0Xuetal. Fixed Point Theory and Applications (2016) 2016:47 DOI 10.1186/s13663-016-0538-z
O Fixed Point Theory and Applications
a SpringerOpen Journal
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Suzuki's type fixed point theorems for generalized mappings in partial cone metric spaces over a solid cone
Wenqing Xu1,2, Chuanxi Zhu
"Correspondence: chuanxizhu@126.com
1 Department of Mathematics, Nanchang University, Nanchang, 330031, P.R. China Full list of author information is available at the end of the article
1 and Chunfang Chen1
Abstract
In this paper, we obtain some Suzuki-type fixed point theorems for generalized mappings in partial cone metric spaces over a solid cone. Our results unify and generalize various known comparable results in the literature. We also provide illustrative examples in support of our new results.
MSC: 47H10; 54H25
Keywords: partial cone metric spaces; solid cone; Suzuki type; fixed point
£ Spri
ringer
1 Introduction and preliminaries
In 1994, Matthews [1] introduced the notion of a partial metric space as a part of the study of denotational semantics of data for networks, showing that the Banach contraction mapping principle can be generalized to the partial metric context for applications in program verification. After that, many fixed point theorems for mappings satisfying different contractive conditions in (ordered) partial metric spaces have been proved (see [2-4]).
In 2007, Huang and Zhang [5] introduced the concept of cone metric spaces and extended the Banach contraction principle to cone spaces over a normal solid cone. Moreover, they defined the convergence via interior points of the cone. Such an approach allows the investigation of the case that the cone is not necessarily normal. Since then, there were many references concerned with fixed point results in (ordered) cone spaces (see [6-15]). In 2012, based on the definition of cone metric spaces and partial metric spaces, Sonmez [16, 17] defined a partial cone metric space and proved some fixed point theorems for contractive type mappings in complete partial cone metric spaces.
Recently, without using the normality of the cone, Malhotra et al. [18] and Jiang and Li [19] extended the results of [16,17] to 0-complete partial cone metric spaces. First, we recall the definition of partial metric spaces (see [1]).
Definition 1.1 ([1]) Let X be a nonempty set. A function p : X x X ^ R+ is said to be a partial metric if for all x, y, z e X, the following conditions are satisfied:
(p1) p(x, x) = p(x,y) = p(y,y) if and only if x = y; (p2) p(x,x) <p(x,y); (p3) p(x, y)=p(y, x);
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(p4) p(x,y) < p(x,z)+p(z,y)-p(z, z).
The pair (X,p) is called a partial metric space. Ifp(x,y) = 0, then the (pi) and (p2) imply that x = y, but the converse does not hold in general. A trivial example of a partial metric space is the pair (R+,p), where p : R+ x R+ — R+ is defined as p(x,y) = max{x,y}; see also [1].
Let E be a topological vector space. A cone of E is a nonempty closed subset P of E such that
(i) ax + by e P for all x,y e P and a, b > 0, and
(ii) P n (-P) = {0}, where 0 is the zero element of E.
Each cone P of E determines a partial order ^ on E by x ^ y if and only if y - x e P for all x, y e E. We shall write x -< y if x ^ y and x = y.
A cone P of a topological vector space E is solid if intP = 0, where intP is the interior of P. For all x, y e E with y - x e int P, we write x ^ y. Let P be a solid cone of a topological vector space E. A sequence {un} of E weakly converges [18] to u e E (denoted un -— u) if for each c e int P, there exists a positive integer n0 such that u - c ^ un ^ u + c for all n > n0.A cone P of a normed vector space (E, || • ||) is normal if there exists K >0 such that 0 < x < y implies that ||x|| < K||y|| for all x,y e P, and the minimal K is called a normal constant of P. Next, we state the definitions of cone metric and partial cone metric spaces and some of their properties (see [5,16-19]).
Definition 1.2 ([5]) Let X be a nonempty set, and let P be a cone of a topological vector space E. A cone metric on X is a mapping d: X x X — P such that, for all x, y, z e X:
(di) d(x, y) =0 if and only if x = y;
(d2) d(x, y)=d(y, x);
(d3) d(x, y) ^ d(x, z) + d(z, y).
The pair (X, d) is called a cone metric space over P.
Definition 1.3 ([16, 17]) Let X be a nonempty set, and let P be a cone of a topological vector space E. A partial cone metric on X is a mappingp : X x X — P such that, for all x, y, z e X:
(p1) p(x, x) = p(x, y) = p(y,y) if and only if x = y;
(p2) p(x,x) ^p(x,y);
(p3) p(x,y)=p(y,x);
(p4) p(x,y) ^p(x,z) + p(z,y) -p(z,z).
In this case, the pair (X, p) is called a partial cone metric space over P.
Note that each cone metric is certainly a partial cone metric. The following example shows that there do exist partial cone metrics that are not cone metrics.
Example 1.1 ([19]) Let E = CR [0,1] with the norm ||u|| = ||u||TO + ||u'||TO, and X = P = {u e E: u(t) > 0, t e [0,1]}, which is a nonnormal solid cone. Define the mappingp: X x X — P by
p(x, y) =
x, x = y, x + y otherwise.
Then p is a partial cone metric, but not a cone metric, since p(x, x) = 0 for all x e X with x = 0.
A partial cone metric p on X over a solid cone P generates a topology tp on X, which has a base of the family of open p-balls {Bp(x, c): x e X, 0 ^ c}, where Bp(x, c) = {y e X: p(x, y) ^ p(x, x) + c} for x e X and c e int P.
