SOME FIXED POINT THEOREMS IN METRIC AND 2-METRIC SPACES

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. Theorem 1. Let f be a self-map on a complete metric space (M,ρ) such that ρ2(fx,fy)≤αρ(x,fx)ρ(y,fy)+βρ(x,fy)ρ(y,fx) (1) for all x,y in M for some nonnegative constants α, β with α < 1. Then f has a fixed point in M . If further β < 1, then f has a unique fixed point in M . In this paper we first obtain generalisations of the existence part of the 2-metric space version of Theorem 1 for 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. Definition 2. Let X be a nonempty set. A real-valued function d on X×X×X is said to be a 2-metric on X if (i) given distinct elements x, y of X, there exists an element z of X such that d(x,y,z)≠ 0, (ii) d(x,y,z)= 0 when at least two of x, y , z are equal, (iii) d(x,y,z)= d(x,z,y)= d(y,z,x) for all x, y , z in X, and (iv) d(x,y,z)≤ d(x,y,w)+d(x,w,z)+d(w,y,z) for all x,y,z,w in X. When d is a 2-metric on X, the ordered pair (X,d) is called a 2-metric space. Definition 3. A sequence {xn} in a 2-metric space (X,d) is said to be (i) convergent with limitx in X if limn→∞d(xn,x,a) = 0 for all a in X, (ii) Cauchy if limm,n→∞d(xn,xm,a)= 0 for all a in X. 626 S. VENKATA RATNAM NAIDU Definition 4. A 2-metric space is said to be complete if every Cauchy sequence in it is convergent. Definition 5. A 2-metric on a setX is said to be continuous onX if 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 sequentially 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 if d is continuous on X [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; R is the set of all real numbers; R+ is the set of all nonnegative real numbers; for a self-map θ on R+, θ1 stands for θ and for a positive integer n, θn+1 is the composite of θ and θn; φ is a monotonically increasing map from R+ to R+ with ∑∞ n=1 √ φn(t) < +∞ for all t in R+; ψ is a map from R+ to R+ with ψ(0) = 0; K is an absolute nonnegative real constant; and, depending upon the context, f , g, S, T are self-maps on X or M . We note that φ(t) < t for all t in (0,∞) and that φ(0)=φ(0+)= 0. Remark 6. For a monotonically increasing nonnegative real-valued function θ on R+ the condition “ ∑∞ n=1 √ θn(t) <+∞ for all t in R+” neither implies nor is implied by the condition “θ(t+) < t for all t in (0,∞).” Examples 7 and 8 illustrate this. Example 7. Define θ :R+ →R+ as θ(t) = t2 if 0≤ t ≤ 3/4 and θ(t) = 3/4 if t > 3/4. Then θ is monotonically increasing on R+. For a positive integer n, we have θn(t) = (3/4) n−1 if t > 3/4 and θn(t) = t2 if t ≤ 3/4. Hence ∑∞n=1 √ θn(t) < +∞ for all t in R+. We note that θ((3/4)+)= 3/4. Example 8. Define θ : R+ → R+ as θ(t) = t/(1+t) for all t in R+. Then θ is a strictly increasing continuous function on R+ with θ(t) < t for all t in (0,∞). We have θ(1/n)= 1/(n+1) for all n= 1,2,3, . . . . Hence θn(1)= 1/(n+1) for all n= 1,2,3, . . . . Hence ∑∞ n=1 √ θn(1) is divergent. We need the following lemma of Naidu [3]. Lemma 9 (see [3]). Let {yn}n=0 be a sequence in (X,d). For a∈ X, let dn(a)= d(yn, yn+1,a). Suppose that dn(ym) = 0 for any nonnegative integers m, n with n > m. Then d(yi,yj,yk)= 0 for all nonnegative integers i, j, k. Proposition 10. Suppose that d2(fx,gy,a)≤φ(Kd(fx,Ty,a)d(Sx,gy,a) +max{d2(Sx,Ty,a),d2(Sx,fx,a),d2(Ty,gy,a)}) +Ψ(d(fx,Ty,a)d(Sx,gy,a)) (2) SOME FIXED POINT THEOREMS IN METRIC . . . 