# A Common Fixed Point Theorem for Two Hybrid Pairs of Mappings inb-Metric Spaces

Authors: K. P. R. Rao; K. R. K. Rao
Journal: International Journal of Analysis
 (2013)

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## Abstract

We introduce a condition (W.C.C) and prove a common fixed point theorem for two hybrid pairs of mappings in b-metric spaces.


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## Body

## 1. Introduction and Preliminaries

The study of fixed points for set-valued contraction maps using the Hausdorff metric was initiated by Nadler [1] in 1969, who extended the Banach contraction principle to set-valued mappings. The theory of set-valued maps has many applications in control theory, convex optimization, differential equations, and economics.Recently Amini-Harandi [2] proved a set-valued version of Ciric's [3] theorem as follows.Theorem 1.
Let(X,d) be a complete metric space. Suppose T:X→CB(X) satisfy
(1)H(Tx,Ty)≤qmax{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)},
for all x,y∈X, where 0≤q<1/2. Then T has a fixed point in X.Aydi et al. [4] proved a slight version of the above theorem in b-metric spaces as follows.Theorem 2.
Let(X,d) be a complete b-metric space. Suppose T:X→CB(X) satisfy
(2)H(Tx,Ty)≤qmax{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)},
for all x,y∈X, where 0≤q<1/(s2+s). Then T has a fixed point in X.In this paper, we introduce a new condition, namely, “condition (W.C.C)” of mappings which are not continuous and non commuting pairwise. Using this new condition we prove a common fixed point theorem for two pairs of hybrid mappings in b-metric spaces.For the sequel, we need the following concepts.Czerwik [5] presented a generalization of the well-known Banach fixed point theorem in b-metric spaces by defining the following.Definition 3.
LetX be a nonempty set and s≥1 a given real number. A function d:X×X→ℝ+ is called a b-metric if the following axioms are satisfied for all x,y,z∈X:(b1)
d(x,y)=0 if and only if x=y,(b2)
d(x,y)=d(y,x),(b3)
d(x,y)≤s[d(x,z)+d(z,y)].
Let(X,d) be a b-metric space. Let CB(X) be the collection of all non-empty closed and bounded subsets of X. For A,B∈CB(X), define
(3)H(A,B)=max{δ(A,B),δ(B,A)},
where
(4)δ(A,B)=sup{d(a,B):a∈A},δ(B,A)=sup{d(b,A):b∈B},
with
(5)d(a,B)=inf{d(a,b):b∈B}.
Then H is called the Hausdorff metric induced by the b-metric.
We cite the following lemmas from Czerwik [6]. For more details we refer to [7, 8].Lemma 4.
Let(X,d) be a  b-metric space. For any A,B,C∈CB(X) and any x,y∈X, we have the following:  (i)
d(x,B)≤d(x,b) for any b∈B,(ii)
δ(A,B)≤H(A,B),(iii)
d(x,B)≤H(A,B) for any x∈A,(iv)
H(A,A)=0,(v)
H(A,B)=H(B,A),(vi)
H(A,C)≤s[H(A,B)+H(B,C)],(vii)
d(x,A)≤s[d(x,y)+d(y,A)].Lemma 5.
Let(X,d) be a  b-metric space. Let A,B∈CB(X). Then for each ϵ>0 and for all b∈B there exists a∈A such that d(a,b)≤H(A,B)+ϵ.Lemma 6.
Let(X,d) be a  b-metric space. For A∈CB(X) and x∈X, we have d(x,A)=0⇔x∈A¯=A.In this paper, we introduce the following condition on three mappings without continuity and commutativity(W.C.C).Definition 7.
Let(X,d) be a  b-metric space. The mappings (f,g;S) where f,g:X→X and S:X→CB(X) are said to satisfy the condition (W.C.C) if d(fx,gy)≤d(y,Sx) for all x,y∈X.Remark 8.
The mappings(f,g;S) satisfying the condition (W.C.C) need not be continuous and commutative in view of the following examples.Example 9.
Let(X,d) be a b-metric space, where X=[0,1] and d(x,y)=|x-y|2.
Definef,g,S:X→X by
(6)fx={x12ifx∈[0,1),124ifx=1,(7)gx={x6ifx∈[0,1),112ifx=1,(8)Sx={x2ifx∈[0,1),14ifx=1.
We distinguish the following cases.Case 1.
Consider0≤x<1and0≤y<1(9)Thend(fx,gy)=|x12-y6|2=136|x2-y|2=136d(y,Sx).Case 2.
Consider0≤x<1andy=1(10)Thend(fx,gy)=|x12-112|2=136|12-x2|2<136d(y,Sx).Case 3.
Considerx=1and0≤y<1(11)Thend(fx,gy)=|124-y6|2=136|y-14|2=136d(y,Sx).Case 4.
Considerx=1andy=1(12)Thend(fx,gy)=|124-112|2=1576<916=d(y,Sx).
Thus d(fx,gy)≤d(y,Sx) for all x,y∈X is satisfied.
In this example all the mappings are discontinuous.Example 10.
Letd(x,y)=|x-y|2 be a  b-metric on a set A={x,y}. Define fx=fy=y, gx=y, gy=x, and Sx=Sy=x. Clearly the mappings (f,g;S) satisfy the condition (W.C.C) and the pairs (f,g), (f,S), and (g,S) are not commuting.Now, we prove our main result with the condition(W.C.C) for two hybrid pairs of mappings in b-metric spaces.


