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Suzuki’s type fixed point theorems for generalized mappings in partial cone metric spaces over a solid cone
Fixed Point Theory and Applications volume 2016, Article number: 47 (2016)
Abstract
In this paper, we obtain some Suzukitype 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.
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 θcomplete 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 \(p:X\times X\to\mathbb {R}^{+}\) is said to be a partial metric if for all \(x,y,z\in X\), the following conditions are satisfied:
 \((p1)\) :

\(p(x,x)=p(x,y)=p(y,y)\) if and only if \(x=y\);
 \((p2)\) :

\(p(x,x)\leq p(x,y)\);
 \((p3)\) :

\(p(x,y)=p(y,x)\);
 \((p4)\) :

\(p(x,y)\leq p(x,z)+p(z,y)p(z,z)\).
The pair \((X,p)\) is called a partial metric space. If \(p(x,y)=0\), then the \((p1)\) 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 \((\mathbb{R}^{+},p)\), where \(p:\mathbb{R}^{+}\times\mathbb{R}^{+}\to\mathbb {R}^{+}\) 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)
\(ax+by \in P\) for all \(x,y \in P\) and \(a,b\geq0\), and

(ii)
\(P \cap(P) = \{\theta\}\), where θ is the zero element of E.
Each cone P of E determines a partial order ⪯ on E by \(x \preceq y \) if and only if \(y x \in P\) for all \(x,y \in E\). We shall write \(x\prec y\) if \(x\preceq y\) and \(x\neq y\).
A cone P of a topological vector space E is solid if \(\operatorname {int}P \neq\varnothing\), where intP is the interior of P. For all \(x,y \in E\) with \(yx \in \operatorname {int}P\), we write \(x\ll y\). Let P be a solid cone of a topological vector space E. A sequence \(\{u_{n}\}\) of E weakly converges [18] to \(u\in E\) (denoted \(u_{n}\stackrel{w}{\to}u\)) if for each \(c\in \operatorname {int}P\), there exists a positive integer \(n_{0}\) such that \(uc\ll u_{n} \ll u+c\) for all \(n\geq n_{0}\). A cone P of a normed vector space \((E,\Vert \cdot \Vert )\) is normal if there exists \(K > 0\) such that \(\theta \preceq x\preceq y\) implies that \(\Vert x\Vert \leq K\Vert y\Vert \) for all \(x,y \in P\), 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 \(d:X\times X\to P\) such that, for all \(x,y,z\in X\):
 \((d1)\) :

\(d(x,y)=\theta\) if and only if \(x=y\);
 \((d2)\) :

\(d(x,y)=d(y,x)\);
 \((d3)\) :

\(d(x,y)\preceq d(x,z)+d(z,y)\).
Definition 1.3
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 mapping \(p:X\times X\to P\) such that, for all \(x,y,z\in X\):
 \((p1)\) :

\(p(x,x)=p(x,y)=p(y,y)\) if and only if \(x=y\);
 \((p2)\) :

\(p(x,x)\preceq p(x,y)\);
 \((p3)\) :

\(p(x,y)=p(y,x)\);
 \((p4)\) :

\(p(x,y)\preceq p(x,z)+p(z,y)p(z,z)\).
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 \(E=C_{R}^{1}[0,1]\) with the norm \(\Vert u\Vert =\Vert u\Vert _{\infty}+\Vert u'\Vert _{\infty}\), and \(X=P=\{u\in E:u(t)\geq0,t\in[0,1]\}\), which is a nonnormal solid cone. Define the mapping \(p:X\times X\to P\) by
Then p is a partial cone metric, but not a cone metric, since \(p(x,x)\neq\theta\) for all \(x\in X\) with \(x\neq\theta\).
A partial cone metric p on X over a solid cone P generates a topology \(\tau_{p}\) on X, which has a base of the family of open pballs \(\{B_{p}(x,c):x\in X,\theta\ll c\}\), where \(B_{p}(x,c)=\{y\in X:p(x,y)\ll p(x,x)+c\}\) for \(x\in X\) and \(c\in \operatorname {int}P\).
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 \(\{x_{n}\}\) in X converges to \(x \in X\) (denoted by \(x_{n} \stackrel{\tau_{p}}{\to}x\)) if for each \(c\in \operatorname {int}P\), there exists a positive integer \(n_{0}\) such that \(p(x_{n},x) \ll p(x,x)+c\) for each \(n \geq n_{0}\) (that is, \(p(x_{n},x)\stackrel{w}{\to}p(x,x)\)).

(ii)
A sequence \(\{x_{n}\}\) in X is θCauchy if for each \(c \in \operatorname {int}P\), there exists a positive integer \(n_{0}\) such that \(p(x_{n},x_{m})\ll c\) for all \(m,n\geq n_{0}\). The partial cone metric space \((X,p)\) is θcomplete if each θCauchy sequence \(\{x_{n}\}\) of X converges to a point \(x \in X\) such that \(p(x,x) = \theta\).
Definition 1.5
Let \((X,p)\) be a partial cone metric space over a solid cone P of a topological vector space \((E,\Vert \cdot \Vert )\).

