 such that 𝐻((𝐴𝑥,𝑢),(𝐶𝑥,𝑢))−𝐻𝐴𝑦,𝑢,𝐶𝑦,𝑢,𝑥−𝑦≥𝜇₁𝐴𝑥−𝐴𝑦²+𝛾₁𝑥−𝑦²,∀𝑥,𝑦,𝑢∈𝑋; (15)(ii)𝐻((⋅,𝐵),(⋅,𝐷)) is said to be (𝜇₂,𝛾₂)-relaxed mixed cocoercive with respect to (𝐵,𝐷) if there exist constants 𝜇₂,𝛾₂>0 such that 𝐻((𝑢,𝐵𝑥),(𝑢,𝐷𝑥))−𝐻𝑢,𝐵𝑦,𝑢,𝐷𝑦,𝑥−𝑦≥−𝜇₂𝐵𝑥−𝐵𝑦²+𝛾₂𝑥−𝑦²,∀𝑥,𝑦,𝑢∈𝑋; (16)(iii)𝐻((𝐴,𝐵),(𝐶,𝐷)) is said to be 𝜏-mixed Lipschitz continuous with respect to 𝐴, 𝐵, 𝐶, and 𝐷 if there exists a constant 𝜏>0 such that 𝐻((𝐴𝑥,𝐵𝑥),(𝐶𝑥,𝐷𝑥))−𝐻𝐴𝑦,𝐵𝑦,𝐶𝑦,𝐷𝑦≤𝜏𝑥−𝑦,∀𝑥,𝑦∈𝑋. (17)

Example 5. Let 𝑋=ℝ² with usual inner product. Let 𝐴,𝐵,𝐶,𝐷∶ℝ²→ℝ² be defined by 𝐴𝑥=4𝑥₁4𝑥₂,𝐵𝑦=−3𝑦₁−3𝑦₂,𝐶𝑥=2𝑥₁2𝑥₂𝑦,𝐷𝑦=¹𝑦₂,∀𝑥₁,𝑥₂,𝑦₁,𝑦₂∈ℝ². (18) Suppose that 𝐻∶(ℝ²×ℝ²)×(ℝ²×ℝ²)→ℝ² is defined by 𝐻𝐴𝑥,𝐵𝑦,𝐶𝑥,𝐷𝑦=𝐴𝑥+𝐵𝑦+𝐶𝑥+𝐷𝑦. (19) Then 𝐻((𝐴,𝐵),(𝐶,𝐷)) is (1/4,2)-mixed cocoercive with respect to (𝐴,𝐶), (1/3,1)-relaxed mixed cocoercive with respect to (𝐵,𝐷), and 4-mixed Lipschitz continuous with respect to 𝐴, 𝐵, 𝐶, and 𝐷.

Indeed, let, for any 𝑢∈𝑋, 𝐻),(,==+=((𝐴𝑥,𝑢𝐶𝑥,𝑢))−𝐻𝐴𝑦,𝑢𝐶𝑦,𝑢,𝑥−𝑦𝐴𝑥+𝐶𝑥−𝐴𝑦−𝐶𝑦,𝑥−𝑦𝐴𝑥−𝐴𝑦,𝑥−𝑦𝐶𝑥−𝐶𝑦,𝑥−𝑦4𝑥₁−4𝑦₁,4𝑥₂−4𝑦₂,𝑥₁−𝑦₁,𝑥₂−𝑦₂+2𝑥₁−2𝑦₁,2𝑥₂−2𝑦₂,𝑥₁−𝑦₁,𝑥₂−𝑦₂=4𝑥₁−𝑦₁²+4𝑥₂−𝑦₂²+2𝑥₁−𝑦₁²+2𝑥₂−𝑦₂²=4𝑥−𝑦²+2𝑥−𝑦²,𝐴𝑥−𝐴𝑦²==𝐴𝑥−𝐴𝑦,𝐴𝑥−𝐴𝑦4𝑥₁−4𝑥₂,4𝑦₁−4𝑦₂,4𝑥₁−4𝑥₂,4𝑦₁−4𝑦₂=16𝑥₁−𝑦₁²+16𝑥₂−𝑦₂²,⟨𝐻),(,≥1((𝐴𝑥,𝑢𝐶𝑥,𝑢))−𝐻𝐴𝑦,𝑢𝐶𝑦,𝑢,𝑥−𝑦4𝐴𝑥−𝐴𝑦²+2𝑥−𝑦², (20) that is, 𝐻((𝐴,𝐵),(𝐶,𝐷)) is (1/4,2)-mixed cocoercive with respect to (𝐴,𝐶). Consider 𝐻),(,==+=((𝑢,𝐵𝑥𝑢,𝐷𝑥))−𝐻𝑢,𝐵𝑦𝑢,𝐷𝑦,𝑥−𝑦𝐵𝑥+𝐷𝑥−𝐵𝑦−𝐷𝑦,𝑥−𝑦𝐵𝑥−𝐵𝑦,𝑥−𝑦𝐷𝑥−𝐷𝑦,𝑥−𝑦−3𝑥₁+3𝑦₁,−3𝑥₂+3𝑦₂,𝑥₁−𝑦₁,𝑥₂−𝑦₂+𝑥₁−𝑦₁,𝑥₂−𝑦₂,𝑥₁−𝑦₁,𝑥₂−𝑦₂=−3𝑥₁−𝑦₁²−3𝑥₂−𝑦₂²+𝑥₁−𝑦₁²+𝑥₂−𝑦₂²=−3𝑥−𝑦²+𝑥−𝑦²,𝐵𝑥−𝐵𝑦²==𝐵𝑥−𝐵𝑦,𝐵𝑥−𝐵𝑦−3𝑥₁+3𝑦₁,−3𝑥₂+3𝑦₂,−3𝑥₁+3𝑦₁,−3𝑥₂+3𝑦₂=−9𝑥₁−𝑦₁²+𝑥₂−𝑦₂²,𝐻),(,1((𝑢,𝐵𝑥𝑢,𝐷𝑥))−𝐻𝑢,𝐵𝑦𝑢,𝐷𝑦,𝑥−𝑦≥−3𝐵𝑥−𝐵𝑦²+𝑥−𝑦²; (21) that is, 𝐻((𝐴,𝐵),(𝐶,𝐷)) is (1/3,1)-relaxed mixed cocoercive with respect to (𝐵,𝐷). 𝐻((𝐴𝑥,𝐵𝑥),(𝐶𝑥,𝐷𝑥))−𝐻𝐴𝑦,𝐵𝑦,𝐶𝑦,𝐷𝑦²=𝐻),(,,𝐻),(,=)−,()−=((𝐴𝑥,𝐵𝑥𝐶𝑥,𝐷𝑥))−𝐻𝐴𝑦,𝐵𝑦𝐶𝑦,𝐷𝑦((𝐴𝑥,𝐵𝑥𝐶𝑥,𝐷𝑥))−𝐻𝐴𝑦,𝐵𝑦𝐶𝑦,𝐷𝑦(𝐴𝑥+𝐵𝑥+𝐶𝑥+𝐷𝑥𝐴𝑦+𝐵𝑦+𝐶𝑦+𝐷𝑦𝐴𝑥+𝐵𝑥+𝐶𝑥+𝐷𝑥𝐴𝑦+𝐵𝑦+𝐶𝑦+𝐷𝑦4𝑥₁,4𝑥₂−4𝑦₁,4𝑦₂,4𝑥₁,4𝑥₂−4𝑦₁,4𝑦₂=16𝑥₁−𝑦₁²+𝑥₂−𝑦₂²,𝐻),(,;((𝐴𝑥,𝐵𝑥𝐶𝑥,𝐷𝑥))−𝐻𝐴𝑦,𝐵𝑦𝐶𝑦,𝐷𝑦≤4𝑥−𝑦 (22) that is, 𝐻((𝐴,𝐵),(𝐶,𝐷)) is 4-mixed Lipschitz continuous with respect to 𝐴, 𝐵, 𝐶, and 𝐷.

