# Some Fixed Point Theorems for Prešić-Hardy-Rogers Type Contractions in Metric Spaces

Authors: Satish Shukla; Stojan Radenović; Slaviša Pantelić
Journal: Journal of Mathematics
 (2013)

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## Abstract

We introduce some generalizations of Prešić type contractions and establish some fixed point theorems for mappings satisfying Prešić-Hardy-Rogers type contractive
conditions in metric spaces. Our results generalize and extend several known results in metric spaces. Some examples are included which illustrate the cases when new results can be applied while old ones cannot.


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## Body

## 1. Introduction

The well-known Banach contraction mapping principle states that if(X,d) is a complete metric space and T:X→X is a self-mapping such that
(1)d(Tx,Ty)≤λd(x,y)
for all x,y∈X, where 0≤λ<1, then there exists a unique x∈X such that Tx=x. This point x is called the fixed point of mapping T.On the other hand, for mappingsT:X→X, Kannan [1] introduced the contractive condition:
(2)d(Tx,Ty)≤λ[d(x,Tx)+d(y,Ty)],
for all x,y∈X, where λ∈[0,1/2) is a constant and proved a fixed point theorem using (2) instead of (1). The conditions (1) and (2) are independent, as it was shown by two examples in [2].Reich [3], for mappings T:X→X, generalized Banach and Kannan fixed point theorems, using contractive condition:
(3)d(Tx,Ty)≤αd(x,y)+βd(x,Tx)+γd(y,Ty),
for all x,y∈X, where α,β,γ are nonnegative constants with α+β+γ<1. An example in [3] shows that the condition (3) is a proper generalization of (1) and (2).For mappingT:X→X Chatterjea [4] introduced the contractive condition:
(4)d(Tx,Ty)≤λ[d(x,Ty)+d(y,Tx)],
for all x,y∈X, where λ∈[0,1/2) is a constant and proved a fixed point result using (4).Ćirić [5], for mappings T:X→X, generalized all above mappings, using contractive condition:
(5)d(Tx,Ty)≤αd(x,y)+βd(x,Tx)+γd(y,Ty)+δ[d(x,Ty)+d(y,Tx)],
for all x,y∈X, where α,β,γ,δ are nonnegative constants with α+β+γ+2δ<1. A mapping satisfying (5) is called Generalized contraction.Hardy and Rogers [6], for mappings T:X→X, used the contractive condition:
(6)d(Tx,Ty)≤αd(x,y)+βd(x,Tx)+γd(y,Ty)+δd(x,Ty)+μd(y,Tx),
for all x,y∈X, where α,β,γ,δ,μ are nonnegative constants with α+β+γ+δ+μ<1 and proved fixed point result. Note that condition (6) generalizes all the previous conditions.In 1965, Prešić [7, 8] extended Banach contraction mapping principle to mappings defined on product spaces and proved the following theorem.Theorem 1.
Let(X,d) be a complete metric space, k a positive integer, and f:Xk→X a mapping satisfying the following contractive type condition:
(7)d(f(x1,x2,…,xk),f(x2,x3,…,xk+1))≤∑i=1k‍qid(xi,xi+1),
for every x1,x2,…,xk+1∈X, where q1,q2,…,qk are nonnegative constants such that q1+q2+⋯+qk<1. Then there exists a unique point x∈X such that f(x,x,…,x)=x. Moreover if x1,x2,…,xk are arbitrary points in X and for n∈ℕ,
(8)xn+k=f(xn,xn+1,…,xn+k-1),
then the sequence {xn} is convergent and limxn=f(limxn,limxn,…,limxn).Note that condition (7) in the case k=1 reduces to the well-known Banach contraction mapping principle. So, Theorem 1 is a generalization of the Banach fixed point theorem. Some generalizations and applications of Prešić theorem can be seen in [9–18].Thek-step iterative sequence given by (8) represents a nonlinear difference equation and the solution of this equation can be assumed to be a fixed point of f; that is, solution of (8) is a point x*∈X such that x*=f(x*,x*,…,x*). The Prešić theorem insures the convergence of the sequence {xn} defined by (8) and provides a sufficient condition for the existence of solution of (8) in the case when mapping f satisfies the condition (7). A condition, independent from (7); namely, the Prešić-Kannan condition, is considered in [11] (for the proof of independency of these conditions in case k=1, we refer [1, 2]). In this paper, we introduce some generalizations of Prešić type contractions in metric spaces and use a more general condition; namely, the Prešić-Hardy-Rogers type condition, to prove the existence of fixed point of f in metric spaces. We note that this condition generalizes the result of Prešić [7, 8], Păcurar [11], Hardy and Rogers [6], and several known results in metric spaces. Some examples are included which illustrate the cases when new results can be applied while old ones cannot.


