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A GENERAL EXISTENCE PRINCIPLE FOR FIXED
POINT THEOREMS IN
D-METRIC SPACES
B. C. DHAGE, A. M. PATHAN, and B. E. RHOADES
(Received 13 April 1998 and in revised form 22 July 1998)
Abstract.We establish two general principles for fixed point theorems inD-metric spaces, and then show that several theorems inD-metric spaces follow as corollaries of these gen-eral principles.
Keywords and phrases.α-condensing maps,D-metric spaces, fixed point theorems.
2000 Mathematics Subject Classification. Primary 47H10.
1. Introduction. The concept of aD-metric space was introduced by the first author in [1]. A nonempty setX, together with a functionD:X×X×X→[0,∞)is called a D-metric space, denoted by(X,D)ifDsatisfies
(i) D(x,y,z)=0 if and only ifx=y=z(coincidence),
(ii) D(x,y,z)=D(p{x,y,z}), wherepis a permutation ofx,y,z(symmetry), (iii) D(x,y,z)≤D(x,y,a)+D(x,a,z)+D(a,y,z)for allx,y,z,a∈X(tetrahedral inequality).
The nonnegative real functionDis called aD-metric onX. Some specific examples ofD-metrics appear in [2]. AD-metric is a continuous function onX3in the topology
ofD-metric convergence, which is Hausdorff (see [5]).
In this paper, we establish two general fixed point principles for mappings in aD -metric space, which yield several fixed point theorems as corollaries.
2. Preliminaries. Letf:X→X. The orbit offat the pointx∈Xis the setO(x)= {x,f x,f2x,...}. An orbit ofxis said to be bounded if there exists a constantK >0
such thatD(u,v,w)≤Kfor allu,v,w∈O(x). The constantKis called aD-bound ofO(x). AD-metric spaceXis said to bef-orbitally bounded ifO(x)is bounded for eachx∈X. A sequencexn⊂Xis said to beD-Cauchy if, for eachε >0, there exists
a positive integern0such that, for allm > n, p≥n0, D(xm,xn,xp) < ε. A sequence
{xn} ⊂X is said to beD-convergent to a pointx∈X if, for eachε >0 there exists
a positive integern0such that, for allm,n≥n0, D(xm,xn,x) < ε. An orbitO(x)is
calledf-orbitally complete if everyD-Cauchy sequence inO(x)converges to a point inX.
Lemma2.1[4]. Let{xn} ⊂Xbe a bounded sequence withD-boundKsatisfying
D(xn,xx+1,xm)≤λnK (2.1)
3. Main results
Theorem3.1. Let(X,D)be aD-metric spaces,fa selfmap ofX. Suppose that there exists anx0∈Xsuch thatO(x0)isD-bounded andf-orbitally complete. Suppose also thatf satisfies
D(f x,f y,f z)≤λmaxD(x,y,z), D(x,f x,z) forx,y,z∈O(x0) (3.1) for some0≤λ <1. Thenfhas a unique fixed point inX.
Proof. Suppose there exists annsuch thatxn=xn+1. Thenf hasxnas a fixed
point inX. Therefore we may assume that all of thexnare distinct.
We wish to show that, for any positive integersm,n, m > n, thatD(xn+1,xn+2,xm)
≤λnK, whereKis theD-bound ofO(x
0). The proof is by induction. From (3.1), for
anym,
D(x1,x2,xm)≤λmaxD(x0,x1,xm−1), D(x0,x1,xm−1)≤λK. (3.2)
Again using (3.1),
D(x2,x3,xm)≤λmaxD(x2,x3,xm−1), D(x1,x2,xm−1). (3.3)
Using (3.2),
D(x2,x3,xm)≤λmaxD(x2,x3,xm−1),λK. (3.4)
Inequality (3.4) can be regarded as a recursion formula inm. Therefore
D(x2,x3,xm)≤λmaxλmaxD(x2,x3,xm−2),λK,λK≤λ2K. (3.5)
Assume the induction hypothesis. Then, from (3.1),
D(xn+1,xn+2,xm)≤λmaxD(xn+1,xn+2,xm−1), D(xn,xn+1,xm−1)
≤λmaxD(xn+1,xn+2,xm−1),λnK. (3.6)
Inequality (3.6) can be regarded as a recursion formula inm. Therefore,
D(xn+1,xn+2,xm)≤λmaxλmaxD(xn+1,xn+2,xm−2),λnK,λnK =maxλ2D(x
n+1,xn+2,xm−2),λn+2K,λn+1K =maxλ2D(x
n+1,xn+2,xm−2),λn+1K ≤maxλ2·λmaxD(x
n+1,xn+2,xm−3),λn+1K,λn+1K =maxλ3D(x
n+1,xn+2,xm−3),λn+1K≤ ··· ≤maxλnD(x
n+1,xn+2,xm−n),λn+1K
≤maxλn·λmaxD(x
n+1,xn+2,xm−n−1),λn+1K,λn+1K =λn+1K,
(3.7)
In (3.1) setx=xn, z=pto obtain
D(xn+1,xn+1,f p)≤λmaxD(xn,xn,p),D(xn,xn+1,p). (3.8)
Taking the limit of (3.8) asn→ ∞yieldsD(p,p,f p)≤λD(p,p,p)=0, andp=f p. To prove uniqueness, suppose thatqis also a fixed point off. Then, from (3.1),
D(p,p,q)=D(f p,f p,f q)≤λmaxD(p,p,q), D(p,f p,q)=λD(p,p,q), (3.9)
which implies thatp=q.
