Approximating Fixed Points via Hybrid Enriched Contractions in Convex Metric Space with an Application

Abstract

In the present study, we define hybrid enriched contractions of the Hardy–Rogers type and of the Ćirić–Reich–Rus type in the framework of convex metric space. We demonstrate the presence and the approximation of fixed points for contraction mappings by using Krasnoselskij iteration. The main conclusions of this study are shown to be corollaries or implications of multiple important fixed point theory findings. Some examples have also been provided to show the validity of our results. Towards the end of this paper, we study the solution of the nonlinear equations as an application of our main results.

1. Introduction

In 1922, Banach established the Banach contraction principle [1], which is a fundamental outcome of fixed point theory on metric spaces. Banach contraction mapping can now be improved in a number of ways, including investigating spaces with more structure than metric spaces, constructing new nonlinear mappings and establishing fixed point results for these new mappings [2,3,4,5,6].

In order to investigate the fixed point problem for nonexpansive mappings, Takahashi [7] introduced the concept of convexity in metric space in 1970. This concept of convexity has been used as a fundamental tool for proving various fixed point problems [8,9,10,11,12].

In 2018, Karapinar [13] developed the concept of an interpolative-type contraction by incorporating the metric fixed point theory with interpolation theory. In 2021, Noorwali et al. [14] studied a hybrid contraction [15,16,17,18,19], which is a combination of the Jaggi contraction [20], the Suzuki contraction [21], and an interpolative type of contraction.

In recent studies [22,23,24,25,26], various authors have used the idea of enriched contraction mapping to generalize multiple contraction mappings within the framework of convex metric space.

The primary goal of the current work is to extend some of the above-mentioned findings for more general case, i.e., that of convex metric space.

We first present some definitions and lemmas that will be helpful in the proving of our main results.

Definition 1 

([7]). Let $(H,\mathfrak{h})$ be a metric space. A continuous mapping $W:H\times H\times [0,1]\to H$ is said to have a convex structure if there exists ${\mathfrak{u}}_1,{\mathfrak{u}}_2$ and $\mu \in [0,1]$ such that

$$\mathfrak{h}({\mathfrak{u}}_3,W({\mathfrak{u}}_1,{\mathfrak{u}}_2,\mu ))\le \mu \mathfrak{h}({\mathfrak{u}}_3,{\mathfrak{u}}_1)+(1-\mu )\mathfrak{h}({\mathfrak{u}}_3,{\mathfrak{u}}_2)$$

(1)

for any ${\mathfrak{u}}_3\in H$.

A metric space $(H,\mathfrak{h})$ endowed with a convex structure W is called a convex metric space and is denoted by $(H,\mathfrak{h},W)$.

Zheng et al. [6] introduced the following set of functions to define the concept of $\theta$-$\varphi$ contractions.

Let $\mathrm{\Theta }$ be the set of all mappings $\theta :(0,\infty )\to (1,\infty )$ satisfying the following conditions:

(i)

$\theta$ is non-decreasing.

(ii)

For each sequence $\left\{s_n\right\}$⊂$(0,\infty )$, ${\displaystyle \underset{n\to \infty }{lim}}$$\theta (s_n)$$=1$ iff ${\displaystyle \underset{n\to \infty }{lim}}$$s_n$$=0^{+}.$

(iii)

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

Let $\mathrm{\Phi }$ be the collection of mappings $\varphi :[1,\infty )\to [1,\infty )$ satisfying the following conditions:

(i)

$\varphi$ is non-decreasing.

(ii)

For each $h>1$, ${\displaystyle \underset{n\to \infty }{lim}{\varphi }^n(h)=1.}$

(iii)

$\varphi$ is continuous on $[1,\infty )$.

Lemma 1 

([6]). Let $\varphi \in \mathrm{\Phi }$. Then, $\varphi (1)=1$ and $\varphi (\mathfrak{u})<\mathfrak{u}$ for each $\mathfrak{u}>1.$

In 2018, Karapinar et al. [27] introduced the idea of interpolative Hardy–Rogers-type contraction mapping as:

Definition 2. 

Let $(H,\mathfrak{h})$ be a metric space. A mapping $R:H\to H$ is said to be an interpolative Hardy–Rogers-type contraction mapping if there exist some $\nu \in [0,1)$ and $d_1,d_2,d_3\in (0,1)$ with $d_1+d_2+d_3<1$, such that

$$\mathfrak{h}(R{\mathfrak{u}}_1,R{\mathfrak{u}}_2)\le \nu {[\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)]}^{d_1}.{[\mathfrak{h}({\mathfrak{u}}_1,R{\mathfrak{u}}_1)]}^{d_2}.{[\mathfrak{h}({\mathfrak{u}}_2,R{\mathfrak{u}}_2)]}^{d_3}.{\left[{\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_1,R{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_2,R{\mathfrak{u}}_1))\right]}^{1-d_1-d_2-d_3}$$

for all ${\mathfrak{u}}_1,{\mathfrak{u}}_2\in H\setminus Fix(R)$.

