Some coupled coincidence and common fixed point results for a hybrid pair of mappings in 0-complete partial metric spaces

Abstract

In this paper we extend some coupled coincidence and common fixed point theorems for a hybrid pair of mappings obtained by Abbas et al. (Fixed Point Theory Appl. 2012:4, 2012, doi:10.1186/1687-1812-2012-4) from the complete metric space to 0-complete partial metric spaces. An example showing that this extension is proper is given.

MSC:47H10, 54H25.

1 Introduction

Let A be any nonempty subset of a metric space $(X,d)$. For $x\in X$, define

$$d(x,A)=inf\{d(x,y):y\in A\}.$$

Let $CB(X)$ denote the set of all nonempty closed bounded subset of X. For $A,B\in CB(X)$, define

$$\begin{array}{c}\delta (A,B)=sup\{d(x,B):x\in A\},\hfill \\ H(A,B)=max\{\delta (A,B),\delta (B,A)\}.\hfill \end{array}$$

Then H is a metric on $CB(X)$ and is called a Hausdorff metric.

Nadler [1] generalized the Banach contraction mapping principle to set-valued functions and proved the following fixed point theorem.

Theorem 1 Let $(X,d)$ be a complete metric space and let T be a mapping from X into $CB(X)$ such that for all $x,y\in X$,

$$H(Tx,Ty)\le \lambda d(x,y),$$

where $0\le \lambda <1$. Then T has a fixed point.

Later, an interesting and rich fixed point theory was developed. On the other hand, Matthews [2] introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks, with the interesting property ‘non-zero self-distance’ in space. He showed that the Banach contraction mapping theorem can be generalized to the partial metric context for applications in program verification. Subsequently, several authors (see, e.g., [3–22]) derived fixed point theorems in partial metric spaces. Romaguera [17] introduced the notion of 0-Cauchy sequence, 0-complete partial metric spaces and proved some characterizations of partial metric spaces in terms of completeness and 0-completeness. Recently, Aydi et al. [9] introduced the notion of a partial Hausdorff metric and extended the Nadler’s theorem in partial metric spaces.

Bhaskar and Lakshmikantham [23] introduced the concept of a coupled fixed point and established some coupled fixed point theorems in partially ordered sets. As an application, they studied the existence and uniqueness of a solution for a periodic boundary value problem associated with a first-order ordinary differential equation. Recently Abbas et al. [24] extended these concepts to set-valued mappings and obtained coupled coincidence points and coupled common fixed point theorems involving a hybrid pair of single-valued and multi-valued maps satisfying generalized contractive conditions in the framework of a complete metric space (see also [25, 26]). The study of a coincidence point and common fixed points of a hybrid pair of mappings in Banach spaces and metric spaces is interesting and well developed. For applications of hybrid fixed point theory, we refer to [27–30].

In this paper, we extend and generalize the results of Abbas et al. [24] and Aydi et al. [9] for a hybrid pair of mappings in 0-complete partial metric spaces. Also, some new results are obtained. An example is included to support our results.

2 Preliminaries

Consistent with [2, 8, 9, 16, 17, 19], the following definitions and results will be needed in the sequel.

Definition 1 A partial metric on a nonempty set X is a function $p:X\times X\to {\mathbb{R}}^{+}$ (${\mathbb{R}}^{+}$ stands for nonnegative reals) such that for all $x,y,z\in X$,

• (P1) $x=y\iff p(x,x)=p(x,y)=p(y,y)$,

• (P2) $p(x,x)\le p(x,y)$,

• (P3) $p(x,y)=p(y,x)$,

• (P4) $p(x,y)\le p(x,z)+p(z,y)-p(z,z)$.

A partial metric space is a pair $(X,p)$ such that X is a nonempty set and p is a partial metric on X.

It is clear that if $p(x,y)=0$, then from (P1) and (P2) $x=y$. But if $x=y$, $p(x,y)$ may not be 0. Also, every metric space is a partial metric space, with zero self-distance.

Example 1 If $p:{\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is defined by $p(x,y)=max\{x,y\}$, for all $x,y\in {\mathbb{R}}^{+}$, then $({\mathbb{R}}^{+},p)$ is a partial metric space.

Some more examples of a partial metric space can be seen in [2, 9, 16].

Each partial metric on X generates a $T_0$ topology ${\tau }_p$ on X which has as a base the family of open p-balls $\{B_p(x,ϵ):x\in X,ϵ>0\}$, where $B_p(x,ϵ)=\{y\in X:p(x,y)<p(x,x)+ϵ\}$ for all $x\in X$ and $ϵ>0$.

Theorem 2 [2]

For each partial metric $p:X\times X\to {\mathbb{R}}^{+}$, the pair $(X,d)$, where $d(x,y)=2p(x,y)-p(x,x)-p(y,y)$ for all $x,y\in X$, is a metric space.

Here $(X,d)$ is called an induced metric space and d is an induced metric. In further discussion, unless specified otherwise, $(X,d)$ will represent an induced metric space.

Let $(X,p)$ be a partial metric space.

• (1) A sequence $\{x_n\}$ in $(X,p)$ converges to a point $x\in X$ if and only if $p(x,x)={lim}_{n\to \mathrm{\infty }}p(x_n,x)$.

• (2) A sequence $\{x_n\}$ in $(X,p)$ is called a Cauchy sequence if there exists (and is finite) ${lim}_{n,m\to \mathrm{\infty }}p(x_n,x_m)$.

• (3) $(X,p)$ is said to be complete if every Cauchy sequence $\{x_n\}$ in X converges with respect to ${\tau }_p$ to a point $x\in X$ such that $p(x,x)={lim}_{n,m\to \mathrm{\infty }}p(x_n,x_m)$.

• (4) A sequence $\{x_n\}$ in $(X,p)$ is called 0-Cauchy sequence if ${lim}_{n,m\to \mathrm{\infty }}p(x_n,x_m)=0$. The space $(X,p)$ is said to be 0-complete if every 0-Cauchy sequence in X converges with respect to ${\tau }_p$ to a point $x\in X$ such that $p(x,x)=0$.

