An $(\alpha ,\vartheta )$-admissibility and Theorems for Fixed Points of Self-maps

Abstract

We introduce $(\alpha ,\vartheta )$-admissibility and an $(\varUpsilon ,\wp )$-integral-type contraction with applications to new fixed point theorems for the admissible and continuous mapping $F: X \rightarrow X$ on a complete metric space (X, d). For the application, an interesting example is added which demonstrate our results.

1 Introduction and Preliminaries

Fixed point theory have a lot of applications in different disciplines of pure and applied mathematics, image processing, engineering, nonlinear functional analysis, computer science, economics, dynamical system etc. [11,12,13]. In 1922, Banach [6] provided an outstanding theorem which has led to many follow-up results have been proven. In this area, one can observe a large number of new fixed point theorems which modify the pre-existing theorems. For instance, Agarwal et al. [1] pointed out many important consequences of the new fixed point theorems in multiplicative metric space. Samet and Vetro [5] provided the $(\alpha$-$\psi )$-contractive type mappings and produced new fixed point theorems and provided some interesting applications of their results. Shatanawi and Rawashdeh [4] proved new fixed point theorems in order metric space by $(\psi ,\phi )$-contractive-type mapping and applied their theorems to some functional equations. Hussian et al. [3] established new fixed point theorems by using $\alpha$-admissible contraction in complete metric space and gave some interesting instructive example in the applications of their work. Chandok [2] produced $(\alpha ,\beta )$-admissible Geraghty-type contractive mappings in metric space and for the usability he provided some explanatory examples. Altun et al. [16] produced new fixed point theorems for weakly compatible mappings sustaining integral-type contractions and provided helpful examples. Rhoades [17] published two fixed point theorems for mappings by the use of integral-type contractions and gave some instructive example. Farajzadeh et al. [14] introduced a new $(\alpha ,\eta ,\psi ,\xi )$ contraction for multi-valued mappings and added interesting instructive example for the applications of their fixed point theorems. Bota et al. [10] introduced $(\alpha$-$\psi )$-Ciric-type contraction for the multi valued operator and proved fixed point theorems in b-metric space.

In this paper, we introduce an $(\alpha ,\vartheta )$-admissibility and an $(\varUpsilon ,\wp )$-integral-type contraction for a self-continuous mapping $F: X \rightarrow X$ on a complete metric space (X, d) and produce new fixed point theorems. We also provide an interesting instructive example which demonstrate an application of our results.

The following definitions are given from the available literature.

Definition 1

[8]. Let (X, d) be a metric space and $F : X \rightarrow X$ be a mapping and $\alpha :X\,\times \,X \rightarrow [0,\infty )$. A mapping F is called $\alpha$-admissible mapping if the following condition holds:

$$\begin{aligned} x,y\in X \text { with } \, \alpha (x,y) \ge 1 \Rightarrow \alpha (Fx,Fy)\ge 1. \end{aligned}$$

(1)

Definition 2

Let $F : X \rightarrow X$ and $\alpha :X\,\times \,X \rightarrow [0,\infty )$ be a given mappings. A mappings $F:X\rightarrow X$ is called a triangular $\alpha$-admissible if

• $(\mathfrak {C}_{1})$ F is $\alpha$-admissible;

• $(\mathfrak {C}_{2})$ $\alpha (x,y)\ge 1$ and $\alpha (y,z)\ge 1 \Rightarrow$ $\alpha (x,z)\ge 1$, $x,y,z\in X$.

By $\varPsi$ we mean a class of functions $\wp :[0,\infty ) \rightarrow [0,\infty )$ satisfying the following assumptions:

• (1) $\wp$ is non decreasing function;

• (2) $\sum _{n=1}^{\infty }\wp ^{n}(t)<\infty$ for all $t>0$, where $\wp ^{n}$ is the nth iteration of $\wp$;

• (3) $\lim _{n\rightarrow \infty }\wp ^{n}(t)=0$ for all $t>0$;

• (4) $\wp (t)<t$ for each $t>0$;

• (5) $\wp (0)=0$.

The $\varPsi$ is known as Bianchini-Grandolf gauge function.

Now the $\varXi$ denotes the family of functions $\varUpsilon : [0,\infty )\rightarrow [0,\infty )$ sustain the following assumptions:

• (1) $\varUpsilon$ is continuous;

• (2) $\varUpsilon$ is nondecreasing on $[0,\infty )$;

• (3) $\varUpsilon (t)=0$ if and only if $t=0$;

• (4) $\varUpsilon (t)>0$ for all $t\in (0,\infty )$;

• (4) $\varUpsilon$ is subadditive.

