A Pata-type fixed point theorem

Abstract

We prove a fixed point theorem for a Pata-type map defined on a complete (normal) cone metric space. Our results generalize the recent work of M. Chakraborty and S. K. Samanta. An example demonstrating this fact is also presented.

Mathematics Subject Classification  47H10 · 54H25 (Primary) · 37C25 (Secondary)

Introduction

The classical Banach fixed point theorem states that if $(X,d)$ is a complete metric space and $T:X\to X$ is a contraction map, i.e., $T$ satisfies

$$\begin{array}{c}\hfill d(Tx,Ty)\le \mathit{\alpha }d(x,y),\end{array}$$

(1)

for all $x,y\in X$ and some $\mathit{\alpha }\in [0,1)$, then $T$ has a unique fixed point. i.e., there exists a unique $a\in X$ such that $Ta=a$.

In [4], Pata considered a map $T:X\to X$ on the complete metric space $(X,d)$ that satisfied the condition: for all $x,y\in X$,

$$\begin{array}{c}\hfill d(Tx,Ty)\le (1-\mathit{ϵ})d(x,y)+\mathrm{\Lambda }{\mathit{ϵ}}^{\mathit{\alpha }}\mathit{\psi }(\mathit{ϵ}){[1+|||x|||+|||y|||]}^{\mathit{\beta }},\end{array}$$

(2)

for every $\mathit{ϵ}\in [0,1]$, fixed constants $\mathrm{\Lambda }\ge 0$, $\mathit{\alpha }\ge 1$ and $\mathit{\beta }\in [0,\mathit{\alpha }]$, a fixed element $x_0\in X$, $|||z|||=d(z,x_0)$ and an increasing function $\mathit{\psi }:[0,1]\to [0,\infty )$ which vanishes at and is continuous at $0$. He proved that the map $T$ satisfying (2) has a unique fixed point. Moreover, he also demonstrated that if $T:X\to X$ is a contraction map, then $T$ satisfies condition (2), thereby obtaining a generalization of the Banach fixed point theorem.

Another fixed point theorem that is widely popular, is due to Kannan which states that if $(X,d)$ is a complete metric space and the map $T:X\to X$ is a Kannan contraction, i.e., $T$ satisfies

$$\begin{array}{c}\hfill d(Tx,Ty)\le \frac{\mathit{\gamma }}{2}\{d(x,Tx)+d(y,Ty)\}\end{array}$$

(3)

for all $x,y\in X$ and some $\mathit{\gamma }\in [0,1)$, then $T$ has a unique fixed point.

There are several generalizations of the Kannan fixed point theorem, but the one of particular interest to us is due to Chakraborty and Samanta in [1]. The authors, in [1], consider a map $T:X\to X$ defined on the complete metric space $(X,d)$ that satisfies the condition: for all $x,y\in X$,

$$\begin{array}{cc}\hfill d(Tx,Ty)& \le \frac{1-\mathit{ϵ}}{2}\{d(x,Tx)+d(y,Ty)\}+\hfill \\ \hfill & \mathrm{\Lambda }{\mathit{ϵ}}^{\mathit{\alpha }}\mathit{\psi }(\mathit{ϵ}){[1+|||x|||+|||y|||+|||Tx|||+|||Ty|||]}^{\mathit{\beta }},\hfill \end{array}$$

(4)

for every $\mathit{ϵ}\in [0,1]$, fixed constants $\mathrm{\Lambda }\ge 0$, $\mathit{\alpha }\ge 1$ and $\mathit{\beta }\in [0,\infty )$, a fixed element $x_0\in X$, $|||z|||=d(z,x_0)$ and a function $\mathit{\psi }:[0,1]\to [0,\infty )$ which vanishes at and is continuous at $0$. The authors prove that the map $T$ has a unique fixed point. Moreover, they also demonstrate that if $T:X\to X$ is a Kannan contraction map, then $T$ satisfies condition (4).

This article contains a generalization of the main result in [1]. The setting considered is that of complete cone metric spaces, where the underlying cone is normal (see [3]). It is shown that for a map $T:X\to X$ defined on the complete (normal) cone metric space $(X,d)$, which satisfies a Pata-type condition, that is an improved version of (4), there exists a unique fixed point. An example to illustrate the main result is also provided.

