Research article Special Issues

Iterative algorithm for solving monotone inclusion and fixed point problem of a finite family of demimetric mappings

  • The goal of this study is to develop a novel iterative algorithm for approximating the solutions of the monotone inclusion problem and fixed point problem of a finite family of demimetric mappings in the context of a real Hilbert space. The proposed algorithm is based on the inertial extrapolation step strategy and combines forward-backward and Tseng's methods. We introduce a demimetric operator with respect to M-norm, where M is a linear, self-adjoint, positive and bounded operator. The algorithm also includes a new step for solving the fixed point problem of demimetric operators with respect to the M-norm. We study the strong convergence behavior of our algorithm. Furthermore, we demonstrate the numerical efficiency of our algorithm with the help of an example. The result given in this paper extends and generalizes various existing results in the literature.

    Citation: Anjali, Seema Mehra, Renu Chugh, Salma Haque, Nabil Mlaiki. Iterative algorithm for solving monotone inclusion and fixed point problem of a finite family of demimetric mappings[J]. AIMS Mathematics, 2023, 8(8): 19334-19352. doi: 10.3934/math.2023986

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  • The goal of this study is to develop a novel iterative algorithm for approximating the solutions of the monotone inclusion problem and fixed point problem of a finite family of demimetric mappings in the context of a real Hilbert space. The proposed algorithm is based on the inertial extrapolation step strategy and combines forward-backward and Tseng's methods. We introduce a demimetric operator with respect to M-norm, where M is a linear, self-adjoint, positive and bounded operator. The algorithm also includes a new step for solving the fixed point problem of demimetric operators with respect to the M-norm. We study the strong convergence behavior of our algorithm. Furthermore, we demonstrate the numerical efficiency of our algorithm with the help of an example. The result given in this paper extends and generalizes various existing results in the literature.



    Monotone inclusion problem (MIP) is the problem to identify a zero of the sum of two operators and it is described as

    FindingζHsuchthat0(A+B)ζ, (1.1)

    where A:HH is an M-cocoercive operator, M is a linear, bounded operator on a real Hilbert space H and B:H2H is a maximal monotone operator. Monotone inclusion problem incorporates various problems such as convex optimization, machine learning, statistical regression, signal and image processing, see [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17]. Lions and Mercier [18] defined the forward-backward splitting algorithm which is one of the most effective method for solving the problem (1.1) and it is defined by

    ζn+1=(I+λnA)1(IλnB)ζnforallnN, (1.2)

    where A is monotone, B is 1/L-cocoercive operator and λn(0,2/L) is a step size parameter. But the algorithm defined in (1.2) converges weakly to a solution of the monotone inclusion problem. Later, Tseng [15] improved this forward-backward splitting algorithm and also proved weak convergence of it. After that, Gibali and Thong [19] proposed a modified version of Tseng's splitting algorithm and proved strong convergence of the proposed algorithm. To speed up the convergence rate of the algorithms, Moudafi and Oliny [20] introduced the following algorithm which includes the inertial parameter.

    {φn=ζn+ϵn(ζnζn1),ζn+1=(I+λnA)1(IλnB)ζnforallnN, (1.3)

    where ϵn[0,1) is inertial parameter. They proved the weak convergence of above algorithm (1.3) assuming the conditions n=1ϵnζnζn12< and λn<2/L, where L is Lipschitz constant of operator B. Later on, the following preconditioning algorithm was defined by Lorenz and Pock [10] for solving the monotone inclusion problem and proved weak convergence of it.

    {φn=ζn+ϵn(ζnζn1),ζn+1=(I+λnM1A)1(IλnM1B)φn. (1.4)

    It is clear that the algorithm (1.4) reduces to forward-backward algorithm (1.2) if we take ϵn=0 and M=I. Dixit et al.[4], in 2021, introduced accelerated preconditioning forward-backward normal S-iteration and proved weak convergence under few assumptions in a real Hilbert space H.

    {φn=ζn+ϵn(ζnζn1),ζn+1=JA,Bλ,M((1γn)φn+γnJA,Bλ,M(φn))forallnN, (1.5)

    where JA,Bλ,M=(I+λM1A)1(IλM1B),γn(0,1),ϵn[0,1) and λ[0,1). Next, Altiparmak and Karahan [21] proposed a new preconditioning forward-backward splitting algorithm and proved strong convergence of it.

