# Coupled Fixed Point Theorems for Nonlinear Contractions in Partial Metric Spaces

Authors: Sharifa Al-Sharif; Mohammad Al-Khaleel; Mona Khandaqji
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

We establish some results on the existence and uniqueness of coupled fixed point involving nonlinear contractive conditions in complete-ordered partial metric spaces.


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## Body

## 1. Introduction

The concept of partial metric which is a generalized metric space was introduced by Matthews [1] in 1994, in which the distance between two identical elements needs not be zero. The existence of fixed point for contraction-type mappings on such spaces was considered by many authors [1–12]. A modified version of a Banach contraction mapping principle, more suitable to solve certain problems arising in computer science using the concept of partial metric space is given in [1].Gnana Bhaskar and Lakshmikantham [13] introduced the concept of coupled fixed point of a mapping F:X×X→X and proved some interesting coupled fixed point theorems for mapping satisfying the mixed monotone property. Later in [14], Lakshmikantham and Ćirić investigated some more coupled fixed point theorems in partially ordered sets. For more on coupled fixed point theory, we refer the reader to [2, 14–20].First, we start by recalling some definitions and properties of partial metric spaces.Definition 1.1 (see [9]).
A partial metric on a nonempty setX is a function p:X×X→R+ such that for all x,y,z∈X:(p1)
x=y⇔p(x,x)=p(x,y)=p(y,y),(p2)
p(x,x)≤p(x,y),(p3)
p(x,y)=p(y,x),(p4)
p(x,y)≤p(x,z)+p(z,y)-p(z,z).A partial metric space is a pair(X,p) such that X is a non empty set and p is a partial metric on X. Each partial metric p on X generates a T0 topology τp on X which has as a base the family of open p-balls {Bp(x,ϵ), x∈X,ϵ>0}, where Bp(x,ϵ)={y∈X:p(x,y)<p(x,x)+ϵ} for all x∈X and ϵ>0. Matthews observed in [1, page 187] that a sequence (xn) in a partial metric space (X,p) converges to some x∈X with respect to p if and only if p(x,x)=limn→∞p(x,xn). It is clear that if p(x,y)=0, then from (p1),(p2), and (p3), x=y. But if x=y,p(x,y) may not be 0.Ifp is a partial metric on X, then the function ps:X×X→R+ given by
(1.1)ps(x,y)=2p(x,y)-p(x,x)-p(y,y),
is a metric on X.Example 1.2 (see, e.g., [1, 7]).
ConsiderX=R+ with p(x,y)=max{x,y}. Then, (R+,p) is a partial metric space.It is clear thatp is not a (usual) metric. Note that in this case ps(x,y)=|x-y|.Definition 1.3 (see [1, Definition 5.2]).
Let(X,p) be a partial metric space and let (xn) be a sequence in X. Then, {xn} is called a Cauchy sequence if limn,m→∞p(xm,xn) exists (and is finite).Definition 1.4 (see [1, Definition 5.3]).
A partial metric space(X,p) is said to be complete if every Cauchy sequence {xn} in X converges, with respect to τp, to a point x∈X, such that p(x,x)=logn,m→∞p(xm,xn).Example 1.5 (see [12]).
LetX:=[0,1]∪[2,3] and define p:X×X→[0,∞) by
(1.2)p(x,y)={max{x,y},{x,y}∩[2,3]≠ϕ,|x-y|,{x,y}⊂[0,1],
Then, (X,p) is a complete partial metric space.It is well known (see, e.g., [1, page 194]) that a sequence in a partial metric space (X,p) is a Cauchy sequence in (X,p) if and only if it is a Cauchy sequence in the metric space (X,ps), and that a partial metric space (X,p) is complete if and only the metric space (X,ps) is complete. Furthermore, limn→∞ps(x,xn)=0 if and only if
(1.3)p(x,x)=limn→∞p(x,xn)=limn,m→∞p(xm,xn).Let(X,p) be a partial metric. We endow X×X with the partial metric q defined for (x,y),(u,v)∈X×X by
(1.4)q((x,y),(u,v))=p(x,u)+p(y,v).A mappingF:X×X→X is said to be continuous at (x,y)∈X×X, if for every ϵ>0, there exists δ>0 such that F(Bq((x,y),δ))⊆Bp(F(x,y),ϵ).In this paper, we establish some results on the existence and uniqueness of a coupled fixed point involving nonlinear contractive conditions in complete-ordered partial metric spaces analogous to some other results in [17, 18].Before presenting our main results, we recall some basic concepts.Definition 1.6 (see [8]).
