Some fixed point theorems in b-metric-like spaces

Abstract

In this work, some fixed point and common fixed point theorems are investigated in b-metric-like spaces. Some of our results generalize related results in the literature. Also, some examples and an application to integral equation are given to support our main results.

1 Introduction and preliminaries

There exist many generalizations of the concept of metric spaces in the literature. In [1, 2], Matthews introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks. A lot of fixed point theorems were investigated in partial spaces (see, e.g., [3–11] and references therein). The notions of metric-like spaces [12] and b-metric spaces [13–16] were introduced in the literature, which are generalizations of metric spaces. Recently, the concept of b-metric-like spaces which is a generalization of metric-like spaces and b-metric spaces and partial metric spaces was introduced in [17]. Recently, Hussain et al. [18] discussed topological structure of b-metric-like spaces and proved some fixed point results in b-metric-like spaces.

Definition 1.1

[17]

A b-metric-like on a nonempty set X is a function $D: X\times X\rightarrow[0, +\infty)$ such that for all $x,y,z\in X$ and a constant $s\geq1$, the following three conditions hold true:

(D₁):

if $D(x,y)=0$ then $x=y$;

(D₂):

$D(x,y)=D(y,x)$;

(D₃):

$D(x,z)\leq s(D(x,y)+D(y,z))$.

The pair $(x,D)$ is then called a b-metric-like space.

Example 1.1

Let $X=\{0,1,2\}$, and let

$$ D(x,y)=\left \{ \textstyle\begin{array}{l@{\quad}l} 2, & x=y=0, \\ \frac{1}{2}, & \mbox{otherwise}. \end{array}\displaystyle \right . $$

Then $(X,D)$ is a b-metric-like space with the constant $s=2$.

In [17], some concepts in b-metric-like spaces were introduced as follows.

Each b-metric-like D on X generalizes a topology $\tau_{D}$ on X whose base is the family of open D-balls $B_{D}(x,\varepsilon)=\{y\in X:|D(x,y)-D(x,x)|<\varepsilon\}$, for all $x\in X$ and $\varepsilon>0$.

A sequence $\{x_{n}\}$ in the b-metric-like space $(X,D)$ converges to a point $x\in X$ if and only if $D(x,x)=\lim_{n\rightarrow+\infty}D(x,x_{n})$.

A sequence $\{x_{n}\}$ in the b-metric-like space $(X,D)$ is called a Cauchy sequence if there exists $\lim_{n,m\rightarrow+\infty}D(x_{m},x_{n})$ (and it is finite).

A b-metric-like space is called complete if every Cauchy sequence $\{ x_{n}\}$ in X converges with respect to $\tau_{D}$ to a point $x\in X$ such that $\lim_{n\rightarrow+\infty}D(x,x_{n})=D(x,x)=\lim_{n,m\rightarrow+\infty}D(x_{m},x_{n})$.

Remark 1.1

In Example 1.1, let $x_{n}=2$ for each $n=1,2,\ldots$ , then it is clear that $\lim_{n\rightarrow+\infty}D(x_{n},2)=D(2,2)$ and $\lim_{n\rightarrow+\infty}D(x_{n},1)=D(1,1)$, hence, in b-metric-like spaces, the limit of a convergent sequence is not necessarily unique.

Remark 1.2

It should be noted that in general, a b-metric-like function $D(x,y)$ need not be jointly continuous in both variables. The following example illustrates this fact.

Example 1.2

Let $X=\mathbb{N}\cup\{ +\infty\}$ (where $\mathbb{N}$ is the set of all natural numbers, similarly hereinafter), and let $D: X\times X\rightarrow R$ be defined by

$$ D(x,y)=\left \{ \textstyle\begin{array}{l@{\quad}l} 0, &m \mbox{ and } n \mbox{ are }+\infty, \\ 1, &m=n=1, \\ |\frac{1}{m}-\frac{1}{n}|, &\mbox{if one of } m,n \mbox{ is odd which is larger than }1 \mbox{ and} \\ &\mbox{the other is odd or }+\infty, \\ 7,&\mbox{if one of }m,n \mbox{ is even and the other is even or }+\infty, \\ 2 ,&\mbox{otherwise}. \end{array}\displaystyle \right . $$

Then considering all possible cases, it can be checked that, for all $m,n,p\in X$, we have

$$D(m,n)\leq\frac{7}{2}\bigl[D(m,p)+D(p,n)\bigr]. $$

Thus, $(X,D)$ is a b-metric-like space with $s=\frac{7}{2}$. Let $x_{n}=2n+1$ for each $n\in\mathbb{N}$, then $D(x_{n}, +\infty )=D(2n+1,+\infty)=\frac{1}{2n+1}\rightarrow0$, as $n\rightarrow +\infty$, that is, $x_{n}\rightarrow+\infty$, but $D(x_{n},2)=2\nrightarrow D(+\infty,2)=7$.

