# 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.

## 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  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 ). 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 . Later several authors proved so many results on b-metric spaces (see ). Mustafa and Sims defined the concept of a generalized metric space which is called a G-metric space . 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 . Aghajani, Abbas and Roshan presented a new type of metric which is called $$G_{b}$$-metric and studied some properties of this metric .

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

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.

## Preliminaries

### Definition 2.1



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



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



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



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



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



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



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



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

1. (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. (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



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



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.

## 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}

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.

### 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. □

### 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 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:

1. (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),

2. (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 $$t>0$$,

3. (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

$$\epsilon\le0,$$

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 . 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. (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. (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. (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). □

## 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.

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