SOME FIXED POINT THEOREMS IN METRIC AND 2-METRIC SPACES

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SOME FIXED POINT THEOREMS IN METRIC AND 2-METRIC SPACES

S. VENKATA RATNAM NAIDU (Received 31 July 2000)

Abstract.We introduce the notion of compatibility for a pair of self-maps on a 2-metric space and we have fixed point theorems for pairs as well as quadruples of self-maps on a 2-metric space satisfying certain generalised contraction conditions. Further metric space versions of the same have also been obtained.

2000 Mathematics Subject Classification. 47H10, 54H25.

Brian Fisher [1] proved the following result.

Theorem1. Letfbe a self-map on a complete metric space(M, ρ)such that

ρ²(f x, f y)≤αρ(x, f x)ρ(y, f y)+βρ(x, f y)ρ(y, f x) (1) for allx, y inM for some nonnegative constantsα,βwithα <1. Thenf has a fixed point inM. If furtherβ <1, thenfhas a unique fixed point inM.

In this paper we first obtain generalisations of the existence part of the 2-metric space version ofTheorem 1for a pair of self-maps and the uniqueness part of the same for four self-maps on a 2-metric space. Next we state without proof the metric space versions of some of these results. We also give a number of examples to throw light on the results discussed and the concept of compatibility of a pair of self-maps on a 2-metric space introduced here.

Recall some basic notions and facts for the sake of completeness.

Definition2. LetXbe a nonempty set. A real-valued functiondonX×X×Xis said to be a 2-metric onXif

(i) given distinct elementsx,y ofX, there exists an elementzofX such that d(x, y, z)≠0,

(ii) d(x, y, z)=0 when at least two ofx,y,zare equal, (iii) d(x, y, z)=d(x, z, y)=d(y, z, x)for allx,y,zinX, and

(iv) d(x, y, z)≤d(x, y, w)+d(x, w, z)+d(w, y, z)for allx, y, z, winX.

Whendis a 2-metric onX, the ordered pair(X, d)is called a 2-metric space.

Definition3. A sequence{xn}in a 2-metric space(X, d)is said to be (i) convergent with limitxinXif lim_n→∞d(xn, x, a)= 0 for allainX, (ii) Cauchy if limm,n→∞d(xn, xm, a)=0 for allainX.

Definition4. A 2-metric space is said to be complete if every Cauchy sequence in it is convergent.

Definition5. A 2-metric on a setXis said to be continuous onXif it is sequentially continuous in two of its arguments.

It is known that a 2-metric is a nonnegative real-valued function, that it is sequen- tially continuous in anyone of its arguments and that if it is sequentially continuous in two of its arguments then it is sequentially continuous in all the three arguments. It was observed by Naidu and Prasad that (i) a convergent sequence in a 2-metric space need not be Cauchy (see [4, Remark 0.1 and Example 0.1]) (ii) in a 2-metric space(X, d) every convergent sequence is Cauchy ifdis continuous onX[4, Remark 0.2] and (iii) the converse of (ii) is false [4, Remark 0.2 and Example 0.2].

Throughout this paper, unless otherwise stated,(X, d)is a 2-metric space;(M, ρ) is a metric space;Ris the set of all real numbers;R⁺is the set of all nonnegative real numbers; for a self-mapθonR⁺,θ¹stands forθand for a positive integern,θⁿ⁺¹is the composite ofθandθⁿ;ϕis a monotonically increasing map fromR⁺toR⁺with _∞

n=1

ϕⁿ(t) <+∞for alltinR⁺;ψis a map fromR⁺toR⁺ withψ(0)= 0;Kis an absolute nonnegative real constant; and, depending upon the context,f,g,S,T are self-maps onXorM.

We note thatϕ(t) < tfor alltin(0,∞)and thatϕ(0)=ϕ(0+)=0.

Remark6. For a monotonically increasing nonnegative real-valued functionθ on R⁺the condition “_∞

n=1

θⁿ(t) <+∞for alltinR⁺” neither implies nor is implied by the condition “θ(t+) < tfor alltin(0,∞).” Examples7and8illustrate this.

Example7. Defineθ:R⁺→R⁺asθ(t)=t²if 0≤t≤3/4 andθ(t)= 3/4 ift >3/4.

Thenθis monotonically increasing onR⁺. For a positive integern, we haveθⁿ(t)= (3/4)²ⁿ⁻¹ ift >3/4 andθⁿ(t)=t²ⁿ ift≤3/4. Hence_∞

n=1

θⁿ(t) <+∞for alltin R⁺. We note thatθ((3/4)+)=3/4.

