Coincidence Points And Common Fixed Points For Expansive Type Mappings In Cone b-Metric Spaces
Sushanta Kumar Mohanta
y, Rima Maitra
zReceived 22 January 2014
Abstract
In this paper we prove coincidence point and common …xed point results for mappings satisfying some expansive type contractions in the setting of a cone b-metric space. Our results improve and supplement some recent results in the literature. Some examples are also provided to illustrate our results.
1 Introduction and Preliminaries
Metric …xed point theory is playing an increasing role in mathematics because of its wide range of applications in applied mathematics and sciences. There has been a number of generalizations of the usual notion of a metric space. One such general- ization is a b-metric space introduced and studied by Bakhtin [3] and Czerwik [4]. In [6], Huang and Zhang introduced the concept of cone metric spaces as a generaliza- tion of metric spaces and proved some …xed point theorems for contractive mappings that extend certain results of …xed points in metric spaces. Recently, Hussain and Shah [7] introduced the concept of coneb-metric spaces as a generalization ofb-metric spaces and cone metric spaces. There are many related works about the …xed point of contractive mappings (see, for example [1, 5, 10]). The aim of this work is to obtain su¢ cient conditions for existence of points of coincidence and common …xed points for a pair of self mappings satisfying some expansive type conditions in coneb-metric spaces.
We need to recall some basic notations, de…nitions, and necessary results from existing literature. LetEbe a real Banach space and denote the zero vector ofE. A coneP is a subset ofE such that
(i) P is closed, nonempty andP 6=f g; (ii) ax+by2P for a; b2R; a; b 0; x; y2P;
(iii) P\( P) =f g:
Mathematics Sub ject Classi…cations: 54H25, 47H10.
yDepartment of Mathematics, West Bengal State University, Barasat, 24 Parganas (North), West Bengal, Kolkata 700126, India
zDepartment of Mathematics, West Bengal State University, Barasat, 24 Parganas (North), West Bengal, Kolkata 700126, India
200
For any coneP E, we can de…ne a partial ordering onEwith respect toP by x y(equivalently,y x) if and only ify x2P. We shall writex y(equivalently, y x) ifx y and x6=y, while x y will stand for y x2int(P), where int(P) denotes the interior ofP. The coneP is called normal if there is a numberk >0such that for allx; y2E,
x y implieskxk kkyk:
The least positive number satisfying the above inequality is called the normal constant of P. Throughout this paper, we suppose thatE is a real Banach space,P is a cone in E withint(P)6=; and is a partial ordering onE with respect toP.
DEFINITION 1.1 ([6]). LetE be a real Banach space with coneP and letX be a nonempty set. Suppose the mappingd:X X !E satis…es
(i) d(x; y)for allx; y2X andd(x; y) = if and only ifx=y;
(ii) d(x; y) =d(y; x)for allx; y2X;
(iii) d(x; y) d(x; z) +d(z; y)for all x; y; z2X:
Thendis called a cone metric onX, and(X; d)is called a cone metric space.
DEFINITION 1.2 ([7]). LetX be a nonempty set andEa real Banach space with coneP. A vector valued functiond:X X !Eis said to be a coneb-metric function onX with the constants 1if the following conditions are satis…ed:
(i) d(x; y)for allx; y2X andd(x; y) = if and only ifx=y;
(ii) d(x; y) =d(y; x)for allx; y2X;
(iii) d(x; y) s(d(x; z) +d(z; y))for all x; y; z2X.
The pair (X; d)is called a coneb-metric space.
Observe that ifs= 1, then the ordinary triangle inequality in a cone metric space is satis…ed, however it does not hold true when s >1. Thus the class of coneb-metric spaces is e¤ectively larger than that of the ordinary cone metric spaces. That is, every cone metric space is a cone b-metric space, but its converse need not be true. The following examples illustrate these facts.
EXAMPLE 1.3 ([7]). Let X =f 1;0;1g, E =R2, P =f(x; y) :x 0; y 0g. De…ne d :X X ! P by d(x; y) =d(y; x)for all x; y2 X; d(x; x) = ; x2 X and d( 1;0) = (3;3); d( 1;1) =d(0;1) = (1;1). Then(X; d)is a coneb-metric space, but not a cone metric space since the triangle inequality is not satis…ed. Indeed, we have
d( 1;1) +d(1;0) = (1;1) + (1;1) = (2;2) (3;3) =d( 1;0):
It is easy to verify that s=32.