(i)
A sequence \(\{x_{n}\}\) in X strongly converges to \(x \in X\) (denoted by \(x_{n} \stackrel{s\tau_{p}}{\to}x\)) if
$$\lim_{n\to\infty}p(x_{n},x)=\lim_{n\to\infty}p(x_{n},x_{n})= p(x,x). $$ 
(ii)
A sequence \(\{x_{n}\}\) in X is Cauchy if there exists \(u\in P\) with \(\Vert u\Vert <\infty\) such that \(\lim_{m,n\to\infty}p(x_{m},x_{n})=u\). The partial cone metric space \((X,p)\) is complete if each Cauchy sequence \(\{x_{n}\}\) of X strongly converges to a point \(x \in X\) such that \(p(x,x) = u\).
Note that if P is a normal solid cone of a normed vector space \((E,\Vert \cdot \Vert )\), then each complete partial cone metric space is θcomplete. But the converse is not true. The following example ([16], Example 4) shows that a θcomplete partial cone metric space is not necessarily complete.
Example 1.2
([16])
Let \(X = \{(x_{1},x_{2},\ldots,x_{k}) : x_{i}\geq0,x_{i}\in Q,i = 1,2,\ldots,k\}\), and \(E = R^{k}\) with the norm \(\Vert x\Vert =\sqrt{\sum_{i=1}^{k}x_{i}^{2}}\), \(P=R_{+}^{k}\), where Q denotes the set of rational numbers. Define the mapping \(p : X \times X \to P\) by
where the symbol ∨ denotes the maximum, that is, \(x\vee y=\max\{ x,y\}\). Clearly, \((X,p)\) is a partial cone metric space, \(p(x,x) = x\) for each \(x \in X\), \(p(x,\theta) = \theta\) if and only if \(x = \theta\), and P is normal. On the other hand, \((X,p)\) is θcomplete but not complete.
Let X be a nonempty set, and \(S,T:X\to X\) be two mappings. A point \(x\in X\) is said to be a coincidence point of S and T if \(Sx=Tx\). A point \(y\in X\) is called point of coincidence of S and T if there exists a point \(x\in X\) 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 \((X,\sqsubseteq)\) be a partially ordered set; \(x,y\in X\) are called comparable if \(x\sqsubseteq y\) or \(y \sqsubseteq x\). A mapping \(T:X\to X\) is said to be nondecreasing if for \(x,y\in X\), \(x\sqsubseteq y\) implies \(Tx\sqsubseteq Ty\). Let \(S,T:X\to X\) be two mappings; T is said to be Snondecreasing if \(Sx\sqsubseteq Sy\) implies \(Tx\sqsubseteq Ty\) for all \(x,y\in X\).
Bhasker and Lakshmikantham [20] introduced the concepts of mixed monotone mappings and coupled fixed point.
Definition 1.6
([20])
Let \((X,\sqsubseteq)\) be a partially ordered set, and \(A : X \times X \to X\). 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 \(x,y \in X\),
Definition 1.7
([20])
An element \((x,y) \in X^{2}\) is said to be a coupled fixed point of the mapping \(A : X^{2}\to X\) if \(A(x,y)= x\) and \(A(y,x)=y\).
Lemma 1.1
([21])
Let X be a nonempty set, and \(S:X\to X\) a mapping. Then there exists a subset \(Y\subseteq X\) such that \(SY=SX\) and \(S:Y\to X\) is onetoone.
Paesano and Vetro [22] proved Suzukitype 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 Suzukitype 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 Suzukitype 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.
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 \(a\preceq b\) and \(b\preceq c\), then \(a\preceq c\).

(2)
If \(a\ll b\) and \(b\ll c\), then \(a\ll c\).

(3)
If \(\theta\preceq u\ll c\) for each \(c\in \operatorname {int}P\), then \(u=\theta\).

(4)
If \(a\preceq\lambda a\), where \(0\leq\lambda<1\), then \(a=\theta\).

(5)
If \(a\preceq b +c\) for each \(c\in \operatorname {int}P\), then \(a\preceq b\).
Now, we establish some Suzukitype fixed point theorems for generalized mappings in partially ordered partial cone metric spaces over a solid cone.
Theorem 2.1
Let \((X,p,\sqsubseteq)\) be a θcomplete partially ordered partial cone metric space over a solid cone P of a normed vector space \((E,\Vert \cdot \Vert )\). Let \(T:X\to X\) be a nondecreasing mapping with respect to ⊑. Define the nonincreasing function \(\psi:[0,1)\to(\frac{1}{2},1]\) by
Assume that there exists \(r\in[0,1)\) such that
for all comparable \(x,y\in X\), where \(U(x,y)\in\{ p(x,y),p(x,Tx),p(y,Ty),\frac{p(x,Ty)+p(y,Tx)}{2}\}\). Suppose that the following conditions hold:

(i)
there exists \(x_{0}\in X\) such that \(x_{0}\sqsubseteq Tx_{0}\);

(ii)
for a nondecreasing sequence \(x_{n}\stackrel{\tau_{p}}{\to} x\), we have \(x_{n}\sqsubseteq x\) for all \(n\in N\);

(iii)
for two nondecreasing sequences \(\{x_{n}\}, \{y_{n}\}\subseteq X\) such that \(x_{n}\sqsubseteq y_{n}\), \(x_{n}\stackrel{\tau_{p}}{\to} x\), and \(y_{n} \stackrel{\tau_{p}}{\to} y\) as \(n\to\infty\), we have \(x\sqsubseteq y\).
Then T has a fixed point in X. Moreover, the fixed point of T is unique if

(iv)
for all \(x,y\in X\) that are not comparable, there exists \(u\in X\) comparable with x and y.
Proof
Since \(\psi(r)\leq1\), \(\psi(r)p(x,Tx)\preceq p(x,Tx)\) for every \(x\in X\). By (2.1) and using the triangular inequality, we have
where
Thus, we get the following cases:

Case 1. \(p(Tx,T^{2}x) \preceq r p(x,Tx)\).

