Open Access

# Four mappings and generalized contractions

Fixed Point Theory and Applications20162016:99

DOI: 10.1186/s13663-016-0591-7

Accepted: 10 November 2016

Published: 17 November 2016

## Abstract

In the paper general theorems on common fixed point for four mappings are presented. The results are compact and they extend and unify the respective part of the fixed point theory.

### Keywords

generalized contraction dislocated metric weakly compatible mappings fixed point

47H10 54H25

## 1 Introduction

The present paper was inspired by the advanced and sophisticated article of Liu et al. [1]. Our assumptions are weaker, as the comparison function is much more general, and we do not assume the spaces under consideration to be metric. In addition the general contraction condition is compact and abstract. Also the proofs are relatively simple.

## 2 Theorems

Let us recall (see [2]) that Φ is the family of all mappings $$\varphi : [0,\infty)\rightarrow[0,\infty)$$ such that $$\varphi(\alpha) < \alpha$$, $$\alpha> 0$$, and $$\boldsymbol {\Phi _{0}}$$ consists of mappings $$\varphi\in\Phi$$ such that $$\varphi(0) = 0$$. In turn, $$\boldsymbol {\Phi}_{P}$$ is the family of all mappings $$\varphi : [0,\infty)\rightarrow[0,\infty)$$ for which every sequence $$(a_{n})_{n \in\mathbb{N}}$$ such that $$a_{n+1}\leq\varphi(a_{n})$$, $$n \in\mathbb{N}$$, converges to zero. It is well known ([2], Proposition 16) that $$\Phi_{P}\subset\Phi_{0}$$.

In turn, $$\boldsymbol {\Psi}_{P}$$ (see [3]) consists of all mappings of Φ for which every sequence $$(a_{n})_{n \in\mathbb{N}}$$ such that $$0 < a_{n+1}\leq\varphi (a_{n})$$, $$n \in\mathbb{N}$$ converges to zero. It is known ([4], Corollary 2.4) that $$\Psi_{P}$$ consists of all mappings $$\varphi\in\Phi$$ satisfying
$$\mbox{for each } \alpha> 0, \varphi(\cdot) \leq\alpha \mbox{ on some interval } (\alpha,\alpha+ \epsilon).$$
(1)

Clearly, $$\Phi_{P}\subset\Psi_{P}$$ holds and consequently, all members of $$\Phi_{P}$$ satisfy (1).

### Lemma 2.1

If a $$\varphi\in\Phi_{0}$$ satisfies (1), then $$\varphi\in \Phi_{P}$$.

### Proof

Let $$(a_{n})_{n \in\mathbb{N}}$$ be a sequence such that $$a_{n+1}\leq \varphi(a_{n})$$, $$n \in\mathbb{N}$$ for a $$\varphi\in\Phi_{0}$$. Then we have
$$a_{n+1}\leq\varphi(a_{n}) \leq a_{n}, \quad n \in \mathbb{N}.$$
Therefore, $$(a_{n})_{n \in\mathbb{N}}$$ is nonincreasing and it converges, say to α. Suppose $$\alpha> 0$$. Then from (1) it follows that there exists an interval $$(\alpha,\alpha +\epsilon)$$ on which $$\varphi(\cdot) \leq\alpha$$. For large n all $$a_{n}$$ belong to this interval, as
$$0 < \alpha\leq\varphi(a_{n+1}) \leq a_{n+1}\leq a_{n},$$
and $$\varphi\in\Phi$$ yield $$\alpha\leq\varphi(a_{n+1}) < a_{n+1}$$. Now, we have $$\alpha< a_{n+1}\leq\varphi(a_{n}) \leq\alpha$$, a contradiction. Consequently, $$\alpha= 0$$, i.e. $$\varphi\in\Phi_{P}$$. □

### Corollary 2.2

$$\Phi_{P}$$ consists of all mappings $$\varphi\in\Phi_{0}$$ satisfying (1).

The notion of dislocated metric space presented below is due to Hitzler and Seda [5].