EXAMPLE 1.4 ([8]). LetE =R2, P =f(x; y) : x 0; y 0g E; X =R and d : X X ! E such that d(x; y) = (jx yjp; jx yjp) where 0 and p > 1 are two constants. Then(X; d)is a coneb-metric space withs= 2p 1, but not a cone metric space.
DEFINITION 1.5 ([7]). Let(X; d)be a coneb-metric space,x2X and(xn)be a sequence in X. Then
(i) (xn) converges to x whenever, for every c 2 E with c, there is a natural numbern0such that for alln > n0,d(xn; x) c. We denote this bylimn!1xn= xor xn!x(n! 1);
(ii) (xn) is a Cauchy sequence whenever, for every c 2 E with c, there is a natural numbern0 such thatd(xn; xm) c for alln; m > n0;
(iii) (X; d)is a complete coneb-metric space if every Cauchy sequence is convergent.
REMARK 1.6 ([7]). Let (X; d) be a cone b-metric space over the ordered real Banach space E with a coneP. Then the following properties are often used:
(i) Ifa bandb c, thena c.
(ii) If a bandb c, then a c.
(iii) If u cfor each c2int(P), thenu= .
(iv) Ifc2int(P); an andan ! , then there existsn0 such that for alln > n0
we havean c.
(v) Let c. If d(xn; x) bn andbn! , then eventually d(xn; x) c, where (xn); xare a sequence and a given point inX.
(vi) If an bn andan !a; bn!b, thena b, for each coneP.
(vii) IfEis a real Banach space with conePand ifa awherea2Pand0 <1, thena= .
(viii) int(P) int(P)for >0.
(ix) For each >0 andx2int(P)there is 0< <1such thatk xk< .
(x) For each c1 and c2 2P, there is an element d such thatc1 d and c2 d.
(xi) For each c1 and c2, there is an element e such that e c1 and e c2.
DEFINITION 1.7. Let(X; d) be a cone b-metric space and let T : X ! X be a given mapping. We say that T is continuous atx02X ifT xn !T x0 as n! 1for every sequences (xn) in X satisfyingxn !x0 as n! 1. IfT is continuous at each pointx02X, then we say that T is continuous onX.
DEFINITION 1.8. Let(X; d)be a coneb-metric space with the constants 1. A mapping T :X !X is called expansive if there exists a real constantk > s such that
d(T x; T y) k d(x; y)for allx; y2X:
DEFINITION 1.9 ([2]). LetT andS be self mappings of a setX. Ify=T x=Sx for somexinX, thenxis called a coincidence point ofT andS andy is called a point of coincidence ofT andS.
DEFINITION 1.10 ([9]). The mappings T; S : X ! X are weakly compatible, if for every x2X, the following holds:
T(Sx) =S(T x)wheneverSx=T x:
PROPOSITION 1.11 ([2]). Let S andT be weakly compatible selfmaps of a non- empty set X. IfS andT have a unique point of coincidencey=Sx=T x, then y is the unique common …xed point of S andT.
2 Main Results
In this section, we prove point of coincidence and common …xed point results in cone b-metric spaces.
THEOREM 2.1. Let (X; d) be a cone b-metric space with the constant s 1.
Suppose the mappings f; g : X ! X satisfy g(X) f(X), either f(X) or g(X) is complete, and
d(f x; f y) d(gx; gy) + d(f x; gx) + d(f y; gy)for allx; y2X; (1) where ; ; are nonnegative real numbers with + + > s; < 1 and 6= 0.
Then f andghave a point of coincidence inX. Moreover, if >1, then the point of coincidence is unique. Iff andgare weakly compatible and >1, thenf andghave a unique common …xed point in X.
