Open Journal of Mathematical Sciences

Vol. 4 (2020), Issue 1, pp. 1 – 8
ISSN: 2523-0212 (Online) 2616-4906 (Print)
DOI: 10.30538/oms2020.0087

A fixed point theorem for generalized weakly contractive mappings in $b$-metric spaces

Eliyas Zinab, Kidane Koyas, Aynalem Girma$^1$
Department of Mathematics, College of Natural Sciences, Jimma University, Jimma, Ethiopia.; (E.Z & K.K & A.G)
$^{1}$Corresponding Author: aynalemgirma46@yahoo.com

Copyright © 2020 Eliyas Zinab, Kidane Koyas, Aynalem Girma. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Received: July 11, 2019 – Accepted: September 30, 2019 – Published: January 16, 2019

Abstract

In this paper we establish a fixed point theorem for generalized weakly contractive mappings in the setting of $b$-metric spaces and prove the existence and uniqueness of a fixed point for a self-mappings satisfying the established theorem. Our result extends and generalizes the result of Cho [1]. Finally, we provided an example in the support of our main result.

Keywords:

Fixed point, generalized weak contractive mapping, $b$-metric space.

1. Introduction

In 1993, Czerwik [2] introduced the concept of $ b $-metric spaces and proved the Banach contraction mapping principle in the setting of $ b $-metric spaces. Afterwards, several research papers [3, 4, 5, 6, 7, 8] were published on the existence of fixed point results for single-valued and multi-valued mappings in the setting of $ b $-metric spaces. In 1997, Alber et al. [9] generalized Banach\textquoteright s contraction principle by introducing the concept of weakly contractive mappings and proved the existence of fixed points for weakly contractive and single valued mappings on Hilbert spaces.

Rhoades [10] proved that every weakly contractive mapping has a unique fixed point in complete metric spaces. Then, many authors obtained generalizations and extensions of the weakly contractive mappings.

In particular, Choudhury et al. [11] generalized fixed point results for weakly contractive mappings by using altering distance functions. Very recently, Cho [1] introduced the notion of generalized weakly contractive mappings in metric spaces and proved a fixed point theorem for generalized weakly contractive mappings defined on complete metric spaces.

Inspired and motivated by the results of Cho [1] the purpose of this paper is to establish a fixed point result for generalized weakly contractive mappings in the setting of $b$-metric spaces.

2. Preliminaries

In this section, we give basic definitions of concepts concerning a generalized weakly contractive mappings in the setting of $b$-metric spaces.

Definition 1.[2] Let $X$ be a nonempty set and $ s \geq 1 $ be a given real number. A function $ d:X \times X \rightarrow R^+ $ is a $ b $-metric if and only if for all $ x, y, z \in X $, the following conditions are satisfied:

(a) $ d(x,y) = 0 $ if and only if $ x=y; $
(b) $ d(x,y)=d(y,x) $;
(c) $ d(x,z) \leq s[d(x,y)+d(y,z)]. $
The pair $(X,d)$ is called a $b$-metric space.

It should be noted that the class of $ b $-metric spaces is effectively larger than that of metric spaces, since $b$-metric is metric when $ s = 1. $ But, in general, the converse is not true.

Example 1. [12] Let $ X = R $ and $ d:X \times X\rightarrow R^{+} $ be given by $ d(x,y)=(x - y)^{2} $ for all $ x ,y \in X $, then $ d $ is a $ b $-metric on X with $ s = 2 $ but it is not a metric on X, because for $ x=2,y=4 $ and $ z=6 $, we have $ d(2,6) \nleq 2[ d(2,4)+d(4,6)] $, hence the triangle inequality for a metric does not hold.

Definition 2. A function $f \colon X \rightarrow R^{+}$, where $X$ is $b$-metric space is called lower semicontinuous if for all $ x \in X $ and $ x_{n} \in X $ with $\lim_{n\rightarrow \infty}x_{n}=x$, we have $$ f(x) \leq \liminf_{n\rightarrow \infty} f(x_{n}).$$

Definition 3. [6] Let $ X $ be a $ b $-metric space and $ \{x_{n}\} $ be a sequence in $ X $, we say that

(a) $x_{n}$ is $ b $-converges to $ x \in X $ if $ d(x_{n},x )\rightarrow 0 $ as $ n \rightarrow\infty. $
(b) $ {x_{n} } $ is a $ b$-Cauchy sequence if $ d(x_{n} ,x_{m} ) \rightarrow 0 $ as $ n,m \rightarrow\infty. $
(c) $ (X,d )$ is $ b $-complete if every $ b $-Cauchy sequence in $ X $ is $ b $-convergent.

Definition 4. [1] Let $ X $ be a complete metric space with metric $d,$ and $ T \colon X \rightarrow X $. Also let $ \varphi\colon X\rightarrow R^{+} $ be a lower semicontinuous function, then $T$ is called a generalized weakly contractive mapping if it satisfies the following condition: $$ \psi (d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)) \leq \psi(m(x,y,d,T,\varphi)) -\phi(l(x,y,d,T,\varphi))$$ where, \begin{eqnarray*} m(x,y,d,T,\varphi) &=& max \left\{d(x,y)+\varphi(x)+\varphi(y), d(x,Tx)+\varphi(x)+\varphi(Tx), d(y,Ty)+\varphi(y)+\varphi(Ty),\right.\\&&{}\left.\dfrac{1}{2} \left[d(x,Ty)+\varphi(x)+\varphi(Ty)+ d(y,Tx)+\varphi(y)+\varphi(Tx) \right] \right\} \end{eqnarray*} and $l(x,y,d,T,\varphi )=\max \{ d(x,y)+\varphi(x)+\varphi(y),d(y,Ty)+\varphi(y)+\varphi(Ty) \} $, for all $ x, y \in X, $ where $\psi\colon R^{+}\rightarrow R^{+} $ is a continuous with $\psi(t)=0$ if and only if $ t=0$ and $\phi \colon R^{+} \rightarrow R^{+} $ is a lower semicontinuous function with $\phi(t)=0$ if and only if $ t=0$.