Lemma 1 [2, 17, 19]

Let $(X,p)$ be a partial metric space and $\{x_n\}$ be any sequence in X.

• (i) $\{x_n\}$ is a Cauchy sequence in $(X,p)$ if and only if it is a Cauchy sequence in the metric space $(X,d)$.

• (ii) $(X,p)$ is complete if and only if the metric space $(X,d)$ is complete. Furthermore, ${lim}_{n\to \mathrm{\infty }}d(x_n,x)=0$ if and only if $p(x,x)={lim}_{n\to \mathrm{\infty }}p(x_n,x)={lim}_{n,m\to \mathrm{\infty }}p(x_n,x_m)$.

• (iii) Every 0-Cauchy sequence in $(X,p)$ is Cauchy in $(X,d)$.

• (iv) If $(X,p)$ is complete, then it is 0-complete.

The converse assertions of (iii) and (iv) do not hold. Indeed, the partial metric space $(\mathbb{Q}\cap [0,\mathrm{\infty }),p)$, where ℚ denotes the set of rational numbers and the partial metric p is given by $p(x,y)=max\{x,y\}$ for all $x,y\in X$, provides an easy example of a 0-complete partial metric space which is not complete. It is easy to see that every closed subset of a 0-complete partial metric space is 0-complete.

Let $(X,p)$ be a partial metric space. Let $C{B}^p(X)$ be the family of all nonempty, closed and bounded subsets of the partial metric space $(X,p)$ induced by the partial metric p. Note that closedness is taken from $(X,{\tau }_p)$ (${\tau }_p$ is the topology induced by p) and boundedness is given as follows: A is a bounded subset in $(X,p)$ if there exist $x_0\in X$ and $M\ge 0$ such that for all $a\in A$, we have $a\in B_p(x_0,M)$, that is, $p(x_0,a)<p(a,a)+M$.

For $A,B\in C{B}^p(X)$ and $x\in X$, define

$$p(x,A)=inf\{p(x,a):a\in A\},{\delta }_p(A,B)=sup\{p(a,B):a\in A\}.$$

Lemma 2 [8]

Let $(X,p)$ be a partial metric space, $A\subset X$. Then $a\in \overline{A}$ if and only if $p(a,A)=p(a,a)$.

Proposition 1 [9]

Let $(X,p)$ be a partial metric space. For any $A,B,C\in C{B}^p(X)$, we have the following:

• (i) ${\delta }_p(A,A)=sup\{p(a,a):a\in A\}$;

• (ii) ${\delta }_p(A,A)\le {\delta }_p(A,B)$;

• (iii) ${\delta }_p(A,A)=0$ implies that $A\subseteq B$;

• (iv) ${\delta }_p(A,B)\le {\delta }_p(A,C)+{\delta }_p(C,B)-{inf}_{c\in C}p(c,c)$.

Let $(X,p)$ be a partial metric space. For $A,B\in C{B}^p(X)$, define

$$H_p(A,B)=max\{{\delta }_p(A,B),{\delta }_p(B,A)\}.$$

Proposition 2 [9]

Let $(X,p)$ be a partial metric space. For $A,B,C\in C{B}^p(X)$, we have

• (h1) $H_p(A,A)\le H_p(A,B)$;

• (h2) $H_p(A,B)=H_p(B,A)$;

• (h3) $H_p(A,B)\le H_p(A,C)+H_p(C,B)-{inf}_{c\in C}p(c,c)$.

Corollary 1 [9]

Let $(X,p)$ be a partial metric space. For $A,B\in C{B}^p(X)$, the following holds:

$$H_p(A,B)=0\mathit{\text{implies that}}A=B.$$

In view of Proposition 2 and Corollary 1, we call the mapping $H_p:C{B}^p(X)\times C{B}^p(X)\to [0,\mathrm{\infty })$ a partial Hausdorff metric induced by p.

Lemma 3 [9]

Let $(X,p)$ be a partial metric space, $A,B\in C{B}^p(X)$ and $h>1$. For any $a\in A$, there exists $b=b(a)\in B$ such that $p(a,b)\le h{H}_p(A,B)$.

The following lemma is crucial for the proof of our main result and its proof is similar to Lemma 3.

Lemma 4 Let $(X,p)$ be a partial metric space and $A,B\in C{B}^p(X)$, $a\in A$. Let $ϵ>0$ be arbitrary, then there exists $b=b(a)\in B$ such that

$$p(a,b)\le H_p(A,B)+ϵ.$$

Definition 2 [24]

Let X be a nonempty set, $F:X\times X\to 2^X$ (collection of all nonempty subsets of X) and $g:X\to X$. An element $(x,y)\in X\times X$ is called

• (i) a coupled fixed point of F if $x\in F(x,y)$ and $y\in F(y,x)$;

• (ii) a coupled coincidence point of the hybrid pair $\{F,g\}$ if $gx\in F(x,y)$ and $gy\in F(y,x)$;

• (iii) a coupled point of coincidence if there exists $(u,v)\in X\times X$ such that $x=gu\in F(u,v)$ and $y=gv\in F(v,u)$;

• (iv) a coupled common fixed point of the hybrid pair $\{F,g\}$ if $x=gx\in F(x,y)$ and $y=gy\in F(y,x)$.

Definition 3 Let X be a nonempty set, let $F:X\times X\to 2^X$ and $g:X\to X$ be two mappings. The hybrid pair $\{F,g\}$ is called weakly compatible if $gF(x,y)\subseteq F(gx,gy)$ and $gF(y,x)\subseteq F(gy,gx)$ whenever $(x,y)$ is a coupled coincidence point of the hybrid pair $\{F,g\}$.

Now we can state our main results.

3 Main results

The following result extends and generalizes the main result of [24] in partial metric spaces.

