Common fixed points of a generalized ordered g-quasicontraction in partially ordered metric spaces

Abstract

The concept of a generalized ordered g-quasicontraction is introduced, and some fixed and common fixed point theorems for a g-nondecreasing generalized ordered g-quasicontraction mapping in partially ordered complete metric spaces are proved. We also show the uniqueness of the common fixed point in the case of a generalized ordered g-quasicontraction mapping. Finally, we prove fixed point theorems for mappings satisfying the so-called weak contractive conditions in the setting of a partially ordered metric space. Presented theorems are generalizations of very recent fixed point theorems due to Golubović et al. (Fixed Point Theory Appl. 2012:20, 2012).

MSC:47H10, 47N10.

1 Introduction

The Banach fixed point theorem for contraction mappings has been extended in many directions (cf. [1–15]). Very recently Golubović et al. [16] presented some new results for ordered quasicontractions and g-quasicontractions in partially ordered metric spaces.

Recall that if $(X,⪯)$ is a partially ordered set and $f:X\to X$ is such that for $x,y\in X$, $x⪯y$ implies $fx⪯fy$, then a mapping F is said to be non-decreasing. The main result of Golubović et al. [16] is the following fixed point theorem.

Theorem 1.1 (See [16], Theorem 1)

Let $(X,d,⪯)$ be a partially ordered metric space and let $f,g:X\to X$ be two self-maps on X satisfying the following conditions:

• (i) $fX\subset gX$;

• (ii) gX is complete;

• (iii) f is g-nondecreasing;

• (iv) f is an ordered g-quasicontraction;

• (v) there exists $x_0\in X$ such that $g{x}_0⪯f{x}_0$;

• (vi) if $\{g{x}_n\}$ is a nondecreasing sequence that converges to some $gz\in gX$, then $g{x}_n⪯gz$ for each $n\in \mathbb{N}$ and $gz⪯g(gz)$.

  Then f and g have a coincidence point, i.e., there exists $z\in X$ such that $fz=gz$. If, in addition,

• (vii) f and g are weakly compatible [17, 18], i.e., $fx=gx$ implies $fgx=gfx$ for each $x\in X$, then they have a common fixed point.

An open problem is to find sufficient conditions for the uniqueness of the common fixed point in the case of an ordered g-quasicontraction in Theorem 1.1.

In Section 2 of this article, we introduce generalized ordered g-quasicontractions in partially ordered metric spaces and prove the respective (common) fixed point theorems which generalize the results of Theorem 1.1.

In Section 3 of this article, the uniqueness of a common fixed point theorem is obtained when for all $x,u\in X$, there exists $a\in X$ such that fa is comparable to fx and fu in addition to the hypotheses in Theorem 2.1 of Section 2. Our results are an answer to finding sufficient conditions for the uniqueness of a common fixed point in the case of an ordered g-quasicontraction in Theorem 1.1. Finally, two examples show that the comparability is a sufficient condition for the uniqueness of a common fixed point in the case of an ordered g-quasicontraction, so our results are extensions of known ones.

In Section 4 of this article, we consider weak contractive conditions in the setting of a partially ordered metric space and prove respective common fixed point theorems.

2 Common fixed points of a generalized ordered g-quasicontraction

We start this section with the following definitions. Consider a partially ordered set $(X,⪯)$ and two mappings $f:X\to X$ and $g:X\to X$ such that $f(X)\subset g(X)$.

Definition 2.1 (See [19])

We will say that the mapping f is g-nondecreasing (resp., g-nonincreasing) if

$$gx⪯gy\Rightarrow fx⪯fy$$

(1)

(resp., $gx⪯gy\Rightarrow fx⪯fy$) holds for each $x,y\in X$.

Definition 2.2 (See [16])

We will say that the mapping f is an ordered g-quasicontraction if there exists $\alpha \in (0,1)$ such that for each $x,y\in X$ satisfying $gy⪯gx$, the inequality

$$d(fx,fy)\le \alpha \cdot M(x,y)$$

holds, where

$$M(x,y)=max\{d(gx,gy),d(gx,fx),d(gy,fy),d(gx,fy),d(gy,fx)\}.$$

Definition 2.3 We will say that the mapping f is a generalized ordered g-quasicontraction if there is a continuous and non-decreasing function $\psi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $\psi (s+t)\le \psi (s)+\psi (t)$ for each $s,t>0$, $\psi (t)\ge t$ for $t\ge 0$ and there exists $\alpha \in (0,1)$

$$\begin{array}{rcl}\psi (d(fx,fy))& \le & \alpha max\{\psi (d(gx,gy)),\psi (d(gx,fx)),\psi (d(gy,fy)),\\ \psi (d(gx,fy)),\psi (d(gy,fx))\}\end{array}$$

(2)

for all $x,y\in X$ for which $gx⪰gy$;

It is obvious that if $\psi =I$, then a generalized ordered g-quasicontraction reduces to an ordered g-quasicontraction.

For arbitrary $x_0\in X$, one can construct the so-called Jungck sequence $\{y_n\}$ in the following way: Denote $y_0=f{x}_0\in f(X)\subset g(X)$; there exists $x_1\in X$ such that $g{x}_1=y_0=f{x}_0$; now $y_1=f{x}_1\in f(X)\subset g(X)$ and there exists $x_2\in X$ such that $g{x}_2=y_1=f{x}_1$ and the procedure can be continued.

Theorem 2.1 Let $(X,⪯)$ be a partially ordered set and suppose there is a metric d on X such that $(X,d)$ is a complete metric space. Let $f,g:X\to X$ be two self-maps on X satisfying the following conditions:

• (i) $f(X)\subset g(X)$;

• (ii) $g(X)$ is closed;

• (iii) f is a g-nondecreasing mapping;

• (iv) f is a generalized ordered g-quasicontraction;

• (v) there exists an $x_0\in X$ with $g{x}_0⪯f{x}_0$;

• (vi) $\{g(x_n)\}\subset X$ is a non-decreasing sequence with $g(x_n)\to gz$ in $g(X)$, then $g{x}_n⪯gz$, $gz⪯g(gz)$, ∀n hold.

