Research article

Certain new iteration of hybrid operators with contractive M -dynamic relations

  • This article investigates Wardowski's contraction in the setting of extended distance spaces known as M-metric spaces using hybrid operators based an M -dynamic iterative process. The main purpose is to observe new set-valued Chatterjea type common fixed point theorems for hybrid operators with respect to an M-dynamic iterative process, i.e., ˇD(Ψ1,Ψ2,s0). We realize an application: the existence of a solution for a multistage system and integral equation. Also, we give a graphical interpretation of our obtained theorems. The main results are explicated with the help of a relevant example. Some important corollaries are extracted from the main theorems as well.

    Citation: Amjad Ali, Muhammad Arshad, Eskandar Ameer, Asim Asiri. Certain new iteration of hybrid operators with contractive M -dynamic relations[J]. AIMS Mathematics, 2023, 8(9): 20576-20596. doi: 10.3934/math.20231049

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  • This article investigates Wardowski's contraction in the setting of extended distance spaces known as M-metric spaces using hybrid operators based an M -dynamic iterative process. The main purpose is to observe new set-valued Chatterjea type common fixed point theorems for hybrid operators with respect to an M-dynamic iterative process, i.e., ˇD(Ψ1,Ψ2,s0). We realize an application: the existence of a solution for a multistage system and integral equation. Also, we give a graphical interpretation of our obtained theorems. The main results are explicated with the help of a relevant example. Some important corollaries are extracted from the main theorems as well.



    Throughout this paper, H denotes a real Hilbert space with inner product , and the induced , I the identity operator on H, N the set of all natural numbers and R the set of all real numbers. For a self-operator T on H, F(T) denotes the set of all fixed points of T.

    Let H1 and H2 be real Hilbert spaces and let T:H1H2 be bounded linear operator. Let {Uj}tj=1:H1H1 and {Ti}ri=1:H2H2 be two finite families of operators, where t,rN. The split common fixed point problem (SCFPP) is formulated as finding a point xH1 such that

    xtj=1F(Uj)   such that  Txri=1F(Ti). (1.1)

    In particular, if t=r=1, the SCFPP (1.1) reduces to finding a point xH1 such that

    xF(U)  such that TxF(T). (1.2)

    The above problem is usually called the two-set SCFPP.

    In recent years, the SCFPP (1.1) and the two-set SCFPP (1.2) have been studied and extended by many authors, see for instance [15,20,23,27,36,37,38,39,40,47,48,49]. It is known that the SCFPP includes the multiple-set split feasibility problem and split feasibility problem as a special case. In fact, let {Cj}tj=1 and {Qi}ri=1 be two finite families of nonempty closed convex subsets in H1 and H2, respectively. Let Uj=PCj and Ti=PQi; then SCFPP (1.1) becomes the multiple-set split feasibility problem (MSSFP) as follows:

    find xtj=1Cj  such that Txri=1Qi. (1.3)

    When t=r=1 the MSSFP (1.3) is reduced to the split feasibility problem (SFP) which is described as finding a point xH1 satisfying the following property

    xC   such that  TxQ. (1.4)

    The SFP was first introduced by Censor and Elfving [22] with the aim of modeling certain inverse problems. It has turned out to also play an important role in, for example, medical image reconstruction and signal processing (see [2,4,15,17,21]). Since then, several iterative algorithms for solving (1.4) have been presented and analyzed. See, for instance [1,5,14,15,16,18,19,23,24,27] and references therein.

    The CQ algorithm has been extended by several authors to solve the multiple-set split convex feasibility problem. See, for instance, the papers by Censor and Segal [25], Elfving, Kopf and Bortfeld [23], Masad and Reich [35], and by Xu [53,54].

    In 2020, Reich and Tuyen [45] proposed and analyzied the following split feasibility problem with multiple output sets in Hilbert spaces: let H,Hi,i=1,2,,m be real Hilbert spaces. Let Ti:HHi, i=1,2,,m, be bounded linear operators. Furthermore, let C and Qi be nonempty, closed and convex subsets of H and Hi,i=1,2,,m, respectively. Find an element u, such that:

    uΩSFP=C(mi=1T1i(Qi)); (1.5)

    that is, uC and TuQi, for all i=1,2,,m.

    To solve problem (1.5), Reich et al. [46] proposed the following iterative methods: for any u0,v0C, let {un} and {vn} be two sequence generated by:

    un+1=PC[unγmi=1Ti(IPQi)Tiun], (1.6)
    vn+1=αnf(vn)+(1αn)PC[vnγnmi=1Ti(IPQi)Tivn], (1.7)

    where f:CC is a strict k-contraction with k[0,1), {γn}(0,) and {αn}(0,1). They established the weak and strong convergence of iterative methods (1.6) and (1.7), respectively.

