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): 47​H​10,54​H​2547\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,y∈Xx,y\in X and u,v∈Xu,v\in X (each distinct from xx and yy ), we have that
(1) d​(x,y)=0⟺x=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 xn≠xmx_{n}\neq x_{m}, for each n≠mn\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≠∅,s≥1X\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,y∈Xx,y\in X and u,v∈Xu,v\in X (each distinct from xx and yy ), we have that
(1) d​(x,y)=0⟺x=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, x∈Xx\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 x∈Xx\in X, if for each ε>0\varepsilon>0, there exists n0∈ℕn_{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 limn→∞xn=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 n0∈ℕn_{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 limn→∞d​(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 x∈Xx\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 s≥1s\geq 1. Let (xn)\left(x_{n}\right) be a b-rectangular Cauchy sequence in XX, such that xn≠xmx_{n}\neq x_{m}, for each n≠mn\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 s≥1s\geq 1. Also, let (xn)\left(x_{n}\right) be a sequence for which xn≠xmx_{n}\neq x_{m}, for every n≠mn\neq m, with limn→∞d​(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 k∈ℕk\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
εs≤lim supk→∞d​(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 }
εs≤lim supk→∞d​(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 s≥1s\geq 1.
(a) Consider two sequences (xn)\left(x_{n}\right) and (yn)\left(y_{n}\right), such that xnx_{n} converges to x∈Xx\in X and yny_{n} converges to y∈Xy\in X, with x≠yx\neq y. Also, suppose that for each n∈ℕ,xn≠xn\in\mathbb{N},x_{n}\neq x and yn≠yy_{n}\neq y. Then

1s​d​(x,y)≤lim infn→∞d​(xn,yn)≤lim supn→∞d​(xn,yn)≤s​d​(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 y∈Xy\in X and a b-rectangular Cauchy sequence ( xnx_{n} ), such that xn≠xmx_{n}\neq x_{m}, for each n≠mn\neq m. Moreover, suppose that the sequence ( xnx_{n} ) converges to an element x≠yx\neq y. Then

1s​d​(x,y)≤lim infn→∞d​(xn,y)≤lim supn→∞d​(xn,y)≤s​d​(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​(T​x,T​y)≤λ​[d​(x,T​x)+d​(y,T​y)], 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​(f​x,g​y)≤a​d​(g​x,g​y)+b​[d​(g​x,f​x)+d​(g​y,f​y)]d(fx,gy)\leq ad(gx,gy)+b[d(gx,fx)+d(gy,fy)]

for each x,y∈Xx,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,T​x)+d​(y,T​y)≤p​d​(x,y), where ​p∈[0,2)d(x,Tx)+d(y,Ty)\leq pd(x,y),\text{ where }p\in[0,2)

and

a​d​(T​x,T​y)+b​[d​(x,T​x)+d​(y,T​y)]≤s​d​(x,y), with ​0≤s+|a|−2​b<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​(f​x,f​y)≥q​d​(g​x,g​y), 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​(T​x,T​y)+d​(x,T​x)+d​(y,T​y)≤r​d​(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

a​d​(x,f​x)+b​d​(y,f​y)+c​d​(f​x,f​y)≤k​d​(x,y), with ​2​b−|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

a​d​(S​x,T​x)+b​d​(S​y,T​y)+c​d​(T​x,T​y)≤e​d​(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

2​b−|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

α​d​(T​x,T​y)+β​[d​(S​x,T​x)+d​(S​y,T​y)]+γ​[d​(S​x,T​y)+d​(S​y,T​x)]≤η​d​(S​x,S​y),\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.

2​β+γ−|γ|−α≤η​<α+2​β+3​γ−|​γ| and ​β+γ≤0.2\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

a​d​(x,T​x)+b​d​(y,T​y)+c​d​(T​x,T​y)+e​d​(T​y,x)+f​d​(y,T​x)≤k​d​(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|​(1−λ)−|c|​λ1−λ≤k<a+b+c+e+f−|c|​λ−|f|​(1−λ)1−λ,\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​(T​x,T​y)≥k​d​(x,y)+l​d​(x,T​x)+p​d​(y,T​y)d(Tx,Ty)\geq kd(x,y)+ld(x,Tx)+pd(y,Ty)

where TT satisfies K≥−1,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​(T​x,T​y)≥a​d​(f​x,f​y)+b​d​(T​x,f​x)+c​d​(T​y,f​y)d(Tx,Ty)\geq ad(fx,fy)+bd(Tx,fx)+cd(Ty,fy)

with b<1,a≠0,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​(f​x,f​y)≥α​d​(g​x,g​y)+β​d​(f​x,g​x)+γ​d​(f​y,g​y)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​(f​x,g​y)≥a1​d​(S​x,T​y)+a2​d​(f​x,S​x)+a3​d​(g​y,T​y)+a4​d​(f​x,T​y)+a5​d​(g​y,S​x).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​(T​x,T​y)≤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​(S2​x,T​S​y))≥13​[d​(S​x,S2​x)+d​(T​S​y,S​y)+d​(S​x,S​y)].\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:X→Xf:X\rightarrow X on a b-g.m.s. XX, satisfying the following condition:

