Some applications via fixed point results in partially ordered $S_{b}$-metric spaces

Abstract

In this paper we give some applications to integral equations as well as homotopy theory via fixed point theorems in partially ordered complete $S_{b}$-metric spaces by using generalized contractive conditions. We also furnish an example which supports our main result.

1 Introduction

Banach contraction principle in metric spaces is one of the most important results in fixed theory and nonlinear analysis in general. Since 1922, when Stefan Banach [1] formulated the concept of contraction and posted a famous theorem, scientists around the world have published new results related to the generalization of a metric space or with contractive mappings (see [1–24]). Banach contraction principle is considered to be the initial result of the study of fixed point theory in metric spaces.

In the year 1989, Bakhtin introduced the concept of b-metric spaces as a generalization of metric spaces [6]. Later several authors proved so many results on b-metric spaces (see [13–16]). Mustafa and Sims defined the concept of a generalized metric space which is called a G-metric space [12]. Sedghi, Shobe and Aliouche gave the notion of an S-metric space and proved some fixed point theorems for a self-mapping on a complete S-metric space [22]. Aghajani, Abbas and Roshan presented a new type of metric which is called $G_{b}$-metric and studied some properties of this metric [2].

Recently Sedghi et al. [20] defined $S_{b}$-metric spaces using the concept of S-metric spaces [22].

The aim of this paper is to prove some unique fixed point theorems for generalized contractive conditions in complete $S_{b}$-metric spaces. Also, we give applications to integral equations as well as homotopy theory. Throughout this paper $R, R^{+}$ and N denote the sets of all real numbers, non-negative real numbers and positive integers, respectively.

First we recall some definitions, lemmas and examples.

2 Preliminaries

Definition 2.1

[22]

Let X be a non-empty set. An S-metric on X is a function $S:X^{3} \to[0,+\infty)$ that satisfies the following conditions for each $x,y,z,a \in X$:

$(S1)$::

$0 < S(x, y, z)$ for all $x,y,z \in X$ with $x \neq y \neq z \neq x$,

$(S2)$::

$S(x, y, z) = 0 \mbox{ if and only if } x = y = z$,

$(S3)$::

$S(x,y,z) \leq S(x ,x, a) + S(y, y, a) + S(z, z, a)$ for all $x,y,z,a \in X$.

Then the pair $(X, S)$ is called an S-metric space.

Definition 2.2

[20]

Let X be a non-empty set and $b \geq1$ be a given real number. Suppose that a mapping $S_{b}:X^{3} \to\mathcal{[}0,\infty )$ is a function satisfying the following properties:

$(S_{b} 1)$ :

$0 < S_{b}(x,y,z)$ for all $x,y,z \in X$ with $x \neq y \neq z \neq x$,

$(S_{b} 2)$ :

$S_{b}(x, y, z) = 0 \mbox{ if and only if } x = y = z$,

$(S_{b} 3)$ :

$S_{b}(x,y,z) \leq b(S_{b}(x, x, a) + S_{b}(y, y, a) + S_{b}(z, z, a))$ for all $x,y,z,a \in X$.

Then the function $S_{b}$ is called an $S_{b}$-metric on X and the pair $(X,S_{b})$ is called an $S_{b}$-metric space.

Remark 2.3

[20]

It should be noted that the class of $S_{b}$-metric spaces is effectively larger than that of S-metric spaces. Indeed each S-metric space is an $S_{b}$-metric space with $b=1$.

The following example shows that an $S_{b}$-metric on X need not be an S-metric on X.

Example 2.4

[20]

Let $(X,S)$ be an S-metric space and $S_{*}(x,y,z) = S(x,y,z)^{p}$, where $p>1$ is a real number. Note that $S_{*}$ is an $S_{b}$-metric with $b = 2^{2(p-1)}$. Also, $(X,S_{*})$ is not necessarily an S-metric space.

Definition 2.5

[20]

Let $(X,S_{b})$ be an $S_{b}$-metric space. Then, for $x \in X$, $r > 0$, we define the open ball $B_{S_{b}} (x,r)$ and the closed ball $B_{S_{b}} [x,r]$ with center x and radius r as follows, respectively:

$$\begin{aligned} &B_{S_{b}} (x,r) = \bigl\{ y \in X: S_{b}(y,y,x) < r\bigr\} , \\ &B_{S_{b}} [x,r] = \bigl\{ y \in X: S_{b}(y,y,x) \le r\bigr\} . \end{aligned}$$

Lemma 2.6

[20]

In an $S_{b}$-metric space, we have

$$S_{b}(x,x,y) \le b S_{b}(y,y,x) $$

and

$$S_{b}(y,y,x) \le b S_{b}(x,x,y). $$

Lemma 2.7

[20]

In an $S_{b}$-metric space, we have

$$S_{b}(x,x,z) \le2 b S_{b}(x,x,y)+b^{2} S_{b}(y,y,z). $$

Definition 2.8

[20]

If $(X,S_{b})$ is an $S_{b}$-metric space, a sequence $\{x_{n}\}$ in X is said to be:

• (1) $S_{b}$-Cauchy sequence if, for each $\epsilon> 0$, there exists $n_{0} \in\mathcal {N}$ such that $S_{b}(x_{n},x_{n},x_{m}) < \epsilon$ for each $m,n \geq n_{0}$.

• (2) $S_{b}$-convergent to a point $x \in X$ if, for each $\epsilon> 0$, there exists a positive integer $n_{0}$ such that $S_{b}(x_{n},x_{n},x) < \epsilon$ or $S_{b}(x,x,x_{n}) < \epsilon$ for all $n \geq n_{0}$, and we denote $\mathop{\lim} _{n \rightarrow\infty}x_{n} = x$.

Definition 2.9

[20]

An $S_{b}$-metric space $(X,S_{b})$ is called complete if every $S_{b}$-Cauchy sequence is $S_{b}$-convergent in X.