    In 2021, Reich and Tuyen [44] considered the following split common null point problem with multiple output sets in Hilbert spaces: let H,Hi,i=1,2,,N, be real Hilbert spaces and let Ti:HHi,i=1,2,,N, be bounded linear operators. Let B:H2Hi,i=1,2,,N be maximal monotone operators. Given H,Hi and Ti as defined above, the split common null point problem with multiple output sets is to find a point u such that

    uΩ:=B1(0)(Ni=1T1i(B1i(0))). (1.8)

    To solve problem (1.8), Reich and Tuyen [44] proposed the following iterative method:

    Algorithm 1.1. For any u0H, Let H0=H,T0=IH,B0=B, and let {un} be the sequence generated by:

    vn=Ni=0βi,n[unτi,nTi(IHiJBiri,n)Tiun]un+1=αnf(un)+(1αn)vn, n0,

    where {αn}(0,1), and {βi,n} and {ri,n},i=0,1,,N, are sequences of positive real numbers, such that {βi,n}[a,b](0,1) and Ni=0βi,n=1 for each n0, and τi,n=ρi,n(IHiJBiri,n)Tiun2Ti(IHiJBiri,n)Tiun2+θi,n, where {ρi,n}[c,d](0,2) and {θi,n} is a sequence of positive real numbers for each i=0,1,,N, and f:HH is a strict contraction mapping H into itself with the contraction coefficient k[0,1).

    They established the strong convergence of the sequence {un} generated by Algorithm 1.1 which is a solution of the Problem (1.8)

    Alvarez and Attouch [7] applied the following inertial technique to develop an inertial proximal method for finding the zero of a monotone operator, i.e.,

    find xH  such that 0G(x). (1.9)

    where G:H2H is a set-valued monotone operator. Given xn1, xnH and two parameters θn[0,1), λn>0, find xn+1H such that

    0λnG(xn+1)+xn+1xnθn(xnxn1). (1.10)

    Here, the inertia is induced by the term θn(xnxn1). The equation (1.10) may be thought as coming from the implicit discretization of the second-other differential system

    0d2xdt2(t)+ρdxdt(t)+G(x(t)) a.e. t>0, (1.11)

    where ρ>0 is a damping or a friction parameter. This point of view inspired various numerical methods related to the inertial terminology which has a nice convergence property [6,7,8,28,29,33] by incorporating second order information and helps in speeding up the convergence speed of an algorithm (see, e.g., [3,7,9,10,11,12,13,51,52] and the references therein).

    Recently, Thong and Hieu [50] introduced an inertial algorithm to solve split common fixed point problem (1.1). The algorithm is of the form

    {x0,x1H1,yn=xn+αn(xnxn1),xn+1=(1βn)yn+βnrj=1wjUj(I+ti=1ηiγT(TiI)T)yn. (1.12)

    Under approximate conditions, they show that the sequence {xn} generated by (1.12) converges weakly to some solution of SCFPP (1.1).

    It was shown in [43, Section 4] by example that one-step inertial extrapolation wn=xn+θ(xnxn1), θ[0,1) may fail to provide acceleration. It was remarked in [32, Chapter 4] that the use of inertial of more than two points xn,xn1 could provide acceleration. For example, the following two-step inertial extrapolation

    yn=xn+θ(xnxn1)+δ(xn1xn2) (1.13)

    with θ>0 and δ<0 can provide acceleration. The failure of one-step inertial acceleration of ADMM was also discussed in [42, Section 3] and adaptive acceleration for ADMM was proposed instead. Polyak [41] also discussed that the multi-step inertial methods can boost the speed of optimization methods even though neither the convergence nor the rate result of such multi-step inertial methods was established in [41]. Some results on multi-step inertial methods have recently be studied in [26].

    Our Contributions. Motivated by [44,50], in this paper, we consider the following split common null point problem with multiple output sets in Hilbert spaces: Let H1 and H2 be real Hilbert spaces. Let {Uj}rj=1:H1H1 be a finite family of quasi-nonexpansive operators and Bi:H22H2,i=1,2,,t. be maximal monotone operators and {Ti}ti=1:H1H2 be a bounded linear operator. The split common null point problem with multiple output set is to find a point xH1 such that

    xrj=1F(Uj)(ti=1T1i(B1i0)). (1.14)

    Let Υ be the solution set of (1.14). We propose a two-step inertial extrapolation algorithm with self-adaptive step sizes for solving problem (1.14) and give the weak convergence result of our problem in real Hilbert spaces. We give numerical computations to show the efficiency of our proposed method.