Case 2. \(p(Tx,T^{2}x) \preceq r p(Tx,T^{2}x)\), which implies that \(p(Tx,T^{2}x)=\theta\).

Case 3. \(p(Tx,T^{2}x) \preceq r \frac{p(x,T^{2}x)+p(Tx,Tx)}{2}\preceq r \frac{p(x,Tx)+p(Tx,T^{2}x)}{2}\), which implies that \(p(Tx,T^{2}x) \preceq r p(x,Tx)\).
Then, in all cases, we have
Let \(x_{0}\in X\) be such that \(x_{0}\sqsubseteq Tx_{0}\). Since T is nondecreasing, we get
Define the sequence \(\{x_{n}\}\) by \(x_{n}=T^{n}x_{0}\), so that \(x_{n+1}=Tx_{n}\). If \(x_{n}=x_{n+1}=Tx_{n}\) for some n, then \(x_{n}\) becomes a fixed point of T. Now, suppose that \(x_{n}\neq x_{n+1}\) for all \(n\in N\). Then \(p(x_{n},x_{n+1})\succ\theta\).
Note that \(\psi(r)p(x_{n1},Tx_{n1})\preceq p(x_{n1},Tx_{n1})\) for all \(n\in Z^{+}\), where \(Z^{+}\) is the set of positive integers. Since \(x_{n1}\) and \(Tx_{n1}\) are comparable for all \(n>0\), by (2.1) we have
where
Thus, we get the following cases:

Case 1. If \(U(x_{n1},Tx_{n1})=p(x_{n1},x_{n})\), then \(p(x_{n},x_{n+1})\preceq r p(x_{n1},x_{n})\).

Case 2. \(p(x_{n},x_{n+1})\preceq r p(x_{n},x_{n+1}) \), which implies that \(p(x_{n},x_{n+1})=\theta\).

Case 3. \(p(x_{n},x_{n+1})\preceq r \cdot\frac {p(x_{n1},x_{n+1})+p(x_{n},x_{n})}{2}\preceq r(\frac {p(x_{n1},x_{n})+p(x_{n},x_{n+1})}{2})\), which implies that \(p(x_{n},x_{n+1})\preceq r p(x_{n1},x_{n})\).
Then, in all cases, we have
Continuing this process, it follows that
Thus, for any \(m,n\in Z^{+}\) with \(m>n\), we get
Let \(\theta\ll c\) be given, Choose \(\delta>0\) such that \(c+N_{\delta}(\theta)\subseteq P\), where \(N_{\delta}(\theta)=\{y\in E: \Vert y\Vert <\delta\}\). Also, choose a natural number \(N_{1}\) such that \(\frac{r^{n}}{1r}p(x_{0},x_{1})\in N_{\delta}(\theta)\) for all \(n\geq N_{1}\). Then \(\frac {r^{n}}{1r}p(x_{0},x_{1})\ll c\) for all \(n\geq N_{1}\). Thus,
for all \(m>n\geq N_{1}\). Hence, \(\{x_{n}\}\) is a θCauchy sequence in \((X,p)\). Since \((X,p)\) is a θcomplete partial cone metric space, there exists \(z\in X\) such that \(x_{n}\stackrel{\tau_{p}}{\to} z\) and \(p(z,z)=\theta\).
First, we show that there exists \(j\in Z^{+}\) such that \(T^{j}z=z\). Arguing by contradiction, we assume that \(T^{j}z\neq z\) for all \(j\in Z^{+}\).
Note that, by condition (ii), if \(\{x_{n}\}\) is nondecreasing, then \(x_{n}\sqsubseteq z\). Since T is nondecreasing, we get \(x_{n+1}=Tx_{n}\sqsubseteq Tz\) for all \(n\in N\). Taking the limit as \(n\to\infty\), by (iii) we obtain that \(z\sqsubseteq Tz\), which implies that \(\{T^{n}z\}\) is a nondecreasing sequence. So, we have shown that for \(\{x_{n}\}\), \(\{T^{n}z\}\) also is a θCauchy sequence. We also have \(T^{j}z\) is comparable with \(x_{n}\) for all \(j,n\in N\).
Now, we prove that
Since \(p(T^{j}z,z)\succ\theta\), \(p(z,z)=\theta\), and \(x_{n}\stackrel{\tau _{p}}{\to} z\), there exists \(N_{2}\in N\) such that \(p(x_{n},z)\preceq\frac {p(T^{j}z,z)}{3} \) for all \(n\geq N_{2}\). We have
By (2.1) and using the triangle inequality, we get that
where
Thus,
where
Since \(x_{n}\stackrel{\tau_{p}}{\to} z\) for every \(c\gg\theta\), there exists \(n_{0}\in N\) such that \(p(x_{n},z)\ll\frac{c}{2}\) and \(p(x_{n},x_{n+1})\ll\frac{c}{2}\) for all \(n> n_{0}\). Now, for \(n>n_{0}\), we consider the following cases:
Case 1. \(p(z,T^{j+1}z)\preceq p(x_{n+1},z) +r p(x_{n},T^{j}z)\preceq p(x_{n+1},z) +r( p(x_{n},z)+p(z,T^{j}z))\preceq rp(z, T^{j}z)+c\). Then it follows from Remark 2.1(5) that \(p(z,T^{j+1}z)\preceq rp(z,T^{j}z)\), and we get
Case 2. \(p(z,T^{j+1}z)\preceq p(x_{n+1},z) +r p(x_{n},x_{n+1})\ll c\), which implies that \(p(z,T^{j+1}z)=\theta\).
Case 3.
which implies that \(p(z,T^{j+1}z)\preceq r p(T^{j}z,T^{j+1}z)\). Then from (2.2) we have
Case 4.
which implies that \(p(z,T^{j+1}z)\preceq rp(z,T^{j}z)\). Then from (2.4) we get \(p(z,T^{j+1}z)\preceq r^{j}p(z,Tz)\).
Thus, in all cases, we obtain \(p(z,T^{j+1}z)\preceq r^{j}p(z,Tz)\nonumber \), that is, (2.3) holds.
Now, we consider the following three cases:

(1)
\(0\leq r<\frac{\sqrt{5}1}{2}\);

(2)
\(\frac{\sqrt{5}1}{2} \leq r< \frac{\sqrt{2}}{2}\);

(3)
\(\frac{\sqrt{2}}{2}\leq r< 1\).
In case (1), we note that \(r^{2}+r<1\) and \(\psi(r)=1\). If we assume that \(p(T^{2}z,z)\prec p(T^{2}z,T^{3}z)\), then by (2.2) we have
This is a contradiction. So, we have \(p(T^{2}z,z)\succeq p(T^{2}z,T^{3}z)= \psi(r)p(T^{2}z,T^{3}z)\). By (2.1)(2.3) we deduce that
where
Thus, we get the following cases:

Case 1. \(p(T^{3}z,Tz)\preceq r p(T^{2}z,z)\preceq r^{2} p(Tz,z)\preceq r p(Tz,z)\).

Case 2. \(p(T^{3}z,Tz)\preceq r p(T^{2}z,T^{3}z) \preceq r^{3}p(Tz,z)\preceq r p(Tz,z)\).

Case 3. \(p(T^{3}z,Tz)\preceq r p(Tz,z)\).

Case 4. \(p(T^{3}z,Tz)\preceq r \frac{p(T^{3}z,z)+p(Tz,T^{2}z)}{2}\preceq r[\frac{r^{2}p(Tz,z)+rp(z,Tz)}{2}]\preceq r^{2} p(Tz,z)\preceq r p(Tz,z)\).
Then, in all cases, we have \(p(T^{3}z,Tz)\preceq r p(Tz,z)\). Hence,
which is a contradiction.
In case (2), we note that \(2r^{2}<1\) and \(\psi(r)=\frac{1r}{r^{2}}\). Now, we show by induction that
for \(n\geq2\). By (2.2), (2.5) holds for \(n=2\). Assume that (2.5) holds for some n with \(n\geq2\). Since
we have
and so
Therefore, by the hypotheses we have
where
Thus, we get the following cases:

Case 1. \(p(T^{n+1}z,Tz)\preceq r p(T^{n}z,z)\preceq r^{n} p(Tz,z)\preceq r p(Tz,z)\).

Case 2. \(p(T^{n+1}z,Tz)\preceq r p(T^{n}z,T^{n+1}z) \preceq r^{n+1}p(Tz,z)\preceq r p(Tz,z)\).

Case 3. \(p(T^{n+1}z,Tz)\preceq r p(Tz,z)\).

Case 4.
$$\begin{aligned}p \bigl(T^{n+1}z,Tz \bigr)&\preceq r \frac {p(T^{n+1}z,z)+p(Tz,T^{n}z)}{2}\preceq r \biggl(\frac{r^{n} p(Tz,z)+rp(Tz,z)}{2} \biggr) \\ &\preceq r^{2} p(Tz,z)\preceq r p(Tz,z). \end{aligned} $$
Then, in all cases, we have \(p(T^{n+1}z,Tz)\preceq r p(Tz,z)\). Therefore, (2.5) holds. Now, from (2.3) we have
for \(n\geq1\). Since \(0\leq r<1\), for every \(c\gg\theta\), there exists \(n_{0}\in N\) such that \(p(Tz,z)\ll c\) for all \(n\geq n_{0}\). Hence,
which implies that \(p(Tz,z)=\theta\). Thus, \(Tz=z\). This is a contradiction.
In case (3), we note that for \(x,y\in X\), either
Indeed, if
then we have
This is a contradiction. Now, since either
for all \(n\in N\), by (2.1) it follows that either
where
Hence, we deduce that either
or
Since \(x_{n}\stackrel{\tau_{p}}{\to} z\) for every \(c\gg\theta\), there exists \(n_{0}\in N\) such that \(p(x_{n},z)\ll\frac{c}{2}\) and \(p(x_{n},x_{n+1})\ll\frac{c}{2}\) for all \(n> n_{0}\). Now, by (2.6) we get the following cases:

Case 1. \(p(Tz,z)\preceq p(x_{2n+1},z)+r p(z,x_{2n})\ll c\) implies \(p(Tz,z)=\theta\).

