# Extensions of almost-F and F-Suzuki contractions with graph and some applications to fractional calculus

## Abstract

In this paper, we introduce the two new concepts of an α-type almost-F-contraction and an α-type F Suzuki contraction and prove some fixed point theorems for such mappings in a complete metric space. Some examples and an application to a nonlinear fractional differential equation are given to illustrate the usability of the new theory.

## Introduction

In recent years two interesting but different generalizations of the Banach-contraction theorem have been given by Samet et al.  and Wardowski . These two results have become of recent interest of many authors (see  and references therein).

Most recently, Piri and Kumam  (respectively, Minak et al. ) extended the results of Wardowski  by introducing the concept of an F-Suzuki contraction (respectively, almost-F-contraction) and obtained some interesting fixed point results. Following this direction of research, we introduce the new concepts of an α-type almost-F-contraction and an α-type F-Suzuki contraction and prove some fixed point theorems concerning such contractions. Moreover, some examples and an application to a nonlinear fractional differential equation are given to illustrate the usability of the new theory.

## Preliminaries

The aim of this section is to present some notions and results used in the paper. Throughout the article $$\mathbb{N}$$, $$\mathbb{R}^{+}$$, and $$\mathbb{R}$$ will denote the set of natural numbers, positive real numbers, and real numbers, respectively.

### Definition 2.1



Let $$F: \mathbb{R}^{+} \rightarrow\mathbb{R}$$ be a mapping satisfying:

1. (F1)

F is strictly increasing, that is, $$\alpha< \beta \Rightarrow F(\alpha) < F(\beta)$$ for all $$\alpha,\beta\in\mathbb{R}^{+}$$,

2. (F2)

for every sequence $$\{\alpha_{n}\}$$ in $$\mathbb{R}^{+}$$ we have $$\lim_{n \rightarrow\infty} \alpha_{n} =0$$ iff $$\lim_{n \rightarrow \infty} F(\alpha_{n}) = -\infty$$,

3. (F3)

there exists a number $$k \in(0,1)$$ such that $$\lim_{\alpha \rightarrow0^{+}} \alpha^{k} F(\alpha)= 0$$.

We denote with $$\mathcal{F}$$ the family of all functions F that satisfy the conditions (F1)-(F3).

### Example 2.1

The following function $$F: \mathbb{R}^{+} \rightarrow\mathbb{R}$$ belongs to $$\mathcal{F}$$:

$$F(\alpha) = \ln\alpha,\quad\quad F(\alpha) = \ln\alpha+ \alpha, \qquad F(\alpha )=- \frac{1}{\sqrt{\alpha}}\quad \text{where } \alpha>0.$$

### Definition 2.2



Let $$(X,d)$$ be a metric space. A mapping $$T:X \rightarrow X$$ is called an F-contraction on X if there exist $$F \in\mathcal{F}$$ and $$\tau> 0$$ such that, for all $$x,y \in X$$ with $$d(Tx, Ty) > 0$$, we have

$$\tau+ F \bigl(d(Tx,Ty)\bigr) \leq F\bigl(d(x,y)\bigr).$$

### Definition 2.3



Let $$(X,d)$$ be a metric space, $$T:X\to X$$ be a mapping. Then the mapping T is said to be an almost-F-contraction if there exist $$F\in\mathcal{F}$$ and $$\tau>0$$, $$L\ge0$$ such that

\begin{aligned}& d(Tx,Ty)>0 \quad \implies\quad \tau+F\bigl(d(Tx,Ty)\bigr)\le F\bigl(d(x,y)+ Ld(y,Tx)\bigr)\quad \text{and} \\& d(Tx,Ty)>0\quad \implies\quad \tau+F\bigl(d(Tx,Ty)\bigr)\le F\bigl(d(x,y)+ Ld(x,Ty)\bigr) \end{aligned}

for all $$x,y\in X$$.

### Definition 2.4

[13, 14]

Let us denote by $$\mathcal{G}$$ the set of all functions $$F:\mathbb{R}^{+}\to\mathbb{R}$$ satisfying the following conditions:

1. (G1)

F is strictly increasing, that is, $$\alpha< \beta \Rightarrow F(\alpha) < F(\beta)$$ for all $$\alpha,\beta\in\mathbb{R}^{+}$$,

2. (G2)

there is a sequence $$\{\alpha_{n}\}$$ of positive real numbers such that $$\lim_{n \rightarrow\infty} F(\alpha_{n}) =-\infty$$, or $$\inf F=-\infty$$,

3. (G3)

F is continuous on $$(0,\infty)$$.

### Example 2.2

The following function $$F: \mathbb{R}^{+} \rightarrow\mathbb{R}$$ belongs to $$\mathcal{G}$$:

$$F(\alpha) = -\frac{1}{\alpha},\qquad F(\alpha) = -\frac{1}{\alpha} + \alpha,\quad\quad F(\alpha)=\ln\alpha \quad \text{where } \alpha>0.$$

### Lemma 2.1



Let $$F:\mathbb{R}^{+}\to\mathbb{R}$$ be an increasing function and $$\{ \alpha_{n}\}$$ be a sequence of positive real numbers. Then the following holds:

1. (a)

if $$\lim_{n\to\infty} F(\alpha_{n})=-\infty$$, then $$\lim_{n\to \infty}\alpha_{n}=0$$;

2. (b)

if $$\inf F=-\infty$$, and $$\lim_{n\to\infty}\alpha_{n}=0$$, then $$\lim_{n\to\infty} F(\alpha_{n})=-\infty$$.

### Definition 2.5



Let $$(X,d)$$ be a metric space. A mapping $$T:X\to X$$ is said to be an F-Suzuki contraction if there exists $$\tau>0$$ such that for all $$x,y\in X$$ with $$Tx\ne Ty$$

$$\frac{1}{2}d(x,Tx)< d(x,y)\quad \implies\quad \tau+F\bigl(d(Tx,Ty)\bigr)\le F \bigl(d(x,y)\bigr),$$

where $$F\in\mathcal{G}$$.