Definition 1.4 ([19]) Let (X,p) be a partial cone metric space over a solid cone P of a topological vector space E.
(i) A sequence {xn} in X converges to x e X (denoted by xn — x) if for each c e int P, there exists a positive integer n0 such that p(xn,x) ^ p(x,x) + c for each n > n0 (that is, p(xn,x) -— p(x,x)).
(ii) A sequence {xn} in X is 0-Cauchy if for each c e int P, there exists a positive integer n0 such that p(xn,xm) ^ c for all m, n > n0. The partial cone metric space (X,p) is 0-complete if each 0-Cauchy sequence {xn} of X converges to a point x e X such that p(x, x) = 0.
Definition 1.5 ([16,17]) Let (X,p) be a partial cone metric space over a solid cone P of a topological vector space (E, || • ||).
(i) A sequence {xn} in X strongly converges to x e X (denoted by xn — x) if
lim p(xn,x) = lim p(xn,xn) = p(x,x).
n—TO n—TO
(ii) A sequence {xn} in X is Cauchy if there exists u e P with ||u|| < to such that limm,n—TO p(xm
, xn) — u. The partial cone metric space (X,p) is complete if each Cauchy sequence {xn} of X strongly converges to a point x e X such that p(x,x) = u.
Note that if P is a normal solid cone of a normed vector space (E, || • ||), then each complete partial cone metric space is 0-complete. But the converse is not true. The following example ([16], Example 4) shows that a 0-complete partial cone metric space is not necessarily complete.
Example 1.2 ([16]) Let X = {(x1,x2,...,xk):xi > 0,xi e Q,i = 1,2,...,k}, andE = Rk with the norm ||x|| = ^ x2, P = R+, where Q denotes the set of rational numbers. Define the mapping p : X x X — P by
p(x,y) = (x1 vy1,x2 vy2,...,xk vyk) for allx,y e X,
where the symbol v denotes the maximum, that is, x v y = max{x,y}. Clearly, (X,p) is a partial cone metric space, p(x, x) =x for each x e X, p(x, 0) =0 if and only if x = 0, andP is normal. On the other hand, (X,p) is 0-complete but not complete.
Let X be a nonempty set, and S, T : X — X be two mappings. A point x e X is said to be a coincidence point of S and T if Sx = Tx. A point y e X is called point of coincidence
of S and T if there exists a point x e X such that y = Sx = Tx. The mappings S, T are said to be weakly compatible if they commute at their coincidence points (that is, TSz = STz whenever Sz = Tz).
Let (X, c) be a partially ordered set; x,y e X are called comparable if x c y or y c x. A mapping T: X ^ X is said to be nondecreasing if for x,y e X, x c y implies Tx c Ty. Let S, T : X ^ X be two mappings; T is said to be S-nondecreasing if Sx c Sy implies Tx c Ty for all x, y e X.
Bhasker and Lakshmikantham [20] introduced the concepts of mixed monotone mappings and coupled fixed point.
Definition 1.6 ([20]) Let (X, c) be a partially ordered set, and A : X x X ^ X. The mapping A is said to have the mixed monotone property if A is monotone nondecreasing in its first argument and is monotone nonincreasing in its second argument, that is, for any x, y e X,
x1,x2 e X, xi c x2 A(xi,y) c A(x2,y),
y1,y2 e X, y2 c y1 A(x,y1) c A(x,y2).
Definition 1.7 ([20]) An element (x,y) e X2 is said to be a coupled fixed point of the mapping A : X2 ^ X if A(x,y) = x and A(y,x) =y.
Lemma 1.1 ([21]) LetX be a nonempty set, and S: X ^ X a mapping. Then there exists a subset Y c X such that SY = SX and S: Y ^ X is one-to-one.
Paesano and Vetro [22] proved Suzuki-type characterizations of completeness for partial metric spaces and fixed points for partially ordered metric spaces. Note that if in Theorem 2 of [22], we assume that p is a metric, then we obtain Theorem 2 of [23]. Recently, also, some Suzuki-type fixed point and coupled fixed point results for mappings or generalized multivalued mappings in different metric spaces were investigated (see [24-29]). The aim of the paper is to give a generalized version of Theorems 2 and 3 of [22] in partially ordered partial cone metric spaces over a solid cone. Meantime, we also establish the corresponding Suzuki-type coupled fixed point results for generalized mappings in partially ordered partial cone metric spaces. It is worth pointing out that some examples are presented to verify the effectiveness and applicability of our results.
2 Fixed point theorems in partial cone metric spaces
In this section, we first give some properties of partial cone metric spaces. The following properties are used (particularly when dealing with cone metric spaces in which the cone need not be normal).
Remark 2.1 Let P be a solid cone. Then the following properties are used:
(1) If a < b and b < c, then a < c.
(2) If a «: b and b «: c, then a «: c.
(3) If 0 < u « c for each c e int P, then u = 0.
(4) If a < ka, where 0 < X <1, then a = 0.
(5) If a < b + c for each c e int P, then a < b.
Now, we establish some Suzuki-type fixed point theorems for generalized mappings in partially ordered partial cone metric spaces over a solid cone.
Theorem 2.1 Let (X, p, c) be a 0-complete partially ordered partial cone metric space over a solid cone P of a normed vector space (E, || • ||). LetT: X — X be a nondecreasing mapping with respect to c. Define the nonincreasingfunction ty : [0,1) — (5,1] by
ty (r) =
1 if 0 < r <
J- f^1 < r <¿?,
1+r f^T < r <J.
