1 Introduction and preliminaries
In this paper, \(\mathbb{N}\), \(\mathbb{N}^{+}\) and \(\mathbb{R}\) are used to denote the set of all nonnegative integer numbers, the set of all positive integer numbers, and the set of all real numbers, respectively.
The Banach contraction mapping principle in metric spaces is an important tool in nonlinear analysis, many authors have been devoting in generalizing metric spaces and the Banach contraction mapping principle. And then, many generalized metric spaces were introduced. In 2013, Alghamdi et al. [1] proposed the concept of b-metric-like spaces which is considered to be an interesting generalization of metric spaces, b-metric spaces [2] and metric-like spaces [3]. After that, some fixed point theorems were investigated by many authors [4‐8]. Firstly, let us recall some definitions about b-metric-like spaces.
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Definition 1.1
([1])
A b-metric-like on a nonempty set X is a function \(b: X\times X\rightarrow [0, +\infty)\) such that, for all \(x,y,z\in X\) and a constant \(s\geq1\), the following three conditions hold true:
The pair \((X,b)\) is then called a b-metric-like space with coefficient s.
\((b_{1})\)
if \(b(x,y)=0 \) then \(x=y\);
\((b_{2})\)
\(b(x,y)=b(y,x)\);
\((b_{3})\)
\(b(x,z)\leq s[b(x,y)+b(y,z)]\).
Some concepts in b-metric-like spaces were introduced as follows.
Each b-metric-like b on X generalizes a topology \(\tau_{b}\) on X whose base is the family of open b-balls \(B_{b}(x,\varepsilon)=\{y\in X:|b(x,y)-b(x,x)|<\varepsilon\}\) for all \(x\in X\) and \(\varepsilon>0\).
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A sequence \(\{x_{n}\}\) in a b-metric-like space \((X,b)\) converges to a point \(x\in X\) if and only if \(b(x,x)=\lim_{n\rightarrow+\infty}b(x,x_{n})\).
A sequence \(\{x_{n}\}\) in a b-metric-like space \((X,b)\) is called a Cauchy sequence if \(\lim_{n,m\rightarrow+\infty}b(x_{m},x_{n})\) exists and is finite.
A b-metric-like space is called to be complete if every Cauchy sequence \(\{x_{n}\}\) in X converges with respect to \(\tau_{b}\) to a point \(x\in X\) such that \(\lim_{n\rightarrow+\infty}b(x,x_{n})=b(x,x)=\lim_{n,m\rightarrow+\infty}b(x_{m},x_{n})\).
Let \(\mathbb{B}\) be the family of all functions β:\([0,\infty)\rightarrow[0,1)\) which satisfy the condition: \(\lim_{n\rightarrow+\infty}\beta(t_{n})=1\) implies \(\lim_{n\rightarrow+\infty}t_{n}=0\).
On the other hand, in [9], Geraghty extended the Banach contraction mapping principle in metric spaces and obtained the following theorem.
Theorem 1.1
([9])
Let
\((X,d)\)
be a complete metric space and
\(T:X\rightarrow X\)
be a mapping. If
T
satisfies
\(d(Tx,Ty)\leq \beta(d(x,y))d(x,y)\)
for any
\(x,y\in X\), where
\(\beta\in \mathbb{B}\), then
T
has a unique fixed point.
Recently, many papers about generalization of Geraghty contraction appeared [10‐13]. In 2017, Fulga et al. [12] introduced the concept of \(\varphi_{E}\)-Geraghty contraction and established a fixed point theorem for such contraction in complete metric spaces. The new Geraghty type contraction was studied on metric-like spaces in [14], and the following theorem was obtained.
Theorem 1.2
([14])
Let
\((X,\sigma)\)
be a complete metric-like space and
\(T:X\rightarrow X\)
be a mapping. If there exists
\(\beta\in \mathbb{B}\)
such that
\(\sigma(Tx,Ty)\leq \beta(F(x,y))F(x,y)\)
for all
\(x,y\in X\), where
\(F(x,y)=\sigma(x,y)+|\sigma(x,Tx)-\sigma(y,Ty)|\), then
T
has a unique fixed point.
In this paper, we define the new concept of \((T,g)_{F}\)-contraction of Geraghty type and investigate common fixed point theorems for such contraction in b-metric-like spaces. An application about the unique solution of an integral equation is given.