627 for all x,y,a in X. Let {xn}n=0 be a sequence in X such that fx2n = Tx2n+1(=y2n, say), gx2n+1 = Sx2n+2(=y2n+1, say) (n= 0,1,2, . . .). (3) Then {yn}n=0 is Cauchy. Proof. Let dn(a)= d(yn,yn+1,a). By taking x = x2n+2 and y = x2n+1 in inequality (2) we obtain d2n+1(a)≤φ ( max { d2n(a),d 2 2n+1(a) }) . (4) By taking x = x2n and y = x2n+1 in inequality (2) we obtain d2n(a)≤φ ( max { d2n−1(a),d 2 2n(a) }) . (5) From the above two inequalities we have dn+1(a)≤φ ( max { dn(a),d 2 n+1(a) }) (n= 0,1,2, . . .). (6) Since φ is nonnegative and φ(t) < t for all t in (0,∞), from the above inequality we have dn+1(a)≤φ ( dn(a) ) (n= 0,1,2, . . .). (7) By repeatedly using inequality (7) and the monotonic increasing nature ofφ we obtain dn(a)≤φ ( d0(a) ) (n= 0,1,2, . . .). (8) From inequality (7) we see that dn+1(a)= 0 if dn(a)= 0. Since dm(ym)= 0 for every nonnegative integer m, it follows that dn(ym) = 0 for any nonnegative integers m, n with n > m. Hence from Lemma 9 we have d(yi,yj,yk) = 0 for all nonnegative integers i, j, k. Hence for any nonnegative integersm andnwithn<m, by repeatedly using the triangle type inequality for 2-metrics, we obtain d ( yn,ym,a )≤ m−1 ∑

Definition 2. Let X be a nonempty set.A real-valued function d on X × X × X is said to be a 2-metric on X if (i) given distinct elements x, y of X, there exists an element z of X such that d(x,y,z) ≠ 0, (ii) d(x,y,z) = 0 when at least two of x, y, z are equal, (iii) d(x,y,z) = d(x,z,y) = d(y,z,x) for all x, y, z in X, and (iv) d(x,y,z) ≤ d(x,y,w)+ d(x,w,z)+ d(w,y,z) for all x, y, z, w in X.
When d is a 2-metric on X, the ordered pair (X, d) is called a 2-metric space.Definition 3. A sequence {x n } in a 2-metric space (X, d) is said to be (i) convergent with limit x in X if lim n→∞ d(x n ,x,a) = 0 for all a in X, (ii) Cauchy if lim m,n→∞ d(x n ,x m ,a) = 0 for all a in X. Definition 4. A 2-metric space is said to be complete if every Cauchy sequence in it is convergent.Definition 5. A 2-metric on a set X is said to be continuous on X if 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 sequentially 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,  Throughout this paper, unless otherwise stated, (X, d) is a 2-metric space; (M, ρ) is a metric space; R is the set of all real numbers; R + is the set of all nonnegative real numbers; for a self-map θ on R + , θ 1 stands for θ and for a positive integer n, θ n+1 is the composite of θ and θ n ; ϕ is a monotonically increasing map from R + to R + with ∞ n=1 ϕ n (t) < +∞ for all t in R + ; ψ is a map from R + to R + with ψ(0) = 0; K is an absolute nonnegative real constant; and, depending upon the context, f , g, S, T are self-maps on X or M.
Remark 6.For a monotonically increasing nonnegative real-valued function θ on R + the condition " ∞ n=1 θ n (t) < +∞ for all t in R + " neither implies nor is implied by the condition "θ(t+) < t for all t in (0, ∞)." Examples 7 and 8 illustrate this.