Definition 6. Let 𝐻∶(𝑋×𝑋)×(𝑋×𝑋)→𝑋,𝑃∶𝑋×𝑋→𝑋, and 𝐴,𝐵,𝐶,𝐷∶𝑋→𝑋 be the single valued mappings and 𝑆,𝑇∶𝑋→𝐶𝐵(𝑋) the set-valued mappings. The mapping 𝐻((𝐴,𝐵),(𝐶,𝐷)) is said to be 𝜐-mixed strongly monotone with respect to 𝑃, 𝑆, and 𝑇 if and only if there exists a constant 𝜐>0 such that 𝐻𝐴𝑢_𝑛,𝐵𝑢_𝑛,𝐶𝑢_𝑛,𝐷𝑢_𝑛−𝐻𝐴𝑢_𝑛−1,𝐵𝑢_𝑛−1,𝐶𝑢_𝑛−1,𝐷𝑢_𝑛−1,𝑃𝑥_𝑛,𝑦_𝑛−𝑃𝑥_𝑛−1,𝑦_𝑛−1≤𝜐𝑢_𝑛−𝑢_𝑛−1²,∀𝑢_𝑛,𝑢_𝑛−1∈𝑋,𝑥_𝑛∈𝑆𝑢_𝑛,𝑦_𝑛∈𝑇𝑢_𝑛,𝑥_𝑛−1∈𝑆𝑢_𝑛−1,𝑦_𝑛−1∈𝑇𝑢_𝑛−1. (23)

3. 𝐻((⋅,⋅),(⋅,⋅))-Mixed Cocoercive Operators

This section deals with a new concept and properties of 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive mappings, which provides a unifying framework for the existing cocoercive operators, monotone operators in Hilbert spaces, and accretive operators in Banach space.

Definition 7. Let 𝐻∶(𝑋×𝑋)×(𝑋×𝑋)→𝑋 and 𝐴,𝐵,𝐶,𝐷∶𝑋→𝑋 be five single-valued mappings. Let 𝐻((𝐴,𝐵),(𝐶,𝐷)) be (𝜇₁,𝛾₁)-mixed cocoercive with respect to (𝐴,𝐶) and (𝜇₂,𝛾₂)-relaxed mixed cocoercive with respect to (𝐵,𝐷). Then the set-valued mapping 𝑀∶𝑋→2^𝑋 is said to be 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive with respect to (𝐴,𝐶) and (𝐵,𝐷) (or simply 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive in the sequel) if(i)𝑀 is 𝑚-relaxed monotone;(ii)(𝐻((⋅,⋅),(⋅,⋅))+𝜌𝑀)(𝑋)=𝑋, for all 𝜌>0.

Example 8. Let 𝑋, 𝐻, 𝐴, 𝐵, 𝐶, and 𝐷 be the same as in Example 5, and let 𝑀∶ℝ²→ℝ² be defined by 𝑀(𝑥)=(−2𝜋,−2𝑥₂),∀𝑥=(𝑥₁,𝑥₂)∈ℝ².
We claim that 𝑀 is 2-relaxed monotone mapping. Indeed, for any 𝑥=(𝑥₁,𝑥₂),𝑦=(𝑦₁,𝑦₂)∈ℝ², we have =𝑀𝑥−𝑀𝑦,𝑥−𝑦−2𝜋,−2𝑥₂−−2𝜋,−2𝑦₂,𝑥₁−𝑦₁,𝑥₂−𝑦₂=0,−2𝑥₂−𝑦₂,𝑥₁−𝑦₁,𝑥₂−𝑦₂=−2𝑥₂−𝑦₂²≥−2𝑥₁−𝑦₁²+𝑥₂−𝑦₂²=−2𝑥−𝑦²𝑀𝑥−𝑀𝑦,𝑥−𝑦≥−2𝑥−𝑦². (24) Furthermore, 𝑀 is also an 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator since (𝐻((⋅,⋅),(⋅,⋅))+𝜌𝑀)(ℝ²)=ℝ² for any 𝜌>0.

Remark 9. (i)If 𝐻((𝐴,𝐵),(𝐶,𝐷))=𝐻(𝐴,𝐵) and 𝑀 is cocoercive, then 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator reduces to 𝐻(⋅,⋅)-cocoercive operator which was studied in [19].(ii)If 𝐻((𝐴,𝐵),(𝐶,𝐷))=𝐻(𝐴,𝐵) and 𝑀 is 𝑚-𝜂-relaxed monotone, then 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator reduces to (𝐻(⋅,⋅),𝜂)-monotone operator which was studied in [18].(iii)If 𝐻((𝐴,𝐵),(𝐶,𝐷))=𝐻(𝐴,𝐵) and 𝑀 is accretive, then 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator reduces to 𝐻(⋅,⋅)-accretive operator which was studied in [14].(iv)If 𝐻((𝐴,𝐵),(𝐶,𝐷))=𝐻, then 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator reduces to 𝐻-monotone operator which was studied in [8, 9].
Since 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator is a generalization of the maximal monotone operator, it is sensible that there are similar properties between them. The following result confirms this expectation.

Proposition 10. Let set-valued mapping 𝑀∶𝑋→2^𝑋 be a 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator with respect to (𝐴,𝐶) and (𝐵,𝐷). If 𝐴 is 𝛼-expansive, 𝐵 is 𝛽-Lipschitz continuous, and 𝜇₁>𝜇₂,𝛼>𝛽, 𝛾₁,𝛾₂>0 with 𝑟=(𝜇₁𝛼²−𝜇₂𝛽²)+(𝛾₁+𝛾₂)>𝑚, then the following inequality 𝑢−𝑣,𝑥−𝑦≥0, (25) holds for all (𝑦,𝑣)∈Graph(𝑀), implies 𝑢∈𝑀𝑥, where Graph(𝑀)={(𝑎,𝑏)∈𝑋×𝑋∶𝑏∈𝑀(𝑎)}. (26)

