Throughout the rest of this paper, we denote by (X,\u2aaf,\sigma ) a complete partially ordered metriclike space, i.e., ⪯ is a partial order on the set X and σ is a complete metriclike on X.
A mapping F:X\to X is said to be nondecreasing if x,y\in X, x\u2aafy\Rightarrow Fx\u2aafFy.
Theorem 1 Let (X,\u2aaf,\sigma ) be a complete partially ordered metriclike space. Let F:X\to X be a continuous and nondecreasing mapping such that for all comparable x,y\in X
\psi (\sigma (Fx,Fy))\le \psi (M(x,y))\varphi (M(x,y)),
(2.1)
where
M
is given by
M(x,y)=max\{\sigma (x,y),\sigma (x,Fx),\sigma (y,Fy),\sigma (x,x),\sigma (y,y),[\sigma (x,Fy)+\sigma (Fx,y)]/2\},
and

(a)
\psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is a continuous monotone nondecreasing function with \psi (t)=0 if and only if t=0;

(b)
\varphi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is a lower semicontinuous function with \varphi (t)=0 if and only if t=0.
If there exists {x}_{0}\in X with {x}_{0}\u2aafF{x}_{0}, then F has a fixed point.
Proof Since F is a nondecreasing function, we obtain by induction that
{x}_{0}\u2aafF{x}_{0}\u2aaf{F}^{2}{x}_{0}\u2aaf\cdots \u2aaf{F}^{n}{x}_{0}\u2aaf{F}^{n+1}{x}_{0}\u2aaf\cdots .
Put {x}_{n+1}=F{x}_{n}. Then, for each integer n=0,1,2,\dots , as the elements {x}_{n+1} and {x}_{n} are comparable, from (2.1) we get
\begin{array}{rcl}\psi (\sigma ({x}_{n+1},{x}_{n}))& =& \psi (\sigma (F{x}_{n},F{x}_{n1}))\\ \le & \psi (M({x}_{n},{x}_{n1}))\varphi (M({x}_{n},{x}_{n1})),\end{array}
(2.2)
which implies \psi (\sigma ({x}_{n+1},{x}_{n}))\le \psi (M({x}_{n},{x}_{n1})). Using the monotone property of the ψfunction, we get
\sigma ({x}_{n+1},{x}_{n})\le M({x}_{n},{x}_{n1}).
(2.3)
Now, from the triangle inequality, for σ we have
\begin{array}{rcl}M({x}_{n},{x}_{n1})& =& max\{\sigma ({x}_{n},{x}_{n1}),\sigma ({x}_{n},F{x}_{n}),\sigma ({x}_{n1},F{x}_{n1}),\sigma ({x}_{n},{x}_{n}),\\ \sigma ({x}_{n1},{x}_{n1}),[\sigma ({x}_{n},F{x}_{n1})+\sigma (F{x}_{n},{x}_{n1})]/2\}\\ =& max\{\sigma ({x}_{n},{x}_{n1}),\sigma ({x}_{n},{x}_{n+1}),\sigma ({x}_{n1},{x}_{n}),\sigma ({x}_{n},{x}_{n}),\\ \sigma ({x}_{n1},{x}_{n1}),[\sigma ({x}_{n},{x}_{n})+\sigma ({x}_{n+1},{x}_{n1})]/2\}\\ \le & max\{\sigma ({x}_{n},{x}_{n1}),\sigma ({x}_{n+1},{x}_{n}),[\sigma ({x}_{n},{x}_{n+1})+\sigma ({x}_{n},{x}_{n1})]/2\}\\ =& max\{\sigma ({x}_{n},{x}_{n1}),\sigma ({x}_{n+1},{x}_{n})\}.\end{array}
If \sigma ({x}_{n+1},{x}_{n})>\sigma ({x}_{n},{x}_{n1}), then M({x}_{n},{x}_{n1})=\sigma ({x}_{n+1},{x}_{n})>0. By (2.2) it furthermore implies that
\psi (\sigma ({x}_{n+1},{x}_{n}))\le \psi (\sigma ({x}_{n+1},{x}_{n}))\varphi (\sigma ({x}_{n+1},{x}_{n})),
which is a contradiction. So, we have
\sigma ({x}_{n+1},{x}_{n})\le M({x}_{n},{x}_{n1})\le \sigma ({x}_{n},{x}_{n1}).
