On new faster fixed point iterative schemes for contraction operators and comparison of their rate of convergence in convex metric spaces

Abstract

In this paper we present new iterative algorithms in convex metric spaces. We show that these iterative schemes are convergent to the fixed point of a single-valued contraction operator. Then we make the comparison of their rate of convergence. Additionally, numerical examples for these iteration processes are given.

Authors

Cristian Daniel Alecsa
Babes-Bolyai University, Department of Mathematics, Cluj-Napoca, Romania,
Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy, Cluj-Napoca, Romania

Keywords

convex metric space; fixed point; iterative algorithm; rate of convergence; convex combination.

Paper coordinates

C.-D. Alecsa, On new faster fixed point iterative schemes for contraction operators and comparison of their rate of convergence in convex metric spaces, International Journal Of Nonlinear Analysis And Applications, 8 (2019) no. 1, pp. 353-388.
doi: 10.22075/ijnaa.2017.11144.1543

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International Journal of Nonlinear Analysis and Applications

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2008-6822

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[4] V. Berinde, Picard iteration converges faster than Man iteration for a class of quasi-contractive operators, Fixed Point Theory Appl. 2014 (2004):1.

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[6] H. Fukhar-ud-din and V. Berinde, Iterative methods for the class of quasi-contractive type operators and comparsion of their rate of convergence in convex metric spaces, Filomat 30 (2016) 223–230.

[7] K. Goebel and W. Kirk, Iteration processes for nonexpansive mappings, Topological Methods in Nonlinear Functional Analysis, Contemporary Mathematics 21, Amer. Math. Soc. Providence (1983), 115–123.

[8] F. Gursoy and V.Karakaya, A Picard-S hybrid type iteration method for solving a differential equation with retarded argument, http://arxiv.org/abs/1403.2546
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[11] W. Phuengrattana and S. Suantai, Comparison of the rate of convergence of various iterative methods for the class of weak contractions in Banach Spaces, Thai J. Math. 11 (2013) 217–226.

[12] S. Reich and I. Safrir, Nonexpansive iteration in hyperbolic spaces, Nonlinear. Anal. 15 (1990) 537–558.

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On new faster fixed point iterative schemes for contraction operators and comparison of their rate of convergence in convex metric spaces

Cristian Daniel Alecsa
Babeş-Bolyai University, Department of Mathematics, Cluj-Napoca, Romania, Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy, Cluj-Napoca, Romania
Abstract

In this paper we present new iterative algorithms in convex metric spaces. We show that these iterative schemes are convergent to the fixed point of a single-valued contraction operator. Then we make the comparison of their rate of convergence. Additionally, numerical examples for these iteration processes are given.

(Communicated by Themistocles M. Rassias)

Keywords: convex metric space; fixed point; iterative algorithm; rate of convergence; convex combination.
2010 MSC: Primary 47H10; Secondary 54H25.

1. Introduction and Preliminaries

Most of the real world problems of applied sciences are, in general, functional equations. Such equations can be written as fixed point equations. Then, it is necessary to develop an iterative process which approximate the solution of these equations that has a good rate of convergence.

Many studies in the field of fixed point theory concerning the existence and uniqueness of fixed points of singlevalued contractions have been developed using basic iterative algorithms, such as : Picard iteration, Krasnoselksii, Mann and Ishikawa iterative processes. Over the years the interest regarding the speed of convergence of such iterations grew very fast. For example, many authors

00footnotetext: *Corresponding author
Email address: cristian.alecsa@math.ubbcluj.ro, cristian.alecsa@ictp.acad.ro (Cristian Daniel Alecsa)

considered numerous iteration processes and studied their rate of convergence. For this see [1]-[4, [6][8] and [10]-14. Some iterations were introduced to study the fixed points of the contractions. Also, others [12] were introduced for the context of nonexpansive mappings. Furthermore, some authors [6] compared the rate of convergence for some iterative algorithms for the class of quasi-contractions. Finally, since the class of convex metric spaces is larger than the well-known class of linear normed spaces, we shall work in the context of convex metric spaces introduced by W. Takahashi.

Our aim is to introduce new iteration processes and prove that these are faster than most of the classical iterations found in literature, in suitable circumstances. We support analytic proof by some numerical examples.

In the present research paper, we work on a nonlinear domain, more explicitly on a convex metric space. Following [15], let ( X,dX,d ) a metric space and W:X×X×[0,1]XW:X\times X\times[0,1]\rightarrow X a mapping called a convexity structure. If

d(u,W(x,y,λ))λd(u,x)+(1λ)d(u,y), for all u,x,yX and λ[0,1],d(u,W(x,y,\lambda))\leq\lambda d(u,x)+(1-\lambda)d(u,y),\text{ for all }u,x,y\in X\text{ and }\lambda\in[0,1],

then ( X,d,WX,d,W ) is called a convex metric space. Additionally, following [16], we have that W(x,y,0)=yW(x,y,0)=y, for all x,yXx,y\in X.
A nonempty subset CC of a convex metric space XX is convex, if W(x,y,λ)CW(x,y,\lambda)\in C, for all x,yCx,y\in C.
We remind the reader of two important basic example of convex metric spaces : CAT(0)CAT(0) spaces and linear normed spaces. For details, we let the reader follow [5] and [15]. Other important examples are : hyperbolic spaces introduced by Goebel and Kirk and hyperbolic spaces in the sense of Reich and Safrir. For details one can follow : [7] and [12].

Simplifying some existing iteration processes in the literature, we recall the definition of Machado from [9] of general convex combinations defined on convex metric spaces:
For a1,,anXa_{1},\ldots,a_{n}\in X and φ1,,φn[0,1]\varphi_{1},\ldots,\varphi_{n}\in[0,1] with i=1nφi=1\sum_{i=1}^{n}\varphi_{i}=1, we define the multiple convex combination of a1,,ana_{1},\ldots,a_{n}

W(a1,,an;φ1,,φn)=W(W(a1,,an1;φ11φn,,φn11φn),an;1φn), if φn1W\left(a_{1},\ldots,a_{n};\varphi_{1},\ldots,\varphi_{n}\right)=W\left(W\left(a_{1},\ldots,a_{n-1};\frac{\varphi_{1}}{1-\varphi_{n}},\ldots,\frac{\varphi_{n-1}}{1-\varphi_{n}}\right),a_{n};1-\varphi_{n}\right),\text{ if }\varphi_{n}\neq 1

and W(a1,,an;0,,1)=anW\left(a_{1},\ldots,a_{n};0,\ldots,1\right)=a_{n}, if φn=1\varphi_{n}=1.
In the next section, we will work in the cases when n=2n=2 and n=3n=3. For the simplicity of this remark, we consider that φn1\varphi_{n}\neq 1. The other case is obvious and follows from the above definition.

We make the convention that, for n=2n=2, we have:
W(a1,a2;φ1,φ2)=W(W(a1,a1;φ11φ2),a2;1φ2)=W(a1,a2;1φ2)=W(a1,a2,φ1)W\left(a_{1},a_{2};\varphi_{1},\varphi_{2}\right)=W\left(W\left(a_{1},a_{1};\frac{\varphi_{1}}{1-\varphi_{2}}\right),a_{2};1-\varphi_{2}\right)=W\left(a_{1},a_{2};1-\varphi_{2}\right)=W\left(a_{1},a_{2},\varphi_{1}\right), where φ1+φ2=1\varphi_{1}+\varphi_{2}=1.

Furthermore, we remind that we have used the following property of convex metric spaces : W(x,x,λ)=x,xXW(x,x,\lambda)=x,\forall x\in X and λ[0,1]\lambda\in[0,1]. For n=3n=3, we have that
W(a1,a2,a3;φ1,φ2,φ3)=W(W(a1,a2;φ11φ3,φ21φ3),a3;1φ3)=W(b3,a3;1φ3)W\left(a_{1},a_{2},a_{3};\varphi_{1},\varphi_{2},\varphi_{3}\right)=W\left(W\left(a_{1},a_{2};\frac{\varphi_{1}}{1-\varphi_{3}},\frac{\varphi_{2}}{1-\varphi_{3}}\right),a_{3};1-\varphi_{3}\right)=W\left(b_{3},a_{3};1-\varphi_{3}\right), where b3=W(a1,a2;φ11φ3,φ21φ3)=W(a1,a2;1φ21φ3)b_{3}=W\left(a_{1},a_{2};\frac{\varphi_{1}}{1-\varphi_{3}},\frac{\varphi_{2}}{1-\varphi_{3}}\right)=W\left(a_{1},a_{2};1-\frac{\varphi_{2}}{1-\varphi_{3}}\right), as in the case when n=2n=2. Also, we have that φ1+φ2+φ3=1\varphi_{1}+\varphi_{2}+\varphi_{3}=1.

In the next sections, we will work under the definition of convex metric spaces and with the notions of single-valued contractions. We recall here this concept.

Definition 1.1. Let ( X,dX,d ) be a metric space and T:XXT:X\rightarrow X an operator. We say that TT is a δ\delta-contraction if there exists δ[0,1)\delta\in[0,1), such that :
d(T(x),T(y))δd(x,y)d(T(x),T(y))\leq\delta d(x,y), for each x,yXx,y\in X.
For the simplicity of notations, we will use TxTx instead of T(x)T(x), for each xXx\in X.
Also, if ( X,dX,d ) is a complete metric space, then TT has a unique fixed point in XX.

As we shall present some iterative algorithms defined by multiple convex combinations and compare their rate of convergence, we shall remind some defintions of convergence suitable for the case of metric spaces. For details, see [6], [10] and [1].

Definition 1.2. Let ana_{n} and bnb_{n} be two sequences of positive numbers that converge to aa, respectively b. Assume that there exist the following limit

limn|ana||bnb|=l\lim_{n\rightarrow\infty}\frac{\left|a_{n}-a\right|}{\left|b_{n}-b\right|}=l

(i) If l=0l=0, then it is said that {an}\left\{a_{n}\right\} converge faster to aa than {bn}\left\{b_{n}\right\} to bb.
(ii) If 0<l<0<l<\infty, then it is said that {an}\left\{a_{n}\right\} and {bn}\left\{b_{n}\right\} have the same rate of convergence.

Definition 1.3. Suppose that we have two iteration sequences {un}\left\{u_{n}\right\} and {vn}\left\{v_{n}\right\} both converging to a fixed point pp.
Let {an}\left\{a_{n}\right\} and {bn}\left\{b_{n}\right\} be two sequences of positive numbers, such that:

d(un,p)an, for all n,\displaystyle d\left(u_{n},p\right)\leq a_{n},\text{ for all }n\in\mathbb{N},
d(vn,p)bn, for all n,\displaystyle d\left(v_{n},p\right)\leq b_{n},\text{ for all }n\in\mathbb{N},

where {an}\left\{a_{n}\right\} and {bn}\left\{b_{n}\right\} converging to 0 . If {an}\left\{a_{n}\right\} converge faster than {bn}\left\{b_{n}\right\} in the sense of (Definition 1.2), then {un}\left\{u_{n}\right\} is said to converge faster than {vn}\left\{v_{n}\right\} to pp.

In this article, we use as references the articles of Abbas, Nazir, Gursoy, Karakaya and Berinde. In [1, [8] and [4]. The authors used for comparing the rate of convergence of new iterations with Picard iteration, the definitions (1.2) and (1.3).

In [11, [6] and [4], Suantai, Berinde et al. made the following remark: that the original definition for comparison of rate of convergence depends on the sequences {an},{bn},{cn},{βn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{c_{n}\right\},\left\{\beta_{n}\right\} and {αn}\left\{\alpha_{n}\right\}, where the already presented sequences appear as auxiliary sequences in some iterative processes. Therefore, the definitions (1.2) and (1.3) are not consistent and this method of comparing the rate of convergence of two iterative algorithms is unclear.

In [11], Phuengrattana and Suantai proposed a new definition in convex metric spaces (see [6]).
Definition 1.4. If {xn}\left\{x_{n}\right\} and {un}\left\{u_{n}\right\} are two iterative sequences that converge to the unique fixed point pp of TT, then {xn}\left\{x_{n}\right\} converges faster than {un}\left\{u_{n}\right\}, if

limnd(xn,p)d(un,p)=0.\lim_{n\rightarrow\infty}\frac{d\left(x_{n},p\right)}{d\left(u_{n},p\right)}=0.

In the case of convex metric spaces, if we use the above definition for comparing the rate of convergence of two iterative schemes, then we need the following property (see [6] and [11]).

Remark 1.5. For each x,y,zXx,y,z\in X and λ[0,1]\lambda\in[0,1], we have that

d(z,W(x,y,λ))(1λ)d(z,y)λd(z,x)d(z,W(x,y,\lambda))\geq(1-\lambda)d(z,y)-\lambda d(z,x)

In the entire fixed point literature, there are a lot of classical iteration schemes defined on normed linear spaces and on metric spaces endowed with a convexity structure. Following [6] and [10], we shall remind some of them, but with the remark that, in the research article [10], the authors use a modified version of convex metric space, that is the hyperbolic space in the sense of Goebel and Kirk. So, the iterative schemes will be defined with the inverse order of the two sequence terms appearing in the convexity structure WW :

Let CC be a convex subset of the convex metric space ( X,d,WX,d,W ) and T:CCT:C\rightarrow C be a contraction mapping. Moreover, let αn,bnan\alpha_{n},b_{n}a_{n} be sequences in ( 0,1 ). The classical Noor iteration is

{xn+1=W(Tyn,xn,αn)yn=W(Tzn,xn,bn)zn=W(Txn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Ty_{n},x_{n},\alpha_{n}\right)\\ y_{n}=W\left(Tz_{n},x_{n},b_{n}\right)\\ z_{n}=W\left(Tx_{n},x_{n},a_{n}\right)\end{array}\right.

Putting an=0a_{n}=0 we have that zn=xnz_{n}=x_{n}, for each nn\in\mathbb{N}, we get the well know Ishikawa iteration in convex metric spaces:

{xn+1=W(Tyn,xn,αn)yn=W(Txn,xn,bn)\left\{\begin{array}[]{l}x_{n+1}=W\left(Ty_{n},x_{n},\alpha_{n}\right)\\ y_{n}=W\left(Tx_{n},x_{n},b_{n}\right)\end{array}\right.

Putting an=bn=0a_{n}=b_{n}=0, then yn=zn=xny_{n}=z_{n}=x_{n}, for each nn\in\mathbb{N}, we get the well know Mann iteration in convex metric spaces:

xn+1=W(Txn,xn,αn)x_{n+1}=W\left(Tx_{n},x_{n},\alpha_{n}\right) (1.3)

Furthermore, we remind that we can employ a condition from hyperbolic spaces, which is satisfied in linear normed spaces, i.e. : W(x,y,λ)=W(y,x,1λ)W(x,y,\lambda)=W(y,x,1-\lambda), for each x,yXx,y\in X and λ[0,1]\lambda\in[0,1]. This conditions is not at all restrictive and it has the advantage that the iteration terms in the convexity structure WW can be swapped one with another and this does not affect convergence of the fixed point iteration.

Moreover, we recall the basic fixed point iteration which appears in Banach contraction principle, that is Picard iteration:

xn+1=Txn, for each nx_{n+1}=Tx_{n},\text{ for each }\mathrm{n}\in\mathbb{N} (1.4)

Other interesting iteration algorithms are the implicit iterations. Following [10], we recall: The implicit Noor iteration

{xn+1=W(Txn+1,yn,αn)yn=W(Tyn,zn,bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Tx_{n+1},y_{n},\alpha_{n}\right)\\ y_{n}=W\left(Ty_{n},z_{n},b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Putting an=0a_{n}=0, then zn=xnz_{n}=x_{n}, for each nn\in\mathbb{N}, we get the implicit Ishikawa iteration in convex metric spaces:

{xn+1=W(Txn+1,yn,αn)yn=W(Tyn,xn,bn)\left\{\begin{array}[]{l}x_{n+1}=W\left(Tx_{n+1},y_{n},\alpha_{n}\right)\\ y_{n}=W\left(Ty_{n},x_{n},b_{n}\right)\end{array}\right.

Additionally, putting an=bn=0a_{n}=b_{n}=0, it follows that yn=zn=xny_{n}=z_{n}=x_{n}, for each nn\in\mathbb{N}; we get the implicit Mann iteration:

xn+1=W(Txn+1,xn,αn)x_{n+1}=W\left(Tx_{n+1},x_{n},\alpha_{n}\right) (1.7)

Now, we recall sufficient conditions for the convergence to the fixed point of a contraction mapping of Noor iteration, respectively implicit Noor iteration.

Remark 1.6. Since Noor iteration is more general than Ishikawa and Mann iterations, we shall remind that, the classical Noor iteration (1.1) is convergent to the fixed point pp of the contraction mapping TT, if k=0αk=\sum_{k=0}^{\infty}\alpha_{k}=\infty. In a similar way, since implicit Noor iteration is more general than implicit Mann and implicit Ishikawa iterations, we remind that implicit Noor algorithm (1.5) is convergent when k=0(1αk)=\sum_{k=0}^{\infty}\left(1-\alpha_{k}\right)=\infty.

Following [3], we remind that
Definition 1.7. Let ( X,dX,d ) be a metric space and T:XXT:X\rightarrow X a map for which there exists the real numbers a,ba,b and cc satisfying 0<a<1,0<b,c<1/20<a<1,0<b,c<1/2 such that for each pair x,yXx,y\in X, at least one of the following is true
(z1)d(Tx,Ty)ad(x,y)(z1)d(Tx,Ty)\leq ad(x,y),
(z2) d(Tx,Ty)b[d(x,Tx)+d(y,Ty)]d(Tx,Ty)\leq b[d(x,Tx)+d(y,Ty)],
(z3) d(Tx,Ty)c[d(x,Ty)+d(y,Tx)]d(Tx,Ty)\leq c[d(x,Ty)+d(y,Tx)].
Then TT is called a Zamfirescu operator. Morevorer, by [17], if ( X,dX,d ) is a complete metric space, TT has a unique fixed point.

Regarding contraction mappings and Zamfirescu operators, we have the following
Remark 1.8. In [3], in the context of normed linear spaces, it is shown that Picard iteration converges faster than Noor iteration for Zamfirescu operators. The same remark can be applied in the context of convex metric spaces as well. Since, any contraction is a Zamfirescu operator by condition ( z1z1 ), the above property remains true for the contraction mappings. Moreover, because of the complex computations of convergence in the case of some iterative schemes, we will use in the next sections the condition that TT must be a contraction with the contraction constant δ\delta. The same proofs can be applied in the same way to Zamfirescu operators. We let this as an open problem.

We present our three goals that we will gain throughout the next sections
(A.) Let CC be a nonempty convex subset of a normed space EE and T:CCT:C\rightarrow C a δ\delta-contraction map.
In 2005 Suantai [14] introduced a modified Noor iterative method with sequences
{αn},{βn},{an},{bn},{cn}[0,1],x1=xC\left\{\alpha_{n}\right\},\left\{\beta_{n}\right\},\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{c_{n}\right\}\subseteq[0,1],x_{1}=x\in C and

{xn+1=αnTyn+βnTzn+(1αnβn)xn,n1yn=bnTzn+cnTxn+(1bncn)xnzn=anTxn+(1an)xn.\left\{\begin{array}[]{l}x_{n+1}=\alpha_{n}Ty_{n}+\beta_{n}Tz_{n}+\left(1-\alpha_{n}-\beta_{n}\right)x_{n},n\geq 1\\ y_{n}=b_{n}Tz_{n}+c_{n}Tx_{n}+\left(1-b_{n}-c_{n}\right)x_{n}\\ z_{n}=a_{n}Tx_{n}+\left(1-a_{n}\right)x_{n}.\end{array}\right.