    {φn=ζn+ϵn(ζnζn1),υn=JA,Bλ,M((1βn)φn+βnJA,Bλ,M(φn)),ζn+1=(1γn)JA,Bλ,M(υn)+γnf(υn)forallnN, (1.6)

    where ϵn[0,θ] with θ[0,1) and βn,γn(0,1) and f:HH is a k- contraction mapping with respect to M-norm. Recently, in 2021, the same authors [22] proved strong convergence of a modified preconditioning algorithm for solving the monotone inclusion problem in the following manner:

    {φn=ζn+ϵn(ζnζn1),υn=(1αn)φn+αnJA,Bλ,M(φn),κn=JA,Bλ,M((1βn)υn+βnJA,Bλ,M(υn)),ζn+1=(1γn)JA,Bλ,M(κn)+γnf(κn)forallnN, (1.7)

    where JA,Bλ,M=(I+λM1B)1(IλM1A),αn,βn,γn(0,1),ϵn[0,θ] with θ[0,1) and h:HH is k-contraction with respect to M-norm.

    On the other hand, the theory of fixed points has been an appealing topic of research. Several researchers have worked in this direction, see [23,24,25,26,27].

    We have considered the Monotone inclusion problem and fixed point problem of a finite family of demimetric mappings in the context of Hilbert space in this paper as we defined the demimetric operator in the notion of inner product with respect to M-norm, where M is the linear, self-adjoint, positive and bounded operator. In the setting of a Hilbert space, that is, in the complete inner product space, the existence of a solution to the monotone inclusion problem is guaranteed under certain conditions. The authors defined contraction, nonexpansive, quasi-nonexpansive operators with respect to M-norm in these algorithms and proved weak and strong convergence of those algorithms under suitable assumptions. Our main contributions to this research are as follows:

    ● We introduce an operator namely, a demimetric operator with respect to M-norm;

    ● We define a new algorithm which is the combination of forward-backward and Tseng method for solving the monotone inclusion problem together with a new step for solving the fixed point problem of the finite family of demimetric operators;

    ● We also utilize the inertial extrapolation step strategy due to Polyak [29] to enhance the convergence rate of the proposed algorithm.

    In this section, we list some definitions and lemmas which contribute significantly to our main result. Throughout the study, suppose H is a real Hilbert space, Q is a nonempty, closed and convex subset of H, M is a linear, self-adjoint, positive and bounded operator on H and Fix(U) denotes the set of all fixed points of the mapping U.

    Definition 2.1. [30] Assume that A:H2H is a set-valued operator. It is called monotone if

    ζφ,υκ0forallζ,φH,υAζandκAφ.

    The operator A is called maximal monotone if the graph of the operator A is not properly contained in the graph of any other monotone operator.

    Definition 2.2. [31] Assume that Q is nonempty subset of H and ζH. If for any υH, there exists a unique point φQ such that φζυζforallυH then φ is called metric projection of ζ onto Q. It is symbolized by φ=PQζ. If PQζ exists and it can be uniquely obtained for all ζH, then PQ:HQ is called metric projection operator. The operator PQ is nonexpansive and it satisfies the following inequality

    ζPQζ,ψPQζ0forallψQ.

    Definition 2.3. [32] Assume that M is a bounded and linear operator on H. The operator M is said to be self-adjoint if M=M where M denotes the adjoint of M. If M(ζ),ζ)>0 for all ζ(0)H, then M is called positive definite operator. The M-inner product is given by ζ,ψM=ζ,M(ψ) for all ζ,ψH and the corresponding M-norm is also given by ζ2M=ζ,M(ζ) for all ζH by using the self-adjoint, linear and bounded operator M.

    Definition 2.4. [4] Assume that Q is a nonempty subset of H, M is a positive definite operator on H and U:QH is an operator. Then U is called

    (i) M-cocoercive operator if

    UζUψ2M1ζψ,UζUψ,forallζ,ψH,

    (ii) nonexpansive operator with respect to M -norm if

    UζUψMζψMforallζ,ψH,

    (iii) quasi-nonexpansive operator with respect to M -norm if

    UζUψMζψMforallζH,ψFix(U),

    (iv) h-contraction with respect to M-norm if there exists h[0,1) such that

    UζUψMhζψMforallζ,ψH.

    Lemma 2.1. [30] For ζ,ψ,ω,ιH and α,β,γ[0,1], where α+β+γ=1, we have

    (i) ζ+ψ2ζ2+2ψ,ζ+ψ,

    (ii) βζ+(1β)ψ2=βζ2+(1β)ψ2β(1β)ζψ2,

    (iii) αζ+βψ+γω2=αζ2+βψ2+γω2αβζψ2αγζω2βγψω2,

    (iv) 2ζψ,ωι=ζι2+ψω2ζω2ψι2,

    (v) ζ±ψ2=ζ2±2ζ,ψ+ψ2.

