THEOREMS IN D-METRIC SPACES
S. V. R. NAIDU, K. P. R. RAO, AND N. SRINIVASA RAO Received 24 March 2004 and in revised form 7 January 2005
Examples of complete D-metric spaces are given in which every convergent sequence has at most two limits and in which there are convergent sequences with exactly two limits. Also an example of a completeD-metric space having a convergent sequence with infinitely many limits is given and, using the example, several fixed point theorems in D-metric spaces are shown to be false. Modifications of some of these theorems and their generalizations are obtained either by imposing restrictions on the number of limits of certain convergent sequences in the space or by assuming the sequential continuity of the D-metric in any two variables and the theorems so obtained are illustrated by means of examples.
1. Introduction
In this paper, we give examples (Examples3.1and3.2) of completeD-metric spaces in which every convergent sequence has at most two limits and in which there are convergent sequences having exactly two limits. We also provide a simple example (Example 3.3) of a completeD-metric space in which there is a convergent sequence, which converges to ev- ery element of the space. Using the last one, we show that several fixed point theorems in D-metric spaces are false. In particular, we observe that Dhage [2, Theorem 2.1], Ahmad et al. [1, Theorem 4.2], and Dhage et al. [3, Theorem 3.1] are false and we modify them.
We provide examples to illustrate these modifications. Further, we obtain generalizations of these modifications.
2. Definitions
Definition 2.1(see [2]). LetX be a nonempty set. A functionρ:X×X×X→[0,∞) is called aD-metric onXif
(i)ρ(x,y,z)=0 if and only ifx=y=z(coincidence);
(ii)ρ(x,y,z)=ρ(p(x,y,z)) for allx,y,z∈Xand for each permutationp(x,y,z) ofx, y,z(symmetry);
(iii)ρ(x,y,z)≤ρ(x,y,a) +ρ(x,a,z) +ρ(a,y,z) for allx,y,z,a∈X(tetrahedral inequa- lity).
Copyright©2005 Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences 2005:12 (2005) 1969–1988 DOI:10.1155/IJMMS.2005.1969
IfXis a nonempty set andρis aD-metric onX, then the ordered pair (X,ρ) is called aD-metric space. When theD-metricρis understood,Xitself is called aD-metric space.
Definition 2.2(see [2]). A sequence{xn}in aD-metric space (X,ρ) is said to be con- vergent (orρ-convergent) if there exists an elementxofXwith the following property.
Given thatε >0, there exists anN∈Nsuch thatρ(xm,xn,x)< εfor allm,n≥N. In such a case,{xn}is said to converge toxandxis said to be a limit of{xn}.
Definition 2.3(see [2]). A sequence{xn}in aD-metric space (X,ρ) is said to be a Cauchy (orρ-Cauchy) sequence if given thatε >0, there exists anN∈Nsuch thatρ(xm,xn,xp)< ε for allm,n,p≥N.
Definition 2.4(see [2]). AD-metric spaceX is said to be complete if everyρ-Cauchy sequence inXconverges to a pointxinX.
Definition 2.5(see [2]). AD-metric space (X,ρ) is said to be bounded if there exists a positive real numberM such thatρ(x,y,z)≤M for allx,y,z∈X. In such a case,M is said to be aD-bound for theD-metricρ.
Definition 2.6(see [5]). A subsetEof aD-metric space (X,ρ) is said to beρ-bounded if there exists a positive real numberMsuch thatρ(x,y,z)≤Mfor allx,y,z∈E.
Definition 2.7(see [6]). A pair (f1,f2) of self-maps on an arbitrary setEis said to be weakly compatible, partially commuting, or coincidentally commuting if (f1◦f2)u= (f2◦f1)uwheneveru∈Eand f1u=f2u.
Remark 2.8. The definition ofρ-Cauchy sequence as given by Dhage [2] appears to be slightly different fromDefinition 2.3but it is actually equivalent to it. It can be shown that in aD-metric space every convergent sequence is a Cauchy sequence.
3. Examples and theorems
Throughout this section, unless otherwise stated, Ris the set of all real numbers,R+ is the set of all nonnegative real numbers,Nis the set of all positive integers and for a monotonically increasing functionϕ:R+→R+andt∈R+,ϕ(t+) stands for the right- hand limit ofϕatt.
We now give examples of complete D-metric spaces in which every convergent se- quence has at most two limits and in which there are convergent sequences having exactly two limits.