Preliminaries and the main result

Let E be a real Banach space. A non-empty closed subset P of E is said to be a $cone$ if

• (a) $\mathit{\alpha }P+\mathit{\beta }P\subset P$ for all $\mathit{\alpha },\mathit{\beta }\in [0,\infty )$.

• (b) $P\cap (-P)=\{\mathit{\theta }\}$, where $\mathit{\theta }\in E$ is the zero vector.

The cone $P$ is said to be $solid$ if the interior of $P$, which we will denote by $intP$, is non-empty.

Examples of solid cones

• (1) Let $E=\mathbb{R}$ and $P=[0,\infty )$.

• (2) Let $E={\mathbb{R}}^2$ and $P=\{(x,y):x,y\ge 0\}$.

• (3) Let $E={\ell }^2$ and $P=\{{(x_n)}_{n\ge 1}:x_n\ge 0\}$.

The norms on $E$ in the examples above are the usual norms.

A cone $P$ in a real Banach space E, induces the following partial order $⪯$ on E. For $x,y\in E$,

$$\begin{array}{c}\hfill x⪯y\iff y-x\in P.\end{array}$$

In the case of a solid cone $P$, we will use the notation $x\ll y$ to denote $y-x\in intP$.

A cone $P$ is said to be $normal$ if for all $x,y\in P$ such that $x⪯y$, there exists a constant $\mathit{\kappa }\ge 1$ such that $\Vert x\Vert \le \mathit{\kappa }\Vert y\Vert$. The examples (1), (2) and (3) above are normal cones with $\mathit{\kappa }=1$.

Let $X$ be a nonempty set, $E$ be a real Banach space and $P\subset E$ be a solid normal cone with normal constant $\mathit{\kappa }\ge 1$. A map $d:X\times X\to E$ is said to be a cone metric if for all $x,y,z\in E$,

• (a) $d(x,y)⪰\mathit{\theta }$, i.e., $d(x,y)\in P$.

• (b) $d(x,y)=\mathit{\theta }$ if and only if $y=x$.

• (c) $d(x,y)⪯d(x,z)+d(z,y)$.

The pair $(X,d)$ is called acone metric space. It is indeed the case that every metric space is a cone metric space.

Examples of cone metric spaces

• (1) ([3]) Let $E$ and $P$ be as in example (2) above, $d:\mathbb{R}\times \mathbb{R}\to E$ be the map $d(x,y)=(|x-y|,\mathit{\alpha }|x-y|)$, where $\mathit{\alpha }\ge 0$ is a constant. The pair $(\mathbb{R},d)$ is a cone metric space.

• (2) ([2]) Let $(X,\mathit{\rho })$ be a metric space, $E$ and $P$ be as in example (3) above, $d:X\times X\to E$ be the map $d(x,y)={\left(\sqrt{2^{-n}\mathit{\rho }(x,y)}\right)}_{n\ge 1}.$ It can be verified that $(X,d)$ is a cone metric space.

Let $(X,d)$ be a cone metric space. A sequence $(x_n)$ of points in $X$ is said to be Cauchy if for any given $c\gg 0$, there exists $N\in \mathbb{N}$ such that $d(x_m,x_n)\ll c$, for all $m,n\ge N$, or equivalently, there exists $M\in \mathbb{N}$ which is independent of $p$ such that $d(x_n,x_{n+p})\ll c$ for all $n\ge M$.

The sequence $(x_n)$ is said to beconvergent, if there exists $x\in X$ such that for any given $c\gg 0$, there exists $N\in \mathbb{N}$ such that $d(x_n,x)\ll c$ for all $n\ge N$.

The cone metric space $(X,d)$ is said to becomplete if every Cauchy sequence in $X$ converges.

The proof of the following lemma can be found in [3].

Lemma 1

Let $(X,d)$ be a cone metric space and $(x_n)$ be a sequence in $X$.