Then f and g have a coincidence point. Further, if f and g are weakly compatible, then f and g have a common fixed point.

Proof Let $x_0\in X$ be such that $g{x}_0⪯f{x}_0$. Since $f(X)\subset g(X)$, we can choose $x_1\in X$ so that $g{x}_1=f{x}_0$. Again from $f(X)\subset g(X)$, we can choose $x_2\in X$ such that $g{x}_2=f{x}_1$. Continuing this process, we can construct a Jungck sequence $\{y_n\}$ in X such that

$$g{x}_{n+1}=f{x}_n=y_n,\mathrm{\forall }n\ge 0.$$

(3)

Since $g{x}_0⪯f{x}_0$ and $g{x}_1=f{x}_0$, we have $g{x}_0⪯g{x}_1$. Then by (1),

$$f{x}_0⪯f{x}_1.$$

(4)

Thus, by (3), $g{x}_1⪯g{x}_2$. Again by (1),

$$f{x}_1⪯f{x}_2,$$

(5)

that is, $g{x}_2⪯g{x}_3$. Continuing this process, we obtain

$$f{x}_0⪯f{x}_1⪯f{x}_2⪯f{x}_3⪯\cdots ⪯f{x}_n⪯f{x}_{n+1}.$$

(6)

Let $O(y_k,n)=\{y_k,y_{k+1},\dots ,y_{k+n}\}$. We will claim that $\{y_n\}$ is a Cauchy sequence. To prove our claim, we follow the arguments of Das and Naik [20]. Fix $k\ge 0$ and $n\in \{1,2,\dots \}$. If $diam[O(y_k;n)]=0$, then $y_k=y_{k+1}$, which yields that $\{y_n\}$ is a constant sequence and also a Cauchy sequence. Then our claims holds. Thus we suppose that $diam[O(y_k;n)]>0$. Now, for i, j with $1\le i<j$ , by (2), we have

and so

$$\psi (d(y_i,y_j))\le \alpha \psi (diam[O(y_{i-1};j-i+1)]).$$

(7)

Now, for some i, j with $k\le i<j\le k+n$, $diam[O(y_k;n)]=d(y_i,y_j)$. If $i>k$ by (2) and (7), then we have

$$\begin{array}{rcl}\psi (diam[O(y_k;n)])& \le & \alpha \psi (diam[O(y_{i-1};j-i+1)])\\ \le & \alpha \psi (diam[O(y_k;n)]).\end{array}$$

(8)

It follows that $\psi (diam[O(y_k;n)])=0$, as $diam[O(y_k;n)]\le \psi (diam[O(y_k;n)])=0$, then $diam[O(y_k;n)]=0$. It is a contradiction! Thus,

$$diam[O(y_k;n)]=d(y_k,y_j)\text{for }j\text{ with }k<j\le k+n.$$

(9)

Also, by (7) and (9), we have

$$\begin{array}{rcl}\psi (diam[O(y_k;n)])& =& \psi (d(y_k,y_j))\\ \le & \alpha \psi (diam[O(y_{k-1};j-k+1)])\\ \le & \alpha \psi (diam[O(y_{k-1};n+1)]).\end{array}$$

(10)

Using the triangle inequality, by (7), (9) and (10), we obtain that

$$\begin{array}{rcl}\psi (diam[O(y_l;m)])& =& \psi (d(y_l,y_j))\\ \le & \psi (d(y_l,y_{l+1})+d(y_{l+1},y_j))\\ \le & \psi (d(y_l,y_{l+1}))+\psi (d(y_{l+1},y_j))\\ \le & \psi (d(y_l,y_{l+1}))+\alpha \psi (diam[O(y_{l+1};m-1)])\\ \le & \psi (d(y_l,y_{l+1}))+\alpha \psi (diam[O(y_l;m)]),\end{array}$$

(11)

and so

$$\psi (diam[O(y_l;m)])\le \frac{1}{1-\alpha }\psi (d(y_l,y_{l+1})).$$

(12)

As a result, we have

$$\begin{array}{rcl}\psi (diam[O(y_k;n)])& \le & \alpha \psi (diam[O(y_{k-1};n+1)])\\ \le & \alpha \cdot \alpha \psi (diam[O(y_{k-2};n+2)])\\ \le & {\alpha }^k\psi (diam[O(y_0;n+k)])\\ \le & \frac{{\alpha }^k}{1-\alpha }\psi (d(y_0,y_1)).\end{array}$$

(13)

Now let $ϵ>0$, there exists an integer $n_0$ such that

$${\alpha }^k\psi (d(y_0,y_1))<(1-\alpha )ϵ\text{for all }k>n_0.$$

(14)

For $m>n>n_0$, we have

$$\begin{array}{rcl}\psi (d(y_m,y_n))& \le & \psi (diam[O(y_{n_0};m-n_0)])\\ \le & \frac{{\alpha }^{n_0}}{1-\alpha }\psi (d(y_0,y_1))\\ <& ϵ.\end{array}$$

(15)

Since $\psi (t)\ge t$ as $t>0$, then $d(y_m,y_n)\le \psi (d(y_m,y_n))<ϵ$. Therefore, $\{y_n\}$ is a Cauchy sequence.

Since by (3) we have $\{f{x}_n=g{x}_{n+1}\}\subseteq g(X)$ and $g(X)$ is closed, then there exists $z\in X$ such that

$$\underset{n\to \mathrm{\infty }}{lim}g{x}_n=gz.$$

(16)

Now we show that z is a coincidence point of f and g. Since from condition (iv) and (9) we have $g{x}_n⪯gz$ for all n, then by the triangle inequality and (2), we have that

$$\begin{array}{rcl}\psi (d(fz,gz))& \le & \psi (d(gz,f{x}_n)+d(f{x}_n,fz))\\ \le & \psi (d(gz,f{x}_n))+\psi (d(f{x}_n,fz))\\ \le & \psi (d(gz,f{x}_n))+\alpha max\{\psi (d(g{x}_n,gz)),\psi (d(g{x}_n,f{x}_n)),\\ \psi (d(gz,fz)),\psi (d(g{x}_n,fz)),\psi (d(gz,f{x}_n))\}.\end{array}$$

(17)

So, letting $n\to \mathrm{\infty }$ yields $\psi (d(fz,gz))\le \alpha \psi (d(fz,gz))$. Hence $\psi (d(fz,gz))=0$, hence $d(fz,gz)=0$, which yields $fz=gz$. Thus we have proved that f and g have a coincidence point.

