We prove the existence of common fixed points for three relatively asymptotically regular mappings defined on an orbitally complete ordered metric space using orbital continuity of one of the involved maps. We furnish a suitable example to demonstrate the validity of the hypotheses of our results.

1. Introduction and Preliminaries

Browder and Petryshyn introduced the concept of asymptotic regularity of a self-map at a point in a metric space.

Definition 1.1 (see [<xref ref-type="bibr" rid="B6">1</xref>]).

A self-map 𝒯 on a metric space (𝒳,d) is said to be asymptotically regular at a point x𝒳 if limnd(𝒯nx,𝒯n+1x)=0.

Recall that the set 𝒪(x0;𝒯)={𝒯nx0:n=0,1,2,} is called the orbit of the self-map 𝒯 at the point x0𝒳.

Definition 1.2 (see [<xref ref-type="bibr" rid="B8">2</xref>]).

A metric space (𝒳,d) is said to be 𝒯-orbitally complete if every Cauchy sequence contained in 𝒪(x;𝒯) (for some x in 𝒳) converges in 𝒳.

Here, it can be pointed out that every complete metric space is 𝒯-orbitally complete for any 𝒯, but a 𝒯-orbitally complete metric space need not be complete.

Definition 1.3 (see [<xref ref-type="bibr" rid="B6">1</xref>]).

A self-map 𝒯 defined on a metric space (𝒳,d) is said to be orbitally continuous at a point z in 𝒳 if for any sequence {xn}𝒪(x;𝒯) (for some x𝒳), xnz as n implies 𝒯xn𝒯z as n.

Clearly, every continuous self-mapping of a metric space is orbitally continuous, but not conversely.

Sastry et al.  extended the above concepts to two and three mappings and employed them to prove common fixed point results for commuting mappings. In what follows, we collect such definitions for three maps.

Definition 1.4 (see [<xref ref-type="bibr" rid="B25">3</xref>]).

Let 𝒮,𝒯, be three self-mappings defined on a metric space (𝒳,d).

If for a point x0𝒳, there exists a sequence {xn} in 𝒳 such that x2n+1=𝒮x2n, x2n+2=𝒯x2n+1, n=0,1,2,, then the set 𝒪(x0;𝒮,𝒯,)={xn:n=1,2,} is called the orbit of (𝒮,𝒯,) at x0.

The space (𝒳,d) is said to be (𝒮,𝒯,)-orbitally complete at x0 if every Cauchy sequence in 𝒪(x0;𝒮,𝒯,) converges in 𝒳.

The map is said to be orbitally continuous at x0 if it is continuous on 𝒪(x0;𝒮,𝒯,).

The pair (𝒮,𝒯) is said to be asymptotically regular (in short a.r.) with respect to at x0 if there exists a sequence {xn} in 𝒳 such that x2n+1=𝒮x2n, x2n+2=𝒯x2n+1, n=0,1,2,, and d(xn,xn+1)0 as n.

On the other side, Khan et al.  introduced the notion of an altering distance function, which is a control function that alters distance between two points in a metric space. This notion has been used by several authors to establish fixed point results in a number of subsequent works, some of which are noted in . In , Choudhury introduced the concept of a generalized altering distance function in three variables which was further generalized by Rao et al.  to four variables and is defined as follows.

Definition 1.5 (see [<xref ref-type="bibr" rid="B24">10</xref>]).

A function ψ:[0,+)4[0,+) is said to be a generalized altering distance function if

ψ is continuous,

ψ is nondecreasing in each variable,

ψ(t1,t2,t3,t4)=0t1=t2=t3=t4=0.

4 will denote the set of all functions ψ satisfying conditions (i)–(iii).

Simple examples of generalized altering distance functions with four variables areψ(t1,t2,t3,t4)=kmax{t1,t2,t3,t4},k>0;ψ(t1,t2,t3,t4)=max{t1,t2,t3,t4}1+max{t1,t2,t3,t4};ψ(t1,t2,t3,t4)=t1p+t2q+t3r+t4s,p,q,r,s1.