In the case when CC is a nonempty convex subset of a convex metric space EE, Berinde modified the above iteration with the use of the convexity structure WW and defined the iteration as follows

{xn+1=W(Tyn,W(Tzn,xn,βn1αn),αn)yn=W(Tzn,W(Txn,xn,cn1bn),bn)zn=W(Txn,xn,an).\left\{\begin{array}[]{l}x_{n+1}=W\left(Ty_{n},W\left(Tz_{n},x_{n},\frac{\beta_{n}}{1-\alpha_{n}}\right),\alpha_{n}\right)\\ y_{n}=W\left(Tz_{n},W\left(Tx_{n},x_{n},\frac{c_{n}}{1-b_{n}}\right),b_{n}\right)\\ z_{n}=W\left(Tx_{n},x_{n},a_{n}\right).\end{array}\right.

For the convergence of the above iteration to the fixed point of the nonlinear contraction mapping, we remind the following

Remark 1.9. Following the same article of Berinde [6], the above iteration is convergent when the next assumptions are satisfied
for each n,{αn+βn}[0,1]n\in\mathbb{N},\left\{\alpha_{n}+\beta_{n}\right\}\in[0,1] and k=0(αk+βk)=\sum_{k=0}^{\infty}\left(\alpha_{k}+\beta_{k}\right)=\infty.
Moreover, we have that d(xn+1,p)[1(1δ)(αn+βn)]d(xn,p)d\left(x_{n+1},p\right)\leq\left[1-(1-\delta)\left(\alpha_{n}+\beta_{n}\right)\right]d\left(x_{n},p\right), for each nn\in\mathbb{N}.

Our first goal of the present research article is to find at least a faster iteration that (1.9) with some assumptions on the sequences {αn},{βn},{an},{bn}\left\{\alpha_{n}\right\},\left\{\beta_{n}\right\},\left\{a_{n}\right\},\left\{b_{n}\right\} and {cn}\left\{c_{n}\right\} that makes usage of the definition of multiple convex combinations introduced by Machado in [9].
(B.) In [2], Agarwal et al. presented a new iteration defined on nonempty convex subset CC on normed spaces, that can be adapted easily on convex metric spaces. This iteration is defined by x1=xCx_{1}=x\in C and

{xn+1=αnTyn+(1αn)Txnyn=bnTxn+(1bn)xn\left\{\begin{array}[]{l}x_{n+1}=\alpha_{n}Ty_{n}+\left(1-\alpha_{n}\right)Tx_{n}\\ y_{n}=b_{n}Tx_{n}+\left(1-b_{n}\right)x_{n}\end{array}\right.

The above iteration (1.10) was introduced as an example of an iteration that is faster than Picard iteration (1.4), with respect to (Definition 1.2 ) and (Definition 1.3 ).

In the context of nonempty convex subset CC of a convex metric space, the above iteration is defined by x1=xCx_{1}=x\in C and

{xn+1=W(Tyn,Txn,αn)yn=W(Txn,xn,bn)\left\{\begin{array}[]{l}x_{n+1}=W\left(Ty_{n},Tx_{n},\alpha_{n}\right)\\ y_{n}=W\left(Tx_{n},x_{n},b_{n}\right)\end{array}\right.

In 1 Abbas and Nazir improved the above iteration and they presented a three-step iteration. We will present it in the context of the convex metric space

{xn+1=W(Tzn,Tyn,αn)yn=W(Tzn,Txn,bn)zn=W(Txn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Tz_{n},Ty_{n},\alpha_{n}\right)\\ y_{n}=W\left(Tz_{n},Tx_{n},b_{n}\right)\\ z_{n}=W\left(Tx_{n},x_{n},a_{n}\right)\end{array}\right.

From the same paper [1], we recall the following convergence concept.

Remark 1.10. The above iteration (1.12) is convergent when the next assumptions are satisfied: ak[a,1a](0,1)a_{k}\in[a,1-a]\in(0,1) and k=0αkbkak=\sum_{k=0}^{\infty}\alpha_{k}b_{k}a_{k}=\infty.
In this case, we have that d(xn+1,p)δ[1(1δ)αnbnan]d(xn,p)d\left(x_{n+1},p\right)\leq\delta\left[1-(1-\delta)\alpha_{n}b_{n}a_{n}\right]d\left(x_{n},p\right), for each nn\in\mathbb{N}.
In the fixed point literature we can find other classical iterations. From [8], we will recall them in the context of convex metric spaces
SPSP iteration, with x0=xCx_{0}=x\in C and

{xn+1=W(Tyn,yn,αn)yn=W(Tzn,zn,bn)zn=W(Txn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Ty_{n},y_{n},\alpha_{n}\right)\\ y_{n}=W\left(Tz_{n},z_{n},b_{n}\right)\\ z_{n}=W\left(Tx_{n},x_{n},a_{n}\right)\end{array}\right.

SS iteration, with x0=xCx_{0}=x\in C and

{xn+1=W(Tyn,Txn,αn)yn=W(Txn,xn,bn)\left\{\begin{array}[]{l}x_{n+1}=W\left(Ty_{n},Tx_{n},\alpha_{n}\right)\\ y_{n}=W\left(Tx_{n},x_{n},b_{n}\right)\end{array}\right.

CRCR iteration, with x0=xCx_{0}=x\in C and

{xn+1=W(Tyn,yn,αn)yn=W(Tzn,Txn,bn)zn=W(Txn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Ty_{n},y_{n},\alpha_{n}\right)\\ y_{n}=W\left(Tz_{n},Tx_{n},b_{n}\right)\\ z_{n}=W\left(Tx_{n},x_{n},a_{n}\right)\end{array}\right.

Additionally, in [8] Gursoy and Karakaya presented a modified Picard-S hybrid iteration, that is faster than all of the classical iterations (1.3), (1.2), (1.1), (1.13), (1.14) and (1.15) :

{xn+1=Tynyn=W(Tzn,Txn,bn)zn=W(Txn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=Ty_{n}\\ y_{n}=W\left(Tz_{n},Tx_{n},b_{n}\right)\\ z_{n}=W\left(Tx_{n},x_{n},a_{n}\right)\end{array}\right.

We recall from 8 that the above iteration (1.16) is faster than S iteration (1.14) and the last one is faster than Picard iteration (1.4). So this iteration answers the question of Agarwal et al., i.e. it is indeed an example of an iterative process that is faster with respect to convergence than Picard’s.

Moreover, we have the following results concerning iteration (1.16)
Remark 1.11. The iteration (1.16) is convergent when k=0bkak=\sum_{k=0}^{\infty}b_{k}a_{k}=\infty. Also, we have that d(xn+1,p)δ2[1(1δ)anbn]d(xn,p)d\left(x_{n+1},p\right)\leq\delta^{2}\left[1-(1-\delta)a_{n}b_{n}\right]d\left(x_{n},p\right), for each nn\in\mathbb{N}.

We let the reader get into details for the following remark.
Remark 1.12. When the sequence {αn}\left\{\alpha_{n}\right\} satisfies limnαn=0\lim_{n\rightarrow\infty}\alpha_{n}=0, iteration (1.16) is faster than iteration (1.12), in the sense of (Definition 1.2 ) and (Definition 1.3 ).

Also, regarding the question of Agarwal, Sintunavarat and Pitea in [13] introduced a new iteration better than that of Agarwal’s and of Picard. That is SnS_{n} iteration, with x0=xCx_{0}=x\in C and

{xn+1=W(Tyn,Tzn,αn)yn=W(Txn,xn,bn)zn=W(yn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Ty_{n},Tz_{n},\alpha_{n}\right)\\ y_{n}=W\left(Tx_{n},x_{n},b_{n}\right)\\ z_{n}=W\left(y_{n},x_{n},a_{n}\right)\end{array}\right.

From the same paper, we recall the assumptions under which the iteration (1.17) is convergent to the unique fixed point of the contraction mapping.

Remark 1.13. If {αn}[α,1α],{bn}[b,1b],{an}[a,1a]\left\{\alpha_{n}\right\}\in[\alpha,1-\alpha],\left\{b_{n}\right\}\in[b,1-b],\left\{a_{n}\right\}\in[a,1-a] and α,b,a(0,12)\alpha,b,a\in\left(0,\frac{1}{2}\right), with α(2a)<a\alpha(2-a)<a, then the iteration (1.17) is faster to the fixed point of TT than iteration (1.11).

It is clearly obvious that this iteration requires stronger conditions that the above ones and so we can eliminate from our discussion. So, another goal of this paper is to find iterations with better rate of convergence than (1.16), which implies that the iteration we seek is faster than (1.12),(1.16) and (1.4), also regarding (Definition 1.2) and (Definition 1.3).
(C.) The last goal of this paper is to present a faster implicit-type Noor iteration, faster than the already existing in literature implicit Noor iteration (1.5). Then, we want to modify this one through multiple convex combinations and analyze the rate of convergence.

2. Convergence Analysis

Let impose in the convex metric space the property of hyperbolic spaces in the sense of Goebel and Kirk, that is W(x,y,λ)=W(y,x,1λ)W(x,y,\lambda)=W(y,x,1-\lambda), for each x,yXx,y\in X and λ[0,1]\lambda\in[0,1]. This property is easily satisfied in a linear normed space.

The first main result of this section improve Suantai’s iteration (1.9) on convex metric spaces. The next iteration is an implicit algorithm made by multiple convex combinations. Let’s call it Suantai implicit

{xn+1=W(yn,Tyn,Txn+1;1αnβn,βn,αn)yn=W(zn,Tzn,Tyn;1bncn,cn,bn)zn=W(Tzn,xn,an).\left\{\begin{array}[]{l}x_{n+1}=W\left(y_{n},Ty_{n},Tx_{n+1};1-\alpha_{n}-\beta_{n},\beta_{n},\alpha_{n}\right)\\ y_{n}=W\left(z_{n},Tz_{n},Ty_{n};1-b_{n}-c_{n},c_{n},b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right).\end{array}\right.

In terms of simple convex combinations, this iteration is

{xn+1=W(Txn+1,W(Tyn,yn,βn1αn);αn)yn=W(Tyn,W(Tzn,zn,cn1bn);bn)zn=W(Tzn,xn,an).\left\{\begin{array}[]{l}x_{n+1}=W\left(Tx_{n+1},W\left(Ty_{n},y_{n},\frac{\beta_{n}}{1-\alpha_{n}}\right);\alpha_{n}\right)\\ y_{n}=W\left(Ty_{n},W\left(Tz_{n},z_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right).\end{array}\right.

Our first results of this section concerns under what condition iteration (2.1) is convergent to the unique fixed point of a δ\delta-contraction.

Theorem 2.1. Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{cn},{αn},{βn},{bn+cn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{c_{n}\right\},\left\{\alpha_{n}\right\},\left\{\beta_{n}\right\},\left\{b_{n}+c_{n}\right\} and {αn+βn}\left\{\alpha_{n}+\beta_{n}\right\} sequences in [0,1][0,1] such that k=0(αk+βk)=\sum_{k=0}^{\infty}\left(\alpha_{k}+\beta_{k}\right)=\infty. Then {xn}\left\{x_{n}\right\} in (2.1) is convergent to the unique fixed point pp of TT.

Proof. We evaluate
d(zn,p)=d(W(Tzn,xn;an),p)and(Tzn,p)+(1an)d(xn,p)δand(zn,p)+(1an)d(xn,p)d\left(z_{n},p\right)=d\left(W\left(Tz_{n},x_{n};a_{n}\right),p\right)\leq a_{n}d\left(Tz_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n,p}\right)\leq\delta a_{n}d\left(z_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right).
Then, we get that d(zn,p)1an1δand(xn,p)d\left(z_{n},p\right)\leq\frac{1-a_{n}}{1-\delta a_{n}}d\left(x_{n},p\right).
In a similar way, we evaluate
d(yn,p)=d(W(Tyn,W(Tzn,zn,cn1bn);bn),p)d\left(y_{n},p\right)=d\left(W\left(Ty_{n},W\left(Tz_{n},z_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right),p\right)
bnd(Tyn,p)+(1bn)d(W(Tzn,zn;cn1bn),p)\leq b_{n}d\left(Ty_{n},p\right)+\left(1-b_{n}\right)d\left(W\left(Tz_{n},z_{n};\frac{c_{n}}{1-b_{n}}\right),p\right)\leq
δbnd(yn,p)+(1bn)[(cn1bn)d(Tzn,p)+(1cn1bn)d(zn,p)]=δbnd(yn,p)+δcnd(zn,p)+(1bncn)d(zn,p)\delta b_{n}d\left(y_{n},p\right)+\left(1-b_{n}\right)\left[\left(\frac{c_{n}}{1-b_{n}}\right)d\left(Tz_{n},p\right)+\left(1-\frac{c_{n}}{1-b_{n}}\right)d\left(z_{n},p\right)\right]=\delta b_{n}d\left(y_{n},p\right)+\delta c_{n}d\left(z_{n},p\right)+(1-\left.b_{n}-c_{n}\right)d\left(z_{n},p\right).

Then, we get that d(yn,p)1bncn(1δ)1δbnd(zn,p)d\left(y_{n},p\right)\leq\frac{1-b_{n}-c_{n}(1-\delta)}{1-\delta b_{n}}d\left(z_{n},p\right).
For {xn+1}\left\{x_{n+1}\right\}, we have that
d(xn+1,p)=d(W(Txn+1,W(Tyn,yn,βn1αn);αn),p)d\left(x_{n+1},p\right)=d\left(W\left(Tx_{n+1},W\left(Ty_{n},y_{n},\frac{\beta_{n}}{1-\alpha_{n}}\right);\alpha_{n}\right),p\right)\leq
αnd(Txn+1,p)+(1αn)d(W(Tyn,yn;βn1αn),p)\alpha_{n}d\left(Tx_{n+1},p\right)+\left(1-\alpha_{n}\right)d\left(W\left(Ty_{n},y_{n};\frac{\beta_{n}}{1-\alpha_{n}}\right),p\right)\leq
δαnd(xn+1,p)+(1αn)[(βn1αn)d(Tyn,p)+(1βn1αn)d(yn,p)]\delta\alpha_{n}d\left(x_{n+1},p\right)+\left(1-\alpha_{n}\right)\left[\left(\frac{\beta_{n}}{1-\alpha_{n}}\right)d\left(Ty_{n},p\right)+\left(1-\frac{\beta_{n}}{1-\alpha_{n}}\right)d\left(y_{n},p\right)\right]\leq
δαnd(xn+1,p)+(δβn+1βnαn)d(yn,p)\delta\alpha_{n}d\left(x_{n+1},p\right)+\left(\delta\beta_{n}+1-\beta_{n}-\alpha_{n}\right)d\left(y_{n},p\right).
Then, we get that d(xn+1,p)1αn(1δ)βn1δαnd(yn,p)d\left(x_{n+1},p\right)\leq\frac{1-\alpha_{n}-(1-\delta)\beta_{n}}{1-\delta\alpha_{n}}d\left(y_{n},p\right).
Combining the above results, we have the estimation
d(xn+1,p)AnBnCnd(xn,p)d\left(x_{n+1},p\right)\leq A_{n}\cdot B_{n}\cdot C_{n}d\left(x_{n},p\right), where An:=1αn(1δ)βn1δαn,Bn:=1bn(1δ)cn1δbnA_{n}:=\frac{1-\alpha_{n}-(1-\delta)\beta_{n}}{1-\delta\alpha_{n}},B_{n}:=\frac{1-b_{n}-(1-\delta)c_{n}}{1-\delta b_{n}} and Cn:=1an1δanC_{n}:=\frac{1-a_{n}}{1-\delta a_{n}}, for each nn\in\mathbb{N}.
It is easy to see that, because of δ<1\delta<1, results that Cn<1C_{n}<1.
Also,
An=1αn(1δ)βn1δαn1An=11αn(1δ)βn1δαn=(1δ)(αn+βn)1δαn1An(1δ)(αn+βn)An1(1δ)(αn+βn)A_{n}=\frac{1-\alpha_{n}-(1-\delta)\beta_{n}}{1-\delta\alpha_{n}}\Longrightarrow 1-A_{n}=1-\frac{1-\alpha_{n}-(1-\delta)\beta_{n}}{1-\delta\alpha_{n}}=\frac{(1-\delta)\left(\alpha_{n}+\beta_{n}\right)}{1-\delta\alpha_{n}}\Longrightarrow 1-A_{n}\geq(1-\delta)\left(\alpha_{n}+\beta_{n}\right)\Longrightarrow A_{n}\leq 1-(1-\delta)\left(\alpha_{n}+\beta_{n}\right).
In a similar manner Bn=1bn(1δ)cn1δbn1(1δ)(bn+cn)B_{n}=\frac{1-b_{n}-(1-\delta)c_{n}}{1-\delta b_{n}}\leq 1-(1-\delta)\left(b_{n}+c_{n}\right). Since δ<1\delta<1 and bn+cn0b_{n}+c_{n}\geq 0, we have that Bn<1B_{n}<1, for each nn\in\mathbb{N}.
So, d(xn+1,p)[1(1δ)(αn+βn)]d(xn,p)d\left(x_{n+1},p\right)\leq\left[1-(1-\delta)\left(\alpha_{n}+\beta_{n}\right)\right]d\left(x_{n},p\right). This means that :

d(xn+1,p)k=0n[1(1δ)(αk+βk)]d(x0,p)d\left(x_{n+1},p\right)\leq\prod_{k=0}^{n}\left[1-(1-\delta)\left(\alpha_{k}+\beta_{k}\right)\right]d\left(x_{0},p\right) (2.2)

In view of the fact that 1xex1-x\leq e^{-x}, for x[0,1]x\in[0,1], the above inequality (2.2) reduces to

d(xn+1,p)k=0ne(1δ)(αk+βk)d(x0,p)=e(1δ)k=0n(αk+βk)d(x0,p)d\left(x_{n+1},p\right)\leq\prod_{k=0}^{n}e^{-(1-\delta)\left(\alpha_{k}+\beta_{k}\right)}d\left(x_{0},p\right)=e^{-(1-\delta)\sum_{k=0}^{n}\left(\alpha_{k}+\beta_{k}\right)}d\left(x_{0},p\right)

Since k=0(αk+βk)=\sum_{k=0}^{\infty}\left(\alpha_{k}+\beta_{k}\right)=\infty, letting nn\rightarrow\infty, we get

d(xn+1,p)0d\left(x_{n+1},p\right)\rightarrow 0

where pp is the unique fixed point of the δ\delta-contraction operator TT.
As particular cases of iteration (2.1) we get classical iterations, such as implicit Noor, respectively implicit Ishikawa iterative processes.

Remark 2.2. In (2.1), taking βn=cn=0\beta_{n}=c_{n}=0, we get Implicit Noor iteration (1.5) and taking βn=cn=an=0\beta_{n}=c_{n}=a_{n}=0, we get Implicit Ishikawa iteration (1.6).

In the following we present a Noor-type implicit iteration which is faster than the original implicit Noor (1.5). Let’s call it Noor implicit II

{xn+1=W(Txn+1,Tyn;αn)yn=W(Tyn,Tzn;bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Tx_{n+1},Ty_{n};\alpha_{n}\right)\\ y_{n}=W\left(Ty_{n},Tz_{n};b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Now, we present our second result regarding the conditions under which iteration (2.3) is convergent to the unique fixed point of a δ\delta-contraction.

Theorem 2.3. Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{αn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{\alpha_{n}\right\} sequences in ( 0,1 ). Then {xn}\left\{x_{n}\right\} in (2.3) is convergent to the unique fixed point pp of TT.

