Fixed point theorems for generalized contraction mappings on b-rectangular metric spaces

Abstract

In the present article, we study some fixed point theorems for a hybrid class of generalized contractive operators in the context of b-rectangular metric spaces. Examples justifying theorems and an open problem regarding to further generalizations for this type of operators are also given.

Authors

Cristian Daniel Alecsa
Babes-Bolyai University Faculty of Mathematics and Computer Sciences Cluj-Napoca, Romania
”Tiberiu Popoviciu” Institute of Numerical Analysis Romanian Academy Cluj-Napoca, Romania

Keywords

Generalized contraction; b-rectangular metric space; expansive mappings; fixed point.

Paper coordinates

Cristian-Daniel Alecsa, Fixed point theorems for generalized contraction mappings on b-rectangular metric spaces, Stud. Univ. Babes-Bolyai Math. 62(2017), No. 4, 495–520 DOI: 10.24193/subbmath.2017.4.08

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Fixed point theorems for generalized contraction mappings on b-rectangular metric spaces

Cristian Daniel Alecsa
Abstract

In the present article, we study some fixed point theorems for a hybrid class of generalized contractive operators in the context of b-rectangular metric spaces. Examples justifying theorems and an open problem regarding to further generalizations for this type of operators are also given.

Mathematics Subject Classification (2010): 47H10,54H2547\mathrm{H}10,54\mathrm{H}25.
Keywords: Generalized contraction, b-rectangular metric space, expansive mappings, fixed point.

1. Introduction and preliminaries

In this section we shall present some useful lemmas and definitions regarding rectangular and b-rectangular metric spaces. Also, we shall present some recent results in the field of fixed point theory concerning expansive operators and some generalized contraction mappings.
In [6], A. Branciari introduced a new metric-type space, when triangle inequality is replaced by an inequality which involves four different elements. This is called a rectangular metric space or a generalized metric space (g.m.s.)

Definition 1.1. Let X,d:X×X[0,)X\neq\emptyset,d:X\times X\rightarrow[0,\infty), such that for each x,yXx,y\in X and u,vXu,v\in X (each distinct from xx and yy ), we have that
(1) d(x,y)=0x=yd(x,y)=0\Longleftrightarrow x=y,
(2) d(x,y)=d(y,x)d(x,y)=d(y,x),
(3) d(x,y)d(x,u)+d(u,v)+d(v,y)d(x,y)\leq d(x,u)+d(u,v)+d(v,y).

Furthermore, from [10] we mention that convergent sequences and Cauchy sequences can be introduced in a similar manner as in metric spaces.
Also, from the same paper, we know that if ( X,dX,d ) is a rectangular metric space and if ( xnx_{n} ) is a b-rectangular Cauchy sequence with the property that xnxmx_{n}\neq x_{m}, for each nmn\neq m, then (xn)\left(x_{n}\right) converge to at most one point, i.e. the property that ( X,dX,d ) is

Haussdorf becomes superfluous.
Moreover, from [8], [9], [22], we recall the definition of b-rectangular metric spaces (or b-generalized metric spaces), briefly b-g.m.s.

Definition 1.2. Let X,s1X\neq\emptyset,s\geq 1 be a given real number and d:X×X[0,)d:X\times X\rightarrow[0,\infty), such that for each x,yXx,y\in X and u,vXu,v\in X (each distinct from xx and yy ), we have that
(1) d(x,y)=0x=yd(x,y)=0\Longleftrightarrow x=y,
(2) d(x,y)=d(y,x)d(x,y)=d(y,x),
(3) d(x,y)s[d(x,u)+d(u,v)+d(v,y)]d(x,y)\leq s[d(x,u)+d(u,v)+d(v,y)].

As in metric spaces, we recall the basic notions regarding sequences in b-g.m.s:
Definition 1.3. Let ( X,dX,d ) be a b-g.m.s, xXx\in X and (xn)X\left(x_{n}\right)\subset X be a given sequence. Then
(a) ( xnx_{n} ) is convergent in ( X,dX,d ) to an element xXx\in X, if for each ε>0\varepsilon>0, there exists n0n_{0}\in\mathbb{N}, such that d(xn,x)<εd\left(x_{n},x\right)<\varepsilon, for each n>n0n>n_{0}. We denote this by limnxn=x\lim_{n\rightarrow\infty}x_{n}=x.
(b) ( xnx_{n} ) is Cauchy in ( X,dX,d ) (or b-rectangular Cauchy, briefly b-g.m.s.), if for each ε>0\varepsilon>0, there exists n0n_{0}\in\mathbb{N}, such that d(xn,xn+p)<εd\left(x_{n},x_{n+p}\right)<\varepsilon, for each n>n0n>n_{0} and for each p>0p>0. We denote this by limnd(xn,xn+p)=0\lim_{n\rightarrow\infty}d\left(x_{n},x_{n+p}\right)=0, for each p>0p>0.
(c) ( X,dX,d ) is said to be complete b-g.m.s, if every Cauchy sequence in X converges to some xXx\in X.

We recall the following important remark from [8]:
Remark 1.4. (1) Every metric space and every rectangular metric space (g.m.s) is b-g.m.s.
(2) The limit of a sequence in a b-rectangular metric space is not unique.
(3) Every convergent sequence in a b-g.m.s is not necessarily a b-g.m.s Cauchy.

For this, we recall a crucial lemma from [8], i.e. (Lemma 1.5), that specify when a b-rectangular Cauchy sequence can’t have two limits in a b-g.m.s.

Lemma 1.5. Let (X,d)(X,d) be a bb-rectangular metric space, with the coefficient s1s\geq 1. Let (xn)\left(x_{n}\right) be a b-rectangular Cauchy sequence in XX, such that xnxmx_{n}\neq x_{m}, for each nmn\neq m. Then ( xnx_{n} ) can converge to at most one point.

Also, we recall from [12] and [8] the following crucial lemma.
Lemma 1.6. Let (X,d)(X,d) be a bb-rectangular metric space, with the coefficient s1s\geq 1. Also, let (xn)\left(x_{n}\right) be a sequence for which xnxmx_{n}\neq x_{m}, for every nmn\neq m, with limnd(xn,xn+1)=0\lim_{n\rightarrow\infty}d\left(x_{n},x_{n+1}\right)=0. If (xn)\left(x_{n}\right) is not a b-rectangular Cauchy sequence, then there exists ε>0\varepsilon>0, such that for each kk\in\mathbb{N}, there exists ( m(k)m(k) ) and ( n(k)n(k) ) two sequences of positive integers, such that

d(xm(k),xn(k))ε\displaystyle d\left(x_{m(k)},x_{n(k)}\right)\geq\varepsilon
εslim supkd(xm(k),xn(k)2)ε and\displaystyle\frac{\varepsilon}{s}\leq\limsup_{k\rightarrow\infty}d\left(x_{m(k)},x_{n(k)-2}\right)\leq\varepsilon\text{ and }
εslim supkd(xm(k)+1,xn(k)1)\displaystyle\frac{\varepsilon}{s}\leq\limsup_{k\rightarrow\infty}d\left(x_{m(k)+1},x_{n(k)-1}\right)

In [22], another crucial lemma regarding sequences in b-rectangular metric spaces was presented. For convenience, we remind it below.

Lemma 1.7. Let (X,d)(X,d) be a b-g.m.s., with coefficient s1s\geq 1.
(a) Consider two sequences (xn)\left(x_{n}\right) and (yn)\left(y_{n}\right), such that xnx_{n} converges to xXx\in X and yny_{n} converges to yXy\in X, with xyx\neq y. Also, suppose that for each n,xnxn\in\mathbb{N},x_{n}\neq x and ynyy_{n}\neq y. Then

1sd(x,y)lim infnd(xn,yn)lim supnd(xn,yn)sd(x,y)\frac{1}{s}d(x,y)\leq\liminf_{n\rightarrow\infty}d\left(x_{n},y_{n}\right)\leq\limsup_{n\rightarrow\infty}d\left(x_{n},y_{n}\right)\leq sd(x,y)

(b) Consider an element yXy\in X and a b-rectangular Cauchy sequence ( xnx_{n} ), such that xnxmx_{n}\neq x_{m}, for each nmn\neq m. Moreover, suppose that the sequence ( xnx_{n} ) converges to an element xyx\neq y. Then

1sd(x,y)lim infnd(xn,y)lim supnd(xn,y)sd(x,y)\frac{1}{s}d(x,y)\leq\liminf_{n\rightarrow\infty}d\left(x_{n},y\right)\leq\limsup_{n\rightarrow\infty}d\left(x_{n},y\right)\leq sd(x,y)

Finally, for the convenience of the reader, we recall some important results in brectangular metric spaces. In [9], George et.al.studied basic contraction-type mappings in b-rectangular metric spaces, like Kannan operators, i.e.

d(Tx,Ty)λ[d(x,Tx)+d(y,Ty)], with λ[0,1s+1]d(Tx,Ty)\leq\lambda[d(x,Tx)+d(y,Ty)],\text{ with }\lambda\in\left[0,\frac{1}{s+1}\right]

In [8], Radenovic et.al. extended the results to mappings satisfying

d(fx,gy)ad(gx,gy)+b[d(gx,fx)+d(gy,fy)]d(fx,gy)\leq ad(gx,gy)+b[d(gx,fx)+d(gy,fy)]

for each x,yXx,y\in X and studied unique coincidence and common fixed points for the pair of operators ( f,gf,g ) that satisfies some additional assumptions.
Also, for more results in b-rectangular metric spaces and for a consistent survey on different generalized metric-type spaces, we recommend [11] and [12].
Now, regarding generalized contraction mappings we recall some recent advances in this subfield of fixed point theory.
In [13], Karapinar studied unique fixed points for some generalized contractions on cone Banach spaces satisfying the following contractive-type conditions

d(x,Tx)+d(y,Ty)pd(x,y), where p[0,2)d(x,Tx)+d(y,Ty)\leq pd(x,y),\text{ where }p\in[0,2)

and

ad(Tx,Ty)+b[d(x,Tx)+d(y,Ty)]sd(x,y), with 0s+|a|2b<2(a+b).ad(Tx,Ty)+b[d(x,Tx)+d(y,Ty)]\leq sd(x,y),\text{ with }0\leq s+|a|-2b<2(a+b).

Moreover, in 2009, Kumar [14] presented some theorems for two maps satisfying the following

d(fx,fy)qd(gx,gy), with q>1d(fx,fy)\geq qd(gx,gy),\text{ with }q>1

where ff is onto and gg is one-to-one.
Moosaei, Azizi, Asadi and Wang generalized the results of Karapinar as follows In [15], Moosaei used Krasnoselskii’s iteration defined in convex metric spaces, for the following mappings, that satisfy

d(Tx,Ty)+d(x,Tx)+d(y,Ty)rd(x,y), where r[2,5),d(Tx,Ty)+d(x,Tx)+d(y,Ty)\leq rd(x,y),\text{ where }r\in[2,5),

respectively

ad(x,fx)+bd(y,fy)+cd(fx,fy)kd(x,y), with 2b|c|k<2(a+b+c)|c|.ad(x,fx)+bd(y,fy)+cd(fx,fy)\leq kd(x,y),\text{ with }2b-|c|\leq k<2(a+b+c)-|c|.

In [17], Moosaei and Azizi extended the results to generalized contraction-type operators, studying coincidence points for various mappings, such as

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

where T(K)S(K),KT(K)\subset S(K),K and S(K)S(K) are closed and convex subsets of a convex metric space and the coefficients satisfy

2b|c|e<2(a+b+c)|c|.2b-|c|\leq e<2(a+b+c)-|c|.

Nevertheless, in 2014, Moosaei [16] studied a more generalized pair of contractions (S,T)(S,T), where

αd(Tx,Ty)+β[d(Sx,Tx)+d(Sy,Ty)]+γ[d(Sx,Ty)+d(Sy,Tx)]ηd(Sx,Sy),\alpha d(Tx,Ty)+\beta[d(Sx,Tx)+d(Sy,Ty)]+\gamma[d(Sx,Ty)+d(Sy,Tx)]\leq\eta d(Sx,Sy),

with some assumptions on contractive-coefficients, i.e.

2β+γ|γ|αη<α+2β+3γ|γ| and β+γ0.2\beta+\gamma-|\gamma|-\alpha\leq\eta<\alpha+2\beta+3\gamma-|\gamma|\text{ and }\beta+\gamma\leq 0.

Asadi in [3], using the same iteration (Krasnoselskii) on convex metric spaces, studied fixed points for generalized Hardy-Rogers type-mappings, as follows

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

where

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},

and λ[0,1]\lambda\in[0,1] is the coefficient of Krasnoselskii’s iteration.
Furthermore, Wang and Zhang, in [23] extended the above results for pairs of generalized Hardy-Rogers type contractions.
Now, expansive and expansive-type mappings can be considered a particular case of generalized contractions. Regarding the former ones, we recall some recent development into the study of this type of operators.
In 2011, Aage [1] considered expansive mappings in cone metric spaces. The more general form of these mappings, with some underlying assumptions, are

d(Tx,Ty)kd(x,y)+ld(x,Tx)+pd(y,Ty)d(Tx,Ty)\geq kd(x,y)+ld(x,Tx)+pd(y,Ty)

where TT satisfies K1,p<1,l>1K\geq-1,p<1,l>1 and k+l+p>1k+l+p>1.
Aydi et.al. studied in [4] some interesting fixed point theorems for pairs of expansive mappings for spaces endowed with c-distances. We recall them using the standard notations for metric spaces, i.e.

d(Tx,Ty)ad(fx,fy)+bd(Tx,fx)+cd(Ty,fy)d(Tx,Ty)\geq ad(fx,fy)+bd(Tx,fx)+cd(Ty,fy)

with b<1,a0,f(X)T(X)b<1,a\neq 0,f(X)\subseteq T(X) and (T(X),d)(X,d)(T(X),d)\subset(X,d) complete.
Also, in cone rectangular metric spaces, some fixed point theorems were developed. For example, in [20], pair of mappings satisfying

d(fx,fy)αd(gx,gy)+βd(fx,gx)+γd(fy,gy)d(fx,fy)\geq\alpha d(gx,gy)+\beta d(fx,gx)+\gamma d(fy,gy)

were studied, with some assumptions on the coefficients α,β\alpha,\beta and γ\gamma and on the range of gg and ff.
These pairs of generalized mappings were extended by Olaoluwa and Olaleru in [18], but in the framework of b-metric spaces and for a pair of four mappings, as follows

d(fx,gy)a1d(Sx,Ty)+a2d(fx,Sx)+a3d(gy,Ty)+a4d(fx,Ty)+a5d(gy,Sx).d(fx,gy)\geq a_{1}d(Sx,Ty)+a_{2}d(fx,Sx)+a_{3}d(gy,Ty)+a_{4}d(fx,Ty)+a_{5}d(gy,Sx).

