# Common Fixed Point of TwoR-Weakly Commuting Mappings in b-Metric Spaces

Authors: Poom Kumam; Wutiphol Sintunavarat; Shaban Sedghi; Nabi Shobkolaei
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We prove some common fixed point results for two mappings satisfying generalized contractive condition inb-metric space. Note that b-metric of main results in this work are not necessarily continuous. So our results extend and improve several previous works. We also present one example that shows the applicability and usefulness of our results.


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## Body

## 1. Introduction

In 1998, Czerwik [1] introduced the concept of b-metric space. Since then, several papers deal with fixed point theory for single-valued and multivalued operators in b-metric spaces (see also [1–19]). For example, Pacurar [16] proved results on sequences of almost contractions and fixed points in b-metric spaces. Also, Hussain and Shah [11] obtained results on KKM mappings in cone b-metric spaces. Furthermore, Khamsi ([12, 13]) also showed that each cone metric space has a b-metric structure.The aim of this paper is to present some common fixed point results for two mappings under generalized contractive condition inb-metric space, where the b-metric is not necessarily continuous. Because many of the authors in their works have used the b-metric spaces in which the b-metric is continuous, the techniques used in this paper can be used for many of the results on the context of b-metric space. From this point of view the results obtained in this paper generalize and extend several ones obtained earlier in a lot of papers concerning b-metric space.


## 2. Preliminaries

Throughout this paper, we denote byN, R+, and R the sets of positive integers, nonnegative real numbers, and real numbers, respectively.Consistent with [1] and [18, page 264], the following definition and results will be needed in the sequel.Definition 1 (see [1]).
LetX be a nonempty set and let b≥1 be a given real number. A function d:X×X→R+ is a b-metric if and only if, for all x,y,z∈X, the following conditions are satisfied: (b1)
d(x,y)=0 if and only if x=y;(b2)
d(x,y)=d(y,x);(b3)
d(x,z)≤b[d(x,y)+d(y,z)].The pair (X,d) is called a b-metric space with coefficient b.It should be noted that the class ofb-metric spaces is effectively larger than that of metric spaces, since a b-metric is a metric when b=1.We present an example which shows that ab-metric on X need not be a metric on X (see also [18, page 264]).Example 2.
Let(X,d) be a metric space and ρ(x,y)=(d(x,y))p, where p>1 is a real number. We show that ρ is a b-metric with b=2p-1.
Obviously conditions (b1) and (b2) in Definition1 are satisfied. Now we show that condition (b3) holds for ρ.
It is easy to see that if1<p<∞, then the convexity of the function f(x)=xp, where x≥0, implies (1)a+c2p≤12ap+cp,and hence(2)a+cp≤2p-1(ap+cp).Therefore, for each x,y,z∈X, we obtain that (3)ρx,y=dx,yp≤dx,z+dz,yp≤2p-1dx,zp+dz,yp=2p-1(ρ(x,z)+ρ(z,y)).So condition (b3) in Definition 1 holds and then ρ is a b-metric coefficient s=2p-1.It should be noted that in Example2, if (X,d) is a metric space, then (X,ρ) is not necessarily a metric space (see Example 3).Example 3.
For example, ifX=R and the usual Euclidean metric d:X×X→R+ is defined by (4)d(x,y)=x-yfor all x,y∈X, then ρ(x,y)=|x-y|2 is a b-metric on R with b=2 but is not a metric on R, because the triangle inequality does not hold.Before stating and proving our results, we present some definition and proposition inb-metric space. We recall first the notions of convergence, closedness, and completeness in a b-metric space.Definition 4 (see [6]).
Let(X,d) be a b-metric space. Then a sequence {xn} in X is called (i)
convergent if and only if there existsx∈X such that d(xn,x)→0 as n→+∞, and in this case, we write limn→∞xn=x,(ii)
Cauchy if and only ifd(xn,xm)→0 as n,m→+∞.Proposition 5 (Remark  2.1 in [6]).
In ab-metric space (X,d) the following assertions hold: (i)
a convergent sequence has a unique limit;(ii)
each convergent sequence is Cauchy;(iii)
in general, ab-metric is not continuous.Definition 6 (see [6]).
Theb-metric space (X,d) is complete if every Cauchy sequence in X converges.It should be noted that in general ab-metric function d for b>1 is not jointly continuous in all the two of its variables.Since in general ab-metric is not continuous, we need the following simple lemma about the b-convergent sequences.Lemma 7 (see [20]).
Let(X,d) be a b-metric space with coefficient b≥1, and suppose that {xn} and {yn} are b-convergent to x,y, respectively; then one has (5)1b2dx,y≤liminf⁡n→∞d(xn,yn)≤limsup⁡n→∞d(xn,yn)≤b2d(x,y).In particular, if x=y, then one has limn→∞d(xn,yn)=0. Moreover, for each z∈X one has (6)1bd(x,z)≤liminf⁡n→∞d(xn,z)≤limsup⁡n→∞d(xn,z)≤bd(x,z).