Proof . First, we evaluate
d(zn,p)=d(W(Tzn,xn,an),p)and(Tzn,p)+(1an)d(xn,p)δand(zn,p)+(1an)d(xn,p)d\left(z_{n},p\right)=d\left(W\left(Tz_{n},x_{n},a_{n}\right),p\right)\leq a_{n}d\left(Tz_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right)\leq\delta a_{n}d\left(z_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right).
That is, d(zn,p)1an1δand(xn,p)d\left(z_{n},p\right)\leq\frac{1-a_{n}}{1-\delta a_{n}}d\left(x_{n},p\right).
In a similar way, we estimate
d(yn,p)=d(W(Tyn,Tzn;bn),p)bnd(Tyn,p)+(1bn)d(Tzn,p)δbnd(yn,p)+δ(1bn)d(zn,p)d\left(y_{n},p\right)=d\left(W\left(Ty_{n},Tz_{n};b_{n}\right),p\right)\leq b_{n}d\left(Ty_{n},p\right)+\left(1-b_{n}\right)d\left(Tz_{n},p\right)\leq\delta b_{n}d\left(y_{n},p\right)+\delta\left(1-b_{n}\right)d\left(z_{n},p\right).
We get that, d(yn,p)δ1bn1δbnd(zn,p)d\left(y_{n},p\right)\leq\delta\cdot\frac{1-b_{n}}{1-\delta b_{n}}d\left(z_{n},p\right).
For {xn+1}\left\{x_{n+1}\right\}, we have
d(xn+1,p)=d(W(Txn+1,Tyn;αn),p)αnd(Txn+1,p)+(1αn)d(Tyn,p)δαnd(xn+1,p)+δ(1αn)d(yn,p)d\left(x_{n+1},p\right)=d\left(W\left(Tx_{n+1},Ty_{n};\alpha_{n}\right),p\right)\leq\alpha_{n}d\left(Tx_{n+1},p\right)+\left(1-\alpha_{n}\right)d\left(Ty_{n},p\right)\leq\delta\alpha_{n}d\left(x_{n+1},p\right)+\delta(1-\left.\alpha_{n}\right)d\left(y_{n},p\right).
We get that, d(xn+1,p)δ1αn1δαnd(yn,p)d\left(x_{n+1},p\right)\leq\delta\cdot\frac{1-\alpha_{n}}{1-\delta\alpha_{n}}d\left(y_{n},p\right).
Combining above results, results that :

d(xn+1,p)δ2AnBnCnd(xn,p), with An:=1αn1δαn,Bn:=1bn1δbn,Cn:=1an1δand\left(x_{n+1},p\right)\leq\delta^{2}\cdot A_{n}\cdot B_{n}\cdot C_{n}\cdot d\left(x_{n},p\right),\text{ with }A_{n}:=\frac{1-\alpha_{n}}{1-\delta\alpha_{n}},B_{n}:=\frac{1-b_{n}}{1-\delta b_{n}},C_{n}:=\frac{1-a_{n}}{1-\delta a_{n}} (2.4)

From the assumption of contraction operator TT that δ[0,1)\delta\in[0,1) and from the assumption that the sequences {an},{bn}\left\{a_{n}\right\},\left\{b_{n}\right\} and {αn}\left\{\alpha_{n}\right\} are positive, it implies that:
An,Bn,Cn<1A_{n},B_{n},C_{n}<1, for each nn\in\mathbb{N}. So, we obtain
d(xn+1,p)<δ2d(xn,p)δ2(n+1)d(x0,p)d\left(x_{n+1},p\right)<\delta^{2}\cdot d\left(x_{n},p\right)\leq\delta^{2(n+1)}d\left(x_{0},p\right).
Letting pp\rightarrow\infty, because of δ<1\delta<1, we get d(xn+1,p)0d\left(x_{n+1},p\right)\rightarrow 0, so the sequence {xn}\left\{x_{n}\right\} is convergent to the unique fixed point pp of TT.

Now, we present an useful remark concerning iteration (2.3).
Remark 2.4. The above iteration (2.3) has weak hypotheses, because, without any other assumptions on the sequences {an},{bn}\left\{a_{n}\right\},\left\{b_{n}\right\} and {αn},d(xn+1,p)<δ2d(xn,p)\left\{\alpha_{n}\right\},d\left(x_{n+1},p\right)<\delta^{2}d\left(x_{n},p\right), for each nn\in\mathbb{N}, so is a contraction sequence.

Iteration (2.3) is faster than Picard iteration (1.4) in the sense of definitions 1.2 and 1.3, as follows

Remark 2.5. In the case of Picard iteration (1.4), the sequence {d(xn,p)}\left\{d\left(x_{n},p\right)\right\} has the term δ\delta between two consecutive elements. The above iteration (2.3) is evidently faster convergent than Picard iteration, because of the term δ2\delta^{2} in the approximation of the sequence {d(xn,p)}\left\{d\left(x_{n},p\right)\right\}, in the sense of definitions (1.2) and (1.3).

Now, we can combine Gursoy-Karakaya iteration (1.16) with implicit Noor II iteration (2.3). Let’s call it GKN implicit II :

{xn+1=Tynyn=W(Tyn,Tzn;bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=Ty_{n}\\ y_{n}=W\left(Ty_{n},Tz_{n};b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Theorem 2.6. Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn}\left\{a_{n}\right\},\left\{b_{n}\right\}, sequences in ( 0,1 ). Then {xn}\left\{x_{n}\right\} in (2.5) is convergent to the unique fixed point pp of TT.

Proof . First, we estimate
d(zn,p)=d(W(Tzn,xn,an),p)and(Tzn,p)+(1an)d(xn,p)δand(zn,p)+(1an)d(xn,p)d\left(z_{n},p\right)=d\left(W\left(Tz_{n},x_{n},a_{n}\right),p\right)\leq a_{n}d\left(Tz_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right)\leq\delta a_{n}d\left(z_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right).
That is, d(zn,p)1an1δand(xn,p)d\left(z_{n},p\right)\leq\frac{1-a_{n}}{1-\delta a_{n}}d\left(x_{n},p\right).
In a similar way, we evaluate
d(yn,p)=d(W(Tyn,Tzn;bn),p)bnd(Tyn,p)+(1bn)d(Tzn,p)δbnd(yn,p)+δ(1bn)d(zn,p)d\left(y_{n},p\right)=d\left(W\left(Ty_{n},Tz_{n};b_{n}\right),p\right)\leq b_{n}d\left(Ty_{n},p\right)+\left(1-b_{n}\right)d\left(Tz_{n},p\right)\leq\delta b_{n}d\left(y_{n},p\right)+\delta\left(1-b_{n}\right)d\left(z_{n},p\right).
We get that, d(yn,p)δ1bn1δbnd(zn,p)d\left(y_{n},p\right)\leq\delta\cdot\frac{1-b_{n}}{1-\delta b_{n}}d\left(z_{n},p\right).
For {xn+1}\left\{x_{n+1}\right\}, we have
d(xn+1,p)=d(Tyn,p)δd(yn,p)d\left(x_{n+1},p\right)=d\left(Ty_{n},p\right)\leq\delta d\left(y_{n},p\right).
Combining above results, we obtain that :

d(xn+1,p)δ2AnBnd(xn,p), with An:=1bn1δbn,Bn:=1an1δand\left(x_{n+1},p\right)\leq\delta^{2}\cdot A_{n}\cdot B_{n}\cdot d\left(x_{n},p\right),\text{ with }A_{n}:=\frac{1-b_{n}}{1-\delta b_{n}},B_{n}:=\frac{1-a_{n}}{1-\delta a_{n}} (2.6)

From the assumption of contraction operator TT that δ[0,1)\delta\in[0,1) and from the assumption that the sequences {an},{bn}\left\{a_{n}\right\},\left\{b_{n}\right\} and {αn}\left\{\alpha_{n}\right\} are positive, it implies that:
An,Bn<1A_{n},B_{n}<1, for each nn\in\mathbb{N}. So, we obtain
d(xn+1,p)<δ2d(xn,p)δ2(n+1)d(x0,p)d\left(x_{n+1},p\right)<\delta^{2}\cdot d\left(x_{n},p\right)\leq\delta^{2(n+1)}d\left(x_{0},p\right).
Letting pp\rightarrow\infty, because of δ<1\delta<1, we get d(xn+1,p)0d\left(x_{n+1},p\right)\rightarrow 0, so the sequence {xn}\left\{x_{n}\right\} is convergent to the unique fixed point pp of TT.

Remark 2.7. The above iteration (2.5) has weak hypotheses, because, without any other assumptions on the sequences {an},{bn}\left\{a_{n}\right\},\left\{b_{n}\right\} and {αn},d(xn+1,p)<δ2d(xn,p)\left\{\alpha_{n}\right\},d\left(x_{n+1},p\right)<\delta^{2}d\left(x_{n},p\right), for each nn\in\mathbb{N}.

In the spirit of definition (1.2) and (1.3), iteration 2.5) is faster than Picard iteration, as follows
Remark 2.8. In the case of Picard iteration (1.4), the sequence {d(xn,p)}\left\{d\left(x_{n},p\right)\right\} has the term δ\delta between two consecutive elements. The above iteration (2.5) is evidently a faster convergent iteration than Picard, because of the term δ2\delta^{2} in the approximation of the sequence with the general term {d(xn,p)}\left\{d\left(x_{n},p\right)\right\}, following definitions (1.2) and (1.3).

The next two iterations are modified version of (2.5) through multiple convex combinations (or simply m.c.c). The iterative scheme presented below will be called GKN implicit II with multiple convex combinations 1, or simply GKN implicit II m.c.c. 1
{xn+1=Tynyn=(Tzn,xn,Tyn;1bncn,cn,bn)zn=W(Tzn,xn,an),\left\{\begin{array}[]{l}x_{n+1}=Ty_{n}\\ y_{n}=\left(Tz_{n},x_{n},Ty_{n};1-b_{n}-c_{n},c_{n},b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right),\end{array}\right.
and in the case of simple convex combinations

{xn+1=Tynyn=W(Tyn,W(xn,Tzn,cn1bn);bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=Ty_{n}\\ y_{n}=W\left(Ty_{n},W\left(x_{n},Tz_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

For our new iteration (2.7), we present the conditions in which our iteration convergences to the unique fixed point of a δ\delta-contraction, as follows

Theorem 2.9. Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{bn+cn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{b_{n}+c_{n}\right\} sequences in ( 0,1 ). Then {xn}\left\{x_{n}\right\} in (2.7) is convergent to the unique fixed point pp of TT.

Proof. We evaluate the following distance
d(zn,p)=d(W(Tzn,xn,an),p)and(Tzn,p)+(1an)d(xn,p)δand(zn,p)+(1an)d(xn,p)d\left(z_{n},p\right)=d\left(W\left(Tz_{n},x_{n},a_{n}\right),p\right)\leq a_{n}d\left(Tz_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right)\leq\delta a_{n}d\left(z_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right).
That is, d(zn,p)1an1δand(xn,p)d\left(z_{n},p\right)\leq\frac{1-a_{n}}{1-\delta a_{n}}d\left(x_{n},p\right).
We estimate the following
d(yn,p)=d(W(Tyn,W(xn,Tzn,cn1bn);bn),p)bnd(Tyn,p)+d\left(y_{n},p\right)=d\left(W\left(Ty_{n},W\left(x_{n},Tz_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right),p\right)\leq b_{n}d\left(Ty_{n},p\right)+
(1bn)d(W(xn,Tzn;cn1bn),p)δbnd(yn,p)+\left(1-b_{n}\right)d\left(W\left(x_{n},Tz_{n};\frac{c_{n}}{1-b_{n}}\right),p\right)\leq\delta b_{n}d\left(y_{n},p\right)+
(1bn)[(cn1bn)d(xn,p)+(1cn1bn)d(Tzn,p)]\left(1-b_{n}\right)\left[\left(\frac{c_{n}}{1-b_{n}}\right)d\left(x_{n},p\right)+\left(1-\frac{c_{n}}{1-b_{n}}\right)d\left(Tz_{n},p\right)\right]\leq
δbnd(yn,p)+cnd(xn,p)+δ(1bncn)d(zn,p)\delta b_{n}d\left(y_{n},p\right)+c_{n}d\left(x_{n},p\right)+\delta\left(1-b_{n}-c_{n}\right)d\left(z_{n},p\right).
That is, d(yn,p)cn+δ(1bncn)1an1δan1δbnd(xn,p)d\left(y_{n},p\right)\leq\frac{c_{n}+\delta\left(1-b_{n}-c_{n}\right)\cdot\frac{1-a_{n}}{1-\delta a_{n}}}{1-\delta b_{n}}d\left(x_{n},p\right).
For {xn+1}\left\{x_{n+1}\right\}, we have that :
d(xn+1,p)=d(Tyn,p)δd(yn,p)d\left(x_{n+1},p\right)=d\left(Ty_{n},p\right)\leq\delta d\left(y_{n},p\right).
Combining these results, we get that

d(xn+1,p)δAnd(xn,p), with An:=cn+δ(1bncn)1an1δan1δbnd(xn,p)d\left(x_{n+1},p\right)\leq\delta\cdot A_{n}\cdot d\left(x_{n},p\right),\text{ with }A_{n}:=\frac{c_{n}+\delta\left(1-b_{n}-c_{n}\right)\cdot\frac{1-a_{n}}{1-\delta a_{n}}}{1-\delta b_{n}}d\left(x_{n},p\right) (2.8)

Since δ<1\delta<1 and {an}\left\{a_{n}\right\} is a sequence of positive numbers, we get that 1an1δan<1\frac{1-a_{n}}{1-\delta a_{n}}<1, so d(xn+1,p)<Bnd(xn,p)d\left(x_{n+1},p\right)<B_{n}\cdot d\left(x_{n},p\right), where Bn:=cn+δ(1cn)δbn1δbnB_{n}:=\frac{c_{n}+\delta\left(1-c_{n}\right)-\delta b_{n}}{1-\delta b_{n}}.
Because of cn<1c_{n}<1, for each n(1δ)cn+δ<1n\in\mathbb{N}\Longrightarrow(1-\delta)c_{n}+\delta<1, so Bn<1B_{n}<1.
Finally, it implies that d(xn+1,p)δd(xn,p)δn+1d(x0,p)d\left(x_{n+1},p\right)\leq\delta\cdot d\left(x_{n},p\right)\leq\delta^{n+1}d\left(x_{0},p\right).
Letting nn\rightarrow\infty, because δ<1\delta<1, we get that {xn}\left\{x_{n}\right\} converges to the unique fixed point pp of the δ\delta-contraction operator TT.

The next remark concerns some particular cases of iteration (2.7).
Remark 2.10. If we put cn=0c_{n}=0 in the above iteration (2.7), we get iteration (2.5).
Additionally, putting αn=0\alpha_{n}=0 in iteration (2.3), we get iteration (2.5).

Let’s call the next iterative scheme by GKN implicit II with multiple convex combinations 2, or simply GKNGKN implicit II m.c.c. 2

{xn+1=Tynyn=(Tzn,zn,Tyn;1bncn,cn,bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=Ty_{n}\\ y_{n}=\left(Tz_{n},z_{n},Ty_{n};1-b_{n}-c_{n},c_{n},b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

In the case of simple convex combinations, we have

{xn+1=Tynyn=W(Tyn,W(zn,Tzn,cn1bn);bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=Ty_{n}\\ y_{n}=W\left(Ty_{n},W\left(z_{n},Tz_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

For the convergence of iteration (2.9) to the unique fixed point of a contraction mapping, we have the following.

Theorem 2.11. Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{bn+cn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{b_{n}+c_{n}\right\} sequences in ( 0,1 ). Then {xn}\left\{x_{n}\right\} in (2.9) is convergent to the unique fixed point pp of TT.

Proof . We evaluate
d(zn,p)=d(W(Tzn,xn,an),p)and(Tzn,p)+(1an)d(xn,p)δand(zn,p)+(1an)d(xn,p)d\left(z_{n},p\right)=d\left(W\left(Tz_{n},x_{n},a_{n}\right),p\right)\leq a_{n}d\left(Tz_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right)\leq\delta a_{n}d\left(z_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right).
That is, d(zn,p)1an1δand(xn,p)d\left(z_{n},p\right)\leq\frac{1-a_{n}}{1-\delta a_{n}}d\left(x_{n},p\right).
Now, we estimate the following distance
d(yn,p)=d(W(Tyn,W(zn,Tzn,cn1bn);bn),p)bnd(Tyn,p)+d\left(y_{n},p\right)=d\left(W\left(Ty_{n},W\left(z_{n},Tz_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right),p\right)\leq b_{n}d\left(Ty_{n},p\right)+
(1bn)d(W(zn,Tzn;cn1bn),p)δbnd(yn,p)+\left(1-b_{n}\right)d\left(W\left(z_{n},Tz_{n};\frac{c_{n}}{1-b_{n}}\right),p\right)\leq\delta b_{n}d\left(y_{n},p\right)+
(1bn)[(cn1bn)d(zn,p)+(1cn1bn)d(Tzn,p)]\left(1-b_{n}\right)\left[\left(\frac{c_{n}}{1-b_{n}}\right)d\left(z_{n},p\right)+\left(1-\frac{c_{n}}{1-b_{n}}\right)d\left(Tz_{n},p\right)\right]\leq
δbnd(yn,p)+cnd(zn,p)+δ(1bncn)d(zn,p)\delta b_{n}d\left(y_{n},p\right)+c_{n}d\left(z_{n},p\right)+\delta\left(1-b_{n}-c_{n}\right)d\left(z_{n},p\right).
That is, d(yn,p)cn+δ(1bncn)1δbnd(zn,p)d\left(y_{n},p\right)\leq\frac{c_{n}+\delta\left(1-b_{n}-c_{n}\right)}{1-\delta b_{n}}d\left(z_{n},p\right).
For {xn+1}\left\{x_{n+1}\right\}, we have
d(xn+1,p)=d(Tyn,p)δd(yn,p)d\left(x_{n+1},p\right)=d\left(Ty_{n},p\right)\leq\delta d\left(y_{n},p\right).
Combining the above results, we get that

d(xn+1,p)δAnBnd(xn,p), with An:=cn+δ(1bncn)1δbn,Bn:=1an1δan.d\left(x_{n+1},p\right)\leq\delta\cdot A_{n}\cdot B_{n}\cdot d\left(x_{n},p\right),\text{ with }A_{n}:=\frac{c_{n}+\delta\left(1-b_{n}-c_{n}\right)}{1-\delta b_{n}},B_{n}:=\frac{1-a_{n}}{1-\delta a_{n}}. (2.10)

Since δ<1\delta<1 and {an}\left\{a_{n}\right\} is a sequence of positive numbers, we get that 1an1δan<1\frac{1-a_{n}}{1-\delta a_{n}}<1, so d(xn+1,p)<And(xn,p)d\left(x_{n+1},p\right)<A_{n}\cdot d\left(x_{n},p\right).
Moreover, 1An=(1δ)(1cn)1δbn>(1δ)(1cn)1-A_{n}=\frac{(1-\delta)\left(1-c_{n}\right)}{1-\delta b_{n}}>(1-\delta)\left(1-c_{n}\right), this means that An<1(1δ)(1cn)<1A_{n}<1-(1-\delta)\left(1-c_{n}\right)<1.
So, d(xn+1,p)<δd(xn,p)δn+1d(x0,p)d\left(x_{n+1},p\right)<\delta d\left(x_{n},p\right)\leq\delta^{n+1}d\left(x_{0},p\right).
Letting nn\rightarrow\infty, because δ<1\delta<1, we get that {xn}\left\{x_{n}\right\} converges to the unique fixed point pp of the δ\delta-contraction operator TT.

A particular case of the iterative process (2.9) is presented in the next remark.