• (1) $(x_n)$ is Cauchy if and only if $\Vert d(x_m,x_n)\Vert \to 0$ as $m,n\to \infty$. i.e., Given $\mathit{\eta }>0$ and $p\in \mathbb{N}$, there exists $N\in \mathbb{N}$ which is independent of $p$ such that $\Vert d(x_n,x_{n+p})\Vert <\mathit{\eta }$, for all $n\ge N$.

• (2) $(x_n)$ is convergent to $x$ if and only if $\Vert d(x_n,x)\Vert \to 0$ as $n\to \infty$.

• (3) If $(x_n)$ converges to $x,y\in X$, then $x=y$.

The following is our main result which generalizes Theorem 2.2 in [1].

Theorem 1

Let $(X,d)$ be a complete cone metric space with normal constant $\mathit{\kappa }\ge 1$, $x_0\in X$, $\mathrm{\Lambda }\ge 0$, $\mathit{\alpha }\ge 1$ and $\mathit{\beta }\in [0,\infty )$ be fixed constants and $\mathit{\psi }:[0,1]\to [0,\infty )$ be such that ${\displaystyle \underset{\mathit{ϵ}\to 0^{+}}{lim}\mathit{\psi }(\mathit{ϵ})=0}$. If for every $x,y\in X$, the map $T:X\to X$ satisfies

$$\begin{array}{c}\hfill \Vert d(Tx,Ty)\Vert \le \frac{(1-\mathit{ϵ})}{\mathit{\kappa }}M(x,y)+\mathrm{\Lambda }{\mathit{ϵ}}^{\mathit{\alpha }}\mathit{\psi }(\mathit{ϵ}){[1+|||x|||+|||y|||+|||Tx|||+|||Ty|||]}^{\mathit{\beta }},\end{array}$$

(5)

for every $\mathit{ϵ}\in [0,1]$, where $M(x,y)=max\left\{\Vert d(x,Tx)\Vert ,\Vert d(y,Ty)\Vert ,\frac{1}{2\mathit{\kappa }}\Vert d(x,y)\Vert \right\}$ and $|||z|||$ denotes $\Vert d(z,x_0)\Vert$, then $T$ has a unique fixed point.

The proof of the above Theorem is given in Sect. 3. We point out that there is no loss of generality in choosing any such $x_0$ in (5), simply because a change in $x_0$ can essentially be absorbed by assigning a different value to $\mathrm{\Lambda }$, thanks to the triangle inequality of the cone metric $d$ and the sub-additivity of the norm.

Proof of The main result

This section contains a proof of our main result. Although the proof follows a similar pattern as that of Theorem 2.2 in [1], the arguments provided here are different and sometimes simpler.

Proof of Theorem 1

First we prove uniqueness of the fixed point. Suppose that $x,y\in X$ are such that $x\ne y$, $Tx=x$ and $Ty=y$. Letting $\mathit{ϵ}=0$ in inequality (5) yields, $\Vert d(x,y)\Vert \le \frac{1}{2\mathit{\kappa }}\Vert d(x,y)\Vert$, a contradiction to the fact that $x\ne y$. The uniqueness follows.

To show the existence of a fixed point, consider the sequence $(T^n{x}_0)$. Without loss of generality, we assume that $T^n{x}_0\ne T^{n+1}{x}_0$, for all $n=0,1,2,\cdots .$ Since for each $n\in \mathbb{N}$, and $\mathit{ϵ}\in [0,1]$,

$$\begin{array}{cc}\hfill \Vert d(T^{n+1}{x}_0,T^n{x}_0)\Vert & \le (1-\mathit{ϵ})M(T^n{x}_0,T^{n-1}{x}_0)\hfill \\ \hfill & +\mathrm{\Lambda }{\mathit{ϵ}}^{\mathit{\alpha }}\mathit{\psi }(\mathit{ϵ})[1+2|||T^n{x}_0|||+|||T^{n-1}{x}_0|||+|||T^{n+1}{x}_0{|||]}^{\mathit{\beta }},\hfill \end{array}$$