Let X be a nonempty set, and $$p : X \times X \rightarrow [0,\infty)$$ a mapping satisfying
\begin{aligned} &p(x,y)= 0 \mbox{ yields } x = y, \quad x,y \in X, \end{aligned}
(2a)
\begin{aligned} &p(x,y)=p(y,x), \quad x,y \in X, \end{aligned}
(2b)
\begin{aligned} &p(x,z) \leq p(x,y)+ p(y,z), \quad x,y,z \in X . \end{aligned}
(2c)
Then p is called a dislocated metric (briefly a d-metric), and $$(X,p)$$ is called a dislocated metric space (briefly a d-metric space). The topology of $$(X,p)$$ is generated by balls.
Many authors applied sophisticated contraction conditions. To present a general idea let us consider a mapping $$h : X^{4} \rightarrow[0,\infty)$$ satisfying the following requirements for each $$\alpha> 0$$:
\begin{aligned} &(a = d \mbox{ or } b = c) \mbox{ yields } h(a,b,c,d) \leq\max \bigl\{ p(a,b),p(c,a),p(d,b) \bigr\} , \end{aligned}
(3a)
\begin{aligned} & \bigl(p(a,b) \rightarrow\alpha\mbox{ and } p(c,a),p(d,b) \rightarrow0 \bigr) \mbox{ yields } \limsup h(a,b,c,d) \leq\alpha, \end{aligned}
(3b)
\begin{aligned} &\mbox{if } p(d,b) \geq\alpha\mbox{ and } p(a,b),p(c,a) \mbox{ are small, then } h(a,b,c,d) \leq p(d,b), \end{aligned}
(3c)
\begin{aligned} &\mbox{if } p(c,a) \geq\alpha\mbox{ and } p(a,b),p(d,b) \mbox{ are small, then } h(a,b,c,d) \leq p(c,a), \end{aligned}
(3d)
\begin{aligned} &h(a,b,a,b) \leq p(a,b). \end{aligned}
(3e)

In order to present an example let us recall the notion of partial metric due to Matthews ([6], Definition 3.1).

A partial metric is a mapping $$p : X \times X \rightarrow [0,\infty)$$ such that
\begin{aligned} &x = y \mbox{ iff } p(x,x)=p(x,y)=p(y,y),\quad x,y \in X, \end{aligned}
(4a)
\begin{aligned} &p(x,x)\leq p(x,y), \quad x,y \in X, \end{aligned}
(4b)
\begin{aligned} &p(x,y)= p(y,x), \quad x,y \in X, \end{aligned}
(4c)
\begin{aligned} &p(x,z) \leq p(x,y)+ p(y,z) - p(y,y), \quad x,y,z \in X. \end{aligned}
(4d)

Conditions (4b), (4a) yield (2a) and therefore each partial metric is a d-metric. As regards the respective spaces, the situation is more complicated because their topologies are different.

### Example 2.3

Let p be a partial metric on X and let
$$h(a,b,c,d) = \bigl[p(a,d) + p(b,c) \bigr]/2.$$
(5)
Then for $$a = d$$ we have
\begin{aligned} \bigl[p(a,d) + p(b,c) \bigr]/2 &\leq \bigl[p(a,d) + p(b,d) + p(d,c) - p(d,d) \bigr]/2 \\ &= \bigl[p(b,d) + p(a,c) \bigr]/2 \leq\max \bigl\{ p(c,a),p(d,b) \bigr\} . \end{aligned}
Similarly, for $$b = c$$ we obtain
\begin{aligned} \bigl[p(a,d) + p(b,c) \bigr]/2& \leq \bigl[p(a,b) + p(b,d) - p(b,b) + p(b,c) \bigr]/2 \\ &= \bigl[p(a,b) + p(b,d) \bigr]/2 \leq\max \bigl\{ p(a,b),p(d,b) \bigr\} . \end{aligned}
Consequently, (3a) holds. From
\begin{aligned} & \bigl[p(a,d) + p(b,c) \bigr]/2 \\ &\quad \leq \bigl[p(a,b) + p(b,d) - p(b,b) + p(b,a) + p(a,c) - p(a,a) \bigr]/2 \\ &\quad = p(a,b) + \bigl[p(d,b) - p(b,b) + p(c,a) - p(a,a) \bigr]/2 \end{aligned}
we obtain (3b), (3c), (3d), and (3e).
The notion of a 0-complete d-metric space (or a set) was presented in [3], Definition 2.3 (condition (2.5)). Let us note that if $$\lim_{n \rightarrow\infty}p(y,x_{n}) = \lim_{n \rightarrow\infty}p(x,x_{n})= 0$$ holds, then from $$p(x,y) \leq p(x,x_{n}) + p(y,x_{n})$$ it follows that $$x = y$$. In addition, $$p(x,x)\leq2p(x,x_{n})$$ means that $$p(x,x)= 0$$. Therefore, condition (2.5) of [3] is equivalent to
\begin{aligned} \begin{aligned}& \mbox{for each sequence } (x_{n})_{n \in\mathbb{N}}\mbox{ in } X \mbox{ with } \lim _{m,n \rightarrow\infty} p(x_{n},x_{m})=0 \\ &\mbox{there exists a unique } x \in X \mbox{ for which } \lim _{n \rightarrow\infty}p(x,x_{n})= p(x,x)= 0. \end{aligned} \end{aligned}
(6)