    Let C be a nonempty, closed, and convex subset of a real Hilbert spaces H. We know that for each point uH, there is a unique element PCuC, such that:

    uPCu=infvCuv. (2.1)

    We recall that the mapping PC:HC defined by (2.1) is said to be metric projection of H onto C. Moreover, we have (see, for instance, Section 3 in [31]):

    uPCu,vPCu0,   uH, vC. (2.2)

    Definition 2.1. Let T:HH be an operator with F(T). Then

    T:HH is called nonexpansive if

    TuTvuv,   u,vH, (2.3)

    T:HH is quasi-nonexpansive if

    Tuvuv,   vF(T), uH. (2.4)

    We denote by F(T) the set of fixed points of mapping T; that is, F(T)={uC:Tu=u}. Given an operator E:H2H, its domain, range, and graph are defined as follows:

    D(E):={uH:E(u)},R(E):={E(u):uD(E)}

    and

    G(E):={(u,v)H×H:uD(E), vE(u)}.

    The inverse operator E1 of E is defined by:

    uE1(v) if and only if vE(u).

    Recall that the operator E is said to be monotone if, for each u,vD(E), we have fg,uv0 for all fE(u) and gE(v). We denote by IH the identity mapping on H. A monotone operator E is said to be maximal monotone if there is no proper monotone extension of E or, equivalently, by Minty's theorem, if R(IH+λE)=H, for all λ>0. If E is maximal monotone, then we can define, for each λ>0, a nonexpansive single-valued operator JEλ:R(IH+λE)D(E) by

    JEλ=(IH+λE)1.

    This operator is called the resolvent of E. It is easy to see that E1(0)=F(JEλ), for all λ>0.

    Lemma 2.2. [45] Suppose that E:D(E)H2H is a monotone operator. Then, we have the following statements:

    (i) For rs>0, we have:

    uJEsu2uJEru,

    for all elements uR(IH+rE)R(IH+sE).

    (ii) For all numbers r>0 and for all points u,vR(IH+rE), we have:

    uv,JEruJErvJEruJErv2.

    (iii) For all numbers r>0 and for all points u,vR(IH+rE), we have:

    (IHJEr)u(IHJEr)v,uv(IHJEr)u(IHJEr)v2.

    (iv) If S=E1(0), then for all elements uS and uR(IH+rE), we have:

    JEruu2uu2uJEru2.

    Lemma 2.3. [30] Suppose that T is a nonexpansive mapping from a closed and convex subset of a Hilbert space H into H. Then, the mapping IHT is demiclosed on C; that is, for any {un}C, such that unuC and the sequence (IHT)(un)v, we have (IHT)(u)=v.

    Lemma 2.4. [34] Given an integer N1. Assume that for each i=1,,N, Ti:HH is a ki-demicontractive operator such that Ni=1F(Ti). Assume that {wi}Ni=1 is a finite sequence of positive numbers such that Ni=1wi=1. Setting U=Ni=1wiTi, then the following results hold:

    (i) F(U)=Ni=1F(Ti).

    (ii) U is λ-demicontractive operator, where λ=max{ki|i=1,,N}.

    (iii) xUx,xz1λ2Ni=1wixTix2 for all xH and zNi=1F(Ti).

    Lemma 2.5. Let x,y,zH and α,βR. Then

    (1+α)x(αβ)yβz2=(1+α)x2(αβ)y2βz2+(1+α)(αβ)xy2+β(1+α)xz2β(αβ)yz2.

    We give the following assumptions in order to obtain our convergence analysis.

    Assumptions 3.1. We assume that the inertial parameters θ[0,12), ρ(0,1) and δ(,0] satisfies the following conditions.

    (a)

    0θ<1ρ1+ρ;

    (b)

    max{2θρ1ρ(1θ),θ1θ+1[ρ(2θ+1)(θ1)]}<δ0;

    (c)

    (2ρ1)(θ2+δ2)+(2ρ)(θδ)+ρ2δθ1<0.