An elementx,y∈X×X is said to be a coupled fixed point of the mapping F:X×X→X if F(x,y)=x and F(y,x)=y.Definition 1.7 (see, Gnana Bhashkar and Lakshmikantham [13]).
Let(X,≤) be a partially ordered set and F:X×X→X. The mapping F is said to has the mixed monotone property if
(1.5)x1,x2∈X,x1≤x2⇒F(x1,y)≤F(x2,y)foranyy∈X,y1,y2∈X,y1≤y2⇒F(x,y1)≥F(x,y2)foranyx∈X.Now, let us denote byΦ the set of all nondecreasing continuous functions φ:[0,∞)→[0,∞) that satisfy(i)
φ(t)=0 if and only if t=0,(ii)
φ(t+s)≤φ(t)+φ(s), for all t,s∈[0,∞).Again, letΨ denote all functions ψ:[0,∞)→[0,∞) which satisfy limt→rψ(t)>0 for all r>0 and limt→0+ψ(t)=0. It is an easy matter to see that Φ⊆Ψ and (1/2)φ(t)≤φ(t/2)  for any t∈[0,∞).


## 2. Main Results

The aim of this work is to prove the following theorem.Theorem 2.1.
Let(X,≤) be a partially ordered set and suppose that there is a partial metric p on X such that (X,p) is a complete partial metric space. Let F:X×X→X be a mapping having the mixed monotone property on X and assume that there exist φ∈Φ and ψ∈Ψ such that
(2.1)φ(d(F(x,y),F(u,v)))≤φ(αp(x,u)+βp(y,v))-ψ(αp(x,u)+βp(y,v)),
for all x,y,u,v∈X with x≥u,y≤v, and α+β<1. Suppose either F is continuous or X has the following properties:(i)
if a nondecreasing sequencexn→x, then xn≤x for all n,(ii)
if a nonincreasing sequencexn→x, then x≤xn for all n.
If there existx0,y0∈X such that x0≤F(x0,y0), and F(y0,x0)≤y0, then there exist x,y∈X such that x=F(x,y) and y=F(y,x), that is, F has a coupled fixed point. Furthermore, p(x,x)=p(y,y)=0.Proof.
Choosex0,y0∈X and set x1=F(x0,y0) and F(y0,x0)=y1. Since x0≤F(x0,y0) and y0≥F(y0,x0), letting x2=F(x1,y1) and y2=F(y1,x1), we denote
(2.2)F2(x0,y0)=F(F(x0,y0),F(y0,x0))=F(x1,y1)=x2,F2(y0,x0)=F(F(y0,x0),F(x0,y0))=F(y1,x1)=y2,
and due to the mixed monotone property of F, we have
(2.3)x2=F(x1,y1)≥F(x0,y0)=x1,y2=F(y1,x1)≤F(y0,x0)=y1.
Further, for n=1,2,…, we can easily verify that
(2.4)x0≤F(x0,y0)=x1≤F2(x0,y0)=x2≤⋯≤F(Fn(x0,y0),Fn(y0,x0))=xn+1,y0≥F(y0,x0)=y1≥F2(y0,x0)=y2≥⋯≥F(Fn(y0,x0),Fn(x0,y0))=yn+1.
