# Quadruple  Fixed Point Theorems for Weakϕ-Contractions

Authors: Erdal Karapınar
Journal: ISRN Mathematical Analysis
 (2011)

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## Abstract

The notion of coupled fixed point is introduced in by Gnana Bhaskar and Lakshmikantham (2006). Very recently, the concept of tripled fixed point is introduced by Berinde and Borcut (2011). In this paper, quadruple fixed point is introduced, and some new fixed point theorems are obtained.


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## Body

## 1. Introduction and Preliminaries

Very recently, Berinde and Borcut [1] introduced the concept of triple fixed point. Their contributions are inspired from the remarkable paper of Gnana Bhaskar and Lakshmikantham [2] in which they introduced the notion of coupled fixed point and proved some fixed point theorems under certain condition. Later, Lakshmikantham and Ljubomir Ćirić in [3] extended these results by defining of g-monotone property. Many authors focused on coupled fixed point theory and proved remarkable results (see, e.g., [4–9]).Here we recall the basic definitions and results from which triple and quadruple fixed point [10] notions are inspired. Let (X,d) be a metric space and X2:=X×X. Then the mapping ρ:=X2×X2:→E such that ρ((x1,y1),(x2,y2)):=d(x1,x2)+d(y1,y2) forms a metric on X2. A sequence ({xn},{yn})∈X2 is said to be a double sequence of X.Definition 1.1 (see [2]).
Let(X,≤) be partially ordered set and F:X×X→X. F is said to have mixed monotone property if F(x,y) is monotone nondecreasing in x and is monotone nonincreasing in y, that is, for any x,y∈X,
(1.1)x1≤x2⟹F(x1,y)≤F(x2,y),forx1,x2∈X,y1≤y2⟹F(x,y2)≤F(x,y1),fory1,y2∈X.Definition 1.2 (see [2]).
An element(x,y)∈X×X is said to be a couple fixed point of the mapping F:X×X→X if
(1.2)F(x,y)=x,F(y,x)=y.Throughout this paper, let(X,≤) be partially ordered set and d a metric on X such that (X,d) is a complete metric space. Further, the product spaces X×X satisfy the following:(1.3)(u,v)≤(x,y)⟺u≤x,y≤v;∀(x,y),(u,v)∈X×X.The following two results of Gnana Bhaskar and Lakshmikantham in [2] were extended to class of cone metric spaces in [7].Theorem 1.3.
LetF:X×X→X be a continuous mapping having the mixed monotone property on X. Assume that there exists a k∈[0,1) with
(1.4)d(F(x,y),F(u,v))≤k2[d(x,u)+d(y,v)],∀u≤x,y≤v.
If there exists x0,y0∈X such that x0≤F(x0,y0) and F(y0,x0)≤y0, then, there exists x,y∈X such that x=F(x,y) and y=F(y,x).Theorem 1.4.
LetF:X×X→X be a mapping having the mixed monotone property on X. Suppose that X has the following properties:  (i)
if a nondecreasing sequence{xn}→x, then xn≤x, for all n;(ii)
if a nonincreasing sequence{yn}→y, then y≤yn, for all n. Assume that there exists a k∈[0,1) with
(1.5)d(F(x,y),F(u,v))≤k2[d(x,u)+d(y,v)],∀u≤x,y≤v.
If there exists x0,y0∈X such that x0≤F(x0,y0) and F(y0,x0)≤y0, then, there exists x,y∈X such that x=F(x,y) and y=F(y,x).Inspired by Definition1.1, Berinde and Borcut [1] introduced the following partial order on the product space X3=X×X×X: (1.6)(u,v,w)≤(x,y,z)iffx≥u,y≤v,z≥w,
where (u,v,w),(x,y,z)∈X3. Regarding this partial order, we state the definition of the following mapping.Definition 1.5 (see [1]).
Let(X,≤) be partially ordered set and F:X3→X. We say that F has the mixed monotone property if F(x,y,z) is monotone nondecreasing in x and y, and it is monotone nonincreasing in y, that is, for any x,y,z∈X(1.7)x1,x2∈X,x1≤x2⟹F(x1,y,z)≤F(x2,y,z),y1,y2∈X,y1≤y2⟹F(x,y1,z)≥F(x,y2,z),z1,z2∈X,z1≤z2⟹F(x,y,z1)≤F(x,y,z2).Definition 1.6 (see [1]).
