The purpose of this paper is to prove a fixed point theorem for \(C\)-class functions in complete \(b\)-metric spaces. Moreover, the solution of the integral equation is obtained using our main result.
The concept of \(b\)-metric spaces was first introduced by Bakhtin [1] and Czerwik [2] and utilized for \(s=2\) and for an arbitrary \(s\geq 1\) to prove some generalizations of Banach’s fixed point theorem [3]. In 2010, Khamsi and Hussain [4] reintroduced the notion of \(b\) -metric and called it a metric-type. Afterwards, several authors proved fixed and common fixed point theorems for single-valued mappings in \(b\)-metric spaces, see [5,6,7,8,9,10,11,12,13,14,15,16].
In this paper, we introduce the definition of \(C\)-class functions and \(% \left( \psi ,\varphi ,F\right) \)-contraction type mappings where \(\psi \) is the altering distance function and \(\varphi \) is the ultra altering distance function. The unique fixed point theorem for self mapping in the setting of \(% b\)-complete metric spaces is proven. In the end of paper, we apply our main result to approximating the solution of the Fredholm integral equation.
In the sequel, we always denote by \(\mathbb{N}\), \(\mathbb{R}\), and \(\mathbb{R}_{+}\) the set of positive integers, real numbers, and nonnegative real numbers, respectively. The following definitions, notations, basic lemma and remarks will be needed in the sequel.
Definition 1. [1] Let \(X\) be a nonempty set and \(s\geq 1\) a given real number. A function \( d:X\times X\rightarrow\mathbb{R}_{+}\) is called a \(b\)-metric on \(X\) if for all \(x,y,z\in X\), the following conditions are satisfied;
Every metric space is a \(b\)-metric space with \(s=1\), but the converse is not true in general as it is shown by the following example.
Example 1. [17] Let \(X=\{0,1,2\}\) and \(d:X\times X\rightarrow\mathbb{R}_{+}\) defined by \begin{align*} d(0,0) &=d(1,1)=d(2,2)=0\text{,} \\ d(1,0) &=d(0,1)=d(2,1)=d(1,2)=1\text{,} \\ d(0,2) &=d(2,0)=m\text{,} \end{align*} where, \(m\) is given real number such that \(m\geq 2\). It is easy to check that for all \(x,y,z\in X\) \begin{equation*} d(x,y)\leq \frac{m}{2}(d(x,z)+d(z,y))\text{.} \end{equation*} Therefore, \((X,d)\) is a \(b\)-metric space with a coefficient \(s=\dfrac{m}{2}\). The ordinary triangle inequality does not hold if \(m>2\) and so \((X,d)\) is not a metric space.
Example 2. [13] Let \((X,d)\) be a metric space and \(\rho (x,y)=(d(x,y))^{p}\), where \(p>1\) is a real number. Then \(\rho \) is a \(b\)-metric with \(s=2^{p-1}\).
For other examples of a \(b\)-metric, see [1].
Definition 2. [18] Let \((X,d)\) be a \(b\)-metric space and \(\left\{ x_{n}\right\} \) a sequence in \(X\). The sequence \(\left\{ x_{n}\right\} \) is said to be
Remark 1. In general, a \(b\)-metric need not be continuous in each variable [13].
The following lemma was established by [12].
Lemma 1. Let \((X,d)\) be a \(b\)-metric space with a coefficient \(s\geq 1\) and \(\left\{ x_{n}\right\} \) a sequence in \(X\) such that \begin{equation*} d(x_{n},x_{n+1})\leq \lambda d(x_{n-1},x_{n}),\text{ }n=1,2,…, \end{equation*} where \(0\leq \lambda < 1\). Then \(\{x_{n}\}\) is a Cauchy sequence.
Recently, Ansari [19] introduced the concept of following \(C\)-class functions which covers a large class of contractive conditions.
Definition 3. [19] A continuous function \(F:[0,+\infty )\rightarrow\mathbb{R}\) is called \(C\)-class function if for any \(s,t\in \lbrack 0,+\infty )\); the following conditions hold
Example 3. The following examples show that the class \(C\) is nonempty;
Definition 4. [20] A function \(\psi :[0,+\infty )\rightarrow \lbrack 0,+\infty )\) is called an altering distance function if the following properties are satisfied;
Definition 5. A tripled \((\psi ,\varphi ,F)\) where \(\psi \in \) \(\Psi \); \(\varphi \) \(\in \Phi _{u}\) and \(F\in C\) is said to be a monotone if for any \(x,y\in \lbrack 0,+\infty )\), we have \begin{equation*} x\leq y\Rightarrow F(\psi (x),\varphi (x))\leq F(\psi (y),\varphi (y)). \end{equation*}
Example 4. Let \(F(s,t)=s-t\), \(\varphi (x)=\sqrt{x}\) and \( \psi (x)=\left\{ \begin{array}{c} \sqrt{x} \;\;\;\;\;\; \text{if}\;\;0\leq x\leq 1 \\ x^{2}\;\;\text{if}\;\;x>1 \end{array} \right., \) then \((\psi ,\varphi ,F)\) is monotone.
