1 Introduction

In [1], Dass and Gupta proved the following fixed point theorem.

Theorem 1

Let \((X,d)\) be a complete metric space and \(T:X\rightarrow X\) a mapping such that there exist \(\alpha ,\beta \ge 0\) with \(\alpha +\beta <1\) satisfying

$$\begin{aligned} d(Tx,Ty)\le \frac{\alpha \, d(y,Ty)[1+d(x,Tx)]}{1+d(x,y)}+\beta \, d(x,y) \end{aligned}$$

for all \(x,y\in X\).

Then \(T\) has a unique fixed point.

The aim of this paper is to give a version of Theorem 1 in the context of partially ordered metric spaces.

Existence of fixed point in partially ordered metric spaces has been considered recently by many authors (see, [2–13], for example).

2 Main result

Definition 1

Let \((X,\le )\) be a partially ordered set and \(T:X\rightarrow X\). \(T\) is said to be a nondecreasing mapping if for \(x,y\in X\)

$$\begin{aligned} x\le y \Rightarrow Tx\le Ty. \end{aligned}$$

Theorem 2

Let \((X,\le )\) be a partially ordered set and suppose that there exists a metric \(d\) in \(X\) such that \((X,d)\) is a complete metric space. Let \(T:X\rightarrow X\) be a continuous and nondecreasing mapping such that there exists \(\alpha ,\beta \ge 0\) with \(\alpha +\beta <1\) satisfying

$$\begin{aligned} d(Tx,Ty)\le \frac{\alpha \, d(y,Ty)[1+d(x,Tx)]}{1+d(x,y)}+\beta \, d(x,y)\quad \text{ for} x,y\in X \text{ with} x\le y.\nonumber \\ \end{aligned}$$
(1)

If there exist \(x_0\in X\) such that \(x_0\le Tx_0\) then \(T\) has a fixed point.

Proof

If \(Tx_0=x_0\) then the proof is finished.

Suppose that \(x_0<Tx_0\). Since \(T\) is a nondecreasing mapping, by using induction, we obtain

$$\begin{aligned} x_0<Tx_0\le T^{2}x_0\le \ldots \le T^{n}x_0\le T^{n+1}x_0\le \ldots . \end{aligned}$$

Put \(x_{n+1}=Tx_n\).

If there exists \(n\ge 1\) such that \(x_{n+1}=x_n\) then \(x_{n+1}=T x_n=x_n\), and \(x_n\) is a fixed point of \(T\) and the proof is finished.

Suppose that \(x_{n+1}\not =x_n\) for \(n\ge 0\).

Since \(x_{n+1}\le x_n\) for any \(n\in \mathbb N \), using the contractive condition (1), we have for \(n\ge 1\)

$$\begin{aligned} d(x_n,x_{n+1})&= d(Tx_{n-1},Tx_n)\\&\le \frac{\alpha \, d(x_n,Tx_n)[1+d(x_{n-1},Tx_{n-1})]}{1+d(x_{n-1},x_n)}+\beta \, d(x_{n-1},x_n)\\&= \frac{\alpha \, d(x_n,x_{n+1})[1+d(x_{n-1},x_n)]}{1+d(x_{n-1},x_n)}+\beta \, d(x_{n-1},x_n)\,, \end{aligned}$$

and, the last inequality implies

$$\begin{aligned} (1-\alpha )\, d(x_n,x_{n+1})\le \beta \, d(x_{n-1},x_n)\quad \text{ for} \text{ any} n\in \mathbb N , \end{aligned}$$

or, equivalently,

$$\begin{aligned} d(x_n,x_{n+1})\le \frac{\beta }{1-\alpha } d(x_{n-1},x_n)\quad \text{ for} \text{ any} n\in \mathbb N . \end{aligned}$$

Using mathematical induction, we have

$$\begin{aligned} d(x_n,x_{n+1})\le \left(\frac{\beta }{1-\alpha }\right)^n d(x_{0},x_1)\quad \text{ for} \text{ any} n\in \mathbb N . \end{aligned}$$

Notice that \(\displaystyle r=\frac{\beta }{1-\alpha }<1\).