a​d​(x,f​x)+b​d​(y,f​y)+c​d​(f​x,f​y)≤k​d​(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=f​xnx_{n+1}=fx_{n}, for each n∈ℕn\in\mathbb{N} and the ’inverse’ Picard iteration, i.e. xn=f​xn+1x_{n}=fx_{n+1}, for each n∈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:X→Xf:X\rightarrow X, satisfying the following contractive condition

a​d​(x,f​x)+b​d​(y,f​y)+c​d​(f​x,f​y)≤k​d​(x,y), where ​0≤k−b<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 k≥0k\geq 0, then kc<1s\frac{k}{c}<\frac{1}{s},
(B) If c>0c>0 and k≤0k\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=f​xnx_{n+1}=fx_{n}, for each n∈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=f​xnx_{n+1}=fx_{n}, for each n∈ℕn\in\mathbb{N}. Applying the contractive condition for the pair (xn−1,xn)\left(x_{n-1},x_{n}\right), we get that

a​d​(xn,f​xn)+b​d​(xn−1,f​xn−1)+c​d​(f​xn−1,f​xn)≤k​d​(xn−1,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)
a​d​(xn,xn+1)+b​d​(xn−1,xn)+c​d​(xn,xn+1)≤k​d​(xn−1,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)≤(k−b)​d​(xn−1,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)≤δ​d​(xn−1,xn)d\left(x_{n},x_{n+1}\right)\leq\delta d\left(x_{n-1},x_{n}\right), where δ:=k−ba+c∈[0,1s)\delta:=\frac{k-b}{a+c}\in\left[0,\frac{1}{s}\right) from the theorem’s assumptions, since 0≤k−b<a+cs0\leq k-b<\frac{a+c}{s}.
So d​(xn,xn+1)≤δn​d​(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 limn→∞d​(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 xn≠xn+1x_{n}\neq x_{n+1}, for each n∈ℕn\in\mathbb{N} and that xn≠xmx_{n}\neq x_{m}, for each n≠mn\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
a​d​(xm​(k),xm​(k)+1)+b​d​(xn​(k)−2,xn​(k)−1)+c​d​(xm​(k)+1,xn​(k)−1)≤k​d​(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)
c​d​(xm​(k)+1,xn​(k)−1)≤k​d​(xm​(k),xn​(k)−2)−a​d​(xm​(k),xm​(k)+1)−b​d​(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)≤kc​d​(xm​(k),xn​(k)−2)−ac​d​(xm​(k),xm​(k)+1)−bc​d​(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 a≥0a\geq 0, then −ac≤0-\frac{a}{c}\leq 0, so −ac​d​(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 a≤0a\leq 0, then −ac≥0-\frac{a}{c}\geq 0, so −ac​d​(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 supk→∞d​(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)​limk→∞d​(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,xn−1))\left(d\left(x_{n},x_{n-1}\right)\right).
Case (A): When k≥0k\geq 0.
Since k≥0k\geq 0, we have that kc≥0\frac{k}{c}\geq 0. We know that lim supk→∞d​(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)
=kc​lim supk→∞d​(xm​(k),xn​(k)−2)≤kc​ε\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 εs≤lim supk→∞d​(xm​(k)+1,xm​(k)−1)≤kc​ε\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 1s≤kc\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 k≤0k\leq 0.
In this case we have that kc≤0\frac{k}{c}\leq 0, so kc​d​(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 εs≤0\frac{\varepsilon}{s}\leq 0. Since ε>0\varepsilon>0 and s≥1s\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 xn≠xmx_{n}\neq x_{m}, for each n≠mn\neq m.
Since ( X,dX,d ) is complete, it implies that there exists u∈Xu\in X, such that xn→ux_{n}\rightarrow u, i.e.

limn→∞d​(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,f​u)\displaystyle d(u,fu) ≤s​[d​(u,xn)+d​(xn,xn+1)+d​(xn+1,f​u)]\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​(f​xn,f​u)]\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​(f​xn,f​u)≤kc​d​(xn,u)−bc​d​(xn,xn+1)−ac​d​(u,f​u)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,f​u)≤s​[d​(u,xn)+d​(xn,xn+1)+kc​d​(xn,u)−bc​d​(xn,xn+1)−ac​d​(u,f​u)]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 n→∞n\rightarrow\infty, we get

(1+s​ac)​d​(u,f​u)≤0\left(1+s\frac{a}{c}\right)d(u,fu)\leq 0

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

a​d​(x,f​x)+b​d​(y,f​y)+c​d​(f​x,f​y)≤k​d​(x,y)\displaystyle ad(x,fx)+bd(y,fy)+cd(fx,fy)\leq kd(x,y)
c​d​(f​x,f​y)≤k​d​(x,y)−a​d​(x,f​x)−b​d​(y,f​y)\displaystyle cd(fx,fy)\leq kd(x,y)-ad(x,fx)-bd(y,fy)