The idea of 0-completeness for partial metric spaces is due to Romaguera ([7], Definition 2.1). A partial metric space $$(X,p)$$ is called 0-complete if any 0-Cauchy sequence $$(x_{n})_{n \in\mathbb{N}}$$ in X (i.e. such that $$\lim_{m,n \rightarrow\infty}p(x_{n},x_{m})=0$$), converges (in the topology of $$(X,p)$$) to a point $$x \in X$$ for which $$p(x,x)= 0$$.

It is well known (e.g. [6]) that $$x \in\lim_{n \rightarrow\infty}x_{n}$$ in a partial metric space $$(X,p)$$ iff $$\lim_{n \rightarrow\infty}p(x,x_{n})= p(x,x)$$. In addition, as it was noticed before, each partial metric is a d-metric. Hence we obtain the following.

### Corollary 2.4

Any partial metric space $$(X,p)$$ is 0-complete iff $$(X,p)$$ treated as a d-metric space is 0-complete, and iff (6) is satisfied.

### Proposition 2.5

Let $$(X,p)$$ be a d-metric space and let $$(x_{n})_{n \in\mathbb{N}}$$ be a sequence of points of X such that
$$p(x_{n+2},x_{n+1}) \leq\varphi \bigl(p(x_{n+1},x_{n}) \bigr), \quad n \in\mathbb{N},$$
holds for a $$\varphi\in\Phi_{P}$$. Then we have
$$\lim_{n \rightarrow\infty}p(x_{n+1},x_{n}) = \lim_{n \rightarrow \infty}p(x_{n},x_{n}) = 0.$$
(7)

### Proof

Let us adopt $$a_{n}= p(x_{n+1},x_{n})$$. Clearly, $$a_{n+1}\leq\varphi(a_{n})$$, $$n \in\mathbb{N}$$ holds, and $$\lim_{n \rightarrow\infty}a_{n}= 0$$, as $$\varphi\in\Phi_{P}$$. Now, $$p(x_{n},x_{n}) \leq2p(x_{n+1},x_{n}) = 2a_{n}$$ completes (7). □

In the following $$f(X)$$ is replaced by fX, $$f(x)$$ is replaced by fx, etc. The precise order in $$\max\{p(a,b),p(c,a),p(d,b)\}$$ informs on the variables of the mapping h, and they are not shown in the proofs.

### Lemma 2.6

Let $$(X,p)$$ be a d-metric space and let $$f,g,i,j$$ be self mappings in X satisfying the following conditions:
\begin{aligned} & fX \subset jX \textit{, } gX \subset iX \textit{ and at least one of these sets is } 0\textit{-complete}, \end{aligned}
(8)
\begin{aligned} &p(fx,gy) \leq\varphi \bigl(\max \bigl\{ p(ix,jy),p(fx,ix),p(gy,jy),h(ix,jy,fx,gy) \bigr\} \bigr), \end{aligned}
(9)
for a $$h : X^{4} \rightarrow[0,\infty)$$ such that (3a), (3b) hold, and a $$\varphi\in\Phi_{P}$$ (see Corollary  2.2). Then there exist sequences $$(x_{n})_{n \in \mathbb{N}}$$, $$(y_{n})_{n \in\mathbb{N}}$$ in X such that
\begin{aligned} \begin{aligned} &x_{2k+1}= gy_{2k} \quad \textit{for } x_{2k}= jy_{2k} \quad\textit{and} \\ & x_{2k}= fy_{2k-1} \quad \textit{ for } x_{2k-1}= iy_{2k-1}, \quad k \in\mathbb{N}, \end{aligned} \end{aligned}
(10)
and $$(x_{n})_{n \in\mathbb{N}}$$ converges to a point x such that $$p(x,x)= 0$$, and
\begin{aligned} \begin{aligned} x& = \lim_{n \rightarrow\infty}x_{n}= \lim_{k \rightarrow\infty }fy_{2k-1}= \lim_{k \rightarrow\infty}iy_{2k-1} \\ &= \lim_{k \rightarrow\infty}gy_{2k}= \lim_{k \rightarrow\infty}jy_{2k}. \end{aligned} \end{aligned}
(11)