### Definition 2.6



Let $$T:X\rightarrow X$$ and $$\alpha:X\times X\rightarrow[0, \infty)$$ be two given mappings. Then T is called an α-admissible if

$$x,y\in X, \quad \alpha(x,y)\geq1\quad \Longrightarrow\quad \alpha(Tx,Ty)\geq1.$$

## Fixed point results for α-type almost-F-contraction and α-type F-Suzuki contraction

In this section, we first introduce the concepts of an α-type almost-F-contraction and an α-type F-Suzuki contraction and then we prove some fixed point theorems for these contractions in a complete metric space.

We begin with the following definitions.

### Definition 3.1

Let $$(X,d)$$ be a metric space, $$T:X\to X$$ be a mapping, and $$\alpha :X\times X\to\{{-\infty}\}\cup(0,\infty)$$ be a symmetric function. Then the mapping T is said to be an α-type almost-F-contraction if there exist $$F\in\mathcal{F}$$ and $$\tau>0$$ and $$L\ge0$$ such that

\begin{aligned}& d(Tx,Ty)>0 \quad \implies\quad \tau+\alpha(x,y)F\bigl(d(Tx,Ty)\bigr)\le F\bigl(d(x,y)+ Ld(y,Tx)\bigr)\quad \text{and} \\& d(Tx,Ty)>0 \quad \implies\quad \tau+\alpha(x,y)F\bigl(d(Tx,Ty)\bigr)\le F \bigl(d(x,y)+ Ld(x,Ty)\bigr) \end{aligned}

for all $$x,y\in X$$.

### Example 3.1

Let $$X=[0,3]\cup[5,6]$$ with the usual metric, $$T:X\to X$$ be defined as

$$Tx =\textstyle\begin{cases}0 & \text{if }x\in[0,3], \\ 3 &\text{if }x\in[5,6], \end{cases}$$

and $$F(\alpha)=\ln\alpha$$.

Then T is not an almost-F-contraction. Since at $$x=3$$ and $$y=5$$, $$d(Tx,Ty)>0$$ but $$\tau+F(d(Tx,Ty))=\tau+F(3)$$, whereas $$F(d(x,y)+ Ld(x,Ty))=F(2)$$.

Define

$$\alpha(x,y) =\textstyle\begin{cases} 1 &\text{if } x,y\in[0,3]\text{ or }x,y\in[5,6],\\ 0.1 &\text{otherwise}. \end{cases}$$

Then T is an α-type almost-F-contraction with $$\tau=0.5$$ and $$L=3$$.

### Definition 3.2

Let $$(X,d)$$ be a metric space $$T:X\to X$$ be a mapping and $$\alpha :X\times X\to\{{-\infty}\}\cup(0,\infty)$$ be a symmetric function. A map $$T:X\to X$$ is said to be an α-type F-Suzuki contraction if there exists $$\tau>0$$ such that for all $$x,y\in X$$ with $$Tx\ne Ty$$

$$\frac{1}{2}d(x,Tx)< d(x,y)\quad \implies\quad \tau+\alpha(x,y)F\bigl(d(Tx,Ty)\bigr)\le F\bigl(d(x,y)\bigr),$$

where $$F\in\mathcal{G}$$.

### Example 3.2

Let $$X=[1,3]\cup[5,9]$$ with the usual metric and $$F:\mathbb{R}^{+}\to \mathbb{R}$$ be defined as $$F(\alpha)=-\frac{1}{\alpha}$$. Define a mapping $$T:X\to X$$ as

$$Tx =\textstyle\begin{cases}5, &x\in[1,3],\\ 8, &x\in[5,9]. \end{cases}$$

Then T is not F-Suzuki contraction as the condition

$$\frac{1}{2}d(x,Tx)< d(x,y)\quad \implies\quad \tau+F\bigl(d(Tx,Ty)\bigr)\le F \bigl(d(x,y)\bigr)$$

fails for $$x=3$$ and $$y=5$$.

Define $$\alpha:X\times X \to\{-\infty\} \cup(0,\infty)$$ as

$$\alpha(x,y)=2,\quad \mbox{for all }x,y\in X.$$

Then T is α-type F-Suzuki contraction i.e.

$$\frac{1}{2}d(x,Tx)< d(x,y)\quad \implies\quad \tau+\alpha(x,y)F\bigl(d(Tx,Ty)\bigr)\le F\bigl(d(x,y)\bigr)$$

holds for all $$x,y\in X$$ with $$\tau=\frac{1}{6}$$.

Now, we prove our first result.

### Theorem 3.1

Let $$(X,d)$$ be a complete metric space and $$T:X\to X$$ be an α-type almost-F-contraction where $$F\in\mathcal{F}$$, satisfying the following conditions:

1. (i)

2. (ii)

there exists $$x_{0}\in X$$ such that $$\alpha(x_{0},Tx_{0})\ge1$$,

3. (iii)

if $$\{x_{n}\}$$ is a sequence in X such that $$\alpha (x_{n},x_{n+1})\ge1$$ for all $$n\in\mathbb{N}\cup\{0\}$$ and $$x_{n}\to x\in X$$ as $$n\to\infty$$ then $$\alpha(x_{n},x)\ge1$$ for all $$n\in\mathbb {N}\cup\{0\}$$.

Then T has a fixed point $$x^{\ast}\in X$$.

### Proof

Let $$x_{0}\in X$$ be such that $$\alpha(x_{0},Tx_{0})\ge1$$. Define the sequence $$\{x_{n}\}$$ in X by $$x_{n+1}=Tx_{n}$$, for all $$n\in\mathbb {N}\cup\{0\}$$. If $$x_{n+1}=x_{n}$$ for some $$n\in\mathbb{N}$$, then $$x^{*}=x_{n}$$ is a fixed point of T. Let us assume that $$x_{n+1}\neq x_{n}$$ for all $$n\in\mathbb{N}\cup\{0\}$$.

Since T is α-admissible, we have $$\alpha(x_{0},x_{1})=\alpha(x_{0},Tx_{0})\ge1$$, which implies $$\alpha (Tx_{0},Tx_{1})=\alpha(x_{1},x_{2})\ge1$$. Continuing in this way we have in general

$$\alpha(x_{n},x_{n+1})\ge1 \quad \text{for all } n\in\mathbb{N}.$$
(3.1)

Now, $$F(d(x_{n+1},x_{n}))=F(d(Tx_{n},Tx_{n-1}))\le\alpha (x_{n},x_{n-1})F(d(Tx_{n},Tx_{n-1}))$$.