## 2. Some Generalizations of Presić Type Contractions

In this section, we introduced some Prešić type contractions in metric spaces.Let(X,d) be a metric space, k a positive integer, and f:Xk→X be a mapping. (i)
f is said to be a Prešić contraction if f satisfies the condition (7).(ii)
f is said to be a Prešić-Kannan contraction (see [11] for detail) if f satisfies following condition:
(9)d(f(x1,x2,…,xk),f(x2,x3,…,xk+1))≤β∑i=1k+1‍d(xi,f(xi,xi,…,xi))for allx1,x2,…,xk,xk+1∈X, where
(10)0≤βk(k+1)<1.(iii)
f is said to be a Prešić-Reich contraction if f satisfies following condition:
(11)d(f(x1,x2,…,xk),f(x2,x3,…,xk+1))≤∑i=1k‍αid(xi,xi+1)+∑i=1k+1‍βid(xi,f(xi,xi,…,xi))for allx1,x2,…,xk,xk+1∈X, where αi,βi are nonnegative constants such that
(12)∑i=1k‍αi+k∑i=1k+1‍βi<1.(iv)
f is said to be a Prešić-Chatterjea contraction if f satisfies following condition:
(13)d(f(x1,x2,…,xk),f(x2,x3,…,xk+1))≤γ∑i=1,i≠jk+1‍∑j=1k+1‍d(xi,f(xj,xj,…,xj))for allx1,x2,…,xk,xk+1∈X, where
(14)0≤γk2(k+1)<1.(v)
f is said to be a Generalized-Prešić contraction if f satisfies following condition:
(15)d(f(x1,x2,…,xk),f(x2,x3,…,xk+1))≤∑i=1k‍αid(xi,xi+1)+∑i=1k+1‍βid(xi,f(xi,xi,…,xi))+β∑i=1,i≠jk+1‍∑j=1k+1‍d(xi,f(xj,xj,…,xj))for allx1,x2,…,xk,xk+1∈X, where αi,βi,β are nonnegative constants such that
(16)∑i=1k‍αi+k∑i=1k+1‍βi+βk2(k+1)<1.(vi)
f is said to be a Prešić-Hardy-Rogers contraction if f satisfies following condition:
(17)d(f(x1,x2,…,xk),f(x2,x3,…,xk+1))≤∑i=1k‍αid(xi,xi+1)+∑i=1k+1‍∑j=1k+1‍βi,jd(xi,f(xj,xj,…,xj))for allx1,x2,…,xk,xk+1∈X, where αi,βi,j are nonnegative constants such that
(18)∑i=1k‍αi+k∑i=1k+1‍∑j=1k+1‍βi,j<1.Remark 2.
Note that forβi,j=β for all i,j∈{1,2,…,k,k+1} with i≠j and βi,i=βi for all i∈{1,2,…,k,k+1}, the Prešić-Hardy-Rogers contraction reduces into the Generalized-Prešić contraction. With β=0, the Generalized-Prešić contraction reduces into the Prešić-Reich contraction and with αi=0 for all i∈{1,2,…,k},  βi=0 for all i∈{1,2,…,k,k+1}, and β=γ, the Generalized-Prešić contraction reduces into the Prešić-Chatterjea contraction. With αi=0 for all i∈{1,2,…,k}, the Prešić-Reich contraction reduces into the Prešić-Kannan contraction and with βi=0 for all i∈{1,2,…,k,k+1}, the Prešić-Reich contraction reduces into the Prešić contraction. Therefore among all above definitions, the Prešić-Hardy-Rogers contraction is the most general contraction.Remark 3.
It is easy to see that fork=1, Prešić-Hardy-Rogers contraction reduces into Hardy-Rogers contraction and for k=1, Generalized-Prešić contraction reduces into Generalized contraction and so forth; therefore, the comparison as considered in [19] shows that the above generalization is proper.Now, we shall prove some fixed point results for Prešić-Hardy-Rogers type contractions in metric spaces.