Theorem 3 Let $(X,p)$ be a 0-complete partial metric space, let $F:X\times X\to C{B}^p(X)$ and $g:X\to X$ be mappings satisfying

$$\begin{array}{rl}{H}_p(F(x,y),F(u,v))\le & a_{1}p(gx,gu)+a_{2}p(gy,gv)+a_{3}p(F(x,y),gx)\\ +a_{4}p(F(x,y),gu)+a_{5}p(F(u,v),gx)\\ +a_{6}p(F(u,v),gu)\end{array}$$

(1)

for all $x,y,u,v\in X$, where $a_i$ are nonnegative reals such that ${\sum }_{i=1}^6{a}_i<1$. If $F(X\times X)\subseteq g(X)$ and $g(X)$ is a closed subset of X, then F and g have a coupled point of coincidence $(w_c,z_c)\in X\times X$ and $p(w_c,w_c)=p(z_c,z_c)=0$.

Proof Let $x_0,y_0\in X$ be arbitrary, then $F(x_0,y_0),F(y_0,x_0)\in C{B}^p(X)$. As $F(X\times X)\subseteq g(X)$, we can choose $g{x}_1\in F(x_0,y_0)$ and $g{y}_1\in F(y_0,x_0)$ for some $x_1,y_1\in X$. Again, as $F(x_1,y_1),F(y_1,x_1)\in C{B}^p(X)$ and $F(X\times X)\subseteq g(X)$, so by Lemma 4, for any $ϵ>0$, there exist $g{x}_2\in F(x_1,y_1)$ and $g{y}_2\in F(y_1,x_1)$ such that

$$\begin{array}{c}p(g{x}_1,g{x}_2)\le H_p(F(x_0,y_0),F(x_1,y_1))+ϵ,\hfill \\ p(g{y}_1,g{y}_2)\le H_p(F(y_0,x_0),F(y_1,x_1))+ϵ.\hfill \end{array}$$

Continuing this process, we obtain two sequences $\{x_n\}$ and $\{y_n\}$ in X such that

$$\begin{array}{c}g{x}_{n+1}\in F(x_n,y_n)\text{and}g{y}_n\in F(y_n,x_n),\hfill \\ p(g{x}_n,g{x}_{n+1})\le H_p(F(x_{n-1},y_{n-1}),F(x_n,y_n))+{ϵ}^n,\hfill \\ p(g{y}_n,g{y}_{n+1})\le H_p(F(y_{n-1},x_{n-1}),F(y_n,x_n))+{ϵ}^n.\hfill \end{array}$$

From the above inequalities and (1), we obtain

$$\begin{array}{rcl}p(g{x}_n,g{x}_{n+1})& \le & H_p(F(x_{n-1},y_{n-1}),F(x_n,y_n))+{ϵ}^n\\ \le & a_{1}p(g{x}_{n-1},g{x}_n)+a_{2}p(g{y}_{n-1},g{y}_n)\\ +a_{3}p(F(x_{n-1},y_{n-1}),g{x}_{n-1})+a_{4}p(F(x_{n-1},y_{n-1}),g{x}_n)\\ +a_{5}p(F(x_n,y_n),g{x}_{n-1})+a_{6}p(F(x_n,y_n),g{x}_n)+{ϵ}^n\\ \le & a_{1}p(g{x}_{n-1},g{x}_n)+a_{2}p(g{y}_{n-1},g{y}_n)+a_{3}p(g{x}_n,g{x}_{n-1})\\ +a_{4}p(g{x}_n,g{x}_n)+a_{5}p(g{x}_{n+1},g{x}_{n-1})+a_{6}p(g{x}_{n+1},g{x}_n)+{ϵ}^n\\ \le & a_{1}p(g{x}_{n-1},g{x}_n)+a_{2}p(g{y}_{n-1},g{y}_n)+a_{3}p(g{x}_n,g{x}_{n-1})\\ +a_{4}p(g{x}_n,g{x}_n)+a_{5}p(g{x}_{n+1},g{x}_n)+a_{5}p(g{x}_n,g{x}_{n-1})\\ -a_{5}p(g{x}_n,g{x}_n)+a_{6}p(g{x}_{n+1},g{x}_n)+{ϵ}^n,\end{array}$$

that is,

$$\begin{array}{rl}(1-a_5-a_6)p(g{x}_n,g{x}_{n+1})\le & (a_1+a_3+a_5)p(g{x}_n,g{x}_{n-1})+a_{2}p(g{y}_{n-1},g{y}_n)\\ +(a_4-a_5)p(g{x}_n,g{x}_n)+{ϵ}^n.\end{array}$$

(2)

Interchanging the roles of $x_n$ and $x_{n+1}$ and using the symmetries of p and $H_p$, we obtain

$$\begin{array}{rl}(1-a_4-a_3)p(g{x}_n,g{x}_{n+1})\le & (a_1+a_6+a_4)p(g{x}_n,g{x}_{n-1})+a_{2}p(g{y}_{n-1},g{y}_n)\\ +(a_5-a_4)p(g{x}_n,g{x}_n)+{ϵ}^n.\end{array}$$

(3)

It follows from (2) and (3) that

$$\begin{array}{rl}(2-a_3-a_4-a_5-a_6)p(g{x}_n,g{x}_{n+1})\le & (2a_1+a_3+a_4+a_5+a_6)p(g{x}_n,g{x}_{n-1})\\ +2a_{2}p(g{y}_{n-1},g{y}_n)+2{ϵ}^n.\end{array}$$

(4)

Similarly, it can be obtained that

$$\begin{array}{rl}(2-a_3-a_4-a_5-a_6)p(g{y}_n,g{y}_{n+1})\le & (2a_1+a_3+a_4+a_5+a_6)p(g{y}_n,g{y}_{n-1})\\ +2a_{2}p(g{x}_{n-1},g{x}_n)+2{ϵ}^n.\end{array}$$

(5)

For simplicity, set $p_n=p(g{x}_n,g{x}_{n+1})+p(g{y}_n,g{y}_{n+1})$, then it follows from (4) and (5) that

$$(2-a_3-a_4-a_5-a_6)p_n\le (2a_1+2a_2+a_3+a_4+a_5+a_6)p_{n-1}+4{ϵ}^n,$$

that is,

$$p_n\le \frac{2a_1+2a_2+a_3+a_4+a_5+a_6}{2-a_3-a_4-a_5-a_6}{p}_{n-1}+\frac{4{ϵ}^n}{2-a_3-a_4-a_5-a_6}.$$