(2.4)
Therefore, the sequence \{\sigma ({x}_{n+1},{x}_{n})\} is monotone nonincreasing and bounded. Thus, there exists r\ge 0 such that
\underset{n\to \mathrm{\infty}}{lim}\sigma ({x}_{n+1},{x}_{n})=\underset{n\to \mathrm{\infty}}{lim}M({x}_{n},{x}_{n1})=r.
(2.5)
We suppose that r>0. Then, letting n\to \mathrm{\infty} in the inequality (2.2), we get
\psi (r)\le \psi (r)\varphi (r),
which is a contradiction unless r=0. Hence,
\underset{n\to \mathrm{\infty}}{lim}\sigma ({x}_{n+1},{x}_{n})=0.
(2.6)
Next we prove that \{{x}_{n}\} is a σCauchy sequence. Suppose that \{{x}_{n}\} is not a σCauchy sequence. Then, there exists \epsilon >0 for which we can find subsequences \{{x}_{{m}_{k}}\} and \{{x}_{{n}_{k}}\} of \{{x}_{n}\} with {n}_{k}>{m}_{k}>k such that
\sigma ({x}_{{n}_{k}},{x}_{{m}_{k}})\ge \epsilon .
(2.7)
Further, corresponding to {m}_{k}, we can choose {n}_{k} in such a way that it is the smallest integer with {n}_{k}>{m}_{k} satisfying (2.7). Then
\sigma ({x}_{{n}_{k}1},{x}_{{m}_{k}})<\epsilon .
(2.8)
Using (2.7), (2.8) and the triangle inequality, we have
\epsilon \le \sigma ({x}_{{n}_{k}},{x}_{{m}_{k}})\le \sigma ({x}_{{n}_{k}},{x}_{{n}_{k}1})+\sigma ({x}_{{n}_{k}1},{x}_{{m}_{k}})<\sigma ({x}_{{n}_{k}},{x}_{{n}_{k}1})+\epsilon .
Letting k\to \mathrm{\infty} and using (2.6), we obtain
\underset{k\to \mathrm{\infty}}{lim}\sigma ({x}_{{n}_{k}},{x}_{{m}_{k}})=\epsilon .
(2.9)
Again, the triangle inequality gives us
Then we have
\sigma ({x}_{{n}_{k}1},{x}_{{m}_{k}})\sigma ({x}_{{n}_{k}},{x}_{{m}_{k}})\le \sigma ({x}_{{n}_{k}},{x}_{{n}_{k}1}).
Letting k\to \mathrm{\infty} in the above inequality and using (2.6) and (2.9), we get
\underset{k\to \mathrm{\infty}}{lim}\sigma ({x}_{{n}_{k}1},{x}_{{m}_{k}})=\epsilon .
(2.10)
Similarly, we can show that
\begin{array}{rcl}\underset{k\to \mathrm{\infty}}{lim}\sigma ({x}_{{n}_{k}},{x}_{{m}_{k}1})& =& \underset{k\to \mathrm{\infty}}{lim}\sigma ({x}_{{n}_{k}1},{x}_{{m}_{k}1})\\ =& \underset{k\to \mathrm{\infty}}{lim}\sigma ({x}_{{n}_{k}},{x}_{{m}_{k}+1})\\ =& \underset{k\to \mathrm{\infty}}{lim}\sigma ({x}_{{n}_{k}+1},{x}_{{m}_{k}})=\epsilon .\end{array}
(2.11)
As
\begin{array}{rcl}M({x}_{{n}_{k}1},{x}_{{m}_{k}1})& =& max\{\sigma ({x}_{{n}_{k}1},{x}_{{m}_{k}1}),\sigma ({x}_{{n}_{k}1},{x}_{{n}_{k}}),\\ \sigma ({x}_{{m}_{k}1},{x}_{{m}_{k}}),\sigma ({x}_{{n}_{k}1},{x}_{{n}_{k}1}),\sigma ({x}_{{m}_{k}1},{x}_{{m}_{k}1}),\\ [\sigma ({x}_{{n}_{k}1},{x}_{{m}_{k}})+\sigma ({x}_{{n}_{k}},{x}_{{m}_{k}1})]/2\}\end{array}
using (2.6) and (2.9)(2.11), we have
\underset{k\to \mathrm{\infty}}{lim}M({x}_{{n}_{k}1},{x}_{{m}_{k}1})=max\{\epsilon ,0,0,0,0,\epsilon \}=\epsilon .