We denote by CL(X) the class of all nonempty closed subsets of X.

Lemma 1

[14]. Let (X, d) be a metric space. $\varUpsilon \in \varXi$ and ${\mathscr {B}}\in CL(X)$. If there exist $x\in X$ such that $\varUpsilon (d(x,{\mathscr {B}}))\ge 0$, then there exists $y\in {\mathscr {B}}$, such that

$$\begin{aligned} \varUpsilon (d(x,y))<q\varUpsilon (d (x,{\mathscr {B}})), \end{aligned}$$

(2)

where $q>1$.

2 Main results

Here we give definition of $(\alpha ,\vartheta )$-admissibility.

Definition 3

Let (X, d) be a metric space and $F :X \rightarrow X$ be a self mapping and $\alpha , \vartheta : X\,\times \,X \rightarrow [0,\infty )$. A mapping F is said to be $(\alpha , \vartheta )$-admissible mapping if the following condition holds; for all $x,y \in X$

$$\begin{aligned} \alpha (x,y)\vartheta (x,y)\ge 1 \text { implies }\alpha (F x,F y)\ge 1\text { and } \vartheta (F x,F y)\ge 1. \end{aligned}$$

If for all $x,y \in X$, we have $\vartheta (x,y)=1$ then $(\alpha , \vartheta )$-admissibility becomes $\alpha$-admissibility given in Definition 1.

Example 1

Let $(X=[0,\infty ),d)$ be a metric space. Define

$$\begin{aligned} F x= {\left\{ \begin{array}{ll} x+2,\,\,\,\text { if } x\ne 0;\\ 0,\,\,\, \text { if } x=0, \end{array}\right. } \end{aligned}$$

(3)

and

$$\begin{aligned} \alpha (x,y)= {\left\{ \begin{array}{ll} 1.2,\,\,\, \text { if } x,y\ne 0;\\ 0,\,\,\, \text { if } x=0 \text { or } y=0, \end{array}\right. } \vartheta (x,y)= {\left\{ \begin{array}{ll} 1.5,\,\,\, \text { if } x,y\ne 0;\\ 0,\,\,\, \text { if } x=0 \text { or } y=0. \end{array}\right. } \end{aligned}$$

(4)

We discuss two cases.

Case I. When x or $y=0 \Rightarrow$ $\alpha (x,y)=0<1$, so we omit the case.

Case II. When $x,y\ne 0$, $\alpha (x,y)=1.2$. Now we have to check whether $\alpha (F x,F y)\ge 1$, or not. Since

$$\begin{aligned} F x=x+2, \quad F y=y+2, \end{aligned}$$

(5)

for $x\ne 0,\,y\ne 0$, then we have

$$\begin{aligned} \alpha (F x,F y)=\alpha (x+2,y+2)=1.2\ge 1. \end{aligned}$$

Now we check whether $\vartheta (F x,F y)\ge 1$ or not. Where $F x=x+2 ,\,\, F y=y+2$, which implies

$$\begin{aligned} \vartheta (x+2,y+2)=1.5\ge 1. \end{aligned}$$

(6)

Thus the self-mapping F is an $(\alpha , \vartheta )$-admissible map.

Definition 4

Let (X, d) be a complete metric space and $F :X \rightarrow X$ be an $(\alpha ,\vartheta )$-admissible mapping. The mapping F satisfies $(\varUpsilon ,\wp )$-integral-type contraction if there exist $\alpha ,\vartheta :X\times X\rightarrow [0,\infty )$ such that for all $x,y \in X$, such that $\alpha (x,y)\vartheta (x,y)\ge 1$ implies

$$\begin{aligned} \varUpsilon \Big (\int _0^{d(y,Fy)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\le \wp \Big ( \varUpsilon \max \Big \{\int _0^{ M_{1}(x,y)}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{M_{2}(x,y)}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{M_{3}(x,y)}\zeta (\mathfrak {j})d\mathfrak {j}\Big \}\Big ), \end{aligned}$$

(7)

where $M_{i}(x,y)$ for $i=1,2,3$ are

$$\begin{aligned} M_{1}(x,y)= & {} \max \Big \{ d(x,y),d(x,Fx),d(y,Fy),\nonumber \frac{d(x,F y)+d(y,F x)}{2}\Big \},\nonumber \\ M_{2}(x,y)= & {} \max \Big \{d(x,y),d(x,Fx),d(y,Fy)\Big \},\nonumber \\ M_{3}(x,y)= & {} \max \Big \{d(x,y),d(y,Fy)\Big \},\nonumber \\ M(x,y)= & {} \max \Big \{M_{1}(x,y), M_{2}(x,y), M_{3}(x,y)\Big \}, \end{aligned}$$

for $\wp \in \varPsi$, $\varUpsilon \in \varXi$ and $\mathfrak {P}\ge 1$, $\zeta :\mathbb {R}^{+}\rightarrow \mathbb {R}^{+}$ Lebesgue integrable with finite integral such that $\int _{0}^{\epsilon }\zeta (\mathfrak {j})d\mathfrak {j}>0$, for each $\epsilon >0$.