Example 3.1. LetX= {0, 1, 1/2, 1/3,...}. Defineρ:X×X×X→R+as follows:
ρ(x,y,z)=
|x−y|+|y−z|+|z−x| ifx,y,z∈X\{1},
0 ifx=y=z=1,
1 if both 0 and 1 occur amongx,y,z, or
if 1 occurs exactly twice amongx,y,z, minmax{x,y},
max{y,z}, max{z,x}
if 1 occurs exactly once and 0 does not occur amongx,y,z.
(3.1)
Then (X,ρ) is a completeD-metric space. Further{1/n}converges to both 0 and 1. If a sequence{xn}inXconverges to an elementx0∈X\{0, 1}, thenx0is the unique limit of {xn}. In particular, a convergent sequence inXhas at most two limits.
Proof. Clearlyρis symmetric in all the three variablesx,y,zandρ(x,y,z)=0⇔x=y= z. Letx,y,z,u∈X.
Case 1. x,y,z∈X\{1}. Ifu∈X\{1}, then
ρ(x,y,z)= |x−y|+|y−z|+|z−x|
≤ |u−y|+|y−z|+|z−u| +|x−u|+|u−z|+|z−x| +|x−y|+|y−u|+|u−x|
=ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u).
(3.2)
Ifu=1, then
ρ(x,y,z)= |x−y|+|y−z|+|z−x|
≤max{y,z}+ max{z,x}+ max{x,y}
≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u)
(3.3)
sinceρ(u,y,z)=ρ(1,y,z)=1 or max{y,z}, and so forth.
Case 2. 1 occurs amongx,y,z.
Subcase 1. x=y=z=1.
Then
ρ(x,y,z)=0≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u). (3.4) Subcase 2. 1 occurs exactly twice amongx,y,z.
Without loss of generality, we may assume thatx=y=1. Thenz∈X\{1}. Ifu=1, then
ρ(x,y,z)=ρ(1, 1,z)=ρ(u,y,z)
≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u). (3.5) Ifu=1, then
ρ(x,y,z)=ρ(1, 1,z)=1=ρ(1, 1,u)=ρ(x,y,u)
≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u). (3.6) Subcase 3. 1 occurs exactly once amongx,y,z.
Without loss of generality, we may assume thatx=1. Then y,z∈X\{1}. We may assume thatz≤y.
Ifu=1, then
ρ(x,y,z)=ρ(1,y,z)=ρ(u,y,z)
≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u). (3.7) Ifz=0, then
ρ(x,y,z)=ρ(1,y, 0)=1=ρ(1,u, 0)=ρ(x,u,z), (3.8) and hence
ρ(x,y,z)≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u). (3.9) Ifu=1 andz=0, then
ρ(x,y,z)=ρ(1,y,z)=max{y,z} =y
≤max{u,y} ≤ρ(1,y,u)=ρ(x,y,u)
≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u).
(3.10)
Thus for allx,y,z,u∈X, we have
ρ(x,y,z)≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u). (3.11) Henceρis aD-metric onX.
We are going to show that (X,ρ) is complete. Let{xn}be a Cauchy sequence inX.
Case 1. There existsN∈Nsuch thatxn=xNfor alln≥N.
In this case, evidently,{xn}converges toxN.
Case 2. There does not existN∈Nsuch thatxn=xNfor alln≥N.
In this case, givenN∈N, there existi,j∈Nsuch thati > j > Nandxi=xj.
Subcase 1. There existsα∈Xsuch thatxn=αfor infinitely manyn. Then givenN∈N, there existi,j,k∈Nsuch thatk > j > i > N,xi=xj=α, andxk=α. We haveρ(xi,xj,xk)= 1 ifα=1. Since{xn}is a Cauchy sequence, it follows thatα=1. Hence, there existsN0∈ Nsuch thatxn∈X\{1}for alln≥N0. Hence,ρ(xi,xj,xk)= |xi−xj|+|xj−xk|+|xk−xi| for alli,j,k≥N0. Since{xn}is a Cauchy,ρ(xi,xj,xk)→0 asi,j,k→ ∞. Sincexn=αfor in- finitely manyn, giveni∈N, there existj,k∈Nsuch thatk > j > iandxj=xk=α. Hence given an integeri≥N0, there exist j,k∈Nsuch thatk > j > iand 2|xi−α| =ρ(xi,xj,xk).