Debnath and de La Sen [28], in 2019, introduced the notion of interpolative Ćirić–Reich–Rus contraction mapping as:

Definition 3. 

Let $(H,\mathfrak{h})$ be a metric space. A mapping $R:H\to H$ is said to be an interpolative Ćirić–Reich–Rus type of contraction mapping, if there exist some $\nu \in [0,1)$ and $d_1,d_2\in (0,1)$ with $d_1+d_2<1$, such that

$$\mathfrak{h}(R{\mathfrak{u}}_1,R{\mathfrak{u}}_2)\le \nu {[\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)]}^{d_1}.{[\mathfrak{h}({\mathfrak{u}}_1,R{\mathfrak{u}}_1)]}^{d_2}.{[\mathfrak{h}({\mathfrak{u}}_2,R{\mathfrak{u}}_2)]}^{1-d_1-d_2}$$

for all ${\mathfrak{u}}_1,{\mathfrak{u}}_2\in H\setminus Fix(R).$

2. Main Results

This section aims to define the concepts of hybrid enriched contractions in the framework of convex metric space.

2.1. Hybrid Enriched Contraction of a Ćirić–Reich–Rus Type

In the present subsection, we define the concept of the hybrid enriched contraction of the Ćirić–Reich–Rus type. We use a Krasnoselskij-type iterative process [29] to approximate the fixed point of such a kind of hybrid enriched contraction.

Definition 4. 

Let $(H,\mathfrak{h},W)$ be a convex metric space. A mapping $R:H\to H$ is said to be a hybrid enriched contraction of the Ćirić–Reich–Rus type if there exist θ∈Θ, ϕ∈Φ and $\mu \in [0,1)$ such that

$$\theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2))\le \varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))],$$

(2)

for all distinct ${\mathfrak{u}}_1,{\mathfrak{u}}_2\in H$, $\widehat{R}\mathfrak{u}=W(\mathfrak{u},R\mathfrak{u},\mu )$ and

$$N({\mathfrak{u}}_1,{\mathfrak{u}}_2)=\left\{\begin{array}{cc}{[d_1{(\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1))}^c+d_3{(\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2))}^c]}^{{\displaystyle \frac{1}{c}}},\hfill & forc>0\hfill \\ \\ ({(\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2))}^{d_1}.{(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1))}^{d_2}.{(\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2))}^{d_3},\hfill & forc=0,{\mathfrak{u}}_1,{\mathfrak{u}}_2\in H\setminus F(\widehat{R})\hfill \end{array}\right.$$

where $c\ge 0$, $d_i>0$, $i=1,2,3$ with $d_1+d_2+d_3=1$ and $F(\widehat{R})$ = $\{\mathfrak{u}\in H:\widehat{R}\mathfrak{u}=\mathfrak{u}\}$.

Theorem 1. 

Let $(H,\mathfrak{h},W)$ be a complete convex metric space and let $R:H\to H$ be a mapping such that R is a hybrid enriched contraction of the Ćirić–Reich–Rus type. Then, R has a fixed point.

Proof. 

Let us choose an element ${\mathfrak{u}}_0$ from the set H. To generate the sequence ${\mathfrak{u}}_n$, let $\widehat{R}{\mathfrak{u}}_n={\mathfrak{u}}_{n+1}$, for every $n\in \mathbb{N}$. If there exists an $n\in \mathbb{N}$ such that ${\mathfrak{u}}_n={\mathfrak{u}}_{n+1}$, then essentially $\mathfrak{u}={\mathfrak{u}}_n$ becomes a fixed point of $\widehat{R}$. However, for the rest of our discussion, we assume that ${\mathfrak{u}}_{n+1}\ne {\mathfrak{u}}_n$ for every $n\in \mathbb{N}$, which implies that $\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})>0$ for all $n\in \mathbb{N}$.

We will examine the two distinct instances of $c=0$ and $c>0$ to demonstrate the assumption of the theorem.

