We start this section with the following definition, which can be seen in [9, 16, 17, 30].
Definition 2.1 Let (x,P) be a partial metric space. A mapping T:X\to X is said to be continuous at {x}_{0}\in X if for every \epsilon >0, there exists \delta >0 such that T({B}_{p}({x}_{0},\delta ))\subset {B}_{p}(T{x}_{0},\epsilon ).
Definition 2.2 [25]
The function \phi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) is called an altering distance function, if the following properties are satisfied:

(1)
φ is continuous and nondecreasing;

(2)
\phi (t)=0 if and only if t=0.
Lemma 2.1 [31]
Let (X,p) be a partial metric space, T:X\to X be a given mapping. Suppose that T is continuous at {x}_{0}\in X. Then, for each sequence \{{x}_{n}\} in X, {x}_{n}\to {x}_{0} in {\tau}_{p}\Rightarrow T{x}_{n}\to T{x}_{0} in {\tau}_{p} holds.
Theorem 2.1 Let (X,\u2aaf) be a partially ordered set and suppose that there exists a partial metric p on X such that (X,p) is complete. Let f:X\to X be a continuous nondecreasing mapping. Suppose that for comparable x,y\in X, we have
\psi (p(fx,fy))\le \phi \left(\frac{p(x,fy)+p(fx,y)}{2}\right)\varphi (p(x,fy),p(fx,y)),
(2.1)
where
ψ
and
φ
are altering distance functions with
\psi (t)\phi (t)\ge 0
(2.2)
for all t\ge 0, and \varphi :[0,+\mathrm{\infty})\times [0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) is a continuous function with \varphi (x,y)=0 if and only if x=y=0. If there exists {x}_{0}\in X such that {x}_{0}\u2aaff{x}_{0}, then f has a fixed point.
Proof If {x}_{0}=f{x}_{0}, then {x}_{0} is a fixed point of f. Suppose that {x}_{0}\prec f{x}_{0}, we can choose {x}_{1}\in X such that f{x}_{0}={x}_{1}. Since f is a nondecreasing function, we have
{x}_{0}\prec {x}_{1}=f{x}_{0}\u2aaf{x}_{2}=f{x}_{1}\u2aaf{x}_{3}=f{x}_{2}.
Continuing this process, we can construct a sequence \{{x}_{n}\} in X such that {x}_{n+1}=f{x}_{n} with
{x}_{0}\prec {x}_{1}\u2aaf{x}_{2}\u2aaf\cdots \u2aaf{x}_{n}\u2aaf{x}_{n+1}\u2aaf\cdots .
It is clear that if p({x}_{n},{x}_{n+1})=0 for some {n}_{0}\in N, then f has a fixed point. Taking p({x}_{n},{x}_{n+1})>0 for all n\in N, now let us prove the following inequality:
p({x}_{n},{x}_{n+1})\le p({x}_{n1},{x}_{n}),\phantom{\rule{1em}{0ex}}n\in {N}^{+}.
(2.3)
Suppose this is not true, then p({x}_{n},{x}_{n+1})>p({x}_{n1},{x}_{n}) for some {n}_{0}, that is,
p({x}_{{n}_{0}},{x}_{{n}_{0}+1})>p({x}_{{n}_{0}1},{x}_{{n}_{0}}).
(2.4)
From (2.1) and (2.4), we obtain that
this together with (2.2) shows that
\varphi (p({x}_{{n}_{0}1},{x}_{{n}_{0}+1}),p({x}_{{n}_{0}},{x}_{{n}_{0}}))=0.
Using the property of ϕ, we have
p({x}_{{n}_{0}1},{x}_{{n}_{0}+1})=0,\phantom{\rule{2em}{0ex}}p({x}_{{n}_{0}},{x}_{{n}_{0}})=0.
(2.5)
Since
applying (2.5), we get
\psi (p({x}_{{n}_{0}},{x}_{{n}_{0}+1}))=0.
(2.6)
From the property of ψ, we have p({x}_{{n}_{0}},{x}_{{n}_{0}+1})=0, which contradicts with p({x}_{n},{x}_{n+1})>0 for all n\in N; hence (2.3) holds. Therefore, \{p({x}_{n},{x}_{n+1})\} is a nonincreasing sequence, and thus there exists r\ge 0 such that
\underset{n\to +\mathrm{\infty}}{lim}p({x}_{n},{x}_{n+1})=r.
Using (2.1), we obtain
it means that
\varphi (p({x}_{n},{x}_{n+2}),p({x}_{n+1},{x}_{n+1}))\le \phi \left(\frac{p({x}_{n},{x}_{n+1})+p({x}_{n+1},{x}_{n+2})}{2}\right)\psi (p({x}_{n+1},{x}_{n+2})).
