Approximating fixed points for nonlinear generalized mappings using Ishikawa iteration

Abstract

We obtain a contractive condition for the existence and uniqueness of fixed points for a generalized contraction-type mapping. The present study focuses on providing a method for the existence of fixed points for nonlinear mappings. Sufficient conditions for the existence and uniqueness of such points are obtained using Ishikawa iteration process. Moreover, an example is given.

Authors

C. D. Alecsa
Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy, Cluj-Napoca, Romania

Keywords

Fixed point;  Generalized contraction;  Ishikawa; Convergence; Convex metric space;

Paper coordinates

Cristian-Daniel Alecsa, Approximating fixed points for nonlinear generalized mappings using Ishikawa iteration, Rendiconti Del Circolo Matematico Di Palermo, 68 (2019) no. 1, pp. 163-191.
https://doi.org/10.1007/s12215-018-0349-7

PDF

About this paper

Journal

Rendiconti del Circolo Matematico di Palermo Series 2

Publisher Name

Springer Milan

DOI

https://doi.org/10.1007/s12215-018-0349-7

Print ISSN
Online ISSN

google scholar link

[1]  Abbas, M., Khan, S.H., Rhoades, B.E., Simpler is also better in approximating fixed points. Appl. Math. Comput. 205(1), 428–431 (2008)

[2] Agarwal, R.P., O’Regan, D., Sahu, D.R., Fixed Point Theory for Lipschitzian-Type Mappings with Applications. Springer, Heidelberg (2003)

[3] Asadi, M., Some results of fixed point theorems in convex metric spaces. Nonlinear Funct. Anal. Appl. 19(2), 171–175 (2014)

[4] Fukhar-ud-dina, H., Berinde, V., Iterative methods for the class of quasi-contractive type operators and comparison of their rate of convergence in convex metric spaces. Filomat 30(1), 223–230 (2016)

[5] Karapinar, E., Fixed point theorems in cone banach spaces. Fixed point theory and applications. Article ID 609281, pp. 1–9 (2009). https://doi.org/10.1155/2009/60928

[6] Moosaei, M., Fixed point theorems in convex metric spaces. Fixed Point Theory Appl. 2012, 164 (2012). https://doi.org/10.1186/1687-1812-2012-164

[7] Moosaei, M., Common fixed points for some generalized contraction pairs in convex metric spaces. Fixed Point Theory Appl. 2014:98 (2014). http://www.fixedpointtheoryandapplications.com/content/2014/1/

[8] Moosaei, M., Azizi, A., On coincidence points of generalized contractive pair mappings in convex metric spaces. J. Hyperstruct. 2(4), 136–141 (2015)

[9] Shimizu, T., Takahashi, W., Fixed point theorems in certain convex metric spaces. Math. Jpn. 37, 855–859 (1992)

[10] Takahashi, W., A convexity in metric spaces and nonexpansive mappings I. Kodai Math. Semin. Rep. 22, 142–149 (1970)

[11] Wang, C., Zhang, T., Approximating common fixed points for a pair of generalized nonlinear mappings in convex metric spaces. J. Nonlinear Sci. Appl. 9, 1–7 (2016)

Paper (preprint) in HTML form

Approximating fixed points for nonlinear generalized mappings using Ishikawa iteration

Cristian Daniel Alecsa 1,2
Abstract

We obtain a contractive condition for the existence and uniqueness of fixed points for a generalized contraction-type mapping. The present study focuses on providing a method for the existence of fixed points for nonlinear mappings. Sufficient conditions for the existence and uniqueness of such points are obtained using Ishikawa iteration process. Moreover, an example is given.

Received: 6 May 2017 / Accepted: 30 April 2018 / Published online: 9 May 2018
© Springer-Verlag Italia S.r.l., part of Springer Nature 2018

Keywords Fixed point \cdot Generalized contraction \cdot Ishikawa \cdot Convergence \cdot Convex metric space

Mathematics Subject Classification 47H1054H2547\mathrm{H}10\cdot 54\mathrm{H}25

1 Introduction and preliminaries

In [10], Takahashi introduced a new concept of convexity in metric spaces and proved that all normed convex spaces and their convex subsets are convex metric spaces. Moreover, he gave some examples of convex metric spaces. We recall the basic definitions and properties of convex metric spaces. For details, we let the reader see [2,4,9][2,4,9] and [10].

Definition 1 Let ( X,dX,d ) be a metric space.
A function W:X×X×[0,1]XW:X\times X\times[0,1]\rightarrow X is called a convexity structure, if

d(u,W(x,y,λ))λd(u,x)+(1λ)d(u,y),d(u,W(x,y,\lambda))\leq\lambda d(u,x)+(1-\lambda)d(u,y),

for each x,yXx,y\in X and λ[0,1]\lambda\in[0,1].

00footnotetext: Cristian Daniel Alecsa
cristian.alecsa@math.ubbcluj.ro; cristian.alecsa@ictp.acad.ro
1 Department of Mathematics, Babeş-Bolyai University, Cluj-Napoca, Romania
2 Tiberiu Popoviciu Institute of Numerical Analysis, Cluj-Napoca, Romania

A metric space ( X,dX,d ) endowed with a convexity structure WW is called a convex metric space. A nonempty subset KK of XX is convex, if W(x,y,λ)KW(x,y,\lambda)\in K, for each x,yKx,y\in K and λ[0,1]\lambda\in[0,1].

Let XX be a convex metric space. Takahashi showed that the open balls and the closed balls are convex subsets of XX.

If {Kα}αJ\left\{K_{\alpha}\right\}_{\alpha\in J} is a family of nonempty, convex subsets of XX, then the intersection αJKα\bigcap_{\alpha\in J}K_{\alpha} is a convex subset of XX.

Also, in [4], Berinde has mentioned an important property of convex metric spaces, i.e.:

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

for each x,yXx,y\in X and λ[0,1]\lambda\in[0,1].
Also, from [6-8] and [11], we recall some important lemmas regarding convex metric spaces:

Lemma 1 Let ( X,d,WX,d,W ) be a convex metric space. Then the following statements hold:
(i) d(x,y)=d(x,W(x,y,λ))+d(y,W(x,y,λ))d(x,y)=d(x,W(x,y,\lambda))+d(y,W(x,y,\lambda)), for each (x,y)X×X(x,y)\in X\times X and λ[0,1]\lambda\in[0,1].
(ii) d(x,W(x,y,λ))=(1λ)d(x,y)d(x,W(x,y,\lambda))=(1-\lambda)d(x,y), for each x,yXx,y\in X.
(iii) d(y,W(x,y,λ))=λd(x,y)d(y,W(x,y,\lambda))=\lambda d(x,y), for each x,yXx,y\in X.

Lemma 2 Let (X,d,W)(X,d,W) be a convex metric space. Then:

d(x,W(x,y,12))=d(y,W(x,y,12))=12d(x,y),d\left(x,W\left(x,y,\frac{1}{2}\right)\right)=d\left(y,W\left(x,y,\frac{1}{2}\right)\right)=\frac{1}{2}d(x,y),

for each x,yXx,y\in X.
Also, for the sake of completeness we remind the classical iterative processes, such as Krasnoselskii, Mann and Ishikawa in convex metric spaces.

In [3], Asadi used Krasnoselskij iteration for a contractive-type nonlinear mapping, i.e.:

xn=W(xn1Txn1,λ),x_{n}=W\left(x_{n-1}\cdot Tx_{n-1},\lambda\right),

with λ[0,1]\lambda\in[0,1]. Also, the Mann iteration is defined as:

xn=W(xn1Txn1,αn),x_{n}=W\left(x_{n-1}\cdot Tx_{n-1},\alpha_{n}\right),

with αn[0,1]\alpha_{n}\in[0,1], for each nn\in\mathbb{N}.
Moreover, the classical Ishikawa iteration is as follows

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

with αn,βn[0,1]\alpha_{n},\beta_{n}\in[0,1], for each nn\in\mathbb{N}.
We make the following remark: that although Berinde, Assadi and Moosaei used the same definition of convex metric spaces, in [4], Berinde defined Mann iteration as

xn=W(Txn1,xn1,αn),x_{n}=W\left(Tx_{n-1},x_{n-1},\alpha_{n}\right),

with αn[0,1]\alpha_{n}\in[0,1], for each nn\in\mathbb{N}.
Since many authors use the property that W(x,y,λ)=W(y,x,1λ)W(x,y,\lambda)=W(y,x,1-\lambda), the Mann iterations can be transformed one to another, so it lies no confusion.

Now, since we work with generalized type of contractype mappings, we recall some generalization of Banach contraction principle in convex metric spaces.

In [5] Karapinar developed existence and uniqueness theorems for mappings satisfying certain contractive conditions in cone Banach spaces, such as

d(Tx,Ty)ad(x,y)\displaystyle d(Tx,Ty)\geq ad(x,y)
d(x,Tx)+d(y,Ty)pd(x,y)\displaystyle d(x,Tx)+d(y,Ty)\leq pd(x,y)
ad(Tx,Ty)+b[d(x,Tx)+d(y,Ty)]sd(x,y)\displaystyle ad(Tx,Ty)+b[d(x,Tx)+d(y,Ty)]\leq sd(x,y)

Karapinar used the following conditions on this contractive type-operators, such as:
a>1,p[0,2)a>1,p\in[0,2), respectively 0s+|a|2b<2(a+b)0\leq s+|a|-2b<2(a+b).
In [3], Asadi generalized the above conditions for mappings endowed with more coefficients, i.e.

ad(x,Tx)+bd(y,Ty)+cd(Tx,Ty)+ed(x,Ty)+fd(y,Tx)kd(x,y).ad(x,Tx)+bd(y,Ty)+cd(Tx,Ty)+ed(x,Ty)+fd(y,Tx)\leq kd(x,y).

Asadi, developed existence and uniqueness for above contractions, using the condition:

b+e|f|(1λ)|c|λ1λk<a+b+c+e+f|c|λ|f|(1λ)1λ\frac{b+e-|f|(1-\lambda)-|c|\lambda}{1-\lambda}\leq k<\frac{a+b+c+e+f-|c|\lambda-|f|(1-\lambda)}{1-\lambda}

Independently, in [11] Wang and Zhang developed a theorem for the existence of fixed points of a pair of contractive mappings, such as

kd(Tx,Sy)ad(x,y)+b[d(x,Tx)+d(y,Sy)]+c[d(x,Sy)+d(y,Tx)],kd(Tx,Sy)\leq ad(x,y)+b[d(x,Tx)+d(y,Sy)]+c[d(x,Sy)+d(y,Tx)],

using a Jungk-Krasnoselskii iteration, as follows

{x2n+1=W(x2n,Tx2n,λ)x2n+2=W(x2n+1,Tx2n+1,λ),\left\{\begin{array}[]{l}x_{2n+1}=W\left(x_{2n},Tx_{2n},\lambda\right)\\ x_{2n+2}=W\left(x_{2n+1},Tx_{2n+1},\lambda\right)\end{array},\right.

and give an exemple of a pointwise mapping satysfing the above condition.
Also, Wang and Zhang imposed the conditions a+2c<ka+2c<k and (k,a,b,c)Γρ(k,a,b,c)\in\Gamma_{\rho}, with ρ{1,2,3,4}\rho\in\{1,2,3,4\}, where Γρ\Gamma_{\rho} were defined in terms of the coefficients a,b,ca,b,c and kk.

In [6] Moosaei developed a theorem for a Banach operator pair ( f,gf,g ) satysfing

ad(gx,fx)+bd(gy,fy)+cd(fx,fy)kd(gx,gy),ad(gx,fx)+bd(gy,fy)+cd(fx,fy)\leq kd(gx,gy),

with 2b|c|k<2(a+b+c)|c|2b-|c|\leq k<2(a+b+c)-|c| and in [8] the same author presented a theorem for existence and uniqueness of a common fixed point of a pair of mappings ( S,TS,T ), where (S,T)(S,T) weakly compatible pair, i.e.:

ad(Sx,Tx)+bd(Sy,Ty)+cd(Tx,Ty)ed(Sx,Sy).ad(Sx,Tx)+bd(Sy,Ty)+cd(Tx,Ty)\leq ed(Sx,Sy).

Likewise, in [7], Moosaei presented sufficient conditions for the existence of a coincidence point of a generalized contraction pair ( S,TS,T ), as follows
αd(Tx,TY)+β[d(Sx,Ty)+d(Sy,Tx)]+γ[d(Sx,Ty)+d(Sy,Tx)]ηd(Sx,Sy)\alpha d(Tx,TY)+\beta[d(Sx,Ty)+d(Sy,Tx)]+\gamma[d(Sx,Ty)+d(Sy,Tx)]\leq\eta d(Sx,Sy),
using as Wang and Zhang a Jungk-Krasnoselskii iteration.
In [1], Abbas et al. discussed the simplicity of the modified Krasnoselskij iteration

xn+1=(1λ)xn+λTnxn,x_{n+1}=(1-\lambda)x_{n}+\lambda T^{n}x_{n},

where λ(0,1)\lambda\in(0,1) and the operator TT is defined on a nonempty, closed convex subset of a uniformly convex Banach space. Furthermore, one can easily extend this iteration to the case of convex metric spaces in the sense of Takahashi. Abbas, Khan and Rhoades suggested that a number of fixed point iteration problems can be solved using the above iteration, which is
much simpler than Ishikawa iteration. However, the simplification principle does not directly apply to the setting of our research since we are dealing with a generalized type of nonlinear contractions in contrast with the case when the operator TT is defined by it’s iterates, for example when TT is an asymptotically nonexpansive mapping. In this case it satisfies the following condition: there exists a sequence (kn)n[1,)\left(k_{n}\right)_{n\in\mathbb{N}}\subset[1,\infty), with n(kn1)<\sum_{n\in\mathbb{N}}\left(k_{n}-1\right)<\infty, such that

TnxTnyknxy.\left\|T^{n}x-T^{n}y\right\|\leq k_{n}\|x-y\|.

So, in the above case, the condition would act on the iterates of the operator TT, contrary to the case when TT is a nonlinear generalized contraction-type mapping. In our case, as we do not assume an asymptotic condition for the nonlinear operator of interest, especially from a computational point of view, the modified Krasnoselskij iteration process would be less cost-efficient than the well-known Ishikawa iteration. The cause of this relies on the fact that the modified Krasnoselskij iteration requires the computation of TnT^{n} in each step.

2 Main results

In this section the existence and uniqueness of fixed points of a generalized contraction mapping are established.