On the other hand, fixed point theory has developed rapidly in metric spaces endowed with a partial ordering. The first result in this direction was given by Ran and Reurings  who presented its applications to matrix equations. Subsequently, Nieto and Rodríguez-López  extended this result for nondecreasing mappings and applied it to obtain a unique solution for a first order ordinary differential equation with periodic boundary conditions. Thereafter, several authors obtained many fixed point theorems in ordered metric spaces. For more details see  and the references cited therein.

In this paper, an attempt has been made to derive some common fixed point theorems for three relatively asymptotically regular mappings defined on an orbitally complete ordered metric space, using orbital continuity of one of the involved maps and conditions involving a generalized altering distance function. The presented theorems generalize, extend, and improve some recent results given in [7, 14, 21, 22]. In the hypotheses, we have considered the space as not necessarily complete, the maps ,𝒮, and 𝒯 as not necessarily continuous and the range of 𝒮 and 𝒯 may not be contained in the range of .

2. Results2.1. Notations and Definitions

First, we introduce some further notations and definitions that will be used later.

If (𝒳,) is a partially ordered set, then x,y𝒳 are called comparable if xy or yx holds. A subset 𝒦 of 𝒳 is said to be well ordered if every two elements of 𝒦 are comparable. If 𝒯:𝒳𝒳 is such that, for x,y𝒳, xy implies 𝒯x𝒯y, then the mapping 𝒯 is said to be nondecreasing.

Definition 2.1.

Let (𝒳,) be a partially ordered set and 𝒮,𝒯:𝒳𝒳.

 The pair (𝒮,𝒯) is called weakly increasing if 𝒮x𝒯𝒮x and 𝒯x𝒮𝒯x for all x𝒳.

 The pair (𝒮,𝒯) is called partially weakly increasing if 𝒮x𝒯𝒮x for all x𝒳.

 The mapping 𝒮 is called a weak annihilator of 𝒯 if 𝒮𝒯xx for all x𝒳.

 The mapping 𝒮 is called dominating if x𝒮x for each x𝒳.

Note that none of two weakly increasing mappings need to be nondecreasing. There exist some examples to illustrate this fact in . Obviously, the pair (𝒮,𝒯) is weakly increasing if and only if the ordered pairs (𝒮,𝒯) and (𝒯,𝒮) are partially weakly increasing. Following is an example of an ordered pair (𝒮,𝒯) which is partially weakly increasing but not weakly increasing.

Example 2.2 (see [<xref ref-type="bibr" rid="B1">24</xref>]).

Let 𝒳=[0,1] be endowed with usual ordering.

Let 𝒮,𝒯:𝒳𝒳 be defined by 𝒮x=x2 and 𝒯x=x. Clearly, (𝒮,𝒯) is partially weakly increasing. But 𝒯x=xx=𝒮𝒯x for x(0,1) implies that (𝒯,𝒮) is not partially weakly increasing.

Let 𝒮,𝒯:𝒳𝒳 be defined by 𝒮x=x2 and 𝒯x=x3. Obviously, 𝒮𝒯x=x6x for all x𝒳. Thus 𝒮 is a weak annihilator of 𝒯.

Let 𝒮:𝒳𝒳 be defined by 𝒮x=xn. Since xxn=𝒮x for all x𝒳, 𝒮 is a dominating map.

Definition 2.3 (see [<xref ref-type="bibr" rid="B11">25</xref>, <xref ref-type="bibr" rid="B12">26</xref>]).

Let (𝒳,d) be a metric space and f,g:𝒳𝒳. The mappings f and g are said to be compatible if limnd(fgxn,gfxn)=0, whenever {xn} is a sequence in 𝒳 such that limnfxn=limngxn=t for some t𝒳.

Definition 2.4.