    Now, we present our proposed method and our convergence analysis as follows:

    Algorithm 3.2. Two-Step Inertial for Split Common Null Point Problem
      Step 1. Choose δ(,0] and θ[0,1/2) such that Assumption 3.1 is fulfilled. Pick x1,x0,x1H1 and set n=1.
      Step 2. Given xn2,xn1 and xn, compute xn+1 as follows
        {yn=xn+θ(xnxn1)+δ(xn1xn2),xn+1=(1ρ)yn+ρrj=1wjUj(Iti=1δi,nτi,nTi(IH2JBiri,n)Ti)yn,(3.1)
    where {δi,n} and {ri,n}, i=1,2,t, are sequences of positive real numbers, such {δi,n}[a,b](0,1) and ti=1δi,n=1, for each n1 and
          τi,n=ρi,n(IH2JBiri,n)Tiyn2Ti(IH2JBiri,n)Tiyn2+θi,n,(3.2)
    where {ρi,n}[c,d](0,2) and {θi,n} are sequences of positive real numbers for each i=1,2,,t, and {Uj}rj=1 is a finite family of quasi-nonexpansive operators.
      Step 3. Set nn+1 and go to Step 2.

    Lemma 3.3. For tN, let {Bi}ti=1:H22H2 be a finite family maximal monotone operators. Let {Ti}ti=1:H1H2, be a finite family of bounded linear operators. Define the operator V:H1H1 by

    V:=Iti=1δi,nτi,nTi(IH2JBiri,n)Ti, (3.3)

    where τi,n is as defined in (3.2), {δi,n}ti=1(0,1) and ti=1δi,n=1. Then we have the following results:

    (1)

    Vxz2xz2ti=1δi,nρi,n(2ρi,n)(IH2JBiri,n)Tix4Ti(IH2JBiri,n)Tix2+θi,n.

    (2) xF(V) if and only if Tixti=1F(JBiri,n), for i=1,2,,t.

    Proof. (1) Given a point zΥ, it follows from the convexity of the function 2 that:

    Vxz2=xti=1δi,nτi,nTi(IH2JBiri,n)Tixz2=ti=1δi,n(xτi,nTi(IH2JBiri,n)Tixz)2ti=1δi,nxτi,nTi(IH2JBiri,n)Tixz2. (3.4)

    Using JBiri,n(Tiz)=Tiz and Lemma 2.2(iii), for each i=1,2,,t, we see that

    xτi,nTi(IH2JBiri,n)Tixz2=xz22τi,nTi(IH2JBiri,n)Tix,xz+τ2i,nTi(IH2JBiri,n)Tix2=xz22τi,n(IH2JBiri,n)Tix,TixTiz+τ2i,nTi(IH2JBiri,n)Tix2=xz22τi,n(IH2JBiri,n)Tix(IH2JBiri,n)Tiz,TixTiz+τ2i,nTi(IH2JBiri,n)Tix2xz22τi,n(IH2JBiri,n)Tix2+τ2i,n(Ti(IH2JBiri,n)Tix2+θi,n)=xz2ρi,n(2ρi,n)(IH2JBiri,n)Tix4Ti(IH2JBiri,n)Tix2+θi,n. (3.5)

    Hence, from (3.4) and (3.5), we get

    Vxz2xz2ti=1δi,nρi,n(2ρi,n)(IH2JBiri,n)Tix4Ti(IH2JBiri,n)Tix2+θi,n.

    (2) It is obvious that if Tixti=1F(JBiri,n) then xF(V). We show the converse, let xF(V) and zT1i(F(JBiri,n)) we have

    xz2=Vxz2xz2ti=1δi,nρi,n(2ρi,n)(IH2JBiri,n)Tix4Ti(IH2JBiri,n)Tix2+θi,n. (3.6)

    Since ρi,n(0,2), we obtain

    (IH2JBiri,n)Tix=0,     i=1,2,,t.

    That is, Tixti=1F(JBiri,n).

    Lemma 3.4. For t,rN, let {Bi}ti=1:H2H2 be maximal monotone operators such that ti=1F(JBiri,n) and {Uj}rj=1:H1H1 be a finite family of quasi-nonexpansive operators such that rj=1F(Uj). Assume that {IUj}rj=1 and {IJBiri,n}ti=1 are demiclosed at zero. Let {Ti}ti=1:H1H2,  be bounded linear operators suppose Υ. Let S:H1H1 be defined by

    Sx=rj=1wjUj(Iti=1δi,nτi,nTi(IH2JBiri,n)Ti)x,

    where {τi,n}, is as defined in (3.2), {wj}rj=1 and {δi,n}ti=1 are in (0,1) with rj=1wj=1 and ti=1δi,n=1. Assume that the following conditions are satisfied

    (A1) mini=1,2,,t{infn{ri,n}}=r>0;

    (A2) maxi=1,2,,t{supn{θi,n}}=K<.