Theorem 1. Let \((X,d)\) be a \(b\)-complete metric space and \(T\) be a self-mapping on \(X\) that satisfies the following contractive condition;
Proof. Let \(x\) in \(X\) and \(\left\{ x_{n}\right\} _{n}\) be a sequence in \(X\) defined as \begin{equation*} Tx_{n}=x_{n+1},y_{n}=x_{n-1}\text{ }n=0,1,2\ldots . \end{equation*} Applying the inequality (1), we obtain \begin{equation*} \psi \left( d(Tx_{n},Tx_{n-1})\right) \leq F\left( \psi (M(x_{n},x_{n-1})),\varphi (M(x_{n},x_{n-1}))\right) , \end{equation*} where \begin{align*} M(x_{n},x_{n-1}) &=\max \left\{ d(x_{n},x_{n-1}),\frac{d^{2}(x_{n},x_{n-1})% }{1+d(x_{n-1},Tx_{n-1})},\frac{d^{2}(x_{n-1},Tx_{n-1})}{1+d(x_{n},x_{n-1})},% \frac{d(x_{n},Tx_{n})d(x_{n-1},Tx_{n-1})}{1+d(Tx_{n-1},Tx_{n})}\right\} \\ &\leq d(x_{n-1},x_{n}). \end{align*} Thus \begin{align*} \psi \left( d(Tx_{n},Tx_{n-1})\right) &\leq F\left( \psi (d(x_{n},x_{n-1})),\varphi (d(x_{n},x_{n-1}))\right) \\ &\leq \psi (d(x_{n},x_{n-1})). \end{align*} Since \(\psi \) is non-decreasing, then \(d(Tx_{n},Tx_{n+1})\leq d(x_{n},x_{n-1}).\) This means \(\left\{ d(x_{n},x_{n+1})\right\} \) is a decreasing sequence. Thus it converges and there exists \(r\geq 0\) such that \(\lim\limits_{n% \rightarrow +\infty }d(x_{n},x_{n+1})=r\). Taking \(n\rightarrow +\infty \), then contractive condition implies \( \psi \left( r\right) \leq F\left( \psi (r),\varphi (r)\right) \leq \psi \left( r\right) . \) So, \(\psi (r)=0\). Therefore \(r=0\), that is \( \lim\limits_{n\rightarrow +\infty }d(x_{n},x_{n+1})=0. \)
Now, we prove that the sequence \(\left\{ x_{n}\right\} \) is a Cauchy sequence. Suppose that \(\left\{ x_{n}\right\} \) is not a Cauchy sequence, then there exists an \(\varepsilon >0\) for which we can nd two sequences of positive integers \(m(k)\) and \(n(k)\) such that for all positive integers \(k\), \(n(k)>m(k)>k\) and \(d(x_{m(k)},x_{n(k)})\geq \varepsilon .\) Let \(n(k)\) be the smallest positive integer \(n(k)>m(k)>k\), such that
\begin{equation*} d(x_{m(k)},x_{n(k)})\geq \varepsilon ,\text{ }d(x_{m(k)},x_{n(k)-1})\leq \varepsilon . \end{equation*} Then, we find \(\psi (\varepsilon )=0\) which is a contradiction. Thus \(% \left\{ x_{n}\right\} \) is a \(b\)-Cauchy sequence in \(X\). Since \((X,d)\) is a complete \(b\)-metric space, so there exists \(u\in X,\) such that \(% \lim\limits_{n\rightarrow +\infty }x_{n}=u.\)The following example supports our Theorem 1.
Example 5. Let the complete \(b\)-metric space \((X,d)\) with \(X=\left[0,\frac{1}{2% }\right]\) and \begin{equation*} d(x,y)=|x-y|\text{ }for\text{ }all\text{ }x,y\in X. \end{equation*} Consider \(T:X\rightarrow X\) be given by \(Tx=\frac{x}{4}\) for all \(x\in X\). Then, for \(\psi \left( t\right) =t\) and \(F(s,t)=ms\) for some \(m\in (0,1),\) we have \begin{align*} d(Tx,Ty) &=\frac{1}{4}\left\vert x-y\right\vert \leq \frac{1}{2}\left\vert x-y\right\vert \\ &\leq \frac{1}{2}d(x,y)\leq \frac{1}{2}M(x,y). \end{align*} Thus, \(T\) is satisfying all the conditions of Theorem 1 and \(0\) is its fixed point, which is unique.
The following results can be obtained immediately from Theorem 1.