Also, for the sake of convenience, we recall other studies in metric-type spaces and for expansive-type mappings, as follows: in [24] generalized mappings were studied on cone rectangular metric spaces using the technique of scalarizing, in [21] mappings that satisfy

d(Tx,Ty)φ(d(x,y))d(Tx,Ty)\leq\varphi(d(x,y))

were studied on cone rectangular metric spaces and in [19], fixed point theorems for a general type of expansive mappings were developed, satisfying

ϕ(d(S2x,TSy))13[d(Sx,S2x)+d(TSy,Sy)+d(Sx,Sy)].\phi\left(d\left(S^{2}x,TSy\right)\right)\geq\frac{1}{3}\left[d\left(Sx,S^{2}x\right)+d(TSy,Sy)+d(Sx,Sy)\right].

Also, in the context of dislocated metric spaces, Daheriya et.al. [7] studied rationaltype expansive mappings, and in [2] Alghamdi studied fixed points for generalized expansive mappings in b-metric like spaces.
The purpose of this work is to extend some fixed results for a hybrid class of generalized contractive-type mappings and for some expansive-type operators in the context of b-rectangular metric spaces. Moreover, at the end of the second section, we shall let and open problem.

2. Main results

Moosaei in [15] used Krasnoselskii iteration to develop fixed point theorems for generalized contractions on convex metric spaces. It is easily seen that we can use Picard instead of Krasnoselkii sequences in metric spaces.
In this section, our aim is to extend the results of Moosaei [15] for generalized contraction mappings from metric spaces to b-rectangular metric spaces. Also, we extend and develop the fixed point results of Aage [1] from cone metric spaces to b-g.m.s. Furthermore, we extend results from [20] of Patil, from rectangular metric spaces to b-rectangular ones (b-g.m.s).
Also, examples similar to those in [1], [12] and [20] justifying our theorems are given. Now, let’s consider generalized contractions f:XXf:X\rightarrow X on a b-g.m.s. XX, satisfying the following condition:

ad(x,fx)+bd(y,fy)+cd(fx,fy)kd(x,y).ad(x,fx)+bd(y,fy)+cd(fx,fy)\leq kd(x,y).

We will analyze two separate cases: when c>0c>0 and c<0c<0. Also, for expansive-type mappings, i.e. when c<0c<0, we consider two types of sequence, namely the classical Picard iteration xn+1=fxnx_{n+1}=fx_{n}, for each nn\in\mathbb{N} and the ’inverse’ Picard iteration, i.e. xn=fxn+1x_{n}=fx_{n+1}, for each nn\in\mathbb{N}, for which we require that the operator ff is onto.
Our first result is a theorem for the existence and uniqueness of the fixed point of a mapping satisfying the contractive condition from above. The technique we will use is based on the (Lemma 1.6).

Theorem 2.1. Let ( X,dX,d ) be a complete bb-rectangular metric space (b-gms), with coefficient s>1s>1. Consider a mapping f:XXf:X\rightarrow X, satisfying the following contractive condition

ad(x,fx)+bd(y,fy)+cd(fx,fy)kd(x,y), where 0kb<a+csad(x,fx)+bd(y,fy)+cd(fx,fy)\leq kd(x,y),\text{ where }0\leq k-b<\frac{a+c}{s}

Also, suppose the following assumptions are satisfied
(A) If c>0c>0 and k0k\geq 0, then kc<1s\frac{k}{c}<\frac{1}{s},
(B) If c>0c>0 and k0k\leq 0, then we have no additional conditions,
(C) If c<0c<0 and k<0k<0, then kc>s2\frac{k}{c}>s^{2}.

Then, the Picard sequence ( xnx_{n} ), defined as xn+1=fxnx_{n+1}=fx_{n}, for each nn\in\mathbb{N} converges to a fixed point of the mapping ff.

Proof. We consider the Picard iterative process ( xnx_{n} ), defined as xn+1=fxnx_{n+1}=fx_{n}, for each nn\in\mathbb{N}. Applying the contractive condition for the pair (xn1,xn)\left(x_{n-1},x_{n}\right), we get that

ad(xn,fxn)+bd(xn1,fxn1)+cd(fxn1,fxn)kd(xn1,xn)\displaystyle ad\left(x_{n},fx_{n}\right)+bd\left(x_{n-1},fx_{n-1}\right)+cd\left(fx_{n-1},fx_{n}\right)\leq kd\left(x_{n-1},x_{n}\right)
ad(xn,xn+1)+bd(xn1,xn)+cd(xn,xn+1)kd(xn1,xn)\displaystyle ad\left(x_{n},x_{n+1}\right)+bd\left(x_{n-1},x_{n}\right)+cd\left(x_{n},x_{n+1}\right)\leq kd\left(x_{n-1},x_{n}\right)
(a+c)d(xn,xn+1)(kb)d(xn1,xn)\displaystyle(a+c)d\left(x_{n},x_{n+1}\right)\leq(k-b)d\left(x_{n-1},x_{n}\right)

So d(xn,xn+1)δd(xn1,xn)d\left(x_{n},x_{n+1}\right)\leq\delta d\left(x_{n-1},x_{n}\right), where δ:=kba+c[0,1s)\delta:=\frac{k-b}{a+c}\in\left[0,\frac{1}{s}\right) from the theorem’s assumptions, since 0kb<a+cs0\leq k-b<\frac{a+c}{s}.
So d(xn,xn+1)δnd(x0,x1)d\left(x_{n},x_{n+1}\right)\leq\delta^{n}d\left(x_{0},x_{1}\right). Since δ[0,1s)\delta\in\left[0,\frac{1}{s}\right), it follows that limnd(xn,xn+1)=0\lim_{n\rightarrow\infty}d\left(x_{n},x_{n+1}\right)=0.
Also, by a routine argument (by reductio ad absurdum), it follows easily that xnxn+1x_{n}\neq x_{n+1}, for each nn\in\mathbb{N} and that xnxmx_{n}\neq x_{m}, for each nmn\neq m.
The next step is to show that the sequence ( xnx_{n} ) is b-rectangular Cauchy. We will use (Lemma 1.6) and we shall apply it on three different cases
(1) Case c>0c>0 : Let’s suppose that the sequence ( xnx_{n} ) is not b-rectangular Cauchy. Then, there exists ε>0\varepsilon>0 and two sequences of nonnegative real numbers ( m(k)m(k) ) and (n(k))(n(k)), such that the assumptions from (Lemma 1.6) are satisfied.
Now, we will apply the contraction condition for x=xm(k)x=x_{m(k)} and y=xn(k)2y=x_{n(k)-2}. It follows that
ad(xm(k),xm(k)+1)+bd(xn(k)2,xn(k)1)+cd(xm(k)+1,xn(k)1)kd(xm(k),xn(k)2)ad\left(x_{m(k)},x_{m(k)+1}\right)+bd\left(x_{n(k)-2},x_{n(k)-1}\right)+cd\left(x_{m(k)+1},x_{n(k)-1}\right)\leq kd\left(x_{m(k)},x_{n(k)-2}\right)
cd(xm(k)+1,xn(k)1)kd(xm(k),xn(k)2)ad(xm(k),xm(k)+1)bd(xn(k)2,xn(k)1)cd\left(x_{m(k)+1},x_{n(k)-1}\right)\leq kd\left(x_{m(k)},x_{n(k)-2}\right)-ad\left(x_{m(k)},x_{m(k)+1}\right)-bd\left(x_{n(k)-2},x_{n(k)-1}\right).
Because c>0c>0, we have that
d(xm(k)+1,xn(k)1)kcd(xm(k),xn(k)2)acd(xm(k),xm(k)+1)bcd(xn(k)2,xn(k)1)d\left(x_{m(k)+1},x_{n(k)-1}\right)\leq\frac{k}{c}d\left(x_{m(k)},x_{n(k)-2}\right)-\frac{a}{c}d\left(x_{m(k)},x_{m(k)+1}\right)-\frac{b}{c}d\left(x_{n(k)-2},x_{n(k)-1}\right).
Now, we want to apply the limit superior. We make the following necessary remark and consider the following cases
If a0a\geq 0, then ac0-\frac{a}{c}\leq 0, so acd(xm(k),xm(k)+1)0-\frac{a}{c}d\left(x_{m(k)},x_{m(k)+1}\right)\leq 0, so an upper bound for this element is 0 .

If a0a\leq 0, then ac0-\frac{a}{c}\geq 0, so acd(xm(k),xm(k)+1)0-\frac{a}{c}d\left(x_{m(k)},x_{m(k)+1}\right)\geq 0. Applying the limit superior, we get that

lim supk(ac)d(xm(k),xm(k)+1)\displaystyle\limsup_{k\rightarrow\infty}\left(-\frac{a}{c}\right)d\left(x_{m(k)},x_{m(k)+1}\right) =(ac)lim supkd(xm(k),xm(k)+1)\displaystyle=\left(-\frac{a}{c}\right)\limsup_{k\rightarrow\infty}d\left(x_{m(k)},x_{m(k)+1}\right)
=(ac)limkd(xm(k),xm(k)+1)=0\displaystyle=\left(-\frac{a}{c}\right)\lim_{k\rightarrow\infty}d\left(x_{m(k)},x_{m(k)+1}\right)=0

The same reasoning can be made about the sign of the coefficient bb and about the limit superior of the sequence (d(xn(k)2,xn(k)1))\left(d\left(x_{n(k)-2},x_{n(k)-1}\right)\right) as a subsequence of (d(xn,xn1))\left(d\left(x_{n},x_{n-1}\right)\right).
Case (A): When k0k\geq 0.
Since k0k\geq 0, we have that kc0\frac{k}{c}\geq 0. We know that lim supkd(xm(k),xn(k)2)ε\limsup_{k\rightarrow\infty}d\left(x_{m(k)},x_{n(k)-2}\right)\leq\varepsilon. Multiplying by (kc)\left(\frac{k}{c}\right) and taking the limit superior, we get that

lim supk(kc)d(xm(k),xn(k)2)\displaystyle\limsup_{k\rightarrow\infty}\left(\frac{k}{c}\right)d\left(x_{m(k)},x_{n(k)-2}\right) =lim supk|kc|d(xm(k),xn(k)2)\displaystyle=\limsup_{k\rightarrow\infty}\left|\frac{k}{c}\right|d\left(x_{m(k)},x_{n(k)-2}\right)
=kclim supkd(xm(k),xn(k)2)kcε\displaystyle=\frac{k}{c}\limsup_{k\rightarrow\infty}d\left(x_{m(k)},x_{n(k)-2}\right)\leq\frac{k}{c}\varepsilon

From (Lemma 1.6), it follows that εslim supkd(xm(k)+1,xm(k)1)kcε\frac{\varepsilon}{s}\leq\limsup_{k\rightarrow\infty}d\left(x_{m(k)+1},x_{m(k)-1}\right)\leq\frac{k}{c}\varepsilon, so 1skc\frac{1}{s}\leq\frac{k}{c}. This is a contradiction with the assumption that in this case we have kc<1s\frac{k}{c}<\frac{1}{s}.
Case (B): When k0k\leq 0.
In this case we have that kc0\frac{k}{c}\leq 0, so kcd(xm(k),xn(k)2)0\frac{k}{c}d\left(x_{m(k)},x_{n(k)-2}\right)\leq 0, then we can take 0 as an upper bound for it. By (Lemma 1.6), we have that εs0\frac{\varepsilon}{s}\leq 0. Since ε>0\varepsilon>0 and s1s\geq 1, we got a contradiction.
Now, in the two cases from above, we have shown that (xn)\left(x_{n}\right) is b-rectangular Cauchy. Moreover, we have said that xnxmx_{n}\neq x_{m}, for each nmn\neq m.
Since ( X,dX,d ) is complete, it implies that there exists uXu\in X, such that xnux_{n}\rightarrow u, i.e.

limnd(xn,u)=0\lim_{n\rightarrow\infty}d\left(x_{n},u\right)=0

Now, we shall show that uu is a fixed point for ff

d(u,fu)\displaystyle d(u,fu) s[d(u,xn)+d(xn,xn+1)+d(xn+1,fu)]\displaystyle\leq s\left[d\left(u,x_{n}\right)+d\left(x_{n},x_{n+1}\right)+d\left(x_{n+1},fu\right)\right]
=s[d(u,xn)+d(xn,xn+1)+d(fxn,fu)]\displaystyle=s\left[d\left(u,x_{n}\right)+d\left(x_{n},x_{n+1}\right)+d\left(fx_{n},fu\right)\right]

Since c>0c>0, then

d(fxn,fu)kcd(xn,u)bcd(xn,xn+1)acd(u,fu)d\left(fx_{n},fu\right)\leq\frac{k}{c}d\left(x_{n},u\right)-\frac{b}{c}d\left(x_{n},x_{n+1}\right)-\frac{a}{c}d(u,fu)