So

d​(f​x,f​y)≥kc​d​(x,y)−ac​d​(x,f​x)−bc​d​(y,f​y)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 k∈ℕk\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]
s​d​(xm​(k)−1,xn​(k)−3)≥d​(xn​(k)−2,xm​(k))−s​d​(xn​(k)−3,xn​(k)−2)−s​d​(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 s≥1s\geq 1, we obtain the following

d​(xm​(k)−1,xn​(k)−3)≥1s​d​(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 kc≥0\frac{k}{c}\geq 0. Multiplying by (kc)\left(\frac{k}{c}\right), it implies that

kc​d​(xm​(k)−1,xn​(k)−3)\displaystyle\frac{k}{c}d\left(x_{m(k)-1},x_{n(k)-3}\right) ≥kc​s​d​(xn​(k)−2,xm​(k))−kc​d​(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)
−kc​d​(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)≥kc​d​(xm​(k)−1,xn​(k)−3)−ac​d​(xm​(k)−1,xm​(k))−bc​d​(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) ≥kc​s​d​(xn​(k)−2,xm​(k))−kc​d​(xn​(k)−3,xn​(k)−2)−kc​d​(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)
−ac​d​(xm​(k)−1,xm​(k))−bc​d​(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) =kc​lim supk→∞−d​(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)
=−kc​lim infk→∞d​(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)​limk→∞d​(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 a≥0a\geq 0, then −ac≥0-\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 a≤0a\leq 0, then −ac≤0-\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) =ac​lim supk→∞−d​(xm​(k)−1,xm​(k))\displaystyle=\frac{a}{c}\limsup_{k\rightarrow\infty}-d\left(x_{m(k)-1},x_{m(k)}\right)
=−ac​lim infk→∞d​(xm​(k)−1,xm​(k))\displaystyle=-\frac{a}{c}\liminf_{k\rightarrow\infty}d\left(x_{m(k)-1},x_{m(k)}\right)
=−ac​limk→∞d​(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

ε≥lim supk→∞d​(xm​(k),xn​(k)−2)≥kc​s​lim supk→∞d​(xm​(k),xn​(k)−2)≥ε​kc​s2.\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 1s2≤ck\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 xn≠xmx_{n}\neq x_{m}, for each n≠m,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 u∈Xu\in X, such that xn→ux_{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

a​d​(u,f​u)+b​d​(xn,f​xn)+c​d​(f​u,f​xn)\displaystyle ad(u,fu)+bd\left(x_{n},fx_{n}\right)+cd\left(fu,fx_{n}\right) ≤k​d​(u,xn)\displaystyle\leq kd\left(u,x_{n}\right)
a​d​(u,f​u)+b​d​(xn,xn+1)+c​d​(f​u,xn+1)\displaystyle ad(u,fu)+bd\left(x_{n},x_{n+1}\right)+cd\left(fu,x_{n+1}\right) ≤k​d​(u,xn)\displaystyle\leq kd\left(u,x_{n}\right)

Letting n→∞n\rightarrow\infty, we have (a+c)​d​(u,f​u)≤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=A∪BX=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)=(x−y)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:X→Xf:X\rightarrow X, such as

f​(x)={13,x∈A15,x∈Bf(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

1⋅d​(f​x,f​y)≤152​d​(x,y)+14​d​(x,f​x)+23100​d​(y,f​y)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,y∈Xx,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.

    x∈Ax\in A and y∈A:d​(f​x,f​y)=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.

    x∈Bx\in B and y∈B:d​(f​x,f​y)=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) x∈Ax\in A and y∈By\in B :

d​(f​x,f​y)=(13,15)=6100d​(x,f​x)=d​(x,13)≥minx∈A⁡d​(x,13)=1200d​(y,f​y)=d​(y,15)=(y−15)2=y2−25​y+125≥miny∈[1,2]=1−14+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)=(y−x)2=|y−x|2d(x,y)=(y-x)^{2}=|y-x|^{2}.
We have that

d​(f​x,f​y)≤k​d​(x,y)+(−a)​minx∈A⁡d​(x,f​x)+(−b)​miny∈B⁡d​(y,f​y)d(fx,fy)\leq kd(x,y)+(-a)\min_{x\in A}d(x,fx)+(-b)\min_{y\in B}d(y,fy)

So

6100≤152​|y−x|2+14⋅1200+23100⋅625\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​|y−x|2≥−661912000\frac{1}{52}|y-x|^{2}\geq\frac{-6619}{12000}, which is obviously true.
4) x∈Bx\in B and y∈Ay\in A

d​(f​x,f​y)=(13,15)=6100d​(x,f​x)≥minx∈B⁡d​(x,f​x)=625d​(y,f​y)≥miny∈A=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)=(y−x)2=|y−x|2.d(x,y)=(y-x)^{2}=|y-x|^{2}.

We have that

6100≤152​|y−x|2+14⋅625+23100⋅1200,\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​|y−x|2≥−4196000\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>0⇔1>0\displaystyle c>0\Leftrightarrow 1>0
k>0⇔152>0\displaystyle k>0\Leftrightarrow\frac{1}{52}>0
a+c=1−14=34>0\displaystyle a+c=1-\frac{1}{4}=\frac{3}{4}>0
b≤k⇔−23100≤152\displaystyle b\leq k\Leftrightarrow-\frac{23}{100}\leq\frac{1}{52}
kc<1s⇔k<13⇔3<52\displaystyle\frac{k}{c}<\frac{1}{s}\Leftrightarrow k<\frac{1}{3}\Leftrightarrow 3<2
k<b+a+cs⇔152+23100<14⇔324<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 x≠yx\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,v∈I1={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​[minu∈I1⁡d​(1,u)+d​(3,4)+minv∈I1⁡d​(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 ​s≥32\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,v∈I2={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​[minu∈I2⁡d​(3,u)+d​(2,4)+minv∈I2⁡d​(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 ​s≥14\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,v∈I3={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​[minu∈I3⁡d​(4,u)+d​(2,3)+minv∈I3⁡d​(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 ​s≥12\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,v∈I4={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​[minu∈I4⁡d​(2,u)+d​(1,3)+minv∈I4⁡d​(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 ​s≥12\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,v∈I5={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​[minu∈I5⁡d​(3,u)+d​(1,2)+minv∈I5⁡d​(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 ​s≥29\displaystyle\leq s\left[\frac{1}{10}+\frac{6}{10}+\frac{2}{10}\right],\text{ so }s\geq\frac{2}{9}

