# Quadruple Fixed Point Theorems under Nonlinear Contractive Conditions in Partially Ordered Metric Spaces

Authors: Erdal Karapınar; Wasfi Shatanawi; Zead Mustafa
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We prove a number of quadruple fixed point theorems underϕ-contractive conditions for a mapping F:X4→X in ordered metric spaces. Also, we introduce an example to illustrate the effectiveness of our results.


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## Body

## 1. Introduction and Preliminaries

The notion of coupled fixed point was initiated by Gnana Bhaskar and Lakshmikantham [1] in 2006. In this paper, they proved some fixed point theorems under a set of conditions and utilized their theorems to prove the existence of solutions to some ordinary differential equations. Recently, Berinde and Borcut [2] introduced the notion of tripled fixed point and extended the results of Gnana Bhaskar and Lakshmikantham [1] to the case of contractive operator F:X×X×X→X, where X is a complete ordered metric space. For some related works in coupled and tripled fixed point, we refer readers to [3–32].For simplicity we will denote the cross product ofk∈ℕ copies of the space X by Xk.Definition 1.1 (see [2]).
LetX be a nonempty set and F:X3→X a given mapping. An element (x,y,z)∈X3 is called a tripled fixed point of F if
(1.1)F(x,y,z)=x,F(y,x,y)=y,F(z,y,x)=z.
Let (X,d) be a metric space. The mapping d¯:X3→X, given by
(1.2)d¯((x,y,z),(u,v,w))=d(x,y)+d(y,v)+d(z,w),
defines a metric on X3, which will be denoted for convenience by d.Definition 1.2 (see [2]).
Let(X,≤) be a partially ordered set and F:X3→X a mapping. One says that F has the mixed monotone property if F(x,y,z) is monotone nondecreasing in x and z and is monotone nonincreasing in y; that is, for any x,y,z∈X,
(1.3)x1,x2∈X,x1≤x2,impliesF(x1,y,z)≤F(x2,y,z),y1,y2∈X,y1≤y2,impliesF(x,y2,z)≤F(x,y1,z),z1,z2∈X,z1≤z2,impliesF(x,y,z1)≤F(x,y,z2).Let us recall the main results of [2] to understand our motivation toward our results in this paper.Theorem 1.3 (see [2]).
Let(X,≤) be a partially ordered set and (X,d) a complete metric space. Let F:X3→X be a continuous mapping such that F has the mixed monotone property. Assume that there exist j,k,l∈[0,1) with j+k+l<1 such that
(1.4)d(F(x,y,z),F(u,v,w))≤jd(x,u)+kd(y,v)+ld(z,w)
for all x,y,z,u,v,w∈X with x≥u, y≤v, and z≥w. If there exist x0,y0,z0∈X such that x0≤F(x0,y0,z0), y0≥F(y0,x0,y0), and z0≤F(z0,y0,x0), then F has a tripled fixed point.Theorem 1.4 (see [2]).
Let(X,≤) be a partially ordered set and (X,d) a complete metric space. Let F:X3→X be a mapping having the mixed monotone property. Assume that there exist j,k,l∈[0,1) with j+k+l<1 such that
(1.5)d(F(x,y,z),F(u,v,w))≤jd(x,u)+kd(y,v)+ld(z,w)
for all x,y,z,u,v,w∈X with x≥u, y≤v, and z≥w. Assume that X has the following properties:  (i)
if a nondecreasing sequencexn→x, then xn≤x for all n∈N,(ii)
if a nonincreasing sequenceyn→y, then yn≥y for all n∈N. If there exist x0,y0,z0∈X such that x0≤F(x0,y0,z0), y0≥F(y0,x0,y0), and z0≤F(z0,y0,x0), then F has a tripled fixed point.Very recently, Karapınar introduced the notion of quadruple fixed point and obtained some fixed point theorems on the topic [33]. Extending this work, quadruple fixed point is developed and related fixed point theorems are proved in [34–39].Definition 1.5 (see [34]).