Let 𝒳 be a nonempty set. Then (𝒳,d,) is called an ordered metric space if

(𝒳,d) is a metric space,

(𝒳,) is a partially ordered set.

2.2. Main Results

The first main result is as follows.

Theorem 2.5.

Let (𝒳,d,) be an ordered metric space. Let 𝒮,𝒯,:𝒳𝒳 be given mappings satisfying Ψ1(d(Sx,Ty))ψ1(M[S,T,R](x,y))-ψ2(M[S,T,R](x,y)), for all x,y𝒪(x0;𝒮,𝒯,)¯ (for some x0) such that x and y are comparable, where M[S,T,R](x,y)=(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Sx)]), and  ψ1 and ψ2 are generalized altering distance functions (in 4) and Ψ1(t)=ψ1(t,t,t,t). We assume the following hypotheses:

(𝒮,𝒯) is a.r. with respect to at x0𝒳;

𝒳 is (𝒮,𝒯,)-orbitally complete at x0;

(,𝒮) and (,𝒯) are partially weakly increasing;

𝒮 and 𝒯 are dominating maps;

𝒮 and 𝒯 are weak annihilators of ;

for a nondecreasing sequence {xn}, xnyn for all n and ynu as n imply that xnu for all n.

Assume either

𝒮 and are compatible; 𝒮 or is orbitally continuous at x0 or

𝒯 and are compatible; 𝒯 or is orbitally continuous at x0.

Then 𝒮,𝒯 and have a common fixed point. Moreover, the set of common fixed points of 𝒮,𝒯, and in 𝒪(x0;𝒮,𝒯,)¯ is well ordered if and only if it is a singleton.

Proof.

Since (𝒮,𝒯) is a.r. with respect to at x0 in 𝒳, there exists a sequence {xn} in 𝒳 such that Sx2n-2=Rx2n-1,Tx2n-1=Rx2n,nN. By the given assumptions, x2n-2𝒮x2n-2=x2n-1𝒮x2n-1x2n-1, and x2n-1𝒯x2n-1=x2n𝒯x2nx2n. Thus, for all n1, we have xnxn+1. In view of (i), we have limnd(Rxn,Rxn+1)=0. Now, we assert that {xn} is a Cauchy sequence in the metric space 𝒪(x0;𝒮,𝒯,).