Lemma 2.10

[20]

If $(X,S_{b})$ is an $S_{b}$-metric space with $b\geq1$, and suppose that $\{x_{n}\}$ is $S_{b}$-convergent to x, then we have

$$\mathrm{(i)}\quad \frac{1}{2b}S_{b}(y,x,x) \le\mathop{\lim} _{n \rightarrow\infty} \inf S_{b}(y,y,x_{n}) \le\mathop{\lim} _{n \rightarrow\infty} \sup S_{b}(y,y,x_{n}) \le2b S_{b}(y,y,x) $$

and

$$\mathrm{(ii)}\quad \frac{1}{b^{2}}S_{b}(x,x,y) \le\mathop{\lim} _{n \rightarrow\infty} \inf S_{b}(x_{n},x_{n},y) \le\mathop{\lim} _{n \rightarrow\infty} \sup S_{b}(x_{n},x_{n},y) \le b^{2} S_{b}(x,x,y) $$

for all $y \in X$.

In particular, if $x=y$, then we have $\mathop{\lim} _{n \rightarrow \infty} S_{b}(x_{n},x_{n},y) = 0$.

Now we prove our main results.

3 Results and discussions

Definition 3.1

Let $(X, S_{b}, \preceq)$ be a partially ordered complete $S_{b}$-metric space which is said to be regular if every two elements of X are comparable,

$$\mbox{i.e., if } x, y \in X \Rightarrow\mbox{ either } x \preceq y \mbox{ or } y \preceq x. $$

Definition 3.2

Let $(X, S_{b}, \preceq)$ be a partially ordered complete $S_{b}$-metric space which is also regular; let $f: X \to X$ be a mapping. We say that f satisfies $(\psi, \phi)$-contraction if there exist $\psi, \phi : [0, \infty) \to[0, \infty)$ such that

(3.2.1):

f is non-decreasing,

(3.2.2):

ψ is continuous, monotonically non-decreasing and ϕ is lower semi-continuous,

(3.2.3):

$\psi(t) = 0 = \phi(t)$ if and only if $t= 0$,

(3.2.4):

$\psi (4b^{4} {S_{b} ( {fx, fx, fy} )} ) \le \psi ( { M_{f}^{i} ( {x,y} )} ) - \phi ( {M_{f}^{i} ( {x,y} )} )$, $\forall x, y \in X$, $x \preceq y$, $i = 3, 4, 5$ and

$$\begin{aligned} &M_{f}^{5} ( {x,y} ) = \max \left \{ {S_{b}(x, x,y), S_{b}(x, x,fx),S_{b}(y, y, fy),S_{b}(x, x,fy), S_{b}(y, y, fx) } \right \}, \\ &M_{f}^{4} ( {x,y} ) = \max \left \{ {S_{b}(x, x,y), S_{b}(x, x,fx),S_{b}(y, y, fy), \frac{1}{4b^{4}} \bigl[S_{b}(x, x,fy)+S_{b}(y, y, fx) \bigr] } \right \}, \\ &M_{f}^{3} ( {x,y} ) = \max\biggl\{ S_{b}(x, x,y), \frac{1}{4b^{4}} \bigl[S_{b}(x, x,fx)+S_{b}(y, y, fy) \bigr],\\ &\phantom{M_{f}^{3} ( {x,y} ) =}\frac{1}{4b^{4}} \bigl[S_{b}(x, x,fy)+S_{b}(y, y, fx) \bigr] \biggr\} . \end{aligned}$$

Definition 3.3

Suppose that $(X, \preceq)$ is a partially ordered set and f is a mapping of X into itself. We say that f is non-decreasing if for every $x, y \in X$,

$$ x \preceq y \mbox{ implies that } f x \preceq f y. $$

(1)

Theorem 3.4

Let $(X,S_{b}, \preceq)$ be an ordered complete $S_{b}$ metric space, which is also regular, and let $f : X \to X$ satisfy $(\psi, \phi )$-contraction with $i = 5$. If there exists $x_{0} \in X$ with $x_{0} \preceq f x_{0}$, then f has a unique fixed point in X.

Proof

Since f is a mapping from X into X, there exists a sequence $\{x_{n} \}$ in X such that

$$x_{n+1} = f x_{n}, \quad n = 0, 1, 2, 3, \ldots. $$

Case (i): If $x_{n} = x_{n+1}$, then $x_{n}$ is a fixed point of f.

Case (ii): Suppose $x_{n} \neq x_{n+1}\ \forall n$.

Since $x_{0} \preceq fx_{0} = x_{1}$ and f is non-decreasing, it follows that

$$x_{0} \preceq f x_{0} \preceq f^{2} x_{0} \preceq f^{3} x_{0} \preceq\cdots \preceq f^{n} x_{0} \preceq f^{n+1} x_{0} \preceq\cdots. $$

Now

$$\begin{aligned} \psi \bigl(4b^{4} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \bigr) &= \psi \bigl(4b^{4} S_{b} ( fx_{0}, fx_{0}, fx_{1} ) \bigr) \\ & \leq\psi \bigl(M_{f}^{5} ( x_{0},x_{1} ) \bigr) - \phi \bigl(M_{f}^{5} ( x_{0},x_{1} ) \bigr), \end{aligned}$$

where

$$\begin{aligned} M_{f}^{5} ( x_{0},x_{1} ) & = \max \left \{ \begin{matrix} S_{b} ( x_{0}, x_{0}, x_{1} ), S_{b} ( x_{0}, x_{0}, fx_{0} ), S_{b} ( x_{1}, x_{1}, fx_{1} )\\ S_{b} ( x_{0}, x_{0}, f^{2}x_{0} ), S_{b} ( fx_{0}, fx_{0}, f x_{0} ) \end{matrix} \right \} \\ & = \max \left \{ S_{b} ( x_{0}, x_{0}, fx_{0} ), S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr), S_{b} \bigl( x_{0}, x_{0}, f^{2}x_{0} \bigr) \right \} . \end{aligned}$$

Therefore

$$\begin{aligned} &\psi \bigl(4b^{4} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \bigr) \\ &\quad \leq \psi \left (\max \left \{ S_{b} ( x_{0}, x_{0}, fx_{0} ), S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr), S_{b} \bigl( x_{0}, x_{0}, f^{2}x_{0} \bigr) \right \} \right ) \\ &\qquad{} - \phi \left (\max \left \{ S_{b} ( x_{0}, x_{0}, fx_{0} ), S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr), S_{b} \bigl( x_{0}, x_{0}, f^{2}x_{0} \bigr) \right \} \right ) \\ &\quad \leq \psi \left (\max \left \{ S_{b} ( x_{0}, x_{0}, fx_{0} ), S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr), S_{b} \bigl( x_{0}, x_{0}, f^{2}x_{0} \bigr) \right \} \right ). \end{aligned}$$

