# Unique Fixed Point Theorems for Generalized Contractive Mappings in Partial Metric Spaces

Authors: Lakshmi Narayan Mishra; Shiv Kant Tiwari; Vishnu Narayan Mishra; Idrees A. Khan
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We establish some unique fixed point theorems in complete partial metric spaces for generalized weaklyS-contractive mappings, containing two altering distance functions under certain assumptions. Also, we discuss some examples in support of our main results.


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## Body

## 1. Introduction and Preliminaries

An abstract metric space was first introduced and studied by the French mathematician Frechet [1] in 1906. Many researchers have generalized the concept of metric space as cone metric space, semimetric space, quasimetric space, and so forth, along with the generalization of contraction mappings with applications (see [2–7]). The best approximations of functions in locally convex spaces were discussed by Mishra et al. [8] and Mishra [9]. The degree of approximation of signals in Lp-space is established in [10].Matthews [11, 12] initiated the concept of partial metric space as another generalization of metric space to study the denotational semantics of dataflow networks. Also, Matthews [11] generalized the Banach contraction principle to the class of partial metric spaces as follows: let (X,p) be a complete partial metric space, and then a self-mapping T on X, satisfying (1)pTx,Ty≤kpx,y∀x,y∈X,where 0≤k<1, has a unique fixed point.After the Matthews [11] historical contribution, several researchers have established some more fixed point theorems in partial metric spaces and also discussed its topological properties (see [13–15] and references therein).First, we recall some useful definitions and results, which is useful throughout the paper.Definition 1 (see [11, 12]).
LetX be a nonempty set, and a mapping p:X×X→[0,∞) satisfying the following conditions is called a partial metric space on X:(P1)
p(x,y)=p(y,x),(P2)
p(x,x)=p(x,y)=p(y,y)⇔x=y,(P3)
p(x,x)≤p(x,y),(P4)
p(x,y)≤p(x,z)+p(z,y)-p(z,z),for all x,y,z∈X, and the pair (X,p) is called a partial metric space. In the rest of the paper, (X,p) represents a partial metric space equipped with a partial metric p, unless otherwise stated. Let (X,p) be a partial metric space and then let a function dp:X×X→[0,∞) be defined as (2)dp(x,y)=2p(x,y)-p(y,y)-p(x,x)which is a metric on X. Consider the function dm:X×X→[0,∞) such that (3)dmx,y=max⁡⁡px,y-px,x,px,y-py,y=p(x,y)-min⁡⁡px,x,py,y;then dm is a metric on X, and both of the above metrics dp and dm are equivalent [16].Remark 2 (see [17]).
In a partial metric space(X,p), (1)
p(x,y)=0⇒x=y but if x=y, then p(x,y) may not be zero,(2)
p(x,y)>0 for all x≠y, for all x,y∈X.Example 3 (see [16]).
Consider a mappingp:[0,∞)×[0,∞)→[0,∞) such that p(x,y)=max⁡⁡{x,y} for all x,y∈[0,∞). Then p will satisfy all the property of partial metric, and hence ([0,∞),p) is a partial metric space but fails to be the condition of p(x,x)=0 for all nonzero x∈[0,∞). Therefore ([0,∞),p) is not a metric space.Example 4 (see [16, 18]).
Letpi:X×X→[0,∞)  (i=1,2,3) be three mappings and for any arbitrary mapping f:X→[0,∞) such thatp1x,y=d(x,y)+p(x,y),p2x,y=d(x,y)+max⁡⁡{f(x),f(y)},p3x,y=d(x,y)+r,for all r≥0, where (X,d) and (X,p) are a metric space and a partial metric space, respectively. Then each pi is the partial metric on X.Definition 5 (see [19]).
In a partial metric space(X,p), (1) a sequence {xn} is said to be convergent to a point x∈X if and only if limn→∞p(xn,x)=p(x,x).
(2) A sequence{xn} is called Cauchy sequence if and only if limn,m→∞p(xn,xm) is finite.
(3) If every Cauchy sequence{xn} converges to a point x∈X such that (4)limn,m→∞p(xn,xm)=p(x,x),then (X,p) is known as complete partial metric space.Definition 6 (see [20, 21]).