An element(x,y,z,w)X3 is called a quadruple fixed point of F:X3→X if
(1.8)F(x,y,z)=x,F(y,z,y)=y,F(z,y,x)=z.For a metric space(X,d), the function ρ:X3→[0,∞), given by
(1.9)ρ((x,y,z),(u,v,w)):=d(x,u)+d(y,v)+d(z,w),
forms a metric space on X3; that is, (X3,ρ) is a metric induced by (X,d).Theorem 1.7.
Let(X,≤) be partially ordered set, and let (X,d) be a complete metric space. Let F:X×X×X→X be a continuous mapping having the mixed monotone property on X. Assume that there exist constants a,b,c∈[0,1) such that a+b+c<1 for which
(1.10)d(F(x,y,z),F(u,v,w))≤ad(x,u)+bd(y,v)+cd(z,w)
for all x≥u,y≤v,z≥w. If there exist x0,y0,z0∈X such that
(1.11)x0≤F(x0,y0,z0),y0≥F(y0,x0,y0),z0≤F(x0,y0,z0),
then there exist x,y,z∈X such that
(1.12)F(x,y,z)=x,F(y,z,y)=y,F(z,y,x)=z.The aim of this paper is to introduce the concept of quadruple fixed point and prove the related fixed point theorems.


## 2. Quadruple Fixed Point Theorems

Let(X,≤) be partially ordered set and (X,d) a complete metric space. We consider the following partial order on the product space X4=X×X×X×X: (2.1)(u,v,r,t)≤(x,y,z,w)iffx≥u,y≤v,z≥r,t≤w,
where (u,v,r,t),(x,y,z,w)∈X4. Regarding this partial order, we state the definition of the following mapping.Definition 2.1.
Let(X,≤) be partially ordered set and F:X4→X. We say that F has the mixed monotone property if F(x,y,z,w) is monotone nondecreasing in x and z, and it is monotone nonincreasing in y and w, that is, for any x,y,z,w∈X(2.2)x1,x2∈X,x1≤x2⟹F(x1,y,z,w)≤F(x2,y,z,w),y1,y2∈X,y1≤y2⟹F(x,y1,z,w)≥F(x,y2,z,w),z1,z2∈X,z1≤z2⟹F(x,y,z1,w)≤F(x,y,z2,w),w1,w2∈X,w1≤w2⟹F(x,y,z,w1)≥F(x,y,z,w2).Definition 2.2.
An element(x,y,z)∈X4 is called a triple fixed point of F:X4→X if
(2.3)F(x,y,z,w)=x,F(x,w,z,y)=y,F(z,y,x,w)=z,F(z,w,x,y)=w.For a metric space(X,d), the function ρ:X4→[0,∞), given by
(2.4)ρ((x,y,z,w),(u,v,r,t)):=d(x,u)+d(y,v)+d(z,r)+d(w,t),
forms a metric space on X4; that is, (X4,ρ) is a metric induced by (X,d). Let Φ denote the all functions ϕ:[0,∞)→[0,∞) which satisfies that limt→rϕ(t)>0 for all r>0 and limt→0+ϕ(t)=0.The aim of this paper is to prove the following theorem.Theorem 2.3.
Let(X,≤) be partially ordered set and (X,d) a complete metric space. Let F:X4→X be a mapping having the mixed monotone property on X. Assume that for all x≥u,y≤v,z≥w,
(2.5)d(F(x,y,z,w),F(u,v,r,t))≤14[d(x,u)+d(y,v)+d(z,r)+d(w,t)]-ϕ(14[d(x,u)+d(y,v)+d(z,r)+d(w,t)]),
where ϕ∈Φ. Suppose there exist x0,y0,z0,w0∈X such that
(2.6)x0≤F(x0,y0,z0,w0),y0≥F(x0,w0,z0,y0),z0≤F(z0,y0,x0,w0),w0≥F(z0,w0,x0,y0).
Suppose either  (a)
F is continuous, or(b)
X has the following property:
(i)
if nondecreasing sequencexn→x (resp., zn→z), then xn≤x (resp., zn≤z) for all n,(ii)
if nonincreasing sequenceyn→y (resp., wn→w), then yn≥y (resp., wn≥w) for all n.  Then there exist x,y,z,w∈X such that
(2.7)F(x,y,z,w)=x,F(x,w,z,y)=y,F(z,y,x,w)=z,F(z,w,x,y)=w.Proof.
