# Quadruple Fixed Point Theorems in Partially Ordered Metric Spaces Depending on Another Function

Authors: Hassen Aydi; Erdal Karapınar; İlker Şavas Yüce
Journal: ISRN Applied Mathematics
 (2012)

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## Abstract

We prove quadruple fixed point theorems in partially ordered metric spaces depending on another function. Also, we state some examples showing that our results are real generalization of known ones in quadruple fixed point theory.


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## Body

## 1. Introduction and Preliminaries

Basic topological properties of an ordered set like convergence were introduced by Wolk [1]. In 1981, Monjardet [2] considered metrics on partially ordered sets. Ran and Reurings [3] proved an analog of Banach contraction mapping principle in partially ordered metric spaces. In their pioneering work, they also provide applications to matrix equations. As an extension, Nieto and Rodríguez-López [4] discovered further fixed point theorems in partially ordered metric spaces. For some other related results in ordered metric spaces, see, for example, [5–7].Bhaskar and Lakshmikantham in [8] introduced the concept of coupled fixed point of a mapping F:X×X→X and investigated the existence and uniqueness of a coupled fixed point theorem in partially ordered complete metric spaces. Lakshmikantham and Ćirić in [9] defined mixed g-monotone property and coupled coincidence point in partially ordered metric spaces. They also proved related fixed point theorems. Later, various results on coupled fixed point have been obtained, see, for example, [9–20].Following this trend, Berinde and Borcut [21] introduced the concept of tripled fixed point in ordered sets. The following two definitions are from [21].Definition 1.1.
Let(X,≤) be a partially ordered set and F:X×X×X→X. The mapping F is said to have the mixed monotone property if, for any x,y,z∈X,
(1.1)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 1.2.
LetF:X×X×X→X. An element (x,y,z) is called a tripled fixed point of F if
(1.2)F(x,y,z)=x,F(y,x,y)=y,F(z,y,x)=z.Also, Berinde and Borcut [21] proved the following theorem.Theorem 1.3.
Let(X,≤,d) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Suppose also that F:X×X×X→X be a mapping such that it has the mixed monotone property and there exist j,r,l≥0 with j+r+l<1 such that
(1.3)d(F(x,y,z),F(u,v,w))≤jd(x,u)+rd(y,v)+ld(z,w),
for anyx,y,z∈X for which x≤u, v≤y, and z≤w. Additionally suppose that either F is continuous or X has the following properties:(1)
if a nondecreasing sequencexn→x, then   xn≤x  for all  n,(2)
if a nonincreasing sequenceyn→y, then  y≤yn  for all  n.
If there existx0,y0,z0∈X such that  x0≤F(x0,y0,z0),  y0≥F(y0,x0,z0), and  z0≤F(z0,y0,x0), then there exist  x,y,z∈X such that
(1.4)F(x,y,z)=x,F(y,x,y)=y,F(z,y,x)=z,
that is,F has a tripled fixed point.The notion of fixed point of orderN≥3 was first introduced by Samet and Vetro [22]. Very recently, Karapınar used the notion of quadruple fixed point and obtained some quadruple fixed point theorems [23] in partially ordered metric spaces. This work motivated the following studies [24–27] which provide further fixed point theorems on quadruple fixed points.From now on, we denoteX4=X×X×X×X.Definition 1.4 (see [24]).
LetX be a nonempty set and let F:X4→X be a given mapping. An element (x,y,z,w)∈X×X3 is called a quadruple fixed point of F if
(1.5)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
(1.6)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 1.5.
In [23, 24, 27], the notion of quadruple fixed point is called quartet fixed point.Definition 1.6 (see [24]).