So s≥32>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,x≠41,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​(f​x,f​y)≥(−3)​d​(x,y)−5​d​(x,f​x)+3​d​(y,f​y)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 c​d​(f​x,f​y)≥k​d​(x,y)−a​d​(x,f​x)−b​d​(y,f​y)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​(f​x,f​y)=d​(f​x,f​x)=0d(fx,fy)=d(fx,fx)=0, so the left hand side is 0 . Now, the right hand side is k⋅0−a​d​(x,f​x)−b​d​(x,f​x)=−(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,f​x)≥0(a+b)d(x,fx)\geq 0. We have two sub-cases:
    If x=3x=3, then d​(x,f​x)=d​(3,3)=0d(x,fx)=d(3,3)=0, so the inequality is valid. Also, if ≠3\neq 3, then d​(x,f​x)>0d(x,fx)>0, so we have the condition that −b≤a-b\leq a.

  2. 2.

    If ≠y\neq y, we have the following sub-cases
    a) For x=4x=4 and y≠4y\neq 4, it follows that d​(f​y,f​x)=d​(f​y,1)d(fy,fx)=d(fy,1). Since y≠4y\neq 4, then f​y=3fy=3, so d​(f​x,f​y)=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 y≠4y\neq 4,
    d​(x,f​x)=d​(4,f​x)=210d(x,fx)=d(4,fx)=\frac{2}{10}, for each x∈Xx\in X and d​(y,f​y)=d​(y,3)≤maxy≠4⁡d​(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 x≠4x\neq 4, it follows that d​(f​x,f​y)=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 x≠4x\neq 4,
d(x,fx)=d(x,3)=≥minx≠4d(x,3)=110d(x,fx)=d(x,3)=\geq\min_{x\neq 4}d(x,3)=\frac{1}{10} and d​(y,f​y)=d​(4,f​y)=210d(y,fy)=d(4,fy)=\frac{2}{10}, for each value of f​yfy.
c) For y≠y≠4y\neq y\neq 4 (simultaneously), it follows that d​(f​x,f​y)=d​(3,3)=0d(fx,fy)=d(3,3)=0. Also

k​d​(x,y)−a​d​(x,f​x)−b​d​(y,f​y)≤0, so k​d​(x,y)−b​d​(x,y)≤a​d​(x,f​x).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,y∈X⁡d​(x,y)=110d(x,y)\geq\min_{x,y\in X}d(x,y)=\frac{1}{10}.
Furthermore, we have that

d​(y,f​y)=d​(y,3)≤maxy≠4⁡d​(y,3)=210 and d​(x,f​x)=d​(x,3)≥minx≠4⁡d​(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 −b≤a-b\leq a. For the case (2a), we get that

d​(f​x,f​y)\displaystyle d(fx,fy) =110≥k​d​(x,y)−a​d​(x,f​x)−b​maxy≠4⁡d​(y,f​y)\displaystyle=\frac{1}{10}\geq kd(x,y)-ad(x,fx)-b\max_{y\neq 4}d(y,fy)
=2​k10−2​a10−2​b10≥2​k10−2​a10−b​d​(y,f​y)\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​(f​x,f​y)\displaystyle d(fx,fy) =110≥k​d​(x,y)−a​minx≠4⁡d​(x,f​x)−b​d​(y,f​y)\displaystyle=\frac{1}{10}\geq kd(x,y)-a\min_{x\neq 4}d(x,fx)-bd(y,fy)
=2​k10−a10−2​b10≥2​k10−2​b10−a​d​(x,f​x)\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​(f​x,f​y)\displaystyle d(fx,fy) =0≥k​minx≠y≠4⁡d​(x,y)−a​minx≠4⁡d​(x,f​x)−b​maxy≠4⁡d​(y,f​y)\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)
=k10−a10−2​b10​k​d​(x,y)−a​d​(x,f​x)−b​d​(y,f​y)\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 k−s​b≤ak-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

{−b≤a⇔3≤5,k<a+b+12⇔−3<2+12k<a2+b+12⇔10+12>0,k−2​b≤a⇔3>1b≤k⇔−3≤−3,kc>s2⇔12>9k<b+a+cs⇔6>0,a+c>0⇔6>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​(f​x,f​y)≤kc​d​(x,y)−ac​d​(x,f​x)−bc​d​(y,f​y), for each ​x,y∈Xd(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∈ℝ\lambda\in\mathbb{R} and x,y,z,wx,y,z,w arbitrary elements of XX, each distinct from each other. Then