Definition 1.2

Suppose that $(X,D)$ is a b-metric-like space. A mapping $T:X\rightarrow X$ is said to be continuous at $x\in X$, if for every $\varepsilon>0$ there exists $\delta>0$ such that $T(B_{D}(x,\delta ))\subseteq B_{D}(Tx,\varepsilon)$. We say that T is continuous on X if T is continuous at all $x\in X$.

Let $(X,D)$ be a b-metric-like space, and let $f: X\rightarrow X$ be a continuous mapping. Then

$$\lim_{n\rightarrow+\infty}D(x_{n},x)=D(x,x)\quad \Rightarrow\quad \lim_{n\rightarrow+\infty} D(fx_{n},fx)=D(fx,fx). $$

In this paper, we investigate some new fixed point and common fixed point theorems in b-metric-like spaces. Some of our results generalize and improve related results in the literature. Some examples and an application are presented to support our main results.

2 Main results

In this section, we begin with the following definitions and lemma which will be needed in the sequel.

Definition 2.1

[19]

Let f and g be two self-mappings on a set X. If $\omega=fx=gx$ for some x in X, then x is called a coincidence point of f and g, where ω is called a point of coincidence of f and g.

Definition 2.2

[19]

Let f and g be two self-mappings defined on a set X. Then f and g are said to be weakly compatible if they commute at every coincidence point, i.e., if $fx=gx$ for some $x\in X$, then $fgx=gfx$.

Lemma 2.1

[17]

Let $(X,D)$ be a b-metric-like space with the constant $s\geq1$. Let $\{y_{n}\}$ be a sequence in $(X,D)$ such that

$$ D(y_{n},y_{n+1})\leq\lambda D(y_{n-1},y_{n}) $$

(2.1)

for some λ, $0<\lambda<\frac{1}{s}$, and each $n=1,2,\ldots$ .

Then $\lim_{m,n\rightarrow+\infty}D(y_{m},y_{n})=0$.

Let Φ denote the set of all functions $\phi: [0,+\infty )\rightarrow[0,+\infty)$ satisfying:

• (1) ϕ is continuous and nondecreasing;

• (2) $\phi(t)=0$ if and only if $t=0$.

Now we prove our main results.

Theorem 2.1

Let $(X,D)$ be a complete b-metric-like space with the constant $s\geq1$ and let $T: X \rightarrow X$ be a mapping such that

$$ D(Tx,Ty)\leq\frac{D(x,y)}{s}-\varphi\bigl(D(x,y)\bigr) $$

(2.2)

for all $x,y\in X$, where $\varphi\in\Phi$. Then T has a unique fixed point.

Proof

Let $x_{0}$ be an arbitrary point in X. Define $x_{n+1}=Tx_{n}$ for $n=0,1,2,\ldots$ , then we can claim that

$$ \lim_{n\rightarrow+\infty}D(x_{n},x_{n+1})=0. $$

(2.3)

In fact, by (2.2), we have

$$ D(x_{n+1},x_{n+2})=D(Tx_{n},Tx_{n+1}) \leq\frac {D(x_{n},x_{n+1})}{s}-\varphi\bigl(D(x_{n},x_{n+1})\bigr)\leq D(x_{n},x_{n+1}), $$

(2.4)

it means that sequence $\{D(x_{n},x_{n+1})\}$ is non-increasing and hence there exists some nonnegative number $r_{0}$ such that

$$ \lim_{n\rightarrow+\infty}D(x_{n},x_{n+1})=r_{0}. $$

(2.5)

Since

$$\begin{aligned} D(x_{n+1},x_{n+2}) =&D(Tx_{n},Tx_{n+1})\leq \frac {D(x_{n},x_{n+1})}{s}-\varphi\bigl(D(x_{n},x_{n+1})\bigr) \\ \leq& D(x_{n},x_{n+1})-\varphi\bigl(D(x_{n},x_{n+1}) \bigr), \end{aligned}$$

taking $n\rightarrow+\infty$ in the above inequalities, the continuity of φ and (2.5) shows that $r_{0}\leq r_{0}-\varphi(r_{0})$, yielding $r_{0}=0$, hence we conclude our claim.