Example 8. Define θ:R⁺→R⁺ asθ(t)=t/(1+t) for all t in R⁺. Then θ is a strictly increasing continuous function onR⁺withθ(t) < tfor alltin (0,∞). We have θ(1/n)=1/(n+1)for alln=1,2,3, . . .. Henceθⁿ(1)=1/(n+1)for alln=1,2,3, . . .. Hence_∞

n=1

θⁿ(1)is divergent.

We need the following lemma of Naidu [3].

Lemma9(see [3]). Let{yn}^∞n=0be a sequence in(X, d). Fora∈X, letdn(a)=d(yn, y_n+1, a). Suppose thatdn(ym)=0for any nonnegative integersm, nwithn > m.

Thend(yi, yj, yk)=0for all nonnegative integersi,j,k.

Proposition10. Suppose that d²(f x, gy, a)≤ϕ

Kd(f x, T y, a)d(Sx, gy, a) +max

d²(Sx, T y, a), d²(Sx, f x, a), d²(T y, gy, a) +Ψ

d(f x, T y, a)d(Sx, gy, a) (2)

for allx, y, ainX. Let{xn}^∞n=0be a sequence inXsuch that

f x2n=T x_2n+1(=y2n,say), gx_2n+1=Sx_2n+2(=y_2n+1,say) (n=0,1,2, . . . ). (3) Then{yn}^∞n=0is Cauchy.

Proof. Letdn(a)=d(yn, y_n+1, a). By takingx=x_2n+2andy=x_2n+1in inequal- ity (2) we obtain

d²_2n+1(a)≤ϕ max

d²_2n(a), d²_2n+1(a)

. (4)

By takingx=x2nandy=x_2n+1in inequality (2) we obtain d²_2n(a)≤ϕ

max

d²_2n−1(a), d²_2n(a)

. (5)

From the above two inequalities we have d²_n+1(a)≤ϕ

max

d²_n(a), d²_n+1(a)

(n=0,1,2, . . . ). (6) Sinceϕis nonnegative andϕ(t) < tfor alltin(0,∞), from the above inequality we have

d²_n+1(a)≤ϕ d²_n(a)

(n=0,1,2, . . . ). (7)

By repeatedly using inequality (7) and the monotonic increasing nature ofϕwe obtain d²_n(a)≤ϕⁿ

d²₀(a)

(n=0,1,2, . . . ). (8)

From inequality (7) we see thatd_n+1(a)=0 ifdn(a)=0. Sincedm(ym)=0 for every nonnegative integerm, it follows thatdn(ym)=0 for any nonnegative integersm, nwith n > m. Hence fromLemma 9we haved(yi, yj, yk)=0 for all nonnegative integersi,j,k. Hence for any nonnegative integersmandnwithn < m, by repeatedly using the triangle type inequality for 2-metrics, we obtain

d

yn, ym, a

≤

m−1

k=n

dk(a). (9)

Hence from inequality (8) we have d

yn, ym, a

≤

m−1

k=n

ϕ^k

t0

, (10)

wheret0=d²₀(a). Since_∞

k=1

ϕ^k(t) <+∞for alltinR⁺,m−1

k=nϕ^k(t0)tends to zero as bothmandntend to+∞. Henced(yn, ym, a)tends to zero as bothmandntend to+∞. Since this is true for anyainX, it follows that{yn}is Cauchy.

Theorem11. Suppose thatΨis right continuous at zero and d²(f x, gy, a)≤ϕ

max

d²(x, y, a), d²(x, f x, a),

d²(y, gy, a), Kd(f x, y, a)d(x, gy, a) +Ψ

d(f x, y, a)d(x, gy, a) (11)

for allx, y, ainX. For anyx0inX, let{xn}^∞n=1be defined iteratively as

x2n+1=f x2n, x2n+2=gx2n+1 (n=0,1,2, . . . ). (12) Then{xn}is Cauchy. If{xn}converges to an elementzofX, thenzis a common fixed point offandg. Further the fixed point sets offandgare the same.