Theorem 5. [1] Let $X$ be complete. If $T$ is a generalized weakly contractive mapping, then there exists a unique $z \in X $ such that $z = Tz$ and $\varphi(z) = 0.$

Lemma 6. [12] Suppose $(X, d)$ is a $b$-metric space and $\{x_{n}\}$ be a sequence in $X$ such that $$\lim _{{n\rightarrow\infty}}d(x_{n}, x_{n+1})\rightarrow 0.$$ If $\{x_{n}\}$ is not a $b$-Cauchy sequence, then there exists $\epsilon>0$ and two sequences of positive integers $\{m(k)\}$ and $\{n(k)\}$ with $n(k)>m(k)\geq k$ such that for all positive intiger $k$, $d(x_{m(k)} ,x_{n(k)} ) \geq \epsilon,\ \ d(x_{m(k)} ,x_{n(k-1)} )< \epsilon$ and

(a) $\epsilon \leq \lim _{{n\rightarrow\infty}}infd(x_{m(k)} ,x_{n(k)} ) \leq\lim _{{n\rightarrow\infty}}sup d(x_{m(k)} ,x_{n(k)} ) \leq s\epsilon.$
(b) $\frac{\epsilon}{s} \leq \lim _{{n\rightarrow\infty}}infd(x_{m(k)} ,x_{n(k)} ) \leq\lim _{{n\rightarrow\infty}}supd(x_{m(k+1)} ,x_{n(k)} ) \leq s^{2}\epsilon.$
(c) $\frac{\epsilon}{s^{2}} \leq \lim _{{n\rightarrow\infty}}infd(x_{m(k+1)} ,x_{n(k)} ) \leq\lim _{{n\rightarrow\infty}}supd(x_{m(k)} ,x_{n(k+1)} ) \leq s^{2}\epsilon.$
(d) $\dfrac{\epsilon}{s^{2}} \leq \lim _{{n\rightarrow\infty}}infd(x_{m(k+1)} ,x_{n(k+1)} ) \leq\lim _{{n\rightarrow\infty}}supd(x_{m(k+1)} ,x_{n(k+1)} ) \leq s^{3}\epsilon$

holds.

3. Results and discussion

In this section, we introduce a generalized weakly contractive mappings in the setting of $b$-metric spaces and prove a fixed point result.

Definition 5. Let $X$ be a $b$-metric space with metric $d$ and parameter $s \geq 1 $, $T : X \rightarrow X$, and let $\varphi: X \rightarrow R^{+}$ be a lower semicontinuous function, then $T$ is called a generalized weakly contractive mapping if satisfies the following condition:

\begin{equation}\label{Equa 1} \psi(S^{3}d(Tx, Ty)+ \varphi(Tx)+ \varphi(Ty)) \leq \psi(m(x,y,d,T,\varphi)) - \phi(l(x,y,d,T,\varphi)) \end{equation}

(1)

for all $x,y \in X$, where,

\begin{eqnarray}\label{Equa2} %\begin{equation*} m(x,y,d,T,\varphi)&=&max\left\{d(x,y)+\varphi(x)+\varphi(y), d(x,Tx)+\varphi(x)+\varphi(Tx),d(y,Ty)+\varphi(y)+\varphi(Ty),\right.\nonumber\\&& \left.\frac{1}{2s^{2}}\left\{d(x,Ty)+\varphi(x)+\varphi(Ty)+d(y,Tx)+\varphi(y)+\varphi(Tx)\right\}\right\} \end{eqnarray}

(2)

and

\begin{equation}\label{Equa3} l(x,y,d,T,\varphi)=max\{d(x,y)+\varphi(x)+\varphi(y), d(y,Ty)+\varphi(y)+\varphi(Ty)\}. \end{equation}

(3)

for all $ x, y \in X, $ where $\psi\colon R^{+}\rightarrow R^{+} $ is a continuous with $\psi(t)=0$ if and only if $ t=0$ and $\phi \colon R^{+} \rightarrow R^{+} $ is a lower semicontinuous function with $\phi(t)=0$ if and only if $ t=0$.

Theorem 2. \ Let $ X$ be a complete $b$-metric space with metric $d$ and $ s\geq 1$. If $T$ is a generalized weakly contractive mapping then $T$ has a unique fixed point $u\in X$ such that $u=Tu$ and $\varphi(u)=0$.