## 3. Common Fixed Point Results

Definition 8.
Letf and g be mappings from a b-metric space (X,d) into itself. The mappings f and g are said to be weakly commuting if (7)d(fgx,gfx)≤d(fx,gx)for each x in X.Definition 9.
Letf and g be mappings from a b-metric space (X,d) into itself. The mappings f and g are said to be R-weakly commuting if there exists some positive real number R such that (8)d(fgx,gfx)≤Rd(fx,gx)for each x in X.Remark 10.
Weak commutativity impliesR-weak commutativity in b-metric space. However, R-weak commutativity implies weak commutativity only when R≤1.Example 11.
LetX=R and d:X×X→R+ defined as follows: (9)d(x,y)=x-y2for all x,y∈X. Then (X,d) is a b-metric space. Define fx=2x-1 and gx=x2. Then (10)d(fgx,gfx)=4x-14=4d(fx,gx)>d(fx,gx).Therefore, for R=4, f and g are R-weakly commuting. But f and g are not weakly commuting.Theorem 12.
Let(X,d) be a complete b-metric space and let f and g be R-weakly commuting self-mappings on X satisfying the following conditions: (a)
f(X)⊆g(X);(b)
f or g is continuous;(c)
d(fx,fy)≤γ((1/b4)d(gx,gy)) for all x,y∈X, where γ:[0,∞)→[0,∞) is a continuous and nondecreasing function such that γ(a)<a for each a>0 and γ(0)=0.Then f and g have a unique common fixed point.Proof.
Letx0 be an arbitrary point in X. By (a), choose a point x1 in X such that fx0=gx1. In general choose xn+1 such that fxn=gxn+1 for all n∈N∪{0}. Now we observe that for each n∈N, we have(11)dfxn,fxn+1≤γ1b4dgxn,gxn+1=γ(1b4d(fxn-1,fxn))≤1b4d(fxn-1,fxn)≤d(fxn-1,fxn).This implies that (12){cn}:=dfxn,fxn+1is nonincreasing sequence in [0,∞). Therefore, it tends to a limit a≥0. Next, we claim that a=0. Suppose that a>0. Making n→∞ in the inequality (11), we get (13)a≤γ1b4a<1b4a≤a,which is a contradiction. Hence a=0; that is, (14)limn→∞cn=limn→∞d(fxn,fxn+1)=0.
Now, we prove that{fxn} is a Cauchy sequence in f(X). Suppose that {fxn} is not a Cauchy sequence in f(X). For convenience, let yn=fxn for n=1,2,3,…. Then there is an ε>0 such that, for each integer k, there exist integers m(k) and n(k) with m(k)>n(k)≥k such that(15)dk∶=d(yn(k),ym(k))≥εfork=1,2,….We may assume that(16)dynk,ymk-1<ε,by choosing m(k) as the smallest number exceeding n(k) for which (15) holds. Using (11), we have(17)ε≤dk≤b[d(ynk,ymk-1)+d(ymk-1,ymk)]<bε+bcm(k)-1≤bε+bck.Hence, ε≤limk→∞dk≤bε. Also notice (18)dk=d(yn(k),ym(k))≤bd(ynk,ynk+1)+b2dynk+1,ymk+1+dymk+1,ymk≤bck+b2γ1b4dynk,ymk+b2ck=bck+b2γ(1b4dk)+b2ckfor each k∈N. Thus, as k→∞ in the above inequality we have(19)ε≤limk→∞dk≤blimk→∞ck+b2γ1b4limk→∞dk+b2limk→∞ck,and thus (20)ε≤b2γ1b3ε<1bε,which is a contradiction. Thus, {fxn} is Cauchy sequence in X and by the completeness of X, {fxn} converges to z in X. Also {gxn} converges to z in X. Let us suppose that the mapping f is continuous. Then limn→∞ffxn=fz and