### Proof

From (8), (9), and (3a) (for $$b = c$$) it follows that
\begin{aligned} \begin{aligned} &\mbox{for each } x \in X \mbox{ there exists a } y \in X \mbox{ such that } fx = jy \mbox{ and} \\ &p(gy,fx) = p(fx,gy) \leq\varphi \bigl(\max \bigl\{ p(ix,fx),p(gy,fx) \bigr\} \bigr), \end{aligned} \end{aligned}
and (for $$a = d$$)
\begin{aligned} \begin{aligned}& \mbox{for each } x \in X \mbox{ there exists a } y \in X \mbox{ such that } gx = iy \mbox{ and} \\ &p(fy,gx) \leq\varphi \bigl(\max \bigl\{ p(gx,jx),p(fy,gx) \bigr\} \bigr). \end{aligned} \end{aligned}
If e.g. $$p(fx,ix) \leq p(gy,fx)$$ holds, then $$p(gy,fx) \leq \varphi(p(gy,fx))$$ means that $$p(gy,fx) = 0$$, as $$\varphi\in\Phi_{0}$$. Consequently, (8) and (9) yield
\begin{aligned} \begin{aligned} &\mbox{for each } x \in X \mbox{ there exists a } y \in X \mbox{ such that } \\ &fx = jy \mbox{ and } p(gy,fx) \leq\varphi \bigl(p(fx,ix) \bigr) \end{aligned} \end{aligned}
(12)
and
\begin{aligned} \begin{aligned}& \mbox{for each } x \in X \mbox{ there exists a } y \in X \mbox{ such that } \\ &gx = iy \mbox{ and } p(fy,gx) \leq\varphi \bigl(p(gx,jx) \bigr). \end{aligned} \end{aligned}
(13)
For an $$x_{0}\in X$$ let us adopt $$x_{1}= gx_{0}= iy_{1}$$, $$x_{2} = fy_{1}$$, where $$p(x_{2},x_{1}) \leq\varphi(p(x_{1},jx_{0}))$$ (see (13)). Now, we define $$x_{3} = gy_{2}$$ for $$y_{2}$$ such that $$x_{2} = fy_{1}= jy_{2}$$ and $$p(x_{3},x_{2}) \leq \varphi (p(x_{2},x_{1}))$$ (see (12)). In turn $$x_{4} = fy_{3}$$ for $$y_{3}$$ such that $$x_{3} = gy_{2} = iy_{3}$$ and $$p(x_{4},x_{3}) \leq\varphi(p(x_{3},x_{2}))$$ (see (13)). By induction we define two sequences $$(x_{n})_{n \in\mathbb{N}}$$, $$(y_{n})_{n \in\mathbb{N}}$$ satisfying (10) and
$$p(x_{n+2},x_{n+1}) \leq\varphi \bigl(p(x_{n+1},x_{n}) \bigr), \quad n \in\mathbb{N}.$$
In view of Proposition 2.5 we have $$\lim_{n \rightarrow \infty}p(x_{n+1},x_{n}) = 0$$.
Suppose that there exists an infinite set $$\mathbb{K}\subset \mathbb{N}$$ such that for each $$k \in\mathbb{K}$$ there exists an $$n \in\mathbb{N}$$ for which $$0 < \alpha< p(x_{2k},x_{2n+1+2k})$$ holds. Let $$n = n(k) > 0$$ be the smallest numbers satisfying this inequality for $$k \in\mathbb{K}$$. We have (see (2c))