## 3. Main Results

The following theorem is the fixed point result for Prešić-Hardy-Rogers type contractions and the main result of this paper.Theorem 4.
Let(X,d) be any complete metric space, k a positive integer. Let f:Xk→X be a Prešić-Hardy-Rogers contraction, then f has a unique fixed point in X.Proof.
Letx0∈X be arbitrary. Define a sequence {xn} in X by
(19)xn+1=f(xn,…,xn),n≥0.
If xn=xn+1 for any n then xn is a fixed point of f. Therefore we assume xn≠xn+1 for all n.
We shall show that this sequence is a Cauchy sequence inX.
For simplicity, set(20)di=d(xi,xi+1),Di,j=d(xi,f(xj,…,xj))∀i,j≥1.
For any n≥0, we obtain
(21)dn+1=d(xn+1,xn+2)=d(f(xn,…,xn),f(xn+1,…,xn+1))≤d(f(xn,…,xn),f(xn,…,xn,xn+1))+d(f(xn,…,xn,xn+1),f(xn,…,xn,xn+1,xn+1))+⋯+d(f(xn,xn+1,…,xn+1),f(xn+1,…,xn+1)),
using (17), it follows from above inequality that
(22)dn+1≤{αkdn+[∑j=1k‍β1,j+∑j=1k‍β2,j+⋯+∑j=1k‍βk,j]Dn,n+[∑i=1k‍βi,k+1]Dn,n+1+[∑j=1k‍βk+1,j]Dn+1,n+βk+1,k+1Dn+1,n+1∑i=1k‍}+{αk-1dn∑j=1k-1‍+[∑j=1k-1‍β1,j+∑j=1k-1‍β2,j+⋯+∑j=1k-1‍βk-1,j]Dn,n+[∑i=1k-1‍βi,k+∑i=1k-1‍βi,k+1]Dn,n+1+[∑j=1k-1‍βk,j+∑j=1k-1‍βk+1,j]Dn+1,n+[∑j=kk+1‍βk,j+∑j=kk+1‍βk+1,j]Dn+1,n+1}+⋯+{∑j=2k+1‍α1dn+β1,1Dn,n+[∑j=2k+1‍β1,j]Dn,n+1+[∑i=2k+1‍βi,1]Dn+1,n+[∑j=2k+1‍β2,j+∑j=2k+1‍β3,j+⋯+∑j=2k+1‍βk+1,j]Dn+1,n+1},
that is,
(23)dn+1≤[∑i=1k‍αi]dn+{[∑i=1k‍∑j=1k‍βi,j]Dn,n+[∑i=1k‍βi,k+1]Dn,n+1+[∑j=1k‍βk+1,j]Dn+1,n+βk+1,k+1Dn+1,n+1}+{[∑i=1k-1‍∑j=1k-1‍βi,j]Dn,n+[∑i=1k-1‍∑j=kk+1‍βi,j]Dn,n+1+[∑i=kk+1‍∑j=1k-1‍βi,j]Dn+1,n+[∑i=kk+1‍∑j=kk+1‍βi,j]Dn+1,n+1}+⋯+{β1,1Dn,n+[∑j=2k+1‍β1,j]Dn,n+1+[∑i=2k+1‍βi,1]Dn+1,n+[∑i=2k+1‍∑j=2k+1‍βi,j]Dn+1,n+1},
that is,
(24)dn+1≤[∑i=1k‍αi]dn+[∑i=1k‍∑j=1k‍βi,j+∑i=1k-1‍∑j=1k-1‍βi,j+⋯+∑i=12‍∑j=12‍βi,j+β1,1]Dn,n+[∑i=1k‍βi,k+1+∑i=1k-1‍∑j=kk+1‍βi,j+⋯+∑i=12‍∑j=3k+1‍βi,j+∑j=2k+1‍β1,j]Dn,n+1+[∑j=1k‍βk+1,j+∑i=kk+1‍∑j=1k-1‍βi,j+⋯+∑i=3k+1‍∑j=12‍βi,j+∑i=2k+1‍βi,1]Dn+1,n+[∑i=2k+1‍∑j=2k+1‍βi,j+∑i=3k+1‍∑j=3k+1‍βi,j+⋯+∑i=kk+1‍∑j=kk+1‍βi,j+βk+1,k+1]Dn+1,n+1=Adn+BDn,n+CDn,n+1+EDn+1,n+FDn+1,n+1,
where A,B,C,E, and F are the coefficients of dn,Dn,n,Dn,n+1,Dn+1,n, and Dn+1,n+1, respectively, in the above inequality.
By definition,Dn,n=d(xn,f(xn,…,xn))=d(xn,xn+1)=dn,Dn,n+1=d(xn,f(xn+1,…,xn+1))=d(xn,xn+2),Dn+1,n=d(xn+1,f(xn,…,xn))=d(xn+1,xn+1)=0,Dn+1,n+1=d(xn+1,f(xn+1,…,xn+1))=d(xn+1,xn+2)=dn+1, therefore
(25)dn+1≤Adn+Bdn+Cd(xn,xn+2)+Fdn+1≤Adn+Bdn+Cd(xn,xn+1)+Cd(xn+1,xn+2)+Fdn+1=(A+B+C)dn+(C+F)dn+1,
that is,
(26)(1-C-F)dn+1≤(A+B+C)dn.
Again, as dn+1=d(xn,xn+1)=d(xn+1,xn), interchanging the role of xn and xn+1, and repeating above process, we obtain
(27)(1-E-B)dn+1≤(A+F+E)dn.