    Lemma 2.2. [33] Let U:QH be the nonexpansive operator with Fix(U)ϕ. Then the mapping IU is demiclosed at origin, that is, for any sequence {ζn}H such that ζnζH and ζnUζn0 as n, we have ζFix(U).

    Lemma 2.3. [34] Assume that S:QH is ξ-demimetric operator with ξ(,1). Then, Fix(S) is closed and convex.

    Lemma 2.4. [4] Suppose that A is M-cocoercive operator on H, B:H2H is maximal monotone operator. Then, JA,Bλ,M=(I+λM1B)1(IλM1A) is nonexpansive with respect to M-norm for λ(0,1].

    Lemma 2.5. [4] Suppose that A is M-cocoercive operator on H, B:H2H is a maximal monotone operator and λ is a non-negative real number. Then, ζH is a solution of monotone inclusion problem (1.1) if and only if (I+λM1B)1(IλM1A)(ζ)=ζ.

    Lemma 2.6. [35] Suppose that {tn}[0,) is a sequence of real numbers. Let

    sn+1(1pn)sn+pntnforallnN,

    where {pn}[0,1] and {tn}(,) satisfying the following assumptions:

    (1) n=1pn=,

    (2) lim supntn0.

    Then limnsn=0.

    Lemma 2.7. [36] Assume that {ζn}[0,) such that there exists a subsequence {ζnj} of {ζn} such that ζnj<ζnj+1. Then, there exists a nondecreasing sequence {nk} of natural numbers satisfying limknk=, ζnkζnk+1 and ζkζnk+1 for all kN. Also, nk is the greatest number n in the set {1,2,...,k} satisfying ζn<ζn+1.

    In this part, we will prove the strong convergence of the following algorithm for finding a common solution of monotone inclusion and fixed point problem of a finite family of demimetric operators in the setting of a real Hilbert space H. Assume that A is M-cocoercive operator on H, B:H2H is a maximally monotone operator. Suppose Si is a finite family of ξ-demimetric operators with ξ(,1) such that ISi is demiclosed at origin for all i=0,1,2,...N1, h:HH is a contraction mapping with respect to M-norm with constant k(0,1].

    Algorithm 3.1. Let ζ0,ζ1H. Compute {φn},{υn},{κn}and{ζn} using

    {φn=ζn+ϵn(ζnζn1),υn=(1αn)φn+αnJA,Bλ,M(φn),κn=JA,Bλ,M((1βn)υn+βnJA,Bλ,M(υn)),ζn+1=γnh(ζn)+(1γnδn)JA,Bλ,M(κn)+δnSnκn, (3.1)

    where Sn=1NN1i=0(1qn)I+qnSi

    Definition 3.1. A mapping U:QH, where Q is closed, convex and nonempty subset of H is called ξ-demimetric with respect to M-norm, where ξ(,1) if Fix(U)ϕ such that

    ζζ,(IU)ζM12(1ξ)(IU)ζ2M,forallζQ,ζFix(U).

    Example 3.1. Let H=R3,Q=H and M(ζ)=(5ζ1,4ζ2,5ζ3) for all ζ=(ζ1,ζ2,ζ3)H. The mapping U:HH defined by U(ζ)=2ζ for all ζ=(ζ1,ζ2,ζ3) is 13-demimetric mapping with respect to M-norm.

    Now, we prove a result for ξ-demimetric operator with respect to M-norm which is motivated by Lemma (2.2) of [37].

    Lemma 3.1. Suppose S:QH is ξ-demimetric operator with respect to M-norm, where ξ(,1) and Fix(S) is nonempty. Let P=(1γ)I+γS, where γ(,) with γ(0,1ξ], then P:QH is a quasi-nonexpansive operator.

    Proof. It is direct that Fix(S)=Fix(P). Now, since S:QH is demimetric operator, we have for any ζQ and yFix(P),

    ζy,ζPζM=ζy,γ(ζSζ)M=γζy,ζSζMγ2(1ξ)(IS)ζ2M=1ξ2γζPζ2Mγ2γζPζ2M=12ζPζ2M.

    This implies that P is 0-demimetric mapping. Also, for ζQ, yFix(P) and using Lemma (2.1), we deduce

    12ζPζ2Mζy,ζPζMζPζ2M2ζy,ζPζMζPζ2MζPζ2M+ζy2MPζy2MPζy2Mζy2MPζyMζyM.

    Hence, P is a quasi-nonexpansive operator.

    Lemma 3.2. The mapping Sn defined by Sn=1NN1i=0(1qn)I+qnSi is quasi-nonexpansive. \proof Let ζΩ.