λ​d​(x,z)\displaystyle\lambda d(x,z) ≥[1+s22​s​λ+1−s22​s​|λ|]​d​(x,y)+[s−12​λ−s+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)
+[s−12​λ−s+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∈ℝ\lambda\in\mathbb{R} :
Case (1): Let λ≥0\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)]
s​d​(x,z)≥d​(x,y)−s​d​(z,w)−s​d​(w,y)\displaystyle sd(x,z)\geq d(x,y)-sd(z,w)-sd(w,y)
d​(x,z)≥1s​d​(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)
λ​d​(x,z)≥λs​d​(x,y)−λ​d​(z,w)−λ​d​(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 λ≤0\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)]
λ​d​(x,z)≥λ​s​d​(x,y)+λ​s​d​(y,w)+λ​s​d​(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

{λ​d​(x,z)≥λs​d​(x,y)−λ​d​(z,w)−λ​d​(w,y),λ≥0λ​d​(x,z)≥λ​s​d​(x,y)+λ​s​d​(y,w)+λ​s​d​(w,z),λ≤0\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
λ​d​(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,λ≥0s​λ,λ≤0​ and ​ψ​(λ):={−λ,λ≥0s​λ,λ≤0\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+s22​s​λ+1−s22​s​|λ|ψ​(λ):=s−12​λ−s+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∈ℝ\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

a​d​(x,f​x)+b​d​(y,f​y)+c​d​(f​x,f​y)≤k​d​(x,y)\displaystyle ad(x,fx)+bd(y,fy)+cd(fx,fy)\leq kd(x,y)
c​d​(f​x,f​y)≤k​d​(x,y)−a​d​(x,f​x)−b​d​(y,f​y)\displaystyle cd(fx,fy)\leq kd(x,y)-ad(x,fx)-bd(y,fy)
d​(f​x,f​y)≥kc​d​(x,y)−ac​d​(x,f​x)−bc​d​(y,f​y)\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​(f​x,f​y)≥α​d​(x,y)+β​d​(x,f​x)+γ​d​(y,f​y)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 α≤0\alpha\leq 0 and α≥0\alpha\geq 0, so, when k≥0,c<0k\geq 0,c<0, respectively k≤0,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 s≥1s\geq 1. Also, consider f:X→Xf:X\rightarrow X a mapping satisfying

d​(f​x,f​y)≥α​d​(x,y)+β​d​(x,f​x)+γ​d​(y,f​y), for each ​x,y∈Xd(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<1−s,γ>s,α+γ<1−βs\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 1−α<γ​(1s−1)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)≤δ​d​(xn−1,xn)d\left(x_{n},x_{n+1}\right)\leq\delta d\left(x_{n-1},x_{n}\right), for each n∈ℕn\in\mathbb{N}, where δ=k−ba+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 0≤k−b<a+cs0\leq k-b<\frac{a+c}{s}.
In our case,

δ=k−ba+c=kc−bcac+1=α+γ1−β\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<1−βs\alpha+\gamma<\frac{1-\beta}{s}. By the contractive-type condition, we have that

d​(f​x,f​y)≥α​d​(x,y)+β​d​(x,f​x)+γ​d​(y,f​y)d(fx,fy)\geq\alpha d(x,y)+\beta d(x,fx)+\gamma d(y,fy)

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

d​(xn,xn+2)≥α​d​(xn−1,xn+1)+β​d​(xn−1,xn)+γ​d​(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 n∈ℕn\in\mathbb{N}, i.e. using (Lemma 2.7), we obtain that

γ​d​(xn+1,xn+2)≥φ​(γ)​d​(xn,xn+2)+ψ​(γ)​d​(xn,xn−1)+ψ​(γ)​d​(xn−1,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​(xn−1,xn)d_{n}:=d\left(x_{n-1},x_{n}\right), for each n∈ℕn\in\mathbb{N}.
From (2.1) we have

dn∗≥α​dn−1∗+β​dn+φ​(γ)​dn∗+ψ​(γ)​dn+ψ​(γ)​dn−1∗.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]​dn−1∗+[−ψ​(γ)−β]​dn≤|ψ​(γ)+α|​dn−1∗+|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=|α−γ|γs−1a_{2}=\frac{|\alpha-\gamma|}{\frac{\gamma}{s}-1}. For a2<1a_{2}<1, we get that |α−γ|<γs−1|\alpha-\gamma|<\frac{\gamma}{s}-1. So, we have two cases:

  • •

    When α>γ\alpha>\gamma, i.e. α−γ>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)<2​γ\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<1−βs\alpha+\gamma<\frac{1-\beta}{s}, we employ two sub-cases If β>0\beta>0, then 1−β<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 1−β>01-\beta>0 (the denominator in δ\delta is positive, so δ\delta is positive). Since β<0\beta<0, then 1−βs>1s\frac{1-\beta}{s}>\frac{1}{s}. Moreover, since α+γ>1\alpha+\gamma>1, then we get β<1−s\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 1−s<11-s<1, then s>0s>0, which are evidently true.