Assume that there exists r e [0,1) such that
ty(r)p(x, Tx) < p(x,y) implies p(Tx, Ty) < rU(x,y) (2.1)
for all comparable x,y e X, where U(x,y) e {p(x,y),p(x, Tx),p(y, Ty), p(x,Ty)+p(y,Tx)}. Suppose that the following conditions hold:
(i) there exists x0 e X such that x0 — Tx0;
(ii) for a nondecreasing sequence xn — x, we have xn — x for all n e N;
(iii) for two nondecreasing sequences {xn}, {yn }cX such that xn —— yn, xn x, and yn -— y as n — to, we have x — y.
Then T has a fixed point in X. Moreover, the fixed point ofT is unique if
(iv) for all x, y e X that are not comparable, there exists u e X comparable with x and y.
Proof Since ty (r) < 1, ty (r)p(x, Tx) < p(x, Tx) for every x e X .By (2.1) and using the triangular inequality, we have
p(Tx, T2x) ^ rU(x, Tx),
p(x, T2x) + p(Tx, Tx)
2 N p(x, T2x) + p(Tx, Tx)
Thus, we get the following cases: Case 1. p(Tx, T2x) ^ rp(x, Tx).
Case 2.p(Tx, T2x) ^ rp(Tx, T2x), which implies thatp(Tx, T2x) = 0. Case 3. p(Tx, T2x) ^ rp(xT2x)+p(Tx,Tx) ^ rp(x,Tx)+p(Tx,T2x), which implies that p(Tx, T2x) ^ rp(x, Tx). Then, in all cases, we have
p(Tx, T2x) ^ rp(x, Tx). (2.2)
Let x0 e X be such that x0 c Tx0. Since T is nondecreasing, we get
x0 c Tx0 c T2x0 c ••• c Tnx0 c •••.
U(x, Tx) e |p(x, Tx),p(x, Tx),p(Tx, T2x), = |p(x, Tx), p(Tx, T2x),
Define the sequence {xn} by xn = Tnx0, so that xn+i = Txn. If xn = xn+i = Txn for some n, then xn becomes a fixed point of T. Now, suppose that xn = xn+i for all n e N. Then p(xn,xn+i) ^ 0.
Note that ty (r)p(xn-i, Txn-i) ^ p(xn-i, Txn-i) for all n e Z+, where Z+ is the set of positive integers. Since xn-i and Txn-i are comparable for all n > 0, by (2.i) we have
p(xn,xn+i) = p(Txn_i, T2xn-i) < rU(xn-i, Txn-i), where
a(xn-i, Txn-i) e Txn-i),p(xn-i, Txn-i),p(Txn-i, T2xn-i),
p(Xn-l, T2xn-i) + p(Txn-i, Txn-i) '
p(xn-i, xn+i) + p(xn, xn)
— ïp(xn-i, xn),p(xn, xn+i),
Thus, we get the following cases:
Case I. If L/(xn-1, = p(xn-1,xn), thenp(xn,xn+i) < rp(xn-1,xn). Case 2. p(xn,xn+1) ^ rp(xn,xn+1), which implies that p(xn,xn+1) = 0. Case 3.p(x„,x„+i) ^ r • p(x„-1,x„+i)+p(x„,x„) ^ r(p(x„-1'x„);p(x„'x„+i)), which implies that
p(xn, x„+1) ^ rp(x„-1, x„).
Then, in all cases, we have
p(x„,x„+1) ^ rp(x n-1, x„). Continuing this process, it follows that
p(x„,x„+1) ^ rp(x„_1,x„) ^ r2p(x„_2,x„_1) ^ • • • ^ r„p(x0,x1). Thus, for any m, „ e Z+ with m > n,we get
p(x„, xm) ^ p(x„, x„+1) + p(x„+1, x„+2) + ••• + p(xm-1, xm)
^ (r„ + r„+1 + ••• + rm-1)p(xo,x1) r„
-p(xo, x1).
Let 0 « c be given, Choose S > 0 such that c + NS(0) ç P, where N (0) — {y e £ :
< 5}. Also, choose a natural number N such that -f-p(xo,xi) e Ns(0) for all n > Ni.
Then 1-rp(xo,xi) « c for all « > N. Thus,
p(Xn, Xm) < --P(Xo, Xi) « C
for all m > n > N.Hence, {xn} is a 0-Cauchy sequence in (X,p). Since (X,p) isa0-complete
partial cone metric space, there exists z e X such that xn ^ z and p(z, z) = 0.
First, we show that there exists j e Z+ such that Tjz = z. Arguing by contradiction, we assume that Tjz = z for all j e Z+.
Note that, by condition (ii), if {xn} is nondecreasing, then xn ç z. Since T is nondecreas-ing, we get xn+i = Txn ç Tz for all n e N. Taking the limit as n — to, by (iii) we obtain that z ç Tz, which implies that {Tnz} is a nondecreasing sequence. So, we have shown that for {xn}, {Tnz} also is a 0-Cauchy sequence. We also have T'z is comparable with xn for all j, n e N. Now, we prove that
p(T'+1z, z) ^ rjp(Tz, z) for all j e N. (.3)
Since p(Tjz, z) > 0, p(z,z) = 0, and xn — z, there exists N2 e N such that p(xn, z) < pT'3z,z) for all n > N2.We have
ty (r)p(x n, Txn) ^ p(xn, Txn) ^ p(xn, z)+p(xn+l, z)
2 . . 1 ^ 3p(T'z,z) = p(T'z,z) - 3p(T'z,z)
< p(Tjz, z) -p(xn, z) < p(xm Tj z). By (2.1) and using the triangle inequality, we get that
p(xn+i, Tj+1z) ^ rU(xn, Tjz), where
U(xn, T'z) e jp(xn, T'z),p(xn,xn+i),p(T'z, Tj+1z),p(Xn, T pT zxn+^j. Thus,
p(z, T'+1z) < p(xn+1,z) + p(xn+1, T'+1z) ^ p(xn+1, z) + rU(xm T'z),
U(xn, T'z) e jp(xn, T'z),p(xn,xn+1),p(T'z, T'+1z),p(Xn, T +p(T z,xn+^j.