2 Main results
In this section, we begin with the following definitions.
Definition 2.1
([15])
Let X be a nonempty set, f and g be self-mappings on X and \(C(f,g)=\{x\in X: fx=gx\}\). The pair f and g are called to be weakly compatible if \(fgx=gfx\) for all \(x\in C(f,g)\). If \(w=fx=gx\) for some \(x\in X\), then x is called to be a coincidence of f and g, and w is called to be a point of coincidence of f and g.
Definition 2.2
Let \((X,b)\) be a b-metric-like space with coefficient \(s\geq 1\) and \(T,g: X\rightarrow X\) be two mappings. We say that the pair \((T,g)\) is a \((T,g)_{F}\)-contraction of Geraghty type if there exists \(\beta\in \mathbb{B}\) such that
for all \(x,y\in X\), where \(F_{g}(x,y)=\frac{1}{s^{2}}[b(gx,gy)+|b(gx,Tx)-b(gy,Ty)|]\).
$$\begin{aligned} b(Tx,Ty)\leq \beta\bigl(F_{g}(x,y)\bigr)F_{g}(x,y) \end{aligned}$$
(1)
Lemma 2.1
Let
\((X,b)\)
be a b-metric-like space, T
and
g
be self-mappings on
X
such that
\((T,g)\)
is a
\((T,g)_{F}\)-contraction of Geraghty type. If
\(v\in X\)
is a point of coincidence of
T
and
g, then
\(b(v,v)=0\).
Proof
Suppose that \(v\in X\) is a point of coincidence of T and g, then there exists \(u\in X\) such that \(Tu=gu=v\). Assume \(b(v,v)>0\), we get \(b(v,v)=b(Tu,Tu)\leq \beta(F_{g}(u,u))F_{g}(u,u)\), since \(F_{g}(u,u)=\frac{1}{s^{2}}[b(gu,gu)+|b(gu,Tu)-b(gu,Tu)|]=\frac{1}{s^{2}}b(v,v)\), then we have \(b(v,v)<\frac{1}{s^{2}}b(v,v)\), which is a contradiction, hence \(b(v,v)=0\). □
Theorem 2.1
Let
\((X,b)\)
be a
b-metric-like space with coefficient
\(s\geq 1\), \(T,g:X\times X\rightarrow X\)
be two mappings with
\(TX\subseteq gX\)
and
gX
is complete. If the pair
\((T,g)\)
is a
\((T,g)_{F}\)-contraction of Geraghty type, then
T
and
g
have a unique point of coincidence. In addition, if
T
and
g
are weakly compatible, then
T
and
g
have a unique common fixed point.
Proof
For an arbitrary \(x_{0}\in X\), since \(TX\subseteq gX\), we can construct a sequence \(\{y_{n}\}\) by
for all \(n\in \mathbb{N^{+}}\). Now, we prove that T and g have a point of coincidence. If there exists some \(n_{0}\in \mathbb{N^{+}}\) such that \(b(y_{n_{0}},y_{n_{0}+1})=0\), then \(y_{n_{0}}=y_{n_{0}+1}\), which implies \(gx_{n_{0}}=Tx_{n_{0}}\), thus, \(x_{n_{0}}\) is a coincidence point of T and g, so \(w_{0}=gx_{n_{0}}=Tx_{n_{0}}\) is a point of coincidence of T and g. We assume that \(b(y_{n},y_{n+1})>0\) for all \(n\in \mathbb{N^{+}}\). From (1), we have
where
Assume that there exists \(n_{0}\in \mathbb{N^{+}}\) such that \(b(y_{n_{0}-1},y_{n_{0}})\leq b(y_{n_{0}},y_{n_{0}+1})\). By (3), we get
which is a contradiction. Thus, we obtain
for all \(n\in \mathbb{N^{+}}\). Therefore, there exists \(a\geq 0\) such that
(3) and (4) yield that
Taking \(n\rightarrow+\infty\) in (6), we get \(\lim_{n\rightarrow+\infty}\beta[\frac{2b(y_{n-1},y_{n})-b(y_{n},y_{n+1})}{s^{2}}]=1\), hence \(\lim_{n\rightarrow+\infty}\frac{2b(y_{n-1},y_{n})-b(y_{n},y_{n+1})}{s^{2}}=0\). On the other hand, \(\lim_{n\rightarrow+\infty}\frac{2b(y_{n-1},y_{n})-b(y_{n},y_{n+1})}{s^{2}}=\frac{a}{s^{2}}\), therefore \(a=0\). Hence,
Now, we prove
If (8) does not hold, then there exists \(\varepsilon>0\), for which we can find two subsequences \(\{y_{m(k)}\}\) and \(\{y_{n(k)}\}\) of \(\{y_{n}\}\), where \(m(k)\) is the smallest index for which \(m(k)>n(k)>k\) with
Applying (1) and (9), we have
where
Next, we discuss two cases.