Therefore,

\begin{aligned} \tau+F\bigl(d(Tx_{n},Tx_{n-1})\bigr) \le& \tau+\alpha (x_{n},x_{n-1})F\bigl(d(Tx_{n},Tx_{n-1}) \bigr) \\ \le& F\bigl(d(x_{n},x_{n-1})+Ld(x_{n},Tx_{n-1}) \bigr). \end{aligned}

So $$F(d(x_{n+1},x_{n}))=F(d(Tx_{n},Tx_{n-1}))\le F(d(x_{n},x_{n-1}))-\tau$$. In general we get

$$F\bigl(d(x_{n+1},x_{n})\bigr)=F\bigl(d(Tx_{n},Tx_{n-1}) \bigr)\le F\bigl(d(x_{1},x_{0})\bigr)-n\tau.$$
(3.2)

Thus as $$n\to\infty$$, we have $$\lim_{n\to\infty }F(d(x_{n+1},x_{n}))=-\infty$$, then by (F2) we have $$\lim_{n\to\infty }d(x_{n+1}, x_{n})=0$$. Now, from (F3), there exists $$k\in(0,1)$$ such that $$\lim_{n\rightarrow\infty} (d(x_{n+1},x_{n}))^{k}F(d(x_{n+1}, x_{n}))=0$$. From (3.2) it follows that

$$\bigl(d(x_{n+1},x_{n})\bigr)^{k}F \bigl(d(x_{n+1},x_{n})\bigr)\le \bigl(d(x_{n+1},x_{n}) \bigr)^{k}\bigl(F\bigl(d(x_{1},x_{0})\bigr)-n\tau \bigr).$$

Then as $$n\to\infty$$ we get

$$\lim_{n\to\infty}n\bigl(d(x_{n+1},x_{n}) \bigr)^{k}=0.$$

Therefore, there exists $$n_{0}\in\mathbb{N}$$ such that

$$n\bigl(d(x_{n+1},x_{n})\bigr)^{k}\le1, \quad \forall n\ge n_{0},$$

i.e.

$$d(x_{n+1},x_{n})\le\frac{1}{n^{1/k}}, \quad \forall n\ge n_{0}.$$

Now, for $$m>n>n_{0}$$,

\begin{aligned} d(x_{n},x_{m}) \le& d(x_{n},x_{n+1})+d(x_{n+1},x_{n+2})+ \cdots +d(x_{m-1},x_{m}) \\ \le& \sum_{n\ge n_{0}}\frac{1}{n^{1/k}}, \end{aligned}

which is convergent as $$k\in(0,1)$$. Therefore as $$m,n\to\infty$$ we get $$d(x_{n},x_{m})\to0$$. Hence $$\{x_{n}\}$$ is a Cauchy sequence. From the completeness of X we then have $$x^{\ast}\in X$$ such that $$x_{n}\to x^{\ast}$$.

Now we claim that $$d(x_{n+1},Tx^{*})=d(Tx_{n},Tx^{*})\to0$$ as $$n\to\infty$$. If $$x^{*}=Tx^{*}$$, then the proof is finished. Assume that $$x^{*}\ne Tx^{*}$$. If $$x_{n+1}=Tx_{n}=Tx^{*}$$ for infinite values of $$n\in\mathbb{N}\cup\{0\}$$, then the sequence has a subsequence that converges to $$Tx^{*}$$ and the uniqueness of the limit implies $$x^{*}=Tx^{*}$$. Then we can assume that $$Tx_{n}\ne Tx^{*}$$ for all $$n\in\mathbb{N}\cup\{0\}$$. Now using (iii), we have

$$\tau+F\bigl(d\bigl(Tx_{n},Tx^{*}\bigr)\bigr)\le\tau+\alpha \bigl(x_{n},x^{*}\bigr)F\bigl(d\bigl(Tx_{n},Tx^{*}\bigr)\bigr)\le F\bigl(d\bigl(x_{n},x^{*}\bigr)+Ld\bigl(Tx_{n},x^{*}\bigr)\bigr).$$

Then, as $$n\to\infty$$ we get

$$\tau+\lim_{n\to\infty}F\bigl(d\bigl(Tx_{n},Tx^{*}\bigr)\bigr) \le-\infty,$$

which will lead to a contradiction of the assumption that $$\lim_{n\to \infty}d(Tx_{n},Tx^{*})>0$$ (in respect of (F2)). Thus we have $$x_{n+1}=Tx_{n}\to Tx^{*}$$ as $$n\to\infty$$ and hence $$Tx^{*}=x^{*}$$. □

### Theorem 3.2

We further assume that $$\alpha(x,y)\ge1$$ for all $$x,y\in \operatorname {Fix}(T)$$ and suppose T also satisfies the following condition: there exist $$G\in\mathcal{F}$$ and some $$L\ge0$$, $$\tau>0$$ such that for all $$x,y\in X$$

$$\tau+\alpha(x,y)G\bigl(d(Tx,Ty)\bigr)\le G\bigl(d(x,y)+Ld(x,Tx)\bigr)$$

holds. Then the fixed point in the above result is unique.

### Proof

Let $$y^{*}\in X$$, $$y^{*}\ne x^{*}$$ such that $$Ty^{*}=y^{*}$$. Then $$d(Tx^{*},Ty^{*})>0$$, which implies

$$\tau+G\bigl(d\bigl(Tx^{*},Ty^{*}\bigr)\bigr)\le\tau+\alpha\bigl(x^{*},y^{*}\bigr)G \bigl(d\bigl(Tx^{*},Ty^{*}\bigr)\bigr)\le G\bigl(d\bigl(x^{*},y^{*}\bigr)+L\bigl(x^{*},Tx^{*} \bigr)\bigr).$$

Therefore we have

$$\tau+G\bigl(d\bigl(x^{*},y^{*}\bigr)\bigr)=\tau+G\bigl(d\bigl(Tx^{*},Ty^{*}\bigr)\bigr) \le G\bigl(d\bigl(x^{*},y^{*}\bigr)\bigr),$$

which is a contradiction as $$\tau>0$$. □

The following example is illustrative of Theorem 3.2.