Hence|xi−α| →0 as i→ ∞. Hence forn,m≥N0,ρ(xn,xm,α)= |xn−xm|+|xm−α|+
|α−xn| →0 asn,m→ ∞. Hence{xn}converges toα.
Subcase 2. For anyα∈X,xn=αfor only finitely manyn.
Letεbe a positive real number. Choose a positive integerM such that (1/M)< ε. Let N1=sup{n∈N:xn∈ {1, 1/2, 1/3,..., 1/M}}, where we adopt the convention that the supremum of the empty subset ofRis zero. Since for anyk∈N, 1/k∈X andxn=1/k for only finitely manyn, it follows thatN1is a nonnegative integer. LetN=N1+ 1. Then
0≤xn<1/M < εfor alln≥NsinceX= {0} ∪ {1/k:k∈N}. Hence{xn}converges to 0 in the usual sense.
Forn,m≥N, we have ρ0,xn,xm
= 0−xn + xn−xm + xm−0
≤2 xn + 2 xm
−→0 asn,m−→ ∞.
(3.12)
Hence{xn}converges to 0 with respect toρ. Thus in any case,{xn}is convergent inX with respect toρ. Hence (X,ρ) is complete.
Evidently{1/n}converges to both 0 and 1.
Let{xn}be a sequence inXsuch that it converges to an elementx0ofX\{0, 1}. Since ρ(1, 1,x0)=1 andρ(xn,xm,x0)→0 asn,m→ ∞, it follows that there existsN0∈Nsuch thatxn∈X\{1}for alln≥N0. Forn,m≥N0, we have
ρxn,xm,x0
= xn−xm + xm−x0 + x0−xn . (3.13) Since ρ(xn,xm,x0)→0 asn,m→ ∞, it follows that|xn−x0| →0 asn→ ∞. Sincex0 is positive, it follows that there existsN1∈Nsuch that xn>0 for alln≥N1. Form,n≥ max{N0,N1}, we have
ρ1,xn,xm
=minmax1,xn
, maxxn,xm
, maxxm, 1=maxxn,xm
−→x0 asn,m−→ ∞. (3.14)
Hence{xn}does not converge to 1. Lety∈X\{1}. Form,n≥N0, we have ρy,xn,xm
= y−xn + xn−xm + xm−y
−→2 y−x0 asn,m−→ ∞, 2 y−x0 =0⇐⇒y=x0.
(3.15)
Hence{xn}does not converge toyfor any y∈X\{1,x0}. Hencex0 is the only limit of
{xn}.
Example 3.2. LetX= {0, 1, 1/2, 1/3,...}. Defineρ:X×X×X→R+as
ρ(x,y,z)=
|x−y|+|y−z|+|z−x| ifx,y,z∈X\{1},
0 ifx=y=z=1,
1 if both 0 and 1 occur amongx,y,z,
2 minmax{x,y},
max{y,z}, max{z,x}
if 1 occurs exactly once and 0 does not occur amongx,y,z,
2 min{x,y,z} if 1 occurs exactly twice amongx,y,zand 0 does not occur amongx,y,z.
(3.16) Then (X,ρ) is a completeD-metric space. Further{1/n}converges to both 0 and 1.
If a sequence{xn}inXconverges to an elementx0∈X\{0, 1}, thenx0 is the unique limit of{xn}. In particular, a convergent sequence inXhas at most two limits.
Proof. Clearly,ρis symmetric in all the three variablesx,y,zandρ(x,y,z)=0⇔x=y= z. Letx,y,z,u∈X.
Case 1. x,y,z∈X\{1}. Ifu∈X\{1}, then
ρ(x,y,z)= |x−y|+|y−z|+|z−x|
≤ |u−y|+|y−z|+|z−u| +|x−u|+|u−z|+|z−x| +|x−y|+|y−u|+|u−x|
=ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u).
(3.17)
Ifu=1, then
ρ(x,y,z)= |x−y|+|y−z|+|z−x|
≤max{y,z}+ max{z,x}+ max{x,y}
≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u)
(3.18)
sinceρ(u,y,z)=ρ(1,y,z)=1 or 2 max{y,z}, and so forth.
Case 2. 1 occurs amongx,y,z.
Subcase 1. x=y=z=1.
Then
ρ(x,y,z)=0≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u). (3.19) Subcase 2. 1 occurs exactly twice amongx,y,z.
