OBSERVATIONS ON A VARIANT OF COMPATIBILITY

We consider a variation of the concept of compatible maps introduced by Hicks and Saliga [1], and obtain generalizations ofresults by Hicks and Saliga and others.

x for all x E X and for any >0.A map S:D( c_ X) -, X is w-continuous at p E D iff whenever {xn is a sequence in D such that Xn p 6 D, then Sxn Sp.For further discussion of d-complete topological spaces and symmetdcs/semi-metdcs see [6].
In this paper we shall focus on the Hicks and Saliga compatibility concept.However, the concept was introduced in [1] for functions S,T :D X (D C_ X) and used on subsets C of the domains D, even though the precise meaning of this compatibility on subsets C was not made clear by the definition.We shall therefore begin by defining what we mean by S being compatible with T so as to (hopefully) preserve the sense intended by Hicks and Saliga.
DEFINITION 1.1.Let X be a topological space, let C c_ D _C X, and let T,S:D X. S is compatible with T on C iff whenever {Xn is a sequence in C such that Sxn is in D and Tx, SXn p D, then TSxn Sp.If C D, we say S is compatible with T. It follows that if C D X, S is compatible with T iffwhenever {xn} is a sequence in X such that Sn,Tn p X, then TSxn Sp.
As we shall see, the relation, "compatible with ", is not necessarily commutative, whereas the concept of compatible pairs {S, T} introduced in [2] is.Although the metric space definition of compatible pairs extends naturally to a topological space having a symmetric, in this paper we are more interested in a property shared by these two compatibility concepts, namely, weak compatibility.Weak compatibility was defined in [7] for semi-metric spaces.We now define it for any set X. DEFINITION 1.2.Let X be any set, let C C_ D C_ X, and let T,S" D X. S and T are weakly compatible on C iff x E C and Tx Sx E D) = (STx TSx).If C D, we say that S and T are weakly compatible or the pair S,T} is weakly compatible.
In the following N will denote the set of positive integers, and for k N, Nk is the set of all n N such that n < k.If S is a map, we shall write Sx for S(x) when convenient and the meaning is clear.Moreover, we require that the topological spaces (X, t) be Hausdorff (which we designate "T2") to ensure that converging sequences have unique limits (See example 2.2 in [7]).
2. TltEOREMS AND RESULTS.PROPOSITION 2.1.Let (X, t) be a T2 topological space and let C C_ D C_ X.Let S,T:D X and suppose that S is compatible with T on C.
1.If x C and Sx Tx D, then T2x TSx STx S2x.
2. If {x} is a sequence in C such that Sxn E D for n N and Sxn, Txn p D, and if T is w- continuous at p, then Sp Tp.
3. If C=D=X and both S and T are w-continuous, then T is compatible with S.
PROOF.Suppose that x C and Sx Tx D. Let xn x for n E N. Then Sxn,Txn Tx (//D, and TSx --n InooTSxn STx since S is compatible with T on C; i.e., the conclusion 1. holds.To see that 2. is true, note that TSxn "-+ Sp since S is compatible with T on C.But since T is w-continuous at p and {Sxn} is a sequence in D convergent to p, TSxn -+ Tp.But (X, t) is T2 and therefore limits of sequences are unique; i.e., Tp Sp.
And to prove 3., suppose Txn, Sxn "-p X.Since S is w-continuous, STxn Sp.But Sp Tp by 2. and therefore STxn Tp; i.e., T is compatible with S. [2 Note that Proposition 2.1.1 tells us that ifS is compatible with T (on C), then the pair {S, T} is weakly compatible (on C), even though-as the next example shows-T is not compatible with S. The following example also shows us that the conclusion of Proposition 2.1.2need not hold if S is compatible with T and T is no_.!t continuous at p, even though S is continuous at p. EXAMPLE 2.1.Let X [0, 1] with d(x, y) [x Y I, let S (the identity map) and define -) and Tx 0 if x , 1]. ,Now, as will be shown T:X X by Tx--(2x+3) $' if x[0, momentarily, is compatible with (any) T and certainly is continuous.But if x from the left, Tx , ITx , whereas T(1/2) 0; i.e., T is not compatible with I.And T() 0 # 1 / 2 I(!)2 ,so the conclusion 2 in Proposition 2.1 does not hold.
In that which follows, we shall use the collapsing principle for series.Thus, if ak is a sequence of numbers, then (a a/) a a for n E N. TREOREM 2.1.Let (X, t) be a d-complete T2 topological space and let D be a closed subset of X.Let S,T:D X where S(D) C_ D N T(D).Suppose there is a map c:D [0, oo) such that d(Tx, Sx) < c(Tx) c(Sx) for x Tq(D).
