On the existence of fixed points that belong to the zero set of a certain function

Abstract

Let $T: X\to X$ be a given operator and $F_{T}$ be the set of its fixed points. For a certain function $\varphi: X\to[0,\infty)$, we say that $F_{T}$ is φ-admissible if $F_{T}$ is nonempty and $F_{T}\subseteq Z_{\varphi}$, where $Z_{\varphi}$ is the zero set of φ. In this paper, we study the φ-admissibility of a new class of operators. As applications, we establish a new homotopy result and we obtain a partial metric version of the Boyd-Wong fixed point theorem.

1 Introduction

Let $(X,d)$ be a metric space. For a given function $\varphi: X\to[0,\infty)$, we define the set

$$Z_{\varphi}=\bigl\{ x\in X: \varphi(x)=0\bigr\} . $$

Let $T: X\to X$ be a given operator. The set of fixed points of T is denoted by $F_{T}$, that is,

$$F_{T}=\{x\in X: Tx=x\}. $$

Definition 1.1

We say that the set $F_{T}$ is φ-admissible if and only if $F_{T}\neq\emptyset$ and $F_{T}\subseteq Z_{\varphi}$.

Let $\mathcal{F}$ be the set of functions $F:[0,\infty)^{3}\to[0,\infty )$ satisfying the following conditions:

(F1):

$\max\{a,b\} \leq F(a,b,c)$, for all $a, b, c \geq0$;

(F2):

$F(a,0,0)=a$, for all $a\geq0$;

(F3):

F is continuous.

As examples, the following functions belong to $\mathcal{F}$:

• 1. $F(a,b,c)=a+b+c$,

• 2. $F(a,b,c)=\max\{a,b\}+\ln(c+1)$,

• 3. $F(a,b,c)= a+b+c(c+1)$,

• 4. $F(a,b,c)=(a+b)e^{c}$,

• 5. $F(a,b,c)=(a+b)(c+1)^{n}$, $n\in\mathbb{N}$.

Let Ψ be the set of functions $\psi:[0,\infty)\to[0,\infty)$ satisfying the following conditions:

(Ψ1):

ψ is upper semi-continuous from the right;

(Ψ2):

$\psi(t)< t$, for all $t>0$.

For given functions $\varphi:X\to[0,\infty)$, $F\in\mathcal{F}$, and $\psi\in\Psi$, we denote by $\mathcal{T}(\varphi,F,\psi)$ the class of operators $T: X\to X$ satisfying

$$ F \bigl(d(Tx,Ty),\varphi(Tx),\varphi(Ty) \bigr)\leq\psi \bigl(F \bigl(d(x,y),\varphi(x),\varphi(y) \bigr) \bigr),\quad (x,y)\in X\times X. $$

(1.1)

The aim of this paper is to study the φ-admissibility of the set $F_{T}$, where T belongs to the class of operators $\mathcal{T}(\varphi,F,\psi)$, $(F,\psi)\in\mathcal{F}\times\Psi$. As applications, we obtain an homotopy result and a partial metric version of the Boyd-Wong fixed point theorem.

2 Main result

Our main result is given in the following theorem.

Theorem 2.1

Let $(X,d)$ be a complete metric space and $T: X\to X$ be a given operator. Suppose that the following conditions hold:

• (i) there exist $\varphi: X\to[0,\infty)$, $F\in\mathcal{F}$, and $\psi\in\Psi$ such that $T\in\mathcal{T}(\varphi,F,\psi)$;

• (ii) φ is lower semi-continuous.

Then the set $F_{T}$ is φ-admissible. Moreover, the operator T has a unique fixed point.

Proof

Let ξ be an arbitrary element of the set $F_{T}$. Take $x=y=\xi$ in (1.1), and we get

$$ F\bigl(0,\varphi(\xi),\varphi(\xi)\bigr)\leq\psi \bigl(F \bigl(0, \varphi(\xi),\varphi (\xi) \bigr) \bigr). $$

(2.1)

If $F(0,\varphi(\xi),\varphi(\xi))\neq0$, from (Ψ2), we get

$$\psi \bigl(F \bigl(0,\varphi(\xi),\varphi(\xi) \bigr) \bigr)< F\bigl(0,\varphi( \xi ),\varphi(\xi)\bigr), $$

which is impossible from (2.1). Consequently, we have

$$F\bigl(0,\varphi(\xi),\varphi(\xi)\bigr)=0. $$

Using the above equality and (F1), we obtain

$$\varphi(\xi)\leq F\bigl(0,\varphi(\xi),\varphi(\xi)\bigr)=0, $$

which yields

$$\varphi(\xi)=0. $$

Consequently, we have

$$ F_{T}\subseteq Z_{\varphi}. $$

(2.2)

