http://www.uab.ro/auajournal/ doi: 10.17114/j.aua.2015.43.08
A GENERAL FIXED POINT THEOREM FOR A PAIR OF MAPPINGS IN PARTIAL METRIC SPACES
V. Popa, A.-M. Patriciu
Abstract. In this paper a general fixed point theorem for a pair of mappings satisfying a new type of implicit relation, different than the type from [8], which gen- eralize the results from Theorem 3.5 [4], Theorem 4.5 [4], Theorem 2.3 [2], Corollary 2.4 [2], Corollary 2.8 [2] and other results from [1] is proved.
2010Mathematics Subject Classification: 54H25, 47H10.
Keywords: fixed point, partial metric spaces, implicit relation.
1. Introduction
In 1994, Matthews [5] introduced the concept of partial metric space as a part of the study of denotional semantics of dataflow networks and proved the Banach contraction principle in such spaces.
Many authors studied some contractive conditions in complete partial metric spaces.
Quite recently, in [1], [2], [4], some fixed point theorems under various contractive conditions in partial metric spaces are proved.
Several classical fixed point theorems and common fixed point theorems have been unified considering a general condition by an implicit relation in [6], [7] and in other papers. Recently, the method is used in the study of fixed points in metric spaces, symmetric spaces, quasi - metric spaces, ultra - metric spaces, convex metric spaces, compact metric spaces, paracompact metric spaces, in two and three metric spaces, for single - valued mappings, hybrid pairs of mappings and set - valued mappings.
Quite recently, the method is used in the study of fixed points for mappings satisfying contractive/extensive conditions of integral type, in fuzzy metric spaces, probabilistic metric spaces, intuitionistic metric spaces andG- metric spaces. Also, this method allows the study of local and global properties of fixed point structures.
The study of fixed point for self mappings in complete partial metric spaces satisfying an implicit relation is initiated in [8].
The purpose of this paper is to prove a general fixed point theorem for a pair of mappings satisfying a new type of implicit relation, different than the type from [8], which generalize Theorem 3.5 [4], Theorem 4.5 [4], Theorem 2.3 [2], Corollary 2.4 [2], Corollary 2.8 [2] and other results from [1].
2. Preliminaries
Definition 1 ([5]). Let X be a nonempty set. A functionp:X×X →R+ is said to be a partial metric on X if for anyx, y, z ∈X, the following conditions hold:
(P1) :p(x, x) =p(y, y) =p(x, y) if and only if x=y, (P2) :p(x, x)≤p(x, y),
(P3) :p(x, y) =p(y, x),
(P4) :p(x, z)≤p(x, y) +p(y, z)−p(y, y).
The pair (X, p) is called a partial metric space.
Ifp(x, y) = 0, then x=y, but the converse does not always hold.
Each partial metric p on X generates a T0 - topology τp which has as a base the family of open p - balls {Bp(x, ε) : x ∈ X, ε >0}, where Bp(x, ε) = {y ∈ X : p(x, y)≤p(x, x) +ε, for all x∈X and ε >0}.
A sequence {xn} in a partial metric space (X, p) converges to a limit x ∈ X, with respect to τp if and only ifp(x, x) = limn→∞p(x, xn).
Ifpis a partial metric onX, then the functionps(x, y) = 2p(x, y)−p(x, x)−p(y, y) define a metric on X.
Further more, a sequence{xn}converges in (X, ps) to a pointx∈X if and only if
n,m→∞lim p(xn, xm) = lim
n→∞p(xn, x) =p(x, x). (1)
Definition 2 ([5]). Let (X, p) be a partial metric space.
a) A sequence{xn}inXis said to be a Cauchy sequence iflimn,m→∞p(xn, xm) exists and is finite.
b) (X, p) is said to be complete if every Cauchy sequence inX converges with respect to τp to a point x∈X such that limn→∞p(xn, x) =p(x, x).
Lemma 1 ([5], [2]). Let (X, p) be a partial metric space. Then:
a) A sequence in X is a Cauchy sequence in(X, p) if and only if is a Cauchy sequence in (X, ps).
b) A partial metric space (X, p) is complete if and only if the metric space (X, ps) is complete.
The following results are obtained in [4].