We construct a sequence{(xn,yn,zn,wn)} in the following way. Set(2.8)x1=F(x0,y0,z0,w0)≥x0,y1=F(x0,w0,z0,y0)≤y0,z1=F(z0,y0,x0,w0)≥z0,w1=F(z0,w0,x0,y0)≤w0,
and by the mixed monotone property of F, for n≥1, inductively we get
(2.9)xn=F(xn-1,yn-1,zn-1,wn-1)≥xn-1≥⋯≥x0,yn=F(xn-1,wn-1,zn-1,yn-1)≤yn-1≤⋯≤y0,zn=F(zn-1,yn-1,xn-1,wn-1)≥zn-1≥⋯≥z0,wn=F(zn-1,wn-1,xn-1,yn-1)≤wn-1≤⋯≤w0.
Due to (2.5) and (2.9), we have
(2.10)d(x1,x2)=d(F(x0,y0,z0,w0),F(x1,y1,z1,w1))≤14[d(x0,x1)+d(y0,y1)+d(z0,z1)+d(w0,w1)]-ϕ(14[d(x0,x1)+d(y0,y1)+d(z0,z1)+d(w0,w1)])≤14[d(x0,x1)+d(y0,y1)+d(z0,z1)+d(w0,w1)],(2.11)d(y1,y2)=d(F(x0,w0,z0,y0),F(x1,w1,z1,y1))≤14[d(x0,x1)+d(w0,w1)+d(z0,z1)+d(y0,y1)]-ϕ(14[d(x0,x1)+d(w0,w1)+d(z0,z1)+d(y0,y1)])≤14[d(x0,x1)+d(w0,w1)+d(z0,z1)+d(y0,y1)],(2.12)d(z1,z2)=d(F(z0,y0,x0,w0),F(z1,y1,x1,w1))≤14[d(z0,z1)+d(y0,y1)+d(x0,x1)+d(w0,w1)]-ϕ(14[d(z0,z1)+d(y0,y1)+d(x0,x1)+d(w0,w1)])≤14[d(z0,z1)+d(y0,y1)+d(x0,x1)+d(w0,w1)],(2.13)d(w1,w2)=d(F(z0,w0,x0,y0),F(z1,w1,x1,y1))≤14[d(z0,z1)+d(w0,w1)+d(x0,x1)+d(y0,y1)]-ϕ(14[d(z0,z1)+d(w0,w1)+d(x0,x1)+d(y0,y1)])≤14[d(z0,z1)+d(w0,w1)+d(x0,x1)+d(y0,y1)].
Regarding (2.5) together with (2.14) we have
(2.14)d(x2,x3)=d(F(x1,y1,z1,w1),F(x2,y2,z2,w2))≤14[d(x1,x2)+d(y1,y2)+d(z1,z2)+d(w1,w2)]-ϕ(14[d(x1,x2)+d(y1,y2)+d(z1,z2)+d(w1,w2)])≤14[d(x1,x2)+d(y1,y2)+d(z1,z2)+d(w1,w2)],d(y2,y3)=d(F(x1,w1,z1,y1),F(x2,w2,z2,y2))≤14[d(x1,x2)+d(w1,w2)+d(z1,z2)+d(y1,y2)]-ϕ(14[d(x1,x2)+d(w1,w2)+d(z1,z2)+d(y1,y2)])≤14[d(x1,x2)+d(w1,w2)+d(z1,z2)+d(y1,y2)],d(z2,z3)=d(F(z1,y1,x1,w1),F(z2,y2,x2,w2))≤14[d(z1,z2)+d(y1,y2)+d(x1,x2)+d(w1,w2)]-ϕ(14[d(z1,z2)+d(y1,y2)+d(x1,x2)+d(w1,w2)])≤14[d(z1,z2)+d(y1,y2)+d(x1,x2)+d(w1,w2)],d(w2,w3)=d(F(z1,w1,x1,y2),F(z2,w2,x2,y2))≤14[d(z1,z2)+d(w1,w2)+d(x1,x2)+d(y1,y2)]-ϕ(14[d(z1,z2)+d(w1,w2)+d(x1,x2)+d(y1,y2)])≤14[d(z1,z2)+d(w1,w2)+d(x1,x2)+d(y1,y2)].