Because of the heavy computations, we work with contraction type mappings developed by Karapinar, i.e

cd(Tx,Ty)+a[d(x,Tx)+d(y,Ty)]kd(x,y),cd(Tx,Ty)+a[d(x,Tx)+d(y,Ty)]\leq kd(x,y),

although the same method can be applied to all the above nonlinear mappings.
Since Karapinar, Moosaei and the already mentioned authors have used Krasnoselskii iteration (when λ[0,1]\lambda\in[0,1] and on the particular case when λ=12\lambda=\frac{1}{2} ), we give a new method of proof through Ishikawa iteration.

We also give an example to illustrate the theorem we develop.
Theorem 1 Let ( X,dX,d ) be a complete convex metric space.
Let KK a nonempty, closed and convex subset of XX and T:KKT:K\rightarrow K be a map satisfying the following contractive condition

cd(Tx,Ty)+a[d(x,Tx)+d(y,Ty)]kd(x,y),cd(Tx,Ty)+a[d(x,Tx)+d(y,Ty)]\leq kd(x,y),

for each x,yKx,y\in K.
Let (αn)\left(\alpha_{n}\right) and (βn)\left(\beta_{n}\right) be two sequences in (0,1)(0,1) and let a1,a2,b1,b2[0,1]a_{1},a_{2},b_{1},b_{2}\in[0,1], such that αn[a1,a2][0,1)\alpha_{n}\in\left[a_{1},a_{2}\right]\subset[0,1) and βn[b1,b2][0,1)\beta_{n}\in\left[b_{1},b_{2}\right]\subset[0,1), for each nn\in\mathbb{N}.

Let:

ε(0,12),w1:={a1a2,a02a3(1a1),a<0,w2:={a(1+b2)1b2,a0a(1+b1)1b1,a<0\displaystyle\varepsilon\in\left(0,\frac{1}{2}\right),w_{1}=\left\{\begin{array}[]{ll}\frac{a}{1-a_{2}},&a\geq 0\\ \frac{2a}{3\left(1-a_{1}\right)},&a<0\end{array},w_{2}:=\begin{cases}\frac{a\left(1+b_{2}\right)}{1-b_{2}},&a\geq 0\\ \frac{a\left(1+b_{1}\right)}{1-b_{1}},&a<0\end{cases}\right.
w3:={12(a+c)|a|(1+b2)1b1,k>012(a+c)|a|(1+b2)1b2,k<0,τ:={13,a2<1213a21a2,a1>12,\displaystyle w_{3}=\left\{\begin{array}[]{ll}\frac{\frac{1}{2}(a+c)-|a|\left(1+b_{2}\right)}{1-b_{1}},&k>0\\ \frac{\frac{1}{2}(a+c)-|a|\left(1+b_{2}\right)}{1-b_{2}},&k<0\end{array},\tau:=\left\{\begin{array}[]{ll}\frac{1}{3},&a_{2}<\frac{1}{2}\\ \frac{1}{3}\cdot\frac{a_{2}}{1-a_{2}},&a_{1}>\frac{1}{2}\end{array},\right.\right.
s:={3(1a1)2(a+c),a0 and c0,a2(1a1)+cε(1+b2)(1a2)(1b2),a0 and c<0,aε(1+b2)(1a2)(1b2)+c2(1a1),a<0 and c0s:=\begin{cases}\frac{3\left(1-a_{1}\right)}{2(a+c)},&a\geq 0\text{ and }c\geq 0,\\ \frac{a}{2\left(1-a_{1}\right)}+\frac{c}{\varepsilon}\frac{\left(1+b_{2}\right)}{\left(1-a_{2}\right)\left(1-b_{2}\right)},&a\geq 0\text{ and }c<0,\\ \frac{a}{\varepsilon}\frac{\left(1+b_{2}\right)}{\left(1-a_{2}\right)\left(1-b_{2}\right)}+\frac{c}{2\left(1-a_{1}\right)},&a<0\text{ and }c\geq 0\end{cases}

Suppose the following conditions hold:
(1) c0c\neq 0,
(2) a+c>0a+c>0,
(3) kw1k\geq w_{1},
(4) kw2k\geq w_{2},
(5) k<w3k<w_{3},
(6) s>0s>0,

|c|τ<sk|a|ε(1+b2)(1a2)(1b2).|c|\tau<s-k-\frac{|a|}{\varepsilon}\frac{\left(1+b_{2}\right)}{\left(1-a_{2}\right)\left(1-b_{2}\right)}. (7)

Then, the operator TT admits a fixed point.
Moreover, if k<ck<c, then the fixed point is unique.
Proof For each nn\in\mathbb{N}, define the Ishikawa iteration

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

for αn\alpha_{n} and βn[0,1)\beta_{n}\in[0,1), for each nn\in\mathbb{N}.
From the contractive condition for the operator TT applied on the pair ( xn,xn1x_{n},x_{n-1} ), we obtain that

cd(Txn1,Txn)+ad(xn,Txn)+ad(xn1,Txn1)kd(xn1,xn)cd\left(Tx_{n-1},Tx_{n}\right)+ad\left(x_{n},Tx_{n}\right)+ad\left(x_{n-1},Tx_{n-1}\right)\leq kd\left(x_{n-1},x_{n}\right)

Moreover, from now on, let’s denote by DD the distance d(xn,Txn)d\left(x_{n},Tx_{n}\right), for each nn\in\mathbb{N} fixed, unless we specify it explicitly.

We divide our proof in the following steps, as follows:
(I) We make the following estimations:

d(xn,xn+1)=d(xn,W(xn,Tyn,αn))=(1αn)d(xn,Tyn)\displaystyle d\left(x_{n},x_{n+1}\right)=d\left(x_{n},W\left(x_{n},Ty_{n},\alpha_{n}\right)\right)=\left(1-\alpha_{n}\right)d\left(x_{n},Ty_{n}\right)
d(xn,yn)=d(xn,W(xn,Txn,βn))=(1βn)d(xn,Txn)=(1βn)D\displaystyle d\left(x_{n},y_{n}\right)=d\left(x_{n},W\left(x_{n},Tx_{n},\beta_{n}\right)\right)=\left(1-\beta_{n}\right)d\left(x_{n},Tx_{n}\right)=\left(1-\beta_{n}\right)D
d(Txn,yn)=d(Txn,W(xn,Txn,βn))=βnd(xn,Txn)=βnD\displaystyle d\left(Tx_{n},y_{n}\right)=d\left(Tx_{n},W\left(x_{n},Tx_{n},\beta_{n}\right)\right)=\beta_{n}d\left(x_{n},Tx_{n}\right)=\beta_{n}D

Also, we have that

d(Tyn,xn+1)=d(Tyn,W(xn,Tyn,αn))=αnd(xn,Tyn).d\left(Ty_{n},x_{n+1}\right)=d\left(Ty_{n},W\left(x_{n},Ty_{n},\alpha_{n}\right)\right)=\alpha_{n}d\left(x_{n},Ty_{n}\right).

So

d(xn,xn+1)=1αnαnd(Tyn,xn+1)d\left(x_{n},x_{n+1}\right)=\frac{1-\alpha_{n}}{\alpha_{n}}d\left(Ty_{n},x_{n+1}\right)

(II) Now, we estimate d(Txn,Tyn)d\left(Tx_{n},Ty_{n}\right) :

Case (A): a0,c0a\geq 0,c\geq 0 :

d(xn,xn+1)\displaystyle d\left(x_{n},x_{n+1}\right) =(1αn)d(xn,Tyn)(1αn)[d(xn,yn)+d(yn,Tyn)]\displaystyle=\left(1-\alpha_{n}\right)d\left(x_{n},Ty_{n}\right)\leq\left(1-\alpha_{n}\right)\left[d\left(x_{n},y_{n}\right)+d\left(y_{n},Ty_{n}\right)\right]
=(1αn)d(xn,yn)+(1αn)d(yn,Tyn)\displaystyle=\left(1-\alpha_{n}\right)d\left(x_{n},y_{n}\right)+\left(1-\alpha_{n}\right)d\left(y_{n},Ty_{n}\right)
=(1αn)(1βn)D+(1αn)d(yn,Tyn)\displaystyle=\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)D+\left(1-\alpha_{n}\right)d\left(y_{n},Ty_{n}\right)

By the definition of a convex metric space, we have that

d(yn,Tyn)\displaystyle d\left(y_{n},Ty_{n}\right) =d(W(xn,Txn,βn),Tyn)βnd(xn,Tyn)+(1βn)d(Txn,Tyn)\displaystyle=d\left(W\left(x_{n},Tx_{n},\beta_{n}\right),Ty_{n}\right)\leq\beta_{n}d\left(x_{n},Ty_{n}\right)+\left(1-\beta_{n}\right)d\left(Tx_{n},Ty_{n}\right)
=βn1αnd(xn,xn+1)+(1βn)d(Txn,Tyn)\displaystyle=\frac{\beta_{n}}{1-\alpha_{n}}d\left(x_{n},x_{n+1}\right)+\left(1-\beta_{n}\right)d\left(Tx_{n},Ty_{n}\right)

So

d(xn,xn+1)(1αn)(1βn)D+βnd(xn,xn+1)+(1αn)(1βn)d(Txn,Tyn)\displaystyle d\left(x_{n},x_{n+1}\right)\leq\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)\cdot D+\beta_{n}d\left(x_{n},x_{n+1}\right)+\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)d\left(Tx_{n},Ty_{n}\right)
d(xn,xn+1)(1αn)D+(1αn)d(Txn,Tyn).\displaystyle\Longrightarrow d\left(x_{n},x_{n+1}\right)\leq\left(1-\alpha_{n}\right)D+\left(1-\alpha_{n}\right)d\left(Tx_{n},Ty_{n}\right).

From the contraction-type condition for the pair (xn,yn)\left(x_{n},y_{n}\right), we obtain that

cd(Txn,Tyn)+ad(xn,Txn)+ad(yn,Tyn)kd(xn,yn)\displaystyle cd\left(Tx_{n},Ty_{n}\right)+ad\left(x_{n},Tx_{n}\right)+ad\left(y_{n},Ty_{n}\right)\leq kd\left(x_{n},y_{n}\right)
cd(Txn,Tyn)kd(xn,yn)ad(xn,Txn)ad(yn,Tyn)\displaystyle cd\left(Tx_{n},Ty_{n}\right)\leq kd\left(x_{n},y_{n}\right)-ad\left(x_{n},Tx_{n}\right)-ad\left(y_{n},Ty_{n}\right)
cd(Txn,Tyn)k(1βn)DaDad(yn,Tyn)\displaystyle cd\left(Tx_{n},Ty_{n}\right)\leq k\left(1-\beta_{n}\right)\cdot D-a\cdot D-ad\left(y_{n},Ty_{n}\right)
cd(Txn,Tyn)[k(1βn)a]Dad(yn,Tyn).\displaystyle cd\left(Tx_{n},Ty_{n}\right)\leq\left[k\left(1-\beta_{n}\right)-a\right]\cdot D-ad\left(y_{n},Ty_{n}\right).

Since c>0c>0, we obtain that

d(Txn,Tyn)[k(1βn)ac]D+(ac)d(yn,Tyn)d\left(Tx_{n},Ty_{n}\right)\leq\left[\frac{k\left(1-\beta_{n}\right)-a}{c}\right]\cdot D+\left(-\frac{a}{c}\right)d\left(y_{n},Ty_{n}\right)

Since a0a\geq 0 and c>0c>0, we have that

(ac)0 and k(1βn)ac0\left(-\frac{a}{c}\right)\leq 0\text{ and }\frac{k\left(1-\beta_{n}\right)-a}{c}\geq 0

because ka1βnk\geq\frac{a}{1-\beta_{n}} from the hypothesis.
We have that (ac)d(yn,Tyn)0\left(-\frac{a}{c}\right)d\left(y_{n},Ty_{n}\right)\leq 0, so, by triangle inequality, it follows that

d(yn,Tyn)d(Txn,Tyn)d(Txn,yn)d(Txn,Tyn)βnD.d\left(y_{n},Ty_{n}\right)\geq d\left(Tx_{n},Ty_{n}\right)-d\left(Tx_{n},y_{n}\right)\geq d\left(Tx_{n},Ty_{n}\right)-\beta_{n}\cdot D.

So

(ac)d(yn,Tyn)(ac)d(Txn,Tyn)+acβnD.\left(-\frac{a}{c}\right)d\left(y_{n},Ty_{n}\right)\leq\left(-\frac{a}{c}\right)d\left(Tx_{n},Ty_{n}\right)+\frac{a}{c}\beta_{n}\cdot D.

From the above relations, we get

d(Txn,Tyn)[k(1βn)ac]Dacd(Txn,Tyn)+acβnD.\displaystyle d\left(Tx_{n},Ty_{n}\right)\leq\left[\frac{k\left(1-\beta_{n}\right)-a}{c}\right]\cdot D-\frac{a}{c}d\left(Tx_{n},Ty_{n}\right)+\frac{a}{c}\beta_{n}\cdot D.
(1+ac)d(Txn,Tyn)[k(1βn)a+aβnc]D\displaystyle\left(1+\frac{a}{c}\right)d\left(Tx_{n},Ty_{n}\right)\leq\left[\frac{k\left(1-\beta_{n}\right)-a+a\beta_{n}}{c}\right]D
a+ccd(Txn,Tyn)[k(1βn)a(1βn)c]D.\displaystyle\Longrightarrow\frac{a+c}{c}d\left(Tx_{n},Ty_{n}\right)\leq\left[\frac{k\left(1-\beta_{n}\right)-a\left(1-\beta_{n}\right)}{c}\right]D.

Since c>0c>0, we have (a+c)d(Txn,Tyn)(1βn)(ka)D(a+c)d\left(Tx_{n},Ty_{n}\right)\leq\left(1-\beta_{n}\right)(k-a)\cdot D. Moreover, since a+c>0a+c>0 from the hypothesis, we obtain:

d(Txn,Tyn)(1βn)(ka)a+cD.d\left(Tx_{n},Ty_{n}\right)\leq\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\cdot D.

We observe that we have used the following conditions:

c>0,a+c>0,k>ac>0,a+c>0,k>a (3)

Now

d(xn,xn+1)(1αn)D+(1αn)d(Txn,Tyn)\displaystyle d\left(x_{n},x_{n+1}\right)\leq\left(1-\alpha_{n}\right)\cdot D+\left(1-\alpha_{n}\right)d\left(Tx_{n},Ty_{n}\right)\Longrightarrow
d(xn,xn+1)(1αn)[1+(1βn)(ka)a+c]D.\displaystyle d\left(x_{n},x_{n+1}\right)\leq\left(1-\alpha_{n}\right)\left[1+\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\right]D.