Let(X,≤) be a partially ordered set and F:X4→X be a mapping. We say 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,
(1.7)x1,x2∈X,x1≤x2impliesF(x1,y,z,w)≤F(x2,y,z,w),y1,y2∈X,y1≤y2impliesF(x,y2,z,w)≤F(x,y1,z,w),z1,z2∈X,z1≤z2impliesF(x,y,z1,w)≤F(x,y,z2,w),w1,w2∈X,w1≤w2impliesF(x,y,z,w2)≤F(x,y,z,w1).In this paper, we prove some quadruple fixed point theorems in partially ordered metric spaces depended on another functionT:X→X.


## 2. Main Results

We start with the following definition (see, e.g., [28–31]).Definition 2.1.
Let(X,d) be a metric space. A mapping T:X→X is said to be ICS if T is injective, continuous, and it has the property: for every sequence {xn} in X, if {Txn} is convergent then, {xn} is also convergent.LetΦ be the set of all functions ϕ:[0,∞)→[0,∞) such that(1)
ϕ(t)<t for all t∈(0,+∞),(2)
limr→t+ϕ(r)<t for all t∈(0,+∞).Our first result is given by the following theorem.Theorem 2.2.
Let(X,≤) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Suppose also that T:X→X is an ICS mapping and F:X4→X is such that F has the mixed monotone property. Assume that there exists ϕ∈Φ such that
(2.1)d(TF(x,y,z,w),TF(u,v,r,s))≤ϕ(max⁡{d(Tx,Tu),d(Ty,Tv),d(Tz,Tr),d(Tw,Ts)})
for anyx,y,z,w,u,v,r,s∈X for which x≤u, v≤y, z≤r, and s≤w. Additionally assume that either(a)
F is continuous, or(b)
X has the following properties:
(i)
if nondecreasing sequencexn→x (respectively, zn→z), then xn≤x (respectively, zn≤z) for all n,(ii)
if nonincreasing sequenceyn→y (respectively, wn→w), then yn≥y (respectively, wn≥w) for all 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 there exist x,y,z,w∈X such that
(2.2)F(x,y,z,w)=x,F(y,z,w,x)=y,F(z,w,x,y)=z,F(w,x,y,z)=w.
that is,F has a quadruple fixed point.Proof.
Letx0,y0,z0,w0∈X such that
(2.3)x0≤F(x0,y0,z0,w0),y0≥F(y0,z0,w0,x0),(2.4)z0≤F(z0,w0,x0,y0),w0≥F(w0,x0,y0,z0).
Set(2.5)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. By continuing this process, we can construct four sequences {xn}, {yn}, {zn}, and {wn} in X such that
(2.6)xn+1=F(xn,yn,zn,wn),yn+1=F(yn,zn,wn,xn),zn+1=F(z1,w1,x1,y1),wn+1=F(wn,xn,yn,zn).
SinceF has the mixed monotone property, by using a mathematical induction it is easy to see that
(2.7)xn≤xn+1,yn+1≤yn,zn≤zn+1,wn+1≤wn,forn=0,1,2,…,
Assume that, for somen∈ℕ,
(2.8)xn=xn+1,yn=yn+1,zn=zn+1,wn=wn+1.
Then, by (2.6), (xn,yn,zn,wn) is a quadruple fixed point of F. Therefore, in the rest of the proof, for any n∈ℕ we will assume that
(2.9)xn≠xn+1oryn≠yn+1orzn≠zn+1orwn≠wn+1.
SinceT is injective, for any n∈ℕ,
(2.10)0<max⁡{d(Txn,Txn+1),d(Tyn,Tyn+1),d(Tzn,Tzn+1),d(Twn,Twn+1)}.
