(2009) pp. 3–14
http://ami.ektf.hu
A common fixed point theorem via a generalized contractive condition
Abdelkrim Aliouche
a, Faycel Merghadi
baDepartment of Mathematics, University of Larbi Ben M’Hidi Oum-El-Bouaghi, Algeria
bDepartment of Mathematics, University of Tebessa, Algeria
Submitted 8 January 2009; Accepted 12 June 2009
Abstract
We prove a common fixed point theorem for mappings satisfying a gener- alized contractive condition which generalizes the results of [3, 4, 12, 15, 19, 20, 24] and we correct the errors of [7, 12, 20].
Keywords: Metric space, weakly compatible mappings, common fixed point.
MSC:47H10, 54H25
1. Introduction
Sessa [21] defined S and T to be weakly commuting as a generalization of commuting if for allx∈X.
d(ST x, T Sx)6d(T x, Sx).
Jungck [9] defined S and T to be compatible as a generalization of weakly com- muting if
n→∞lim d(ST xn, T Sxn) = 0
whenever {xn} is a sequence in X such thatlimn→∞Sxn = limn→∞T xn =t for somet∈X. It is easy to show that commuting implies weakly commuting implies compatible and there are examples in the literature verifying that the inclusions are proper, see [9, 21]. Jungck et al [10] definedS andT to be compatible mappings of type (A) if
n→∞lim d ST xn, T2xn
= 0 and lim
n→∞d T Sxn, S2xn
= 0, 3
whenever {xn} is a sequence in X such thatlimn→∞Sxn = limn→∞T xn =t for some t ∈ X. Example are given to show that the two concepts of compatibility are independent, see [10]. Recently, Pathak and Khan [16] definedS andT to be compatible mappings of type (B) as a generalization of compatible mappings of type (A) if
n→∞lim d T Sxn, S2xn 61
2 hlim
n→∞d(T Sxn, T t) + lim
n→∞d T t, T2xni ,
n→∞lim d ST xn, T2xn 61
2 hlim
n→∞d(ST xn, St) + lim
n→∞d St, S2xni ,
whenever {xn} is a sequence in X such thatlimn→∞Sxn = limn→∞T xn =t for somet∈X. Clearly compatible mappings of type (A) are compatible mappings of type (B), but the converse is not true, see [16]. However, compatible mappings of type (A) and compatibility of type (B) are equivalent if S andT are continuous, see [16]. Pathak et al [17] definedS andT to be compatible mappings of type (P) if
n→∞lim d S2xn, T2xn
= 0,
whenever {xn} is a sequence in X such thatlimn→∞Sxn = limn→∞T xn =t for somet ∈X. However, compatibility, compatibility of type (A) and compatibility of type (P) are equivalent if S and T are continuous, see [17]. Pathak et al [18]
defined S and T to be compatible mappings of type (C) as a generalization of compatible mappings of type (A) if
lim
n→∞
d T Sxn, S2xn
6 1 3
hlim
n→∞
d(T Sxn, T t) + lim
n→∞
d T t, S2xn
+ lim
n→∞
d T t, T2xn
i ,
lim
n→∞
d ST xn, T2xn
6 1 3
hlim
n→∞
d(ST xn, St) + lim
n→∞
d St, T2xn
+ lim
n→∞
d St, S2xn
i , whenever {xn} is a sequence in X such thatlimn→∞Sxn = limn→∞T xn =t for some t ∈ X. Compatibility, compatibility of type (A) and compatibility of type (C) are equivalent if S and T are continuous, see [18]. Pant [15] definedS andT to be reciprocally continuous if
n→∞lim ST xn=St and lim
n→∞T Sxn=T t,
whenever {xn} is a sequence in X such that limn→∞Sxn = limn→∞T xn = t for some t ∈ X. It is clear that if S and T are both continuous, then they are reciprocally continuous, but the converse is not true. Moreover, it was proved in [15] that in the setting of common fixed point theorem for compatible mappings satisfying contractive conditions, the continuity of one of the mappings S and T implies their reciprocal continuity, but not conversely.
2. Preliminaries
Definition 2.1 (See [11]). S andT are said to be weakly compatible if they com- mute at their coincidence points; i.e., ifSu=T ufor someu∈X, thenST u=T Su.
Lemma 2.2 (See [9, 10, 16, 17, 18]). If S andT are compatible, or compatible of type (A), or compatible of type (P), or compatible of type (B), or compatible of type (C), then they are weakly compatible.
The converse is not true in general, see [4].