Here, we observe, that we need the same condition, i.e. k>ak>a. In a similar way, we estimate a lower bound for d(xn,xn+1)d\left(x_{n},x_{n+1}\right) :

We know that

d(xn,xn+1)\displaystyle d\left(x_{n},x_{n+1}\right) =(1αn)d(xn,Tyn)(1αn)[d(xn,yn)d(yn,Tyn)]\displaystyle=\left(1-\alpha_{n}\right)d\left(x_{n},Ty_{n}\right)\geq\left(1-\alpha_{n}\right)\left[d\left(x_{n},y_{n}\right)-d\left(y_{n},Ty_{n}\right)\right]
=(1αn)(1βn)D(1αn)d(yn,Tyn).\displaystyle=\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)\cdot D-\left(1-\alpha_{n}\right)d\left(y_{n},Ty_{n}\right).

But we have shown that

d(yn,Tyn)βn1αnd(xn,xn+1)+(1βn)d(Txn,Tyn)\displaystyle d\left(y_{n},Ty_{n}\right)\leq\frac{\beta_{n}}{1-\alpha_{n}}d\left(x_{n},x_{n+1}\right)+\left(1-\beta_{n}\right)d\left(Tx_{n},Ty_{n}\right)\Longrightarrow
(1αn)d(yn,Tyn)βnd(xn,xn+1)(1αn)(1βn)d(Txn,Tyn)\displaystyle-\left(1-\alpha_{n}\right)d\left(y_{n},Ty_{n}\right)\geq-\beta_{n}d\left(x_{n},x_{n+1}\right)-\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)d\left(Tx_{n},Ty_{n}\right)

So

d(xn,xn+1)(1αn)(1βn)Dβnd(xn,xn+1)(1αn)(1βn)d(Txn,Tyn)d\left(x_{n},x_{n+1}\right)\geq\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)D-\beta_{n}d\left(x_{n},x_{n+1}\right)-\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)d\left(Tx_{n},Ty_{n}\right)

Then

d(xn,xn+1)(1αn)(1βn)1+βnD(1αn)(1βn)1+βnd(Txn,Tyn).d\left(x_{n},x_{n+1}\right)\geq\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\cdot D-\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\cdot d\left(Tx_{n},Ty_{n}\right).

Since

d(Txn,Tyn)(1βn)(ka)a+cD, we get\displaystyle d\left(Tx_{n},Ty_{n}\right)\leq\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\cdot D,\text{ we get }
d(xn,xn+1)(1αn)(1βn)1+βn[1(1βn)(ka)a+c]D.\displaystyle d\left(x_{n},x_{n+1}\right)\geq\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\cdot\left[1-\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\right]\cdot D.

We observe that we have used the following condition, such that the right hand side is positive:

(1βn)(ka)a+c<1\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}<1 (4)

Case (D): a<0,c<0a<0,c<0 :
This case is not possible, since from the hypothesis we have that a+c>0a+c>0 and this leads to a contradiction.

Case (B): a<0,c>0a<0,c>0 :
We have that

d(yn,Tyn)d(Txn,Tyn)+d(yn,Txn)=d(Txn,Tyn)+βnD.d\left(y_{n},Ty_{n}\right)\leq d\left(Tx_{n},Ty_{n}\right)+d\left(y_{n},Tx_{n}\right)=d\left(Tx_{n},Ty_{n}\right)+\beta_{n}\cdot D.

We have shown that

cd(Txn,Tyn)[k(1βn)a]Dad(yn,Tyn).cd\left(Tx_{n},Ty_{n}\right)\leq\left[k\left(1-\beta_{n}\right)-a\right]\cdot D-ad\left(y_{n},Ty_{n}\right).

Since a<0a<0, this means that a>0-a>0, we have

ad(yn,Tyn)ad(Txn,Tyn)aβnD.-ad\left(y_{n},Ty_{n}\right)\leq-ad\left(Tx_{n},Ty_{n}\right)-a\beta_{n}D.

So

cd(Txn,Tyn)[k(1βn)a]Dad(Txn,Tyn)aβnDcd\left(Tx_{n},Ty_{n}\right)\leq\left[k\left(1-\beta_{n}\right)-a\right]D-ad\left(Tx_{n},Ty_{n}\right)-a\beta_{n}D\Longrightarrow
(a+c)d(Txn,Tyn)[k(1βn)a(1+βn)]D(a+c)d\left(Tx_{n},Ty_{n}\right)\leq\left[k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)\right]D

Since c>0c>0 and a+c>0a+c>0, it follows that

d(Txn,Tyn)[k(1βn)a(1+βn)a+c]Dd\left(Tx_{n},Ty_{n}\right)\leq\left[\frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}\right]D

Now, we have used the following condition:

k>a(1+βn1βn)k>a\left(\frac{1+\beta_{n}}{1-\beta_{n}}\right) (5)

Also, we know that

d(xn,xn+1)(1αn)D+(1αn)d(Txn,Tyn)d\left(x_{n},x_{n+1}\right)\leq\left(1-\alpha_{n}\right)D+\left(1-\alpha_{n}\right)d\left(Tx_{n},Ty_{n}\right)

So

d(xn,xn+1)(1αn)[1+k(1βn)a(1+βn)a+c]D.d\left(x_{n},x_{n+1}\right)\leq\left(1-\alpha_{n}\right)\cdot\left[1+\frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}\right]D.

We recall that this estimation does not depend on the sign of the coefficients aa and cc. Also, we notice that we have k(1βn)a(1+βn)a+c>0\frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}>0, which is the condition imposed above. Now, for a lower bound of d(xn,xn+1)d\left(x_{n},x_{n+1}\right), we proceed in a similar way:

We have shown that

d(Txn,Tyn)[k(1βn)a(1+βn)a+c]D.d\left(Tx_{n},Ty_{n}\right)\leq\left[\frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}\right]D.

Moreover, we know that

d(xn,xn+1)(1αn)(1βn)1+βnD(1αn)(1βn)1+βnd(Txn,Tyn)d\left(x_{n},x_{n+1}\right)\geq\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}D-\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}d\left(Tx_{n},Ty_{n}\right)

We mention that this estimation does not depend on the coefficients from the contractivecondition. From these two relations, we obtain that

d(xn,xn+1)(1αn)(1βn)1+βn[1k(1βn)a(1+βn)a+c]D.d\left(x_{n},x_{n+1}\right)\geq\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\left[1-\frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}\right]D.

We observe that we have used the condition:

k(1βn)a(1+βn)a+c<1\frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}<1 (6)

Case (C): a0,c<0a\geq 0,c<0 :
From the contraction condition, we have shown that

cd(Txn,Tyn)[k(1βn)a]Dad(yn,Tyn)cd\left(Tx_{n},Ty_{n}\right)\leq\left[k\left(1-\beta_{n}\right)-a\right]D-ad\left(y_{n},Ty_{n}\right)

Since c<0c<0, we get

d(Txn,Tyn)[k(1βn)ac]D+(ac)d(yn,Tyn)d\left(Tx_{n},Ty_{n}\right)\geq\left[\frac{k\left(1-\beta_{n}\right)-a}{c}\right]D+\left(-\frac{a}{c}\right)d\left(y_{n},Ty_{n}\right)

Now, we estimate the following:

d(yn,Tyn)d(Txn,Tyn)d(Txn,yn)=d(Txn,Tyn)βnDd\left(y_{n},Ty_{n}\right)\geq d\left(Tx_{n},Ty_{n}\right)-d\left(Tx_{n},y_{n}\right)=d\left(Tx_{n},Ty_{n}\right)-\beta_{n}D\Longrightarrow

Because ac0-\frac{a}{c}\geq 0, we have that

d(Txn,Tyn)[k(1βn)ac]Dacd(Txn,Tyn)+acβnD\displaystyle d\left(Tx_{n},Ty_{n}\right)\geq\left[\frac{k\left(1-\beta_{n}\right)-a}{c}\right]D-\frac{a}{c}d\left(Tx_{n},Ty_{n}\right)+\frac{a}{c}\beta_{n}D\Longrightarrow
(1+ac)d(Txn,Tyn)[k(1βn)ac+acβn]D.\displaystyle\left(1+\frac{a}{c}\right)d\left(Tx_{n},Ty_{n}\right)\geq\left[\frac{k\left(1-\beta_{n}\right)-a}{c}+\frac{a}{c}\beta_{n}\right]D.

This means that

(a+cc)d(Txn,Tyn)[k(1βn)ac+acβn]D.\left(\frac{a+c}{c}\right)d\left(Tx_{n},Ty_{n}\right)\geq\left[\frac{k\left(1-\beta_{n}\right)-a}{c}+\frac{a}{c}\beta_{n}\right]D.

Whereas c<0c<0 the right hand side is negative and the left hand side is also negative, because c<0c<0 and a+c>0a+c>0. Moreover, we notice that k(1βn)a0k\left(1-\beta_{n}\right)-a\geq 0, which means that

k(1βn)a.k\left(1-\beta_{n}\right)\geq a.

What with cc cannot be 0 , we divide by cc and obtain:
(a+c)d(Txn,Tyn)[k(1βn)a+aβn]Dd(Txn,Tyn)[(1βn)(ka)a+c]D(a+c)d\left(Tx_{n},Ty_{n}\right)\leq\left[k\left(1-\beta_{n}\right)-a+a\beta_{n}\right]D\Longrightarrow d\left(Tx_{n},Ty_{n}\right)\leq\left[\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\right]D.
Furthermore, we notice that we used the following condition:

k>ak>a (7)

such that the right hand side is strict positive.
The above inequalities imply that

d(xn,xn+1)\displaystyle d\left(x_{n},x_{n+1}\right) (1αn)D+(1αn)d(Txn,Tyn)\displaystyle\leq\left(1-\alpha_{n}\right)D+\left(1-\alpha_{n}\right)d\left(Tx_{n},Ty_{n}\right)\Longrightarrow
d(xn,xn+1)\displaystyle d\left(x_{n},x_{n+1}\right) (1αn)[1+(1βn)(ka)a+c]D.\displaystyle\leq\left(1-\alpha_{n}\right)\left[1+\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\right]D.

Also, we observe that the condition k>ak>a is satisfied by hypothesis assumptions. In a similar way, for a lower bound of d(xn,xn+1)d\left(x_{n},x_{n+1}\right), we have shown that

d(xn,xn+1)(1αn)(1βn)1+βn[Dd(Txn,Tyn)]d\left(x_{n},x_{n+1}\right)\geq\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\left[D-d\left(Tx_{n},Ty_{n}\right)\right]

This means that

d(xn,xn+1)(1αn)(1βn)1+βn[1(1βn)(ka)a+c]D.d\left(x_{n},x_{n+1}\right)\geq\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\left[1-\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\right]D.

Given that the right hand side must pe positive, we use the following condition

(1βn)(ka)a+c<1\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}<1 (8)

(III) Now we find θn\theta_{n}, such that d(xn,xn+1)θnd(xn,Txn)d\left(x_{n},x_{n+1}\right)\cdot\theta_{n}\leq d\left(x_{n},Tx_{n}\right) :

We have a major bound for d(xn,xn+1):d(xn,xn+1)ηd(xn,Txn)d\left(x_{n},x_{n+1}\right):d\left(x_{n},x_{n+1}\right)\leq\eta^{\prime}d\left(x_{n},Tx_{n}\right). This means that

d(xn,Txn)1ηd(xn,xn+1)θnd(xn,xn+1)d\left(x_{n},Tx_{n}\right)\geq\frac{1}{\eta^{\prime}}d\left(x_{n},x_{n+1}\right)\geq\theta_{n}d\left(x_{n},x_{n+1}\right)

Let us define

η={η1, in the case (A)η2, in the case (B)η3, in the case (C)\eta^{\prime}=\left\{\begin{array}[]{l}\eta_{1},\text{ in the case }(A)\\ \eta_{2},\text{ in the case }(B)\\ \eta_{3},\text{ in the case }(C)\end{array}\right.

where

η1=(1αn)[1+(1βn)(ka)a+c]=(1αn)[1+k(1βn)a+c+a(1βn)a+c],\displaystyle\eta_{1}=\left(1-\alpha_{n}\right)\left[1+\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\right]=\left(1-\alpha_{n}\right)\left[1+\frac{k\left(1-\beta_{n}\right)}{a+c}+\frac{-a\left(1-\beta_{n}\right)}{a+c}\right],
η2=(1αn)[1+k(1βn)a+c+a(1+βn)a+c] and η3 is equal to η1.\displaystyle\eta_{2}=\left(1-\alpha_{n}\right)\left[1+\frac{k\left(1-\beta_{n}\right)}{a+c}+\frac{-a\left(1+\beta_{n}\right)}{a+c}\right]\text{ and }\eta_{3}\text{ is equal to }\eta_{1}.

We observe that η1,η2,η3\eta_{1},\eta_{2},\eta_{3} and η\eta^{\prime} depend on the index nn\in\mathbb{N}. Moreover, we observe that :
For the cases (A)(A) and (C)(C), since a+c>0a+c>0 and a0a\geq 0, we have that

a(1βn)a+c0a(1+βn)a+c=|a|1+βna+c.\frac{-a\left(1-\beta_{n}\right)}{a+c}\leq 0\leq\frac{a\left(1+\beta_{n}\right)}{a+c}=|a|\frac{1+\beta_{n}}{a+c}.

Also, for the case ( BB ), since a+c>0a+c>0 and a<0a<0, we get

a(1+βn)a+c=|a|1+βna+c.\frac{-a\left(1+\beta_{n}\right)}{a+c}=|a|\frac{1+\beta_{n}}{a+c}.

So 0<ηη0<\eta^{\prime}\leq\eta, where

η:=(1αn)[1+k(1βn)a+c+|a|1+βna+c].\eta:=\left(1-\alpha_{n}\right)\left[1+\frac{k\left(1-\beta_{n}\right)}{a+c}+|a|\frac{1+\beta_{n}}{a+c}\right].

Simplifying, it follows that

η=1αna+c[k(1βn)+|a|(1+βn)+(a+c)].\eta=\frac{1-\alpha_{n}}{a+c}\left[k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)+(a+c)\right].