Since xn≥xn-1 and yn≤yn-1, from (2.1), we have
(2.5)φ(p(xn,xn+1))=φ(p(F(xn-1,yn-1),F(xn,yn)))≤φ(αp(xn-1,xn)+βp(yn-1,yn))-ψ(αp(xn-1,xn)+βp(yn-1,yn))≤φ(αp(xn-1,xn)+βp(yn-1,yn)).
Similarly, since yn-1≥yn and xn-1≤xn, from (2.1), we also have
(2.6)φ(p(yn,yn+1))=φ(p(F(yn-1,xn-1),F(yn,xn)))≤φ(αp(yn-1,yn)+βp(xn-1,xn))-ψ(αp(yn-1,yn)+βp(xn-1,xn))≤φ(αp(yn-1,yn)+βp(xn-1,xn)).
Consequently, since φ is nondecreasing, using (2.5) and (2.6), we get
(2.7)p(xn,xn+1)≤αp(xn-1,xn)+βp(yn-1,yn),(2.8)p(yn,yn+1)≤αp(yn-1,yn)+βp(xn-1,xn).
By adding (2.7) to (2.8), we have
(2.9)p(xn,xn+1)+p(yn,yn+1)≤(α+β)(p(xn-1,xn)+p(yn-1,yn)).
Now, we will show that both xn and yn are Cauchy sequences. Note that
(2.10)p(xn,xn+1)+p(yn,yn+1)≤(α+β)(p(xn-1,xn)+p(yn-1,yn))≤(α+β)2(p(xn-2,xn-1)+p(yn-2,yn-1))⋮≤(α+β)n(p(x0,x1)+p(y0,y1)).
Consequently, if p(x0,x1)+p(y0,y1)=0, then x1=x0=F(x0,y0) and y1=y0=F(y0,x0), that means (x0,y0) is a coupled fixed point of F. If p(x0,x1)+p(y0,y1)>0,  for each m>n, combining (2.7) and (2.8) using property p4, we have
(2.11)p(xn,xm)≤p(xn,xn+1)+p(xn+1,xn+2)+⋯+p(xm-1,xm)-∑k=n+1m-1p(xk,xk)≤p(xn,xn+1)+p(xn+1,xn+2)+⋯+p(xm-1,xm),p(yn,ym)≤p(yn,yn+1)+p(yn+1,yn+2)+⋯+p(ym-1,ym)-∑k=n+1m-1p(yk,yk)≤p(yn,yn+1)+p(yn+1,yn+2)+⋯+p(ym-1,ym).
Thus,
(2.12)p(xn,xm)+p(yn,ym)≤p(xn,xn+1)+p(yn,yn+1)+p(xn+1,xn+2)+p(yn+1,yn+2)+⋯+p(xm-1,xm)+p(ym-1,ym)≤((α+β)n+(α+β)n+1+⋯+(α+β)m-1)(p(x0,x1)+p(y0,y1))≤(α+β)n1-(α+β)(p(x0,x1)+p(y0,y1)).
By definition of ps, we have ps(x,y)≤2p(x,y), so
(2.13)ps(xn,xm)+ps(yn,ym)≤2p(xn,xm)+2p(yn,ym)≤2(α+β)n1-(α+β)(p(x0,x1)+p(y0,y1)).
Consequently,
(2.14)ps(xn,xm)≤ps(xn,xm)+ps(yn,ym)≤2(α+β)n1-(α+β)(p(x0,x1)+p(y0,y1)),ps(yn,ym)≤ps(xn,xm)+ps(yn,ym)≤2(α+β)n1-(α+β)(p(x0,x1)+p(y0,y1)),
which implies that both (xn) and (yn) are Cauchy sequences in the metric space (X,ps). Since the metric space (X,ps) is complete, it follows that there exist x,y∈X such that
(2.15)limn→∞ps(xn,x)=limn→∞ps(yn,y)=0.
Therefore, using property p2 and the fact that (X,p) is complete if and only if (X,ps) is complete, using (1.3), we have
(2.16)p(x,x)=limn→∞p(xn,x)=limn→∞p(xn,xn),p(y,y)=limn→∞p(yn,y)=limn→∞p(yn,yn).