Now, we have to prove that $F_{T}$ is a nonempty set. Let $x_{0}$ be an arbitrary element of X. Consider the Picard sequence $\{x_{n}\}\subset X$ defined by

$$x_{n}=T^{n}x_{0},\quad n\in\mathbb{N}=\{0,1,2,\ldots\}, $$

where $T^{n}$ is the nth iterate of T. If for some $N\in \mathbb{N}$ we have $x_{N}=x_{N+1}$, then $x_{N}$ will be an element of $F_{T}$. As a result we can suppose that

$$ d(x_{n},x_{n+1})>0,\quad n\in\mathbb{N}. $$

(2.3)

Using (1.1), we have

$$\begin{aligned}& F \bigl(d(Tx_{n},Tx_{n-1}),\varphi(Tx_{n}), \varphi(Tx_{n-1}) \bigr) \\& \quad\leq \psi \bigl(F \bigl(d(x_{n},x_{n-1}), \varphi(x_{n}),\varphi(x_{n-1}) \bigr) \bigr),\quad n\in\mathbb{N}^{*}; \end{aligned}$$

(2.4)

here $\mathbb{N}^{*}=\{1,2,\ldots\}$. If for some $N\in\mathbb{N}^{*}$, we have

$$F \bigl(d(x_{N},x_{N-1}),\varphi(x_{N}), \varphi(x_{N-1}) \bigr)=0, $$

then property (F1) yields

$$d(x_{N},x_{N-1})\leq F \bigl(d(x_{N},x_{N-1}), \varphi(x_{N}),\varphi (x_{N-1}) \bigr)=0, $$

which is a contradiction with (2.3). Thus

$$ F \bigl(d(x_{n},x_{n-1}),\varphi(x_{n}), \varphi(x_{n-1}) \bigr)>0,\quad n\in \mathbb{N}^{*}. $$

(2.5)

Using (2.4), (2.5), the definition of the sequence $\{x_{n}\}$, and (Ψ2), we have

$$\begin{aligned} \textstyle\begin{cases} F (d(x_{n+1},x_{n}),\varphi(x_{n+1}),\varphi(x_{n}) )\leq\psi (F (d(x_{n},x_{n-1}),\varphi(x_{n}),\varphi(x_{n-1}) ) ),\\ \psi (F (d(x_{n},x_{n-1}),\varphi(x_{n}),\varphi(x_{n-1}) ) )< F (d(x_{n},x_{n-1}),\varphi(x_{n}),\varphi(x_{n-1}) ), \end{cases}\displaystyle \displaystyle \quad n\in\mathbb{N}^{*}. \end{aligned}$$

(2.6)

It follows immediately from (2.6) that there exists some $c\geq 0$ such that

$$\begin{aligned}& \lim_{n\to\infty} F \bigl(d(x_{n+1},x_{n}), \varphi(x_{n+1}),\varphi (x_{n}) \bigr) \\& \quad=\lim_{n\to\infty}\psi \bigl(F \bigl(d(x_{n},x_{n-1}), \varphi(x_{n}),\varphi (x_{n-1}) \bigr) \bigr)=c. \end{aligned}$$

(2.7)

Suppose now that $c>0$. Using the properties (Ψ1)-(Ψ2), we deduce from (2.7) that

$$c=\limsup_{n\to\infty}\psi \bigl(F \bigl(d(x_{n},x_{n-1}), \varphi (x_{n}),\varphi(x_{n-1}) \bigr) \bigr)\leq\psi(c)< c, $$

which is a contradiction. As a consequence, we have $c=0$, that is,

$$\begin{aligned}& \lim_{n\to\infty} F \bigl(d(x_{n+1},x_{n}), \varphi(x_{n+1}),\varphi (x_{n}) \bigr) \\& \quad=\lim_{n\to\infty}\psi \bigl(F \bigl(d(x_{n},x_{n-1}), \varphi(x_{n}),\varphi (x_{n-1}) \bigr) \bigr)=0. \end{aligned}$$

(2.8)

Using (F1) and (2.8), we get

$$ \lim_{n\to\infty} d(x_{n+1},x_{n})= \lim_{n\to\infty}\varphi(x_{n})=0. $$

(2.9)

Next, we show that $\{x_{n}\}$ is a Cauchy sequence in the metric space $(X,d)$. Suppose that $\{x_{n}\}$ is not a Cauchy sequence. Then there exists $\varepsilon>0$ for which we can find two sequences of positive integers $\{m(k)\}$ and $\{n(k)\}$ such that, for all $k\in\mathbb{N}$,

$$ n(k)>m(k)>k,\qquad d(x_{m(k)}, x_{n(k)}) \geq \varepsilon,\quad\quad d(x_{m(k)},x_{n(k)-1})< \varepsilon. $$

(2.10)