Proof. Let $x_{0}\in X$ be fixed and define a sequence $\{x_{n}\}$ by $x_{1}$ = $Tx_{0}$, $x_{2}$ = $Tx_{1}$,\ldots, $x_{n+1}$= $Tx_{n}$ for all $n = 0,1,2,\ldots.$ If $x_{n}$ = $x_{n+1}$ for some $n$, $x_{n}= x_{n+1}=Tx_{n}$, $x_{n}$ is fixed point of $T$. Assume $x_{n} \neq x_{n+1}$ for all $n = 0,1,2,\ldots.$ From (2) by using $x=x_{n-1}$ and $y=x_{n}$, we have\\ $ m(x_{n-1},x_{n},d,T,\varphi)=\max\{d(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n}),d(x_{n-1},Tx_{n-1})+\varphi(x_{n-1})+\varphi(Tx_{n-1}), d(x_{n},Tx_{n})+\varphi(x_{n})+\varphi(Tx_{n}), \frac{1}{2s^{2}}\{d(x_{n-1},Tx_{n})+\varphi(x_{n-1})+\varphi(Tx_{n})+d(x_{n},Tx_{n-1})+\varphi(x_{n})+\varphi(Tx_{n-1})\}\}= \max\{d(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n}),d(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n})d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1}), \frac{1}{2s^{2}}\{d(x_{n-1},x_{n+1})+\varphi(x_{n-1})+\varphi(x_{n+1})+d(x_{n},x_{n})+\varphi(x_{n})+\varphi(x_{n})\}\}.$ Since\\ $ \frac{1}{2s^{2}}\{d(x_{n-1},x_{n+1})+\varphi(x_{n-1})+\varphi(x_{n+1})+d(x_{n},x_{n})+\varphi(x_{n})+\varphi(x_{n})\}\leq \frac{1}{2s^{2}}\{sd(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n})+sd(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1})\}\leq \frac{1}{2s^{2}}\{s[d(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n})+d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1})]\}= \frac{1}{2s}\{d(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n})+d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1})\}\leq \frac{1}{2}\{d(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n})+d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1})\}\leq max\{d(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n}), d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1})\}. $ So, we obtain

\begin{equation}\label{*}m(x_{n-1},x_{n},d,T,\varphi)= max\{d(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n}),d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1})\}. \end{equation}

(4)

Similarly from (3)

\begin{equation}\label{**} l(x_{n-1},x_{n},d,T,\varphi)= max\{d(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n}),d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1})\}. \end{equation}

(5)

Then (1) becomes

\begin{eqnarray}\label{Equa 4} \psi(s^{3}d (x_{n},x_{n+1})+ \varphi(x_{n})+ \varphi(x_{n+1})\leq \psi(m(x_{n-1},x_{n},d,T,\varphi)) -\phi(l(x_{n-1},x_{n},d,T,\varphi)). \end{eqnarray}

(6)

Now, if $d(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n})\leq d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1})$, for some positive integer $n$\ then (6) becomes\\ $ \psi(s^{3}d (x_{n},x_{n+1})+ \varphi(x_{n})+ \varphi(x_{n+1})\leq \psi( d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1})) -\phi( d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1})).$ It follows $ \psi( d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1}))\leq \psi(s^{3}d (x_{n},x_{n+1})+ \varphi(x_{n})+ \varphi(x_{n+1})\leq \psi( d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1}))-\phi( d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1}))\psi( d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1})), $ which is a contradiction. Thus,

\begin{equation}\label{Equa 5} d(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n})> d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1}). \end{equation}

(7)

From (4), (5) and (7), we obtain

\begin{equation}\label{ver} m(x_{n-1},x_{n},d,T,\varphi)= l(x_{n-1},x_{n},d,T,\varphi)= d(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n}), \end{equation}

(8)

So (6) becomes:

\begin{eqnarray}\label{Equa 6} \psi(s^{3}d (x_{n},x_{n+1})+ \varphi (x_{n})+\varphi(x_{n+1})\leq\psi(m(x_{n-1},x_{n},d,T,\varphi))-\phi(l(x_{n-1},x_{n},d,T,\varphi)). \end{eqnarray}

(9)

From (7), the sequence $(d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1}))$ is decreasing and bounded below. Hence $d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1}) \rightarrow r$ as $ n \rightarrow \infty $ for some $ r \geq 0$. Assume $r>0$ and letting $n\rightarrow\infty$ in (\ref{Equa 6}) and using the continuity of $\psi$ and the lower semicontinuity of $\phi$, we have \begin{eqnarray*} \psi(s^{3}r)&\leq&\psi(r)-\lim _{{n\rightarrow\infty}}inf\phi(d(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n}))\\&\leq &\psi(r)-\lim _{{n\rightarrow\infty}}\phi(d(x_{n-1},x_{n})+\varphi(x_{n-1})+\varphi(x_{n}))\\&=&\psi(r)-\phi(r). \end{eqnarray*} It follows that $\psi(r)\leq\psi(s^{3}r)\leq\psi(r)-\phi(r)< \psi(r)$, which is a contradiction, hence we have $ r=0$ and consequently, $\lim _{n\rightarrow\infty}[d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1})] =0.$ Implies

\begin{equation}\label{Equa 7} \lim _{n\rightarrow\infty}d(x_{n},x_{n+1})=0. \end{equation}

(10)

\begin{equation} \lim _{n\rightarrow\infty}\varphi(x_{n})=\lim _{n\rightarrow\infty}\varphi(x_{n+1}) =0. \end{equation}

(11)