From p2 and (2.10), we have
(2.17)p(xn,xn)≤p(xn,xn+1)≤(α+β)n(p(x0,x1)+p(y0,y1)).
Since α+β<1, we get, limn→∞p(xn,xn)=0. Similarly, one can show that limn→∞p(yn,yn). Therefore,
(2.18)p(x,x)=limn→∞p(xn,x)=limn→∞p(xn,xn)=0,p(y,y)=limn→∞p(yn,y)=limn→∞p(yn,yn)=0.
Finally, we will show that x=F(x,y) and y=F(y,x).
(a) Assume thatF is continuous on X. In particular, F is continuous at (x,y), hence for any ϵ>0, there exists δ>0 such that if (u,v)∈X×X verifying q((x,y),(u,v))<q((x,y),(x,y))+δ, meaning that
(2.19)p(x,u)+p(y,v)<p(x,x)+p(y,y)+δ=δ,
because p(x,x)=p(y,y)=0, then we have
(2.20)p(F(x,y),F(u,v))<p(F(x,y),F(x,y))+ϵ2.
Since limn→∞p(xn,x)=limn→∞p(yn,y)=0, for γ=min{δ/2,ϵ/2}>0, there exist n0,m0∈N such that, for n≥n0,m≥m0,p(xn,x)<γ and p(yn,y)<γ. Therefore, for n∈N,n≥max(n0,m0), we have
(2.21)p(xn,x)+p(yn,y)<2γ<δ,
so we get
(2.22)p(F(x,y),F(xn,yn))<p(F(x,y),F(x,y))+ϵ2.
Now, for any n≥max(n0,m0),
(2.23)p(F(x,y),x)≤p(F(x,y),xn+1)+p(xn+1,x)=p(F(x,y),F(xn,yn))+p(xn+1,x)<p(F(x,y),F(x,y))+ϵ2+γ<p(F(x,y),F(x,y))+ϵ.
On the other hand, inserting p(x,x)=p(y,y)=0 in (2.1), we get
(2.24)φ(p(F(x,y),F(x,y)))≤φ(αp(x,x)+βp(y,y))-ψ(αp(x,x)+βp(y,y))=φ(0)-ψ(0)=-ψ(0)≤0,
which implies p(F(x,y),F(x,y))=0, so for any ϵ>0,p(F(x,y),x)<ϵ. This implies that F(x,y)=x. Similarly, we can show that F(y,x)=y.
(b) Assume thatX satisfies the two conditions given by (i) and (ii). Since (xn),(yn) are a nondecreasing sequences and xn→x,yn→y, we have xn≤x and yn≥y  for all n. By the condition (p4), we have
(2.25)p(x,F(x,y))≤p(x,xn+1)+p(xn+1,F(x,y))=p(x,xn+1)+p(F(xn,yn),F(x,y)).
Therefore,
(2.26)φ(p(x,F(x,y)))≤φ(p(x,xn+1))+φ(p(F(xn,yn),F(x,y)))≤φ(p(x,xn+1))+φ(αp(xn,x)+βp(yn,y))-ψ(αp(xn,x)+βp(yn,y)).
Taking the limit as n→+∞ in the above inequality, using (2.18), and the properties of φ and ψ, we get φ(p(x,F(x,y)))=0. Thus, p(x,F(x,y))=0. Hence, x=F(x,y). Similarly, one can show that y=F(y,x). Thus, we proved that F has a coupled fixed point.Corollary 2.2.
Let(X,≤) be a partially ordered set and suppose that there is a partial metric p on X such that (X,p) is a complete partial metric space. Let F:X×X→X be a mapping having the mixed monotone property on X. Supposed that
(2.27)p(F(x,y),F(u,v))≤αp(x,u)+βp(y,v)
for all x,y,u,v∈X with x≥u,y≤v and α+β<1. Suppose either F is continuous or X has the following properties:(i)
if a nondecreasing sequencexn→x, then xn≤x for all n,(ii)
if a nonincreasing sequencexn→x, then x≤xn for all n. If there exist x0,y0∈X such that x0≤F(x0,y0) and F(y0,x0)≤y0 then, there exist x,y∈X such that x=F(x,y) and y=F(y,x), that is, F has a coupled fixed point. Also, p(x,x)=p(y,y)=0.Proof.