So

d(u,fu)s[d(u,xn)+d(xn,xn+1)+kcd(xn,u)bcd(xn,xn+1)acd(u,fu)]d(u,fu)\leq s\left[d\left(u,x_{n}\right)+d\left(x_{n},x_{n+1}\right)+\frac{k}{c}d\left(x_{n},u\right)-\frac{b}{c}d\left(x_{n},x_{n+1}\right)-\frac{a}{c}d(u,fu)\right]

Taking the limit when nn\rightarrow\infty, we get

(1+sac)d(u,fu)0\left(1+s\frac{a}{c}\right)d(u,fu)\leq 0

so (c+sa)d(u,fu)0(c+sa)d(u,fu)\leq 0. Furthermore, since c>0c>0 and 0<(a+c)<(a+cs)0<(a+c)<(a+cs), then uu is a fixed point for ff.
(2) Case c<0c<0 : We have that

ad(x,fx)+bd(y,fy)+cd(fx,fy)kd(x,y)\displaystyle ad(x,fx)+bd(y,fy)+cd(fx,fy)\leq kd(x,y)
cd(fx,fy)kd(x,y)ad(x,fx)bd(y,fy)\displaystyle cd(fx,fy)\leq kd(x,y)-ad(x,fx)-bd(y,fy)

So

d(fx,fy)kcd(x,y)acd(x,fx)bcd(y,fy)d(fx,fy)\geq\frac{k}{c}d(x,y)-\frac{a}{c}d(x,fx)-\frac{b}{c}d(y,fy)

This is a case of expansive-type mapping. By (Lemma 1.6), there exists ε>0\varepsilon>0, such that for every kk\in\mathbb{N}, there exists (m(k)),(n(k))(m(k)),(n(k)) two sequences of nonnegative real numbers such that the assumptions in the already mentioned lemma are true. By b-rectangular inequality, we have that

d(xn(k)2,xm(k))s[d(xm(k)1,xn(k)3)+d(xn(k)3,xn(k)2)+d(xm(k)1,xm(k))]\displaystyle d\left(x_{n(k)-2},x_{m(k)}\right)\leq s\left[d\left(x_{m(k)-1},x_{n(k)-3}\right)+d\left(x_{n(k)-3},x_{n(k)-2}\right)+d\left(x_{m(k)-1},x_{m(k)}\right)\right]
sd(xm(k)1,xn(k)3)d(xn(k)2,xm(k))sd(xn(k)3,xn(k)2)sd(xm(k)1,xm(k))\displaystyle sd\left(x_{m(k)-1},x_{n(k)-3}\right)\geq d\left(x_{n(k)-2},x_{m(k)}\right)-sd\left(x_{n(k)-3},x_{n(k)-2}\right)-sd\left(x_{m(k)-1},x_{m(k)}\right)

Dividing by s1s\geq 1, we obtain the following

d(xm(k)1,xn(k)3)1sd(xn(k)2,xm(k))d(xn(k)3,xn(k)2)d(xm(k)1,xm(k))d\left(x_{m(k)-1},x_{n(k)-3}\right)\geq\frac{1}{s}d\left(x_{n(k)-2},x_{m(k)}\right)-d\left(x_{n(k)-3},x_{n(k)-2}\right)-d\left(x_{m(k)-1},x_{m(k)}\right)

Case (C): When k<0k<0 : Here we have that kc0\frac{k}{c}\geq 0. Multiplying by (kc)\left(\frac{k}{c}\right), it implies that

kcd(xm(k)1,xn(k)3)\displaystyle\frac{k}{c}d\left(x_{m(k)-1},x_{n(k)-3}\right) kcsd(xn(k)2,xm(k))kcd(xn(k)3,xn(k)2)\displaystyle\geq\frac{k}{cs}d\left(x_{n(k)-2},x_{m(k)}\right)-\frac{k}{c}d\left(x_{n(k)-3},x_{n(k)-2}\right)
kcd(xm(k)1,xm(k))\displaystyle-\frac{k}{c}d\left(x_{m(k)-1},x_{m(k)}\right)

Now, we apply the contractive condition for x=xm(k)1x=x_{m(k)-1} and y=xn(k)3y=x_{n(k)-3}, i.e.
d(xm(k),xn(k)2)kcd(xm(k)1,xn(k)3)acd(xm(k)1,xm(k))bcd(xn(k)3,xn(k)2)d\left(x_{m(k)},x_{n(k)-2}\right)\geq\frac{k}{c}d\left(x_{m(k)-1},x_{n(k)-3}\right)-\frac{a}{c}d\left(x_{m(k)-1},x_{m(k)}\right)-\frac{b}{c}d\left(x_{n(k)-3},x_{n(k)-2}\right).
So, combining the above inequalities, we get that

d(xm(k),xn(k)2)\displaystyle d\left(x_{m(k)},x_{n(k)-2}\right) kcsd(xn(k)2,xm(k))kcd(xn(k)3,xn(k)2)kcd(xm(k)1,xm(k))\displaystyle\geq\frac{k}{cs}d\left(x_{n(k)-2},x_{m(k)}\right)-\frac{k}{c}d\left(x_{n(k)-3},x_{n(k)-2}\right)-\frac{k}{c}d\left(x_{m(k)-1},x_{m(k)}\right)
acd(xm(k)1,xm(k))bcd(xn(k)3,xn(k)2)\displaystyle-\frac{a}{c}d\left(x_{m(k)-1},x_{m(k)}\right)-\frac{b}{c}d\left(x_{n(k)-3},x_{n(k)-2}\right)

From the limit superior, we have get the following

lim supk(kc)d(xn(k)3,xn(k)2)\displaystyle\limsup_{k\rightarrow\infty}\left(-\frac{k}{c}\right)d\left(x_{n(k)-3},x_{n(k)-2}\right) =kclim supkd(xn(k)3,xn(k)2)\displaystyle=\frac{k}{c}\limsup_{k\rightarrow\infty}-d\left(x_{n(k)-3},x_{n(k)-2}\right)
=kclim infkd(xn(k)3,xn(k)2)\displaystyle=-\frac{k}{c}\liminf_{k\rightarrow\infty}d\left(x_{n(k)-3},x_{n(k)-2}\right)
=(kc)limkd(xn(k)3,xn(k)2)=0\displaystyle=\left(-\frac{k}{c}\right)\lim_{k\rightarrow\infty}d\left(x_{n(k)-3},x_{n(k)-2}\right)=0

We have the same reasoning for d(xm(k)1,xm(k))d\left(x_{m(k)-1},x_{m(k)}\right), with coefficient kc-\frac{k}{c}. Also, for coefficients aa and bb, we have that
If a0a\geq 0, then ac0-\frac{a}{c}\geq 0, so (ac)d(xm(k)1,xm(k))0\left(-\frac{a}{c}\right)d\left(x_{m(k)-1},x_{m(k)}\right)\geq 0, so we can make the lower bound 0 .
If a0a\leq 0, then ac0-\frac{a}{c}\leq 0, so (ac)d(xm(k)1,xm(k))0\left(-\frac{a}{c}\right)d\left(x_{m(k)-1},x_{m(k)}\right)\leq 0, so taking the limit superior, it follows that:

lim supk(ac)d(xm(k)1,xm(k))\displaystyle\limsup_{k\rightarrow\infty}\left(-\frac{a}{c}\right)d\left(x_{m(k)-1},x_{m(k)}\right) =aclim supkd(xm(k)1,xm(k))\displaystyle=\frac{a}{c}\limsup_{k\rightarrow\infty}-d\left(x_{m(k)-1},x_{m(k)}\right)
=aclim infkd(xm(k)1,xm(k))\displaystyle=-\frac{a}{c}\liminf_{k\rightarrow\infty}d\left(x_{m(k)-1},x_{m(k)}\right)
=aclimkd(xm(k)1,xm(k))=0\displaystyle=-\frac{a}{c}\lim_{k\rightarrow\infty}d\left(x_{m(k)-1},x_{m(k)}\right)=0

Same remarks can be made about the coefficient bb and for d(xn(k)3,xn(k)2)d\left(x_{n(k)-3},x_{n(k)-2}\right).
By (Lemma 1.6), we get that

εlim supkd(xm(k),xn(k)2)kcslim supkd(xm(k),xn(k)2)εkcs2.\varepsilon\geq\limsup_{k\rightarrow\infty}d\left(x_{m(k)},x_{n(k)-2}\right)\geq\frac{k}{cs}\limsup_{k\rightarrow\infty}d\left(x_{m(k)},x_{n(k)-2}\right)\geq\frac{\varepsilon k}{cs^{2}}.

So 1s2ck\frac{1}{s^{2}}\leq\frac{c}{k}. This is a contradiction with the fact that in this case kc>s2\frac{k}{c}>s^{2}.
Now, since xnxmx_{n}\neq x_{m}, for each nm,d(xn,xn+1)0,(xn)n\neq m,d\left(x_{n},x_{n+1}\right)\rightarrow 0,\left(x_{n}\right) Cauchy b-rectangular and ( X,dX,d ) is complete, then there exists uXu\in X, such that xnux_{n}\rightarrow u. We shall show that uu is a fixed point for the mapping ff.
Applying the contractive condition on the pair ( u,xnu,x_{n} ), we get

ad(u,fu)+bd(xn,fxn)+cd(fu,fxn)\displaystyle ad(u,fu)+bd\left(x_{n},fx_{n}\right)+cd\left(fu,fx_{n}\right) kd(u,xn)\displaystyle\leq kd\left(u,x_{n}\right)
ad(u,fu)+bd(xn,xn+1)+cd(fu,xn+1)\displaystyle ad(u,fu)+bd\left(x_{n},x_{n+1}\right)+cd\left(fu,x_{n+1}\right) kd(u,xn)\displaystyle\leq kd\left(u,x_{n}\right)

Letting nn\rightarrow\infty, we have (a+c)d(u,fu)0(a+c)d(u,fu)\leq 0 and since we know that a+c>0a+c>0, it follows that uu is a fixed point for the mapping ff.

Relative to (Theorem 2.1), we give two examples that validate cases (A) and(C): From [12], we recall an example of a complete b-rectangular metric space.

Example 2.2. Let X=ABX=A\cup B, where A={1n|n=2,5¯}A=\left\{\left.\frac{1}{n}\right\rvert\,n=\overline{2,5}\right\} and B=[1,2]B=[1,2]. We define d:X×X[0,)d:X\times X\rightarrow[0,\infty), such that d(x,y)=d(y,x)d(x,y)=d(y,x) and

d(12,13)=d(14,15)=3100,d\left(\frac{1}{2},\frac{1}{3}\right)=d\left(\frac{1}{4},\frac{1}{5}\right)=\frac{3}{100},
d(12,15)=d(13,14)=2100\displaystyle d\left(\frac{1}{2},\frac{1}{5}\right)=d\left(\frac{1}{3},\frac{1}{4}\right)=\frac{2}{100}
d(14,13)=d(15,13)=6100\displaystyle d\left(\frac{1}{4},\frac{1}{3}\right)=d\left(\frac{1}{5},\frac{1}{3}\right)=\frac{6}{100}

d(x,y)=(xy)2d(x,y)=(x-y)^{2}, otherwise.
Then ( X,dX,d ) is a complete b-rectangular metric space, with coefficient s=3s=3. Furthermore, ( X,dX,d ) is not a metric space or a rectangular metric space.

Regarding case (A) of (Theorem 2.1), we give the following example.
Example 2.3. Let (X,d)(X,d) be the b-rectangular metric space defined above, with s=3s=3. Also, define f:XXf:X\rightarrow X, such as

f(x)={13,xA15,xBf(x)=\begin{cases}\frac{1}{3},&x\in A\\ \frac{1}{5},&x\in B\end{cases}

It is easy to observe that ff has a unique fixed point 13\frac{1}{3}. Moreover, we shall show that ff satisfies

1d(fx,fy)152d(x,y)+14d(x,fx)+23100d(y,fy)1\cdot d(fx,fy)\leq\frac{1}{52}d(x,y)+\frac{1}{4}d(x,fx)+\frac{23}{100}d(y,fy)

for each x,yXx,y\in X.
Let’s define: a=14,b=23100,k=152,c=1a=\frac{-1}{4},b=\frac{-23}{100},k=\frac{1}{52},c=1 and s=3s=3.
We have the following cases

  1. 1.

    xAx\in A and yA:d(fx,fy)=d(13,13)=0y\in A:d(fx,fy)=d\left(\frac{1}{3},\frac{1}{3}\right)=0, so the above inequality is valid.

  2. 2.

    xBx\in B and yB:d(fx,fy)=d(15,15)=0y\in B:d(fx,fy)=d\left(\frac{1}{5},\frac{1}{5}\right)=0, so the inequality of ff is true.

Now, for the non-trivial cases, it follows that:
3) xAx\in A and yBy\in B :

d(fx,fy)=(13,15)=6100d(x,fx)=d(x,13)minxAd(x,13)=1200d(y,fy)=d(y,15)=(y15)2=y225y+125miny[1,2]=114+125=625\begin{gathered}d(fx,fy)=\left(\frac{1}{3},\frac{1}{5}\right)=\frac{6}{100}\\ d(x,fx)=d\left(x,\frac{1}{3}\right)\geq\min_{x\in A}d\left(x,\frac{1}{3}\right)=\frac{1}{200}\\ d(y,fy)=d\left(y,\frac{1}{5}\right)=\left(y-\frac{1}{5}\right)^{2}=y^{2}-\frac{2}{5}y+\frac{1}{25}\geq\min_{y\in[1,2]}=1-\frac{1}{4}+\frac{1}{25}=\frac{6}{25}\end{gathered}

Also d(x,y)=(yx)2=|yx|2d(x,y)=(y-x)^{2}=|y-x|^{2}.
We have that

d(fx,fy)kd(x,y)+(a)minxAd(x,fx)+(b)minyBd(y,fy)d(fx,fy)\leq kd(x,y)+(-a)\min_{x\in A}d(x,fx)+(-b)\min_{y\in B}d(y,fy)

So

6100152|yx|2+141200+23100625\frac{6}{100}\leq\frac{1}{52}|y-x|^{2}+\frac{1}{4}\cdot\frac{1}{200}+\frac{23}{100}\cdot\frac{6}{25}

so 152|yx|2661912000\frac{1}{52}|y-x|^{2}\geq\frac{-6619}{12000}, which is obviously true.
4) xBx\in B and yAy\in A

d(fx,fy)=(13,15)=6100d(x,fx)minxBd(x,fx)=625d(y,fy)minyA=1200\begin{gathered}d(fx,fy)=\left(\frac{1}{3},\frac{1}{5}\right)=\frac{6}{100}\\ d(x,fx)\geq\min_{x\in B}d(x,fx)=\frac{6}{25}\\ d(y,fy)\geq\min_{y\in A}=\frac{1}{200}\end{gathered}

and

d(x,y)=(yx)2=|yx|2.d(x,y)=(y-x)^{2}=|y-x|^{2}.