Without loss of generality, we may assume thatx=y=1. Thenz∈X\{1}. Ifu=1, then
ρ(x,y,z)=ρ(1,y,z)=ρ(u,y,z)
≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u). (3.20) Ifz=0, then
ρ(x,y,z)=ρ(1, 1, 0)=1=ρ(u, 1, 0)=ρ(u,y,z)
≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u). (3.21) Ifz=0 andu=1, then
ρ(x,y,z)=ρ(1, 1,z)=2z
≤2 max{u,z} ≤ρ(u, 1,z)=ρ(u,y,z)
≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u).
(3.22)
Subcase 3. 1 occurs exactly once amongx,y,z.
Without loss of generality, we may assume thatx=1. Theny,z∈X\{1}. We may assume thatz≤y.
Ifu=1, then
ρ(x,y,z)=ρ(1,y,z)=ρ(u,y,z)≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u). (3.23) Ifz=0, then
ρ(x,y,z)=ρ(1,y, 0)=1=ρ(1,u, 0)=ρ(x,u,z), (3.24) and hence
ρ(x,y,z)≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u). (3.25) Ifu=1 andz=0, then
ρ(x,y,z)=ρ(1,y,z)=2 max{y,z}
=2y≤2 max{u,y} ≤ρ(1,y,u)
=ρ(x,y,u)≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u).
(3.26)
Thus for allx,y,z,u∈X, we have
ρ(x,y,z)≤ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u). (3.27) Henceρis aD-metric onX.
We are going to show that (X,ρ) is complete. Let{xn}be a Cauchy sequence inX.
Case 1. xn=1 for infinitely manyn.
Giveni∈N, we can choosej,k∈Nsuch thatk > j > iandxk=xj=1 so that
ρxi,xj,xk
=ρxi, 1, 1=
0 ifxi=1, 1 ifxi=0, 2xi ifxi∈X\{0, 1}.
(3.28)
Since{xn}is a Cauchy sequence,ρ(xp,xq,xr)→0 asp,q,r→ ∞.
Hence,ρ(xi, 1, 1)→0 asi→ ∞. Hence,xn=0 for only finitely manyn. Hence, there existsN∈Nsuch thatxn=0 for alln≥N.
Ifxn=1 for infinitely manyn, thenxn→0 in the usual sense asn→ ∞among thosen for whichxn=1.
Form,n≥N, we have
ρ1,xm,xn
=
0 ifxm=xn=1,
2xm ifxn=1,xm=1, 2xn ifxm=1,xn=1, 2 maxxm,xn
ifxm=1,xn=1.
(3.29)
Hence,ρ(1,xm,xn)→0 asm,n→ ∞. Hence,{xn}converges to 1 with respect toρ.
Case 2. xn=1 for only finitely manynand there existsα∈X\{1}such thatxn=αfor infinitely manyn.
Sincexn=1 for only finitely manyn, there existsN0∈Nsuch thatxn∈X\{1}for alln≥N0. Hence, ρ(xi,xj,xk)= |xi−xj|+|xj−xk|+|xk−xi|for alli,j,k≥N0. Since xn=α for infinitely manyn, given i∈N, there exist j,k∈N such thatk > j > i and xj=xk=α.
Hence given an integeri≥N0, there existj,k∈Nsuch thatk > j > iand 2 xi−α =ρxi,xj,xk
. (3.30)
Since {xn} is a Cauchy sequence,ρ(xi,xj,xk)→0 asi,j,k→ ∞. Hence,|xi−α| →0 as i→ ∞. Hence forn,m≥N0,
ρxn,xm,α= xn−xm + xm−α + α−xn
≤2 xm−α + 2 xn−α
−→0 asm,n−→ ∞.
(3.31)
Hence{xn}converges toα.
Case 3. For anyα∈X,xn=αfor only finitely manyn.
Letεbe a positive real number. Choose a positive integerMsuch that (1/M)< ε.
LetN1=sup{n∈N:xn∈ {1, 1/2, 1/3,..., 1/M}}, where we adopt the convention that the supremum of the empty subset ofRis zero. Since for anyk∈N, 1/k∈Xandxn=1/k for only finitely manyn, it follows thatN1is a nonnegative integer.
LetN=N1+ 1. SinceX= {0, 1, 1/2, 1/3,...}, it follows that 0≤xn<1/M < εfor all n≥N.
Hence{xn}converges to 0 in the usual sense.