Using (2.10), for all $k\in\mathbb{N}$ we have

$$\begin{aligned} \varepsilon \leq& d(x_{m(k)}, x_{n(k)}) \\ \leq& d(x_{m(k)},x_{n(k)-1})+d(x_{n(k)-1},x_{n(k)}) \\ < & \varepsilon+ d(x_{n(k)-1},x_{n(k)}), \end{aligned}$$

which yields

$$ \varepsilon\leq d(x_{m(k)}, x_{n(k)})< \varepsilon+ d(x_{n(k)-1},x_{n(k)}), \quad k\in\mathbb{N}. $$

(2.11)

Letting $k\to\infty$ in the above inequality and using (2.9), we obtain

$$ \begin{aligned}&\lim_{k\to\infty} d(x_{m(k)}, x_{n(k)})=\varepsilon^{+} \quad \textit{i.e.} \\ & \lim_{k\to\infty} d(x_{m(k)}, x_{n(k)})=\varepsilon\quad \hbox{and}\quad d(x_{m(k)}, x_{n(k)}) \geq \epsilon\quad \mbox{for } k\in\mathbb{N}. \end{aligned} $$

(2.12)

Using the properties (F2)-(F3), (2.9), and (2.12), we get

$$\lim_{k\to\infty}F \bigl(d(x_{n(k)},x_{m(k)}), \varphi(x_{n(k)}),\varphi (x_{m(k)}) \bigr)=F(\varepsilon,0,0)= \varepsilon^{+}. $$

Using the above limit and (Ψ1), we obtain

$$ \limsup_{k\to\infty} \psi \bigl(F \bigl(d(x_{n(k)},x_{m(k)}), \varphi (x_{n(k)}),\varphi(x_{m(k)}) \bigr) \bigr)\leq\psi( \varepsilon). $$

(2.13)

On the other hand, using (1.1) and (F1), for all $k\in\mathbb {N}$ we have

$$\begin{aligned} \varepsilon \leq& d(x_{n(k)},x_{m(k)})\leq d(x_{n(k)},x_{n(k)+1})+d(x_{n(k)+1},x_{m(k)+1})+d(x_{m(k)+1},x_{m(k)}) \\ \leq& d(x_{n(k)},x_{n(k)+1})+F \bigl(d(x_{n(k)+1},x_{m(k)+1}), \varphi (x_{n(k)+1}),\varphi(x_{m(k)+1}) \bigr)+d(x_{m(k)+1},x_{m(k)}) \\ \leq& d(x_{n(k)},x_{n(k)+1})+\psi \bigl(F \bigl(d(x_{n(k)},x_{m(k)}), \varphi(x_{n(k)}),\varphi(x_{m(k)}) \bigr) \bigr)+d(x_{m(k)+1},x_{m(k)}). \end{aligned}$$

Passing to the limit superior as $k\to\infty$, using (2.9), (2.13), and (Ψ2), we obtain

$$\varepsilon\leq\psi(\varepsilon)< \varepsilon, $$

which is a contradiction. As a consequence, $\{x_{n}\}$ is a Cauchy sequence. Since $(X,d)$ is a complete metric space, there is a $z\in X$ such that

$$ \lim_{n\to\infty}d(x_{n},z)=0. $$

(2.14)

Since φ is lower semi-continuous, it follows from (2.14) and (2.9) that

$$0\leq\varphi(z)\leq\liminf_{n\to\infty}\varphi(x_{n})=0, $$

which yields

$$ z\in Z_{\varphi}. $$

(2.15)

Now we show that $z\in F_{T}$. Using (1.1), (F1), and (2.15), we have

$$ d(x_{n+1},Tz)\leq \psi \bigl(F \bigl(d(x_{n},z), \varphi(x_{n}),0 \bigr) \bigr),\quad n\in\mathbb{N}. $$

(2.16)

Also using the continuity of F, (F2), (2.14), and (2.9), we have

$$\lim_{n\to\infty}F \bigl(d(x_{n},z),\varphi(x_{n}),0 \bigr)=F(0,0,0)=0. $$

Note that from (Ψ2), we have

$$\lim_{t\to0^{+}}\psi(t)=0. $$

Then

$$ \lim_{n\to\infty} \psi \bigl(F \bigl(d(x_{n},z), \varphi(x_{n}),0 \bigr) \bigr)=\lim_{t\to0^{+}}\psi(t)=0. $$

(2.17)

Now, passing $n\to\infty$ in (2.16) and using (2.17), we get

$$\lim_{n\to\infty}d(x_{n+1},Tz)=0. $$

The uniqueness of the limit yields $z=Tz$. Thus $F_{T}$ is a nonempty set, and the φ-admissibility of $F_{T}$ is proved. Finally, in order to prove the uniqueness of the fixed point, let us assume that $w\in F_{T}$ with $d(z,w)>0$. Since $F_{T}$ is φ-admissible, we know that $z,w\in Z_{\varphi}$. Now, applying (1.1) with $(x,y)=(z,w)$, we obtain

$$F \bigl(d(z,w),0,0 \bigr)\leq\psi \bigl(F \bigl(d(z,w),0,0 \bigr) \bigr). $$

Using the properties (F2) and (Ψ2), we get

$$d(z,w)\leq\psi\bigl(d(z,w)\bigr)< d(z,w), $$

which is a contradiction. Thus T has a unique fixed point. □

Remark 2.2

Take $\varphi\equiv0$ and $F(a,b,c)=a+b+c$ in Theorem 2.1, and we recover the Boyd-Wong fixed point theorem [1].