Proof. By taking S = T =I, the identity map on X, inProposition 10, we can conclude that{xn}is a Cauchy sequence inX. Suppose that it converges to an element zofX. By takingx=x2nandy=zin inequality (11) we obtain

d²

x_2n+1, gz, a

≤ϕ max

d²

x2n, z, a , d²

x2n, x_2n+1, a , d²(z, gz, a), Kd

x_2n+1, z, a d

x2n, gz, a +Ψ

d

x2n+1, z, a d

x2n, gz, a .

(13)

The limit of the first term on the right-hand side of inequality (13) asntends to+∞is ϕ(d²(z, gz, a))ifd(z, gz, a)is positive. Otherwise, it isϕ(0)orϕ(0+). Sinceϕ(0+)= ϕ(0)=0, in either case it can be written asϕ(d²(z, gz, a)). SinceΨ(0+)=Ψ(0)=0, by taking limits on both sides of the above inequality asntends to+∞we obtain

d²(z, gz, a)≤ϕ

d²(z, gz, a)

. (14)

Sinceϕ(t) < tfor alltin(0,∞), we haved²(z, gz, a)=0. Since this is true for alla inX,gz=z. Similarly it can be shown thatf z=z. Ifx is a fixed point off, then by takingy=x in inequality (11) we obtain d²(x, gx, a)≤ϕ(d²(x, gx, a)). Hence gx=x. Similarly it can be shown that any fixed point ofgis also a fixed point off.

Hencef andghave the same fixed point sets.

Remark12. The hypothesis ofTheorem 11does not ensure the uniqueness of the common fixed point forfandg. This can be seen by takingfandgas identity maps onX,K=2,ϕ(t)=(1/2)tandΨ(t)=0 for alltinR⁺. We can also takeK=0 and ϕ(t)=Ψ(t)=(1/2)tfor alltinR⁺orϕ(t)=0 andΨ(t)=tfor alltinR⁺.Theorem 11 is an improvement over the existence part of Theorem 3 of Naidu [3] in which the first Ψ occurring in the governing inequality is to be read asϕ.Proposition 10is also an improvement over that of Naidu [3].

Corollary13. Suppose that(X, d)is complete and

d²(f x, f y, a)≤αd(x, f x, a)d(y, f y, a)+βd(x, f y, a)d(f x, y, a) (15) for allx, yinXfor some nonnegative constantsα,βwithα <1. Thenf has a fixed point inX. If furtherβ <1, thenf has a unique fixed point inX.

Proof. The existence part of the corollary follows from Theorem 11by taking g=f,K=0,ϕ(t)=αt, andΨ(t)=βtfor alltinR⁺. The rest of it is evident.

Remark14. Corollary 13is the 2-metric space version ofTheorem 1.

A perusal of the proof ofTheorem 11leads to the following variant.

Theorem15. Suppose thatϕ(t+) < tfor alltin(0,∞),Ψis right continuous at zero and

d²(f x, gy, a)≤ϕ

Kd(f x, y, a)d(x, gy, a) +max

d²(x, y, a), d²(x, f x, a), d²(y, gy, a) +Ψ

d(f x, y, a)d(x, gy, a) (16)

for allx, y, ainX. For anyx0inX, let{xn}^∞n=1be defined iteratively as inTheorem 11.

Then{xn}is Cauchy. If{xn}converges to an elementzofX, thenzis a common fixed point offandg. Further the fixed point sets offandgare the same.

Remark16. The hypothesis ofTheorem 15does not ensure the uniqueness of a common fixed point forfandg. This can be seen by takingfandgas identity maps onX,K=1,ϕ(t)=(1/2)t, andΨ(t)=0 for alltinR⁺.

The concept of weak continuity of a 2-metric and that of weak commutativity for a pair of self-maps on a 2-metric space were introduced by Naidu and Prasad [4]. The notion of compatibility for a pair of self-maps on a metric space and that of weak compatibility for a pair of self-maps on an arbitrary set can be found in Jeong and Rhoades [2]. We state them below for the sake of completeness.

Definition 17 (see [4]). We say that d is weakly continuous at z∈X if every convergent sequence inXwith limitzis Cauchy.

Definition18(see [4]). A pair(f1, f2)of self-maps on(X, d)is said to be a weakly commuting pair (w.c.p.) ifd(f1f2x, f2f1x, a)≤d(f2x, f1x, a)for allx, y, ainX.

Definition19(see [2]). A pair(f1, f2)of self-maps on (M, ρ) is said to be a compat- ible pair (co.p.) if{ρ(f1f2xn, f2f1xn)}converges to zero whenever{xn}is a sequence inMsuch that{f1xn}and{f2xn}are convergent inMand have the same limit.