Inductively we have(2.15)d(xn+1,xn+2)=d(F(xn,yn,zn,wn),F(xn+1,yn+1,zn+1,wn+1))≤14[d(xn,xn+1)+d(yn,yn+1)+d(zn,zn+1)+d(wn,wn+1)]-ϕ(14[d(xn,xn+1)+d(yn,yn+1)+d(zn,zn+1)+d(wn,wn+1)])≤14[d(xn,xn+1)+d(yn,yn+1)+d(zn,zn+1)+d(wn,wn+1)],d(yn+1,yn+2)=d(F(xn,wn,zn,yn),F(xn+1,wn+1,zn+1,yn+1))≤14[d(xn,xn+1)+d(wn,wn+1)+d(zn,zn+1)+d(yn,yn+1)]-ϕ(14[d(xn,xn+1)+d(wn,wn+1)+d(zn,zn+1)+d(yn,yn+1)])≤14[d(xn,xn+1)+d(wn,wn+1)+d(zn,zn+1)+d(yn,yn+1)],d(zn+1,zn+2)=d(F(zn,yn,xn,wn),F(zn+1,yn+1,xn+1,wn+1))≤14[d(zn,zn+1)+d(yn,yn+1)+d(xn,xn+1)+d(wn,wn+1)]-ϕ(14[d(zn,zn+1)+d(yn,yn+1)+d(xn,xn+1)+d(wn,wn+1)])≤14[d(zn,zn+1)+d(yn,yn+1)+d(xn,xn+1)+d(wn,wn+1)],d(wn+1,wn+2)=d(F(zn,wn,xn,yn),F(zn+1,wn+1,xn+1,yn+1))≤14[d(zn,zn+1)+d(wn,wn+1)+d(xn,xn+1)+d(yn,yn+1)]-ϕ(14[d(zn,zn+1)+d(wn,wn+1)+d(xn,xn+1)+d(yn,yn+1)])≤14[d(zn,zn+1)+d(wn,wn+1)+d(xn,xn+1)+d(yn,yn+1)].
Set δn=d(xn,xn-1)+d(yn,yn-1)+d(zn,zn-1)+d(wn,wn-1). Due to (2.15), we conclude that {δn} is a nonincreasing sequence. Since it is bounded below, there is some δ≥0 such that
(2.16)lim⁡n→∞δn=δ.
We shall show that δ=0. Suppose, to the contrary, that δ>0.
Again by (2.15) and (2.9) together with (2.5), we have
(2.17)δn≤δn-4ϕ(14[δn]).
Lettingn→∞ in (2.17) and having in mind that we suppose limt→rϕ(t)>0 for all r>0 and limt→0+ϕ(t)=0, we have
(2.18)δ≤δ-4ϕ(14δ),
which is a contradiction. Thus, δ=0, that is,
(2.19)lim⁡n→∞δn=lim⁡n→∞[d(xn,xn-1)+d(yn,yn-1)+d(zn,zn-1)+d(wn,wn-1)]=0.
Now, we shall prove that{xn},{yn},{zn}, and {wn} are Cauchy sequences. Suppose, to the contrary, that at least one of {xn},{yn},{zn}, and {wn} is not Cauchy. So, there exists an ɛ>0 for which we can find subsequences {xn(k)}, {xn(k)} of {xn} and {yn(k)}, {yn(k)} of {yn} and {zn(k)}, {zn(k)} of {zn} and {wn(k)}, {wn(k)} of {wn} with n(k)>m(k)≥k such that
(2.20)d(xn(k),xm(k))+d(yn(k),ym(k))+d(zn(k),zm(k))+d(wn(k),wm(k))≥ɛ.
Additionally, corresponding to m(k), we may choose n(k) such that it is the smallest integer satisfying (2.20) and n(k)>m(k)≥k. Thus,
(2.21)d(xn(k)-1,xm(k))+d(yn(k)-1,ym(k))+d(zn(k)-1,zm(k))+d(wn(k)-1,wm(k))<ɛ.