Now, we prove that the sequence $\{x_{n}\}$ is a $b$-Cauchy sequence. If $\{x_{n}\}$ is not a $b$-Cauchy sequence, then by Lemma 1 there exists $\epsilon>0$ and sequences of positive integers ${m(k)}$ and ${n(k)}$ such that for all positive integer $k$, $n(k)>m(k)\geq k$, $d(x_{m(k)} ,x_{n(k)} ) \geq \epsilon$ and $ d(x_{m(k)} ,x_{n(k-1)} )< \epsilon $ and conditions from (a)-(d) of 1 hold. From (2) and by setting $x=x_{m(k)}$ and $y=x_{n(k)}$ we have:
$m(x_{m(k)},x_{n(k)},d,T,\varphi)= \max\{(d(x_{m(k)},x_{n(k)})+\varphi(x_{m(k)})+\varphi(x_{n(k)}){},d(x_{m(k)},Tx_{m(k)})+\varphi(x_{m(k)})+\varphi(Tx_{m(k)}), d(x_{n(k)},Tx_{n(k)})+\varphi(x_{n(k)})+\varphi(Tx_{n(k)}), \frac{1}{2s^{2}}\{d(x_{m(k)},Tx_{n(k)})+\varphi(x_{m(k)})+\varphi(Tx_{n(k)})+d(x_{n(k)},Tx_{m(k)})+\varphi(x_{n(k)})+\varphi(Tx_{m(k)})\}\}= \max\{d(x_{m(k)},x_{n(k)})+\varphi(x_{m(k)})+\varphi(x_{n(k)}),d(x_{m(k)},x_{m(k)+1})+\varphi(x_{m(k)})+\varphi(x_{m(k)+1}), d(x_{n(k)},x_{n(k)+1})+\varphi(x_{n(k)})+\varphi(x_{n(k)+1}), \frac{1}{2S^{2}}\{d(x_{m(k)},x_{n(k)+1})+\varphi(x_{m(k)})+\varphi(x_{n(k)+1})+d(x_{n(k)},x_{m(k)+1})+\varphi(x_{n(k)})+\varphi(x_{m(k)+1})\}\}. $ Taking the limit as $ k\rightarrow\infty$ and using (10), (11) and Lemma 1, we have

\begin{eqnarray} \label{Equa 9} && \lim _{k\rightarrow\infty} m(x_{m(k)},x_{n(k)},d,T,\varphi)= \max\lim _{k\rightarrow\infty}\{d(x_{m(k)},x_{n(k)})+\varphi(x_{m(k)})+\varphi(x_{n(k)}),d(x_{m(k)},x_{m(k)+1})+\varphi(x_{m(k)})\nonumber\\&& +\varphi(x_{m(k)+1}), d(x_{n(k)},x_{n(k)+1})+\varphi(x_{n(k)})+\varphi(x_{n(k)+1}), \frac{1}{2S^{2}} \{d(x_{m(k)},x_{n(k)+1})+\varphi(x_{m(k)})+\varphi(x_{n(k)+1})\nonumber\\&&+d(x_{n(k)},x_{m(k)+1})+\varphi(x_{n(k)})+\varphi(x_{m(k)+1})\}\} \leq max\{s\epsilon,0,0,\dfrac{1}{2s^{2}}(s^{2}\epsilon+s^{2}\epsilon)\}=s\epsilon. \end{eqnarray}

(12)

Similarly from (3), we have
$ l(x_{m(k)},x_{n(k)},d,T,\varphi)= max\{d(x_{m(k)},x_{n(k)})+\varphi(x_{m(k)})+\varphi(x_{n(k)}), d(x_{n(k)},Tx_{n(k)})+\varphi(x_{n(k)})+\varphi(Tx_{n(k)})\}= max\{d(x_{m(k)},x_{n(k)})+\varphi(x_{m(k)})+\varphi(x_{n(k)}), d(x_{n(k)},x_{n(k)+1})+\varphi(x_{n(k)})+\varphi(x_{n(k)+1})\}$

\begin{eqnarray}\label{Equa 10} &&\lim _{k\rightarrow\infty} l(x_{m(k)},x_{n(k)},d,T,\varphi)=\lim _{k\rightarrow\infty} max\{d(x_{m(k)},x_{n(k)})+\varphi(x_{m(k)})+\varphi(x_{n(k)}),d(x_{n(k)},x_{n(k)+1})+\varphi(x_{n(k)})\nonumber\\&&+\varphi(x_{n(k)+1})\}\leq \max\{s\epsilon,0\}=s\epsilon.\end{eqnarray}

(13)

Now from (1), we have
$ \psi(s^{3}d(Tx, Ty)+ \varphi(Tx)+ \varphi(Ty))= \psi(s^{3}d(x_{m(k)+1}, x_{n(k)+1})+ \varphi(x_{m(k)+1})+ \varphi(x_{n(k)+1}))\leq \psi(m(x_{m(k)},x_{n(k)},d,T,\varphi))-\phi(l(x_{m(k)},x_{n(k)},d,T,\varphi)).$ Letting $k\rightarrow\infty$, using (11), (12), (13), applying the continuity of $\psi$ and lower semicontinuity of $\phi$, we have, \begin{eqnarray*} \lim _{k\rightarrow\infty}\psi(s^{3}d(x_{m(k)+1}, x_{n(k)+1}))\leq\psi(s\varepsilon)-\phi(s\epsilon). \end{eqnarray*} This implies that \begin{eqnarray*} \psi(s\epsilon)= \psi(s^{3}\dfrac{\varepsilon}{s^{2}})\leq \psi( s^{3}\lim _{k\rightarrow\infty}d(x_{m(k)+1}, x_{n(k)+1}))\leq\psi(s\varepsilon)-\phi(s\epsilon) < \psi(s\epsilon), \end{eqnarray*} which is a contradiction. Therefore $\{x_{n}\}$ is a $b$-Cauchy sequence. Now since $ \{x_{n}\}$ is a $b$-Cauchy and $X$ is $b$-complete we have, \begin{equation*} \lim_{n\rightarrow\infty}x_{n}=u\in X. \end{equation*} Since $\varphi$ is lower semicontinuous, \begin{equation*} \varphi(u)\leq\lim_{n\rightarrow\infty}inf\varphi(x_{n})\leq \lim_{n\rightarrow\infty}\varphi(x_{n})=0, \end{equation*} which implies

\begin{equation}\label{Equa 11} \varphi(u)=0. \end{equation}

(14)