LetX be a nonempty set and F:X4→X a given mapping. An element (x,y,z,w)∈X×X3 is called a quadruple fixed point of F if
(1.6)F(x,y,z,w)=x,F(y,z,w,x)=y,F(z,w,x,y)=z,F(w,x,y,z)=w.
Let (X,d) be a metric space. The mapping d¯:X4→X, given by
(1.7)d¯((x,y,z,w),(u,v,h,l))=d(x,y)+d(y,v)+d(z,h)+d(w,l),
defines a metric on X4, which will be denoted for convenience by d.Remark 1.6.
In [33, 34, 38], the notion of quadruple fixed point is called quartet fixed point.Definition 1.7 (see [34]).
Let(X,≤) be a partially ordered set and F:X4→X a mapping. One says that F has the mixed monotone property if F(x,y,z,w) is monotone nondecreasing in x and z and is monotone nonincreasing in y and w; that is, for any x,y,z,w∈X,
(1.8)x1,x2∈X,x1≤x2,impliesF(x1,y,z,w)≤F(x2,y,z,w),y1,y2∈X,y1≤y2,impliesF(x,y2,z,w)≤F(x,y1,z,w),z1,z2∈X,z1≤z2,impliesF(x,y,z1,w)≤F(x,y,z2,w),w1,w2∈X,w1≤w2,impliesF(x,y,z,w2)≤F(x,y,z,w1).By following Matkowski [40], we let Φ be the set of all nondecreasing functions ϕ:[0,+∞)→[0,+∞) such that limn→+∞ϕn(t)=0 for all t>0. Then, it is an easy matter to show that(1)
ϕ(t)<t for all t>0,(2)
ϕ(0)=0.In this paper, we prove some quadruple fixed point theorems for a mappingF:X4→X satisfying a contractive condition based on some ϕ∈Φ.


## 2. Main Results

Our first result is the following.Theorem 2.1.
Let(X,≤) be a partially ordered set and (X,d) a complete metric space. Let F:X4→X be a continuous mapping such that F has the mixed monotone property. Assume that there exists ϕ∈Φ such that
(2.1)d(F(x,y,z,w),F(u,v,h,l))≤ϕ(max⁡{d(x,u),d(y,v),d(z,h),d(w,l)})
for all x,y,z,w,u,v,h,l∈X with x≥u, y≤v, z≥h, and w≤l. If there exist x0,y0,z0,w0∈X such that x0≤F(x0,y0,z0,w0), y0≥F(y0,z0,w0,x0), z0≤F(z0,w0,x0,y0) and w0≥F(w0,x0,y0,z0), then F has a quadruple fixed point.Proof.
Supposex0,y0,z0,w0∈X are such that x0≤F(x0,y0,z0,w0), y0≥F(y0,z0,w0,x0), z0≤F(z0,w0,x0,y0), and w0≥F(w0,x0,y0,z0). Define
(2.2)x1=F(x0,y0,z0,w0),y1=F(y0,z0,w0,x0),z1=F(z0,w0,x0,y0),w1=F(w0,x0,y0,z0).
Then, x0≤x1, y0≥y1, z0≤z1, and w0≥w1. Again, define x2=F(x1,y1,z1,w1), y2=F(y1,z1,w1,x1), z2=F(z1,w1,x1,y1), and w2=F(w1,x1,y1,z1). Since F has the mixed monotone property, we have x0≤x1≤x2, y2≤y1≤y0, z0≤z1≤z2, and w2≤w1≤w0. Continuing this process, we can construct four sequences (xn), (yn), (zn), and (wn) in X such that
(2.3)xn=F(xn-1,yn-1,zn-1,wn-1)≤xn+1=F(xn,yn,zn,wn),yn+1=F(yn,zn,wn,xn)≤yn=F(yn-1,zn-1,wn-1,xn-1),zn=F(zn-1,wn-1,xn-1,yn-1)≤zn+1=F(zn,wn,xn,yn),wn+1=F(wn,xn,yn,zn)≤wn=F(wn-1,xn-1,yn-1,zn-1).