Now, we give some examples to illustrate our main result given by Theorem 2.1.

Example 2.3

We endow the set $X=[0,\infty)$ with the standard metric

$$d(x,y)=|x-y|, \quad(x,y)\in X\times X. $$

Let $T: X\to X$ be the mapping defined by

$$ Tx=\textstyle\begin{cases} 0 &\mbox{if } 0\leq x\leq1,\\ \frac{x}{2} &\mbox{if } 1< x. \end{cases} $$

Observe that T is not continuous in X. So, there is no $\psi\in\Psi$ such that

$$d(Tx,Ty)\leq\psi\bigl(d(x,y)\bigr),\quad (x,y)\in X\times X. $$

Then the Boyd-Wong fixed point theorem cannot be applied in this case. Let $\varphi: X\to[0,\infty)$ be the function defined by

$$\varphi(x)=x^{n},\quad x\in X, \mbox{ for some } n\in\mathbb{N}^{*}. $$

Let $F: [0,\infty)^{3}\to[0,\infty)$ be the function defined by

$$F(a,b,c)=a+b+c, \quad a,b,c\geq0. $$

Let $\psi:[0,\infty)\to[0,\infty)$ be the function defined by

$$\psi(t)=\frac{t}{2},\quad t\geq0. $$

Observe that F belongs to the set $\mathcal{F}$ and ψ belongs to the set Ψ. We claim that $T\in\mathcal{T}(\varphi,F,\psi)$, that is,

$$ d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)\leq\frac{1}{2}\bigl[d(x,y)+ \varphi (x)+\varphi(y)\bigr],\quad (x,y)\in X\times X. $$

(2.18)

In order to prove our claim, we distinguish three cases.

Case 1. $(x,y)\in[0,1]\times[0,1]$.

In this case, we have

$$d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)=0\leq\frac{1}{2}\bigl[d(x,y)+\varphi (x)+ \varphi(y)\bigr]. $$

Case 2. $(x,y)\in[0,1]\times(1,\infty)$.

In this case, we have

$$d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)=\frac{y}{2}+ \biggl(\frac{y}{2} \biggr)^{n}, $$

while

$$\frac{1}{2}\bigl[d(x,y)+\varphi(x)+\varphi(y)\bigr]= \frac{1}{2} \bigl[y-x+x^{n}+y^{n}\bigr]. $$

Then we have to prove that

$$y^{n} \biggl(\frac{1}{2^{n-1}}-1 \biggr)\leq x^{n}-x. $$

Observe that the function $h: [0,1]\to\mathbb{R}$ defined by

$$h(x)=x^{n}-x,\quad x\in[0,1], $$

has a global minimum at $x_{n}= (\frac{1}{n} )^{\frac{1}{n-1}}$ which is equal to $x_{n} (\frac{1-n}{n} )\geq\frac{1-n}{n}$. So, we have just to check that

$$y^{n} \biggl(\frac{1}{2^{n-1}}-1 \biggr)\leq\frac{1}{n}-1. $$

Since $y>1$, we have

$$y^{n} \biggl(\frac{1}{2^{n-1}}-1 \biggr)\leq\frac{1}{2^{n-1}}-1\leq \frac{1}{n}-1. $$

Then our claim holds in this case.

Case 3. $x,y>1$.

In this case, we have

$$d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)=\frac{|x-y|}{2}+ \biggl(\frac{x}{2} \biggr)^{n}+ \biggl(\frac{y}{2} \biggr)^{n} $$

and

$$\frac{1}{2}\bigl[d(x,y)+\varphi(x)+\varphi(y)\bigr]=\frac{|x-y|}{2}+ \frac{x^{n}+y^{n}}{2}. $$

Obviously, we have

$$d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)\leq\frac{1}{2}\bigl[d(x,y)+\varphi (x)+ \varphi(y)\bigr]. $$

Finally, in all cases our claim (2.18) holds, which yields $T\in \mathcal{T}(\varphi,F,\psi)$. By Theorem 2.1, the set $F_{T}$ is φ-admissible and T has a unique fixed point. In this example, $F_{T}=\{0\}$ and $\varphi(0)=0$.