Now from (2) by putting $x=x_{n}$ and $y=u$, we have
$ m(x_{n},u,d,T,\varphi)= max\{d(x_{n},u)+\varphi(x_{n})+\varphi(u),d(x_{n},Tx_{n})+\varphi(x_{n})+\varphi(Tx_{n}), {}d(u,Tu)+\varphi(u)+\varphi(Tu), \frac{1}{2s^{2}}\{d(x_{n},Tu)+\varphi(x_{n})+\varphi(Tu)+d(u,Tx_{n})+\varphi(u)+\varphi(Tx_{n})\}\}= max\{d(x_{n},u)+\varphi(x_{n})+\varphi(u),d(x_{n},x_{n+1})+\varphi(x_{n})+\varphi(x_{n+1}),{} d(u,Tu)+\varphi(u)+\varphi(Tu), \frac{1}{2s^{2}}\{d(x_{n},Tu)+\varphi(x_{n})+\varphi(Tu)+d(u,x_{n+1})+\varphi(u)+\varphi(x_{n+1})\}\}. $ Applying the limit as $ n\rightarrow\infty$ and using (10), (11) and(14) we have

\begin{eqnarray}\label{Equa 12} \lim _{n\rightarrow\infty} m(x_{n},u,d,T,\varphi) = d(u,Tu)+\varphi(Tu). \end{eqnarray}

(15)

Similarly

\begin{equation}\label{Equa 13} \lim _{n\rightarrow\infty}l(x_{n},u,d,T,\varphi)=d(u,Tu)+\varphi(Tu). \end{equation}

(16)

Then using (1), we have \begin{eqnarray*} \psi(s^{3}d(Tx_{n}, Tu)+ \varphi(Tx_{n})+ \varphi(Tu))&=&\psi(s^{3}d(x_{n+1}, Tu)+ \varphi(x_{n+1})+ \varphi(Tu))\\ &\leq& \psi(m(x_{n},u,d,T,\varphi))-\phi(l(x_{n},u,d,T,\varphi)). \end{eqnarray*} Letting $n\rightarrow\infty$, using (14),(15), (16) and by using the continuity of $\psi$ and lower semicontinuity of $\phi$, we have \begin{eqnarray*} \psi(s^{3}d(x_{n+1}, Tu)+ \varphi(x_{n+1})+ \varphi(Tu))&=&\psi(s^{3}d(u, Tu)+ \varphi(Tu))\\ &\leq& \psi(m(x_{n},u,d,T,\varphi))-\phi(l(x_{n},u,d,T,\varphi))\\&=&\psi (d(u,Tu)+\varphi(Tu))-\phi(d(u,Tu)+\varphi(Tu)). \end{eqnarray*} This implies \begin{eqnarray*} \psi(d(u, Tu)+ \varphi(Tu))&\leq& \psi(s^{3}d(u, Tu)+ \varphi(Tu)) \\&\leq& \psi(d(u,Tu)+\varphi(Tu)) -\phi(d(u,Tu)+\varphi(Tu)). \end{eqnarray*} This holds if and only if, $\phi(d(u,Tu)+\varphi(Tu))=0$ and then from the property of $\phi$ we have, $$d(u,Tu)+\varphi(Tu)=0.$$ Hence, $d(u,Tu)=0 $ so that u = Tu and $\varphi(Tu)=0.$ Since $u= Tu$ this implies $\varphi(u)=0. $
Therefore $u$ is fixed point of $T$.
Uniqueness
Suppose $v$ is another fixed point of $T$. Then $Tv=v$ and $\varphi(v)=0$.\\By (1) with $x=u$ and $y=v$ \begin{eqnarray*} \psi(s^{3}d(Tu, Tv)+ \varphi(Tu)+ \varphi(Tv))=\psi(s^{3}d(u, v))\leq \psi(m(Tu,Tv,d,T,\varphi))-\phi(l(Tu,Tv,d,T,\varphi)). \end{eqnarray*} From (2) we have
$ m(Tu,Tv,d,T,\varphi)= max\{(d(Tu,Tv)+\varphi(Tu)+\varphi(Tv),d(Tu,T^{2}u)+\varphi(Tu)+\varphi(T^{2}u), d(Tv,T^{2}v)+\varphi(Tv)+\varphi(T^{2}v), \frac{1}{2s^{2}}\{d(Tu,T^{2}v)+\varphi(Tu)+\varphi(T^{2}v)+d(Tv,T^{2}u)+\varphi(Tv)+\varphi(T^{2}u)\}\}= max\{(d(u,v)+\varphi(u)+\varphi(v),d(u,u)+\varphi(u)+\varphi(u), d(v,v)+\varphi(v)+\varphi(v), \frac{1}{2s^{2}}\{d(u,v)+\varphi(u)+\varphi(v)+d(v,u)+\varphi(v)+\varphi(u)\}\}= d(u,v). $ Similarly from (3), we have \begin{eqnarray*} l(Tu,Tv,d,T,\varphi)&=& max \{d(Tu,Tv)+\varphi(Tu)+\varphi(Tv),d(Tv,T^{2}v)+\varphi(Tv)+\varphi(T^{2}v)\\&=& d(u,v)+\varphi(u)+\varphi(v),d(v,v)+\varphi(v)+\varphi(v)\}=d(u,v). \end{eqnarray*} Then using (1) and the continuity of $\psi$, we have \begin{eqnarray*} \psi(d(u,v))&\leq &\psi(s^{3}d(u, v))\leq \psi(d(u,v))-\phi(d(u,v)). \end{eqnarray*} This holds if $\phi(d(u,v))=0$ and then we have $d(u,v)=0.$ Hence $u=v.$ Therefore, $T$ has a unique fixed point.