Suppose now that f and g commute at z. Set $w=fz=gz$. Then

$$fw=f(gz)=g(fz)=gw.$$

(18)

Since from (vi) we have that $gz⪯g(gz)=gw$ and as $fz=gz$ and $fw=gw$, from (2) we have that

$$\begin{array}{rcl}\psi (d(fz,fw))& \le & \alpha max\{\psi (d(gz,gw)),\psi (d(gz,fz)),\psi (d(gw,fw)),\\ \psi (d(gz,fw)),\psi (d(gw,fz))\}\\ =& \alpha \psi (d(gz,gw)).\end{array}$$

(19)

Hence, $\psi (d(fz,fw))=0$, that is, $d(w,fw)=0$. Therefore,

$$fw=gw=w.$$

(20)

Thus, we have proved that f and g have a common fixed point. □

Accordingly, we can also obtain the results similar to Theorem 2 in [16].

Theorem 2.2 Let the conditions of Theorem  2.1 be satisfied, except that (iii), (v) and (vi) are, respectively, replaced by:

• (iii′) f is a g-nonincreasing mapping;

• (v′) there exists $x_0\in X$ such that $f{x}_0$ and $g{x}_0$ are comparable;

• (vi′) if $\{g{x}_n\}$ is a sequence in $g(X)$ which has comparable adjacent terms and that converges to some $gz\in gX$, then there exists a subsequence $g{x}_{n_k}$ of $\{g{x}_n\}$ having all the terms comparable with gz and gz is comparable with $ggz$. Then all the conclusions of Theorem  2.1 hold.

Proof Regardless of whether $f{x}_0⪯g{x}_0$ or $g{x}_0⪯f{x}_0$ (condition (v′)), Lemma 1 of [16] implies that the adjacent terms of the Jungck sequence $\{y_n\}$ are comparable. This is again sufficient to imply that $\{y_n\}$ is a Cauchy sequence. Hence, it converges to some $gz\in gX$.

By (vi′), there exists a subsequence $y_{n_k}=f{x}_{n_k}=g{x}_{n_k+1}$, $k\in \mathbb{N}$, having all the terms comparable with gz. Hence, we can apply the contractive condition to obtain

$$\begin{array}{rcl}\psi (d(fz,gz))& \le & \psi (d(gz,f{x}_{n_k}))+\psi (d(fz,f{x}_{n_k}))\\ \le & \psi (d(gz,f{x}_{n_k}))+\alpha max\{\psi (d(gz,g{x}_{n_k})),\psi (d(gz,fz)),\\ \psi (d(g{x}_{n_k},f{x}_{n_k})),\psi (d(gz,f{x}_{n_k})),\psi (d(g{x}_{n_k},fz))\}.\end{array}$$

Letting $k\to \mathrm{\infty }$, it yields that $\psi (d(fz,gz))\le \alpha \psi (d(gz,fz))$, then $\psi (d(fz,gz))=0$. Thus $d(fz,gz)=0$. It follows that $fz=gz=w$. The rest of conclusions follow in the same way as in Theorem 2.1. □

Corollary 2.1 (a) Let $(X,⪯)$ be a partially ordered set and suppose there is a metric d on X such that $(X,d)$ is a complete metric space. Let $f:X\to X$ be a nondecreasing self-map such that for some $\alpha \in (0,1)$

$$d(fx,fy)\le \alpha max\{d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)\}$$

for all $x,y\in X$ for which $x⪰y$. Suppose also that either

• (i) $\{x_n\}\subset X$ is a non-decreasing sequence with $x_n\to u$ in X, then $x_n⪯u$, ∀n hold, or

• (ii) f is continuous.

If there exists an $x_0\in X$ with $x_0⪯f{x}_0$, then f has a fixed point.

(b) The same holds if f is nonincreasing, there exists $x_0$ comparable with $f{x}_0$ and (i) is replaced by

(i′) if a sequence $\{x_n\}$ converging to some $u\in X$ has every two adjacent terms comparable, then there exists a subsequence $\{x_{n_k}\}$ having each term comparable with x.

Proof (a) If (i) holds, then take $\psi =I$ and $g=I$ (I= the identity mapping) in Theorem 2.1.

If (ii) holds, then from (16) with $g=I$, we get

$$z=\underset{n\to \mathrm{\infty }}{lim}{x}_{n+1}=\underset{n\to \mathrm{\infty }}{lim}f{x}_n=f\left(\underset{n\to \mathrm{\infty }}{lim}{x}_n\right)=fz.$$

(21)

(b) Let u be the limit of the Picard sequence $\{f^n{x}_0\}$ and let $f^{n_k}{x}_0$ be a subsequence having all the terms comparable with u. Then we can apply the contractivity condition to obtain

$$\begin{array}{rcl}\psi (d(fu,u))& \le & \psi (d(u,f^{n_k+1}{x}_0)+d(fu,f^{n_k+1}{x}_0))\\ \le & \psi \left(d(u,f^{n_k+1}{x}_0)\right)+\psi \left(d(fu,f^{n_k+1}{x}_0)\right)\\ \le & \psi \left(d(u,f^{n_k+1}{x}_0)\right)+\alpha max\{\psi \left(d(u,f^{n_k}{x}_0)\right),\psi (d(u,fu)),\\ \psi \left(d(f^{n_k}{x}_0,f^{n_k+1}{x}_0)\right),\psi \left(d(u,f^{n_k+1}{x}_0)\right),\psi \left(d(fu,f^{n_k}{x}_0)\right)\}.\end{array}$$

Letting $k\to \mathrm{\infty }$, we have that

$$\begin{array}{rcl}\psi (d(fu,u))& \le & \alpha max\{0,\psi (d(u,fu)),0,0,\psi (d(u,fu))\}\\ =& \alpha \psi (d(u,fu)).\end{array}$$

It follows that $\psi (d(fu,u))=0$. Thus $d(fu,u)=0$ as $d(fu,u)\le \psi (d(fu,u))=0$. Therefore, $fu=u$.