  • •

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

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

Moreover, since 1s−1<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<1−βs\alpha+\gamma<\frac{1-\beta}{s}, if β>0\beta>0, then α+γ<1−βs<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<1−s\beta<1-s means that the right hand side 1−βs>1\frac{1-\beta}{s}>1, so 1<α+γ<1−βs1<\alpha+\gamma<\frac{1-\beta}{s}, which is valid.
So dn∗≤a2​dn−1∗+a1​dnd_{n}^{*}\leq a_{2}d_{n-1}^{*}+a_{1}d_{n}, for each n∈ℕn\in\mathbb{N}. We know that

dn=d​(xn−1,xn)≤δ​d​(xn−1,xn−2)≤…≤δn−1​D0d_{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 dn∗≤a2​dn−1∗+a1​δn−1​D0d_{n}^{*}\leq a_{2}d_{n-1}^{*}+a_{1}\delta^{n-1}D_{0}.
We take a major bound for dn∗d_{n}^{*} :

dn∗\displaystyle d_{n}^{*} ≤a2​dn−1∗+a1​δn−1​D0≤a2​(a2​dn−2∗+a1​δn−2​D0)+a1​δn−1​D0\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}
=a22​dn−2∗+a2​a1​δn−2​D0+a1​δn−1​D0\displaystyle=a_{2}^{2}d_{n-2}^{*}+a_{2}a_{1}\delta^{n-2}D_{0}+a_{1}\delta^{n-1}D_{0}
≤a22​(a2​dn−3∗+a1​δn−3​D0)+a1​a2​δn−2​D0+a1​δn−1​D0\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}
=a23​dn−3∗+D0​a1​(δn−1+a2​δn−2+a22​δn−3)​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
≤a2k​dn−k∗+a1​(δn−1+a2​δn−2+…+a2k−1​δn−k)​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 n−k=0⟹k=nn-k=0\Longrightarrow k=n. This means that

dn∗≤a2n​d0∗+a1​D0​(a20​δn−1+a2​δn−2+…+a2n−1​δ0)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:=a20​δn−1+a2​δn−2+…+a2n−1​δ0S:=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 δn−1\delta^{n-1}. This is a geometric progression, with general term bnb_{n} and b3b2=a2​δn−3δn−2=a2δ\frac{b_{3}}{b_{2}}=a_{2}\frac{\delta^{n-3}}{\delta^{n-2}}=\frac{a_{2}}{\delta}, so

S=δn−1⋅(1−(a2δ)n)1−(a2δ)=δn−a2nδ−a2S=\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 dn∗≤a2n​d0∗+δn−a2nδ−a2​a1​D0d_{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 n∈ℕn\in\mathbb{N} and p>0p>0 fixed. We divide in two cases: the first one, when p=2​mp=2m, with m≥2m\geq 2 and the second one, when p=2​m+1p=2m+1, with m≥1m\geq 1 :
Case (i): When p=2​m+1p=2m+1, with m≥1m\geq 1. We evaluate

d​(xn,xn+p)=d​(xn,xn+2​m+1)≤s​[d​(xn,xn+1)+d​(xn+1,xn+1)+d​(xn+2,xn+2​m+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+2​m+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]+…+sm​dn+2​m\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+2​m=d​(xn+2​m,xn+2​m+1)d_{n+2m}=d\left(x_{n+2m},x_{n+2m+1}\right). So, we get the following estimation

d​(xn,xn+2​m+1)\displaystyle d\left(x_{n},x_{n+2m+1}\right) ≤s​[δn​D0+δn+1​D0]+s2​[δn+2​D0+δn+3​D0]\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+4​D0+δn+5​D0]+…+sm​δn+2​m​D0\displaystyle+s^{3}\left[\delta^{n+4}D_{0}+\delta^{n+5}D_{0}\right]+\ldots+s^{m}\delta^{n+2m}D_{0}
≤s​δn​[1+s​δ2+s2​δ4+…+]​D0+s​δn+1​[1+s​δ2+s2​δ4+…+]​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+δ1−s​δ2​s​δn​D0\displaystyle=\frac{1+\delta}{1-s\delta^{2}}s\delta^{n}D_{0}

and by hypothesis we know that s​δ2<1s\delta^{2}<1 is satisfied. So, d​(xn,xn+2​m+1)→0d\left(x_{n},x_{n+2m+1}\right)\rightarrow 0, when n→∞n\rightarrow\infty and m≥1m\geq 1 fixed.
Case (ii): When p=2​mp=2m, with m≥2m\geq 2. We evaluate

d​(xn,xn+2​m)\displaystyle d\left(x_{n},x_{n+2m}\right) ≤s​[d​(xn,xn+1)+d​(xn+1,xn+2)+d​(xn+2,xn+2​m)]\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]+s​d​(xn+2,xn+2​m)\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+
+sm−1​[d2​m−3+d2​m−2]+sm−1​d​(xn+2​m−2,xn+2​m)\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​[δn​D0+δn+1​D0]+s2​[δn+2​D0+δn+3​D0]+…+\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+
+sm−1​[δ2​m−4​D0+δ2​m−3​D0]+sm−1​d​(xn+2​m−2,xn+2​m)\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)
≤s​δn​[1+s​δ2+s2​δ4+…]​D0\displaystyle\leq s\delta^{n}\left[1+s\delta^{2}+s^{2}\delta^{4}+\ldots\right]D_{0}
+s​δn+1​[1+s​δ2+s2​δ4+…]​D0+sm−1​dn+2​m∗\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+δ1−s​δ2​s​δn​D0+sm−1​dn+2​m∗\displaystyle=\frac{1+\delta}{1-s\delta^{2}}s\delta^{n}D_{0}+s^{m-1}d_{n+2m}^{*}