By the definition of ψ, we have that

$$ S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \leq \max \left \{ \begin{matrix} {\frac{1}{4b^{4}} S_{b} ( x_{0}, x_{0}, fx_{0} )} \\ {\frac{1}{4b^{4}} S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} )}\\ {\frac{1}{4b^{4}} S_{b} ( x_{0}, x_{0}, f^{2}x_{0} )} \end{matrix} \right \}. $$

(2)

But

$$\begin{aligned} \frac{1}{4b^{4}} S_{b} \bigl( x_{0}, x_{0}, f^{2}x_{0} \bigr)&\leq \frac{1}{4b^{4}} \bigl[2b S_{b} ( x_{0}, x_{0}, fx_{0} ) + b^{2} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \bigr] \\ &\leq \max \biggl\{ \frac{1}{b^{3}} S_{b} ( x_{0}, x_{0}, fx_{0} ) , \frac{1}{2b^{2}} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \biggr\} . \end{aligned}$$

From (2) we have that

$$\begin{aligned} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \leq \max \biggl\{ \frac{1}{b^{3}} S_{b} ( x_{0}, x_{0}, fx_{0} ) , \frac{1}{2b^{2}} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \biggr\} . \end{aligned}$$

If $\frac{1}{2b^{2}} S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} )$ is maximum, we get a contradiction. Hence

$$ S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \leq \frac{1}{b^{3}} S_{b} ( x_{0}, x_{0}, fx_{0} ). $$

(3)

Also

$$\begin{aligned} \psi \bigl(4b^{4} S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \bigr) &= \psi \bigl(4b^{4} S_{b} ( fx_{1}, fx_{1}, fx_{2} ) \bigr) \\ & \leq\psi \bigl(M_{f}^{5} ( x_{1},x_{2} ) \bigr) - \phi \bigl(M_{f}^{4} ( x_{1},x_{2} ) \bigr), \end{aligned}$$

where

$$\begin{aligned} M_{f}^{5} ( x_{1},x_{2} ) & = \max \left \{ \begin{matrix} S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} ), S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} ), S_{b} ( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} )\\ S_{b} ( fx_{0}, fx_{0}, f^{3}x_{0} ), S_{b} ( f^{2}x_{0}, f^{2}x_{0}, f^{2} x_{0} ) \end{matrix} \right \} \\ & = \max \left \{ S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr), S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr), S_{b} \bigl( fx_{0}, fx_{0}, f^{3}x_{0} \bigr) \right \} . \end{aligned}$$

Therefore

$$\begin{aligned} &\psi \bigl(4b^{4} S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \bigr)\\ &\quad\leq \psi \left (\max \left \{ \begin{matrix} S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} ), S_{b} ( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} )\\ S_{b} ( fx_{0}, fx_{0}, f^{3}x_{0} ) \end{matrix} \right \} \right ) \\ &\qquad{} - \phi \left (\max \left \{ \begin{matrix} S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} ), S_{b} ( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} )\\ S_{b} ( fx_{0}, fx_{0}, f^{3}x_{0} ) \end{matrix} \right \} \right ) \\ &\quad\leq \psi \left (\max \left \{ \begin{matrix} S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} ), S_{b} ( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} )\\ S_{b} ( fx_{0}, fx_{0}, f^{3}x_{0} ) \end{matrix} \right \} \right ). \end{aligned}$$

By the definition of ψ, we have that

$$ S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \leq \max \left \{ \begin{matrix} {\frac{1}{4b^{4}} S_{b} ( fx_{0}, fx_{0}, f^{2}x_{0} )} \\ {\frac{1}{4b^{4}} S_{b} ( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} )} \\ {\frac{1}{4b^{4}} S_{b} ( fx_{0}, fx_{0}, f^{3}x_{0} )} \end{matrix} \right \}. $$

(4)

But

$$\begin{aligned} &\frac{1}{4b^{4}} S_{b} \bigl( fx_{0}, fx_{0}, f^{3}x_{0} \bigr)\\ &\quad\leq \frac {1}{4b^{4}} \bigl[2b S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) + b^{2} S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \bigr] \\ &\quad\leq \max \biggl\{ \frac{1}{b^{3}} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) , \frac{1}{2b^{2}} S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \biggr\} . \end{aligned}$$

From (4) we have that

$$\begin{aligned} S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \leq \max \biggl\{ \frac {1}{b^{3}} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) , \frac{1}{2 b^{2}} S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \biggr\} . \end{aligned}$$

If $\frac{1}{2b^{2}} S_{b} ( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} )$ is maximum, we get a contradiction. Hence

$$\begin{aligned} S_{b} \bigl( f^{2}x_{0}, f^{2}x_{0}, f^{3}x_{0} \bigr) \leq& \frac{1}{b^{3}} S_{b} \bigl( fx_{0}, fx_{0}, f^{2}x_{0} \bigr) \\ \le&\frac{1}{(b^{3})^{2}} S_{b} ( x_{0}, x_{0}, fx_{0} ) . \end{aligned}$$

Continuing this process, we can conclude that

$$\begin{aligned} S_{b} \bigl( f^{n}x_{0}, f^{n}x_{0}, f^{n+1}x_{0} \bigr) &\leq \frac {1}{(b^{3})^{n}} S_{b} ( x_{0}, x_{0}, f x_{0} ) \\ &\to 0 \quad\mbox{as } n \to\infty. \end{aligned}$$

That is,

$$\begin{aligned} \mathop{\lim} _{n \rightarrow\infty} S_{b} \bigl( f^{n}x_{0}, f^{n}x_{0}, f^{n+1}x_{0} \bigr) = 0. \end{aligned}$$

(5)

Now we prove that $\{ f^{n} x_{0} \}$ is an $S_{b}$-Cauchy sequence in $(X, S_{b})$. On the contrary, we suppose that $\{ f^{n} x_{0} \}$ is not $S_{b}$-Cauchy. Then there exist $\epsilon> 0$ and monotonically increasing sequences of natural numbers $\{m_{k}\}$ and $\{ n_{k}\}$ such that $n_{k} > m_{k}$.

$$ S_{b} \bigl( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k}}x_{0} \bigr) \geq\epsilon $$