Forα+β<1, taking φ(t)=t and ψ(t)=0,  we get the result.The following main theorem for Gnana Bhaskar and Lakshmikantham in [13] proved the next theorem.Theorem 2.3 (see Gnana Bhaskar and Lakshmikantham [13]).
Let(X,≤) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Let F:X×X→X be a mapping having the mixed monotone property on X. Assume that there exists a k∈[0,1) with
(2.28)d(F(x,y),F(u,v))≤k2(d(x,u)+d(y,v)),
for all x,y,u,v∈X with x≤u and v≤y. suppose either F is continuous or X  has the following properties:(i)
if a nondecreasing sequencexn→x, then xn≤x  for all n,(ii)
if a nonincreasing sequencexn→x, then x≤xn for all n.
If there existx0,y0∈X such that x0≤F(x0,y0) and F(y0,x0)≤y0 then there exist x,y∈X such that x=F(x,y) and y=F(y,x), that is, F has a coupled fixed point.Note that fork∈[0,1), and α=β=k/2 in Corollary 2.2, we get analogous to Theorem 2.3 in complete-ordered partial metric space.Theorem 2.4.
Let(X,≤) be a partially ordered set having the property that for every (x,y),(z,t)∈X×X, there exists a(u,v) in X×X such that u≤x,z and v≤y,t and suppose that there is a partial metric p on X such that (X,p) is a complete partial metric space. Let F:X×X→X be a mapping having the mixed monotone property on X and assume that there exist φ∈Φ and ψ∈Ψ such that
(2.29)φ(p(F(x,y),F(u,v)))≤φ(αp(x,u)+βp(y,v))-ψ(αp(x,u)+βp(y,v))
for all x,y,u,v∈X with x≥u,y≤v, and α+β<1. Suppose either F is continuous or X has the following properties:(i)
if a nondecreasing sequencexn→x, then xn≤x for all n,(ii)
if a nonincreasing sequencexn→x, then x≤xn for all n.
If there existx0,y0∈X such that x0≤F(x0,y0)  and F(y0,x0)≤y0 then F has a unique coupled fixed point.Proof.
From Theorem2.1, the set of coupled fixed points of F is non-empty. Suppose (x,y) and (z,t) are coupled fixed points of F, that is, x=F(x,y),y=F(y,x),z=F(z,t)  and  t=F(t,z). We shall show that x=z and y=t.
By assumption, there exists(u,v)∈X×X such that u≤x,z and v≤y,t. We define sequences (un),(vn) as follows:
(2.30)u0=u,v0=v,un+1=F(un,vn),vn+1=F(vn,un)∀n.
Since we may assume that (u0,v0)=(u,v). By using the mathematical induction, it is easy to prove that un≤x and y≤vn for any n∈N. From (2.1), we have
(2.31)φ(p(x,un+1))=φ(p(F(x,y),F(un,vn)))≤φ(αp(x,un)+βp(y,vn))-ψ(αp(x,un)+βp(y,vn))≤φ(αp(x,un)+βp(y,vn)),φ(p(vn+1,y))=φ(p(F(vn,un),F(y,x)))≤φ(αp(vn,y)+βp(un,x))-ψ(αp(vn,y)+βp(un,x))≤φ(αp(vn,y)+βp(un,x)).
Since φ is nondecreasing, from the above inequalities, we have
(2.32)p(x,un+1)≤αp(x,un)+βp(y,vn),(2.33)p(vn+1,y)≤αp(vn,y)+βp(un,x).