Proof. Suppose on contrary that there exists (𝑥₀,𝑢₀)∉Graph(𝑀) such that 𝑢₀−𝑣,𝑥₀−𝑦≥0,∀𝑦,𝑣∈Graph(𝑀). (27) Since 𝑀 is 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive, we know (𝐻((⋅,⋅),(⋅,⋅))+𝜌𝑀)(𝑋)=𝑋 that holds for all 𝜌>0, and so there exists (𝑥₁,𝑢₁)∈Graph(𝑀) such that 𝐻𝐴𝑥₀,𝐵𝑥₀,𝐶𝑥₀,𝐷𝑥₀+𝜌𝑢₀=𝐻𝐴𝑥₁,𝐵𝑥₁,𝐶𝑥₁,𝐷𝑥₁+𝜌𝑢₁∈𝑋. (28)
Now, 𝜌𝑢₀−𝜌𝑢₁=𝐻𝐴𝑥₁,𝐵𝑥₁,𝐶𝑥₁,𝐷𝑥₁−𝐻𝐴𝑥₀,𝐵𝑥₀,𝐶𝑥₀,𝐷𝑥₀∈𝑋.𝜌𝑢₀−𝜌𝑢₁,𝑥₀−𝑥₁=−𝐻𝐴𝑥₀,𝐵𝑥₀,𝐶𝑥₀,𝐷𝑥₀−𝐻𝐴𝑥₁,𝐵𝑥₁,𝐶𝑥₁,𝐷𝑥₁,𝑥₀−𝑥₁. (29) Setting (𝑦,𝑣)=(𝑥₁,𝑢₁) in (27) and then from the resultant, (28) and 𝑚-relaxed monotonicity of 𝑀, we obtain −𝑚𝑥₀−𝑥₁²≤𝜌𝑢₀−𝑢₁,𝑥₀−𝑥₁=−𝐻𝐴𝑥₀,𝐵𝑥₀,𝐶𝑥₀,𝐷𝑥₀−𝐻𝐴𝑥₁,𝐵𝑥₁,𝐶𝑥₁,𝐷𝑥₁,𝑥₀−𝑥₁=−𝐻𝐴𝑥₀,𝐵𝑥₀,𝐶𝑥₀,𝐷𝑥₀−𝐻𝐴𝑥₁,𝐵𝑥₀,𝐶𝑥₁,𝐷𝑥₀,𝑥₀−𝑥₁−𝐻𝐴𝑥₁,𝐵𝑥₀,𝐶𝑥₁,𝐷𝑥₀−𝐻𝐴𝑥₁,𝐵𝑥₁,𝐶𝑥₁,𝐷𝑥₁,𝑥₀−𝑥₁. (30) Since 𝐻((𝐴,𝐵),(𝐶,𝐷)) is (𝜇₁,𝛾₁)-mixed cocoercive with respect to (𝐴,𝐶) and (𝜇₂,𝛾₂)-relaxed mixed cocoercive with respect to (𝐵,𝐷) and 𝐴 is 𝛼-expansive and 𝐵 is 𝛽-Lipschitz continuous; thus, (30) becomes −𝑚𝑥₀−𝑥₁²≤−𝜇₁𝐴𝑥₀−𝐴𝑥₁²−𝛾₁𝑥₀−𝑥₁²+𝜇₂𝐵𝑥₀−𝐵𝑥₁²−𝛾₂𝑥₀−𝑥₁²≤−𝜇₁𝛼²𝑥₀−𝑥₁²−𝛾₁𝑥₀−𝑥₁²+𝜇₂𝛽²𝑥₀−𝑥₁²−𝛾₂𝑥₀−𝑥₁²=−𝜇₁𝛼²−𝜇₂𝛽²+𝛾₁+𝛾₂𝑥₀−𝑥₁²=−𝑟𝑥₀−𝑥₁²,where𝑟=𝜇₁𝛼²−𝜇₂𝛽²+𝛾₁+𝛾₂≤−(𝑟−𝑚)𝑥₀−𝑥₁²≤0, (31) which gives 𝑥₀=𝑥₁ since 𝑟>𝑚. By (27), we have 𝑢₀=𝑢₁, a contradiction. This completes the proof.

Theorem 11. Let the set-valued mapping 𝑀∶𝑋→2^𝑋 be a 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator with respect to (𝐴,𝐶) and (𝐵,𝐷). If 𝐴 is 𝛼-expansive, 𝐵 is 𝛽-Lipschitz continuous, and 𝜇₁>𝜇₂,𝛼>𝛽, 𝛾₁,𝛾₂>0 with 𝑟=(𝜇₁𝛼²−𝜇₂𝛽²)+(𝛾₁+𝛾₂)>𝜌𝑚, then (𝐻((𝐴,𝐵),(𝐶,𝐷))+𝜌𝑀)⁻¹ is single valued.

Proof. For any given 𝑥∈𝑋, let 𝑢,𝑣∈(𝐻((𝐴,𝐵),(𝐶,𝐷))+𝜌𝑀)⁻¹(𝑥). It follows that ),(),(−𝐻((𝐴𝑢,𝐵𝑢𝐶𝑢,𝐷𝑢))+𝑥∈𝜌𝑀𝑢,−𝐻((𝐴𝑣,𝐵𝑣𝐶𝑣,𝐷𝑣))+𝑥∈𝜌𝑀𝑣. (32) Since 𝑀 is 𝑚-relaxed monotone, we have −𝑚‖𝑢−𝑣‖²≤1𝜌⟨),((),()⟩‖‖−𝐻((𝐴𝑢,𝐵𝑢𝐶𝑢,𝐷𝑢))+𝑥−−𝐻((𝐴𝑣,𝐵𝑣𝐶𝑣,𝐷𝑣))+𝑥,𝑢−𝑣−𝜌𝑚𝑢−𝑣²≤⟨),((),()⟩⟨𝐻),(),(⟩⟨𝐻),(),(⟩−⟨𝐻),(),(⟩.−𝐻((𝐴𝑢,𝐵𝑢𝐶𝑢,𝐷𝑢))+𝑥−−𝐻((𝐴𝑣,𝐵𝑣𝐶𝑣,𝐷𝑣))+𝑥,𝑢−𝑣=−((𝐴𝑢,𝐵𝑢𝐶𝑢,𝐷𝑢))−𝐻((𝐴𝑣,𝐵𝑣𝐶𝑣,𝐷𝑣)),𝑢−𝑣=−((𝐴𝑢,𝐵𝑢𝐶𝑢,𝐷𝑢))−𝐻((𝐴𝑣,𝐵𝑢𝐶𝑣,𝐷𝑢)),𝑢−𝑣((𝐴𝑣,𝐵𝑢𝐶𝑣,𝐷𝑢))−𝐻((𝐴𝑣,𝐵𝑣𝐶𝑣,𝐷𝑣)),𝑢−𝑣 (33)
Since 𝐻((𝐴,𝐵),(𝐶,𝐷)) is (𝜇₁,𝛾₁)-mixed cocoercive with respect to (𝐴,𝐶) and (𝜇₂,𝛾₂)-relaxed mixed cocoercive with respect to (𝐵,𝐷) and 𝐴 is 𝛼-expansive and 𝐵 is 𝛽-Lipschitz continuous; thus, (33) becomes −𝜌𝑚‖𝑢−𝑣‖²≤−𝜇₁‖𝐴𝑢−𝐴𝑣‖²−𝛾₁‖𝑢−𝑣‖²+𝜇₂‖𝐵𝑢−𝐵𝑣‖²−𝛾₂‖𝑢−𝑣‖²≤−𝜇₁𝛼²‖𝑢−𝑣‖²−𝛾₁‖𝑢−𝑣‖²+𝜇₂𝛽²‖𝑢−𝑣‖²−𝛾₂‖𝑢−𝑣‖²=−𝜇₁𝛼²−𝜇₂𝛽²+𝛾₁+𝛾₂‖𝑢−𝑣‖²=−𝑟‖𝑢−𝑣‖²,where𝑟=𝜇₁𝛼²−𝜇₂𝛽²+𝛾₁+𝛾₂=−𝑟−𝜌𝑚‖𝑢−𝑣‖²≤0, (34) since 𝑟>𝜌𝑚. Hence, it follows that ‖𝑢−𝑣‖≤0. This implies that 𝑢=𝑣, and so (𝐻((𝐴,𝐵),(𝐶,𝐷))+𝜌𝑀)⁻¹ is single valued.

Definition 12. Let set-valued mapping 𝑀∶𝑋→2^𝑋 be a 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator with respect to (𝐴,𝐶) and (𝐵,𝐷). If 𝐴 is 𝛼-expansive, 𝐵 is 𝛽-Lipschitz continuous, and 𝜇₁>𝜇₂,𝛼>𝛽, 𝛾₁,𝛾₂>0 with 𝑟=(𝜇₁𝛼²−𝜇₂𝛽²)+(𝛾₁+𝛾₂)>𝜌𝑚, then the resolvent operator 𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀∶𝑋→𝑋 is defined by 𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀(𝑢)=𝐻((𝐴,𝐵),(𝐶,𝐷))+𝜌𝑀⁻¹(𝑢),∀𝑢∈𝑋. (35)

Now we prove that the resolvent operator defined by (35) is Lipschitz continuous.