(6)

and

$$ S_{b} \bigl( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k} - 1}x_{0} \bigr) < \epsilon . $$

(7)

From (6) and (7), we have

$$\begin{aligned} \epsilon\le{}& S_{b} \bigl( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k}}x_{0} \bigr) \\ \le{}& 2b S_{b} \bigl( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{m_{k}+1}x_{0} \bigr) \\ &{}+ b^{2} S_{b} \bigl( f^{m_{k}+1}x_{0}, f^{m_{k}+1}x_{0}, f^{n_{k}}x_{0} \bigr). \end{aligned}$$

So that

$$\begin{aligned} 4 b^{2} \epsilon\le{}& 8 b^{3} S_{b} \bigl( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{m_{k}+1}x_{0} \bigr)\\ &{} + 4b^{4} S_{b} \bigl( f^{m_{k}+1}x_{0}, f^{m_{k}+1}x_{0}, f^{n_{k}}x_{0} \bigr). \end{aligned}$$

Letting $k \to\infty$ and applying ψ on both sides, we have that

$$\begin{aligned} \psi \bigl(4 b^{2} \epsilon \bigr) &\le \mathop{\lim} _{k \rightarrow \infty} \psi \bigl(4b^{4} S_{b} \bigl( f^{m_{k}+1}x_{0}, f^{m_{k}+1}x_{0}, f^{n_{k}}x_{0} \bigr) \bigr) \\ &= \mathop{\lim} _{k \rightarrow\infty} \psi \bigl(4b^{4} S_{b} ( fx_{m_{k}}, fx_{m_{k}}, fx_{n_{k}-1} ) \bigr) \\ &\leq \mathop{\lim} _{k \rightarrow\infty} \psi \bigl( M_{f}^{5} ( x_{m_{k}}, x_{n_{k}-1} ) \bigr) - \mathop{\lim} _{k \rightarrow \infty} \phi \bigl( M_{f}^{5} ( x_{m_{k}}, x_{n_{k}-1} ) \bigr), \end{aligned}$$

(8)

where

$$\begin{aligned} & \mathop{\lim} _{k \rightarrow\infty} M_{f}^{5} ( x_{m_{k}}, x_{n_{k}-1} ) \\ &\quad= \mathop{\lim} _{k \rightarrow\infty} \max \left \{ \begin{matrix} {S_{b} ( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k}-1}x_{0} ), S_{b} ( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{m_{k}+1}x_{0} ) }\\ {S_{b} ( f^{n_{k}-1}x_{0} , f^{n_{k}-1}x_{0} , f^{n_{k}}x_{0} )}, S_{b} ( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k}}x_{0} ) \\ S_{b} ( f^{n_{k}-1}x_{0},f^{n_{k}-1}x_{0}, f^{m_{k}+1}x_{0} ) \end{matrix} \right \} \\ &\quad < \mathop{\lim} _{k \rightarrow\infty} \max \left \{ {\epsilon, 0, 0, } S_{b} \bigl( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k}}x_{0} \bigr), S_{b} \bigl( f^{n_{k}-1}x_{0},f^{n_{k}-1}x_{0}, f^{m_{k}+1}x_{0} \bigr) \right \}. \end{aligned}$$

But

$$\begin{aligned} \mathop{\lim} _{k \rightarrow\infty} S_{b} \bigl( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k}}x_{0} \bigr)&\le \mathop{\lim} _{k \rightarrow \infty} \left [ \begin{matrix} 2b S_{b} ( f^{m_{k}}x_{0}, f^{m_{k}}x_{0}, f^{n_{k}-1}x_{0} )\\ {}+ b^{2} S_{b} ( f^{n_{k}-1}x_{0}, f^{n_{k}-1}x_{0}, f^{n_{k}}x_{0} ) \end{matrix} \right ] < 2 b \epsilon. \end{aligned}$$

Also

$$\begin{aligned} \mathop{\lim} _{k \rightarrow\infty} S_{b} \bigl( f^{n_{k}-1}x_{0},f^{n_{k}-1}x_{0}, f^{m_{k}+1}x_{0} \bigr) &\le \mathop {\lim} _{k \rightarrow\infty} \left [ \begin{matrix} 2b S_{b} ( f^{n_{k}-1}x_{0},f^{n_{k}-1}x_{0}, f^{m_{k}}x_{0} )\\ {}+ b^{2}S_{b} ( f^{m_{k}}x_{0},f^{m_{k}}x_{0}, f^{m_{k}+1}x_{0} ) \end{matrix} \right ] < 2 b^{2} \epsilon. \end{aligned}$$

Therefore

$$\begin{aligned} \mathop{\lim} _{k \rightarrow\infty} M_{f}^{5} ( x_{m_{k}}, x_{n_{k}-1} ) \le& \max \left \{ {\epsilon, 2 b \epsilon, 2 b^{2} \epsilon} \right \} \\ =& 2 b^{2} \epsilon. \end{aligned}$$

From (8), by the definition of ψ, we have that

$$4 b ^{2} \epsilon\le 2 b^{2} \epsilon, $$

which is a contradiction. Hence $\{ f^{n} x_{0} \}$ is an $S_{b}$-Cauchy sequence in complete regular $S_{b}$-metric spaces $(X, S_{b}, \preceq)$. By the completeness of $(X, S_{b})$, it follows that the sequence $\{ f^{n} x_{0} \}$ converges to α in $( X, S_{b} )$. Thus

$$\mathop{\lim} _{k \rightarrow\infty} f^{n} x_{0} = \alpha= \mathop{\lim } _{k \rightarrow\infty} f^{n+1} x_{0}. $$

Since $x_{n}, \alpha\in X$ and X is regular, it follows that either $x_{n} \preceq\alpha$ or $\alpha\preceq x_{n}$.

Now we have to prove that α is a fixed point of f.