Letting n\to +\mathrm{\infty} in the above inequality, we get
\underset{n\to +\mathrm{\infty}}{lim\hspace{0.17em}inf}\varphi (p({x}_{n},{x}_{n+2}),p({x}_{n+1},{x}_{n+1}))=0,
the continuity of ϕ guarantees that
\varphi (\underset{n\to +\mathrm{\infty}}{lim\hspace{0.17em}inf}p({x}_{n},{x}_{n+2}),\underset{n\to +\mathrm{\infty}}{lim\hspace{0.17em}inf}p({x}_{n+1},{x}_{n+1}))=0,
and the property of ϕ gives that
\underset{n\to +\mathrm{\infty}}{lim\hspace{0.17em}inf}p({x}_{n},{x}_{n+2})=0,\phantom{\rule{2em}{0ex}}\underset{n\to +\mathrm{\infty}}{lim\hspace{0.17em}inf}p({x}_{n+1},{x}_{n+1})=0.
(2.8)
Since
on taking inferior limit in the above inequalities and using (2.8), we obtain that \psi (r)=0 and so r=0, therefore,
\underset{n\to +\mathrm{\infty}}{lim}p({x}_{n},{x}_{n+1})=0,
moreover, we have
\underset{n\to +\mathrm{\infty}}{lim}p({x}_{n},{x}_{n})=0.
Now, we claim that \{{x}_{n}\} is a Cauchy sequence in the metric space (X,{d}_{p}) (and so also in the space (X,p) by Lemma 1.1). For this, it is sufficient to show that \{{x}_{2n}\} is a Cauchy sequence in (X,{d}_{p}). Suppose that this is not the case, then using Lemma 1.1 we have that \{{x}_{2n}\} is not a Cauchy sequence in (X,p). By Lemma 1.3, we obtain that there exist \epsilon >0 and two sequences \{m(k)\} and \{n(k)\} of positive integers such that n(k)>m(k)>k and sequences in (1.2) tend to ε when k\to +\mathrm{\infty}. For two comparable elements y={x}_{2n(k)+1} and x={x}_{2m(k)}, we can obtain, from (2.1), that
Taking k\to +\mathrm{\infty} in (2.9), we get
\psi (\epsilon )\le \phi (\epsilon )\varphi (\epsilon ,\epsilon ),
which implies that \varphi (\epsilon ,\epsilon )=0, hence \epsilon =0, a contradiction. Thus, \{{x}_{2n}\} is a Cauchy sequence in (X,{d}_{p}) and so \{{x}_{n}\} is a Cauchy sequence both in (X,{d}_{p}) and in (X,p). Since (X,p) is complete then the sequence \{{x}_{n}\} converges to some z\in X, that is
p(z,z)=\underset{n\to +\mathrm{\infty}}{lim}p({x}_{n},z)=\underset{n,m\to +\mathrm{\infty}}{lim}p({x}_{n},{x}_{m}).
(2.10)
Moreover, since \{{x}_{n}\} is a Cauchy sequence in (X,{d}_{p}), we have {lim}_{n\to +\mathrm{\infty}}{d}_{p}({x}_{n},{x}_{m})=0. By {d}_{p}({x}_{n},{x}_{m})=2p({x}_{n},{x}_{m})p({x}_{n},{x}_{n})p({x}_{m},{x}_{m}) and {lim}_{n\to +\mathrm{\infty}}p({x}_{n},{x}_{n})=0, we have {lim}_{n\to +\mathrm{\infty}}p({x}_{n},{x}_{m})=0. Then (2.10) yields that
p(z,z)=\underset{n\to +\mathrm{\infty}}{lim}p({x}_{n},z)=0.
(2.11)
Applying the triangular inequality, we have
p(z,fz)\le p(z,{x}_{n})+p({x}_{n},fz)p({x}_{n},{x}_{n})\le p(z,{x}_{n})+p({x}_{n},fz)=p(z,{x}_{n})+p(f{x}_{n1},fz),
taking n\to +\mathrm{\infty} in the above inequalities, then the continuity of f and Lemma 2.1 give that
hence
By combining (2.1) and (2.12), we have
\begin{array}{rcl}\psi (p(z,fz))& =& \psi (p(fz,fz))\\ \le & \phi \left(\frac{p(z,fz)+p(fz,z)}{2}\right)\varphi (p(z,fz),p(fz,z))\\ =& \phi (p(z,fz))\varphi (p(z,fz),p(fz,z)),\end{array}
which yields that \varphi (p(z,fz),p(fz,z))=0, and thus p(z,fz)=0, that is z=fz. Therefore, z is a fixed point of f. □
Theorem 2.2 Suppose that X, f, ψ, φ, and ϕ are the same as in Theorem 2.1 except the continuity of f. Suppose that for a nondecreasing sequence \{{x}_{n}\} in X with {x}_{n}\to x\in X, we have {x}_{n}\u2aafx for all n\in N. If there exists {x}_{0}\in X such that {x}_{0}\u2aaff{x}_{0}, then f has a fixed point.