    Then the following hold:

    (a) The operator S is quasi-nonexpansive.

    (b) F(S)=Υ.

    (c) IS is demiclosed at zero.

    Proof. From the definition of V we can rewrite the operator S as

    Sx=rj=1wjUjVx.

    We show the following

    (i) {UjV}rj=1 is a finite family of quasi-nonexpansive operator,

    (ii) rj=1F(UjV)=Υ,

    (iii) for each j=1,2,,r then IUjV is demiclosed at zero.

    By Lemma 3.3, V is quasi-nonexpansive. Therefore, for each j=1,2,,r the operator UjV is quasi-nonexpansive. Next, we show that for each j=1,2,,r, then

    F(UjV)=F(Uj)F(V).

    Indeed, it suffices to show that for each j=1,2,,r F(UjV)F(Uj)F(V). Let pF(UjV). It is enough to show that pF(V). Now, taking zF(Uj)F(V); we have

    pz2=UjVpz2Vpz2pz2ti=1δi,nρi,n(2ρi,n)(IH2JBiri,n)Tix4Ti(IH2JBiri,n)Tix2+θi,n.

    This implies that

    ti=1δi,nρi,n(2ρi,n)(IH2JBiri,n)Tix4Ti(IH2JBiri,n)Tix2+θi,n=0.

    That is JBiri,n(Tip)=Tip,     i=1,2,,t. This implies that Tipti=1F(JBiri,n). Thus, pF(V).

    Therefore, F(Uj)F(V)=F(UjV),  j=1,,r. We now show that

    Υ={prj=1F(Uj)    such that  Tipti=1F(JBiri,n)}=tj=1F(UjV).

    By Lemma 3.3, we have

    Υ={xrj=1F(Uj)|Tixri=1F(JBiri,n)}={xri=1F(Uj)|xF(V)}=rj=1F(Uj)F(V)=rj=1F(UjV).

    Finally, we show that for each j=1,..r, IUjV is demiclosed at zero. Let {xn}H1 be a sequence such that xnzH1 and UjVxnxn0 we have

    0xnzUjVxnzxnUjVxn0.

    This implies that

    xnz2UjVxnz20.

    By Lemma 3.3, we have

    UjVxnz2Vxnz2xnz2ti=0δi,nρi,n(2ρi,n)(IH2JBiri,n)Tixn4Ti(IH2JBiri,n)Tixn2+θi,n. (3.7)

    This implies that

    ti=0δi,nρi,n(2ρi,n)(IH2JBiri,n)Tix4Ti(IH2JBiri,n)Tix2+θi,nxnz2UjVxnz2.

    Since {δi,n}[a,b](0,1), {ρi,n}[c,d](0,2), and (3.7) implies

    (IH2JBiri,n)Tixn4Ti(IH2JBiri,n)Tixn2+θi,n0,   as   n, (3.8)

       i=0,1,2t. It follows from the boundedness of the sequence {xn} that L:=maxi=0,1,,N{sup{Ti(IH2JBiri,n)Tixn2}}<. Thus from Condition (A2), It follows that

    (IH2JBiri,n)Tixn4Ti(IH2JBiri,n)Tixn2+θi,n(IH2JBiri,n)Tixn4L+K.

    Combining this with (3.8), we deduce that

    (IH2JBiri,n)Tixn0 (3.9)

     i=0,1,,N, Lemma 2.2(i) and Condition (A1) now imply that

    (IH2JBir)Tixn2(IH2JBiri,n)Tixn, (3.10)

     i=0,1,,N. Thus using (3.9) and (3.10), we are able to deduce that

    (IH2JBir)Tixn0, (3.11)

    i=0,1,,N.

    From Vxnxn=ti=0δi,nτi,nTi(IH2JBiri,n)Tixn, the assumptions on {δi,n} and {τi,n} and (3.10), it follows that

    Vxnxn0.

    On the other hand

    UjVxnVxnUjVxnxn+Vxnxn0. (3.12)

    Since xnz, we have Vxnz and by the demiclosedness of Uj we have zF(Uj). Since, for each i=1,2,,N, Ti is a bounded linear operator, it follows that TixnkTiz. Thus by Lemma 2.3 and (3.11) implies that TizF(JBir)  i=1,,t that is Tizti=1F(JBir). By Lemma 3.3 we get zF(V). Therefore, zF(Uj)F(V)=F(UjV).

    By Claim (i) and Lemma 2.4, we obtain Sx=rj=1wjUjVx is quasi-nonexpansive and F(S)=tj=1F(UjV)=Υ.