limn→∞fgxn=fz. Further we have that since f and g are R-weakly commuting (21)d(fgxn,gfxn)≤Rd(fxn,gxn)for all n∈N. Taking the upper limit as n→∞ in the above inequality, we get (22)1b2dfz,limsup⁡n→∞gfxn≤limsup⁡n→∞d(fgxn,gfxn)≤Rlimsup⁡n→∞d(fxn,gxn)≤Rb2d(z,z)=0.Similarly,(23)1b2d(fz,liminf⁡n→∞gfxn)=0,and hence we get limn→∞gfxn=fz. We now prove that z=fz. Suppose that z≠fz and then d(z,fz)>0. By (c), we have (24)1bdfz,fxm≤limsup⁡n→∞dffxn,fxm≤γlimsup⁡n→∞1b4dgfxn,gxm≤γ1b3dfz,gxmfor each m∈N. On taking the upper limit as m→∞ in the above inequality we get (25)1b2dfz,z≤limsup⁡m→∞1bd(fz,fxm)≤γlimsup⁡m→∞1b3dfz,gxm≤γ(1b2d(fz,z))<1b2d(fz,z),which is a contradiction. Therefore, z=fz. Since f(X)⊆g(X) we can find z1 in X such that z=fz=gz1. Now, we have (26)d(ffxn,fz1)≤γ1b4dgfxn,gz1for all n∈N. Taking limit sup⁡ as n→∞ we get (27)1bdfz,fz1≤limsup⁡n→∞dffxn,fz≤γlimsup⁡n→∞1b4dgfxn,gz1≤γ(1b3d(fz,gz1))=0since γ(0)=0, which implies that fz=fz1; that is, z=fz=fz1=gz1. Also, (28)d(fz,gz)=d(fgz1,gfz1)≤Rd(fz1,gz1)=0which again implies that fz=gz. Thus z is a common fixed point of f and g.
Now to prove uniqueness let if possiblez′≠z be another common fixed point of f and g. Then d(z,z′)>0 and so (29)dz,z′=dfz,fz′≤γ1b4dgz,gz′=γ1b4dz,z′<1b4dz,z′≤dz,z′,which is a contradiction. Therefore, z=z′, that is, z is a unique common fixed point of f and g. This completes the proof.Now we give an example to support Theorem12.Example 13.
LetX=R and d:X×X→R+ defined by (30)d(x,y)=x-y2for all x,y∈X. Then (X,d) is a b-metric space for b=2. Define fx=1 and gx=2x-1 on X. It is evident that f(X)⊆g(X) and f is continuous. Now we observe that (31)d(fx,fy)≤γ116dgx,gyfor all x,y in X and γ:(0,∞)→(0,∞) defined by γ(t)=kt for 0<k<1. Moreover, it is easy to see that f and g are R-weakly commuting. Thus all the conditions of Theorem 12 are satisfied and 1 is a common fixed point of f and g.Corollary 14.
Let(X,d) be a complete b-metric space and let f be a self-mapping on X satisfying the following condition: (32)d(fx,fy)≤γ1b4dx,yfor all x,y∈X, where γ:[0,∞)→[0,∞) is a continuous and nondecreasing function such that γ(a)<a for each a>0 and γ(0)=0. Then f has a unique fixed point.Proof.
If we takeg as identity mapping on X, then Theorem 12 follows that f has a unique fixed point.Corollary 15.
Let(X,d) be a complete metric space and let f and g be R-weakly commuting self-mappings of X satisfying the following conditions: (a)
f(X)⊆g(X);(b)
f or g is continuous;(c)
d(fx,fy)≤γ(d(gx,gy)) for all x,y∈X, where γ:[0,∞)→[0,∞) is a continuous and nondecreasing function such that γ(a)<a for each a>0 and γ(0)=0.Then f and g have a unique common fixed point.Proof.
If we takeb=1, then Theorem 12 follows that f and g have a unique common fixed point.



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