Theorem 2 (Theorem 3.5 [4]). Let (X, p) be a complete partial metric space and f, g :X→Xbe two self mappings. If there exist nonnegative constantsA, B, C, D, E with A+B+C+ 2D+E <1 and A+B+C+D+ 2E <1 such that
p(f x, gy)≤Ap(x, y) +Bp(x, f x) +Cp(y, gy) +Dp(x, gy) +Ep(y, f x) (2) for each x ∈ X, then f and g have an unique common fixed point z such that p(z, z) = 0.
Theorem 3 (Theorem 4.3 [4]). Let (X, p) be a complete partial metric space and f :X→X be a self map. If there exists k∈
0,12
such that
p(f x, f y)≤kmax{p(x, y), p(x, f x), p(y, f y), p(x, f y), p(y, f x)} (3) for all x, y∈X, then f has an unique fixed pointz such thatp(z, z) = 0.
Definition 3 ([3]). Let X be a nonempty set. Two mappings f, g : X → X are said to have property Q if F(f)∩F(g) = F(fn) ∩F(gn) for each n ∈ N, where F(f) ={x∈X:x=f x}.
The following theorem is proved in [4].
Theorem 4 (Theorem 3.9 [4]). Let (X, p) be a complete metric space and let f, g: X →X be two self mappings. Suppose that
p(f x, gy)≤Ap(x, y) +B[p(x, f x) +p(y, gy)] +D[p(x, gy) +p(y, f x)] (4) holds for all x, y∈X, where A, B, D≥0 are such thatA+ 2B+ 3D <1.
3. Implicit relations
Definition 4. Let Fp be the set of all continuous functions F(t1, ..., t6) : R6+ → R satisfying
(F1) : F is non - increasing in variablet2, t3, t4, t5, t6,
(F2) : There exists h1, h2 ∈[0,1) such that for all u, v≥0 with (F2a) : F(u, v, v, u, u+v, v)≤0 implies u≤h1v,
(F2b) :F(u, v, u, v, v, u+v)≤0 implies u≤h2v.
In the following examples, property (F1) is obviously.
Example 1. F(t1, ..., t6) = t1−at2−bt3−ct4 −dt5 −et6, where a, b, c, d, e ≥ 0, a+b+c+ 2d+e <1 and a+b+c+d+ 2e <1.
(F2) :Let u, v≥0andF(u, v, v, u, u+v, v) =u−av−bv−cu−d(u+v)−ev≤0.
Then u≤h1v, where 0≤h1= a+b+d+e1−(c+d) <1.
Similarly, F(u, v, u, v, v, u+v)≤0implies u≤h2v, where 0≤h2= a+c+d+e1−(b+e) <
1.
Example 2. F(t1, ..., t6) =t1−kmax{t2, ..., t6}, where k∈ 0,12
.
(F2) :Let u, v≥0 and F(u, v, v, u, u+v, v) =u−k(u+v)≤0. Thenu≤h1v, where 0≤h1 = 1−kk <1.
Similarly, F(u, v, u, v, v, u+v) ≤ 0 implies u ≤ h2v, where 0 ≤ h2 = h1 =
k 1−k <1.
Example 3. F(t1, ..., t6) = t21−t1(at2+bt3+ct4)−dt5t6, where a, b, c, d≥0 and a+b+c+ 2d <1.
(F2) :Letu, v≥0andF(u, v, v, u, u+v, v) =u2−u(av+bv+cu)−dv(u+v)≤0.
If u > v, then u2[1−(a+b+c+ 2d)] ≤ 0, a contradiction. Hence u ≤ v, which implies u≤h1v, where 0≤h1 =√
a+b+c+ 2d <1.
Similarly, F(u, v, u, v, v, u+v) ≤ 0 implies u ≤ h2v, where 0 ≤ h2 = h1 =
√a+b+c+ 2d <1.
Example 4. F(t1, ..., t6) =t21−amax{t22, t23, t24}−bt5t6, wherea, b≥0anda+2b <1.
(F2) :Let u, v≥0andF(u, v, v, u, u+v, v) =u2−amax{u2, v2} −bv(u+v)≤0.
If u > v, then u2[1−(a+ 2b)] ≤ 0, a contradiction. Hence u ≤ v, which implies u≤h1v, where 0≤h1 =√
a+ 2b <1.