Proof As in the proof of Theorem 2.1, we have a Cauchy sequence \{{x}_{n}\} in X. Since (X,p) is complete, there exists z\in X such that {x}_{n}\to z, that is,
p(z,z)=\underset{n\to +\mathrm{\infty}}{lim}p({x}_{n},z)=\underset{n,m\to +\mathrm{\infty}}{lim}p({x}_{n},{x}_{m}),
due to the hypothesis, we get {x}_{n}\u2aafz. Similar to the proof of Theorem 2.1, we have that
p(z,z)=\underset{n\to +\mathrm{\infty}}{lim}p({x}_{n},z)=0.
From (2.1), we obtain that
\begin{array}{rcl}\psi (p({x}_{n},fz))& =& \psi (p(f{x}_{n1},fz))\\ \le & \phi \left(\frac{p({x}_{n1},fz)+p(f{x}_{n1},z)}{2}\right)\varphi (p({x}_{n1},fz),p({x}_{n},z))\\ =& \phi \left(\frac{p({x}_{n1},fz)+p({x}_{n},z)}{2}\right)\varphi (p({x}_{n1},fz),p({x}_{n},z)).\end{array}
Letting n\to +\mathrm{\infty} in the above inequalities, and by Lemma 1.2, we have
\psi (p(z,fz))\le \phi (p(z,fz))\varphi (p(z,fz),0),
which implies, from (2.2), that \varphi (p(z,fz),0)=0, hence p(z,fz)=0, and thus z=fz. Therefore, f has a fixed point. □
Theorem 2.3 Let (X,p) be a complete partial metric space, f and g be selfmappings on X. Suppose that for all x,y\in X
\psi (p(fx,gy))\le \phi \left(\frac{p(x,gy)+p(fx,y)}{2}\right)\varphi (p(x,gy),p(fx,y)),
(2.13)
where
ψ
and
φ
are altering distance functions with
\psi (t)\phi (t)\ge 0
(2.14)
for all t\ge 0, and \varphi :[0,+\mathrm{\infty})\times [0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) is a continuous function with \varphi (x,y)=0 if and only if x=y=0.
Then f and g have a unique common fixed point.
Proof Let {x}_{0} be an arbitrary point in X. One can choose {x}_{1}\in X such that f{x}_{0}={x}_{1}. Also, one can choose {x}_{2}\in X such that g{x}_{1}={x}_{2}. Continuing this process, one can construct a sequence \{{x}_{n}\} in X such that
{x}_{2n+1}=f{x}_{2n},\phantom{\rule{2em}{0ex}}{x}_{2n+2}=g{x}_{2n+1},\phantom{\rule{1em}{0ex}}n\in N.
(2.15)
Now, we discuss the following two cases.
Case 1. If p({x}_{n},{x}_{n+1})=0 for some {n}_{0}\in N, then f and g have at least one common fixed point. In fact, if p({x}_{n},{x}_{n+1})=0 for some {n}_{0}\in N, that is p({x}_{{n}_{0}},{x}_{{n}_{0}+1})=0, which implies that {x}_{{n}_{0}}={x}_{{n}_{0}+1}. If {n}_{0}=2k (k\in N), then {x}_{2k}={x}_{2k+1}. Using (2.13), we have
\begin{array}{r}\psi (p({x}_{2k+1},{x}_{2k+2}))\\ \phantom{\rule{1em}{0ex}}=\psi (p(f{x}_{2k},g{x}_{2k+1}))\\ \phantom{\rule{1em}{0ex}}\le \phi \left(\frac{p({x}_{2k},g{x}_{2k+1})+p(f{x}_{2k},{x}_{2k+1})}{2}\right)\varphi (p({x}_{2k},g{x}_{2k+1}),p(f{x}_{2k},{x}_{2k+1}))\\ \phantom{\rule{1em}{0ex}}=\phi \left(\frac{p({x}_{2k},{x}_{2k+2})+p({x}_{2k+1},{x}_{2k+1})}{2}\right)\varphi (p({x}_{2k},{x}_{2k+2}),p({x}_{2k+1},{x}_{2k+1}))\\ \phantom{\rule{1em}{0ex}}\le \phi \left(\frac{p({x}_{2k},{x}_{2k+1})+p({x}_{2k+1},{x}_{2k+2})}{2}\right)\varphi (p({x}_{2k},{x}_{2k+2}),p({x}_{2k+1},{x}_{2k+1}))\\ \phantom{\rule{1em}{0ex}}\le \phi (max\{p({x}_{2k},{x}_{2k+1}),p({x}_{2k+1},{x}_{2k+2})\})\varphi (p({x}_{2k},{x}_{2k+2}),p({x}_{2k+1},{x}_{2k+1}))\\ \phantom{\rule{1em}{0ex}}=\phi (max\{p({x}_{2k+1},{x}_{2k+1}),p({x}_{2k+1},{x}_{2k+2})\})\varphi (p({x}_{2k+1},{x}_{2k+2}),p({x}_{2k+1},{x}_{2k+1}))\\ \phantom{\rule{1em}{0ex}}=\phi (p({x}_{2k+1},{x}_{2k+2}))\varphi (p({x}_{2k+1},{x}_{2k+2}),p({x}_{2k+1},{x}_{2k+1})).\end{array}
(2.16)
With the help of (2.14) and (2.16), we conclude that \varphi (p({x}_{2k+1},{x}_{2k+2}),p({x}_{2k+1},{x}_{2k+1}))=0, hence, using the property of ϕ, we get p({x}_{2k+1},{x}_{2k+2})=0, that is {x}_{2k+1}={x}_{2k+2}. By similar arguments, we obtain {x}_{2k+2}={x}_{2k+3}, {x}_{2k+3}={x}_{2k+4} and so on. Thus, \{{x}_{n}\} becomes a constant from n=2k, that is,
{x}_{2k}={x}_{2k+1}={x}_{2k+2}=\cdots .