By using triangle inequality and having (2.20), (2.21) in mind
(2.22)ɛ≤tk=d(xn(k),xm(k))+d(yn(k),ym(k))+d(zn(k),zm(k))+d(wn(k),wm(k))≤d(xn(k),xn(k)-1)+d(xn(k)-1,xm(k))+d(yn(k),yn(k)-1)+d(yn(k)-1,ym(k))+d(zn(k),zn(k)-1)+d(zn(k)-1,zm(k))+d(wn(k),wn(k)-1)+d(wn(k)-1,wm(k))<d(xn(k),xn(k)-1)+d(yn(k),yn(k)-1)+d(zn(k),zn(k)-1)+d(wn(k),wn(k)-1)+ɛ.
Letting k→∞ in (2.22) and using (2.16)
(2.23)lim⁡k→∞tk=lim⁡k→∞d(xn(k),xm(k))+d(yn(k),ym(k))+d(zn(k),zm(k))+d(wn(k),wm(k))=ɛ.
Again by triangle inequality,
(2.24)tk=d(xn(k),xm(k))+d(yn(k),ym(k))+d(zn(k),zm(k))+d(wn(k),wm(k))≤d(xn(k),xn(k)+1)+d(xn(k)+1,xm(k)+1)+d(xm(k)+1,xm(k))+d(yn(k),yn(k)+1)+d(yn(k)+1,ym(k)+1)+d(ym(k)+1,ym(k))+d(zn(k),zn(k)+1)+d(zn(k)+1,zm(k)+1)+d(zm(k)+1,zm(k))+d(wn(k),wn(k)+1)+d(wn(k)+1,wm(k)+1)+d(wm(k)+1,wm(k))≤δn(k)+1+δm(k)+1+d(xn(k)+1,xm(k)+1)+d(yn(k)+1,ym(k)+1)+d(zn(k)+1,zm(k)+1)+d(wn(k)+1,wm(k)+1).
Since n(k)>m(k), then
(2.25)xn(k)≥xm(k),yn(k)≤ym(k),zn(k)≥zm(k),wn(k)≤wm(k).
Hence from (2.25), (2.9) and (2.5), we have
(2.26)d(xn(k)+1,xm(k)+1)=d(F(xn(k),yn(k),zn(k),wn(k)),F(xm(k),ym(k),zm(k),wm(k)))≤14[d(xn(k),xm(k))+d(yn(k),ym(k))+d(zn(k),zm(k))+d(wn(k),wm(k))]-ϕ(14[d(xn(k),xm(k))+d(yn(k),ym(k))+d(zn(k),zm(k))+d(wn(k),wm(k))]14),d(yn(k)+1,ym(k)+1)=d(F(xn(k),wn(k),zn(k),yn(k)),F(xm(k),wm(k),zm(k),ym(k)))≤14[d(xn(k),xm(k))+d(wn(k),wm(k))+d(zn(k),zm(k))+d(yn(k),ym(k))]-ϕ(14[d(xn(k),xm(k))+d(wn(k),wm(k))+d(zn(k),zm(k))+d(yn(k),ym(k))]14),d(zn(k)+1,zm(k)+1)=d(F(zn(k),yn(k),xn(k),wn(k)),F(zm(k),ym(k),xm(k),wm(k)))≤14[d(zn(k),zm(k))+d(yn(k),ym(k))+d(xn(k),xm(k))+d(wn(k),wm(k))]-ϕ(14[d(zn(k),zm(k))+d(yn(k),ym(k))+d(xn(k),xm(k))+d(wn(k),wm(k))]14),d(wn(k)+1,wm(k)+1)=d(F(zn(k),wn(k),xn(k),yn(k)),F(zm(k),wm(k),xm(k),ym(k)))≤14[d(zn(k),zm(k))+d(wn(k),wm(k))+d(xn(k),xm(k))+d(yn(k),ym(k))]-ϕ(14[d(zn(k),zm(k))+d(wn(k),wm(k))+d(xn(k),xm(k))+d(yn(k),ym(k))]14).
Combining (2.24) with (2.26), we obtain that
(2.27)tk≤δn(k)+1+δm(k)+1+d(xn(k)+1,xm(k)+1)+d(yn(k)+1,ym(k)+1)+d(zn(k)+1,zm(k)+1)+d(wn(k)+1,wm(k)+1)≤δn(k)+1+δm(k)+1+tk-4ϕ(14tk).