$$\begin{aligned} y_{n}=gx_{n}=Tx_{n-1} \end{aligned}$$
(2)
$$\begin{aligned} b(y_{n},y_{n+1}) =&b(Tx_{n-1},Tx_{n}) \\ \leq& \beta\bigl(F_{g}(x_{n-1},x_{n}) \bigr)F_{g}(x_{n-1},x_{n}), \end{aligned}$$
(3)
$$\begin{aligned} F_{g}(x_{n-1},x_{n}) =&\frac{1}{s^{2}} \bigl[b(gx_{n-1},gx_{n})+ \bigl\vert b(gx_{n-1},Tx_{n-1})-b(gx_{n},Tx_{n}) \bigr\vert \bigr] \\ =&\frac{1}{s^{2}}\bigl[b(y_{n-1},y_{n})+ \bigl\vert b(y_{n-1},y_{n})-b(y_{n},y_{n+1}) \bigr\vert \bigr]. \end{aligned}$$
$$\begin{aligned} b(y_{n_{0}},y_{n_{0}+1}) =&b(Tx_{n_{0}-1},Tx_{n_{0}}) \\ \leq&\beta\bigl(F_{g}(x_{n_{0}-1},x_{n_{0}}) \bigr)F_{g}(x_{n_{0}-1},x_{n_{0}}) \\ < &F_{g}(x_{n_{0}-1},x_{n_{0}}) \\ =&\frac{1}{s^{2}}\bigl[b(gx_{n_{0}-1},gx_{n_{0}})+ \bigl\vert b(gx_{n_{0}-1},Tx_{n_{0}-1})-b(gx_{n_{0}},Tx_{n_{0}}) \bigr\vert \bigr] \\ =&\frac{1}{s^{2}}\bigl[b(y_{n_{0}-1},y_{n_{0}})+ \bigl\vert b(y_{n_{0}-1},y_{n_{0}})-b(y_{n_{0}},y_{n_{0}+1}) \bigr\vert \bigr] \\ =&\frac{1}{s^{2}}b(y_{n_{0}},y_{n_{0}+1})\leq b(y_{n_{0}},y_{n_{0}+1}), \end{aligned}$$
$$\begin{aligned} b(y_{n-1},y_{n})>b(y_{n},y_{n+1}) \end{aligned}$$
(4)
$$\begin{aligned} \lim_{n\rightarrow+\infty}b(y_{n-1},y_{n})=a. \end{aligned}$$
(5)
$$\begin{aligned} b(y_{n},y_{n+1}) =&b(Tx_{n-1},Tx_{n}) \\ \leq&\beta\bigl(F_{g}(x_{n-1},x_{n})\bigr)F_{g}(x_{n-1},x_{n}) \\ =&\beta\biggl[\frac{1}{s^{2}}\bigl(2b(y_{n-1},y_{n})-b(y_{n},y_{n+1}) \bigr)\biggr]\cdot\frac{1}{s^{2}}\bigl(2b(y_{n-1},y_{n})-b(y_{n},y_{n+1}) \bigr) \\ \leq&\beta\biggl[\frac{1}{s^{2}}\bigl(2b(y_{n-1},y_{n})-b(y_{n},y_{n+1}) \bigr)\biggr]\cdot\bigl(2b(y_{n-1},y_{n})-b(y_{n},y_{n+1}) \bigr) \\ < & 2b(y_{n-1},y_{n})-b(y_{n},y_{n+1}). \end{aligned}$$
(6)
$$\begin{aligned} \lim_{n\rightarrow +\infty}b(y_{n-1},y_{n})=0. \end{aligned}$$
(7)
$$\begin{aligned} \lim_{m,n\rightarrow+\infty}b(y_{m},y_{n})=0. \end{aligned}$$
(8)