Definition 2.3 (See [13]). S andT are said to be R−weakly commuting if there exists anR >0such that
d(ST x, T Sx)6Rd(T x, Sx) for allx∈X. (2.1) Definition 2.4 (See [14]). S and T are pointwiseR−weakly commuting if for all x∈X, there exists anR >0 such that (2.1) holds.
It was proved in [14] that R-weakly commutativity is equivalent to commuta- tivity at coincidence points; i.e.,SandT are pointwiseR-weakly commuting if and only if they are weakly compatible.
Lemma 2.5 (See [22]). For any t∈(0,∞),ψ(t)< t ifflimn→∞ψn(t) = 0, where ψn denotes the n-times repeated composition ofψwith itself.
Several authors proved fixed point and common fixed point theorem for map- pings satisfying contractive conditions of integral type, see [1, 3, 4, 5, 6, 7, 12, 19, 20]. The following theorem was proved by [3].
Theorem 2.6 (See [3]). Let A, B, S and T be self-mappings of a metric space (X, d)satisfying
S(X)⊂B(X) and T(X)⊂A(X), Z d(Sx,T y)
0
ϕ(t) dt6ψ
Z M(x,y) 0
ϕ(t) dt
!
for all x, y ∈X, ψ:R+ →R+ is a right continuous function such that ψ(0) = 0 andψ(s)< sfor alls >0andϕ:R+→R+is a Lebesgue integrable mapping which is summable and satisfies
Z ǫ 0
ϕ(t) dt >0, M(x, y) = max
d(Ax, By), d(Sx, Ax), d(T y, By),1
2[d(Sx, By) +d(T y, Ax)]
. If one of A(X), B(X), S(X)andT(X)is a complete subspace ofX, thenAand S have a coincidence point and B and T have a coincidence point. Further, if S andAas well asT andB are weakly compatible, thenA, B, SandT have a unique common fixed point inX.
Recently, Zhang [24] and Aliouche [2] proved common fixed point theorems using generalized contractive conditions in metric spaces.
LetA∈(0,∞],R+A= [0, A)and F:RA+→Rsatisfying
(i)F(0) = 0andF(t)>0 for eacht∈(0, A), (ii)F is nondecreasing onR+A,
(iii)F is continuous.
Define̥[0, A) ={F :F satisfies (i)–(iii)}.
Lemma 2.7 (See [24]). Let A ∈ (0,∞], F ∈ ̥[0, A). If limn→∞F(ǫn) = 0 for ǫn ∈R+A, thenlimn→∞ǫn= 0.
The following examples were given in [24].
(i) LetF(t) =t, thenF ∈̥[0, A)for eachA∈(0,∞].
(ii) Suppose thatϕis nonnegative, Lebesgue integrable on[0, A)and satisfies Z ǫ
0
ϕ(t) dt >0for eachǫ∈(0, A).
LetF(t) =Rt
0ϕ(s) ds, thenF ∈[0, A).
(iii) Suppose thatψis nonnegative, Lebesgue integrable on[0, A)and satisfies Z ǫ
0
ψ(t) dt >0 for eachǫ∈(0, A) andϕis nonnegative, Lebesgue integrable onh
0,RA
0 ψ(s) ds
and satisfies Z ǫ
0
ϕ(t) dt >0for each0< ǫ <
Z A 0
ψ(s) ds.
LetF(t) =RR0tψ(s) ds
0 ϕ(u) du, thenF ∈̥[0, A).
(iv) If G ∈ [0, A) and F ∈ ̥[0, G(A −0)), then a composition mapping F ◦G∈̥[0, A). For instance, letH(t) =RF(t)
0 ϕ(s) ds, then H ∈ ̥[0, A)when- everF∈̥[0, A)andϕis nonnegative, Lebesgue integrable on̥[0, F(A−0))and satisfies
Z ǫ 0
ϕ(t) dt >0for eachǫ∈(0, F(A−0)).
LetA∈(0,∞]andψ:RA+→R+satisfying (i)ψ(t)< tfor allt∈(0, A)
(ii)ψis upper semi-continuous.
(iii)ψis nondecreasing onR+A,
DefineΨ[0, A) ={ψ:ψsatisfies (i)-(iii)}.
3. Main results
Theorem 3.1. Let (X, d)be a metric space and D= sup{d(x, y) :x, y∈X}. Set A=D if D=∞ andA > D ifD <∞. LetA1, A2, S andT be self-mappings of (X, d)satisfying
A1(X)⊂T(X)andA2(X)⊂S(X),
F(d(A1x, A2y))6ψ(F(L(x, y))) (3.1) for allx, y inX, where
L(x, y) = maxn
d(Sx, T y), d(A1x, Sx), d(T y, A2y),1
2[d(Sx, A2y) +d(A1x, T y)]o , F ∈ ̥[0, A) and ψ ∈ Ψ[0, F(A−0)) for all A ∈ (0,∞]. Suppose that the pair (A1, S)is weakly compatible and there exists w∈C(A2, T): the set of coincidence points ofA2 andT such thatA2T w=T A2w. If one ofA1(X), A2(X), S(X)and T(X) is a complete subspace ofX, then A1, A2, S and T have a unique common fixed point in X.