(6)

it follows that there exists no $n\in \mathbb{N}$, for which $M(T^n{x}_0,T^{n-1}{x}_0)=\Vert d(T^{n+1}{x}_0,T^n{x}_0)\Vert$. For otherwise, it would mean that there exists some $m\in \mathbb{N}$ such that for every $\mathit{ϵ}\in (0,1]$,

$$\begin{array}{c}\hfill \Vert d(T^{m+1}{x}_0,T^m{x}_0)\Vert \le \mathrm{\Lambda }{\mathit{ϵ}}^{\mathit{\alpha }-1}\mathit{\psi }(\mathit{ϵ})[1+2|||T^m{x}_0|||+|||T^{m-1}{x}_0|||+|||T^{m+1}{x}_0{|||]}^{\mathit{\beta }}.\end{array}$$

Letting $\mathit{ϵ}\to 0^{+}$ yields, $\Vert d(T^{m+1}{x}_0,T^m{x}_0)\Vert =0$, i.e., $T^{m+1}{x}_0=T^m{x}_0$, a contradiction. Thus, for each $n\in \mathbb{N}$, letting $\mathit{ϵ}=0$ in inequality (6) yields,

$$\begin{array}{c}\hfill \Vert d(T^{n+1}{x}_0,T^n{x}_0)\Vert \le \Vert d(T^n{x}_0,T^{n-1}{x}_0)\Vert .\end{array}$$

(7)

Iterating we obtain

$$\begin{array}{c}\hfill \Vert d(T^{n+1}{x}_0,T^n{x}_0)\Vert \le \Vert d(T{x}_0,x_0)\Vert ,\end{array}$$

(8)

for all $n=0,1,2,\cdots$

Next, we show that the sequence $(\Vert d(T^n{x}_0,x_0)\Vert )$ is bounded above by $c=2\mathit{\kappa }\Vert d(x_0,T{x}_0)\Vert$. This is certainly the case when $n=1$. Assume that $\Vert d(T^{m-1}{x}_0,x_0)\Vert \le c$. The claim follows from induction, if we show that $\Vert d(T^m{x}_0,x_0)\Vert \le c$. Using (8), it follows that

$$\begin{array}{cc}\hfill \Vert d(T^m{x}_0,x_0)\Vert & \le \mathit{\kappa }\{\Vert d(T^m{x}_0,T{x}_0)\Vert +\Vert d(T{x}_0,x_0)\Vert \}\hfill \\ \hfill & \le \mathit{\kappa }M(T^{m-1}{x}_0,x_0)+\frac{c}{2}\hfill \\ \hfill & =\mathit{\kappa }max\left\{\Vert d(T^{m-1}{x}_0,T^m{x}_0)\Vert ,\Vert d(x_0,T{x}_0)\Vert ,\frac{1}{2\mathit{\kappa }}\Vert d(T^{m-1}{x}_0,x_0)\Vert \right\}+\frac{c}{2}\hfill \\ \hfill & \le \mathit{\kappa }max\left\{\Vert d(x_0,T{x}_0)\Vert ,\frac{1}{2\mathit{\kappa }}\Vert d(T^{m-1}{x}_0,x_0)\Vert \right\}+\frac{c}{2}\hfill \\ \hfill & \le \mathit{\kappa }\left\{\frac{c}{2\mathit{\kappa }}\right\}+\frac{c}{2}\hfill \\ \hfill & =c.\hfill \end{array}$$

Thus, for all $n\in \mathbb{N}$,

$$\begin{array}{c}\hfill \Vert d(T^n{x}_0,x_0)\Vert \le c.\end{array}$$

(9)