(6)

As $ϵ>0$ was arbitrary, choose $ϵ=\frac{2a_1+2a_2+a_3+a_4+a_5+a_6}{2-a_3-a_4-a_5-a_6}$; also, as ${\sum }_{i=1}^6{a}_i<1$, we have $ϵ<1$. Therefore, from (6) we have

$$p_n\le ϵp_{n-1}+\frac{4{ϵ}^n}{1+a_1+a_2}.$$

From a successive application of the above inequality, we obtain

$$\begin{array}{r}{p}_n\le ϵp_{n-1}+\frac{4{ϵ}^n}{1+a_1+a_2},\\ p_n\le ϵ[ϵp_{n-2}+\frac{4{ϵ}^{n-1}}{1+a_1+a_2}]+\frac{4{ϵ}^n}{1+a_1+a_2},\\ p_n\le {ϵ}^2{p}_{n-2}+\frac{8{ϵ}^n}{1+a_1+a_2},\\ ⋮\\ p_n\le {ϵ}^n{p}_0+\frac{4n{ϵ}^n}{1+a_1+a_2}.\end{array}$$

(7)

For $m,n\in \mathbb{N}$ with $m>n$, using (7) we obtain

$$\begin{array}{rcl}p(g{x}_n,g{x}_m)+p(g{y}_n,g{y}_m)& \le & p(g{x}_n,g{x}_{n+1})+p(g{y}_n,g{y}_{n+1})+p(g{x}_{n+1},g{x}_{n+2})\\ +p(g{y}_{n+1},g{y}_{n+2})+\cdots +p(g{x}_{m-1},g{x}_m)+p(g{y}_{m-1},g{y}_m)\\ =& p_n+p_{n+1}+\cdots +p_{m-1}\\ \le & {ϵ}^n{p}_0+\frac{4n{ϵ}^n}{1+a_1+a_2}+{ϵ}^{n+1}{p}_0+\frac{4(n+1){ϵ}^{n+1}}{1+a_1+a_2}\\ +\cdots +{ϵ}^{m-1}{p}_0+\frac{4(m-1){ϵ}^{m-1}}{1+a_1+a_2}\\ =& p_0\sum _{i=0}^{m-n-1}{ϵ}^{n+i}+\frac{4}{1+a_1+a_2}\sum _{i=0}^{m-n-1}(n+i){ϵ}^{n+i}.\end{array}$$

As $ϵ<1$, it follows from the above inequality that

$$\underset{n,m\to \mathrm{\infty }}{lim}p(g{x}_n,g{x}_m)=\underset{n,m\to \mathrm{\infty }}{lim}p(g{y}_n,g{y}_m)=0.$$

So, $\{g{x}_n\}$ and $\{g{y}_n\}$ are 0-Cauchy sequences in $g(X)$; therefore, by the closedness of $g(X)$, there exists $w,z\in X$ such that

(8)

(9)

Using (1) we obtain

$$\begin{array}{rcl}p(F(w,z),gw)& \le & p(F(w,z),g{x}_{n+1})+p(g{x}_{n+1},gw)\\ \le & H_p(F(w,z),F(x_n,y_n))+p(g{x}_{n+1},gw)\\ \le & a_{1}p(gw,g{x}_n)+a_{2}p(gz,g{y}_n)+a_{3}p(F(w,z),gw)\\ +a_{4}p(F(w,z),g{x}_n)+a_{5}p(F(x_n,y_n),gw)\\ +a_{6}p(F(x_n,y_n),g{x}_n)+p(g{x}_{n+1},gw),\end{array}$$

that is,

$$\begin{array}{rcl}(1-a_3-a_4)p(F(w,z),gw)& \le & a_{1}p(gw,g{x}_n)+a_{2}p(gz,g{y}_n)+a_{4}p(gw,g{x}_n)\\ +a_{5}p(g{x}_{n+1},gw)+a_{6}p(g{x}_{n+1},g{x}_n)+p(g{x}_{n+1},gw)\\ =& (a_1+a_4)p(gw,g{x}_n)+a_{2}p(gz,g{y}_n)\\ +(1+a_5)p(g{x}_{n+1},gw)+a_{6}p(g{x}_{n+1},g{x}_n).\end{array}$$

Using (8) and (9) and the fact that $1-a_3-a_4>0$ in the above inequality, we obtain

$$p(F(w,z),gw)=p(gw,gw)=0.$$

Therefore, by Lemma 2, $gw\in F(w,z)$. Similarly, $gz\in F(z,w)$. Thus $(w,z)$ is a coupled coincidence point and $(gw,gz)=(w_c,z_c)$ (say) is a point of coincidence of the mappings F and g with $p(gw,gw)=p(gz,gz)=p(w_c,w_c)=p(z_c,z_c)=0$. □

The following is a coupled fixed point result for a set-valued mapping and it can be obtained by taking $g=I_X$ (that is an identity mapping of X) in the above theorem.

Corollary 2 Let $(X,p)$ be a 0-complete partial metric space, let $F:X\times X\to C{B}^p(X)$ be a mapping satisfying

$$\begin{array}{rl}{H}_p(F(x,y),F(u,v))\le & a_{1}p(x,u)+a_{2}p(y,v)+a_{3}p(F(x,y),x)+a_{4}p(F(x,y),u)\\ +a_{5}p(F(u,v),x)+a_{6}p(F(u,v),u)\end{array}$$

for all $x,y,u,v\in X$, where $a_i$ are nonnegative reals such that ${\sum }_{i=1}^6{a}_i<1$. Then F has a coupled fixed point $(w,z)\in X\times X$ and $p(w,w)=p(z,z)=0$.

With suitable values of control constants in Theorem 3, one can obtain the following corollaries.