From (2.5), it will be sufficient to prove that {x2n} is a Cauchy sequence. We proceed by negation and suppose that {x2n} is not a Cauchy sequence. Then, there exists ε>0 for which we can find two sequences of positive integers {m(i)} and {n(i)} such that for all positive integers i, n(i)>m(i)>i,d(Rx2m(i),Rx2n(i))ε,d(Rx2m(i),Rx2n(i)-2)<ε. From (2.6) and using the triangular inequality, we get εd(Rx2m(i),Rx2n(i))d(Rx2m(i),Rx2n(i)-2)+d(Rx2n(i)-2,Rx2n(i)-1)+d(Rx2n(i)-1,Rx2n(i))<ε+d(Rx2n(i)-2,Rx2n(i)-1)+d(Rx2n(i)-1,Rx2n(i)). Letting i in the above inequality and using (2.5), we obtain limid(Rx2m(i),Rx2n(i))=ε. Again, the triangular inequality gives us |d(Rx2n(i),Rx2m(i)-1)-d(Rx2n(i),Rx2m(i))|d(Rx2m(i)-1,Rx2m(i)). Letting i in the above inequality and using (2.5) and (2.8), we get limid(Rx2n(i),Rx2m(i)-1)=ε. Similarly, we have limid(Rx2n(i)+1,Rx2m(i)-1)=ε. On the other hand, we have d(Rx2n(i),Rx2m(i))d(Rx2n(i),Rx2n(i)+1)+d(Rx2n(i)+1,Rx2m(i))=d(Rx2n(i),Rx2n(i)+1)+d(Tx2n(i),Sx2m(i)-1). Then, from (2.5), (2.8), and the continuity of Ψ1, we get by letting i in (2.1) Ψ1(ε)limiΨ1(d(Sx2m(i)-1,Tx2n(i))). Now, using the considered contractive condition (2.1) for x=x2m(i)-1 and y=x2n(i), we have Ψ1(d(Sx2m(i)-1,Tx2n(i)))ψ1(12d(Rx2m(i)-1,Rx2n(i)),d(Rx2m(i)-1,Rx2m(i)),d(Rx2n(i),Rx2n(i)+1),12[d(Rx2m(i)-1,Rx2n(i)+1)+d(Rx2n(i),Rx2m(i))])-ψ2(12d(Rx2m(i)-1,Rx2n(i)),d(Rx2m(i)-1,Rx2m(i)),d(Rx2n(i),Rx2n(i)+1),12[d(Rx2m(i)-1,Rx2n(i)+1)+d(Rx2n(i),Rx2m(i))]). Then, from (2.5), (2.10), (2.11), and the continuity of ψ1 and ψ2, we get by letting i in the above inequality limiΨ1(d(Sx2m(i)-1,Tx2n(i)))ψ1(ε,0,0,ϵ)-ψ2(ε,0,0,ϵ)Ψ1(ε)-ψ2(ε,0,0,ϵ). Now, combining (2.13) with the above inequality, we get Ψ1(ε)Ψ1(ε)-ψ2(ε,0,0,ϵ), which implies that ψ2(ε,0,0,ϵ)=0, which is a contradiction since ε>0. Hence {xn} is a Cauchy sequence in 𝒪(x0;𝒮,𝒯,). Since 𝒳 is (𝒮,𝒯,)-orbitally complete at x0, there exists some z𝒳 such that xnz as n.

Finally, we prove the existence of a common fixed point of the three mappings 𝒮,𝒯, and .

We have Rx2n+1=Sx2nzas  n,Rx2n+2=Tx2n+1zas  n. Suppose that (a) holds. Since {𝒮,} are compatible, we have limnSRx2n+2=limnRSx2n+2=Rz. Also, x2n+1𝒯x2n+1=x2n+2. Now Ψ1(d(SRx2n+2,Tx2n+1))ψ1(12d(RRx2n+2,Rx2n+1),d(RRx2n+2,SRx2n+2),d(Rx2n+1,Tx2n+1),12[d(RRx2n+2,Tx2n+1)+d(SRx2n+2,Rx2n+1)])-ψ2(12d(RRx2n+2,Rx2n+1),d(RRx2n+2,SRx2n+2),d(Rx2n+1,Tx2n+1),  12[d(RRx2n+2,Tx2n+1)+d(SRx2n+2,Rx2n+1)]). Assume that is orbitally continuous. Passing to the limit as n, we obtain Ψ1(d(Rz,z))ψ1(d(Rz,z),0,0,d(Rz,z))-ψ2(d(Rz,z),0,0,d(Rz,z))Ψ1(d(Rz,z))-ψ2(d(Rz,z),0,0,d(Rz,z)), so ψ2(d(z,z),0,0,d(z,z))=0, which implies that Rz=z. Now, x2n+1𝒯x2n+1 and 𝒯x2n+1z as n, so by the assumption we have x2n+1z and (2.1) becomes Ψ1(d(Sz,Tx2n+1))ψ1(12d(Rz,Rx2n+1),d(Sz,Rz),d(Tx2n+1,Rx2n+1),12[d(Rz,Tx2n+1)+d(Sz,Rx2n+1)])-ψ2(12d(Rz,Rx2n+1),d(Sz,Rz),d(Tx2n+1,Rx2n+1),12[d(Rz,Tx2n+1)+d(Sz,Rx2n+1)]). Passing to the limit as n in the above inequality and using (2.21), it follows that Ψ1(d(Sz,z))ψ1(0,d(Sz,z),0,12d(Sz,z))-ψ2(0,d(Sz,z),0,12d(Sz,z))Ψ1(d(Sz,z))-ψ2(0,d(Sz,z),0,12d(Sz,z)), which holds unless ψ2(0,d(𝒮z,z),0,(1/2)d(𝒮z,z))=0, so Sz=z. Now, since x2n𝒮x2n and 𝒮x2nz as n implies that x2nz, from (2.1) Ψ1(d(Sx2n,Tz))ψ1(d(Rx2n,Rz),d(Rx2n,Sx2n),d(Rz,Tz),12(d(Rx2n,Tz)+d(Sx2n,Rz)))-ψ2(d(Rx2n,Rz),d(Rx2n,Sx2n),d(Rz,Tz),12(d(Rx2n,Tz)+d(Sx2n,Rz))). Passing to the limit as n, we have Ψ1(d(z,Tz))ψ1(0,0,d(z,Tz),d(z,Tz))-ψ2(d(z,Tz),0,0,d(z,Tz))Ψ1(d(z,Tz))-ψ2(0,0,d(z,Tz),d(z,Tz)), which gives that z=Tz. Therefore, 𝒮z=𝒯z=z=z, hence z is a common fixed point of ,𝒮, and 𝒯. The proof is similar when 𝒮 is orbitally continuous.