### Example 3.3

Let $$X=[1,3]\cup[5,9]$$ with usual metric and $$F:\mathbb{R}^{+}\to\mathbb {R}$$ be defined as $$F(\alpha)=\ln\alpha$$. Define a continuous map $$T:X\to X$$ as

$$Tx =\textstyle\begin{cases}2x+3, &x\in[1,3],\\ 9, &x\in[5,9]. \end{cases}$$

Then T satisfies all the conditions of Theorem 3.2 for $$\tau\le 1.9408$$ and $$L=1$$, and hence T has a unique fixed point $$x^{\ast}=9$$.

If $$\alpha(x,y)=1$$ for all $$x,y\in X$$ then we have following result as in .

### Corollary 3.1

Let $$(X,d)$$ be a complete metric space and $$T:X\to X$$ be an almost-F-contraction. Then T has a fixed point $$x^{*}$$ in X.

If $$\alpha(x,y)=1$$ for all $$x,y\in X$$ and $$L=0$$ then we have following result of Wardowski’s .

### Corollary 3.2

Let $$(X,d)$$ be a complete metric space and $$T:X\to X$$ be an F-contraction. Then T has a unique fixed point $$x^{*}$$ in X.

To prove our next result, we first give the following.

### Definition 3.3



An α-admissible map T is said to have the K-property whenever for each sequence $$\{x_{n}\}\subset X$$ with $$\alpha (x_{n},x_{n+1})\ge1$$ for all $$n\in\mathbb{N}\cup\{0\}$$, then there exists a natural number k such that $$\alpha(Tx_{m},Tx_{n})\ge1$$ for all $$m>n\ge k$$.

### Theorem 3.3

Let $$(X,d)$$ be a complete metric space and $$T:X\to X$$ be an α-type F-Suzuki contraction satisfying the following conditions:

1. (i)

2. (ii)

there exists $$x_{0}\in X$$ such that $$\alpha(x_{0},Tx_{0})\ge1$$,

3. (iii)

T has the K-property,

4. (iv)

if $$\{x_{n}\}$$ is a sequence in X such that $$\alpha (x_{n},x_{n+1})\ge1$$ for all $$n\in\mathbb{N}\cup\{0\}$$ and $$x_{n}\to x\in X$$ as $$n\to\infty$$, then $$\alpha(x_{n},x)\ge1$$ for all $$n\in\mathbb {N}\cup\{0\}$$.

Then T has a fixed point $$x^{\ast}\in X$$.

### Proof

Let $$x_{0}\in X$$ be such that $$\alpha(x_{0},Tx_{0})\ge1$$. Define the sequence $$\{x_{n}\}\subseteq X$$ by $$x_{n+1}=Tx_{n}$$, for all $$n\in\mathbb {N}\cup\{0\}$$. Since T is α-admissible we have $$\alpha(x_{0},x_{1})=\alpha(x_{0},Tx_{0})\ge1$$, which implies $$\alpha (Tx_{0},Tx_{1})=\alpha(x_{1},x_{2})\ge1$$. Continuing in this way we have in general

$$\alpha(x_{n},x_{n+1})\ge1 \quad \text{for all } n\in\mathbb{N}\cup \{0\}.$$
(3.3)

If $$x_{n+1}=x_{n}$$ for some $$n\in\mathbb{N}\cup\{0\}$$, then $$x^{*}=x_{n}$$ is a fixed point of T. Let us assume that $$x_{n+1}\neq x_{n}$$ for all $$n\in \mathbb{N}\cup\{0\}$$. Therefore $$\frac{1}{2}d(x_{n},Tx_{n})< d(x_{n},Tx_{n})$$ for all $$n\in\mathbb{N}\cup\{0\}$$ and hence

\begin{aligned} \tau+F\bigl(d\bigl(Tx_{n},T^{2}x_{n}\bigr)\bigr) &\le\tau+\alpha(x_{n},Tx_{n})F\bigl(d\bigl(Tx_{n},T^{2}x_{n} \bigr)\bigr)\\ &\le F\bigl(d(x_{n},Tx_{n})\bigr) \quad \text{for all } n \in\mathbb{N}\cup\{0\}. \end{aligned}

So $$F(d(Tx_{n},T^{2}x_{n}))\le F(d(x_{n},Tx_{n}))-\tau$$ and repeating this process in general we get

$$F\bigl(d\bigl(Tx_{n},T^{2}x_{n}\bigr)\bigr)\le F \bigl(d(x_{0},x_{1})\bigr)-n\tau.$$

As $$n\to\infty$$ we obtain

$$\lim_{n\to\infty}F\bigl(d(x_{n+1},x_{n+2})\bigr)=- \infty,$$

which together with (G2) and by Lemma 2.1, gives

$$\lim_{n\to\infty}d(x_{n+1},x_{n+2})=0.$$

Suppose $$\{x_{n}\}$$ is not a Cauchy sequence. Then there exist $$\varepsilon>0$$ and $$p(n)>q(n)>n\ge k$$ such that $$d(x_{p(n)},x_{q(n)})\ge\varepsilon$$ and $$d(x_{(p(n)-1)},x_{q(n)})< \varepsilon$$.

Now

$$\varepsilon\le d(x_{p(n)},x_{q(n)})\le d(x_{p(n)},x_{p(n)-1})+d(x_{p(n)-1},x_{q(n)})< d(x_{p(n)},x_{p(n)-1})+ \varepsilon.$$

Therefore

$$\lim_{n\to\infty}d(x_{p(n)},x_{q(n)})=\varepsilon.$$
(3.4)

Again we have

$$d(x_{p(n)},x_{q(n)})\le d(x_{p(n)},x_{p(n)+1})+d(x_{p(n)+1},x_{q(n)+1})+d(x_{q(n)+1},x_{q(n)})$$

and

$$d(x_{p(n)+1},x_{q(n)+1})\le d(x_{p(n)+1},x_{p(n)})+d(x_{p(n)},x_{q(n)})+d(x_{q(n)},x_{q(n)+1}).$$

So as $$n\to\infty$$, from the above two inequalities we have

$$\lim_{n\to\infty}d(x_{p(n)+1},x_{q(n)+1})=\varepsilon.$$
(3.5)