    Finally, we show that IS is demiclosed at zero. Indeed, Let {xn}H1 be a sequence such that xnzH1 and xnSxn0. Let pF(S) by Lemma 2.4, we have

    xnSxn,xnp12tj=1xnUjVxn2.

    This imples that, for each j=1,,t we have

    xnUjVxn0   as  n.

    By the demiclosedness of IUjV we have zF(UjV). Therefore ztj=1F(UjV)=F(S).

    Theorem 3.5. For t,rN. Let {Bi}ti=1:H2H2 be a finite family of maximal monotone operators such that ti=1F(JBir) and {Uj}rj=1:H1H1 be a finite family of quasi-nonexpansive operators such that rj=1F(Uj). Assume that {IUj}rj=1 and {IJBir}ti=1 are demiclosed at zero. Let Ti:H1H2, i=1,2,,N be bounded linear operators. Suppose Υ. Let {xn} be a sequence generated by Algorithm 3.2. and suppose that Assumptions (3.1) (a)-(c) are fulfilled. Then {xn} converges weakly to an element of Υ.

    Proof. Let S=rj=1wjUj(Iti=1δi,nτi,nTi(IH2JBiri,n)Ti), then the sequence {xn+1} can be rewritten as follows

    xn+1=(1ρ)yn+ρSyn. (3.13)

    By Lemma 3.3, we have that S is quasi-nonexpansive. Let zΥ, from (3.13), we get

    xn+1z2=(1ρ)(ynz)+ρ(Synz)2=(1ρ)ynz2+ρSynz2ρ(1ρ)ynSyn2ynz2ρ(1ρ)ynSyn2. (3.14)

    Observe that

    ynz=xn+θ(xnxn1)+δ(xn1xn2)z=(1+θ)(xnz)(θδ)(xn1z)δ(xn2z).

    Hence by Lemma 2.5, we have

    ynz2=(1+θ)(xnz)(θδ)(xn1z)δ(xn2z)2=(1+θ)xnz2(θδ)xn1z2δxn2z2+(1+θ)(θδ)xnxn12+δ(1+θ)xnxn22δ(θδ)xn1xn22. (3.15)

    Note also that

    2θxn+1xn,xnxn1=2θ(xn+1xn),xnxn12|θ|xn+1xnxnxn1=2θxn+1xnxnxn1,

    and so,

    2θxn+1xn,xnxn12θxn+1xnxnxn1. (3.16)

    Also,

    2δxn+1xn,xn1xn2=2δ(xn+1xn),xn1xn22|δ|xn+1xnxn1xn2,

    which implies that

    2δxn+1xn,xn1xn22|δ|xn+1xnxn1xn2. (3.17)

    Similarly, we have

    2δθxn1xn,xn1xn22|δ|θxnxn1xn1xn2,

    and thus

    2δθxnxn1,xn1xn22|δ|θxnxn1xn1xn2. (3.18)

    By (3.16)–(3.18) and Cauchy-Schwartz inequality one has

    xn+1yn2=xn+1(xn+θ(xnxn1)+δ(xn1xn2))2=xn+1xnθ(xnxn1)δ(xn1xn2)2=xn+1xn22θxn+1xn,xnxn12δxn+1xn,xn1xn2+θ2xnxn12+2δθxnxn1,xn1xn2+δ2xn1xn22xn+1xn22θxn+1xnxnxn12|δ|xn+1xnxn1xn2+θ2xnxn122|δ|θxnxn1xn1xn2+δ2xn1xn22xn+1xn2θxn+1xn2θxnxn12|δ|xn+1xn2|δ|xn1xn22+θ2xnxn12|δ|θxnxn12|δ|θxn1xn22+δ2xn1xn22=(1|δ|θ)xn+1xn2+(θ2θ|δ|θ)xnxn12+(δ2|δ||δ|θ)xn1xn22. (3.19)

    Observe that

    Synyn2=1ρ2xn+1yn2. (3.20)

    Putting (3.20) in (3.14), we get

    xn+1z2ynz2ρ(1ρ)Synyn2=ynz21ρρxn+1yn2. (3.21)