Example 2. Let $ X = [0, 1] $ and $d(x, y) = ( x - y)^{2}$. Then $(X,d)$ is $b$-metric space with $ s = 2.$ Define $T:X\rightarrow X$, $\varphi:X\rightarrow R^{+} \ and \ \psi,\phi:R^{+}\rightarrow R^{+} $ by $T(x)=\left\{ \begin{array}{ll} 0, & \text{if}\; x\leq \frac{1}{4}; \\ \frac{1}{16}, & \text{if}\; x\in (\frac{1}{4},1]. \end{array} \right.$, $\psi(t)=\frac{5t}{4}$, $\phi(t)=\left\{ \begin{array}{ll} \frac{t}{8}, & \text{if}\; t\leq 3; \\ \frac{t}{4}, & \text{if}\; t\geq 3. \end{array} \right.$ and $\varphi(t)=\left\{ \begin{array}{ll} 2t, & \text{if}\; t\geq 1; \\ t, & \text{if}\; 0\leq t\leq 1. \end{array} \right.$ Now we verify condition (1).
Case I: If $ x,y \in [0,\frac{1}{4}]$ and $x\geq y$. Then $ \psi[s^{3}d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)]=\psi[2^{3}(Tx-Ty)^{2}+\varphi(Tx)+\varphi(Ty)]=\psi[8(0)+\varphi(0)+\varphi(0)]= 0.$ Also $ d(x,y)+\varphi(x)+\varphi(y)=(x-y )^{2}+\varphi(x)+\varphi(y)=(x-y )^{2}+x+y,$
$ d(x,Tx)+\varphi(x)+\varphi(Tx)=(x-Tx)^{2}+\varphi(x)+\varphi(Tx)= x^{2}+x, $
$ d(y,Ty)+\varphi(y)+\varphi(Ty)=(y-Ty)^{2}+\varphi(y)+\varphi(Ty)= y^{2}+y,$
$\frac{1}{s^{2}}[d(x,Ty)+\varphi(x)+\varphi(Ty)+d(y,Tx)+\varphi(y)+\varphi(Tx)]=\frac{1}{8}(x^{2}+x+y^{2}+y).$ And $m(x,y,d,T,\varphi)= max\{(x-y )^{2}+x+y, x^{2}+x,y^{2}+y, \frac{1}{8}(x^{2}+x+y^{2}+y)\},$
$\frac{1}{8}(x^{2}+x+y^{2}+y)\leq max\{x^{2}+x,y^{2}+y\}= x^{2}+x,$
$m(x,y,d,T,\varphi)=max\{(x-y )^{2} + x+y, x^{2}+x\}$
and $l(x,y,d,T,\varphi)= max\{(x-y )^{2} + x+y, y^{2}+y\}.$ But $(x-y )^{2} + x+y \geq x+y\geq y+y \geq y^{2}+y,$ so , $ l(x,y,d,T,\varphi)=(x-y )^{2} + x+y$ and $ \phi[l(x,y,d,T,\varphi)]= \frac{1}{8}[(x-y)^{2} + x+y].$ If $m(x,y,d,T,\varphi)= (x-y )^{2} + x+y$ then (1) becomes $0\leq \frac{5}{4} [(x-y )^{2} + x+y]-\frac{1}{8}[(x-y )^{2} + x+y]=\frac{9}{8} [(x-y )^{2} + x+y]$ and if $m(x,y,d,T,\varphi)= x^{2} + x $ then by (1), we have $\frac{5}{4} (x^{2} + x)-\frac{1}{8}[(x-y )^{2} + x+y]\geq \frac{5}{4} (x^{2} + x)-\frac{1}{8}(x^{2} + x)=\frac{9}{8} (x^{2} + x)\geq 0.$ Let $ x,y \in [0,\frac{1}{4}]$ and $x < y$. Then $m(x,y,d,T,\varphi)= max\{(x-y )^{2}+x+y,x^{2}+x,y^{2}+y,\frac{1}{8}(x^{2}+x+y^{2}+y)\},$
$\frac{1}{8}(x^{2}+x+y^{2}+y)\leq max\{x^{2}+x,y^{2}+y\}= y^{2}+y, $ implies $m(x,y,d,T,\varphi)= max\{(x-y )^{2} + x+y, y^{2}+y\}.$ Similarly $l(x,y,d,T,\varphi) = max \{(x-y )^{2} + x+y, y^{2}+y\}.$ Now, if $m(x,y,d,T,\varphi)=(x-y )^{2} + x+y=l(x,y,d,T,\varphi)$, (1) becomes $ 0\leq\frac{9}{8} [(x-y )^{2} + x+y].$ If $m(x,y,d,T,\varphi)=y^{2}+y=l(x,y,d,T,\varphi)$, (1) becomes $0\leq\frac{9}{8} (y^{2}+y).$
Case II: If $ x \in [0,\frac{1}{4}], y \in (\frac{1}{4},1]$. This implies $x< y$. Then $\psi[s^{3}d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)]=\psi[2^{3}(Tx-Ty)^{2}+\varphi(Tx)+\varphi(Ty)]=\frac{5}{4}[8(0-\frac{1}{16})^{2}+\varphi(0)+\varphi(\frac{1}{16})]=\frac{5}{4}(\frac{1}{32}+\frac{1}{16}) =\dfrac{15}{128}.$ Also $ d(x,y)+\varphi(x)+\varphi(y)=(x-y )^{2}+\varphi(x)+\varphi(y)=(x-y )^{2}+x+y, $