Forn,m≥N, we have ρ0,xn,xm
= 0−xn + xn−xm + xm−0
≤2 xn + 2 xm
−→0 asn,m−→ ∞.
(3.32)
Hence{xn}converges to 0 with respect toρ. Thus in any case,{xn}is convergent inX with respect toρ. Hence (X,ρ) is complete.
Evidently,{1/n}converges to both 0 and 1.
If a sequence{xn}inXconverges to an elementx0∈X\{0, 1}, then it can be shown as inExample 3.1thatx0is a unique limit of{xn}. We now give a simple example of a completeD-metric space having a convergent sequence, which converges to every element of the space.
Example 3.3. LetX= {1/2n:n∈N}. Defineρ:X×X×X→R+as follows:
ρ(x,y,z)=
0 ifx=y=z,
minmax{x,y}, max{y,z}, max{z,x}
otherwise. (3.33)
Then (X,ρ) is a completeD-metric space. Further{1/2n}∞n=1is aρ-Cauchy sequence and converges with respect toρtoxfor allx∈X.
Proof. We are going to prove thatρis a D-metric on X. Clearly ρis symmetric in all the three variables and ρ(x,y,z)=0⇔x=y=z. We note that ρ(x,y,z)≤1/2 for all x,y,z∈X. Letx,y,z∈Xand at least let two amongx,y,zbe distinct. We may assume thatx≥y≥z. Thenρ(x,y,z)=y.
Letu∈X. Whenu≥y,ρ(u,y,z)=y; and whenu < y,ρ(x,y,u)=y. Henceρ(x,y,z)≤ ρ(u,y,z) +ρ(x,u,z) +ρ(x,y,u). Henceρis aD-metric onX.
We are going to show that (X,ρ) is complete. Let{xn}be a Cauchy sequence inX.
Case 1. There existsN∈Nsuch thatxn=xNfor alln≥N.
In this case, evidently{xn}converges toxN.
Case 2. GivenN∈N, there existi,j∈Nsuch thati > N,j > N, andxi=xj. Claim3.4. xn→0asn→ ∞in the usual sense.
Suppose not. Then there exists a positive real number εsuch thatxn≥εfor infinitely manyn∈N. GivenN∈N, we can choosei,j,k∈Nsuch thatk > j > i > N,xi≥ε,xj≥ε, andxk=xj. Then
ρxi,xj,xk
=minmaxxi,xj
, maxxj,xk
, maxxk,xi
≥ε. (3.34)
This is a contradiction, since{xn}is a Cauchy sequence.
Form,n∈Nandx∈X, we have ρx,xm,xn
≤minmaxx,xm
, maxxm,xn
, maxxn,x
≤maxxm,xn
−→0 asm,n−→ ∞.
(3.35)
Hence{xn}converges toxfor anyx∈Xwith respect to theD-metricρ.
Hence (X,ρ) is a completeD-metric space.
Dhage [2] proved the following theorem.
Theorem3.5 (see [2, Theorem 2.1]). Let f be a self-map of a complete and bounded D-metric spaceXsuch that
ρ(f x,f y,f z)≤αρ(x,y,z) (3.36) for allx,y,z∈X, and someα∈[0, 1). Then f has a unique fixed point.
Remark 3.6. Example 3.7shows that the existence part of fixed point ofTheorem 3.5is false.
Example 3.7. Let (X,ρ) be as inExample 3.3. Define f :X→X as f x=x/2 for allx∈ X. Thenρ(f x,f y,f z)≤(1/2)ρ(x,y,z) for allx,y,z∈Xbut f has no fixed point inX.
Further for allx∈X,{fnx}converges to every element ofX, and hence has infinitely many limits.
Proof. Letx,y,z∈X Case 1. x=y=z.
Then
ρ(f x,f y,f z)=ρ(f x,f x,f x)=0=1
2ρ(x,x,x)=1
2ρ(x,y,z). (3.37) Case 2. At least two amongx,y,zare distinct.
We may assume thatx≥y≥z. Then ρ(f x,f y,f z)=ρx
2,y 2,z
2
= y 2 =
1
2ρ(x,y,z). (3.38)
Thus
ρ(f x,f y,f z)≤1
2ρ(x,y,z) ∀x,y,z∈X. (3.39) Evidentlyf has no fixed point and for anyx∈X,{fnx}inXconverges to every point of
X.