If, for some integer n, we have (xn+1,yn+1,zn+1,wn+1)=(xn,yn,zn,wn), then F(xn,yn,zn,wn)=xn, F(yn,zn,wn,xn)=yn, F(zn,wn,xn,yn)=zn, and F(wn,xn,yn,zn)=wn; that is, (xn,yn,zn,wn) is a quadruple fixed point of F. Thus, we will assume that (xn+1,yn+1,zn+1,wn+1)≠(xn,yn,zn,wn) for all n∈ℕ; that is, we assume that xn+1≠xn,yn+1≠yn, or zn+1≠zn or wn+1≠wn. For any n∈ℕ, we have
(2.4)d(xn+1,xn):=d(F(xn,yn,zn,wn),F(xn-1,yn-1,zn-1,wn-1))≤ϕ(max⁡{d(xn,xn-1),d(yn,yn-1),d(zn,zn-1),d(wn,wn-1)}),d(yn,yn+1):=d(F(yn-1,zn-1,wn-1,xn-1),F(yn,zn,wn,xn))≤ϕ(max⁡{d(yn-1,yn),d(zn,zn-1),d(wn,wn-1),d(xn-1,xn)}),d(zn+1,zn):=d(F(zn,wn,xn,yn),F(zn-1,wn-1,xn-1,yn-1))≤ϕ(max⁡{d(zn,zn-1),d(wn,wn-1),d(xn,xn-1),d(yn,yn-1)}),d(wn,wn+1):=d(F(wn-1,xn-1,yn-1,zn-1),F(wn,xn,yn,zn))≤ϕ(max⁡{d(yn-1,yn),d(zn,zn-1),d(wn,wn-1),d(xn-1,xn)}).
From (2.4), it follows that
(2.5)max⁡{d(xn+1,xn),d(yn,yn+1),d(zn+1,zn),d(wn,wn+1)}≤ϕ(max⁡{d(xn,xn-1),d(yn,yn-1),d(zn,zn-1),d(wn,wn-1)}).
By repeating (2.5) n times, we get that
(2.6)max⁡{d(xn+1,xn),d(yn,yn+1),d(zn+1,zn),d(wn,wn+1)}≤ϕ(max⁡{d(xn,xn-1),d(yn,yn-1),d(zn,zn-1),d(wn,wn-1)})≤ϕ2(max⁡{d(xn-1,xn-2),d(yn-1,yn-2),d(zn-1,zn-2),d(wn-1,wn-1)})⋮≤ϕn(max⁡{d(x1,x0),d(y1,y0),d(z1,z0),d(w1,w0)}).
Now, we will show that (xn), (yn), (zn), and (wn) are Cauchy sequences in X. Let ϵ>0. Since
(2.7)lim⁡n→+∞ϕn(max⁡{d(x1,x0),d(y1,y0),d(z1,z0),d(w1,w0)})=0
and ϵ>ϕ(ϵ), there exist n0∈ℕ such that
(2.8)ϕn(max⁡{d(x1,x0),d(y1,y0),d(z1,z0),d(w1,w0)})<ϵ-ϕ(ϵ)∀n≥n0.
This implies that
(2.9)max⁡{d(xn+1,xn),d(yn,yn+1),d(zn+1,zn),d(wn,wn+1)}<ϵ-ϕ(ϵ)∀n≥n0.
For m,n∈ℕ, we will prove by induction on m that
(2.10)max⁡{d(xn,xm),d(yn,ym),d(zn,zm),d(wn,wm)}<ϵ∀m≥n≥n0.
Since ϵ-ϕ(ϵ)<ϵ, then by using (2.9) we conclude that (2.10) holds when m=n+1. Now suppose that (2.10) holds for m=k. For m=k+1, we have
(2.11)d(xn,xk+1)≤d(xn,xn+1)+d(xn+1,xk+1)≤ϵ-ϕ(ϵ)+d(F(xn,yn,zn,wn),F(xk,yk,zk,wk))≤ϵ-ϕ(ϵ)+ϕ(max⁡⁡{d(xn,xk),d(yn,yk),d(zn,zk),d(wn,wk)})<ϵ-ϕ(ϵ)+ϕ(ϵ)=ϵ.