Due to (2.1), (2.6), and (2.7), we have
(2.11)d(Txn,Txn+1)=d(TF(xn-1,yn-1,zn-1,wn-1),TF(xn,yn,zn,wn))≤ϕ(max⁡{d(Txn-1,Txn),d(Tyn-1,Tyn),d(Tzn-1,Tzn),d(Twn-1,Twn)}),d(Tyn+1,Tyn)=d(TF(yn,zn,wn,xn),TF(yn-1,zn-1,wn-1,xn-1))≤ϕ(max⁡{d(Tyn-1,Tyn),d(Tzn-1,Tzn),d(Twn-1,Twn),d(Txn-1,Txn)}),d(Tzn,Tzn+1)=d(TF(zn-1,wn-1,xn-1,yn-1),TF(zn,wn,xn,yn))≤ϕ(max⁡{d(Tzn-1,Tzn),d(Twn-1,Twn),d(Txn-1,Tzn),d(Tyn-1,Tyn)}),d(Twn+1,Twn)=d(TF(wn,xn,yn,zn),TF(wn-1,xn-1,yn-1,zn-1))≤ϕ(max⁡{d(Twn-1,Twn),d(Txn-1,Txn),d(Tyn-1,Tyn),d(Tzn-1,Tzn)}).
Using the fact thatϕ(t)<t for all t>0 together with (2.11), we obtain that
(2.12)0<max⁡{d(Txn,Txn+1),d(Tyn,Tyn+1),d(Tzn,Tzn+1),d(Twn,Twn+1)}≤ϕ(max⁡{d(Txn-1,Txn),d(Tyn-1,Tyn),d(Tzn-1,Tzn),d(Twn-1,Twn)})<max⁡{d(Txn-1,Txn),d(Tyn-1,Tyn),d(Tzn-1,Tzn),d(Twn-1,Twn)}.
It follows that(2.13)max⁡{d(Txn,Txn+1),d(Tyn,Tyn+1),d(Tzn,Tzn+1),d(Twn,Twn+1)}<max⁡{d(Txn-1,Txn),d(Tyn-1,Tyn),d(Tzn-1,Tzn),d(Twn-1,Twn)}.
Thus,max{d(Txn,Txn+1),d(Tyn,Tyn+1),d(Tzn,Tzn+1),d(Twn,Twn+1)} is a positive decreasing sequence. Hence, there exists r≥0 such that
(2.14)lim⁡n→+∞max⁡{d(Txn,Txn+1),d(Tyn,Tyn+1),d(Tzn,Tzn+1),d(Twn,Twn+1)}=r.
Suppose thatr>0. Letting n→+∞ in (2.12), we obtain that
(2.15)0<r≤lim⁡n→+∞ϕ(max⁡{d(Txn,Txn+1),d(Tyn,Tyn+1),d(Tzn,Tzn+1),d(Twn,Twn+1)})=lim⁡t→r+ϕ(t)<r,
which is a contradiction. Therefore, we deduce that(2.16)lim⁡n→+∞max⁡{d(Txn,Txn+1),d(Tyn,Tyn+1),d(Tzn,Tzn+1),d(Twn,Twn+1)}=0.
We will show that{Txn}, {Tyn}, {Tzn}, and {Twn} are Cauchy sequences. Assume the contrary, that is, either {Txn} or {Tyn} or {Tzn} or {Twn} is not a Cauchy sequence, consequently,
(2.17)lim⁡n,m→+∞d(Txm,Txn)≠0orlim⁡n,m→+∞d(Tym,Tyn)≠0,
orlimn,m→+∞d(Tzm,Tzn)≠0 or limn,m→+∞d(Twm,Twn)≠0. This means that there exists ε>0 for which we can find subsequences of integers (mk) and (nk) with nk>mk>k such that
(2.18)max⁡{d(Txmk,Txnk),d(Tymk,Tynk),d(Tzmk,Tznk),d(Twmk,Twnk)}≥ε.
Furthermore, corresponding tomk, we can choose nk in such a way that it is the smallest integer with nk>mk and satisfying (2.18). Then,
(2.19)max⁡{d(Txmk,Txnk-1),d(Tymk,Tynk-1),d(Tzmk,Tznk-1),d(Twmk,Twnk-1)}<ε.
By the triangle  inequality and (2.19), we have
(2.20)d(Txmk,Txnk)≤d(Txmk,Txnk-1)+d(Txnk-1,Txnk)<ϵ+d(Txnk-1,Txnk).