Moreover

1η1η\frac{1}{\eta}\leq\frac{1}{\eta^{\prime}}

So

d(xn,xn+1)ηd(xn,Txn)ηd(xn,Txn)\displaystyle d\left(x_{n},x_{n+1}\right)\leq\eta^{\prime}d\left(x_{n},Tx_{n}\right)\leq\eta d\left(x_{n},Tx_{n}\right)\Longrightarrow
d(xn,Txn)θnd(xn,xn+1)\displaystyle d\left(x_{n},Tx_{n}\right)\geq\theta_{n}d\left(x_{n},x_{n+1}\right)
where θn=a+c1αn1k(1βn)+|a|(1+βn)+(a+c)=1η\displaystyle\text{ where }\theta_{n}=\frac{a+c}{1-\alpha_{n}}\cdot\frac{1}{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)+(a+c)}=\frac{1}{\eta}
and η depends on the index n.\displaystyle\text{ and }\eta\text{ depends on the index }n\in\mathbb{N}\text{. }

(IV) Now we find θn\theta_{n}^{\prime}, such that d(xn,Txn)θnd(xn,xn+1)d\left(x_{n},Tx_{n}\right)\leq\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right) :

We have that

d(xn,xn+1)μd(xn,Txn)d(xn,Txn)1μd(xn,xn+1)θnd(xn,xn+1)d\left(x_{n},x_{n+1}\right)\geq\mu^{\prime}d\left(x_{n},Tx_{n}\right)\Longrightarrow d\left(x_{n},Tx_{n}\right)\leq\frac{1}{\mu^{\prime}}d\left(x_{n},x_{n+1}\right)\leq\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)

We denote by:

μ={μ1, in the case (A),μ2, in the case (B),μ3, in the case (C)\mu^{\prime}=\left\{\begin{array}[]{l}\mu_{1},\text{ in the case }(A),\\ \mu_{2},\text{ in the case }(B),\\ \mu_{3},\text{ in the case }(C)\end{array}\right.

where

μ1=(1αn)(1βn)1+βn[1(1βn)(ka)a+c],\mu_{1}=\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\left[1-\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\right],
μ2=(1αn)(1βn)1+βn[1k(1βn)a(1+βn)a+c] and\displaystyle\mu_{2}=\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\left[1-\frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}\right]\text{ and }
μ3 is equal to μ1\displaystyle\mu_{3}\text{ is equal to }\mu_{1}

Moreover, we observe that

μ3=μ1=(1αn)(1βn)1+βn[1+k(1βn)a+c+a(1βn)a+c].\mu_{3}=\mu_{1}=\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\left[1+\frac{-k\left(1-\beta_{n}\right)}{a+c}+\frac{a\left(1-\beta_{n}\right)}{a+c}\right].

For the cases (A)(A) and (C)(C), we have that

a(1βn)=|a|(1βn)|a|(1βn)>|a|(1+βn)a\left(1-\beta_{n}\right)=|a|\left(1-\beta_{n}\right)\geq-|a|\left(1-\beta_{n}\right)>-|a|\left(1+\beta_{n}\right)

, since 1βn<1+βn1-\beta_{n}<1+\beta_{n}.
Also, for the case ( BB ), we have

a(1+βn)=|a|(1+βn)a\left(1+\beta_{n}\right)=-|a|\left(1+\beta_{n}\right)

From all the inequalities presented above, we derive

μμ\mu^{\prime}\geq\mu

where

μ:=(1αn)(1βn)1+βn[1k(1βn)a+c|a|(1+βn)a+c].\mu:=\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\left[1-\frac{k\left(1-\beta_{n}\right)}{a+c}-\frac{|a|\left(1+\beta_{n}\right)}{a+c}\right].

So

d(xn,xn+1)\displaystyle d\left(x_{n},x_{n+1}\right) μd(xn,Txn)μd(xn,Txn)\displaystyle\geq\mu^{\prime}d\left(x_{n},Tx_{n}\right)\geq\mu d\left(x_{n},Tx_{n}\right)\Longrightarrow
d(xn,Txn)\displaystyle d\left(x_{n},Tx_{n}\right) 1μd(xn,xn+1)=θnd(xn,xn+1)\displaystyle\leq\frac{1}{\mu}d\left(x_{n},x_{n+1}\right)=\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)

where θn=1μ\theta_{n}^{\prime}=\frac{1}{\mu}, where μ\mu depends on the index nn\in\mathbb{N}.
If μ>0\mu>0, then μ>0\mu^{\prime}>0, where μ\mu^{\prime} is μ1,μ2,μ3>0\mu_{1},\mu_{2},\mu_{3}>0, depending on the cases (A),(B)(A),(B) and (C). For μ>0\mu>0, we have that

k(1βn)+|a|(1+βn)a+c<1.\frac{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)}{a+c}<1.

So, we impose the following condition:

k(1βn)+|a|(1+βn)<(a+c)k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)<(a+c) (9)

Now, we have that

θnd(xn,xn+1)d(xn,Txn)θnd(xn,xn+1)\theta_{n}d\left(x_{n},x_{n+1}\right)\leq d\left(x_{n},Tx_{n}\right)\leq\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)

If there exists nn\in\mathbb{N} such that xn=xn+1:=xx_{n}=x_{n+1}:=x^{*}, then, by the inequality above d(x,Tx)=d\left(x^{*},Tx^{*}\right)= 0 , so the sequence (xn)\left(x_{n}\right) is convergent to the fixed point xx^{*}.

So, let’s suppose that xnxn+1x_{n}\neq x_{n+1}. On the other hand, because μ>0\mu>0 and η>0η>0\eta^{\prime}>0\Longrightarrow\eta>0, we have that

θn,θn>0, which implies θnθn.\theta_{n},\theta_{n}^{\prime}>0,\text{ which implies }\theta_{n}\leq\theta_{n}^{\prime}.

(V) At this step, we can show that θn<11αn\theta_{n}<\frac{1}{1-\alpha_{n}}, for each nn\in\mathbb{N} :

We recall the following inequalities

For the case (A):

d(Txn1,Tyn1)(1βn1)(ka)a+cd(xn1,Txn1)d\left(Tx_{n-1},Ty_{n-1}\right)\leq\frac{\left(1-\beta_{n-1}\right)(k-a)}{a+c}d\left(x_{n-1},Tx_{n-1}\right)

For the case ( BB ):

d(Txn1,Tyn1)k(1βn1)a(1+βn1)a+cd(xn1,Txn1) and d\left(Tx_{n-1},Ty_{n-1}\right)\leq\frac{k\left(1-\beta_{n-1}\right)-a\left(1+\beta_{n-1}\right)}{a+c}d\left(x_{n-1},Tx_{n-1}\right)\text{ and }

For the case (C)(C) :

d(Txn1,Tyn1)(1βn1)(ka)a+cd(xn1,Txn1)d\left(Tx_{n-1},Ty_{n-1}\right)\leq\frac{\left(1-\beta_{n-1}\right)(k-a)}{a+c}d\left(x_{n-1},Tx_{n-1}\right)

Moreover, for all three cases we have the same conditions for positivity. Also, for (A),(B)(A),(B) and (C), we get

d(Txn1,Tyn1)[k(1βn1)a+c+|a|(1+βn1)a+c]D, because d\left(Tx_{n-1},Ty_{n-1}\right)\leq\left[\frac{k\left(1-\beta_{n-1}\right)}{a+c}+\frac{|a|\left(1+\beta_{n-1}\right)}{a+c}\right]D,\text{ because }

In the cases (A)(A) and (C)(C), when a0a\geq 0, we have

a(1βn1)a(1βn1)a(1+βn1)=|a|(1+βn1)-a\left(1-\beta_{n-1}\right)\leq a\left(1-\beta_{n-1}\right)\leq a\left(1+\beta_{n-1}\right)=|a|\left(1+\beta_{n-1}\right)

and for the case (B)(B), when a<0a<0, we have

a(1+βn1)=|a|(1+βn1)-a\left(1+\beta_{n-1}\right)=|a|\left(1+\beta_{n-1}\right)

At the same time

η=1αna+c[k(1βn)+|a|(1+βn)+(a+c)]\eta=\frac{1-\alpha_{n}}{a+c}\left[k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)+(a+c)\right]

We can denote by ηn\eta_{n} the right hand side since η\eta depends on the index nn\in\mathbb{N}. So

k(1βn)+|a|(1+βn)a+c+1=ηn1αn=η1αn, for each n\frac{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)}{a+c}+1=\frac{\eta_{n}}{1-\alpha_{n}}=\frac{\eta}{1-\alpha_{n}},\text{ for each }n\in\mathbb{N}

This means that

k(1βn)+|a|(1+βn)a+c=η1αn1.\frac{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)}{a+c}=\frac{\eta}{1-\alpha_{n}}-1.

It implies that

d(Txn1,Tyn1)[η1αn11]d(xn1,Txn1)=[1θn1(1αn1)1]d(xn1,Txn1)d\left(Tx_{n-1},Ty_{n-1}\right)\leq\left[\frac{\eta}{1-\alpha_{n-1}}-1\right]d\left(x_{n-1},Tx_{n-1}\right)=\left[\frac{1}{\theta_{n-1}\left(1-\alpha_{n-1}\right)}-1\right]d\left(x_{n-1},Tx_{n-1}\right)

From the condition of positivity from above, the right hand side is positive, since the upper bound for d(Txn,Tyn)d\left(Tx_{n},Ty_{n}\right) is positive, we deduce that

1θn1(1αn1)>1, i.e. θn<11αn, for each n\frac{1}{\theta_{n-1}\left(1-\alpha_{n-1}\right)}>1\text{, i.e. }\theta_{n}<\frac{1}{1-\alpha_{n}}\text{, for each }n\in\mathbb{N}\text{. }

(VI) Now we evaluate d(Txn1,Txn)d\left(Tx_{n-1},Tx_{n}\right) :

  • The case when c0c\geq 0 :

We have that

d(xn,Txn)d(xn,Txn1)+d(Txn1,Txn)d\left(x_{n},Tx_{n}\right)\leq d\left(x_{n},Tx_{n-1}\right)+d\left(Tx_{n-1},Tx_{n}\right)

So

d(Txn1,Txn)\displaystyle d\left(Tx_{n-1},Tx_{n}\right) d(xn,Txn)d(xn,Txn1)d(Txn1,Txn)\displaystyle\geq d\left(x_{n},Tx_{n}\right)-d\left(x_{n},Tx_{n-1}\right)\Longrightarrow d\left(Tx_{n-1},Tx_{n}\right)
θnd(xn,xn+1)d(xn,Txn1)\displaystyle\geq\theta_{n}d\left(x_{n},x_{n+1}\right)-d\left(x_{n},Tx_{n-1}\right)

Moreover, we know that

d(xn,Txn1)=\displaystyle d\left(x_{n},Tx_{n-1}\right)= d(W(xn1,Tyn1,αn1),Txn1)\displaystyle d\left(W\left(x_{n-1},Ty_{n-1},\alpha_{n-1}\right),Tx_{n-1}\right)
αn1d(xn1,Txn1)+(1αn1)d(Tyn1,Txn1)\displaystyle\leq\alpha_{n-1}d\left(x_{n-1},Tx_{n-1}\right)+\left(1-\alpha_{n-1}\right)d\left(Ty_{n-1},Tx_{n-1}\right)
αn1d(xn1,Txn1)+(1αn1)[1θn1(1αn1)1]d(xn1,Txn1)\displaystyle\leq\alpha_{n-1}d\left(x_{n-1},Tx_{n-1}\right)+\left(1-\alpha_{n-1}\right)\left[\frac{1}{\theta_{n-1}\left(1-\alpha_{n-1}\right)}-1\right]d\left(x_{n-1},Tx_{n-1}\right)
=[αn1+1θn11+αn1]d(xn1,Txn1)\displaystyle=\left[\alpha_{n-1}+\frac{1}{\theta_{n-1}}-1+\alpha_{n-1}\right]d\left(x_{n-1},Tx_{n-1}\right)

So

d(xn,Txn1)[2αn11+1θn1]d(xn1,Txn1).d\left(x_{n},Tx_{n-1}\right)\leq\left[2\alpha_{n-1}-1+\frac{1}{\theta_{n-1}}\right]d\left(x_{n-1},Tx_{n-1}\right).

We show that the condition 2αn11+1θn102\alpha_{n-1}-1+\frac{1}{\theta_{n-1}}\geq 0 is true for each nn\in\mathbb{N}, so the right hand side is positive:

If

αn<12, i.e. 12αn>0,\alpha_{n}<\frac{1}{2},\text{ i.e. }1-2\alpha_{n}>0,

then

θn<112αn, since θn<11αn<112αn.\theta_{n}<\frac{1}{1-2\alpha_{n}},\text{ since }\theta_{n}<\frac{1}{1-\alpha_{n}}<\frac{1}{1-2\alpha_{n}}.

Moreover, if

αn>12, i.e. 12αn<0,\alpha_{n}>\frac{1}{2},\text{ i.e. }1-2\alpha_{n}<0,

then

12αn<0<1θn.1-2\alpha_{n}<0<\frac{1}{\theta_{n}}.

So

d(Txn1,Txn)θnd(xn,xn+1)[2αn11+1θn1]d(xn1,Txn1).d\left(Tx_{n-1},Tx_{n}\right)\geq\theta_{n}d\left(x_{n},x_{n+1}\right)-\left[2\alpha_{n-1}-1+\frac{1}{\theta_{n-1}}\right]d\left(x_{n-1},Tx_{n-1}\right).