Example 2.4

We endow the set $X=[\sqrt{2},\infty)$ with the standard metric

$$d(x,y)=|x-y|, \quad(x,y)\in X\times X. $$

Let $T: X\to X$ be the mapping defined by

$$ Tx=\textstyle\begin{cases} \sqrt{2} &\mbox{if } \sqrt{2}\leq x\leq2\sqrt{2},\\ \frac{x}{2} &\mbox{if } 2\sqrt{2}< x. \end{cases} $$

As in the previous example, the Boyd-Wong fixed point theorem cannot be applied in this case. Let $\varphi: X\to[0,\infty)$ be the function defined by

$$\varphi(x)=x^{2}-2,\quad x\in X. $$

Let $F: [0,\infty)^{3}\to[0,\infty)$ be the function defined by

$$F(a,b,c)=a+b+c, \quad a,b,c\geq0. $$

Let $\psi:[0,\infty)\to[0,\infty)$ be the function defined by

$$\psi(t)=\frac{t}{2},\quad t\geq0. $$

We distinguish three cases.

Case 1. $(x,y)\in[\sqrt{2}, 2\sqrt{2}]\times [\sqrt{2}, 2\sqrt{2}]$.

In this case, we have

$$d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)=0\leq\frac{1}{2}\bigl[d(x,y)+\varphi (x)+ \varphi(y)\bigr]. $$

Case 2. $(x,y)\in[\sqrt{2}, 2\sqrt{2}]\times (2\sqrt{2},\infty)$.

In this case, we have

$$d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)=\frac{y}{2}-\sqrt{2} +\frac{y^{2}}{4}-2, $$

while

$$\frac{1}{2}\bigl[d(x,y)+\varphi(x)+\varphi(y)\bigr]=\frac{y}{2}- \frac{x}{2}+\frac {x^{2}}{2}+\frac{y^{2}}{2}-2. $$

Clearly, we have

$$d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)\leq\frac{1}{2}\bigl[d(x,y)+\varphi (x)+ \varphi(y)\bigr]. $$

Case 3. $(x,y)\in(2\sqrt{2},\infty)$.

In this case, we have

$$d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)=\frac{|x-y|}{2}+\frac{x^{2}}{4}+ \frac{y^{2}}{4}-4, $$

while

$$\frac{1}{2}\bigl[d(x,y)+\varphi(x)+\varphi(y)\bigr]=\frac{|x-y|}{2}+ \frac {x^{2}}{2}+\frac{y^{2}}{2}-2. $$

Also, we have

$$d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)\leq\frac{1}{2}\bigl[d(x,y)+\varphi (x)+ \varphi(y)\bigr]. $$

As a consequence, the mapping T belongs to $\mathcal{T}(\varphi,F,\psi )$. By Theorem 2.1, the set $F_{T}$ is φ-admissible and T has a unique fixed point. In this example, $F_{T}=\{\sqrt{2}\}$ and $\varphi(\sqrt{2})=0$.

Example 2.5

Let $(X,d)$ be the metric space considered in Example 2.4. We take the functions φ, F, and ψ defined in Example 2.4. Let $T: X\to X$ be the mapping defined by

$$\begin{aligned} Tx=\textstyle\begin{cases} \sqrt{2} &\mbox{if } \sqrt{2}\leq x\leq2\sqrt{2},\\ \frac{\sin x}{2} &\mbox{if } 2\sqrt{2}< x. \end{cases}\displaystyle \displaystyle \end{aligned}$$

Similarly, we have $T\in\mathcal{T}(\varphi,F,\psi)$. By Theorem 2.1, the set $F_{T}$ is φ-admissible and T has a unique fixed point. In this example, $F_{T}=\{\sqrt{2}\}$ and $\varphi(\sqrt{2})=0$.

3 Applications

3.1 An homotopy result

Let us denote by $\mathcal{F}^{*}$ the set of functions $F\in \mathcal{F}$ satisfying the following property:

(F4):

for all $a,b,c,d\geq0$,

$$a\leq c+d \quad\Longrightarrow\quad F(a,b,0)\leq F(c,b,0)+d. $$

As examples, the following functions belong to $\mathcal{F}^{*}$:

• 1. $F(a,b,c)=(a+b)e^{c}$,

• 2. $F(a,b,c)=(a+b)(c+1)^{n}$, $n\in\mathbb{N}$.

Observe that $\mathcal{F}^{*}\subsetneq\mathcal{F}$. To see this, let us consider the function

$$F(a,b,c)=ae^{c+b}+be^{a+c},\quad a,b,c\geq0. $$

It is not difficult to check that $F\in\mathcal{F}$ but $F\notin \mathcal{F}^{*}$.

We have the following homotopy result.