Theorem 1

Let $F :X \rightarrow X$ be a self-mapping on a complete metric space (X, d) and F satisfies $(\varUpsilon ,\wp )$-integral-contraction with the following assumptions:

• $(\mathfrak {C}_{1})~F$ is $(\alpha ,\vartheta )$-admissible mapping;

• $(\mathfrak {C}_{2})$ there exist $x_{0},y_{0}\in X$ such that $\alpha (x_{0},y_{0})\ge 1$;

• $(\mathfrak {C}_{3})~F$ satisfies $(\varUpsilon ,\wp )$-integral-type contraction.

Then F has a unique fixed point in (X, d).

Proof

Since F is $(\alpha ,\vartheta )$-admissible self mapping. Then there exist $x_{0},y_{0}\in X$ such that $\alpha (x_{0},y_{0})\ge 1$ which implies $\alpha (F x_{0},F y_{0})\ge 1$ implies $\vartheta (F x_{0},F y_{0})\ge 1$. Since $F :X\rightarrow X$, again there exist some $x_1 \in X$ such that $F x_{0}=x_{1}$, similarly $F y_{0}=y_{1}$ for some $y_{1}\in X$, thus $\alpha (x_{1},y_{1})\ge 1$. By $( C_{1})$, $\alpha (F x_{1},F y_{1})\ge 1$ which implies $\vartheta (F x_{1},F y_{1})\ge 1$. By continuing this process and using mathematical induction, we may have $\alpha (x_{n},y_{n})\ge 1$ implies $\alpha (F x_{n},F y_{n})\ge 1$ which give us $\vartheta (F x_{n},F y_{n})\ge 1$. Ultimately, we have

$$\begin{aligned} \alpha (x_{n},y_{n})\vartheta (x_{n},y_{n})\ge 1. \end{aligned}$$

(8)

Now by $(\mathfrak {C}_{3})$, we may use the inequality (7). By putting $x=x_0$ and $y=x_1$, in the inequality (7), we have

$$\begin{aligned} \varUpsilon \Big (\int _0^{d(x_{1},F x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\le \wp \Big ( \varUpsilon \max \Big \{\int _0^{ M_{1}(x_{0},x_{1})}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{M_{2}(x_{0},x_{1})}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{M_{3}(x_{0},x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big \}\Big ), \end{aligned}$$

(9)

where $M_{i}(x_{1},x_{2})$ for $i=1, 2, 3$

$$\begin{aligned} M_{1}(x_{0},x_{1})= & {} \max \Big \{d(x_{0},x_{1}),d(x_{0},F x_{0}),d(x_{1},F x_{1}),\nonumber \\&\frac{d(x_{0},F x_{1})+d(x_{1},F x_{0})}{2}\Big \}\nonumber \\\le & {} \max \Big \{d(x_{0},x_{1}),d(x_{0},x_{1}),d(x_{1}, F x_{1}),\nonumber \\&\frac{d(x_{0},F x_{1})+d(x_{1},x_{1})}{2}\Big \}\nonumber \\\le & {} \max \Big \{d(x_{0},x_{1}),d(x_{1},F x_{1}),\nonumber \frac{d(x_{0},x_{1})+d(x_{1},x_{1})}{2}\Big \}\nonumber \\\le & {} \max \Big \{d(x_{0},x_{1}),d(x_{1},F x_{1})\Big \}, \end{aligned}$$

$$\begin{aligned} M_{2}(x_{0},x_{1})= & {} \max \Big \{d(x_{0},x_{1}),d(x_{0},F x_{0}),d(x_{1},F x_{1})\Big \}\nonumber \\\le & {} \max \Big \{d(x_{0},x_{1}),d(x_{0},x_{1}),d(x_{1},F x_{1})\Big \}\nonumber \\\le & {} \max \Big \{d(x_{0},x_{1}),d(x_{1},F x_{1})\Big \}, \end{aligned}$$

$$\begin{aligned} M_{3}(x_{0},x_{1})= & {} \max \Big \{d(x_{0},x_{1}),d(x_{1},F x_{1})\Big \}\nonumber \\\le & {} \max \Big \{d(x_{0},x_{1}),d(x_{1},F x_{1})\Big \}. \end{aligned}$$