Similarly, we show that
(2.12)d(yn,yk+1)<ϵ,d(zn,zk+1)<ϵ,d(wn,wk+1)<ϵ.
Hence, we have
(2.13)max⁡⁡{d(xn,xk+1),d(yn,yk+1),d(zn,zk+1),d(wn,wk+1)}<ϵ.
Thus, (2.10) holds for all m≥n≥n0. Hence, (xn), (yn), (zn), and (wn) are Cauchy sequences in X.
SinceX is a complete metric space, there exist x,y,z,w∈X such that (xn), (yn), (zn) and (wn) converge to x, y, z, and w, respectively. Finally, we show that (x,y,z,w) is a quadruple fixed point of F. Since F is continuous and (xn,yn,zn,wn)→(x,y,z,w), we have xn+1=F(xn,yn,zn,wn)→F(x,y,z,w). By the uniqueness of limit, we get that x=F(x,y,z,w). Similarly, we show that y=F(y,z,w,x), z=F(z,w,x,y), and w=F(w,x,y,z). So, (x,y,z,w) is a quadruple fixed point of F.By takingϕ(t)=kt, where k∈[0,1), in Theorem 2.1, we have the following.Corollary 2.2.
Let(X,≤) be a partially ordered set and (X,d) a complete metric space. Let F:X4→X be a continuous mapping such that F has the mixed monotone property. Assume that there exists k∈[0,1) such that
(2.14)d(F(x,y,z,w),F(u,v,h,l))≤kmax⁡⁡{d(x,u),d(y,v),d(z,h),d(w,l)}
for all x,y,z,w,u,v,h,l∈X with x≥u, y≤v, z≥h, and w≤l. If there exist x0,y0,z0,w0∈X such that x0≤F(x0,y0,z0,w0), y0≥F(y0,z0,w0,x0), z0≤F(z0,w0,x0,y0), and w0≥F(w0,x0,y0,z0), then F has a quadruple fixed point.As a consequence of Corollary2.2, we have the following.Corollary 2.3.
Let(X,≤) be a partially ordered set and (X,d) a complete metric space. Let F:X4→X be a continuous mapping such that F has the mixed monotone property. Assume that there exist a1,a2,a3,a4∈[0,1) with a1+a2+a3+a4<1 such that
(2.15)d(F(x,y,z,w),F(u,v,h,l))≤a1d(x,u)+a2d(y,v)+a3d(z,h)+a4d(w,l)
for all x,y,z,w,u,v,h,l∈X with x≥u, y≤v, z≥h, and w≤l. If there exist x0,y0,z0,w0∈X such that x0≤F(x0,y0,z0,w0), y0≥F(y0,z0,w0,x0), z0≤F(z0,w0,x0,y0) and w0≥F(w0,x0,y0,z0), then F has a quadruple fixed point.By adding an additional hypothesis, the continuity ofF in Theorem 2.1 can be dropped.Theorem 2.4.
Let(X,≤) be a partially ordered set and (X,d) a complete metric space. Let F:X4→X be a mapping having the mixed monotone property. Assume that there exists ϕ∈Φ such that
(2.16)d(F(x,y,z,w),F(u,v,h,l))≤ϕ(max⁡{d(x,u),d(y,v),d(z,h),d(w,l)})
for all x,y,z,w,u,v,h,l∈X with x≥u, y≤v, z≥h, and w≤l. Assume also that X has the following properties:  (i)
if a nondecreasing sequencexn→x, then xn≤x for all n∈N,(ii)
if a nonincreasing sequenceyn→y, then yn≥y for all n∈N.
If there existx0,y0,z0,w0∈X such that x0≤F(x0,y0,z0,w0), y0≥F(y0,z0,w0,x0), z0≤F(z0,w0,x0,y0), and w0≥F(w0,x0,y0,z0), then F has a quadruple fixed point.Proof.