Theorem 13. Let set-valued mapping 𝑀∶𝑋→2^𝑋 be a 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator with respect to (𝐴,𝐶) and (𝐵,𝐷). If 𝐴 is 𝛼-expansive, 𝐵 is 𝛽-Lipschitz continuous, and 𝜇₁>𝜇₂,𝛼>𝛽, 𝛾₁,𝛾₂>0 with 𝑟=(𝜇₁𝛼²−𝜇₂𝛽²)+(𝛾₁+𝛾₂)>𝜌𝑚, then the resolvent operator 𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀∶𝑋→𝑋 is 1/(𝑟−𝜌𝑚)-Lipschitz continuous; that is, 𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀(𝑢)−𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀(𝑣≤1)‖‖𝑟−𝜌𝑚𝑢−𝑣,∀𝑢,𝑣∈𝑋. (36)

Proof. Let 𝑢,𝑣∈𝑋 be any given points. It follows from (35) that 𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀(𝑢)=𝐻((𝐴,𝐵),(𝐶,𝐷))+𝜌𝑀⁻¹(𝑢),𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀(𝑣)=𝐻((𝐴,𝐵),(𝐶,𝐷))+𝜌𝑀⁻¹(𝑣),1𝜌𝑢−𝐻𝐴𝑅^{𝐻((⋅,⋅)(⋅,⋅))}_𝜌,𝑀(𝑢),𝐵𝑅^{𝐻((⋅,⋅)(⋅,⋅))}_𝜌,𝑀(𝑢),𝐶𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀(𝑢),𝐷𝑅^{𝐻((⋅,⋅)(⋅,⋅))}_𝜌,𝑀(𝑢𝑅)∈𝑀^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀(𝑢),1𝜌𝑣−𝐻𝐴𝑅^{𝐻((⋅,⋅)(⋅,⋅))}_𝜌,𝑀(𝑣),𝐵𝑅^{𝐻((⋅,⋅)(⋅,⋅))}_𝜌,𝑀(𝑣),𝐶𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀(𝑣),𝐷𝑅^{𝐻((⋅,⋅)(⋅,⋅))}_𝜌,𝑀(𝑣)∈𝑀𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀(𝑣). (37) Let 𝑧₁=𝑅^{𝐻((⋅,⋅)(⋅,⋅))}_𝜌,𝑀(𝑢) and 𝑧₂=𝑅^{𝐻((⋅,⋅)(⋅,⋅))}_𝜌,𝑀(𝑣). Since 𝑀 is 𝑚-relaxed monotone, we have 1𝜌𝑢−𝐻𝐴𝑧₁,𝐵𝑧₁,𝐶𝑧₁,𝐷𝑧₁−𝑣−𝐻𝐴𝑧₂,𝐵𝑧₂,𝐶𝑧₂,𝐷𝑧₂,𝑧₁−𝑧₂≥−𝑚𝑧₁−𝑧₂²,1𝜌𝑢−𝑣−𝐻𝐴𝑧₁,𝐵𝑧₁,𝐶𝑧₁,𝐷𝑧₁+𝐻𝐴𝑧₂,𝐵𝑧₂,𝐶𝑧₂,𝐷𝑧₂,𝑧₁−𝑧₂≥−𝑚𝑧₁−𝑧₂², (38) which implies that 𝑢−𝑣,𝑧₁−𝑧₂≥𝐻𝐴𝑧₁,𝐵𝑧₁,𝐶𝑧₁,𝐷𝑧₁−𝐻𝐴𝑧₂,𝐵𝑧₂,𝐶𝑧₂,𝐷𝑧₂,𝑧₁−𝑧₂−𝜌𝑚𝑧₁−𝑧₂². (39) Further, we have ‖𝑢−𝑣‖𝑧₁−𝑧₂≥𝑢−𝑣,𝑧₁−𝑧₂≥𝐻𝐴𝑧₁,𝐵𝑧₁,𝐶𝑧₁,𝐷𝑧₁−𝐻𝐴𝑧₂,𝐵𝑧₂,𝐶𝑧₂,𝐷𝑧₂,𝑧₁−𝑧₂−𝜌𝑚𝑧₁−𝑧₂²=𝐻𝐴𝑧₁,𝐵𝑧₁,𝐶𝑧₁,𝐷𝑧₁−𝐻𝐴𝑧₂,𝐵𝑧₁,𝐶𝑧₂,𝐷𝑧₁,𝑧₁−𝑧₂+𝐻𝐴𝑧₂,𝐵𝑧₁,𝐶𝑧₂,𝐷𝑧₁−𝐻𝐴𝑧₂,𝐵𝑧₂,𝐶𝑧₂,𝐷𝑧₂,𝑧₁−𝑧₂−𝜌𝑚𝑧₁−𝑧₂²≥𝜇₁𝐴(𝑧₁)−𝐴(𝑧₂)²+𝛾₁𝑧₁−𝑧₂²−𝜇₂𝐵(𝑧₁)−𝐵(𝑧₂)²+𝛾₂𝑧₁−𝑧₂²−𝜌𝑚𝑧₁−𝑧₂²≥𝜇₁𝛼²−𝜇₂𝛽²+𝛾₁+𝛾₂𝑧₁−𝑧₂²−𝜌𝑚𝑧₁−𝑧₂²≥𝑟−𝜌𝑚𝑧₁−𝑧₂²,where𝑟=𝜇₁𝛼²−𝜇₂𝛽²+𝛾₁+𝛾₂, (40) and hence ‖𝑢−𝑣‖𝑧₁−𝑧₂≥𝑟−𝜌𝑚𝑧₁−𝑧₂², (41) that is 𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀(𝑢)−𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀(𝑣≤1)‖‖𝑟−𝜌𝑚𝑢−𝑣,∀𝑢,𝑣∈𝑋. (42) This completes the proof.

4. An Application of 𝐻((⋅,⋅),(⋅,⋅))-Mixed Cocoercive Operators for Solving Variational Inclusions

In this section, we shall show that, under suitable assumptions, the 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator can also play important roles for solving the generalized mixed variational inclusions in Hilbert spaces.

Let 𝐺,𝑆,𝑇∶𝑋→𝐶𝐵(𝑋) be set-valued mappings, and let 𝑔∶𝑋→𝑋,𝐴,𝐵,𝐶,𝐷∶𝑋→𝑋, 𝑃,𝐹∶𝑋×𝑋→𝑋 and 𝐻∶(𝑋×𝑋)×(𝑋×𝑋)→𝑋 be single-valued mappings. Suppose that 𝑀∶𝑋×𝑋→2^𝑋 is a set-valued mapping such that for each fixed 𝑧∈𝐺(𝑥),𝑀(⋅,𝑧)∶𝑋→2^𝑋 is 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator with respect to (𝐴,𝐶) and (𝐵,𝐷), and 𝑔(𝑋)∩dom(𝑀(⋅,𝑧))≠∅. We consider the following generalized mixed variational inclusion: for given 𝜔∈𝑋, find 𝑢∈𝑋, 𝑥∈𝑆(𝑢), 𝑦∈𝑇(𝑢), and 𝑧∈𝐺(𝑢) such that 𝜔∈𝐹(𝑢,𝑧)+𝑃𝑥,𝑦+𝑀𝑔(𝑢),𝑧. (43) The problem of type (43) was studied by Xu and Wang [18] in the setting of Banach spaces when 𝑀 is (𝐻(⋅,⋅),𝜂)-monotone. Problem (43) includes many variational inequalities (inclusions) and complementarity problems as special cases as follows.

If 𝐹=𝜔=0, then problem (43) reduces to a generalized mixed quasi-variational inclusion with 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operators in Hilbert spaces: find 𝑢∈𝑋, 𝑥∈𝑆(𝑢), 𝑦∈𝑇(𝑢), and 𝑧∈𝐺(𝑢) such that 0∈𝑃𝑥,𝑦+𝑀𝑔(𝑢),𝑧. (44) If 𝑀 is 𝑃-𝜂-proximal point mapping, then the problem of type (44) was studied by Kazmi and Khan [10].

If 𝐹(𝑢,𝑧)=𝐹(𝑢), 𝑀(𝑔(𝑢),𝑧)=𝑀(𝑢), and 𝑆, 𝑇, 𝐺=0, 𝑃=0, 𝑔=𝐼, then problem (43) reduces to generalized variational inclusion problem: find 𝑢∈𝑋 such that 0∈𝐹(𝑢)+𝑀(𝑢), (45) which was studied by many authors in the setting of Hilbert spaces when 𝑀 is maximal monotone and 𝐹 is strongly monotone operator.