    Consider

    SnζζM=1NN1i=0((1qn)I+qnSi)ζζM1NN1i=0((1qn)I+qnSi)ζζM1NN1i=0ζζM=ζζM.

    Hence, Sn is quasi-nonexpansive.

    Lemma 3.3. Assume that {ζn} is bounded sequence of real numbers and ζΩ=(A+B)1(0)N1i=0Fix(Si). If limnJA,Bλ,MζnζnM=0 and limnκnSnκnM=0. Then lim supnh(ζ)ζ,ζn+1ζM0.

    Proof. Since the sequence {ζn} is bounded, so there exists a subsequence {ζnk} of {ζn} such that ζnkζ

    Consider

    lim supnh(ζ)ζ,ζn+1ζM=lim supkh(ζ)ζ,ζnk+1ζM=h(ζ)ζ,ζζM. (3.2)

    Using the condition JA,Bλ,MζnζnM0 as n and by using Lemma (2.5), we deduce that ζ(A+B)1(0). Also, since limnζnκn=0 and ζnkζ, so we obtain κnkζ. By using the condition limnκnSnκn=0 and using Lemma (2.2), we obtain that ζFix(Si). Thus ζΩ.

    From Eq (3.2) and by the property of metric projection, we have

    lim supnh(ζ)ζ,ζn+1ζMh(ζ)ζ,ζζ0.

    Hence the lemma is proved.

    Theorem 3.1. Suppose that the solution set Ω=(A+B)1(0)N1i=0Fix(Si) is non-empty and the sequence {ζn} is generated by algorithm (3.1), where {ϵn}[0,θ] with θ[0,1),{αn},{βn},{γn},{δn}(0,1) such that the following conditions hold:

    (i) 0<aαnb<1forsomea,bR,

    (ii) 0<cβnd<1forsomec,dR,

    (iii) limnγn=0,n=1γn=,

    (iv) n=1ϵnζnζn1<,

    (v) for any nN,0<a<lim infnδnlim supnδn<b<1γn, where a,bR+.

    Then the sequence {ζn} converges strongly to a point ζΩ=PΩh(ζ).

    Proof. First we prove that the sequence {ζn} is bounded. Let ζΩ.

    Consider

    φnζM=ζn+ϵn(ζnζn1)ζMζnζM+ϵn(ζnζn1)M=ζnζM+ϵnζnζn1M. (3.3)

    Since, JA,Bλ,M is nonexpansive with respect to M-norm, so using nonexpansiveness of JA,Bλ,M, we have

    υnζM=(1αn)φn+αnJA,Bλ,M(φn)ζM(1αn)φnζM+αnJA,Bλ,M(φn)ζM(1αn)φnζM+αnφnζM=φnζM, (3.4)

    and

    κnζM=JA,Bλ,M((1βn)υn+βnJA,Bλ,M(υn))ζM(1βn)υn+βnJA,Bλ,M(υn)ζM(1βn)υnζM+βnJA,Bλ,M(υn)ζM(1βn)υnζM+βnυnζM=υnζM. (3.5)

    From Eqs (3.1)–(3.5) and using the fact that h is k-contraction and Sn is quasi-nonexpansive with respect to M-norm, we have

    ζn+1ζM=γnh(ζn)+(1γnδn)JA,Bλ,M(κn)+δnSnκnζMγnh(ζn)ζM+(1γnδn)JA,Bλ,M(κn)ζM+δnSnκnζMγnh(ζn)ζM+(1γnδn)κnζM+δnκnζM=γnh(ζn)ζM+(1γn)κnζMγnh(ζn)h(ζ)M+γnh(ζ)ζM+(1γn)υnζMγnh(ζn)h(ζ)M+γnh(ζ)ζM+(1γn)φnζMγnh(ζn)h(ζ)M+γnh(ζ)ζM+(1γn)[ζnζM+ϵnζnζn1M]γnkζnζM+γnh(ζ)ζM+(1γn)ζnζM+ϵn(1γn)ζnζn1M=(1γn+γnk)ζnζM+γnh(ζ)ζM+ϵn(1γn)ζnζn1M[1γn(1k)]ζnζM+γnh(ζ)ζM+ϵnζnζn1M=[1γn(1k)]ζnζM+γnh(ζ)ζM+ϵnγnγnζnζn1M. (3.6)

    From assumptions (iii) and (iv), we have limnϵnγnζnζn1M=0. So, there exists a positive integer N1>0 such that ϵnγnζnζn1N1. By using Eq (3.6), we obtain

    ζn+1ζM[1γn(1k)]ζnζM+γn[N1+h(ζ)ζM][1γn(1k)]ζnζM+γn(1k)[N1+h(ζ)ζM1k]max.{ζnζM,N1+h(ζ)ζM1k}.