A self-mappingψ on a positive real number is said to be an altering distance function, if it holds for all t∈[0,∞) such that(1)
ψ is continuous and nondecreasing,(2)
ψ(t)=0⇔t=0.The generalization of contractive mappings intoC-contractive mappings has been introduced by Chatterjea [6].Definition 7 (see [2, 21]).
A self-mappingT on a metric space (X,d), satisfying (5)dTx,Ty≤12dx,Ty+dTx,y-ϕdx,Ty,dTx,y,for all x,y∈X and ϕ:[0,∞)2→[0,∞) is a continuous mapping with ϕ(x,y)=0 if and only if x=y=0 is called weakly C-contractive mapping or a weak C-contraction.Shukla and Tiwari [3] have introduced the concept of weakly S-contractive mappings.Definition 8 (see [3]).
A self-mappingT on a complete metric space (X,d) is said to be weakly S-contractive mapping or a weak S-contraction, if the following inequality holds: (6)dTx,Ty≤13dx,Ty+dTx,y+dx,y-ϕdx,Ty,dTx,y,dx,y,for all x,y∈X and ϕ:[0,∞)3→[0,∞) is a continuous function with ϕ(x,y,z)=0 if and only if x=y=z=0.Lemma 9 (see [7, 14]).
In a partial metric space(X,p), if a sequence {xn} is convergent to a point x∈X, then limn→∞p(xn,x)≤p(x,z) for all z∈X. Also, if p(x,x)=0, then (7)limn→∞p(xn,z)=p(x,z)∀z∈X.Lemma 10 (see [13]).
If{x2n} is not a Cauchy sequence in (X,p) and two sequences {m(k)} and {n(k)} of positive integers such that n(k)>m(k)>k, then the four sequences (8)px2mk,x2nk+1,px2mk,x2nk,px2mk-1,x2nk+1,px2mk-1,x2nktend to ε>0, when k→∞.Lemma 11 (see [13, 16]).
I
n a partial metric space (X,p):(1)
a sequence{xn} is a Cauchy if and only if it is a Cauchy in (X,dp),(2)
X is complete if and only if it is complete in (X,dp).
In addition,limn→∞dp(xn,x)=0 if and only if (9)limn,m→∞p(xn,xm)=limn→∞p(xn,x)=p(x,x).If{xn} is a Cauchy sequence in the metric space (X,dp), we have(10)limn,m→∞dp(xn,xm)=0and therefore, by definition of dp, we have (11)limn,m→∞p(xn,xm)=0.


## 2. Main Results

Theorem 12.
Let(X,p) be a complete partial metric space and ψ and φ be two altering distance functions such that ψ(t)-φ(t)≥0∀t≥0. Then the self-continuous nondecreasing mapping T on X, satisfying the condition(12)ψpTx,Ty≤φpx,Ty+pTx,y+px,y3-ϕpx,Ty,pTx,y,px,y,for all x,y∈X and ϕ:[0,∞)3→[0,∞) is a continuous function such that ϕ(x,y,z)=0 if and only if x=y=z=0, has a unique fixed point in X.Proof.
First we prove that if fixed point ofT exists, then it will be unique. On the contrary, we consider two fixed points z,u∈X of T such that z≠u. Then by (12), we have (13)ψpz,u=ψpTz,Tu≤φpz,Tu+pTz,u+pz,u3-ϕpz,Tu,pTz,u,pz,u⟹0≤(ψ-φ)(p(z,u))≤-ϕ(p(z,u),p(z,u),p(z,u)).By the property of ϕ, we obtain (14)ϕpz,u,pz,u,pz,u=0⟹pz,u=0.Using Remark 2, we obtain z=u, which is a contradiction with respect to z≠u. Thus, we conclude that T has a unique fixed point in X.