Moreover, for \(m>n\), we have

$$\begin{aligned} d(x_n,x_m)&\le d(x_n,x_{n+1})+\ldots +d(x_{m-1},x_m)\\&\le (r^n+\ldots +r^{m-1}) d(x_0,x_1)\\&< \frac{r^n}{1-r}d(x_0,x_1). \end{aligned}$$

Letting \(n,m\rightarrow \infty \) and, since \(r<1\), we obtain \(\displaystyle \lim _{n,m\rightarrow \infty }d(x_n,x_m)=0\).

This proves that \((x_n)\) is a Cauchy sequence.

Since \((X,d)\) is a complete metric space, \(\displaystyle \lim _{n\rightarrow \infty }x_n=x\) for certain \(x\in X\).

The continuity of \(T\) gives us

$$\begin{aligned} T x=T\left(\lim _{n\rightarrow \infty }x_n\right)=\lim _{n\rightarrow \infty }T x_n=\lim _{n\rightarrow \infty }x_{n+1}=x. \end{aligned}$$

Therefore, \(x\) is a fixed point.

This finishes the proof. \(\square \)

In the sequel, we will prove that Theorem 2 is still valid for \(T\) not necessarily continuous, assuming the following hypothesis in \(X\):

$$\begin{aligned} \text{ if} (x_n) \text{ is} \text{ a} \text{ nondecreasing} \text{ sequence} \text{ in} X \text{ such} \text{ that} x_n\rightarrow x \text{ then} x_n\le x \text{ for} \text{ all} n\in \mathbb N .\nonumber \\ \end{aligned}$$
(2)

Theorem 3

If in Theorem 2 we replace the condition of continuity of \(T\) by (2), the same conclusion holds.

Proof

In fact, following the proof of Theorem 2, we only have to check that \(x\) is a fixed point.

Since \((x_n)\) is a nondecreasing sequence in \(X\) and \(x_n\rightarrow x\) then, by (2), we have \(x_n\le x\) for all \(n\in \mathbb N \).

Using the contractive condition, for any \(n\in \mathbb N \), we get

$$\begin{aligned} d(x_{n+1},T x)&= d(Tx_{n},Tx)\\&\le \frac{\alpha \, d(x,Tx)[1+d(x_{n},Tx_{n})]}{1+d(x_{n},x)}+\beta \, d(x_{n},x)\\&= \frac{\alpha \, d(x,Tx)[1+d(x_{n},x_{n+1})]}{1+d(x_{n},x)}+\beta \, d(x_{n},x)\,. \end{aligned}$$

Taking into account that if \(x_n\rightarrow x\) then \(d(x_n,x_{n+1})\xrightarrow [n\rightarrow \infty ]{}0\), letting \(n\rightarrow \infty \) in the last inequality, it follows

$$\begin{aligned} d(x,Tx)\le \alpha \, d(x,Tx) \end{aligned}$$

and since \(\alpha <1\), this is imposible unless that \(d(x,Tx)=0\) this proves that \(Tx=x\).

Thus, the proof is complete. \(\square \)

Now, we present an example where it can be appreciated that assumptions in Theorem 2 do not guarantee the uniqueness of the fixed point. This example appears in [9].

Example 1

Let \(X=\{(1,0)\, \ (0,1)\}\subset \mathbb R ^2\) and consider the usual order given by

$$\begin{aligned} (x,y)\le (z,t) \Leftrightarrow x\le z \text{ and} y\le t. \end{aligned}$$

\((X,\le )\) is a partially ordered set whose different elements are not comparable. Besides, \((X,d_2)\), where \(d_2\) is the euclidean distance, is a complete metric space. The identity map \(T(x,y)=(x,y)\) is obviously continuous and nondecreasing and the contractive condition appearing in Theorem 2 is satisfied since elements in \(X\) are only comparable to themselves. Moreover \((1,0)\le T(1,0)\) and \(T\) has two fixed points.

In what follows, we present a sufficient condition for the uniqueness of the fixed point in Theorems 2 and 3.

The condition is:

$$\begin{aligned} \text{ for} x,y\in X \quad \text{ there} \text{ exists} \text{ a} \text{ lower} \text{ bound}. \end{aligned}$$
(3)

Theorem 4

Adding assumption (3) to the hypotheses of Theorem 2 (or Theorem 3) we obtain uniqueness of the fixed point.