Case 2. \(p(Tz,z)\preceq p(x_{2n+1},z)+rp(x_{2n},x_{2n+1})\ll c \) implies \(p(Tz,z)=\theta\).

Case 3. \(p(Tz,z)\preceq p(x_{2n+1},z)+rp(z,Tz) \ll c+ rp(z,Tz) \) implies \(p(Tz,z)=\theta\).

Case 4.
$$\begin{aligned}p(Tz,z)&\preceq p(x_{2n+1},z)+r \biggl[ \frac {p(x_{2n},Tz)+p(z,x_{2n+1})}{2} \biggr] \\ &\preceq p(x_{2n+1},z)+r \biggl[\frac {p(x_{2n},z)+p(z,Tz)+p(z,x_{2n+1})}{2} \biggr]\ll c+ rp(z,Tz), \end{aligned} $$which implies that \(p(Tz,z)=\theta\). Then, in all cases, we have \(p(Tz,z)=\theta\). Similarly, by (2.7) we also have \(p(Tz,z)=\theta\). Thus, \(Tz=z\). This is a contradiction.
Therefore, in all cases, there exists \(j\in N\) such that \(T^{j}z=z\). Since \(\{T^{n}z\}\) is a θCauchy sequence, we obtain \(z=Tz\). If not, that is, if \(z\neq Tz\), from \(p(T^{nj}z,T^{nj+1}z)=p(z,Tz)\) for all \(n\in N\) it follows that \(\{T^{n}z\}\) is not a θCauchy sequence. Hence, z is a fixed point of T.
Finally, we prove the uniqueness of the fixed point. Suppose that there exist \(z_{1},z_{2}\in X\) with \(z_{1}\neq z_{2}\) such that \(Tz_{1}=z_{1}\) and \(Tz_{2}=z_{2}\). We have two possible cases:
Case (a). If \(z_{1}\) and \(z_{2}\) are comparable, using (2.1) with \(x=z_{1}\) and \(y=z_{2}\), we get that
where
Thus, we get the following cases:

Case 1. \(p(z_{1},z_{2})\preceq rp(z_{1},z_{2})\) implies \(p(z_{1},z_{2})=\theta\).

Case 2. \(p(z_{1},z_{2})\preceq rp(z_{1},z_{1})\preceq rp(z_{1},z_{2})\) implies \(p(z_{1},z_{2})=\theta\).

Case 3. \(p(z_{1},z_{2})\preceq rp(z_{2},z_{2})\preceq rp(z_{1},z_{2})\) implies \(p(z_{1},z_{2})=\theta\).
Thus, in all cases, we have \(p(z_{1},z_{2})=\theta\), that is, \(z_{1}=z_{2}\). This is a contradiction. Hence, \(z_{1}=z_{2}\).
Case (b). If \(z_{1}\) and \(z_{2}\) are not comparable, then there exists \(x\in X\) comparable with \(z_{1}\) and \(z_{2}\). First, we note that for each \(x\in X\) comparable with \(z_{1}\), we have that \(T^{n}z_{1}\) and \(T^{n}x\) are comparable and \(\psi (r)p(T^{n1}z_{1},T^{n}z_{1})\preceq p(T^{n1}z_{1},T^{n}z_{1})=p(z_{1},z_{1})\preceq p(T^{n1}z_{1},T^{n1}x)\). By (2.1) we obtain
where
Thus, we get the following cases:

Case 1. \(p(z_{1},T^{n}x)\preceq rp(z_{1},T^{n1}x)\preceq r^{2}p(z_{1},T^{n2}x) \preceq\cdots\preceq r^{n} p(z_{1},x) \).

Case 2. \(p(z_{1},T^{n}x)\preceq rp(z_{1},z_{1})\preceq rp(z_{1},T^{n1}x)\preceq\cdots\preceq r^{n} p(z_{1},x) \).

Case 3. \(p(z_{1},T^{n}x)\preceq r p(T^{n1}x,T^{n}x)\preceq r^{2}p(T^{n2}x,T^{n1}x) \preceq\cdots\preceq r^{n} p(x,Tx)\).

Case 4. \(p(z_{1},T^{n}x)\preceq r[\frac{p(z_{1},T^{n1}x)+p(z_{1},T^{n}x)}{2}] \) implies \(p(z_{1},T^{n}x)\preceq rp(z_{1},T^{n1}x)\preceq r^{n} p(z_{1},x) \).
Thus, in all cases, we have
Similarly, \(p(z_{2},T^{n}x)\preceq r^{n} p(z_{2},x)\) or \(p(z_{2},T^{n}x)\preceq r^{n} p(x,Tx)\). Let \(\theta\ll c\) and choose a natural number \(N_{3}\) such that \(r^{n} p(z_{1},x) \ll\frac{c}{2}\), or \(r^{n} p(x,Tx)\ll\frac{c}{2}\) and \(r^{n} p(z_{2},x) \ll\frac{c}{2}\) for all \(n\geq N_{3}\). Thus,
which is again a contradiction. Hence, \(z_{1}=z_{2}\). □
Now, in order to support the usability of Theorem 2.1, we present the following example.
Example 2.1
Let \(E=C_{R}^{1}[0,1]\) with the norm \(\Vert u\Vert =\Vert u\Vert _{\infty}+\Vert u'\Vert _{\infty}\), and \(X=P=\{u\in E:u(t)\geq0,t\in[0,1]\}\), which is a nonnormal solid cone. Define the mapping \(p:X\times X\to P\) by
Then \((X,p)\) is a θcomplete partial cone metric space. We can define a partial order on X as follows:
Then \((X,p,\sqsubseteq)\) is a θcomplete partially ordered partial cone metric space. For every fixed \(r\in[0,1)\), define \(T:X\to X\) by
Thus, for all \(x\in X\), we consider the following three cases:

Case 1. If \(t\in[0,1]\) such that \(x(t)=0\), then \(Tx(t)=x(t)\).