It follows from (26) and (27) that
(28)(2-B-C-E-F)dn+1≤(2A+B+C+E+F)dn,dn+1≤2A+B+C+E+F2-B-C-E-Fdn,dn+1≤λdn,
where λ=(2A+B+C+E+F)/(2-B-C-E-F).
Using (18), we obtain
(29)A+B+C+E+F=∑i=1k‍αi+∑i=1k‍∑j=1k‍βi,j+∑i=1k-1‍∑j=1k-1‍βi,j+⋯+∑i=12‍∑j=12‍βi,j+β1,1+∑i=1k‍βi,k+1+∑i=1k-1‍∑j=kk+1‍βi,j+⋯+∑i=12‍∑j=3k+1‍βi,j+∑j=2k+1‍β1,j+∑j=1k‍βk+1,j+∑i=kk+1‍∑j=1k-1‍βi,j+⋯+∑i=3k+1‍∑j=12‍βi,j+∑i=2k+1‍βi,1+∑i=2k+1‍∑j=2k+1‍βi,j+∑i=3k+1‍∑j=3k+1‍βi,j+⋯+∑i=kk+1‍∑j=kk+1‍βi,j+βk+1,k+1=∑i=1k‍αi+k∑i=1k+1‍∑j=1k+1‍βi,j<1.
So 0≤λ<1. By (28), we obtain
(30)dn+1≤λn+1d0∀n≥0.
Suppose n,m∈ℕ with m>n. Then
(31)d(xn,xm)≤d(xn,xn+1)+d(xn+1,xn+2)+⋯+d(xm-1,xm)=dn+dn+1+⋯+dm-1≤λnd0+λn+1d0+⋯+λm-1d0≤λn1-λd0,
as 0≤λ<1, it follows from the above inequality that limn→∞d(xn,xm)=0. Therefore {xn} is a Cauchy sequence. By completeness of X, there exists u∈X such that limn→∞xn=u.
We shall show thatu is the fixed point of f. Note that
(32)d(u,f(u,…,u))≤d(u,xn+1)+d(xn+1,f(u,…,u))=d(u,xn+1)+d(f(xn,…,xn),f(u,…,u)),
using a similar process as used in the calculation of dn+1, we obtain
(33)d(u,f(u,…,u))≤d(u,xn+1)+Ad(xn,u)+Bd(xn,f(xn,…,xn))+Cd(xn,f(u,…,u))+Ed(u,f(xn,…,xn))+Fd(u,f(u,…,u))≤d(u,xn+1)+Ad(xn,u)+Bd(xn,xn+1)+Cd(xn,u)+Cd(u,f(u,…,u))+Ed(u,xn+1)+Fd(u,f(u,…,u)),
that is,
(34)d(u,f(u,…,u))≤A+B+C1-C-Fd(xn,u)+1+B+E1-C-Fd(xn+1,u).
Using the fact that limn→∞xn=u, it follows from the above inequality that
(35)d(u,f(u,…,u))=0thatis,f(u,…,u)=u.
Thus u is a fixed point of f. For uniqueness, let v be another fixed point of f, that is, f(v,…,v)=v. Again using a similar process as used in the calculation of dn+1, we obtain
(36)d(u,v)≤Ad(u,v)+Bd(u,f(u,…,u))+Cd(u,f(v,…,v))+Ed(v,f(u,…,u))+Fd(v,f(v,…,v))=(A+C+E)d(u,v),
as A+B+C+E+F<1, we obtain d(u,v)=0, that is, u=v. Thus fixed point is unique.Remark 5.
Fork=1 in the above theorem, we obtain the result of Hardy and Rogers [6]. For βi,j=0 for all i,j∈{1,2,…,k,k+1}, we obtain the fixed point result of Prešić. Therefore, above theorem is a generalization of the results of Hardy and Rogers and Prešić.With Remark2, the following corollaries are obtained.Corollary 6.
Let(X,d) be any complete metric space, k a positive integer, and f:Xk→X a Generalized Prešić contraction. Then f has a unique fixed point in X.Fork=1 in above corollary, we obtain the fixed point result of Ćirić [5].Corollary 7.
Let(X,d) be any complete metric space, k a positive integer, and f:Xk→X a Prešić-Reich contraction. Then f has a unique fixed point in X.Fork=1 in the above corollary, we obtain the fixed point result of Reich [3].Corollary 8.