    Continuing like this,

    ζn+1ζMmax.{ζ1ζM,N1+h(ζ)ζM1k}.

    Thus the sequence {ζn} is bounded and therefore, the sequences {φn},{υn},{κn} are also bounded.

    Now, we show that ζnζ. Using Lemma (2.1), we obtain

    φnζ2M=ζn+ϵn(ζnζn1)ζ2Mζnζ2M+2ϵnζnζMζnζn1M+ϵ2nζnζn12M, (3.7)
    υnζ2M=(1αn)φn+αnJA,Bλ,M(φn)ζ2M=(1αn)φnζ2M+αnJA,Bλ,M(φn)ζ2Mαn(1αn)φnJA,Bλ,M(φn)2M(1αn)φnζ2M+αnφnζ2Mαn(1αn)φnJA,Bλ,M(φn)2M=φnζ2M(1αn)αnφnJA,Bλ,M(φn)2Mφnζ2M, (3.8)

    and

    κnζ2M=JA,Bλ,M((1βn)υn+βnJA,Bλ,M(υn))ζ2M(1βn)υn+βnJA,Bλ,M(υn)ζ2M=(1βn)(υnζ)+βn(JA,Bλ,M(υn)ζ)2M=(1βn)υnζ2M+βnJA,Bλ,M(υn)ζ2M(1βn)βnυnJA,Bλ,M(υn)2M(1βn)υnζ2M+βnυnζ2(1βn)βnυnJA,Bλ,M(υn)2Mυnζ2M. (3.9)

    Using Eqs (3.7)(3.9), Lemma (2.1) and using the fact that JA,Bλ,M is nonexpansive and Sn is quasi-nonexpansive, we obtain

    ζn+1ζ2M=γnh(ζn)+(1γnδn)JA,Bλ,M(κn)+δnSnκnζ2M=γn(h(ζn)ζ)+(1γnδn)(JA,Bλ,M(κn)ζ)+δn(Snκnζ)2M=γn(h(ζn)h(ζ))+γn(h(ζ)ζ)+(1γnδn)(JA,Bλ,M(κn)ζ)+δn(Snκnζ)2Mγn(h(ζn)h(ζ))+(1γnδn)(JA,Bλ,M(κn)ζ)+δn(Snκnζ)2M+2γn(h(ζ)ζ),ζn+1ζM=γn(h(ζn)h(ζ))+(1γnδn)(JA,Bλ,M(κn)ζ)+δn(Snκnζ)2M+2γnh(ζ)ζ,ζn+1ζMγnh(ζn)h(ζ)2M+(1γnδn)JA,Bλ,M(κn)ζ2M+δnSnκnζ2M+2γnh(ζ)ζ,ζn+1ζMγnh(ζn)h(ζ)2M+(1γnδn)κnζ2M+δnκnζ2M+2γnh(ζ)ζ,ζn+1ζM=(1γn)κnζ2M+γnh(ζn)h(ζ)2M+2γnh(ζ)ζ,ζn+1ζM(1γn)[ζnζ2M+2ϵnζnζMζnζn1M+ϵ2nζnζn12M]+γnh(ζn)h(ζ)2M+2γnh(ζ)ζ,ζn+1ζM(1γn)ζnζ2M+2ϵn(1γn)ζnζMζnζn1M+ϵ2n(1γn)ζnζn12M+γnkζnζ2M+2γnh(ζ)ζ,ζn+1ζM[1γn(1k)]ζnζ2M+2ϵnζnζMζnζn1M+ϵ2nζnζn12M+2γnh(ζ)ζ,ζn+1ζM, (3.10)

    which implies that

    ζn+1ζ2M[1γn(1k)]ζnζ2M+γn(1k)[21kh(ζ)ζ,ζn+1ζ+2ϵnγn(1k)ζnζMζnζn1M+ϵ2nγn(1k)ζnζn12M]. (3.11)

    Equation (3.11) is equivalent to

    sn+1(1pn)sn+pntn, (3.12)

    where, sn=ζnζ2M, pn=γn(1k) and tn=21kh(ζ)ζ,ζn+1ζ+2ϵnγn(1k)ζnζMζnζn1M+ϵ2nγn(1k)ζnζn12M

    Now, we consider two possible cases on the sequence {ζnζM}.

    Case 1. Suppose that there exists some positive integer n0 such that the sequence {ζnζM} is nonincreasing sequence for any nn0. Also, the sequence {ζnζM} is bounded below by zero. So, it is convergent.