Corollary3.2[2, Theorem 2.1]. Letf be a selfmap of a complete and bounded D-metric spaceXsatisfying
D(f x,f y,f z)≤λD(x,y,z) (3.10)
for allx,y,z∈X, for some0≤λ <1. Thenf has a unique fixed pointp, andf is continuous atp.
Proof. In (3.10) sety=f xto obtain (3.1). Then, from Theorem 3.1,fhas a unique fixed pointp.
To prove continuity, let{zn} ⊂Xwith limzn=p. From (3.10),
D(p,p,f zn)=D(f p,f p,f zn)≤λD(p,p,zn). (3.11)
Taking the limit asn→ ∞gives limsupD(p,p,f zn)=0, and liminfD(p,p,f zn)=0
which implies that limf zn=p=f p, andfis continuous atp.
Corollary3.3[2, Corollary 1.1]. Letf be a selfmap of a complete and bounded D-metric space satisfying the condition that there exists a positive integerqsuch that
D(fqx,fqy,fqz)≤λD(x,y,z) (3.12)
for allx,y,z∈X, for some0≤λ <1. Thenf has a unique fixed pointp, andf is f-orbitally continuous atp.
Proof. DefineT=fq. Then (3.12) reduces to (3.10), andThas a unique fixed point
pby Corollary 3.2; i.e.,p=T p=fqp. Thusf p=fq+1p=T (f p), andf pis also a
fixed point ofT. Uniqueness implies thatf p=p, andpis a fixed point off. Condition (3.12) implies the uniqueness ofpas a fixed point off.
For the continuity, let{zn} ⊂O(f ), with limzn=p. From (3.12),
D(fqp,fqp,fqz
n)≤λD(p,p,zn). (3.13)
Taking the limit asn→ ∞shows that limfqzn=p=fqp, andfqisf-orbitally
con-tinuous atp. But, since eachzn∈O(f ), limfqzn=limf zn+q−1, andfisf-orbitally
continuous atp.
Corollary3.4. Letf be a selfmap ofX,Xanf-orbitally bounded and complete D-metric space satisfying
D(f x,f y,f z)≤α
1+D(x,f x,z) 1+D(x,y,z)
for allx,y,z∈X, α,β≥0, α+β <1. Then f has a unique fixed pointp andf is continuous atp.
Proof. In (3.14) sety=f xto obtain
D(f x,f2x,f z)≤αD(f x,f2x,z)+βD(x,f x,z)
≤λmaxD(f x,f2x,z), D(x,f x,z), (3.15)
whereλ=α+β <1, and (3.1) is satisfied. The conclusion follows from Theorem 3.1. To prove the continuity off atp, let{zn} ⊂Xwith limzn=p. In (3.14) setx=z
=p, y=zn, to obtain
D(p,f zn,p)=D(f p,f zn,f p)
≤α
1+D(p,f p,p) 1+D(p,zn,p)
D(zn,f zn,p)+βD(p,zn,p)
≤αD(zn,f zn,p)+βD(p,zn,p).
(3.16)
Taking the limsup of both sides of (3.16) asn→ ∞yields
D(p,limsupf zn,p)≤αD(p,limsupf zn,p), (3.17)
which implies that limsupf zn=p. Similarly, taking the liminf of both sides of (3.16)
asn→ ∞yields
D(p,liminff zn,p)≤αD(p,liminff zn,p), (3.18)
which implies that liminff zn=p. Therefore limf zn=p=f p, andfis continuous
atp.