Proof. Letx0be arbitrary point inX. Inductively, we can define a sequence{yn} in X such that
y2n =A1x2n=T x2n+1andy2n+1=Sx2n+2=A2x2n+1
for all n = 0,1,2, . . .. As in the proof of [2], {yn} is a Cauchy sequence in X. Assume thatS(X)is complete. Therefore
n→∞lim A1x2n= lim
n→∞T x2n+1= lim
n→∞A2x2n+1= lim
n→∞Sx2n+1=z=Su for some u∈X. IfA1u6=z using (3.1) we obtain
F(d(A1u, A2x2n+1))6ψ(F(L(u, x2n))) where
L(u, x2n) = maxn
d(Su, T x2n+1), d(A1u, Su), d(T x2n+1, A2x2n+1), 1
2[d(Su, A2x2n+1) +d(A1u, T x2n+1)]o . Letting n→ ∞, we get
F(d(A1u, z))6ψ(F(d(A1u, z)))< F(d(A1u, z))
which is a contradiction and so z = A1u = Su. If z 6= A2w, applying (3.1) we obtain
F(d(A1u, A2w))6ψ(F(d(A1u, A2w))) where
L(u, v) = maxn
d(Su, T w), d(A1u, Su), d(T w, A2w),1
2[d(Su, A2w) +d(A1u, T w)]o . Hence
F(d(z, A2w))6ψ(F(d(z, A2w)))< F(d(z, A2w)).
which is a contradiction and soz=A1u=Su=A2w=T w.
Since the pairs(A1, S)is weakly compatible and there existsw∈C(A2, T)such that A2T w=T A2w, we haveSz=A1z andT z=A2z.
IfA1z6=z we have by (3.1)
F(d(A1z, A2w))6ψ(F(L(z, w))) where
L(z, w) = maxn
d(Sz, T w), d(A1z, Sz), d(Bw, A2w),1
2[d(Sz, A2w) +d(A1z, T w)]o . Therefore
F(d(A1z, z))6ψ(F(d(A1z, z)))< F(d(A1z, z)) and so A1z=Sz=z. Similarly, we can prove that A2z=T z=z.
The proof is similar when T(X) is assumed to be a complete subspace of X.
The case in whichA1(X)or A2(X)is a complete subspace ofX is similar to the case in which T(X)or S(X) respectively is complete since A1(X)⊂T(X) and A2(X)⊂S(X). The uniqueness ofz follows from (3.1).
Theorem 3.1 generalizes Theorem 2.6 of [3].
Corollary 3.2. Let (X, d) be a metric space and D = sup{d(x, y) : x, y ∈ X}.
Set A=D if D =∞ andA > D if D <∞. Let {Ai},i = 1,2, . . . , S andT be self-mappings of (X, d)satisfying
A1(X)⊂T(X)andAi(X)⊂S(X),i>2 and
F(d(A1x, Aiy))6ψ(F(Li(x, y))),i>2 for allx, y inX, where
Li(x, y) = maxn
d(Sx, T y), d(A1x, Sx), d(Aiy, T y),1
2[d(Sx, Aiy) +d(A1x, T y)]o , F ∈ ̥[0, A) and ψ ∈ Ψ[0, F(A−0)) for all A ∈ (0,∞]. Suppose that the pair (A1, S)is weakly compatible and there existsw∈C(Ai, T): the set of coincidence points ofAiandT such thatAiT w=T Aiwfor somei>2. If one ofAi(X), S(X) andT(X)is a complete subspace ofX. ThenAi, S andT have a unique common fixed point in X.
Ifϕ(t) = 1in Corollary 3.2, we get a generalization of a theorem of [15]. The following example illustrates our corollary 3.2.
Example 3.3. LetX = [0,10]be endowed with the metricd(x, y) =|x−y|,
Sx=
0, ifx= 0, x+ 8, ifx∈(0,2], x−2, ifx∈(2,10],
T x=
0, ifx= 0, x+ 5, ifx∈(0,2], x−2, ifx∈(2,10],
A1x=
(3, ifx∈(0,2],
0, ifx∈ {0} ∪(2,10], A2x=
(0, ifx∈[0,2], 4, ifx∈(2,10], A3x=
(0, ifx∈[0,2],
5, ifx∈(2,10], A4x=
(0, ifx∈[0,2], 6, ifx∈(2,10], Aix=
(2 + 2i, ifx∈(0,2],
0, ifx∈ {0} ∪(2,10], for alli >4.