Now if $\max \Big \{d(x_{0},x_{1}),d(x_{1},F x_{1})\Big \}=d(x_{1},F x_{1})$. From (9), we proceed

$$\begin{aligned} 0< & {} \varUpsilon \Big (\int _0^{d(x_{1},F x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\nonumber \\\le & {} \wp \Big (\varUpsilon \nonumber \max \Big \{\int _0^{d(x_{1},F x_{1})}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{d(x_{1},F x_{1})}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{d(x_{1},F x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big \}\Big )\\\nonumber\le & {} \wp \Big ( \varUpsilon \Big (\int _0^{d(x_{1},F x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big )\\\nonumber< & {} \varUpsilon \Big (\int _0^{d(x_{1},F x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big ). \end{aligned}$$

This is contradiction. If we consider $\max \Big \{d(x_{0},x_{1}),d(x_{1},F x_{1})\Big \}=d(x_{0},x_{1})$, then we have

$$\begin{aligned} 0<\varUpsilon \Big (\int _0^{d(x_{1},F x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\le \wp \Big ( \varUpsilon \Big (\int _0^{d(x_{0},x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big ). \end{aligned}$$

(10)

From Lemma 1, we have

$$\begin{aligned} \varUpsilon \Big (\int _0^{d(x_{1},x_{2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )< q\varUpsilon \Big (\int _0^{d(x_{1},F x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big ), \end{aligned}$$

for some $x_{2}= Fx_1$ and $q>1$. If $x_{2}=Fx_{2}$, then $x_{2}$ is the fixed point of F, we assume $x_{2}\ne Fx_{2}$ then from Eqs. (10) and (11), we have

$$\begin{aligned} 0<\varUpsilon \Big (\int _0^{d(x_{1},x_{2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )<q\wp \Big ( \varUpsilon \Big (\int _0^{d(x_{0},x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big ). \end{aligned}$$

(11)

Applying $\wp$ is non-decreasing to the inequality (11), we obtain

$$\begin{aligned} 0<\wp \Big (\varUpsilon \Big (\int _0^{d(x_{1},x_{2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big )<\wp \Big (q\wp \Big ( \varUpsilon \Big (\int _0^{d(x_{0},x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big )\Big ). \end{aligned}$$

This implies

$$\begin{aligned} q_{1}=\frac{\wp \Big (q\wp \Big ( \varUpsilon \Big (\int _0^{d(x_{0},x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big )\Big )}{\wp \Big (\varUpsilon \Big (\int _0^{d(x_{1},x_{2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big )}>1. \end{aligned}$$

(12)

Next, by putting $x=x_2, y=x_2$ in the inequality (7), we have

$$\begin{aligned} 0< & {} \varUpsilon \Big (\int _0^{d(x_{2},F x_{2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\nonumber \\\le & {} \wp \Big ( \varUpsilon \max \Big \{\int _0^{ M_{1}(x_{1},x_{2})}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{M_{2}(x_{1},x_{2})}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{M_{3}(x_{1},x_{2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big \}\Big ), \end{aligned}$$

(13)

where $\phi (M_{i}(x_{1},x_{2}))$ for $i=1,\,2,\,3$

$$\begin{aligned} M_{1}(x_{1},x_{2})= & {} \max \Big \{d(x_{1},x_{2}),d(x_{1},F x_{1}),d(x_{2},F x_{2}),\nonumber \\&\frac{d(x_{1},F x_{2})+d(x_{2},F x_{1})}{2}\Big \},\nonumber \\\le & {} \max \Big \{d(x_{1},x_{2}),d(x_{1},x_{1}),d(x_{2},F x_{2})\nonumber ,\nonumber \\&\frac{d(x_{1},F x_{2})+d(x_{2},x_{1})}{2}\Big \},\nonumber \\\le & {} \max \Big \{d(x_{1},x_{2}),d(x_{2},F x_{2})\nonumber ,\frac{d(x_{1},x_{2})+d(x_{2},x_{2})}{2}\Big \},\nonumber \\\le & {} \max \Big \{d(x_{1},x_{2}),d(x_{2},F x_{2})\Big \}, \end{aligned}$$