(2.17)
Equations (2.15) and (2.17) yield that
{x}_{2k}=g{x}_{2k}=f{x}_{2k},
(2.18)
which implies that {x}_{2k} is the common fixed point of f and g. Similarly, one can show that if {n}_{0}=2k+1 (k\in N), then f and g have at least one common fixed point. Therefore, we have proved that if p({x}_{n},{x}_{n+1})=0 for some {n}_{0}\in N, then f and g have at least one common fixed point.
Case 2. If p({x}_{n},{x}_{n+2})=0 for some {n}_{0}\in N, then f and g have at least one common fixed point. Indeed, if {n}_{0}=2k (k\in N), then p({x}_{2k},{x}_{2k+2})=0. Hence, {x}_{2k}={x}_{2k+2}, due to (2.13), we have
\begin{array}{r}\psi (p({x}_{2k+1},{x}_{2k+2}))\\ \phantom{\rule{1em}{0ex}}=\psi (p(f{x}_{2k},g{x}_{2k+1}))\\ \phantom{\rule{1em}{0ex}}\le \phi \left(\frac{p({x}_{2k},g{x}_{2k+1})+p(f{x}_{2k},{x}_{2k+1})}{2}\right)\varphi (p({x}_{2k},g{x}_{2k+1}),p(f{x}_{2k},{x}_{2k+1}))\\ \phantom{\rule{1em}{0ex}}=\phi \left(\frac{p({x}_{2k},{x}_{2k+2})+p({x}_{2k+1},{x}_{2k+1})}{2}\right)\varphi (p({x}_{2k},{x}_{2k+2}),p({x}_{2k+1},{x}_{2k+1}))\\ \phantom{\rule{1em}{0ex}}\le \phi \left(\frac{p({x}_{2k},{x}_{2k+1})+p({x}_{2k+1},{x}_{2k+2})}{2}\right)\varphi (p({x}_{2k},{x}_{2k+2}),p({x}_{2k+1},{x}_{2k+1}))\\ \phantom{\rule{1em}{0ex}}=\phi \left(\frac{p({x}_{2k+2},{x}_{2k+1})+p({x}_{2k+1},{x}_{2k+2})}{2}\right)\varphi (p({x}_{2k},{x}_{2k+2}),p({x}_{2k+1},{x}_{2k+1}))\\ \phantom{\rule{1em}{0ex}}=\phi (p({x}_{2k+2},{x}_{2k+1}))\varphi (p({x}_{2k},{x}_{2k+2}),p({x}_{2k+1},{x}_{2k+1})).\end{array}
(2.19)
Applying (2.14) and (2.19), we obtain \varphi (p({x}_{2k},{x}_{2k+2}),p({x}_{2k+1},{x}_{2k+1}))=0. Using the property of ϕ, we have
p({x}_{2k+1},{x}_{2k+1})=0.
(2.20)
From (2.20) and using p({x}_{2k},{x}_{2k+2})=0, we get that
\begin{array}{r}\psi (p({x}_{2k+1},{x}_{2k+2}))\\ \phantom{\rule{1em}{0ex}}=\psi (p(f{x}_{2k},g{x}_{2k+1}))\\ \phantom{\rule{1em}{0ex}}\le \phi \left(\frac{p({x}_{2k},g{x}_{2k+1})+p(f{x}_{2k},{x}_{2k+1})}{2}\right)\varphi (p({x}_{2k},g{x}_{2k+1}),p(f{x}_{2k},{x}_{2k+1}))\\ \phantom{\rule{1em}{0ex}}=\phi \left(\frac{p({x}_{2k},{x}_{2k+2})+p({x}_{2k+1},{x}_{2k+1})}{2}\right)\varphi (p({x}_{2k},{x}_{2k+2}),p({x}_{2k+1},{x}_{2k+1}))\\ \phantom{\rule{1em}{0ex}}=\phi (0)\varphi (0,0)\\ \phantom{\rule{1em}{0ex}}=0,\end{array}
(2.21)
which implies that \psi (p({x}_{2k+1},{x}_{2k+2}))=0, and thus p({x}_{2k+1},{x}_{2k+2})=0. Hence we obtain that f and g have at least one common fixed point from case 1. Similarly, it is easy to show that if p({x}_{n},{x}_{n+2})=0 for some n=2k+1 (k\in N), then f and g have at least one common fixed point, this completes the proof of case 2.