(2.12)
As {n}_{k}>{m}_{k} and {x}_{{n}_{k}1} and {x}_{{m}_{k}1} are comparable, setting x={x}_{{n}_{k}1} and y={x}_{{m}_{k}1} in (2.1), we obtain
\begin{array}{rcl}\psi (\sigma ({x}_{{n}_{k}},{x}_{{m}_{k}}))& =& \psi (\sigma (F{x}_{{n}_{k}1},F{x}_{{m}_{k}1}))\\ \le & \psi (M({x}_{{n}_{k}1},{x}_{{m}_{k}1}))\varphi (M({x}_{{n}_{k}1},{x}_{{m}_{k}1})).\end{array}
Letting k\to \mathrm{\infty} in the above inequality and using (2.9) and (2.12), we get
\psi (\epsilon )\le \psi (\epsilon )\varphi (\epsilon ),
which is a contradiction as \epsilon >0. Hence \{{x}_{n}\} is a σCauchy sequence. By the completeness of X, there exists z\in X such that {lim}_{n\to \mathrm{\infty}}{x}_{n}=z, that is,
\underset{n\to \mathrm{\infty}}{lim}\sigma ({x}_{n},z)=\sigma (z,z)=\underset{m,n\to \mathrm{\infty}}{lim}\sigma ({x}_{m},{x}_{n})=0.
(2.13)
Moreover, the continuity of F implies that
\underset{n\to \mathrm{\infty}}{lim}\sigma ({x}_{n+1},z)=\underset{n\to \mathrm{\infty}}{lim}\sigma (F{x}_{n},z)=\sigma (Fz,z)=0
and this proves that z is a fixed point. □
Notice that the continuity of F in Theorem 1 is not necessary and can be dropped.
Theorem 2 Under the same hypotheses of Theorem 1 and without assuming the continuity of F, assume that whenever \{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x\in X implies {x}_{n}\u2aafx for all n\in \mathbb{N}, then F has a fixed point in X.
Proof Following similar arguments to those given in Theorem 1, we construct a nondecreasing sequence \{{x}_{n}\} in X such that {x}_{n}\to z for some z\in X. Using the assumption of X, we have {x}_{n}\u2aafz for every n\in \mathbb{N}. Now, we show that Fz=z. By (2.1), we have
\begin{array}{rcl}\psi (\sigma (Fz,{x}_{n+1}))& =& \psi (\sigma (Fz,F{x}_{n}))\\ \le & \psi (M(z,{x}_{n}))\varphi (M(z,{x}_{n})),\end{array}
(2.14)
where
\begin{array}{rcl}\sigma (Fz,z)& \le & M(z,{x}_{n})=max\{\sigma (z,{x}_{n}),\sigma (Fz,z),\sigma ({x}_{n},{x}_{n+1}),\\ \sigma (z,z),\sigma ({x}_{n},{x}_{n}),[\sigma (z,{x}_{n+1})+\sigma (Fz,{x}_{n})]/2\}\\ \le & max\{\sigma (z,{x}_{n}),\sigma (Fz,z),\sigma ({x}_{n},{x}_{n+1}),\\ \sigma (z,z),\sigma ({x}_{n},{x}_{n}),[\sigma (z,{x}_{n+1})+\sigma (Fz,z)+\sigma (z,{x}_{n})]/2\}.\end{array}
Taking limit as n\to \mathrm{\infty}, by (2.13), we obtain
\underset{n\to \mathrm{\infty}}{lim}M(z,{x}_{n})=\sigma (Fz,z).