Thus, by (2.16), we obtain
(2.21)lim⁡k→+∞d(Txmk,Txnk)≤lim⁡k→+∞d(Txmk,Txnk-1)≤ε.
Similarly, we have(2.22)lim⁡k→+∞d(Tymk,Tynk)≤lim⁡k→+∞d(Tymk,Tynk-1)≤ε,lim⁡k→+∞d(Tzmk,Tznk)≤lim⁡k→+∞d(Tzmk,Tznk-1)≤ε,lim⁡k→+∞d(Twmk,Twnk)≤lim⁡k→+∞d(Twmk,Twnk-1)≤ε.
Again, by (2.19), we have
(2.23)d(Txmk,Txnk)≤d(Txmk,Txmk-1)+d(Txmk-1,Txnk-1)+d(Txnk-1,Txnk)≤d(Txmk,Txmk-1)+d(Txmk-1,Txmk)+d(Txmk,Txnk-1)+d(Txnk-1,Txnk)<d(Txmk,Txmk-1)+d(Txmk-1,Txmk)+ε+d(Txnk-1,Txnk).
Lettingk→+∞ and using (2.16), we get
(2.24)lim⁡k→+∞d(Txmk,Txnk)≤lim⁡k→+∞d(Txmk-1,Txnk-1)≤ε,lim⁡k→+∞d(Tymk,Tynk)≤lim⁡k→+∞d(Tymk-1,Tynk-1)≤ε,lim⁡k→+∞d(Tzmk,Tznk)≤lim⁡k→+∞d(Tzmk-1,Tznk-1)≤ε,lim⁡k→+∞d(Twmk,Twnk)≤lim⁡k→+∞d(Twmk-1,Twnk-1)≤ε.
Using (2.18) and (2.24), we have
(2.25)lim⁡k→+∞max⁡{d(Txmk,Txnk),d(Tymk,Tynk),d(Tzmk,Tznk),d(Twmk,Twnk)}=lim⁡k→+∞max⁡{d(Txmk-1,Txnk-1),d(Tymk-1,Tynk-1),d(Tzmk-1,Tznk-1),d(Twmk-1,Twnk-1)}=ε.
Now, using inequality (2.1), we obtain
(2.26)d(Txmk,Txnk)=d(TF(xmk-1,ymk-1,zmk-1,wmk-1),TF(xnk-1,ynk-1,znk-1,wnk-1))≤ϕ(max⁡{d(Txmk-1,Txnk-1),d(Tymk-1,Tynk-1),d(Tzmk-1,Tznk-1),d(Twmk-1,Twnk-1)})d(Tymk,Tynk)=d(TF(ymk-1,zmk-1,wmk-1,xmk-1),TF(ynk-1,znk-1,wnk-1,xnk-1))≤ϕ(max⁡{d(Tymk-1,Tynk-1),d(Tzmk-1,Tznk-1),d(Twmk-1,Twnk-1),d(Txmk-1,Txnk-1)}),d(Tymk,Tynk)=d(TF(zmk-1,wmk-1,xmk-1,ymk-1),TF(znk-1,wnk-1,xnk-1,ynk-1))≤ϕ(max⁡{d(Tzmk-1,Tznk-1),d(Twmk-1,Twnk-1),d(Txmk-1,Txnk-1),d(Tymk-1,Tynk-1)}),d(Tymk,Tynk)=d(TF(wmk-1,xmk-1,ymk-1,zmk-1),TF(wnk-1,xnk-1,ynk-1,znk-1))≤ϕ(max⁡{d(Twmk-1,Twnk-1),d(Txmk-1,Txnk-1),d(Tymk-1,Tynk-1),d(Tzmk-1,Tznk-1)})
From (2.26), we deduce that
(2.27)max⁡{d(Txmk,Txnk),d(Tymk,Tynk),d(Tzmk,Tznk),d(Twmk,Twnk)}≤ϕ(max⁡{d(Txmk-1,Txnk-1),d(Tymk-1,Tynk-1),d(Tzmk-1,Tznk-1),d(Twmk-1,Twnk-1)}).