Since xn — z for every c » 0, there exists no e N such thatp(xn, z) ^ § andp(xn,xn+1) ^ | for all n > no. Now, for n > no, we consider the following cases:
Case 1. p(z, T'+1z) ^ p(xn+1, z) + rp(xn, T'z) ^ p(xn+1, z) + r(p(xn, z) + p(z, T'z)) ^ rp(z, T'z) + c. Then it follows from Remark 2.1(5) that p(z, T'+1z) ^ rp(z, T'z), and we get
p(z, T'+1z) ^ rp(z, T'z) < r2p(z, T''-1z) < ••• ^ r'p(z, Tz). (2.4)
Case 2. p(z, T'+1z) ^ p(xn+1, z) + rp(xn,xn+1) ^ c, which implies thatp(z, T'+1z) = 0. Case 3.
p(z, T'+1z) ^p(xn+1,z) + rp(T'z, T'+1z) ^ c + rp(T'z, T''+1z), which implies thatp(z, T'+1z) ^ rp(T'z, T'+1z). Then from (2.2) we have
p(z, T^z) ^ rp(Tjz, T^z) < r2p(Tj-iz, Tjz) < ■ ■ ■ ^ rj+1p(z, Tz) < rjp(z, Tz). Case 4.
p(x«, Tj+1 z) + p(Tjz,x«+i)
p(z, T,+1z) < p(x«+1, z)+r ^ p(x«+1, z)+r
p(x«, z) + p(z, Tj+1z) + p(T'z, z) + p(z, x«+1)
« c + r
p(z, Tj+1z) + p(Tjz, z)
which implies thatp(z, Tj+1z) ^ rp(z, Tjz). Then from (2.4) we getp(z, Tj+1 z) ^ rjp(z, Tz).
Thus, in all cases, we obtainp(z, Tj+1z) ^ rjp(z, Tz), that is, (2.3) holds.
Now, we consider the following three cases:
(1) o< r <4-^
(2) 4-1 < r ;
(3) f < r <1.
Incase (1), we note that r2 + r <1 and ft (r) = 1. If we assume that p(T2z, z) -< p(T2z, T 3z), then by (2.2) we have
p(z, Tz) <p(z, T2z) + p(Tz, T2z)
^ p(T 2z, T 3z) + p(Tz, T2z) ^ r2p(z, Tz) + rp(z, Tz) <p(z, Tz).
This is a contradiction. So, we havep(T2z,z) h p(T2z, T3z) = ft(r)p(T2z, T3z). By (2.1)-(2.3) we deduce that
p(T3z, Tz) ^ rtf(T2z,z), where
p(T 3z, z)+p(Tz, T 2z)
a(T2z,z) e |p(T2z,z),p(T2z, T3z),p(z, Tz),
Thus, we get the following cases:
Casel.p(T3z, Tz) ^ rp(T2z,z) ^ r2p(Tz,z) ^ rp(Tz,z). Case 2.p(T3z, Tz) < rp(T2z, T3z) ^ r3p(Tz,z) ^ rp(Tz,z). Case 3. p(T3z, Tz) ^ rp(Tz, z).
Case 4.p(T3z, Tz) ^ rpTzz+pTTz) < r[r2p(Tz^rp(zT)] ^ r2p(Tz,z) ^ rp(Tz,z). Then, in all cases, we havep(T3z, Tz) < rp(Tz,z). Hence,
p(z, Tz) < p(z, T3z) + p(T3z, Tz)
^ r2p(z, Tz) + rp(z, Tz) = ( r2 + 1p(z, Tz) -< p(z, Tz),
which is a contradiction.
In case (2), we note that 2r2 < 1 and ty (r) = ^ • Now, we show by induction that
p(Tnz,z) < rp(z, Tz) (2.5)
for n > 2. By (2.2), (2.5) holds for n = 2. Assume that (2.5) holds for some n with n > 2. Since
p(z, Tz) <p(z, Tnz) + p(Tnz, Tz) <p(z, Tnz) + rp(z, Tz),
we have
p(z, Tz) <~^p(z, Tnz), 1-r '
and so
1_r 1_r
f (r)p(Tnz, Tn+1^ = —p(Tnz, Tn+1z) ^ ~r-^p(Tnz, Tn+1z)
^ (1 - r)p(z, Tz) < p(z, Tnz). Therefore, by the hypotheses we have
p(Tn+iz, Tz) ^ rU(Tnz, Tz), where
U(Tnz,z) e jp(Tnz,z),p(Tnz, Tn+1z),p(z, Tz),p(тn+1z,z)+p(Tz Tnz)
Thus, we get the following cases:
Case I.p(Tn+1z, Tz) < rp(Tnz,z) < rnp(Tz,z) < rp(Tz,z). Case 2.p(Tn+1z, Tz) < rp(Tnz, Tn+1z) < rn+1p(Tz,z) < rp(Tz,z). Case 3. p(Tn+1z, Tz) < rp(Tz, z). Case 4.