Therefore, letting n\to \mathrm{\infty} in (2.14), we get
\psi (\sigma (Fz,z))\le \psi (\sigma (Fz,z))\varphi (\sigma (Fz,z)),
which is a contradiction unless \sigma (Fz,z)=0. Thus Fz=z. □
Next theorem gives a sufficient condition for the uniqueness of the fixed point.
Theorem 3 Let all the conditions of Theorem 1 (resp. Theorem 2) be fulfilled and let the following condition be satisfied: For arbitrary two points x,y\in X, there exists z\in X which is comparable with both x and y. Then the fixed point of F is unique.
Proof Suppose that there exist z,x\in X which are fixed points. We distinguish two cases.
Case 1. If x is comparable to z, then {F}^{n}x=x is comparable to {F}^{n}z=z for n=0,1,2,\dots and
\begin{array}{rcl}\psi (\sigma (z,x))& =& \psi \left(\sigma ({F}^{n}z,{F}^{n}x)\right)\\ \le & \psi \left(M({F}^{n1}z,{F}^{n1}x)\right)\varphi \left(M({F}^{n1}z,{F}^{n1}x)\right)\\ \le & \psi (M(z,x))\varphi (M(z,x)),\end{array}
(2.15)
where
\begin{array}{rcl}M(z,x)& =& max\{\sigma (z,x),\sigma (z,Fz),\sigma (x,Fx),\sigma (z,z),\sigma (x,x),[\sigma (z,Fx)+\sigma (Fz,x)]/2\}\\ =& max\{\sigma (z,x),\sigma (z,z),\sigma (x,x),[\sigma (z,x)+\sigma (z,x)]/2\}=\sigma (z,x).\end{array}
(2.16)
Using (2.15) and (2.16), we have
\psi (\sigma (z,x))\le \psi (\sigma (z,x))\varphi (\sigma (z,x)),
which is a contradiction unless \sigma (z,x)=0. This implies that z=x.
Case 2. If x is not comparable to z, then there exists y\in X comparable to x and z. The monotonicity of F implies that {F}^{n}y is comparable to {F}^{n}x=x and {F}^{n}z=z, for n=0,1,2,\dots . Moreover,
\begin{array}{rcl}\psi \left(\sigma (z,{F}^{n}y)\right)& =& \psi \left(\sigma ({F}^{n}z,{F}^{n}y)\right)\\ \le & \psi \left(M({F}^{n1}z,{F}^{n1}y)\right)\varphi \left(M({F}^{n1}z,{F}^{n1}y)\right),\end{array}
(2.17)
where
\begin{array}{rcl}M({F}^{n1}z,{F}^{n1}y)& =& max\{\sigma ({F}^{n1}z,{F}^{n1}y),\sigma ({F}^{n1}z,{F}^{n}z),\sigma ({F}^{n1}y,{F}^{n}y),\\ \sigma ({F}^{n1}z,{F}^{n1}z),\sigma ({F}^{n1}y,{F}^{n1}y),\\ [\sigma ({F}^{n1}z,{F}^{n}y)+\sigma ({F}^{n}z,{F}^{n1}y)]/2\}\\ =& max\{\sigma (z,{F}^{n1}y),\sigma (z,z),\sigma ({F}^{n1}y,{F}^{n}y),\\ \sigma ({F}^{n1}y,{F}^{n1}y),[\sigma (z,{F}^{n}y)+\sigma (z,{F}^{n1}y)]/2\}\\ \le & max\{\sigma (z,{F}^{n1}y),\sigma (z,{F}^{n}y)\}\end{array}
(2.18)
for n sufficiently large, because \sigma ({F}^{n1}y,{F}^{n1}y)\to 0 and \sigma ({F}^{n1}y,{F}^{n}y)\to 0 when n\to \mathrm{\infty}. Similarly as in the proof of Theorem 1, it can be shown that \sigma (z,{F}^{n}y)\le M(z,{F}^{n1}y)\le \sigma (z,{F}^{n1}y). It follows that the sequence \{\sigma (z,{F}^{n}y)\} is nonnegative decreasing and, consequently, there exists \alpha \ge 0 such that
\underset{n\to \mathrm{\infty}}{lim}\sigma (z,{F}^{n}y)=\underset{n\to \mathrm{\infty}}{lim}M(z,{F}^{n1}y)=\alpha .