But

d(xn1,Txn1)d(xn1,xn)+d(xn,Txn1)=d(xn1,xn)+d(W(xn1,Tyn1,αn1),Txn1)\displaystyle d\left(x_{n-1},Tx_{n-1}\right)\leq d\left(x_{n-1},x_{n}\right)+d\left(x_{n},Tx_{n-1}\right)=d\left(x_{n-1},x_{n}\right)+d\left(W\left(x_{n-1},Ty_{n-1},\alpha_{n-1}\right),Tx_{n-1}\right)
d(xn1,xn)+αn1d(xn1,Txn1)+(1αn1)d(Tyn1,Txn1)\displaystyle\quad\leq d\left(x_{n-1},x_{n}\right)+\alpha_{n-1}d\left(x_{n-1},Tx_{n-1}\right)+\left(1-\alpha_{n-1}\right)d\left(Ty_{n-1},Tx_{n-1}\right)
d(xn1,xn)+αn1d(xn1,Txn1)+(1αn1)[1θn1(1αn1)1]d(xn1,Txn1)\displaystyle\quad\leq d\left(x_{n-1},x_{n}\right)+\alpha_{n-1}d\left(x_{n-1},Tx_{n-1}\right)+\left(1-\alpha_{n-1}\right)\left[\frac{1}{\theta_{n-1}\left(1-\alpha_{n-1}\right)}-1\right]d\left(x_{n-1},Tx_{n-1}\right)

This implies that

(1αn1)d(xn1,Txn1)d(xn1,xn)+(1αn1)[1θn1(1αn1)1]\displaystyle\left(1-\alpha_{n-1}\right)d\left(x_{n-1},Tx_{n-1}\right)\leq d\left(x_{n-1},x_{n}\right)+\left(1-\alpha_{n-1}\right)\left[\frac{1}{\theta_{n-1}\left(1-\alpha_{n-1}\right)}-1\right]
d(xn1,Txn1)\displaystyle d\left(x_{n-1},Tx_{n-1}\right)\Longrightarrow
(1αn1)[11θn1(1αn1)+1]d(xn1,Txn1)d(xn1,xn)\displaystyle\left(1-\alpha_{n-1}\right)\left[1-\frac{1}{\theta_{n-1}\left(1-\alpha_{n-1}\right)}+1\right]d\left(x_{n-1},Tx_{n-1}\right)\leq d\left(x_{n-1},x_{n}\right)\Longrightarrow
[2(1αn1)1θn1]d(xn1,Txn1)d(xn1,xn), with θn>12(1αn)\displaystyle{\left[2\left(1-\alpha_{n-1}\right)-\frac{1}{\theta_{n-1}}\right]d\left(x_{n-1},Tx_{n-1}\right)\leq d\left(x_{n-1},x_{n}\right),\text{ with }\theta_{n}>\frac{1}{2\left(1-\alpha_{n}\right)}}

This means that
d(xn1,Txn1)12(1αn1)1θn1d(xn1,xn)=θn12θn1(1αn1)1d(xn1,xn)d\left(x_{n-1},Tx_{n-1}\right)\leq\frac{1}{2\left(1-\alpha_{n-1}\right)-\frac{1}{\theta_{n-1}}}d\left(x_{n-1},x_{n}\right)=\frac{\theta_{n-1}}{2\theta_{n-1}\left(1-\alpha_{n-1}\right)-1}d\left(x_{n-1},x_{n}\right).
We notice that the right hand side is positive, so

2θn1(1αn1)1>0, i.e. θn12(1αn).2\theta_{n-1}\left(1-\alpha_{n-1}\right)-1>0,\text{ i.e. }\theta_{n}\geq\frac{1}{2\left(1-\alpha_{n}\right)}.

We imposed the condition that:

θn>12(1αn), for each n\theta_{n}>\frac{1}{2\left(1-\alpha_{n}\right)},\text{ for each }\mathrm{n}\in\mathbb{N} (10)

This does not contradict the fact that

θn11αn, since αn<1, for each n.\theta_{n}\leq\frac{1}{1-\alpha_{n}},\text{ since }\alpha_{n}<1,\text{ for each }n\in\mathbb{N}.

It implies that

d(Txn1,Txn)θnd(xn,xn+1)[2αn11+1θn1][θn12θn1(1αn1)1]d(xn1,xn)d\left(Tx_{n-1},Tx_{n}\right)\geq\theta_{n}d\left(x_{n},x_{n+1}\right)-\left[2\alpha_{n-1}-1+\frac{1}{\theta_{n-1}}\right]\cdot\left[\frac{\theta_{n-1}}{2\theta_{n-1}\left(1-\alpha_{n-1}\right)-1}\right]d\left(x_{n-1},x_{n}\right)

Now, let’s denote by

τn:=[2αn1+1θn][θn2θn(1αn)1]=θn(2αn1)+12θn(1αn)1.\tau_{n}:=\left[2\alpha_{n}-1+\frac{1}{\theta_{n}}\right]\cdot\left[\frac{\theta_{n}}{2\theta_{n}\left(1-\alpha_{n}\right)-1}\right]=\frac{\theta_{n}\left(2\alpha_{n}-1\right)+1}{2\theta_{n}\left(1-\alpha_{n}\right)-1}.

Then

d(Txn1,Txn)θnd(xn,xn+1)τn1d(xn1,xn)\displaystyle d\left(Tx_{n-1},Tx_{n}\right)\geq\theta_{n}d\left(x_{n},x_{n+1}\right)-\tau_{n-1}d\left(x_{n-1},x_{n}\right)\Longrightarrow
cd(Txn1,Txn)cθnd(xn,xn+1)cτn1d(xn1,xn).\displaystyle cd\left(Tx_{n-1},Tx_{n}\right)\geq c\theta_{n}d\left(x_{n},x_{n+1}\right)-c\tau_{n-1}d\left(x_{n-1},x_{n}\right).

From the definition of τn\tau_{n}, we have that τn>0\tau_{n}>0, for each nn\in\mathbb{N}. Also, we have seen that the inequality θn<11αn\theta_{n}<\frac{1}{1-\alpha_{n}} holds. The condition that 12(1αn)<θn\frac{1}{2\left(1-\alpha_{n}\right)}<\theta_{n} can be written as follows

12<a+ck(1βn)+|a|(1+βn)+(a+c)\displaystyle\frac{1}{2}<\frac{a+c}{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)+(a+c)}\Longrightarrow
k(1βn)+|a|(1+βn)+(a+c)<2(a+c), i.e.\displaystyle k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)+(a+c)<2(a+c),\text{ i.e. }
k(1βn)<(a+c)|a|(1+βn).\displaystyle k\left(1-\beta_{n}\right)<(a+c)-|a|\left(1+\beta_{n}\right).

So, we impose the condition that:

k(1βn)+|a|(1+βn)a+c<1\frac{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)}{a+c}<1 (11)
  • The case when c<0c<0 :

We have

d(Txn1,Txn)d(xn,Txn)+d(xn,Txn1).d\left(Tx_{n-1},Tx_{n}\right)\leq d\left(x_{n},Tx_{n}\right)+d\left(x_{n},Tx_{n-1}\right).

Moreover, we have shown that

d(xn,Txn1)[2αn11+1θn1]d(xn1,Txn1)d\left(x_{n},Tx_{n-1}\right)\leq\left[2\alpha_{n-1}-1+\frac{1}{\theta_{n-1}}\right]d\left(x_{n-1},Tx_{n-1}\right)

Also, we mention that this inequality is universal, i.e. it does not depend on the cases defined by the condition of positivity of aa and cc.

So

d(Txn1,Txn)d(xn,Txn)+[2αn11+1θn1]d(xn1,Txn1)d\left(Tx_{n-1},Tx_{n}\right)\leq d\left(x_{n},Tx_{n}\right)+\left[2\alpha_{n-1}-1+\frac{1}{\theta_{n-1}}\right]d\left(x_{n-1},Tx_{n-1}\right)

But

d(xn1,Txn1)θn12θn1(1αn1)1d(xn1,xn)\displaystyle d\left(x_{n-1},Tx_{n-1}\right)\leq\frac{\theta_{n-1}}{2\theta_{n-1}\left(1-\alpha_{n-1}\right)-1}d\left(x_{n-1},x_{n}\right)\Longrightarrow
d(Txn1,Txn)θnd(xn,xn+1)+τn1d(xn1,xn)\displaystyle d\left(Tx_{n-1},Tx_{n}\right)\leq\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)+\tau_{n-1}d\left(x_{n-1},x_{n}\right)\Longrightarrow
cd(Txn1,Txn)cθnd(xn,xn+1)+cτn1d(xn1,xn)\displaystyle cd\left(Tx_{n-1},Tx_{n}\right)\geq c\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)+c\tau_{n-1}d\left(x_{n-1},x_{n}\right)
  • Now, for these two cases we have that:

If c0c\geq 0, then

cd(Txn1,Txn)cθnd(xn,xn+1)|c|τn1d(xn1,xn)cd\left(Tx_{n-1},Tx_{n}\right)\geq c\theta_{n}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right)

If c<0c<0, then

cd(Txn1,Txn)cθnd(xn,xn+1)|c|τn1d(xn1,xn)cd\left(Tx_{n-1},Tx_{n}\right)\geq c\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right)

(VII) Now we analyze cases (A),(B)(A),(B) and (C)(C) on the contraction-type condition:

  • Case (A)(A), when a0a\geq 0 and c>0c>0 :

We get that

ad(xn,Txn)aθnd(xn,xn+1)\displaystyle ad\left(x_{n},Tx_{n}\right)\geq a\theta_{n}d\left(x_{n},x_{n+1}\right)
ad(xn1,Txn1)aθn1d(xn1,xn) and\displaystyle ad\left(x_{n-1},Tx_{n-1}\right)\geq a\theta_{n-1}d\left(x_{n-1},x_{n}\right)\text{ and }
cd(Txn1,Txn)cθnd(xn,xn+1)|c|τn1d(xn1,xn)\displaystyle cd\left(Tx_{n-1},Tx_{n}\right)\geq c\theta_{n}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right)

By the contraction-type condition on the pair ( xn1,xnx_{n-1},x_{n} ), it follows that

aθnd(xn,xn+1)+aθn1d(xn1,xn)+cθnd(xn,xn+1)|c|τn1d(xn1,xn)\displaystyle a\theta_{n}d\left(x_{n},x_{n+1}\right)+a\theta_{n-1}d\left(x_{n-1},x_{n}\right)+c\theta_{n}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right)
ad(xn,Txn)+ad(xn1,Txn1)+cd(Txn1,Txn)kd(xn1,xn)\displaystyle\leq ad\left(x_{n},Tx_{n}\right)+ad\left(x_{n-1},Tx_{n-1}\right)+cd\left(Tx_{n-1},Tx_{n}\right)\leq kd\left(x_{n-1},x_{n}\right)

So

d(xn,xn+1)kaθn1+|c|τn1aθn+cθnδ1d(xn1,xn).d\left(x_{n},x_{n+1}\right)\leq\underbrace{\frac{k-a\theta_{n-1}+|c|\tau_{n-1}}{a\theta_{n}+c\theta_{n}}}_{\delta_{1}}d\left(x_{n-1},x_{n}\right).
  • Case (C)(C), when a0a\geq 0 and c<0c<0 :

We get that

ad(xn,Txn)aθnd(xn,xn+1)\displaystyle ad\left(x_{n},Tx_{n}\right)\geq a\theta_{n}d\left(x_{n},x_{n+1}\right)
ad(xn1,Txn1)aθn1d(xn1,xn) and\displaystyle ad\left(x_{n-1},Tx_{n-1}\right)\geq a\theta_{n-1}d\left(x_{n-1},x_{n}\right)\text{ and }
cd(Txn1,Txn)cθnd(xn,xn+1)|c|τn1d(xn1,xn)\displaystyle cd\left(Tx_{n-1},Tx_{n}\right)\geq c\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right)

By the contraction-type condition on the pair ( xn1,xnx_{n-1},x_{n} ), it follows that

aθnd(xn,xn+1)+aθn1d(xn1,xn)+cθnd(xn,xn+1)|c|τn1d(xn1,xn)kd(xn1,xn)a\theta_{n}d\left(x_{n},x_{n+1}\right)+a\theta_{n-1}d\left(x_{n-1},x_{n}\right)+c\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right)\leq kd\left(x_{n-1},x_{n}\right)

So

d(xn,xn+1)kaθn1+|c|τn1aθn+cθnδ2d(xn,xn1).d\left(x_{n},x_{n+1}\right)\leq\underbrace{\frac{k-a\theta_{n-1}+|c|\tau_{n-1}}{a\theta_{n}+c\theta_{n}^{\prime}}}_{\delta_{2}}d\left(x_{n},x_{n-1}\right).
  • Case (B), when a<0a<0 and c0c\geq 0 :

We have that

ad(xn,Txn)aθnd(xn,xn+1),\displaystyle\operatorname{ad}\left(x_{n},Tx_{n}\right)\geq a\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right),
ad(xn1,Txn1)aθn1d(xn1,xn) and\displaystyle\operatorname{ad}\left(x_{n-1},Tx_{n-1}\right)\geq a\theta_{n-1}^{\prime}d\left(x_{n-1},x_{n}\right)\text{ and }
cd(Txn1,Txn)cθnd(xn,xn+1)|c|τn1d(xn1,xn).\displaystyle\operatorname{cd}\left(Tx_{n-1},Tx_{n}\right)\geq c\theta_{n}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right).

By the contraction-type condition on the pair ( xn1,xnx_{n-1},x_{n} ), it follows that

aθnd(xn,xn+1)+aθn1d(xn1,xn)+cθnd(xn,xn+1)|c|τn1d(xn1,xn)kd(xn1,xn).a\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)+a\theta_{n-1}^{\prime}d\left(x_{n-1},x_{n}\right)+c\theta_{n}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right)\leq kd\left(x_{n-1},x_{n}\right).

So

d(xn,xn+1)kaθn1+|c|τn1aθn+cθnδ3d(xn1,xn).d\left(x_{n},x_{n+1}\right)\leq\underbrace{\frac{k-a\theta_{n-1}^{\prime}+|c|\tau_{n-1}}{a\theta_{n}^{\prime}+c\theta_{n}}}_{\delta_{3}}d\left(x_{n-1},x_{n}\right).
  • Now we impose conditions on the already defined δ1,δ2\delta_{1},\delta_{2} and δ3\delta_{3}, which depend on the index nn\in\mathbb{N} and give some bounds that do not depend on nn for these numbers such that the sequence ( xnx_{n} ) will be Cauchy:

We have shown that

θn[12(1αn),11αn][12(1a1),11a2].\theta_{n}\in\left[\frac{1}{2\left(1-\alpha_{n}\right)},\frac{1}{1-\alpha_{n}}\right]\subset\left[\frac{1}{2\left(1-a_{1}\right)},\frac{1}{1-a_{2}}\right].

Moreover, we know that

τn=(2αn1+1θn)(θn2θn(1αn)1).\tau_{n}=\left(2\alpha_{n}-1+\frac{1}{\theta_{n}}\right)\cdot\left(\frac{\theta_{n}}{2\theta_{n}\left(1-\alpha_{n}\right)-1}\right).

Now, we impose a little more restrictive condition on θn\theta_{n} :
Let

12(1αn)<1q(1αn)<θn<11αn, so q<2\frac{1}{2\left(1-\alpha_{n}\right)}<\frac{1}{q\left(1-\alpha_{n}\right)}<\theta_{n}<\frac{1}{1-\alpha_{n}},\text{ so }q<2\text{. }

Also

1q<θn(1αn)<1, so q>1.\frac{1}{q}<\theta_{n}\left(1-\alpha_{n}\right)<1,\text{ so }q>1.