Consider the monotonically decreasing sequence $(\Vert d(T^n{x}_0,T^{n+1}{x}_0)\Vert )$ [see (7)]. Since it is bounded below by $0$, it is convergent. Let ${\displaystyle \ell =\underset{n\to \infty }{lim}\Vert d(T^n{x}_0,T^{n+1}{x}_0)\Vert }$. Clearly $\ell \ge 0$. We will in fact, prove that $\ell =0$. For each $\mathit{ϵ}\in (0,1]$, it follows from (6) and (9) that

$$\begin{array}{c}\hfill \Vert d(T^n{x}_0,T^{n+1}{x}_0)\Vert \le (1-\mathit{ϵ})\Vert d(T^{n-1}{x}_0,T^n{x}_0)\Vert +K\mathrm{\Lambda }{\mathit{ϵ}}^{\mathit{\alpha }}\mathit{\psi }(\mathit{ϵ}),\end{array}$$

where $K={(1+4c)}^{\mathit{\beta }}$. Rearranging the above inequality and using the fact that the sequence $(\Vert d(T^n{x}_0,T^{n+1}{x}_0)\Vert )$ is monotonically decreasing yields,

$$\begin{array}{c}\hfill \Vert d(T^n{x}_0,T^{n+1}{x}_0)\Vert \le \frac{\Vert d(T^{n-1}{x}_0,T^n{x}_0)\Vert }{(1+\mathit{ϵ})}+\frac{K\mathrm{\Lambda }{\mathit{ϵ}}^{\mathit{\alpha }}\mathit{\psi }(\mathit{ϵ})}{(1+\mathit{ϵ})},\end{array}$$

(10)

for each $\mathit{ϵ}\in (0,1]$. Fixing such an $\mathit{ϵ}$ and letting $n\to \infty$ in (10), we obtain

$$\begin{array}{c}\hfill \ell \le \frac{\ell }{(1+\mathit{ϵ})}+\frac{K\mathrm{\Lambda }{\mathit{ϵ}}^{\mathit{\alpha }}\mathit{\psi }(\mathit{ϵ})}{(1+\mathit{ϵ})}\end{array}$$

Multiplying both sides by $(1+\mathit{ϵ})$ and simplifying yields, for each $\mathit{ϵ}\in (0,1]$,

$$\begin{array}{c}\hfill \ell \le K\mathrm{\Lambda }{\mathit{ϵ}}^{\mathit{\alpha }-1}\mathit{\psi }(\mathit{ϵ}).\end{array}$$

Letting $\mathit{ϵ}\to 0^{+}$ yields, $\ell \le 0$. Thus, in fact,

$$\begin{array}{c}\hfill \ell =0.\end{array}$$

(11)

Next, we show that the sequence $(T^n{x}_0)$ is Cauchy. In view of Lemma 1 (i), it suffices to show that, given $\mathit{\eta }>0$ and $p\in \mathbb{N}$, there exists $M\in \mathbb{N}$ which is independent of $p$, such that $\Vert d(T^n{x}_0,T^{n+p}{x}_0)\Vert <\mathit{\eta }$, for all $n\ge M$.

By (11), choose $N\in \mathbb{N}$ such that

$$\begin{array}{c}\hfill \Vert d(T^n{x}_0,T^{n+1}{x}_0)\Vert <\frac{\mathit{\eta }}{2}\end{array}$$

(12)

for all $n\ge N$.

Consider $\Vert d(T^n{x}_0,T^{n+p}{x}_0)\Vert$ for all $n\ge N+1$. Letting $\mathit{ϵ}=0$ in (5) yields,

$$\begin{array}{c}\hfill \Vert d(T^n{x}_0,T^{n+p}{x}_0)\Vert \le M(T^{n-1}{x}_0,T^{n+p-1}{x}_0).\end{array}$$

If $M(T^{n-1}{x}_0,T^{n+p-1}{x}_0)=\Vert d(T^{n-1}{x}_0,T^n{x}_0)\Vert$ or $\Vert d(T^{n+p-1}{x}_0,T^{n+p}{x}_0)\Vert$, then it follows from (12) that

$$\begin{array}{c}\hfill \Vert d(T^n{x}_0,T^{n+p}{x}_0)\Vert <\frac{\mathit{\eta }}{2}.\end{array}$$

(13)