Remark 2.12. If we put cn=0c_{n}=0 in the above iteration (2.9), we get iteration (2.5).

Let’s call the next iteration GKN implicit II with multiple convex combinations 3, or simply GKN implicit II m.c.c. 3
{xn+1=Tynyn=(Tzn,Txn,Tyn;1bncn,cn,bn)zn=W(Tzn,xn,an).\left\{\begin{array}[]{l}x_{n+1}=Ty_{n}\\ y_{n}=\left(Tz_{n},Tx_{n},Ty_{n};1-b_{n}-c_{n},c_{n},b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right).\end{array}\right.
, and in the case of simple convex combinations :

{xn+1=Tynyn=W(Tyn,W(Txn,Tzn,cn1bn);bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=Ty_{n}\\ y_{n}=W\left(Ty_{n},W\left(Tx_{n},Tz_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Now, we shall present a theorem for the convergence of iteration (2.11) to the unique fixed point of the contraction mapping, as follows.

Theorem 2.13. Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{bn+cn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{b_{n}+c_{n}\right\} sequences in ( 0,1 ). Then {xn}\left\{x_{n}\right\} in (2.11) is convergent to the unique fixed point pp of TT.

Proof . We have that
d(zn,p)=d(W(Tzn,xn,an),p)and(Tzn,p)+(1an)d(xn,p)δand(zn,p)+(1an)d(xn,p)d\left(z_{n},p\right)=d\left(W\left(Tz_{n},x_{n},a_{n}\right),p\right)\leq a_{n}d\left(Tz_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right)\leq\delta a_{n}d\left(z_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right).
That is, d(zn,p)1an1δand(xn,p)d\left(z_{n},p\right)\leq\frac{1-a_{n}}{1-\delta a_{n}}d\left(x_{n},p\right).
In a similar way, it follows that
d(yn,p)=d(W(Tyn,W(Txn,Tzn,cn1bn);bn),p)bnd(Tyn,p)+d\left(y_{n},p\right)=d\left(W\left(Ty_{n},W\left(Tx_{n},Tz_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right),p\right)\leq b_{n}d\left(Ty_{n},p\right)+
(1bn)d(W(Txn,Tzn;cn1bn),p)δbnd(yn,p)+\left(1-b_{n}\right)d\left(W\left(Tx_{n},Tz_{n};\frac{c_{n}}{1-b_{n}}\right),p\right)\leq\delta b_{n}d\left(y_{n},p\right)+
(1bn)[(cn1bn)d(Txn,p)+(1cn1bn)d(Tzn,p)]\left(1-b_{n}\right)\left[\left(\frac{c_{n}}{1-b_{n}}\right)d\left(Tx_{n},p\right)+\left(1-\frac{c_{n}}{1-b_{n}}\right)d\left(Tz_{n},p\right)\right]\leq
δbnd(yn,p)+δcnd(xn,p)+δ(1bncn)d(zn,p)\delta b_{n}d\left(y_{n},p\right)+\delta c_{n}d\left(x_{n},p\right)+\delta\left(1-b_{n}-c_{n}\right)d\left(z_{n},p\right).
That is, d(yn,p)δcn+(1bncn)1an1δan1δbnd(xn,p)d\left(y_{n},p\right)\leq\delta\frac{c_{n}+\left(1-b_{n}-c_{n}\right)\cdot\frac{1-a_{n}}{1-\delta a_{n}}}{1-\delta b_{n}}d\left(x_{n},p\right).
For {xn+1}\left\{x_{n+1}\right\}, we have
d(xn+1,p)=d(Tyn,p)δd(yn,p)d\left(x_{n+1},p\right)=d\left(Ty_{n},p\right)\leq\delta d\left(y_{n},p\right).
From the above results, we get that

d(xn+1,p)δ2And(xn,p), with An:=cn+(1bncn)1an1δan1δbnd(xn,p)d\left(x_{n+1},p\right)\leq\delta^{2}\cdot A_{n}\cdot d\left(x_{n},p\right),\text{ with }A_{n}:=\frac{c_{n}+\left(1-b_{n}-c_{n}\right)\cdot\frac{1-a_{n}}{1-\delta a_{n}}}{1-\delta b_{n}}d\left(x_{n},p\right) (2.12)

Since δ<1\delta<1 and {an}\left\{a_{n}\right\} is a sequence of positive numbers, we get that 1an1δan<1\frac{1-a_{n}}{1-\delta a_{n}}<1, so d(xn+1,p)<Bnd(xn,p)d\left(x_{n+1},p\right)<B_{n}\cdot d\left(x_{n},p\right), where Bn:=cn+(1cn)bn1δbn=1bn1δbnB_{n}:=\frac{c_{n}+\left(1-c_{n}\right)-b_{n}}{1-\delta b_{n}}=\frac{1-b_{n}}{1-\delta b_{n}}.

Because of bn<1b_{n}<1, for each n,Bn<1n\in\mathbb{N},B_{n}<1.
Finally, it implies that d(xn+1,p)<δ2d(xn,p)δ2(n+1)d(x0,p)d\left(x_{n+1},p\right)<\delta^{2}\cdot d\left(x_{n},p\right)\leq\delta^{2(n+1)}d\left(x_{0},p\right).
Letting nn\rightarrow\infty, because δ<1\delta<1, we get that {xn}\left\{x_{n}\right\} converges to the unique fixed point pp of the δ\delta-contraction operator TT.

Now we present a particular case of the iteration algorithm (2.11).
Remark 2.14. If we put cn=0c_{n}=0 in the above iteration (2.11), we get iteration (2.5).

Finally, we will study two iterations which are modified implicit Noor II-type iterations through multiple convex combinations.

Let’s call the first one Implicit Noor II with multiple convex combinations, or simply, IN II m.c.c. This is defined, through multiple convex combinations, as

{xn+1=W(Tyn,Tzn,Txn+1;1αnβn,βn,αn)yn=W(Tzn,Txn,Tyn;1bncn,cn,bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Ty_{n},Tz_{n},Tx_{n+1};1-\alpha_{n}-\beta_{n},\beta_{n},\alpha_{n}\right)\\ y_{n}=W\left(Tz_{n},Tx_{n},Ty_{n};1-b_{n}-c_{n},c_{n},b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

With simple convex combinations, the iteration become

{xn+1=W(Txn+1,W(Tzn,Tyn;βn1αn);αn)yn=W(Tyn,W(Txn,Tzn;cn1bn);bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Tx_{n+1},W\left(Tz_{n},Ty_{n};\frac{\beta_{n}}{1-\alpha_{n}}\right);\alpha_{n}\right)\\ y_{n}=W\left(Ty_{n},W\left(Tx_{n},Tz_{n};\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Concerning the convergence of the iteration (2.13), we have the following theorem.
Theorem 2.15. Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{αn},{bn+cn},{αn+βn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{\alpha_{n}\right\},\left\{b_{n}+c_{n}\right\},\left\{\alpha_{n}+\beta_{n}\right\} sequences in (0,1)(0,1). Then {xn}\left\{x_{n}\right\} in (2.13) is convergent to the unique fixed point pp of TT.

Proof . We evaluate the following distance
d(zn,p)=d(W(Tzn,xn,an),p)and(Tzn,p)+(1an)d(xn,p)δand(zn,p)+(1an)d(xn,p)d\left(z_{n},p\right)=d\left(W\left(Tz_{n},x_{n},a_{n}\right),p\right)\leq a_{n}d\left(Tz_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right)\leq\delta a_{n}d\left(z_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right).
So, we have d(zn,p)1an1δand(xn,p)d\left(z_{n},p\right)\leq\frac{1-a_{n}}{1-\delta a_{n}}d\left(x_{n},p\right).
In the same manner, we get that
d(yn,p)=d(W(Tyn,W(Txn,Tzn;cn1bn);bn),p)d\left(y_{n},p\right)=d\left(W\left(Ty_{n},W\left(Tx_{n},Tz_{n};\frac{c_{n}}{1-b_{n}}\right);b_{n}\right),p\right)\leq
bnd(Tyn,p)+(1bn)d(W(Txn,Tzn;cn1bn),p)b_{n}d\left(Ty_{n},p\right)+\left(1-b_{n}\right)d\left(W\left(Tx_{n},Tz_{n};\frac{c_{n}}{1-b_{n}}\right),p\right)\leq
δbnd(yn,p)+(1bn)[(cn1bn)d(Txn,p)+(1cn1bn)d(Tzn,p)]\delta b_{n}d\left(y_{n},p\right)+\left(1-b_{n}\right)\left[\left(\frac{c_{n}}{1-b_{n}}\right)d\left(Tx_{n},p\right)+\left(1-\frac{c_{n}}{1-b_{n}}\right)d\left(Tz_{n},p\right)\right]\leq
δbnd(yn,p)+δcnd(xn,p)+δ(1bncn)d(zn,p)\delta b_{n}d\left(y_{n},p\right)+\delta c_{n}d\left(x_{n},p\right)+\delta\left(1-b_{n}-cn\right)d\left(z_{n},p\right).
So, d(yn,p)δcn+(1bncn)1an1δan1δbnd(xn,p)d\left(y_{n},p\right)\leq\delta\cdot\frac{c_{n}+\left(1-b_{n}-c_{n}\right)\cdot\frac{1-a_{n}}{1-\delta a_{n}}}{1-\delta b_{n}}d\left(x_{n},p\right).
For {xn}\left\{x_{n}\right\}, it follows that
d(xn+1,p)=d(W(Txn+1,W(Tzn,Tyn;βn1αn);αn),p)d\left(x_{n+1},p\right)=d\left(W\left(Tx_{n+1},W\left(Tz_{n},Ty_{n};\frac{\beta_{n}}{1-\alpha_{n}}\right);\alpha_{n}\right),p\right)\leq
αnd(Txn+1,p)+(1αn)d(W(Tzn,Tyn;βn1αn),p)\alpha_{n}d\left(Tx_{n+1},p\right)+\left(1-\alpha_{n}\right)d\left(W\left(Tz_{n},Ty_{n};\frac{\beta_{n}}{1-\alpha_{n}}\right),p\right)\leq
δαnd(xn+1,p)+(1αn)[(βn1αn)d(Tzn,p)+(1βn1αn)d(Tyn,p)]\delta\alpha_{n}d\left(x_{n+1},p\right)+\left(1-\alpha_{n}\right)\left[\left(\frac{\beta_{n}}{1-\alpha_{n}}\right)d\left(Tz_{n},p\right)+\left(1-\frac{\beta_{n}}{1-\alpha_{n}}\right)d\left(Ty_{n},p\right)\right]\leq
δαnd(xn+1,p)+δ(βnd(zn,p)+(1αnβn)d(yn,p))\delta\alpha_{n}d\left(x_{n+1},p\right)+\delta\left(\beta_{n}d\left(z_{n},p\right)+\left(1-\alpha_{n}-\beta_{n}\right)d\left(y_{n},p\right)\right).
So, it follows that

d(xn+1,p)δβn1an1δan+δ(1αnβn)cn+(1bncn)1an1δan1δbn1δαnd(xn,p).d\left(x_{n+1},p\right)\leq\delta\cdot\frac{\beta_{n}\cdot\frac{1-a_{n}}{1-\delta a_{n}}+\delta\cdot\left(1-\alpha_{n}-\beta_{n}\right)\cdot\frac{c_{n}+\left(1-b_{n}-c_{n}\right)\cdot\frac{1-a_{n}}{1-\delta a_{n}}}{1-\delta b_{n}}}{1-\delta\alpha_{n}}d\left(x_{n},p\right). (2.14)

Because δ<1\delta<1, we get that 1an1δan<1\frac{1-a_{n}}{1-\delta a_{n}}<1 and cn+(1bncn)1δbn=1bn1δbn<1\frac{c_{n}+\left(1-b_{n}-c_{n}\right)}{1-\delta b_{n}}=\frac{1-b_{n}}{1-\delta b_{n}}<1.
The above computations imply that :
d(xn+1,p)<δAnd(xn,p)d\left(x_{n+1},p\right)<\delta A_{n}\cdot d\left(x_{n},p\right), with An:=βn+δ(1αnβn)1δαnA_{n}:=\frac{\beta_{n}+\delta\left(1-\alpha_{n}-\beta_{n}\right)}{1-\delta\alpha_{n}}.
With the assumption that βn<1\beta_{n}<1, for each nn\in\mathbb{N}, it follows that An<1A_{n}<1.
Then, d(xn+1,p)<δd(xn,p)δn+1d(x0,p)d\left(x_{n+1},p\right)<\delta d\left(x_{n},p\right)\leq\delta^{n+1}d\left(x_{0},p\right).
Since δ<1\delta<1, letting nn\rightarrow\infty, we have that d(xn+1,p)0d\left(x_{n+1},p\right)\rightarrow 0, that means the sequence {xn}\left\{x_{n}\right\} is convergent to the unique fixed point pp of TT.

As a particular case of iteration (2.13), we have the following.
Remark 2.16. Putting βn=cn=0\beta_{n}=c_{n}=0, we get iteration (2.3).

We present the last iteration. Let’s call it Double Implicit Noor II with multiple convex combinations, or simply, DIN II m.c.c. This is defined, through multiple convex combinations, as

{xn+1=W(Tyn,Txn+1,Txn+1;1αnβn,βn,αn)yn=W(Tzn,Tyn,Tyn;1bncn,cn,bn)zn=W(Tzn,xn,an).\left\{\begin{array}[]{l}x_{n+1}=W\left(Ty_{n},Tx_{n+1},Tx_{n+1};1-\alpha_{n}-\beta_{n},\beta_{n},\alpha_{n}\right)\\ y_{n}=W\left(Tz_{n},Ty_{n},Ty_{n};1-b_{n}-c_{n},c_{n},b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right).\end{array}\right.

With simple convex combinations, the iteration become

{xn+1=W(Txn+1,W(Txn+1,Tyn;βn1αn);αn)yn=W(Tyn,W(Tyn,Tzn;cn1bn);bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Tx_{n+1},W\left(Tx_{n+1},Ty_{n};\frac{\beta_{n}}{1-\alpha_{n}}\right);\alpha_{n}\right)\\ y_{n}=W\left(Ty_{n},W\left(Ty_{n},Tz_{n};\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

In our last theorem of this section, sufficient conditions for the convergence of the iterative process (2.15) are presented.

Theorem 2.17. Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{αn},{bn+cn},{αn+βn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{\alpha_{n}\right\},\left\{b_{n}+c_{n}\right\},\left\{\alpha_{n}+\beta_{n}\right\} sequences in (0,1)(0,1). Then {xn}\left\{x_{n}\right\} in (2.15) is convergent to the unique fixed point pp of TT.

Proof . As in the above proofs, we estimate the following distance
d(zn,p)=d(W(Tzn,xn,an),p)and(Tzn,p)+(1an)d(xn,p)δand(zn,p)+(1an)d(xn,p)d\left(z_{n},p\right)=d\left(W\left(Tz_{n},x_{n},a_{n}\right),p\right)\leq a_{n}d\left(Tz_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right)\leq\delta a_{n}d\left(z_{n},p\right)+\left(1-a_{n}\right)d\left(x_{n},p\right).

So, we have d(zn,p)1an1δand(xn,p)d\left(z_{n},p\right)\leq\frac{1-a_{n}}{1-\delta a_{n}}d\left(x_{n},p\right).
In the same manner, it follows that
d(yn,p)=d(W(Tyn,W(Tyn,Tzn;cn1bn);bn),p)d\left(y_{n},p\right)=d\left(W\left(Ty_{n},W\left(Ty_{n},Tz_{n};\frac{c_{n}}{1-b_{n}}\right);b_{n}\right),p\right)\leq
bnd(Tyn,p)+(1bn)d(W(Tyn,Tzn;cn1bn),p)b_{n}d\left(Ty_{n},p\right)+\left(1-b_{n}\right)d\left(W\left(Ty_{n},Tz_{n};\frac{c_{n}}{1-b_{n}}\right),p\right)\leq
δbnd(yn,p)+(1bn)[(cn1bn)d(Tyn,p)+(1cn1bn)d(Tzn,p)]\delta b_{n}d\left(y_{n},p\right)+\left(1-b_{n}\right)\left[\left(\frac{c_{n}}{1-b_{n}}\right)d\left(Ty_{n},p\right)+\left(1-\frac{c_{n}}{1-b_{n}}\right)d\left(Tz_{n},p\right)\right]\leq
δbnd(yn,p)+δ[cnd(yn,p)+(1bncn)d(zn,p)]\delta b_{n}d\left(y_{n},p\right)+\delta\left[c_{n}d\left(y_{n},p\right)+\left(1-b_{n}-c_{n}\right)d\left(z_{n},p\right)\right].
So, d(yn,p)δ(bn+cn)d(yn,p)+δ(1bncn)d(zn,p)d\left(y_{n},p\right)\leq\delta\cdot\left(b_{n}+c_{n}\right)d\left(y_{n},p\right)+\delta\cdot\left(1-b_{n}-c_{n}\right)d\left(z_{n},p\right).
This means that d(yn,p)δ1(bn+cn)1δ(bn+cn)d(zn,p)d\left(y_{n},p\right)\leq\delta\cdot\frac{1-\left(b_{n}+c_{n}\right)}{1-\delta\left(b_{n}+c_{n}\right)}\cdot d\left(z_{n},p\right).
For {xn}\left\{x_{n}\right\}, we have
d(xn+1,p)=d(W(Txn+1,W(Txn+1,Tyn;βn1αn);αn),p)d\left(x_{n+1},p\right)=d\left(W\left(Tx_{n+1},W\left(Tx_{n+1},Ty_{n};\frac{\beta_{n}}{1-\alpha_{n}}\right);\alpha_{n}\right),p\right)\leq
αnd(Txn+1,p)+(1αn)d(W(Txn+1,Tyn;βn1αn),p)\alpha_{n}d\left(Tx_{n+1},p\right)+\left(1-\alpha_{n}\right)d\left(W\left(Tx_{n+1},Ty_{n};\frac{\beta_{n}}{1-\alpha_{n}}\right),p\right)\leq
δαnd(xn+1,p)+(1αn)[(βn1αn)d(Txn+1,p)+(1βn1αn)d(Tyn,p)]\delta\alpha_{n}d\left(x_{n+1},p\right)+\left(1-\alpha_{n}\right)\left[\left(\frac{\beta_{n}}{1-\alpha_{n}}\right)d\left(Tx_{n+1},p\right)+\left(1-\frac{\beta_{n}}{1-\alpha_{n}}\right)d\left(Ty_{n},p\right)\right]\leq
δ(αn+βn)d(xn+1,p)+δ(1αnβn)d(yn,p)\delta\left(\alpha_{n}+\beta_{n}\right)d\left(x_{n+1},p\right)+\delta\left(1-\alpha_{n}-\beta_{n}\right)d\left(y_{n},p\right).
So, d(xn+1,p)δ1(αn+βn)1δ(αn+βn)d(yn,p)d\left(x_{n+1},p\right)\leq\delta\cdot\frac{1-\left(\alpha_{n}+\beta_{n}\right)}{1-\delta\left(\alpha_{n}+\beta_{n}\right)}d\left(y_{n},p\right).
This means that

d(xn+1,p)δ2AnBnCnd(xn,p), with An:=1(αn+βn)1δ(αn+βn),Bn:=1(bn+cn)1δ(bn+cn),Cn:=1an1δan.d\left(x_{n+1},p\right)\leq\delta^{2}A_{n}B_{n}C_{n}d\left(x_{n},p\right),\text{ with }A_{n}:=\frac{1-\left(\alpha_{n}+\beta_{n}\right)}{1-\delta\left(\alpha_{n}+\beta_{n}\right)},B_{n}:=\frac{1-\left(b_{n}+c_{n}\right)}{1-\delta\left(b_{n}+c_{n}\right)},C_{n}:=\frac{1-a_{n}}{1-\delta a_{n}}. (2.16)

Since δ<1\delta<1, we get that An,Bn,Cn<1A_{n},B_{n},C_{n}<1.
Then, d(xn+1,p)<δ2d(xn,p)δ2(n+1)d(xo,p)d\left(x_{n+1},p\right)<\delta^{2}d\left(x_{n},p\right)\leq\delta^{2(n+1)}d\left(x_{o},p\right).
Since δ<1\delta<1, letting nn\rightarrow\infty, we have that d(xn+1,p)0d\left(x_{n+1},p\right)\rightarrow 0, that means the sequence {xn}\left\{x_{n}\right\} is convergent to the unique fixed point pp of TT.