Lettingk→+∞ in (2.27) and by using (2.25), we get that
(2.28)0<ε≤lim⁡t→ε+ϕ(t)<ε,
which is a contradiction. Thus,{Txn}, {Tyn}, {Tzn}, and {Twn} are Cauchy sequences in (X,d). Since X is a complete metric space, {Txn}, {Tyn}, {Tzn}, and {Twn} are convergent sequences.
SinceT is an ICS mapping, there exist x,y,z,w∈X such that
(2.29)lim⁡n→+∞xn=x,lim⁡n→+∞yn=y,lim⁡n→+∞zn=z,lim⁡n→+∞wn=w.
SinceT is continuous, we have
(2.30)lim⁡n→+∞Txn=Tx,lim⁡n→+∞Tyn=Ty,lim⁡n→+∞Tzn=Tz,lim⁡n→+∞Twn=Tw.
Suppose now the assumption(a) holds, that is, F is continuous. By (2.6), (2.29), and (2.30) we obtain
(2.31)x=lim⁡n→+∞xn+1=lim⁡n→+∞F(xn,yn,zn,wn)=F(lim⁡n→+∞xn,lim⁡n→+∞yn,lim⁡n→+∞zn,lim⁡n→+∞wn)=F(x,y,z,w),y=lim⁡n→+∞yn+1=lim⁡n→+∞F(yn,zn,wn,xn)=F(lim⁡n→+∞yn,lim⁡n→+∞zn,lim⁡n→+∞wn,lim⁡n→+∞xn)=F(y,z,w,x),z=lim⁡n→+∞zn+1=lim⁡n→+∞F(zn,wn,xn,yn)=F(lim⁡n→+∞zn,lim⁡n→+∞wn,lim⁡n→+∞xn,lim⁡n→+∞yn)=F(z,w,x,y),w=lim⁡n→+∞wn+1=lim⁡n→+∞F(wn,xn,yn,zn)=F(lim⁡n→+∞wn,lim⁡n→+∞xn,lim⁡n→+∞yn,lim⁡n→+∞zn)=F(w,x,y,z).
We have proved thatF has a quadruple fixed point.
Suppose now the assumption(b) holds. Since {xn} and {zn} are nondecreasing with xn→x and zn→z and also {yn} and {wn} are nonincreasing, with yn→y and wn→w, we have
(2.32)xn≤x,yn≥y,zn≤z,wn≥w
for alln. Consider now
(2.33)d(Tx,TF(x,y,z,w))≤d(Tx,Txn+1)+d(Txn+1,TF(x,y,z,w))=d(Tx,Txn+1)+d(TF(xn,yn,zn,wn),TF(x,y,z,w))≤d(Tx,Txn+1)+ϕ(max⁡{d(Txn,Tx),d(Tyn,Ty),d(Tzn,Tz),d(Twn,Tw)}).
Takingn→∞ and using (2.30), the right-hand side of (2.33) tends to 0, so we get that d(Tx,TF(x,y,z,w))=0. Thus, Tx=TF(x,y,z,w), and since T is injective, we get that x=F(x,y,z,w). Analogously, one finds that
(2.34)F(y,z,w,x)=y,F(z,w,x,y)=z,F(w,x,y,z)=w.
Thus, we proved thatF has a quadrupled fixed point. This completes the proof of Theorem 2.2.Repeating the same proof of Theorem2.2, we may state the following corollary.Corollary 2.3.