We have that

6100152|yx|2+14625+231001200,\frac{6}{100}\leq\frac{1}{52}|y-x|^{2}+\frac{1}{4}\cdot\frac{6}{25}+\frac{23}{100}\cdot\frac{1}{200},

so 152|yx|24196000\frac{1}{52}|y-x|^{2}\geq\frac{-419}{6000}, which is also true.
Moreover, we show that the conditions from (Theorem 2.1) - case ( AA ) on the coefficients are satisfied

c>01>0\displaystyle c>0\Leftrightarrow 1>0
k>0152>0\displaystyle k>0\Leftrightarrow\frac{1}{52}>0
a+c=114=34>0\displaystyle a+c=1-\frac{1}{4}=\frac{3}{4}>0
bk23100152\displaystyle b\leq k\Leftrightarrow-\frac{23}{100}\leq\frac{1}{52}
kc<1sk<133<52\displaystyle\frac{k}{c}<\frac{1}{s}\Leftrightarrow k<\frac{1}{3}\Leftrightarrow 3<2
k<b+a+cs152+23100<14324<325\displaystyle k<b+\frac{a+c}{s}\Leftrightarrow\frac{1}{52}+\frac{23}{100}<\frac{1}{4}\Leftrightarrow 24<25

Now, we construct an example of a complete b-rectangular metric space, which will be used further in this section.

Example 2.4. Let X={1,2,3,4}X=\{1,2,3,4\} and define d:X×X[0,)d:X\times X\rightarrow[0,\infty), such as

d(1,2)=d(2,1)=610\displaystyle d(1,2)=d(2,1)=\frac{6}{10}
d(1,3)=d(3,1)=110\displaystyle d(1,3)=d(3,1)=\frac{1}{10}
d(2,3)=d(3,2)=110\displaystyle d(2,3)=d(3,2)=\frac{1}{10}
d(1,4)=d(4,1)=d(2,4)=d(4,2)=d(3,4)=d(4,3)=210\displaystyle d(1,4)=d(4,1)=d(2,4)=d(4,2)=d(3,4)=d(4,3)=\frac{2}{10}

We will prove that (X,d)(X,d) is a b-rectangular metric space with coefficient s=32s=\frac{3}{2}, which is not a rectangular metric space.

For a b-rectangular metric space, we have that d(x,y)s[d(x,u)+d(u,v)+d(v,y)]d(x,y)\leq s[d(x,u)+d(u,v)+d(v,y)], for each u,v{x,y}u,v\notin\{x,y\}, with u,vu,v being distinct. We have the following cases.

  • When x=yx=y, the right hand side is 0 , so the above inequality remains valid.

  • When xyx\neq y, we employ the following sub-cases

Case (1): If x=1x=1 and y=2(x=2y=2(x=2 and y=1y=1 by symmetry):

610\displaystyle\frac{6}{10} s[d(1,u)+d(u,v)+d(v,2)], for u,v{1,2}, i.e. u,vI1={3,4}\displaystyle\leq s[d(1,u)+d(u,v)+d(v,2)],\text{ for }u,v\notin\{1,2\},\text{ i.e. }u,v\in I_{1}=\{3,4\}
610\displaystyle\frac{6}{10} =d(1,2)s[minuI1d(1,u)+d(3,4)+minvI1d(v,2)]\displaystyle=d(1,2)\leq s\left[\min_{u\in I_{1}}d(1,u)+d(3,4)+\min_{v\in I_{1}}d(v,2)\right]
610\displaystyle\frac{6}{10} s[110+210+110], so s32\displaystyle\leq s\left[\frac{1}{10}+\frac{2}{10}+\frac{1}{10}\right],\text{ so }s\geq\frac{3}{2}

Case (2): If x=3x=3 and y=1y=1 ( x=1x=1 and y=3y=3 by symmetry):

110\displaystyle\frac{1}{10} s[d(3,u)+d(u,v)+d(v,1)], for u,v{1,3}, i.e. u,vI2={2,4}\displaystyle\leq s[d(3,u)+d(u,v)+d(v,1)],\text{ for }u,v\notin\{1,3\},\text{ i.e. }u,v\in I_{2}=\{2,4\}
110\displaystyle\frac{1}{10} =d(3,1)s[minuI2d(3,u)+d(2,4)+minvI2d(v,1)]\displaystyle=d(3,1)\leq s\left[\min_{u\in I_{2}}d(3,u)+d(2,4)+\min_{v\in I_{2}}d(v,1)\right]
110\displaystyle\frac{1}{10} s[110+210+110], so s14\displaystyle\leq s\left[\frac{1}{10}+\frac{2}{10}+\frac{1}{10}\right],\text{ so }s\geq\frac{1}{4}

Case (3): If x=4x=4 and y=1y=1 ( x=1x=1 and y=4y=4 by symmetry):

210\displaystyle\frac{2}{10} s[d(3,u)+d(u,v)+d(v,1)], for u,v{1,4}, i.e. u,vI3={2,3}\displaystyle\leq s[d(3,u)+d(u,v)+d(v,1)],\text{ for }u,v\notin\{1,4\},\text{ i.e. }u,v\in I_{3}=\{2,3\}
210\displaystyle\frac{2}{10} =d(4,1)s[minuI3d(4,u)+d(2,3)+minvI3d(v,1)]\displaystyle=d(4,1)\leq s\left[\min_{u\in I_{3}}d(4,u)+d(2,3)+\min_{v\in I_{3}}d(v,1)\right]
210\displaystyle\frac{2}{10} s[210+110+110], so s12\displaystyle\leq s\left[\frac{2}{10}+\frac{1}{10}+\frac{1}{10}\right],\text{ so }s\geq\frac{1}{2}

Case (4): If x=2x=2 and y=4(x=4y=4(x=4 and y=2y=2 by symmetry):

210\displaystyle\frac{2}{10} s[d(2,u)+d(u,v)+d(v,4)], for u,v{2,4}, i.e. u,vI4={1,3}\displaystyle\leq s[d(2,u)+d(u,v)+d(v,4)],\text{ for }u,v\notin\{2,4\},\text{ i.e. }u,v\in I_{4}=\{1,3\}
210\displaystyle\frac{2}{10} =d(4,2)s[minuI4d(2,u)+d(1,3)+minvI4d(v,4)]\displaystyle=d(4,2)\leq s\left[\min_{u\in I_{4}}d(2,u)+d(1,3)+\min_{v\in I_{4}}d(v,4)\right]
210\displaystyle\frac{2}{10} s[110+110+210], so s12\displaystyle\leq s\left[\frac{1}{10}+\frac{1}{10}+\frac{2}{10}\right],\text{ so }s\geq\frac{1}{2}

Case (5): If x=3x=3 and y=4(x=4y=4(x=4 and y=3y=3 by symmetry):

210\displaystyle\frac{2}{10} s[d(3,u)+d(u,v)+d(v,4)], for u,v{3,4}, i.e. u,vI5={1,2}\displaystyle\leq s[d(3,u)+d(u,v)+d(v,4)],\text{ for }u,v\notin\{3,4\},\text{ i.e. }u,v\in I_{5}=\{1,2\}
210\displaystyle\frac{2}{10} =d(3,4)s[minuI5d(3,u)+d(1,2)+minvI5d(4,v)]\displaystyle=d(3,4)\leq s\left[\min_{u\in I_{5}}d(3,u)+d(1,2)+\min_{v\in I_{5}}d(4,v)\right]
210\displaystyle\frac{2}{10} s[110+610+210], so s29\displaystyle\leq s\left[\frac{1}{10}+\frac{6}{10}+\frac{2}{10}\right],\text{ so }s\geq\frac{2}{9}

So s32>1s\geq\frac{3}{2}>1,so we can take s=32s=\frac{3}{2}.
Furthermore, ( X,dX,d ) is not a b-g.m.s., because

610=d(1,2)>d(1,3)+d(3,u)+d(u,2)=110+210+210=510,\frac{6}{10}=d(1,2)>d(1,3)+d(3,u)+d(u,2)=\frac{1}{10}+\frac{2}{10}+\frac{2}{10}=\frac{5}{10},

so 6>56>5, which is valid.
Now, we construct an example, justifying case ( CC ) of (Theorem 2.1).
Example 2.5. Let X={1,2,3,4}X=\{1,2,3,4\} the b-rectangular metric space defined above, with coefficient s=32s=\frac{3}{2}.

Let f(x)={3,x41,x=4f(x)=\left\{\begin{array}[]{ll}3,&x\neq 4\\ 1,&x=4\end{array}\right. a self-mapping defined on XX.
We shall show that ff satisfies

d(fx,fy)(3)d(x,y)5d(x,fx)+3d(y,fy)d(fx,fy)\geq(-3)d(x,y)-5d(x,fx)+3d(y,fy)

and also the conditions from case ( CC ) of (Theorem 2.1).
Let ff satisfy cd(fx,fy)kd(x,y)ad(x,fx)bd(y,fy)cd(fx,fy)\geq kd(x,y)-ad(x,fx)-bd(y,fy). Let’s normalize the contractive condition, by taking c=1<0c=-1<0 We shall determine the coefficients k,a,bk,a,b, with k<0,a>0k<0,a>0 and b<0b<0. We have the following cases

  1. 1.

    If x=yx=y, then d(fx,fy)=d(fx,fx)=0d(fx,fy)=d(fx,fx)=0, so the left hand side is 0 . Now, the right hand side is k0ad(x,fx)bd(x,fx)=(a+b)d(x,f)k\cdot 0-ad(x,fx)-bd(x,fx)=-(a+b)d(x,f). This implies that (a+b)d(x,fx)0(a+b)d(x,fx)\geq 0. We have two sub-cases:
    If x=3x=3, then d(x,fx)=d(3,3)=0d(x,fx)=d(3,3)=0, so the inequality is valid. Also, if 3\neq 3, then d(x,fx)>0d(x,fx)>0, so we have the condition that ba-b\leq a.

  2. 2.

    If y\neq y, we have the following sub-cases
    a) For x=4x=4 and y4y\neq 4, it follows that d(fy,fx)=d(fy,1)d(fy,fx)=d(fy,1). Since y4y\neq 4, then fy=3fy=3, so d(fx,fy)=d(1,3)=110d(fx,fy)=d(1,3)=\frac{1}{10}.
    Moreover, one can easily verify that d(x,y)=d(4,y)=210d(x,y)=d(4,y)=\frac{2}{10}, for each y4y\neq 4,
    d(x,fx)=d(4,fx)=210d(x,fx)=d(4,fx)=\frac{2}{10}, for each xXx\in X and d(y,fy)=d(y,3)maxy4d(y,3)=210d(y,fy)=d(y,3)\leq\max_{y\neq 4}d(y,3)=\frac{2}{10}.
    b) For y=4y=4 and x4x\neq 4, it follows that d(fx,fy)=110d(fx,fy)=\frac{1}{10}.

Moreover, we have that d(x,y)=d(4,x)=210d(x,y)=d(4,x)=\frac{2}{10}, for each x4x\neq 4,
d(x,fx)=d(x,3)=minx4d(x,3)=110d(x,fx)=d(x,3)=\geq\min_{x\neq 4}d(x,3)=\frac{1}{10} and d(y,fy)=d(4,fy)=210d(y,fy)=d(4,fy)=\frac{2}{10}, for each value of fyfy.
c) For yy4y\neq y\neq 4 (simultaneously), it follows that d(fx,fy)=d(3,3)=0d(fx,fy)=d(3,3)=0. Also

kd(x,y)ad(x,fx)bd(y,fy)0, so kd(x,y)bd(x,y)ad(x,fx).kd(x,y)-ad(x,fx)-bd(y,fy)\leq 0,\quad\text{ so }\quad kd(x,y)-bd(x,y)\leq ad(x,fx).