Now, we show that $\{x_{n}\}$ is a Cauchy sequence. For arbitrary $\varepsilon>0$, we choose $N\in\mathbb{N}$ such that

$$ D(x_{n},x_{n+1})< \min\biggl\{ \frac{\varepsilon}{2s},\varphi\biggl(\frac {\varepsilon}{2s}\biggr)\biggr\} $$

(2.6)

for $n\geq N$.

We claim that if $D(x,x_{N_{0}})\leq\varepsilon$ for $N_{0}>N$, then $D(Tx,x_{N_{0}})\leq\varepsilon$. For this, we distinguish two cases.

Case 1. If $D(x,x_{N_{0}})\leq\frac{\varepsilon}{2s}$, then

$$\begin{aligned} D(Tx,x_{N_{0}}) \leq&s \bigl(D(Tx,Tx_{N_{0}})+D(Tx_{N_{0}},x_{N_{0}}) \bigr) \\ =&sD(Tx,Tx_{N_{0}})+sD(Tx_{N_{0}},x_{N_{0}}) \\ \leq&D(x,x_{N_{0}})-s\varphi\bigl(D(x,x_{N_{0}}) \bigr)+sD(x_{N_{0}+1},x_{N_{0}}) \\ < &\frac{\varepsilon}{2s}+\frac{\varepsilon}{2} \\ \leq&\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{aligned}$$

Case 2. If $\frac{\varepsilon}{2s}< D(x,x_{N_{0}})\leq\varepsilon$, then $\varphi(D(x,x_{N_{0}}))\geq\varphi(\frac{\varepsilon}{2s})$, from which we obtain

$$\begin{aligned} D(Tx,x_{N_{0}}) \leq&s \bigl(D(Tx,Tx_{N_{0}})+D(Tx_{N_{0}},x_{N_{0}}) \bigr) \\ = &sD(Tx,Tx_{N_{0}})+sD(Tx_{N_{0}},x_{N_{0}}) \\ \leq&D(x,x_{N_{0}})-s\varphi \bigl(D(x,x_{N_{0}}) \bigr)+sD(x_{N_{0}+1},x_{N_{0}}) \\ \leq&\varepsilon-s\varphi\biggl(\frac{\varepsilon}{2s}\biggr)+s\varphi\biggl( \frac {\varepsilon}{2s}\biggr)=\varepsilon. \end{aligned}$$

(2.7)

By the above two cases, we show that our claim is true. From (2.6), we have $D(x_{N_{0}+1},x_{N_{0}})<\varepsilon$, which together with our claim implies that $D(Tx_{N_{0}+1},x_{N_{0}})\leq\varepsilon$, that is, $D(x_{N_{0}+2},x_{N_{0}})\leq\varepsilon$. Continue this process, one can deduce that $D(x_{n},x_{N_{0}})<\varepsilon$ for each $n> N_{0}$. Therefore, for any $m,n>N$, we have $D(x_{n},x_{m})\leq s(D(x_{n},x_{N_{0}})+D(x_{N_{0}},x_{m}))<2s\varepsilon$, it follows that $\lim_{n,m\rightarrow+\infty}D(x_{m},x_{n})=0$ and $\{x_{n}\}$ is a Cauchy sequence in $(X,D)$. Since $(X,D)$ is complete, there exists some $u\in X$ such that

$$ \lim_{n\rightarrow+\infty}D(x_{n},u)=D(u,u)=\lim _{m,n\rightarrow+\infty}D(x_{m},x_{n})=0. $$

(2.8)

Since

$$ D(x_{n+1},Tu)=D(Tx_{n},Tu)\leq \frac{D(x_{n},u)}{s}-\varphi\bigl(D(x_{n},u)\bigr), $$

(2.9)

the continuity of φ shows, from (2.8) and (2.9), that $\lim_{n\rightarrow+\infty}D(x_{n},Tu)=0$, which together with the inequality $D(u,Tu)\leq sD(x_{n},u)+sD(x_{n},Tu)$ and (2.8) yields $D(u,Tu)=0$, hence $u=Tu$. Let v be a fixed point of T, that is, $Tv=v$, we have

$$D(u,v)=D(Tu,Tv)\leq\frac{D(u,v)}{s}-\varphi\bigl(D(u,v)\bigr)\leq D(u,v)- \varphi\bigl(D(u,v)\bigr), $$

it implies that $D(u,v)=0$ and so $u=v$, this means that T has a unique fixed point. □

In Theorem 2.1, taking $\varphi(t)=\frac{t}{s}-\lambda t$ with $0<\lambda<\frac{1}{s}$, we can get the following corollary.