Then, if xo D _q sequences {x },{y in D such that y Sxa.t Tx for n N and y p D.
Moreover, ifT is w-continuous at p and S is compatible with T on T " (D), then Sp=Tp.
PROOF.Since S(D) _C T(X), given xo E D, we can choose x D such that Tx Sxo.We can then choose x2 D such that Tx2 Sx.In general, given x, D for Nk such that Tx,= Sxi., we can choose xk+ E D such that Txk+ Sx.Thus, by induction, a sequence {Yn} ofthe type cited in the statement of the theorem exists.Since Tx Sxk.D for k N, x T (D) for k N, so for k E N we can write: d(yk, Yk+l) d(Txk, Sxk) _< a(Tx)-a(Sxk) a(yk)-c(yk+) which by the collapsing principle implies d(yk, Yk+l) a(yl) c(y.+l) _< a(yl), for n e N.
k=l Thus d(Yk, Yk+l) < oo.Therefore, Yk P E X since (X, t) is d-complete.
k=l Consequently, Sxk, Txk -p.But SXk E D for k E N and D is closed, so that p E D. Moreover, if S is compatible with T on T "l (D) and T is w-continuous at p, since Xk 6 T "I (D) for k 6 N, Proposition 2. I.
(with C T l (D)) implies that Tp Sp.El Example 2.1 shows us that even though the identity map on a space X is compatible with any map on X, 3 maps S on X not compatible with I.The following proposition says that nice things happen when a function S is compatible with I.
PROPOSITION 2.2.Let X be a T2 topological space and let D C_ X. Suppose S:D X and is the identity map.Then is compatible with S.Moreover, if S is compatible with and Sx. and {x. are sequences in D which converge to p 6 D, then Sp p.And if S is w-continuous, then S is compatible with I. PROOF.To see that is compatible with S, let {x.be a sequence in D such that Ix. x, Sx. p 6 D. Then SIxn Sx. -p Ip, so is (trivially) compatible with S.
If S is compatible with and {x.}, Sx. are sequences in D which converge to p E D, then p= Sp by Proposition 2. (with T I).Now suppose that S is w-continuous and let {x.}, SXn be sequences in D such that Ix., Sx.Sp by continuity.Thus, Sxn ISxn Sp, and hence S is compatible with I. El The last sentence in Proposition 2.2 prompts the question, "If S:D X is compatible with (the identity map), is S continuous on D?" The next example tells us that the answer is "no", even if S:X X. EXAMPLE 2.2.Let X [0, 1] with the usual metric and define S: X X by Sx if x6[0,] andSx=0 ifx6(,l].Now ifSx.p, then p6{0,1}.But ifIxn(=Xn) 0, Sx.
COROLLARY 2.1.Let (X, t) be a d-complete T topological space and let S:X -X.Ifthere exists a map a: X [0, oo) such that for x X d(x, Sx) < a(x)-a(Sx), then for any x X, Sn(x) p for some p Px X.If S is compatible with the identity map, then Sp -----p.
(To see that P=Px need not be unique, let S I, the identity map.) PROOF.If we let T I, the identity map, in Theorem 2.1, then Yn xn SXn-l for n E N. Thus, Yn--Snxo for n E N. Since I(X) X, the conclusion follows.El NOTE 2.1.Corollary 2.1 is the topological version of Caristi's Theorem [8] for complete metric spaces.Caristi required that c be lower semi-continuous, whereas we required that S be compatible with the identity map.Dien [I0] noted -as Browder [9] had already known in 1975 -that for metric spaces, the lower semi-continuity requirement on c can be dropped by requiring that S be continuous.In view ofExample 2.2, Dien's comment suggests that Corollary 2. is of interest.COROLLARY 2.2.Let (X, t) be a d-complete T2 topological space and let T:D X, where D is a closed subset ofX.IfD C_ T(D) and if3 a map c: X [0, oo) such that d(Tx, x) _< c(Tx)-c(x), for x /T "l (D), then any xo E D determines a sequence ix.} with x,.= Txn such that xn p E D. IfT is w-continuous at p, then Tp p. PROOF.Let S I, the idemity map in Theorem 2.1., and note that is compatible with T by Proposition 2.2.C! TREOREM 2.2.Let (X, t) be a d-complete Hausdorff topological space and let D be a closed subset of X. Suppose S, T: D X and that S(D) _C D f T(D).If 3 maps a, :X [0,oo) and r (0, 1) such that (*) d(Sx, Sy) _< r d(Tx, Ty) + (a(Tx) a(Sx) + ((Ty) -/(Sy)) for x,y T'I(D), then for any xo D, S sequences ix=}, {y,} such that y= Tx, Sx.l for n N and y,p p E D. If S is compatible with T on T " (D) and T is w-continuous at p, then Tp Sp p, and p is the only common fixed point of S and T.