Proof

Suppose that \(x,y\in X\) are fixed point of \(T\). We distinguish two cases:

  1. Case 1:

    \(x\) and \(y\) are comparable. Suppose \(x\le y\) (the same argument works for \(y\le x\)). By using the contractive condition, we get

    $$\begin{aligned} d(x,y)&= d(Tx,Ty)\\&\le \frac{\alpha \, d(y,Ty)[1+d(x,Tx)]}{1+d(x,y)}+\beta \, d(x,y)\\&= \beta \, d(x,y)\,. \end{aligned}$$

    Since \(\beta <1\), this is only posible when \(d(x,y)=0\). Thus \(x=y\).

  2. Case 2:

    \(x\) and \(y\) are not comparable. By (3), there exists \(z\in X\) with \(z\le x\) and \(z\le y\). Since \(z\le x\), the nondecreasing character of \(T\) gives us

    $$\begin{aligned} T^n z\le T^n x=x\quad \text{ for} \text{ any} n\in \mathbb N . \end{aligned}$$

    By using the contractive condition, for any \(n\in \mathbb N \), we have

    $$\begin{aligned} d(T^{n}z,x)&= d(T^{n}z,T^{n}x)\\&\le \frac{\alpha \, d(T^{n-1}x,T^{n}x)[1+d(T^{n-1}z,T^{n}z)]}{1+d(T^{n-1}z,T^{n-1}x)}+\beta \, d(T^{n-1}z,T^{n-1}x)\\&= \frac{\alpha \, d(x,x)[1+d(T^{n-1}z,T^{n}z)]}{1+d(T^{n-1}z,x)}+\beta \, d(T^{n-1}z,x)\\&= \beta \, d(T^{n-1}z,x)\,. \end{aligned}$$

    Using mathematical induction, we obtain

    $$\begin{aligned} d(T^n z,x)\le \beta ^n \, d(z,x) \end{aligned}$$

    and, since \(\beta <1, \,\displaystyle \lim _{n\rightarrow \infty } d(T^n z,x)=0\). This means that \(\displaystyle \lim _{n\rightarrow \infty }T^n z=x\). Using a similar argument, we can get \(\displaystyle \lim _{n\rightarrow \infty }T^n z=y\). Finally, the uniqueness of the limit gives us \(x=y\). This finishes the proof. \(\square \)

3 Some remarks

Remark 1

In [9] instead condition (3), the authors use the following weaker condition:

$$\begin{aligned} \text{ For} x,y\in X \text{ there} \text{ exists} z\in X \text{ which} \text{ is} \text{ comparable} \text{ to} x \text{ and} y. \end{aligned}$$
(4)

We have not been able to prove Theorem 4 using condition (4).

The reason is that the contractive condition appearing in Theorem 2 is not symmetric in the following sense, it adopts distinct forms depending \(x\le y\) or \(y\le x\).

Remark 2

If in Theorems 2, 3 and 4, we put \(\alpha =0\), then Theorems 2.1, 2.2 and 2.3 of [9] are obtained.

If in Theorems 2, 3 and 4, \(\beta \) is equal to zero, we have the following fixed point theorem in ordered metric spaces.

Theorem 5

Let \((X,\le )\) be a partially ordered set and suppose that there exists a metric \(d\) in \(X\) such that \((X,d)\) is a complete metric space. Let \(T:X\rightarrow X\) be a nondecreasing mapping such that there exists \(\alpha \in [0,1)\) satisfying

$$\begin{aligned} d(Tx,Ty)\le \frac{\alpha \, d(y,Ty)[1+d(x,Tx)]}{1+d(x,y)}\quad \text{ for} x,y\in X \text{ with} x\le y. \end{aligned}$$
(5)

Suppose also that either \(T\) is continuous or \(X\) satisfies condition (2).

If there exist \(x_0\in X\) such that \(x_0\le Tx_0\) then \(T\) has a fixed point.

Besides, if \((X,\le )\) satisfies condition (3), then the fixed point is unique.

Remark 3

If in Theorems 2, 3 and 4, \(\alpha =0\) and \(0<\beta <\frac{1}{3}\) then the contractive condition (1) appearing in Theorem 2 implies that

$$\begin{aligned} d(Tx,Ty)\le \gamma \, [d(x,Tx)+d(y,Ty)]\quad \text{ for} x,y\in X \text{ with} x\le y, \end{aligned}$$

where \(\gamma \in (0,\frac{1}{2})\) (this is a Kannan type condition).