## 2.   Main Result

Theorem 11.
Let(X,d) be a complete b-metric space. Let f,g:X→X and S,T:X→CB(X) be mappings satisfying  the following:(2.1.1)
H(Sx,Ty)≤qM(x,y), where M(x,y)=max{d(fx,gy),d(fx,Sx), d(gy,Ty), d(fx,Ty), d(gy,Sx)} for all x,y∈X, where 0≤q<1/(s2+s),(2.1.2)
⋃x∈XSx⊆g(X)and⋃x∈XTx⊆f(X),(2.1.3)
the mappings(f,g;S) or the mappings (f,g;T) satisfy the condition (W.C.C).Then either the pairs
(
f
,
S
),and(g,T)have coincidence points or the mappingsS,T,fandghave a common fixed point inX.Proof.
SupposeM(x,y)=0 for some x,y∈X.
Thenfx∈Sx and gy∈Ty. Thus x is a coincidence point of f and S, and y is a coincidence point of g and T.
Assume thatM(x,y)>0 for all x,y∈X.
Letϵ=(1/2)(1/(s2+s)-q) and β=q+ϵ=(1/2)(1/(s2+s)+q).
Thenϵ>0 and 0<β<1.
Letx0∈X. From (2.1.2), there exist x1,y1∈X such that y1=gx1∈Sx0.
From (2.1.2), there existx2,y2∈X such that y2=fx2∈Tx1 and from Lemma 5, we have
(13)d(y1,y2)≤H(Sx0,Tx1)+ϵM(x0,x1)≤qM(x0,x1)+ϵM(x0,x1),from(2.1.1)=βM(x0,x1).
Again from (2.1.2), there exist x3,y3∈X such that y3=gx3∈Sx2 and from Lemma 5, we have
(14)d(y2,y3)≤H(Sx2,Tx1)+ϵM(x2,x1)≤qM(x2,x1)+ϵM(x2,x1),from(2.1.1)=βM(x2,x1).
Continuing in this way, we get the sequences {xn} and {yn} in X such that  y2n+1=gx2n+1∈Sx2n, y2n+2=fx2n+2∈Tx2n+1 for all n=0,1,2,3,… and
(15)d(y2n+1,y2n)≤βM(x2n,x2n-1),d(y2n+1,y2n+2)≤βM(x2n,x2n+1)forn=1,2,3,….
For simplification, writedn=d(yn,yn+1).
Now(16)d2n+1=d(y2n+1,y2n+2)≤βM(x2n,x2n+1)=βmax{d(y2n,y2n+1),d(y2n,y2n+1),d(y2n+1,y2n+2),d(y2n,y2n+2),0}≤βmax{d2n,d2n+1,s[d2n+d2n+1]}
which in turn yields
(17)d2n+1≤γd2n,
where γ=max{β,sβ/(1-sβ)}.
Similarly, we can show that(18)d2n≤γd2n-1.
Thus, from (17) and (18), we have dn≤γdn-1 for all n=1,2,3,…, so that
(19)dn≤γnd0∀n=1,2,3,….
If γ=β, then γs=βs=(s/2)(1/(s2+s)+q)<s/(s2+s)=1/(s+1)≤1/2<1.
Ifγ=sβ/(1-sβ), then γs=s(sβ/(1-sβ))<s[1/(s+1)/1-1/(s+1)]=1.
Thus(20)γs<1.
Since γ<1, from (19), we have
(21)dn⟶0asn⟶∞.
Now for m,n∈ℕ with m>n, we have
(22)d(yn,ym)≤sd(yn,yn+1)+s2d(yn+1,yn+2)+⋯+sm-nd(ym-1,ym)≤(sγn+s2γn+1+⋯+sm-nγm-1)d0,from(19)≤sγn(1+sγ+⋯+sm-n-1γm-n-1+⋯)d0=γns1-γsd0→0asn,m→∞.
Hence {yn} is a Cauchy sequence in X. Since X is complete, there exists z∈X such that yn→z which in turn yields fx2n→z and gx2n+1→z as n→∞.
Suppose the mappings(f,g;S) satisfy the condition (W.C.C). Then
(23)d(fx,gy)≤d(y,Sx)∀x,y∈X.
From (23), we have
(24)d(fx2n,gz)≤d(z,Sx2n)≤d(z,gx2n+1).
Letting n→∞, we get
(25)d(z,gz)≤0sothatgz=z.
Now(26)1sd(z,Tz)≤d(z,gx2n+1)+d(gx2n+1,Tz)≤d(z,gx2n+1)+H(Sx2n,Tz)≤d(z,y2n+1)+qmax{d(fx2n,gz),d(fx2n,Sx2n),d(gz,Tz),d(fx2n,Tz),d(gz,Sx2n)}≤d(z,y2n+1)+qmax{d(y2n,z),d(y2n,y2n+1),d(z,Tz),d(y2n,Tz),d(z,y2n+1)}.
Using (vii) of Lemma 4 and letting n→∞ and using (21), we get
(27)1sd(z,Tz)≤0+qsd(z,Tz).
Thus we have
(28)d(z,Tz)≤qs2d(z,Tz),
so that
(29)d(z,Tz)=0sinceqs2<1.
From Lemma 6, we have z∈Tz. Thus
(30)gz=z∈Tz.
From (23), we get
(31)d(fz,z)=d(fz,gz)≤d(z,Sz).
Now from (2.1.1), we have
(32)d(z,Sz)≤H(Sz,Tz)≤qmax{d(fz,gz),d(fz,Sz),d(gz,Tz),d(fz,Tz),d(gz,Sz)}≤qmax{d(z,Sz),s[d(z,Sz)+d(z,Sz)],0,sd(z,Sz),d(z,Sz)}from(30)and(31)=2qsd(z,Sz).
Hencez∈Sz, since 2qs<1.
From (31), we have fz=z. Thus
(33)fz=z∈Sz.
From (30) and (33), it follows that z is a common fixed point of f,g,S, and T.



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