Case 2. If \(t\in[0,1]\) such that \(2n1\leq x(t)\leq2n\), then
$$Tx(t)=\frac{2n1}{2}rt+\frac{2n1}{4n}rx(t)\leq\frac{2n1}{2}r+ \frac {2n1}{2}r\leq rx(t). $$ 
Case 3. If \(t\in[0,1]\) such that \(2n\leq x(t)\leq2n+1\), then
$$Tx(t)=nrt+\frac{n}{2n+1}rx(t)\leq nr+ nr\leq rx(t). $$
In all cases, for all \(x\in X\) and \(t\in[0,1]\), we have \(Tx(t)\leq rx(t)\). Hence, for all \(x,y\in X\), \(x\sqsubseteq y\), we have
where \(U(x,y)\in\{p(x,y),p(x,Tx),p(y,Ty),\frac{p(x,Ty)+p(y,Tx)}{2}\}\), 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, \(x=\theta\) is a fixed point of T.
Theorem 2.2
Let \((X,p,\sqsubseteq)\) be a partially ordered partial cone metric space over a solid cone P of a normed vector space \((E,\Vert \cdot \Vert )\). Let \(S,T:X\to X\) be such that T is an Snondecreasing mapping with respect to ⊑, \(TX\subseteq SX\), and SX is a θcomplete subset of X. Define \(\psi:[0,1)\to(\frac{1}{2},1]\) as in Theorem 2.1. Suppose that there exists \(r\in[0,1)\) such that
for all comparable \(Sx,Sy\in X\), where \(U(Sx,Sy)\in\{ p(Sx,Sy),p(Sx,Tx),p(Sy,Ty),(p(Sx,Ty)+p(Sy,Tx))/{2}\}\). Suppose that the following conditions hold:

(i)
there exists \(x_{0}\in X\) such that \(Sx_{0}\sqsubseteq Tx_{0}\);

(ii)
for a nondecreasing sequence \(x_{n}\stackrel{\tau_{p}}{\to} x\), we have \(x_{n}\sqsubseteq x\) for all \(n\in N\);

(iii)
for two nondecreasing sequences \(\{x_{n}\}, \{y_{n}\}\subseteq X\) such that \(x_{n}\sqsubseteq y_{n}\), \(x_{n}\stackrel{\tau_{p}}{\to} x\), and \(y_{n} \stackrel{\tau_{p}}{\to} y\) as \(n\to\infty\), we have \(x\sqsubseteq y\);