$$\begin{aligned} M_{2}(x_{1},x_{2})= & {} \max \Big \{d(x_{1},x_{2}),d(x_{1},F x_{1}),d(x_{2},F x_{2})\Big \},\nonumber \\\le & {} \max \Big \{d(x_{1},x_{2}),d(x_{1},x_{2}),d(x_{2},F x_{2})\Big \},\nonumber \\\le & {} \max \Big \{d(x_{1},x_{2}),d(x_{2},F x_{2})\Big \}, \end{aligned}$$

$$\begin{aligned} M_{3}(x_{1},x_{2})= & {} \max \Big \{d(x_{1},x_{2}),d(x_{2},F x_{2})\Big \},\nonumber \\\le & {} \max \Big \{d(x_{1},x_{2}),d(x_{2},F x_{2})\Big \}. \end{aligned}$$

Now if $\max \Big \{d(x_{1},x_{2}),d(x_{2},F x_{2})\Big \}=d(x_{1},x_{2})$. Then from (13) and $\wp (t)<t$, we have

$$\begin{aligned} 0<\varUpsilon \Big (\int _0^{d(x_{2},F x_{2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\le & {} \wp \Big ( \varUpsilon \nonumber \max \Big \{\int _0^{d(x_{1},x_{2})}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{d(x_{1},x_{2})}\zeta (\mathfrak {j})d\mathfrak {j},\\&\int _0^{d(x_{1},x_{2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big \}\Big )\\\nonumber\le & {} \wp \Big ( \varUpsilon \Big (\int _0^{d(x_{1},x_{2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big )<\varUpsilon \Big (\int _0^{d(x_{1},x_{2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big ).\nonumber \end{aligned}$$

(14)

Therefore, we get

$$\begin{aligned} 0<\varUpsilon \Big (\int _0^{d(x_{2},F x_{2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\le \wp \Big ( \varUpsilon \Big (\int _0^{d(x_{1},x_{2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big ). \end{aligned}$$

(15)

As $q_{1}>1$ from the Lemma 1 and there exists some $x_3\in X$, such that $x_{3}=F x_2$, which gives us

$$\begin{aligned} \varUpsilon \Big (\int _0^{d(x_{2},x_{3})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )< q_{1}\varUpsilon \Big (\int _0^{d(x_{2},F x_{2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big ). \end{aligned}$$

(16)

From (13), (15) and (16), we have

$$\begin{aligned} 0<\varUpsilon \Big (\int _0^{d(x_{2},x_{3})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\le & {} q_{1}\wp \Big ( \varUpsilon \Big (\int _0^{d(x_{1},x_{2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big )\nonumber \\ {}= & {} \wp \Big (q\wp \Big ( \varUpsilon \Big (\int _0^{d(x_{0},x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big )\Big ). \end{aligned}$$

(17)

Applying $\wp$ on (17), we have

$$\begin{aligned} 0<\wp \Big (\varUpsilon \Big (\int _0^{d(x_{2},x_{3})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big )<\wp ^{2} \Big (q\wp \Big ( \varUpsilon \Big (\int _0^{d(x_{0},x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big )\Big ). \end{aligned}$$

(18)

Continuing the same process upto $x_n$ with the assumption that $x_{n}\ne x_{n+1}= F x_{n}$, we have

$$\begin{aligned} 0<\varUpsilon \Big (\int _0^{d(x_{n+1},x_{n+2})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )<\wp ^{n} \Big (q\wp \Big ( \varUpsilon \Big (\int _0^{d(x_{0},x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big )\Big ), \end{aligned}$$

for all $n\in N_{0}$. Now we show that $\{x_{n}\}$ in X is a Cauchy sequence. For this, let $m,n \in N$ such that $m>n$, and triangle inequality then we have

$$\begin{aligned} 0<\varUpsilon \Big (\int _0^{d(x_{m},x_{n})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\le & {} \sum _{i=n}^{m-1}\varUpsilon \Big (\int _0^{d(x_{i},x_{i+1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\\< & {} \sum _{i=n}^{m-1}\wp ^{i-1} \Big (q\wp \Big ( \nonumber \varUpsilon \Big (\int _0^{d(x_{0},x_{1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big )\Big ). \end{aligned}$$

(19)

Applying $\lim _{n,m \rightarrow \infty }$ to (19), and $\wp \rightarrow 0$ as $n \rightarrow \infty$, therefore, we have

$$\begin{aligned} \lim _{n,m\rightarrow \infty }\varUpsilon \Big (\int _0^{d(x_{m},x_{n})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )=0. \end{aligned}$$

(20)