Suppose $f\alpha\neq\alpha$, by Lemma (2.10), we have that

$$\frac{1}{2b} S_{b}(f\alpha, f\alpha, \alpha) \le\mathop{\lim} _{n \rightarrow\infty} \inf S_{b} (f\alpha, f\alpha, f^{n+1}x_{0} ). $$

Now from (3.2.4) and applying ψ on both sides, we have that

$$\begin{aligned} \psi \bigl(2b^{3} S_{b}(f\alpha, f\alpha, \alpha) \bigr) \le{}& \mathop {\lim} _{n \rightarrow\infty} \inf \psi \bigl(4 b^{4} S_{b} \bigl(f\alpha, f\alpha, f^{n+1}x_{0} \bigr) \bigr) \\\le{}& \mathop{\lim} _{n \rightarrow\infty} \inf \psi \bigl( M_{f}^{5} ( \alpha, x_{n} ) \bigr) - \mathop{\lim} _{n \rightarrow\infty} \inf \phi \bigl( M_{f}^{5} ( \alpha, x_{n} ) \bigr). \end{aligned}$$

(9)

Here

$$\begin{aligned} \mathop{\lim} _{n \rightarrow\infty} \inf M_{f}^{5} ( \alpha, x_{n} ) &= \mathop{\lim} _{n \rightarrow\infty} \inf\max \left \{ \begin{matrix} {S_{b} (\alpha, \alpha, x_{n} ), S_{b} (\alpha, \alpha, f\alpha ), S_{b} (x_{n}, x_{n}, fx_{n} ) }\\ {S_{b} (\alpha, \alpha, fx_{n} ), S_{b} (x_{n}, x_{n}, f\alpha ) } \end{matrix} \right \} \\ &\le \mathop{\lim} _{n \rightarrow\infty} \sup\max \left \{ {0, S_{b} ( \alpha, \alpha, f\alpha ), 0, 0, S_{b} (x_{n}, x_{n}, f\alpha ) } \right \} \\ & \le\max \left \{ S_{b} (\alpha, \alpha, f\alpha ), b^{2} S_{b} (\alpha, \alpha, f\alpha ) \right \} \\ &\le b^{3} S_{b} (f\alpha, f\alpha, \alpha ). \end{aligned}$$

Hence from (9) we have that

$$\begin{aligned} \psi \bigl(2 b^{3} S_{b}(f\alpha, f\alpha, \alpha) \bigr) & \le \psi \bigl(b^{3}S_{b} (\alpha, \alpha, f\alpha ) \bigr) - \mathop {\lim} _{n \rightarrow\infty} \inf \phi \bigl( M_{f}^{5} ( \alpha, x_{n} ) \bigr) \\ &\le \psi \bigl(b^{3} S_{b} (f\alpha, f\alpha, \alpha ) \bigr) , \end{aligned}$$

which is a contradiction. So that α is a fixed point of f.

Suppose that $\alpha^{\ast}$ is another fixed point of f such that $\alpha\neq\alpha^{\ast}$.

Consider

$$\begin{aligned} \psi \bigl( 4 b^{4} S_{b} \bigl(\alpha, \alpha, \alpha^{\ast} \bigr) \bigr) \le{}& \psi \bigl( M_{f}^{5} \bigl(\alpha, \alpha^{\ast} \bigr) \bigr) - \phi \bigl( M_{f}^{5} \bigl( \alpha, \alpha^{\ast} \bigr) \bigr) \\ = {}& \psi \bigl(\max \bigl\{ S_{b} \bigl(\alpha, \alpha, \alpha^{\ast} \bigr), S_{b} \bigl(\alpha^{\ast}, \alpha^{\ast}, \alpha \bigr) \bigr\} \bigr) \\ &{}- \phi \bigl(\max \bigl\{ S_{b} \bigl(\alpha, \alpha, \alpha^{\ast } \bigr), S_{b} \bigl(\alpha^{\ast}, \alpha^{\ast}, \alpha \bigr) \bigr\} \bigr) \\ \le{} & \psi \bigl(b S_{b} \bigl(\alpha, \alpha, \alpha^{\ast} \bigr) \bigr), \end{aligned}$$

which is a contradiction.

Hence α is a unique fixed point of f in $(X, S_{b} )$. □

Example 3.5

Let $X = [0, 1]$ and $S : X \times X \times X \to\mathbb {R}^{+}$ by $S_{b}(x,y,z) = ( \vert y+z-2x \vert + \vert y-z \vert )^{2}$ and ⪯ by $a \preceq b \iff a\le b$, then $(X, S_{b} , \preceq )$ is a complete ordered $S_{b}$-metric space with $b = 4$. Define $f: X \rightarrow X$ by $f(x) = \frac{x}{32\sqrt{2}}$. Also define $\psi,\phi:\mathbb{R}^{+} \to\mathbb{R}^{+}$ by $\psi(t) = t$ and $\phi(t) = \frac{t}{2}$.

$$\begin{aligned} \psi \bigl(4 b^{4} S_{b}(fx, fx, fy) \bigr) &= 4 b^{4} \bigl( \vert fx + fy - 2 fx \vert + \vert fx - fy \vert \bigr)^{2} \\ &= 4 b^{4} \biggl(2 \biggl\vert \frac{x}{32\sqrt{2}} - \frac{y}{32\sqrt {2}} \biggr\vert \biggr)^{2} \\ &= \frac{4 b^{4}}{8 b^{4}}S_{b}(x,x, y) \\ &\le \frac{1}{2}M_{f}^{5}(x, y) \\ &\le \psi \bigl(M_{f}^{5}(x, y) \bigr) - \phi \bigl(M_{f}^{5}(x, y) \bigr), \end{aligned}$$

where

$$M_{f}^{5} ( {x,y} ) = \max \left \{ S_{b}(x, x,y), S_{b}(x, x,fx),S_{b}(y, y, fy),S_{b}(x, x,fy), S_{b}(y, y, fx) \right \}. $$

Hence, all the conditions of Theorem 3.4 are satisfied and 0 is a unique fixed point of f.

Theorem 3.6

Let $(X,S_{b}, \preceq)$ be an ordered complete $S_{b}$ metric space, and let $f : X \to X$ satisfy $(\psi, \phi)$-contraction with $i = 3\textit{ or }4$. If there exists $x_{0} \in X$ with $x_{0} \preceq f x_{0}$, then f has a unique fixed point in X.

Proof

Follows along similar lines of Theorem 3.4 if we take $M_{f}^{3} ( {x,y} )$ or $M_{f}^{4} ( {x,y} )$ in place of $M_{f}^{5} ( {x,y} )$ in Theorem 3.4. □

Theorem 3.7

Let $(X,S_{b}, \preceq)$ be an ordered complete $S_{b}$ metric space, and let $f : X \to X$ satisfy

$$4 b^{4} S_{b} ( {fx, fx, fy} ) \le M_{f}^{i} ( {x,y} ) - \varphi \bigl( {M_{f}^{i} ( {x,y} )} \bigr), $$

where $\varphi:[0, \infty) \to[0, \infty)$ and i= 3 or 4 or 5. If there exists $x_{0} \in X$ with $x_{0} \preceq f x_{0}$, then f has a unique fixed point in X.