Let(X,≤) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Suppose also T:X→X is an ICS mapping and F:X4→X is such that F has the mixed monotone property. Assume that there exists ϕ∈Φ such that
(2.35)d(TF(x,y,z,w),TF(u,v,t,s))≤ϕ(d(Tx,Tu)+d(Ty,Tv)+d(Tz,Tt)+d(Tw,Ts)4)
for anyx,y,z,w,u,v,t,s∈X for which x≤u, v≤y, z≤t, and s≤w. Additionally suppose that either(a)
F is continuous, or(b)
X has the following property:
(i)
if non-decreasing sequencexn→x (resp., zn→z), then xn≤x (resp., zn≤z) for all n,(ii)
if non-increasing sequenceyn→y (resp., wn→w), then yn≥y (resp., wn≥w) for all 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,w0,y0,z0), then there exist x,y,z,w∈X such that
(2.36)F(x,y,z,w)=x,F(y,z,w,x)=y,F(z,w,x,y)=z,F(w,x,y,z)=w,
that is,F has a quadruple fixed point.Corollary 2.4.
Let(X,≤) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Suppose also that T:X→X is an ICS mapping and F:X4→X is such that F has the mixed monotone property. Assume that there exists k∈[0,1) such that
(2.37)d(TF(x,y,z,w),TF(u,v,t,s))≤kmax⁡{d(Tx,Tu),d(Ty,Tv),d(Tz,Tt),d(Tw,Ts)}
for anyx,y,z,w,u,v,t,s∈X for which  x≤u,   v≤y,   z≤t, and  s≤w. Suppose that 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.
If there existx0,y0,z0,w0∈X such that  x0≤F(x0,y0,w0,z0),  y0≥F(y0,z0,w0,x0),  z0≤F(z0,w0,x0,y0), and  w0≥F(w0,w0,y0,z0) then, there exist  x,y,z,w∈X such that
(2.38)F(x,y,z,w)=x,F(y,z,w,x)=y,F(z,w,x,y)=z,F(w,x,y,z)=w,
that is,F has a quadruple fixed point.Proof.
It suffices to remark thatϕ(t)=kt in Theorem 2.2.Corollary 2.5.
Let(X,≤) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Suppose also that T:X→X is an ICS mapping and F:X4→X is such that F has the mixed monotone property. Assume that there exists k∈[0,1) such that
(2.39)d(TF(x,y,z,w),TF(u,v,t,s))≤k4(d(Tx,Tu),d(Ty,Tv),d(Tz,Tt),d(Tw,Ts))
for anyx,y,z,w,u,v,t,s∈X for which x≤u, v≤y, z≤t and s≤w. Suppose that 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.
If there existx0,y0,z0,w0∈X such that x0≤F(x0,y0,w0,z0), y0≥F(y0,z0,w0,x0), z0≤F(z0,w0,x0,y0) and w0≥F(w0,w0,y0,z0) then there exist x,y,z,w∈X such that
(2.40)F(x,y,z,w)=x,F(y,z,w,x)=y,F(z,w,x,y)=z,F(w,x,y,z)=w
that is,F has a quadruple fixed point.Proof.
It suffices to takeϕ(t)=kt in Corollary 2.3.Now, we shall prove the existence and uniqueness of a 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,w),(u,v,t,s)∈X4,
(2.41)(x,y,z,w)≤(u,v,t,s)⟺x≤u,y≥v,z≤t,w≥s.We say that(x,y,z,w) and (u,v,t,s) are comparable if
(2.42)(x,y,z,w)≤(u,v,t,s)or(u,v,s,t)≤(x,y,z,w).Also, we say that(x,y,z,w) is equal to (u,v,t,s) if and only if x=u,y=v,z=t,w=s.Theorem 2.6.
In addition to hypotheses of Theorem2.2, suppose that that for all (x,y,z,w),(u,v,t,s)∈X4, there exists (a,b,c,d)∈X4 such that (F(a,b,c,d), F(b,c,d,a), F(c,d,a,b), F(d,a,b,c)) is comparable to (F(x,y,z,w), F(y,z,w,x), F(z,w,x,y), F(w,x,y,z)) and (F(u,v,t,s), F(v,t,s,u), F(t,s,u,v), F(s,u,v,t)). Then, F has a unique quadruple fixed point (x,y,z,w).Proof.