By the continuity of $\varUpsilon$, we get

$$\begin{aligned} \lim _{n,m\rightarrow \infty } \int _0^{d(x_{m},x_{n})}\zeta (\mathfrak {j})d\mathfrak {j}=0. \end{aligned}$$

Thus $\{x_{n}\}$ is a Cauchy sequence in X. Since (X, d) is complete, therefore there exists $x^*\in X$ such that $x_{n}\rightarrow x^*$ as $n \rightarrow \infty$, thus we have $\lim _{n,m\rightarrow \infty }\int _0^{d(x_{n},x^*)}\zeta (\mathfrak {j})d\mathfrak {j}=0$ from the continuity of F we have

$$\begin{aligned} \lim _{n,m\rightarrow \infty }\int _0^{d(F x_{n},F x^*)}\zeta (\mathfrak {j})d\mathfrak {j}=0, \end{aligned}$$

and

$$\begin{aligned} \int _0^{d(x^*,F x^*)}\zeta (\mathfrak {j})d\mathfrak {j}= \lim _{n\rightarrow \infty }\int _0^{d(x_{n+1},F x^*)}\zeta (\mathfrak {j})d\mathfrak {j}= \lim _{n\rightarrow \infty }\int _0^{d(F x_{n},F x^*)}\zeta (\mathfrak {j})d\mathfrak {j}=0. \end{aligned}$$

This implies $d(x^*,F x^*)=0$ or $x^*=F x^*$ and therefore, $x^*$ is a fixed point of F in (X, d).

Theorem 2

Let $F :X \rightarrow X$ be a self-mapping on a complete metric soace (X, d) and F satisfies $(\varUpsilon ,\wp )$-integral-contraction with the following assumptions:

• $(\mathfrak {C}^*_{1})$ F is $(\alpha ,\vartheta )$-admissible self mapping;

• $(\mathfrak {C}^*_{2})$ there exist $x_{0},x_{1}\in X$ such that $\alpha (x_{0},x_{1})\ge 1$ and $\alpha (x_{0},x_{1})\vartheta (x_{0},x_{1})\ge 1$;

• $(\mathfrak {C}^*_{3})$ f $\{x_n\}$ is a sequence in X with $x_{n+1}\in F x_n$, $x_n \rightarrow x \in X$ as $n\rightarrow \infty$ and $\alpha (x_n,x_{n+1})\ge 1$ for all $n\in N_0$.

Then we have

$$\begin{aligned} \varUpsilon \Big (\int _0^{d(x_{n+1},F x)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\le \wp \Big ( \varUpsilon \max \Big (\int _0^{ M{1}(x_n,x)}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{M{2}(x_n,x)}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{M{3}(x_n,x)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big ), \end{aligned}$$

for all $n\in N_0$. Then F has a fixed point in X.

Proof

Let $\{x_n\}$ be a Cauchy sequence in X, such that $x_n\rightarrow x^*$ as $n\rightarrow \infty$, then

$$\begin{aligned} \alpha (x_n,x_{n+1})\ge 1, \end{aligned}$$

and

$$\begin{aligned} \alpha (x_n,x_{n+1})\vartheta (x_n,x_{n+1})\ge 1, \end{aligned}$$

for all $n\in N$. Then from $(\mathfrak {C}^*_{3})$, we have

$$\begin{aligned} \varUpsilon \Big (\int _0^{d(x_{n+1},F x^*)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )&\le \wp \Big ( \varUpsilon \max \Big (\int _0^{ M{1}(x_n,x^*)}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{M{2}(x_n,x^*)}\zeta (\mathfrak {j})d\mathfrak {j}, \int _0^{M{3}(x_n,x^*)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big ), \end{aligned}$$

(21)

where $M_{i}$ for $i=1,2,3,$ are

$$\begin{aligned} M_{1}(x_n,x^*)= & {} \max \Big \{ d(x_n,x^*),d(x_n,F x_n),d(x^*,F x^*)\nonumber ,\nonumber \\&\frac{d(x_n,F x^*)+d(x^*,F x_n)}{2}\Big \}, \end{aligned}$$

$$\begin{aligned} M_{2}(x_n,x^*)= & {} \max \Big \{d(x_n,x^*),d(x_n,F x_n),d(x^*,F x^*)\Big \}, \end{aligned}$$

$$\begin{aligned} M_{3}(x_n,x^*)= & {} \max \Big \{d(x_n,x^*),d(x^*,F x^*)\Big \}, \end{aligned}$$

for all $n\in N$. Here, we assume that $d(x^*,F x^*)>0$ and let $\epsilon :=\frac{d(x_n,F x^*)}{2}$. Since $x_n\rightarrow x^*$ as $n\rightarrow \infty$, so we can find $N_1\in N_0$ such that

$$\begin{aligned} d(x^*,F x_n)<\frac{d(x^*,F x^*)}{2}, \end{aligned}$$

for all $n\ge N_1$. Furthermore, we obtain

$$\begin{aligned} d(x^*,F x_n)\le d(x^*,x_{n+1})\le \frac{d(x_n,F x^*)}{2}, \end{aligned}$$