(_+1 p(Tn+1z,z) +p(Tz, Tnz) , /rnp(Tz, z) + rp(Tz, z) p(тn+1z, Tz) ^ r-2-^ rl-2-
^ r2p(Tz, z) ^ rp(Tz, z).
Then, in all cases, we havep(Tn+1 z, Tz) ^ rp(Tz, z). Therefore, (2.5) holds. Now, from (2.3) we have
p(z, Tn+1z) < rp(z, Tnz) < ••• ^ rnp(z, Tz)
for n > 1. Since 0 < r < 1, for every c » 0, there exists no e N such thatp(Tz, z) ^ c for all n > n0. Hence,
p(z, Tz) ^p(z, Tn+1^ + p(Tn+1z, Tz)
< rnp(z, Tz) + rp(z, Tz) < c + rp(z, Tz),
which implies that p(Tz,z) = 0. Thus, Tz = z. This is a contradiction. In case (3), we note that for x, y e X, either
ft(r)p(x, Tx) <p(x,y) or ft(r)p(Tx, T2x) <p(Tx,y).
Indeed, if
ft(r)p(x, Tx) >p(x,y) and ft(r)p(Tx, T2x) >p(Tx,y),
then we have
p(x, Tx) <p(x,y) + p(Tx,y)
< ft(r) [p(x, Tx) +p(2x, T2x)] -< ft (r) [p(x, Tx) + rp(x, Tx)] 1
, (1 + r)p(x, Tx) = p(x, Tx). 1 + r'
This is a contradiction. Now, since either
ft (r)p(x2n, Tx2n) < p(x2n, z) or ft (r)p(x2«+1, Tx2«+1) ^ p(x2«+1, z) for all « e N,by (2.1) it follows that either
p(Tx2n, Tz) < rU(x2n, z) or p(Tx2«+1, Tz) ^ rU(x2n+1, z), where
U(x2n,z) e |p(z,x2n),p(x2n,x2«+1),p(z, Tz),p(x2n, Tz) + p(z,x2n+1^,
, i , w w \ P(x2n+1, T^+p^ x2n+2) U(x2«+1, z) e jp(z,x2«+1),p(x2n+1,x2«+2),p(z, Tz),-2-
Hence, we deduce that either
p(Tz, z) ^ p(Tx2«, z) + p(Tx2«, Tz) (.6)
p(Tz, z) ^ p(Tx2«+1, z) + p(Tx2«+1, Tz). (2.7)
Since xn ^ z for every c » 0 ,thereexists no e N suchthat p(xn, z) ^ | and p(xn, xn+i) ^ | for all n > no. Now, by (2.6) we get the following cases:
Case I. p(Tz, z) ^ p(x2n+i, z) + rp(z,x2n) ^ c implies p(Tz, z) = 0. Case 2. p(Tz, z) ^ p(x2n+i, z) + rp(x2n,x2n+i) ^ c implies p(Tz, z) = 0. Case 3. p(Tz, z) < p(x2n+i, z) + rp(z, Tz) ^ c + rp(z, Tz) implies p(Tz, z) = 0.
Case 4.
p(Tz, z) < p(x2n+1, z) + r ^ p(x2n+1, z) + r
p(x2n, Tz)+p(z, x2n+1)
p(x2n, z) + p(z, Tz) + p(z, x2n+1)
« c + rp(z, Tz),
which implies thatp(Tz,z) = Q. Then, in all cases, we havep(Tz,z) = 0. Similarly, by (2.7) we also have p(Tz, z) = Q. Thus, Tz = z. This is a contradiction. Therefore, in all cases, there exists j e N such that Tj z = z. Since {Tnz} is a Q-Cauchy sequence, we obtain z = Tz. If not, that is, if z = Tz, from p(Tnj z, Tnj+1z) = p(z, Tz) for all n e N it follows that {Tnz} is not a Q-Cauchy sequence. Hence, z is a fixed point of T.
Finally, we prove the uniqueness of the fixed point. Suppose that there exist zi, z2 e X with zi = z2 such that Tzi = zi and Tz2 = z2. We have two possible cases: Case (a). If zi and z2 are comparable, using (2.1) with x = zi and y = z2, we get that
p(z1, z2) = p(Tz1, Tz2) ^ rU(z1, z2),
U(z1, z2) e |p(z1, z2),p(z1, Tz1),p(z2, Tz2),p(z^ Tz2) 2 p(z2, Tz1) j
p(z1, z2)+p(z2, z1)
= |p(z1, z2), p(z1, z1), p(z2, z2), = {p(z1, z2), p(z1, z1), p(z2, z2^.
Thus, we get the following cases:
Casel. p(z1, z2) < rp(z1, z2) implies p(z1, z2) = 0. Case 2. p(z1, z2) < rp(z1, z1) < rp(z1,z2) implies p(z1, z2) = 0. Case 3. p(z1, z2) < rp(z2, z2) < rp(z1, z2) implies p(z1, z2) = 0. Thus, in all cases, we have p(z1,z2) = 0, that is, z1 = z2. This is a contradiction. Hence, z1 = z2.