Next, we show that the mappingsT, satisfying (12), have a fixed point. We choose an arbitrary point x0 in X. If x0=Tx0, then the theorem follows trivially. Now, we suppose that x0≤Tx0 and we choose x1∈X such that Tx0=x1. Since T is a nondecreasing function, then we have x0≤x1=Tx0≤Tx1. Again, let x2=Tx1. Then we get (15)x0≤x1=Tx0≤Tx1=x2≤Tx2.Proceeding with this work, we obtained a sequence {xn} in X such that xn+1=Txn and (16)x0≤x1≤x2≤x3≤⋯≤xn≤xn+1⋯.Supposing that p(xn0,xn0+1)=0 for some n0≥0, then by Remark 2 we have (17)xn0=xn0+1=Txn0,thatis,xn0isafixedpointofT.Again, we suppose that p(x2n,x2n+1)>0∀n≥0. Firstly, we prove that the sequence {p(x2n,x2n+1)} is nonincreasing. Suppose this is not true, and then(18)p(x2n,x2n+1)≥p(x2n-1,x2n)∀n≥0.Putting x=x2n-1 and y=x2n in (12) and using (P4), we have (19)ψpx2n,x2n+1=ψpTx2n-1,Tx2n≤φpx2n-1,Tx2n+pTx2n-1,x2n+px2n-1,x2n3-ϕpx2n-1,Tx2n,pTx2n-1,x2n,px2n-1,x2n.Using (P4) above, we get (20)ψpx2n,x2n+1≤φ2px2n-1,x2n+px2n,x2n+13-ϕpx2n-1,x2n+1,px2n,x2n,px2n-1,x2n.Using (18) above, we have (21)0≤ψ-φpx2n,x2n+1≤-ϕ(p(x2n-1,x2n+1),p(x2n,x2n),p(x2n-1,x2n)).⟹ϕ(p(x2n-1,x2n+1),p(x2n,x2n),p(x2n-1,x2n))=0⟹p(x2n-1,x2n+1)=0,p(x2n,x2n)=0,p(x2n-1,x2n)=0llllllllllllllllllllll∀n≥0⟹p(x2n,x2n+1)=0∀n≥0,which contradicts our assumption that p(x2n-1,x2n+1)>0 for all n≥0. Thus, we deduce that {p(x2n,x2n+1)} is a nonincreasing sequence. Therefore(22)p(x2n,x2n+1)≤p(x2n-1,x2n)∀n≥0.Since {p(x2n,x2n+1)} is a monotonically decreasing and bounded below sequence in X, then there exists r≥0 such that(23)limn→∞p(x2n,x2n+1)=r.Using (23) and letting n→∞ in (20), we get (24)0≤ψ-φr≤-ϕlimn→∞px2n-1,x2n+1,limn→∞px2n,x2n,r⟹ϕlimn→∞px2n-1,x2n+1,limn→∞px2n,x2n,r=0⟹limn→∞p(x2n-1,x2n+1)=0,limn→∞px2n,x2n=0,r=0.Then (23) reduces to(25)limn→∞p(x2n,x2n+1)=0∀n≥0.Now, we have required proving that the sequence {xn} is a Cauchy sequence in the metric space (X,dp) and so in (X,p) by Lemma 11. On the contrary, that is, the sequence {x2n} not being a Cauchy sequence in (X,dp), sequences in Lemma 10 tend to ε, when k→∞. Now, we put x=x2n(k)+1 and y=x2m(k) in (12). We have (26)ψpx2nk+1,x2mk=ψpTx2nk,Tx2mk-1≤φpx2nk,Tx2mk-1+pTx2nk,x2mk-13-1+px2nk,x2mk-13-1-ϕpx2nk,Tx2mk-1,pTx2nk,x2mk-1,px2nk,x2mk-1=φpx2nk,x2mk+px2nk+1,x2mk-13-1+px2nk,x2mk-13-1-ϕpx2nk,x2mk,px2nk+1,x2mk-1,px2nk,x2mk-1.Taking k→∞ and applying Lemma 10 in the above inequality, we have (27)0≤ψ-φε≤-ϕε,ε,ε⟹ϕε,ε,ε=0⟹ε=0,which is a contradiction with respect to ε>0. Thus {x2n} is a Cauchy sequence in (X,dp) and so in (X,p). Since (X,p) is complete, (X,dp) is also complete (by Lemma 11). Therefore, the Cauchy sequence {xn} converges in (X,dp); that is, limn→∞dp(xn,z)=0; then by Lemma 11, we have(28)p(z,z)=limn→∞p(xn,z)=limn,m→∞p(xn,xm).By Lemma 11, we get limn,m→∞dp(xn,xm)=0. So, by definition of dp, we get (29)dp(xn,xm)=2p(xn,xm)-p(xn,xn)-p(xm,xm).Using (24) and taking n,m→∞ in the above inequality, we obtain(30)limn,m→∞p(xn,xm)=0.From (28) and (30), we get(31)p(z,z)=limn→∞p(xn,z)=0.By (P4), we obtain (32)p(z,Tz)≤p(z,xn)+p(xn,Tz)-p(xn,xn).Taking n→∞ and using (31), (24), and Lemma 9 in the above inequality, we have(33)p(z,Tz)≤p(Tz,Tz).From (P2), we have(34)p(Tz,Tz)≤p(z,Tz).By (33) and (34), we get(35)p(z,Tz)=p(Tz,Tz).From (35) and (12), we obtain (36)ψpz,Tz=ψpTz,Tz≤φpz,Tz+pTz,z+pz,z3-ϕpz,Tz,pTz,z,pz,z.Using (31) and property of φ in the above inequality, we obtain (37)0≤ψ-φpz,Tz≤-ϕpz,Tz,pTz,z,0⟹ϕ(p(z,Tz),p(Tz,z),0)=0⟹p(z,Tz)=0⟹Tz=z.Thus, z is a unique fixed point of T in X.Example 13.