Since q(1,2)q\in(1,2), taking q=32q=\frac{3}{2}, it implies that

2θn(1αn)1>2q1=13, so θn>23(1αn).2\theta_{n}\left(1-\alpha_{n}\right)-1>\frac{2}{q}-1=\frac{1}{3},\text{ so }\theta_{n}>\frac{2}{3\left(1-\alpha_{n}\right)}.

This means that

θn[23(1αn),11αn].\theta_{n}\in\left[\frac{2}{3\left(1-\alpha_{n}\right)},\frac{1}{1-\alpha_{n}}\right].

Now, the condition becomes
k(1βn)+|a|(1+βn)+(a+c)<3(a+c)2k(1βn)<(a+c)2|a|(1+βn)k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)+(a+c)<\frac{3(a+c)}{2}\Longrightarrow k\left(1-\beta_{n}\right)<\frac{(a+c)}{2}-|a|\left(1+\beta_{n}\right).
So, we imposed the following condition:

k(1βn)<(a+c)2|a|(1+βn)k\left(1-\beta_{n}\right)<\frac{(a+c)}{2}-|a|\left(1+\beta_{n}\right) (12)

Also θn(2αn1)+10\theta_{n}\left(2\alpha_{n}-1\right)+1\geq 0 is equivalent to θn(12αn)1\theta_{n}\left(1-2\alpha_{n}\right)\leq 1. In a simiar manner, for θn(2αn1)+1\theta_{n}\left(2\alpha_{n}-1\right)+1, we have two cases:

When

a2<12αn1<12,a_{2}<\frac{1}{2}\Longrightarrow\alpha_{n-1}<\frac{1}{2},

it implies that

2αn11<0(2αn11)θn1+1<1, i.e.\displaystyle 2\alpha_{n-1}-1<0\Longrightarrow\left(2\alpha_{n-1}-1\right)\theta_{n-1}+1<1,\text{ i.e. }
τn1<113=3, for each n,\displaystyle\tau_{n-1}<\frac{1}{\frac{1}{3}}=3,\text{ for each }n\in\mathbb{N},
since θn>23(1αn)12θn(1αn)1<3.\displaystyle\text{ since }\theta_{n}>\frac{2}{3\left(1-\alpha_{n}\right)}\Longrightarrow\frac{1}{2\theta_{n}\left(1-\alpha_{n}\right)-1}<3.

When

a1>12αn1>12,a_{1}>\frac{1}{2}\Longrightarrow\alpha_{n-1}>\frac{1}{2},

it implies that

2αn11>0θn1(2αn11)+1<1+2a211a2=a21a2.2\alpha_{n-1}-1>0\Longrightarrow\theta_{n-1}\left(2\alpha_{n-1}-1\right)+1<1+\frac{2a_{2}-1}{1-a_{2}}=\frac{a_{2}}{1-a_{2}}.

So

τn1<a21a213.\tau_{n-1}<\frac{a_{2}}{1-a_{2}}\cdot\frac{1}{3}.

Taking these two cases into account, we have that τn1<τ\tau_{n-1}<\tau, where

τ:={13a2<1213a21a2a1>12\tau:=\begin{cases}\frac{1}{3}&a_{2}<\frac{1}{2}\\ \frac{1}{3}\cdot\frac{a_{2}}{1-a_{2}}&a_{1}>\frac{1}{2}\end{cases}

Now, let ε(0,1)\varepsilon\in(0,1). Furthermore, we impose the condition that

θn1+βnε(1αn)(1βn)\theta_{n}^{\prime}\leq\frac{1+\beta_{n}}{\varepsilon\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}

So

(a+c)(a+c)k(1βn)|a|(1+βn)<1ε\displaystyle\frac{(a+c)}{(a+c)-k\left(1-\beta_{n}\right)-|a|\left(1+\beta_{n}\right)}<\frac{1}{\varepsilon}\Longrightarrow
ε(a+c)(a+c)k(1βn)|a|(1+βn)<1\displaystyle\frac{\varepsilon(a+c)}{(a+c)-k\left(1-\beta_{n}\right)-|a|\left(1+\beta_{n}\right)}<1\Longrightarrow
(ε1)(a+c)<k(1βn)|a|(1+βn)(ε1)(a+c)\displaystyle(\varepsilon-1)(a+c)<-k\left(1-\beta_{n}\right)-|a|\left(1+\beta_{n}\right)\Longrightarrow(\varepsilon-1)(a+c)
+|a|(1+βn)<k(1βn), i.e.\displaystyle\quad+|a|\left(1+\beta_{n}\right)<-k\left(1-\beta_{n}\right),\text{ i.e. }
(1ε)(a+c)|a|(1+βn)>k(1βn).(1-\varepsilon)(a+c)-|a|\left(1+\beta_{n}\right)>k\left(1-\beta_{n}\right).

We observe that we have imposed the following condition:

k(1βn)<(1ε)(a+c)|a|(1+βn)k\left(1-\beta_{n}\right)<(1-\varepsilon)(a+c)-|a|\left(1+\beta_{n}\right) (13)

Now, for δ1,δ2\delta_{1},\delta_{2} and δ3\delta_{3}, we analyze all three cases:

  • Case (A)(A), when a0a\geq 0 and c>0c>0 :

We have that

δ1=kaθn1+|c|τn1(a+c)θnk+|c|τn1(a+c)θnk+|c|τ(a+c)θn.\delta_{1}=\frac{k-a\theta_{n-1}+|c|\tau_{n-1}}{(a+c)\theta_{n}}\leq\frac{k+|c|\tau_{n-1}}{(a+c)\theta_{n}}\leq\frac{k+|c|\tau}{(a+c)\theta_{n}}.

At the same time

(a+c)θn>2(a+c)3(1αn)2(a+c)3(1a1)1(a+c)θn<3(1a1)2(a+c)\displaystyle(a+c)\theta_{n}>\frac{2(a+c)}{3\left(1-\alpha_{n}\right)}\geq\frac{2(a+c)}{3\left(1-a_{1}\right)}\Longrightarrow\frac{1}{(a+c)\theta_{n}}<\frac{3\left(1-a_{1}\right)}{2(a+c)}\Longrightarrow
δ13(k+|c|τ)(1a1)2(a+c), i.e. δ13(1a1)2(a+c)[k+|c|τ].\displaystyle\delta_{1}\leq\frac{3(k+|c|\tau)\left(1-a_{1}\right)}{2(a+c)},\text{ i.e. }\delta_{1}\leq\frac{3\left(1-a_{1}\right)}{2(a+c)}\cdot[k+|c|\tau].
  • Case (C)(C), when a0a\geq 0 and c<0c<0 :

We know that

δ2=kaθn1+|c|τn1aθn+cθn.\delta_{2}=\frac{k-a\theta_{n-1}+|c|\tau_{n-1}}{a\theta_{n}+c\theta_{n}^{\prime}}.

But

kaθn1+|c|τn1ka1a2+|c|τk+|c|τk-a\theta_{n-1}+|c|\tau_{n-1}\leq k-\frac{a}{1-a_{2}}+|c|\tau\leq k+|c|\tau

Now

θn1+βnε(1αn)(1βn)1+b2ε(1a2)(1b2)\theta_{n}^{\prime}\leq\frac{1+\beta_{n}}{\varepsilon\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}\leq\frac{1+b_{2}}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}

where a2a_{2} and b2b_{2} are upper bounds for the sequences ( αn\alpha_{n} ), respectively ( βn\beta_{n} ).
Furthermore

aθn+cθna2(1a1)+c(1+b2)ε(1a2)(1b2).a\theta_{n}+c\theta_{n}^{\prime}\geq\frac{a}{2\left(1-a_{1}\right)}+\frac{c\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}.

Since

θn23(1a1)>12(1a1),\theta_{n}\geq\frac{2}{3\left(1-a_{1}\right)}>\frac{1}{2\left(1-a_{1}\right)},

in this case it is appropriate to use the above inequality with 12(1a1)\frac{1}{2\left(1-a_{1}\right)}. Also, we notice that we have the condition :

a2(1a1)+c(1+b2)ε(1a2)(1b2)>0\frac{a}{2\left(1-a_{1}\right)}+\frac{c\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}>0 (14)

Then

δ2k+|c|τa2(1a1)+c(1+b2)(1a2)(1b2)\delta_{2}\leq\frac{k+|c|\tau}{\frac{a}{2\left(1-a_{1}\right)}+\frac{c\left(1+b_{2}\right)}{\left(1-a_{2}\right)\left(1-b_{2}\right)}}
  • Case (B)(B), when a<0a<0 and c0c\geq 0 :

We know that

δ3=kaθn1+|c|τn1aθn+cθn\delta_{3}=\frac{k-a\theta_{n-1}^{\prime}+|c|\tau_{n-1}}{a\theta_{n}^{\prime}+c\theta_{n}}

But

kaθn1+|c|τn1ka(1+b2)ε(1a2)(1b2)+|c|τ and\displaystyle k-a\theta_{n-1}^{\prime}+|c|\tau_{n-1}\leq k-a\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+|c|\tau\text{ and }
cθnc2(1a1)\displaystyle c\theta_{n}\geq\frac{c}{2\left(1-a_{1}\right)}

As in the previous case it is more convenient to use the inequality for θn\theta_{n} with 12(1a1)\frac{1}{2\left(1-a_{1}\right)}, than with 23(1a1)\frac{2}{3\left(1-a_{1}\right)}. Moreover

aθn+cθnaθn+c2(1a1)a\theta_{n}^{\prime}+c\theta_{n}\geq a\theta_{n}^{\prime}+\frac{c}{2\left(1-a_{1}\right)}

It follows that

aθn+cθna(1+b2)ε(1a2)(1b2)+c2(1a1)a\theta_{n}^{\prime}+c\theta_{n}\geq\frac{a\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+\frac{c}{2\left(1-a_{1}\right)}

Also, we imposed the condition:

a(1+b2)ε(1a2)(1b2)+c2(1a1)>0\frac{a\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+\frac{c}{2\left(1-a_{1}\right)}>0 (15)

Finally, we get the upper bound for δ3\delta_{3}, i.e.

δ3ka(1+b2)ε(1a2)(1b2)+c|τ|a(1+b2)ε(1a2)(1b2)+c2(1a1).\delta_{3}\leq\frac{k-\frac{a\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+c|\tau|}{\frac{a\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+\frac{c}{2\left(1-a_{1}\right)}}.
  • Now, we can get an upper bound for all δi\delta_{i} ’s:

k+|a|θn1+|c|τn1k+|a|(1+b2)ε(1a2)(1b2)+|c|τ.k+|a|\theta_{n-1}^{\prime}+|c|\tau_{n-1}\leq k+|a|\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+|c|\tau.

Now, we denote the denominator by

s:={3(1a1)2(a+c), in the case (A), a2(1a1)+c(1+b2)ε(1a2)(1b2), in the case (C), a(1+b2)ε(1a2)(1b2)+c2(1a1), in the case (B) s:=\begin{cases}\frac{3\left(1-a_{1}\right)}{2(a+c)},&\text{ in the case (A), }\\ \frac{a}{2\left(1-a_{1}\right)}+c\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)},&\text{ in the case (C), }\\ \frac{a\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+\frac{c}{2\left(1-a_{1}\right)},&\text{ in the case (B) }\end{cases}

Also, let’s denote

t:=k+|a|(1+b2)ε(1a2)(1b2)+|c|τs.t:=\frac{k+|a|\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+|c|\tau}{s}.

In the step ( IXIX ), we shall show that the numerators of all δi\delta_{i} are positive, so that δi0\delta_{i}\geq 0. For the moment, since δi0\delta_{i}\geq 0, for each i{1,2,3}i\in\{1,2,3\}, then t0t\geq 0. Moreover, we want that t<1t<1, that is

|c|τ<sk|a|(1+b2)ε(1a2)(1b2).|c|\tau<s-k-|a|\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}.

We now have the condition:

τ<1|c|[sk|a|(1+b2)ε(1a2)(1b2)]\tau<\frac{1}{|c|}\cdot\left[s-k-|a|\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}\right] (16)

(VIII) Now we show that the sequence ( xnx_{n} ) is Cauchy:

Let’s take t[0,1)t\in[0,1) as before. For all the three cases, (A),(B)(A),(B) and (C)(C), we get

d(xn,xn+1)td(xn1,xn)d(xn,xn+1)tnd(x1,x0),\displaystyle\quad d\left(x_{n},x_{n+1}\right)\leq td\left(x_{n-1},x_{n}\right)\Longrightarrow d\left(x_{n},x_{n+1}\right)\leq t^{n}d\left(x_{1},x_{0}\right),
where {x1=W(x0,Ty0,α0)y0=W(x0,Tx0,β0), with x0K is a fixed arbitrary element\displaystyle\text{ where }\left\{\begin{array}[]{l}x_{1}=W\left(x_{0},Ty_{0},\alpha_{0}\right)\\ y_{0}=W\left(x_{0},Tx_{0},\beta_{0}\right),\end{array}\text{ with }x_{0}\in\mathrm{~K}\right.\text{ is a fixed arbitrary element }

Let m>nm>n. We now have that

d(xm,xn)[tn+tn+1++tm1]d(x1,x0)tn1td(x1,x0)d\left(x_{m},x_{n}\right)\leq\left[t^{n}+t^{n+1}+\cdots+t^{m-1}\right]d\left(x_{1},x_{0}\right)\leq\frac{t^{n}}{1-t}d\left(x_{1},x_{0}\right)

Letting mm and nn go to \infty, it follows that ( xnx_{n} ) is a Cauchy sequence, because ( xnx_{n} ) KX\subset K\subset X and KK is a convex subset of the convex metric space ( X,d,WX,d,W ). Since ( xnx_{n} ) is Cauchy and KK is closed in the complete metric space ( X,d,WX,d,W ), then ( xnx_{n} ) is convergent, i.e. there exists pKp\in K, such that limnxn=p\lim_{n\rightarrow\infty}x_{n}=p. Taking in the contraction-type conditions p instead of x and xnx_{n} instead of y , it follows

cd(Txn,Tp)+ad(p,Tp)+ad(xn,Txn)kd(xn,p)cd\left(Tx_{n},Tp\right)+ad(p,Tp)+ad\left(x_{n},Tx_{n}\right)\leq kd\left(x_{n},p\right) (*)

But

d(Txn,p)d(p,xn)+d(xn,Txn)d(xn,p)+θnd(xn,xn+1)d\left(Tx_{n},p\right)\leq d\left(p,x_{n}\right)+d\left(x_{n},Tx_{n}\right)\leq d\left(x_{n},p\right)+\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)

Also, we now that

d(xn,p)0,d(xn,xn+1)0 and that limnθn<d\left(x_{n},p\right)\rightarrow 0,d\left(x_{n},x_{n+1}\right)\rightarrow 0\text{ and that }\lim_{n\rightarrow\infty}\theta_{n}^{\prime}<\infty

because

θn(a+c)(1+b2)ε(1a2)(1b2).\theta_{n}^{\prime}\leq\frac{(a+c)\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}.