Definition20(see [2]). A pair(f1, f2)of self-maps on an arbitrary setE is said to be a weakly compatible pair (w.co.p.) iff1f2x=f2f1xwheneverx∈Eis such that f1x=f2x.

In analogy withDefinition 19we introduce the concept of compatibility for a pair of self-maps on a 2-metric space.

Definition21. A pair (f1, f2)of self-maps on (X, d)is called a compatible pair (co.p.) if{d(f1f2xn, f2f1xn, a)}converges to zero for eachainXwhenever{xn}is a sequence inXsuch that{f1xn}and{f2xn}are convergent sequences inXhaving the same limit and{d(f2xn, f1xn, a)}converges to zero for eachainX.

Remark22. The notion of asymptotic weak commutativity for a pair of self-maps on a 2-metric space introduced by Naidu [3] is slightly more stringent than the no- tion of compatibility introduced here. In 2-metric spaces, weak commutativity implies compatibility. But the converse is false. The following example illustrates it.

Example23. DefinedonR⁺×R⁺×R⁺asd(x, y, z)=min{|x−y|,|y−z|,|z−x|}. Thend is a 2-metric onR⁺. Definef1, f2 fromR⁺ to R⁺ as f1x =x/(1+x)and f2x =2x for all x in R⁺. Let {xn}be a sequence in R⁺. Then {d(f2xn, f1xn, a)}

converges to zero for eachainR⁺if and only if{xn}converges to zero inR⁺in the usual sense. We have

d

f1f2xn, f2f1xn, a

= 2xn

1+2xn

, 2xn

1+xn

, a

≤2xn

1

1+xn− 1 1+2xn

→0 asn → +∞

(17)

when{xn}converges to zero inR⁺in the usual sense. Hence(f1, f2)is a co.p. For any positive real numberx, whiled(f1f2x, f2f1x,2x)is positived(f2x, f1x,2x)is zero.

Hencef1andf2do not commute weakly.

Theorem24. Suppose thatΨis monotonically increasing onR⁺,Ψ(t+) < tfor allt in(0,∞),

d²(f x, gy, a)≤ϕ

max{d²(Sx, f x, a), d²(T y, gy, a)} +Ψ

d(f x, T y, a)d(Sx, gy, a) (18) for allx, y, a in X and that there are sequences {xn}^∞n=0and {yn}^∞n=0 as stated in Proposition 10. Then{yn}is Cauchy. Suppose that it converges to an elementzofX.

Then the following statements are true.

(I) Neither the pair of mapsf and S nor the pair of maps gand T can have a common fixed point other thanz. IfSz=z, thenf z=z. IfT z=z, thengz=z.

(II) IfSz=T z, thenzis a common fixed point offandSif and only if it is a common fixed point ofgandT.

(III) The pointzis a unique common fixed point off andS if one of the following five groups of conditions is true.

(i) (f , S)is a w.co.p. andz∈S(X).

(ii) (f , S)is a co.p., andf andS are continuous atz.

(iii) (f , S)is a co.p.,S is continuous atzanddis weakly continuous atSz.

(iv) (f , S)is a w.co.p. and, for some positive integerk,f S^k=S^kf,S^kis contin- uous atzanddis weakly continuous atS^kz.

(v) (f , S)is a co.p. and, for some positive integerk,S^kis continuous atzand commutes with each of the mapsf,g, andT.

(IV) The pointzis a fixed point off if one of the following two groups of conditions is true.

(i) (f , S)is a co.p.,f is continuous atzanddis weakly continuous atf z.

(ii) f is continuous atzand commutes with each of the mapsg,S, andT. (V) Statements(III) and (IV) withf,g,S, andT replaced byg,f,T, andS, respec- tively.

Proof. That{yn}is Cauchy follows from Proposition 10. Suppose that it con- verges to an elementzofX. By takingy=x_2n+1in inequality (18) we obtain

d²(f x, y_2n+1, a)≤ϕ max

d²(Sx, f x, a), d²

y2n, y_2n+1, a +Ψ

d(f x, y2n, a)d(Sx, y_2n+1, a)

. (19)

By taking limits on both sides of inequality (19) asn→ +∞and using the facts that

{yn}is a Cauchy sequence converging toz,ϕ(0)=ϕ(0+)=0 andΨis monotonically increasing onR⁺we obtain

d²(f x, z, a)≤ϕ

d²(Sx, f x, a) +Ψ

d(f x, z, a)d(Sx, z, a)⁺

(20) for allainX.