In the next remark, a particular case of the iteration (2.15) is presented.
Remark 2.18. Putting βn=cn=0\beta_{n}=c_{n}=0, we get iteration (2.3).

In our last two remarks of this section, we refer to the convergence of iteration of iteration (2.1), respectively to the case ( 𝐁\mathbf{B} ) of our first section.

Remark 2.19. It is important to observe the strong property that except the first iteration introduced by us, that is (2.1), depend on convergence only from δ\delta.

Remark 2.20. In the remarks (2.5) and (2.8), we answered the question posed by Agarwal in 2 . and found iterations with better rate of convergence than iteration (1.10). The points (A) and (C) from the first section are solved in the next section.

3. Rate of Convergence

  • Using the definition for rate of convergence given by Berinde, Suantai et. al. in [6] and [11], we provide sufficient conditions for some iterations given by us to converge better than iteration (1.9). More exactly, the first three iterations are compared with iteration (1.9), using definition (1.4), since iteration (1.9) was considered by Berinde et.al. in convex metric spaces and compared with various classical iterations with the already mentioned definition.

Our first main result of this section relates to the comparing of the reate of convergence of iterations (2.1) and (1.9).

Theorem 3.1. Let {xn}\left\{x_{n}\right\} be the sequence defined by iteration (2.1), that is

{xn+1=W(Txn+1,W(Tyn,yn,βn1αn);αn)yn=W(Tyn,W(Tzn,zn,cn1bn);bn)zn=W(Tzn,xn,an).\left\{\begin{array}[]{l}x_{n+1}=W\left(Tx_{n+1},W\left(Ty_{n},y_{n},\frac{\beta_{n}}{1-\alpha_{n}}\right);\alpha_{n}\right)\\ y_{n}=W\left(Ty_{n},W\left(Tz_{n},z_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right).\end{array}\right.

Let {un}\left\{u_{n}\right\} be the sequence defined by iteration (1.9), that is

{un+1=W(Tvn,W(Twn,un,βn1αn),αn)vn=W(Twn,W(Tun,un,cn1bn),bn)wn=W(Tun,un,an).\left\{\begin{array}[]{l}u_{n+1}=W\left(Tv_{n},W\left(Tw_{n},u_{n},\frac{\beta_{n}}{1-\alpha_{n}}\right),\alpha_{n}\right)\\ v_{n}=W\left(Tw_{n},W\left(Tu_{n},u_{n},\frac{c_{n}}{1-b_{n}}\right),b_{n}\right)\\ w_{n}=W\left(Tu_{n},u_{n},a_{n}\right).\end{array}\right.

Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{cn},{αn},{βn},{bn+cn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{c_{n}\right\},\left\{\alpha_{n}\right\},\left\{\beta_{n}\right\},\left\{b_{n}+c_{n}\right\} and {αn+βn}\left\{\alpha_{n}+\beta_{n}\right\} sequences in ( 0,1 ) such that k=0(αk+βk)=\sum_{k=0}^{\infty}\left(\alpha_{k}+\beta_{k}\right)=\infty.
Additionally, let’s suppose that the following assumptions are satisfied :
(C1) 0<a<an<1a<10<a<a_{n}<1-a<1,
(C2) limn=αn=0,limn=βn=0,limn=bn=0\lim_{n=\infty}\alpha_{n}=0,\lim_{n=\infty}\beta_{n}=0,\lim_{n=\infty}b_{n}=0 and limn=cn=0\lim_{n=\infty}c_{n}=0,
(C3) 11+δ>1a\frac{1}{1+\delta}>1-a,
(C)[1(αn+βn)]>δ{αn[1bncn(1δ)]+[1an(1δ)][βn+δαnbn]}\left(C^{*}\right)\left[1-\left(\alpha_{n}+\beta_{n}\right)\right]>\delta\left\{\alpha_{n}\left[1-b_{n}-c_{n}(1-\delta)\right]+\left[1-a_{n}(1-\delta)\right]\cdot\left[\beta_{n}+\delta\alpha_{n}b_{n}\right]\right\}, for each nn\in\mathbb{N}.
Then, iteration (2.1) converges faster than (1.9).
Proof. We know that d(xn+1,p)And\left(x_{n+1},p\right)\leq A_{n} and d(un+1,p)Bnd\left(u_{n+1},p\right)\leq B_{n}, with
An=k=0n(1αkβk(1δ)1δαk1bkck(1δ)1δbk1ak1δak)d(x0,p)A_{n}=\prod_{k=0}^{n}\left(\frac{1-\alpha_{k}-\beta_{k}(1-\delta)}{1-\delta\alpha_{k}}\cdot\frac{1-b_{k}-c_{k}(1-\delta)}{1-\delta b_{k}}\cdot\frac{1-a_{k}}{1-\delta a_{k}}\right)\cdot d\left(x_{0},p\right),
We make the following evaluation
d(un+1,p)(1αn)d(W(Twn,un,βn1αn),p)αnd(Tvn,p)d\left(u_{n+1},p\right)\geq\left(1-\alpha_{n}\right)d\left(W\left(Tw_{n},u_{n},\frac{\beta_{n}}{1-\alpha_{n}}\right),p\right)-\alpha_{n}d\left(Tv_{n},p\right).
Since d(Tvn,p)δd(vn,p)d\left(Tv_{n},p\right)\leq\delta d\left(v_{n},p\right), we get that
d(un+1,p)(1αn)[(1βn1αn)d(un,p)βn1αnd(Twn,p)]δαnd(vn,p)d\left(u_{n+1},p\right)\geq\left(1-\alpha_{n}\right)\left[\left(1-\frac{\beta_{n}}{1-\alpha_{n}}\right)d\left(u_{n},p\right)-\frac{\beta_{n}}{1-\alpha_{n}}d\left(Tw_{n},p\right)\right]-\delta\alpha_{n}d\left(v_{n},p\right).
Since d(Twn,p)δd(wn,p)d\left(Tw_{n},p\right)\leq\delta d\left(w_{n},p\right), we obtain
d(un+1,p)(1αnβn)d(un,p)δβnd(wn,p)δαnd(vn,p)d\left(u_{n+1},p\right)\geq\left(1-\alpha_{n}-\beta_{n}\right)d\left(u_{n},p\right)-\delta\beta_{n}d\left(w_{n},p\right)-\delta\alpha_{n}d\left(v_{n},p\right).

We know that d(wn,p)=d(W(Tun,un,an),p)and(Tun,p)+(1an)d(un,p)(1an(1δ))d(un,p)d\left(w_{n},p\right)=d\left(W\left(Tu_{n},u_{n},a_{n}\right),p\right)\leq a_{n}d\left(Tu_{n},p\right)+\left(1-a_{n}\right)d\left(u_{n},p\right)\leq\left(1-a_{n}(1-\right.\delta))d\left(u_{n},p\right).
In a similar manner, we have that
d(vn,p)bnd(Twn,p)+(1bn)[(cn1bn)d(Tun,p)+(1cn1bn)d(un,p)]δbnd(wn,p)+δcnd(un,p)+(1bncn)d(un,p)=δbnd(wn,p)+(1bncn(1δ))d(un,p)d\left(v_{n},p\right)\leq b_{n}d\left(Tw_{n},p\right)+\left(1-b_{n}\right)\left[\left(\frac{c_{n}}{1-b_{n}}\right)d\left(Tu_{n},p\right)+\left(1-\frac{c_{n}}{1-b_{n}}\right)d\left(u_{n},p\right)\right]\leq\delta b_{n}d\left(w_{n},p\right)+\delta c_{n}d\left(u_{n},p\right)+\left(1-b_{n}-c_{n}\right)d\left(u_{n},p\right)=\delta b_{n}d\left(w_{n},p\right)+\left(1-b_{n}-c_{n}(1-\delta)\right)d\left(u_{n},p\right).
So :
d(un+1,p)(1αnβn)d(un,p)δβn[1an(1δ)]d(un,p)d\left(u_{n+1},p\right)\geq\left(1-\alpha_{n}-\beta_{n}\right)d\left(u_{n},p\right)-\delta\beta_{n}\left[1-a_{n}(1-\delta)\right]d\left(u_{n},p\right)-
δαn[δbnd(wn,p)+(1bncn+δcn)d(un,p)]\delta\alpha_{n}\left[\delta b_{n}d\left(w_{n},p\right)+\left(1-b_{n}-c_{n}+\delta c_{n}\right)d\left(u_{n},p\right)\right].
This means that d(un+1,p)[(1αnβn)δαn(1bncn(1δ))]δ[1an(1δ)][βn+δαnbn]d\left(u_{n+1},p\right)\geq\left[\left(1-\alpha_{n}-\beta_{n}\right)-\delta\alpha_{n}\left(1-b_{n}-c_{n}(1-\delta)\right)\right]-\delta\left[1-a_{n}(1-\delta)\right]\cdot\left[\beta_{n}+\delta\alpha_{n}b_{n}\right]. d(un,p)d\left(u_{n},p\right).
So, let’s denote by
Bn=k=0n((1αkβk)δαk[(1bkck(1δ))]δ[1ak(1δ)][βk+δαkbk])d(u0,p)B_{n}=\prod_{k=0}^{n}\left(\left(1-\alpha_{k}-\beta_{k}\right)-\delta\alpha_{k}\left[\left(1-b_{k}-c_{k}(1-\delta)\right)\right]-\delta\left[1-a_{k}(1-\delta)\right]\cdot\left[\beta_{k}+\delta\alpha_{k}b_{k}\right]\right)\cdot d\left(u_{0},p\right),
Let’s denote θn:=AnBn\theta_{n}:=\frac{A_{n}}{B_{n}}.
We have that
θn+1θn=1αn+1βn+1(1δ)1δαn+11bn+1cn+1(1δ)1δbn+11an+11δan+1[(1αn+1βn+1)δαn+1(1bn+1cn+1(1δ))]δ[1an+1(1δ)][βn+1+δαn+1bn+1]\frac{\theta_{n+1}}{\theta_{n}}=\frac{\frac{1-\alpha_{n+1}-\beta_{n+1}(1-\delta)}{1-\delta\alpha_{n+1}}\cdot\frac{1-b_{n+1}-c_{n+1}(1-\delta)}{1-\delta b_{n+1}}\cdot\frac{1-a_{n+1}}{1-\delta a_{n+1}}}{\left[\left(1-\alpha_{n+1}-\beta_{n+1}\right)-\delta\alpha_{n+1}\left(1-b_{n+1}-c_{n+1}(1-\delta)\right)\right]-\delta\left[1-a_{n+1}(1-\delta)\right]\cdot\left[\beta_{n+1}+\delta\alpha_{n+1}b_{n+1}\right]}
From assumption (C1), we get that
θn+1θn1αn+1βn+1(1δ)1δαn+11bn+1cn+1(1δ)1δbn+11a1δ(1a)[(1αn+1βn+1)δαn+1(1bn+1cn+1(1δ))]δ[1a(1δ)][βn+1+δαn+1bn+1]\frac{\theta_{n+1}}{\theta_{n}}\leq\frac{\frac{1-\alpha_{n+1}-\beta_{n+1}(1-\delta)}{1-\delta\alpha_{n+1}}\cdot\frac{1-b_{n+1}-c_{n+1}(1-\delta)}{1-\delta b_{n+1}}\cdot\frac{1-a}{1-\delta(1-a)}}{\left[\left(1-\alpha_{n+1}-\beta_{n+1}\right)-\delta\alpha_{n+1}\left(1-b_{n+1}-c_{n+1}(1-\delta)\right)\right]-\delta[1-a(1-\delta)]\cdot\left[\beta_{n+1}+\delta\alpha_{n+1}b_{n+1}\right]}
From assumptions (C2), we have that, limnθn+1θn1a1(1a)δ\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}\leq\frac{1-a}{1-(1-a)\delta}.
Moreover, from assumption (C3), we get limnθn+1θn<1\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}<1.
Then, we get that n=1θn<\sum_{n=1}^{\infty}\theta_{n}<\infty, which implies that limnθn=0\lim_{n\rightarrow\infty}\theta_{n}=0.
This means that limnAnBn=0\lim_{n\rightarrow\infty}\frac{A_{n}}{B_{n}}=0.
In conclusion, iteration (2.1) converges faster than (1.9).
Now, our first remark of this section, useful in the last section regarding numerical examples, refers to the condition ( CC^{*} ) of the previous theorem.

Remark 3.2. The condition ( C\mathrm{C}^{*} ) from the above theorem represent the condition such that the denominator is positive.
Also, we observe that a<12<1aa<\frac{1}{2}<1-a, so if in our next theorems we have a relation between a coefficient, for example q<1qq<1-q, then, in numerical examples, we must take q<12q<\frac{1}{2}.

In our second theorem of this section we compare the rate of convergence of the iterations (2.15) and (1.9).

Theorem 3.3. Let {xn}\left\{x_{n}\right\} be the sequence defined by the iteration (2.15), that is

{xn+1=W(Txn+1,W(Txn+1,Tyn;βn1αn);αn)yn=W(Tyn,W(Tyn,Tzn;cn1bn);bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Tx_{n+1},W\left(Tx_{n+1},Ty_{n};\frac{\beta_{n}}{1-\alpha_{n}}\right);\alpha_{n}\right)\\ y_{n}=W\left(Ty_{n},W\left(Ty_{n},Tz_{n};\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Let {un}\left\{u_{n}\right\} be the sequence defined by iteration (1.9), that is

{un+1=W(Tvn,W(Twn,un,βn1αn),αn)vn=W(Twn,W(Tun,un,cn1bn),bn)wn=W(Tun,un,an)\left\{\begin{array}[]{l}u_{n+1}=W\left(Tv_{n},W\left(Tw_{n},u_{n},\frac{\beta_{n}}{1-\alpha_{n}}\right),\alpha_{n}\right)\\ v_{n}=W\left(Tw_{n},W\left(Tu_{n},u_{n},\frac{c_{n}}{1-b_{n}}\right),b_{n}\right)\\ w_{n}=W\left(Tu_{n},u_{n},a_{n}\right)\end{array}\right.

Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{cn},{αn},{βn},{bn+cn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{c_{n}\right\},\left\{\alpha_{n}\right\},\left\{\beta_{n}\right\},\left\{b_{n}+c_{n}\right\} and {αn+βn}\left\{\alpha_{n}+\beta_{n}\right\} sequences in ( 0,1 ) such that k=0(αk+βk)=\sum_{k=0}^{\infty}\left(\alpha_{k}+\beta_{k}\right)=\infty.
Additionally, let’s suppose that the following assumptions are satisfied :
(C1) an(a,1a),bn(b,1b)a_{n}\in(a,1-a),b_{n}\in(b,1-b) and cn(c,1c)c_{n}\in(c,1-c),
(C2) limnβn=0\lim_{n\rightarrow\infty}\beta_{n}=0 and limnαn=0\lim_{n\rightarrow\infty}\alpha_{n}=0
(C)[1(αn+βn)]>δ{αn[1bncn(1δ)]+[1an(1δ)][βn+δαnbn]}\left(C^{*}\right)\left[1-\left(\alpha_{n}+\beta_{n}\right)\right]>\delta\left\{\alpha_{n}\left[1-b_{n}-c_{n}(1-\delta)\right]+\left[1-a_{n}(1-\delta)\right]\cdot\left[\beta_{n}+\delta\alpha_{n}b_{n}\right]\right\}, for each nn\in\mathbb{N}.
Then, iteration (2.15) converges faster than iteration (1.9).
Proof. We know that d(xn+1,p)And\left(x_{n+1},p\right)\leq A_{n} and d(un+1,p)Bnd\left(u_{n+1},p\right)\leq B_{n}, with
An=k=0n(δ21(αk+βk)1δ(αk+βk)1(bk+ck)1δ(bk+ck)1ak1δak)d(x0,p)A_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot\frac{1-\left(\alpha_{k}+\beta_{k}\right)}{1-\delta\left(\alpha_{k}+\beta_{k}\right)}\cdot\frac{1-\left(b_{k}+c_{k}\right)}{1-\delta\left(b_{k}+c_{k}\right)}\cdot\frac{1-a_{k}}{1-\delta a_{k}}\right)\cdot d\left(x_{0},p\right),
Bn=k=0n((1αkβk)δαk[(1bkck(1δ))]δ[1ak(1δ)][βk+αkbk])d(u0,p)B_{n}=\prod_{k=0}^{n}\left(\left(1-\alpha_{k}-\beta_{k}\right)-\delta\alpha_{k}\left[\left(1-b_{k}-c_{k}(1-\delta)\right)\right]-\delta\left[1-a_{k}(1-\delta)\right]\cdot\left[\beta_{k}+\alpha_{k}b_{k}\right]\right)\cdot d\left(u_{0},p\right).
Let’s denote θn:=AnBn\theta_{n}:=\frac{A_{n}}{B_{n}}. We have that θn+1θn\frac{\theta_{n+1}}{\theta_{n}}\leq
δ21(1αn+1βn+1)δαn+1[(1bn+1cn+1(1δ))]δ[1an+1(1δ)][βn+1+αn+1bn+1]\delta^{2}\cdot\frac{1}{\left(1-\alpha_{n+1}-\beta_{n+1}\right)-\delta\alpha_{n+1}\left[\left(1-b_{n+1}-c_{n+1}(1-\delta)\right)\right]-\delta\left[1-a_{n+1}(1-\delta)\right]\cdot\left[\beta_{n+1}+\alpha_{n+1}b_{n+1}\right]}. 1(αn+1+βn+1)1δ(αn+1+βn+1)\cdot\frac{1-\left(\alpha_{n+1}+\beta_{n+1}\right)}{1-\delta\left(\alpha_{n+1}+\beta_{n+1}\right)}, because 1(bn+cn)1δ(bn+cn)<1\frac{1-\left(b_{n}+c_{n}\right)}{1-\delta\left(b_{n}+c_{n}\right)}<1 and 1an1δan<1\frac{1-a_{n}}{1-\delta a_{n}}<1, for each nn\in\mathbb{N}.
From assumptions (C1) and (C2), we have that
limnθn+1θn=δ2<1\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}=\delta^{2}<1.
Then, we get that n=1θn<\sum_{n=1}^{\infty}\theta_{n}<\infty, which implies that limnθn=0\lim_{n\rightarrow\infty}\theta_{n}=0.
This means that limnAnBη=0\lim_{n\rightarrow\infty}\frac{A_{n}}{B_{\eta}}=0.
In conclusion, iteration (2.15) converges faster than iteration (1.9).
In our third main result, the comparison of the rate of convergence between the iterations (2.13) and (1.9) is presented.