Since $\{x_n\}$ is Cauchy sequence, therefore there exists $N_2 \in N_0$ such that

$$\begin{aligned} d(x_n,F x_n)\le d(x_n,x_{n+1})\le \frac{d(x^*,F x^*)}{2}, \end{aligned}$$

for all $n\ge N_2$. Since $d(x_n,F x^*)\rightarrow d(x^*,F x^*)$ as $n\rightarrow \infty$, it follows that there exists $N_3\in N_0$ such that

$$\begin{aligned} d(x_n,F x^*)<\frac{3d(x^*,F x^*)}{2}, \end{aligned}$$

for all $n\ge N_3$, and we get

$$\begin{aligned} M_{1}(x_n,x^*)= & {} \max \Big \{ d(x_n,x^*),d(x_n,F x_n),d(x^*,F x^*)\nonumber ,\nonumber \\&\frac{d(x_n,F x^*)+d(x^*,F x_n)}{2}\Big \}\nonumber \\= & {} d(x^*,F x^*), \end{aligned}$$

$$\begin{aligned} M_{2}(x_n,x^*)= & {} \max \Big \{d(x_n,x^*),d(x_n,F x_n),d(x^*,F x^*)\Big \}=d(x^*,F x^*), \end{aligned}$$

$$\begin{aligned} M_{3}(x_n,x^*)= & {} \max \Big \{d(x_n,x^*),d(x^*,F x^*)\Big \}=d(x^*,F x^*), \end{aligned}$$

for all $n\ge N:=\max \{N_1,N_2,N_3\}$. For each $n\ge N$, from (21) by using the triangular inequality we have

$$\begin{aligned} \varUpsilon \Big (\int _0^{d(x^*,F x^*)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\le & {} \nonumber \varUpsilon \Big (\int _0^{d(x^*,x_{n+1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )+\varUpsilon \Big (\int _0^{d(x_{n+1},F x^*)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\\ {}\le & {} \varUpsilon \Big (\int _0^{d(x^*,x_{n+1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )+\wp \Big ( \nonumber \varUpsilon \max \Big \{\int _0^{ M_{1}(x_n,x)}\zeta (\mathfrak {j})d\mathfrak {j},\nonumber \\&\int _0^{M_{2}(x_n,x)}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{ M_{3}(x_n,x)}\zeta (\mathfrak {j})d\mathfrak {j}\Big \}\Big )\nonumber \\ {}= & {} \varUpsilon \Big (\int _0^{d(x^*,x_{n+1})}\zeta (\mathfrak {j})d\mathfrak {j}\Big )+\wp \Big (\varUpsilon \nonumber \Big (\int _0^{d(x^*,F x^*)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big ). \end{aligned}$$

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

$$\begin{aligned} \varUpsilon \Big (\int _0^{d(x^*,F x^*)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\le \wp \Big (\varUpsilon \Big (\int _0^{d(x^*,F x^*)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big ). \end{aligned}$$

This is a contradiction of $\wp (t)\le t$. This implies $\varUpsilon \Big (\int _0^{d(x^*,F x^*)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )=0$, which further implies that $\int _0^{d(x^*,F x^*)}\zeta (\mathfrak {j})d\mathfrak {j}=0$. Consequently, we have $x^*= F x^*$.

3 Applications

Example 2

Let $(X=[0,10],d)$ be a metric space with $d(x,y)=|x-y|$, for $x,\,y\in X$. Defining $F :X\rightarrow X$ and $\alpha ,\vartheta :X\times X\rightarrow [0,\infty )$ as

$$\begin{aligned} F (x)= {\left\{ \begin{array}{ll} 2.5x, \text { if }x\in [0,5);\\ \frac{x}{64}, \text { if }x\in [5,10], \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \alpha (x,y)= {\left\{ \begin{array}{ll} 6,\,\,\, \text { if }x\in [5,10];\\ 2.5,\,\,\, \text { otherwise} , \end{array}\right. } \qquad \vartheta (x,y)= {\left\{ \begin{array}{ll} 7,\,\,\, \text { if }x\in [5,10];\\ 3,\,\,\, \text { otherwise} . \end{array}\right. } \end{aligned}$$