The following is a modification of Dhage [2, Theorem 2.1].
Theorem3.8. Let (X,ρ) be a bounded completeD-metric space and let f :X→Xbe such that
ρ(f x,f y,f z)≤αρ(x,y,z) (3.40) for allx,y,z∈X, and someα∈[0, 1). Then for anyx∈X,{fnx}is aD-Cauchy sequence.
If there existsx0∈Xsuch that{fnx0}has only a finite number of limits, then f has a unique fixed point.
We now illustrateTheorem 3.8using the completeD-metric spaces given in Examples 3.1and3.2.
Example 3.9. Let (X,ρ) be as inExample 3.1. Define f :X→Xas f x=
0 ifx=1, x
3 otherwise. (3.41)
Then ρ(f x,f y,f z)≤(2/3)ρ(x,y,z) for all x,y,z∈X. Evidently, 0 is the unique fixed point of f.
Proof. Letx,y,z∈X. In view of the symmetry ofρ, it is enough to consider the following five cases.
Case 1. x=y=z=1.
Then
ρ(f x,f y,f z)=ρ(f1,f1,f1)=ρ(0, 0, 0)=0≤2
3ρ(x,y,z). (3.42)
Case 2. x=y=1 andz=1.
Then
ρ(f x,f y,f z)=ρ(f1,f1,f z)
=ρ0, 0,z 3
=2z 3 <2
3(1)=2
3ρ(x,y,z). (3.43) Case 3. x=1,y=0 andz=1.
Then
ρ(f x,f y,f z)=ρ(f1,f0,f z)
=ρ0, 0,z 3
=2z 3 <2
3(1)=2
3ρ(x,y,z). (3.44) Case 4. x=1 and 0< z≤y <1.
Then
ρ(f x,f y,f z)=ρ(f1,f y,f z)
=ρ0,y 3,z
3
=2y 3 =
2
3ρ(x,y,z). (3.45) Case 5. z≤y≤x <1.
Then
ρ(f x,f y,f z)=ρx 3,y
3,z 3
= x
3− y 3
+ y 3−
z 3
+ z 3−
x 3
=2
3(x−z)≤2
32(x−z)=2
3ρ(x,y,z).
(3.46)
Thusρ(f x,f y,f z)≤(2/3)ρ(x,y,z) for allx,y,z∈X.
Example 3.10. Let (X,ρ) be as inExample 3.2. Definef :X→Xandg:X→Xas follows:
f x=
0 ifx=1, x
3 otherwise.
gx=
1 ifx∈ {0, 1}, x
3 otherwise.
(3.47)
It can be verified that
ρ(f x,f y,f z)≤1
3ρ(x,y,z), ρ(gx,g y,gz)≤1
3ρ(x,y,z) (3.48)
for allx,y,z∈X.
For anyx∈X\{0, 1}, the sequences{fnx}and{gnx}are the same and converge to 0 and 1 only. While 0 is the fixed point of f, 1 is the fixed point ofginX.
Theorem 3.8is a corollary of the following theorem.
Theorem3.11. Let (X,ρ) be aD-metric space, letf be a self-map onX, and letϕ:R+→R+ be a monotonically increasing map such thatϕn(t)→0asn→ ∞for eacht∈R+. Suppose that
ρ(f x,f y,f z)≤ϕρ(x,y,z) (3.49) for allx,y,z∈Xand that there existsx0∈X such that{ρ(x0,fix0,fjx0) :i,j∈ {0} ∪N}
is bounded. Then{fnx0}is aρ-Cauchy sequence. Suppose that{fnx0}is convergent and has only a finite number of limits. Then f has a unique fixed point inX, which is a limit of {fnx0}.
Proof. Letxn= fnx0for alln∈N.
For nonnegative integersrands, andn∈N, we have ρxn,xn+r,xn+s
=ρfnx0,fnxr,fnxs
≤ϕρfn−1x0,fn−1xr,fn−1xs
(from (3.49))
≤ϕϕρfn−2x0,fn−2xr,fn−2xs
(from (3.49) and the monotonic increasing nature ofϕ)
...
≤ϕnρx0,xr,xs .
(3.50) Since{ρ(x0,fix0,fjx0) :i,j∈ {0} ∪N}is bounded, there exists a positive real numberM such thatρ(x0,fix0,fjx0)≤Mfor alli,j∈ {0} ∪N. Hence from inequality (3.50) and the monotonic increasing nature ofϕ, we haveρ(xn,xn+r,xn+s)≤ϕn(M). Sinceϕn(M)→0 as n→ ∞, from the above inequality, it follows that{xn}is aρ-Cauchy sequence.