Now d(x,y)minx,yXd(x,y)=110d(x,y)\geq\min_{x,y\in X}d(x,y)=\frac{1}{10}.
Furthermore, we have that

d(y,fy)=d(y,3)maxy4d(y,3)=210 and d(x,fx)=d(x,3)minx4d(x,3)=110d(y,fy)=d(y,3)\leq\max_{y\neq 4}d(y,3)=\frac{2}{10}\quad\text{ and }\quad d(x,fx)=d(x,3)\geq\min_{x\neq 4}d(x,3)=\frac{1}{10}

Now, we analyze the conditions on ff.
For the case (1), we get ba-b\leq a. For the case (2a), we get that

d(fx,fy)\displaystyle d(fx,fy) =110kd(x,y)ad(x,fx)bmaxy4d(y,fy)\displaystyle=\frac{1}{10}\geq kd(x,y)-ad(x,fx)-b\max_{y\neq 4}d(y,fy)
=2k102a102b102k102a10bd(y,fy)\displaystyle=\frac{2k}{10}-\frac{2a}{10}-\frac{2b}{10}\geq\frac{2k}{10}-\frac{2a}{10}-bd(y,fy)

because b<0b<0. So k<a+b+12k<a+b+\frac{1}{2}.
For the case (2b), we obtain

d(fx,fy)\displaystyle d(fx,fy) =110kd(x,y)aminx4d(x,fx)bd(y,fy)\displaystyle=\frac{1}{10}\geq kd(x,y)-a\min_{x\neq 4}d(x,fx)-bd(y,fy)
=2k10a102b102k102b10ad(x,fx)\displaystyle=\frac{2k}{10}-\frac{a}{10}-\frac{2b}{10}\geq\frac{2k}{10}-\frac{2b}{10}-ad(x,fx)

because a>0a>0. So k<a2+b+12k<\frac{a}{2}+b+\frac{1}{2}.
For the case (2c), it follows that

d(fx,fy)\displaystyle d(fx,fy) =0kminxy4d(x,y)aminx4d(x,fx)bmaxy4d(y,fy)\displaystyle=0\geq k\min_{x\neq y\neq 4}d(x,y)-a\min_{x\neq 4}d(x,fx)-b\max_{y\neq 4}d(y,fy)
=k10a102b10kd(x,y)ad(x,fx)bd(y,fy)\displaystyle=\frac{k}{10}-\frac{a}{10}-\frac{2b}{10}kd(x,y)-ad(x,fx)-bd(y,fy)

because b,k<0b,k<0 and a>0a>0, so ksbak-sb\leq a.
Additionally, ff satisfies the conditions from (Theorem 2.1) - Case (C).
Let’s take k=3,c=1,a=5,b=3k=-3,c=-1,a=5,b=-3, with s=32s=\frac{3}{2}. We verify that the coefficients a,b,c,ka,b,c,k verify all of the above conditions

{ba35,k<a+b+123<2+12k<a2+b+1210+12>0,k2ba3>1bk33,kc>s212>9k<b+a+cs6>0,a+c>06>0\left\{\begin{array}[]{l}-b\leq a\Leftrightarrow 3\leq 5,k<a+b+\frac{1}{2}\Leftrightarrow-3<2+\frac{1}{2}\\ k<\frac{a}{2}+b+\frac{1}{2}\Leftrightarrow 10+\frac{1}{2}>0,k-2b\leq a\Leftrightarrow 3>1\\ b\leq k\Leftrightarrow-3\leq-3,\frac{k}{c}>s^{2}\Leftrightarrow 12>9\\ k<b+\frac{a+c}{s}\Leftrightarrow 6>0,a+c>0\Leftrightarrow 6>0\end{array}\right.

Remark 2.6. We observe that the contractive condition when c>0c>0, can be written as:

d(fx,fy)kcd(x,y)acd(x,fx)bcd(y,fy), for each x,yXd(fx,fy)\leq\frac{k}{c}d(x,y)-\frac{a}{c}d(x,fx)-\frac{b}{c}d(y,fy),\text{ for each }x,y\in X

Taking k>0,a<0k>0,a<0 and b<0b<0, it follows that the operator ff is of Reich-type, so the above theorem (when k>0k>0 ) is similar with the results of [8].

Now, we present an useful lemma for expansive-type mappings in b-rectangular metric spaces, following the technique used in [18].

Lemma 2.7. Let ( X,dX,d ) a b-rectangular metric space. Also, consider λ\lambda\in\mathbb{R} and x,y,z,wx,y,z,w arbitrary elements of XX, each distinct from each other. Then

λd(x,z)\displaystyle\lambda d(x,z) [1+s22sλ+1s22s|λ|]d(x,y)+[s12λs+12|λ|]d(z,w)\displaystyle\geq\left[\frac{1+s^{2}}{2s}\lambda+\frac{1-s^{2}}{2s}|\lambda|\right]d(x,y)+\left[\frac{s-1}{2}\lambda-\frac{s+1}{2}|\lambda|\right]d(z,w)
+[s12λs+12|λ|]d(w,y)\displaystyle+\left[\frac{s-1}{2}\lambda-\frac{s+1}{2}|\lambda|\right]d(w,y)

Proof. Let x,y,z,wx,y,z,w arbitrary points from XX, each distinct from each other. We analyze two cases for the parameter λ\lambda\in\mathbb{R} :
Case (1): Let λ0\lambda\geq 0. From the b-rectangular inequality, we get that:

d(x,y)s[d(x,z)+d(z,w)+d(w,y)]\displaystyle d(x,y)\leq s[d(x,z)+d(z,w)+d(w,y)]
sd(x,z)d(x,y)sd(z,w)sd(w,y)\displaystyle sd(x,z)\geq d(x,y)-sd(z,w)-sd(w,y)
d(x,z)1sd(x,y)d(z,w)d(w,y)\displaystyle d(x,z)\geq\frac{1}{s}d(x,y)-d(z,w)-d(w,y)
λd(x,z)λsd(x,y)λd(z,w)λd(w,y)\displaystyle\lambda d(x,z)\geq\frac{\lambda}{s}d(x,y)-\lambda d(z,w)-\lambda d(w,y)

Case (2): Let λ0\lambda\leq 0. From the b-rectangular inequality, it follows that:

d(x,z)s[d(x,y)+d(y,w)+d(w,z)]\displaystyle d(x,z)\leq s[d(x,y)+d(y,w)+d(w,z)]
λd(x,z)λsd(x,y)+λsd(y,w)+λsd(w,z)\displaystyle\lambda d(x,z)\geq\lambda sd(x,y)+\lambda sd(y,w)+\lambda sd(w,z)

So, from the above inequality, we have that

{λd(x,z)λsd(x,y)λd(z,w)λd(w,y),λ0λd(x,z)λsd(x,y)+λsd(y,w)+λsd(w,z),λ0\begin{cases}\lambda d(x,z)\geq\frac{\lambda}{s}d(x,y)-\lambda d(z,w)-\lambda d(w,y),&\lambda\geq 0\\ \lambda d(x,z)\geq\lambda sd(x,y)+\lambda sd(y,w)+\lambda sd(w,z),&\lambda\leq 0\end{cases}

Combining these cases, it follows that
λd(x,z)φ(λ)d(x,y)+ψ(λ)d(z,w)+ψ(λ)d(w,y)\lambda d(x,z)\geq\varphi(\lambda)d(x,y)+\psi(\lambda)d(z,w)+\psi(\lambda)d(w,y), where

φ(λ):={λs,λ0sλ,λ0 and ψ(λ):={λ,λ0sλ,λ0\varphi(\lambda):=\left\{\begin{array}[]{ll}\frac{\lambda}{s},&\lambda\geq 0\\ s\lambda,&\lambda\leq 0\end{array}\text{ and }\psi(\lambda):=\begin{cases}-\lambda,&\lambda\geq 0\\ s\lambda,&\lambda\leq 0\end{cases}\right.

Similar to [18], we get that

{φ(λ):=1+s22sλ+1s22s|λ|ψ(λ):=s12λs+12|λ|\left\{\begin{array}[]{l}\varphi(\lambda):=\frac{1+s^{2}}{2s}\lambda+\frac{1-s^{2}}{2s}|\lambda|\\ \psi(\lambda):=\frac{s-1}{2}\lambda-\frac{s+1}{2}|\lambda|\end{array}\right.

Also, as a final remark, we observe that ψ(λ)0\psi(\lambda)\leq 0, for each λ\lambda\in\mathbb{R}.

For expansive-type mappings, i.e. when c<0c<0, we make the following important remark.

Remark 2.8. We have studied contraction-type mappings, that satisfied

ad(x,fx)+bd(y,fy)+cd(fx,fy)kd(x,y)\displaystyle ad(x,fx)+bd(y,fy)+cd(fx,fy)\leq kd(x,y)
cd(fx,fy)kd(x,y)ad(x,fx)bd(y,fy)\displaystyle cd(fx,fy)\leq kd(x,y)-ad(x,fx)-bd(y,fy)
d(fx,fy)kcd(x,y)acd(x,fx)bcd(y,fy)\displaystyle d(fx,fy)\geq\frac{k}{c}d(x,y)-\frac{a}{c}d(x,fx)-\frac{b}{c}d(y,fy)

By some substitutions we can make the mapping ff satisfy

d(fx,fy)αd(x,y)+βd(x,fx)+γd(y,fy)d(fx,fy)\geq\alpha d(x,y)+\beta d(x,fx)+\gamma d(y,fy)

where

{α=kcβ=acγ=bc\left\{\begin{array}[]{l}\alpha=\frac{k}{c}\\ \beta=-\frac{a}{c}\\ \gamma=-\frac{b}{c}\end{array}\right.

We will analyze the cases when α0\alpha\leq 0 and α0\alpha\geq 0, so, when k0,c<0k\geq 0,c<0, respectively k0,c<0k\leq 0,c<0.

Now, involving rate of convergence, we present a constructive fixed point theorem for expansive-type mappings in b-rectangular metric spaces, using Picard iterative process.

Theorem 2.9. Let ( X,dX,d ) a complete bb-rectangular metric space, endowed with coefficient s1s\geq 1. Also, consider f:XXf:X\rightarrow X a mapping satisfying

d(fx,fy)αd(x,y)+βd(x,fx)+γd(y,fy), for each x,yXd(fx,fy)\geq\alpha d(x,y)+\beta d(x,fx)+\gamma d(y,fy),\text{ for each }x,y\in X

Moreover, suppose the following conditions are satisfied
(i) β<1s,γ>s,α+γ<1βs\beta<1-s,\gamma>s,\alpha+\gamma<\frac{1-\beta}{s},
(ii) If α>γ\alpha>\gamma, then we have the additional assumptions α+1<γ(1+1s)\alpha+1<\gamma\left(1+\frac{1}{s}\right).

If α<γ\alpha<\gamma, then we have the additional assumptions α>1\alpha>1 and 1α<γ(1s1)1-\alpha<\gamma\left(\frac{1}{s}-1\right). Then, the mapping ff has a fixed point.

Proof. In the proof of (Theorem 2.1), we have shown that the Picard sequence for generalized contraction satisfy d(xn,xn+1)δd(xn1,xn)d\left(x_{n},x_{n+1}\right)\leq\delta d\left(x_{n-1},x_{n}\right), for each nn\in\mathbb{N}, where δ=kba+c\delta=\frac{k-b}{a+c}. This is also valid for the situation of expansive-type mappings, when c<0c<0. The condition that the Picard sequence is asymptotically regular was that 0kb<a+cs0\leq k-b<\frac{a+c}{s}.
In our case,

δ=kba+c=kcbcac+1=α+γ1β\delta=\frac{k-b}{a+c}=\frac{\frac{k}{c}-\frac{b}{c}}{\frac{a}{c}+1}=\frac{\alpha+\gamma}{1-\beta}

Now δ[0,1s)\delta\in\left[0,\frac{1}{s}\right), by hypothesis assumptions: β<1,α+γ>0\beta<1,\alpha+\gamma>0 and α+γ<1βs\alpha+\gamma<\frac{1-\beta}{s}. By the contractive-type condition, we have that

d(fx,fy)αd(x,y)+βd(x,fx)+γd(y,fy)d(fx,fy)\geq\alpha d(x,y)+\beta d(x,fx)+\gamma d(y,fy)

and applying it for the pair ( xn1,xn+1x_{n-1},x_{n+1} ), we obtain

d(xn,xn+2)αd(xn1,xn+1)+βd(xn1,xn)+γd(xn+1,xn+2)d\left(x_{n},x_{n+2}\right)\geq\alpha d\left(x_{n-1},x_{n+1}\right)+\beta d\left(x_{n-1},x_{n}\right)+\gamma d\left(x_{n+1},x_{n+2}\right) (2.1)

Now, we will try to evaluate an upper bound for d(xn,xn+2)d\left(x_{n},x_{n+2}\right), for each nn\in\mathbb{N}, i.e. using (Lemma 2.7), we obtain that

γd(xn+1,xn+2)φ(γ)d(xn,xn+2)+ψ(γ)d(xn,xn1)+ψ(γ)d(xn1,xn+1).\gamma d\left(x_{n+1},x_{n+2}\right)\geq\varphi(\gamma)d\left(x_{n},x_{n+2}\right)+\psi(\gamma)d\left(x_{n},x_{n-1}\right)+\psi(\gamma)d\left(x_{n-1},x_{n+1}\right).

Now, let’s denote by dn:=d(xn,xn+2)d_{n}^{*}:=d\left(x_{n},x_{n+2}\right) and by dn:=d(xn1,xn)d_{n}:=d\left(x_{n-1},x_{n}\right), for each nn\in\mathbb{N}.
From (2.1) we have

dnαdn1+βdn+φ(γ)dn+ψ(γ)dn+ψ(γ)dn1.d_{n}^{*}\geq\alpha d_{n-1}^{*}+\beta d_{n}+\varphi(\gamma)d_{n}^{*}+\psi(\gamma)d_{n}+\psi(\gamma)d_{n-1}^{*}.