Case I: Let $c>0$. Using (2) with ${\mathfrak{u}}_1={\mathfrak{u}}_{n-1}$ and ${\mathfrak{u}}_2={\mathfrak{u}}_n,$ we get

$$\begin{array}{cc}\theta (\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))\hfill & =\theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_{n-1},\widehat{R}{\mathfrak{u}}_n))\hfill \\ \hfill & \le \varphi [\theta (N({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n))]\hfill \\ \hfill & =\varphi [\theta ({[d_1{(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_{n-1},\widehat{R}{\mathfrak{u}}_{n-1}))}^c+d_3{(\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}{\mathfrak{u}}_n))}^c]}^{{\displaystyle \frac{1}{c}}})]\hfill \\ \hfill & =\varphi [\theta ({[d_1{(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n))}^c+d_3{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c]}^{{\displaystyle \frac{1}{c}}})].\hfill \end{array}$$

(3)

Assume that $\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})>\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)$, then the use of (3) gives

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \theta (\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))& \le \varphi [\theta ({[d_1{(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n))}^c+d_3{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c]}^{{\displaystyle \frac{1}{c}}})]\hfill \\ \hfill & <\varphi [\theta ({[d_1{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c+d_3{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c]}^{{\displaystyle \frac{1}{c}}})]\hfill \\ \hfill & =\varphi [\theta ({[(d_1+d_2+d_3){(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c]}^{{\displaystyle \frac{1}{c}}})]\hfill \\ \hfill & =\varphi [\theta ({[{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c]}^{{\displaystyle \frac{1}{c}}})]\hfill \\ \hfill & =\varphi [\theta ((\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})))]\hfill \\ \hfill & <\theta ((\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))),\hfill \end{array}\end{array}$$

which is a contradiction. Hence, $\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\le \mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n).$

Returning to inequality (3), it is easily observed that

$$\theta (\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))\le \varphi [\theta ((\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)))].$$

(4)

This leads us to the inductive conclusion that

$$\theta (\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))\le {\varphi }^n[\theta ((\mathfrak{h}({\mathfrak{u}}_0,{\mathfrak{u}}_1)))].$$

(5)

The definition of $\mathrm{\Theta }$ and $\mathrm{\Phi }$ gives us

$${\displaystyle \underset{n\to \infty }{lim}{\varphi }^n[\theta (\mathfrak{h}({\mathfrak{u}}_0,{\mathfrak{u}}_1))]=1,}$$

which implies

$${\displaystyle \underset{n\to \infty }{lim}\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})=0.}$$

Now, our goal is to establish that $\left\{{\mathfrak{u}}_n\right\}$ is a Cauchy sequence. Assuming, on the contrary, that there exist $ϵ>0$ and corresponding subsequences $n_k$ and $m_k$ of $\mathbb{N}$ such that $n_k>m_k>k$ and:

$$\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k})\ge ϵ,$$

(6)

where $n_k$ and $m_k$ are selected as the smallest integers that satisfy Equation (6), that is

$$\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k-1})<ϵ.$$

(7)

By (6) and (7), and the triangular inequality, it is easily observed that

$$\begin{array}{cc}ϵ\le \mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k})\hfill & \le \mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k-1})+\mathfrak{h}({\mathfrak{u}}_{n_k-1},{\mathfrak{u}}_{n_k})\hfill \\ \hfill & <ϵ+\mathfrak{h}({\mathfrak{u}}_{n_k-1},{\mathfrak{u}}_{n_k}).\hfill \end{array}$$

(8)

Using (8), we have

$${\displaystyle \underset{k\to \infty }{lim}\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k})=ϵ.}$$

(9)

By using (2), we get

$$\begin{array}{cc}\hfill \theta (\mathfrak{h}({\mathfrak{u}}_{m_k+1},{\mathfrak{u}}_{n_k+1}))& =\theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_{m_k},\widehat{R}{\mathfrak{u}}_{n_k}))\hfill \\ \hfill & \le \varphi [\theta ((N({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k})))]\hfill \\ \hfill & =\varphi [\theta ({[d_1{(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k}))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{m_k+1}))}^c+d_3{(\mathfrak{h}({\mathfrak{u}}_{n_k},{\mathfrak{u}}_{n_k+1}))}^c]}^{{\displaystyle \frac{1}{c}}})]\hfill \\ \hfill & <\theta ({[d_1{(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k}))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{m_k+1}))}^c+d_3{(\mathfrak{h}({\mathfrak{u}}_{n_k},{\mathfrak{u}}_{n_k+1}))}^c]}^{{\displaystyle \frac{1}{c}}})\hfill \\ \hfill \mathfrak{h}({\mathfrak{u}}_{m_k+1},{\mathfrak{u}}_{n_k+1})& <{[d_1{(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k}))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{m_k+1}))}^c+d_3{(\mathfrak{h}({\mathfrak{u}}_{n_k},{\mathfrak{u}}_{n_k+1}))}^c]}^{{\displaystyle \frac{1}{c}}}.\hfill \end{array}$$

(10)

Using limit $k\to \infty$ in Equation (10),

$$ϵ<d_1^{{\displaystyle \frac{1}{c}}}.ϵ$$

which is a contradiction as $d_1^{{\displaystyle \frac{1}{c}}}<1.$ Hence, $ϵ=0.$ Therefore, $\left\{{\mathfrak{u}}_n\right\}$ is a Cauchy sequence.