The pair (A1, S) is weakly compatible, but it is not compatible of type (A), (B), (P) and (C), see [6].
A1(X)⊂T(X)andAi(X)⊂S(X).
The pair(Ai, T), i >4, is weakly compatible because Ai and T commute at their coincidence point x= 0, but it is not compatible of type (A), (B), (P) and (C).
Letxn= 2 + 1n. We haveT xn= n1 andAixn= 0,hence
n→∞lim T xn= lim
n→∞Aixn =t= 0.
In the other hand, AiT xn = Ai(n1) = 2 + 2i and T Aixn = T0 = 0 and so limn→∞d(AiT xn, T Aixn) = 2 +2i 6= 0.Therefore, the pair(Ai, T)is not compat- ible.
A2ixn =Ai0 = 0and T2xn =T n1
= 5 +n1, so limn→∞
T Aixn−A2ixn = 0 andlimn→∞
AiT xn−T2xn
= limn→∞(3 +n1−2i)6= 0for alli >3. Then,(Ai, T) is not compatible of type (A).
n→∞lim
AiT xn−T2xn
= 3−2 i > 1
2 hlim
n→∞|AiT xn−Ai0|+ lim
n→∞
Ai0 −A2ixn i
=1 2
2 +2 i
=1 i + 1, hence (Ai, T)is not compatible of type (B).
limn→∞
A2ixn−T2xn
= limn→∞(5 + n1) = 5 6= 0. Therefore, (Ai, T)is not compatible of type (P).
n→∞lim
AiT xn−T2xn
= 3−2 i
>1 3
h lim
n→∞|AiT xn−Ai0|+ lim
n→∞
Ai0−T2xn
+ lim
n→∞
Ai0 −A2ixn
i
=1 3
7 + 2
i
fori >4. So, the pair(Ai, T)is not compatible of type (C).
It can be verified that the pairs (A2, T) , (A3, T) and(A4, T)are not weakly compatible because x = 6 is a coincidence point ofA2 and T, but A2T6 = 4 6=
T A26 = 2,x= 7is a coincidence point ofA3andT, butA3T(7) = 56=T A3(7) = 3 andx= 8is a point of coincidence for A4andT, but A4T(8) = 66=T A4(8) = 4.
Now, we begin to verify the rest of conditions of Corollary 3.2. Let F(t) = ln(1 +t)and ψ(t) =ht, where06h <1 andt >0. Set
R= ln(1 +|A1x−Aiy|)−hmax
ln(1 +|Sx−T y|),ln(1 +|A1x−Sx|), ln(1 +|Aiy−T y|),
1
2[ln(1 +|A1x−T y|) + ln(1 +|Sx−Aiy|)]
We have the following cases. Ifx= 0andy= 0 we getR60 for all06h <1.If x= 0andy∈(0,2], we get
R= ln
3 +2 i
−hmax
ln (y+ 6),ln y+ 4−2i ,
1 2
ln (y+ 6) + ln 3 +2i
60
forh> ln(3+2i)
3 ln 2 and so there exists0 6h <1.Ifx= 0 andy∈(2,10],we get R=−hmax{ln (y−1),ln (y−1),ln (y−1)}60
for all06h <1. Ifx∈(0,2]andy= 0, we get R= ln 4−hmax
ln (x+ 9),ln (x+ 6),
1
2[ln 4 + ln (x+ 9)]
60 forh> ln 4
ln 11 and so there exists06h <1.Ifx∈(0,2]andy∈(0,2],we get R= ln
2−2
i
−hmax
ln(x−y+ 4),ln(x+ 6),ln y+ 4−2i ,
1 2
ln (y+ 3) + ln x+ 7−2i
60
forh> ln(3−2i)
ln 8 .Hence, there exists06h <1. Ifx∈(0,2]andy∈(2,10],we get R= ln 4−hmax
ln (x+ 11−y),ln (x+ 6),ln (y−1),
1
2[ln (|5−y|+ 1) + ln (x+ 9)]
60
forh> ln 11ln 4.Hence, there exists06h <1.Ifx∈(2,10]andy= 0, we get R=−hmax
ln (x−1),ln (x−1),0,1
2ln (x−1)
60
for allh>0.Hence, there exists06h <1.Ifx∈(2,10]andy∈(0,2],we get R= ln
3 +2
i
−hmax
ln (|x−(y+ 7)|+ 1),ln (x−1), ln y+ 3−2i
,12
ln (y+ 5) + ln
x−4−2i + 1
60
forh> ln(3+2i)
ln 9 .Hence, there exists06h <1.Ifx, y∈(2,10]we get R=−hmax
ln (|x−y|+ 1),ln (x−1),ln (y−1),
1
2[ln (y−1) + ln (x−1)]
60
for all06h <1.