$$\begin{aligned} b(y_{m(k)},y_{n(k)})\geq \varepsilon,\qquad b(y_{m(k)-1},y_{n(k)})< \varepsilon. \end{aligned}$$
(9)
$$\begin{aligned} \varepsilon \leq& b(y_{m(k)},y_{n(k)}) \\ =& b(Tx_{m(k)-1},Tx_{n(k)-1}) \\ \leq& \beta\bigl(F_{g}(x_{m(k)-1},x_{n(k)-1}) \bigr)F_{g}(x_{m(k)-1},x_{n(k)-1}) \\ < &F_{g}(x_{m(k)-1},x_{n(k)-1}), \end{aligned}$$
(10)
$$\begin{aligned} &F_{g}(x_{m(k)-1},x_{n(k)-1}) \\ &\quad =\frac{1}{s^{2}}\bigl[b(gx_{m(k)-1},gx_{n(k)-1})+ \bigl\vert b(gx_{m(k)-1},Tx_{m(k)-1})-b(gx_{n(k)-1},Tx_{n(k)-1}) \bigr\vert \bigr] \\ &\quad =\frac{1}{s^{2}}\bigl[b(y_{m(k)-1},y_{n(k)-1})+ \bigl\vert b(y_{m(k)-1},y_{m(k)})-b(y_{n(k)-1},y_{n(k)}) \bigr\vert \bigr]. \end{aligned}$$
(11)
Case I: Case of \(s>1\). Applying (7), (10), and (11), we obtain
Moreover, from (9), we have
Taking \(n\rightarrow +\infty\) in the above inequalities, we have
(12) and (13) imply \(\varepsilon\leq \frac{\varepsilon}{s}\), which is a contradiction.
$$\begin{aligned} \varepsilon\leq \liminf_{n\rightarrow+\infty} \frac{1}{s^{2}}b(y_{m(k)-1},y_{n(k)-1}). \end{aligned}$$
(12)
$$\begin{aligned} b(y_{m(k)-1},y_{n(k)-1})\leq sb(y_{m(k)-1},y_{n(k)})+sb(y_{n(k)},y_{n(k)-1})< s \varepsilon+sb(y_{n(k)},y_{n(k)-1}). \end{aligned}$$
$$\begin{aligned} \liminf_{n\rightarrow+\infty}b(y_{m(k)-1},y_{n(k)-1}) \leq s\varepsilon. \end{aligned}$$
(13)
Case II: Case of \(s=1\). From (9), we have
By (7), taking \(n\rightarrow +\infty\) in (14), we have
Since \(s=1\), by (10) and (11), we have
(7), (15) and (16) yield
From (11) and (15), and taking \(s=1\) into account, we get \(\lim_{n\rightarrow+\infty}F_{g}(x_{m(k)-1},x_{n(k)-1})= \varepsilon\), which together with (17) implies
thus \(\lim_{n\rightarrow+\infty}F_{g}(x_{m(k)-1},x_{n(k)-1})=0\), which is contradictive with \(\lim_{n\rightarrow+\infty}F_{g}(x_{m(k)-1}, x_{n(k)-1})= \varepsilon\).