As $(H,\mathfrak{h},W)$ is a complete convex metric space, so there exists $\mathfrak{u}\in H$ such that the sequence $\left\{{\mathfrak{u}}_n\right\}$ converges to $\mathfrak{u}$ and

$${\displaystyle \underset{n\to \infty }{lim}\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u})=0.}$$

(11)

Using Equation (2) with ${\mathfrak{u}}_1={\mathfrak{u}}_n$ and ${\mathfrak{u}}_2=\mathfrak{u}$.

$$\begin{array}{cc}\hfill \theta (\mathfrak{h}({\mathfrak{u}}_{n+1},\widehat{R}\mathfrak{u}))& =\theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_n,\widehat{R}\mathfrak{u}))\hfill \\ \hfill & =\varphi [\theta ({[d_1{(\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u}))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c+d_3{(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u}))}^c]}^{{\displaystyle \frac{1}{c}}})]\hfill \\ \hfill & <\theta ({[d_1{(\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u}))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c+d_3{(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u}))}^c]}^{{\displaystyle \frac{1}{c}}})\hfill \\ \hfill \mathfrak{h}({\mathfrak{u}}_{n+1},\widehat{R}\mathfrak{u})& <{[d_1{(\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u}))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c+d_3{(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u}))}^c]}^{{\displaystyle \frac{1}{c}}}.\hfill \end{array}$$

(12)

Assume that $\mathfrak{u}\ne \widehat{R}\mathfrak{u}$.

$$\begin{array}{cc}0<\mathfrak{h}(\widehat{R}\mathfrak{u},\mathfrak{u})\hfill & \le \mathfrak{h}(\widehat{R}\mathfrak{u},{\mathfrak{u}}_{n+1})+\mathfrak{h}({\mathfrak{u}}_{n+1},\mathfrak{u})\hfill \\ \hfill & \le {[d_1{(\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u}))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c+d_3{(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u}))}^c]}^{{\displaystyle \frac{1}{c}}}+\mathfrak{h}({\mathfrak{u}}_{n+1},\mathfrak{u}).\hfill \end{array}$$

(13)

Take $n\to \infty$ in the above equation,

$$\mathfrak{h}(\widehat{R}\mathfrak{u},\mathfrak{u})\le {(d_3)}^{{\displaystyle \frac{1}{c}}}\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u}),$$

which is a contradiction as ${(d_3)}^{{\displaystyle \frac{1}{c}}}<1.$ Hence, $\mathfrak{u}=\widehat{R}\mathfrak{u}$.

Case II: Let c = 0. Using (2) with ${\mathfrak{u}}_1={\mathfrak{u}}_{n-1}$ and ${\mathfrak{u}}_2={\mathfrak{u}}_n,$ we get

$$\begin{array}{cc}\hfill \theta (\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))& =\theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_{n-1},\widehat{R}{\mathfrak{u}}_n))\hfill \\ \hfill & \le \varphi [\theta (N({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n))]\hfill \\ \hfill & =\varphi [\theta ({(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n))}^{d_1}.{(\mathfrak{h}({\mathfrak{u}}_{n-1},\widehat{R}{\mathfrak{u}}_{n-1}))}^{d_2}.{(\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}{\mathfrak{u}}_n))}^{d_3})]\hfill \\ \hfill & <\theta ({(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n))}^{d_1}.{(\mathfrak{h}({\mathfrak{u}}_{n-1},\widehat{R}{\mathfrak{u}}_{n-1}))}^{d_2}.{(\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}{\mathfrak{u}}_n))}^{d_3})\hfill \\ \hfill \mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})& <{(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n))}^{d_1}.{(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n))}^{d_2}.{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^{d_3}\hfill \\ \hfill {(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^{1-d_3}& <{(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n))}^{d_1+d_2}\hfill \\ \hfill \mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})& <\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n).\hfill \end{array}$$

(14)

By employing the same techniques as those in the case of $c>0,$ we quickly see that $\left\{{\mathfrak{u}}_n\right\}$ forms a Cauchy sequence. Consequently, there exist $\mathfrak{u}\in H$ such that ${\displaystyle \underset{n\to \infty }{lim}\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u})=0.}$

Use of Equation (2) with ${\mathfrak{u}}_1={\mathfrak{u}}_n$ and ${\mathfrak{u}}_2=\mathfrak{u}$, gives

$$\begin{array}{cc}\hfill \theta (\mathfrak{h}({\mathfrak{u}}_{n+1},\widehat{R}\mathfrak{u}))& =\theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_n,\widehat{R}\mathfrak{u}))\hfill \\ \hfill & =\varphi [\theta ([{(\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u}))}^{d_1}.{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^{d_2}.{(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u}))}^{d_3}])]\hfill \\ \hfill & <\theta ([{(\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u}))}^{d_1}.{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^{d_2}.{(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u}))}^{d_3}])\hfill \\ \hfill \mathfrak{h}({\mathfrak{u}}_{n+1},\widehat{R}\mathfrak{u})& <{(\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u}))}^{d_1}.{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^{d_2}.{(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u}))}^{d_3}.\hfill \end{array}$$