Also, we have shown that dn∗≤a2n​d0∗+δn−a2nδ−a2​a1​D0d_{n}^{*}\leq a_{2}^{n}d_{0}^{*}+\frac{\delta^{n}-a_{2}^{n}}{\delta-a_{2}}a_{1}D_{0}. So dn+2​m∗≤a2n+2​m​d0∗+Q​a1​D0d_{n+2m}^{*}\leq a_{2}^{n+2m}d_{0}^{*}+Qa_{1}D_{0}, where Q:=δn+2​m−a2n+2​mδ−a2Q:=\frac{\delta^{n+2m}-a_{2}^{n+2m}}{\delta-a_{2}}.
Now, we have two cases: if δ−a2>0\delta-a_{2}>0, then Q=δn+2​m−a2n+2​mδ−a2≤δn+2​mδ−a2Q=\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 n→∞n\rightarrow\infty. In a similar manner, if δ−a2<0\delta-a_{2}<0, then

Q=a2n+2​m−δn+2​ma2−δ≤a2n+2​ma2−δQ=\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 n→∞n\rightarrow\infty. This reasoning is valid, since, from the theorem’s assumptions, we know that 0≤a2<10\leq a_{2}<1 and δ<1s<1\delta<\frac{1}{s}<1. So, in this case, since Q→0Q\rightarrow 0, then d​(xn,xn+2​m)→0d\left(x_{n},x_{n+2m}\right)\rightarrow 0, as n→∞n\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 xn≠xmx_{n}\neq x_{m}, for each n≠mn\neq m and that ( X,dX,d ) is complete. This means that there exists u∈Xu\in X, such that limn→∞xn=u\lim_{n\rightarrow\infty}x_{n}=u.
Moreover, since the contractive condition can be reduced to the original form, i.e. a​d​(x,f​x)+b​d​(y,f​y)+c​d​(f​x,f​y)≤k​d​(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 4∈X4\in X. We will determine the coefficients α,β\alpha,\beta and γ\gamma, such that ff satisfies d​(f​x,f​y)≥α​d​(x,y)+β​d​(x,f​x)+γ​d​(y,f​y)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 ​110≥α​610+β​110+γ​610\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 ​110≥α​610+β​610+γ​110\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 ​610≥α​110+β​610+γ​110\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 ​610≥α​110+β​110+γ​610\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 ​210≥α​210+β​610+γ​210\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 ​210≥α​210+β​210+γ​610\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 ​110≥α​110+β​110+γ​110\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 ​110≥α​110+β​110+γ​110\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 ​210≥α​210+β​210+γ​110\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 ​210≥α​210+β​110+γ​210\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 ​210≥α​210+β​210+γ​110\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 ​210≥α​210+β​110+γ​210\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+γ≤0\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 1≥6​α+β+2​γ1\geq 6\alpha+\beta+2\gamma, instead of (2.7) and (2.8) we require only (2.7) and instead of 1≥6​α+β+2​γ1\geq 6\alpha+\beta+2\gamma and (2.5), we require 1≥6​α+β+6​γ1\geq 6\alpha+\beta+6\gamma. We mention that all of the above reasoning was made under the assumptions that β≤0\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+γ≤0,1≥6​α+β+6​γβ<1−s,γ>s,α​γα+γ<1−βs,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+1≤α1​an+α2​an−1+…+αk​an−k+1+bn, where ​n≥k−1.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 limn→∞bn=0\lim_{n\rightarrow\infty}b_{n}=0, then it follows that limn→∞an=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)≤a2n​d0∗+δn−a2nδ−a2​a1​D0d_{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+1≤α1​an+bna_{n+1}\leq\alpha_{1}a_{n}+b_{n}, with α1∈[0,1)\alpha_{1}\in[0,1) and limn→∞bn=0\lim_{n\rightarrow\infty}b_{n}=0. Then limn→∞an=0\lim_{n\rightarrow\infty}a_{n}=0.
Now, we have proved that dn∗≤a2​dn−1∗+a1​δn−1​D0d_{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:=a1​D0​δn−1b_{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 limn→∞dn∗=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:X→Xf:X\rightarrow X a mapping satisfying

d​(f​x,f​y)≥α​d​(x,y)+β​d​(x,f​x)+γ​d​(y,f​y)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 1−β<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​(1−αγ)<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​(f​x,f​y)≥α​d​(x,y)+β​d​(x,f​x)+γ​d​(y,f​y)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 y∈Xy\in X, there exists x∈Xx\in X, such that y=f​xy=fx.
Now, for x0∈Xx_{0}\in X, there exists x1∈Xx_{1}\in X, such that x0=f​x1x_{0}=fx_{1}. Also, for x1∈Xx_{1}\in X, there exists x2∈Xx_{2}\in X, such that x1=f​x2x_{1}=fx_{2}. Inductively, we get that xn=f​xn+1x_{n}=fx_{n+1}, for each n∈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​(f​xn+1,f​xn)≥α​d​(xn,xn+1)+β​d​(xn,f​xn)+γ​d​(xn+1,f​xn+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,xn−1)≥α​d​(xn,xn+1)+β​d​(xn,xn−1)+γ​d​(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)≤(1−β)​d​(xn−1,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)≤θ​d​(xn−1,xn)\displaystyle\Longrightarrow d\left(x_{n},x_{n+1}\right)\leq\theta d\left(x_{n-1},x_{n}\right)

