# A Unique Common Triple Fixed Point Theorem for  Hybrid Pair of Maps

Authors: K. P. R. Rao; G. N. V. Kishore; Kenan Tas
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We obtain a unique common triple fixed point theorem for hybrid pair of mappings in metric spaces. Our result extends the recent results of B. Samet and C. Vetro (2011). We also introduced a suitable example supporting our result.


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## Body

## 1. Introduction

The study of fixed points for multivalued contraction mappings using the Hausdorff metric was initiated by Nadler [1].Let(X,d) be a metric space. We denote CB(X) the family of all nonempty closed and bounded subsets of X and CL(X) the set of all nonempty closed subsets of X. For A,B∈CB(X) and x∈X, we denote D(x,A)=inf{d(x,a):a∈A}. Let H be the Hausdorff metric induced by the metric d on X, that is,
(1.1)H(A,B)=max{supx∈Ad(x,B),supy∈Bd(y,A)},
for every A,B∈CB(X).It is clear that forA,B∈CB(X) and a∈A, we have d(a,B)≤H(A,B).Definition 1.1.
An elementx∈X is said to be a fixed point of a set-valued mapping T:X→CB(X) if and only if x∈Tx.In 1969, Nadler [1] extended the famous Banach contraction principle [2] from single-valued mapping to multivalued mapping and proved the following fixed point theorem for the multivalued contraction.Theorem 1.2 (see, Nadler [1]).
Let(X,d) be a complete metric space and let T be a mapping from X into CB(X). Assume that there exists c∈[0,1) such that
(1.2)H(Tx,Ty)≤cd(x,y),
for all x,y∈X. Then, T has a fixed point.Lemma 1.3   (see, Nadler [1]).
LetA,B∈CB(X) and α>1. Then for every a∈A, there exists b∈B such that d(a,b)≤αH(A,B).Lemma 1.4   (see, Nadler [1]).
Letα>0. If A,B∈CB(X) with H(A,B)≤α, then for each a∈A, there exists b∈B such that d(a,b)≤α.Lemma 1.5   (see, Nadler [1]).
Let{An} be a sequence in CB(X) with limn→+∞H(An,A)=0, for A∈CB(X). If xn∈An and limn→+∞d(xn,x)=0, then x∈A.The existence of fixed points for various multivalued contractive mappings has been studied by many authors under different conditions. For details, we refer the reader to [1, 3–11] and the references therein.The concept of coupled fixed point for multivalued mapping was introduced by Samet and Vetro [12], and later several authors, namely, Hussain and Alotaibi [13], Aydi et al. [14], and Abbas et al. [15], proved coupled coincidence point theorems in partially ordered metric spaces.Definition 1.6   (see, Samet and Vetro [12]).
LetF:X×X→CL(X) be a given mapping. We say that (x,y)∈X×X is a coupled fixed point of F if and only if(1.3)x∈F(x,y),y∈F(y,x).Definition 1.7   (see, Hussain and Alotaibi [13]).
Let the mappingsF:X×X→CB(X) and g:X→X be given. An element (x,y)∈X×X is called(1)
a coupled coincidence point of a pair{F,g} if gx∈F(x,y) and gy∈F(y,x);(2)
a coupled common fixed point of a pair{F,g} if x=gx∈F(x,y) and y=gy∈F(y,x).
Berinde and Borcut [16] introduced the concept of triple fixed points and obtained a tripled fixed point theorem for single valued map.
Now we give the following.Definition 1.8.
LetX be a nonempty set, T:X×X×X→2X (collection of all nonempty subsets of X). f:X→X.(i)
The point(x,y,z)∈X×X×X is called a tripled fixed point of T if
(1.4)x∈T(x,y,z),y∈T(y,x,y),z∈T(z,y,x).(ii)
The point(x,y,z)∈X×X×X is called a tripled coincident point of T and f if
(1.5)fx∈T(x,y,z),fy∈T(y,x,y),fz∈T(z,y,x).(iii)
The point(x,y,z)∈X×X×X is called a tripled common fixed point of T and f if
(1.6)x=fx∈T(x,y,z),y=fy∈T(y,x,y),z=fz∈T(z,y,x).Definition 1.9.
LetT:X×X×X→2X be a multivalued map and f be a self map on X. The Hybrid pair {T,f} is called w-compatible if f(T(x,y,z))⊆T(fx,fy,fz) whenever (x,y,z) is a tripled coincidence point of T and f.