    By using Lemma (2.1), Eqs (3.7) and (3.9), we have

    ζn+1ζ2M=γn(h(ζn)ζ)+(1γnδn)(JA,Bλ,M(κn)ζ)+δn(Snκnζ)2Mγnh(ζn)ζ2M+(1γnδn)JA,Bλ,M(κn)ζ2M+δnSnκnζ2Mγnh(ζn)ζ2M+(1γnδn)κnζ2M+δnκnζ2M=γnh(ζn)ζ2M+(1γn)κnζ2Mγnh(ζn)ζ2M+(1γn)υnζ2M, (3.13)

    and

    υnζ2M=(1αn)(φnζ)+αn(JA,Bλ,M(φn)ζ)2M=(1αn)φnζ2M+αnJA,Bλ,M(φn)ζ2Mαn(1αn)JA,Bλ,M(φn)φn2M=φnζ2Mαn(1αn)JA,Bλ,M(φn)φn2Mζnζ2M+2ϵnζnζMζnζn1M+ϵ2nζnζn12Mαn(1αn)JA,Bλ,M(φn)φn2M. (3.14)

    From Eqs (3.13) and (3.14), we obtain

    ζn+1ζ2Mγnh(ζn)ζ2M+(1γn)[ζnζ2M+2ϵnζnζMζnζn1M+ϵ2nζnζn12Mαn(1αn)JA,Bλ,M(φn)φn2M],

    which implies that

    αn(1αn)JA,Bλ,M(φn)φn2Mγnh(ζn)ζ2M+(1γn)ζnζ2M+2ϵnζnζMζnζn1M+ϵ2nζnζn12Mζn+1ζ2Mγnh(ζn)ζ2M+ζnζ2M+2ϵnζnζMζnζn1M+ϵ2nζnζn12Mζn+1ζ2M. (3.15)

    Since, n=1ϵnζnζn1<, so we have

    limnϵnζnζn1M=0. (3.16)

    Using the condition limnγn=0, Eq (3.16) and taking limit n in Eq (3.15), we get

    limnJA,Bλ,M(φn)φn2M=0. (3.17)

    Now, consider

    υnφnM=(1αn)φn+αnJA,Bλ,M(φn)φnM=αn(JA,Bλ,M(φn)φn)M=αnJA,Bλ,M(φn)φnM. (3.18)

    Taking limit n in Eq (3.18) and using Eq (3.17), we obtain

    limnυnφnM=0. (3.19)

    By using triangle inequality, we have

    JA,Bλ,M(φn)υnMJA,Bλ,M(φn)φnM+φnυnM. (3.20)

    Taking limit n in Eq (3.20) and using Eqs (3.17) and (3.19), we deduce

    limnJA,Bλ,M(φn)υnM=0. (3.21)

    Also,

    JA,Bλ,M(υn)υnM=JA,Bλ,M(υn)JA,Bλ,M(φn)+JA,Bλ,M(φn)υnMJA,Bλ,M(υn)JA,Bλ,M(φn)M+JA,Bλ,M(φn)υnMυnφnM+JA,Bλ,M(φn)υnM, (3.22)

    Taking limit n in Eq (3.22) and using Eqs (3.19) and (3.21) in Eq (3.22), we have

    limnJA,Bλ,M(υn)υnM=0, (3.23)

    and

    κnφnM=κnJA,Bλ,M(φn)+JA,Bλ,M(φn)φnMκnJA,Bλ,M(φn)M+JA,Bλ,M(φn)φnM=JA,Bλ,M((1βn)υn+βnJA,Bλ,M(υn))JA,Bλ,M(φn)M+JA,Bλ,M(φn)φnM(1βn)υn+βnJA,Bλ,M(υn)φnM+JA,Bλ,M(φn)φnM=(υnφn)+βn(JA,Bλ,M(υn)υn)M+JA,Bλ,M(φn)φnMυnφnM+βnJA,Bλ,M(υn)υnM+JA,Bλ,M(φn)φnM. (3.24)

    Taking limit n in Eq (3.24) and using Eqs (3.17), (3.19) and (3.23), we obtain

    limnκnφnM=0. (3.25)

    Again using Eq (3.1) and Lemma (2.1), we deduce that

    ζn+1ζ2M=γn(h(ζn)ζ)+(1γnδn)(JA,Bλ,M(κn)ζ)+δn(Snκnζ)2Mγnh(ζn)ζ2M+(1γnδn)JA,Bλ,M(κn)ζ2M+δnSnκnζ2M(1γnδn)δnJA,Bλ,M(κn)Snκn2M.