Note also that instead of the completeness of X, its f-orbitally completeness is sufficient to obtain the conclusion of the corollary. □

3 Uniqueness of a common fixed point of f and g

The following theorem gives the sufficient condition for the uniqueness of a common fixed point of f and g.

Theorem 3.1 In addition to the hypotheses of Theorem  2.1, suppose that for all $x,u\in X$, there exists $a\in X$ such that

$$fa\mathit{\text{is comparable to}}fx\mathit{\text{and}}fu.$$

(22)

Then f and g have a unique common fixed point.

Proof Since a set of common fixed points of f and g is not empty due to Theorem 2.1, assume now that x and u are two common fixed points of f and g, i.e.,

$$fx=gx=x,fu=gu=u.$$

(23)

We claim that $gx=gu$.

By assumption, there exists $a\in X$ such that fa is comparable to fx and fu. Define a sequence $\{g{a}_n\}$ such that $a_0=a$ and

$$g{a}_n=f{a}_{n-1}\text{for all }n.$$

(24)

Further, set $x_0=x$ and $u_0=u$ and in the same way define $\{g{x}_n\}$ and $\{g{u}_n\}$ such that

$$g{x}_n=f{x}_{n-1},g{u}_n=f{u}_{n-1}\text{for all }n.$$

(25)

Since fx ($=g{x}_1=gx$) is comparable to fa ($=f{a}_0=g{a}_1$) and f is g-nondecreasing, it is easy to show

$$gx⪰g{a}_1.$$

(26)

Recursively, we can get that

$$g{a}_n⪯gx\text{for all }n.$$

(27)

By (27), we have that

$$\begin{array}{rcl}\psi (d(g{a}_{n+1},gx))& =& \psi (d(f{a}_n,fx))\\ \le & \alpha max\{\psi (d(g{a}_n,gx)),\psi (d(g{a}_n,f{a}_n)),\psi (d(gx,fx)),\\ \psi (d(g{a}_n,fx)),\psi (d(gx,f{a}_n))\}.\end{array}$$

(28)

By the proof of Theorem 2.1, we obtain that $\{g{a}_n\}$ is a convergent sequence, and there exists $g\overline{a}$ such that $g{a}_n\to g\overline{a}$. Letting $n\to \mathrm{\infty }$ in (28) and ψ is continuous, we can obtain that

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}\psi (d(g{a}_{n+1},gx))& =& \psi (d(g\overline{a},gx))\\ \le & \alpha max\{\psi (d(g\overline{a},gx)),0,0,\psi (d(g\overline{a},fx)),\psi (d(gx,g\overline{a}))\}\\ =& \alpha \psi (d(g\overline{a},gx)).\end{array}$$

Therefore, we obtain

$$\psi (d(g\overline{a},gx))=0.$$

Since $\psi (t)\ge t$ as $t\ge 0$, then $d(g\overline{a},gx)=0$ and hence

$$g\overline{a}=gx.$$

(29)

Similarly, we can show that

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}\psi (d(g{a}_{n+1},gu))& =& \psi (d(g\overline{a},gu))\\ \le & \alpha max\{\psi (d(g\overline{a},gu)),0,0,\psi (d(g\overline{a},fu)),\psi (d(gu,g\overline{a}))\}\\ =& \alpha \psi (d(g\overline{a},gu)).\end{array}$$

Therefore, we obtain

$$\psi (d(g\overline{a},gu))=0.$$

Since $\psi (t)\ge t$ as $t\ge 0$, then $d(g\overline{a},gu)=0$ and hence

$$g\overline{a}=gu.$$

(30)

Thus, from (29) and (30), we have $gx=gu$. It follows that

$$x=fx=gx=gu=fu=u.$$

(31)

It means that x is the unique common fixed point of f and g. □

Remark 3.1 Theorem 3.1 can be considered as an answer to Theorem 3 in [16]. We find the sufficient conditions for the uniqueness of the common fixed point in the case of an ordered g-quasicontraction. In this paper, condition (vi) in Theorem 2.1 is weaker than the ordered g-quasicontraction in [16]. When $\psi =I$ (I= the identity mapping), our condition (vi) reduces to the ordered g-quasicontraction in [16].

Example 3.1 Let $X=\{(0,2),(2,3)\}$, let $(a,b)⪯(c,d)$ if and only if $a\le c$ and $b\ge d$, and let d be the Euclidean metric. We define the functions as follows:

$$f((x,y))=(x^2,5y-8),g((x,y))=(2x,y^2-2)\text{for all }(x,y)\in X.$$

Let $\varphi ,\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ be given by

$$\psi (t)=\frac{2}{5}t\text{for all }t\in [0,\mathrm{\infty }).$$

Obviously, for $(0,2)$ and $(2,3)\in X$, but $f((0,2))=(0,2)$ is not comparable to $g((2,3))=(2,3)$. However, f and g have two common fixed points $(0,2)$ and $(2,3)$ since

$$f((0,2))=g(0,2)=(0,2),f((2,3))=g((2,3))=(2,3).$$

Example 3.2 Let $X=[-\mathrm{\infty },+\mathrm{\infty })$ with the usual metric $d(x,y)=|x-y|$ for all $x,y\in X$. Let $f:X\to X$ and $g:X\to X$ be given by

$$f(x)=\frac{x}{16},g(x)=\frac{3}{4}x$$

for all $x,y,z,w\in X$. Let $\varphi ,\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ be given by

$$\psi (t)=3t\text{for all }t\in [0,\mathrm{\infty }).$$

It is easy to check that all the conditions of Theorem 2.1 are satisfied.

$$\begin{array}{rcl}\psi (d(fx,fy))& =& \frac{3}{16}|x-y|\\ \le & 3\cdot \alpha \cdot \frac{3}{4}|x-y|\\ \le & max\{3\cdot \frac{3}{4}|x-y|,3\cdot |\frac{3}{4}x-\frac{x}{16}|,3\cdot |\frac{3}{4}y-\frac{y}{16}|,\\ 3\cdot |\frac{3}{4}x-\frac{y}{16}|,3\cdot |\frac{3}{4}y-\frac{x}{16}|\}\\ =& max\{\psi (d(gx,gy)),\psi (d(gx,fx)),\psi (d(gy,fy)),\psi (d(gx,fy)),\psi (d(gy,fx))\}.\end{array}$$

It holds when $\alpha =\frac{1}{12}$ and $gx\ge gy$, i.e., $\frac{3}{4}x\ge \frac{3}{4}y$, i.e., $x\ge y$.