$ d(x,Tx)+\varphi(x)+\varphi(Tx)=(x-Tx)^{2}+\varphi(x)+\varphi(Tx)=x^{2}+x, $
$d(y,Ty)+\varphi(y)+\varphi(Ty)=(y-\frac{1}{16})^{2}+y+\frac{1}{16}=y^{2}+\frac{7y}{8}+\frac{17}{256},$
$ \frac{1}{s^{2}}[d(x,Ty)+\varphi(x)+\varphi(Ty)+d(y,Tx)+\varphi(y)+\varphi(Tx)] =\frac{1}{8}[x^{2}+\frac{7x}{8}+y^{2}+y+\frac{17}{256}].$ And $m(x,y,d,T,\varphi)= max\{(x-y )^{2}+x+y,x^{2}+x,y^{2}+\frac{7y}{8}+\frac{17}{256},\frac{1}{8}[x^{2}+\frac{7x}{8}+y^{2}+y+\frac{17}{256}]\}, $
$\frac{1}{8}[x^{2}+\frac{7x}{8}+y^{2}+y+\frac{17}{256}]= \frac{1}{4}[\frac{x^{2}}{2}+\frac{7x}{16}+\frac{y^{2}}{2}+\frac{y}{2}+\frac{17}{512}] \leq \frac{1}{4}[x^{2}+x +y^{2}+\frac{7y}{8}+\frac{17}{256}] \leq max\{x^{2}+x,y^{2}+\frac{7y}{8}+\frac{17}{256}\} = y^{2}+\frac{7y}{8}+\frac{17}{256}.$ Therefore $m(x,y,d,T,\varphi)= max\{(x-y )^{2} + x+y,y^{2}+\frac{7y}{8}+\frac{17}{256}\}$. Similarly, $l(x,y,d,T,\varphi)= max\{(x-y )^{2} + x+y,y^{2}+\frac{7y}{8}+\frac{17}{256}\}.$ Now if, $m(x,y,d,T,\varphi)= (x-y )^{2} + x+y= l(x,y,d,T,\varphi)$, then $\psi[m(x,y,d,T,\varphi)]=\frac{5}{4} [(x-y )^{2} + x+y]$ and $\phi[m(x,y,d,T,\varphi)]=\frac{1}{8} [(x-y )^{2} + x+y].$ So (1) becomes $ \frac{15}{128} \leq \frac{9}{8} [(x-y )^{2} + x+y].$ Also if, $m(x,y,d,T,\varphi)= y^{2}+\frac{7y}{8}+\frac{17}{256} = l(x,y,d,T,\varphi),$ then $\psi[m(x,y,d,T,\varphi)]=\frac{5}{4} [y^{2}+\frac{7y}{8}+\frac{17}{256}]$ and $\phi[m(x,y,d,T,\varphi)]=\frac{1}{8} [y^{2}+\frac{7y}{8}+\frac{17}{256}].$ So (1) becomes $ \frac{15}{128} \leq \frac{9}{8}[ y^{2}+\frac{7y}{8}+\frac{17}{256}]$.
Case III: If $ x,y \in (\frac{1}{4},1]$ and $x\geq y$. Then $\psi[s^{3}d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)]=\psi[8(\frac{1}{16}-\frac{1}{16})^{2}+\frac{1}{16}+\frac{1}{16}]=\psi[\dfrac{1}{8}]= \frac{5}{32}.$ Also $d(x,y)+\varphi(x)+\varphi(y)=(x-y )^{2}+x+y,$ $ d(x,Tx)+\varphi(x)+\varphi(Tx) =(x-\frac{1}{16})^{2}+x+\frac{1}{16},$
$d(y,Ty)+\varphi(y)+\varphi(Ty)=(y-\frac{1}{16})^{2}+y+\frac{1}{16},$
$\frac{1}{s^{2}}[d(x,Ty)+\varphi(x)+\varphi(Ty)+d(y,Tx)+\varphi(y)+\varphi(Tx)]=\frac{1}{8}[(x-\frac{1}{16})^{2}+x+\frac{1}{16}+(y-\frac{1}{16})^{2}+y+\frac{1}{16})] \leq max[(x-\frac{1}{16})^{2}+x+\frac{1}{16},(y-\frac{1}{16})^{2}+y+\frac{1}{16}]=(x-\frac{1}{16})^{2}+x+\frac{1}{16}.$ This implies that $m(x,y,d,T,\varphi)=max\{(x-y)^{2}+x+y,(x-\frac{1}{16})^{2}+x+\frac{1}{16}\}$ But $(y-\frac{1}{16})^{2}+y+\frac{1}{16}= (\frac{1}{16}-y)^{2}+y+\frac{1}{16}< (x-y)^{2}+y+x.$ So $l(x,y,d,T,\varphi)=(x-y)^{2}+x+y $. Now if, $ m(x,y,d,T,\varphi)= (x-y)^{2}+x+y$ \ (1) becomes, $\frac{5}{32}\leq \frac{5}{4} [(x-y )^{2} + x+y]-\frac{1}{8}[(x-y )^{2} + x+y]=\frac{9}{8} [(x-y )^{2} + x+y].$ If $m(x,y,d,T,\varphi)=(x-\frac{1}{16})^{2}+x+\frac{1}{16}$