    Using nonexpansiveness of JA,Bλ,M, quasi-nonexpansiveness of Sn, Eqs (3.7)–(3.9), we have

    (1γnδn)δnJA,Bλ,M(κn)Snκn2Mγnh(ζn)ζ2M+ζnζ2M+2ϵnζnζMζnζn1M+ϵ2nζnζn12Mζn+1ζ2M. (3.26)

    Taking limit n in Eq (3.26) and using Eq (3.16), condition limnγn=0, we obtain

    limnJA,Bλ,M(κn)SnκnM=0. (3.27)

    By using triangle inequality, we have

    JA,Bλ,M(κn)κnM=JA,Bλ,M(κn)JA,Bλ,M(φn)+JA,Bλ,M(φn)φn+φnκnMJA,Bλ,M(κn)JA,Bλ,M(φn)M+JA,Bλ,M(φn)φn+φnκnMκnφnM+JA,Bλ,M(φn)φnM+φnκnM. (3.28)

    Taking limit n in Eq (3.28) and using Eqs (3.17) and (3.25), we get

    limnJA,Bλ,M(κn)κnM=0. (3.29)

    Also,

    SnκnκnM=SnκnJA,Bλ,M(κn)+JA,Bλ,M(κn)κnMSnκnJA,Bλ,M(κn)M+JA,Bλ,M(κn)κnM. (3.30)

    Taking limit n in Eq (3.30) and using Eqs (3.27) and (3.29), we obtain

    limnSnκnκnM=0. (3.31)

    From Eqs (3.19) and (3.25) and using triangle inequality, we obtain

    limnκnυnM=0. (3.32)

    From Eq (3.1),

    φnζnM=ζn+ϵn(ζnζn1)ζnM=ϵn(ζnζn1)M. (3.33)

    Taking limit n in Eq (3.33) and using Eq (3.16), we get

    limnφnζnM=0. (3.34)

    Similarly, we can prove

    limnκnζnM=0. (3.35)

    Now, from Eqs (3.31) and (3.35), we can prove

    SnκnζnM=Snκnκn+κnζnMSnκnκnM+κnζnM,

    which implies

    limnSnκnζnM=0. (3.36)

    Again consider

    ζn+1ζnM=ζn+1Snκn+SnκnζnMζn+1SnκnM+SnκnζnM=γnh(ζn)+(1γnδn)JA,Bλ,M(κn)+δnSnκnSnκnM+SnκnζnM=(JA,Bλ,M(κn)Snκn)+γn(h(ζn)JA,Bλ,M(κn))+δn(SnκnJA,Bλ,M(κn))M+SnκnζnM. (3.37)

    Taking limit n in Eq (3.37) and using the condition limnγn=0, Eqs (3.27) and (3.36), we obtain

    limnζn+1ζnM=0. (3.38)

    Also, using Eqs (3.34) and (3.38), we have

    limnζn+1φnM=0. (3.39)

    From Eqs (3.17) and (3.34), we can prove

    JA,Bλ,M(ζn)ζnM=JA,Bλ,M(ζn)JA,Bλ,M(φn)(ζnφn)+(JA,Bλ,M(φn)φn)M2ζnφnM+JA,Bλ,M(φn)φnM.

    Taking limit n in above equation, we have

    limnJA,Bλ,M(ζn)ζnM=0. (3.40)

    Since, the sequence {ζn} is a bounded sequence of real numbers, so using Lemma (3.3), we obtain

    lim supnh(ζ)ζ,ζn+1ζM0. (3.41)

    Thus, by making use of Eq (3.16), we have lim supntn0. Hence from Lemma (2.6), ζnζ as n.

    Case 2. There exists a subsequence {ζnjζ2M} of {ζnζ2M} such that ζnjζ2Mζnj+1ζ2M for any jN. By using Lemma (2.7), we see that there exists a nondecreasing sequence {ζnk} of N such that limknk= and the following inequalities hold for all kN.

    ζnkζ2Mζnk+1ζ2M, (3.42)

    and

    ζkζ2Mζnk+1ζ2M. (3.43)

    Similar to Eq (3.15), we have

    αnk(1αnk)JA,Bλ,M(φnk)φnk2Mγnkh(ζnk)ζ2M+ζnkζ2M+2ϵnkζnkζMζnkζnk1M+ϵ2nkζnkζnk12Mζnk+1ζ2M. (3.44)

    Using Eqs (3.16), (3.42) and condition limkγnk=0 in Eq (3.44), we obtain

    limkJA,Bλ,M(φnk)φnkM=0. (3.45)

    Similarly as in Case 1, we have

    limkυnkφnkM=0,limkφnkκnkM=0,limkζnk+1φnkM=0,limkζnk+1ζnkM=0,limkJA,Bλ,M(ζnk)ζnkM=0.