In addition, $\mathrm{\forall }x,u\in X$, there exists $a\in X$ such that $fa=\frac{a}{16}$ is comparable to $fx=\frac{x}{16}$ and $fu=\frac{u}{16}$. So, all the conditions of Theorem 3.1 are satisfied.

By applying Theorem 3.1, we conclude that f and g have a unique common fixed point. In fact, f and g have only one common fixed point. It is $x=0$.

4 Weak ordered contractions

We denote by Ψ the set of functions $\psi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ satisfying the following hypotheses:

(${\psi }_1$) ψ is continuous and nondecreasing,

(${\psi }_2$) $\psi (t)=0$ if and only if $t=0$.

We denote by Φ the set of functions $\varphi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ satisfying the following hypotheses:

(${\varphi }_1$) ${lim}_{s\to t+}\varphi (s)>0$ for all $t>0$,

(${\varphi }_2$) $\varphi (t)=0$ if and only if $t=0$.

Let $(X,d)$ be a metric space and let $f,g:X\to X$. In the article [16] (in the setting of partially ordered metric spaces), the authors obtained contractive conditions of the form

$$\psi (d(fx,fy))\le \psi (M(x,y))-\varphi (M(x,y)),$$

(32)

where

$$M(x;y)=max\{d(gx,gy),d(gx,fx),d(gy,fy),\frac{d(gx,fy)+d(gy,fx)}{2}\}.$$

(33)

We will use here the following more general contractive condition:

$$M(x,y)=max\{d(gx,gy),d(gx,fx),d(gy,fy),d(gx,fy),d(gy,fx)\}.$$

(34)

We begin with the following result.

Theorem 4.1 Let $(X,d,⪯)$ be a partially ordered metric space and let f and g be self-mappings of X satisfying the following conditions:

• (i) $f(X)\subset g(X)$;

• (ii) $g(X)$ is complete;

• (iii) f is g-nondecreasing;

• (iv) f and g satisfy the following condition:

  $$\psi (d(fx,fy))\le \psi (M(x,y))-\varphi (M(x,y))$$

  (35)

  for all $x,y\in X$ such that $gy⪯gx$, where $\psi \in \mathrm{\Psi }$, $\varphi \in \mathrm{\Phi }$ and

  $$M(x,y)=max\{d(gx,gy),d(gx,fx),d(gy,fy),d(gx,fy),d(gy,fx)\}.$$

  (36)

  Suppose that, in addition,

• (v) $\psi (t)-\varphi (t)$ is nondecreasing;

• (vi) $\psi (s+t)\le \psi (s)+\psi (t)$ for each $s,t>0$;

• (vii) ${lim}_{t\to +\mathrm{\infty }}\varphi (t)=\mathrm{\infty }$;

• (viii) there exists $x_0\in X$ such that $g{x}_0⪯f{x}_0$;

• (ix) if $\{g{x}_n\}$ is a nondecreasing sequence that converges to some $gz\in gX$, then $g{x}_n⪯gz$ for each $n\in N$ and $gz⪯g(gz)$.

  Then f and g have a coincidence point. If, in addition,

• (x) f and g are weakly compatible, then they have a common fixed point.

  Further, if

• (xi) for arbitrary $v,w\in X$, there exists $y_0\in X$ such that $f{y}_0$ is comparable to fv and fw, then f and g have a unique common fixed point.

Proof As in the proof of Theorem 2.1, we can construct a nondecreasing Jungck sequence $\{y_n\}$ with

$$y_n=f{x}_n=g{x}_{n+1}$$

for all $n\ge 0$. Denote

(37)

(38)

We will prove that the Jungck sequence $\{y_n\}$ is bounded, that is,

$$diam(O(y_0))=diam(\{y_0,y_1,y_2,\dots ,y_n,\dots \})\le K$$

(39)

for some $K\in \mathbb{R}$. Let $k<n$ be any fixed positive integer and let $diam(O(y_k,n))=d(y_i,y_j)$ for some i, j with $k\le i<j\le k+n$. We will show that

$$\psi (diam(O(y_k,n)))\le \psi (diam(O(y_{i-1},j-i+1)))-\varphi (diam(O(y_{i-1},j-i+1))).$$

(40)

Since $diam(O(y_k,n))=d(y_i,y_j)$, $y_i=f{x}_i$, $y_j=f{x}_j$ and $g{x}_i⪯g{x}_j$, then from (35) we have

$$\psi (diam(O(y_k,n)))=\psi (d(f{x}_i,f{x}_j))\le \psi (M(x_i,x_j))-\varphi (M(x_i,x_j)),$$

(41)

where

$$\begin{array}{rcl}M(x_i,x_j)& =& max\{d(g{x}_i,g{x}_j),d(g{x}_i,f{x}_i),d(g{x}_j,f{x}_j),d(g{x}_i,f{x}_j),d(g{x}_j,f{x}_i)\}\\ =& max\{d(y_{i-1},y_{j-1}),d(y_{i-1},y_i),d(y_{j-1},y_j),d(y_{i-1},y_j),d(y_{j-1},y_i)\}.\end{array}$$

Since $y_{i-1},y_i,y_{j-1},y_j\in O(y_{i-1},j-i+1)$, then

$$M(x_i,x_j)\le diam(\{y_{i-1},y_i,y_{j-1},y_j\})\le diam(O(y_{i-1},j-i+1)).$$

So, from (v),

$$\psi (M(x_i,x_j))-\varphi (M(x_i,x_j))\le \psi (diam(O(y_{i-1},j-i+1)))-\varphi (diam(O(y_{i-1},j-i+1))).$$

Hence from (41) we obtain (40).