If $M(T^{n-1}{x}_0,T^{n+p-1}{x}_0)=\frac{1}{2\mathit{\kappa }}\Vert d(T^{n-1}{x}_0,T^{n+p-1}{x}_0)\Vert$, then the normality of the underlying cone and an application of the triangle inequality yield,

$$\begin{array}{cc}\hfill \Vert d(T^n{x}_0,T^{n+p}{x}_0)\Vert & \le \frac{1}{2\mathit{\kappa }}\Vert d(T^{n-1}{x}_0,T^{n+p-1}{x}_0)\Vert \hfill \\ \hfill & \le \frac{1}{2}\{\Vert d(T^{n-1}{x}_0,T^n{x}_0)\Vert +\Vert d(T^n{x}_0,T^{n+p}{x}_0)\Vert \hfill \\ \hfill & +\Vert d(T^{n+p}{x}_0,T^{n+p-1}{x}_0)\Vert \}.\hfill \end{array}$$

From a rearrangement of terms, it follows from (12) that

$$\begin{array}{cc}\hfill \Vert d(T^n{x}_0,T^{n+p}{x}_0)\Vert & \le \Vert d(T^{n-1}{x}_0,T^n{x}_0)\Vert +\Vert d(T^{n+p}{x}_0,T^{n+p-1}{x}_0)\Vert <\mathit{\eta }.\hfill \end{array}$$

(14)

Thus, it follows from (13) and (14) that given $\mathit{\eta }>0$ and $p\in \mathbb{N}$, by choosing $M=N+1$, which is independent of $p$, one has that

$$\begin{array}{c}\hfill \Vert d(T^n{x}_0,T^{n+p}{x}_0)\Vert <\mathit{\eta },\end{array}$$

for all $n\ge M$. i.e., the sequence $(T^n{x}_0)$ is Cauchy in the cone metric space $(X,d)$. By completeness, there exists a unique $u\in X$ such that $T^n{x}_0\to u$ as $n\to \infty$.

We complete the proof by showing that this $u$ is a fixed point of $T$. Let $\mathit{\eta }>0$ be arbitrary, $B=\mathrm{\Lambda }\mathit{\kappa }{(1+|||u|||+|||Tu|||+2c)}^{\mathit{\beta }}$. Choose ${\mathit{ϵ}}_0\in (0,1]$ such that

$$\begin{array}{c}\hfill B{\mathit{ϵ}}_0^{\mathit{\alpha }-1}\mathit{\psi }({\mathit{ϵ}}_0)<\frac{\mathit{\eta }}{2}.\end{array}$$

(15)

Choose $L\in \mathbb{N}$ such that

$$\begin{array}{c}\hfill \Vert d(T^L{x}_0,u)\Vert +\Vert d(T^L{x}_0,T^{L+1}{x}_0)\Vert <\frac{{\mathit{ϵ}}_0\mathit{\eta }}{4\mathit{\kappa }}.\end{array}$$

(16)

Consider $\Vert d(Tu,u)\Vert$. From (5), it follows that

$$\begin{array}{cc}\hfill \Vert d(Tu,u)\Vert & \le \mathit{\kappa }\{\Vert d(Tu,T^{L+1}{x}_0)\Vert +\Vert d(T^{L+1}{x}_0,T^L{x}_0)\Vert +\Vert d(T^L{x}_0,u)\Vert \}\hfill \\ \hfill & \le (1-{\mathit{ϵ}}_0)M(u,T^L{x}_0)+\mathit{\kappa }\mathrm{\Lambda }{\mathit{ϵ}}_0^{\mathit{\alpha }}\mathit{\psi }({\mathit{ϵ}}_0)(1+|||u|||+|||T^L{x}_0|||+|||Tu|||+|||T^{L+1}{x}_0{|||)}^{\mathit{\beta }}\hfill \\ \hfill & +\mathit{\kappa }\{\Vert d(T^{L+1}{x}_0,T^L{x}_0)\Vert +\Vert d(T^L{x}_0,u)\Vert \}\hfill \\ \hfill & \le (1-{\mathit{ϵ}}_0)\{\Vert d(u,Tu)\Vert +\Vert d(T^{L+1}{x}_0,T^L{x}_0)\Vert +\Vert d(u,T^L{x}_0)\Vert \}+B{\mathit{ϵ}}_0^{\mathit{\alpha }}\mathit{\psi }({\mathit{ϵ}}_0)\hfill \\ \hfill & +\mathit{\kappa }\{\Vert d(T^{L+1}{x}_0,T^L{x}_0)\Vert +\Vert d(T^L{x}_0,u)\Vert \}\hfill \\ \hfill & \le (1-{\mathit{ϵ}}_0)\Vert d(u,Tu)\Vert +2\mathit{\kappa }\{\Vert d(T^{L+1}{x}_0,T^L{x}_0)\Vert +\Vert d(u,T^L{x}_0)\Vert \}+B{\mathit{ϵ}}_0^{\mathit{\alpha }}\mathit{\psi }({\mathit{ϵ}}_0).\hfill \end{array}$$