Let([0,1],p) be a complete partial metric space defined by p(x,y)=max⁡⁡{x,y}  ∀x,y∈[0,1]. Consider a self-map T on [0,1] such that Tx=3x2+2x3. Also, we define ψ,φ:[0,∞)→[0,∞) such that ψ(t)=t+t2/2, φ(t)=3t2+t3, respectively, and ϕ:[0,∞)3→[0,∞) such that ϕ(p,q,r)=(p+q+r)2/54.Ifx≥y, then (38)pTx,Ty=max⁡⁡Tx,Ty=Tx=3x2+2x3,ψpTx,Ty=3x2+2x3+92x4+6x5+2x6,φpx,y+pTx,y+px,Ty3=2x+3x2+2x323+2x+3x2+2x3,ϕpx,y,pTx,y,px,Ty=2x+3x2+2x3254.We observe that, for all x,y∈[0,1], (39)ψpTx,Ty≤φpx,y+pTx,y+px,Ty3-ϕpx,y,pTx,y,px,Ty.Similarly, we can show the result for y≥x. Thus, (12) holds for all x,y∈[0,1] and satisfies all the requirements of Theorem 12. So, 0 is the unique fixed point of T.Corollary 14.
Let(X,p) be a complete partial metric space. Then the self-continuous nondecreasing mapping T on X, satisfying the condition(40)ψpTx,Ty≤ψpx,Ty+pTx,y+px,y3-ϕpx,Ty,pTx,y,px,yfor all x,y∈X and ψ and ϕ which are the same as in Theorem 12, has a unique fixed point in X.Corollary 15.
In Corollary14, if partial metric space (X,p) is replaced by usual metric space (X,d), then it reduces to the result of [21].Corollary 16.
In Theorem12, if we take ψ(t)=ϕ(t)=t and partial metric space (X,p) is replaced by usual metric space (X,d), then we obtain the main result of [3], which unifies the main result of [2].Corollary 17.
If we putp(x,y)=0 in (12) and let ϕ:[0,∞)×[0,∞)→[0,∞) be a function, such that ϕ(x,y)=0 if and only if x=y=0, then Theorem 12 reduces to Theorem 2.1 of [13].Theorem 18.
Let(X,p) be a complete partial metric space and ψ and φ be two altering distance functions such that ψ(t)-φ(t)≥0  ∀t≥0. Then the two self-continuous nondecreasing mappings S and T on X, satisfying the condition(41)ψpTx,Sy≤φpx,Sy+pTx,y+px,y3-ϕ(p(x,Sy),p(Tx,y),p(x,y))for all x,y∈X and ϕ:[0,∞)3→[0,∞) is a continuous function such that ϕ(x,y,z)=0 if and only if x=y=z=0, having a unique common fixed point in X.Proof.