Therefore there exists $$k\in\mathbb{N}$$ such that $$\frac {1}{2}d(x_{p(n)},x_{q(n)})< d(x_{p(n)},x_{q(n)})$$ for all $$n\ge k$$. Using the K-property we have

\begin{aligned} \tau+F\bigl(d(Tx_{p(n)},Tx_{q(n)})\bigr) \le& \tau+\alpha (x_{p(n)},x_{q(n)})F\bigl(d(Tx_{p(n)},Tx_{q(n)}) \bigr) \\ \le& F\bigl(d(x_{p(n)},x_{q(n)})\bigr). \end{aligned}

So as $$n\to\infty$$ and by (G3), we get $$\tau+F(\varepsilon)\le F(\varepsilon)$$, which is a contradiction. Hence $$\{x_{n}\}$$ is a Cauchy sequence in X and so it converges to some $$x^{\ast}$$ in X.

Next, we claim that

\begin{aligned}& \frac{1}{2}d(x_{n},Tx_{n})< d\bigl(x_{n},x^{\ast}\bigr)\quad \mbox{or} \\& \frac{1}{2}d\bigl(Tx_{n},T^{2}x_{n}\bigr)< d \bigl(Tx_{n},x^{\ast}\bigr)\quad \text{for all } n\in\mathbb{N.} \end{aligned}

Assume there exists $$m\in\mathbb{N}$$ such that

$$\frac{1}{2}d(x_{m},Tx_{m})\ge d \bigl(x_{m},x^{\ast}\bigr) \quad \text{and}\quad \frac {1}{2}d \bigl(Tx_{m},T^{2}x_{m}\bigr)\ge d \bigl(Tx_{m},x^{\ast}\bigr).$$

Then,

$$2d\bigl(x_{m},x^{\ast}\bigr)\le d(x_{m},Tx_{m}) \le d\bigl(x_{m},x^{\ast}\bigr)+d\bigl(x^{\ast},Tx_{m} \bigr)$$

and hence

$$d\bigl(x_{m},x^{\ast}\bigr)\le d\bigl(x^{\ast},Tx_{m} \bigr) \le\frac{1}{2}d\bigl(Tx_{m},T^{2}x_{m} \bigr).$$
(3.6)

Since $$\frac{1}{2}d(x_{m},Tx_{m}) < d(x_{m},Tx_{m})$$, we have

$$\tau+F\bigl(d\bigl(Tx_{m},T^{2}x_{m}\bigr)\bigr) \le\tau+\alpha(x_{m},Tx_{m})F\bigl(d\bigl(Tx_{m},T^{2}x_{m} \bigr)\bigr)\le F\bigl(d(x_{m},Tx_{m})\bigr),$$

which implies $$F(d(Tx_{m},T^{2}x_{m}))< F(d(x_{m},Tx_{m}))$$ and so $$d(Tx_{m},T^{2}x_{m})< d(x_{m},Tx_{m})$$. Now

\begin{aligned} d\bigl(Tx_{m},T^{2}x_{m}\bigr) < &d(x_{m},Tx_{m}) \\ \le& d\bigl(x_{m},x^{\ast}\bigr)+d\bigl(x^{\ast},Tx_{m} \bigr) \\ \le& \frac{1}{2}d\bigl(Tx_{m},T^{2}x_{m} \bigr)+\frac{1}{2}d\bigl(Tx_{m},T^{2}x_{m} \bigr)=d\bigl(Tx_{m},T^{2}x_{m}\bigr). \end{aligned}

Thus, for every $$n\in\mathbb{N}$$ either

\begin{aligned}& \tau+F\bigl(d\bigl(Tx_{n},Tx^{\ast}\bigr)\bigr)\le\tau+\alpha \bigl(x_{n},x^{\ast}\bigr)F\bigl(d\bigl(Tx_{n},Tx^{\ast}\bigr)\bigr)\le F\bigl(d\bigl(x_{n},x^{\ast}\bigr)\bigr) \quad \text{or} \\& \tau+F\bigl(d\bigl(T^{2}x_{n},Tx^{\ast}\bigr)\bigr) \le\tau+\alpha\bigl(x_{n+1},x^{\ast}\bigr)F\bigl(d \bigl(T^{2}x_{n},Tx^{\ast}\bigr)\bigr)\le F\bigl(d \bigl(x_{n+1},x^{\ast}\bigr)\bigr). \end{aligned}

As $$n\to\infty$$ we get from the above

\begin{aligned}& \lim_{n\to\infty}F\bigl(d\bigl(Tx_{n},Tx^{\ast}\bigr)\bigr)=-\infty \quad \text{and} \\& \lim_{n\to\infty}F\bigl(d\bigl(T^{2}x_{n},Tx^{\ast}\bigr)\bigr)=-\infty, \end{aligned}

respectively. This further gives

$$\lim_{n\to\infty}d\bigl(Tx_{n},Tx^{\ast}\bigr)=0 \quad \text{and}\quad \lim_{n\to\infty }d\bigl(T^{2}x_{n},Tx^{\ast}\bigr)=0.$$

Then

$$0\le d\bigl(x^{\ast},Tx^{\ast}\bigr)\le d\bigl(x^{\ast},Tx_{n} \bigr)+d\bigl(Tx_{n},Tx^{\ast}\bigr),$$

which as $$n\to\infty$$ gives $$d(x^{\ast},Tx^{\ast})=0$$ and hence $$Tx^{\ast}=x^{\ast}$$. □

### Theorem 3.4

If we further assume that $$\alpha(x,y)\ge1$$ for all $$x,y\in \operatorname {Fix}(T)$$, then the fixed point is unique in the above result.