Similarly, the result follows when condition (b) holds.

Now, suppose that the set of common fixed points of 𝒮,𝒯, and in 𝒪(x0;𝒮,𝒯,)¯ is well ordered. We claim that it cannot contain more than one point. Assume to the contrary that 𝒮u=𝒯u=u=u and 𝒮v=𝒯v=v=v but uv. By supposition, we can replace x by u and y by v in (2.1) to obtain Ψ1(d(u,v))=Ψ1(d(Su,Tv))ψ1(d(Ru,Rv),d(Ru,Su),d(Rv,Tv),12[d(Ru,Tv)+d(Su,Rv)])-ψ2(d(Ru,Rv),d(Ru,Su),d(Rv,Tv),12[d(Ru,Tv)+d(Su,Rv)])=ψ1(d(u,v),0,0,d(u,v))-ψ2(d(u,v),0,0,d(u,v))<Ψ1(d(u,v)), a contradiction. Hence, u=v. The converse is trivial.

Now, it is easy to state a corollary of Theorem 2.5 involving a contraction of integral type.

Corollary 2.6.

Let 𝒮,𝒯, and satisfy the conditions of Theorem 2.5, except that condition (2.1) is replaced by the following: there exists a positive Lebesgue integrable function u on + such that 0εu(t)dt>0 for each ε>0 and that 0Ψ1(d(Sx,Ty))u(t)dt0ψ1(M[S,T,R](x,y))u(t)dt-0ψ2(M[S,T,R](x,y))u(t)dt. Then, 𝒮,𝒯, and have a common fixed point. Moreover, the set of common fixed points of 𝒮,𝒯, and in 𝒪(x0;𝒮,𝒯,)¯ is well ordered if and only if it is a singleton.

Remark 2.7.

If we take ψ1(t1,t2,t3,t4)=max{t1,t2,t3,t4},ψ2(t1,t2,t3,t4)=(1-k)max{t1,t2,t3,t4}, for k(0,1), then Ψ1(t)=t for all t0, and the contractive condition (2.1) becomes d(Sx,Ty)kmax{d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Sx)]}, which corresponds to the contraction given by Theorem  2.1 in  by taking ψ(t)=t and φ(t)=(1-k)t. Hence, the result of Abbas et al.  is covered by Theorem 2.5 for three maps.

Other results could be derived for other choices of ψ1 and ψ2.

As consequences of Theorem 2.5, we may state the following corollaries.

Corollary 2.8.