(15)

Assume that $\mathfrak{u}\ne \widehat{R}\mathfrak{u}.$

$$\begin{array}{cc}0<\mathfrak{h}(\widehat{R}\mathfrak{u},\mathfrak{u})\hfill & \le \mathfrak{h}(\widehat{R}\mathfrak{u},{\mathfrak{u}}_{n+1})+\mathfrak{h}({\mathfrak{u}}_{n+1},\mathfrak{u})\hfill \\ \hfill & \le {(\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u}))}^{d_1}.{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^{d_2}.{(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u}))}^{d_3}+\mathfrak{h}({\mathfrak{u}}_{n+1},\mathfrak{u}).\hfill \end{array}$$

(16)

Take $n\to \infty$ in the above equation

$$\mathfrak{h}(\widehat{R}\mathfrak{u},\mathfrak{u})\le 0,$$

Hence, $\mathfrak{u}=\widehat{R}\mathfrak{u}$.

By the use of Lemma 4 of [24], we have $R\mathfrak{u}=\mathfrak{u}.$ □

We now present an example to show the validity of this result.

Example 1. 

Let H = $[0,2]$ with the metric $\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)$ = $|{\mathfrak{u}}_1-{\mathfrak{u}}_2|$ and let $W({\mathfrak{u}}_1,{\mathfrak{u}}_2,\mu )=\mu {\mathfrak{u}}_1+(1-\mu ){\mathfrak{u}}_2.$ Assume that the mapping R is defined as

$$R(\mathfrak{u})=\left\{\begin{array}{cc}{\displaystyle \frac{3-\mathfrak{u}}{8}}\hfill & for\mathfrak{u}\in [0,1]\hfill \\ \\ 0\hfill & for\mathfrak{u}\in (1,2]\hfill \end{array}\right.$$

Since the mapping R is not continuous, it is therefore neither a Banach contraction nor is it an enriched contraction. But the mapping R satisfies the conditions of Theorem 2 with $\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)=|{\mathfrak{u}}_1-{\mathfrak{u}}_2|$, $c=1$,$d_1=d_2=d_3={\displaystyle \frac{1}{3}}$, $\theta (\mathfrak{u})=2^{\mathfrak{u}}$ for all $\mathfrak{u}\in [0,+\infty ),$ and $\varphi (\mathfrak{u})={\mathfrak{u}}^{0.9}$ for all $\mathfrak{u}\in [1,+\infty ).$

For $\mu ={\displaystyle \frac{1}{9}}$, $\widehat{R}$ is defined as

$$\widehat{R}(\mathfrak{u})=\left\{\begin{array}{cc}{\displaystyle \frac{1}{3}}\hfill & for\mathfrak{u}\in [0,1]\hfill \\ \\ {\displaystyle \frac{\mathfrak{u}}{9}}\hfill & for\mathfrak{u}\in (1,2]\hfill \end{array}\right.$$

Case I For ${\mathfrak{u}}_1,{\mathfrak{u}}_2\in [0,1]$,

$$\theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2))=\theta \left(\mathfrak{h}\left({\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{3}}\right)\right)=1$$

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))]& =\varphi \left[\theta \left(\left[{\displaystyle \frac{1}{3}}\left(\mathfrak{h}\left({\mathfrak{u}}_1,{\mathfrak{u}}_2\right)\right)+{\displaystyle \frac{1}{3}}\left(\mathfrak{h}\left({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1\right)\right)+{\displaystyle \frac{1}{3}}\left(\mathfrak{h}\left({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2\right)\right)\right]\right)\right]\hfill \\ \hfill & =\varphi \left[\theta \left(\left[{\displaystyle \frac{1}{3}}\left(|{\mathfrak{u}}_1-{\mathfrak{u}}_2|\right)+{\displaystyle \frac{1}{3}}\left(|{\mathfrak{u}}_1-{\displaystyle \frac{1}{3}}|\right)+{\displaystyle \frac{1}{3}}\left(|{\mathfrak{u}}_2-{\displaystyle \frac{1}{3}}|\right)\right]\right)\right]\hfill \\ \hfill & >\varphi [\theta (0)]=1\hfill \end{array}\end{array}$$

which implies

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))]& >1\hfill \\ \hfill -\varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))]& <-1\hfill \\ \hfill \theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2))-\varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))]& <0\hfill \\ \hfill \theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2))& <\varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))].\hfill \end{array}\end{array}$$