Case (b). If z1 and z2 are not comparable, then there exists x e X comparable with z1 and z2. First, we note that for each x e X comparable with z1, we have that Tnz1 and Tnx are comparable and ty(r)p(Tn-1z1, Tnz1) < p(Tn-1z1, Tnz1) = p(z1, z1) ^ p(Tn-1z1, Tn-1x). By (2.1) we obtain
p(z1, Tnx) = p(Tnz1, Tnx) < rU(Tn-1zi, Tn-1x),
U(Tn-1z1, Tn-1x) e Jp(Tn-1z1, Tn-1x,p(Tn-1 z1, Tnz^,p(Tn-1x, Tnx), p(Tn-1x, Tnz1) + p(Tn-1z1, Tnx) '
= p(z1, Tn-1^, p(z1, z1), p(Tn-1x, Tnx),
p(z1, Tn-1x) + p(z1, Tnx) '
Thus, we get the following cases:
Case 1.p(z1, T«x) < rp(zh Tn-1x) < r2p(z1, Tn-2x) ^ ••• ^ r«p(z1,x). Case 2.p(z1, T«x) < rp(z1,z1) ^ rp(z1, Tn-1x) ^ • • • ^ r«p(z1,x). Case 3.p(z1, T«x) < rp(Tn-1x, T«x) < r2p(Tn-2x, Tn-1x) ^ • • • ^ r«p(x, Tx). Case 4.p(zu T«x) < r[p(z1,r«-1x2+p(z1,r«x)] impliesp(z1, T«x) < rp(zh Tn-1x) < r«p(zi,x). Thus, in all cases, we have
p(z1, T«x) < r«p(z1,x)orp(z1, T«x) < r«p(x, Tx).
Similarly, p(z2, T«x) < r«p(z2,x) orp(z2, T«x) < r«p(x, Tx). Let 0 « c and choose a natural number N3 such that r«p(z1, x) « | ,orr«p(x, Tx) « | and r«p(z2, x) « | forall n > N3. Thus,
p(z1,z2) ^p(z1, T«x) + p(z2, T«x « 2 + 2 = c,
which is again a contradiction. Hence, z1 = z2. □
Now, in order to support the usability of Theorem 2.1, we present the following example.
Example 2.1 Let E = CR [0,1] with the norm \\u\\ = \\u\\œ + \\u'\\œ, and X = P = {u e E : u(t) > 0, t e [0,1]}, which is a nonnormal solid cone. Define the mapping p : X x X P by
p(x, y) =
x, x = y, x + y otherwise.
Then (X,p) is a 0-complete partial cone metric space. We can define a partial order on X as follows:
x c y if and only if x(t) < y(t) for all t e [0,1].
Then (X,p, c) is a 0-complete partially ordered partial cone metric space. For every fixed r e [0,1), define T: X ^ X by
Tx(t) =
if t e [0,1] such that x(t) = 0,
^rt + ^rx(t) if t e [0,1] such that 2« -1 < x(t) < 2« for n e Z+, nrt + 2«+Trx(t) if t e [0,1] such that 2n < x(t) < 2n +1 for n e Z+.
Thus, for all x e X, we consider the following three cases: Case l.ft e [0,1] such that x(t) = 0, then Tx(t) = x(t). Case 2.1ft e [0,1] such that 2n -1 < x(t) < 2«, then
^ , x 2« -1 2« -1 , x 2« -1 2« -1
Tx(t) =-rt +-rx(t) <-r +-r < rx(t).
w 2 4«w- 2 2"w
Case 3.1ft e [0,1] such that 2« < x(t) < 2« + 1, then
Tx(t) = nrt +
2« + 1
rx(t) < nr + nr < rx(t).
In all cases, for allx e X and t e [0,1],wehave Tx(t) < rx(t). Hence, for allx,y e X, x ç y, we have
p(Tx, Ty) < rp(x,y) < rU(x,y),
where U(x,y) e {p(x,y),p(x, Tx),p(y, Ty), Pi-x,Ty')+P(y,Tx')}, which ensures that condition (2.1) is satisfied. Also, conditions (i)-(iii) of Theorem 2.1 are satisfied. Following the Theorem 2.1, we deduce that T has a fixed point in X; indeed, x = 0 is a fixed point of T.
Theorem 2.2 Let (X,p, ç) be a partially ordered partial cone metric space over a solid cone P of a normed vector space (E, || ■ ||). Let S, T : X — X be such that T is an S-nondecreasing mapping with respect to ç, TX ç SX, and SX is a 0-complete subset ofX. Define ty : [0,1) — (2,1] as in Theorem 2.1. Suppose that there exists r e [0,1) such that
ty (r)p(Sx, Tx) < p(Sx, Sy) implies p(Tx, Ty) < rU (Sx, Sy) (2.8)
for all comparable Sx, Sy e X, where U (Sx, Sy) e {p(Sx, Sy), p(Sx, Tx), p(Sy, Ty), (p(Sx, Ty) + p(Sy, Tx))/2}. Suppose that the following conditions hold:
(i) there exists x0 e X such that Sx0 ç Tx0;
(ii) for a nondecreasing sequence xn — x, we have xn ç x for all n e N ;
(iii) for two nondecreasing sequences {xn}, {yn }ç X such that xn ^— yn, xn x, and yn — y as n — to, we have x ç y;
(iv) the set of points of coincidence of S and T is totally ordered, and S, T are weakly compatible.
Then S and T have a unique common fixed point in X.
Proof By Lemma 1.1 there exists Y ç X, such that SY = SX and S : Y — X is one-to-one. Define f : SY — SX by fSx = Tx for all Sx e SY. Since S is one-to-one on Y, f is well defined. Note that, for all comparable Sx, Sy e SY,
ty (r)p(Sx,fSx) < p(Sx, Sy) implies p(fSx,fSy) < rU (Sx, Sy),
where U (Sx, Sy) e {p(Sx, Sy), p(Sx,fSx), p(Sy,fSy), P(SxfSy^+P(SyfSx}.