Note that $\varphi (diam(O(y_{i-1},j-i+1)))>0$, and so from (40),

$$diam(O(y_k,n))<diam(O(y_{i-1},j-i+1)).$$

(42)

Now we will show that if $diam(O(y_k,n))=d(y_i,y_j)$, then $i=k$, that is,

$$diam(O(y_k,n))=d(y_k,y_j)\text{for some }k<j\le k+n.$$

(43)

Suppose, to the contrary, that $i>k$. Then $\{y_{i-1},y_i,\dots ,y_j\}\subseteq \{y_k,y_{k+1},\dots ,y_i,\dots ,y_j\}$ and hence we conclude that

$$\begin{array}{rcl}diam(O(y_k,n))& =& d(y_i,y_j)=diam(O(y_{i-1},j-i+1))\\ =& diam(O(y_i,j-i))=diam(O(y_k,j-k)).\end{array}$$

This contradicts (42). Therefore, $i=k$ and so we have proved (43).

We will prove that the Jungck sequence $\{y_n\}$ is bounded. From (43) it follows that $diam(O(y_0,n))=d(y_0,y_j)$ for some $y_j\in \{y_1,y_2,\dots ,y_n\}$. By the triangle inequality,

$$diam(O(y_0,n))=d(y_0,y_j)\le d(y_0,y_1)+d(y_1,y_j).$$

Now, from (${\psi }_1$) and (${\psi }_3$), we get

$$\begin{array}{rcl}\psi (diam(O(y_0,n)))& \le & \psi [d(y_0,y_1)+d(y_1,y_j)]\\ \le & \psi (d(y_0,y_1))+\psi (d(y_1,y_j)).\end{array}$$

(44)

Since $d(y_1,y_j)=d(f{x}_1,f{x}_j)$ and as $g{x}_1⪯g{x}_j$, from (35) we have

$$\psi (d(y_1,y_j))\le \psi (M(x_1,x_j))-\varphi (M(x_1,x_j)),$$

where

$$M(x_1,x_j)=max\{d(y_0,y_{j-1}),d(y_0,y_1),d(y_{j-1},y_j),d(y_0,y_j),d(y_{j-1},y_1)\}.$$

Clearly, $M(x_1,x_j)\le diam\{y_0,y_1,y_{j-1},y_j\}\le diam(O(y_0,n))$. Thus by (v), we get

$$\psi (M(x_1,x_j))-\varphi (M(x_1,x_j))\le \psi (diam(O(y_0,n)))-\varphi (diam(O(y_0,n))).$$

Now, by (44),

$$\psi (diam(O(y_0,n)))\le \psi (d(y_0,y_1))+\psi (diam(O(y_0,n)))-\varphi (diam(O(y_0,n))).$$

Hence

$$\varphi (diam(O(y_0,n)))\le \psi (d(y_0,y_1)).$$

(45)

Since $diam(\{y_0,y_1,\dots ,y_n\})\le diam(\{y_0,y_1,\dots ,y_{n+1}\})$, the sequence ${\{diam(O(y_0,n))\}}_{n=1}^{\mathrm{\infty }}$ is nondecreasing, and so there exists its limit $diam(O(y_0))$, which is finite or infinite. Suppose that ${lim}_{n\to \mathrm{\infty }}diam(O(y_0,n))=+\mathrm{\infty }$. Then (vii) implies that the left-hand side of (45) becomes unbounded when n tends to infinity, but the right-hand side is bounded, a contradiction. Therefore, ${lim}_{n\to \mathrm{\infty }}diam(O(y_0,n))=diam(O(y_0))<+\mathrm{\infty }$. Thus we have proved (39).

Now we show that $\{y_n\}$ is a Cauchy sequence. For all $n\ge 1$, set similarly as in (38),

$$O(y_n)=\{y_n,y_{n+1},\dots \}.$$

Clearly, $O(y_{n+1})\subset O(y_n)$ and so $diam(O(y_{n+1}))\le diam(O(y_n))$. Therefore, ${\{diam(O(y_n))\}}_{n=0}^{\mathrm{\infty }}$ is the monotone decreasing sequence of finite nonnegative numbers and converges to some $\delta \ge 0$.

We will prove that $\delta =0$. Let $n\ge 1$ and $s\ge n+2$. Since $g{x}_{n+1}⪯g{x}_s$, from (35),

$$\psi (d(y_{n+1},y_s))=\psi (d(f{x}_{n+1},f{x}_s))\le \psi (M(x_{n+1},x_s))-\varphi (M(x_{n+1},x_s)),$$

where

$$M(x_{n+1},x_s)=max\{d(y_n,y_{s-1}),d(y_n,y_{n+1}),d(y_{s-1},y_s),d(y_n,y_s),d(y_{s-1},y_{n+1})\}.$$

Since $y_n,y_{n+1},y_{s-1},y_s\in \{y_n,y_{n+1},\dots \}=O(y_n)$, we conclude that $M(x_{n+1},x_s)\le diam(O(y_n))$, and so by (v), we get

$$\psi (d(y_{n+1},y_s))\le \psi (diam(O(y_n)))-\varphi (diam(O(y_n))).$$

(46)

Since ${lim}_{s\to +\mathrm{\infty }}d(y_{n+1},y_s)=diam(O(y_{n+1}))$ and ψ is continuous, we have ${lim}_{s\to +\mathrm{\infty }}\psi (d(y_{n+1},y_s))=\psi (diam(O(y_{n+1})))$. Thus, taking the limit in (46) when $s\to +\mathrm{\infty }$, we get

$$\psi (diam(O(y_{n+1})))\le \psi (diam(O(y_n)))-\varphi (diam(O(y_n))).$$

(47)

Suppose that ${lim}_{n\to \mathrm{\infty }}diam(O(y_n))=\delta >0$. Since $diam(O(y_n))\to \delta +$ as $n\to \mathrm{\infty }$, then from (${\varphi }_1$), we have ${lim}_{n\to \mathrm{\infty }}\varphi (diam(O(y_n)))=q>0$. Therefore, taking the limits as $n\to +\mathrm{\infty }$ in (47) and using the continuity of ψ, we get

$$\psi (\delta )\le \psi (\delta )-q<\psi (\delta ),$$

a contradiction. Therefore, $\delta =0$ and so we have proved that

$$\underset{n\to \mathrm{\infty }}{lim}diam(\{y_n,y_{n+1},\dots \})=0.$$

Hence we conclude that $\{y_n\}$ is a Cauchy sequence.