(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 \(Y\subseteq X\), such that \(SY=SX\) and \(S:Y\to X\) is onetoone. Define \(f:SY\to SX\) by \(fSx=Tx\) for all \(Sx\in SY\).
Since S is onetoone on Y, f is well defined. Note that, for all comparable \(Sx,Sy\in SY\),
where \(U(Sx,Sy)\in\{p(Sx,Sy),p(Sx,fSx),p(Sy,fSy),\frac {p(Sx,fSy)+p(Sy,fSx)}{2}\}\).
Since T is Snondecreasing, we have that f is nondecreasing. In fact, \(Sx\sqsubseteq Sy\) implies \(Tx\sqsubseteq Ty\), and hence \(fSx=Tx\sqsubseteq Ty=fSy\). Since SY is θcomplete, 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 \(z\neq w\). Then
and by (2.8) we have
where \(U(Sz,Sw)\in\{p(Sz,Sw),p(Sz,Tz),p(Sw,Tw),\frac {p(Sz,Tw)+p(Sw,Tz)}{2}\}\). Since \(Sz=Tz\) and \(Sw=Tw\), it follows from (2.9) that
where \(U(Tz,Tw)\in\{p(Tz,Tw),p(Tz,Tz),p(Tw,Tw),\frac {p(Tz,Tw)+p(Tw,Tz)}{2}\}= \{p(Tz,Tw),p(Tz,Tz), p(Tw,Tw)\}\), 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. □
Coupled point theorems in partial cone metric spaces
In this section, we will apply the results obtained in Section 2 to establish the corresponding Suzukitype coupled fixed point theorems for generalized mappings in partially ordered partial cone metric spaces over a nonnormal cone.
For \(\tilde{a}=(x,y)\), \(\tilde{b}=(u,v)\in X^{2}\), we introduce the mapping \(\tilde{p}:X^{2}\times X^{2}\to P\) defined by \(\tilde{p}(\tilde{a},\tilde{b})=p(x,u)+p(y,v)\).
The following conclusion is valid, and for its proof, we refer to [30].
Lemma 3.1
If \((X,p )\) is a partial cone metric space over a solid cone P of a normed vector space \((E,\Vert \cdot \Vert )\), then \((X^{2},\tilde{p})\) is also a θcomplete partial cone metric space.
Proof
It suffices to prove that, for \(\tilde{a}=(x,y)\), \(\tilde {b}=(u,v)\), \(\tilde{c}=(z,w)\in X^{2}\),
In fact, for \(\tilde{a}=(x,y)\), \(\tilde{b}=(u,v)\), \(\tilde{c}=(z,w)\in X^{2}\), we have
Suppose that the sequence \(\{\tilde{x}_{n}\}=\{(x_{n},y_{n})\}\) is a θCauchy sequence in \((X^{2},\tilde{p})\). Then, for every \(c\gg\theta\), there exists a positive integer \(n_{0}\in N\) such that \(\tilde{p}(\tilde{x}_{n},\tilde {x}_{m})=p(x_{n},x_{m})+p(y_{n},y_{m})\ll c\) for all \(n,m>n_{0}\). Then \(p(x_{n},x_{m})\ll c\) and \(p(y_{n},y_{m})\ll c\). Thus, \(\{x_{n}\}\) and \(\{y_{n}\}\) are θCauchy sequences in \((X,p)\). Since \((X,p)\) is θcomplete, there exist \(x,y\in X\) such that \(x_{n}\stackrel{\tau_{p}}{\to} x\), \(y_{n}\stackrel{\tau_{p}}{\to} y\), and \(p(x,x)=\theta\), \(p(y,y)=\theta\).
Thus, for every \(c\gg\theta\), there exists \(n_{1}\in N\) such that \(p(x_{n},x)\ll\frac{c}{2}\) and \(p(y_{n},y)\ll\frac{c}{2}\) for all \(n>n_{1}\). Then \(\tilde{p}((x_{n},y_{n}),(x,y))=p(x_{n},x)+p(y_{n},y)\ll\frac{c}{2}+\frac {c}{2}=c\), and \(\tilde{p}((x,y),(x,y))=\theta\). Thus, \(\{(x_{n},y_{n})\}\stackrel{\tau_{\tilde{p}}}{\to} (x,y) \).
Therefore, \((X^{2},\tilde{p})\) is a θcomplete partial cone metric space. □
Theorem 3.1
Let \((X,p,\sqsubseteq)\) be a θcomplete partially ordered partial cone metric space over a solid cone P of a normed vector space \((E,\Vert \cdot \Vert )\). Let \(A:X\times X\to X\) be a mapping satisfying the mixed monotone property on X with respect to ⊑. Define \(\psi:[0,1)\to (\frac{1}{2},1]\) as in Theorem 2.1. Assume that there exists \(r\in[0,1)\) such that
for all \(x,y\in X\) such that \(x\sqsubseteq u\) and \(v\sqsubseteq y\), where
Suppose that the following conditions hold:

(i)
there exists \(x_{0},y_{0}\in X\) such that \(x_{0}\sqsubseteq A(x_{0},y_{0})\) and \(A(y_{0},x_{0})\sqsubseteq y_{0}\);

(ii)
for a nondecreasing sequence \(x_{n}\stackrel{\tau_{p}}{\to} x\), we have \(x_{n}\sqsubseteq x\) for all \(n\in N\);

(iii)
for a nonincreasing sequence \(y_{n}\stackrel{\tau_{p}}{\to} x\), we have \(y\sqsubseteq y_{n}\) for all \(n\in N\);

(iv)
for two nondecreasing sequences \(\{x_{n}\}, \{u_{n}\}\subseteq X\) such that \(x_{n}\sqsubseteq u_{n}\) for all \(n\in N\), \(x_{n}\stackrel{\tau_{p}}{\to} x\), and \(u_{n} \stackrel{\tau_{p}}{\to} u\) as \(n\to\infty\), we have \(x\sqsubseteq u\);