Corollary 2.1

Let $(X,D)$ be a complete b-metric-like space with the constant $s\geq1$ and let $T: X \rightarrow X$ be a mapping such that

$$ D(Tx,Ty)\leq\lambda D(x,y) $$

(2.10)

for all $x,y\in X$, where $0<\lambda<\frac{1}{s}$. Then T has a unique fixed point in X.

Remark 2.1

By taking $s=1$ in Theorem 2.1, we get Theorem 2.7 in [12].

Theorem 2.2

Let $(X,D)$ be a complete b-metric-like space with the constant $s\geq1$ and let $T: X \rightarrow X$ be a surjection such that

$$ D(Tx,Ty)\geq a_{1}D(x,y)+a_{2}D(x,Tx)+a_{3}D(y,Ty)+a_{4}D(x,Ty) $$

(2.11)

for all $x,y\in X$, where $a_{i}\geq0$ ($i=1,2,3,4$) satisfy $s(a_{1}+a_{2})+a_{4}+s^{2}(a_{3}-a_{4})>s^{2}$ and $1-a_{3}+a_{4}>0$. Then T has a fixed point.

Proof

Let $x_{0}\in X$. Since T is surjective, choose $x_{1}\in X$ such that $Tx_{1}=x_{0}$. Continuing this process, we can define a sequence $\{x_{n}\}$ such that $x_{n-1}=Tx_{n}$, $n\geq1$, $n\in\mathbb{N}$. Without loss of generality, we assume that $x_{n-1}\neq x_{n}$ for all $n\geq 1$, $n\in\mathbb{N}$. Due to (2.11), we have

$$\begin{aligned} D(x_{n}, x_{n-1}) =&D(Tx_{n+1},Tx_{n}) \\ \geq& a_{1}D(x_{n+1},x_{n})+a_{2}D(x_{n+1},Tx_{n+1})+a_{3}D(x_{n},Tx_{n})+a_{4}D(x_{n+1},Tx_{n}) \\ =& a_{1}D(x_{n+1},x_{n})+a_{2}D(x_{n+1},x_{n})+a_{3}D(x_{n},x_{n-1})+a_{4}D(x_{n+1},x_{n-1}). \end{aligned}$$

(2.12)

By $D(x_{n+1},x_{n-1})\geq\frac {D(x_{n},x_{n+1})-sD(x_{n},x_{n-1})}{s}$, (2.12) implies that

$$ D(x_{n+1},x_{n})\leq\frac{s-sa_{3}+sa_{4}}{sa_{1}+sa_{2}+a_{4}}D(x_{n},x_{n-1}). $$

(2.13)

Letting $\lambda=\frac{s-sa_{3}+sa_{4}}{sa_{1}+sa_{2}+a_{4}}$, by $s(a_{1}+a_{2})+a_{4}+s^{2}(a_{3}-a_{4})>s^{2}$, we have $0<\lambda<\frac{1}{s}$. Applying Lemma 2.1, we see that $\lim_{m,n\rightarrow+\infty}D(x_{m},x_{n})=0$ and $\{x_{n}\}$ is a Cauchy sequence. Since $(X,D)$ is complete, there exists $z\in X$ such that

$$ \lim_{n\rightarrow+\infty}D(x_{n},z)=D(z,z)=\lim _{m,n\rightarrow+\infty}D(x_{m},x_{n})=0. $$

(2.14)