This means that
[φ(γ)1]dn[ψ(γ)α]dn1+[ψ(γ)β]dn|ψ(γ)+α|dn1+|ψ(γ)+β|dn[\varphi(\gamma)-1]d_{n}^{*}\leq[-\psi(\gamma)-\alpha]d_{n-1}^{*}+[-\psi(\gamma)-\beta]d_{n}\leq|\psi(\gamma)+\alpha|d_{n-1}^{*}+|\psi(\gamma)+\beta|d_{n}.
Let’s denote by a2:=|α+ψ(γ)|φ(γ)1a_{2}:=\frac{|\alpha+\psi(\gamma)|}{\varphi(\gamma)-1} and by a1:=|β+ψ(γ)|φ(γ)1a_{1}:=\frac{|\beta+\psi(\gamma)|}{\varphi(\gamma)-1}.
From the hypothesis, we know that φ(γ)>1\varphi(\gamma)>1, i.e. γ>s>0\gamma>s>0, since φ(γ)=γs\varphi(\gamma)=\frac{\gamma}{s}. Then it follows that a1a_{1} and a2a_{2} are positive.
Furthermore, since γ>0\gamma>0, we have that ψ(γ)=γ<0\psi(\gamma)=-\gamma<0. So a2=|αγ|γs1a_{2}=\frac{|\alpha-\gamma|}{\frac{\gamma}{s}-1}. For a2<1a_{2}<1, we get that |αγ|<γs1|\alpha-\gamma|<\frac{\gamma}{s}-1. So, we have two cases:

  • When α>γ\alpha>\gamma, i.e. αγ>0\alpha-\gamma>0 :

Then, the condition that a2<1a_{2}<1 becomes α+1<γs+γ\alpha+1<\frac{\gamma}{s}+\gamma, i.e. α+1<γ(1+1s)\alpha+1<\gamma\left(1+\frac{1}{s}\right).
Now, since γ+1<α+1<γ(1+1s)\gamma+1<\alpha+1<\gamma\left(1+\frac{1}{s}\right), then s<γs<\gamma, which is true. Also, since γ+1<α+1<γ(1+1s)<2γ\gamma+1<\alpha+1<\gamma\left(1+\frac{1}{s}\right)<2\gamma, then 1<γ1<\gamma, which is a valid assumption.
Moreover, from the hypothesis condition that α+γ<1βs\alpha+\gamma<\frac{1-\beta}{s}, we employ two sub-cases If β>0\beta>0, then 1β<11-\beta<1, i.e. α+γ<1s<1\alpha+\gamma<\frac{1}{s}<1, so α+γ<1\alpha+\gamma<1. Since α,γ>s>1\alpha,\gamma>s>1, this is obviously not true.
If β<0\beta<0, then β<1\beta<1, so 1β>01-\beta>0 (the denominator in δ\delta is positive, so δ\delta is positive). Since β<0\beta<0, then 1βs>1s\frac{1-\beta}{s}>\frac{1}{s}. Moreover, since α+γ>1\alpha+\gamma>1, then we get β<1s\beta<1-s, which is valid from hypothesis (ii).
Finally, we can verify easily that since s>1s>1, then β<1\beta<1 and since 1s<11-s<1, then s>0s>0, which are evidently true.

  • We know verify the case when α<γ\alpha<\gamma, i.e. αγ<0\alpha-\gamma<0 :

Since |αγ|=γα<γs1|\alpha-\gamma|=\gamma-\alpha<\frac{\gamma}{s}-1, then 1α<γ(1s1)1-\alpha<\gamma\left(\frac{1}{s}-1\right), which is true by hypothesis (ii).

Moreover, since 1s1<0\frac{1}{s}-1<0, then α>1\alpha>1 is obviously true, also by hypothesis. Also, since γ>α>1\gamma>\alpha>1, then γ>1\gamma>1, which is valid by the fact that γ>s\gamma>s.
Also, as in previous case, by the assumption on δ\delta that α+γ<1βs\alpha+\gamma<\frac{1-\beta}{s}, if β>0\beta>0, then α+γ<1βs<1\alpha+\gamma<\frac{1-\beta}{s}<1, which contradicts the fact that α,γ>1\alpha,\gamma>1.
So β<0\beta<0 and from the assumption that β<1s\beta<1-s means that the right hand side 1βs>1\frac{1-\beta}{s}>1, so 1<α+γ<1βs1<\alpha+\gamma<\frac{1-\beta}{s}, which is valid.
So dna2dn1+a1dnd_{n}^{*}\leq a_{2}d_{n-1}^{*}+a_{1}d_{n}, for each nn\in\mathbb{N}. We know that

dn=d(xn1,xn)δd(xn1,xn2)δn1D0d_{n}=d\left(x_{n-1},x_{n}\right)\leq\delta d\left(x_{n-1},x_{n-2}\right)\leq\ldots\leq\delta^{n-1}D_{0}

where D0:=d1=d(x0,x1)D_{0}:=d_{1}=d\left(x_{0},x_{1}\right), with x0x_{0} an arbitrary fixed element.
So dna2dn1+a1δn1D0d_{n}^{*}\leq a_{2}d_{n-1}^{*}+a_{1}\delta^{n-1}D_{0}.
We take a major bound for dnd_{n}^{*} :

dn\displaystyle d_{n}^{*} a2dn1+a1δn1D0a2(a2dn2+a1δn2D0)+a1δn1D0\displaystyle\leq a_{2}d_{n-1}^{*}+a_{1}\delta^{n-1}D_{0}\leq a_{2}\left(a_{2}d_{n-2}^{*}+a_{1}\delta^{n-2}D_{0}\right)+a_{1}\delta^{n-1}D_{0}
=a22dn2+a2a1δn2D0+a1δn1D0\displaystyle=a_{2}^{2}d_{n-2}^{*}+a_{2}a_{1}\delta^{n-2}D_{0}+a_{1}\delta^{n-1}D_{0}
a22(a2dn3+a1δn3D0)+a1a2δn2D0+a1δn1D0\displaystyle\leq a_{2}^{2}\left(a_{2}d_{n-3}^{*}+a_{1}\delta^{n-3}D_{0}\right)+a_{1}a_{2}\delta_{n-2}D_{0}+a_{1}\delta^{n-1}D_{0}
=a23dn3+D0a1(δn1+a2δn2+a22δn3)D0\displaystyle=a_{2}^{3}d_{n-3}^{*}+D_{0}a_{1}\left(\delta^{n-1}+a_{2}\delta^{n-2}+a_{2}^{2}\delta^{n-3}\right)D_{0}\leq\ldots
a2kdnk+a1(δn1+a2δn2++a2k1δnk)D0\displaystyle\leq a_{2}^{k}d_{n-k}^{*}+a_{1}\left(\delta^{n-1}+a_{2}\delta^{n-2}+\ldots+a_{2}^{k-1}\delta^{n-k}\right)D_{0}

The last term is d0=d(x2,x0)d_{0}^{*}=d\left(x_{2},x_{0}\right), so nk=0k=nn-k=0\Longrightarrow k=n. This means that

dna2nd0+a1D0(a20δn1+a2δn2++a2n1δ0)d_{n}^{*}\leq a_{2}^{n}d_{0}^{*}+a_{1}D_{0}\left(a_{2}^{0}\delta^{n-1}+a_{2}\delta^{n-2}+\ldots+a_{2}^{n-1}\delta^{0}\right)

Let’s denote by S:=a20δn1+a2δn2++a2n1δ0S:=a_{2}^{0}\delta^{n-1}+a_{2}\delta^{n-2}+\ldots+a_{2}^{n-1}\delta^{0}. The first term in the sum is δn1\delta^{n-1}. This is a geometric progression, with general term bnb_{n} and b3b2=a2δn3δn2=a2δ\frac{b_{3}}{b_{2}}=a_{2}\frac{\delta^{n-3}}{\delta^{n-2}}=\frac{a_{2}}{\delta}, so

S=δn1(1(a2δ)n)1(a2δ)=δna2nδa2S=\frac{\delta^{n-1}\cdot\left(1-\left(\frac{a_{2}}{\delta}\right)^{n}\right)}{1-\left(\frac{a_{2}}{\delta}\right)}=\frac{\delta^{n}-a_{2}^{n}}{\delta-a_{2}}

So dna2nd0+δna2nδa2a1D0d_{n}^{*}\leq a_{2}^{n}d_{0}^{*}+\frac{\delta^{n}-a_{2}^{n}}{\delta-a_{2}}a_{1}D_{0}. Now we can show that the sequence ( xnx_{n} ) is b-rectangular Cauchy. We shall evaluate d(xn,xn+p)d\left(x_{n},x_{n+p}\right), for each nn\in\mathbb{N} and p>0p>0 fixed. We divide in two cases: the first one, when p=2mp=2m, with m2m\geq 2 and the second one, when p=2m+1p=2m+1, with m1m\geq 1 :
Case (i): When p=2m+1p=2m+1, with m1m\geq 1. We evaluate

d(xn,xn+p)=d(xn,xn+2m+1)s[d(xn,xn+1)+d(xn+1,xn+1)+d(xn+2,xn+2m+1)]\displaystyle d\left(x_{n},x_{n+p}\right)=d\left(x_{n},x_{n+2m+1}\right)\leq s\left[d\left(x_{n},x_{n+1}\right)+d\left(x_{n+1},x_{n+1}\right)+d\left(x_{n+2},x_{n+2m+1}\right)\right]
s[dn+2+dn+1]+s2[d(xn+2,xn+3)+d(xn+3,xn+4)+d(xn+4,xn+2m+1)]\displaystyle\quad\leq s\left[d_{n+2}+d_{n+1}\right]+s^{2}\left[d\left(x_{n+2},x_{n+3}\right)+d\left(x_{n+3},x_{n+4}\right)+d\left(x_{n+4},x_{n+2m+1}\right)\right]
s[dn+2+dn+1]+s2[dn+3+dn+4]+s3[dn+5+dn+6]++smdn+2m\displaystyle\quad\leq s\left[d_{n+2}+d_{n+1}\right]+s^{2}\left[d_{n+3}+d_{n+4}\right]+s^{3}\left[d_{n+5}+d_{n+6}\right]+\ldots+s^{m}d_{n+2m}

where dn+2m=d(xn+2m,xn+2m+1)d_{n+2m}=d\left(x_{n+2m},x_{n+2m+1}\right). So, we get the following estimation

d(xn,xn+2m+1)\displaystyle d\left(x_{n},x_{n+2m+1}\right) s[δnD0+δn+1D0]+s2[δn+2D0+δn+3D0]\displaystyle\leq s\left[\delta^{n}D_{0}+\delta^{n+1}D_{0}\right]+s^{2}\left[\delta^{n+2}D_{0}+\delta^{n+3}D_{0}\right]
+s3[δn+4D0+δn+5D0]++smδn+2mD0\displaystyle+s^{3}\left[\delta^{n+4}D_{0}+\delta^{n+5}D_{0}\right]+\ldots+s^{m}\delta^{n+2m}D_{0}
sδn[1+sδ2+s2δ4++]D0+sδn+1[1+sδ2+s2δ4++]D0\displaystyle\leq s\delta^{n}\left[1+s\delta^{2}+s^{2}\delta^{4}+\ldots+\right]D_{0}+s\delta^{n+1}\left[1+s\delta^{2}+s^{2}\delta^{4}+\ldots+\right]D_{0}
=1+δ1sδ2sδnD0\displaystyle=\frac{1+\delta}{1-s\delta^{2}}s\delta^{n}D_{0}

and by hypothesis we know that sδ2<1s\delta^{2}<1 is satisfied. So, d(xn,xn+2m+1)0d\left(x_{n},x_{n+2m+1}\right)\rightarrow 0, when nn\rightarrow\infty and m1m\geq 1 fixed.
Case (ii): When p=2mp=2m, with m2m\geq 2. We evaluate

d(xn,xn+2m)\displaystyle d\left(x_{n},x_{n+2m}\right) s[d(xn,xn+1)+d(xn+1,xn+2)+d(xn+2,xn+2m)]\displaystyle\leq s\left[d\left(x_{n},x_{n+1}\right)+d\left(x_{n+1},x_{n+2}\right)+d\left(x_{n+2},x_{n+2m}\right)\right]
s[dn+2+dn+1]+sd(xn+2,xn+2m)\displaystyle\leq s\left[d_{n+2}+d_{n+1}\right]+sd\left(x_{n+2},x_{n+2m}\right)
s[dn+2+dn+1]+s2[dn+4+dn+3]+s3[dn+6+dn+5]++\displaystyle\leq s\left[d_{n+2}+d_{n+1}\right]+s^{2}\left[d_{n+4}+d_{n+3}\right]+s^{3}\left[d_{n+6}+d_{n+5}\right]+\ldots+
+sm1[d2m3+d2m2]+sm1d(xn+2m2,xn+2m)\displaystyle+s^{m-1}\left[d_{2m-3}+d_{2m-2}\right]+s^{m-1}d\left(x_{n+2m-2},x_{n+2m}\right)
s[δnD0+δn+1D0]+s2[δn+2D0+δn+3D0]++\displaystyle\leq s\left[\delta^{n}D_{0}+\delta^{n+1}D_{0}\right]+s^{2}\left[\delta^{n+2}D_{0}+\delta^{n+3}D_{0}\right]+\ldots+
+sm1[δ2m4D0+δ2m3D0]+sm1d(xn+2m2,xn+2m)\displaystyle+s^{m-1}\left[\delta^{2m-4}D_{0}+\delta^{2m-3}D_{0}\right]+s^{m-1}d\left(x_{n+2m-2},x_{n+2m}\right)
sδn[1+sδ2+s2δ4+]D0\displaystyle\leq s\delta^{n}\left[1+s\delta^{2}+s^{2}\delta^{4}+\ldots\right]D_{0}
+sδn+1[1+sδ2+s2δ4+]D0+sm1dn+2m\displaystyle+s\delta^{n+1}\left[1+s\delta^{2}+s^{2}\delta^{4}+\ldots\right]D_{0}+s^{m-1}d_{n+2m}^{*}
=1+δ1sδ2sδnD0+sm1dn+2m\displaystyle=\frac{1+\delta}{1-s\delta^{2}}s\delta^{n}D_{0}+s^{m-1}d_{n+2m}^{*}

Also, we have shown that dna2nd0+δna2nδa2a1D0d_{n}^{*}\leq a_{2}^{n}d_{0}^{*}+\frac{\delta^{n}-a_{2}^{n}}{\delta-a_{2}}a_{1}D_{0}. So dn+2ma2n+2md0+Qa1D0d_{n+2m}^{*}\leq a_{2}^{n+2m}d_{0}^{*}+Qa_{1}D_{0}, where Q:=δn+2ma2n+2mδa2Q:=\frac{\delta^{n+2m}-a_{2}^{n+2m}}{\delta-a_{2}}.
Now, we have two cases: if δa2>0\delta-a_{2}>0, then Q=δn+2ma2n+2mδa2δn+2mδa2Q=\frac{\delta^{n+2m}-a_{2}^{n+2m}}{\delta-a_{2}}\leq\frac{\delta^{n+2m}}{\delta-a_{2}} and this converge to 0 as nn\rightarrow\infty. In a similar manner, if δa2<0\delta-a_{2}<0, then

Q=a2n+2mδn+2ma2δa2n+2ma2δQ=\frac{a_{2}^{n+2m}-\delta^{n+2m}}{a_{2}-\delta}\leq\frac{a_{2}^{n+2m}}{a_{2}-\delta}

and this converge to 0 as nn\rightarrow\infty. This reasoning is valid, since, from the theorem’s assumptions, we know that 0a2<10\leq a_{2}<1 and δ<1s<1\delta<\frac{1}{s}<1. So, in this case, since Q0Q\rightarrow 0, then d(xn,xn+2m)0d\left(x_{n},x_{n+2m}\right)\rightarrow 0, as nn\rightarrow\infty.
So, from both cases, we have shown that (xn)\left(x_{n}\right) is a b-rectangular Cauchy sequence. Also, we know that xnxmx_{n}\neq x_{m}, for each nmn\neq m and that ( X,dX,d ) is complete. This means that there exists uXu\in X, such that limnxn=u\lim_{n\rightarrow\infty}x_{n}=u.
Moreover, since the contractive condition can be reduced to the original form, i.e. ad(x,fx)+bd(y,fy)+cd(fx,fy)kd(x,y)ad(x,fx)+bd(y,fy)+cd(fx,fy)\leq kd(x,y), then, as in the proof of (Theorem 2.1), there exists a unique point uu of ff, as long as a+c>0a+c>0 and c<kc<k.