The set of quadruple fixed points ofF is not empty due to Theorem 2.2. Assume, now, (x,y,z,w) and (u,v,t,s) are two quadrupled fixed points of F, that is,
(2.43)F(x,y,z,w)=x,F(u,v,t,s)=u,F(y,z,w,x)=y,F(v,t,s,u)=v,F(z,w,x,y)=z,F(t,s,u,v)=t,F(w,x,y,z)=w,F(s,u,v,t)=s.
We shall show that(x,y,z,w) and (u,v,t,s) are equal. By assumption, there exists (a,b,c,d)∈X4 such that (F(a,b,c,d), F(b,c,d,a), F(c,d,a,b), F(d,a,b,c)) is comparable to (F(x,y,z,w), F(y,z,w,x), F(z,w,x,y), F(w,x,y,z)) and (F(u,v,t,s), F(v,t,s,u), F(t,s,u,v), F(s,u,v,t)).
Define sequences{an},{bn}, {cn}, and {dn} such that
(2.44)a0=a,b0=b,c0=c,d0=d,foranyn≥1,an=F(an-1,bn-1,cn-1,dn-1),bn=F(bn-1,cn-1,dn-1,an-1),cn=F(cn-1,dn-1,an-1,bn-1),dn=F(dn-1,an-1,bn-1,cn-1),
for alln. Further, set x0=x, y0=y, z0=z,  w0=w and u0=u, v0=v, t0=t, s0=s and on the same way define the sequences {xn},{yn}, {zn}, and {wn} and {un},{vn}, {tn}, and {sn}. Then, it is easy that
(2.45)xn=F(x,y,z,w),yn=F(y,z,w,x,),zn=F(z,w,x,y),wn=F(w,x,y,z),un=F(u,v,t,s),vn=F(v,t,s,u),tn=F(t,s,u,v),sn=F(s,u,v,t),
for alln≥1. Since (F(x,y,z,w),F(y,z,w,x),F(z,w,x,y),F(w,x,y,z))=(x1,y1,z1,w1)=(x,y,z,w) is comparable to (F(a,b,c,d),F(b,c,d,a),F(c,d,a,b))=(a1,b1,c1,d1), then it is easy to show (x,y,z,w)≥(a1,b1,c1,d1). Recursively, we get that
(2.46)(x,y,z,w)≥(an,bn,cn,dn)∀n.
By (2.46) and (2.1), we have
(2.47)d(Tx,Tan+1)=d(TF(x,y,z,w),TF(an,bn,cn,dn))≤ϕ(max⁡{d(Tx,Tan),d(Ty,Tbn),d(Tz,Tcn),d(Tw,Tdn)}),d(Tbn+1,Ty)=d(TF(bn,cn,dn,an),TF(y,z,w,x))≤ϕ(max⁡{d(Ty,Tbn),d(Tz,Tcn),d(Tw,Tdn),d(Tx,Tan)}),d(Tz,Tcn+1)=d(TF(z,w,x,y),TF(cn,dn,an,bn))≤ϕ(max⁡⁡{d(Tz,Tcn),d(Tw,Tdn),d(Tx,Tan),d(Ty,Tbn)}),d(Tw,Tdn+1)=d(TF(w,x,y,z),TF(dn,an,bn,cn))≤ϕ(max⁡{d(Tw,Tdn),d(Tx,Tan),d(Ty,Tbn),d(Tz,Tcn)}).
It follows from (2.47) that
(2.48)max⁡{d(Tz,Tcn+1),d(Tw,Tdn+1),d(Tx,Tan+1),d(Ty,Tbn+1)}≤ϕ(max⁡{d(Tz,Tcn),d(Tw,Tdn),d(Tx,Tan),d(Ty,Tbn)}).
Therefore, for eachn≥1,
(2.49)max⁡{d(Tz,Tcn),d(Tw,Tdn),d(Tx,Tan),d(Ty,Tbn)}≤ϕn(max⁡{d(Tz,Tc0),d(Tw,Td0),d(Tx,Ta0),d(Ty,Tb0)}).