Taking p({x}_{n},{x}_{n+1})>0 and p({x}_{n},{x}_{n+2})>0 for all n\in N. Now we prove that for every k\in N, we have
p({x}_{2k+2},{x}_{2k+1})\le p({x}_{2k+1},{x}_{2k}).
(2.22)
Suppose this is not true, then p({x}_{2k+2},{x}_{2k+1})>p({x}_{2k+1},{x}_{2k}) for some k={k}_{0}, that is,
p({x}_{2{k}_{0}+2},{x}_{2{k}_{0}+1})>p({x}_{2{k}_{0}+1},{x}_{2{k}_{0}}).
Using (2.13) and (2.15), we obtain that
\begin{array}{r}\psi (p({x}_{2{k}_{0}+1},{x}_{2{k}_{0}+2}))\\ \phantom{\rule{1em}{0ex}}=\psi (p(f{x}_{2{k}_{0}},g{x}_{2{k}_{0}+1}))\\ \phantom{\rule{1em}{0ex}}\le \phi \left(\frac{p({x}_{2{k}_{0}},g{x}_{2{k}_{0}+1})+p(f{x}_{2{k}_{0}},{x}_{2{k}_{0}+1})}{2}\right)\varphi (p({x}_{2{k}_{0}},g{x}_{2{k}_{0}+1}),p(f{x}_{2{k}_{0}},{x}_{2{k}_{0}+1}))\\ \phantom{\rule{1em}{0ex}}=\phi \left(\frac{p({x}_{2{k}_{0}},{x}_{2{k}_{0}+2})+p({x}_{2{k}_{0}+1},{x}_{2{k}_{0}+1})}{2}\right)\varphi (p({x}_{2{k}_{0}},{x}_{2{k}_{0}+2}),p({x}_{2{k}_{0}+1},{x}_{2{k}_{0}+1}))\\ \phantom{\rule{1em}{0ex}}\le \phi \left(\frac{p({x}_{2{k}_{0}},{x}_{2{k}_{0}+1})+p({x}_{2{k}_{0}+1},{x}_{2{k}_{0}+2})}{2}\right)\varphi (p({x}_{2{k}_{0}},{x}_{2{k}_{0}+2}),p({x}_{2{k}_{0}+1},{x}_{2{k}_{0}+1}))\\ \phantom{\rule{1em}{0ex}}\le \phi (max\{p({x}_{2{k}_{0}},{x}_{2{k}_{0}+1}),p({x}_{2{k}_{0}+1},{x}_{2{k}_{0}+2})\})\varphi (p({x}_{2{k}_{0}},{x}_{2{k}_{0}+2}),p({x}_{2{k}_{0}+1},{x}_{2{k}_{0}+1}))\\ \phantom{\rule{1em}{0ex}}=\phi (p({x}_{2{k}_{0}+1},{x}_{2{k}_{0}+2}))\varphi (p({x}_{2{k}_{0}},{x}_{2{k}_{0}+2}),p({x}_{2{k}_{0}+1},{x}_{2{k}_{0}+1})).\end{array}
(2.23)
Equations (2.14) and (2.23) give that \varphi (p({x}_{2{k}_{0}},{x}_{2{k}_{0}+2}),p({x}_{2{k}_{0}+1},{x}_{2{k}_{0}+1}))=0. Using the property of ϕ, we get p({x}_{2{k}_{0}},{x}_{2{k}_{0}+2})=0, which contradicts with p({x}_{n},{x}_{n+2})>0 for n\in N, hence (2.22) holds.
Similarly, one can show that for every k\in {N}^{+}, the following inequality holds.
p({x}_{2k+1},{x}_{2k})\le p({x}_{2k},{x}_{2k1}).
(2.24)
Equations (2.22) and (2.24) imply that the sequence \{p({x}_{n},{x}_{n+1})\} is nonincreasing, and consequently there exists some r\ge 0 such that
\underset{n\to +\mathrm{\infty}}{lim}p({x}_{n},{x}_{n+1})=r.
(2.25)
By (2.25) and the following inequalities,
\begin{array}{rcl}p({x}_{2n},{x}_{2n+2})& \le & p({x}_{2n},{x}_{2n+1})+p({x}_{2n+1},{x}_{2n+2})p({x}_{2n+1},{x}_{2n+1})\\ \le & p({x}_{2n},{x}_{2n+1})+p({x}_{2n+1},{x}_{2n+2}),\end{array}
we get that \{p({x}_{2n},{x}_{2n+2})\} is bounded, and hence it has some subsequence \{p({x}_{2n(k)},{x}_{2n(k)+2})\} converging to some {r}_{0}, that is,
\underset{k\to +\mathrm{\infty}}{lim}p({x}_{2n(k)},{x}_{2n(k)+2})={r}_{0}.