We now prove the following statements.

(1) Iff x=Sxfor somex∈X, thenf x=Sx=z.

(2) IfSx=zfor somex∈X, thenf x=z.

Proof of(1). Suppose thatf x=Sxfor somex∈X. Then, sinceϕ(0)=0, from inequality (20) we have d²(f x, z, a)≤Ψ(d²(f x, z, a)⁺). SinceΨ(t+) < t for alltin (0,∞), we haved²(f x, z, a)=0. Since this is true for allainX, we havef x=z.

Proof of(2). Suppose thatSx=zfor somex∈X. Then, sinceΨ(0+)=0, from inequality (20) we haved²(f x, z, a)≤ϕ(d²(z, f x, a)). Sinceϕ(t) < tfor alltin(0,∞), we haved²(f x, z, a)=0. Since this is true for allainX, we havef x=z.

Since inequality (18) remains unaffected if we interchangef,g,S,Twithg,f,T,S, respectively, in analogy with statements (1) and (2) we have the following statements.

(3) Ifgx=T xfor somex∈X, thengx=T x=z.

(4) IfT x=zfor somex∈X, thengx=z.

Statement (I) is evident from statements (1), (2), (3), and (4). Statement (II) is evident from statement (I). We now prove the following statement (5).

(5) IfS is continuous atzand(f , S)is a co.p., then{f Sx2n}converges toSz.

Suppose thatS is continuous at z and (f , S)is a co.p. Since d(Sx2n, f x2n, a)= d(y2n−1, y2n, a) → 0 as n → +∞, {yn} converges to z and (f , S) is a co.p., d(f Sx2n, Sf x2n, a)→0 asn→ +∞. SinceSis continuous atzand{f x2n}converges toz,{Sf x2n}converges toSz. We have

d

f Sx2n, Sz, a

≤d

f Sx2n, Sf x2n, a +d

Sf x2n, Sz, a +d

f Sx2n, Sf x2n, Sz

→0 (21) asn→ +∞. Hence{f Sx2n}converges toSz.

We now prove statement (III).

(i) Suppose thatz∈S(X). Then there exists an x∈X such that z=Sx. From statement (2) it follows thatf x=z. Suppose that(f , S)is a w.co.p. Then, sinceSx= f x, we have f Sx=Sf x, that is, f z=Sz. Hence from statement (1) we have f z= Sz=z.

(ii) Suppose thatf andSare continuous atzand(f , S)is a co.p. From statement (5) it follows that{f Sx2n}converges toSz. Sincef is continuous atz and{Sx2n} converges toz,{f Sx2n}converges tof z. Hencef z=Sz. Hence from statement (1) we havef z=Sz=z.

(iii) Suppose thatdis weakly continuous atSz,S is continuous atzand(f , S)is a co.p. From statement (5) it follows that{f Sx2n}converges toSz. SinceSis contin- uous at zand {Sx2n}converges to z, {SSx2n}converges toSz. Since d is weakly continuous at Sz and both{f Sx2n} and {SSx2n} converge to Sz, it follows that

{d(f Sx2n, SSx2n, a)}converges to zero. By takingx=Sx2nin inequality (20) we obtain d²

f Sx2n, z, a

≤ϕ d²

SSx2n, f Sx2n, a +Ψ

d

f Sx2n, z, a d

SSx2n, z, a+

. (22)

By taking limits on both sides of (22) as n → +∞, we obtain d²(Sz, z, a) ≤ Ψ(d²(Sz, z, a)⁺). SinceΨ(t+) < t for allt in(0,∞), we haved²(Sz, z, a)=0. Since this is true for allainX, we haveSz=z. Hence from statement (I) we havef z=z.

We now establish the following statement (6), which is needed to complete the proof of statement (III).

(6) If(f , S)is a w.co.p. and, for some positive integerk,f S^k=S^kf,S^kis continuous atz, and{d(S^ky2n−1, S^ky2n, a)}converges to zero for eachainX, thenf z=Sz=z.