Theorem 3.1

Let $(X,d)$ be a complete metric space, U be an open subset of X, and V be a closed subset of X with $U\subset V$. Suppose that $H: V\times[0,1]\to X$ has the following properties:

(C1):

$x\neq H(x,\lambda)$ for every $x\in V\backslash U$ and $\lambda\in[0,1]$;

(C2):

there exist a continuous function $\varphi: X\to[0,\infty )$, $L\in(0,1)$, and $F\in\mathcal{F}^{*}$ such that for all $x,y\in V$ and $\lambda\in[0,1]$,

$$F \bigl(d\bigl(H(x,\lambda),H(y,\lambda)\bigr),\varphi\bigl(H(x,\lambda)\bigr), \varphi \bigl(H(y,\lambda) \bigr)\bigr)\leq L F \bigl(d(x,y),\varphi(x), \varphi(y) \bigr); $$

(C3):

there exists a continuous function $\eta:[0,1]\to\mathbb {R}$ such that for all $x\in V$ and $\lambda,\mu\in[0,1]$,

$$F \bigl(d\bigl(H(x,\lambda),H(x,\mu)\bigr),\varphi\bigl(H(x,\lambda)\bigr), \varphi \bigl(H(x,\mu ) \bigr)\bigr)\leq \bigl|\eta(\lambda)-\eta(\mu)\bigr|. $$

Then $H(\cdot,0)$ has a fixed point if and only if $H(\cdot,1)$ has a fixed point.

Proof

Suppose that $H(\cdot,0)$ has a fixed point. Consider the set

$$Q=\bigl\{ t\in[0,1]: x= H(x,t)\mbox{ for some }x \in U\bigr\} . $$

From (C1), clearly 0 is an element of Q, so Q is a nonempty set. We will show that Q is both closed and open in $[0,1]$, and so by the connectedness of $[0,1]$, we are finished since $Q=[0,1]$. First, let us prove that Q is open in $[0,1]$. Let $t_{0}\in Q$ and $x_{0}\in U$ with $x_{0}= H(x_{0},t_{0})$. Using (C2) with $x=y=x_{0}$ and $\lambda=t_{0}$, we obtain

$$F\bigl(0,\varphi(x_{0}), \varphi(x_{0})\bigr)\leq L F \bigl(0,\varphi(x_{0}), \varphi(x_{0})\bigr), $$

which implies since $L\in(0,1)$ that

$$F\bigl(0,\varphi(x_{0}), \varphi(x_{0})\bigr)=0. $$

Then (F1) yields

$$\varphi(x_{0})=0. $$

Moreover, observe that, for all $t\in[0,1]$, if $x\in U$ is a fixed point of $H(\cdot,t)$, then $\varphi(x)=0$. On the other hand, since U is open in $(X,d)$, there exists $r>0$ such that $B(x_{0},r) \subseteq U$, where

$$B(x_{0},r)=\bigl\{ z\in X: d(x_{0},z)< r\bigr\} . $$

Consider the set

$$\Lambda(x_{0},\varphi)=\bigl\{ z\in X: F\bigl(d(z,x_{0}), \varphi(z),0\bigr) < r\bigr\} . $$

Clearly $\Lambda(x_{0},\varphi)$ is nonempty (since $x_{0}\in \Lambda (x_{0},\varphi)$) and $\Lambda(x_{0},\varphi)\subseteq B(x_{0},r)$. Let $\varepsilon= (1-L)r>0$. Since η is continuous on $t_{0}$, there exists $\alpha(\varepsilon)>0$ such that

$$t\in\bigl(t_{0}-\alpha(\varepsilon),t_{0}+\alpha( \varepsilon)\bigr)\cap [0,1]\quad\Longrightarrow\quad\bigl|\eta(t)-\eta(t_{0})\bigr|< \varepsilon. $$

Let $t\in(t_{0}-\alpha(\varepsilon),t_{0}+\alpha(\varepsilon))\cap[0,1]$. For $x\in\overline{\Lambda(x_{0},\varphi)}$ (the closure of $\Lambda(x_{0},\varphi)$), we have

$$F \bigl(d\bigl(H(x,t),x_{0}\bigr),\varphi\bigl(H(x,t)\bigr),0 \bigr) =F \bigl(d\bigl(H(x,t),H(x_{0},t_{0})\bigr),\varphi \bigl(H(x,t)\bigr),0 \bigr). $$

Also since

$$d\bigl(H(x,t),H(x_{0},t_{0})\bigr)\leq d \bigl(H(x,t),H(x,t_{0})\bigr)+d\bigl(H(x,t_{0}),H(x_{0},t_{0}) \bigr), $$

using the properties (F1), (F4) we get

$$\begin{aligned}& F \bigl(d\bigl(H(x,t),H(x_{0},t_{0})\bigr),\varphi \bigl(H(x,t)\bigr),0 \bigr) \\& \quad\leq F \bigl(d\bigl(H(x,t),H(x,t_{0})\bigr),\varphi\bigl(H(x,t) \bigr),0 \bigr)+d\bigl(H(x,t_{0}),H(x_{0},t_{0}) \bigr) \\& \quad\leq F \bigl(d\bigl(H(x,t),H(x,t_{0})\bigr),\varphi\bigl(H(x,t) \bigr),0 \bigr)+F \bigl(d\bigl(H(x,t_{0}),H(x_{0},t_{0}) \bigr),\varphi\bigl(H(x,t_{0})\bigr),0 \bigr) \\& \quad\leq\bigl|\eta(t)-\eta(t_{0})\bigr|+L F \bigl(d(x,x_{0}), \varphi(x),0 \bigr) \\& \quad< \varepsilon+Lr=r. \end{aligned}$$