Case II For ${\mathfrak{u}}_1,{\mathfrak{u}}_2\in (1,2]$,

$$\theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2))=\theta \left(|{\displaystyle \frac{{\mathfrak{u}}_1}{9}}-{\displaystyle \frac{{\mathfrak{u}}_2}{9}}|\right)<\theta \left({\displaystyle \frac{1}{9}}\right)=1.080$$

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))]& =\varphi \left[\theta \left(\left[{\displaystyle \frac{1}{3}}\left(\mathfrak{h}\left({\mathfrak{u}}_1,{\mathfrak{u}}_2\right)\right)+{\displaystyle \frac{1}{3}}\left(\mathfrak{h}\left({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1\right)\right)+{\displaystyle \frac{1}{3}}\left(\mathfrak{h}\left({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2\right)\right)\right]\right)\right]\hfill \\ \hfill & =\varphi \left[\theta \left(\left[{\displaystyle \frac{1}{3}}\left(|{\mathfrak{u}}_1-{\mathfrak{u}}_2|\right)+{\displaystyle \frac{1}{3}}\left(|{\mathfrak{u}}_1-{\displaystyle \frac{{\mathfrak{u}}_1}{9}}|\right)+{\displaystyle \frac{1}{3}}\left(|{\mathfrak{u}}_2-{\displaystyle \frac{{\mathfrak{u}}_2}{9}}|\right)\right]\right)\right]\hfill \\ \hfill & =\varphi \left[\theta \left(\left[{\displaystyle \frac{1}{3}}\left(|{\mathfrak{u}}_1-{\mathfrak{u}}_2|\right)+{\displaystyle \frac{1}{3}}\left(8.{\displaystyle \frac{{\mathfrak{u}}_1}{9}}\right)+{\displaystyle \frac{1}{3}}\left(8.{\displaystyle \frac{{\mathfrak{u}}_2}{9}}\right)\right]\right)\right]\hfill \\ \hfill & >\varphi \left[\theta \left(\left[{\displaystyle \frac{1}{3}}(0)+{\displaystyle \frac{1}{3}}\left(8.{\displaystyle \frac{1}{9}}\right)+{\displaystyle \frac{1}{3}}\left(8.{\displaystyle \frac{1}{9}}\right)\right]\right)\right]\hfill \\ \hfill & =\varphi \left[\theta \left(\left[{\displaystyle \frac{1}{3}}\left({\displaystyle \frac{8}{9}}\right)+{\displaystyle \frac{1}{3}}\left({\displaystyle \frac{8}{9}}\right)\right]\right)\right]\hfill \\ \hfill & =1.447\hfill \end{array}\end{array}$$

which implies

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))]& >1.447\hfill \\ \hfill -\varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))]& <-1.447\hfill \\ \hfill \theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2))-\varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))]& <-0.367<0\hfill \\ \hfill \theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2))& <\varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))]\hfill \end{array}\end{array}$$

Case III For ${\mathfrak{u}}_1\in [0,1]$ and ${\mathfrak{u}}_2\in (1,2]$,

$$\theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2))=\theta \left(\mathfrak{h}\left({\displaystyle \frac{1}{3}},{\displaystyle \frac{{\mathfrak{u}}_2}{9}}\right)\right)=\theta \left(|{\displaystyle \frac{1}{3}}-{\displaystyle \frac{{\mathfrak{u}}_2}{9}}|\right)<\theta \left({\displaystyle \frac{2}{9}}\right)=1.166$$

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))]& =\varphi \left[\theta \left(\left[{\displaystyle \frac{1}{3}}\left(\mathfrak{h}\left({\mathfrak{u}}_1,{\mathfrak{u}}_2\right)\right)+{\displaystyle \frac{1}{3}}\left(\mathfrak{h}\left({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1\right)\right)+{\displaystyle \frac{1}{3}}\left(\mathfrak{h}\left({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2\right)\right)\right]\right)\right]\hfill \\ \hfill & =\varphi \left[\theta \left(\left[{\displaystyle \frac{1}{3}}\left(|{\mathfrak{u}}_1-{\mathfrak{u}}_2|\right)+{\displaystyle \frac{1}{3}}\left(|{\mathfrak{u}}_1-{\displaystyle \frac{1}{3}}|\right)+{\displaystyle \frac{1}{3}}\left(|{\mathfrak{u}}_2-{\displaystyle \frac{{\mathfrak{u}}_2}{9}}|\right)\right]\right)\right]\hfill \\ \hfill & >\varphi \left[\theta \left(\left[{\displaystyle \frac{1}{3}}(0)+{\displaystyle \frac{1}{3}}(0)+{\displaystyle \frac{1}{3}}\left({\displaystyle \frac{8}{9}}\right)\right]\right)\right]\hfill \\ \hfill & =1.203\hfill \end{array}\end{array}$$

which implies

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))]& >1.203\hfill \\ \hfill -\varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))]& <-1.203\hfill \\ \hfill \theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2))-\varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))]& <-0.037<0\hfill \\ \hfill \theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2))& <\varphi [\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))].\hfill \end{array}\end{array}$$