Similarly, F(u, v, u, v, v, u+v) ≤ 0 implies u ≤ h2v, where 0 ≤ h2 = h1 =
√
a+ 2b <1.
Example 5. F(t1, ..., t6) = t31 −at21t2 −ct2t3t4−dt1t5t6, where a, b, c, d ≥ 0 and a+b+c+ 2d <1.
(F2) :Letu, v≥0andF(u, v, v, u, u+v, v) =u3−auv2−buv2−cv2u−dv(u+v)v≤ 0. If u > v, then u3[1−(a+b+c+ 2d)]≤0, a contradiction. Henceu≤v, which implies u≤h1v, where 0≤h1 =√3
a+b+c+ 2d <1.
Similarly, F(u, v, u, v, v, u+v) ≤ 0 implies u ≤ h2v, where 0 ≤ h2 = h1 =
√3
a+b+c+ 2d <1.
Example 6. F(t1, ..., t6) =t1−at2−bt3−cmax{2t4, t5+t6}, wherea, b, c≥0 and a+b+ 3c <1.
(F2) :Letu, v≥0andF(u, v, v, u, u+v, v) =u−av−bv−cmax{2u, u+2v} ≤0.
If u > v, then u[1−(a+b+ 3c)]≤0, a contradiction. Hence u≤v, which implies u≤h1v, where 0≤h1 =a+b+ 3c <1.
Similarly,F(u, v, u, v, v, u+v)≤0impliesu≤h2v, where 0≤h2 = 1−(b+c)a+2c <1.
4. Fixed point theorems
Theorem 5. Let (X, p) be a complete partial metric space and let f, g:X →X be two self mappings. If
F(p(f x, gy), p(x, y), p(x, f x), p(y, gy), p(x, gy), p(y, f x))≤0 (5) for all x, y∈X, where F ∈Fp. Then, f and g have an unique common fixed point z such thatp(z, z) = 0.
Proof. Letx0 ∈X be an arbitrary point inX. Iff x0=x0 and gx06=x0, by (5) we have successively
F(p(f x0, gx0), p(x0, x0), p(x0, f x0), p(x0, gx0), p(x0, gx0), p(x0, f x0))≤0 F(p(x0, gx0), p(x0, x0), p(x0, x0), p(x0, gx0), p(x0, gx0), p(x0, x0))≤0.
By (F1) and (P2) we obtain
F(p(x0, gx0), p(x0, gx0), p(x0, gx0), p(x0, gx0), p(x0, gx0), p(x0, gx0))≤0.
By (F2a) we obtain
p(x0, gx0)≤h1p(x0, gx0)< p(x0, gx0),
a contradiction. Hence,p(x0, gx0) = 0 which impliesx0=gx0. Thusx0is a common fixed point of f and g. So we assume thatx06=f x0 and x0 6=gx0. Then, we define the sequence{xn} inX such that
x2n+1=f x2n and x2n+2 =gx2n+1, forn∈N (6) By (5) forx=x2n and y=x2n+1 we get that
F(p(f x2n, gx2n+1), p(x2n, x2n+1), p(x2n, f x2n), p(x2n+1, gx2n+1), p(x2n, gx2n+1), p(x2n+1, f x2n))≤0.
By (6) we obtain
F(p(x2n+1, x2n+2), p(x2n, x2n+1), p(x2n, x2n+1),
p(x2n+1, x2n+2), p(x2n, x2n+2), p(x2n+1, x2n+1))≤0. (7) Since by (P4)
p(x2n, x2n+2) ≤ p(x2n, x2n+1) +p(x2n+1, x2n+2)−p(x2n+1, x2n+1)
≤ p(x2n, x2n+1) +p(x2n+1, x2n+2)
and by (P2)
p(x2n+1, x2n+1)≤p(x2n, x2n+1), by (7) and (F1) we obtain
F(p(x2n+1, x2n+2), p(x2n, x2n+1), p(x2n, x2n+1), p(x2n+1, x2n+2), p(x2n, x2n+1) +p(x2n+1, x2n+2), p(x2n, x2n+1))≤0.
By (F2a) we obtain
p(x2n+1, x2n+2)≤hp(x2n, x2n+1), where h= max{h1, h2}.