By following the same process in Theorem2.1, we construct four Cauchy sequences (xn), (yn), (zn), and (wn) in X with
(2.17)x1≤x2≤⋯≤xn≤…,y1≥y2≥⋯≥yn≥⋯,z1≤z2≤⋯≤zn≤⋯,w1≥w2≥⋯≥wn≥⋯,
such that xn→x∈X, yn→y∈X, zn→z∈X, and wn→w∈X. By the hypotheses on X, we have xn≤x, yn≥y, zn≤z, and wn≥w for all n∈N. From (2.16), we have
(2.18)d(F(x,y,z,w),xn+1):=d(F(x,y,z,w),F(xn,yn,zn,wn))≤ϕ(max⁡{d(x,xn),d(y,yn),d(z,zn),d(w,wn)}),d(yn+1,F(y,z,w,x)):=d(F(yn,zn,wn,xn),F(y,z,w,x))≤ϕ(max⁡{d(yn,y),d(zn,z),d(wn,w),d(xn,x)}),d(F(z,w,x,y),zn+1):=d(F(z,w,x,y),F(zn,wn,xn,yn))≤ϕ(max⁡{d(x,xn),d(y,yn),d(z,zn),d(w,wn)}),d(wn+1,F(w,x,y,z)):=d(F(wn,xn,yn,zn),F(w,x,y,z))≤ϕ(max⁡⁡{d(yn,y),d(zn,z),d(wn,w),d(xn,x)}).
From (2.18), we have
(2.19)max⁡{d(F(x,y,z,w),xn+1),d(yn+1,F(y,z,w,x)),d(F(z,w,x,y),zn+1),d(wn+1,F(w,x,y,z))}≤ϕ(max⁡{d(x,xn),d(y,yn),d(z,zn),d(w,wn)}).
Letting n→+∞ in (2.19), it follows that x=F(x,y,z,w), y=F(y,z,w,x), z=F(z,w,x,y), and w=F(w,x,y,z). Hence, (x,y,z,w) is a quadruple fixed point of F.By takingϕ(t)=kt, where k∈[0,1), in Theorem 2.4, we have the following result.Corollary 2.5.
Let(X,≤) be a partially ordered set and (X,d) a complete metric space. Let F:X4→X be a mapping having the mixed monotone property. Assume that there exists k∈[0,1) such that
(2.20)d(F(x,y,z,w),F(u,v,h,l))≤kmax⁡{d(x,u),d(y,v),d(z,h),d(w,l)}
for all x,y,z,w,u,v,h,l∈X with x≥u, y≤v, z≥h, and w≤l. Assume also that X has the following properties:  (i)
if a nondecreasing sequencexn→x, then xn≤x for all n∈N,(ii)
if a nonincreasing sequenceyn→y, then yn≥y for all n∈N. If there exist x0,y0,z0,w0∈X such that x0≤F(x0,y0,z0,w0), y0≥F(y0,z0,w0,x0), z0≤F(z0,w0,x0,y0), and w0≥F(w0,x0,y0,z0), then F has a quadruple fixed point.As a consequence of Corollary2.5, we have the following.Corollary 2.6.
Let(X,≤) be a partially ordered set and (X,d) a complete metric space. Let F:X4→X be a mapping having the mixed monotone property. Assume that there exist a1,a2,a3,a4∈[0,1) with a1+a2+a3+a4<1 such that
(2.21)d(F(x,y,z,w),F(u,v,h,l))≤a1d(x,u)+a2d(y,v)+a3d(z,h)+a4d(w,l)
for all x,y,z,w,u,v,h,l∈X with x≥u, y≤v, z≥h, and w≤l. Assume that X has the following properties:  (i)
if a nondecreasing sequencexn→x, then xn≤x for all n∈N,(ii)
if a nonincreasing sequenceyn→y, then yn≥y for all n∈N. If there exist x0,y0,z0,w0∈X such that x0≤F(x0,y0,z0,w0), y0≥F(y0,z0,w0,x0), z0≤F(z0,w0,x0,y0), and w0≥F(w0,x0,y0,z0), then F has a quadruple fixed point.Now we prove the following result.Theorem 2.7.