Let(X,d) be any complete metric space, k a positive integer, and f:Xk→X a Prešić-Kannan contraction. Then f has a unique fixed point in X.Fork=1 in above the corollary, we obtain the fixed point result of Kannan [2].Corollary 9.
Let(X,d) be any complete metric space, k a positive integer, and f:Xk→X a Prešić-Chatterjea contraction. Then f has a unique fixed point in X.Fork=1 in above corollary, we obtain the fixed point result of Chatterjea [4].The following are some examples which illustrate the cases when known results are not applicable, while our new results can be used to conclude the existence of fixed point of mapping.Example 10.
LetX=[0,1] with usual metric. For k=2 define f:X2→X by
(37)f(x,y)={15,ifx=y=1;x+y5,otherwise.
Then (i)
f is a Prešić-Reich contraction with α1=α2=1/5,β1=β2=β3=1/11;(ii)
f is not a Prešić contraction;(iii)
f is not a Prešić-Kannan contraction.Proof.
(i) Note that forx1,x2,x3∈[0,1) with x1≤x2≤x3,
(38)d(f(x1,x2),f(x2,x3))=d(x1+x25,x2+x35)=x3-x15=15[(x2-x1)+(x3-x2)]=15[d(x1,x2)+d(x2,x3)]=15∑i=12‍d(xi,xi+1).
Therefore conditions (11) and (12) are satisfied for α1=α2=1/5 and β1,β2,β3 with β1+β2+β3∈[0,3/10).
If any one ofx1,x2,x3 is 1 then proof is similar. If any two of x1,x2,x3 are 1, for example, if x1∈[0,1) and x2=x3=1, then
(39)d(f(x1,x2),f(x2,x3))=d(f(x1,1),f(1,1))=d(x1+15,15)=x1+15-15=x15111∑i=13‍d(xi,f(xi,xi))=111[d(x1,f(x1,x1))+d(x2,f(x2,x2))=111+d(x3,f(x3,x3))]=111[3x15+45+45]=155[3x1+8].
As x1∈[0,1), so conditions (11) and (12) are satisfied for β1=β2=β3=1/11 and α1,α2 with α1+α2∈[0,5/11).
Similarly in all possible cases conditions (11) and (12) are satisfied with α1=α2=1/5, β1=β2=β3=1/11. Therefore f is a Prešić-Reich contraction. All other conditions of Corollary 7 are satisfied and 0 is the unique fixed point of f.
(ii) Note that forx1=9/10 and x2=x3=1(40)d(f(x1,x2),f(x2,x3))=d(1950,15)=950∑i=12‍αid(xi,xi+1)=α1d(x1,x2)+α2d(x2,x3)=α1d(910,1)+α2d(1,1)=110α1.
Therefore, we cannot find nonnegative constants α1,α2 such that condition (7) is satisfied with α1+α2<1. So f is not a Prešić contraction.
(iii) Again forx1=x2=0, x3=1(41)d(f(x1,x2),f(x2,x3))=d(0,15)=15β∑i=13‍d(xi,f(xi,xi))=β[d(x1,f(x1,x1))+d(x2,f(x2,x2))+d(x3,f(x3,x3))]=β[d(0,0)+d(0,0)+d(1,15)]=45β.
Therefore, we cannot find nonnegative constant β such that conditions (9) and (10) are satisfied. So f is not a Prešić-Kannan contraction.Remark 11.
In the above example, we cannot apply the result of Prešić [7, 8] and Păcurar [11] to conclude the existence of fixed point of f. But Corollary 7 is applicable which insures the existence of unique fixed point of f.Example 12.