Letting k→∞ and having in mind (2.19) we get a contradiction. This shows that {xn}, {yn}, {zn}, and {wn} are Cauchy sequences. Since X is complete metric space, there exists x,y,z,w∈X such that
(2.28)lim⁡n→∞xn=x,lim⁡n→∞yn=y,lim⁡n→∞zn=z,lim⁡n→∞wn=w.
Suppose now the assumption(a) holds. Then by (2.9) and (2.28), we have
(2.29)x=lim⁡n→∞xn=lim⁡n→∞F(xn-1,yn-1,zn-1,wn-1)=F(lim⁡n→∞xn-1,lim⁡n→∞yn-1,lim⁡n→∞zn-1,lim⁡n→∞wn-1)=F(x,y,z,w).
Analogously, we also observe that
(2.30)y=lim⁡n→∞yn=lim⁡n→∞F(xn-1,wn-1,zn-1,yn-1)=F(x,w,z,y),z=lim⁡n→∞zn=lim⁡n→∞F(zn-1,yn-1,xn-1,wn-1)=F(z,y,x,w),w=lim⁡n→∞wn=lim⁡n→∞F(zn-1,wn-1,xn-1,yn-1)=F(z,w,x,y).
Thus, we have
(2.31)F(x,y,z,w)=x,F(x,w,z,y)=y,F(z,y,x,w)=z,F(z,w,x,y)=w.
Suppose now the assumption(b) holds. Since {xn},{zn} are nondecreasing and xn→x, zn→z, and also {yn},{wn} are nonincreasing and yn→y,wn→w, then by assumption (b) we have
(2.32)xn≥x,yn≤y,zn≥z,wn≤w
for all n. Consider now
(2.33)d(x,F(x,y,z,w))≤d(x,xn+1)+d(xn+1,F(x,y,z,w))=d(x,xn+1)+d(F(xn,yn,zn,wn),F(x,y,z,w))≤14[d(xn,x)+d(yn,y)+d(zn,z)+d(wn,w)]-ϕ(14[d(xn,x)+d(yn,y)+d(zn,z)+d(wn,w)]).
Taking n→∞ in (2.33) and using (2.28), we get that d(x,F(x,y,z,w))=0. Thus, x=F(x,y,z,w). Analogously, we get that
(2.34)F(x,w,z,y)=y,F(z,y,x,w)=z,F(z,w,x,y)=w.
Thus, we proved that F has a quadruple fixed point.Corollary 2.4.
Let(X,≤) be partially ordered set and (X,d) a complete metric space. Let F:X4→X be a mapping having the mixed monotone property on X. Assume that there exists k∈[0,1) such that
(2.35)d(F(x,y,z,w),F(u,v,r,t))≤k4[d(x,u)+d(y,v)+d(z,r)+d(w,t)]
for all x≥u, y≤v, z≥w. Suppose there exist x0,y0,z0,w0∈X such that
(2.36)x0≤F(x0,y0,z0,w0),y0≥F(x0,w0,z0,y0),z0≤F(z0,y0,x0,w0),w0≥F(z0,w0,x0,y0).
Suppose either  (a)
F is continuous, or(b)
X has the following property:
(i)
if nondecreasing sequencexn→x (resp., zn→z), then xn≤x (resp., zn≤z) for all n,(ii)
if nonincreasing sequenceyn→y (resp., wn→w), then yn≥y (resp., wn≥w) for all n,  then there exist x,y,z,w∈X such that
(2.37)F(x,y,z,w)=x,F(x,w,z,y)=y,F(z,y,x,w)=z,F(z,w,x,y)=w.Proof.
It is sufficient to takeϕ(t)=((1-k)/2)t in previous theorem.


## 3. Uniqueness of Quadruple Fixed Point

In this section we shall prove the uniqueness of quadruple fixed point. For a productX4 of a partial ordered set (X,≤) we define a partial ordering in the following way. For all (x,y,z,t),(u,v,r,t)∈X×X×X×X(3.1)(x,y,z,w)≤(u,v,r,t)⟺x≤u,y≥v,z≤r,w≥r.
We say that (x,y,z,w) is equal (u,v,r,t) if and only if x=u,y=v,z=r, and w=t.Theorem 3.1.
In addition to hypothesis of Theorem2.3, suppose that for all (x,y,z,t),(u,v,r,t)∈X×X×X×X; there exists (a,b,c,d)∈X×X×X×X that is comparable to (x,y,z,t) and (u,v,r,t); then F has a unique quadruple fixed point.Proof.