Theorem 3.4. Let {xn}\left\{x_{n}\right\} be the sequence defined by the iteration (2.13), that is

{xn+1=W(Txn+1,W(Tzn,Tyn;βn1αn);αn)yn=W(Tyn,W(Txn,Tzn;cn1bn);bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Tx_{n+1},W\left(Tz_{n},Ty_{n};\frac{\beta_{n}}{1-\alpha_{n}}\right);\alpha_{n}\right)\\ y_{n}=W\left(Ty_{n},W\left(Tx_{n},Tz_{n};\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Let {un}\left\{u_{n}\right\} be the sequence defined by iteration (1.9), that is

{un+1=W(Tvn,W(Twn,un,βn1αn),αn)vn=W(Twn,W(Tun,un,cn1bn),bn)wn=W(Tun,un,an)\left\{\begin{array}[]{l}u_{n+1}=W\left(Tv_{n},W\left(Tw_{n},u_{n},\frac{\beta_{n}}{1-\alpha_{n}}\right),\alpha_{n}\right)\\ v_{n}=W\left(Tw_{n},W\left(Tu_{n},u_{n},\frac{c_{n}}{1-b_{n}}\right),b_{n}\right)\\ w_{n}=W\left(Tu_{n},u_{n},a_{n}\right)\end{array}\right.

Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{cn},{αn},{βn},{bn+cn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{c_{n}\right\},\left\{\alpha_{n}\right\},\left\{\beta_{n}\right\},\left\{b_{n}+c_{n}\right\} and {αn+βn}\left\{\alpha_{n}+\beta_{n}\right\} sequences in ( 0,1 ) such that k=0(αk+βk)=\sum_{k=0}^{\infty}\left(\alpha_{k}+\beta_{k}\right)=\infty.
Additionally, let’s suppose that the following assumptions are satisfied :
(C1) limnβn=0\lim_{n\rightarrow\infty}\beta_{n}=0 and limncn=0\lim_{n\rightarrow\infty}c_{n}=0,
(C2) limnαn=0,limnbn=0\lim_{n\rightarrow\infty}\alpha_{n}=0,\lim_{n\rightarrow\infty}b_{n}=0 and limnan=0\lim_{n\rightarrow\infty}a_{n}=0
(C)[1(αn+βn)]>δ{αn[1bncn(1δ)]+[1an(1δ)][βn+δαnbn]}\left(C^{*}\right)\left[1-\left(\alpha_{n}+\beta_{n}\right)\right]>\delta\left\{\alpha_{n}\left[1-b_{n}-c_{n}(1-\delta)\right]+\left[1-a_{n}(1-\delta)\right]\cdot\left[\beta_{n}+\delta\alpha_{n}b_{n}\right]\right\}, for each nn\in\mathbb{N}. Then, iteration (2.13) converges faster than iteration (1.9).

Proof. We know that d(xn+1,p)And\left(x_{n+1},p\right)\leq A_{n} and d(un+1,p)Bnd\left(u_{n+1},p\right)\leq B_{n}, with
An=k=0n(δβk1ak1δak+δ(1αkβk)ck+(1bkck)1ak1δak1δbk1δαk)d(x0,p)A_{n}=\prod_{k=0}^{n}\left(\delta\cdot\frac{\beta_{k}\cdot\frac{1-a_{k}}{1-\delta a_{k}}+\delta\cdot\left(1-\alpha_{k}-\beta_{k}\right)\cdot\frac{c_{k}+\left(1-b_{k}-c_{k}\right)\cdot\frac{1-a_{k}}{1-\delta a_{k}}}{1-\delta b_{k}}}{1-\delta\alpha_{k}}\right)\cdot d\left(x_{0},p\right),
Bn=k=0n((1αkβk)δαk[(1bkck(1δ))]δ[1ak(1δ)][βk+αkbk])d(u0,p)B_{n}=\prod_{k=0}^{n}\left(\left(1-\alpha_{k}-\beta_{k}\right)-\delta\alpha_{k}\left[\left(1-b_{k}-c_{k}(1-\delta)\right)\right]-\delta\left[1-a_{k}(1-\delta)\right]\cdot\left[\beta_{k}+\alpha_{k}b_{k}\right]\right)\cdot d\left(u_{0},p\right).
Let’s denote θn:=AnBn\theta_{n}:=\frac{A_{n}}{B_{n}}. We factorize and simplify the terms in AnA_{n} and we get
θn+1θnδ1(1δan+1)(1δαn+1)(1δbn)\frac{\theta_{n+1}}{\theta_{n}}\leq\delta\cdot\frac{1}{\left(1-\delta a_{n+1}\right)\left(1-\delta\alpha_{n+1}\right)\left(1-\delta b_{n}\right)}.
βn+1(1an+1)(1δbn+1)+δ(1an+1bn+1)[cn+1(1δan+1)+(1an+1)(1bn+1cn+1)](1αn+1βn+1)δαn+1[(1bn+1cn+1(1δ))]δ[1an+1(1δ)][βn+1+δαn+1bn+1]\frac{\beta_{n+1}\left(1-a_{n+1}\right)\left(1-\delta b_{n+1}\right)+\delta\left(1-a_{n+1}-b_{n+1}\right)\left[c_{n+1}\left(1-\delta a_{n+1}\right)+\left(1-a_{n+1}\right)\left(1-b_{n+1}-c_{n+1}\right)\right]}{\left(1-\alpha_{n+1}-\beta_{n+1}\right)-\delta\alpha_{n+1}\left[\left(1-b_{n+1}-c_{n+1}(1-\delta)\right)\right]-\delta\left[1-a_{n+1}(1-\delta)\right]\cdot\left[\beta_{n+1}+\delta\alpha_{n+1}b_{n+1}\right]} From assumptions (C1) and (C2), we get that
limnθn+1θnδ2<1\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}\leq\delta^{2}<1.
So we get that n=1θn<\sum_{n=1}^{\infty}\theta_{n}<\infty, which implies that limnθn=0\lim_{n\rightarrow\infty}\theta_{n}=0.
This means that limnAnBn=0\lim_{n\rightarrow\infty}\frac{A_{n}}{B_{n}}=0.
In conclusion, iteration (2.13) converges faster than iteration (1.9).

  • Now, we compare the rate of convergence between the iterations given by Gursoy, Karakaya, Abbas, Nazir et. al, and our improved iterations with convex combinations or based on improved implicit Noor iteration, following (1.2) and (1.3), using the technique present in the articles (1) and [8] and [10]. So from now until the next section we use the already mentioned definitions as in the articles of Gursoy, Karakaya, Abbas et. al.

A comparison between the rate of convergence of the iterations (2.3) and (2.5) is presented below.
Theorem 3.5. Let {xn}\left\{x_{n}\right\} be the sequence defined by iteration (2.3), that is

{xn+1=W(Txn+1,Tyn;αn)yn=W(Tyn,Tzn;bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Tx_{n+1},Ty_{n};\alpha_{n}\right)\\ y_{n}=W\left(Ty_{n},Tz_{n};b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Let {un}\left\{u_{n}\right\} be the sequence defined by iteration (2.5), that is

{un+1=Tvnvn=W(Tvn,Twn;bn)wn=W(Twn,un,an)\left\{\begin{array}[]{l}u_{n+1}=Tv_{n}\\ v_{n}=W\left(Tv_{n},Tw_{n};b_{n}\right)\\ w_{n}=W\left(Tw_{n},u_{n},a_{n}\right)\end{array}\right.

Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{αn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{\alpha_{n}\right\} sequences in (0,1)(0,1).
Additionally, let’s suppose that the following assumptions are satisfied :
(C1) 0<α<αn<1α<10<\alpha<\alpha_{n}<1-\alpha<1,
(C2) α>δ1+δ\alpha>\frac{\delta}{1+\delta}.
Then, iteration (2.3) converges faster than (2.5).
Proof. We know that d(xn+1,p)And\left(x_{n+1},p\right)\leq A_{n} and d(un+1,p)Bnd\left(u_{n+1},p\right)\leq B_{n}, with
An=k=0n(δ21αk1δαk1bk1δbk1ak1δak)d(x0,p)A_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot\frac{1-\alpha_{k}}{1-\delta\alpha_{k}}\cdot\frac{1-b_{k}}{1-\delta b_{k}}\cdot\frac{1-a_{k}}{1-\delta a_{k}}\right)\cdot d\left(x_{0},p\right),
Bn=k=0n(δ21bk1δbk1ak1δak)d(u0,p)B_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot\frac{1-b_{k}}{1-\delta b_{k}}\cdot\frac{1-a_{k}}{1-\delta a_{k}}\right)\cdot d\left(u_{0},p\right).
Let’s denote θn:=AnBn\theta_{n}:=\frac{A_{n}}{B_{n}}.
We have that θn+1θn=δ21αn+11δαn+11bn+11δbn+11an+11δan+1δ21bn+11δbn+11an+11δan+1=1αn+11δαn+1\frac{\theta_{n+1}}{\theta_{n}}=\frac{\delta^{2}\cdot\frac{1-\alpha_{n+1}}{1-\delta\alpha_{n+1}}\cdot\frac{1-b_{n+1}}{1-\delta b_{n+1}}\cdot\frac{1-a_{n+1}}{1-\delta a_{n+1}}}{\delta^{2}\cdot\frac{1-b_{n+1}}{1-\delta b_{n+1}}\cdot\frac{1-a_{n+1}}{1-\delta a_{n+1}}}=\frac{1-\alpha_{n+1}}{1-\delta\alpha_{n+1}}.
From assumption (C1), we get that 1αn+11δαn+1<1α1δ(1α)\frac{1-\alpha_{n+1}}{1-\delta\alpha_{n+1}}<\frac{1-\alpha}{1-\delta(1-\alpha)}.
From assumption (C2), we get that 1α1δ(1α)<1\frac{1-\alpha}{1-\delta(1-\alpha)}<1.
Then, we have that limnθn+1θn<1\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}<1.
Then, we get that n=1θn<\sum_{n=1}^{\infty}\theta_{n}<\infty, which implies that limnθn=0\lim_{n\rightarrow\infty}\theta_{n}=0.
This means that limnAnBη=0\lim_{n\rightarrow\infty}\frac{A_{n}}{B_{\eta}}=0.
In conclusion, iteration (2.3) converges faster than (2.5).

Regarding iterations (2.5) and (1.16), sufficient conditions are presented such that iteration (2.5) is faster than iteration (1.16).

Theorem 3.6. Let {xn}\left\{x_{n}\right\} be the sequence defined by the iteration (2.5), that is

{xn+1=Tynyn=W(Tyn,Tzn;bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=Ty_{n}\\ y_{n}=W\left(Ty_{n},Tz_{n};b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Let {un}\left\{u_{n}\right\} be the sequence defined by the iteration (1.16), that is

{un+1=Tvnvn=W(Twn,Tun,bn)wn=W(Tun,un,an)\left\{\begin{array}[]{l}u_{n+1}=Tv_{n}\\ v_{n}=W\left(Tw_{n},Tu_{n},b_{n}\right)\\ w_{n}=W\left(Tu_{n},u_{n},a_{n}\right)\end{array}\right.

Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn}\left\{a_{n}\right\},\left\{b_{n}\right\}, sequences in ( 0,1 ), with k=0akbk=\sum_{k=0}^{\infty}a_{k}b_{k}=\infty.
Additionally, let’s suppose that the following assumptions are satisfied :
(C1) 0<b<bn<1b<10<b<b_{n}<1-b<1,
(C2) limnan=0\lim_{n\rightarrow\infty}a_{n}=0,
(C3) b>δ1+δb>\frac{\delta}{1+\delta}.
Then, iteration (2.5) converges faster than iteration (1.16).
Proof. We know that d(xn+1,p)And\left(x_{n+1},p\right)\leq A_{n} and d(un+1,p)Bnd\left(u_{n+1},p\right)\leq B_{n}, with
An=k=0n(δ21bk1δbk1ak1δak)d(x0,p)A_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot\frac{1-b_{k}}{1-\delta b_{k}}\cdot\frac{1-a_{k}}{1-\delta a_{k}}\right)\cdot d\left(x_{0},p\right),
Bn=k=0n[δ2(1akbk(1δ))]d(u0,p)B_{n}=\prod_{k=0}^{n}\left[\delta^{2}\cdot\left(1-a_{k}b_{k}(1-\delta)\right)\right]\cdot d\left(u_{0},p\right).
Let’s denote θn:=AnBn\theta_{n}:=\frac{A_{n}}{B_{n}}.
We have that θn+1θn=δ21bn+11δbn+11an+11δan+1δ2(1an+1bn+1(1δ))=1bn+11δbn+11an+11δan+1(1an+1bn+1(1δ))\frac{\theta_{n+1}}{\theta_{n}}=\frac{\delta^{2}\cdot\frac{1-b_{n+1}}{1-\delta b_{n+1}}\cdot\frac{1-a_{n+1}}{1-\delta a_{n+1}}}{\delta^{2}\cdot\left(1-a_{n+1}b_{n+1}(1-\delta)\right)}=\frac{\frac{1-b_{n+1}}{1-\delta b_{n+1}}\cdot\frac{1-a_{n+1}}{1-\delta a_{n+1}}}{\left(1-a_{n+1}b_{n+1}(1-\delta)\right)}.
We know that 1an+11δan+1<1\frac{1-a_{n+1}}{1-\delta a_{n+1}}<1.
Also, from assumptions (C1) and (C3), we have that 1bn+11δbn+11b1δ(1b)<1\frac{1-b_{n+1}}{1-\delta b_{n+1}}\leq\frac{1-b}{1-\delta(1-b)}<1.
Then, θn+1θn1b1δ(1b)11an+1bn+1(1δ)\frac{\theta_{n+1}}{\theta_{n}}\leq\frac{1-b}{1-\delta(1-b)}\cdot\frac{1}{1-a_{n+1}b_{n+1}(1-\delta)}.
From assumption (C2), we have that limnθn+1θn1b1δ(1b)<1\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}\leq\frac{1-b}{1-\delta(1-b)}<1.
Then, we get that n=1θn<\sum_{n=1}^{\infty}\theta_{n}<\infty, which implies that limnθn=0\lim_{n\rightarrow\infty}\theta_{n}=0.
This means that limnAnBn=0\lim_{n\rightarrow\infty}\frac{A_{n}}{B_{n}}=0.
Now, by definitions (1.2), respectively (1.3), since AnA_{n} and BnB_{n} converge to 0 in the hypothesis assumptions, it follows that : iteration (2.5) converges faster than iteration (1.16).

Comparing the rate of convergence between the iterations (2.5) and (2.11), we have the following.
Theorem 3.7. Let {xn}\left\{x_{n}\right\} be the sequence defined by the iteration (2.5), that is

{xn+1=Tynyn=W(Tyn,Tzn;bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=Ty_{n}\\ y_{n}=W\left(Ty_{n},Tz_{n};b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Let {un}\left\{u_{n}\right\} be the sequence defined by the iteration (2.11), that is

{un+1=Tvnvn=W(Tvn,W(Tun,Twn,cn1bn);bn)wn=W(Twn,un,an)\left\{\begin{array}[]{l}u_{n+1}=Tv_{n}\\ v_{n}=W\left(Tv_{n},W\left(Tu_{n},Tw_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ w_{n}=W\left(Tw_{n},u_{n},a_{n}\right)\end{array}\right.

Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{bn+cn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{b_{n}+c_{n}\right\} sequences in ( 0,1 ).
Suppose the following assumptions are satisfied:
(C1) 0<a<an<1a<10<a<a_{n}<1-a<1 and 0<b<bn<1b<10<b<b_{n}<1-b<1,
(C2) limncn=0\lim_{n\rightarrow\infty}c_{n}=0,
Then, iteration (2.11) converges faster than (2.5), with respect to definitions (1.2) and (1.3).
Proof. We know that d(xn+1,p)And\left(x_{n+1},p\right)\leq A_{n} and d(un+1,p)Bnd\left(u_{n+1},p\right)\leq B_{n}, with
An=k=0n(δ21bk1δbk1ak1δak)d(x0,p)A_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot\frac{1-b_{k}}{1-\delta b_{k}}\cdot\frac{1-a_{k}}{1-\delta a_{k}}\right)\cdot d\left(x_{0},p\right),
Bn=k=0n(δ2ck+(1bkck)1ak1δak1δbk)d(u0,p)B_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot\frac{c_{k}+\left(1-b_{k}-c_{k}\right)\cdot\frac{1-a_{k}}{1-\delta a_{k}}}{1-\delta b_{k}}\right)\cdot d\left(u_{0},p\right)
Then, Bn=k=0n(δ2ck(1δak)+(1bkck)(1ak)(1δak)(1δbk))d(u0,p)B_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot\frac{c_{k}\left(1-\delta a_{k}\right)+\left(1-b_{k}-c_{k}\right)\left(1-a_{k}\right)}{\left(1-\delta a_{k}\right)\left(1-\delta b_{k}\right)}\right)\cdot d\left(u_{0},p\right).
Let’s denote θn:=AnBn\theta_{n}:=\frac{A_{n}}{B_{n}}.
Then, we have that θn+1θn=(1an+1)(1bn+1)cn+1(1δan+1)+(1bn+1cn+1)(1an+1)\frac{\theta_{n+1}}{\theta_{n}}=\frac{\left(1-a_{n+1}\right)\left(1-b_{n+1}\right)}{c_{n+1}\left(1-\delta a_{n+1}\right)+\left(1-b_{n+1}-c_{n+1}\right)\left(1-a_{n+1}\right)}.
From assumption (C1), we have that θn+1θn(1a)(1b)cn+1(1δ+δa)+a(bcn+1)\frac{\theta_{n+1}}{\theta_{n}}\leq\frac{(1-a)(1-b)}{c_{n+1}(1-\delta+\delta a)+a\left(b-c_{n+1}\right)}.
From assumption (C2), we get limnθn+1θn1aa1bb\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}\leq\frac{1-a}{a}\cdot\frac{1-b}{b}.
It is easy to see that limnθn+1θn>1\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}>1, because a<1aa<1-a from (C1).
We will define ϕn:=1θn\phi_{n}:=\frac{1}{\theta_{n}}. Then ϕn+1ϕn=θnθn+1\frac{\phi_{n+1}}{\phi_{n}}=\frac{\theta_{n}}{\theta_{n+1}}.
Moreover, limnϕn+1ϕn=limnθnθn+1=1limnθn+1θn<1\lim_{n\rightarrow\infty}\frac{\phi_{n+1}}{\phi_{n}}=\lim_{n\rightarrow\infty}\frac{\theta_{n}}{\theta_{n+1}}=\frac{1}{\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}}<1.
So we get that n=1ϕn<\sum_{n=1}^{\infty}\phi_{n}<\infty, which implies that limnϕn=0\lim_{n\rightarrow\infty}\phi_{n}=0.

This means that limnBnAn=0\lim_{n\rightarrow\infty}\frac{B_{n}}{A_{n}}=0.
In conclusion, iteration (2.11) converges faster than (2.5).
Now, in our next theorem we make a comparison between the rate of convergence of the iterations (2.11) and (2.9) in a complete convex metric space, under the assumptions of convergence of both iterative processes.

Theorem 3.8. Let {xn}\left\{x_{n}\right\} be the sequence defined by the iteration (2.11), that is

{xn+1=Tynyn=W(Tyn,W(Txn,Tzn,cn1bn);bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=Ty_{n}\\ y_{n}=W\left(Ty_{n},W\left(Tx_{n},Tz_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Let {un}\left\{u_{n}\right\} be the sequence defined by the iteration (2.9), that is

{un+1=Tvnvn=W(Tvn,W(wn,Twn,cn1bn);bn)wn=W(Twn,un,an)\left\{\begin{array}[]{l}u_{n+1}=Tv_{n}\\ v_{n}=W\left(Tv_{n},W\left(w_{n},Tw_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ w_{n}=W\left(Tw_{n},u_{n},a_{n}\right)\end{array}\right.

Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{bn+cn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{b_{n}+c_{n}\right\} sequences in ( 0,1 ).
Let’s suppose the following assumptions are satisfied:
(C1) 0<a<an<1a<10<a<a_{n}<1-a<1 and 0<b<bn<1b<10<b<b_{n}<1-b<1,
(C2) limncn=0\lim_{n\rightarrow\infty}c_{n}=0,
(C3) 1a2a<b\frac{1-a}{2-a}<b.
Then, iteration (2.11) converges faster than iteration (2.9).
Proof. We know that d(xn+1,p)And\left(x_{n+1},p\right)\leq A_{n} and d(un+1,p)Bnd\left(u_{n+1},p\right)\leq B_{n}, with
An=k=0n(δ2ck+(1bkck)1ak1δak1δbk)d(x0,p)A_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot\frac{c_{k}+\left(1-b_{k}-c_{k}\right)\cdot\frac{1-a_{k}}{1-\delta a_{k}}}{1-\delta b_{k}}\right)\cdot d\left(x_{0},p\right).
This means that An=k=0n(δ2ck(1δak)+(1bkck)(1ak)(1δak)(1δbk))d(x0,p)A_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot\frac{c_{k}\left(1-\delta a_{k}\right)+\left(1-b_{k}-c_{k}\right)\left(1-a_{k}\right)}{\left(1-\delta a_{k}\right)\left(1-\delta b_{k}\right)}\right)d\left(x_{0},p\right),
Bn=k=0n(δck+δ(1bkck)1δbk1ak1δak)d(u0,p)B_{n}=\prod_{k=0}^{n}\left(\delta\cdot\frac{c_{k}+\delta\left(1-b_{k}-c_{k}\right)}{1-\delta b_{k}}\cdot\frac{1-a_{k}}{1-\delta a_{k}}\right)\cdot d\left(u_{0},p\right).
Also let Bn=k=0n(δck+δ(1bkck)1δbk)d(u0,p)B_{n}^{\prime}=\prod_{k=0}^{n}\left(\delta\cdot\frac{c_{k}+\delta\left(1-b_{k}-c_{k}\right)}{1-\delta b_{k}}\right)\cdot d\left(u_{0},p\right), with Bn<BnB_{n}<B_{n}^{\prime}. and d(un+1,p)<Bnd(un,p)d\left(u_{n+1},p\right)<B_{n}^{\prime}d\left(u_{n},p\right).
Let’s denote θn:=AnBn\theta_{n}:=\frac{A_{n}}{B_{n}^{\prime}}.
Then, we have that θn+1θn=δcn+1(1δan+1)+(1bn+1cn+1)(1an+1)cn+1+δ(1bn+1cn+1)\frac{\theta_{n+1}}{\theta_{n}}=\delta\cdot\frac{c_{n+1}\left(1-\delta a_{n+1}\right)+\left(1-b_{n+1}-c_{n+1}\right)\left(1-a_{n+1}\right)}{c_{n+1}+\delta\left(1-b_{n+1}-c_{n+1}\right)}.
Assumption (C1) implies that θn+1θnδcn+1(1δa)+(1bcn+1)(1a)cn+1+δ(bcn+1)\frac{\theta_{n+1}}{\theta_{n}}\leq\delta\cdot\frac{c_{n+1}(1-\delta a)+\left(1-b-c_{n+1}\right)(1-a)}{c_{n+1}+\delta\left(b-c_{n+1}\right)}.

From assumption (C2), we get limnθn+1θnδ(1b)(1a)δb=(1a)(1b)b\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}\leq\delta\cdot\frac{(1-b)(1-a)}{\delta b}=(1-a)\cdot\frac{(1-b)}{b}.
From assumption (C3), since 1a2a<b\frac{1-a}{2-a}<b, it implies that (1a)(1b)<b(1-a)(1-b)<b, that is limnθn+1θn<1\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}<1.
So we get that n=1θn<\sum_{n=1}^{\infty}\theta_{n}<\infty, which implies that limnθn=0\lim_{n\rightarrow\infty}\theta_{n}=0.
This means that limnAnBn=0\lim_{n\rightarrow\infty}\frac{A_{n}}{B_{n}}=0.
In conclusion, iteration (2.11) converges faster than (2.9).

For the comparison between the iterative algorithms (2.9) and (2.7) in a complete convex metric space, under the asssumptions of convergence of iterations, we have the following.

Theorem 3.9. Let {xn}\left\{x_{n}\right\} be the sequence defined by the iteration (2.9), that is

{xn+1=Tynyn=W(Tyn,W(zn,Tzn,cn1bn);bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=Ty_{n}\\ y_{n}=W\left(Ty_{n},W\left(z_{n},Tz_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Let {un}\left\{u_{n}\right\} be the sequence defined by the iteration (2.7), that is

{un+1=Tvnvn=W(Tvn,W(un,Twn,cn1bn);bn)wn=W(Twn,un,an)\left\{\begin{array}[]{l}u_{n+1}=Tv_{n}\\ v_{n}=W\left(Tv_{n},W\left(u_{n},Tw_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ w_{n}=W\left(Tw_{n},u_{n},a_{n}\right)\end{array}\right.

Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{bn+cn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{b_{n}+c_{n}\right\} sequences in ( 0,1 ).
Let’s suppose the following assumptions are satisfied:
(C1) 0<c<cn<1c<10<c<c_{n}<1-c<1,
(C2) limnan=1\lim_{n\rightarrow\infty}a_{n}=1 and limnbn=0\lim_{n\rightarrow\infty}b_{n}=0.
(C3) c>1δ2c>\frac{1-\delta}{2}.
Then, iteration (2.7) converges faster than iteration (2.9).
Proof. We know that d(xn+1,p)And\left(x_{n+1},p\right)\leq A_{n} and d(un+1,p)Bnd\left(u_{n+1},p\right)\leq B_{n}, with
An=k=0n(δck+δ(1bkck)1δbk1ak1δak)d(x0,p)A_{n}=\prod_{k=0}^{n}\left(\delta\cdot\frac{c_{k}+\delta\left(1-b_{k}-c_{k}\right)}{1-\delta b_{k}}\cdot\frac{1-a_{k}}{1-\delta a_{k}}\right)\cdot d\left(x_{0},p\right)
Bn=k=0n(δck+δ(1bkck)1ak1δak1δbk)d(u0,p)B_{n}=\prod_{k=0}^{n}\left(\delta\cdot\frac{c_{k}+\delta\left(1-b_{k}-c_{k}\right)\cdot\frac{1-a_{k}}{1-\delta a_{k}}}{1-\delta b_{k}}\right)\cdot d\left(u_{0},p\right).
This means that Bn=k=0n(δck(1δak)+δ(1bkck)(1ak)(1δak)(1δbk))d(u0,p)B_{n}=\prod_{k=0}^{n}\left(\delta\cdot\frac{c_{k}\left(1-\delta a_{k}\right)+\delta\left(1-b_{k}-c_{k}\right)\left(1-a_{k}\right)}{\left(1-\delta a_{k}\right)\left(1-\delta b_{k}\right)}\right)d\left(u_{0},p\right),
Let’s denote θn:=AnBn\theta_{n}:=\frac{A_{n}}{B_{n}}.
Then, we have that θn+1θn=cn+1+δ(1bn+1cn+1)cn+1(1δan+1)+δ(1bn+1cn+1)(1an+1)\frac{\theta_{n+1}}{\theta_{n}}=\frac{c_{n+1}+\delta\left(1-b_{n+1}-c_{n+1}\right)}{c_{n+1}\left(1-\delta a_{n+1}\right)+\delta\left(1-b_{n+1}-c_{n+1}\right)\left(1-a_{n+1}\right)}.

From assumption (C1), we get θn+1θnc+δ(cbn+1)(1c)(1δan+1)+δ(1cbn+1)(1an+1)\frac{\theta_{n+1}}{\theta_{n}}\geq\frac{c+\delta\left(c-b_{n+1}\right)}{(1-c)\left(1-\delta a_{n+1}\right)+\delta\left(1-c-b_{n+1}\right)\left(1-a_{n+1}\right)}.
From assumption (C2), we get limnθn+1θnc+δ(c)(1c)(1δ)\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}\leq\frac{c+\delta(c)}{(1-c)(1-\delta)}.
This means that the limit is c+δc(1c)(1δ)\frac{c+\delta c}{(1-c)(1-\delta)}, i.e. 1>c>1δ21>c>\frac{1-\delta}{2}. By assumption (C3), we get that the limit is greater than 1 .
Denote φn=1θn\varphi_{n}=\frac{1}{\theta_{n}}. Then limnφn+1φn=limnθnθn+1\lim_{n\rightarrow\infty}\frac{\varphi_{n+1}}{\varphi_{n}}=\lim_{n\rightarrow\infty}\frac{\theta_{n}}{\theta_{n+1}}. We get limnφn+1φn<1\lim_{n\rightarrow\infty}\frac{\varphi_{n+1}}{\varphi_{n}}<1.
So we get that n=1φn<\sum_{n=1}^{\infty}\varphi_{n}<\infty, which implies that limnφn=0\lim_{n\rightarrow\infty}\varphi_{n}=0.
This means that limnBnAn=0\lim_{n\rightarrow\infty}\frac{B_{n}}{A_{n}}=0.
In conclusion, iteration (2.7) converges faster than (2.9).

For comparison of the rate of convergence of iterative processes (2.15) and (2.3) under the assumptions of convergence, we present the following theorem.

Theorem 3.10. Let {xn}\left\{x_{n}\right\} be the sequence defined by the iteration (2.15), that is

{xn+1=W(Txn+1,W(Txn+1,Tyn;βn1αn);αn)yn=W(Tyn,W(Tyn,Tzn;cn1bn);bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Tx_{n+1},W\left(Tx_{n+1},Ty_{n};\frac{\beta_{n}}{1-\alpha_{n}}\right);\alpha_{n}\right)\\ y_{n}=W\left(Ty_{n},W\left(Ty_{n},Tz_{n};\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Let {un}\left\{u_{n}\right\} be the sequence defined by the iteration (2.3), that is

{un+1=W(Tun+1,Tvn;αn)vn=W(Tvn,Twn;bn)wn=W(Twn,un,an)\left\{\begin{array}[]{l}u_{n+1}=W\left(Tu_{n+1},Tv_{n};\alpha_{n}\right)\\ v_{n}=W\left(Tv_{n},Tw_{n};b_{n}\right)\\ w_{n}=W\left(Tw_{n},u_{n},a_{n}\right)\end{array}\right.

Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{αn},{bn+cn},{αn+βn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{\alpha_{n}\right\},\left\{b_{n}+c_{n}\right\},\left\{\alpha_{n}+\beta_{n}\right\} sequences in ( 0,1 ).
Let’s suppose the following assumptions are satisfied:
(C1) 0<β<βn<1β<10<\beta<\beta_{n}<1-\beta<1,
(C2) limnαn=0\lim_{n\rightarrow\infty}\alpha_{n}=0,
(C3) δ<β1β\delta<\frac{\beta}{1-\beta}. Then, iteration 2.15 converges faster than iteration 2.3 .
Proof. We know that d(xn+1,p)And\left(x_{n+1},p\right)\leq A_{n} and d(un+1,p)Bnd\left(u_{n+1},p\right)\leq B_{n}, with
An=k=0n(δ21(αk+βk)1δ(αk+βk)1(bk+ck)1δ(bk+ck)1ak1δak)d(x0,p)A_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot\frac{1-\left(\alpha_{k}+\beta_{k}\right)}{1-\delta\left(\alpha_{k}+\beta_{k}\right)}\cdot\frac{1-\left(b_{k}+c_{k}\right)}{1-\delta\left(b_{k}+c_{k}\right)}\cdot\frac{1-a_{k}}{1-\delta a_{k}}\right)\cdot d\left(x_{0},p\right)
Bn=k=0n(δ21αk1δαk1bk1δbk1ak1δak)d(u0,p)B_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot\frac{1-\alpha_{k}}{1-\delta\alpha_{k}}\cdot\frac{1-b_{k}}{1-\delta b_{k}}\cdot\frac{1-a_{k}}{1-\delta a_{k}}\right)\cdot d\left(u_{0},p\right).
Let’s denote θn:=AnBn\theta_{n}:=\frac{A_{n}}{B_{n}}.
We have that θn+1θn=1(αn+1+βn+1)1δ(αn+1+βn+1)1δαn+11αn+11(bn+1+cn+1)1δ(bn+1+cn+1)1δbn+11bn+1\frac{\theta_{n+1}}{\theta_{n}}=\frac{1-\left(\alpha_{n+1}+\beta_{n+1}\right)}{1-\delta\left(\alpha_{n+1}+\beta_{n+1}\right)}\cdot\frac{1-\delta\alpha_{n+1}}{1-\alpha_{n+1}}\cdot\frac{1-\left(b_{n+1}+c_{n+1}\right)}{1-\delta\left(b_{n+1}+c_{n+1}\right)}\cdot\frac{1-\delta b_{n+1}}{1-b_{n+1}}.

Since δ<1\delta<1, it is easy to see that 1(bn+1+cn+1)1δ(bn+1+cn+1)1δbn+11bn+11\frac{1-\left(b_{n+1}+c_{n+1}\right)}{1-\delta\left(b_{n+1}+c_{n+1}\right)}\cdot\frac{1-\delta b_{n+1}}{1-b_{n+1}}\leq 1.
Then, we have θn+1θn1(αn+1+βn+1)1δ(αn+1+βn+1)1δαn+11αn+1\frac{\theta_{n+1}}{\theta_{n}}\leq\frac{1-\left(\alpha_{n+1}+\beta_{n+1}\right)}{1-\delta\left(\alpha_{n+1}+\beta_{n+1}\right)}\cdot\frac{1-\delta\alpha_{n+1}}{1-\alpha_{n+1}}.
Using the assumption (C1), we get θn+1θn1δαn+11αn+11(αn+1+β)1δ(αn+1+1β)\frac{\theta_{n+1}}{\theta_{n}}\leq\frac{1-\delta\alpha_{n+1}}{1-\alpha_{n+1}}\cdot\frac{1-\left(\alpha_{n+1}+\beta\right)}{1-\delta\left(\alpha_{n+1}+1-\beta\right)}.
From assumption (C2), we get limnθn+1θn1β1δ(1β)<1\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}\leq\frac{1-\beta}{1-\delta(1-\beta)}<1, because of assumption (C3), i.e. δ<β1β<1\delta<\frac{\beta}{1-\beta}<1, since β<12\beta<\frac{1}{2}.
Since δ<1\delta<1, we get that n=1θn<\sum_{n=1}^{\infty}\theta_{n}<\infty, which implies that limnθn=0\lim_{n\rightarrow\infty}\theta_{n}=0.
This means that limnAnBn=0\lim_{n\rightarrow\infty}\frac{A_{n}}{B_{n}}=0.
In conclusion, iteration (2.15) converges faster than iteration (2.3).

Now, if iteration algorithms (2.11) and (1.16) are convergent to the fixed of the contraction in a complete convex metric space, for comparison of their rate of convergence, we have the following.

Theorem 3.11. Let {xn}\left\{x_{n}\right\} be the sequence defined by the iteration (2.11), that is

{xn+1=Tynyn=W(Tyn,W(Txn,Tzn,cn1bn);bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=Ty_{n}\\ y_{n}=W\left(Ty_{n},W\left(Tx_{n},Tz_{n},\frac{c_{n}}{1-b_{n}}\right);b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Let {un}\left\{u_{n}\right\} be the sequence defined by the iteration (1.16), that is

{un+1=Tvnvn=W(Twn,Tun,bn)wn=W(Tun,un,an)\left\{\begin{array}[]{l}u_{n+1}=Tv_{n}\\ v_{n}=W\left(Tw_{n},Tu_{n},b_{n}\right)\\ w_{n}=W\left(Tu_{n},u_{n},a_{n}\right)\end{array}\right.

Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{bn+cn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{b_{n}+c_{n}\right\} sequences in ( 0,1 ), with k=0akbk=\sum_{k=0}^{\infty}a_{k}b_{k}=\infty.
Let’s suppose the following assumptions are satisfied:
(C1) 0<a<an<1a<10<a<a_{n}<1-a<1,
(C2) limnbn=0\lim_{n\rightarrow\infty}b_{n}=0 and limncn=0\lim_{n\rightarrow\infty}c_{n}=0,
(C3) a>δ1+δa>\frac{\delta}{1+\delta}.
Then, iteration (2.11) converges faster than iteration (1.16).
Proof. We know that d(xn+1,p)And\left(x_{n+1},p\right)\leq A_{n} and d(un+1,p)Bnd\left(u_{n+1},p\right)\leq B_{n}, with
An=k=0n(δ2ck+(1bkck)1ak1δak1δbk)d(x0,p)A_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot\frac{c_{k}+\left(1-b_{k}-c_{k}\right)\cdot\frac{1-a_{k}}{1-\delta a_{k}}}{1-\delta b_{k}}\right)\cdot d\left(x_{0},p\right).
This means that An=k=0n(δ2ck(1δak)+(1ak)(1bkck)(1δak)(1δbk))d(x0,p)A_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot\frac{c_{k}\left(1-\delta a_{k}\right)+\left(1-a_{k}\right)\left(1-b_{k}-ck\right)}{\left(1-\delta a_{k}\right)\left(1-\delta b_{k}\right)}\right)d\left(x_{0},p\right).
and Bn=k=0n(δ21akbk(1δ))d(u0,p)B_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot 1-a_{k}b_{k}(1-\delta)\right)\cdot d\left(u_{0},p\right).
Let’s denote θn:=AnBn\theta_{n}:=\frac{A_{n}}{B_{n}}.
Then, θn+1θn=cn+1(1δan+1)+(1an+1)(1bn+1cn+1)1an+1bn+1(1δ)1(1δan+1)(1δbn+1)\frac{\theta_{n+1}}{\theta_{n}}=\frac{c_{n+1}\left(1-\delta a_{n+1}\right)+\left(1-a_{n+1}\right)\left(1-b_{n+1}-c_{n+1}\right)}{1-a_{n+1}b_{n+1}(1-\delta)}\cdot\frac{1}{\left(1-\delta a_{n+1}\right)\left(1-\delta b_{n+1}\right)}.
From assumption (C1), we have that
θn+1θncn+1(1δa)+(1a)(1bn+1cn+1)1(1a)bn+1(1δ)1(1δ(1a))(1δbn+1)\frac{\theta_{n+1}}{\theta_{n}}\leq\frac{c_{n+1}(1-\delta a)+(1-a)\left(1-b_{n+1}-c_{n+1}\right)}{1-(1-a)b_{n+1}(1-\delta)}\cdot\frac{1}{(1-\delta(1-a))\left(1-\delta b_{n+1}\right)}.
From assumption (C2), we get limnθn+1θn1a1δ(1a)\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}\leq\frac{1-a}{1-\delta(1-a)}.
From assumption (C3), since a>δ1+δa>\frac{\delta}{1+\delta}, we get 1a1δ(1a)<1\frac{1-a}{1-\delta(1-a)}<1.
This implies that limnθn+1θn<1\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}<1.
So we get that n=1θn<\sum_{n=1}^{\infty}\theta_{n}<\infty, which implies that limnθn=0\lim_{n\rightarrow\infty}\theta_{n}=0.
This means that limnAnBn=0\lim_{n\rightarrow\infty}\frac{A_{n}}{B_{n}}=0.
In conclusion, iteration (2.11) converges faster than iteration (1.16), by definitions (1.2) and (1.3).

Finally, our last theoretical result regarding the comparison of the rate of convergence of iterations (2.3) and (1.12), we have the following theorem.