$$\begin{aligned} \varepsilon \leq &b(y_{m(k)},y_{n(k)}) \\ \leq& b(y_{m(k)},y_{m(k)-1})+ b(y_{m(k)-1},y_{n(k)-1})+b(y_{n(k)-1},y_{n(k)}) \\ \leq& b(y_{m(k)},y_{m(k)-1})+ b(y_{m(k)-1},y_{n(k)})+2b(y_{n(k)-1},y_{n(k)}) \\ < &b(y_{m(k)},y_{m(k)-1})+\varepsilon+2b(y_{n(k)-1},y_{n(k)}). \end{aligned}$$
(14)
$$\begin{aligned} \lim_{n\rightarrow+\infty}b(y_{m(k)-1},y_{n(k)-1})= \varepsilon. \end{aligned}$$
(15)
$$\begin{aligned} \varepsilon \leq &\beta\bigl(F_{g}(x_{m(k)-1},x_{n(k)-1}) \bigr)F_{g}(x_{m(k)-1},x_{n(k)-1}) \\ < &b(y_{m(k)-1},y_{n(k)-1})+ \bigl\vert b(y_{m(k)-1},y_{m(k)})-b(y_{n(k)-1},y_{n(k)}) \bigr\vert . \end{aligned}$$
(16)
$$\begin{aligned} \lim_{n\rightarrow+\infty}\beta\bigl(F_{g}(x_{m(k)-1},x_{n(k)-1}) \bigr)F_{g}(x_{m(k)-1},x_{n(k)-1})= \varepsilon. \end{aligned}$$
(17)
$$\begin{aligned} \lim_{n\rightarrow+\infty}\beta\bigl(F_{g}(x_{m(k)-1},x_{n(k)-1}) \bigr)=1, \end{aligned}$$
From the above discussions, we get that (8) holds. Therefore, the sequence \(\{y_{n}\}=\{gx_{n}\}\) is a Cauchy sequence in gX. Since gX is complete, then there exist \(v,u\in X\) such that \(v=gu\), and the following equalities hold:
$$\begin{aligned} \lim_{n,m\rightarrow+\infty}b(y_{n},v)=b(v,v)=\lim _{n,m\rightarrow+\infty}b(y_{n},y_{m})=\lim _{n\rightarrow+\infty}b(y_{n},gu)=0. \end{aligned}$$
(18)
By (1), we have
where
$$\begin{aligned} b(y_{n},Tu)=b(Tx_{n-1},Tu)\leq\beta \bigl(F_{g}(x_{n-1},u)\bigr)F_{g}(x_{n-1},u)< F_{g}(x_{n-1},u), \end{aligned}$$
(19)
$$\begin{aligned} F_{g}(x_{n-1},u) =&\frac{1}{s^{2}} \bigl[b(gx_{n-1},gu)+ \bigl\vert b(gx_{n-1},Tx_{n-1})-b(gu,Tu) \bigr\vert \bigr] \\ =&\frac{1}{s^{2}}\bigl[b(y_{n-1},v)+ \bigl\vert b(y_{n-1},y_{n})-b(v,Tu) \bigr\vert \bigr]. \end{aligned}$$
(20)
Next, we prove \(b(Tu,v)=0\) in two cases:
Case I. \(s>1\). Suppose \(b(Tu,v)>0\). Letting \(n\rightarrow +\infty\) in (19), applying (20), we obtain
By the triangle inequality, we get \(b(v,Tu)\leq sb(y_{n},v)+sb(y_{n},Tu)\), which yields
Applying (22), we have \(\liminf_{n\rightarrow\infty}b(y_{n},Tu)\geq \frac{1}{s}b(v,Tu)>0\). From (21) and (22), we get \(b(v,Tu)\leq\frac{1}{s}b(v,Tu)< b(v,Tu)\), this is a contradiction, therefore \(b(Tu,v)=0\).
$$\begin{aligned} \liminf_{n\rightarrow+\infty}b(y_{n},Tu)\leq \frac{1}{s^{2}}b(v,Tu). \end{aligned}$$
(21)
$$\begin{aligned} b(v,Tu)\leq s\liminf_{n\rightarrow+\infty}b(y_{n},Tu). \end{aligned}$$
(22)
Case II. \(s=1\). Taking \(n\rightarrow +\infty\) in (20), and taking \(s=1\) into account, we obtain
On the other hand, from (1), we have
Letting \(n\rightarrow +\infty\) in (24), by (23), we get \(\lim_{n\rightarrow+\infty}\beta(F_{g}(x_{n-1},u))=1\), hence \(\lim_{n\rightarrow+\infty}F_{g}(x_{n-1},u)=0\), by (23), we get \(b(Tu,v)=0\). The above two cases mean \(b(Tu,v)=0\), which implies \(Tu=v\), thus \(Tu=v=gu\). Therefore, T and g have a coincidence point u, and v is a point of coincidence of T and g. By Lemma 2.1, we get \(b(v,v)=0\). Suppose that \(v_{1}\) is also a point of coincidence of T and g, then we can find \(u_{1}\in X\) such that \(Tu_{1}=v_{1}=gu_{1}\) and \(b(v_{1},v_{1})=0\). Now, we prove \(b(v,v_{1})=0\) by contradiction. Suppose \(b(v,v_{1})>0\), applying (1), we have