Thus we proved that, for all $t\in(t_{0}-\alpha(\varepsilon),t_{0}+\alpha (\varepsilon))\cap[0,1]$, the operator

$$H(\cdot,t): \overline{\Lambda(x_{0},\varphi)}\to\overline{\Lambda (x_{0},\varphi)} $$

is well defined. Now, using (C2) and Theorem 2.1, we deduce that, for all $t\in (t_{0}-\alpha(\varepsilon), t_{0}+\alpha(\varepsilon))\cap[0,1]$, the operator $H(\cdot,t)$ has a fixed point in V. However, such a fixed point should be in U from (C1). As a consequence,

$$\bigl(t_{0}-\alpha(\varepsilon),t_{0}+\alpha(\varepsilon) \bigr)\cap[0,1]\subseteq Q, $$

which proves that Q is open in $[0,1]$. Next, we show that Q is closed in $[0,1]$. To see this, let $\{t_{n}\}$ be a sequence in Q with $t_{n}\to t\in[0,1]$ as $n\to\infty$. We have to prove that $t\in Q$. From the definition of Q, for all $n\in\mathbb{N}$, there exists $x_{n}\in U$ with

$$x_{n} =H(x_{n},t_{n}) \quad\mbox{and}\quad \varphi(x_{n})=0. $$

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

$$\begin{aligned} d(x_{n},x_{m}) =&d\bigl(H(x_{n},t_{n}),H(x_{m},t_{m}) \bigr)\\ \leq& d\bigl(H(x_{n},t_{n}),H(x_{n},t_{m}) \bigr)+d\bigl(H(x_{n},t_{m}),H(x_{m},t_{m}) \bigr) \\ \leq& F \bigl(d\bigl(H(x_{n},t_{n}),H(x_{n},t_{m}) \bigr),\varphi\bigl(H(x_{n},t_{n})\bigr),\varphi \bigl(H(x_{n},t_{m})\bigr) \bigr) \\ &{}+F \bigl(d\bigl(H(x_{n},t_{m}),H(x_{m},t_{m}) \bigr),\varphi\bigl(H(x_{n},t_{m})\bigr),\varphi \bigl(H(x_{m},t_{m})\bigr) \bigr) \\ \leq&\bigl|\eta(t_{n})-\eta(t_{m})\bigr|+LF \bigl(d(x_{n},x_{m}),0,0 \bigr) \\ =&\bigl|\eta(t_{n})-\eta(t_{m})\bigr|+Ld(x_{n},x_{m}), \end{aligned}$$

which yields

$$d(x_{n},x_{m})\leq\frac{|\eta(t_{n})-\eta(t_{m})|}{1-L},\quad m,n\in\mathbb{N}. $$

Letting $m,n\to\infty$ in the above inequality and using the continuity of η, we get $d(x_{n},x_{m})\to0$ as $m,n\to\infty$, which implies that $\{x_{n}\}$ is a Cauchy sequence in the complete metric space $(X,d)$. Then there is some $z\in V$ (since V is closed) such that

$$\lim_{n\to\infty} d(x_{n},z)=0 \quad\mbox{and}\quad \varphi(z)=0, $$

since φ is lower semi-continuous. Now, for all $n\in\mathbb {N}$ we have

$$\begin{aligned} d\bigl(x_{n},H(z,t)\bigr) =& d\bigl(H(x_{n},t_{n}),H(z,t) \bigr) \leq d\bigl(H(x_{n},t_{n}),H(x_{n},t)\bigr)+d \bigl(H(x_{n},t),H(z,t)\bigr) \\ \leq& F \bigl(d\bigl(H(x_{n},t_{n}),H(x_{n},t) \bigr),\varphi\bigl(H(x_{n},t_{n})\bigr),\varphi \bigl(H(x_{n},t)\bigr) \bigr)\\ &{}+ F \bigl(d\bigl(H(x_{n},t),H(z,t) \bigr),\varphi\bigl(H(x_{n},t)\bigr),\varphi\bigl(H(z,t)\bigr) \bigr) \\ \leq&\bigl|\eta(t_{n})-\eta(t)\bigr|+LF \bigl(d(x_{n},z),0,0 \bigr) \\ =&\bigl|\eta(t_{n})-\eta(t)\bigr|+Ld(x_{n},z). \end{aligned}$$

Letting $n\to\infty$ in the above inequality, we obtain

$$\lim_{n\to\infty}d\bigl(x_{n},H(z,t)\bigr)=0. $$

The uniqueness of the limit yields $z=H(z,t)$. Using (C1), we deduce that $z\in U$ and $t\in Q$. Thus Q is closed in $[0,1]$.