Corollary3.5. Letf be a selfmap of an f-orbitally bounded and complete D -metric spaceX,qa fixed positive integer. Suppose thatf satisfies
D(fqx,fqy,fqz)≤α
1+D(x,fqx,z)
1+D(x,y,z)
D(y,fqy,z)+βD(x,y,z) (3.19)
for allx,y,z∈X, whereα,β≥0, α+β <1. Thenf has a unique fixed pointpandf isf-orbitally continuous atp.
Proof. SetT =fq. ThenT satisfies (3.14). ThereforeT has a unique fixed point
atp, and is continuous atp. A standard argument then verifies thatfhaspas a unique fixed point. As in the proof of Corollary 3.3,f isf-orbitally continuous atp.
4. α-condensing maps. For any set A in aD-metric space X, the D-diameter of A,δ(A), is defined byδ(A)=supx,y,z∈AD(x,y,z). The measure of noncompactness
of a bounded setAin aD-metric spaceXis a nonnegative real numberα(A)defined by
α(A)=infγ >0 :A= ∪n
Definition4.1. A selfmapfofXis calledα-condensing if, for any bounded set AinX,f (A)is bounded andα(f (A)) < α(A)ifα(A) >0.
Some authors refer toα-condensing maps as densifying maps.
Lemma4.2. Letf:X→X,Xanf-orbitally bounded and completeD-metric space, beα-condensing. ThenO(x)is compact for eachx∈X.
Proof. Letx∈Xand defineA⊂XbyA= {xn}, wherexn=fnx. Then
A= {x,f x,f2x,...} = {x}∪{f x,f2x,...} = {x}∪f (A). (4.2)
Therefore, if A is not precompact, then α(A)=α(f (A)) < α(A), a contradiction. Therefore ¯A=O(x)is compact, since ¯Ais a completeD-metric space.
Defineδ(x,y,z)=δ(O(x)∪O(y)O(z))
Theorem4.3. Letfbe a continuous compact selfmap of a boundedD-metric space X, satisfying
D(frx,fsy,ftz) < δ(x,y,z)for eachx,y,z∈X, with two of{x,y,z}distinct,
(4.3)
wherer ,s, andtare fixed positive integers. Thenf has a unique fixed point inX.
Proof. Sincef is compact, there exists a compact subsetY ofXcontainingf X. Thenf Y⊂YandA:= ∩∞
n=1fnYis a nonempty compactf-invariant subset ofXwhich
is mapped byf onto itself.Ahas the same properties with respect tofr,fs, andft.
Suppose thatδ(A) >0. SinceAis compact there existx,y,z∈Asuch thatδ(A)= D(x,y,z). Sincef A=A, there existx,y, andzinAsuch thatx=frx, y=fsy, andz=ftz. Then, from (4.3),
δ(A)=D(x,y,z)=D(frx,fsy,ftz) < δ(x,y,z)=δ(A), (4.4)
a contradiction. ThereforeAconsists of a single point, which is a fixed point off. Supposepandqare fixed points off,p=q. Then, from (4.3),
0< D(p,p,q)=D(frp,fsp,ftq) < D(p,p,q), (4.5)
a contradiction. Therefore the fixed point is unique.
Corollary4.4[8, Theorem 2]. LetXbe a compactD-metric space,fa continuous selfmap ofXsatisfying
D(f x,f y,f z)
<maxD(x,y,z), D(x,f x,z), D(y,f y,z), D(x,f y,z), D(y,f x,z)D(p,p,q) (4.6)
Proof. Inequality (4.6) implies thatD(f x,f y,f z) < δ(x,y,z), and the existence and uniqueness of a fixed pointpfollows from Theorem 4.3.
For continuity, let{zn} ⊂Xwithzn=pfor eachnand limzn=p. From (4.6)
D(p,p,f zn)=D(f p,f p,f zn) < D(p,f p,zn). (4.7)
Taking the limit asn→ ∞implies thatf is continuous atp.
Theorem4.5. Letf be anf-orbitally continuousα-condensing selfmap of a com-plete boundedD-metric spaceX. Leta∈X. If (4.3) holds onO(a), thenfhas a unique fixed pointp∈O(a), andlimnfnx=pfor eachx∈O(a).
Proof. From Lemma 4.2,O(a)is compact. Sincef is a continuousα-condensing selfmap ofO(a),fis compact. Now apply Theorem 4.3.
Corollary4.6. Letfbe a continuousα-condensing selfmap of a complete bounded D-metric spaceXsatisfying (4.6) for allx,y,z∈Xwithx≠f x, y≠f y, orz≠f z. Thenf has a unique fixed pointpinX.