Finally, we give an example regarding (Theorem 2.9).
Example 2.10. Let ( X,dX,d ), with X={1,2,3,4}X=\{1,2,3,4\} be the b-rectangular metric space, endowed with the b-rectangular metric from (Example 2.2). Define a self-mapping ff, by: f(1)=2,f(2)=3,f(3)=1f(1)=2,f(2)=3,f(3)=1 and f(4)=4f(4)=4. It is obviously that ff has as a unique fixed point the element 4X4\in X. We will determine the coefficients α,β\alpha,\beta and γ\gamma, such that ff satisfies d(fx,fy)αd(x,y)+βd(x,fx)+γd(y,fy)d(fx,fy)\geq\alpha d(x,y)+\beta d(x,fx)+\gamma d(y,fy) :

By x=2 and y=1, we get that 110α610+β110+γ610\displaystyle\text{ By }x=2\text{ and }y=1,\text{ we get that }\frac{1}{10}\geq\alpha\frac{6}{10}+\beta\frac{1}{10}+\gamma\frac{6}{10} (2.2)
By x=1 and y=2, we get that 110α610+β610+γ110\displaystyle\text{ By }x=1\text{ and }y=2,\text{ we get that }\frac{1}{10}\geq\alpha\frac{6}{10}+\beta\frac{6}{10}+\gamma\frac{1}{10} (2.3)
By x=1 and y=3, we get that 610α110+β610+γ110\displaystyle\text{ By }x=1\text{ and }y=3,\text{ we get that }\frac{6}{10}\geq\alpha\frac{1}{10}+\beta\frac{6}{10}+\gamma\frac{1}{10} (2.4)
By x=3 and y=1, we get that 610α110+β110+γ610\displaystyle\text{ By }x=3\text{ and }y=1,\text{ we get that }\frac{6}{10}\geq\alpha\frac{1}{10}+\beta\frac{1}{10}+\gamma\frac{6}{10} (2.5)
By x=1 and y=4, we get that 210α210+β610+γ210\displaystyle\text{ By }x=1\text{ and }y=4,\text{ we get that }\frac{2}{10}\geq\alpha\frac{2}{10}+\beta\frac{6}{10}+\gamma\frac{2}{10} (2.6)
By x=4 and y=1, we get that 210α210+β210+γ610\displaystyle\text{ By }x=4\text{ and }y=1,\text{ we get that }\frac{2}{10}\geq\alpha\frac{2}{10}+\beta\frac{2}{10}+\gamma\frac{6}{10} (2.7)
By x=3 and y=2, we get that 110α110+β110+γ110\displaystyle\text{ By }x=3\text{ and }y=2,\text{ we get that }\frac{1}{10}\geq\alpha\frac{1}{10}+\beta\frac{1}{10}+\gamma\frac{1}{10} (2.8)
By x=2 and y=3, we get that 110α110+β110+γ110\displaystyle\text{ By }x=2\text{ and }y=3,\text{ we get that }\frac{1}{10}\geq\alpha\frac{1}{10}+\beta\frac{1}{10}+\gamma\frac{1}{10} (2.9)
By x=4 and y=2, we get that 210α210+β210+γ110\displaystyle\text{ By }x=4\text{ and }y=2,\text{ we get that }\frac{2}{10}\geq\alpha\frac{2}{10}+\beta\frac{2}{10}+\gamma\frac{1}{10} (2.10)
By x=2 and y=4, we get that 210α210+β110+γ210\displaystyle\text{ By }x=2\text{ and }y=4,\text{ we get that }\frac{2}{10}\geq\alpha\frac{2}{10}+\beta\frac{1}{10}+\gamma\frac{2}{10} (2.11)
By x=4 and y=3, we get that 210α210+β210+γ110\displaystyle\text{ By }x=4\text{ and }y=3,\text{ we get that }\frac{2}{10}\geq\alpha\frac{2}{10}+\beta\frac{2}{10}+\gamma\frac{1}{10} (2.12)
By x=3 and y=4, we get that 210α210+β110+γ210\displaystyle\text{ By }x=3\text{ and }y=4,\text{ we get that }\frac{2}{10}\geq\alpha\frac{2}{10}+\beta\frac{1}{10}+\gamma\frac{2}{10} (2.13)
By x=y, we get that β+γ0\displaystyle\text{ By }x=y,\text{ we get that }\beta+\gamma\leq 0 (2.14)

Now, we observe that (2.11) and (2.14) are equivalent relations. Also, we shall employ the more restrictive conditions on the coefficients α,β\alpha,\beta and γ\gamma, i.e. inequalities (2.11), (2.3), (2.5), (2.7), (2.8) and (2.14). Furthermore, we shall impose more restrictive conditions such that the number of inequalities is reduced: instead of (2.11) and (2.3), we impose that 16α+β+2γ1\geq 6\alpha+\beta+2\gamma, instead of (2.7) and (2.8) we require only (2.7) and instead of 16α+β+2γ1\geq 6\alpha+\beta+2\gamma and (2.5), we require 16α+β+6γ1\geq 6\alpha+\beta+6\gamma. We mention that all of the above reasoning was made under the assumptions that β0\beta\leq 0 and γ>0\gamma>0. Now, we have only two conditions, along with the conditions from (Theorem 2.9), when α>γ\alpha>\gamma

{β+γ0,16α+β+6γβ<1s,γ>s,αγα+γ<1βs,α+1<γ(1+1s)\left\{\begin{array}[]{l}\beta+\gamma\leq 0,1\geq 6\alpha+\beta+6\gamma\\ \beta<1-s,\gamma>s,\alpha\gamma\\ \alpha+\gamma<\frac{1-\beta}{s},\alpha+1<\gamma\left(1+\frac{1}{s}\right)\end{array}\right.

Now, taking account of the fact that s=32s=\frac{3}{2}, we can find some values for the coefficients α,β\alpha,\beta and γ\gamma. For example, the inequalities are satisfied when α=950,β=1015\alpha=\frac{9}{50},\beta=-\frac{101}{5} and γ=17100\gamma=\frac{17}{100}.

Now, we recall (Lemma 2) from [5], that is crucial for inequalities involving difference inequations.

Lemma 2.11. Let ( ana_{n} ) and ( bnb_{n} ) be two sequences of nonnegative real numbers, such that

an+1α1an+α2an1++αkank+1+bn, where nk1.a_{n+1}\leq\alpha_{1}a_{n}+\alpha_{2}a_{n-1}+\ldots+\alpha_{k}a_{n-k+1}+b_{n},\text{ where }n\geq k-1.

If α1,,αk[0,1),i=1kαi<1\alpha_{1},\ldots,\alpha_{k}\in[0,1),\sum_{i=1}^{k}\alpha_{i}<1 and limnbn=0\lim_{n\rightarrow\infty}b_{n}=0, then it follows that limnan=0\lim_{n\rightarrow\infty}a_{n}=0.
Remark 2.12. In the previous proof, we have shown that the following estimation is valid

dn=d(xn+2,xn)a2nd0+δna2nδa2a1D0d_{n}^{*}=d\left(x_{n+2},x_{n}\right)\leq a_{2}^{n}d_{0}^{*}+\frac{\delta^{n}-a_{2}^{n}}{\delta-a_{2}}a_{1}D_{0}

So, based on this lemma, we give a nonconstructive approach for evaluating ( xnx_{n} ) as a Cauchy sequence.
In the above lemma, let’s take k=1k=1. Then, we get that an+1α1an+bna_{n+1}\leq\alpha_{1}a_{n}+b_{n}, with α1[0,1)\alpha_{1}\in[0,1) and limnbn=0\lim_{n\rightarrow\infty}b_{n}=0. Then limnan=0\lim_{n\rightarrow\infty}a_{n}=0.
Now, we have proved that dna2dn1+a1δn1D0d_{n}^{*}\leq a_{2}d_{n-1}^{*}+a_{1}\delta^{n-1}D_{0}.
Let’s define the following: α1:=a2\alpha_{1}:=a_{2} and bn:=a1D0δn1b_{n}:=a_{1}D_{0}\delta^{n-1}. Since δ<1s<1\delta<\frac{1}{s}<1 and a2[0,1)a_{2}\in[0,1), then apply (Lemma 2) from [5] with the particular case when k=1k=1, we get that limndn=0\lim_{n\rightarrow\infty}d_{n}^{*}=0.

Now, we give a proof for expansive-type mappings under the new assumption such that the mapping ff is onto and we shall use the ’inverse’ Picard iterative process.
Theorem 2.13. Let ( X,dX,d ) be a complete b-rectangular metric space and f:XXf:X\rightarrow X a mapping satisfying

d(fx,fy)αd(x,y)+βd(x,fx)+γd(y,fy)d(fx,fy)\geq\alpha d(x,y)+\beta d(x,fx)+\gamma d(y,fy)

Let ff continuous and onto. Suppose that
(i) β<1,α+γ>0\beta<1,\alpha+\gamma>0 and 1β<α+γs1-\beta<\frac{\alpha+\gamma}{s}.

Also, suppose the following additional assumptions
Case (E1), i.e. α>0\alpha>0 : Suppose that the following assumptions are satisfied:
(ii) α>1\alpha>1

Case (E2), i.e. α<0\alpha<0 : Suppose the following assumptions are satisfied:
(ii) α<1,γ>0\alpha<-1,\gamma>0
(iii) s(1αγ)<1+1αs\left(1-\frac{\alpha}{\gamma}\right)<1+\frac{1}{\alpha}

Then, the mapping ff has a fixed point in XX.

Proof. Here, we know that ff is continuous and onto. Let x0x_{0} be an arbitrary point. As we have shown in the previous theorem, i.e. (Theorem 2.9), we reduce the contractive condition to

d(fx,fy)αd(x,y)+βd(x,fx)+γd(y,fy)d(fx,fy)\geq\alpha d(x,y)+\beta d(x,fx)+\gamma d(y,fy)

Because ff is an onto mapping, by definition, we have that for each yXy\in X, there exists xXx\in X, such that y=fxy=fx.
Now, for x0Xx_{0}\in X, there exists x1Xx_{1}\in X, such that x0=fx1x_{0}=fx_{1}. Also, for x1Xx_{1}\in X, there exists x2Xx_{2}\in X, such that x1=fx2x_{1}=fx_{2}. Inductively, we get that xn=fxn+1x_{n}=fx_{n+1}, for each nn\in\mathbb{N}.
Applying the contractive condition on the pair (xn+1,xn)\left(x_{n+1},x_{n}\right), it follows that:

d(fxn+1,fxn)αd(xn,xn+1)+βd(xn,fxn)+γd(xn+1,fxn+1)\displaystyle d\left(fx_{n+1},fx_{n}\right)\geq\alpha d\left(x_{n},x_{n+1}\right)+\beta d\left(x_{n},fx_{n}\right)+\gamma d\left(x_{n+1},fx_{n+1}\right)
d(xn,xn1)αd(xn,xn+1)+βd(xn,xn1)+γd(xn+1,xn)\displaystyle d\left(x_{n},x_{n-1}\right)\geq\alpha d\left(x_{n},x_{n+1}\right)+\beta d\left(x_{n},x_{n-1}\right)+\gamma d\left(x_{n+1},x_{n}\right)
(α+γ)d(xn+1,xn)(1β)d(xn1,xn)\displaystyle\Longrightarrow(\alpha+\gamma)d\left(x_{n+1},x_{n}\right)\leq(1-\beta)d\left(x_{n-1},x_{n}\right)
d(xn,xn+1)θd(xn1,xn)\displaystyle\Longrightarrow d\left(x_{n},x_{n+1}\right)\leq\theta d\left(x_{n-1},x_{n}\right)

where θ:=1βα+γ\theta:=\frac{1-\beta}{\alpha+\gamma}. From the hypothesis,we know that θ[0,1s)\theta\in\left[0,\frac{1}{s}\right), because β<1\beta<1, α+γ>0\alpha+\gamma>0 and 1β<α+γs1-\beta<\frac{\alpha+\gamma}{s}. Furthermore, we have that dn+1:=d(xn+1,xn)θnd1d_{n+1}:=d\left(x_{n+1},x_{n}\right)\leq\theta^{n}d_{1}. For simplicity, let’s denote by D0:=d1=d(x1,x0)D_{0}:=d_{1}=d\left(x_{1},x_{0}\right).
Furthermore, as in the previous theorem, let dn:=d(xn,xn+2)d_{n}^{*}:=d\left(x_{n},x_{n+2}\right), for each nn\in\mathbb{N}.
Now, we shall analyze two different cases for estimation of d(xn,xn+2)d\left(x_{n},x_{n+2}\right)
Case (E1): When α>0\alpha>0, or with the original notation, kc>0\frac{k}{c}>0. Since c<0c<0, we get that k<0k<0.
Applying the expansive-type condition on the pair ( xn,xn+2x_{n},x_{n+2} ), it follows that

d(xn1,xn+1)=d(fxn,fxn+2)αd(xn,xn+2)+βd(xn,fxn)+γd(xn+2,fxn+2)\displaystyle d\left(x_{n-1},x_{n+1}\right)=d\left(fx_{n},fx_{n+2}\right)\geq\alpha d\left(x_{n},x_{n+2}\right)+\beta d\left(x_{n},fx_{n}\right)+\gamma d\left(x_{n+2},fx_{n+2}\right)
=αd(xn,xn+2)+βd(xn,xn1)+γd(xn+1,xn+2)\displaystyle=\alpha d\left(x_{n},x_{n+2}\right)+\beta d\left(x_{n},x_{n-1}\right)+\gamma d\left(x_{n+1},x_{n+2}\right)\Longrightarrow
αd(xn,xn+2)d(xn1,xn+1)βd(xn1,xn)γd(xn+1,xn+2)\displaystyle\alpha d\left(x_{n},x_{n+2}\right)\leq d\left(x_{n-1},x_{n+1}\right)-\beta d\left(x_{n-1},x_{n}\right)-\gamma d\left(x_{n+1},x_{n+2}\right)
d(xn,xn+2)1αdn1+(βα)dn+(γα)dn+2\displaystyle d\left(x_{n},x_{n+2}\right)\leq\frac{1}{\alpha}d_{n-1}^{*}+\left(-\frac{\beta}{\alpha}\right)d_{n}+\left(-\frac{\gamma}{\alpha}\right)d_{n+2}
d(xn,xn+2)1αdn1+(|βα|)dn+(|γα|)dn+2\displaystyle d\left(x_{n},x_{n+2}\right)\leq\frac{1}{\alpha}d_{n-1}^{*}+\left(\left|\frac{\beta}{\alpha}\right|\right)d_{n}+\left(\left|\frac{\gamma}{\alpha}\right|\right)d_{n+2}