Adding (2.32) to (2.33), we get
(2.34)p(x,un+1)+p(y,vn+1)≤(α+β)(p(y,vn)+p(x,un)).
Therefore,
(2.35)p(x,un+1),p(y,vn+1)≤p(x,un+1)+p(y,vn+1)≤(α+β)(p(x,un)+p(y,vn))≤(α+β)2(p(x,un-1)+p(y,vn-1))⋮≤(α+β)n(p(x,u1)+p(y,v1)),
that is, since α+β<1, the sequences {p(y,vn) and p(x,un)} are convergent and
(2.36)limn→∞p(x,un)=limn→∞p(y,vn)=0.
Similarly, one can show that limn→∞p(z,un)=limn→∞p(t,vn)=0.  Since
(2.37)p(x,z)≤p(x,un)+p(un,z),p(y,t)≤p(y,vn)+p(vn,t),
letting n→∞, we obtain p(x,z)=p(y,t)=0, so x=z and y=t.Theorem 2.5.
Let(X,≤) be a partially ordered set such that for every (x,y),(z,t)∈X×X, there exists a(u,v) in X×X such that u≤x,z and v≤y,t and suppose there is a partial metric p on X such that (X,p) is a complete partial metric space. Let F:X×X→X be a mapping having the mixed monotone property on X and assume that there exist φ∈Φ and ψ∈Ψ such that
(2.38)φ(p(F(x,y),F(u,v)))≤φ(αp(x,u)+βp(y,v))-ψ(αp(x,u)+βp(y,v)),
for all x,y,u,v∈X with x≥u,y≤v  and α+β<1. Suppose either F is continuous or X has the following properties:(i)
if a nondecreasing sequencexn→x, then xn≤x for all n,(ii)
if a nonincreasing sequencexn→x, then x≤xn for all n.
If there existx0,y0∈X such that x0≤F(x0,y0)  and F(y0,x0)≤y0, then F has a unique coupled fixed point. In addition, if x0≤y0 or y0≤x0, then x=F(x,y)=F(y,x)=y where (x,y) is a coupled fixed point of F.Proof.
Following the proof of Theorem2.4, F has a unique coupled fixed point (x,y). We only have to show that x=y. Assume y0≤x0. Using the mathematical induction, one can show that xn≥yn for any n∈N. Note that, by (p4),
(2.39)p(x,y)≤p(x,xn+1)+p(xn+1,yn+1)+p(yn+1,y)=p(x,xn+1)+p(yn+1,y)+p(F(xn,yn),F(yn,xn)).
Therefore, using the condition (p3), (2.1), and a property of φ,
(2.40)φ(p(x,y))≤φ(p(x,xn+1))+φ(p(yn+1,y))+φ(p(F(xn,yn),F(yn,xn)))≤φ(p(x,xn+1)+φp(yn+1,y))+φ((α+β)p(xn,yn))-ψ((α+β)p(xn,yn)).
From limn→∞p(xn,x)=limn→∞p(yn,y)=0, we have limn→∞p(xn,yn)=p(x,y). Assume that p(x,y)≠0. Letting n→+∞ in (2.40) we get
(2.41)φ(p(x,y))≤2φ(0)+φ((α+β)p(x,y))-limn→∞ψ((α+β)p(xn,yn))=φ((α+β)p(x,y))-limp(xn,yn)→p(x,y)ψ((α+β)p(xn,yn)).
Since α+β<1, and φ is nondecreasing it follows that
(2.42)φ(p(x,y))≤φ(p(x,y))-limp(xn,yn)→p(x,y)ψ((α+β)p(xn,yn)),
that is, limp(xn,yn)→p(x,y)ψ((α+β)p(xn,yn))≤0, which is a contradiction. Thus, p(x,y)=0, so x=y.Example 2.6 (see [17]).