Lemma 14. Let 𝑋, 𝜔, 𝑔, 𝑃, 𝐹, 𝑆, 𝑇, 𝐺, 𝐴, 𝐵, 𝐶, 𝐷, 𝐻, and 𝑀 be the same as in problem (43). Then (𝑢,𝑥,𝑦,𝑧) is a solution of problem (43) if and only if (𝑢,𝑥,𝑦,𝑧) satisfies the following relation: 𝑔(𝑢)=𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀𝐻𝐴𝑔(𝑢)𝑔(𝑢),𝐶𝑔(𝑢)𝑔(𝑢)(,,𝐵,𝐷+𝜌𝜔−𝜌𝑃𝑥,𝑦−𝜌𝐹𝑢,𝑧) (46) where 𝜌>0 is a constant and 𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_𝜌,𝑀 is the resolvent operator defined by (35).

Proof. Observe that, for 𝜌>0, ()𝑔(𝑢)⟺𝐻𝐴𝑔(𝑢)𝑔(𝑢),𝐶𝑔(𝑢)𝑔(𝑢)(∈𝐻𝐴𝑔(𝑢)𝑔(𝑢),𝐶𝑔(𝑢)𝑔(𝑢)𝑔(𝑢)⟺𝐻𝐴𝑔(𝑢)𝑔(𝑢),𝐶𝑔(𝑢)𝑔(𝑢)(∈𝐻),(()𝑔(𝑢)(𝑢)=𝐻),(()𝜔∈𝐹𝑢,𝑧+𝑃𝑥,𝑦+𝑀,𝑧,𝐵,𝐷+𝜌𝜔−𝜌𝑃𝑥,𝑦−𝜌𝐹𝑢,𝑧),𝐵,𝐷+𝜌𝑀,𝑧,𝐵,𝐷+𝜌𝜔−𝜌𝑃𝑥,𝑦−𝜌𝐹𝑢,𝑧)((𝐴,𝐵𝐶,𝐷))+𝜌𝑀⋅,𝑧⟺𝑔((𝐴,𝐵𝐶,𝐷))+𝜌𝑀⋅,𝑧⁻¹𝐻𝐴𝑔(𝑢)𝑔(𝑢),𝐶𝑔(𝑢)𝑔(𝑢)((𝑢),𝐵,𝐷+𝜌𝜔−𝜌𝑃𝑥,𝑦−𝜌𝐹𝑢,𝑧)⟺𝑔=𝑅^{(𝐻(⋅,⋅),(⋅,⋅))}_{𝜌𝑀(⋅,𝑧)}𝐻𝐴𝑔(𝑢)𝑔(𝑢),𝐶𝑔(𝑢)𝑔(𝑢)(.,𝐵,𝐷+𝜌𝜔−𝜌𝑃𝑥,𝑦−𝜌𝐹𝑢,𝑧) (47)

Remark 15. To develop a fixed point algorithm for (43), we rewrite (46) as follows: 𝑢=𝑢−𝑔(𝑢)+𝑅₍₎^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀⋅,𝑧}𝐻𝐴𝑔(𝑢𝑔(𝑢,𝐶𝑔(𝑢)𝑔(𝑢)(,),𝐵),𝐷+𝜌𝜔−𝜌𝑃𝑥,𝑦−𝜌𝐹𝑢,𝑧)𝜌>0. (48) This fixed point formulation allows us to suggest the following iterative algorithm.

Algorithm 16.  
Step 0. For given 𝜔∈𝑋 and 𝜌>0, choose 𝑢₀∈𝑋, 𝑥₀∈𝑆(𝑢₀), 𝑦₀∈𝑇(𝑢₀), and 𝑧₀∈𝐺(𝑢₀). Set 𝑛=0.
Step 1. Let 𝑢_𝑛+1=𝑢_𝑛−𝑔𝑢_𝑛+𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀(⋅,𝑧𝑛)}𝐻𝐴𝑔𝑢_𝑛,𝐵𝑔𝑢_𝑛,𝐶𝑔𝑢_𝑛,𝐷𝑔𝑢_𝑛+𝜌𝜔−𝜌𝑃𝑥_𝑛,𝑦_𝑛−𝜌𝐹𝑢_𝑛,𝑧_𝑛. (49)
Step 2. Let 𝑢_𝑛+1∈𝑋, 𝑥_𝑛+1∈𝑆(𝑢_𝑛+1), 𝑦_𝑛+1∈𝑇(𝑢_𝑛+1), and 𝑧_𝑛+1∈𝐺(𝑢_𝑛+1)   satisfy 𝑥_𝑛∈𝑆𝑢_𝑛,𝑥_𝑛−𝑥_𝑛+1≤11+𝒟𝑆𝑢𝑛+1^𝑛,𝑆𝑢_𝑛+1,𝑦_𝑛∈𝑇𝑢_𝑛,𝑦_𝑛−𝑦_𝑛+1≤11+𝒟𝑇𝑦𝑛+1^𝑛,𝑇𝑦_𝑛+1,𝑧_𝑛∈𝐺𝑢_𝑛,𝑧_𝑛−𝑧_𝑛+1≤11+𝒟𝐺𝑢𝑛+1^𝑛,𝐺𝑢_𝑛+1, (50) for all 𝑛=0,1,2,….
Step 3. If 𝑢_𝑛+1,𝑥_𝑛+1,𝑦_𝑛+1 and 𝑧_𝑛+1 satisfy (50) to sufficient accuracy, stop; otherwise, set 𝑛=𝑛+1 and return to Step 1.

We need the following definitions which will be used to state and prove the main result.

Definition 17. A set-valued mapping 𝐺∶𝑋→𝐶𝐵(𝑋) is said to be 𝒟-Lipschitz continuous if there exists a constant 𝑙>0 such that 𝒟𝐺𝑥,𝐺𝑦≤ł𝑥−𝑦,∀𝑥,𝑦∈𝑋. (51)

Definition 18. Let 𝐹∶𝑋×𝑋→𝑋 be a single-valued mapping. Then(i)𝐹 is said to be 𝛿₁-Lipschitz continuous in the first argument if there exists a constant 𝛿₁>0 such that 𝐹𝑢₁,⋅−𝐹𝑣₁,⋅≤𝛿₁𝑢₁−𝑣₁,∀𝑢₁,𝑣₁∈𝑋; (52)(ii)𝐹 is said to be 𝛿₂-Lipschitz continuous in the second argument if there exists a constant 𝛿₂>0 such that 𝐹⋅,𝑢₂−𝐹⋅,𝑣₂≤𝛿₂𝑢₂−𝑣₂,∀𝑢₂,𝑣₂∈𝑋. (53)

Definition 19. Let 𝑃∶𝑋×𝑋→𝑋 be a single-valued mapping. Then(i)𝑃 is said to be 𝜖₁-Lipschitz continuous in the first argument if there exists a constant 𝜖₁>0 such that 𝑃𝑢₁,𝑣^′−𝑃𝑢₂,𝑣^′≤𝜖₁𝑢₁−𝑢₂,∀𝑢₁,𝑢₂∈𝑋,𝑣^′∈𝑋; (54)(ii)𝑃 is said to be 𝜖₂-Lipschitz continuous in the second argument if there exists a constant 𝜖₂>0 such that 𝑃𝑢^′,𝑣₁−𝑃𝑢^′,𝑣₂≤𝜖₂𝑣₁−𝑣₂,∀𝑣₁,𝑣₂∈𝑋,𝑢^′∈𝑋. (55)