Lemma 9 (see [3]).Let {y n } ∞ n=0 be a sequence in (X, d).For a ∈ X, let d n (a) = d(y n , y n+1 ,a).Suppose that d n (y m ) = 0 for any nonnegative integers m, n with n > m.Then d(y i ,y j ,y k ) = 0 for all nonnegative integers i, j, k. for all x, y, a in X.Let {x n } ∞ n=0 be a sequence in X such that
Proof.Let d n (a) = d(y n ,y n+1 ,a).By taking x = x 2n+2 and y = x 2n+1 in inequality (2) we obtain By taking x = x 2n and y = x 2n+1 in inequality (2) we obtain From the above two inequalities we have Since ϕ is nonnegative and ϕ(t) < t for all t in (0, ∞), from the above inequality we have By repeatedly using inequality (7) and the monotonic increasing nature of ϕ we obtain From inequality (7) we see that d n+1 (a) = 0 if d n (a) = 0. Since d m (y m ) = 0 for every nonnegative integer m, it follows that d n (y m ) = 0 for any nonnegative integers m, n with n > m.Hence from Lemma 9 we have d(y i ,y j ,y k ) = 0 for all nonnegative integers i, j, k.Hence for any nonnegative integers m and n with n < m, by repeatedly using the triangle type inequality for 2-metrics, we obtain Hence from inequality (8) we have where k=n ϕ k (t 0 ) tends to zero as both m and n tend to +∞.Hence d(y n ,y m ,a) tends to zero as both m and n tend to +∞.Since this is true for any a in X, it follows that {y n } is Cauchy.for all x, y, a in X.For any x 0 in X, let {x n } ∞ n=1 be defined iteratively as Then {x n } is Cauchy.If {x n } converges to an element z of X, then z is a common fixed point of f and g.Further the fixed point sets of f and g are the same.
Proof.By taking S = T = I, the identity map on X, in Proposition 10, we can conclude that {x n } is a Cauchy sequence in X. Suppose that it converges to an element z of X.By taking x = x 2n and y = z in inequality (11) we obtain The limit of the first term on the right-hand side of inequality (13) as n tends to +∞ is ϕ(d 2 (z, gz, a)) if d(z, gz, a) is positive.Otherwise, it is ϕ(0) or ϕ(0+).Since ϕ(0+) = ϕ(0) = 0, in either case it can be written as ϕ(d 2 (z, gz, a)).Since Ψ (0+) = Ψ (0) = 0, by taking limits on both sides of the above inequality as n tends to +∞ we obtain Since ϕ(t) < t for all t in (0, ∞), we have d 2 (z, gz, a) = 0. Since this is true for all a in X, gz = z.Similarly it can be shown that f z = z.If x is a fixed point of f , then by taking y = x in inequality (11) we obtain d 2 (x, gx, a) ≤ ϕ(d 2 (x, gx, a)).Hence gx = x.Similarly it can be shown that any fixed point of g is also a fixed point of f .Hence f and g have the same fixed point sets.
Remark 12.The hypothesis of Theorem 11 does not ensure the uniqueness of the common fixed point for f and g.This can be seen by taking f and g as identity maps on X, K = 2, ϕ(t) = (1/2)t and Ψ (t) = 0 for all t in R + .We can also take K = 0 and ϕ(t) = Ψ (t) = (1/2)t for all t in R + or ϕ(t) = 0 and Ψ (t) = t for all t in R + .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 as ϕ.Proposition 10 is also an improvement over that of Naidu [3].
Corollary 13.Suppose that (X, d) is complete and for all x, y in X for some nonnegative constants α, β with α < 1.Then f has a fixed point in X.If further β < 1, then f has a unique fixed point in X.
Proof.The existence part of the corollary follows from Theorem 11 by taking g = f , K = 0, ϕ(t) = αt, and Ψ (t) = βt for all t in R + .The rest of it is evident.Remark 14. Corollary 13 is the 2-metric space version of Theorem 1.
A perusal of the proof of Theorem 11 leads to the following variant.
Theorem 15.Suppose that ϕ(t+) < t for all t in (0, ∞), Ψ is right continuous at zero and for all x, y, a in X.For any x 0 in X, let {x n } ∞ n=1 be defined iteratively as in Theorem 11.Then {x n } is Cauchy.If {x n } converges to an element z of X, then z is a common fixed point of f and g.Further the fixed point sets of f and g are the same.
Remark 16.The hypothesis of Theorem 15 does not ensure the uniqueness of a common fixed point for f and g.This can be seen by taking f and g as identity maps on X, K = 1, ϕ(t) = (1/2)t, and Ψ (t) = 0 for all t in R + .
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 in X with limit z is Cauchy.
Definition 18 (see [4]).A pair (f 1 ,f 2 ) of self-maps on (X, d) is said to be a weakly commuting pair (w.c.p. Definition 19 (see [2]).A pair (f 1 ,f 2 ) of self-maps on (M, ρ) is said to be a compatible pair (co.p.) if {ρ(f 1 f 2 x n ,f 2 f 1 x n )} converges to zero whenever {x n } is a sequence in M such that {f 1 x n } and {f 2 x n } are convergent in M and have the same limit.