LetX=[0,1] with usual metric. For k=2, define f:X2→X by
(42)f(x,y)={415,ifx=y=1;0,otherwise.
Then (i)
f is a Prešić-Chatterjea contraction with γ∈[1/13,1/12);(ii)
f is not a Prešić contraction;(iii)
f is not a Prešić-Kannan contraction.Proof.
(i) Note that ifx1,x2,x3∈[0,1) or any one of x1,x2,x3 is 1, then conditions (17) and (18) are satisfied trivially.
If any two ofx1,x2,x3 are 1, for example, if x1∈[0,1), x2=x3=1, then
(43)d(f(x1,x2),f(x2,x3))=d(f(x1,1),f(1,1))=d(0,415)=415γ∑i=1,i≠j3‍∑j=13‍d(xi,f(xj,xj))=γ[d(x1,415)+d(x1,415)+d(1,0)+d(1,415)+d(1,0)+d(1,415)]=γ[2|x1-415|+2+2215]=γ[2|x1-415|+5215]≤γ5215.
Therefore conditions (13) and (14) are satisfied with γ∈[1/13,1/12). Also all other conditions of Corollary 9 are satisfied and f has a unique fixed point 0.
(ii) Forx1=9/10, x2=1, x2=1, we have
(44)d(f(x1,x2),f(x2,x3))=d(f(910,1),f(1,1))=d(0,415)=415∑i=12‍αid(xi,xi+1)=α1d(x1,x2)+α2d(x2,x3)=α1110.
Therefore we cannot find nonnegative constants α1,α2 such that condition (7) is satisfied with α1+α2<1. So f is not a Prešić contraction.
(iii) Forx1=0, x2=x3=1, we have
(45)d(f(x1,x2),f(x2,x3))=d(f(0,1),f(1,1))=d(0,415)=415β∑i=13‍d(xi,f(xi,xi))=β[d(x1,f(x1,x1))+d(x2,f(x2,x2))=β+d(x3,f(x3,x3))]=β2215.
Therefore we cannot find nonnegative constant β such that conditions (9) and (10) are satisfied. So f is not a Prešić-Kannan contraction.Remark 13.
In the above example, we cannot apply the result of Prešić [7, 8] and Păcurar [11] to conclude the existence of fixed point of f. But Corollary 9 is applicable which insures the existence of unique fixed point of f.The following theorem is a consequence of Theorem4 and the recent result of Aydi et al. [20].Theorem 14.
Let(X,d) be any complete metric space and k a positive integer. Let f:Xk→X and T:X→X be two mappings such that the following condition holds:
(46)d(Tf(x1,x2,…,xk),Tf(x2,x3,…,xk+1))≤∑i=1k‍αid(Txi,Txi+1)+∑i=1k+1‍∑j=1k+1‍βi,jd(Txi,Tf(xj,xj,…,xj))
for all x1,x2,…,xk,xk+1∈X, where αi,βi,j are nonnegative constants such that
(47)∑i=1k‍αi+k∑i=1k+1‍∑j=1k+1‍βi,j<1
and T is continuous, injective, and sequentially convergent. Then f has a unique fixed point in X.Proof.
Define a mappingρ:X×X→[0,∞) by
(48)ρ(x,y)=d(Tx,Ty)∀x,y∈X.
Then (X,ρ) is a complete metric space (see [20]). Note that condition (46) reduces to the condition (17); that is, mapping f reduces to Prešić-Hardy-Rogers contraction with respect to metric ρ. So the rest of the proof followed Theorem 4.



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*Source: 295093-2013-03-13.xml*