k=-I
We drop the second tcfm {n the {eR member ofthe last inequality above to obtain, (I r) d(Yk, Yk+l) < a(yo) +//(y, + r d(yo, y, M, a constant _> 0. k=l Thus, for n E N: 'd(yk, Yk+l) _< M(I r) "l a nonnegative re,a{, since 0 < r < I. Therefore, k=l d(yk, Yk+l) < oo, and {Yk converges to p 6 X since X is d-complete.So Txk, Sxk -+ p Pxo.But Sxk 6 D for k 6 N and D is closed, which implies that p 6 D. By the above, {Xk is a sequence in C T "l (D) such that Sxk 6 D for k 6 N, and Txk,Sxk p 6 D. Therefore, if S is compatible with T on T q (D) and T is w-continuous at such a p, Proposition 2. (2).implies that Sp Tp.But since p D, Sp(=Tp) E D, so that p 6 T "l (D).Thus Pfoposition 2. l(1) with C T "l (D) tel{s us that S2p STp TSp T2p (2.1)Moreover, since Tp Sp E D and TSp SSp D, we know p, Sp T "I (D).So (*) implies d(Sp, SSp) _< r d(Tp, TSp) + (a(Tp) a(Sp)) + ((TSp) (STp)) d(Tp, TSp) + 0; i.e., d(Sp, SSp) < r d(Sp, SSp) by (2.1).Since 6 (0, I), and d(x, y) 0 implies x=y, we have Sp SSP.Then Sp SSp TSp by (2.I), so that Sp is a common fixed point S and T. That Sp is the only common fixed point of S and T follows easily from (*). {"{ If we let , be identically 0 in Theorem 2.2, we obtain the sufficency portion of Theorem 3. in [I], with the assumption that the phrase, "g is compatible with f on f(C)" in the statement of the Theorem 3. conforms to our definition.Note also that the argument given in the proof of Theorem 3. to prove the necessity portion could be used to obtain a necessary and sufficient condition that the T in Theorem 9_.2 have a fixed point.Our final result appeals to the following lemma.LEMMA 2.1.Let (X, t) be a d-complete topological space with d symmetric, and let D C_ X.Let A,B,S,T: D --.X, such that A(D) C_ T(D) and B(D) C S(D).If   (i) d(Ax, By) < Q(m(x, y)) for x,y 6 X, where m(x, y)= max d(Ax, Sx), d(By, Ty), d(Sx, Ty) }, then the sequences {y.defined below in (ii) exist for any x 6 D and converge to a point p 6 X.
(ii) y2.Sx2.Bx2.-l, y2..l Ax2._ Tx2.., and x. 6 D for n 6 N tJ {0}.Lcmma 2. is proved in [7].It is proved for a semi-metric d, but is valid if d is a symmetric.In [7] the sequence {y, is proven to be "d-Cauchy" which justifies the conclusion above that, with the above hypothesis of"d-completeness", {y.converges to a point p X. THEOREM 2.3.Let (X, t) be a d-complete Hausdorff topological space with d symmetric and let D be a closed subset of X. Suppose A,B,S,T:D --, X, A(D) C_ D N T(X) and B(D) C_ D N SO:)).IfS and T are w-continuous, if (i) in Lemma 2.1 holds, and if A(B) is compatible with S(T), then A,B,S, and T have a unique common fixed poim.
PROOF.Let Xo 6 D and let {y.}, {x.be sequences assured by Lemma 2. I, so that we have Yn, AX2n, BX2n-], SX2n, Tx.. p 6 X.We know that p 6 D since Ax2, 6 D for all n and D is closed.Since A is compatible with S and S is w- cominuous, and since Ax.,Sx2n---* p 6 D, Proposition 2.1 implies Ap Sp, and A2p ASp SAp S2p.Similarly, (2.2) Bp Tp, and Bp BTp TBp T2p._< Q(max{ 0, 0, d(Sp, Tp)}) Q(d(Ap, Bp)), by (2.2) and (2.3).Therefore, Ap Bp by Note 2.2, and hence Ap Bp Sp Tp. (2.4) , (2.3), and (2.4) above.Therefore, Ap BAp.By symmetry we also have Bp ABp.Therefore, the above equalities yield Ap AAp BAp SAp TAp; i.e., Ap is a common fixed point of A,B,S, and T. That x Ap is the only common fixed point of A,B,S and T follows from the contraction property (i).I-I We now state a result described in the final paragraph in [l l] which is a variation of their Theorem 5., page 793.THEOREM 5B.[l Let A, B, S, and T be mappings of a d-topological space (X,t) into itself satisfying the following conditions: (a) A(X) C_ T(X) and B(X) C_ T(X), (b) d(Ax, By _< b( max{ d(Ax, Sx), d(By, Ty), d(Sx, Ty) }) for all x,y 6 X, where b: [o, oo)  [0, oo), b(0) 0, b is nondecreasing, b is upper continuous, and b(t) < for > 0, (c) S and T are w-continuous, (d) A, S and B,T satisfy the Hicks and Saliga definition of compatibility, and (e) d is a continuous symmetric.