Corollary 3 Let $(X,p)$ be a 0-complete partial metric space, let $F:X\times X\to C{B}^p(X)$ and $g:X\to X$ be mappings satisfying

$$H_p(F(x,y),F(u,v))\le a_{1}p(gx,gu)+a_{2}p(gy,gv)$$

(10)

for all $x,y,u,v\in X$, where $a_1$ and $a_2$ are nonnegative reals such that $a_1+a_2<1$. If $F(X\times X)\subseteq g(X)$ and $g(X)$ is a closed subset of X, then F and g have a coupled point of coincidence $(w_c,z_c)\in X\times X$ and $p(w_c,w_c)=p(z_c,z_c)=0$.

Corollary 4 Let $(X,p)$ be a 0-complete partial metric space, let $F:X\times X\to C{B}^p(X)$ and $g:X\to X$ be mappings satisfying

$$\begin{array}{rl}{H}_p(F(x,y),F(u,v))\le & a_{1}p(F(x,y),gx)+a_{2}p(F(x,y),gu)+a_{3}p(F(u,v),gx)\\ +a_{4}p(F(u,v),gu)\end{array}$$

for all $x,y,u,v\in X$, where $a_i$ are nonnegative reals such that ${\sum }_{i=1}^4{a}_i<1$. If $F(X\times X)\subseteq g(X)$ and $g(X)$ is a closed subset of X, then F and g have a coupled point of coincidence $(w_c,z_c)\in X\times X$ and $p(w_c,w_c)=p(z_c,z_c)=0$.

The following example illustrates the case when the results in partial metric spaces are applicable while the same results in usual metric spaces are not.

Example 2 Let $X=[0,1]\cap \mathbb{Q}$, and let $p:X\times X\to {\mathbb{R}}^{+}$ be defined by

$$p(x,y)=|x-y|+max\{x,y\}\text{for all }x,y\in X.$$

Then the metric induced by p is given by $d(x,y)=3|x-y|$ for all $x,y\in X$ and $(X,d)$ is not complete, therefore $(X,p)$ is not complete. Now, it is easy to see that $(X,p)$ is a 0-complete partial metric space and every singleton subset of X is closed with respect to p. Define $F:X\times X\to C{B}^p(X)$ and $g:X\to X$ by

$$F(x,y)=\{\begin{array}{cc}\{0\}\hfill & \text{if }x=y=1;\hfill \\ \{0,\frac{x+y}{8}\}\hfill & \text{otherwise}\hfill \end{array}\text{and}gx=x\text{for all }x\in X.$$

We shall show that F and g satisfy all the conditions of Corollary 3, with $a_1=a_2=\alpha \in [\frac{1}{4},\frac{1}{2})$, while the metric versions of Corollary 3 are not applicable. We consider the following cases.

Case (i) If $x,y,u,v\in X\setminus \{1\}$ and $x+y\ne u+v$, then suppose $u+v<x+y$, so

$$\begin{array}{c}{H}_p(F(x,y),F(u,v))\hfill \\ =H_p(\{0,\frac{x+y}{8}\},\{0,\frac{u+v}{8}\})\hfill \\ =max\{\underset{a\in \{0,\frac{x+y}{8}\}}{sup}p(a,\{0,\frac{u+v}{8}\}),\underset{a\in \{0,\frac{u+v}{8}\}}{sup}p(a,\{0,\frac{x+y}{8}\})\}\hfill \\ =max\{p(\frac{x+y}{8},\{0,\frac{u+v}{8}\}),p(\frac{u+v}{8},\{0,\frac{x+y}{8}\})\}\hfill \\ =max\{min\{\frac{x+y}{4},\frac{1}{8}|x-u+y-v|+\frac{1}{8}max\{x+y,u+v\}\},\hfill \\ min\{\frac{u+v}{4},\frac{1}{8}|x-u+y-v|+\frac{1}{8}max\{x+y,u+v\}\}\}\hfill \\ =max\{\frac{1}{8}|x-u+y-v|+\frac{x+y}{8},min\{\frac{u+v}{4},\frac{1}{8}|x-u+y-v|+\frac{x+y}{8}\}\}\hfill \\ =\frac{1}{8}|x-u+y-v|+\frac{x+y}{8}\le \frac{1}{8}[|x-u|+|y-v|]+\frac{x+y}{8}\hfill \\ \le \alpha [p(gx,gu)+p(gy,gv)],\hfill \end{array}$$

where $\frac{1}{8}\le \alpha$. Similarly, we obtain the same result for $u+v>x+y$.

Case (ii) If $x,y,u,v\in X\setminus \{1\}$ and $x+y=u+v$, then

$$\begin{array}{rcl}{H}_p(F(x,y),F(u,v))& =& H_p(\{0,\frac{x+y}{8}\},\{0,\frac{x+y}{8}\})\\ =& \underset{a\in \{0,\frac{x+y}{8}\}}{sup}p(a,a)\\ =& \frac{x+y}{8}\le \frac{max\{x,u\}+max\{y,v\}}{8}\\ \le & \alpha [p(gx,gu)+p(gy,gv)],\end{array}$$

where $\frac{1}{8}\le \alpha$. Similarly, if any one of x, y, u, v is equal to 1, then we obtain the same result.

Case (iii) If any one of $(x,y)$, $(u,v)$ is equal to $(1,1)$, for example, let $(u,v)=(1,1)$ and $(x,y)\ne (1,1)$, then we have

$$\begin{array}{rcl}{H}_p(F(x,y),F(u,v))& =& H_p(\{0,\frac{x+y}{8}\},\{0\})\\ =& max\{\underset{a\in \{0,\frac{x+y}{8}\}}{sup}p(a,\{0\}),\underset{a\in \{0\}}{sup}p(a,\{0,\frac{x+y}{8}\})\}\\ =& max\{\frac{x+y}{4},0\}=\frac{x+y}{4}\\ \le & \alpha [p(gx,gu)+p(gy,gv)],\end{array}$$

where $\frac{1}{4}\le \alpha$. Similarly, the condition (10) is satisfied for $a_1=a_2=\alpha \in [\frac{1}{4},\frac{1}{2})$ in all possible cases and $0=g0\in F(0,0)$, that is, $(0,0)$ is a coupled coincidence point of F and g (here it is the unique common fixed point of F and g).