Since T is S-nondecreasing, we have that f is nondecreasing. In fact, Sx ç Sy implies Tx ç Ty, and hence fSx = Tx ç Ty = fSy. Since SY is 0-complete, by Theorem 2.1 we get that f has a fixed point on SY, say Sz. Then z = y is a coincidence point of S and T, that is, Tz = fSz = Sz.
Now, we prove that S and T have a unique coincidence point. Suppose that w is another coincidence point of S and T with z = w. Then
ty(r)p(Sz, Tz) < p(Sz, Sw),
and by (2.8) we have
p(Tz, Tw ) ^ rU (Sz, Sw),
where U(Sz, Sw) e {p(Sz, Sw),p(Sz, Tz),p(Sw, Tw), p(Sz,Tw)^p(Sw,Tz)}. Since Sz = Tz and Sw = Tw, it follows from (2.9) that
p(Tz, Tw) ^ rU(Tz, Tw),
where U(Tz, Tw) e {p(Tz, Tw),p(Tz, Tz),p(Tw, Tw), p(Tz,Tw);p(Tw,Tz)} = {p(Tz, Tw),p(Tz, Tz), p(Tw, Tw)}, which is a contradiction. Hence, z = w. Let v = Sz = Tz. Since S and T are weakly compatible, we have Sv = STz = TSz = Tv. Then v is also a coincidence point of S and T, Thus, v = z by uniqueness. Therefore, z is the unique common fixed point of S and T. □
3 Coupled point theorems in partial cone metric spaces
In this section, we will apply the results obtained in Section 2 to establish the corresponding Suzuki-type coupled fixed point theorems for generalized mappings in partially ordered partial cone metric spaces over a nonnormal cone.
For a = (x,y), b = (u, v) e X2, we introduce the mapping p : X2 x X2 — P defined by p (a, b) = p(x, u) + p(y, v). The following conclusion is valid, and for its proof, we refer to [30].
Lemma 3.1 If (X, p) is a partial co«e metric space over a solid co«e P of a «ormed vector space (E, || • ||), the« (X2,p) is also a 0-complete partial co«e metric space.
Proof It suffices to prove that, for a = (x,y), b = (u, v), c = (z, w) e X2,
p (a, b) ^ p (a, c) + p (b, c) - p (c, c).
In fact, for a = (x, y), b = (u, v), C = (z, w) e X2, we have
p (a, b) = p(x, u) + p(y, v)
^ p(x, z) + p(u, z) -p(z, z) + p(y, w) + p(v, w) -p(w, w) = p(x, z) + p(y, w) + p(u, z) + p(v, w) - [p(z, z) + p(w, w)]
=p (a, c) + p (b, c) - p (c, c).
Suppose that the sequence {x«} = {(x«,y«)} is a 0-Cauchy sequence in (X2,p). Then, for every c » 0, there exists a positive integer «0 e N such that p(c«,xm) = p(x«,xm) + p(y«,ym) ^ c for all «, m > «0. Thenp(x«,xm) ^ c and p(y«,ym) ^ c. Thus, {x«} and {y«} are 0-Cauchy sequences in (X,p). Since (X,p) is 0-complete, there exist x, y e X such that x« -- x, y« y, and p(x, x) = 0, p(y, y) = 0.
Thus, for every c » 0, there exists « e N such that p(x«,x) ^ | andp(y«,y) ^ f for all « > «1. Then p((x«,y«), (x,y)) = p(x«,x) + p(y«,y) « 2 + § = c, and p((x,y), (x,y)) = 0. Thus, {(x«,y«)} --- (x,y).
Therefore, (X2,p) is a 0-complete partial cone metric space. □
Theorem 3.1 Let (X,p, c) be a 0-complete partially ordered partial co«e metric space over a solid co«e P of a «ormed vector space (E, || • ||). Let A : X x X — X be a mappi«g
satisfying the mixed monotone property on X with respect to c. Define ty : [0,1) — (2, 1] as in Theorem 2.1. Assume that there exists r e [0,1) such that
ty(r)p(x,A(x,y)) + p(y,A(y,x))] ^p(x,u) + p(y,v) implies
p(A(x,y),A(u, v)) + p(A(y,x),A(v,u)) < rU((x,y),(u, v)) for all x, y e X such thatx c u and vc y, where
U((x,y), (u, v)) e |p(x, u) + p(y, v),p(x,A(x,y)) + p(y,A(y,x)), p(u, A(u, v)) + p(v, A(v, u)),
p(x, A(u, v)) + p(y, A(v, u)) + p(u, A(x, y)) + p(v, A(v, u)) 1
Suppose that the following conditions hold:
(i) there exists x0,y0 e X such that x0 c A(x0,y0) and A(y0,x0) c y0;
(ii) for a nondecreasing sequence xn — x, we have xn c x for all n e N;
(iii) for a nonincreasing sequence yn — x, we have y c yn for all n e N;
(iv) for two nondecreasing sequences {xn}, {un} c X such that xn c un for all n e N, xn — x, and un — u as n — to, we have x c u;
(v) for two nonincreasing sequences {yn}, {vn}c X such that vn c yn for all n e N, vn — v, andyn — y as n — to, we have v c y.
Then A has a coupledfixed point in X, that is, there exist z, w e X such thatA(z, w)= z and A(w, z) = w.
Proof Let X = X x X. For a = (x,y), b = (u, v) e X, we introduce the order -< as
a < b if and only if x c u, v c y.
It follows from Lemma 3.1 that (X,p, -<) is also a 0-complete partially ordered partial cone metric space, where
p (a, b) = p(x, u) + p(y, v).