Now, we verify that (A2, T) and (A3, T) satisfy all the conditions of Theo- rem 4.2. Set
R1=
Z |A1x−A2y|
0
1 1 +tdt−
−hmax
R|Sx−T y|
0
1
1+tdt,R|A1x−Sx|
0
1
1+tdt,R|A2y−T y|
0
1 1+tdt,
1 2
hR|A1x−T y|
0
1
1+tdt+R|Sx−A2y|
0
1 1+tdti
We have the following cases. If x= 0andy = 0we get R160for all0 6h <1.
Ifx= 0andy∈(0,2], we get R1=−hmax
ln (y+ 6),0,ln (y+ 6),1 2[y+ 6]
60
for all06h <1.Ifx= 0andy∈(2,10],we get R1= ln 5−hmax
ln (y−1),ln (|y−6|+ 1),
1
2[ln (y−1) + ln 5]
60 forh> ln 5
ln 9,hence there exists06h <1.Ifx∈(0,2]andy= 0, we get R1= ln 4−hmax
ln (x+ 9),ln (x+ 6),0,
1
2[ln 4 + ln (x+ 9)]
60 for allh> ln 4
ln 11.Hence, there exists06h <1.Ifx∈(0,2]andy∈(0,2],we get R1= ln 4−hmax
ln (4 +x−y),ln (x+ 6),ln (y+ 6),
1
2[ln (y+ 3) + ln (x+ 9)]
60
forh> ln 4ln 8. Hence there exists06h <1. Ifx∈(0,2]andy∈(2,10],we get R1= ln 2−hmax
ln 11,ln (x+ 6),ln (|y−6|+ 1),
1
2[ln (|5−y|+ 1) + ln (x+ 5)]
60
forh> ln 11ln 2.Hence,there exists06h <1.Ifx∈(2,10]andy= 0, we get R1=−hmax
ln (x−1),ln (x−1),1
2ln (x−1)
60
for all06h <1. In the same manner, ifx∈(2,10]andy ∈(0,2],we getR1 60 for all06h <1.Ifx∈(2,10]andy∈(2,10],we get
R1= ln 5−hmax
ln (|x−y|+ 1),ln (x−1),ln (|y−6|+ 1),
1
2[ln (y−1) + ln|x−6|+ 1]
60
for h> ln 5ln 9. Hence, there exists0 6h < 1.Similarly, we can prove the conditions of Theorem 4.2 if we take the mapping A3 instead ofA2. Finally we remark that all conditions of our theorem are verified and 0is the unique common fixed point ofAi, S andT.
The following example support our Theorem 3.1.
Remark 3.4. In this example, Theorem 2.6 of [3] is not applicable since the pair (A2, T)is not weakly compatible, but Theorem 3.1 is applicable. Also, a theorem of [15] forAi=A2for alli>2is not applicable since the pairs(A1, S)and(A2, T) are not compatible. In the same manner, Theorem 1 of [12] is not applicable.
Remark 3.5. In the proof of Lemma 1 of [20] and Theorem 2.1 of [7], the authors applied the inequality
a6b+c=⇒ Z a
0
ϕ(t)dt6 Z b
0
ϕ(t)dt+ Z c
0
ϕ(t)dt which is false in general as it is shown by the following example.
Example 3.6. Letϕ(t) =t,a= 1,b= 12 andc= 34. Then1< 12+34, but Z 1
0
ϕ(t)dt= 1 2 >
Z 12
0
ϕ(t)dt+ Z 34
0
ϕ(t)dt
= 1 8+ 9
32 =13 32.
To correct these errors, the authors should follow the proof of Theorem 2 of [19].
Remark 3.7. In the proof of Theorem 1 of [12], the authors applied the inequality
n→∞lim d(xn, xn+1) = 0 =⇒ {xn} is a Cauchy sequence
which is false in general. It suffices to take xn =n1,n∈N∗. Thus, To correct this error, the authors should follow the proof of Theorem 2 of [19].
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A. Aliouche
Department of Mathematics
University of Larbi Ben M’Hidi Oum-El-Bouaghi 04000
Algeria
e-mail: alioumath@yahoo.fr F. Merghadi
Department of Mathematics University of Tebessa 12000
Algeria
e-mail: faycel_mr@yahoo.fr