    Combining (3.15) and (3.19) in (3.21) with noting that δ0 we obtain

    xn+1z2=(1+θ)xnz2(θδ)xn1z2δxn2z2+(1+θ)(θδ)xnxn12+δ(1+θ)xnxn22δ(θδ)xn1xn22(1ρ)ρ(1|δ|θ)xn+1xn2(1ρ)ρ(θ2θ|δ|θ)xnxn12(1ρ)ρ(δ2|δ||δ|θ)xn1xn22=(1+θ)xnz2(θδ)xn1z2δxn2z2+δ(1+θ)xnxn22+[(1+θ)(θδ)(1ρ)ρ(θ2θ|δ|θ)]xnxn12[δ(θδ)+(1ρ)ρ(δ2|δ||δ|θ)]xn1xn22(1ρ)ρ(1|δ|θ)xn+1xn2(1+θ)xnz2(θδ)xn1z2δxn2z2+[(1+θ)(θδ)(1ρ)ρ(θ2θ+δθ)]xnxn12[δ(θδ)+(1ρ)ρ(δ2+δ+δθ)]xn1xn22(1ρ)ρ(1+δθ)xn+1xn2. (3.22)

    By rearranging we get

    xn+1z2θxnz2δxn1z2+(1ρ)ρ(1+δθ)xn+xn2xnz2θxn1z2δxn2z2+(1ρ)ρ(1+δθ)xnxn12+[(1+θ)(θδ)(1ρ)ρ(θ22θ+δθ+δ+1)][δ(θδ)+(1ρ)ρ(δ2+δ+δθ)]xn1xn22. (3.23)

    Define

    Υn:=xnz2θxn1z2δxn2z+(1ρ)ρ(1+δθ)xnxn12.

    Let us show that Υn0,n1. Now

    Υn=xnz2θxn1z2δxn2z2+(1ρ)ρ(1+δθ)xnxn12xnz22θxnxn122θxnz2δxn2z2+(1ρ)ρ(1+δθ)xnxn12=(12θ)xnz2+[(1ρ)ρ(1+δθ)2θ]xnxn12δxn2z2. (3.24)

    Since θ<1/2, δ0, 2θρ1ρ(1θ)<δ and 0θ<1ρ1+ρ, it follows from (3.24) that Υn0. Furthermore, we drive from (3.23)

    Υn+1Υn[(1+θ)(θδ)(1ρ)ρ(θ22θ+δθ+δ+1)]xnxn12[δ(θδ)+(1ρ)ρ(δ2+δ+δθ)]xn1xn22=[(1+θ)(θδ)(1ρ)ρ(θ22θ+δθ+δ+1)]×(xn1xn2xnxn12)+[(1+θ)(θδ)(1ρ)ρ(θ22θ+δθ+δ+1)δ(θδ)(1ρ)ρ(δ2+δ+δθ)]xn1xn22=q1[xn1xn22xnxn12]q2xn1xn22, (3.25)

    where

    q1:=[(1+θ)(θδ)(1ρ)ρ(θ22θ+δθ+δ+1)]

    and

    q2:=[(1+θ)(θδ)(1ρ)ρ(θ22θ+δθ+δ+1)δ(θδ)(1ρ)ρ(δ2+δ+δθ)]xn1xn22. (3.26)

    By our assumption, it holds that

    θ1θ+1[ρ(2θ+1)(θ1)]<δ. (3.27)

    As a result q1>0. Also q2>0 by Assumption 3.1 (c). Then by (3.25) we have

    Υn+1+q1xnxn12Υn+q1xn1xn22q2xn1xn22. (3.28)

    Letting ˉΥn:=Υn+q1xn1xn22. Then ˉΥn0, n1. Also, it follows from (3.28) that

    ˉΥn+1ˉΥn. (3.29)

    These facts imply that the sequence {ˉΥn} is decreasing and bounded from below and thus limnˉΥn exists. Consequently, we get from (3.28) and the squeeze theorem that

    limnq1xn1xn22=0. (3.30)

    Hence

    limnxn1xn22=0. (3.31)

    As a result

    xn+1yn=xn+1xnθ(xnxn1)δ(xn1)xn+1xn+θxnxn1+|δ|xn1xn20

    as n. By limnxn+1xn=0, one has

    xnynxnxn+1+xn+1yn0, as n.

    By (3.31) and the existence of limnˉΥn, we have that limnΥn exists and hence {Υn} is bounded. Now, since limnxn+1xn=0, we have from the definition of Υn that

    limn[xnz2θxn1z2δxn2z2] (3.32)

    exists. Using the boundedness of {Υn}, we obtain from (3.24) that {xn} is bounded. Consequently {yn} is bounded. From (3.8), we obtain

    ρ(1ρ)Synynynz2xn+1z2.