Consequently, we can find $u\in X$ such that $z=Tu$. Now, we show that $z=u$. From (2.11), we get

$$\begin{aligned} D(x_{n},z) =&D(Tx_{n+1},Tu) \\ \geq& a_{1}D(x_{n+1},u)+a_{2}D(x_{n+1},Tx_{n+1})+a_{3}D(u,Tu)+a_{4}D(x_{n+1},Tu) \\ =& a_{1}D(x_{n+1},u)+a_{2}D(x_{n+1},x_{n})+a_{3}D(u,z)+a_{4}D(x_{n+1},z) \end{aligned}$$

and

$$\begin{aligned} D(z,x_{n}) =&D(Tu,Tx_{n+1}) \\ \geq& a_{1}D(u,x_{n+1})+a_{2}D(u,Tu)+a_{3}D(x_{n+1},Tx_{n+1})+a_{4}D(u,Tx_{n+1}) \\ =&a_{1}D(u,x_{n+1})+a_{2}D(u,z)+a_{3}D(x_{n+1},x_{n})+a_{4}D(u,x_{n}). \end{aligned}$$

Adding the above inequalities, we have

$$\begin{aligned} 2D(z,x_{n}) \geq& 2a_{1}D(u,x_{n+1})+(a_{2}+a_{3})D(u,z)+(a_{2}+a_{3})D(x_{n+1},x_{n})+a_{4}D(u,x_{n}) \\ &{}+a_{4}D(x_{n+1},z). \end{aligned}$$

(2.15)

Since $D(u,x_{n+1})\geq\frac{D(u,z)-sD(x_{n+1},z)}{s}$ and $D(u,x_{n})\geq\frac{D(u,z)-sD(x_{n},z)}{s}$, (2.15) gives

$$\begin{aligned} 2D(z,x_{n}) \geq&2a_{1}\frac {D(u,z)-sD(x_{n+1},z)}{s}+(a_{2}+a_{3})D(u,z)+(a_{2}+a_{3})D(x_{n+1},x_{n}) \\ &{}+a_{4}\frac{D(u,z)-sD(x_{n},z)}{s}+a_{4}D(x_{n+1},z). \end{aligned}$$

Letting $n\rightarrow+\infty$ in the above inequality, we obtain

$$ 0\geq\biggl(\frac{2a_{1}}{s}+a_{2}+a_{3}+ \frac{a_{4}}{s}\biggr)D(u,z), $$

it implies that $D(u,z)=0$, hence $u=z$, that is, $u=z=Tu$. This shows that u is a fixed point of T. □

Corollary 2.2

Let $(X,D)$ be a complete b-metric-like space with the constant $s\geq1$ and let $T: X \rightarrow X$ be a surjection such that

$$ D(Tx,Ty)\geq kD(x,y) $$

(2.16)

for all $x,y\in X$ and $k>s$. Then T has a unique fixed point.

Proof

Letting $a_{i}=0$ ($i=2,3,4$) and $a_{1}=k$, we find that T has a fixed point from Theorem 2.2. Suppose that u and v are fixed points of T, then we get $D(u,v)=0$ (otherwise $D(u,v)=D(Tu,Tv)\geq kD(u,v)>D(u,v)$, which is a contradiction), hence $u=v$, therefore T has a unique fixed point. □

Lemma 2.2

[20]

Let X be a nonempty set and $T:X\rightarrow X$ a function. Then there exists a subset $E\subseteq X$ such that $T(E)=T(X)$ and $T:E\rightarrow X$ is one-to-one.

Corollary 2.3

Let $(X,D)$ be a complete b-metric-like space with the constant $s\geq1$ and the self-mappings F and T satisfy the following condition:

$$ D(Fx,Fy)\geq k D(Tx,Ty) $$

(2.17)

for all $x,y\in X$, where $k>s$ is a constant. If $F(X)\subseteq T(X)$ and $T(X)$ is complete subset of X, then F and T have a unique point of coincidence in X. Moreover, if F and T are weakly compatible, then F and T have a unique common fixed point.

Proof

By Lemma 2.2, there exists $E\subseteq X$ such that $T(E)=T(X)$ and $T:E\rightarrow X$ is one-to-one. Now, we define a mapping $h:T(E)\rightarrow T(E)$ by $h(Tx)=Fx$. Since T is one-to-one on E, h is well defined. Note that $D(h(Tx),h(Ty))\geq kD(Tx,Ty)$ for all $Tx,Ty\in T(E)$. Since $T(E)=T(X)$ is complete, by using Corollary 2.2, there exists a unique $x_{0}\in X$ such that $h(Tx_{0})=Tx_{0}$, hence $Fx_{0}=Tx_{0}$, which means that F and T have a unique point of coincidence in X. Let $Fx_{0}=Tx_{0}=z$, since F and T are weakly compatible, $Fz=Tz$, which together with the uniqueness of point of coincidence implies that $Fz=Tz=z$, therefore, z is the unique common fixed point of F and T. □

Now, we introduce some examples to illustrate the validity of our main results.