where θ:=1−βα+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 1−β<α+γs1-\beta<\frac{\alpha+\gamma}{s}. Furthermore, we have that dn+1:=d​(xn+1,xn)≤θn​d1d_{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 n∈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​(xn−1,xn+1)=d​(f​xn,f​xn+2)≥α​d​(xn,xn+2)+β​d​(xn,f​xn)+γ​d​(xn+2,f​xn+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)
=α​d​(xn,xn+2)+β​d​(xn,xn−1)+γ​d​(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
α​d​(xn,xn+2)≤d​(xn−1,xn+1)−β​d​(xn−1,xn)−γ​d​(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)≤1α​dn−1∗+(−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)≤1α​dn−1∗+(|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+1≤θn​D0d_{n+1}\leq\theta^{n}D_{0}, so dn≤θn−1​D0d_{n}\leq\theta^{n-1}D_{0}, it follows that

dn∗≤1α​dn−1∗+θn−1​Q​D0, 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 limn→∞dn∗=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 dn∗≤a2​dn−1∗+a1​θn−1​D0d_{n}^{*}\leq a_{2}d_{n-1}^{*}+a_{1}\theta^{n-1}D_{0}, briefly dn∗≤a2n​d0∗+θn−a2nθ−a2​a1​D0d_{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=2​m+1p=2m+1, then d​(xn,xn+2​m+1)≤1+θ1−s​θ2​s​θn​D0d\left(x_{n},x_{n+2m+1}\right)\leq\frac{1+\theta}{1-s\theta^{2}}s\theta^{n}D_{0}, and, by hypothesis, s​θ2<1s\theta^{2}<1, then d​(xn,xn+2​m+1)d\left(x_{n},x_{n+2m+1}\right) converges to 0 .
When p=2​mp=2m, then dn+2​m∗≤a2n+2​m​d0∗+θn+2​m−a2n+2​mθ−a2​a1​D0d_{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+2​m∗d_{n+2m}^{*} converges to 0 .
Moreover, d​(xn,xn+2​m)≤1+θ1−s​θ2​s​θn​D0+sm−1​dn+2​m∗d\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)≤θ​d​(xn−1,xn)d\left(x_{n},x_{n+1}\right)\leq\theta d\left(x_{n-1},x_{n}\right), for each n≥1n\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)≥A​dn−1∗+B​dn+C​dn+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

C​dn+1≥φ​(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}^{*}
dn∗≥A​dn−1∗+B​dn+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+1∗≤dn∗​[1−ψ​(C)]+(−A)​dn−1∗−φ​(C)​dn+3−B​dn\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}^{*} ≤1−ψ​(C)φ​(C)​dn∗−A​dn−1∗−[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}+Bd_{n}\right]
dn+1∗\displaystyle d_{n+1}^{*} ≤1−ψ​(C)φ​(C)​dn∗−A​dn−1∗+[|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}^{*} ≤1−ψ​(C)φ​(C)​dn∗−A​dn−1∗+[|φ​(C)|​θ2+|B|]​θn​D0\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|]​θn​D0,α1:=1−ψ​(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=1−ψ​(C)φ​(C)−A=1+CCs−1α\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|=γ−αC=\frac{\gamma}{|\alpha|}=\frac{\gamma}{-\alpha}, then s​(1−αγ)<1+1αs\left(1-\frac{\alpha}{\gamma}\right)<1+\frac{1}{\alpha}. Now, we have two sub-cases.
If 1−αγ<01-\frac{\alpha}{\gamma}<0, then α−γ>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 1−αγ>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​(1−αγ)<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 limn→∞bn=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 limn→∞dn∗=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→ℝ+d:X\times X\rightarrow\mathbb{R}_{+}, such that d​(x,y)=(x−y)2d(x,y)=(x-y)^{2}, for each x,y∈Xx,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:X→Xf:X\rightarrow X as f​(x)=x+δ1δ2f(x)=\frac{x+\delta_{1}}{\delta_{2}}, with δ1,δ2≥0\delta_{1},\delta_{2}\geq 0. It is easy to see that ff is continuous. Also, for each y∈Xy\in X, there exists x=y​δ2−δ1≥0x=y\delta_{2}-\delta_{1}\geq 0,, since δ1\delta_{1} and δ2\delta_{2} are positive, so ff is onto. Moreover:

d​(f​x,f​y)=(f​x−f​y)2=|x+δ1δ2−y+δ1δ2|=1δ2​|x−y|2=1δ2​d​(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​(f​x,f​y)≥10​d​(x,y)d(fx,fy)\geq 10d(x,y), for each x,y∈Xx,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 s≥1s\geq 1, that satisfy

a​d​(x,f​x)+b​d​(y,f​y)+c​d​(f​x,f​y)+e​d​(x,f​y)+g​d​(y,f​x)≤k​d​(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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