Rearranging the terms in the above inequality yields,

$$\begin{array}{c}\hfill \Vert d(u,Tu)\Vert \le \frac{2\mathit{\kappa }}{{\mathit{ϵ}}_0}\{\Vert d(T^{L+1}{x}_0,T^L{x}_0)\Vert +\Vert d(u,T^L{x}_0)\Vert \}+B{\mathit{ϵ}}_0^{\mathit{\alpha }-1}\mathit{\psi }({\mathit{ϵ}}_0).\end{array}$$

It follows from (15) and (16) that

$$\begin{array}{c}\hfill \Vert d(u,Tu)\Vert <\mathit{\eta }.\end{array}$$

Since $\mathit{\eta }>0$ is arbitrary, the proof is complete. $\square$

Applications and examples

This section contains applications of Theorem 1, presented as corollaries. Recall the definition of $M(x,y)$ and the notation $|||·|||$ from Theorem 1.

Corollary 1

Let $(X,d)$ be a complete cone metric space with normal constant $\mathit{\kappa }\ge 1$ and $\mathit{\delta }\in \left(0,\frac{1}{\mathit{\kappa }}\right)$. If the map $T:X\to X$ satisfies

$$\begin{array}{c}\hfill \Vert d(Tx,Ty)\Vert \le \mathit{\delta }M(x,y)\end{array}$$

for all $x,y\in X$, then $T$ has a unique fixed point.

Proof

It suffices to show that $T$ satisfies condition (5). The result then follows from Theorem 1. Fix $x_0\in X$. Observe that

$$\begin{array}{cc}\hfill M(x,y)& =max\left\{\Vert d(x,Tx)\Vert ,\Vert d(y,Ty)\Vert ,\frac{1}{2\mathit{\kappa }}\Vert d(x,y)\Vert \right\}\hfill \\ \hfill & \le 1+\Vert d(x,x_0)\Vert +\Vert d(x_0,Tx)\Vert +\Vert d(y,x_0)\Vert +\Vert d(x_0,Ty)\Vert \hfill \\ \hfill & =1+|||x|||+|||Tx|||+|||y|||+|||Ty|||.\hfill \end{array}$$

(17)

It follows from (17) and a Bernoulli inequality argument similar to Sect. 3 in [1], that for all $\mathit{ϵ}\in [0,1]$, $x,y\in X$,

$$\begin{array}{cc}\hfill \Vert d(Tx,Ty)\Vert & \le \mathit{\delta }M(x,y)\hfill \\ \hfill & =\frac{(1-\mathit{ϵ})}{\mathit{\kappa }}M(x,y)+\left(\mathit{\delta }+\frac{(\mathit{ϵ}-1)}{\mathit{\kappa }}\right)M(x,y)\hfill \\ \hfill & =\frac{(1-\mathit{ϵ})}{\mathit{\kappa }}M(x,y)+\mathit{\delta }\left(1+\frac{(\mathit{ϵ}-1)}{\mathit{\kappa }\mathit{\delta }}\right)M(x,y)\hfill \\ \hfill & \le \frac{(1-\mathit{ϵ})}{\mathit{\kappa }}M(x,y)+\mathit{\delta }{(1+(\mathit{ϵ}-1))}^{\frac{1}{\mathit{\kappa }\mathit{\delta }}}M(x,y)\hfill \\ \hfill & \le \frac{(1-\mathit{ϵ})}{\mathit{\kappa }}M(x,y)+\mathit{\delta }{\mathit{ϵ}}^{\frac{1}{\mathit{\kappa }\mathit{\delta }}}[1+|||x|||+|||Tx|||+|||y|||+|||Ty|||].\hfill \\ \hfill & =\frac{(1-\mathit{ϵ})}{\mathit{\kappa }}M(x,y)+\mathit{\delta }{\mathit{ϵ}}^{1+\mathit{\eta }}[1+|||x|||+|||Tx|||+|||y|||+|||Ty|||],\hfill \end{array}$$