Example 2.1

Let $X=\{0,1,2\}$. Define $D: X \times X\rightarrow[0,+\infty)$ as follows: $D(0,0)=0$, $D(1,1)=3$, $D(2,2)=1$, $D(0,1)=D(1,0)=8$, $D(0,2)=D(2,0)=1$, $D(1,2)=D(2,1)=4$. Let $\varphi(t)=\frac{t}{1+t}$, and define the mapping $T:X\rightarrow X$ by $T0=0$, $T1=2$, $T2=0$. Then one has the following.

• (1) $(X,D)$ is a complete b-metric-like space with the constant $s=\frac{8}{5}$.

• (2) For all $x,y\in X$, we have $D(Tx,Ty)\leq\frac{D(x,y)}{s}-\varphi (D(x,y))$.

Proof

It is clear that $(X,D)$ is a complete b-metric-like space with the constant $s=\frac{8}{5}$. Now, we show that (2) is true. Since

$$\begin{aligned}& D(T0,T0) = 0=\frac{D(0,0)}{s}-\varphi\bigl(D(0,0)\bigr); \\& D(T0,T1) = 1< 5-\frac {8}{9}=\frac{D(0,1)}{s}-\varphi\bigl(D(0,1)\bigr); \\& D(T0,T2) = 0< \frac{5}{8}-\frac{1}{2}=\frac{D(0,2)}{s}-\varphi \bigl(D(0,2)\bigr); \\& D(T1,T1) = 1< \frac{15}{8}-\frac{3}{4}=\frac {D(1,1)}{s}-\varphi \bigl(D(1,1)\bigr); \\& D(T1,T2) = 1< \frac{5}{2}-\frac{4}{5}=\frac{D(1,2)}{s}-\varphi \bigl(D(1,2)\bigr); \\& D(T2,T2) = 0< \frac{5}{8}-\frac{1}{2}=\frac {D(2,2)}{s}-\varphi \bigl(D(2,2)\bigr), \end{aligned}$$

then, for all $x,y\in X$, we have $D(Tx,Ty)\leq\frac {D(x,y)}{s}-\varphi(D(x,y))$. Hence we conclude that (2) holds, therefore all the required hypotheses of Theorem 2.1 are satisfied, and thus we deduce the existence and uniqueness of the fixed point of T. Here, 0 is the unique fixed point of T. □

Example 2.2

Let $X=[0,+\infty)$ and let a b-metric-like $D: X \times X\rightarrow[0,+\infty)$ by

$$D(x,y)=(x+y)^{2}. $$

Clearly, $(X,D)$ is a complete b-metric-like space with the constant $s=2$. Define self-mappings F and T on X as follows: $Fx=\frac{x}{2}$ and $Tx=\ln(1+\frac{x}{4})$. Since $t\geq \ln(1+t)$ for each $t\in[0,+\infty)$, for all $x,y\in X$, we have

$$\begin{aligned} D(Fx,Fy) =&\biggl(\frac{x}{2}+\frac{y}{2}\biggr)^{2}= \biggl(2\frac{x}{4}+2\frac {y}{4}\biggr)^{2}=4\biggl( \frac{x}{4}+\frac{y}{4}\biggr)^{2} \\ \geq&4\biggl(\ln\biggl(1+ \frac {x}{4}\biggr)+\ln\biggl(1+\frac{y}{4}\biggr) \biggr)^{2}=4D(Tx,Ty), \end{aligned}$$

which means $D(Fx,Fy)\geq K D(Tx,Ty)$, where $K=4>s=2$. Therefore all the required hypotheses of Corollary 2.3 are satisfied, hence F and T have a unique point of coincidence, in fact, 0 is the unique point coincidence. Moreover, by $FT0=TF0$, we find that 0 is the unique common fixed point of F and T.

3 Existence of a solution for an integral equation

Consider the following integral equation:

$$ x(t)=\int_{0}^{T}K\bigl(t,r,x(r) \bigr)\, dr, $$

(3.1)

where $T>0$ and $K:[0,T]\times[0,T]\times R\rightarrow R$.