We suppose that \alpha >0. Then letting n\to \mathrm{\infty} in (2.17), we have
\psi (\alpha )\le \psi (\alpha )\varphi (\alpha )
which is a contradiction. Hence \alpha =0. Similarly, it can be proved that
\underset{n\to \mathrm{\infty}}{lim}\sigma (x,{F}^{n}y)=0.
Now, passing to the limit in \sigma (x,z)\le \sigma (x,{F}^{n}y)+\sigma ({F}^{n}y,z), it follows that \sigma (x,z)=0, so x=z, and the uniqueness of the fixed point is proved. □
Now, we present an example to support the useability of our results.
Example 2 Let X=\{0,1,2\} and a partial order be defined as x\u2aafy whenever y\le x, and define \sigma :X\times X\to {\mathbb{R}}^{+} as follows:
Then (X,\u2aaf,\sigma ) is a complete partially ordered metriclike space.
Let F:X\to X be defined by
F0=1,\phantom{\rule{2em}{0ex}}F1=2\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}F2=2.
Define \psi ,\varphi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) by \psi (t)=t and \varphi (t)=\frac{t}{2}. We next verify that the function F satisfies the inequality (2.1). For that, given x,y\in X with x\u2aafy, so y\le x. Then we have the following cases.
Case 1. If x=1, y=0, then
\sigma (F1,F0)=\sigma (2,1)=5
and
\begin{array}{rcl}M(1,0)& =& max\{\sigma (1,0),\sigma (1,F1),\sigma (0,F0),\\ \sigma (1,1),\sigma (0,0),[\sigma (1,F0)+\sigma (F1,0)]/2\}\\ =& max\{9,5,3,1,\frac{3+4}{2}\}=9.\end{array}
As \psi (\sigma (F1,F0))=5<9\frac{9}{2}=\psi (M(1,0))\varphi (M(1,0)), the inequality (2.1) is satisfied in this case.
Case 2. If x=2, y=0, then
\sigma (F2,F0)=\sigma (2,1)=5
and
\begin{array}{rcl}M(2,0)& =& max\{\sigma (2,0),\sigma (2,F2),\sigma (0,F0),\\ \sigma (2,2),\sigma (0,0),[\sigma (2,F0)+\sigma (F2,0)]/2\}\\ =& max\{4,0,9,1,\frac{5+4}{2}\}=9.\end{array}
As \psi (\sigma (F2,F0))=5<9\frac{9}{2}=\psi (M(2,0))\varphi (M(2,0)), the inequality (2.1) is satisfied in this case.
Case 3. If x=2, y=1, then as \sigma (F2,F1)=0 and M(2,1)=5, the inequality (2.1) is satisfied in this case.
Case 4. If x=0, y=0, then as \sigma (F0,F0)=3 and M(0,0)=9, the inequality (2.1) is satisfied in this case.
Case 5. If x=1, y=1, then as \sigma (F1,F1)=0 and M(1,1)=5, the inequality (2.1) is satisfied in this case.
Case 6. If x=2, y=2, then as \sigma (F2,F2)=0 and M(2,2)=0, the inequality (2.1) is satisfied in this case.
So, F, ψ and ϕ satisfy all the hypotheses of Theorem 1. Therefore F has a unique fixed point. Here 2 is the unique fixed point of F.
If we take \psi (t)=t in Theorem 1, we have the following corollary.