LetX=[0,1] with usual order. Define p:[0,1]×[0,1]→R+ by p(x,y)=max{x,y} and F:[0,1]×[0,1]→[0,1] by F(x,y)=(1/8)x. Then,(i)
(X,≤,p) is a complete partially ordered partial metric space,(ii)
F has the mixed monotone property,(iii)
forx,y,u,v∈X with x≥u and y≤v, we have
(2.43)p(F(x,y),F(u,v))≤18(p(x,u)+p(y,v)),(iv)
F is continuous.Proof.
The proofs of (i), (ii), and (iii) are clear. To prove (iv), letting(x,y)∈X×X and ϵ>0, we claim F(Bq((x,y),8ϵ))⊆Bq(F(x,y),ϵ). To prove our claim, let (s,t)∈Bq((x,y),8ϵ), then
(2.44)q((s,t),(x,y))<q((x,y),(x,y))+8ϵ.
So,
(2.45)p(s,x)+p(t,y)<p(x,x)+p(y,y)+8ϵ.
Since y≤p(t,y),p(x,x)=x and p(y,y)=y, we have p(s,x)<x+8ϵ. Therefore, and hence p(F(s,t),F(x,y))≤p(F(x,y),p(x,y)). So, F(s,t)∈Bp(F(x,y),ϵ). We deduce that all the hypotheses of Theorem 2.1 are satisfied with φ(t)=t, ψ(t)=0 and α=β=1/8. Therefore, F has a coupled fixed point. Here, (0,0) is the coupled fixed point of F.


## 3. Application

In this part, from previous obtained results, we will deduce some coupled fixed point results for mappings satisfying a contraction of integral type in a complete partial metric space.LetΓ be the set of all functions α:[0,+∞)→[0,+∞) satisfying the following conditions:(i)
α is a Lebesgue integrable mapping on each compact subset of [0,+∞),(ii)
for allϵ>0, we have∫0ϵα(s)ds>0,(iii)
α is subadditive on each [a,b]⊂[0,+∞), that is,
(3.1)∫0a+bα(s)ds≤∫0aα(s)ds+∫0bα(s)ds.LetN∈N* be fixed. Let {αi}1≤i≤N be a family of Nfunctions that belong to Γ. For all t≥0, we denote (Ii),i=1,…,Nas follows:
(3.2)I1(t)=∫0tα1(s)ds,I2(t)=∫0I1(t)α2(s)ds,…,IN(t)=∫0IN-1(t)αN(s)ds.
We have the following result.Theorem 3.1.
Let(X,≤) be a partially ordered set and suppose there is a partial metric p on Xsuch that (X,p) is a complete partial metric space. Let F:X×X→X be a mapping having the mixed monotone property on X. Assume that there exist φ∈Φ and ψ∈Ψ such that
(3.3)IN(φ(p(F(x,y),F(u,v))))≤IN(φ((αp(x,u)+βp(y,v))))-IN(ψ(αp(x,u)+βp(y,v))),
for all x,y,u,v∈X with x≥u, y≤v, and α+β<1. Suppose either F is continuous, or X has the following properties:(i)
if a nondecreasing sequencexn→x, then xn≤x for all n,(ii)
if a nonincreasing sequencexn→x, then x≤xn for all n.
If there existx0,y0∈X such that x0≤F(x0,y0) and F(y0,x0)≤y0, then there exist x,y∈X such that x=F(x,y) and y=F(y,x), that is, F has a coupled fixed point.Proof.
Takeφ~=IN∘φ and ψ~=IN∘ψ. Note that the (αi),i=1,…,N are taken to be subadditive on each [a,b]⊂[0,∞) in order to get φ~(a+b)≤φ~(a)+φ~(b). Moreover, it is easy to show that φ~ is continuous, nondecreasing and verifies φ~(t)=0⇔t=0. We get that φ~∈Φ. Also, we can find that ψ~∈Ψ. From (3.3), we have
(3.4)φ~(p(F(x,y),F(u,v)))≤φ~(αp(x,u)+βp(y,v))-ψ~(αp(x,u)+βp(y,v)).
Now, applying Theorem 2.1, we obtain the desired result.



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*Source: 763952-2012-09-20.xml*