In fact, since \(\alpha =0\) and \(0<\beta <\frac{1}{3}\), the condition (1) takes the form

$$\begin{aligned} d(Tx,Ty)\le \beta \, d(x,y)\quad \text{ for} x,y\in X \text{ with} x\le y. \end{aligned}$$

By using the triangular inequality, for \(x,y\in X\) with \(x\le y\) we have

$$\begin{aligned} d(Tx,Ty)\le \beta \, d(x,y)\le \beta \, [d(x,Tx)+d(y,Ty)]+\beta \, d(Tx,Ty), \end{aligned}$$

and from this inequality it follows

$$\begin{aligned} d(Tx,Ty)\le \frac{\beta }{1-\beta }[d(x,Tx)+d(y,Ty)]. \end{aligned}$$

From \(0<\beta <\frac{1}{3}\) it is easily proved that \(\gamma =\frac{\beta }{1-\beta }\in \left(0,\frac{1}{2}\right)\).

Remark 4

If diam\(X\le 1\) and \(2\alpha +\beta <1\) then the contractive condition (1) appearing in Theorem 2 implies the following Reich type condition:

$$\begin{aligned} d(Tx,Ty)\le \alpha \, d(x,Tx)+\alpha \, d(y,Ty)+\beta \, d(x,y)\quad \text{ for} x,y\in X \text{ with} x\le y. \end{aligned}$$

In fact, for \(x,y\in X\) with \(x\le y\) we have

$$\begin{aligned} d(Tx,Ty)&\le \frac{\alpha \, d(y,Ty)[1+d(x,Tx)]}{1+d(x,y)}+\beta \, d(x,y)\\&\le \alpha \, d(y,Ty)[1+d(x,Tx)]+\beta \, d(x,y)\\&\le \alpha \, d(y,Ty)+\alpha \, d(y,Ty)d(x,Tx)+\beta \, d(x,y). \end{aligned}$$

Since diam\(X\le 1, \,d(y,Ty)\le 1\), and therefore, the last inequality implies that

$$\begin{aligned} d(Tx,Ty)\le \alpha \, d(y,Ty)+\alpha \, d(x,Tx)+\beta \, d(x,y). \end{aligned}$$

In the sequel, we present an example where Theorem 2 can be applied and it cannot be treated by Theorem 1.

Example 2

Let \(X=\{(0,1)\, \ (1,0)\, \ (1,1)\}\) and we consider in \(X\) the partial order given by \(\mathcal R =\{(x,x):x\in X\}\). Notice that elements in \(X\) are only comparable to themselves. Moreover, \((X,d_2)\), where \(d_2\) is the euclidean distance, is a complete metric space.

Let \(T:X\rightarrow X\) be defined by \(T(0,1)=(1,0)\, \ T(1,0)=(0,1)\) and \(T(1,1)=(1,1)\).

Obviously, \(T\) is continuous and nondecreasing, and the contractive condition appearing in Theorem 2 is satisfied since elements in \(X\) are only comparable to themselves.

As \((1,1)\le T(1,1)\), Theorem 2 says us that \(T\) has a fixed point [this fixed point is \((1,1)\)].

On the other hand, for \(x=(0,1),\ y=(1,1)\), we have

$$\begin{aligned} d_2(Tx, Ty)&= d_2(x,y)=1\\ d_2(x,Tx)&= \sqrt{2}\\ d_2(y,Ty)&= 0 \end{aligned}$$

and the contractive condition appearing in Theorem 1 is not satisfied since

$$\begin{aligned} 1=d_2(Tx,Ty)&\le \frac{\alpha \, d(y,Ty)[1+d(x,Tx)]}{1+d(x,y)}+\beta \, d(x,y)\\&= \frac{\alpha \cdot 0[1+\sqrt{2}]}{1+1}+\beta \cdot 1=\beta \cdot 1=\beta . \end{aligned}$$

Therefore, this example cannot be treated by Theorem 1.

Moreover, notice that in this example we have uniqueness of the fixed point and \((X,\le )\) does not satisfy condition (3). This proves that condition (3) is not necessary condition for the uniqueness of the fixed point.