### Proof

Let $$y^{*}\in X$$, $$y^{*}\ne x^{*}$$ such that $$Ty^{*}=y^{*}$$. Then $$\frac {1}{2}d(Tx^{*},x^{*})=0< d(x^{*},y^{*})$$, which implies

$$\tau+F\bigl(d\bigl(Tx^{*},Ty^{*}\bigr)\bigr)\le\tau+\alpha\bigl(x^{*},y^{*}\bigr)F \bigl(d\bigl(Tx^{*},Ty^{*}\bigr)\bigr)\le F\bigl(d\bigl(x^{*},y^{*}\bigr)\bigr).$$

Therefore, we have

$$\tau+F\bigl(d\bigl(x^{*},y^{*}\bigr)\bigr)=\tau+F\bigl(d\bigl(Tx^{*},Ty^{*}\bigr)\bigr) \le F\bigl(d\bigl(x^{*},y^{*}\bigr)\bigr),$$

which is a contradiction as $$\tau>0$$. □

The following example illustrates Theorem 3.4.

### Example 3.4

Let $$X=[1,3]\cup[5,9]$$ with the usual metric and $$F:\mathbb{R}^{+}\to \mathbb{R}$$ be defined as $$F(\alpha)=-\frac{1}{\alpha}$$. Define a continuous map $$T:X\to X$$ as

$$Tx =\textstyle\begin{cases}2x+3, &x\in[1,3],\\ 9, &x\in[5,9]. \end{cases}$$

Then T is not a F-Suzuki contraction as the condition

$$\frac{1}{2}d(x,Tx)< d(x,y)\quad \implies\quad \tau+F\bigl(d(Tx,Ty)\bigr)\le F \bigl(d(x,y)\bigr)$$

fails for $$x=1$$ and $$y=5$$.

Now, we distinguish following cases:

Case 1. If $$x,y\in[1,3]$$ then there are no points for which $$\frac {1}{2}d(x,Tx)< d(x,y)$$ holds, so we are through.

Case 2. If $$x,y\in[5,9]$$ then $$Tx=Ty$$ so we are done.

Case 3. Let $$x\in[1,3]$$ and $$y\in[5,9]$$. In this case we have $$d(Tx,Ty)\le4$$ and $$2\le d(x,y)$$. Therefore $$\tau+\alpha(x,y)F(d(Tx,Ty))\le\tau+\alpha(x,y) F(4)=\tau-\frac{\alpha (x,y)}{4}$$ and $$F(2)\le F(d(x,y))$$ by (F2). So for given $$\alpha(x,y)$$ we can choose τ for which $$\tau\le \frac{\alpha(x,y)-2}{4}$$ holds. Then for that $$\alpha(x,y)$$ and $$\tau>0$$ we are done.

In particular, if we define $$\alpha:X\times X \to(0,\infty)\cup\{ -\infty\}$$ as

$$\alpha(x,y)=3 \quad \text{for all } x,y\in X.$$

Then T satisfies all the conditions of the above theorem with $$\tau =\frac{1}{4}$$ and hence T has an unique fixed point $$x^{\ast}=9$$.

If $$\alpha(x,y)=1$$ for all $$x,y\in X$$ then we have following result

### Corollary 3.3

Let $$(X,d)$$ be a complete metric space and $$T:X\to X$$ be an F-Suzuki contraction. Then T has a unique fixed point $$x^{*}$$ in X.

## Consequences

In this section we will show that some existing results in the literature can be deduced easily from our theorems proved in Section 3.

### Fixed point with graph

Following Jachymski , let $$(X,d)$$ be a metric space and $$\Delta= \{(x,x) : x \in X\}$$. Consider a graph G with the set $$V(G)$$ of its vertices equal to X and the set $$E(G)$$ of its edges as a superset of Δ. Assume that G has no parallel edges, that is, $$(x,y), (y,x) \in E(G)$$ implies $$x=y$$. Also, G is directed if the edges have a direction associated with them. Now we can identify the graph G with the pair $$(V(G), E(G) )$$.

Denote

$$\mathcal{G}: = \bigl\{ G : G \text{ is a directed graph with } V(G) = X \text{ and } \Delta\subseteq E(G)\bigr\} .$$

### Definition 4.1

A mapping $$T:X \rightarrow X$$ is called G-continuous, if we have a given $$x \in X$$ and a sequence $$\{x_{n}\}$$ such that $$x_{n} \to x$$, as $$n \to\infty$$, $$(x_{n}, x_{n+1}) \in E(G)$$, $$\forall n \in\mathbb{N}$$ imply $$Tx_{n} \to Tx$$.

### Theorem 4.1

Let $$(X,d)$$ be a metric space endowed with a graph G and let T be a self-mapping on X. Suppose that the following assertions hold:

1. (i)

for all $$x,y \in X$$, $$(x,y) \in E(G) \Rightarrow (Tx, Ty) \in E(G)$$;

2. (ii)

there exists $$x_{0} \in X$$ such that $$(x_{0}, Tx_{0}) \in E(G)$$;

3. (iii)

there exist a number $$\tau> 0$$, $$L \geq0$$ and $$F \in \mathcal{F}$$ such that

\begin{aligned}& d(Tx,Ty)>0 \quad \Rightarrow\quad \tau+F\bigl(d(Tx,Ty)\bigr)\le F\bigl[d(x,y)+ Ld(y,Tx) \bigr]\quad \textit{and} \\& d(Tx,Ty)>0 \quad \Rightarrow\quad \tau+F\bigl(d(Tx,Ty)\bigr)\le F\bigl[d(x,y)+ Ld(x,Ty)\bigr] \end{aligned}

for all $$(x, y) \in E(G)$$;

4. (iv)

for any sequence $$\{x_{n}\} \subset X$$, $$x\in X$$ with $$x_{n} \to x$$ as $$n \rightarrow\infty$$ and $$(x_{n}, x_{n+1}) \in E(G)$$ we have $$(x_{n}, x) \in E(G)$$ or T is G-continuous.

Then T has a fixed point.

### Proof

Define $$\alpha: X \times X \to\{ - \infty\} \cup(0, \infty)$$ by

$$\alpha(x,y) = \textstyle\begin{cases} 1 & \text{if } (x,y) \in E(G),\\ -\infty& \text{otherwise.} \end{cases}$$

First we prove that T is α-admissible. If $$\alpha(x,y) \geq 1$$, then $$(x, y) \in E(G)$$. As, from (i), we have $$(Tx, Ty) \in E(G)$$, $$\alpha(Tx, Ty) \geq1$$. So T is an α-admissible mapping. From (ii) there exists $$x_{0} \in X$$ such that $$(x_{0}, Tx_{0}) \in E(G)$$, i.e. $$\alpha(x_{0}, Tx_{0}) \geq1$$.