\begin{aligned} \begin{aligned} &\alpha- p(x_{2k},x_{2k-1}) - p(x_{2n+2k},x_{2n+1+2k}) \\ &\quad< p(x_{2k},x_{2n+1+2k}) - p(x_{2k},x_{2k-1}) - p(x_{2n+2k},x_{2n+1+2k}) \\ &\quad \leq p(x_{2k-1},x_{2n+2k}) \\ &\quad\leq p(x_{2k-1},x_{2k}) + p(x_{2k},x_{2n-1+2k}) + p(x_{2n-1+2k},x_{2n+2k}) \\ &\quad \leq p(x_{2k-1},x_{2k}) + \alpha+ p(x_{2n-1+2k},x_{2n+2k}) \end{aligned} \end{aligned}
for $$n = n(k)$$, and therefore (see (7)),
$$\lim_{k \in\mathbb{K}}p(x_{2k-1},x_{2n+2k}) = \alpha.$$
Now, (9), (10), (7), and the above equality yield
\begin{aligned} \begin{aligned} p(x_{2k},x_{2n+1+2k})& = p(fy_{2k-1},gy_{2n+2k}) \\ &\leq \varphi \bigl(\max \bigl\{ p(iy_{2k-1},jy_{2n+2k}),p(fy_{2k-1},iy_{2k-1}),p(gy_{2n+2k} ,jy_{2n+2k}),h(\cdot) \bigr\} \bigr) \\ &= \varphi \bigl(\max \bigl\{ p(x_{2k-1},x_{2n+2k}),p(x_{2k},x_{2k-1}),p(x_{2n+1+2k},x_{2n+2k} ),h(\cdot) \bigr\} \bigr) \\ & = \varphi \bigl(\max \bigl\{ p(x_{2k-1},x_{2n+2k}),h(\cdot) \bigr\} \bigr) \end{aligned} \end{aligned}
for large k. In turn, (3b) yields
$$\limsup_{k \in\mathbb{K}} h(\cdot) \leq\lim_{k \in\mathbb {K}}p(x_{2k-1},x_{2n+2k}) = \alpha,$$
and hence we obtain $$\max\{p(x_{2k-1},x_{2n+2k}),h(\cdot)\} < \alpha+ \epsilon$$ for large k. From $$\varphi(\beta) < \alpha$$ for $$\beta\leq\alpha$$, and (1) we get
$$\alpha< p(x_{2k},x_{2n+1+2k}) \leq\varphi \bigl(\max \bigl\{ p(x_{2k-1},x_{2n+2k}),h(\cdot) \bigr\} \bigr) \leq\alpha,$$
for large k, a contradiction. Therefore
$$\lim_{k,n \rightarrow\infty} p(x_{2k},x_{2n+1+2k}) = 0$$
holds, and (7) with the triangle inequality (2c) yield $$\lim_{m,n \rightarrow\infty}p(x_{n},x_{m}) = 0$$. Consequently (see (6)), there exists an $$x \in X$$ such that $$\lim_{n \rightarrow\infty}p(x,x_{n})= p(x,x)= 0$$, and (11) holds. □
Let us recall (see [8]) that a pair $$(f,i)$$ of mappings $$f,i : X \rightarrow X$$ is called weakly compatible if
$$\mbox{for each } x \in X \mbox{, } fx = ix \mbox{ yields } fix=ifx.$$