## 2. Main Results

Theorem 2.1.
Let(X,d) be a metric space and let T:X×X×X→CB(X) and f:X→X mappings satisfying  (2.1.1)
H(T(x,y,z),T(u,v,w))≤jd(fx,fy)+kd(fy,fv)+ld(fz,fw),  for all x,y,z,u,v,w∈X and j,k,l∈[0,1) with j+k+l≤h<1, where h is a fixed number,(2.1.2)
T(X×X×X)⊆f(X) and f(X) is a complete subspace of X.
Then the mapsT and f have a tripled coincidence point.
Further,T and f have a tripled common fixed point if one of the following conditions holds.(2.1.3)
(a){T,f} is w-compatible, there exist u,v,w∈X such that limn→∞fnx=u,limn→∞fny=v and limn→∞fnz=w, whenever (x,y,z) is a tripled coincidence point of {T,f} and f is continuous at u,v,w.(b)  There existu,v,w∈X such that limn→∞fnu=x,  limn→∞fnv=y and limn→∞fnw=z whenever (x,y,z) is a tripled coincidence point of {T,f} and f is continuous at x,y, and z.Proof.
Letx0,y0,z0∈X. From (2.1.2), there exist sequences {xn}, {yn}, and {zn} in X such that fxn+1∈T(xn,yn,zn), fyn+1∈T(yn,xn,yn) and fzn+1∈T(zn,yn,xn), n=0,1,2,3,….
For simplification, denote(2.1)dnx=d(fxn-1,fxn),dny=d(fyn-1,fyn),dnz=d(fzn-1,fzn).
From (2.1.1), we obtain
(i)d2x=d(fx1,fx2)≤H(T(x0,y0,z0),T(x1,y1,z1))+h≤jd(fx0,fx1)+kd(fy0,fy1)+ld(fz0,fz1)+h=jd1x+kd1y+ld1z+h,(ii)d2y=d(fy1,fy2)≤H(T(y0,x0,y0),T(y1,x1,y1))+h≤jd(fy0,fy1)+kd(fx0,fx1)+ld(fy0,fy1)+h=kd1x+(j+l)d1y+h,(iii)d2z=d(fz1,fz2)≤H(T(z0,y0,x0),T(z1,y1,x1))+h≤jd(fz0,fz1)+kd(fy0,fy1)+ld(fx0,fx1)+h=ld1x+kd1y+jd1z+h,(iv)d3x=d(fx2,fx3)≤H(T(x1,y1,z1),T(x2,y2,z2))+h2≤jd(fx1,fx2)+kd(fy1,fy2)+ld(fz1,fz2)+h2=jd2x+kd2y+ld2z+h2≤j(jd1x+kd1y+ld1z+h)+k(kd1x+(j+l)d1y+h)+l(ld1x+kd1y+jd1z+h)+h2=(j2+k2+l2)d1x+(2jk+2lk)d1y+(2jl)d1z+h2+(j+k+l)h=(j2+k2+l2)d1x+(2jk+2lk)d1y+(2jl)d1z+2h2,(v)d3y=d(fy2,fy3)≤H(T(y1,x1,y1),T(y2,x2,y2))+h2≤jd(fy1,fy2)+kd(fx1,fx2)+ld(fy1,fy2)+h2=kd2x+(j+l)d2y+h2≤k(jd1x+kd1y+ld1z+h)+(j+l)(kd1x+(j+l)d1y+h)+h2=(2jk+lk)d1x+[(j+l)2+k2]d1y+kld1z+(j+k+l)h+h2≤(2jk+lk)d1x+[(j+l)2+k2]d1y+kld1z+2h2,(vi)d3z=d(fz2,fz3)≤H(T(z1,y1,x1),T(z2,y2,x2))+h2≤jd(fz1,fz2)+kd(fy1,fy2)+ld(fx1,fx2)+h2=jd2z+kd2y+ld2x+h2=ld2x+kd2y+jd2z+h2≤l(jd1x+kd1y+ld1z+h)+k(kd1x+(j+l)d1y+h)+j(ld1x+kd1y+jd1z+h)+h2=(2jl+k2)d1x+2[jk+lk]d1y+(j2+l2)d1z+(j+k+l)h+h2≤(2jl+k2)d1x+2[jk+lk]d1y+(j2+l2)d1z+2h2.