Let $$\alpha(x,y) \geq1$$, then $$(x,y) \in E(G)$$. Now, from (iii) we have

\begin{aligned}& \tau+ F\bigl(d(Tx, Ty)\bigr) \leq F\bigl[d(x,y) + L d(x, Ty)\bigr] \quad \text{and} \\& \tau+F\bigl(d(Tx,Ty)\bigr)\le F\bigl[d(x,y)+ Ld(y,Tx)\bigr] \quad \textit{i.e.} \\& \alpha(x,y ) \geq1 \quad \Rightarrow\quad \tau+ F\bigl(d(Tx, Ty)\bigr) \leq F\bigl[d(x,y) + L d(x, Ty)\bigr] \quad \text{and} \\& \tau+F\bigl(d(Tx,Ty)\bigr)\le F\bigl[d(x,y)+ Ld(y,Tx)\bigr]. \end{aligned}

Now, let $$\{x_{n}\} \subset X$$ be a sequence such that $$x_{n} \rightarrow x$$ as $$n \to\infty$$ and $$\alpha(x_{n}, x_{n+1}) \geq1$$.

Then, $$(x_{n+1}, x_{n}) \in E(G)$$ and then from (iv) $$(x_{n}, x) \in E(G)$$ i.e. $$\alpha(x_{n}, x) \geq1$$. Thus, all the conditions of Theorem 3.1 are satisfied and hence T has a fixed point in X. □

If $$L =0$$, then Theorem 4.1 reduces to Corollary 2.9 given in .

### Fixed point with partial order

Let $$(X, d, \preceq)$$ be a partially ordered metric space. Define the graph G by

$$E(G) = \bigl\{ (x, y)\in X \times X : x \preceq y\bigr\} .$$

For the graph condition (i) in Theorem 4.1 means that T is nondecreasing with respect to this order . From Theorem 4.1 we derive the following important results in partially ordered metric spaces.

### Theorem 4.2

Let $$(X, d, \preceq)$$ be a partially ordered metric space and let T be a self-mapping on X. Suppose that the following assertions hold:

1. (i)

T is a nondecreasing map.

2. (ii)

There exists $$x_{0} \in X$$ such that $$x_{0} \preceq Tx_{0}$$.

3. (iii)

There exist a number $$\tau> 0$$, $$L \geq0$$, and $$F \in \mathcal{F}$$ such that

\begin{aligned}& d(Tx, Ty) > 0 \quad \Rightarrow\quad \tau+ F\bigl(d(Tx, Ty)\bigr) \leq F\bigl[d(x, y) + L d(y, Tx)\bigr] \quad \textit{and} \\& d(Tx, Ty) > 0\quad \Rightarrow\quad \tau+ F\bigl(d(Tx, Ty)\bigr) \leq F\bigl[d(x, y) + L d(x, Ty)\bigr] \end{aligned}

for all $$x, y \in X$$ with $$x\preceq y$$.

4. (iv)

Either for any sequence $$\{ x_{n}\}\subset X$$ and $$x \in X$$ with

\begin{aligned}& x_{n} \to x, \quad \textit{as } n \to\infty \quad \textit{and} \\& x_{n} \preceq x_{n+1}, \quad \forall n \in\mathbb{N}\cup \{0\}, \quad \textit{we have} \quad x_{n}\preceq x, \end{aligned}

or T is continuous. Then T has a fixed point.

### Corollary 4.1



Let $$(X, d, \preceq)$$ be a partially ordered complete metric space and let $$T: X \to X$$ be a continuous, nondecreasing self-mapping such that $$x_{0} \preceq Tx_{0}$$ for some $$x_{0} \in X$$. Assume that

$$d(Tx, Ty) \leq r d(x,y)$$

holds for all $$x,y \in X$$ with $$x \preceq y$$ where $$0 \leq r < 1$$. Then T has a fixed point.

## Some applications to fractional calculus

First, let us recall some basic definitions of fractional calculus (see in ). For a continuous function $$g: [0, \infty) \rightarrow\mathbb{R}$$, the Caputo derivative of fractional order β is defined as

$$^{C}D^{\beta}\bigl(g(t)\bigr) = \frac{1}{\Gamma ( n-\beta )} \int_{0}^{t} (t-s)^{n-\beta-1} g^{n}(s) \,ds \quad \bigl(n-1< \beta< n, n = [\beta] +1\bigr)$$

where $$[\beta]$$ denotes the integer part of the real number β and Γ is a gamma function.

In this section, we present an application of Theorem 3.2 in establishing the existence of solutions for a nonlinear fractional differential equation:

$${}^{C}D^{\beta}\bigl(x(t)\bigr)+ f\bigl(t,x(t) \bigr)= 0\quad (0\leq t \leq1, \beta< 1)$$
(5.1)

via the boundary conditions $$x(0) = 0 = x(1)$$, where $$x \in C([0,1], \mathbb{R})$$ ($$C([0,1], \mathbb{R})$$ is the set of all continuous functions from $$[0, 1]$$ into $$\mathbb{R}$$), $$^{C}D^{\beta}$$ denotes the Caputo fractional derivative of order β, and $$f: [0,1] \times\mathbb{R} \rightarrow\mathbb{R}$$ is continuous function (see ). Recall that the Green function associated to the problem (5.1) is given by

$$G(t,s) =\textstyle\begin{cases}(t(1-s))^{\alpha-1}-(t-s)^{\alpha-1} &\text{if } 0\le s\le t\le1,\\ \frac{(t(1-s))^{\alpha-1}}{\Gamma(\alpha)} &\text{if } 0\le t\le s\le1. \end{cases}$$

Now, we prove the following existence theorem.

### Theorem 5.1

Consider the nonlinear fractional differential equation (5.1). Let $$\zeta:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$$ be a given function. Suppose that the following conditions hold:

1. (i)

$$\vert f(t,a)-f(t,b)\vert \le e^{-\tau} \vert a-b\vert$$ ($$\tau>0$$) for all $$t\in [0,1]$$ and $$a,b\in\mathbb{R}$$ with $$\zeta(a,b)\ge0$$.