Similarly, by (5) we have
F(p(f x2n, gx2n−1), p(x2n, x2n−1), p(x2n, f x2n), p(x2n−1, gx2n−1), p(x2n, gx2n−1), p(x2n−1, f x2n))≤0.
By (6) we obtain
F(p(x2n+1, x2n), p(x2n, x2n−1), p(x2n, x2n+1),
p(x2n−1, x2n), p(x2n, x2n), p(x2n−1, x2n+1))≤0. (8) Since by (P4)
p(x2n−1, x2n+1) ≤ p(x2n−1, x2n) +p(x2n, x2n+1)−p(x2n, x2n)
≤ p(x2n−1, x2n) +p(x2n, x2n+1), and by (P2),
p(x2n, x2n)≤p(x2n, x2n−1), by (8) we obtain
F(p(x2n+1, x2n), p(x2n, x2n−1), p(x2n, x2n+1), p(x2n−1, x2n), p(x2n, x2n−1), p(x2n−1, x2n) +p(x2n, x2n+1))≤0.
By (F2b) we obtain
p(x2n, x2n+1)≤hp(x2n, x2n−1).
Hence
p(xn, xn+1)≤hp(xn−1, xn)≤...≤hnp(x0, x1).
Forn > m we have
p(xn, xm) ≤ (hm+hm+1+...+hn−1)p(x0, x1)
≤ hm
1−hp(x0, x1)
and so limn,m→∞p(xn, xm) = 0. From the definition of ps we get ps(xn, xm)≤2p(xn, xm)→0 asn, m→ ∞.
This implies that{xn}is a Cauchy sequence in metric space (X, ps). Since (X, p) is complete, then by Lemma 1, (X, ps) is a complete metric space. Therefore, the sequence {xn} converges to some z ∈ X with respect to the metric ps, that is, limn→∞ps(xn, z) = 0. Again, by (1) we have
p(z, z) = lim
n→∞p(xn, z) = lim
m,n→∞p(xn, xm) = 0.
By (5) forx=z and y=x2n+1 we obtain
F(p(f z, gx2n+1), p(z, x2n+1), p(z, f z), p(x2n+1, gx2n+1), p(z, gx2n+1), p(x2n+1, f z))≤0, F(p(f z, x2n+2), p(z, x2n+1), p(z, f z), p(x2n+1, x2n+2), p(z, x2n+2), p(x2n+1, f z))≤0.
Lettingn tends to infinity we obtain
F(p(f z, z),0, p(z, f z),0,0, p(z, f z))≤0, which implies by (F2b) thatp(z, f z) = 0, hencez=f z.
Similarly,
F(p(f x2n, gz), p(x2n, z), p(x2n, f z), p(z, gz), p(x2n, gz), p(z, f x2n))≤0, F(p(x2n+1, gz), p(x2n, z), p(x2n, f z), p(z, gz), p(x2n, gz), p(z, x2n+1))≤0.
Lettingn tends to infinity we obtain
F(p(z, gz),0,0, p(z, gz), p(z, gz),0)≤0,
which implies by (F2b) that p(z, gz) = 0. Hencez =gz and z is the common fixed point of f and g. On the other hand by (P2):
p(z, z)≤p(z, f z) = 0, hence p(z, z) = 0.
Suppose that there exists an other point u such that f u =gu =u. By (5) we obtain
F(p(f z, gu), p(z, u), p(z, f z), p(u, gu), p(z, gu), p(u, f z))≤0, F(p(z, u), p(z, u), p(z, z), p(u, u), p(z, u), p(u, z))≤0, By (P2) and (F1) we obtain
F(p(z, u), p(z, u), p(z, u), p(z, u),2p(z, u), p(z, u))≤0 which implies by (F2a) that
p(z, u)≤h1p(z, u)< p(z, u), a contradiction. Hence p(z, u) = 0, i.e. u=z.
Remark 1. 1) By Theorem 1 and Example 1 we obtain Theorem 2.
2) By Theorem 1 and Example 2 we obtain the following result
Theorem 6. Let (X, p) be a complete partial metric space and let f, g:X →X be two self mappings. If there exists k∈
0,12
such that
p(f x, gy)≤kmax{p(x, y), p(x, f x), p(y, f y), p(x, gy), p(y, f x)}, (9) for all x, y ∈ X, then, f and g have an unique common fixed point z such that p(z, z) = 0.