where
From (25) and (26), we obtain \(b(v,v_{1})<\frac{1}{s^{2}}b(v,v_{1})\), which is a contradiction, thus \(b(v,v_{1})=0\), which implies \(v=v_{1}\), therefore T and g have a unique point of coincidence. Moreover, T and g are weakly compatible, then we have \(Tv=gv\). Let \(Tv=gv=\omega\). From the uniqueness of the point of coincidence, we have \(Tv=gv=\omega=v\), that is, \(Tv=gv=v\). Therefore, T and g have a unique common fixed point. □
$$\begin{aligned} \lim_{n\rightarrow+\infty}F_{g}(x_{n-1},u)=b(v,Tu). \end{aligned}$$
(23)
$$\begin{aligned} b(v,Tu) \leq& b(v,y_{n})+b(y_{n},Tu) \\ =& b(v,y_{n})+b(Tx_{n-1},Tu) \\ \leq& b(v,y_{n})+\beta\bigl(F_{g}(x_{n-1},u) \bigr)F_{g}(x_{n-1},u) \\ < & b(v,y_{n})+F_{g}(x_{n-1},u). \end{aligned}$$
(24)
$$\begin{aligned} b(v,v_{1}) =& b(Tu,Tu_{1}) \leq \beta\bigl(F_{g}(u,u_{1})\bigr)F_{g}(u,u_{1}) < F_{g}(u,u_{1}), \end{aligned}$$
(25)
$$\begin{aligned} F_{g}(u,u_{1}) =&\frac{1}{s^{2}} \bigl[b(gu,gu_{1})+ \bigl\vert b(gu,Tu)-b(gu_{1},Tu_{1}) \bigr\vert \bigr] \\ =&\frac{1}{s^{2}}\bigl[b(v,v_{1})+ \bigl\vert b(v,v)-b(v_{1},v_{1}) \bigr\vert \bigr] \\ =&\frac{1}{s^{2}}b(v,v_{1}). \end{aligned}$$
(26)
Letting \(g=I_{x}\) (identity mapping) in Theorem 2.1, we can get the following corollary.
Corollary 2.1
Let
\((X,b)\)
be a complete b-metric-like space with coefficient
\(s\geq1\), and
\(T:X\rightarrow X\)
be a mapping. If there exists
\(\beta\in \mathbb{B}\)
such that
\(b(Tx,Ty)\leq \beta(F(x,y))F(x,y)\)
for any
\(x,y\in X\), where
\(F(x,y)=\frac{1}{s^{2}}[b(x,y)+|b(x,Tx)-b(y,Ty)|]\), then
T
has a unique fixed point.
Taking \(s=1\) in Corollary 2.1, we have the following corollary.
Corollary 2.2
Let
\((X,b)\)
be a complete metric-like space and
\(T:X\rightarrow X\)
be a mapping. If there exists
\(\beta\in \mathbb{B}\)
such that
\(b(Tx,Ty)\leq \beta(F(x,y))F(x,y)\)
for any
\(x,y\in X\), where
\(F(x,y)=b(x,y)+|b(x,Tx)-b(y,Ty)|\), then
T
has a unique fixed point.
Remark 2.1
Taking \(s=1\) in Theorem 2.1, we have the following corollary.
Corollary 2.3
Let
\((X,b)\)
be a metric-like space and
\(T,g:X\times X\rightarrow X\)
be two mappings with
\(TX\subseteq gX\)
and
gX
is complete. Suppose that there exists
\(\beta\in \mathbb{B}\)
such that
where
\(F_{g}(x,y)=b(gx,gy)+|b(gx,Tx)-b(gy,Ty)|\). Then
T
and
g
have a unique point of coincidence. In addition, if
T
and
g
are weakly compatible, then
T
and
g
have a unique common fixed point.
$$\begin{aligned} b(Tx,Ty)\leq \beta\bigl(F_{g}(x,y)\bigr)F_{g}(x,y), \end{aligned}$$
Now, we use an example to illustrate the validity of our main result.