Note that, the metric spaces $(X,\rho )$ and $(X,d)$ (where ρ is usual and d is metric induced by p) are not complete, therefore metric versions of Corollary 3 are not applicable. Also, this example shows that F and g do not satisfy the metric versions of inequality (10). Indeed, if $H_{\rho }$ is the Hausdorff metric induced by the usual metric ρ, then for $x=y=u=1$, $v=\frac{9}{10}$, we have

$$H_{\rho }(F(x,y),F(u,v))=H_{\rho }(\{0\},\{0,\frac{19}{80}\})=\frac{19}{80}$$

and

$$a_1\rho (gx,gu)+a_2\rho (gy,gv)=\frac{1}{10}{a}_2.$$

Therefore, we cannot find the nonnegative reals $a_1$, $a_2$ such that

$$H_{\rho }(F(x,y),F(u,v))\le a_1\rho (gx,gu)+a_2\rho (gy,gv)$$

for all $x,y,u,v\in X$ with $a_1+a_2<1$. So, F is not a contraction (in view of contraction condition (10)) with respect to the usual metric ρ. Similarly, one can see that F is not a contraction with respect to the induced metric d.

The following theorem provides a sufficient condition for the uniqueness of a coupled point of coincidence and a common fixed point of the hybrid pair $\{F,g\}$.

Theorem 4 Let $(X,p)$ be a 0-complete partial metric space, let $F:X\times X\to C{B}^p(X)$ and $g:X\to X$ be mappings such that all the conditions of Theorem  3 are satisfied and, for any coupled coincidence point $(w,z)$ of F and g, we have $F(w,z)=\{gw\}$ and $F(z,w)=\{gz\}$, then F and g have a unique coupled point of coincidence. Suppose in addition that the hybrid pair $\{F,g\}$ is weakly compatible, then F and g have a unique coupled common fixed point.

Proof The existence of a coupled coincidence point $(w,z)$ and a point of coincidence $(w_c,z_c)$ follows from Theorem 3. Suppose that, for any coupled coincidence point $(w,z)$ of F and g, we have $F(w,z)=\{gw\}=\{w_c\}$ and $F(z,w)=\{gz\}=\{z_c\}$. We shall show that the coupled point of coincidence is unique. Let $(w^{\prime },z^{\prime })$ be another coupled coincidence point and $(w_c^{\prime },z_c^{\prime })$ be the coupled point of coincidence of F and g, that is, $w_c^{\prime }=g{w}^{\prime }\in F(w^{\prime },z^{\prime })$, $z_c^{\prime }=g{z}^{\prime }\in F(z^{\prime },w^{\prime })$ and $F(w^{\prime },z^{\prime })=\{g{w}^{\prime }\}=\{w_c^{\prime }\}$, $F(z^{\prime },w^{\prime })=\{g{z}^{\prime }\}=\{z_c^{\prime }\}$.

Using (1), we obtain

$$\begin{array}{rcl}p(gw,g{w}^{\prime })& =& H_p(\{gw\},\left\{g{w}^{\prime }\right\})\\ =& H_p(F(w,z),F(w^{\prime },z^{\prime }))\\ \le & a_{1}p(gw,g{w}^{\prime })+a_{2}p(gz,g{z}^{\prime })+a_{3}p(F(w,z),gw)+a_{4}p(F(w,z),g{w}^{\prime })\\ +a_{5}p(F(w^{\prime },z^{\prime }),gw)+a_{6}p(F(w^{\prime },z^{\prime }),g{w}^{\prime })\\ =& a_{1}p(gw,g{w}^{\prime })+a_{2}p(gz,g{z}^{\prime })+a_{3}p(gw,gw)+a_{4}p(gw,g{w}^{\prime })\\ +a_{5}p(g{w}^{\prime },gw)+a_{6}p(g{w}^{\prime },g{w}^{\prime }).\end{array}$$

(11)

Again, using (1) we obtain

$$\begin{array}{rcl}p(gz,g{z}^{\prime })& =& H_p(\{gz\},\left\{g{z}^{\prime }\right\})\\ =& H_p(F(z,w),F(z^{\prime },w^{\prime }))\\ \le & a_{1}p(gz,g{z}^{\prime })+a_{2}p(gw,g{w}^{\prime })+a_{3}p(F(z,w),gz)+a_{4}p(F(z,w),g{z}^{\prime })\\ +a_{5}p(F(z^{\prime },w^{\prime }),gz)+a_{6}p(F(z^{\prime },w^{\prime }),g{z}^{\prime })\\ =& a_{1}p(gz,g{z}^{\prime })+a_{2}p(gw,g{w}^{\prime })+a_{3}p(gz,gz)+a_{4}p(gz,g{z}^{\prime })\\ +a_{5}p(g{z}^{\prime },gz)+a_{6}p(g{z}^{\prime },g{z}^{\prime }).\end{array}$$

(12)

It follows from (11) and (12) that

$$\begin{array}{rcl}p(gw,g{w}^{\prime })+p(gz,g{z}^{\prime })& \le & (a_1+a_2+a_4+a_5)p(gw,g{w}^{\prime })+a_{3}p(gw,gw)\\ +a_{6}p(g{w}^{\prime },g{w}^{\prime })+(a_1+a_2+a_4+a_5)p(gz,g{z}^{\prime })\\ +a_{3}p(gz,gz)+a_{6}p(g{z}^{\prime },g{z}^{\prime })\\ =& (a_1+a_2+a_3+a_4+a_5+a_6)[p(gw,g{w}^{\prime })+p(gz,g{z}^{\prime })].\end{array}$$

As ${\sum }_{i=1}^6{a}_i<1$, it follows from the above inequality that $p(gw,g{w}^{\prime })+p(gz,g{z}^{\prime })=0$, that is, $p(gw,g{w}^{\prime })=p(gz,g{z}^{\prime })=0$, so $w_c=gw=g{w}^{\prime }=w_c^{\prime }$ and $z_c=gz=g{z}^{\prime }=z_c^{\prime }$. Therefore, a coupled point of coincidence, that is, $(w_c,z_c)$, of F and g is unique.