Then A, B, S, and T have a unique common fixed point, provided 3 xo,xl e X such that Axo Tx and On(d(Axo, BxI)) < OO.
Our Theorem 2.3 has Theorem 5B as a corollary.In fact, Theorem 5B has the following restrictions in the hypothesis not required by Theorem 2.3.The contractive function is upper semi- continuous.A,S and B,T are compatible in the sense of"Hick's and Saliga's definition", which suggests that A(B) is compatible with S(T) and conversely.They also require that the symmetric d be continuous and that the domains of A,B,S, and T be X.
We should note that Harder and Saliga require that 3 Xo, xl X such that Axo Txl and On(d(Axo, Bx)) < oo.Thus, their requirement for convergence for one appears to be lighter than n0 our requirement that Qn(t) < oo for all [0, oo).However, the next result states that because of n the requirements imposed on in Co) of the hypothesis, the inequality on(t) < oo holds for all if it n=0 holds for one (0,oo).PROPOSITION 2.3.Suppose that :[0, oo)  [0, oo), is upper-semicontinuous, nondecreasing, and that O(t) < for > 0. If #(to) < oo for some to > 0, then #(t) < oo for all > 0.
Then ,(t) < t, and O2(t) O(O(t)) 0n(t).Thus n lira #(t) c > 0, where #(t) is upper semi-continuous, :1 6 > 0 such that O(t) < c for (c 6, c+ 6).Since O"(t) --, c as n--, oo, om(t) e (c-6, c+ 6) for some m.Therefore, we have the contradiction om/(t) < c _ om+l(t), bythe choice Now suppose that to > 0 and On(to) < oo.Let e (0, oo).By (1)', we cam choose m N such that om(t) < to, so om/(t) _ O(to), and in general we have om/n(t) _ O(to) for n N, since is nondecreasing.But then, om/n(t) _ on(to) M < oo, and therefore, n=! rl Ok(t) 'Abk(t) + 'Abm+n(t) <_ bk(t) + M < k=l k=l n=! k=l 3. RETROSPECT.In Definition 1.2 we defined the concept of a weakly compatible pair, and in Proposition 2.1 we proved in part 1. that "S compatible with T" implies that {S, T} is a weakly compatible pair.We close with an example which shows that even though S is not compatible with T and T is not compatible with S, S, T} may be a weakly compatible pair.Moreover, since in this example S and T have a common fixed point and satisfy a contractive condition, d(Sx, Sy) < d(Tx, Ty), we ask, to what extent can weak compatibility be used in lieu of the stronger forms of compatibility and still produce common fixed points.'? (A partial answer can be found in Theorems 3.1 and 3. (3/4)-(x/2) and Sx=2-3x forx6[0,1/2], and Ax= and Sx=0 for xe(,l].Then A,S: D X and A(X) C_ S(X), and d is a semi-metric but not a metric (no triangle inequality).Now {A, S is weakly compatible since Ax Tx iff x , and A() S(1/2) AS() SA(1/2).On the other hand consider x. 1 / 2 ()n for n>l.Then Ax., Sx, ---, 1/2, SAx, 0 but A() 1 / 2 and AS 3/4 and S()=1/2.Thus A is not compatible with S and S is not compatible with A. But 1 / 2 is a common fixed point ofA and S, and a quick check shows that d(Ax, Ay) <_ ] d(Sx, Sy).
We conclude by observing that the expression (*) in our Theorem 2.2 was prompted by the analogous but more general expression (2.1) used by Dien in Theorem 2.1 [9].The left member of Dien's contractive expression (2.1) was d(Sx, Ty) and our expression (*) used d(Sx,Sy).However, Dien's theorem is for metric spaces, and a check of his proof(s) reveals the central role of the triangle inequality, which we did not have for topological spaces with only a symmetric.Note also that the right member of(2.I) [9] contained an expression ofthe form ' [i(Lx) bi(Sx)] (3.I) iffil where qi" X ---, [0, oo) for /Nn.But if we let c bi, then o: X --, [0, oo) and (3.1) can be written [c(Ix) c(Sx)] i.e., no generality is gained by using n functions qi in lieu of one function