PROOF. Let x0 2 X and choose x1 2 X such that gx0 =f x1. This is possible since g(X) f(X). Continuing this process, we can construct a sequence(xn)in X such thatf xn=gxn 1, for alln 1. By (1), we have
d(gxn 1; gxn) = d(f xn; f xn+1)
d(gxn; gxn+1) + d(f xn; gxn) + d(f xn+1; gxn+1)
= d(gxn; gxn+1) + d(gxn 1; gxn) + d(gxn; gxn+1)
which gives that
d(gxn; gxn+1) d(gxn 1; gxn)
where = 1+ . It is easy to see that 2(0;1s). By induction, we get that
d(gxn; gxn+1) nd(gx0; gx1) (2) for alln 0. Letm; n2Nwithm > n. Then, by using condition (2) we have
d(gxn; gxm) s[d(gxn; gxn+1) +d(gxn+1; gxm)]
sd(gxn; gxn+1) +s2d(gxn+1; gxn+2) + +sm n 1[d(gxm 2; gxm 1) +d(gxm 1; gxm)]
s n+s2 n+1+ +sm n 1 m 2+sm n 1 m 1 d(gx0; gx1) s n+s2 n+1+ +sm n 1 m 2+sm n m 1 d(gx0; gx1)
= s n 1 +s + (s )2+ + (s )m n 2+ (s )m n 1 d(gx0; gx1) s n
1 s d(gx0; gx1): (3)
It is to be noted that 1ssn d(gx0; gx1)! as n! 1. Let c be given. Then we can …nd m02Nsuch that
s n
1 s d(gx0; gx1) c for eachn > m0: Therefore, it follows from (3) that
d(gxn; gxm) s n
1 s d(gx0; gx1) cfor allm > n > m0:
So(gxn)is a Cauchy sequence ing(X). Suppose that g(X)is a complete subspace of X. Then there existsy2g(X) f(X)such thatgxn !y and alsof xn!y. In case, f(X) is complete, this holds also withy 2 f(X). Let u2 X be such that f u = y.
For c, one can choose a natural number n0 2 N such that d(y; gxn) 2sc and d(f xn; f u) 2sc for alln > n0. By (1), we have
d(gxn 1; f u) = d(f xn; f u)
d(gxn; gu) + d(f xn; gxn) + d(f u; gu) d(gxn; gu):
If 6= 0, then
d(gxn; gu) 1
d(gxn 1; f u):
Therefore,
d(y; gu) s[d(y; gxn) +d(gxn; gu)]
s[d(y; gxn) + 1
d(gxn 1; f u)]
= s[d(y; gxn) + 1
d(f xn; f u)]
c; f or all n > n0:
This gives that d(y; gu) = , i.e., gu =y and hence f u = gu= y. Therefore, y is a point of coincidence off andg.
Now we suppose that >1. Letv be another point of coincidence off andg. So f x=gx=v for somex2X. Then
d(y; v) =d(f u; f x) d(gu; gx) + d(f u; gu) + d(f x; gx) = d(y; v);
which implies that
d(y; v) 1 d(y; v):
By Remark1:6(vii), we haved(v; y) = i.e.,v=y. Therefore,f andghave a unique point of coincidence inX. Iff andgare weakly compatible, then by Proposition1:11, f andg have a unique common …xed point inX. The proof is complete.
COROLLARY 2.2. Let(X; d) be a coneb-metric space with the constants 1.
Suppose the mappingsf; g:X !X satisfy the condition d(f x; f y) d(gx; gy)for allx; y2X;
where > sis a constant. Ifg(X) f(X)andf(X)or g(X)is complete, thenf and ghave a unique point of coincidence inX. Moreover, iff andgare weakly compatible, thenf andg have a unique common …xed point inX.
PROOF. It follows by taking = = 0in Theorem2:1.
The following corollary is the Theorem2:1[8].
COROLLARY 2.3. Let(X; d)be a complete coneb-metric space with the constant s 1. Suppose the mappingg:X !X satis…es the contractive condition
d(gx; gy) d(x; y)for allx; y2X;
where 2[0;1s)is a constant. Then g has a unique …xed point in X. Furthermore, the iterative sequence(gnx)converges to the …xed point.
PROOF. It follows by taking = = 0and f =I, the identity mapping on X, in Theorem2:1.
COROLLARY 2.4. Let(X; d)be a complete coneb-metric space with the constant s 1. Suppose the mappingf :X!X is onto and satis…es
d(f x; f y) d(x; y)for allx; y2X;
where > sis a constant. Thenf has a unique …xed point inX.
PROOF. Takingg=Iand = = 0in Theorem2:1, we obtain the desired result.
REMARK 2.5. Corollary2:4gives a su¢ cient condition for the existence of unique
…xed point of an expansive mapping in coneb-metric spaces.
COROLLARY 2.6. Let(X; d)be a complete coneb-metric space with the constant s 1. Suppose the mappingf :X!X is onto and satis…es the condition
d(f x; f y) d(x; y) + d(f x; x) + d(f y; y)forx; y2X;
where ; ; are nonnegative real numbers with 6= 0; <1; + + > s. Then f has a …xed point inX. Moreover, if >1, then the …xed point off is unique.