Theorem 3.12. Let {xn}\left\{x_{n}\right\} be the sequence defined by the iteration (2.3), that is

{xn+1=W(Txn+1,Tyn;αn)yn=W(Tyn,Tzn;bn)zn=W(Tzn,xn,an)\left\{\begin{array}[]{l}x_{n+1}=W\left(Tx_{n+1},Ty_{n};\alpha_{n}\right)\\ y_{n}=W\left(Ty_{n},Tz_{n};b_{n}\right)\\ z_{n}=W\left(Tz_{n},x_{n},a_{n}\right)\end{array}\right.

Let {un}\left\{u_{n}\right\} be the sequence defined by the iteration (1.12), that is

{un+1=W(Twn,Tvn,αn)vn=W(Twn,Tun,bn)wn=W(Tun,un,an)\left\{\begin{array}[]{l}u_{n+1}=W\left(Tw_{n},Tv_{n},\alpha_{n}\right)\\ v_{n}=W\left(Tw_{n},Tu_{n},b_{n}\right)\\ w_{n}=W\left(Tu_{n},u_{n},a_{n}\right)\end{array}\right.

Let CC be a nonempty, closed and convex subset of a complete convex metric space XX. Let T:CCT:C\rightarrow C be a δ\delta-contraction. Let {an},{bn},{αn}\left\{a_{n}\right\},\left\{b_{n}\right\},\left\{\alpha_{n}\right\} sequences in ( 0,1 ), with k=0αkakbk=\sum_{k=0}^{\infty}\alpha_{k}a_{k}b_{k}=\infty.
Let’s suppose the following assumptions are satisfied :
(C1) 0<α<αn<1α<10<\alpha<\alpha_{n}<1-\alpha<1 and 0<a<an<1a<10<a<a_{n}<1-a<1,
(C2) limnbn=0\lim_{n\rightarrow\infty}b_{n}=0,
(C3) α>δ1+δ\alpha>\frac{\delta}{1+\delta}.
Then, iteration (2.3) converges faster than iteration (1.12).
Proof. We know that d(xn+1,p)And\left(x_{n+1},p\right)\leq A_{n} and d(un+1,p)Bnd\left(u_{n+1},p\right)\leq B_{n}, with
An=k=0n(δ21αk1δαk1bk1δbk1ak1δak)d(x0,p)A_{n}=\prod_{k=0}^{n}\left(\delta^{2}\cdot\frac{1-\alpha_{k}}{1-\delta\alpha_{k}}\cdot\frac{1-b_{k}}{1-\delta b_{k}}\cdot\frac{1-a_{k}}{1-\delta a_{k}}\right)\cdot d\left(x_{0},p\right).

Also, Bn=k=0n(δ(1αkakbk(1δ)))d(u0,p)B_{n}=\prod_{k=0}^{n}\left(\delta\cdot\left(1-\alpha_{k}a_{k}b_{k}(1-\delta)\right)\right)\cdot d\left(u_{0},p\right).
Let’s denote θn:=AnBn\theta_{n}:=\frac{A_{n}}{B_{n}}.
Then, θn+1θn=δ1αn+11δαn+11bn+11δbn+11an+11δan+11(1αn+1an+1bn+1(1δ))\frac{\theta_{n+1}}{\theta_{n}}=\delta\cdot\frac{1-\alpha_{n+1}}{1-\delta\alpha_{n+1}}\cdot\frac{1-b_{n+1}}{1-\delta b_{n+1}}\cdot\frac{1-a_{n+1}}{1-\delta a_{n+1}}\cdot\frac{1}{\left(1-\alpha_{n+1}a_{n+1}b_{n+1}(1-\delta)\right)}.
Since 1bn+11δbn+1<1\frac{1-b_{n+1}}{1-\delta b_{n+1}}<1 and 1an+11δan+1<1\frac{1-a_{n+1}}{1-\delta a_{n+1}}<1 and using assumption (C1), we get that:
θn+1θnδ1α1δ(1α)1(1(1α)an+1bn+1(1δ))\frac{\theta_{n+1}}{\theta_{n}}\leq\delta\cdot\frac{1-\alpha}{1-\delta(1-\alpha)}\cdot\frac{1}{\left(1-(1-\alpha)a_{n+1}b_{n+1}(1-\delta)\right)}.
Assumption (C2) implies that limnθn+1θn1α1δ(1α)\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}\leq\frac{1-\alpha}{1-\delta(1-\alpha)}.
From assumption (C3), we get that limnθn+1θn<1\lim_{n\rightarrow\infty}\frac{\theta_{n+1}}{\theta_{n}}<1.
So we get that n=1θn<\sum_{n=1}^{\infty}\theta_{n}<\infty, which implies that limnθn=0\lim_{n\rightarrow\infty}\theta_{n}=0.
This means that limnAnBn=0\lim_{n\rightarrow\infty}\frac{A_{n}}{B_{n}}=0.
In conclusion, iteration (2.3) converges faster than iteration (1.12).

4. Numerical Examples

Throughout this section, we cover with numerical examples the points (A) and (C) from the first section. All of the examples presented below satisfy the conditions for the comparison of rate of convergence and the conditions from the convergence analysis.

Let T:XXT:X\rightarrow X, where Tx=x2Tx=\frac{x}{2}, with δ=12\delta=\frac{1}{2} and X=[0,)X=[0,\infty). Also, let’s take the first iteration x0=100x_{0}=100 and the number of iterations for each comparison of iterative processes be n=15n=15. Regarding Theorem 3.1, we present a numerical example for iterations (2.1) and (1.9).
Example 4.1. Let αk=1k+9,βk=1k+9,bk=1k+3,ck=1k+3\alpha_{k}=\frac{1}{k+9},\beta_{k}=\frac{1}{k+9},b_{k}=\frac{1}{k+3},c_{k}=\frac{1}{k+3} and ak=12a_{k}=\frac{1}{2}.
Also, iteration (2.1) is

{zn=21an2anxnyn=2cn2bn2bnznxn+1=2βn2αn2αnyn.\left\{\begin{array}[]{l}z_{n}=2\frac{1-a_{n}}{2-a_{n}}x_{n}\\ y_{n}=\frac{2-c_{n}-2b_{n}}{2-b_{n}}z_{n}\\ x_{n+1}=\frac{2-\beta_{n}-2\alpha_{n}}{2-\alpha_{n}}y_{n}.\end{array}\right.

and iteration (2.1) is

{zn=1an2xnyn=bn2zn+(1bncn2)xnxn+1=αn2yn+βn2zn+(1αnβn)xn.\left\{\begin{array}[]{l}z_{n}=1-\frac{a_{n}}{2}x_{n}\\ y_{n}=\frac{b_{n}}{2}z_{n}+\left(1-b_{n}-\frac{c_{n}}{2}\right)x_{n}\\ x_{n+1}=\frac{\alpha_{n}}{2}y_{n}+\frac{\beta_{n}}{2}z_{n}+\left(1-\alpha_{n}-\beta_{n}\right)x_{n}.\end{array}\right.

Furthermore, condition (C)\left(C^{*}\right) is : 1>2n+9+14(n+3)(n+9)[2n+3+34(2n+7)]1>\frac{2}{n+9}+\frac{1}{4(n+3)(n+9)}\left[2n+3+\frac{3}{4}(2n+7)\right], which is satisfied for n>3n>-3, so it is a valid assumption.
We have that

ITERATION 1.9 ITERATION (2.1)
100.0 100.0
42.60651629 87.34375
19.98824221 77.51757813
9.95461865 69.6446991
5.16619183 63.1872579
2.76380633 57.79236367
1.51364647 53.21713488
0.84462672 49.288213597
0.47858118 45.87823358
0.27466398 42.89136941
0.1593546 40.25407114
0.09332257 37.90891209
0.05509877 35.81038302
0.03276441 33.92194486
0.01960717 32.21391854
0.01180006 30.6619459

Regarding Theorem 3.3, we present a numerical example for iterations (2.15) and (2.1):
Example 4.2. Let αk=1k+7,βk=1k+7,bk=13,ck=12\alpha_{k}=\frac{1}{k+7},\beta_{k}=\frac{1}{k+7},b_{k}=\frac{1}{3},c_{k}=\frac{1}{2} and ak=13a_{k}=\frac{1}{3}.
Also, iteration (2.15) is

{zn=21an2anxnyn=1bncn2bncnznxn+1=1αnβn2αnβnyn.\left\{\begin{array}[]{l}z_{n}=2\frac{1-a_{n}}{2-a_{n}}x_{n}\\ y_{n}=\frac{1-b_{n}-c_{n}}{2-b_{n}-c_{n}}z_{n}\\ x_{n+1}=\frac{1-\alpha_{n}-\beta_{n}}{2-\alpha_{n}-\beta_{n}}y_{n}.\end{array}\right.

and iteration (2.1) is

{zn=1an2xnyn=bn2zn+(1bncn2)xnxn+1=αn2yn+βn2zn+(1αnβn)xn.\left\{\begin{array}[]{l}z_{n}=1-\frac{a_{n}}{2}x_{n}\\ y_{n}=\frac{b_{n}}{2}z_{n}+\left(1-b_{n}-\frac{c_{n}}{2}\right)x_{n}\\ x_{n+1}=\frac{\alpha_{n}}{2}y_{n}+\frac{\beta_{n}}{2}z_{n}+\left(1-\alpha_{n}-\beta_{n}\right)x_{n}.\end{array}\right.

Moreover, condition (C)\left(C^{*}\right) is 1>2n+7+12[512+3536]1n+71>\frac{2}{n+7}+\frac{1}{2}\left[\frac{5}{12}+\frac{35}{36}\right]\frac{1}{n+7}, which is satisfied for each n>11536n>-\frac{115}{36}, so it is valid.
Now, we have that
Regarding Theorem 3.4, we present a numerical example for iterations (2.13) and (2.1):
Example 4.3. Let αk=1k+10,βk=1k+10,bk=1k+4,ck=1k+4\alpha_{k}=\frac{1}{k+10},\beta_{k}=\frac{1}{k+10},b_{k}=\frac{1}{k+4},c_{k}=\frac{1}{k+4} and ak=22k+7a_{k}=\frac{2}{2k+7}.

ITERATION (2.15) ITERATION (2.1)
100.0 100.0
4.89795918 83.68055555
0.24489796 71.54170953
0.01243926 62.2015419
0.00063973 54.81903566
0.00003323 48.85492762
0.00000174 43.94855668
0.00000009 39.85017937
4871011487\cdot 10^{-11} 36.38173783
2601012260\cdot 10^{-12} 33.41308909
1391013139\cdot 10^{-13} 30.84705120
7491015749\cdot 10^{-15} 28.60968792
4041016404\cdot 10^{-16} 26.64381755
2181017218\cdot 10^{-17} 24.90456835
1191018119\cdot 10^{-18} 23.35626846
6461020646\cdot 10^{-20} 21.97022728

Also, iteration (2.13) is

{zn=21an2anxnyn=21bncn2bnzn+cn2bnxnxn+1=1αnβn2αnyn+βn2αnzn.\left\{\begin{array}[]{l}z_{n}=2\frac{1-a_{n}}{2-a_{n}}x_{n}\\ y_{n}=2\frac{1-b_{n}-c_{n}}{2-b_{n}}z_{n}+\frac{c_{n}}{2-b_{n}}x_{n}\\ x_{n+1}=\frac{1-\alpha_{n}-\beta_{n}}{2-\alpha_{n}}y_{n}+\frac{\beta_{n}}{2-\alpha_{n}}z_{n}.\end{array}\right.

and iteration (2.1) is

{zn=1an2xnyn=bn2zn+(1bncn2)xnxn+1=αn2yn+βn2zn+(1αnβn)xn.\left\{\begin{array}[]{l}z_{n}=1-\frac{a_{n}}{2}x_{n}\\ y_{n}=\frac{b_{n}}{2}z_{n}+\left(1-b_{n}-\frac{c_{n}}{2}\right)x_{n}\\ x_{n+1}=\frac{\alpha_{n}}{2}y_{n}+\frac{\beta_{n}}{2}z_{n}+\left(1-\alpha_{n}-\beta_{n}\right)x_{n}.\end{array}\right.

Also, condition (C)\left(C^{*}\right) is (n+4)(n+10)(2n+7)(8n3+124n2+598n+915)>0(n+4)(n+10)(2n+7)\left(8n^{3}+124n^{2}+598n+915\right)>0, so the assumption is valid.

Now, we have that
Regarding Theorem 3.5, we present a numerical example for iterations (2.3) and (2.5):
Example 4.4. Let αk=55120,bk=1k+2\alpha_{k}=\frac{55}{120},b_{k}=\frac{1}{\sqrt{k+2}} and ak=11k+2a_{k}=1-\frac{1}{\sqrt{k+2}}.
Also, iteration (2.3) is

{zn=21an2anxnyn=1bn2bnznxn+1=1αn2αnyn.\left\{\begin{array}[]{l}z_{n}=2\frac{1-a_{n}}{2-a_{n}}x_{n}\\ y_{n}=\frac{1-b_{n}}{2-b_{n}}z_{n}\\ x_{n+1}=\frac{1-\alpha_{n}}{2-\alpha_{n}}y_{n}.\end{array}\right.
ITERATION (2.13) ITERATION (2.1)
100.0 100.0
33.92857143 89.44444444
12.32425184 81.00256032
4.69269589 74.06975455
1.84976637 68.26101041
0.74877079 63.31617686
0.30955125 59.05170739
0.13018573 55.33372046
0.0555366 52.06192489
0.0239783 49.15952527
0.01045996 46.56662042
0.0046038 44.2357406
0.00204218 42.12874589
0.00091215 40.21461916
0.00040992 38.46786313
0.00018523 36.86731548

and iteration (2.5) is

{zn=21an2anxnyn=1bn2bnznxn+1=12yn.\left\{\begin{array}[]{l}z_{n}=2\frac{1-a_{n}}{2-a_{n}}x_{n}\\ y_{n}=\frac{1-b_{n}}{2-b_{n}}z_{n}\\ x_{n+1}=\frac{1}{2}y_{n}.\end{array}\right.

Finally, we have

ITERATION (2.3) ITERATION (2.5)
100.0 100.0
7.64126855 10.87411293
0.59661556 1.20823477
0.04612052 0.13291663
0.00349281 0.0143248
0.000258176 0.0015068
0.00001861 0.00015453
0.00000131 0.00001545
0.00000009 0.00000151
61096\cdot 10^{-9} 0.00000014
410104\cdot 10^{-10} 0.00000001
25101225\cdot 10^{-12} 11081\cdot 10^{-8}
15101315\cdot 10^{-13} 210102\cdot 10^{-10}
97101597\cdot 10^{-15} 95101395\cdot 10^{-13}
58101658\cdot 10^{-16} 81101481\cdot 10^{-14}
34101734\cdot 10^{-17} 68101568\cdot 10^{-15}

Regarding Theorem 3.6, we present a numerical example for iterations (2.5) and (1.16):

Example 4.5. Let bk=12+1k+2b_{k}=\frac{1}{2}+\frac{1}{k+2} and ak=12(k+3)a_{k}=\frac{1}{2(k+3)}.
Also, iteration (1.16) is

{zn=2(1an)xnyn=bn2zn+(1bn2)xnxn+1=12yn.\left\{\begin{array}[]{l}z_{n}=2\left(1-a_{n}\right)x_{n}\\ y_{n}=\frac{b_{n}}{2}z_{n}+\left(1-\frac{b_{n}}{2}\right)x_{n}\\ x_{n+1}=\frac{1}{2}y_{n}.\end{array}\right.

and iteration (2.5) is

{zn=21an2anxnyn=1bn2bnznxn+1=12yn.\left\{\begin{array}[]{l}z_{n}=2\frac{1-a_{n}}{2-a_{n}}x_{n}\\ y_{n}=\frac{1-b_{n}}{2-b_{n}}z_{n}\\ x_{n+1}=\frac{1}{2}y_{n}.\end{array}\right.

By comparison, we get

ITERATION (1.16) ITERATION (2.5)
100.0 100.0
23.69791667 6.66666667
5.7023112 0.63157895
1.38399845 0.06970604
0.33776153 0.00839054
0.08274403 0.00106841
0.02032688 0.00014153
0.00500408 0.00001931
0.00123396 0.00000269
0.00030469 0.00000038
0.00007532 0.00000006
0.00001864 81098\cdot 10^{-9}
0.00000461 11091\cdot 10^{-9}
0.00000114 110101\cdot 10^{-10}
0.00000028 210112\cdot 10^{-11}
0.00000007 410124\cdot 10^{-12}

Regarding Theorem 3.7, we present a numerical example for iterations (2.7) and (2.5):
Example 4.6. Let bk=38,ak=1730b_{k}=\frac{3}{8},a_{k}=\frac{17}{30} and ck=1k+2c_{k}=\frac{1}{\sqrt{k+2}}.
In our context of normed spaces, iteration (2.7) is

{zn=21an2anxnyn=1bncn2bnzn+cn2bnxnxn+1=12yn.\left\{\begin{array}[]{l}z_{n}=2\frac{1-a_{n}}{2-a_{n}}x_{n}\\ y_{n}=\frac{1-b_{n}-c_{n}}{2-b_{n}}z_{n}+\frac{c_{n}}{2-b_{n}}x_{n}\\ x_{n+1}=\frac{1}{2}y_{n}.\end{array}\right.

and iteration (2.5) is

{zn=21an2anxnyn=1bn2bnznxn+1=12yn.\left\{\begin{array}[]{l}z_{n}=2\frac{1-a_{n}}{2-a_{n}}x_{n}\\ y_{n}=\frac{1-b_{n}}{2-b_{n}}z_{n}\\ x_{n+1}=\frac{1}{2}y_{n}.\end{array}\right.

By comparison, we get

ITERATION (2.7) ITERATION (2.5)
100.0 100.0
18.65113029 11.62790698
3.30315188 1.35208221
0.56378435 0.15721886
0.0935548 0.01828126
0.01517991 0.00212573
0.00241797 0.00024718
0.0003792 0.00002874
0.00005868 0.00000334
0.00000898 0.00000039
0.00000136 0.00000005
0.0000002 51095\cdot 10^{-9}
0.00000003 610106\cdot 10^{-10}
41094\cdot 10^{-9} 710117\cdot 10^{-11}
710107\cdot 10^{-10} 810128\cdot 10^{-12}
9.610119.6\cdot 10^{-11} 97101397\cdot 10^{-13}

5. Conclusions

In the present article, through an exhaustive approach involving iterative processes in the context of convex metric spaces, we introduced numerous iterations which converge faster to the fixed point of a single-valued mapping than some iterations found in fixed point literature. The first three iterations introduced by us through multiple convex combination were compared to iteration (2.1) using definition (1.4) as in the article of Berinde [6]. The other fixed point iterations were compared with well know iterative processes using definition (1.2) and (1.3) from two points of view : the first one consists on the fact that Agarwal et.al. [2] used this type of definitions, respectively the second point of view lies on the idea that in [1, [8] and in the other articles gave as references, Gursoy, Abbas and the rest of the authors compared iterations also with this type of definitions.

Moreover, from (Example 4.1), (Example 4.2 ) and (Example 4.3 ) the definition (1.4) is more precise for iterations given through multiple convex combinations and our new iterative processes converge faster than the iteration given by Suantai, Berinde et. al. in [6] and [14].

Finally, from (Example 4.6), we can easily observe that the definitions (1.2) and (1.3) depend on the auxiliary sequences (bn),(cn)\left(b_{n}\right),\left(c_{n}\right) and (an)\left(a_{n}\right). So, we gave also an example of two iterations justifying the fact that definitions (1.2) and (1.3) are not very useful in practical applications.

Acknowledgments

The author thank the referees and the editors for valuable comments and suggestions that improved the article in its present form.

References

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2017

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