Since dn+1θnD0d_{n+1}\leq\theta^{n}D_{0}, so dnθn1D0d_{n}\leq\theta^{n-1}D_{0}, it follows that

dn1αdn1+θn1QD0, where Q:=|βα|+|γα|θ3d_{n}^{*}\leq\frac{1}{\alpha}d_{n-1}^{*}+\theta^{n-1}QD_{0},\text{ where }Q:=\left|\frac{\beta}{\alpha}\right|+\left|\frac{\gamma}{\alpha}\right|\theta^{3}

Since θ[0,1s)[0,1)\theta\in\left[0,\frac{1}{s}\right)\subset[0,1) and α>1\alpha>1, we get, by (Lemma 2) in [5] and by (Lemma 2.11), that limndn=0\lim_{n\rightarrow\infty}d_{n}^{*}=0. Now, as in the proof of (Theorem 2.9), we give a constructive approach for the upper bound of d(xn,xn+p)d\left(x_{n},x_{n+p}\right). Furthermore, we shall omit the details. We know that dna2dn1+a1θn1D0d_{n}^{*}\leq a_{2}d_{n-1}^{*}+a_{1}\theta^{n-1}D_{0}, briefly dna2nd0+θna2nθa2a1D0d_{n}^{*}\leq a_{2}^{n}d_{0}^{*}+\frac{\theta^{n}-a_{2}^{n}}{\theta-a_{2}}a_{1}D_{0}, where
a1:=Qa_{1}:=Q and a2:=1αa_{2}:=\frac{1}{\alpha}. When p=2m+1p=2m+1, then d(xn,xn+2m+1)1+θ1sθ2sθnD0d\left(x_{n},x_{n+2m+1}\right)\leq\frac{1+\theta}{1-s\theta^{2}}s\theta^{n}D_{0}, and, by hypothesis, sθ2<1s\theta^{2}<1, then d(xn,xn+2m+1)d\left(x_{n},x_{n+2m+1}\right) converges to 0 .
When p=2mp=2m, then dn+2ma2n+2md0+θn+2ma2n+2mθa2a1D0d_{n+2m}^{*}\leq a_{2}^{n+2m}d_{0}^{*}+\frac{\theta^{n+2m}-a_{2}^{n+2m}}{\theta-a_{2}}a_{1}D_{0}. Since θ<1s<1\theta<\frac{1}{s}<1 and a2<1a_{2}<1, by theorem’s assumptions, then dn+2md_{n+2m}^{*} converges to 0 .
Moreover, d(xn,xn+2m)1+θ1sθ2sθnD0+sm1dn+2md\left(x_{n},x_{n+2m}\right)\leq\frac{1+\theta}{1-s\theta^{2}}s\theta^{n}D_{0}+s^{m-1}d_{n+2m}^{*}.
Case (E2): When α<0\alpha<0. We shall use (Lemma 2.7):
We know that d(xn,xn+1)θd(xn1,xn)d\left(x_{n},x_{n+1}\right)\leq\theta d\left(x_{n-1},x_{n}\right), for each n1n\geq 1.
As in the previous case, with the remark that we divide by α<0\alpha<0, we get that

d(xn,xn+2)Adn1+Bdn+Cdn+1, where A:=1α,B:=β|α| and C:=γ|α|.d\left(x_{n},x_{n+2}\right)\geq Ad_{n-1}^{*}+Bd_{n}+Cd_{n+1},\text{ where }A:=\frac{1}{\alpha},B:=\frac{\beta}{|\alpha|}\text{ and }C:=\frac{\gamma}{|\alpha|}.

By (Lemma 2.7), we get that

Cdn+1φ(C)dn+1+ψ(C)dn+3+ψ(C)dn\displaystyle Cd_{n+1}\geq\varphi(C)d_{n+1}^{*}+\psi(C)d_{n+3}+\psi(C)d_{n}^{*}
dnAdn1+Bdn+φ(C)dn+1+ψ(C)dn+3+ψ(C)dn\displaystyle d_{n}^{*}\geq Ad_{n-1}^{*}+Bd_{n}+\varphi(C)d_{n+1}^{*}+\psi(C)d_{n+3}+\psi(C)d_{n}^{*}
φ(C)dn+1dn[1ψ(C)]+(A)dn1φ(C)dn+3Bdn\displaystyle\varphi(C)d_{n+1}^{*}\leq d_{n}^{*}[1-\psi(C)]+(-A)d_{n-1}^{*}-\varphi(C)d_{n+3}-Bd_{n}

Since, by theorem’s assumptions, φ(C)>0\varphi(C)>0, we get that

dn+1\displaystyle d_{n+1}^{*} 1ψ(C)φ(C)dnAdn1[φ(C)dn+3+Bdn]\displaystyle\leq\frac{1-\psi(C)}{\varphi(C)}d_{n}^{*}-Ad_{n-1}^{*}-\left[\varphi(C)d_{n+3}+Bd_{n}\right]
dn+1\displaystyle d_{n+1}^{*} 1ψ(C)φ(C)dnAdn1+[|φ(C)|dn+3+|B|dn]\displaystyle\leq\frac{1-\psi(C)}{\varphi(C)}d_{n}^{*}-Ad_{n-1}^{*}+\left[|\varphi(C)|d_{n+3}+|B|d_{n}\right]
dn+1\displaystyle d_{n+1}^{*} 1ψ(C)φ(C)dnAdn1+[|φ(C)|θ2+|B|]θnD0\displaystyle\leq\frac{1-\psi(C)}{\varphi(C)}d_{n}^{*}-Ad_{n-1}^{*}+\left[|\varphi(C)|\theta^{2}+|B|\right]\theta^{n}D_{0}

On the other hand, let’s denote by bn:=[|φ(C)|θ2+|B|]θnD0,α1:=1ψ(C)φ(C)b_{n}:=\left[|\varphi(C)|\theta^{2}+|B|\right]\theta^{n}D_{0},\alpha_{1}:=\frac{1-\psi(C)}{\varphi(C)} and by α2:=A\alpha_{2}:=-A. Since γ>0\gamma>0 and C=γ|α|>0C=\frac{\gamma}{|\alpha|}>0, then φ(C)=Cs>0\varphi(C)=\frac{C}{s}>0. Also, from C>0C>0, then ψ(C)=C<0\psi(C)=-C<0. Now, α1>0\alpha_{1}>0 requires that C<1-C<1 and this is true since C>0C>0. Moreover, α2=A=1α>0\alpha_{2}=-A=-\frac{1}{\alpha}>0, because α<0\alpha<0 and so 1α<0\frac{1}{\alpha}<0. This means that α1\alpha_{1} and α2\alpha_{2} are positive, so the sum of these two is positive. Now, we want to validate if the sum of α1\alpha_{1} and α2\alpha_{2} is less than 1 .

α1+α2=1ψ(C)φ(C)A=1+CCs1α\alpha_{1}+\alpha_{2}=\frac{1-\psi(C)}{\varphi(C)}-A=\frac{1+C}{\frac{C}{s}}-\frac{1}{\alpha}

So α1+α2<1\alpha_{1}+\alpha_{2}<1 is equivalent to s(1+CC)<1+1αs\left(\frac{1+C}{C}\right)<1+\frac{1}{\alpha}. Since C=γ|α|=γαC=\frac{\gamma}{|\alpha|}=\frac{\gamma}{-\alpha}, then s(1αγ)<1+1αs\left(1-\frac{\alpha}{\gamma}\right)<1+\frac{1}{\alpha}. Now, we have two sub-cases.
If 1αγ<01-\frac{\alpha}{\gamma}<0, then αγ>0\alpha-\gamma>0, i.e. α>γ\alpha>\gamma, so this is false, because α<0\alpha<0 and γ>0\gamma>0.

So, the only valid case is when 1αγ>01-\frac{\alpha}{\gamma}>0, so α<γ\alpha<\gamma. Since α\alpha and γ\gamma have different signs, this is also valid. Now, because s(1αγ)<1+1αs\left(1-\frac{\alpha}{\gamma}\right)<1+\frac{1}{\alpha} and by the fact that the right hand side is positive, it follows that 1+1α>01+\frac{1}{\alpha}>0, i.e. α<1\alpha<-1, which is valid by hypothesis assumptions. Since θ[0,1s)[0,1)\theta\in\left[0,\frac{1}{s}\right)\subset[0,1), then limnbn=0\lim_{n\rightarrow\infty}b_{n}=0.
Also, since α1+α2[0,1),α1[0,1)\alpha_{1}+\alpha_{2}\in[0,1),\alpha_{1}\in[0,1) and α2[0,1)\alpha_{2}\in[0,1), then limndn=0\lim_{n\rightarrow\infty}d_{n}^{*}=0. The rest of the proof follows as usual.

Now, we give an example of a b-rectangular metric space, which is b-rectangular and validate (Theorem 2.13) through another example, showing that the hypotheses and conclusion of the already mentioned theorem are true also in b-metric spaces.

Example 2.14. Let X=[0,)X=[0,\infty), endowed with d:X×X+d:X\times X\rightarrow\mathbb{R}_{+}, such that d(x,y)=(xy)2d(x,y)=(x-y)^{2}, for each x,yXx,y\in X. Then (X,d)(X,d) is a complete b-metric space, with coefficient s=2s=2. Then, it is also a complete b-rectangular metric space, with coefficient s=4s=4.

Example 2.15. Let X=[0,)X=[0,\infty), where dd is the above b-rectangular metric, with s=4s=4. Define f:XXf:X\rightarrow X as f(x)=x+δ1δ2f(x)=\frac{x+\delta_{1}}{\delta_{2}}, with δ1,δ20\delta_{1},\delta_{2}\geq 0. It is easy to see that ff is continuous. Also, for each yXy\in X, there exists x=yδ2δ10x=y\delta_{2}-\delta_{1}\geq 0,, since δ1\delta_{1} and δ2\delta_{2} are positive, so ff is onto. Moreover:

d(fx,fy)=(fxfy)2=|x+δ1δ2y+δ1δ2|=1δ2|xy|2=1δ2d(x,y)d(fx,fy)=(fx-fy)^{2}=\left|\frac{x+\delta_{1}}{\delta_{2}}-\frac{y+\delta_{1}}{\delta_{2}}\right|=\frac{1}{\delta_{2}}|x-y|^{2}=\frac{1}{\delta_{2}}d(x,y)

Let’s take β=0,γ=0\beta=0,\gamma=0 and α=10\alpha=10. Also, let δ<1s\delta<\frac{1}{s}, i.e. δ2<14\delta_{2}<\frac{1}{4}. For example: δ2=110\delta_{2}=\frac{1}{10} and δ1=1\delta_{1}=1.
Then ff satisfies d(fx,fy)10d(x,y)d(fx,fy)\geq 10d(x,y), for each x,yXx,y\in X.
As an open problem with respect to generalized contractions in b-rectangular metric spaces, we give the following.
Open Problem. Following [3], consider a self-mapping ff defined on a complete brectangular space ( X,dX,d ) with coefficient s1s\geq 1, that satisfy

ad(x,fx)+bd(y,fy)+cd(fx,fy)+ed(x,fy)+gd(y,fx)kd(x,y)ad(x,fx)+bd(y,fy)+cd(fx,fy)+ed(x,fy)+gd(y,fx)\leq kd(x,y)

Develop fixed point theorems for the self-mapping above, in the context of brectangular metric spaces, with suitable conditions on the coefficients a,b,c,e,g,ka,b,c,e,g,k.
Acknowledgments. The author is grateful to the referees for their suggestions that contributed to the improvement of the paper.

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Cristian Daniel Alecsa
Babeş-Bolyai University
Faculty of Mathematics and Computer Sciences
Cluj-Napoca, Romania
e-mail: cristian.alecsa@math.ubbcluj.ro
"Tiberiu Popoviciu" Institute of Numerical Analysis
Romanian Academy
Cluj-Napoca, Romania
e-mail: cristian.alecsa@ictp.acad.ro

2017

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