Suppose that there is a positive integerksuch thatf S^k=S^kf,S^kis continuous atz and{d(S^ky_2n−1, S^ky2n, a)}converges to zero for eachainX. SinceS^kis continuous atz and {yn}converges to z, {S^kyn}converges to S^kz. Sincef S^k=S^kf,f S^kx2n

= S^kf x2n = S^ky2n. We have SS^kx2n = S^kSx2n =S^ky_2n−1. By taking x =S^kx2n in inequality (20), we obtain

d²

S^ky2n, z, a

≤ϕ d²

S^ky2n−1, S^ky2n, a +Ψ

d

S^ky2n, z, a d

S^ky_2n−1, z, a₊

. (23)

By taking limits on both sides of (23) as n → +∞, we obtain d²(S^kz, z, a) ≤ Ψ(d²(S^kz, z, a)⁺). SinceΨ(t+) < tfor alltin(0,∞), we haved²(S^kz, z, a)=0. Since this is true for allainX, we haveS^kz=z. Hencez∈S(X).

Hence, if(f , S)is a w.co.p., then conditions (i) of statement (III) are fulfilled so that from what we already proved we havef z=Sz=z.

We now resume the proof of statement (III).

(iv) Suppose thatS^k is continuous atzfor some positive integerk. Then{S^kyn} converges to S^kz. Suppose now that d is weakly continuous at S^kz. Then {d(S^k y2n−1, S^ky2n, a)}converges to zero for eachainX. Hence, if(f , S)is a w.co.p. and f S^k=S^kf, then from statement (6) we havef z=Sz=z.

(v) Suppose thatS^k commutes with each of the mapsf,g, andT for some posi- tive integerk. Then from equation (3) we havef (S^kx2n)=T (S^kx_2n+1)=S^ky2nand g(S^kx2n+1)=S(S^kx2n+2)=S^ky2n+1for alln=0,1,2, . . . .Hence fromProposition 10, it follows that{S^kyn}is Cauchy. In particular,{d(S^ky_2n−1, S^ky2n, a)}converges to zero for eachainX. Hence from statement (6) it follows thatf z=Sz=zif condi- tions (v) of statement (III) are fulfilled.

The proof of the following statement is similar to that of statement (5).

(7) Iff is continuous atzand(f , S)is a co.p., then{Sf x2n}converges tof z.

We now prove statement (IV).

(i) Suppose thatdis weakly continuous atf z,fis continuous atzand(f , S)is a co.p. From statement (7) it follows that{Sf x2n}converges tof z. Sincefis continuous atzand{f x2n}converges toz,{f f x2n}converges tof z. Sincedis weakly contin- uous at f z and both{Sf x2n}and {f f x2n}converge tof z,{d(Sf x2n, f f x2n, a)} converges to zero. By taking x=f x2n in inequality (20) and then taking limits on both sides of the inequality as n→ +∞, we obtain d²(f z, z, a)≤Ψ(d²(f z, z, a)⁺).

SinceΨ(t+) < tfor alltin(0,∞), we haved²(f z, z, a)=0. Since this is true for alla inX, we havef z=z.

We note that if f S = Sf, then d(Sf x2n, f f x2n, a) = d(f Sx2n, f f x2n, a) = d(f y_2n−1, f y2n, a)and(f , S)is a co.p. Hence from a perusal of the above proof we have the following statement.

(8) Iff S=Sf,fis continuous atzand{d(f y_2n−1, f y2n, a)}converges to zero for eachainX, thenf z=z.

We now resume the proof of statement (IV).

(ii) Suppose thatfcommutes with each of the mapsg,S, andT. Then from equa- tions (3) we have

f f x2n

=T

f x_2n+1

=f y2n, g

f x_2n+1

=S

f x_2n+2

=f y_2n+1 ∀n=0,1,2, . . . . (24) Hence fromProposition 10, it follows that{f yn}is Cauchy. Hence{d(f y_2n−1, f y2n, a)}converges to zero for eachainX. Hence from statement (8) it follows thatf z=z if conditions (ii) of statement (IV) are fulfilled.

Statement (V) follows from symmetry considerations.

Remark25. Theorem 24is an improvement over Theorem 2 in [3].

Corollary26. Suppose thatΨis monotonically increasing onR⁺,Ψ(t+) < tfor all tin(0,∞),

d²(f x, gy, a)≤ϕ max

d²(Sx, f x, a), d²(Sy, gy, a) +Ψ

d(f x, Sy, a)d(Sx, gy, a) (25)

for allx,y,ainXand that there are sequences{xn}^∞_n=0and{yn}^∞_n=0inXsuch that

f x2n=Sx_2n+1=y2n, gx_2n+1=Sx_2n+2=y_2n+1 (n=0,1,2, . . . ). (26) Then{yn}is Cauchy. Suppose that it converges to an elementzofX. Thenzis a unique common fixed point off,g, andS if one of the following groups of conditions is true.