(22)

Let $\wp ,\varUpsilon :[0,\infty )\rightarrow [0,\infty )$ by $\wp (t)=\frac{t}{8}$, $\varUpsilon (t)=5\sqrt{t}$, $\zeta (t)=1$ and $\mathfrak {P}=1$. Since $\wp \in \varPsi$ and $\varUpsilon \in \varXi$. To show that F is a $(\varUpsilon ,\wp )$-integral-type contraction. For this, let $x,y \in X$, then we have

$$\begin{aligned} \alpha (x,y)\vartheta (x,y)\ge 1. \end{aligned}$$

Let $x,y\in [5,10]$, then we have

$$\begin{aligned} \varUpsilon \Big (\int _0^{d(y,F y)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )= & {} 5\sqrt{\Big (\int _0^{d(y,F y)}\zeta (\mathfrak {j})d\mathfrak {j}}\Big )\nonumber \\= & {} 5\sqrt{\Big (\int _0^{d(F x,F y)}\zeta (\mathfrak {j})d\mathfrak {j}}\Big )\nonumber \\= & {} 5\sqrt{\Big (\int _0^{{\frac{|x-y|}{64}}}\zeta (\mathfrak {j})d\mathfrak {j}}\Big )\nonumber \\= & {} \frac{5}{8}\sqrt{|x-y|}\nonumber \\\le & {} \frac{5}{8}\sqrt{\int _0^{\phi (M(x,y))}\zeta (\mathfrak {j})d\mathfrak {j}}\nonumber \\= & {} \wp \Big ( \varUpsilon \max \Big (\int _0^{M{1}(x,y)}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{M{2}(x,y)}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{ M{3}(x,y)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big ). \end{aligned}$$

Therefore F is a $(\varUpsilon ,\wp )$-integral-type contractive mapping. Now for the condition $(\mathfrak {C}^*_{3})$, we assume a sequence $\{x_n\}$ in X with $x_{n+1}= F x_n$ where $x_n \rightarrow x \in X$ as $n \rightarrow \infty$ and $\alpha (x_{n},x_{n+1})\ge 1$ which implies $\vartheta (x_{n},x_{n+1})\ge 1$ for all $n\in N$, then

$$\begin{aligned} \varUpsilon \Big (\int _0^{d(x_{n+1},F x)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )= & {} 5\sqrt{\Big (\int _0^{d(x_{n+1},F x)}\zeta (\mathfrak {j})d\mathfrak {j}}\Big )\nonumber \\= & {} 5\sqrt{\Big (\int _0^{d(F x_{n},F x)}\zeta (\mathfrak {j})d\mathfrak {j}}\Big )\nonumber \\= & {} 5\sqrt{\Big (\int _0^{{\frac{|x_{n}-x|}{64}}}\zeta (\mathfrak {j})d\mathfrak {j}}\Big )\nonumber \\= & {} \frac{5}{8}\sqrt{|x_{n}-x|}\nonumber \\\le & {} \frac{5}{8}\sqrt{\int _0^{\phi M(x_{n},x)}\zeta (\mathfrak {j})d\mathfrak {j}}\nonumber \\= & {} \wp \Big ( \varUpsilon \max \Big (\int _0^{ M{1}(x_n,x)}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{ M{2}(x_n,x)}\zeta (\mathfrak {j})d\mathfrak {j},\int _0^{ M{3}(x_n,x)}\zeta (\mathfrak {j})d\mathfrak {j}\Big )\Big ). \end{aligned}$$

Thus $(\mathfrak {C}^*_{3})$ of Theorem 2 holds. Therefore, we conclude that 0 is a fixed point of the self-mapping F.

4 Conclusion

In this paper, the notion of $(\alpha ,\vartheta )$-admissibility, $(\varUpsilon ,\wp )$-integral-type contraction are given and new fixed point theorems are proved for the admissible and continuous self-mapping $F: X \rightarrow X$ on a complete metric space (X, d). For the fixed point of $F :X \rightarrow X$, we further assume that F be an $(\alpha ,\vartheta )$-admissible mapping, there exist $x_{0},y_{0}\in X$ such that $\alpha (x_{0},y_{0})\ge 1$ and the map F sustaining $(\varUpsilon ,\wp )$-integral-type contraction. For the application, we have presented an interesting example which fulfill all the conditions of our Theorem 2 and the self-map X has a fixed point 0.