Definition 20 (see [2]).A pair (f 1 ,f 2 ) of self-maps on an arbitrary set E is said to be a weakly compatible pair (w.co.p.) if In analogy with Definition 19 we introduce the concept of compatibility for a pair of self-maps on a 2-metric space.
a)} converges to zero for each a in X whenever {x n } is a sequence in X such that {f 1 x n } and {f 2 x n } are convergent sequences in X having the same limit and {d(f 2 x n ,f 1 x n ,a)} converges to zero for each a in X.
Remark 22.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 notion of compatibility introduced here.In 2-metric spaces, weak commutativity implies compatibility.But the converse is false.The following example illustrates it.
converges to zero for each a in R + if and only if {x n } converges to zero in R + in the usual sense.We have when {x n } converges to zero in R + in the usual sense.Hence (f 1 ,f 2 ) is a co.p.For any positive real number x, while d(f Hence f 1 and f 2 do not commute weakly. Theorem 24.Suppose that Ψ is monotonically increasing on R + , Ψ (t+) < t for all t in (0, ∞), for all x, y, a in X and that there are sequences {x n } ∞ n=0 and {y n } ∞ n=0 as stated in Proposition 10.Then {y n } is Cauchy.Suppose that it converges to an element z of X.Then the following statements are true.
(I) Neither the pair of maps f and S nor the pair of maps g and T can have a common fixed point other than z.
(II) If Sz = T z, then z is a common fixed point of f and S if and only if it is a common fixed point of g and T .
(III) The point z is a unique common fixed point of f and S if one of the following five groups of conditions is true.
(i) (f , S) is a w.co.p. and z ∈ S(X).
(ii) (f , S) is a co.p., and f and S are continuous at z. (iii) (f , S) is a co.p., S is continuous at z and d is weakly continuous at Sz. (iv) (f , S) is a w.co.p. and, for some positive integer k, f S k = S k f , S k is continuous at z and d is weakly continuous at S k z. (v) (f , S) is a co.p. and, for some positive integer k, S k is continuous at z and commutes with each of the maps f , g, and T .(IV) The point z is a fixed point of f if one of the following two groups of conditions is true.
(i) (f , S) is a co.p., f is continuous at z and d is weakly continuous at f z. (ii) f is continuous at z and commutes with each of the maps g, S, and T .(V) Statements (III) and (IV) with f , g, S, and T replaced by g, f , T , and S, respectively.
Proof.That {y n } is Cauchy follows from Proposition 10.Suppose that it converges to an element z of X.By taking y = x 2n+1 in inequality (18) we obtain By taking limits on both sides of inequality (19) as n → +∞ and using the facts that {y n } is a Cauchy sequence converging to z, ϕ(0) = ϕ(0+) = 0 and Ψ is monotonically increasing on R + we obtain for all a in X.
We now prove the following statements.
(1) If f x = Sx for some x ∈ X, then f x = Sx = z.
(2) If Sx = z for some x ∈ X, then f x = z.
(5) If S is continuous at z and (f , S) is a co.p., then {f Sx 2n } converges to Sz. Suppose that S is continuous at z and (f , S) is a co.p.Since d(Sx 2n ,f x 2n ,a) = d(y 2n−1 ,y 2n ,a) → 0 as n → +∞, {y n } converges to z and (f , S) is a co.p., d(f Sx 2n ,Sf x 2n ,a) → 0 as n → +∞.Since S is continuous at z and {f x 2n } converges to z, {Sf x 2n } converges to Sz.We have as n → +∞.Hence {f Sx 2n } converges to Sz.We now prove statement (III).
(i) Suppose that z ∈ S(X).Then there exists an x ∈ X such that z = Sx.From statement (2) it follows that f x = z.Suppose that (f , S) is a w.co.p.Then, since Sx = 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 that f and S are continuous at z and (f , S) is a co.p.From statement (5) it follows that {f Sx 2n } converges to Sz.Since f is continuous at z and {Sx 2n } converges to z, {f Sx 2n } converges to f z.Hence f z = Sz.Hence from statement (1) we have f z = Sz = z.