Proof

The proof follows from Theorems 3.4 and 3.6 by taking $\psi(t) = t$ and $\phi(t) = \varphi(t)$. □

Theorem 3.8

Let $(X,S_{b}, \preceq)$ be an ordered complete $S_{b}$ metric space, and let $f : X \to X$ satisfy

$$S_{b} ( {fx, fx, fy} ) \le\lambda M_{f}^{i} ( {x,y} ), $$

where $\lambda\in [ {0, \frac{1}{4b^{4}}} )$ and $i = 3, 4, 5$. If there exists $x_{0} \in X$ with $x_{0} \preceq f x_{0}$, then f has a unique fixed point in X.

3.1 Application to integral equations

In this section, we study the existence of a unique solution to an initial value problem as an application to Theorem 3.4.

Theorem 3.9

Consider the initial value problem

$$ x^{1} (t) = T \bigl(t, x(t) \bigr),\quad t \in I= [0, 1], x(0) = x_{0}, $$

(10)

where $T: I \times [ {\frac{{x_{0} }}{4},\infty} ) \to [ {\frac{{x_{0} }}{4},\infty} )$ and $x_{0} \in\mathbb{R}$. Then there exists a unique solution in $C (I, [ {\frac{{x_{0} }}{4},\infty} ) )$ for initial value problem (10).

Proof

The integral equation corresponding to initial value problem (10) is

$$\begin{aligned} x(t) = x_{0} +3 b^{2} \int_{0}^{t} {T \bigl(s, x(s) \bigr)} \,ds . \end{aligned}$$

Let $X =C (I, [ {\frac{{x_{0} }}{4},\infty} ) )$ and $S_{b}(x,y,z) = ( \vert y+z-2x \vert + \vert y-z \vert )^{2} for x, y \in X$. Define $\psi, \phi: [0, \infty) \to[0, \infty)$ by $\psi(t) = t$, $\phi(t)= \frac{5t}{9}$. Define $f: X \to X$ by

$$ f(x) (t) = \frac{x_{0}}{3b^{2}} + \int_{0}^{t} {T \bigl(s, x(s) \bigr)} \,ds . $$

(11)

Now

$$\begin{aligned} &\psi \bigl(4 b^{4} S_{b}\bigl(fx(t), fx(t), fy(t)\bigr) \bigr) \\ &\quad = 4 b^{4} \bigl\{ { \bigl\vert fx(t)+ fy(t)-2fx(t) \bigr\vert + \bigl\vert fx(t)- fy(t) \bigr\vert } \bigr\} ^{2} \\ &\quad = 16 b^{4} \bigl\vert fx(t)- fy(t) \bigr\vert ^{2} \\ &\quad = \frac{16 b^{4}}{9b^{4}} \biggl\vert x_{0} + 3 b^{2} \int_{0}^{t} {T\bigl(s, x(s)\bigr)} \,ds - y_{0} - 3 b^{2} \int_{0}^{t} {T\bigl(s, y(s)\bigr)} \,ds \biggr\vert ^{2} \\ & \quad= \frac{16}{9} \bigl\vert x(t) - y(t) \bigr\vert ^{2} \\ &\quad = \frac{4}{9} S(x,x,y) \\ & \quad\le\frac{4}{9} M_{f}^{5}(x, y) \\ &\quad= \psi \bigl(M_{f}^{5}(x, y) \bigr) - \phi \bigl(M_{f}^{5}(x, y) \bigr) , \end{aligned}$$

where

$$M_{f}^{5} ( {x,y} ) = \max \left \{ {S_{b}(x, x,y), S_{b}(x, x,fx),S_{b}(y, y, fy), S_{b}(x, x,fy), S_{b}(y, y, fx) } \right \}. $$

It follows from Theorem 3.4 that f has a unique fixed point in X. □

3.2 Application to homotopy

In this section, we study the existence of a unique solution to homotopy theory.

Theorem 3.10

Let $(X, S_{b})$ be a complete $S_{b}$-metric space, U be an open subset of X and U̅ be a closed subset of X such that $U \subseteq\overline{U}$. Suppose that $H : \overline{U} \times[0, 1] \to X$ is an operator such that the following conditions are satisfied:

• (i) $x \neq H(x, \lambda)$ for each $x \in\partial{U}$ and $\lambda \in[0, 1]$ (here ∂U denotes the boundary of U in X),

• (ii) $\psi(4b^{4} S_{b}(H(x, \lambda),H(x, \lambda), H(y, \lambda) )) \leq\psi( S_{b}(x, x, y)) - \phi( S_{b}(x, x, y))$ $\forall x, y \in\overline{U}$ and $\lambda\in[0, 1]$, where $\psi :[0,\infty) \to[0,\infty)$ is continuous, non-decreasing and $\phi :[0,\infty) \to[0,\infty)$ is lower semi-continuous with $\phi(t)>0$ for $t>0$,

• (iii) there exists $M\geq0$ such that

  $$S_{b} \bigl(H(x, \lambda),H(x, \lambda), H(x, \mu) \bigr) \leq M \vert \lambda - \mu \vert $$

  for every $x \in\overline{U}$ and $\lambda, \mu\in[0, 1]$.

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

Proof

Consider the set

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

Since $H(\cdot, 0)$ has a fixed point in U, we have that $0 \in A$. So that A is a non-empty set.

We will show that A is both open and closed in $[0, 1]$, and so, by the connectedness, we have that $A = [0, 1]$. As a result, $H(\cdot, 1)$ has a fixed point in U. First we show that A is closed in $[0, 1]$. To see this, let $\{ { \lambda_{n}} \}_{n = 1}^{\infty }\subseteq A$ with $\lambda_{n} \to\lambda\in[0, 1]$ as $n \to\infty$.

We must show that $\lambda\in A$. Since $\lambda_{n} \in A$ for $n = 1, 2, 3, \ldots$ , there exists $x_{n} \in U$ with $x_{n} = H(x_{n}, \lambda _{n})$.