Then, we obtain that limnTxn=p\lim_{n\rightarrow\infty}Tx_{n}=p. Also, we know that d(xn,Txn)θnd(xn,xn+1)0d\left(x_{n},Tx_{n}\right)\leq\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)\rightarrow 0. From (*), we get cd(p,Tp)+ad(p,Tp)0cd(p,Tp)+ad(p,Tp)\leq 0, i.e. (a+c)d(p,Tp)0(a+c)d(p,Tp)\leq 0. Since a+c>0a+c>0, then d(p,Tp)=0p=Tpd(p,Tp)=0\Longleftrightarrow p=Tp, i.e. the limit of the sequence (xn)\left(x_{n}\right) is a fixed point for the operator TT.

  • The uniqueness of the fixed point: Let’s consider p,qp,q two fixed points for TT, i.e. p=Tpp=Tp and q=Tqq=Tq. Applying the contraction-type condition for the pair ( p,qp,q ), it implies that:

cd(Tp,Tq)+ad(p,Tp)+ad(q,Tq)kd(p,q)cd(Tp,Tq)+ad(p,Tp)+ad(q,Tq)\leq kd(p,q)

Since d(Tp,Tq)=d(p,q),d(p,Tp)=0d(Tp,Tq)=d(p,q),d(p,Tp)=0 and d(q,Tq)=0d(q,Tq)=0, we get (ck)d(p,q)0(c-k)d(p,q)\leq 0. Also, because k<ck<c, then we have that d(p,q)=0d(p,q)=0, i.e. p=qp=q.

We have imposed the condition:

k<c, for the uniqueness of the fixed point for Tk<c\text{, for the uniqueness of the fixed point for }\mathrm{T}

(IX) Now we verify that all the conditions we imposed are valid because of the hypothesis assumptions:

We recall the conditions:
For the case ( AA ):

c0,a+c>0,k>a and (1βn)(ka)a+c<1c\neq 0,a+c>0,k>a\text{ and }\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}<1

For the case ( BB ):

k>a1+βn1βn and k(1βn)a(1+βn)a+c<1k>a\frac{1+\beta_{n}}{1-\beta_{n}}\text{ and }\frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}<1

For the case ( CC ):

c0,k>a and (1βn)(ka)a+c<1.c\neq 0,k>a\text{ and }\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}<1.

For θn\theta_{n}^{\prime}, we have

k(1βn)+|a|(1+βn)a+c<1.\frac{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)}{a+c}<1.

The conditions (10) and (11):

k(1βn)+|a|(1+βn)a+c<1.\frac{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)}{a+c}<1.

The condition (12):

k(1βn)<12(a+c)|a|(1+βn)k\left(1-\beta_{n}\right)<\frac{1}{2}(a+c)-|a|\left(1+\beta_{n}\right)

The condition (13):

k(1βn)<(1ε)(a+c)|a|(1+βn), with ε(0,1)k\left(1-\beta_{n}\right)<(1-\varepsilon)(a+c)-|a|\left(1+\beta_{n}\right),\text{ with }\varepsilon\in(0,1)

The conditions (14) and (15):

s>0s>0

The condition (16):

|c|τ<sk|a|(1+b2)ε(1a2)(1b2)|c|\tau<s-k-|a|\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}

We analyze these conditions by cases:
Casew ( AA ) and ( CC ) imply that

k>a,(1βn)(ka)a+c<1, i.e. (1βn)k<(a+c)+a(1βn).\Longrightarrow k>a,\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}<1,\text{ i.e. }\left(1-\beta_{n}\right)k<(a+c)+a\left(1-\beta_{n}\right).

Case (B) implies that

k>a1+βn1βn and k(1βn)<(a+c)+a(1+βn)\Longrightarrow k>a\frac{1+\beta_{n}}{1-\beta_{n}}\text{ and }k\left(1-\beta_{n}\right)<(a+c)+a\left(1+\beta_{n}\right)\text{. }

Also, for a0a\geq 0, for k>ak>a, we take k>a(1+βn)1βn>ak>\frac{a\left(1+\beta_{n}\right)}{1-\beta_{n}}>a. In a similar way, for a<0a<0, we have the same inequality. So, for the both cases involving aa, we can take

k>a(1+βn)(1βn).k>a\frac{\left(1+\beta_{n}\right)}{\left(1-\beta_{n}\right)}.

Now, the second conditions for aa can be written as:

{a0(1βn)k<(a+c)+|a|(1βn),a<0(1βn)k<(a+c)|a|(1+βn)\left\{\begin{array}[]{l}a\geq 0\Longrightarrow\left(1-\beta_{n}\right)k<(a+c)+|a|\left(1-\beta_{n}\right),\\ a<0\Longrightarrow\left(1-\beta_{n}\right)k<(a+c)-|a|\left(1+\beta_{n}\right)\end{array}\right.

Also, the condition for θn\theta_{n}^{\prime} is

k(1βn)<(a+c)|a|(1+βn).k\left(1-\beta_{n}\right)<(a+c)-|a|\left(1+\beta_{n}\right).

In a similar manner, the condition (12) is

k(1βn)<12(a+c)|a|(1+βn).k\left(1-\beta_{n}\right)<\frac{1}{2}(a+c)-|a|\left(1+\beta_{n}\right).

From the above two relations, we impose the restrictive one, i.e.

k(1βn)<12(a+c)|a|(1+βn).k\left(1-\beta_{n}\right)<\frac{1}{2}(a+c)-|a|\left(1+\beta_{n}\right).

Moreover, at condition (13), we have

k(1βn)<(1ε)(a+c)|a|(1+βn).k\left(1-\beta_{n}\right)<(1-\varepsilon)(a+c)-|a|\left(1+\beta_{n}\right).

For the cases involving aa, condition (**) can be restrained as follows:

(1βn)k<(a+c)|a|(1+βn).\left(1-\beta_{n}\right)k<(a+c)-|a|\left(1+\beta_{n}\right).

So, the valid condition remains only condition (12), that is

k(1βn)<12(a+c)|a|(1+βn).k\left(1-\beta_{n}\right)<\frac{1}{2}(a+c)-|a|\left(1+\beta_{n}\right).
  • Now, we impose condition such that numerators of δi\delta_{i} ’s are positive:

We recall the cases reminded above:
Cases (A)(A) and (C)(C) :

kaθn1|c|τn1,k\geq a\theta_{n-1}-|c|\tau_{n-1},

Case (B):

kaθn1|c|τn1.k\geq a\theta_{n-1}^{\prime}-|c|\tau_{n-1}.

At case (B)(B), since a<0a<0 and θn1θn1\theta_{n-1}\leq\theta_{n-1}^{\prime}, we have that

aθn1aθn1.a\theta_{n-1}\geq a\theta_{n-1}^{\prime}.

So, for all the cases we need to impose

kaθn1|c|τn1.k\geq a\theta_{n-1}-|c|\tau_{n-1}.
  • Finally, the conditions that are valid are:
    (a) kaθn1|c|τn1k\geq a\theta_{n-1}-|c|\tau_{n-1},
    (b) k(1βn)<(1ε)(a+c)|a|(1+βn)k\left(1-\beta_{n}\right)<(1-\varepsilon)(a+c)-|a|\left(1+\beta_{n}\right),
    (c) k(1βn)<12(a+c)|a|(1+βn)k\left(1-\beta_{n}\right)<\frac{1}{2}(a+c)-|a|\left(1+\beta_{n}\right),
    (d) k>a(1+βn)(1βn)k>\frac{a\left(1+\beta_{n}\right)}{\left(1-\beta_{n}\right)},
    (e) s0s\geq 0 and |c|τ<sk|a|(1+b2)ε(1a2)(1b2)|c|\tau<s-k-|a|\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}.

Also, now we analyze in detail conditions (a),(c) and (d):
For condition (c), we have that

k<12(a+c)|a|(1+βn)1βn.k<\frac{\frac{1}{2}(a+c)-|a|\left(1+\beta_{n}\right)}{1-\beta_{n}}.

So, for k>0k>0, we impose k(1b1)<12(a+c)|a|(1+b2)k\left(1-b_{1}\right)<\frac{1}{2}(a+c)-|a|\left(1+b_{2}\right) and for k<0k<0 we impose k(1b2)<12(a+c)|a|(1+b2)k\left(1-b_{2}\right)<\frac{1}{2}(a+c)-|a|\left(1+b_{2}\right). Furthermore, for condition (b), we have that

k(1βn)<(1ε)(a+c)|a|(1+βn).k\left(1-\beta_{n}\right)<(1-\varepsilon)(a+c)-|a|\left(1+\beta_{n}\right).

Now, in a similar manner, using the boundaries for the sequence (βn)\left(\beta_{n}\right), we have two cases:
If k>0k>0, then

k(1b1)<(1ε)(a+c)|a|(1+b2)k\left(1-b_{1}\right)<(1-\varepsilon)(a+c)-|a|\left(1+b_{2}\right)

and for k<0k<0 we impose

k(1b2)<(1ε)(a+c)|a|(1+b2).k\left(1-b_{2}\right)<(1-\varepsilon)(a+c)-|a|\left(1+b_{2}\right).

Taking ε(0,12]\varepsilon\in\left(0,\frac{1}{2}\right], the condition (c) implies that:

k(1βn)<12(a+c)|a|(1+βn)(1ε)(a+c)|a|(1+βn),k\left(1-\beta_{n}\right)<\frac{1}{2}(a+c)-|a|\left(1+\beta_{n}\right)\leq(1-\varepsilon)(a+c)-|a|\left(1+\beta_{n}\right),

for each nn\in\mathbb{N}, so condition (c) implies condition (b).
For condition (d), we have two cases:
If a0a\geq 0, then

k>a(1+b2)1b2,k>a\frac{\left(1+b_{2}\right)}{1-b_{2}},

and if a<0a<0, we have

ka1+b11b1k\geq a\frac{1+b_{1}}{1-b_{1}}

For condition (a), it follows that:
Since

aθn1|c|τn1aθn1,a\theta_{n-1}-|c|\tau_{n-1}\leq a\theta_{n-1},

then we can take

kaθn1.k\geq a\theta_{n-1}.

If a0a\geq 0, since aθn1a1a2a\theta_{n-1}\leq\frac{a}{1-a_{2}}, then we need to impose

ka1a2.k\geq\frac{a}{1-a_{2}}.

If a<0a<0, since θn123(1αn1)23(1a1)\theta_{n-1}\geq\frac{2}{3\left(1-\alpha_{n-1}\right)}\geq\frac{2}{3\left(1-a_{1}\right)}, then

aθn12a3(1a1)a\theta_{n-1}\leq\frac{2a}{3\left(1-a_{1}\right)}

We impose that

k2a3(1a1)k\geq\frac{2a}{3\left(1-a_{1}\right)}

Remark 1 For the sake of completeness involving the proof from (Theorem 1), we make the following crucial remark: if a0a\geq 0, then k0k\geq 0.

As a consequence of the above result, we can obtain a fixed point theorem for a certain iterate of a self-operator in a complete convex metric space.

Corollary 1 Let ( X,dX,d ) be a complete convex metric space. Let KK a nonempty, closed and convex subset of XX and T:KKT:K\rightarrow K be a self-mapping. Suppose, there exists p>1p>1, such that TT satisfies:

cd(Tpx,Tpy)+a[d(x,Tpx)+d(y,Tpy)]kd(x,y)cd\left(T^{p}x,T^{p}y\right)+a\left[d\left(x,T^{p}x\right)+d\left(y,T^{p}y\right)\right]\leq kd(x,y)

for each x,yKx,y\in K. Also, suppose the operator Tp:KKT^{p}:K\rightarrow K satisfies conditions (1)-(8) from (Theorem 1). Then, TT has a fixed point. Moreover, if k<ck<c then the fixed point is unique.

Proof Let T:=TpT^{\prime}:=T^{p}. Applying (Theorem 1) for the operator TT^{\prime}, we find that TT^{\prime} has a fixed point uKu\in K, i.e. Tu=uT^{\prime}u=u. Also uu is the unique fixed point of TT^{\prime} if k<ck<c. But Tp(Tu)=T(Tpu)=T(Tu)=T(u)T^{p}(Tu)=T\left(T^{p}u\right)=T\left(T^{\prime}u\right)=T(u). Also, since Tp(Tu)=T(Tu)T^{p}(Tu)=T^{\prime}(Tu), from these two equalities, we find that Tu=uTu=u, so TT has a fixed point. Moreover, if k<ck<c, then uu is the unique element in KK such that Tu=uT^{\prime}u=u. Finally, with the same reasoning we find that uu is the unique fixed point of TT.

Example 1 Let X:=[0,)X:=[0,\infty) and T:XXT:X\rightarrow X, such that :

T(x)={x194,x[0,x0)x250,x[x0,)T(x)=\begin{cases}\frac{x}{194},&x\in\left[0,x_{0}\right)\\ \frac{x}{250},&x\in\left[x_{0},\infty\right)\end{cases}

where x0x_{0} can be any point in [0,)[0,\infty). We remark that because x0x_{0} is a discontinuity point for TT, then T can’t be a contraction type mapping. Then TT satisfies the inequality:

1800d(Tx,Ty)+12d(x,y)10[d(x,Tx)+d(y,Ty)]1800d(Tx,Ty)+\frac{1}{2}d(x,y)\leq 10[d(x,Tx)+d(y,Ty)]

Also, TT has a unique fixed point, i.e. 0X0\in X.