Since $y_n=f{x}_n=g{x}_{n+1}$, by the assumption (ii) that $g(X)$ is complete, there is some $z\in X$ such that

$$\underset{n\to \mathrm{\infty }}{lim}g{x}_n=gz.$$

We show that $fz=gz$. Suppose, to the contrary, that $d(fz,gz)>0$. Condition (ix) implies that $g{x}_n⪯gz$ and we can apply the contractive condition (35) to obtain

$$\psi (d(fz,f{x}_{n+1}))\le \psi (M(z,x_{n+1}))-\varphi (M(z,x_{n+1})),$$

(48)

where

$$\begin{array}{rcl}M(z,x_{n+1})& =& max\{d(gz,g{x}_{n+1}),d(gz,fz),d(g{x}_{n+1},f{x}_{n+1}),d(gz,f{x}_{n+1}),d(g{x}_{n+1},fz)\}\\ =& max\{d(gz,f{x}_n),d(gz,fz),d(f{x}_n,f{x}_{n+1}),d(gz,f{x}_{n+1}),d(f{x}_n,fz)\}.\end{array}$$

By the triangle inequality,

$$d(gz,fz)\le d(gz,f{x}_{n+1})+d(fz,f{x}_{n+1}).$$

Now, from (${\psi }_1$) and (${\psi }_3$),

$$\begin{array}{rcl}\psi (d(gz,fz))& \le & \psi [d(gz,f{x}_{n+1})+d(fz,f{x}_{n+1})]\\ \le & \psi (d(gz,f{x}_{n+1}))+\psi (d(fz,f{x}_{n+1})).\end{array}$$

Hence from (48) we have

$$\psi (d(gz,fz))\le \psi (d(gz,f{x}_{n+1}))+\psi (M(z,x_{n+1}))-\varphi (M(z,x_{n+1})).$$

(49)

Since ${lim}_{n\to \mathrm{\infty }}f{x}_n=gz$, for large enough n, we have

$$M(z,x_{n+1})=max\{d(gz,fz),d(f{x}_n,fz)\}.$$

If $M(z,x_{n+1})=d(gz,fz)$, then from (49)

$$\psi (d(gz,fz))\le \psi (d(gz,f{x}_{n+1}))+\psi (d(gz,fz))-\varphi (d(gz,fz)).$$

Letting n tend to infinity and using the continuity of ψ, we get

$$\psi (d(gz,fz))\le \psi (d(gz,fz))-\varphi (d(gz,fz)).$$

Hence $\varphi (d(gz,fz))=0$, a contradiction with (${\varphi }_2$) and the assumption $d(gz,fz)>0$.

Similarly, if $M(z,x_{n+1})=d(f{x}_n,fz)$, then from (48)

$$\psi (d(gz,fz))\le \psi (d(gz,f{x}_n))+\psi (d(f{x}_n,fz))-\varphi (d(f{x}_n,fz)).$$

Letting n tend to infinity and having in mind that $d(f{x}_n,fz)\to d(gz,fz)+$, we obtain

$$\psi (d(gz,fz))\le \psi (d(gz,fz))-\underset{d(f{x}_n,fz)\to d(gz,fz)+}{lim}\varphi (d(f{x}_n,fz))$$

and hence we get

$$\underset{d(f{x}_n,fz)\to d(gz,fz)+}{lim}\varphi (d(f{x}_n,fz))\le 0,$$

a contradiction with (${\varphi }_1$). Thus our assumption $d(gz,fz)>0$ is wrong. Therefore, $d(gz,fz)=0$. Hence $gz=fz$, that is, z is a coincidence point of f and g.

If the condition (x) is fulfilled, put $w=fz=gz$. We will show that w is a common fixed point of f and g. Since $fz=gz$ and f and g are weakly compatible, we obtain, by the definition of weak compatibility, that $fgz=gfz$. Thus we have $fw=gw$. Using the condition (ix) that $gz⪯ggz=gw$, we can apply the contractive condition (35) to obtain

$$\psi (d(fw,fz))\le \psi (M(w,z))-\varphi (M(w,z)),$$

where

$$M(w,z)=max\{d(gw,gz),d(gw,fw),d(gz,fz),d(gw,fz),d(gz,fw)\}=d(fw,fz).$$

Thus

$$\psi (d(fw,fz))\le \psi (d(fw,fz))-\varphi (d(fw,fz)).$$

Hence $\varphi (d(fw,fz))=0$, and so by (${\varphi }_2$), $d(fw,fz)=0$. Hence $fw=fz$. Therefore

$$w=fz=fw=ffz=gfz=gw.$$

Thus we showed that w is a common fixed point of f and g.

Suppose now that the condition (xi) is fulfilled. Since a set of common fixed points of f and g is not empty, assume that w and v are two common fixed points of f and g, i.e.,

$$fw=gw=w,fv=gv=v.$$

(50)

We claim that $gw=gv$.