In addition to the hypotheses of Theorem2.1 (resp., Theorem 2.4), suppose that
(2.22)[(x0≤y0)∧(z0≤y0)∧(x0≤w0)∧(z0≤w0)]∨[(y0≤x0)∧(y0≤z0)∧(w0≤x0)∧(w0≤z0)].
Then, x=y=z=w.Proof.
Without loss of generality, we may assume thatx0≤y0, z0≤y0, x0≤w0, and z0≤w0. By the mixed monotone property of F, we have xn≤yn, zn≤yn, xn≤wn, and zn≤wn for all n∈N. Thus, by (2.1), we have
(2.23)d(yn+1,xn+1):=d(F(yn,zn,wn,xn),F(xn,yn,zn,wn))≤ϕ(max⁡{d(yn,xn),d(zn,yn),d(wn,zn),d(xn,wn)}),(2.24)d(yn+1,zn+1):=d(F(yn,zn,wn,xn),F(zn,wn,xn,yn))≤ϕ(max⁡⁡{d(yn,zn),d(zn,wn),d(wn,xn),d(xn,yn)}),(2.25)d(wn+1,xn+1)≔d(F(wn,xn,yn,zn),F(xn,yn,zn,wn))≤ϕ(max⁡⁡{d(xn,wn),d(yn,xn),d(zn,yn),d(wn,zn)}),(2.26)d(wn+1,zn+1):=d(F(wn,xn,yn,zn),F(zn,wn,xn,yn))≤ϕ(max⁡⁡{d(zn,wn),d(wn,xn),d(xn,yn),d(yn,zn)}).
By (2.23) and (2.26), we have
(2.27)max⁡⁡{d(yn+1,xn+1),d(yn+1,zn+1),d(wn+1,xn+1),d(wn+1,zn+1)}≤ϕ(max⁡⁡{d(yn,xn),d(yn,zn),d(wn,xn),d(wn,zn)})≤ϕ2(max⁡⁡{d(yn-1,xn-1),d(yn-1,zn-1),d(wn-1,xn-1),d(wn-1,zn-1)})⋮≤ϕn+1(max⁡⁡{d(y0,x0),d(y0,z0),d(w0,x0),d(w0,z0)}).
By letting n→+∞ in (2.27) and using the property of ϕ and the fact that d is continuous on its variable, we get that max{d(y,x),d(y,z),d(w,x),d(w,z)}=0. Hence, y=z=x=w.Corollary 2.8.
In addition to the hypotheses of Corollary2.3 (resp., Corollary 2.5), suppose that
(2.28)[(x0≤y0)∧(z0≤y0)∧(x0≤w0)∧(z0≤w0)]∨[(y0≤x0)∧(y0≤z0)∧(w0≤x0)∧(w0≤z0)].
Then, x=y=z=w.Example 2.9.
LetX=[0,1] with usual order. Define d:X×X→X by d(x,y)=|x-y|. Define F:X4→X by
(2.29)F(x,y,z,w)={0,max⁡{y,w}≥min⁡{x,z},14(min⁡{x,z}-max⁡{y,w}),max⁡{y,w}<min⁡{x,z}.
Then,  (a)
(X,d,≤) is a complete ordered metric space,(b)
forx,y,z,w,u,v,h,l∈X with x≥u, y≤v, z≥h, and w≤l, we have that
(2.30)d(F(x,y,z,w),F(u,v,h,l))≤12max⁡{d(x,u),d(y,v),d(z,h),d(w,l)},(c)
holds for allx≥u, y≤v, z≥h, and w≤l,(d)
F has the mixed monotone property.Proof.
To prove(b), given x,y,z,w,u,v,h,l∈X with x≥u, y≤v, z≥h, and w≤l, we examine the following cases.
Case1.If max{y,w}≥min{x,z}, and max{v,l}≥min{u,w}. Here, we have
(2.31)d(F(x,y,z,w),F(u,v,h,l))=0≤12max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
Case2.If max{y,w}≥min{x,z} and max{v,l}<min{u,h}. This case is impossible since
(2.32)y≤v<min⁡{u,h}≤min⁡{x,z},w≤l<min⁡{u,h}≤min⁡{x,z}.
So,
(2.33)max⁡{y,w}<min⁡{x,z}.