For the reverse implication, we use the same technique. □

3.2 A partial metric version of Boyd-Wong fixed point theorem

In this part, using Theorem 2.1, we obtain a partial metric version of the Boyd-Wong fixed point theorem.

We start by recalling some basic definitions and properties of partial metric spaces. For more details of such spaces, we refer the reader to [2–20].

A partial metric on a nonempty set X is a function $p: X\to X \to[0, \infty)$ such that for all $x,y,z\in X$, we have

• (i) $p(x,x)=p(y,y)=p(x,y)\Longleftrightarrow x=y$;

• (ii) $p(x,x)\leq p(x,y)$;

• (iii) $p(x,y)=p(y,x)$;

• (iv) $p(x,y)\leq p(x,z)+p(z,y)-p(z,z)$.

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 (i)-(ii), $x=y$; but if $x=y$, $p(x,y)$ may not be 0. A basic example of a partial metric space is the pair $([0, \infty),p)$, where $p(x,y)= \max\{x,y\}$.

Each partial metric p on X generates a $T_{0}$ topology $\tau_{p}$ on X which has as a base the family of open p-balls $\{ B_{p}(x,\varepsilon): x\in X, \varepsilon>0\}$, where

$$B_{p}(x,\varepsilon):=\bigl\{ y\in X: p(x,y)< p(x,x)+\varepsilon\bigr\} . $$

Let $(X,p)$ be a partial metric space. A sequence $\{x_{n}\}\subset X$ converges to some $x\in X$ with respect to p if and only if

$$\lim_{n\to\infty} p(x_{n},x)=p(x,x). $$

A sequence $\{x_{n}\}\subset X$ is said to be a Cauchy sequence if and only if $\lim_{m,n\to\infty}p(x_{n},x_{m})$ exists and is finite. The partial metric space $(X,p)$ is said to be complete if and only if every Cauchy sequence $\{x_{n}\}$ in X converges to some $x\in X$ such that $\lim_{n,m\to\infty}p(x_{n},x_{m})=p(x,x)$.

If p is a partial metric on X, then the function $d_{p}: X\to X\to [0,\infty)$ defined by

$$ d_{p}(x,y)=2p(x,y)-p(x,x)-p(y,y),\quad (x,y)\in X^{2}, $$

(3.1)

is a metric on X.

Lemma 3.2

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

• (i) $\{x_{n}\}$ is a Cauchy sequence in $(X,p)$ if and only if $\{x_{n}\}$ is a Cauchy sequence in the metric space $(X,d_{p})$;

• (ii) the partial metric space $(X,p)$ is complete if and only if the metric space $(X,d_{p})$ is complete. Furthermore, $\lim_{n\to\infty} d_{p}(x_{n}, x)=0$ if and only if

  $$\lim_{n\to\infty}p(x_{n},x)=p(x,x)=\lim _{m,n\to\infty}p(x_{n},x_{m}). $$

We have the following result.

Corollary 3.3

Let $(X,p)$ be a complete partial metric space and let $T: X\to X$ be an operator such that

$$ p(Tx,Ty)\leq\psi\bigl(p(x,y)\bigr), \quad(x,y)\in X\times X, $$

(3.2)

where $\psi\in\Psi$. We have the following results:

• (i) if $z\in X$ is a fixed point of T then $p(z,z)=0$;

• (ii) T has a unique fixed point.

Proof

Let $d_{p}$ be the metric on X defined by (3.1). We have

$$p(x,y)=d(x,y)+\varphi(x)+\varphi(y), \quad(x,y)\in X\times X, $$

where

$$d(x,y)=\frac{d_{p}(x,y)}{2}, \quad\quad\varphi(x)=\frac{p(x,x)}{2}. $$

Then (3.2) yields

$$F\bigl(d(Tx,Ty),\varphi(Tx),\varphi(Ty)\bigr)\leq\psi \bigl(F \bigl(d(x,y), \varphi (x),\varphi(y) \bigr) \bigr), \quad(x,y)\in X\times X, $$

where

$$F(a,b,c)=a+b+c, \quad a,b,c\geq0. $$

From (ii) Lemma 3.2, the metric space $(X,d)$ is complete and the function φ is continuous with respect to the topology of d. Finally the desired result follows from Theorem 2.1. □

Remark 3.4

Take in Corollary 3.3, $\psi(t)= kt$ with $k\in(0,1)$, and we recover Matthews fixed point theorem [9].