then we have, $\frac{5}{4} [(x-\frac{1}{16})^{2}+x+\frac{1}{16}]-\frac{1}{8}[(x-y )^{2} + x+y] > \frac{5}{4} [(x-\frac{1}{16})^{2}+x+\frac{1}{16}]-\frac{1}{8}[(x-\frac{1}{16})^{2}+x+\frac{1}{16}] =\frac{9}{8} [(x-\frac{1}{16})^{2}+x+\frac{1}{16}] >\frac{5}{32}.$ Let $ x,y \in (\frac{1}{4},1]$ and $x < y$. Then $m(x,y,d,T,\varphi)= max\{(x-y)^{2}+x+y,(y-\frac{1}{16})^{2}+y+\frac{1}{16}\}=l(x,y,d,T,\varphi).$ Now if $ m(x,y,d,T,\varphi)=l(x,y,d,T,\varphi)=(x-y)^{2}+x+y,$ then\ (1) becomes $\frac{5}{32}\leq \frac{5}{4} [(x-y )^{2} + x+y]-\frac{1}{8}[(x-y )^{2} + x+y]=\frac{9}{8} [(x-y )^{2} + x+y].$ If $ m(x,y,d,T,\varphi)=l(x,y,d,T,\varphi)=(y-\frac{1}{16})^{2}+y+\frac{1}{16},$ (1) becomes $\frac{5}{32}\leq \frac{5}{4} [(y-\frac{1}{16})^{2}+y+\frac{1}{16}]-\frac{1}{8}[(y-\frac{1}{16})^{2}+y+\frac{1}{16}]=\frac{9}{8} [(y-\frac{1}{16})^{2}+y+\frac{1}{16}].$
Case IV: If $ x \in (\frac{1}{4},1], y \in [0,\frac{1}{4}]$. This implies $x>y$. Then $\psi[s^{3}d(Tx,Ty)+\varphi(Tx)+\varphi(Ty)]=\psi[2^{3}(Tx-Ty)^{2}+\varphi(Tx)+\varphi(Ty)]=\frac{5}{4}[8(\frac{1}{16}-0)^{2}+\varphi(\frac{1}{16})++\varphi(0)]=\frac{5}{4}(\frac{1}{32}+\frac{1}{16}) =\dfrac{15}{128}.$ Also $d(x,y)+\varphi(x)+\varphi(y)=(x-y )^{2}+x+y,$
$ d(x,Tx)+\varphi(x)+\varphi(Tx)= x^{2}+\frac{7x}{8}+\frac{17}{256},$
$d(y,Ty)+\varphi(y)+\varphi(Ty)=y^{2}+y,$
$\frac{1}{s^{2}}[d(x,Ty)+\varphi(x)+\varphi(Ty)+d(y,Tx)+\varphi(y)+\varphi(Tx)]=\frac{1}{8}[x^{2}+x + y^{2}+\frac{7y}{8}+\frac{17}{256}].$ And $m(x,y,d,T,\varphi)= max\{(x-y )^{2}+x+y,x^{2}+\frac{7x}{8}+\frac{17}{256},y^{2}+y,\frac{1}{8}[x^{2}+x + y^{2}+\frac{7y}{8}+\frac{17}{256}]\},$
$m(x,y,d,T,\varphi)= max\{(x-y )^{2} + x+y,x^{2}+\frac{7x}{8}+\frac{17}{256}\}.$ Similarly, $l(x,y,d,T,\varphi)=max\{(x-y )^{2} + x+y,y^{2}+y\}= (x-y )^{2} + x+y.$ Now if, $m(x,y,d,T,\varphi)= (x-y )^{2} + x+y= \l(x,y,d,T,\varphi)$, then (1) becomes $ \frac{15}{128} \leq \frac{9}{8} [(x-y )^{2} + x+y].$ If, $m(x,y,d,T,\varphi)= x^{2}+\frac{7x}{8}+\frac{17}{256},$ we have $\frac{5}{4} [x^{2}+\frac{7x}{8}+\frac{17}{256}]-\frac{1}{8}[(x-y )^{2} + x+y] > \frac{5}{4} [x^{2}+\frac{7x}{8}+\frac{17}{256}]-\frac{1}{8}[x^{2}+\frac{7x}{8}+\frac{17}{256}] =\frac{9}{8} [x^{2}+\frac{7x}{8}+\frac{17}{256}] >\frac{15}{128}.$ Then (1) becomes $\frac{15}{128} \leq \frac{9}{8}[ x^{2}+\frac{7x}{8}+\frac{17}{256}].$ Thus all the condition of Theorem (2) are satisfied and $0$ is the unique fixed point of $T$.

Remark 1. If we take s=1 in Theorem (2) we get the result of Cho [1]. Hence Our result generalizes the result of Cho [1] and related results in the literature.

Acknowledgments

The authors would like to thank Jimma University for material support.

Authorcontributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflictofinterests

The authors declare no conflict of interest.

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A fixed point theorem for generalized weakly contractive mappings in \(b\)-metric spaces