Case3.If max{y,w}<min{x,z} and max{v,l}≥min{u,h}.
This case will have different possibilities.
(i) Letmax{y,w}=y and max{v,l}=v. Suppose that h≤v; then h-y≤v-y and hence
(2.34)min⁡{x,z}-max⁡{y,w}=min⁡{x,z}-y≤z-y=z-h+h-y≤z-h+v-y=d(z,h)+d(y,v)≤2max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
Therefore,
(2.35)d(F(x,y,z,w),F(u,v,h,l))=d(14(min⁡{x,z}-max⁡{y,w}),0)=14(min⁡⁡{x,z}-y)≤12max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
Suppose that u≤v; then u-y≤v-y and hence
(2.36)min⁡{x,z}-max⁡{y,w}=min⁡{x,z}-y≤x-y=x-u+u-y≤(x-u)+(v-y)=d(x,u)+d(v,y)≤2max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
Therefore,
(2.37)d(F(x,y,z,w),F(u,v,h,l))=d(14(min⁡{x,z}-max⁡{y,w}),0)=14(min⁡{x,z}-y)≤12max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
(ii) Letmax{y,w}=y and max{v,l}=l. Suppose that h≤l; then h-y≤l-y and (since w≤y) hence
(2.38)min⁡{x,z}-max⁡{y,w}=min⁡{x,z}-y≤z-y=z-h+h-y≤z-h+l-y≤z-h+l-w=d(z,h)+d(w,l)≤2max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
Therefore,
(2.39)d(F(x,y,z,w),F(u,v,h,l))=d(14(min⁡{x,z}-max⁡{y,w}),0)=14(min⁡{x,z}-y)≤12max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
Suppose thatu≤l; then u-y≤l-y and (since w≤y ) hence
(2.40)min⁡{x,z}-max⁡{y,w}=min⁡{x,z}-y≤x-y=x-u+u-y≤(x-u)+(l-y)≤x-u+l-w=d(x,u)+d(w,l)≤2max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
Therefore,
(2.41)d(F(x,y,z,w),F(u,v,h,l))=d(14(min⁡{x,z}-max⁡{y,w}),0)=14(min⁡{x,z}-y)≤12max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
(iii) Letmax{y,w}=w and max{v,l}=v. Suppose that h≤v; then h-w≤v-w, but y≤w, and hence
(2.42)min⁡{x,z}-max⁡{y,w}=min⁡{x,z}-w≤z-w=z-h+h-w≤z-h+v-w≤z-h+v-y=d(z,h)+d(y,v)≤2max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
Therefore,
(2.43)d(F(x,y,z,w),F(u,v,h,l))=d(14(min⁡{x,z}-max⁡{y,w}),0)=14(min⁡{x,z}-w)≤12max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
Suppose thatu≤v; then u-w≤v-w and hence
(2.44)min⁡{x,z}-max⁡{y,w}=min⁡{x,z}-w≤x-w=x-u+u-w≤(x-u)+(v-w)≤x-u+v-y=d(x,u)+d(v,y)≤2max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
Therefore,
(2.45)d(F(x,y,z,w),F(u,v,h,l))=d(14(min⁡{x,z}-max⁡{y,w}),0)=14(min⁡{x,z}-w)≤12max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
(iv) Letmax{y,w}=w and max{v,l}=l. Suppose that h≤l; then h-w≤l-w and hence
(2.46)min⁡{x,z}-max⁡{y,w}=min⁡{x,z}-w≤z-w=z-h+h-w≤z-h+l-w=d(z,h)+d(w,l)≤2max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
Therefore,
(2.47)d(F(x,y,z,w),F(u,v,h,l))=d(14(min⁡{x,z}-max⁡{y,w}),0)=14(min⁡{x,z}-w)≤12max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
Suppose thatu≤l; then u-w≤l-w and hence
(2.48)min⁡{x,z}-max⁡{y,w}=min⁡{x,z}-w≤x-w=x-u+u-w≤(x-u)+(l-w)=d(x,u)+d(w,l)≤2max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
Therefore,
(2.49)d(F(x,y,z,w),F(u,v,h,l))=d(14(min⁡{x,z}-max⁡{y,w}),0)=14(min⁡{x,z}-w)≤12max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
Case4. (i) If max{y,w}<min{x,z} and max{v,l}<min{u,h}.