Example 2.1
Let \(X=\{0,1,2\}\). Define \(b: X \times X\rightarrow \mathbb{R}\) by \(b(0,0)=0\), \(b(1,1)=3\), \(b(2,2)=1\), \(b(0,1)=b(1,0)=8\), \(b(0,2)=b(2,0)=1\), \(b(1,2)=b(2,1)=4\). It is easy to prove that \((X,b)\) is a complete b-metric-like space with coefficient \(s=\frac{8}{5}\). Consider \(T: X\rightarrow X\) as \(T0=0, T1=2, T2=0\). Take
By the following cases, we prove \(b(Tx,Ty)\leq \beta(F(x,y))F(x,y)\) for any \(x,y\in X\), where \(F(x,y)=\frac{1}{s^{2}}[b(x,y)+|b(x,Tx)-b(y,Ty)|]\).
$$\begin{aligned} \beta(t)=\textstyle\begin{cases}\frac{1}{1+\frac{1}{100}t}, & t>0, \\\frac{1}{3}, &t=0.\end{cases}\displaystyle \end{aligned}$$
Case 1: \((x,y)=(0,0)\), \((x,y)=(2,2)\), \((x,y)=(0,2)\). Since \(b(T0,T0)=b(0,0)=0\), \(b(T2,T2)=b(0,0)=0\), \(b(T0,T2)= b(0,0)=0\), then \(b(Tx,Ty)\leq \beta(F(x,y))F(x,y)\) holds for \((x,y)=(0,0)\), \((x,y)=(2,2)\), \((x,y)=(0,2)\).
Case 2: \((x,y)=(0,1)\).
We get \(b(T0,T1)=b(0,2)=1\) and \(F(0,1)=\frac{25}{64}[b(0,1)+|b(0,T0)-b(1,T1)|]=\frac{300}{64}\). Hence \(b(T0,T1)=1<\beta (F(0,1))F(0,1)=\frac{1}{1+\frac{1}{100}\frac{300}{64}}\frac{300}{64}=\frac{300}{94}\).
Case 3: \((x,y)=(1,1)\).
We get \(b(T1,T1)=b(2,2)=1\) and \(F(1,1)=\frac{25}{64}[b(1,1)+|b(1,T1)-b(1,T1)|]=\frac{75}{64}\). Hence \(b(T1,T1)=1<\beta (F(1,1))F(1,1)=\frac{1}{1+\frac{1}{100}\frac{75}{64}}\frac{75}{64}=\frac{750}{715}\).
Case 4: \((x,y)=(1,2)\).
We get \(b(T1,T2)=b(2,0)=1\) and \(F(1,2)=\frac{25}{64}[b(1,2)+|b(1,T1)-b(2,T2)|]=\frac{175}{64}\). Hence \(b(T1,T2)=1<\beta (F(0,1))F(0,1)=\frac{1}{1+\frac{1}{100}\frac{175}{64}}\frac{175}{64}=\frac{1750}{815}\).
From the above discussions, we know that \(b(Tx,Ty)\leq \beta(F(x,y))F(x,y)\) for any \(x,y\in X\), where \(F(x,y)=\frac{1}{s^{2}}[b(x,y)+|b(x,Tx)-b(y,Ty)|]\). By Corollary 2.1, we obtain that T has a unique fixed point, 0 is the unique fixed point of T.
3 Existence of a solution for an integral equation
Consider the following integral equation:
where \(K:[0,1]\times [0,1]\times R\rightarrow R\). The purpose of this section is to present an existence theorem of the solution about (27). Let \(X=C[0,1]\) be the set of continuous real functions defined on \([0,1]\). We endow X with the b-metric-like:
where \(p\geq1\). Obviously, \((X,b)\) is a complete b-metric-like space with coefficient \(s=2^{p-1}\).