The set of quadruple fixed point ofF is not empty due to Theorem 2.3. Assume, now, that (x,y,z,t) and (u,v,r,t) are the quadruple fixed point of F, that is,
(3.2)F(x,y,z,w)=x,F(u,v,r,t)=u,F(x,w,z,y)=y,F(u,t,r,v)=v,F(z,y,x,w)=z,F(r,v,u,t)=r,F(z,w,x,y)=w,F(r,t,u,v)=t.
We shall show that (x,y,z,w) and (u,v,r,t) are equal. By assumption, there exists (a,b,c,d)∈X×X×X×X that is comparable to (x,y,z,t) and (u,v,r,t). Define sequences {an},{bn},{cn}, and {dn} such that
(3.3)a=a0,b=b0,c=c0,d=d0,an=F(an-1,bn-1,zn-1,dn-1),bn=F(an-1,dn-1,cn-1,bn-1),cn=F(cn-1,bn-1,an-1,dn-1),dn=F(cn-1,dn-1,an-1,bn-1)
for all n. Since (x,y,z,w) is comparable with (a,b,c,d), we may assume that (x,y,z,w)≥(a,b,c,d)=(a0,b0,c0,d0). Recursively, we get that
(3.4)(x,y,z,w)≥(an,bn,cn,dn)∀n.
By (3.4) and (2.5), we have
(3.5)d(x,an+1)=d(F(x,y,z,w),F(an,bn,cn,dn))≤14[d(x,an)+d(y,bn)+d(z,cn)+d(w,dn)]-ϕ(14[d(x,an)+d(y,bn)+d(z,cn)+d(w,dn)]),d(bn+1,y)=d(F(an,dn,cn,bn),F(x,w,z,y))≤14[d(an,x)+d(dn,w)+d(cn,z)+d(bn,y)]-ϕ(14[d(an,x)+d(dn,w)+d(cn,z)+d(bn,y)]),d(z,cn+1)=d(F(z,y,x,w),F(cn,bn,an,dn))≤14[d(z,cn)+d(y,bn)+d(x,an)+d(w,dn)]-ϕ(14[d(z,cn)+d(y,bn)+d(x,an)+d(w,dn)]),d(dn+1,w)=d(F(cn,dn,an,bn),F(z,w,x,y))≤14[d(cn,z)+d(dn,w)+d(an,x)+d(bn,y)]-ϕ(14[d(cn,z)+d(dn,w)+d(an,x)+d(bn,y)]).
Set γn=d(x,an)+d(y,bn)+d(z,cn)+d(w,dn). Then, due to (3.5), we have
(3.6)γn+1≤γn,∀n.Hence, the sequence{γn} is decreasing and bounded below. Thus, there exists γ≥0 such that(3.7)lim⁡n→∞γn=γ.Now, we shall show thatγ=0. Suppose, to the contrary, that γ>0. Again by (3.5), we have (3.8)γn+1≤γn-4(14ϕ(γn)),∀n.
Letting n→∞ in (3.8), we obtain that
(3.9)γ≤γ-4(14ϕ(γ)),which is a contradiction. Therefore,γ=0. That is,
(3.10)lim⁡n→∞γn=0.Consequently, we have(3.11)lim⁡n→∞d(x,an)=0,lim⁡n→∞d(y,bn)=0,lim⁡n→∞d(z,cn)=0,lim⁡n→∞d(w,dn)=0.
Analogously, we show that(3.12)lim⁡n→∞d(u,an)=0,lim⁡n→∞d(v,bn)=0,lim⁡n→∞d(r,cn)=0,lim⁡n→∞d(s,dn)=0.
Combining (3.11) and (3.12) yields that (x,y,z,w) and (u,v,r,t) are equal.Example 3.2.
LetX=ℝ with the metric d(x,y)=|x-y|, for all x,y∈X and the usual ordering.
LetF:X4→X be given by
(3.13)F(x,y,z,w)=x-y+z-w16,∀x,y,z,w∈X,
and let ϕ:[0,∞)→[0,∞) be given by ϕ(t)=t/4 for all t∈[0,∞).
It is easy to check that all the conditions of Theorem2.3 are satisfied and (0,0,0,0) is the unique quadruple fixed point of F.



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*Source: 989423-2011-12-05.xml*