Now, we are ready to prove our theorem.

### Theorem 2.7

Let $$(X,p)$$ be a d-metric space such that (4b) holds. Assume that f, g, i, j are self mappings in X satisfying
$$(f,i) \textit{, } (g,j)\quad \textit{are weakly compatible},$$
(14)
(8), and (9) for a $$\varphi\in\Phi_{P}$$ (see Corollary  2.2) and a mapping $$h : X^{4} \rightarrow[0,\infty)$$ for which the system of conditions (3a)-(3e) holds. Then f, g, i, j have a unique common fixed point, say x, and $$p(x,x)= 0$$.

### Proof

Let $$(x_{n})_{n \in\mathbb{N}}$$, $$(y_{n})_{n \in\mathbb{N}}$$, and x be as in Lemma 2.6. Assume e.g. that $$x \in jX$$ (for 0-complete jX or fX). Then there exists a v such that $$x = jv$$. Let us prove that $$gv = x$$. We have
$$p(gv,x) = p(x,gv) \leq p(x,x_{2k}) + p(x_{2k},gv).$$
Suppose $$p(gv,x) > 0$$. Then we obtain
\begin{aligned} \begin{aligned} p(x_{2k},gv) &= p(fy_{2k-1},gv) \\ & \leq \varphi \bigl(\max \bigl\{ p(iy_{2k-1},jv),p(fy_{2k-1},iy_{2k-1}),p(gv,jv),h( \cdot) \bigr\} \bigr) \\ &= \varphi \bigl(\max \bigl\{ p(x_{2k-1},x),p(x_{2k},x_{2k-1}),p(gv,x),h( \cdot) \bigr\} \bigr) = \varphi \bigl(p(gv,x) \bigr), \end{aligned} \end{aligned}
for large k (see (9), (3c). Hence,
$$p(gv,x) \leq\lim_{k \rightarrow\infty}p(x,x_{2k}) + \varphi \bigl(p(gv,x) \bigr)$$
yields $$p(gv,x) = 0$$ ($$\varphi\in\Phi_{0}$$), and $$gv = x$$. Thus we have $$x = gv = jv$$.
From $$gX \subset iX$$ it follows that there exists a w such that $$x = iw$$. Let us show that $$fw = x$$. We have
$$p(fw,x) \leq p(fw,gy_{2k}) + p(gy_{2k},x) = p(fw,gy_{2k}) + p(x_{2k+1},x).$$
Suppose $$p(fw,x) > 0$$. Then we have
\begin{aligned} \begin{aligned} p(fw,gy_{2k})& \leq\varphi \bigl(\max \bigl\{ p(iw,jy_{2k}),p(fw,iw),p(gy_{2k},jy_{2k} ),h( \cdot) \bigr\} \bigr) \\ &= \varphi \bigl(\max \bigl\{ p(x,x_{2k}),p(fw,x),p(x_{2k+1},x_{2k}) \bigr\} \bigr) = \varphi \bigl(p(fw,x) \bigr) \end{aligned} \end{aligned}
for large k (see (9)), (3d)). Hence, we obtain
$$p(fw,x) \leq\varphi \bigl(p(fw,x) \bigr) + \lim_{k \rightarrow\infty }p(x_{2k+1},x) = \varphi \bigl(p(fw,x) \bigr),$$
and $$p(fw,x) = 0$$, i.e. $$x = fw = iw$$.
Now, (14) yields
$$jx = jgv = gjv = gx\quad \mbox{and} \quad ix = ifw = fiw = fx.$$
Let us show that $$fx = gx$$. We have (see (9), (3e), (4b))
\begin{aligned} \begin{aligned} p(fx,gx) &\leq\varphi \bigl(\max \bigl\{ p(ix,jx),p(fx,ix),p(gx,jx),h(\cdot) \bigr\} \bigr) \\ & = \varphi \bigl(\max \bigl\{ p(fx,gx),p(fx,fx),p(gx,gx) \bigr\} \bigr) = \varphi \bigl(p(fx,gx) \bigr), \end{aligned} \end{aligned}
and therefore $$p(fx,gx) = 0$$, i.e. $$fx = gx$$ (let us note that $$p(fx,fx) = 0$$ not necessarily holds if p is a d-metric).

Now, it is clear that $$fx = gx = ix = jx$$ holds.

Let us prove that $$gx = x$$. We have (see (9), (3e), (4b))
\begin{aligned} \begin{aligned} p(x,gx) &= p(fw,gx) \leq\varphi \bigl(\max \bigl\{ p(iw,jx),p(fw,iw),p(gx,jx),h(\cdot) \bigr\} \bigr) \\ &= \varphi \bigl(\max \bigl\{ p(x,gx),p(x,x),p(gx,gx) \bigr\} \bigr) = \varphi \bigl(p(x,gx) \bigr), \end{aligned} \end{aligned}
and consequently, $$p(x,gx) = 0$$, i.e. $$gx = x$$.

We have proved that $$x = fx = gx = ix = jx$$.

If y is a common fixed point of our mappings, then (see (9), (3e), (4b))
$$p(x,y) = p(fx,gy) \leq\varphi \bigl(\max \bigl\{ p(ix,jy),p(fx,ix),p(gy,jy),h( \cdot) \bigr\} \bigr) = \varphi \bigl(p(x,y) \bigr)$$
yields $$p(x,y) = 0$$, i.e. $$x = y$$. □

The authors of [1] consider (in metric spaces) formulas $$M_{1}$$, $$M_{2}$$, and $$M_{3}$$ instead of ‘max…’ from our condition (9). It would be rather exhausting to cite $$M_{1}$$ or $$M_{2}$$, nevertheless all three formulas are applied in conditions that are particular cases of (9) for h satisfying (3a)-(3e).

The next theorem is a consequence of Corollary 2.4, and Theorem 2.7 (see also Example 2.3).

### Theorem 2.8

Let $$(X,p)$$ be a partial metric space. Assume that f, g, i, j are self mappings in X satisfying (14), (8), and (9) for a $$\varphi\in\Phi_{P}$$ (see Corollary  2.2) and a mapping $$h : X^{4} \rightarrow[0,\infty)$$ for which the system of conditions (3a)-(3e) holds. Then f, g, i, j have a unique common fixed point, say x, and $$p(x,x)= 0$$.

The above theorem is a far extension of [9], Theorem 2.8 in its part concerning a unique common fixed point; our mapping φ need not be continuous (see also Example 2.3).

## Declarations

### Acknowledgements

The work has been supported by the Polish Ministry of Science and Higher Education.

## Authors’ Affiliations

(1)
Faculty of Applied Mathematics, AGH University of Science and Technology

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