By assumption, there exists $y_0\in X$ such that $f{y}_0$ is comparable to fw and fv. Define a sequence $\{g{y}_n\}$ such that

$$g{y}_n=f{y}_{n-1}\text{for all }n.$$

(51)

Further, set $w_0=w$ and $v_0=v$ and, in the same way, define $\{g{w}_n\}$ and $\{g{v}_n\}$ such that

$$g{w}_n=f{w}_{n-1},g{v}_n=f{v}_{n-1}\text{for all }n.$$

(52)

From (50) and (52), we have $f{w}_0=g{w}_1=g{w}_0$ and $f{v}_0=g{v}_1=g{v}_0$. Since $f{y}_0$ is comparable to fw and fv, and f is g-nondecreasing, it is easy to show

$$gw⪰g{y}_1.$$

(53)

Recursively, we can get that

$$g{y}_n⪯gw\text{for all }n.$$

(54)

By (35), we have that

(55)

Similarly as in the proof of Theorem 2.1, we can prove that $\{g{y}_n\}$ is a convergent sequence. Thus there exists $\overline{y}\in X$ such that $g{y}_n\to g\overline{y}$. Since also ${lim}_{n\to \mathrm{\infty }}f{y}_n=g\overline{y}$, for large enough n, we have

$$max\{d(g{y}_n,gw),d(g{y}_n,f{y}_n),d(gw,fw),d(g{y}_n,fw),d(gw,f{y}_n)\}=d(g\overline{y},gw).$$

Thus from (55), for large enough n,

$$\psi (d(g{y}_{n+1},gw))\le \psi (d(g\overline{y},gw))-\varphi (d(g\overline{y},gw)).$$

(56)

Letting $n\to \mathrm{\infty }$ in (56), by (${\psi }_1$) we get

$$\underset{n\to \mathrm{\infty }}{lim}\psi (d(g{y}_{n+1},gw))=\psi (d(g\overline{y},gw))\le \psi (d(g\overline{y},gw))-\varphi (d(g\overline{y},gw)).$$

Hence we obtain

$$\psi (d(g\overline{y},gw))=0.$$

Then by (${\psi }_2$), $d(g\overline{y},gw)=0$ and hence

$$g\overline{y}=gw.$$

(57)

Similarly, we can show that

$$\underset{n\to \mathrm{\infty }}{lim}\psi (d(g{y}_{n+1},gv))=\psi (d(g\overline{y},gv))\le \psi (d(g\overline{y},gv))-\varphi (d(g\overline{y},gv)),$$

and hence we obtain

$$g\overline{y}=gv.$$

(58)

Therefore, from (57) and (58), we have $gw=gv$. It follows that

$$w=fw=gw=gv=fv=v.$$

(59)

It means that w is the unique common fixed point of f and g. □

Corollary 4.1 Let $(X,d,⪯)$ be a complete partially ordered metric space and let f be a self-mapping of X satisfying the following condition:

$$d(fx,fy)\le m(x,y)-\varphi (m(x,y))$$

for all $x,y\in X$ such that $gy⪯gx$, where

$$m(x,y)=max\{d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)\}$$

and $\varphi \in \mathrm{\Phi }$. Suppose that, in addition, $t-\varphi (t)$ is non-decreasing, ${lim}_{t\to +\mathrm{\infty }}\varphi (t)=\mathrm{\infty }$, there exists $x_0\in X$ such that $x_0⪯f{x}_0$ and if $\{f{x}_n\}$ is a nondecreasing sequence such that it converges to some $z\in X$, then $f{x}_n⪯z$. Then f has a unique fixed point.

Proof Taking $\psi (t)=t$ and $g(t)=t$ in the proof of Theorem 4.1, we obtain Corollary 4.1. □

Remark 4.1 Theorem 4.1 extends Theorem 1 due to Berinde [21], Theorems 2.1 and 2.5 due to Beg and Abbas [22] and Theorem 3.1 due to Song [23].

We present an example to show that our result is a real generalization of the recent result of Golubović et al. [16] as well as of the existing results in the literature.

Example 4.1 Let $X=[0,\frac{1}{2}]$ be the closed interval with the usual metric and let $f,g:X\to X$ and $\psi ,\varphi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ be mappings defined as follows:

Let x, y in X be arbitrary. We say that $x⪯y$ if $x\le y$. For any $x,y\in X$ such that $x⪯y$, we have

Since $y^2\ge y^2-x^2(1-x^2)$ for all $x\in [0,\frac{1}{2}]$, it follows that

$$-y^4\le -{(y^2-x^2(1-x^2))}^2.$$

Thus we have

$$\begin{array}{rcl}\psi \left(d(f(x),f(y))\right)& =& |y^2-y^4-x^2+x^4|=(y^2-x^2(1-x^2))-y^4\\ \le & (y^2-x^2(1-x^2))-{(y^2-x^2(1-x^2))}^2\\ =& d(g(y),f(x))-{\left[d(g(y),f(x))\right]}^2\\ =& \psi (M(x,y))-\varphi (M(x,y)).\end{array}$$

Therefore, f and g satisfy (35). Also, it is easy to see that the mappings $\psi (t)$ and $\varphi (t)$ possess all properties (${\psi }_1$), (${\psi }_2$) and (${\varphi }_1$), (${\varphi }_2$) respectively, as well as hypotheses (v), (vi) and (vii) in Theorem 4.1. Thus we can apply our Theorem 4.1 and Corollary 4.1.

On the other hand, for $x=0$ and each $y>0$, the contractive condition in Theorems 1 and 2 of Golubović et al. [16]:

$$d(fx,fy)\le \lambda \cdot M(x,y),$$

(60)

where $0<\lambda <1$ and

$$M(x;y)=max\{d(gx;gy);d(gx;fx);d(gy;fy);d(gx;fy),d(gy;fx)\},$$

is not satisfied. Indeed,

$$\begin{array}{rcl}M(0;y)& =& max\{d(g(0);g(y));d(g(0);f(0));d(g(y);f(y));d(g(0);f(y)),d(g(y);f(0))\}\\ =& max\{y^2;0;y^4;(y^2-y^4),y^2\}=y^2.\end{array}$$

Thus, for any fixed λ; $0<\lambda <1$, we have, for $x=0$ and each $y\in X$ with $0<y<\sqrt{1-\lambda }$,

$$\begin{array}{rcl}d(f(0),f(y))& =& y^2-y^4=(1-y^2)y^2>\lambda \cdot y^2\\ =& \lambda \cdot d(g(y),g(0))=\lambda \cdot M(0,y).\end{array}$$

Thus, f does not satisfy (60). Therefore, the theorems of Jungck and Hussain [24], Al-Thagafi and Shahzad [25] and Das and Naik [26] also cannot be applied.