(iii) Suppose that d is weakly continuous at Sz, S is continuous at z and (f , S) is a co.p.From statement (5) it follows that {f Sx 2n } converges to Sz.Since S is continuous at z and {Sx 2n } converges to z, {SSx 2n } converges to Sz.Since d is weakly continuous at Sz and both {f Sx 2n } and {SSx 2n } converge to Sz, it follows that {d(f Sx 2n , SSx 2n ,a)}converges to zero.By taking x = Sx 2n in inequality (20) we obtain By taking limits on both sides of (22) as n → +∞, we obtain d 2 (Sz, z, a) ≤ Ψ (d 2 (Sz, z, a) + ).Since Ψ (t+) < t for all t in (0, ∞), we have d 2 (Sz, z, a) = 0. Since this is true for all a in X, we have Sz = z.Hence from statement (I) we have f 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 integer k, f S k = S k f , S k is continuous at z, and {d(S k y 2n−1 ,S k y 2n ,a)} converges to zero for each a in X, then f z = Sz = z.Suppose that there is a positive integer k such that f S k = S k f , S k is continuous at z and {d(S k y 2n−1 ,S k y 2n ,a)} converges to zero for each a in X.Since S k is continuous at z and {y n } converges to z, {S k y n } converges to S k z.Since f S k = S k f , f S k x 2n = S k f x 2n = S k y 2n .We have SS k x 2n = S k Sx 2n = S k y 2n−1 .By taking x = S k x 2n in inequality (20), we obtain By taking limits on both sides of (23) as n → +∞, we obtain Since this is true for all a in X, we have S k z = z.Hence z ∈ 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 have f z = Sz = z.
We now resume the proof of statement (III).
(iv) Suppose that S k is continuous at z for some positive integer k.Then {S k y n } converges to S k z.Suppose now that d is weakly continuous at S k z.Then {d(S k y 2n−1 ,S k y 2n ,a)} converges to zero for each a in X.Hence, if (f , S) is a w.co.p. and f S k = S k f , then from statement (6) we have f z = Sz = z.
(v) Suppose that S k commutes with each of the maps f , g, and T for some positive integer k.Then from equation (3) we have f (S k x 2n ) = T (S k x 2n+1 ) = S k y 2n and g(S k x 2n+1 ) = S(S k x 2n+2 ) = S k y 2n+1 for all n = 0, 1, 2,.... Hence from Proposition 10, it follows that {S k y n } is Cauchy.In particular, {d(S k y 2n−1 ,S k y 2n ,a)} converges to zero for each a in X. Hence from statement (6) it follows that f z = Sz = z if conditions (v) of statement (III) are fulfilled.
The proof of the following statement is similar to that of statement ( 5).
(7) If f is continuous at z and (f , S) is a co.p., then {Sf x 2n } converges to f z.We now prove statement (IV).
(i) Suppose that d is weakly continuous at f z, f is continuous at z and (f , S) is a co.p.From statement (7) it follows that {Sf x 2n } converges to f z.Since f is continuous at z and {f x 2n } converges to z, {f f x 2n } converges to f z.Since d is weakly continuous at f z and both {Sf x 2n } and {f f x 2n } converge to f z, {d(Sf x 2n ,f f x 2n ,a)} converges to zero.By taking x = f x 2n in inequality (20) and then taking limits on both sides of the inequality as n → +∞, we obtain d 2 (f z, z, a) ≤ Ψ (d 2 (f z, z, a) + ).
Since Ψ (t+) < t for all t in (0, ∞), we have d 2 (f z, z, a) = 0. Since this is true for all a in X, we have f z = z.
We note that if f S = Sf , then d(Sf x 2n ,f f x 2n ,a) = d(f Sx 2n ,f f x 2n ,a) = d(f y 2n−1 ,f y 2n ,a) and (f , S) is a co.p.Hence from a perusal of the above proof we have the following statement.
(8) If f S = Sf , f is continuous at z and {d(f y 2n−1 ,f y 2n ,a)} converges to zero for each a in X, then f z = z.
We now resume the proof of statement (IV).
(ii) Suppose that f commutes with each of the maps g, S, and T .Then from equations (3) we have Hence from Proposition 10, it follows that {f y n } is Cauchy.Hence {d(f y 2n−1 ,f y 2n , a)} converges to zero for each a in X. Hence from statement (8) it follows that f z = z if conditions (ii) of statement (IV) are fulfilled.Statement (V) follows from symmetry considerations.