Example 7. Let X = [0,1] and p(x, y) = max{x, y}. Then (X, p) is a complete partial metric space and p is not a metric. Then for mappingsf, g:X →X, where f(x) = x2
1 + 3x and g(x) = 0 we have
p(f x, gy) = max{f x,0}=f x= x2 1 + 3x ≤ 1
4·x= 1
4p(x, gy).
Hence
p(f x, gy)≤hmax{p(x, y), p(x, f x), p(y, gy), p(x, gy), p(y, , f x)}, where h= 1
4 < 1 2.
Then by Theorem 6, f andg have a unique common fixed point x= 0.
Remark 2. By Theorem 5 and Examples 3 - 6 we obtain new particular results.
Iff =g, then by Theorem 1 we obtain
Theorem 7. Let (X, p) be a complete partial metric space and let f :X→X be a self mapping. If
F(p(f x, f y), p(x, y), p(x, f x), p(y, f y), p(x, f y), p(y, f x))≤0, (10) for all x, y ∈ X and F satisfies properties (F1) and (F2a), then, f has an unique fixed point z such thatp(z, z) = 0.
Remark 3. By Example 2 and Theorem 7 we obtain Theorem 3.
Corollary 8 (Theorem 2.3 [2]). Let (X, p) be a complete partial metric space and let T :X→X be a self mapping such that for all x, y∈X
p(T x, T y)≤max{ap(x, y), bp(x, T x), cp(y, T y), d[p(x, T y)+p(y, T x)], p(x, x), p(y, y)}, wherea, b, c∈[0,1)andd∈
0,12
. ThenT has an unique fixed pointzandp(z, z) = 0.
Proof. By (P2),
p(x, x) ≤ p(x, T x), p(y, y) ≤ p(y, T y) and
max{ap(x, y), bp(x, T x), cp(y, T y),2d[p(x,T y)+p(y,T x)]
2 , p(x, x), p(y, y)} ≤
≤kmax{p(x, y), p(x, T x), p(y, T y), p(x, T y), p(y, T x)},
wherek= max{a, b, c,2d}<1 and the proof follows from Theorem 7 and Examples 2.
Remark 4. A similar result with Corollary 8 we obtain by Example 1 and Theorem 7 (Corollary 2.8 [2]).
5. Q - property in partial metric spaces
Definition 5. Let Fq be the set of all continuous functions F(t1, ..., t6) : R6+ → R satisfying
(Fq) :There exists q ∈[0,1) such that for all u, v≥0, F(u, v,0, u+v, u, v)≤0 implies u≤qv.
Example 8. F(t1, ..., t6) =t1−at2−bt3−ct4−dt5−et6, where a, b, c, d≥0 and a+ 2c+d+e <1.
(Fq) : Letu, v≥0such thatF(u, v,0, u+v, u, v) =u−av−c(u+v)−du−ev≤ 0, which implies u≤qv, where 0< q= 1−(c+d)a+c+e <1.
Example 9. F(t1, ..., t6) =t1−kmax{t2, ..., t6}, where k∈ 0,12 .
(Fq) : Let u, v≥0 such that F(u, v,0, u+v, u, v) =u−k(u+v)≤0, which implies u≤qv, where 0< q= 1−kk <1.
Example 10. F(t1, ..., t6) =t21−t1(at2+bt3+ct4)−dt5t6, where a, b, c, d≥0 and a+ 2c+d <1.
(Fq) : Letu, v≥0such thatF(u, v,0, u+v, u, v) =u2−u[av+c(u+v)]−duv≤ 0. If u > 0, then u −[av +c(u +v)]−dv ≤ 0 which implies u ≤ qv, where 0< q= a+c+d1−c <1. If u= 0 then u≤qv.
Example 11. F(t1, ..., t6) = t21 −amax{t22, t23, t24} −bt5t6, where a, b ≥ 0 and 0 <
4a+b <1.
(Fq) : Let u, v≥0such thatF(u, v,0, u+v, u, v) =u2−amax{v2,(u+v)2}−
buv =u2−a(u+v)2−buv ≤0. If u > v, then u2[1−(4a+b)]<0, a contradiction.