(i) z∈S(X)and either(f , S)or(g, S)is a w.co.p.

(ii) S is continuous atz, and eitherfis continuous atzand(f , S)is a co.p. orgis continuous atzand(g, S)is a co.p.

(iii) Sis continuous atz,dis weakly continuous atSzand either(f , S)or(g, S)is a co.p.

(iv) S^kis continuous atzanddis weakly continuous atS^kzfor some positive integer k, andS commutes with eitherf org.

(v) Scommutes with each of the mapsf andg, andS^kis continuous atzfor some positive integerk.

Remark27. InTheorem 24if inequality (18) is replaced with the following more stringent inequality

d²(f x, gy, a)≤ϕ

d(Sx, f x, a)d(T y, gy, a) +Ψ

d(f x, T y, a)d(Sx, gy, a)

, (27)

then the weak continuity ofdcan be dropped from all those numbered statements in which it appears. A similar remark applies toCorollary 26also.

We now state without proof the metric space versions of some of the results we obtained in 2-metric spaces. Hereafter, unless otherwise stated,f,g, S, T are self- maps onM.

Proposition28. Suppose that ρ²(f x, gy)≤ϕ

Kρ(f x, T y)ρ(Sx, gy) +max

ρ²(Sx, T y), ρ²(Sx, f x), ρ²(T y, gy) +Ψ

ρ(f x, T y)ρ(Sx, gy) (28)

for allx,yinM and that there are sequences{xn}^∞n=0and{yn}^∞n=0inMsatisfying equations (3). Then{yn}^∞n=0is Cauchy.

Remark29. Proposition 28fails if the condition_∞

n=1

ϕⁿ(t) <+∞for alltinR⁺ is replaced by the conditionϕ(t+) < tfor alltin(0,∞).Example 30illustrates this wheng=f,S=T=I(the identity map onM) andΨ(t)=tfor alltinR⁺.

Example30. LetM= {xn:n=1,2,3, . . .}, wherexn=n

k=11/k. Definef:M→M asf xn=x_n+1for alln=1,2,3, . . .. Defineϕ:R⁺→R⁺asϕ(t)=t/(1+t)for alltin R⁺. Thenϕis a strictly increasing continuous function onR⁺withϕ(t) < tfor allt in(0,∞)and

|f x−f y|²≤ϕ

|x−y|²

+|f x−y||x−f y| (29) for allx, y inM. Evidently for anyxinM the sequence{fⁿx}diverges to+∞and hence is not Cauchy.

Theorem31. Suppose thatΨis right continuous at zero and ρ²(f x, gy)≤ϕ

max

ρ²(x, y), ρ²(x, f x), ρ²(y, gy), Kρ(f x, y)ρ(x, gy) +Ψ

ρ(f x, y)ρ(x, gy) (30)

for allx, yinM. For anyx0inM, let{xn}^∞n=1be defined iteratively as inTheorem 11.

Then{xn}is Cauchy. If{xn}converges to an elementzofM, thenzis a common fixed point offandg. Further the fixed point sets offandgare the same.

Theorem32. Suppose thatϕ(t+) < tfor all tin(0,∞),Ψ is right continuous at zero and

ρ²(f x, gy)≤ϕ

Kρ(f x, y)ρ(x, gy) +max

ρ²(x, y), ρ²(x, f x), ρ²(y, gy) +Ψ

ρ(f x, y)ρ(x, gy) (31)

for allx, yinM. For anyx0inM, let{xn}^∞n=1be defined iteratively as inTheorem 11.

Then{xn}is Cauchy. If{xn}converges to an elementzofM, thenzis a common fixed point offandg. Further the fixed point sets offandgare the same.

Remark33. InTheorem 32the conclusion thatzis a common fixed point offand gfails in the absence of the conditionϕ(t+) < tfor alltin(0,∞)even if(M, ρ)is complete,f=g,K=1 andΨis identically zero onR⁺. Examples34and35illustrate this. While inExample 34the functionf has no fixed point, inExample 35it has.