Suppose that F and g are weakly compatible, then we have

$$\begin{array}{c}g\{w_c\}=gF(w,z)\subseteq F(gw,gz)\text{that is}\{g{w}_c\}\subseteq F(w_c,z_c)\text{and}\hfill \\ g\{z_c\}=gF(z,w)\subseteq F(gz,gw)\text{that is}\{g{z}_c\}\subseteq F(z_c,w_c).\hfill \end{array}$$

Therefore, $(g{w}_c,g{z}_c)$ is another coupled point of coincidence of F and g, and by uniqueness we have $w_c=g{w}_c\in F(w_c,z_c)$ and $z_c=g{z}_c\in F(z_c,w_c)$. Thus $(z_c,w_c)$ is the unique coupled common fixed point of F and g. □

The following theorem is a new result for a hybrid pair of mappings in partial metric as well as in metric spaces.

Theorem 5 Let $(X,p)$ be a 0-complete partial metric space, let $F:X\times X\to C{B}^p(X)$ and $g:X\to X$ be mappings satisfying

$$\begin{array}{rl}{H}_p(F(x,y),F(u,v))\le & a_{1}p(F(y,x),gy)+a_{2}p(F(y,x),gv)+a_{3}p(F(v,u),gy)\\ +a_{4}p(F(v,u),gv)\end{array}$$

(13)

for all $x,y,u,v\in X$, where $a_i$ are nonnegative reals such that ${\sum }_{i=1}^4{a}_i<1$. If $F(X\times X)\subseteq g(X)$ and $g(X)$ is a closed subset of X, then F and g have a coupled point of coincidence $(w_c,z_c)\in X\times X$ and $p(g{w}_c,g{w}_c)=p(g{z}_c,g{z}_c)=0$.

Proof By a similar process as used in Theorem 3, we can find two sequences $\{x_n\}$ and $\{y_n\}$ such that

$$\begin{array}{c}g{x}_{n+1}\in F(x_n,y_n)\text{and}g{y}_{n+1}\in F(y_n,x_n),\hfill \\ p(g{x}_n,g{x}_{n+1})\le H_p(F(x_{n-1},y_{n-1}),F(x_n,y_n))+{ϵ}^n,\hfill \\ p(g{y}_n,g{y}_{n+1})\le H_p(F(y_{n-1},x_{n-1}),F(y_n,x_n))+{ϵ}^n,\hfill \end{array}$$

where $ϵ>0$ is arbitrary.

From the above inequality and (13), we obtain

$$\begin{array}{rcl}p(g{x}_n,g{x}_{n+1})& \le & H_p(F(x_{n-1},y_{n-1}),F(x_n,y_n))+{ϵ}^n\\ \le & a_{1}p(F(y_{n-1},x_{n-1}),g{y}_{n-1})+a_{2}p(F(y_{n-1},x_{n-1}),g{y}_n)\\ +a_{3}p(F(y_n,x_n),g{y}_{n-1})+a_{4}p(F(y_n,x_n),g{y}_n)+{ϵ}^n\\ \le & a_{1}p(g{y}_n,g{y}_{n-1})+a_{2}p(g{y}_n,g{y}_n)+a_{3}p(g{y}_{n+1},g{y}_{n-1})\\ +a_{4}p(g{y}_{n+1},g{y}_n)+{ϵ}^n,\end{array}$$

that is,

$$\begin{array}{rl}p(g{x}_n,g{x}_{n+1})\le & (a_1+a_3)p(g{y}_n,g{y}_{n-1})+(a_3+a_4)p(g{y}_{n+1},g{y}_n)+{ϵ}^n\\ +(a_2-a_3)a_{2}p(g{y}_n,g{y}_n).\end{array}$$

(14)

Interchanging the roles of $x_n$ and $x_{n+1}$ and using the symmetries of p and $H_p$, we obtain

$$\begin{array}{rl}p(g{x}_n,g{x}_{n+1})\le & (a_4+a_2)p(g{y}_n,g{y}_{n-1})+(a_2+a_1)p(g{y}_{n+1},g{y}_n)\\ +(a_3-a_2)a_{2}p(g{y}_n,g{y}_n)+{ϵ}^n.\end{array}$$

(15)

It follows from (14) and (15) that

$$\begin{array}{rl}2p(g{x}_n,g{x}_{n+1})\le & (a_1+a_2+a_3+a_4)[p(g{y}_n,g{y}_{n-1})\\ +p(g{y}_{n+1},g{y}_n)]+2{ϵ}^n.\end{array}$$

(16)

Similarly, it can be shown that

$$\begin{array}{rl}2p(g{y}_n,g{y}_{n+1})\le & (a_1+a_2+a_3+a_4)[p(g{x}_n,g{x}_{n-1})\\ +p(g{x}_{n+1},g{x}_n)]+2{ϵ}^n.\end{array}$$

(17)

For simplicity, set $p_n=p(g{x}_n,g{x}_{n+1})+p(g{y}_n,g{y}_{n+1})$, then it follows from (16) and (17) that

$$p_n\le \frac{a_1+a_2+a_3+a_4}{2-a_1-a_2-a_3-a_4}{p}_{n-1}+\frac{4{ϵ}^n}{2-a_1-a_2-a_3-a_4}.$$

As $ϵ>0$ was arbitrary, choose $ϵ=\frac{a_1+a_2+a_3+a_4}{2-a_1-a_2-a_3-a_4}$; also, as ${\sum }_{i=1}^4{a}_i<1$, we have $ϵ<1$. Therefore

$$p_n\le ϵp_{n-1}+\frac{4{ϵ}^n}{2-a_1-a_2-a_3-a_4}\le ϵp_{n-1}+4{ϵ}^n.$$

It follows from a successive application of the above inequality that

$$\begin{array}{r}{p}_n\le ϵp_{n-1}+4{ϵ}^n,\\ p_n\le ϵ[ϵp_{n-2}+4{ϵ}^{n-1}]+4{ϵ}^n,\\ p_n\le {ϵ}^2{p}_{n-2}+8{ϵ}^n,\\ ⋮\\ p_n\le {ϵ}^n{p}_0+4n{ϵ}^n.\end{array}$$