First, we show that the common fixed point ofT and S is unique, if it exists. On the contrary, we assume two common fixed points z,u∈X of T and S such that z≠u. Then by (41), we get (42)ψpz,u=ψpTz,Su≤φ(pz,Su+pTz,u+pz,u3)-ϕ(p(z,Su),p(Tz,u),p(z,u))⟹0≤(ψ-φ)(p(z,u))≤-ϕ(p(z,u),p(z,u),p(z,u)).Property of ϕ implies that (43)ϕ(p(z,u),p(z,u),p(z,u))=0⟹p(z,u)=0⟹z=u,which contradicts our assumption that u≠z. Therefore, we conclude that T and S have a unique common fixed point in X.
Now, we prove that the mappingsS and T, satisfying (41), have a common fixed point in X. We choose an arbitrary point x0 in X. If x0=Sx0 and x0=Tx0, then theorem follows trivially. So, we suppose that x0≠Sx0 and x0≠Tx0. Then we construct a sequence {xn} in X, in such a way that Sx2n+1=x2n+2 and Tx2n=x2n+1  ∀n≥0.
Let us assume thatp(x2n,x2n+1)>0 and p(x2n,x2n+2)>0∀n≥0. Then, we can prove that S and T have a common fixed point in X. Firstly, we show that {p(x2n,x2n+1)} is nonincreasing sequence. Suppose this is not true, and then(44)p(x2n,x2n+1)≥p(x2n-1,x2n)∀n≥0.Putting x=x2n and y=x2n+1 in (41) and using (P4), we get (45)ψpx2n+1,x2n+2=ψpTx2n,Sx2n+1≤φpx2n,Sx2n+1+pTx2n,x2n+1+px2n,x2n+13-ϕpx2n,Sx2n+1,pTx2n,x2n+1,px2n,x2n+1.Using (P4) above, we get (46)ψpx2n+1,x2n+2≤φ2px2n,x2n+1+px2n+1,x2n+23-ϕpx2n,x2n+2,px2n+1,x2n+1,px2n,x2n+1.By (44) and (46), we obtain (47)0≤ψ-φpx2n+1,x2n+2≤-ϕpx2n,x2n+2,px2n+1,x2n+1,px2n,x2n+1⟹ϕpx2n,x2n+2,px2n+1,x2n+1,px2n,x2n+1=0⟹p(x2n+1,x2n+1)=0,px2n,x2n+1=0,p(x2n,x2n+2)=0llllllllllllllllllllll∀n≥0,which is a contradiction with respect to p(x2n,x2n+1)>0 and p(x2n,x2n+2)>0∀n≥0. Therefore {p(x2n,x2n+1)} is a nonincreasing sequence in X. Thus, we have(48)p(x2n,x2n+1)≤p(x2n-1,x2n)∀n≥0.Since {p(x2n,x2n+1)} is a monotonically decreasing sequence in X, then there exists r≥0 such that(49)limn→∞p(x2n,x2n+1)=r.Letting n→∞ in (46) and using (49), consequently we get (50)0≤(ψ-φ)(r)≤-ϕ(limn→∞p(x2n,x2n+2),limn→∞p(x2n+1,x2n+1),r)⟹ϕ(limn→∞p(x2n,x2n+2),limn→∞p(x2n+1,x2n+1),r)=0⟹limn→∞p(x2n,x2n+2)=0,limn→∞p(x2n+1,x2n+1)=0,r=0.Then (49) will get reduced to(51)limn→∞p(x2n,x2n+1)=r=0∀n≥0.