(2.26)
Taking (P_{2}) into account, we have
p({x}_{2n(k)+1},{x}_{2n(k)+1})\le p({x}_{2n(k)+1},{x}_{2n(k)}),
which combining with (2.25) shows that p({x}_{2n(k)+1},{x}_{2n(k)+1}) is bounded, and hence there exists subsequence p({x}_{2n({k}_{i})+1},{x}_{2n({k}_{i})+1}) of p({x}_{2n(k)+1},{x}_{2n(k)+1}) such that p({x}_{2n({k}_{i})+1},{x}_{2n({k}_{i})+1}) converges to some {r}_{1}, that is,
\underset{i\to +\mathrm{\infty}}{lim}p({x}_{2n({k}_{i})+1},{x}_{2n({k}_{i})+1})={r}_{1}.
(2.27)
By (2.13), we have
\begin{array}{rcl}\psi (p({x}_{2n({k}_{i})+1},{x}_{2n({k}_{i})+2}))& =& \psi (p(f{x}_{2n({k}_{i})},g{x}_{2n({k}_{i})+1}))\\ \le & \phi \left(\frac{p({x}_{2n({k}_{i})},g{x}_{2n({k}_{i})+1})+p(f{x}_{2n({k}_{i})},{x}_{2n({k}_{i})+1})}{2}\right)\\ \varphi (p({x}_{2n({k}_{i})},g{x}_{2n({k}_{i})+1}),p(f{x}_{2n({k}_{i})},{x}_{2n({k}_{i})+1}))\\ =& \phi \left(\frac{p({x}_{2n({k}_{i})},{x}_{2n({k}_{i})+2})+p({x}_{2n({k}_{i})+1},{x}_{2n({k}_{i})+1})}{2}\right)\\ \varphi (p({x}_{2n({k}_{i})},{x}_{2n({k}_{i})+2}),p({x}_{2n({k}_{i})+1},{x}_{2n({k}_{i})+1}))\\ \le & \phi \left(\frac{p({x}_{2n({k}_{i})},{x}_{2n({k}_{i})+1})+p({x}_{2n({k}_{i})+1},{x}_{2n({k}_{i})+2})}{2}\right)\\ \varphi (p({x}_{2n({k}_{i})},{x}_{2n({k}_{i})+2}),p({x}_{2n({k}_{i})+1},{x}_{2n({k}_{i})+1})).\end{array}
(2.28)
Letting i\to +\mathrm{\infty} in (2.28), and using (2.25)(2.27), we obtain that
\psi (r)\le \phi (r)\varphi ({r}_{0},{r}_{1}),
(2.29)
which means that \varphi ({r}_{0},{r}_{1})=0, hence {r}_{0}=0 and {r}_{1}=0.
Since
\begin{array}{rcl}\psi (p({x}_{2n({k}_{i})+1},{x}_{2n({k}_{i})+2}))& =& \psi (p(f{x}_{2n({k}_{i})},g{x}_{2n({k}_{i})+1}))\\ \le & \phi \left(\frac{p({x}_{2n({k}_{i})},g{x}_{2n({k}_{i})+1})+p(f{x}_{2n({k}_{i})},{x}_{2n({k}_{i})+1})}{2}\right)\\ \varphi (p({x}_{2n({k}_{i})},g{x}_{2n({k}_{i})+1}),p(f{x}_{2n({k}_{i})},{x}_{2n({k}_{i})+1}))\\ =& \phi \left(\frac{p({x}_{2n({k}_{i})},{x}_{2n({k}_{i})+2})+p({x}_{2n({k}_{i})+1},{x}_{2n({k}_{i})+1})}{2}\right)\\ \varphi (p({x}_{2n({k}_{i})},{x}_{2n({k}_{i})+2}),p({x}_{2n({k}_{i})+1},{x}_{2n({k}_{i})+1})),\end{array}
taking the limit as i\to +\mathrm{\infty}, we have \psi (r)=0, which implies that r=0, that is,
\underset{n\to +\mathrm{\infty}}{lim}p({x}_{n},{x}_{n+1})=0.
(2.30)
Now, we claim that \{{x}_{n}\} is a Cauchy sequence in the metric space (X,{d}_{p}) (and so also in the space (X,p) by Lemma 1.1). For this, it is sufficient to show that \{{x}_{2n}\} is a Cauchy sequence in (X,{d}_{p}). Suppose that this is not the case, then using Lemma 1.1, we have that \{{x}_{2n}\} is not a Cauchy sequence in (X,p). By Lemma 1.3, we obtain that there exist \epsilon >0 and two sequences \{m(k)\} and \{n(k)\} of positive integers such that n(k)>m(k)>k and sequences in (1.2) tend to ε when k\to +\mathrm{\infty}.