Corollary 1. Let \((X,d)\) be a complete \(b\)-metric space and \(T\) be a self-mapping on \(% X\) that satisfies the following contractive condition; \begin{equation*} \psi \left( d(Tx,Ty)\right) \leq \psi (M(x,y))-\varphi (M(x,y)), \end{equation*} for all \(x,y\in X\) where \(\psi \in \Psi \); \(\varphi \in \Phi _{u}\) and \(F\in C\) such that \((\psi ,\varphi ,F)\) is monotone and \begin{equation*} M(x,y)=\max \left\{ d(x,y),\frac{d^{2}(x,y)}{1+d(y,Ty)},\frac{d^{2}(y,Ty)}{% 1+d(x,y)},\frac{d(x,Tx)d(y,Ty)}{1+d(Ty,Tx)}\right\} . \end{equation*} Then, \(T\) has a unique fixed point in \(X.\)
Proof. Taking \(F(s,t)=s-t\), in Theorem 1, we obtain the desired result.
Corollary 2. Let \((X,d)\) be a complete \(b\)-metric space and \(T\) be a self-mapping on \(% X\) that satisfies the following contractive condition; \begin{equation*} \psi \left( d(Tx,Ty)\right) \leq M(x,y)\Phi \left( \varphi (M(x,y)\right) , \end{equation*} for all \(x,y\in X\) where \(\psi \in \Psi \); \(\varphi \in \Phi _{u}\) and \(F\in C\) such that \((\psi ,\varphi ,F)\) is monotone and \begin{equation*} M(x,y)=\max \left\{ d(x,y),\frac{d^{2}(x,y)}{1+d(y,Ty)},\frac{d^{2}(y,Ty)}{% 1+d(x,y)},\frac{d(x,Tx)d(y,Ty)}{1+d(Ty,Tx)}\right\} . \end{equation*} Then, \(T\) has a unique fixed point in \(X\)
Proof. Taking \(\psi \left( t\right) =t\) and \(F(s,t)=\frac{s}{(1+t)^{r}}\) for some \(% r\in (0,1)\) in Theorem 1, we obtain the desired result.
Corollary 3. Let \((X,d)\) be a complete \(b\)-metric space and \(T\) be a self-mapping on \(% X\) that satisfies the following contractive condition; \begin{equation*} d(Tx,Ty)\leq \frac{M(x,y)}{\left( 1+M(x,y)\right) ^{r}}, \end{equation*} for all \(x,y\in X\) where \(\psi \in \Psi \); \(\varphi \in \Phi _{u}\) ,\(% r\in (0,1)\) and \(F\in C\) such that \((\psi ,\varphi ,F)\) is monotone and \begin{equation*} M(x,y)=\max \left\{ d(x,y),\frac{d^{2}(x,y)}{1+d(y,Ty)},\frac{d^{2}(y,Ty)}{% 1+d(x,y)},\frac{d(x,Tx)d(y,Ty)}{1+d(Ty,Tx)}\right\} . \end{equation*} Then, \(T\) has a unique fixed point in \(X\)
Proof. Taking \(F(s,t)=s\Phi \left( t\right) \), (\(s,t>0\)) in Theorem 1, we obtain the desired result.
Theorem 2. We assume the following conditions;
Proof. Define the mapping \(T:X\rightarrow X\) by \begin{equation*} Tx(t)=f(t)+\lambda \int\limits_{a}^{b}K(t,\tau ,x(\tau ))d\tau , \end{equation*} for all \(t\in \lbrack a,b]\). So, the existence of a solution of (3) is equivalent to the existence of fixed point \(T\). Let \(q\in\mathbb{R}\) such that \(\dfrac{1}{p}+\dfrac{1}{q}=1\). Using the Hölder inequality, and conditions (i)-(iii), we have \begin{align*} d(Tx,Ty) &=\max\limits_{t\in \lbrack a,b]}|Tx(t)-Ty(t)|^{p} \\ &\leq \left\vert \lambda \right\vert ^{p}\max\limits_{t\in \lbrack a,b]}\left( \int\limits_{a}^{b}\left\vert K(t,\tau ,x)-K(t,\tau ,y\right\vert ^{p}d\tau \right) \\ &\leq \left[ \max\limits_{t\in \lbrack a,b]}\left( \int\limits_{a}^{b}1^{q}dz\right) ^{\frac{1}{q}}\left( \int\limits_{a}^{b}\left\vert \left( K(t,\tau ,x)-K(t,\tau ,y\right) )\right\vert ^{p}d\tau \right) ^{\frac{1}{p}}\right] ^{p} \\ &\leq \left( b-a\right) ^{\frac{p}{q}}\left[ \max\limits_{t\in \lbrack a,b]}\left( \int\limits_{a}^{b}\psi (t,\tau )\left\vert x-y\right\vert ^{p}d\tau \right) \right] \\ &\leq \left( b-a\right) ^{p-1}\max\limits_{t\in \lbrack a,b]}\left( \int\limits_{a}^{b}\psi (t,\tau )d\tau \right) d(x,y) \\ &\leq \left( b-a\right) ^{p-1}\dfrac{1}{(b-a)^{p-1}}.M\left( x,y\right). \end{align*} Thus \begin{equation*} d(Tx,Ty)\leq M\left( x,y\right). \end{equation*} Hence, all the conditions of Theorem 1 hold. Consequently, the Equation (3) has a solution \(z\in C[a,b]\).