(18)

For $m,n\in \mathbb{N}$ with $m>n$, using (18) we obtain

$$\begin{array}{rcl}p(g{x}_n,g{x}_m)+p(g{y}_n,g{y}_m)& \le & p(g{x}_n,g{x}_{n+1})+p(g{y}_n,g{y}_{n+1})+p(g{x}_{n+1},g{x}_{n+2})\\ +p(g{y}_{n+1},g{y}_{n+2})+\cdots +p(g{x}_{m-1},g{x}_m)\\ +p(g{y}_{m-1},g{y}_m)\\ =& p_n+p_{n+1}+\cdots +p_{m-1}\\ \le & {ϵ}^n{p}_0+4n{ϵ}^n+{ϵ}^{n+1}{p}_0+4(n+1){ϵ}^{n+1}+\cdots +{ϵ}^{m-1}{p}_0\\ +4(m-1){ϵ}^{m-1}\\ =& p_0\sum _{i=0}^{m-n-1}{ϵ}^{n+i}+4\sum _{i=0}^{m-n-1}(n+i){ϵ}^{n+i}.\end{array}$$

As $ϵ<1$, it follows from the above inequality that

$$\underset{n,m\to \mathrm{\infty }}{lim}p(g{x}_n,g{x}_m)=\underset{n,m\to \mathrm{\infty }}{lim}p(g{y}_n,g{y}_m)=0.$$

So, $\{g{x}_n\}$ and $\{g{y}_n\}$ are 0-Cauchy sequences in $g(X)$, therefore by the closedness of $g(X)$, there exists $w,z\in X$ such that

(19)

(20)

We shall show that $p(F(w,z),gw)=p(gw,gw)=0$ and $p(F(z,w),gz)=p(gz,gz)=0$.

For all $n\in \mathbb{N}$, we have

$$\begin{array}{rcl}p(F(w,z),gw)& \le & p(F(w,z),g{x}_{n+1})+p(g{x}_{n+1},gw)\\ \le & H_p(F(w,z),F(x_n,y_n))+p(g{x}_{n+1},gw)\\ \le & a_{1}p(F(z,w),gz)+a_{2}p(F(z,w),g{y}_n)+a_{3}p(F(y_n,x_n),gz)\\ +a_{4}p(F(y_n,x_n),g{y}_n)+p(g{x}_{n+1},gw)\\ \le & (a_1+a_2)p(F(z,w),gz)+a_{2}p(gz,g{y}_n)+a_{3}p(g{y}_{n+1},gz)\\ +a_{4}p(g{y}_{n+1},g{y}_n)+p(g{x}_{n+1},gw).\end{array}$$

Using (19) and (20) in the above inequality, we obtain

$$p(F(w,z),gw)\le (a_1+a_2)p(F(z,w),gz)<p(F(z,w),gz).$$

(21)

Again, for all $n\in \mathbb{N}$, we have

$$\begin{array}{rcl}p(F(z,w),gz)& \le & p(F(z,w),g{y}_{n+1})+p(g{y}_{n+1},gz)\\ \le & H_p(F(z,w),F(y_n,x_n))+p(g{y}_{n+1},gz)\\ \le & a_{1}p(F(w,z),gw)+a_{2}p(F(w,z),g{x}_n)+a_{3}p(F(x_n,y_n),gw)\\ +a_{4}p(F(x_n,y_n),g{x}_n)+p(g{y}_{n+1},gz)\\ \le & (a_1+a_2)p(F(w,z),gw)+a_{2}p(gw,g{x}_n)+a_{3}p(g{x}_{n+1},gw)\\ +a_{4}p(g{x}_{n+1},g{x}_n)+p(g{y}_{n+1},gz).\end{array}$$

Using (20) and (19) in the above inequality, we obtain

$$p(F(z,w),gz)\le (a_1+a_2)p(F(w,z),gw)<p(F(w,z),gw).$$

(22)

Note that if $p(F(w,z),gw)\ne p(gw,gw)=0$ or $p(F(z,w),gz)\ne p(gz,gz)=0$, then (21) and (22) give a contradiction. Therefore, we have $p(F(w,z),gw)=p(gw,gw)=0$ and $p(F(z,w),gz)=p(gz,gz)=0$, and by Lemma 2, $gw\in F(w,z)$ and $gz\in F(z,w)$. Thus $(w,z)$ is a coupled coincidence point and $(gw,gz)=(w_c,z_c)$ (say) is a point of coincidence of the mappings F and g with $p(gw,gw)=p(gz,gz)=p(w_c,w_c)=p(z_c,z_c)=0$. □

The following is a coupled fixed point result for a set-valued mapping and can be obtained by taking $g=I_X$ (that is an identity mapping of X) in the above theorem.

Corollary 5 Let $(X,p)$ be a 0-complete partial metric space, let $F:X\to C{B}^p(X)$ be a mapping satisfying

$$\begin{array}{rl}{H}_p(F(x,y),F(u,v))\le & a_{1}p(F(y,x),y)+a_{2}p(F(y,x),v)+a_{3}p(F(v,u),y)\\ +a_{4}p(F(v,u),v)\end{array}$$

for all $x,y,u,v\in X$, where $a_i$ are nonnegative reals such that ${\sum }_{i=1}^4{a}_i<1$. Then F has a coupled fixed point $(w,z)\in X\times X$ and $p(w,w)=p(z,z)=0$.

Theorem 6 Let $(X,p)$ be a 0-complete partial metric space, let $F:X\times X\to C{B}^p(X)$ and $g:X\to X$ be mappings such that all the conditions of Theorem  5 are satisfied, and for any coupled coincidence point $(w,z)$ of F and g, we have $F(w,z)=\{gw\}$ and $F(z,w)=\{gz\}$. Then F and g have a unique coupled point of coincidence. Suppose in addition that the hybrid pair $\{F,g\}$ is weakly compatible, then F and g have a unique coupled common fixed point.

Proof The proof of this theorem is followed by a similar process as used in Theorem 4. □