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.
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 quoteright 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.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:
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
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
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:
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 \(x_{1}\) = \(Tx_{0}\), \(x_{2}\) = \(Tx_{1}\),…, \(x_{n+1}\)= \(Tx_{n}\) 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
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, \(m(x,y,d,T,\varphi)= (x-y )^{2} + x+y= (x,y,d,T,\varphi)\),
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.