From (2.13), we get that
\begin{array}{rcl}\psi (p({x}_{2n(k)+1},{x}_{2m(k)}))& =& \psi (p(f{x}_{2n(k)},g{x}_{2m(k)1}))\\ \le & \phi \left(\frac{p({x}_{2n(k)},g{x}_{2m(k)1})+p(f{x}_{2n(k)},{x}_{2m(k)1})}{2}\right)\\ \varphi (p({x}_{2n(k)},g{x}_{2m(k)1}),p(f{x}_{2n(k)},{x}_{2m(k)1}))\\ =& \phi \left(\frac{p({x}_{2n(k)},{x}_{2m(k)})+p({x}_{2n(k)+1},{x}_{2m(k)1})}{2}\right)\\ \varphi (p({x}_{2n(k)},{x}_{2m(k)}),p({x}_{2n(k)+1},{x}_{2m(k)1})).\end{array}
Letting k\to +\mathrm{\infty} in the above inequalities and using the continuity of ψ, φ and ϕ, we get that
\psi (\epsilon )\le \phi (\epsilon )\varphi (\epsilon ,\epsilon ),
therefore, we get that \varphi (\epsilon ,\epsilon )=0. Hence, \epsilon =0 which is a contradiction. Thus, \{{x}_{n}\} is a Cauchy sequence in (X,{d}_{p}), and \{{x}_{n}\} is also a Cauchy sequence in (X,p). Since (X,p) is complete, then the sequence \{{x}_{n}\} converges to some z\in X, that is,
p(z,z)=\underset{n\to +\mathrm{\infty}}{lim}p({x}_{n},z)=\underset{n,m\to +\mathrm{\infty}}{lim}p({x}_{n},{x}_{m}).
Moreover, the sequence \{{x}_{2n}\} and \{{x}_{2n+1}\} converge to z\in X, that is,
p(z,z)=\underset{n\to +\mathrm{\infty}}{lim}p({x}_{2n},z)=\underset{n,m\to +\mathrm{\infty}}{lim}p({x}_{2n},{x}_{2m})
and
p(z,z)=\underset{n\to +\mathrm{\infty}}{lim}p({x}_{2n+1},z)=\underset{n,m\to +\mathrm{\infty}}{lim}p({x}_{2n+1},{x}_{2m+1}).
Using the fact that \{{x}_{n}\} is a Cauchy sequence in (X,{d}_{p}), we have {lim}_{n\to +\mathrm{\infty}}{d}_{p}({x}_{n},{x}_{m})=0, which together with {d}_{p}({x}_{n},{x}_{m})=2p({x}_{n},{x}_{m})p({x}_{n},{x}_{n})p({x}_{m},{x}_{m}) yields that {lim}_{n\to +\mathrm{\infty}}p({x}_{n},{x}_{m})=0. Hence, we have
p(z,z)=\underset{n\to +\mathrm{\infty}}{lim}p({x}_{n},z)=\underset{n\to +\mathrm{\infty}}{lim}p({x}_{2n},z)=\underset{n\to +\mathrm{\infty}}{lim}p({x}_{2n+1},z)=0.
By substituting x={x}_{2n(k)+1}, y=z in (2.13), we get that
\begin{array}{rcl}\psi (p({x}_{2n(k)+1},gz))& =& \psi (p(f{x}_{2n(k)},gz))\\ \le & \phi \left(\frac{p({x}_{2n(k)},gz)+p(f{x}_{2n(k)},z)}{2}\right)\\ \varphi (p({x}_{2n(k)},gz),p(f{x}_{2n(k)},z))\\ =& \phi \left(\frac{p({x}_{2n(k)},gz)+p({x}_{2n(k)+1},z)}{2}\right)\\ \varphi (p({x}_{2n(k)},gz),p({x}_{2n(k)+1},z)),\end{array}
letting k\to +\mathrm{\infty} and applying Lemma 1.2, we conclude that
\begin{array}{rcl}\psi (p(z,gz))& \le & \phi \left(\frac{p(z,gz)+p(z,z)}{2}\right)\varphi (p(z,gz),p(z,z))\\ \le & \phi (p(z,gz))\varphi (p(z,gz),0),\end{array}
which yields that \varphi (p(z,gz),0)=0; hence, p(z,gz)=0, and thus z=gz. Similarly, one can easily show that z=fz, therefore, z is the common fixed point of f and g.
Now we prove the uniqueness of common fixed point. Let us suppose that u is also the common fixed point of f and g. Since
\begin{array}{rcl}\psi (p(u,z))& =& \psi (p(fu,gz))\\ \le & \phi \left(\frac{p(u,gz)+p(fu,z)}{2}\right)\varphi (p(u,gz),p(fu,z))\\ =& \phi (p(u,z))\varphi (p(u,z),p(u,z)),\end{array}
which means that \varphi (p(u,z),p(u,z))=0; hence, p(u,z)=0, and so u=z. Thus, the uniqueness of the common fixed point is proved. □
By taking \phi =\psi in Theorems 2.12.3, respectively, we have the following results.