Remark 2 We will construct a self-mapping TT defined on [0,)[0,\infty), which satisfies the above generalized contraction condition with a0,k0a\leq 0,k\leq 0, and c>0c>0. as well as the assumptions from (Theorem 1). For this purpose, let b1=110,b2=910,a1=110b_{1}=\frac{1}{10},b_{2}=\frac{9}{10},a_{1}=\frac{1}{10} and a2=25a_{2}=\frac{2}{5}. Following the notations from (Theorem 1), we have that

a2<12 and τ=13a_{2}<\frac{1}{2}\text{ and }\tau=\frac{1}{3}

Also,

s=a(1+b2)ε(1a2)(1b2)+c12(1a1)s=\frac{a\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+c\frac{1}{2\left(1-a_{1}\right)}

Since τ|c|<sk|a|(1+b2)ε(1a2)(1b2)\tau|c|<s-k--|a|\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}, that is the condition (e) in the proof of (Theorem 1), then we get that

c3<sk|a|(1+b2)(1a2)(1b2), i.e.\displaystyle\frac{c}{3}<s-k-|a|\frac{\left(1+b_{2}\right)}{\left(1-a_{2}\right)\left(1-b_{2}\right)},\text{ i.e. }
c3<2aε(1+b2)(1a2)(1b2)k\displaystyle\frac{c}{3}<\frac{2a}{\varepsilon}\frac{\left(1+b_{2}\right)}{\left(1-a_{2}\right)\left(1-b_{2}\right)}-k

Furthermore,

c3<2aε1906k+5c9\frac{c}{3}<\frac{2a}{\varepsilon}\frac{190}{6}-k+\frac{5c}{9}

Also a+c>0a+c>0, i.e. c>ac>-a and k2a3(1a1)k\geq\frac{2a}{3\left(1-a_{1}\right)}, so k2027ak\geq\frac{20}{27}a, which is the condition kaθn1k\geq a\theta_{n-1}, i.e. which is the condition (a) in the proof of (Theorem 1). Moreover,

(1βn)k<[12(a+c)|a|(1+b2)]\left(1-\beta_{n}\right)k<\left[\frac{1}{2}(a+c)-|a|\left(1+b_{2}\right)\right]

for each nn\in\mathbb{N}. Since k0k\leq 0, then we impose the condition that (1b2)k<12(a+c)|a|(1+b2)\left(1-b_{2}\right)k<\frac{1}{2}(a+c)-|a|\left(1+b_{2}\right), which is the condition (c) in the proof of (Theorem 1):

k<10[a+c2+19a10]k<10\cdot\left[\frac{a+c}{2}+\frac{19a}{10}\right]

Since ka(1+b1)(1b1)k\geq a\frac{\left(1+b_{1}\right)}{\left(1-b_{1}\right)}, then

k119ak\geq\frac{11}{9}a

which is the condition (d) in the proof of (Theorem 1). Furthermore, because s>0s>0, then

196εa+5c9>0\frac{19}{6\varepsilon}a+\frac{5c}{9}>0

which is the condition (e) in the proof of (Theorem 1). Now, let’s take a=10a=-10. Then c>10c>10. The other inequalities become :

k2027(10)k20027=7.4k\geq\frac{20}{27}(-10)\Longrightarrow k\geq-\frac{200}{27}=-7.4
k119ak1109\displaystyle k\geq\frac{11}{9}a\Longrightarrow k\geq-\frac{110}{9}
k10[a2+c2+19a10]=10[c224].\displaystyle k\leq 0\cdot\left[\frac{a}{2}+\frac{c}{2}+\frac{19a}{10}\right]=0\cdot\left[\frac{c}{2}-24\right].

Since 5c919006ε\frac{5c}{9}\geq\frac{1900}{6\varepsilon}, then

c570ε.c\geq\frac{570}{\varepsilon}.

On the other hand, we have that

2c9<1900εk.-\frac{2c}{9}<\frac{-1900}{\varepsilon}-k.

So, it follows that

c570ε,k20027c\geq\frac{570}{\varepsilon},k\geq-\frac{200}{27}

and the following conditions hold:

(1) k<10(c224) and\displaystyle\text{ (1) }k<0\cdot\left(\frac{c}{2}-24\right)\text{ and }
(2) 2c9<1900εk.\displaystyle\text{ (2) }-\frac{2c}{9}<\frac{-1900}{\varepsilon}-k.

Also, taking ε=13(0,12)\varepsilon=\frac{1}{3}\in\left(0,\frac{1}{2}\right), it follows that c1710c\geq 1710. Now, condition (2) becomes k<57002c9k<5700-\frac{2c}{9}. Since k<0k<0, we put the condition that 57002c905700-\frac{2c}{9}\geq 0, i.e. c25650c\geq 25650, so that condition (2) is satisfied. Furthermore, condition (1) is k5c240k\leq 5c-240. We put 5c240>05c-240>0, so c>48c>48. Taking c570c\geq 570, condition (1) is satisfied, since k<0k<0. Now, we can take c=1800c=1800. So, let’s construct TT as a linear piecewise function, such as

T(x)={xp1,x[0,x0)xp2,x[x0,),T(x)=\left\{\begin{array}[]{ll}\frac{x}{p_{1}},&x\in\left[0,x_{0}\right)\\ \frac{x}{p_{2}},&x\in\left[x_{0},\infty\right)\end{array},\right.

where x0x_{0} is a discontinuity point such that TT is no a contraction mapping. Furthermore p1,p2>0p_{1},p_{2}>0 with p1p2p_{1}\neq p_{2} and p1,p2>0p_{1},p_{2}>0. Now, denote

I1:=[0,x0) and I2:=[x0,).I_{1}:=\left[0,x_{0}\right)\text{ and }I_{2}:=\left[x_{0},\infty\right).

We have four cases two analyze the generalized contraction condition:
(I) When x,yI1x,y\in I_{1}, we get, by the contraction-type condition that:

(cp1k)|xy|p11p1(x+y)a\displaystyle\left(\frac{c}{p_{1}}-k\right)|x-y|\leq-\frac{p_{1}-1}{p_{1}}(x+y)a
(cp1k)(p1+1p1)acp1k(1p1)a\displaystyle\left(\frac{c}{p_{1}}-k\right)\leq\left(\frac{-p_{1}+1}{p_{1}}\right)a\Longrightarrow c-p_{1}k\leq\left(1-p_{1}\right)a

(II) When xI1x\in I_{1} and yI2y\in I_{2}, by contraction-type condition we get:

c|xp1yp2|k|xy|=cp1p2|p2xp1y|k|xy|c\left|\frac{x}{p_{1}}-\frac{y}{p_{2}}\right|-k|x-y|=\frac{c}{p_{1}p_{2}}\left|p_{2}x-p_{1}y\right|-k|x-y|
(cp1x+cp2y)kxky\displaystyle\leq\left(\frac{c}{p_{1}}x+\frac{c}{p_{2}}y\right)-kx-ky\Longrightarrow
(cp1k)x+(cp2k)y(p11)xp1(p21)yp2\displaystyle\left(\frac{c}{p_{1}}-k\right)x+\left(\frac{c}{p_{2}}-k\right)y\leq-\frac{\left(p_{1}-1\right)x}{p_{1}}-\frac{\left(p_{2}-1\right)y}{p_{2}}
{cp1k+p11p10cp2k+p21p20\displaystyle\Longrightarrow\left\{\begin{array}[]{l}\frac{c}{p_{1}}-k+\frac{p_{1}-1}{p_{1}}\leq 0\\ \frac{c}{p_{2}}-k+\frac{p_{2}-1}{p_{2}}\leq 0\end{array}\right.

(III) When xI2x\in I_{2} and yI1y\in I_{1} : it’s analogous to the case above.
(IV) When x,yI2x,y\in I_{2} :

Like in case (I), we get that cp2k1p2c-p_{2}k\leq 1-p_{2}. Now, we have shown that k[20027,0)=[7.4,0)k\in\left[-\frac{200}{27},0\right)=[-7.4,0). By the conditions above, we have that

1800p1k(1p1)(10)1810p1(10+k)k101800-p_{1}k\leq\left(1-p_{1}\right)(-10)\Longrightarrow 1810\leq p_{1}(10+k)\Longrightarrow k\geq-10

Let’s take k=12k=-\frac{1}{2}. Then 1810p1(1012)1810\leq p_{1}\left(10-\frac{1}{2}\right), so we can take p1362019=190.52Inp_{1}\geq\frac{3620}{19}=190.52\mathrm{In} a similar way, we obtain that p2190.52p_{2}\geq 190.52. So, we can take, for example p1=194p_{1}=194 and p2=205p_{2}=205.

  • Now, let’s construct the sequences ( αn\alpha_{n} ) and ( βn\beta_{n} ):

We know that

αn[a1,a2]=[110,25]\alpha_{n}\in\left[a_{1},a_{2}\right]=\left[\frac{1}{10},\frac{2}{5}\right]

and that

βn[b1,b2]=[110,910].\beta_{n}\in\left[b_{1},b_{2}\right]=\left[\frac{1}{10},\frac{9}{10}\right].

Let’s take αn=1n+w+a1+a22αn>0\alpha_{n}=\frac{1}{n+w}+\frac{a_{1}+a_{2}}{2}\Longrightarrow\alpha_{n}>0, for each nn\in\mathbb{N}. Then αn>a1\alpha_{n}>a_{1}, which means that 1n+w+a1+a22>1n+w+a1>a1\frac{1}{n+w}+\frac{a_{1}+a_{2}}{2}>\frac{1}{n+w}+a_{1}>a_{1}. Also, αn<a2\alpha_{n}<a_{2} means that 1n+w+a1+a22<a2\frac{1}{n+w}+\frac{a_{1}+a_{2}}{2}<a_{2}, i.e.

1n+w<a2a12=320n+w>203, for each n.\frac{1}{n+w}<\frac{a_{2}-a_{1}}{2}=\frac{3}{20}\Longrightarrow n+w>\frac{20}{3},\text{ for each }n\in\mathbb{N}.

Taking n=0n=0 and observing that the sequence (n+w)n(n+w)_{n\in\mathbb{N}} is increasing, we get that w>203w>\frac{20}{3}. So, we can take

w=213=7αn=1n+7+14w=\frac{21}{3}=7\Longrightarrow\alpha_{n}=\frac{1}{n+7}+\frac{1}{4}

In a similar manner, let βn=1n+w+b1+b22\beta_{n}=\frac{1}{n+w^{\prime}}+\frac{b_{1}+b_{2}}{2}. Analogous as in the previous case: βn>b1\beta_{n}>b_{1}. Furthermore: βn<b2\beta_{n}<b_{2} means that

1n+w<b2b12n+w>52\frac{1}{n+w^{\prime}}<\frac{b_{2}-b_{1}}{2}\Longrightarrow n+w^{\prime}>\frac{5}{2}

As before, the sequence (n+w)n\left(n+w^{\prime}\right)_{n\in\mathbb{N}} is increasing. Taking n=0n=0, it follows that w>52w^{\prime}>\frac{5}{2} and taking w=62=3w^{\prime}=\frac{6}{2}=3, we get that βn=1n+3+12\beta_{n}=\frac{1}{n+3}+\frac{1}{2}. In conclusion, we have constructed the operator TT as:

T(x)={x194,x[0,x0)x250,x[x0,),T(x)=\left\{\begin{array}[]{ll}\frac{x}{194},&x\in\left[0,x_{0}\right)\\ \frac{x}{250},&x\in\left[x_{0},\infty\right)\end{array},\right.

which verifies that

1800d(Tx,Ty)+12d(x,y)10[d(x,Tx)+d(y,Ty)], for each x,y[0,).1800d(Tx,Ty)+\frac{1}{2}d(x,y)\leq 10[d(x,Tx)+d(y,Ty)],\text{ for each }x,y\in[0,\infty).

Finally, the sequences ( αn\alpha_{n} ) and ( βn\beta_{n} ) from the Ishikawa iteration are:

αn=1n+7+14 and βn=1n+3+12.\alpha_{n}=\frac{1}{n+7}+\frac{1}{4}\text{ and }\beta_{n}=\frac{1}{n+3}+\frac{1}{2}.

Conclusion

Our goal throughout the entire article was to use Ishikawa iteration process in the context of convex metric spaces for showing sufficient conditions for the existence and uniqueness of the fixed points of nonlinear generalized contractive-type mappings. Even though there are simpler techniques for showing the convergence of an iterative process to a fixed point of an operator, we wanted to show that the technique of Karapinar from [5] can be successfully applied to Ishikawa iterative process, such that this iterative sequence convergence to a fixed point of the nonlinear mapping under some suitable conditions. In a future paper, we ought to construct theorems for the existence of the fixed points of generalized contractions, using other techniques that can be further used for comparing rate of convergence.

Acknowledgements The author is thankful to his Phd coordinator, professor dr. Adrian Petruşel from Babeş - Bolyai University, for his support and useful suggestions throughout the entire article.

Funding This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.

References

  1. 1.

    Abbas, M., Khan, S.H., Rhoades, B.E.: Simpler is also better in approximating fixed points. Appl. Math. Comput. 205(1), 428-431 (2008)

  2. 2.

    Agarwal, R.P., O’Regan, D., Sahu, D.R.: Fixed Point Theory for Lipschitzian-Type Mappings with Applications. Springer, Heidelberg (2003)

  3. 3.

    Asadi, M.: Some results of fixed point theorems in convex metric spaces. Nonlinear Funct. Anal. Appl. 19(2), 171-175 (2014)

  4. 4.

    Fukhar-ud-dina, H., Berinde, V.: Iterative methods for the class of quasi-contractive type operators and comparison of their rate of convergence in convex metric spaces. Filomat 30(1), 223-230 (2016)

  5. 5.

    Karapinar, E.: Fixed point theorems in cone banach spaces. Fixed point theory and applications. Article ID 609281, pp. 1-9 (2009). https://doi.org/10.1155/2009/60928

  6. 6.

    Moosaei, M.: Fixed point theorems in convex metric spaces. Fixed Point Theory Appl. 2012, 164 (2012). https://doi.org/10.1186/1687-1812-2012-164

  7. 7.

    Moosaei, M.: Common fixed points for some generalized contraction pairs in convex metric spaces. Fixed Point Theory Appl. 2014:98 (2014). http://www.fixedpointtheoryandapplications.com/content/2014/1/ 98

  8. 8.

    Moosaei, M., Azizi, A.: On coincidence points of generalized contractive pair mappings in convex metric spaces. J. Hyperstruct. 2(4), 136-141 (2015)

  9. 9.

    Shimizu, T., Takahashi, W.: Fixed point theorems in certain convex metric spaces. Math. Jpn. 37, 855-859 (1992)

  10. 10.

    Takahashi, W.: A convexity in metric spaces and nonexpansive mappings I. Kodai Math. Semin. Rep. 22, 142-149 (1970)

  11. 11.

    Wang, C., Zhang, T.: Approximating common fixed points for a pair of generalized nonlinear mappings in convex metric spaces. J. Nonlinear Sci. Appl. 9, 1-7 (2016)

2019

Related Posts