Sincex≥u and z≥h, then min{x,z}≥min{u,h}, and also since y≥v and w≥l, then max{v,l}≥max{y,w}. Thus,
(2.50)d(F(x,y,z),F(u,v,w))=d(14(min⁡{x,z}-max⁡{y,w}),14(min⁡{u,h}-max⁡{v,l}))=14|(min⁡{x,z}-min⁡{u,h})+(max⁡{v,l}-max⁡{y,w})|.
(ii) Ifmin{u,h}=u and max{v,l}=v, then min{x,z}-min{u,h}≤x-u and max{v,l}-max{y,w}≤v-y. Thus,
(2.51)d(F(x,y,z,w),F(u,v,h,l))≤14[(x-u)+(v-y)]=14[d(x,u)+d(y,v)]≤12max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
(iii) Ifmin{u,h}=h and max{v,l}=v, then min{x,z}-min{u,h}≤z-h and max{v,l}-max{y,w}≤v-y, hence
(2.52)d(F(x,y,z,w),F(u,v,h,l))≤14[(z-h)+(v-y)]=14[d(z,h)+d(y,v)]≤12max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
(iv) Ifmin{u,h}=u and max{v,l}=l, then min{x,z}-min{u,h}≤x-u and max{v,l}-max{y,w}≤l-w, and hence
(2.53)d(F(x,y,z,w),F(u,v,h,l))≤14[(x-u)+(l-w)]=14[d(x,u)+d(w,l)]≤12max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
(v) Ifmin{u,h}=h and max{v,l}=l, then min{x,z}-min{u,h}≤z-h and max{v,l}-max{y,w}≤l-w, and hence
(2.54)d(F(x,y,z,w),F(u,v,h,l))≤14[(z-h)+(l-w)]=14[d(z,h)+d(w,l)]≤12max⁡{d(x,u),d(y,v),d(z,h),d(w,l)}.
To prove (c), letx,y,z,w∈X. To show that F(x,y,z,w) is monotone nondecreasing in x, let x1,x2∈X with x1≤x2.
Ifmax{y,w}≥min{x1,z}, then F(x1,y,z,w)=0≤F(x2,y,z,w). If max{y,w}<min{x1,z}, then
(2.55)F(x1,y,z,w)=14(min⁡{x1,z}-max⁡{y,w})≤14(min⁡{x2,z}-max⁡{y,w})=F(x2,y,z,w).
Therefore, F(x,y,z,w) is monotone nondecreasing in x. Similarly, we may show that F(x,y,z,w) is monotone nondecreasing in z.
To show thatF(x,y,z,w) is monotone nonincreasing in y, let y1,y2∈X with y1≤y2. If max{y2,w}≥min{x,z}, then F(x,y2,z,w)=0≤F(x1,y,z,w). If max{y2,w}<min{x,z}, then
(2.56)F(x,y2,z,w)=14(min⁡{x,z}-max⁡{y2,w})≤14(min⁡{x,z}-max⁡{y1,w})=F(x,y2,z,w).
Therefore, F(x,y,z,w) is monotone nonincreasing in y. Similarly, we may show that F(x,y,z,w) is monotone nonincreasing in w.
Thus, by Theorem2.1 (let ϕ(t)=(t/2)), F has a unique quadruple fixed point, namely, (0,0,0,0). Since the condition of Theorem 2.7 is satisfied, (0,0,0,0) is the unique quadruple fixed point of F.Remark 2.10.
We notice that for,F:X2n→X,(n∈ℕ), it is very natural to consider the analog of Theorem 2.1–Theorem 2.7 to get fixed points. Moreover, for F:X2n+1→X(n∈ℕ), the analog of Theorem 7–Theorem 11 of Berinde and Borcut [2] yields fixed points.



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*Source: 951912-2012-07-08.xml*