Let A=[jklkj+l0lkj]  denoted by  [a1b1c1d1e1f1g1b1h1].
Clearly,a1+b1+c1=d1+e1+f1=g1+b1+h1=(j+k+l)≤h<1.
Then,(2.2)A2=[j2+k2+l22jk+2lk2jl2jk+lk(j+l)2+k2kl2jl+k22jk+2lkj2+l2]denoteA2by[a2b2c2d2e2f2g2b2h2].
It is clear that a2+b2+c2=d2+e2+f2=g2+b2+h2=(j+k+l)2≤h2<1.
Now we prove by induction that(2.3)An=[anbncndnenfngnbnhn],
where
(2.4)an+bn+cn=dn+en+fn=gn+bn+hn=(j+k+l)n≤hn<1.
Equation (2.3) is true for n=1,2.
Assume that (2.3) is true for some n. Consider
(2.5)An+1=An·A=[anbncndnenfngnbnhn][jklkj+l0lkj]=[jan+kbn+lcnkan+(j+l)bn+kcnlan+jcnjdn+ken+lfnkdn+(j+l)en+kfnldn+jfnjgn+kbn+lhnkgn+(j+l)bn+khnlgn+jhn].
We have
(2.6)an+1+bn+1+cn+1=(j+k+l)(an+bn+cn)=(j+k+l)n+1≤hn+1<1.
Similarly, we have
(2.7)dn+1+en+1+fn+1=gn+1+bn+1+hn+1=(j+k+l)n+1≤hn+1<1.
Thus (2.3) is true for all +ve integer values of n.
Now from (i)–(vi) and continuing this process, we get(2.8)[dn+1xdn+1ydn+1z]≤[anbncndnenfngnbnhn][d1xd1yd1z]+[nhnnhnnhn],
for all n=1,2,3,…. That is,
(2.9)dn+1x≤and1x+bnd1y+cnd1z+nhn,dn+1y≤dnd1x+end1y+fnd1z+nhn,dn+1z≤gnd1x+bnd1y+hnd1z+nhn,∀n=1,2,3,….For m>n, we have(2.10)d(fxm,fxn)≤d(fxm,fxm-1)+d(fxm-1,fxm-2)+⋯+d(fxn+2,fxn+1)+d(fxn+1,fxn)=dmx+dm-1x+⋯+dn+2x+dn+1x≤am-1d1x+bm-1d1y+cm-1d1z+(m-1)hm-1+am-2d1x+bm-2d1y+cm-2d1z+(m-2)hm-2+⋯+an+1d1x+bn+1d1y+cn+1d1z+(n+1)hn+1+and1x+bnd1y+cnd1z+nhn≤(am-1+am-2+⋯+an+1+an)d1x+(bm-1+bm-2+⋯+bn+1+bn)d1y+(cm-1+cm-2+⋯+cn+1+cn)d1z+[(m-1)hm-1+(m-2)hm-2+⋯+(n+1)hn+1+nhn]≤(hm-1+hm-2+⋯+hn+1+hn)(d1x+d1y+d1z)+∑j=nm-1‍jhj≤hn1-h(d1x+d1y+d1z)+∑j=nm-1‍jhj→0asn→∞,because0≤h<1.
Hence {fxn} is a Cauchy. Similarly, we can show that {fyn} and {fzn} are Cauchy.
Supposef(X) is complete, the sequences {fxn},  {fyn}, and {fzn} are convergent to some α,β,γ in f(X), respectively. There exist x,y,z∈X such that α=fx,  β=fy, and γ=fz.
Now, we have(2.11)d(T(x,y,z),α)≤d(T(x,y,z),fxn+1)+d(fxn+1,α)≤H(T(x,y,z),T(xn,yn,zn))+d(fxn+1,α)≤jd(fx,fxn)+kd(fy,fyn)+ld(fz,fzn)+d(fxn+1,α)=jd(α,fxn)+kd(β,fyn)+ld(γ,fzn)+d(fxn+1,α).
Letting n→∞, we get d(T(x,y,z),α)≤0 so that α∈T(x,y,z). That is, fx∈T(x,y,z). Similarly, we can show that fy∈T(y,x,y) and fz∈T(z,y,x). Thus (x,y,z) is a tripled coincidence point of T and f. Suppose (2.1.3) (a) holds.