$$\begin{aligned} x(t)= \int_{0}^{1}K\bigl(t,r,x(r)\bigr)\,dr, \end{aligned}$$
(27)
$$\begin{aligned} b(u,v)= \Vert u-v \Vert _{\infty}=\max_{t\in[0,1]}\bigl( \bigl\vert u(t) \bigr\vert + \bigl\vert v(t) \bigr\vert \bigr)^{p}\quad \mbox{for all }u,v\in X, \end{aligned}$$
Let \(f(x)(t)=\int_{0}^{1}K(t,r,x(r))\,dr\) for all \(x\in X\) and for all \(t\in [0,1]\). Then the existence of a solution to (27) 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)
\(K:[0,1]\times [0,1]\times R\rightarrow R\)
is continuous;
(ii)
For all
\(t,r\in [0,1]\), there exists continuous
\(\xi:[0,1]\times [0,1]\rightarrow [0,+\infty)\)
such that
$$\begin{aligned} \bigl\vert K\bigl(t,r,x(r)\bigr) \bigr\vert \leq\xi(t,r) \bigl\vert x(r) \bigr\vert . \end{aligned}$$
(28)
(iii)
There exists
\(\beta\in \mathbb{B}\)
such that
$$\begin{aligned} &\sup_{t\in [0,1]} \int_{0}^{1}\xi(t,r)\,dr \\ &\quad \leq \sqrt[p]{\frac{1}{2^{2p-2}}\beta\biggl[\frac{1}{2^{2p-2}}\bigl( \Vert x-y \Vert _{\infty}+ \bigl\vert \Vert x-fx \Vert _{\infty}- \Vert y-fy \Vert _{\infty}\bigr\vert \bigr) \biggr]}. \end{aligned}$$
(29)
Proof
From (28) and (29), for all \(t\in [0,1]\), we have
where \(F(x,y)=\frac{1}{2^{2p-2}}(\|x-y\|_{\infty}+|\|x-fx\|_{\infty}-\|y-fy\|_{\infty}|)\). Then we have \(\|f(x)-f(y)\|_{\infty}\leq F(x,y)\beta(F(x,y))\). Therefore, we get \(b(f(x),f(y))\leq F(x,y)\beta(F(x,y))\) for all \(x,y\in X\).
$$\begin{aligned} & \bigl( \bigl\vert f(x) (t) \bigr\vert + \bigl\vert f(y) (t) \bigr\vert \bigr)^{p} \\ &\quad = \biggl( \biggl\vert \int_{0}^{1}K\bigl(t,r,x(r)\bigr)\,dr \biggr\vert + \biggl\vert \int_{0}^{1}K\bigl(t,r,y(r)\bigr)\,dr \biggr\vert \biggr)^{p} \\ &\quad \leq \biggl( \int_{0}^{1} \bigl\vert K\bigl(t,r,x(r)\bigr) \bigr\vert \,dr+ \int_{0}^{1} \bigl\vert K\bigl(t,r,y(r)\bigr) \bigr\vert \,dr \biggr)^{p} \\ &\quad = \biggl( \int_{0}^{1}\bigl( \bigl\vert K\bigl(t,r,x(r) \bigr) \bigr\vert + \bigl\vert K\bigl(t,r,y(r)\bigr) \bigr\vert \bigr)\,dr \biggr)^{p} \\ &\quad \leq \biggl( \int_{0}^{1}\xi(t,r) \bigl( \bigl\vert x(r) \bigr\vert + \bigl\vert y(r) \bigr\vert \bigr)\,dr \biggr)^{p} \\ &\quad = \biggl( \int_{0}^{1}\xi(t,r) \bigl(\bigl( \bigl\vert x(r) \bigr\vert + \bigl\vert y(r) \bigr\vert \bigr)^{p} \bigr)^{\frac{1}{p}}\biggr)\,dr )^{p} \\ &\quad \leq \biggl( \int_{0}^{1}{\xi(t,r)\bigl[b\bigl(x(\cdot),y(\cdot) \bigr)\bigr]^{\frac{1}{p}}\biggr)\,dr} )^{p} \\ &\quad = \Vert x-y \Vert _{\infty}\biggl( \int_{0}^{1}\xi(t,r)\,dr \biggr)^{p} \\ &\quad \leq \Vert x-y \Vert _{\infty}\frac{1}{2^{2p-2}}\beta\biggl[ \frac{1}{2^{2p-2}}\bigl( \Vert x-y \Vert _{\infty}+ \bigl\vert \Vert x-fx \Vert _{\infty}- \Vert y-fy \Vert _{\infty}\bigr\vert \bigr)\biggr] \\ &\quad \leq F(x,y)\beta\bigl(F(x,y)\bigr), \end{aligned}$$
4 Conclusion
In this paper, we introduce a new concept of \((T,g)_{F}\)-contraction in b-metric-like spaces and investigate some fixed point theorems about such contraction. Our results generalize Theorem 2.1 in [14]. At the same time, an application about our theorem is also proposed.
Acknowledgements
The authors are thankful to the referees for their valuable comments and suggestions to improve this paper.
Competing interests
The authors declare that they have no competing interests.
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