It is known thatϕ(t)<t and limr→t+ϕ(r)<t imply limn→∞ϕn(t)=0 for each t>0. Thus, from (2.49),
(2.50)lim⁡n→∞max⁡{d(Tz,Tcn),d(Tw,Tdn),d(Tx,Tan),d(Ty,Tbn)}=0.
This yields that(2.51)lim⁡n→∞d(Tx,Tan)=0,lim⁡n→∞d(Ty,Tbn)=0,lim⁡n→∞d(Tz,Tcn)=0,lim⁡n→∞d(Tw,Tdn)=0.
Analogously, we show that(2.52)lim⁡n→∞d(Tu,Tan)=0,lim⁡n→∞d(Tv,Tbn)=0,lim⁡n→∞d(Tt,Tcn)=0,lim⁡n→∞d(Ts,Tdn)=0.
Combining (2.51) to (2.52) yields that (Tx,Ty,Tz,Tw) and (Tu,Tv,Tt,Ts) are equal. The fact that T is injective gives us x=u, y=v, z=t, and w=s.We state some examples showing that our results are effective.Example 2.7.
LetX=[1,64] with the metric d(x,y)=|x-y|, for all x,y∈X and the usual ordering. Clearly, (X,d) is a complete metric space.
LetT:X→X and F:X4→X be defined by
(2.53)Tx=ln⁡(x)+1,F(x,y,z,w)=8(xy)1/3,∀x,y,z,w∈X.
It is clear thatT is an ICS mapping, F has the mixed monotone property and continuous.
Setϕ(t)=2t/3. Taking x,y,z,w,u,v,s,t∈X for which x≤u, v≤y, z≤s, and t≤w, we have
(2.54)d(TF(x,y,z,w),TF(u,v,s,t))=13|(ln⁡x-ln⁡y)-(ln⁡u-ln⁡v)|≤13|ln⁡x-ln⁡u|+13|ln⁡y-ln⁡v|≤13max⁡{|ln⁡x-ln⁡u|,|ln⁡y-ln⁡v|}=ϕ(max⁡⁡{d(Tx,Tu),d(Ty,Tv),d(Tz,Ts),d(Tw,Tt)}),which is the contractive condition (2.1). Moreover, taking x0=y0=z0=8=w0, we have
(2.55)x0≤F(x0,y0,z0,w0),y0≥F(y0,z0,w0,x0),z0≤F(z0,w0,x0,y0),w0≥F(w0,x0,y0,z0).
Therefore, all the conditions of Theorem2.2 hold and (8,8,8,8) is the unique quadruple fixed point of F, since also the hypotheses of Theorem 2.6 hold.
On the other hand, we can not apply Corollary 15 of Karapınar [27] to this example. Indeed, for x=1=y=v, u=2, 1≤z=s and 1≤t=w, we have
(2.56)d(F(x,y,z,w),F(u,v,s,t))=8|(2)1/3-1|>14>k4d(x,u)=k4[d(x,u)+d(y,v)+d(z,s)+d(w,t)],for any k∈[0,1).Example 2.8.
LetX=ℝ with d(x,y)=|x-y| and natural ordering. Let T:X→X and F:X4→X be defined by Tx=x/12 and F(x,y,z,w)=2/5(x-y+z-w). It is clear that T is an ICS mapping and F has the monotone property and continuous. Set ϕ(t)=2t/3∈Φ. It is clear that all conditions of Theorem 2.2 are satisfied and (0,0,0,0) is the desired quadruple point.
Note that Corollary 15 of Karapınar [27] is not applicable. Indeed, for x=0, u=1 and y=v=z=s=t=w=0, we have
(2.57)d(F(x,y,z,w),F(u,v,s,t))=25>k4d(x,u)=k4[d(x,u)+d(y,v)+d(z,s)+d(w,t)],
for any k∈[0,1).



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*Source: 539125-2012-08-30.xml*