Now, we have to show that{xn} is a Cauchy sequence in the partial metric space (X,p). By similar arguments as used in case of proving Theorem 12 we find that the sequence {x2n} is a Cauchy sequence. Putting x=x2n(k) and y=x2m(k)-1 in (41), we have (52)ψpx2nk+1,x2mk=ψ(p(Tx2nk,Sx2mk-1))≤φpx2nk,Sx2mk-1+pTx2nk,x2mk-13-1+px2nk,x2mk-13-1-ϕpx2nk,Sx2mk-1,pTx2nk,x2mk-1,px2nk,x2mk-1.Taking k→∞ and using Lemma 10 in the above inequality, we obtain (53)0≤(ψ-φ)(ε)≤-ϕ(ε,ε,ε)⟹ϕ(ε,ε,ε)=0⟹ε=0,which contradicts our assumption that ε>0. Thus {x2n} is a Cauchy sequence in (X,dp) and so in (X,p). Further, by similar arguments of Theorem 12, we obtain(54)p(z,z)=limn→∞p(xn,z)=limn,m→∞p(xn,xm)=0.By substituting x=z, y=x2m(k)-1 in (41), we obtain(55)ψpTz,x2mk=ψ(p(Tz,Sx2mk-1))≤φpz,Sx2mk-1+pTz,x2mk-13-1+pz,x2mk-13-1-ϕpz,Sx2mk-1,pTz,x2mk-1,pz,x2mk-1.Letting k→∞ and using (54) with property of nondecreasing function φ in the above inequality, we obtain (56)0≤(ψ-φ)p(Tz,z)≤-ϕp0,pTz,z,0⟹ϕ(0,p(Tz,z),0)=0⟹p(z,Tz)=0⟹Tz=z.Hence z is a fixed point of T. Similarly, if we take x=x2n(k)+1 and y=z in (41) and use (54), we obtain Sz=z. By uniqueness of the fixed point, z is a unique common fixed point of S and T.
Again, ifp(x2n,x2n+1)=0 or p(x2n,x2n+2)=0∀n≥0, then we will show that the mappings S and T have a common fixed point in X.
Here, we suppose thatp(x2n,x2n+2)=0  ∀n≥0. Then by Remark 2, x2n=x2n+2, for all n≥0. Let n=k, and then(57)x2k=x2k+2∀k≥0.From (41), we get (58)ψpx2k+1,x2k+2=ψ(p(Tx2k,Sx2k+1))≤φpx2k,Sx2k+1+pTx2k,x2k+13-1+px2k,x2k+13-1-ϕpx2k,Sx2k+1,pTx2k,x2k+1,px2k,x2k+1.Using (P4), (P1), and (57) above, we obtain (59)0≤ψ-φpx2k+1,x2k+2≤-ϕ(p(x2k,x2k+2),p(x2k+1,x2k+1),p(x2k,x2k+1))⟹ϕp(x2k,x2k+2),p(x2k+1,x2k+1),p(x2k,x2k+1)=0⟹p(x2k,x2k+2)=0,p(x2k+1,x2k+1)=0,p(x2k,x2k+1)=0⟹x2k=x2k+1=x2k+2∀k≥0.Similarly, we can show that (60)x2k=x2k+1=x2k+2=x2k+3=⋯∀k≥0.Thus {xn} becomes a constant sequence. So xn=Txn=Sxn for all n≥0. Hence xn is a common fixed point of T and S.
Finally, we assume thatp(x2n,x2n+1)=0∀n≥0. Then by Remark 2, we have x2n=x2n+1∀n≥0. Let n=k, and then(61)x2k=x2k+1∀k≥0.Using (58), (61), and (P4) with property of nondecreasing function φ, we have (62)0≤ψ-φpx2k+1,x2k+2≤-ϕ(p(x2k,x2k+2),p(x2k+1,x2k+1),p(x2k,x2k+1)).Using similar property of ϕ, as used in first case, we have (63)x2k=x2k+1=x2k+2=x2k+3=⋯∀k≥0.Thus, {xn} becomes a constant sequence. So xn=Txn=Sxn. Hence xn is a common fixed point of T and S.Example 19.
LetT, p, ψ, φ, and ϕ all be the same as in Example 13 and a self-mapping S on [0,1] defined as Sx=x2/2+x3/3. Then 0 is a unique common fixed point of S and T. One can compute the solution similarly as done in Example 13.Corollary 20.
Two self-continuous nondecreasing mappingsS and T on a complete partial metric space (X,p), satisfying the condition(64)ψpTx,Sy≤ψpx,Sy+pTx,y+px,y3-ϕ(p(x,Sy),p(Tx,y),p(x,y))for all x,y∈X and ψ and ϕ, are the same as in Theorem 18, having a unique common fixed point in X.Corollary 21.
In Corollary20, if partial metric space (X,p) is replaced by usual metric space (X,d), then one gets Theorem 2.3 of [21].Corollary 22.
If one putsp(x,y)=0 in (41) and lets ϕ:[0,∞)×[0,∞)→[0,∞) be a function, such that ϕ(x,y)=0 if and only if x=y=0, then Theorem 18 reduces to Theorem 2.3 of [13].



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