Corollary 1 Let (X,\u2aaf,\sigma ) be a complete partially ordered metriclike space. Let F:X\to X be a nondecreasing mapping such that for all comparable x,y\in X with
\sigma (Fx,Fy)\le M(x,y)\varphi (M(x,y)),
where
M
is given by
M(x,y)=max\{\sigma (x,y),\sigma (x,Fx),\sigma (y,Fy),\sigma (x,x),\sigma (y,y),[\sigma (x,Fy)+\sigma (Fx,y)]/2\},
\varphi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is lower semicontinuous, and \varphi (t)=0 if and only if t=0. If there exists {x}_{0}\in X with {x}_{0}\u2aafF{x}_{0} and one of the following two conditions is satisfied:

(a)
F is continuous in (X,\sigma );

(b)
\{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x\in X implies {x}_{n}\u2aafx for all n\in \mathbb{N}.
Then F has a fixed point. Moreover, if the following condition is satisfied: For arbitrary two points x,y\in X, there exists z\in X which is comparable with both x and y, then the fixed point of F is unique.
If we take \varphi (t)=(1k)t for k\in [0,1) in Corollary 1, we have the following corollary.
Corollary 2 Let (X,\u2aaf,\sigma ) be a complete partially ordered metriclike space. Let F:X\to X be a nondecreasing mapping such that for all comparable x,y\in X
\sigma (Fx,Fy)\le kM(x,y),
where
M
is given by
M(x,y)=max\{\sigma (x,y),\sigma (x,Fx),\sigma (y,Fy),\sigma (x,x),\sigma (y,y),[\sigma (x,Fy)+\sigma (Fx,y)]/2\},
and k\in [0,1). If there exists {x}_{0}\in X with {x}_{0}\u2aafF{x}_{0} and one of the following two conditions is satisfied:

(a)
F is continuous in (X,\sigma );

(b)
\{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x\in X implies {x}_{n}\u2aafx for all n\in \mathbb{N}.
Then F has a fixed point. Moreover, if the following condition is satisfied: For arbitrary two points x,y\in X, there exists z\in X which is comparable with both x and y, then the fixed point of F is unique.
The following corollary improves Theorem 2.7 in [33].
Corollary 3 Let (X,\u2aaf,\sigma ) be a complete partially ordered metriclike space. Let F:X\to X be a nondecreasing mapping such that for all comparable x,y\in X
\sigma (Fx,Fy)\le \sigma (x,y)\varphi (\sigma (x,y)),
where \varphi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is a lower semicontinuous, and \varphi (t)=0 if and only if t=0. If there exists {x}_{0}\in X with {x}_{0}\u2aafF{x}_{0} and one of the following two conditions is satisfied:

(a)
F is continuous in (X,\sigma );

(b)
\{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x\in X implies {x}_{n}\u2aafx for all n\in \mathbb{N}.
Then F has a fixed point. Moreover, if the following condition is satisfied: For arbitrary two points x,y\in X, there exists z\in X which is comparable with both x and y, then the fixed point of F is unique.
The following corollary improves Theorem 2.1 in [10].
Corollary 4 Let (X,\u2aaf,p) be a complete partially ordered partial metric space. Let F:X\to X be a nondecreasing mapping such that for all comparable x,y\in X
\psi (p(Fx,Fy))\le \psi (M(x,y))\varphi (M(x,y)),
where
M
is given by
M(x,y)=max\{p(x,y),p(x,Fx),p(y,Fy),p(x,x),p(y,y),[p(x,Fy)+p(Fx,y)]/2\},
\psi ,\varphi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}), ψ is continuous monotone nondecreasing, ϕ is lower semicontinuous, and \psi (t)=\varphi (t)=0 if and only if t=0. If there exists {x}_{0}\in X with {x}_{0}\u2aafF{x}_{0} and one of the following two conditions is satisfied:

(a)
F is continuous in (X,p);

(b)
\{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to x\in X implies {x}_{n}\u2aafx for all n\in \mathbb{N}.
Then F has a fixed point. Moreover, the set of fixed points of F is well ordered if and only if F has one and only one fixed point.