PROOF. It follows by takingg=I in Theorem2:1.
THEOREM 2.7. Let (X; d) be a complete cone b-metric space with the constant s 1. Suppose the mappingsS; T :X !X satisfy the following conditions:
d(T(Sx); Sx) +k
sd(T(Sx); x) d(Sx; x) (4)
and
d(S(T x); T x) +k
sd(S(T x); x) d(T x; x) (5)
for all x2X, where ; ; k are nonnegative real numbers with > s+ (1 +s)k and
> s+ (1 +s)k. If S and T are continuous and surjective, then S and T have a common …xed point in X.
PROOF. Letx02X be arbitrary and choosex12X such thatx0=T x1. This is possible sinceT is surjective. Since S is also surjective, there existsx22X such that x1 =Sx2. Continuing this process, we can construct a sequence (xn) in X such that x2n=T x2n+1 andx2n 1=Sx2n for alln2N. Using (4), we have forn2N[ f0g
d(T(Sx2n+2); Sx2n+2) +k
sd(T(Sx2n+2); x2n+2) d(Sx2n+2; x2n+2) which implies that
d(x2n; x2n+1) +k
sd(x2n; x2n+2) d(x2n+1; x2n+2):
Hence, we have
d(x2n+1; x2n+2) d(x2n; x2n+1) +kd(x2n; x2n+1) +kd(x2n+1; x2n+2):
Therefore,
d(x2n+1; x2n+2) 1 +k
kd(x2n; x2n+1): (6)
Using (5) and by an argument similar to that used above, we obtain that d(x2n; x2n+1) 1 +k
kd(x2n 1; x2n): (7)
Let =max 1+kk; 1+kk . It is easy to see that 2(0;1s). Then, by combining (6) and (7), we get
d(xn; xn+1) d(xn 1; xn) (8)
for alln 1. By repeated application of (8), we obtain d(xn; xn+1) nd(x0; x1):
By an argument similar to that used in Theorem 2.1, it follows that(xn)is a Cauchy sequence inX. SinceX is complete, there existsu2X such thatxn !uasn! 1. Now, x2n+1 ! u and x2n ! u as n ! 1. The continuity of S and T imply that T x2n+1!T uandSx2n!Suas n! 1 i.e.,x2n!T uandx2n 1 !Suasn! 1. The uniqueness of limit yields thatu=Su=T u. Hence,uis a common …xed point of S andT. The proof is complete.
COROLLARY 2.8. Let(X; d)be a complete coneb-metric space with the constant s 1. LetT :X!X be a continuous surjective mapping such that
d(T2x; T x) +k
sd(T2x; x) d(T x; x)for allx2X;
where ; k are nonnegative real numbers with > s+ (1 +s)k. Then T has a …xed point in X.
PROOF. It follows from Theorem2:7by takingS=T and = . We conclude this paper with the following two examples.
EXAMPLE 2.9. LetE=R2, the Euclidean plane andP =f(x; y)2R2:x; y 0g a cone inE. LetX= [0;1]andp >1be a constant. We de…ned:X X !Eas
d(x; y) = (jx yjp;jx yjp) for allx; y2X:
Then (X; d)is a cone b-metric space with the constants= 2p 1. Let us de…nef; g : X !X as f x= x3 andgx= x9 x272 for allx2X. Then, for everyx; y2X one has d(f x; f y) 3pd(gx; gy)i.e., the condition (1) holds for = 3p; = = 0. Thus, we have all the conditions of Theorem2:1and02X is the unique common …xed point of f andg.
EXAMPLE 2.10. Let E =R2 and P =f(x; y)2R2 :x; y 0g a cone in E. Let X = [0;1). We de…ned:X X!E as
d(x; y) = (jx yj2;jx yj2) for allx; y2X:
Then (X; d)is a complete coneb-metric space with the constant s= 2. Let us de…ne S; T :X !X asSx= 3xandT x= 4xfor allx2X. Then, the conditions (4) and (5) hold for = = 3 + 3k > s+ (1 +s)k, wherekis a nonnegative real number. We see that all hypotheses of Theorem2:7are satis…ed and02X is a common …xed point of S andT.
Acknowledgment. The authors would like to express their thanks to the referees for their valuable comments and useful suggestions.
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