The purpose of this section is to present an existence theorem for (3.1). Let $X=C[0,T]$ be the set of continuous real functions defined on $[0,T]$. We endow X with the b-metric-like

$$ D(u,v)=\max_{t\in[0,T]}\bigl(\bigl\vert u(t)\bigr\vert +\bigl\vert v(t)\bigr\vert \bigr)^{p} \quad \mbox{for all } u,v\in X, $$

where $p>1$. Obviously, $(X,D)$ is a complete b-metric-like space with the constant $s=2^{p-1}$.

Let $f(x(t))=\int_{0}^{T}K(t,r,x(r))\, dr$ for all $x\in X$ and for all $t\in[0,T]$. Then the existence of a solution to (3.1) is equivalent to the existence of a fixed point of f. Now, we prove the following result.

Theorem 3.1

Suppose that the following hypotheses hold:

• (i) $K:[0,T]\times[0,T]\times R\rightarrow R$ is continuous;

• (ii) for all $t,r\in[0,T]$, there exists a continuous $\xi :[0,T]\times[0,T]\rightarrow R$ such that

  $$ \bigl\vert K\bigl(t,r,x(r)\bigr)\bigr\vert +\bigl\vert K \bigl(t,r,y(r)\bigr)\bigr\vert < \lambda^{\frac{1}{p}}\xi(t,r) \bigl(\bigl\vert x(r)\bigr\vert +\bigl\vert y(r)\bigr\vert \bigr) $$

  (3.2)

  and

  $$ \sup_{t\in[0,T]}\int_{0}^{T} \xi(t,r)\, dr \leq1, $$

  (3.3)

  where $0<\lambda<\frac{1}{s}$.

Then the integral equation (3.1) has a unique solution $x\in X$.

Proof

From (3.2) and (3.3), for all $t\in[0,T]$, we have

$$\begin{aligned} \bigl(\bigl\vert f\bigl(x(t)\bigr)\bigr\vert +\bigl\vert f\bigl(y(t)\bigr) \bigr\vert \bigr)^{p} =& \biggl(\biggl\vert \int _{0}^{T}K\bigl(t,r,x(r)\bigr)\,dr\biggr\vert + \biggl\vert \int_{0}^{T}K\bigl(t,r,y(r)\bigr)\,dr \biggr\vert \biggr)^{p} \\ \leq& \biggl(\int_{0}^{T}\bigl\vert K \bigl(t,r,x(r)\bigr)\bigr\vert \,dr+\int_{0}^{T} \bigl\vert K\bigl(t,r,y(r)\bigr)\bigr\vert \,dr \biggr)^{p} \\ =& \biggl(\int_{0}^{T}\bigl(\bigl\vert K \bigl(t,r,x(r)\bigr)\bigr\vert +\bigl\vert K\bigl(t,r,y(r)\bigr)\bigr\vert \bigr)\,dr \biggr)^{p} \\ \leq& \biggl(\int_{0}^{T}\bigl( \lambda^{\frac{1}{p}}\xi (t,r) \bigl(\bigl\vert x(r)\bigr\vert +\bigl\vert y(r)\bigr\vert \bigr)\bigr)\,dr \biggr)^{p} \\ =& \biggl(\int_{0}^{T}\bigl(\xi(t,r) \lambda^{\frac{1}{p}} \bigl(\bigl(\bigl\vert x(r)\bigr\vert +\bigl\vert y(r) \bigr\vert \bigr)^{p} \bigr)^{\frac{1}{p}}\bigr)\,dr \biggr)^{p} \\ \leq& \biggl(\int_{0}^{T}\bigl(\xi(t,r) \lambda^{\frac{1}{p}}D^{\frac {1}{p}}\bigl(x(t),y(t)\bigr)\bigr)\,dr \biggr)^{p} \\ =&\lambda D\bigl(x(t),y(t)\bigr) \biggl(\int_{0}^{T} \xi(t,r)\,dr \biggr)^{p} \\ \leq&\lambda D\bigl(x(t),y(t)\bigr), \end{aligned}$$

which implies that $D(f(x(t)),f(y(t)))\leq\lambda D(x(t),y(t))$.

Now, all the conditions of Corollary 2.1 hold and f has a unique fixed point $x\in X$, which means that x is the unique solution for the integral equation (3.1). □