Corollary 26.Suppose that Ψ is monotonically increasing on R + , Ψ (t+) < t for all t in (0, ∞), for all x, y, a in X and that there are sequences {x n } ∞ n=0 and {y n } ∞ n=0 in X such that Then {y n } is Cauchy.Suppose that it converges to an element z of X.Then z is a unique common fixed point of f , g, and S 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 at z, and either f is continuous at z and (f , S) is a co.p. or g is continuous at z and (g, S) is a co.p.
(iii) S is continuous at z, d is weakly continuous at Sz and either (f , S) or (g, S) is a co.p.
(iv) S k is continuous at z and d is weakly continuous at S k z for some positive integer k, and S commutes with either f or g.
(v) S commutes with each of the maps f and g, and S k is continuous at z for some positive integer k.
Remark 27.In Theorem 24 if inequality (18) is replaced with the following more stringent inequality then the weak continuity of d can be dropped from all those numbered statements in which it appears.A similar remark applies to Corollary 26 also.
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 selfmaps on M.
Remark 29.Proposition 28 fails if the condition ∞ n=1 ϕ n (t) < +∞ for all t in R + is replaced by the condition ϕ(t+) < t for all t in (0, ∞).Example 30 illustrates this when g = f , S = T = I (the identity map on M) and Ψ (t) = t for all t in R + .
Example 30.Let M = {x n : n = 1, 2, 3,...}, where x n = n k=1 1/k.Define f : M → M as f x n = x n+1 for all n = 1, 2, 3,.... Define ϕ : R + → R + as ϕ(t) = t/(1 + t) for all t in R + .Then ϕ is a strictly increasing continuous function on R + with ϕ(t) < t for all t in (0, ∞) and for all x, y in M. Evidently for any x in M the sequence {f n x} diverges to +∞ and hence is not Cauchy.
Theorem 31.Suppose that Ψ is right continuous at zero and ρ 2 (f x, gy) ≤ ϕ max ρ 2 (x, y), ρ 2 (x,f x),ρ 2 (y,gy),Kρ(f x,y)ρ(x,gy) + Ψ ρ(f x, y)ρ(x, gy) (30) for all x, y in M. For any x 0 in M, let {x n } ∞ n=1 be defined iteratively as in Theorem 11.Then {x n } is Cauchy.If {x n } converges to an element z of M, then z is a common fixed point of f and g.Further the fixed point sets of f and g are the same.
Theorem 32.Suppose that ϕ(t+) < t for all t in (0, ∞), Ψ is right continuous at zero and ρ 2 (f x, gy) ≤ ϕ Kρ(f x, y)ρ(x, gy) + max ρ 2 (x, y), ρ 2 (x,f x),ρ 2 (y, gy) + Ψ ρ(f x, y)ρ(x, gy) (31) for all x, y in M. For any x 0 in M, let {x n } ∞ n=1 be defined iteratively as in Theorem 11.Then {x n } is Cauchy.If {x n } converges to an element z of M, then z is a common fixed point of f and g.Further the fixed point sets of f and g are the same.
(2) If Sz = T z, then z is a common fixed point of f and S if and only if it is a common fixed point of g and T .
(3) If (f , S) is a w.co.p. and z ∈ S(X), then f z = Sz = z.(4) If (f , S) is a co.p. and S is continuous at z, then f z = Sz = z.
(5) If (f , S) is a w.co.p. and, for some positive integer k, f S k = S k f and S k is continuous at z, then f z = Sz = z.
(6) If f is continuous at z and (f , S) is a co.p., then f z = z.
Finally we conclude the paper with the following open problem.
Open problem.Does Theorem 15 remain valid if the condition ϕ(t+) < t for all t in (0, ∞) is deleted from the hypothesis?Note 40.The results in Naidu and Prasad [4] remain valid if the weak commutativity condition in them is replaced with compatibility condition as introduced in Definition 21.
Remark 0.1 and Example 0.1]) (ii) in a 2-metric space (X, d) every convergent sequence is Cauchy if d is continuous on X [4, Remark 0.2] and (iii) the converse of (ii) is false [4, Remark 0.2 and Example 0.2].