Corollary 2.1 Let (X,\u2aaf) be a partially ordered set and suppose that there exists a partial metric p on X such that (X,p) is complete. Let f:X\to X be a continuous nondecreasing mapping. Suppose that for comparable x,y\in X, we have
\psi (p(fx,fy))\le \psi \left(\frac{p(x,fy)+p(fx,y)}{2}\right)\varphi (p(x,fy),p(fx,y)),
where ψ is an altering distance function and \varphi :[0,+\mathrm{\infty})\times [0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) is a continuous function with \varphi (x,y)=0 if and only if x=y=0. If there exists {x}_{0}\in X such that {x}_{0}\u2aaff{x}_{0}, then f has a fixed point.
Corollary 2.2 Suppose that X, f, ψ, and ϕ are the same as in Corollary 2.1 except the continuity of f. Suppose that for a nondecreasing sequence \{{x}_{n}\} in X with {x}_{n}\to x\in X, we have {x}_{n}\u2aafx for all n\in N. If there exists {x}_{0}\in X such that {x}_{0}\u2aaff{x}_{0}, then f has a fixed point.
Corollary 2.3 Let (X,p) be a complete partial metric space, f and g be selfmappings on X. Suppose that there exist functions ψ and ϕ such that for all x,y\in X
\psi (p(fx,gy))\le \psi \left(\frac{p(x,gy)+p(fx,y)}{2}\right)\varphi (p(x,gy),p(fx,y)),
where ψ is an altering distance function and \varphi :[0,+\mathrm{\infty})\times [0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) is a continuous function with \varphi (x,y)=0 if and only if x=y=0.
Then f and g have a unique common fixed point.
Remark 2.1 If we replace the partial metric p by (usual) metric d in Corollaries 2.12.3, then we get Theorems 2.12.3 of [27].
Now, we introduce an example to support the usability of our results.
Example 2.1 Let X=[0,1] be endowed with the usual partial metric p:X\times X\to [0,+\mathrm{\infty}) defined by p(x,y)=max\{x,y\}. It is easy to show that the partial metric space (X,p) is complete. Also, define the mappings f,g:X\to X by fx=\frac{{x}^{2}}{4} and gx=\frac{{x}^{2}}{5}, respectively. Let us take \psi ,\phi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) such that \psi (t)={t}^{2} and \phi (t)=\frac{{t}^{2}}{2}, respectively, and take \varphi :[0,+\mathrm{\infty})\times [0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) such that \varphi (t,s)=\frac{{(t+s)}^{2}}{16}. If x\ge y, then
p(fx,gy)=max\{\frac{{x}^{2}}{4},\frac{{y}^{2}}{5}\}=\frac{{x}^{2}}{4},
and
p(x,gy)+p(fx,y)=p(x,\frac{{y}^{2}}{5})+p(\frac{{x}^{2}}{4},y)=max\{x,\frac{{y}^{2}}{5}\}+p(\frac{{x}^{2}}{4},y)=x+p(\frac{{x}^{2}}{4},y).
So, we have
\begin{array}{rcl}\psi (p(fx,gy))& =& \frac{{x}^{4}}{16}\le \frac{{x}^{2}}{16}\\ \le & \frac{{(x+p(\frac{{x}^{2}}{4},y))}^{2}}{16}\\ =& \frac{{(x+p(\frac{{x}^{2}}{4},y))}^{2}}{8}\frac{{(x+p(\frac{{x}^{2}}{4},y))}^{2}}{16}\\ =& \phi \left(\frac{p(x,gy)+p(fx,y)}{2}\right)\varphi (p(x,gy),p(fx,y)).\end{array}
If x\le y, then
p(fx,gy)=max\{\frac{{x}^{2}}{4},\frac{{y}^{2}}{5}\}\le \frac{{y}^{2}}{4}
and
p(x,gy)+p(fx,y)=p(x,\frac{{y}^{2}}{5})+p(\frac{{x}^{2}}{4},y)=p(x,\frac{{y}^{2}}{5})+max\{\frac{{x}^{2}}{4},y\}=p(x,\frac{{y}^{2}}{5})+y.
So, we have
\begin{array}{rcl}\psi (p(fx,gy))& =& \frac{{y}^{4}}{16}\le \frac{{y}^{2}}{16}\\ \le & \frac{{(y+p(x,\frac{{y}^{2}}{5}))}^{2}}{16}\\ =& \frac{{(y+p(x,\frac{{y}^{2}}{5}))}^{2}}{8}\frac{{(y+p(x,\frac{{y}^{2}}{5}))}^{2}}{16}\\ =& \phi \left(\frac{p(x,gy)+p(fx,y)}{2}\right)\varphi (p(x,gy),p(fx,y)).\end{array}
From the above arguments, we conclude that (2.13) holds; hence, all the required hypotheses of Theorem 2.3 are satisfied. Thus, we deduce the existence and uniqueness of a common fixed point of f and g. Here, 0 is the unique common fixed point.