The mapping T : XX — XX is given by Ta = (A(x,y), A(y,x)) for all a = (x,y) e XX. Then a coupled point of A is a fixed point of T and vice versa.
If a < lb, then x c u and v c y. Noting the mixed monotone property of A, we see that if A(x,y) c A(u, v) and A(v, u) c A(y,x), then Ta < Tb. Thus, T is a nondecreasing mapping with respect to the order -< on XX.
On the other hand, for all a = (x,y), b = (u, v) e XX with a < b, if ty(r)p(a, Ta) = ty(r)[p(x, A(x,y)) + p(y, A(y,x))] ^ p(x, u) + p(y, v) = p(a, b), then we have
p(Ta, Tb) = p(A(x,y), A(u, v^ + p(A(y,x), A(v, u)) < rU((x,y),(u, v^,
U (a, b) = U ((x, y),(u, v))
e jp(x, u) + p(y, v),p(x,A(x,y)) + p(y,A(y,x)),p(u,A(u, v)) + p(v,A(v, u)), p(x, A(u, v)) + p(y, A(v, u)) + p(u, A(x, y)) + p(v, A(v, u))
= \p (a, b), p (a, Ta), p (b, Tb),
p (a, Tb) + p (Ta, b)
Also, there exists an x0 = (x0,y0) e X such that x0 -< Tx0 = (A(x0,y0), A(y0,x0)).
If a nondecreasing monotone sequence {x«} = {(x«,y«)} in X tp-converges to x = (x,y), then x« = (x«,y«) < (x«+i,y«+i) = x«+i,thatis, x« c x«+i and y«+i c y«. Thus, {x«} is a nondecreasing sequence xp-converging to x, and {y«} is a nonincreasing sequence Tp-converging to y. Thus, x« c— x and y c y« for all« e N. This implies x« ^N x .
If two nondecreasing sequence {x«} = {(x«, y«)}, {y«} = {(u«, v«)} are such that x« < y« for
all« eN,x« — (x,y), andy« — (u, v) as« — to, thenx« c u«, v« cy«,x« cx«+1,y«+1 cy«, and u« c u«+1, v«+1 c v«. Thus, {x«}, {u«} c X are two nondecreasing sequences, x« c u« for all « e N, x« ^ x, and u« — u as« — to, and by condition (iv) we have x c u. Similarly, by condition (v) we have v c y. Thus, (x,y) -< (u, v).
Therefore, all hypotheses of Theorem 2.1 are satisfied. Following Theorem 2.1, we deduce that A has a coupled point, that is, there exist z, w e X such that A(z, w) = z and A(w, z) = w. □
Now, we present the following example.
Example 3.1 Let X = P = {(x1,x2) : x1,x2 e R+} c R2, and E = R2 with the norm ||x|| = ^/xfTx2. Define the mappingp: X x X — P by
p(x, y) = (xi V yi, x2 V y2)
for all x = (x1,x2), y = (y1,y2) e X, where a V b = max{a, b}, a, b e R+. Then (X,p) is a 0-complete partial cone metric space. Define the partial order on X by
x c y if and only if x1 < x2, y1 < y2.
Then (X,p, c) is a 0-complete partially ordered partial cone metric space. In fact, "c" is equal to For any fixed r e [0,1), define A : X x X — X by
A(x, y)=A((xi, x2),(yi, y2)) = ' e^r)
for all x,y e X. It is clear that A satisfies the mixed monotone property on X with respect to c. Define ft : [0,1) — (5,1] as in Theorem 3.1.
Now, for all x, y, u, v G X, x ç u, v ç y, we have
ty(r)[p(x,A(x,y)) + p(y,A(y,x))] ^p(x,A(x,y)) + p(y,A(y,x)),
(rxi rx2 \ ( ry\ ry2 \
xi v-, x2 v — + y1 v , y2 v —
1 i+yi ey2 J V 1 + xi ex2 J
= (xi + yi, x2 + y2)
^ (ui + yi, U2 + y2) = p(x, u) + p(y, v)
and, on the other hand,
p(A(x, y), A(u, v)) + p(A(y, x), A(v, u))
rxi rui rx2 ru2\ ( ryi rvi ry2 rv2
v --, — v — I + I ^— v :-, ^y v
i+yi i + vi ey2 ev2 ) \i + xi i + ui ex2 eu2 rui ryi ru2 ry2 \ i + vi i + xi, ev2 ex2 J
< (rui + ryi, ru2 + ry2) = r[p(x, u) + p(y, v)]
^ rU((x, y), (u, v^,
where U ((x, y), (u, v)) is as in Theorem 3.i.
Also, conditions (i)-(v) of Theorem 3.i are satisfied. From Theorem 2.i we obtain that A has a fixed point in X ; indeed, x = (0,0) is a fixed point of A.
Competing interests
The authors declare that they have no competing interests. Authors' contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript. Author details
1 Department of Mathematics, Nanchang University, Nanchang, 330031, P.R. China. 2Department of Mathematics, Fengcheng No. 2 Senior School, Yichun, Jiangxi 331100, P.R. China.
Acknowledgements
The authors thank the editor and the referees for their valuable comments and suggestions. The research has been supported by the National Natural Science Foundation of China (11071108,11361042,11326099,11461045) and the Provincial Natural Science Foundation of Jiangxi, China (2010GZS0147,20114BAB201007,20142BAB211004) and supported partly by the Provincial Graduate Innovation Foundation of Jiangxi, China (YC2012-B004).
Received: 14 July 2015 Accepted: 31 March 2016 Published online: 06 April 2016
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