Theorem 20. Let 𝑆,𝑇,𝐺∶𝑋→𝐶𝐵(𝑋) be set-valued mappings, let 𝑔, 𝐴, 𝐵, 𝐶, 𝐷∶𝑋→𝑋, 𝑃, 𝐹∶𝑋×𝑋→𝑋 and 𝐻∶(𝑋×𝑋)×(𝑋×𝑋)→𝑋 be single-valued mappings, and let 𝑀∶𝑋×𝑋→2^𝑋 be a set-valued mapping such that for each 𝑧∈𝐺(𝑥), 𝑀(⋅,𝑧)∶𝑋→2^𝑋 is 𝐻((⋅,⋅),(⋅,⋅))-mixed cocoercive operator with respect to (𝐴,𝐶) and (𝐵,𝐷), and 𝑔(𝑋)∩dom(𝑀(⋅,𝑧))≠∅. Assume that(i)𝑆, 𝑇, and 𝐺 are 𝒟-Lipschitz continuous with constants 𝑙₁, 𝑙₂ and 𝑙₃, respectively;(ii)𝐴 is 𝛼-expansive, and 𝐵 is 𝛽-Lipschitz continuous;(iii)𝐻((𝐴,𝐵),(𝐶,𝐷)) is 𝜏-mixed Lipschitz continuous with constant 𝜏>0;(iv)𝑔 is 𝜁-Lipschitz continuous and 𝜉-strongly monotone;(v)𝑃 is 𝜖₁-Lipschitz continuous in the first argument and 𝜖₂-Lipschitz continuous in the second argument;(vi)𝐹 is 𝛿₁-Lipschitz continuous in the first argument and 𝛿₂-Lipschitz continuous in the second argument;(vii)𝐻((𝐴,𝐵),(𝐶,𝐷)) is 𝜐-mixed strongly monotone with respect to 𝑃, 𝑆, and 𝑇 with constant 𝜐>0;(viii)Suppose that there exist constants 𝜅>0, 𝜌>0 such that, for each 𝑢∈𝑋 and (𝑧,𝑧)∈𝑋×𝑋, 𝑅₍₎^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀⋅,𝑧}(𝑢)−𝑅₍^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀⋅,𝑧)}(𝑢)=𝜅‖𝑧−𝑧‖; (56) 0<1−2𝜉+𝜁²+𝜅𝑙₃+𝜏²𝜁²−2𝜌𝜉𝜐+𝜌²𝜖₁𝑙₁+𝜖₂𝑙₂²+𝜌𝛿₁+𝛿₂𝑙₃𝑟−𝜌𝑚⁻¹<1, (57) where 𝑟=(𝜇₁𝛼²−𝜇₂𝛽²)+(𝛾₁+𝛾₂)>𝜌𝑚 and 𝜇₁>𝜇₂,𝛼>𝛽,𝛾₁,𝛾₂>0.
Then generalized mixed variational inclusion problem (43) has a solution (𝑢,𝑥,𝑦,𝑧), where 𝑢∈𝑋,𝑥∈𝑆(𝑢), 𝑦∈𝑇(𝑢), and 𝑧∈𝐺(𝑢), and the iterative sequences {𝑢_𝑛},{𝑥_𝑛},{𝑦_𝑛},{𝑧_𝑛}, generated by Algorithm 16 converges strongly to 𝑢, 𝑥, 𝑦, and 𝑧, respectively.