2. (ii)

There exists $$x_{0}\in C([0,1],\mathbb{R})$$ such that $$\zeta (x_{0}(t),\int_{0}^{1}Tx_{0}(t)\,dt)\ge0$$ for all $$t\in[0,1]$$ where $$T:C([0,1],\mathbb{R})\to C([0,1],\mathbb{R})$$ is defined by

$$Tx(t)= \int_{0}^{1}G(t,s)f\bigl(s,x(s)\bigr)\,ds$$

for all $$t\in[0,1]$$.

3. (iii)

For each $$t\in[0,1]$$ and $$x,y\in C([0,1],\mathbb{R})$$, $$\zeta(x(t),y(t))> 0$$ implies $$\zeta(Tx(t),Ty(t))>0$$.

4. (iv)

For each $$t\in[0,1]$$, if $$\{x_{n}\}$$ is a sequence in $$C([0,1],\mathbb{R})$$ such that $$x_{n}\to x$$ in $$C([0,1],\mathbb{R})$$ and $$\zeta(x_{n}(t),x_{n+1}(t))>0$$ for all $$n\in\mathbb{N}$$, then $$\zeta (x_{n}(t),x(t))>0$$ for all $$n\in\mathbb{N}$$.

Then the problem (5.1) has at least one solution.

### Proof

First of all, we let $$X=C([0,1],\mathbb{R})$$. It is well known that X is a Banach space endowed with the supremum norm $$\Vert x\Vert _{\infty}=\sup_{t\in[0,1]} \vert x(t)\vert$$ for all $$x\in X$$. It is easy to see that $$x\in X$$ is a solution of (5.1) if and only if $$x\in X$$ is a solution of the equation $$x(t)=\int_{0}^{1}G(t,s)f(s,x(s))\,ds$$ for all $$t\in[0,1]$$. Then the problem (5.1) is equivalent to finding $$x^{*}\in X$$ which is a fixed point of T.

Now let $$x,y\in X$$ such that $$\zeta(x(t),y(t))\ge0$$ for all $$t\in [0,1]$$. By (i) we have

\begin{aligned} \bigl\vert Tx(t)-Ty(t)\bigr\vert =&\biggl\vert \int_{0}^{1}G(t,s)\bigl[f\bigl(s,x(s) \bigr)-f\bigl(s,y(s)\bigr)\bigr]\,ds \biggr\vert \\ \le& \int_{0}^{1}G(t,s)\bigl\vert f\bigl(s,x(s) \bigr)-f\bigl(s,y(s)\bigr)\bigr\vert \,ds \\ \le& \int_{0}^{1}G(t,s)\cdot e^{-\tau}\bigl(\bigl\vert x(s)-y(s)\bigr\vert \bigr)\,ds \\ \le& e^{-\tau} \Vert x-y\Vert _{\infty}\sup _{t\in I} \int_{0}^{1}G(t,s)\,ds \\ \le& e^{-\tau} \Vert x-y\Vert _{\infty}. \end{aligned}

Thus for each $$x,y\in X$$, with $$\zeta(x(t),y(t))>0$$ for all $$t\in[0,1]$$ we have

$$\Vert Tx-Ty\Vert _{\infty}\le e^{-\tau} \Vert x-y\Vert _{\infty}\quad \text{or}\quad d(Tx,Ty)\le e^{-\tau}d(x,y).$$

By passing through a logarithm, we write $$\ln d(Tx,Ty)\le\ln(e^{-\tau }d(x,y))$$ and hence $$\tau+\ln d(Tx,Ty)\le\ln d(x,y)$$.

Now consider the function $$F:\mathbb{R^{+}}\to\mathbb{R}$$ defined by $$F(u)=\ln u$$ for each $$u\in X$$, then $$F\in\mathcal{F}$$. Also, define $$\alpha:X\times X \to\{-\infty\}\cup(0,\infty)$$ by

$$\alpha(x,y) =\textstyle\begin{cases}1 &\text{if } \zeta(x(t),y(t))>0, t\in[0,1],\\ -\infty &\mbox{otherwise}. \end{cases}$$

Therefore,

\begin{aligned}& \tau+\alpha(x,y)F\bigl(d(Tx,Ty)\bigr)\le F\bigl(d(x,y)\bigr)\le F \bigl[d(x,y)+Ld(y,Tx)\bigr] \quad \text{and} \\& \tau+\alpha(x,y)F\bigl(d(Tx,Ty)\bigr)\le F\bigl(d(x,y)\bigr)\le F \bigl[d(x,y)+Ld(x,Ty)\bigr] \end{aligned}

for all $$x,y\in X$$ with $$d(Tx,Ty)>0$$ and $$L\ge0$$. This implies that T is an α-type almost-F-contraction. From (ii) there exists $$x_{0}\in X$$ such that $$\alpha(x_{0},Tx_{0})\ge1$$. Next, by using (iii), we get the following assertions holding for all $$x,y\in X$$

\begin{aligned} \alpha(x,y)\ge1 \quad &\implies\quad \zeta\bigl(x(t),y(t)\bigr)>0\quad \text{for all } t\in [0,1] \\ &\implies \quad \zeta\bigl(Tx(t),Ty(t)\bigr)>0\quad \text{for all } t\in[0,1] \\ &\implies \quad \alpha(Tx,Ty)\ge1; \end{aligned}

hence T is α-admissible. Finally, from condition (iv) in the hypothesis, condition (iii) of Theorem 3.1 holds. Therefore, as an application of Theorem 3.1 we conclude to the existence of $$x^{*}\in X$$ such that $$x^{*}=Tx^{*}$$ and so $$x^{*}$$ is a solution of the problem (5.1). This completes the proof. □

## Conclusions

In the present work we introduced the new concepts of an α-type almost-F-contraction and an α-type F-Suzuki contraction, which are generalizations of the concepts given in [14, 15]. Next, we established some fixed point theorems for these contractions. Further, the attached examples and an application to a nonlinear fractional differential equation illustrate the usability of the obtained results.

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## Acknowledgements

The authors thank the referees for many perceptive comments that led to improvement of the work.

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Correspondence to Poom Kumam.

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The authors declare that they have no competing interests.

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All authors contributed equally and significantly in writing this article. All authors read and approved final manuscript.

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