Since $\widehat{R}$ satisfies all the conditions of Theorem 2, $\widehat{R}$ therefore has a fixed point and ${\displaystyle \frac{1}{3}}$ is the fixed point of $\widehat{R}$ (as shown in Figure 1). By using Lemma 4 of [24], we have $R\mathfrak{u}=\widehat{R}\mathfrak{u}={\displaystyle \frac{1}{3}}$ (as shown in Figure 2).

We now present the following results, which are direct consequences of Theorem 1.

Corollary 1. 

Let $(R,\mathfrak{h},W)$ be a complete convex metric space. Consider a function $R:H\to H$ such that $\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)\le aN({\mathfrak{u}}_1,{\mathfrak{u}}_2)$, for $a\in (0,1)$ and

$$N({\mathfrak{u}}_1,{\mathfrak{u}}_2)=\left\{\begin{array}{cc}{[d_1{(\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1))}^c+d_3{(\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2))}^c]}^{{\displaystyle \frac{1}{c}}},\hfill & forc>0\hfill \\ \\ ({(\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2))}^{d_1}.{(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1))}^{d_2}.{(\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2))}^{d_3},\hfill & forc=0,{\mathfrak{u}}_1,{\mathfrak{u}}_2\in H\setminus F(\widehat{R})\hfill \end{array}\right.$$

where $c\ge 0$, $d_i>0$, $i=1,2,3$ with $d_1+d_2+d_3=1$ and $F(\widehat{R})$ = $\{\mathfrak{u}\in H:\widehat{R}\mathfrak{u}=\mathfrak{u}\}$. Then, R has a fixed point.

Proof. 

Assume that $\theta (\mathfrak{u})=exp(\mathfrak{u})$ for all $\mathfrak{u}\in [0,+\infty ),$ and $\varphi (\mathfrak{u})={\mathfrak{u}}^a$ for all $\mathfrak{u}\in [1,+\infty )$ and $a\in (0,1).$ It is clear that $\theta \in \mathrm{\Theta }$, $\varphi \in \mathrm{\Phi }.$

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2))& =e^{\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)}\hfill \\ \hfill & \le e^{aN({\mathfrak{u}}_1,{\mathfrak{u}}_2)}\hfill \\ \hfill & ={\left(e^{\left(N({\mathfrak{u}}_1,{\mathfrak{u}}_2)\right)}\right)}^a\hfill \\ \hfill & =\varphi \left(\theta (N({\mathfrak{u}}_1,{\mathfrak{u}}_2))\right).\hfill \end{array}\end{array}$$

As a result, all requirements of Theorem 1 are met. Therefore, R has a fixed point. □

Corollary 2. 

Let $(R,\mathfrak{h},W)$ be a complete convex metric space. Let $R:H\to H$ be a mapping such that the following equation holds:

$$\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)\le {\displaystyle \frac{a}{3}}\left(\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1)+\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2)\right).$$

Then, R has a unique fixed point.

Proof. 

Put $c=1$ and $d_1=d_2=d_3={\displaystyle \frac{1}{3}}$ in Corollary 1, then the mapping

$$\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)\le {\displaystyle \frac{a}{3}}\left(\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1)+\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2)\right),$$

has a fixed point. Now, suppose that ${\mathfrak{u}}_1$ and ${\mathfrak{u}}_2$ are two distinct fixed points, then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)& \le {\displaystyle \frac{a}{3}}[\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1)+\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2)]\hfill \\ \hfill \mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)& \le {\displaystyle \frac{a}{3}}[\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_1)+\mathfrak{h}({\mathfrak{u}}_2,{\mathfrak{u}}_2)]\hfill \end{array}\end{array}$$

which is a contradiction. Hence, R has a unique fixed point. □

Corollary 3. 

Let $(R,\mathfrak{h},W)$ be a complete convex metric space. Let $R:H\to H$ be a mapping such that

$$\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)\le a{(\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2))}^{\mathfrak{b}}.{(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1))}^{\mathfrak{c}}.{(\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2))}^{1-\mathfrak{b}-\mathfrak{c}},$$

for $a,\mathfrak{b},\mathfrak{c}\in (0,1)$. Then, R has a fixed point.

Proof. 