Since(x,y,z) is a tripled coincidence point of T and f, there exist u,v,w∈X such that limn→∞fnx=u, limn→∞fny=v and limn→∞fnz=w.
Sincef is continuous at u,v and w, we have fu=u,  fv=v and fw=w.
Sincefx∈T(x,y,z), we have f2x∈f(T(x,y,z))⊆T(fx,fy,fz).
Sincefy∈T(y,x,y), we have f2y∈f(T(y,x,y))⊆T(fy,fx,fy).
Sincefz∈T(z,y,x), we have f2z∈f(T(z,y,x))⊆T(fz,fy,fx).
Then(fx,fy,fz) is tripled coincidence point of T and f.
Similarly, we can show that(fnx,fny,fnz) is a tripled coincidence point of T and f.
Also it is clear that(2.12)fnx∈T(fn-1x,fn-1y,fn-1z),fny∈T(fn-1y,fn-1x,fn-1y),fnz∈T(fn-1z,fn-1y,fn-1x).
From (2.1.1), we have
(2.13)d(fu,T(u,v,w))≤d(fu,fnx)+d(fnx,T(u,v,w))≤d(fu,fnx)+H(T(fn-1x,fn-1y,fn-1z),T(u,v,w))≤d(fu,fnx)+jd(fnx,fu)+kd(fny,fv)+ld(fnz,fw).
Letting n→∞, we obtain
(2.14)d(fu,T(u,v,w))≤0,
which implies that
(2.15)fu∈T(u,v,w).
Thus u=fu∈T(u,v,w). Similarly, we can show that v=fv∈T(v,u,v) and w=fw∈T(w,v,u). Thus (u,v,w) is a tripled common fixed point of T and f. Suppose (2.1.3) (b) holds.
Since(x,y,z) is a tripled coincidence point of {T,f}, there exist u,v,w∈X such that limn→∞fnu=x, limn→∞fnv=y and limn→∞fnw=z.
Sincef is continuous at x,y and z, we have fx=x,  fy=y and fz=z.  Thus x=fx∈T(x,y,z), y=fy∈T(y,x,y) and z=fz∈T(z,y,x). Hence (x,y,z) is a tripled common fixed point of {T,f}.
The following example illustrates Theorem2.1.Example 2.2.
LetX=[0,1],T:X×X×X→CB(X) and f:X→X defined as T(x,y,z)=[0,(1/8)sinx+(1/4)siny+(1/3)sinz] and fx=(7/8)x. Then
(2.16)H(T(x,y,z),T(u,v,w))=|(18sinx+14siny+13sinz)-(18sinu+14sinv+13sinw)|≤18|sinx-sinu|+14|siny-sinv|+13|sinz-sinw|≤18|x-u|+14|y-v|+13|z-w|=17|78x-78u|+27|78y-78v|+821|78z-78w|=17d(fx,fu)+27d(fy,fv)+821d(fz,fw).
It is clear that all conditions of Theorem 2.1 are satisfied and (0,0,0) is the tripled common fixed point of T and f.
The following example shows thatT and f have no tripled common fixed point if (2.1.3) (a) or (2.1.3) (b) is not satisfied.Example 2.3.
LetX=[0,4], T(x,y,z)=[1.5,2] and fx=2-(1/2)x. Then (0,1/2,1) is a tripled coincidence point of T and f. Clearly T and f have no tripled common fixed point.



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*Source: 750403-2012-10-02.xml*