    Using Eq (3.12), we have

    snk+1(1pnk)snk+pnktnk. (3.46)

    Using Eq (3.42) in Eq (3.46), we obtain

    ξnksnkξnktnk. (3.47)

    As pnk>0, so Eq (3.47) implies snktnk, that is, we have

    ζnk+1ζ2M21kh(ζ)ζ,ζnk+1ζ+2ϵnkγnk(1k)ζnkζMζnkζnk1M+ϵ2nkγnk(1k)ζnkζnk12M. (3.48)

    Similar to Case 1, we obtain lim supntnk0. So, we have by using Lemma (2.6),

    limkζnk+1ζ2M=0. (3.49)

    From Eq (3.43), we get ζkζ as k. This completes the proof.

    In this section, we provide a numerical example to show the numerical efficiency of the proposed algorithm. We performed all numerical experiments on a Dell computer equipped with a 3.20 GHz Intel (R) Core (TM) i5-3470 CPU and 8 GB of memory. The MATLAB R 2022 a platform was used as the implementation environment.

    Let H=R3. Define the operators h,A,B,M,Si:HH as h(ζ)=ζ2,A(ζ)=(5ζ1,4ζ2,5ζ3),B=(4ζ1,4ζ2,4ζ3),M=A,Siζ=3ζ for i=0,1. for all ζ=(ζ1,ζ2,ζ3)H. Clearly, M is self-adjoint, bounded and positive operator, A is M-cocoercive operator, B is maximally monotone operator, h is 12-contraction operator with respect to M-norm and Si is a finite family of 12-demimetric operators. To find the numerical values of ζn, we choose αn=12n,λ=0.5,qn=0.1,βn=12n,γn=1n,δn=n2(n+1),ϵn=0.8. We compare our algorithm (3.1) with algorithms (1.4) and (1.5) with different choices of ζ0 and ζ1. We consider three cases:

    Case 1. ζ0=[20,20,20] and ζ1=[10,10,10].

    Case 2. ζ0=[10,10,10] and ζ1=[5,5,5].

    Case 3. ζ0=[1,1,1] and ζ1=[2,2,2].

    The values of iteration number (n) for various choices of ζ0 and ζ1 are given in the Tables 13. Comparison of Algorithm (3.1) by taking En=ζnζn1<104 with the algorithms Dixit et al. (1.4) and Lorenz and Pock (1.5) for Cases 1–3 are shown in Figures 13 respectively.

    Table 1.  Iteration number (with cpu time) for Algorithms (3.1), (1.4) and (1.5) for case 1.
    Case 1 Iteration number(n) cpu time (in seconds)
    Algorithm (3.1) 11 0.015498
    Dixit et al. (1.4) 75 0.699281
    Lorenz and Pock (1.5) 15 0.019471

     | Show Table
    DownLoad: CSV
    Table 2.  Iteration number (with cpu time) for Algorithms (3.1), (1.4) and (1.5) for case 2.
    Case 2 Iteration number(n) cpu time (in seconds)
    Algorithm (3.1) 10 0.001373
    Dixit et al. (1.4) 75 0.003053
    Lorenz and Pock (1.5) 14 0.001431

     | Show Table
    DownLoad: CSV
    Table 3.  Iteration number (with cpu time) for Algorithms (3.1), (1.4) and (1.5) for case 3.
    Case 3 Iteration number(n) cpu time (in seconds)
    Algorithm(3.1) 9 0.001283
    Dixit et al. (1.4) 75 0.002931
    Lorenz and Pock (1.5) 13 0.001368

     | Show Table
    DownLoad: CSV
    Figure 1.  Comparison of Algorithm (3.1) for ζ0=[20,20,20] and ζ1=[10,10,10].
    Figure 2.  Comparison of Algorithm (3.1) for ζ0=[10,10,10] and ζ1=[5,5,5].
    Figure 3.  Comparison of Algorithm (3.1) for ζ0=[1,1,1] and ζ1=[2,2,2].

    We introduced an algorithm for finding the common solution of monotone inclusion and fixed point problem of a finite family of demimetric mappings in a real Hilbert space and showed that our algorithm has strong convergence under some conditions. Moreover, we proved the algorithm has a better rate of convergence by giving a numerical example.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The author Anjali delightedly acknowledge the University Grants Commission (UGC), New Delhi and the authors S. Haque and N. Mlaiki would like to thank the Prince Sultan University for paying the publication fees for this work through TAS LAB.

    The authors declare no conflicts of interest.



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