Example34. LetM= {1/2ⁿ|n=0,1,2, . . .} ∪ {0}. Then M is a complete metric space under the metric induced by the modulus function. Definef:M→Masf x= (1/2)xifx≠0 andf0=1. Defineϕ:R⁺→R⁺asϕ(t)=1 ift >1 andϕ(t)=(1/4)t ift≤1. Thenϕis monotonically increasing onR⁺,_∞

n=1

ϕⁿ(t) <+∞for alltinR⁺, ϕ(1+)=1 and

|f x−f y|²≤ϕ

|f x−y||x−f y|+max

|x−y|²,|x−f x|²,|y−f y|²

(32) for allx, yinM. We note that for anyx0inMthe sequence{fⁿx0}converges to zero.

But 0 is not a fixed point off. In fact,f has no fixed point.

Example 35. Let M be as in Example 34. Define f : M →M as f x=(1/2)x if x∉{0,1}andf0 =f1 = 1. Defineϕ:R⁺→R⁺asϕ(t)=1 ift >1 andϕ(t)=(9/10)t ift≤1. Thenϕis monotonically increasing onR⁺,_∞

n=1

ϕⁿ(t) <+∞for alltinR⁺, ϕ(1+)=1 and inequality (32) is satisfied for allx, yinM. We note that for anyx0in M\{0,1}the sequence{fⁿx0}converges to zero. But 0 is not a fixed point off.

Remark36. A pair(f1, f2)of self-maps on (M, ρ) is a w*.c.p. ifρ(f1f2x, f2f1x)≤ γρ(f2x, f1x)for allxinMfor some nonnegative real numberγand a w.c.p. (weakly commuting pair) ifρ(f1f2x, f2f1x)≤ρ(f2x, f1x)for allxinM. (The notion of weak commutativity for a pair of self-maps on a metric space was introduced by Sessa [5].) Clearly a w.c.p. is a w*.c.p. and a w*.c.p. is a co.p. But the converse is false in either case. Examples37and38illustrate this.

Example37. Definef1,f2fromRtoRasf1x=x²andf2x=2x−1 for allxin R. Then|f1f2x−f2f1x| =2(x−1)²=2|f1x−f2x|for allxinR. Hence(f1, f2)is a w*.c.p. but not a w.c.p.

Example38. Definef1, f2fromRtoRasf1x=x²andf2x= −x²for allx inR. Then|f1f2x−f2f1x| =2x⁴and|f1x−f2x| =2x²for allxinR. Clearly there is no nonnegative real numberγsuch that 2x⁴≤γ(2x²)for allxinR. Hence(f1, f2)is not a w*.c.p. Clearly it is a co.p.

Theorem39. Suppose thatΨis monotonically increasing onR⁺,Ψ(t+) < tfor allt in(0,∞),

ρ²(f x, gy)≤ϕ max

ρ²(Sx, f x), ρ²(T y, gy) +Ψ

ρ(f x, T y)ρ(Sx, gy) (33) for allx, y inM and that there are sequences{xn}^∞_n=0and{yn}^∞_n=0 inM satisfying equations (3). Then{yn}^∞_n=0is Cauchy. Suppose that it converges to an elementzofM.

Then the following statements are true.

(1) Neither the pair of mapsf and S nor the pair of mapsg and T can have a common fixed point other thanz. IfSz=z, thenf z=z. IfT z=z, thengz=z.

(2)IfSz=T z, thenzis a common fixed point offandSif and only if it is a common fixed point ofgandT.

(3)If(f , S)is a w.co.p. andz∈S(X), thenf z=Sz=z.

(4)If(f , S)is a co.p. andSis continuous atz, thenf z=Sz=z.

(5)If(f , S)is a w.co.p. and, for some positive integerk,f S^k=S^kfandS^kis continuous atz, thenf z=Sz=z.

(6)Iffis continuous atzand(f , S)is a co.p., thenf z=z.

(7)Statements (3), (4), (5), and (6) withfandSreplaced bygandT, respectively.

Finally we conclude the paper with the following open problem.

Open problem. DoesTheorem 15remain valid if the conditionϕ(t+) < t for all tin(0,∞)is deleted from the hypothesis?

Note40. The results in Naidu and Prasad [4] remain valid if the weak commu- tativity condition in them is replaced with compatibility condition as introduced in Definition 21.

Acknowledgement. The author expresses his heart felt thanks to the referee for his valuable suggestions and to Dr. J. Rajendra Prasad for helping the author in collecting the necessary research material.

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S. Venkata Ratnam Naidu: Department of Applied Mathematics, Andhra University, Visakhapatnam-530003, India

E-mail address:[email protected]