Consider

$$\begin{aligned} S_{b}(x_{n}, x_{n}, x_{n + 1}) ={}& S_{b}\bigl(H(x_{n},\lambda_{n}), H(x_{n},\lambda_{n}), H(x_{n+1}, \lambda_{n+1}) \bigr) \\ \leq{}&2b S_{b}\bigl(H(x_{n},\lambda_{n}), H(x_{n},\lambda_{n}), H(x_{n+1}, \lambda _{n})\bigr) \\ &{} + b^{2}S_{b}\bigl(H(x_{n+1}, \lambda_{n}), H(x_{n+1},\lambda_{n}), H(x_{n+1}, \lambda_{n+1})\bigr) \\ \leq{}& S_{b}\bigl(H(x_{n},\lambda_{n}), H(x_{n},\lambda_{n}), H(x_{n+1}, \lambda_{n})\bigr) + M \vert \lambda_{n} - \lambda_{n+1} \vert . \end{aligned}$$

Letting $n \to\infty$, we get

$$\mathop{\lim} _{n \to\infty} S_{b}(x_{n}, x_{n}, x_{n + 1}) \leq \mathop{\lim} _{n \to\infty} S_{b} \bigl(H(x_{n},\lambda_{n}), H(x_{n},\lambda _{n}), H(x_{n+1}, \lambda_{n}) \bigr) + 0. $$

Since ψ is continuous and non-decreasing, we obtain

$$\begin{aligned} \mathop{\lim} _{n \to\infty} \psi\bigl(4b^{4} S_{b}(x_{n}, x_{n}, x_{n + 1})\bigr) &\leq\mathop{\lim} _{n \to\infty}\psi \bigl( 4b^{4} S_{b}\bigl(H(x_{n}, \lambda_{n}), H(x_{n},\lambda_{n}), H(x_{n+1}, \lambda_{n})\bigr)\bigr) \\ &\leq \mathop{\lim} _{n \to\infty} \bigl[ {\psi\bigl( S_{b}(x_{n}, x_{n}, x_{n+1})\bigr) - \phi\bigl( S_{b}(x_{n}, x_{n}, x_{n+1})\bigr)} \bigr]. \end{aligned}$$

By the definition of ψ, it follows that

$$\mathop{\lim} _{n \to\infty} \bigl(4 b^{4} -1 \bigr)S_{b} (x_{n}, x_{n}, x_{n+1} ) \leq0. $$

So that

$$ \mathop{\lim} _{n \to\infty} S_{b}(x_{n}, x_{n}, x_{n+1}) = 0 . $$

(12)

Now we prove that $\{x_{n}\}$ is an $S_{b}$-Cauchy sequence in $(X, d_{p})$. On the contrary, suppose that $\{x_{n} \}$ is not $S_{b}$-Cauchy.

There exists $\epsilon> 0$ and monotone increasing sequences of natural numbers $\{ m_{k}\}$ and $\{ n_{k}\}$ such that $n_{k} > m_{k}$,

$$ S_{b}(x_{m_{k}},x_{m_{k}}, x_{n_{k}}) \geq \epsilon $$

(13)

and

$$ S_{b}(x_{m_{k}}, x_{m_{k}}, x_{n_{k}-1}) < \epsilon. $$

(14)

From (13) and (14), we obtain

$$\begin{aligned} \epsilon&\leq S_{b}(x_{m_{k}}, x_{m_{k}}, x_{n_{k}}) \\ &\leq2 b S_{b}(x_{m_{k}},x_{m_{k}}, x_{m_{k} + 1}) + b^{2} S_{b}(x_{m_{k}+1}, x_{m_{k}+1}, x_{n_{k}}). \end{aligned}$$

Letting $k \to\infty$ and applying ψ on both sides, we have that

$$ \psi \bigl(2b^{2} \epsilon \bigr) \leq \mathop{\lim} _{n \to\infty } \psi \bigl(4b^{4} S_{b}(x_{m_{k}+1}, x_{m_{k}+1}, x_{n_{k}}) \bigr). $$

(15)

But

$$\begin{aligned} &\mathop{\lim} _{n \to\infty}\psi \bigl( 4b^{4} S_{b}(x_{m_{k}+1}, x_{m_{k}+1}, x_{n_{k}}) \bigr)\\ & \quad= \mathop{\lim} _{n \to\infty}\psi \bigl( S_{b}\bigl( 4b^{4} H(x_{m_{k}+1}, \lambda_{m_{k}+1}), H(x_{m_{k}+1},\lambda_{m_{k}+1}), H(x_{n_{k}}, \lambda_{n_{k}})\bigr)\bigr) \\ &\quad\leq \mathop{\lim} _{n \to\infty} \bigl[ {\psi\bigl( S_{b}(x_{m_{k}+1}, x_{m_{k}+1}, x_{n_{k}})\bigr) - \phi\bigl( S_{b}(x_{m_{k}+1}, x_{m_{k}+1}, x_{n_{k}})\bigr)} \bigr]. \end{aligned}$$

It follows that

$$\mathop{\lim} _{n \to\infty} \bigl( 4b^{4} -1 \bigr) S_{b} (x_{m_{k}+1}, x_{m_{k}+1}, x_{n_{k}}) \leq0. $$

Thus

$$\mathop{\lim} _{n \to\infty} S_{b} (x_{m_{k}+1}, x_{m_{k}+1}, x_{n_{k}}) = 0. $$

Hence from (15) and the definition of ψ, we have that

$$\epsilon\le0, $$

which is a contradiction.

Hence $\{x_{n} \}$ is an $S_{b}$-Cauchy sequence in $(X, S_{b})$ and, by the completeness of $(X, S_{b})$, there exists $\alpha\in U$ with

$$\begin{aligned} &\mathop{\lim} _{n \to\infty}x_{n} = \alpha= \mathop{\lim} _{n \to\infty} x_{n+1}, \\ &\psi \bigl(2b^{3} S_{b}\bigl(H(\alpha, \lambda), H(\alpha, \lambda), \alpha \bigr) \bigr) \leq\mathop{\lim} _{n \to\infty} \inf\psi \bigl(4b^{4} S_{b}\bigl(H(\alpha, \lambda), H(\alpha, \lambda), H(x_{n}, \lambda)\bigr) \bigr) \\ &\phantom{\psi \bigl(2b^{3} S_{b}\bigl(H(\alpha, \lambda), H(\alpha, \lambda), \alpha \bigr) \bigr)}\leq\mathop{\lim} _{n \to\infty} \inf\bigl[\psi\bigl( S_{b}(\alpha, \alpha, x_{n})\bigr) - \phi\bigl( S_{b}(\alpha, \alpha, x_{n})\bigr) \bigr] \\ &\phantom{\psi \bigl(2b^{3} S_{b}\bigl(H(\alpha, \lambda), H(\alpha, \lambda), \alpha \bigr) \bigr)}= 0. \end{aligned}$$

(16)

It follows that $\alpha= H(\alpha, \lambda)$.