Proof. Since 𝑆, 𝑇, and 𝐺 are 𝒟-Lipschitz continuous with constants 𝑙₁, 𝑙₂, and 𝑙₃, respectively, it follows from Algorithm 16 that 𝑥_𝑛+1−𝑥_𝑛≤11+𝒟𝑆𝑢𝑛+1^𝑛+1,𝑆𝑢_𝑛≤11+𝑛+1ł₁𝑢_𝑛+1−𝑢_𝑛,𝑦_𝑛+1−𝑦_𝑛≤11+𝒟𝑇𝑢𝑛+1^𝑛+1,𝑇𝑢_𝑛≤11+𝑛+1ł₂𝑢_𝑛+1−𝑢_𝑛,𝑧_𝑛+1−𝑧_𝑛≤11+𝒟𝐺𝑢𝑛+1^𝑛+1,𝐺𝑢_𝑛≤11+𝑛+1ł₃𝑢_𝑛+1−𝑢_𝑛, (58) for 𝑛=0,1,2,….
Now, we estimate ‖𝑢_𝑛−𝑢_𝑛−1−(𝑔(𝑢_𝑛)−𝑔(𝑢_𝑛−1))‖, by Lipschitz continuity and strong monotonicity of 𝑔, we have 𝑢_𝑛−𝑢_𝑛−1−(𝑔(𝑢_𝑛)−𝑔(𝑢_𝑛−1))²=𝑢_𝑛−𝑢_𝑛−1²−2𝑔𝑢_𝑛−𝑔𝑢_𝑛−1,𝑢_𝑛−𝑢_𝑛−1+𝑔𝑢_𝑛−𝑔𝑢_𝑛−1²≤𝑢_𝑛−𝑢_𝑛−1²−2𝜉𝑢_𝑛−𝑢_𝑛−1²+𝜁²𝑢_𝑛−𝑢_𝑛−1²=1−2𝜉+𝜁²𝑢_𝑛−𝑢_𝑛−1². (59) Now, we estimate ‖𝑢_𝑛+1−𝑢_𝑛‖ by using Algorithm 16 and the Lipschitz continuity of 𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀(⋅,𝑧𝑛)}: 𝑢_𝑛+1−𝑢_𝑛=𝑢_𝑛−𝑔𝑢_𝑛+𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀⋅,𝑧𝑛}𝐻𝐴𝑔𝑢_𝑛,𝐵𝑔𝑢_𝑛,𝐶𝑔𝑢_𝑛,𝐷𝑔𝑢_𝑛+𝜌𝜔−𝜌𝑃𝑥_𝑛,𝑦_𝑛−𝜌𝐹𝑢_𝑛,𝑧_𝑛−𝑢_𝑛−1−𝑔𝑢_𝑛−1+𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀⋅,𝑧𝑛−1}𝐻𝐴𝑔𝑢_𝑛−1,𝐵𝑔𝑢_𝑛−1,𝐶𝑔𝑢_𝑛−1,𝐷𝑔𝑢_𝑛−1+𝜌𝜔−𝜌𝑃𝑥_𝑛−1,𝑦_𝑛−1−𝜌𝐹𝑢_𝑛−1,𝑧_𝑛−1≤𝑢_𝑛−𝑢_𝑛−1−𝑔𝑢_𝑛−𝑔𝑢_𝑛−1+𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀⋅,𝑧𝑛}𝐻𝐴𝑔𝑢_𝑛,𝐵𝑔𝑢_𝑛,𝐶𝑔𝑢_𝑛,𝐷𝑔𝑢_𝑛+𝜌𝜔−𝜌𝑃𝑥_𝑛,𝑦_𝑛−𝜌𝐹𝑢_𝑛,𝑧_𝑛−𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀⋅,𝑧𝑛−1}𝐻𝐴𝑔𝑢_𝑛−1,𝐵𝑔𝑢_𝑛−1,𝐶𝑔𝑢_𝑛−1,𝐷𝑔𝑢_𝑛−1+𝜌𝜔−𝜌𝑃𝑥_𝑛−1,𝑦_𝑛−1−𝜌𝐹𝑢_𝑛−1,𝑧_𝑛−1≤𝑢_𝑛−𝑢_𝑛−1−𝑔𝑢_𝑛−𝑔𝑢_𝑛−1+𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀⋅,𝑧𝑛}𝐻𝐴𝑔𝑢_𝑛,𝐵𝑔𝑢_𝑛,𝐶𝑔𝑢_𝑛,𝐷𝑔𝑢_𝑛+𝜌𝜔−𝜌𝑃𝑥_𝑛,𝑦_𝑛−𝜌𝐹𝑢_𝑛,𝑧_𝑛−𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀⋅,𝑧𝑛}𝐻𝐴𝑔𝑢_𝑛−1,𝐵𝑔𝑢_𝑛−1,𝐶𝑔𝑢_𝑛−1,𝐷𝑔𝑢_𝑛−1+𝜌𝜔−𝜌𝑃𝑥_𝑛−1,𝑦_𝑛−1−𝜌𝐹𝑢_𝑛−1,𝑧_𝑛−1+𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀⋅,𝑧𝑛}𝐻𝐴𝑔𝑢_𝑛−1,𝐵𝑔𝑢_𝑛−1,𝐶𝑔𝑢_𝑛−1,𝐷𝑔𝑢_𝑛−1+𝜌𝜔−𝜌𝑃𝑥_𝑛−1,𝑦_𝑛−1−𝜌𝐹𝑢_𝑛−1,𝑧_𝑛−1−𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀⋅,𝑧𝑛−1}𝐻𝐴𝑔𝑢_𝑛−1,𝐵𝑔𝑢_𝑛−1,𝐶𝑔𝑢_𝑛−1,𝐷𝑔𝑢_𝑛−1+𝜌𝜔−𝜌𝑃𝑥_𝑛−1,𝑦_𝑛−1−𝜌𝐹𝑢_𝑛−1,𝑧_𝑛−1≤𝑢_𝑛−𝑢_𝑛−1−𝑔𝑢_𝑛−𝑔𝑢_𝑛−1+1×𝐻𝐴𝑔𝑢𝑟−𝜌𝑚_𝑛,𝐵𝑔𝑢_𝑛,𝐶𝑔𝑢_𝑛,𝐷𝑔𝑢_𝑛−𝐻𝐴𝑔𝑢_𝑛−1,𝐵𝑔𝑢_𝑛−1,𝐶𝑔𝑢_𝑛−1,𝐷𝑔𝑢_𝑛−1−𝜌𝑃𝑥_𝑛,𝑦_𝑛−𝑃𝑥_𝑛−1,𝑦_𝑛−1+𝜌𝐹𝑢_𝑛,𝑧_𝑛−𝐹𝑢_𝑛−1,𝑧_𝑛−1+𝜅𝑧_𝑛−𝑧_𝑛−1. (60) Now, by Algorithm 16 and conditions (i), (vi), and (vii), we get 𝑃𝑥_𝑛,𝑦_𝑛−𝑃𝑥_𝑛−1,𝑦_𝑛−1≤𝑃𝑥_𝑛,𝑦_𝑛−𝑃𝑥_𝑛−1,𝑦_𝑛+𝑃𝑥_𝑛−1,𝑦_𝑛−𝑃𝑥_𝑛−1,𝑦_𝑛−1≤𝜖₁𝑥_𝑛−𝑥_𝑛−1+𝜖₂𝑦_𝑛−𝑦_𝑛−1≤𝜖₁11+𝑛𝒟𝑆𝑢_𝑛,𝑆𝑢_𝑛−1+𝜖₂11+𝑛𝒟𝑇𝑢_𝑛,𝑇𝑢_𝑛−1≤𝜖₁11+𝑛𝑙₁𝑢_𝑛−𝑢_𝑛−1+𝜖₂11+𝑛𝑙₂𝑢_𝑛−𝑢_𝑛−1≤𝜖₁𝑙₁+𝜖₂𝑙₂11+𝑛𝑢_𝑛−𝑢_𝑛−1,𝐹𝑢_𝑛,𝑧_𝑛−𝐹𝑢_𝑛−1,𝑧_𝑛−1≤𝐹𝑢_𝑛,𝑧_𝑛−𝐹𝑢_𝑛−1,𝑧_𝑛+𝐹𝑢_𝑛−1,𝑧_𝑛−𝐹𝑢_𝑛−1,𝑧_𝑛−1≤𝛿₁𝑢_𝑛−𝑢_𝑛−1+𝛿₂𝑧_𝑛−𝑧_𝑛−1≤𝛿₁𝑢_𝑛−𝑢_𝑛−1+𝛿₂11+𝑛𝒟𝐺𝑢_𝑛,𝐺𝑢_𝑛−1≤𝛿₁+11+𝑛𝛿₂𝑙₃𝑢_𝑛−𝑢_𝑛−1. (61)(62) Since 𝐻((𝐴,𝐵),(𝐶,𝐷)) is 𝜏-mixed Lipschitz continuous and 𝜐-mixed strongly monotone with respect to 𝑃, 𝑆, and 𝑇 and 𝑔 is 𝜉-Lipschitz continuous and from (61), we have 𝐻𝐴𝑔𝑢_𝑛,𝐵𝑔𝑢_𝑛,𝐶𝑔𝑢_𝑛,𝐷𝑔𝑢_𝑛−𝐻𝐴𝑔𝑢_𝑛−1,𝐵𝑔𝑢_𝑛−1,𝐶𝑔𝑢_𝑛−1,𝐷𝑔𝑢_𝑛−1−𝜌𝑃𝑥_𝑛,𝑦_𝑛−𝑃𝑥_𝑛−1,𝑦_𝑛−1²≤𝐻𝐴𝑔𝑢_𝑛,𝐵𝑔𝑢_𝑛,𝐶𝑔𝑢_𝑛,𝐷𝑔𝑢_𝑛−𝐻𝐴𝑔𝑢_𝑛−1,𝐵𝑔𝑢_𝑛−1,𝐶𝑔𝑢_𝑛−1,𝐷𝑔𝑢_𝑛−1²−2𝜌𝐻𝐴𝑔𝑢_𝑛,𝐵𝑔𝑢_𝑛,𝐶𝑔𝑢_𝑛,𝐷𝑔𝑢_𝑛−𝐻𝐴𝑔𝑢_𝑛−1,𝐵𝑔𝑢_𝑛−1,𝐶𝑔𝑢_𝑛−1,𝐷𝑔𝑢_𝑛−1,−𝜌𝑃𝑥_𝑛,𝑦_𝑛−𝑃𝑥_𝑛−1,𝑦_𝑛−1+𝜌²𝑃𝑥_𝑛,𝑦_𝑛−𝑃𝑥_𝑛−1,𝑦_𝑛−1²≤𝜏²𝑔(𝑢_𝑛)−𝑔(𝑢_𝑛−1)²−2𝜌𝜉𝜐𝑢_𝑛−𝑢_𝑛−1²+𝜌²𝜖₁𝑙₁+𝜖₂𝑙₂²11+𝑛²𝑢_𝑛−𝑢_𝑛−1²≤𝜏²𝜁²−2𝜌𝜉𝜐+𝜌²𝜖₁𝑙₁+𝜖₂𝑙₂²11+𝑛²×𝑢_𝑛−𝑢_𝑛−1². (63)
Using (59)–(63), we have 𝑢_𝑛+1−𝑢_𝑛≤1−2𝜉+𝜁²𝑢_𝑛−𝑢_𝑛−1+1×𝑟−𝜌𝑚𝜏²𝜁²−2𝜌𝜉𝜐+𝜌²𝜖₁𝑙₁+𝜖₂𝑙₂²11+𝑛²×𝑢_𝑛−𝑢_𝑛−1+𝜌𝛿₁+11+𝑛𝛿₂𝑙₃𝑢_𝑛−𝑢_𝑛−11+𝜅1+𝑛𝑙₃𝑢_𝑛−𝑢_𝑛−1=1−2𝜉+𝜁²1+𝜅1+𝑛𝑙₃𝑢_𝑛−𝑢_𝑛−1+𝜏²𝜁²−2𝜌𝜉𝜐+𝜌²𝜖₁𝑙₁+𝜖₂𝑙₂²11+𝑛²+𝜌𝛿₁+11+𝑛𝛿₂𝑙₃𝑟−𝜌𝑚⁻¹×𝑢_𝑛−𝑢_𝑛−1=1−2𝜉+𝜁²1+𝜅1+𝑛𝑙₃+𝜏²𝜁²−2𝜌𝜉𝜐+𝜌²𝜖₁𝑙₁+𝜖₂𝑙₂²11+𝑛²+𝜌𝛿₁+11+𝑛𝛿₂𝑙₃𝑟−𝜌𝑚⁻¹×𝑢_𝑛−𝑢_𝑛−1, (64) which implies 𝑢_𝑛+1−𝑢_𝑛≤Λ_𝑛𝑢_𝑛−𝑢_𝑛−1, (65) where Λ_𝑛=1−2𝜉+𝜁²1+𝜅1+𝑛𝑙₃+𝜏²𝜁²−2𝜌𝜉𝜐+𝜌²𝜖₁𝑙₁+𝜖₂𝑙₂²11+𝑛²+𝜌𝛿₁+11+𝑛𝛿₂𝑙₃𝑟−𝜌𝑚⁻¹. (66) Let Λ=1−2𝜉+𝜁²+𝜅𝑙₃+𝜏²𝜁²−2𝜌𝜉𝜐+𝜌²𝜖₁𝑙₁+𝜖₂𝑙₂²+𝜌𝛿₁+𝛿₂𝑙₃.𝑟−𝜌𝑚 (67)
Since 𝑛→∞, so Λ_𝑛→Λ. By assumption (ix), it is easy to see that Λ<1. Therefore, (67) implies that {𝑢_𝑛} is a Cauchy sequence in 𝑋. Since 𝑋 is a Hilbert space, there exists 𝑢∈𝑋 such that 𝑢_𝑛→𝑢 as 𝑛→∞. From (58), we know that {𝑥_𝑛}, {𝑦_𝑛}, and {𝑧_𝑛} are also Cauchy sequences in 𝑋; thus, there exists 𝑥,𝑦,𝑧∈𝑋 such that 𝑥_𝑛→𝑥, 𝑦_𝑛→𝑦, and 𝑧_𝑛→𝑧 as 𝑛→∞, respectively. By the continuity of 𝑔,𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀(⋅,𝑧)},𝐴,𝐵,𝐶,𝐷,𝑆,𝑇, and 𝐺 and Algorithm 16, we have 𝑔(𝑢)=𝑅^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀(⋅,𝑧)}×𝐻𝐴𝑔(𝑢)𝑔(𝑢),𝐶𝑔(𝑢)𝑔(𝑢)(.,𝐵,𝐷−𝜌𝜔−𝜌𝑃𝑥,𝑦−𝜌𝐹𝑢,𝑧) (68)
Now, we prove that 𝑥∈𝑆(𝑢). In fact, since 𝑥_𝑛∈𝑆(𝑢_𝑛), we have 𝑑(𝑥,𝑆(𝑢))≤𝑥−𝑥_𝑛+𝑑𝑥_𝑛(𝑢)≤,𝑆𝑥−𝑥_𝑛+𝒟𝑆𝑢_𝑛(𝑢)≤,𝑆𝑥−𝑥_𝑛+𝜌𝑥_𝑛−𝑥⟶0as𝑛⟶∞, (69) which implies that 𝑑(𝑥,𝑆(𝑢))=0. Since 𝑆(𝑢)∈𝐶𝐵(𝑋), it follows that 𝑥∈𝑆(𝑢). Similarly, it is easy to see that 𝑦∈𝑇(𝑢),𝑧∈𝐺(𝑢). By Lemma 14, (𝑢,𝑥,𝑦,𝑧) is the solution of problem (43).