(18)

where $\mathit{\eta }=\left(\frac{1}{\mathit{\kappa }\mathit{\delta }}-1\right)>0$. Comparing (18) with (5), one sees that $T$ satisfies (5) with $\mathrm{\Lambda }=\mathit{\delta }$, $\mathit{\beta }=\mathit{\alpha }=1$, $\mathit{\psi }(\mathit{ϵ})={\mathit{ϵ}}^{\mathit{\eta }}$. $\square$

Corollary 2

(The Kannan Fixed Point Theorem) Let $(X,d)$ be a complete metric space and $\mathit{\delta }\in (0,1)$. If the map $T:X\to X$ satisfies

$$\begin{array}{c}\hfill d(Tx,Ty)\le \frac{\mathit{\delta }}{2}(d(x,Tx)+d(y,Ty))\end{array}$$

for all $x,y\in X$, then $T$ has a unique fixed point.

Proof

Observe that $\mathit{\kappa }=1$ and $T$ satisfies the hypothesis of Corollary 1. The result follows. $\square$

We end with an example which demonstrates that Theorem 1 indeed generalizes the main result in [1]. Observe that it suffices to produce a complete metric space $(X,d)$ and a map $T:X\to X$ which satisfies

• (A) $d(Tx,Ty)\le \mathit{\delta }max\left\{d(x,Tx),d(y,Ty),\frac{1}{2}d(x,y)\right\}$ for some $\mathit{\delta }\in (0,1)$ and $x,y\in X$,

• (B) $d(Ta,Tb)>\frac{1}{2}\{d(a,Ta)+d(b,Tb)\}$ for some $a,b\in X$.

Example

Let $X=\left[0,\frac{1}{2}\right]\cup \{1,2\}$ with $d$ being the usual metric. It is clear that $(X,d)$ is a complete cone metric space with $\mathit{\kappa }=1$. Define the map $T:X\to X$ by $Tx=2$ if $x\ne 1,2$ and $T1=T2=1$. Observe that for $x\ne 1,2$, $d(Tx,T1)=1$ and

$$\begin{array}{c}\hfill max\left\{d(x,Tx),d(1,T1),\frac{1}{2}d(x,1)\right\}=max\left\{2-x,0,\frac{1}{2}(1-x)\right\}=2-x.\end{array}$$

Similarly, for $x\ne 1,2$, $d(Tx,T2)=1$ and

$$\begin{array}{c}\hfill max\left\{d(x,Tx),d(2,T2),\frac{1}{2}d(x,2)\right\}=max\left\{2-x,1,\frac{1}{2}(2-x)\right\}=2-x.\end{array}$$

Since, by choice, $x\in \left[0,\frac{1}{2}\right]$, it follows that $1\le \frac{2}{3}(2-x)$. Moreover, $d(T1,T2)=d(Tx,Ty)=0$ for all $x,y\in \left[0,\frac{1}{2}\right]$. Putting all this together implies that $T$ satisfies condition (A) above with $\mathit{\delta }=\frac{2}{3}$. Since $\frac{1}{2}\left\{d\left(\frac{1}{4},T\frac{1}{4}\right)+d(1,T1)\right\}=\frac{\left(2-\frac{1}{4}\right)}{2}<1=d\left(T\frac{1}{4},T1\right)$, it follows that $T$ also satisfies condition (B) above with $a=\frac{1}{4}$ and $b=1$. From the proof of Corollary 1, it follows that $T$ satisfies (5). However, $T$ does not satisfy (4), as it satisfies condition (B) above. This can be seen by setting $\mathit{ϵ}=0$ in (4). It is immediate that $T$ has a unique fixed point.