Thus $\lambda\in A$. Hence A is closed in $[0, 1]$.

Let $\lambda_{0} \in A$. Then there exists $x_{0} \in U$ with $x_{0} = H(x_{0}, \lambda_{0})$.

Since U is open, there exists $r > 0$ such that $B_{S_{b}}(x_{0}, r) \subseteq U$.

Choose $\lambda\in(\lambda_{0} - \epsilon, \lambda_{0} + \epsilon)$ such that $\vert \lambda- \lambda_{0} \vert \leq\frac{1}{M^{n}} < \epsilon$.

Then, for $x \in\overline{B_{p} (x_{0}, r)} = \{x \in X / S_{b}(x, x, x_{0}) \leq r + b^{2} S_{b}(x_{0}, x_{0}, x_{0}) \}$,

$$\begin{aligned} &S_{b}\bigl(H(x, \lambda), H(x, \lambda), x_{0}\bigr)\\ &\quad= S_{b}\bigl(H(x,\lambda),H(x,\lambda ), H(x_{0}, \lambda_{0})\bigr) \\ &\quad\leq2 b S_{b}\bigl(H(x,\lambda),H(x,\lambda), H(x, \lambda_{0})\bigr) + b^{2} S_{b}\bigl(H(x, \lambda_{0}), H(x,\lambda_{0}), H(x_{0}, \lambda_{0})\bigr) \\ &\quad\leq2 b M \vert \lambda- \lambda_{0} \vert + b^{2} S_{b}\bigl(H(x, \lambda_{0}), H(x,\lambda_{0}), H(x_{0}, \lambda_{0})\bigr) \\ &\quad\leq\frac{2 b}{M^{n-1}} + b^{2} S_{b}\bigl(H(x, \lambda_{0}), H(x,\lambda_{0}), H(x_{0}, \lambda_{0})\bigr). \end{aligned}$$

Letting $n \to\infty$, we obtain

$$S_{b} \bigl(H(x, \lambda), H(x, \lambda), x_{0} \bigr) \leq b^{2} S_{b} \bigl(H(x, \lambda_{0}), H(x, \lambda_{0}), H(x_{0}, \lambda_{0}) \bigr). $$

Since ψ is continuous and non-decreasing, we have

$$\begin{aligned} \psi\bigl(S_{b}\bigl(H(x, \lambda), H(x, \lambda), x_{0} \bigr)\bigr) &\leq\psi\bigl(4b^{2}S_{b}\bigl(H(x, \lambda), H(x, \lambda), x_{0}\bigr)\bigr) \\ &\leq \psi\bigl( 4b^{4} S_{b}\bigl(H(x, \lambda_{0}), H(x,\lambda_{0}), H(x_{0}, \lambda _{0})\bigr)\bigr) \\ &\leq\psi\bigl( S_{b}(x, x, x_{0})\bigr) - \phi\bigl( S_{b}(x, x, x_{0})\bigr) \\ &\leq \psi\bigl( S_{b}(x, x, x_{0})\bigr). \end{aligned}$$

Since ψ is non-decreasing, we have

$$\begin{aligned} S_{b}\bigl(H(x, \lambda),H(x, \lambda), x_{0}\bigr) &\leq S_{b}(x, x, x_{0}) \\ &\leq r + b^{2} S_{b}(x_{0}, x_{0}, x_{0}). \end{aligned}$$

Thus, for each fixed $\lambda\in(\lambda_{0} - \epsilon, \lambda_{0} + \epsilon)$, $H(\cdot, \lambda): \overline{B_{p} (x_{0}, r)} \to\overline {B_{p} (x_{0}, r)}$.

Since also (ii) holds and ψ is continuous and non-decreasing and ϕ is continuous with $\phi(t)> 0$ for $t > 0$, then all the conditions of Theorem (3.10) are satisfied.

Thus we deduce that $H(\cdot, \lambda)$ has a fixed point in U̅. But this fixed point must be in U since (i) holds.

Thus $\lambda\in A$ for any $\lambda\in(\lambda_{0} - \epsilon, \lambda _{0} + \epsilon)$.

Hence $(\lambda_{0} - \epsilon, \lambda_{0} + \epsilon) \subseteq A$ and therefore A is open in [0, 1].

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

Corollary 3.11

Let $(X, p)$ be a complete partial metric space, U be an open subset of X and $H : \overline{U} \times[0, 1] \to X$ with the following properties:

• (1) $x \neq H(x, t)$ for each $x \in\partial U$ and each $\lambda \in[0, 1]$ (here ∂U denotes the boundary of U in X),

• (2) there exist $x, y \in\overline{U}$ and $\lambda\in[0, 1], L \in [0, \frac{1}{4b^{4}} )$ such that

  $$S_{b} \bigl(H(x, \lambda),H(x, \lambda), H(y, \mu) \bigr) \leq L S_{b}(x, x, y), $$

• (3) there exists $M \geq0$ such that

  $$S_{b} \bigl(H(x, \lambda), H(x, \lambda), H(x, \mu) \bigr) \leq M \vert \lambda- \mu \vert $$

  for all $x \in\overline{U}$ and $\lambda, \mu\in[0, 1]$.

If $H(\cdot, 0)$ has a fixed point in U, then $H(\cdot, 1)$ has a fixed point in U.

Proof

Proof follows by taking $\psi(x) = x, \phi(x) = x - Lx \mbox{ with } L \in [0, \frac{1}{4b^{4}} )$ in Theorem (3.10). □

4 Conclusions

In this paper we conclude some applications to homotopy theory and integral equations by using fixed point theorems in partially ordered $S_{b}$-metric spaces.