Hence u≤v which implies u≤qv, where 0< q=√
4a+b <1.
Example 12. F(t1, ..., t6) =t31−at21t2−bt1t22−ct2t3t4−dt1t5t6, where a, b, c, d≥0 and 0< a+b+d <1.
(Fq) : Let u, v≥0such thatF(u, v,0, u+v, u, v) =u3−au2v−buv2−du2v≤ 0. Letu >0, then u2−auv−bv2−duv≤0. Ifu > v, then u2[1−(a+b+d)]≤0, a contradiction. Hence u≤v which implies u ≤qv, where 0< q=√
a+b+d <1.
If u= 0, then u≤qv.
Example 13. F(t1, ..., t6) = t1−at2−bt3−cmax{2t4, t5+t6}, where a, b, c ≥ 0 and 0< a+ 4c <1.
(Fq) : Let u, v≥0 such that F(u, v,0, u+v, u, v) =u−av−2c(u+v)≤0, which implies u≤qv, where 0< q= a+2c1−2c <1.
Theorem 9. Let (X, p) be a complete partial metric space and let f, g:X →X be satisfying the conditions of Theorem 5. If F ∈Fq, then f and g have property Q.
Proof. By Theorem 5 we have thatF(f)∩F(g) ={z}, wherezis the unique common fixed point of f and g and p(z, z) = 0. So F(fn)∩F(gn) 6=∅, for eachn∈N. Let u ∈F(fn)∩F(gn), where n >1 is arbitrary and suppose that n6= 2. Then (5) we have successively
F(p(fnz, gnu), p(fn−1z, gn−1u), p(fn−1z, fnz), p(gn−1u, gnu), p(fn−1z, gnu), p(gn−1u, fnz))≤0,
F(p(z, gnu), p(z, gn−1u), p(z, z), p(gn−1u, gnu), p(z, gnu), p(z, gn−1u))≤0.
By (P2) and (F1) we obtain
F(p(z, gnu), p(z, gn−1u),0, p(gn−1u, z) +p(z, gnu), p(z, gnu), p(z, gn−1u))≤0,
which implies by (Fq) that
p(z, gnu)≤qp(z, gn−1u)≤...≤qnp(z, u).
Since gnu = gu then p(z, u) ≤ qnp(z, u), so p(z, u)(1−qn) ≤ 0 which implies p(z, u) = 0. Hence u=z, which implies that f and g have property Q.
Remark 5. By Examples 1 and 8 and Theorem 9 we obtain Theorem 4.
References
[1] I. Altun, F. Sola and H. Simsek, Generalized contractive principle on partial metric spaces, Topology Appl. 157 (18) (2010), 2778-2785.
[2] R. Chi, E. Karapinar and T.D. Thanh, A generalized contraction principle in partial metric spaces, Math. Comput. Modelling 55 (2012), 1673-1681.
[3] G. S. Jeong and B. E. Rhoades, Maps for which F(T) = F(Tn), Fixed Point Theory Appl. 6 (2005), 87-131.
[4] Z. Kadelburg, H. K. Nashine and S. Radenovi´c,Fixed point results under various contractive conditions in partial metric spaces, R.A.C.S.A.M. 107 (2013), 241-256.
[5] S. G. Matthews, Partial metric topology and applications, Proc. 8th Summer Conference on General Topology and Applications, Ann. New York Acad. Sci. 728 (1996), 183-197.
[6] V. Popa,Fixed point theorems for implicit contractive mappings, Stud. Cercet.
S¸tiint¸., Ser. Mat., Univ. Bac˘au 7 (1997), 129-133.
[7] V. Popa,Some fixed point theorems for compatible mappings satisfying an im- plicit relation, Demonstr. Math. 32, 1 (1999), 157-163.
[8] C. Vetro and F. Vetro, Common fixed points of mappings satisfying implicit relations in partial metric spaces, J. Nonlinear Sci. Appl. 6 (2013), 152-161.
Valeriu Popa
“Vasile Alecsandri” University of Bac˘au, Bac˘au, Romania
email: [email protected] Alina-Mihaela Patriciu
Department of Mathematics and Computer Sciences, Faculty of Sciences and Environment,
“Dun˘area de Jos” University of Galat¸i, Galat¸i, Romania
email: [email protected]