Suppose now that{fnx0}is convergent and has only a finite number of limits. LetS denote the set of all limits of{fnx0}. ThenSis a nonempty finite subset ofX. Sinceϕ is a nonnegative real-valued monotonically increasing function onR+andϕn(t)→0 as n→ ∞, it follows thatϕ(t)< tfor allt∈(0,∞) andϕ(0)=0. Letz∈S. Then
ρfn+1x0,fm+1x0,f z≤ϕρfnx0,fmx0,z
≤ρfnx0,fmx0,z
−→0 asm,n−→ ∞.
(3.51)
Hence{fnx0}converges to f zalso.
Hence, f z∈S. Thus f(S)⊆S. For any positive integerm, we have fm(S)⊆ fm−1(S), where f0=I, the identity map onX.
Suppose thatm0is a positive integer such that fm0−1(S) has at least two elements. Let β=inf{ρ(x,y,z) :x,y,z∈fm0−1(S) and at least two of x, y, z are different}. Sincefm0−1(S)
is a finite set having at least two elements,{ρ(x,y,z) :x,y,z∈ fm0−1(S) and at least two ofx,y,zare different}is a nonempty finite subset of (0,∞). Hence, there existu,v,z∈ fm0−1(S) such that at least two of them are distinct andρ(u,v,z)=β. We haveρ(f u,f v,f z)
≤ϕ(ρ(u,v,z))=ϕ(β)< β, sinceβis a positive real number andϕ(t)< tfor allt∈(0,∞).
Sinceu,v,z∈fm0−1(S), f u,f v,f z∈fm0(S)⊆fm0−1(S). Hence from the definition ofβ, it follows that f u= f v= f z. Since fm0−1(S) is finite, at least two ofu,v,zare distinct and f u=f v=f z, it follows that fm0(S) is a proper subset of fm0−1(S). SinceSis finite, from the above discussion, it follows that fn0−1(S) is a singleton for somen0∈N.
Let fn0−1(S)= {w}. Since fn0(S)⊆fn0−1(S), we must have f w=w.
Remark 3.12. Examples3.9and3.10illustrateTheorem 3.11.
Remark 3.13. Example 3.7shows that Dhage and Rhoades [4, Theorem 2] and its corol- lary are false.
Ahmad et al. [1] proved the following theorem.
Theorem3.14. Let (X,ρ) be a completeD-metric space. Let f andSbe commutative self- maps onXsuch that f is injective,Sis surjective, and
ρ(f x,f y,f z)≤αρ(Sx,Sy,Sz) (3.52)
for allx,y,z∈X, and someα∈[0, 1). Then there exists a unique common fixed point of f andS.
Remark 3.15. Example 3.7shows thatTheorem 3.14is false since the mapf in the exam- ple is injective, the identity mapIis surjective, f ◦I=I◦f, and inequality (3.52) is true for allx,y,zinXwithα=1/2 andS=I.Theorem 3.14remains valid if its hypothesis is strengthened by imposing the additional condition that (X,ρ) is bounded and that every ρ-Cauhy sequence in f(X) is convergent inS(X) and has a unique limit inS(X).
The following is a generalization of the modification ofTheorem 3.14suggested above.
Theorem3.16. Let (X,ρ) be a bounded completeD-metric space, let f andSbe self-maps onX. Letϕ:R+→R+be a monotonically increasing map such thatϕn(t)→0asn→ ∞, for each t∈(0,∞). Suppose that f andSare partially commuting, f(X)⊆S(X),
ρ(f x,f y,f z)≤ϕρ(Sx,Sy,Sz) (3.53)
for allx,y,zinXand that everyρ-Cauchy sequence in f(X)is convergent inS(X)and has a unique limit inS(X). Then f andShave a unique common fixed point inX.
Proof. Letx0∈X. Since f(X)⊆S(X), there exists a sequence{xn}inXsuch that f xn= Sxn+1forn=0, 1, 2,....
Letyn=f xn. Sinceϕ:R+→R+is monotonically increasing onR+and{ϕn(t)} →0 as n→ ∞for allt∈(0,∞),ϕ(t)< tfor allt∈(0,∞). Henceϕ(0+)=0.