(v)
for two nonincreasing sequences \(\{y_{n}\}, \{v_{n}\}\subseteq X\) such that \(v_{n}\sqsubseteq y_{n}\) for all \(n\in N\), \(v_{n}\stackrel{\tau_{p}}{\to} v\), and \(y_{n} \stackrel{\tau_{p}}{\to} y\) as \(n\to\infty\), we have \(v\sqsubseteq y\).
Then A has a coupled fixed point in X, that is, there exist \(z,w\in X\) such that \(A(z,w)=z\) and \(A(w,z)=w\).
Proof
Let \(\tilde{X}= X \times X\). For \(\tilde{a}=(x,y)\), \(\tilde {b}=(u,v)\in\tilde{X}\), we introduce the order ≺ as
It follows from Lemma 3.1 that \((\tilde{X},\tilde{p},\prec)\) is also a θcomplete partially ordered partial cone metric space, where
The mapping \(T :\tilde{X} \to\tilde{X }\) is given by \(T\tilde{ a} =(A(x,y),A(y,x))\) for all \(\tilde{a} = (x,y) \in\tilde{X}\). Then a coupled point of A is a fixed point of T and vice versa.
If \(\tilde{ a} \prec\tilde{b}\), then \(x \sqsubseteq u\) and \(v \sqsubseteq y\). Noting the mixed monotone property of A, we see that if \(A(x,y) \sqsubseteq A(u,v)\) and \(A(v,u) \sqsubseteq A(y,x)\), then \(T \tilde{a}\prec T\tilde{b}\). Thus, T is a nondecreasing mapping with respect to the order ≺ on X̃.
On the other hand, for all \(\tilde{a} = (x,y)\), \(\tilde{b} = (u,v)\in \tilde{X}\) with \(\tilde{a} \prec\tilde{ b}\), if \(\psi(r)\tilde{p}(\tilde{a},T\tilde{a})=\psi (r)[p(x,A(x,y))+p(y,A(y,x))]\preceq p(x,u)+p(y,v)=\tilde{p}(\tilde {a},\tilde{b})\), then we have
where
Also, there exists an \(\tilde{x}_{0}= (x_{0},y_{0})\in\tilde{X} \) such that \(\tilde{x}_{0} \prec T \tilde{x}_{0}= (A(x_{0},y_{0}),A(y_{0},x_{0}))\).
If a nondecreasing monotone sequence \(\{ \tilde{x}_{n}\} = \{(x_{n},y_{n})\}\) in X̃ \(\tau_{p}\)converges to \(\tilde{ x} = (x,y)\), then \(\tilde{x}_{n}= (x_{n},y_{n})\prec(x_{n+1},y_{n+1}) = \tilde{x}_{n+1}\), that is, \(x_{n}\sqsubseteq x_{n+1}\) and \(y_{n+1}\sqsubseteq y_{n}\). Thus, \(\{x_{n}\}\) is a nondecreasing sequence \(\tau_{p}\)converging to x, and \(\{y_{n}\}\) is a nonincreasing sequence \(\tau_{p}\)converging to y. Thus, \(x_{n}\sqsubseteq x\) and \(y \sqsubseteq y_{n}\) for all \(n\in N\). This implies \(\tilde{x}_{n}\prec \tilde{x}\).
If two nondecreasing sequence \(\{ \tilde{x}_{n}\} = \{(x_{n},y_{n})\}\), \(\{ \tilde{y}_{n}\} = \{(u_{n},v_{n})\}\) are such that \(\tilde{x}_{n}\prec\tilde {y}_{n}\) for all \(n\in N\), \(\tilde{x}_{n}\stackrel{\tau_{\tilde{p}}}{\to} (x,y)\), and \(\tilde {y}_{n}\stackrel{\tau_{\tilde{p}}}{\to} (u,v)\) as \(n\to\infty\), then \(x_{n}\sqsubseteq u_{n}\), \(v_{n}\sqsubseteq y_{n}\), \(x_{n}\sqsubseteq x_{n+1}\), \(y_{n+1}\sqsubseteq y_{n}\), and \(u_{n}\sqsubseteq u_{n+1}\), \(v_{n+1}\sqsubseteq v_{n}\). Thus, \(\{ x_{n}\}, \{u_{n}\}\subseteq X\) are two nondecreasing sequences, \(x_{n}\sqsubseteq u_{n}\) for all \(n\in N\), \(x_{n}\stackrel{\tau_{p}}{\to} x\), and \(u_{n} \stackrel{\tau_{p}}{\to} u\) as \(n\to\infty\), and by condition (iv) we have \(x\sqsubseteq u\). Similarly, by condition (v) we have \(v\sqsubseteq y\). Thus, \((x,y)\prec(u,v)\).
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 \(z,w \in\tilde{X}\) such that \(A(z,w) = z\) and \(A(w,z) = w\). □
Now, we present the following example.
Example 3.1
Let \(X =P =\{(x_{1},x_{2}) : x_{1},x_{2}\in R^{+}\}\subseteq R^{2}\), and \(E = R^{2}\) with the norm \(\Vert x\Vert =\sqrt{x_{1}^{2}+x_{2}^{2}}\). Define the mapping \(p : X \times X \to P\) by
for all \(x=(x_{1},x_{2})\), \(y=(y_{1},y_{2}) \in X\), where \(a\vee b=\max\{a,b\}\), \(a,b\in R^{+}\). Then \((X,p)\) is a θcomplete partial cone metric space. Define the partial order on X by
Then \((X,p,\sqsubseteq)\) is a θcomplete partially ordered partial cone metric space. In fact, “⊑” is equal to “⪯”. For any fixed \(r\in[0,1)\), define \(A:X\times X\to X\) by
for all \(x,y\in X\). It is clear that A satisfies the mixed monotone property on X with respect to ⊑. Define \(\psi:[0,1)\to (\frac{1}{2},1]\) as in Theorem 3.1.
Now, for all \(x,y,u,v\in X\), \(x\sqsubseteq u\), \(v\sqsubseteq y\), we have
and, on the other hand,
where \(U((x,y),(u,v))\) is as in Theorem 3.1.
Also, conditions (i)(v) of Theorem 3.1 are satisfied. From Theorem 2.1 we obtain that A has a fixed point in X; indeed, \(x=(0,0)\) is a fixed point of A.
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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 (YC2012B004).
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Xu, W., Zhu, C. & Chen, C. Suzuki’s type fixed point theorems for generalized mappings in partial cone metric spaces over a solid cone. Fixed Point Theory Appl 2016, 47 (2016). https://doi.org/10.1186/s136630160538z
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MSC
 47H10
 54H25
Keywords
 partial cone metric spaces
 solid cone
 Suzuki type
 fixed point