Let (𝒳,d,) be an ordered metric space. Let 𝒯,:𝒳𝒳 be given mappings satisfying Ψ1(d(Tx,Ty))ψ1(d(Rx,Ry),d(Rx,Tx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Tx)])-ψ2(d(Rx,Ry),d(Rx,Tx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Tx)]), for all x,y𝒪(x0;𝒯,𝒯,)¯ such that x and y are comparable, where ψ1 and ψ2 are generalized altering distance functions (in 4) and Ψ1(t)=ψ1(t,t,t,t). We assume the following hypotheses:

𝒯 is a.r. with respect to at x0;

𝒳 is (𝒯,)-orbitally complete at x0;

𝒯 or is orbitally continuous at x0;

(𝒯,) is partially weakly increasing;

𝒯 is a dominating map;

𝒯 is a weak annihilator of ;

𝒯 and are compatible.

Let for a nondecreasing sequence {xn} with xnyn  for all n, ynu as n imply that xnu for all n.

Then 𝒯 and have a common fixed point. Moreover, the set of common fixed points of 𝒯 and in 𝒪(x0;𝒯,𝒯,)¯ is well ordered if and only if it is a singleton.

Corollary 2.9.

Let (𝒳,d,) be an ordered metric space. Let 𝒯:𝒳𝒳 be a mapping satisfying Ψ1(d(Tx,Ty))ψ1(d(x,y),d(x,Tx),d(y,Ty),12[d(x,Ty)+d(y,Tx)])-ψ2(d(x,y),d(x,Tx),d(y,Ty),12[d(x,Ty)+d(y,Tx)]), for all (x,y)𝒪(x0;𝒯)¯ such that x and y are comparable, where ψ1 and ψ2 are generalized altering distance functions (in 4) and Ψ1(t)=ψ1(t,t,t,t). We assume the following hypotheses:

𝒯 is a.r. at some point x0 of 𝒳;

𝒳 is 𝒯-orbitally complete at x0;

𝒯 is a dominating map.

Let for a nondecreasing sequence {xn} with xnyn  for all n, ynu as n imply that xnu for all n.

Then 𝒯 has a fixed point. Moreover, the set of fixed points of 𝒯 in 𝒪(x0;𝒯)¯ is well ordered if and only if it is a singleton.

We present an example showing the usage of our results.

Example 2.10.

Let the set 𝒳=[0,+) be equipped with the usual metric d and the order defined by xy  xy. Consider the following self-mappings on 𝒳: Rx=6x,Sx={12x,0x12,x,x>12,Tx={13x,0x13,x,x>13. Take x0=1/2. Then it is easy to show that O(x0;S,T,R){12k3l:k,lN},O(x0;S,T,R)¯=O(x0;S,T,R){0} and all the conditions (i)–(vi) and (a)-(b) of Theorem 2.5 are fulfilled. Take ψ1(t1,t2,t3,t4)=max{t1,t2,t3,t4} and ψ2(t1,t2,t3,t4)=(5/6)max{t1,t2,t3,t4}. Then contractive condition (2.1) takes the form |12x-13y|16max{|6x-6y|,112x,173y,12[|6x-13y|+|6y-12x|]}, for x,y𝒪(x0;𝒮,𝒯,). Using substitution y=tx, t>0, the last inequality reduces to |3-2t|max{6|1-t|,112,173t,12[|6-13t|+|6t-12|]}, and can be checked by discussion on possible values for t>0. Hence, all the conditions of Theorem 2.5 are satisfied and 𝒮,𝒯, have a common fixed point (which is 0).

Remark 2.11.

It was shown by examples in  that (in similar situations)

if the contractive condition is satisfied just on 𝒪(x0;𝒮,𝒯,), there might not exist a (common) fixed point;

under the given hypotheses (common), fixed point might not be unique in the whole space 𝒳.

Acknowledgment

The authors thank the referees for their careful reading of the text and for suggestions that helped to improve the exposition of the paper. The second and third authors are thankful to the Ministry of Science and Technological Development of Serbia.

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