Example 21. Let 𝑋=ℝ² with usual inner product.
Let 𝑆,𝑇,𝐺∶ℝ²→2^ℝ2 be defined by 𝑆(𝑥)(𝑥)(𝑥)=1=𝑇=𝐺𝑛𝑥₁1𝑛𝑥₂,∀𝑛∈𝑁,𝑥=𝑥₁,𝑥₂∈ℝ². (70) Then, it is easy to check that(i)𝑆,𝑇, and 𝐺 are (1/10)𝒟-Lipschitz continuous for 𝑛=10. Let 𝐴,𝐵,𝐶,𝐷∶ℝ²→ℝ² be defined by 1𝐴𝑥=𝑥10¹1𝑥10²−1,𝐵𝑦=5𝑦₁−15𝑦₂,𝐶𝑥=2𝑥₁2𝑥₂𝑦,𝐷𝑦=¹𝑦₂,∀𝑥=𝑥₁,𝑥₂,𝑦=𝑦₁,𝑦₂∈ℝ². (71) Suppose that 𝐻∶(ℝ²×ℝ²)×(ℝ²×ℝ²)→ℝ² is defined by 𝐻𝐴𝑥,𝐵𝑦,𝐶𝑥,𝐷𝑦=𝐴𝑥+𝐵𝑦+𝐶𝑥+𝐷𝑦,∀𝑥∈ℝ². (72) Then it is easy to cheek that(ii)𝐻((⋅,⋅),(⋅,⋅)) is (10,2)-mixed cocoercive with respect to (𝐴,𝐶) and (5,1)-relaxed mixed cocoercive with respect to (𝐵,𝐷), 𝐴 is (1/𝑛)-expansive for 𝑛=10,11, and 𝐵 is (1/𝑛)-Lipschitz continuous for 𝑛=4,5;(iii)𝐻((𝐴,𝐵),(𝐶,𝐷)) is (29/𝑛)-mixed Lipschitz continuous with respect to 𝐴,𝐵,𝐶, and 𝐷 for 𝑛=9,10. Let 𝑔∶ℝ²→ℝ² be defined by 𝑔(𝑥)=9𝑥10¹9𝑥10²,∀𝑥∈ℝ². (73) Then it is easy to check that

(iv)𝑔 is (9/𝑛)-strongly monotone for 𝑛=10,11 and (9/𝑛)-Lipschitz continuous for 𝑛=9,10. Let 𝑃,𝐹∶ℝ²×ℝ²→ℝ² be defined by 𝑃=𝑥𝑥,𝑦4+𝑦5=𝑥,𝐹𝑥,𝑦5+𝑦4,∀𝑥,𝑦∈ℝ². (74) Then it is easy to check that(v)𝑃 is (1/𝑛)-Lipschitz continuous in the first argument for 𝑛=3,4 and (1/𝑛)-Lipschitz continuous in the second argument for 𝑛=4,5;(vi)𝐹 is (1/𝑛)-Lipschitz continuous in the first argument for 𝑛=4,5 and (1/𝑛)-Lipschitz continuous in the second argument for 𝑛=3,4;(vii)𝐻((𝐴,𝐵),(𝐶,𝐷)) is (135/𝑛)-mixed strongly monotone with respect to 𝑃, 𝑆, and 𝑇 for 𝑛=999, 1000.  In addition, Let 𝑀∶ℝ²×ℝ²→2^ℝ2 be defined by 𝑀(𝑥,𝑧)=−2𝑥+𝑧,𝑀(𝑥,𝑧)9=−10𝑥+𝑧, (75) where𝑧=(−(9/10)𝑥₁,−(9/10)𝑥₂), 𝑧=(−2𝑥₁,−2𝑥₂)∀𝑥,𝑧,𝑧∈ℝ².

Then it is easy to check that(viii)𝑅₍₎^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀⋅,𝑧}(𝑢)−𝑅₍^{𝐻((⋅,⋅),(⋅,⋅))}_{𝜌,𝑀⋅,𝑧)}(𝑢≤9)𝑛‖𝑧−𝑧‖,for𝑛=28,29. (76)Therefore, for the constants 𝑙₁=𝑙₂=𝑙₃=1,𝜇10₁=10,𝛾₁=2,𝜇₂𝛾=5,₂=1,𝜖₁=14,𝜖₂=15,𝛿₁=15,𝛿₂=141,𝛼=,110𝛽=59,𝜁=910,𝜉=,10𝜏=2.9,𝜐=135,91000𝑚=2.9,𝜅=29,𝜌=0.01,𝑟=2.9, (77) obtained in (i) to (viii), all the conditions of Theorem 20 are satisfied for the generalized mixed variational inclusion problem (43) for 𝜌=0.01.

Remark 22. If the set-valued mapping 𝑀 is 𝜂-monotone, 𝜂(𝑥,𝑦) is Lipschitz continuous, then 𝐻((⋅,⋅),(⋅,⋅)) becomes a new (𝐻((⋅,⋅),(⋅,⋅)),𝜂) monotone operator; see, for example, [18]. We leave the proofs to readers who are interested in this area.

Acknowledgment

The authors are extremely grateful to the referee for the useful suggestions that improved the content of the paper.