    This implies that

    limnSynyn=0. (3.33)

    Finally, we show that the sequence {xn} converges weakly to xΥ. Indeed, since {xn} is bounded we assume that there exists a subsequence {xnj} of {xn} such that xnjxH. Since xnyn0, we also have ynjx. Then by the demiclosedness of IS, we obtain xF(S)=Υ.

    Now, we show that {xn} has unique weak limit point in Υ. Suppose that {xmj} is another subsequence of {xn} such that xmjv as j. Observe that

    2xn,xv=xnv2xnx2v2+x22xn1,xv=xn1v2xn1x2v2+x2 (3.34)

    and

    2xn2,xv=xn2v2xn2x2v2+x2.

    Therefore

    2θxn1,xv=θxn1v2+θxn1x2+θv2θx2. (3.35)

    and

    2δxn2,xv=δxn2v2+δxn2x2+δv2δx2. (3.36)

    Addition of (3.34)–(3.36) gives

    2xnθxn1δxn2,xv=(xnv2θxn1v2δxn2v2) (3.37)
    (xnx2θxn1x2δxn2x2)+(1θδ)(xv2). (3.38)

    According to (3.32), we get

    limn[xnx2θxn1x2δxn2x2] (3.39)

    exists and

    limn[xnv2θxn1v2δxn2v2] (3.40)

    exists. This implies from (3.21) that limnxnθxn1δxn2,xv exists. Consequently,

    vθvδv,xv=limjxnjθxnj1βxnj2,xv=limnxn1θxn1δxn2,xv=limjxmjθxmj1βxmj2,xv=xθxβx,xv.

    Hence

    (1θδ)xv2=0.

    Since δ0<1θ, we obtain that x=v. Therefore, the sequence {xn} converges weakly to xΥ. This completes the proof.

    In this section, we give a numerical description to illustrate how our proposed algorithm can be implemented in the setting of the real Hilbert space R. Furthermore, we shall show the effect of the double inertia in the fast convergence of the sequence generated by our proposed Algorithm 3.2. First, we give the set of parameters that satisfy the conditions given in assumption 3.1. To this end, fix ρ=12 and take

    θ=14,δ=15 and 11000;θ=15,δ=110 and 11000;θ=16,δ=1100 and 11000

    Clearly, these parameters satisfy the conditions given in assumption 3.1. Next, we define the operators to be used in the implementation of Algorithm 3.2. In Algorithm 3.2, fix t=N=r=3. Set H1=H2=H3=R. Let δi,n=13, ρi,n=θi,n=23, ri,n=12 and wj=13, where i,j{1,2,3} and n1. Let Bi,Ti,Uj:RR be defined by

    Bix=2ix, then JBiri,nx=x1+2iri,n,Tix=ix,Ujx=jx.

    Then,

    Ti(IH2JBiri,n)Tix=2i3ri,n1+2iri,n, and τi,n=23(2i2ri,nyn)2(2i3ri,n)2+2(1+2iri,n)2

    With this, we are ready to implement our proposed Algorithm 3.2 on MATLAB. Choosing x0=1,x1=2 and x2=0.5, and setting maximum number of iterations to 150 or 1016, as our stopping criteria, we varied the double inertial parameters as given above. We obtained the following successive approximations:

    From the numerical simulations presented in Table 1 and Figures 13, we saw that in this example, the best choice for the double inertial parameters is θ=14 and δ=11000. Furthermore, we observed that as θ decreases and δ approaches 0, the number of iterations required to satisfy the stopping criteria increases.

    Table 1.  Results of the numerical simulations.
    No. Iter Inertia Para. |xn+1xn|
    120 θ=14 1.11E-16
    δ=-15
    80 θ=14 1.11E-16
    δ=-11000
    107 θ=15 1.11E-16
    δ=-110
    88 θ=15 1.11E-16
    δ=-11000
    95 θ=16 1.11E-16
    δ=-1100
    116 θ=16 1.11E-16
    δ=-11000

     | Show Table
    DownLoad: CSV
    Figure 1.  Graph of the iterates of Algorithm 3.2 when θ=14, δ=-15 and δ=-11000.
    Figure 2.  Graph of the iterates of Algorithm 3.2 when θ=15, δ=-110 and δ=-11000.
    Figure 3.  Graph of the iterates of Algorithm 3.2 when θ=16, δ=-1100 and δ=-11000.

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

    This research was supported by The Science, Research and Innovation Promotion Funding (TSRI) [grant number FRB660012/0168]. This research block grants was managed under Rajamangala University of Technology Thanyaburi [grant number FRB66E0628].

    All authors declare no conflicts of interest in this paper.



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