As in the proof of Corollary 4.4,D(f x,f y,f z) < δ(x,y,z) and the result follows from Theorem 4.5.
Theorem4.7. Letfbe a selfmap of aD-metric spaceX. Suppose that there exists a pointa∈X with O(a)bounded and complete. Suppose that f is continuous and α-condensing onO(a)and satisfies (4.3) for eachx,y,z∈O(a)with two of{x,y,z} distinct, andx≠f x, y≠f y, z≠f z. Thenf has a fixed point inO(a).
Proof. By Lemma 4.2O(a)is compact. If there exists some integernfor which fna=fn+1a, thenfhas a fixed point inO(a). Assume thatfna≠fn+1afor eachn.
Note thatf, restricted toO(a)is a continuous compact selfmap of O(a). Suppose thatu≠f ufor each cluster pointuofO(a). Thenf satisfies condition (4.3) for all x,y,z∈O(a), with two of{x,y,z}distinct. Therefore, by Theorem 4.3,f, restricted toO(a), has a unique fixed pointp∈O(a). This contradicts the assumption that u≠f ufor each cluster pointuofO(a). Thereforeu=f ufor some cluster point u∈O(a).
The proofs of Theorems 4.3, 4.5, and 4.7 are very similar to their metric space coun-terparts in [6] and [7], but have been given here for completeness.
The following results are proved using the proof technique analogous to the corre-sponding metric space theorems.
Theorem4.8. Letf be a selfmap of X, anf-orbitally bounded and completeD -metric space. Suppose thatfisα-condensing,f-orbitally continuous and satisfies
D(f x,f y,f z) < α
1+D(x,f x,z) 1+D(x,y,z)
D(y,f y,z)+βD(x,y,z)=M(x,y,z) (4.8)
for allx,y,z∈Xwithx≠f x, y≠f y, z≠f z, whereα,β >0, α+β≤1. Thenfhas a unique fixed pointp∈Xandf is continuous atp.
Proof. Ifα+β <1, the result follows from Corollary 3.4. Therefore we assume thatα+β=1. Letx0∈Xand definexn+1=f xn, n≥0. From Lemma 4.2 it follows
Case I. There exists somex,y,z∈O(x0)for whichM=0. Theny=f y=z=x,
andyis a fixed point off. Inequality (4.8) implies uniqueness.
Case II. M≠0 for allx,y,z∈O(x0). Define a functionF:(O(x0))3→[0,∞)by
F(x,y,z)=D(f x,f y,f z)M(x,y,z) . (4.9)
The functionFis well defined on(O(x0))3sinceM≠0 onO(x0).
SinceFis continuous onO(x0), it attains its maximum value at some point(u,v,w) ∈O(x0). We call this maximum valuec. From (4.8) it follows that 0< c <1. Therefore
D(f x,f y,f z)≤cM(x,y,z)
≤α
1+D(x,f x,z) 1+D(x,y,z)
D(y,f y,z)+βD(x,y,z) (4.10)
for allx,y,z∈O(x0), whereα=cα >0, β=cβ >0, andα+β=c(α+β) <1. Since O(x0)is compact, it is bounded and complete. The result follows from Corollary 3.4.
Corollary4.9. Letfbe a selfmap of a complete andf-orbitally boundedD-metric space. Suppose thatf isα-condensing andf-orbitally continuous. Letqbe a positive integer. Suppose thatfsatisfies
D(fqx,fqy,fqz) < α
1+D(x,fqx,z)
1+D(x,y,z)
D(y,fqy,z)+βD(x,y,z) (4.11)
for allx,y,z∈Xfor which the right-hand side of (4.11) is not zero, whereα,β >0, α+β≤1. Thenf has a unique fixed pointpandfisf-orbitally continuous atp.
Proof. SetT=fq. ThenTsatisfies (4.8), and the existence and uniqueness of the
fixed pointp, forT, follows from Theorem 4.8. It then follows thatpis the unique fixed point forf. The continuity argument is the same as that used in the proof of Corollary 3.3.
Corollary4.10. Letfbe a continuous selfmap of a compactD-metric space sat-isfying (4.8). Thenf has a unique fixed pointp, andfis continuous atp.
This result is an immediate consequence of Theorem 4.8. Corollary 4.10 includes[3, Theorem 2.2]as a special case.
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Dhage and Pathan: Mathematical Research Centre, Mahatma Gandhi Mahavidy-alaya, Ahmedpur-413 515, India
Rhoades: Department ofMathematics, Indiana University, Bloomington, Indiana 47405-4301, USA