Upon putting $c=0$ in Corollary 1 with $d_1=\mathfrak{b}$, $d_2=\mathfrak{c}$ and $d_3=1-\mathfrak{b}-\mathfrak{c}$, we get the result. □

Corollary 4 

([28]). Let $(R,\mathfrak{h})$ be a complete metric space. Let $R:H\to H$ be an interpolative Ćirić–Reich–Rus type of contraction mapping. Then, R has a fixed point.

Proof. 

By putting $c=0$, $d_3=1-d_1-d_2$ and $\mu =0$ in Corollary 1, we get the result. □

2.2. Hybrid Enriched Contraction of a Hardy–Rogers Type

The present subsection aims to define the concept of a hybrid enriched contraction of a Hardy–Rogers type and to study the fixed point result for such kind of contraction.

Definition 5. 

Let $(H,\mathfrak{h},W)$ be a complete convex metric space. A mapping $R:H\to H$ is said to be a hybrid enriched contraction of a Hardy–Rogers type if there exist θ∈Θ, ϕ∈Φ, and $\mu \in [0,1)$ such that

$$\theta (\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2))\le \varphi [\theta (M({\mathfrak{u}}_1,{\mathfrak{u}}_2))],$$

(17)

for all distinct ${\mathfrak{u}}_1,{\mathfrak{u}}_2\in H$, $\widehat{R}\mathfrak{u}=W(\mathfrak{u},R\mathfrak{u},\mu )$ and

$$M({\mathfrak{u}}_1,{\mathfrak{u}}_2)=\left\{\begin{array}{c}[d_1{(\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1))}^c+d_3{(\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2))}^c+\hfill \\ d_4{\left[{\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_1))\right]}^c{]}^{{\displaystyle \frac{1}{c}}},\hfill & forc>0\hfill \\ \\ {(\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2))}^{d_1}.{(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1))}^{d_2}.{(\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2))}^{d_3}.\hfill \\ {\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_1))\right)}^{d_4},\hfill & forc=0,{\mathfrak{u}}_1,{\mathfrak{u}}_2\in R\setminus F(\widehat{R})\hfill \end{array}\right.$$

where $c\ge 0$, $d_i>0$, $i=1,2,3,4$ with $d_1+d_2+d_3+d_4=1$ and $F(\widehat{R})$ = $\{\mathfrak{u}\in H:\widehat{R}\mathfrak{u}=\mathfrak{u}\}$.

Theorem 2. 

Let $(H,\mathfrak{h},W)$ be a complete convex metric space and let $R:H\to H$ be a mapping such that R is a hybrid enriched contraction of a Hardy–Rogers type mapping. Then, R has a fixed point.

Proof. 

Let us select an element ${\mathfrak{u}}_0$ from the set H. For each $n\in \mathbb{N}$, let $\widehat{R}{\mathfrak{u}}_n={\mathfrak{u}}_{n+1}$ to construct the sequence ${\mathfrak{u}}_n$. If there exist a $n\in \mathbb{N}$ such that ${\mathfrak{u}}_n={\mathfrak{u}}_{n+1}$, then $\mathfrak{u}={\mathfrak{u}}_n$ becomes a fixed point of $\widehat{R}$. For the remaining part of the discussion, we assume that ${\mathfrak{u}}_{n+1}\ne {\mathfrak{u}}_n$ for each $n\in \mathbb{N}$.

The assumptions of the theorem will be proven by looking at the two different cases of $c=0$ and $c>0$.

Case I: Let $c>0$. By making use of (17) with ${\mathfrak{u}}_1={\mathfrak{u}}_{n-1}$ and ${\mathfrak{u}}_2={\mathfrak{u}}_n,$ we get

$$\begin{array}{cc}\theta \left(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\right)\hfill & =\theta \left(\mathfrak{h}(\widehat{R}{\mathfrak{u}}_{n-1},\widehat{R}{\mathfrak{u}}_n)\right)\hfill \\ \hfill & \le \varphi \left[\theta \left(M({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)\right)\right]\hfill \\ \hfill & =\varphi \left[\theta \right([d_1{\left(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)\right)}^c+d_2{\left(\mathfrak{h}({\mathfrak{u}}_{n-1},\widehat{R}{\mathfrak{u}}_{n-1})\right)}^c+d_3{\left(\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}{\mathfrak{u}}_n)\right)}^c+\hfill \\ \hfill & d_4{\left[{\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_{n-1},\widehat{R}{\mathfrak{u}}_n)+\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}{\mathfrak{u}}_{n-1}))\right]}^c{]}^{{\displaystyle \frac{1}{c}}}\left)\right]\hfill \\ \hfill & =\varphi \left[\theta \right([d_1{\left(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)\right)}^c+d_2{\left(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)\right)}^c+d_3{\left(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\right)}^c\hfill \\ \hfill & d_4{\left[{\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_{n+1})+\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_n))\right]}^c{]}^{{\displaystyle \frac{1}{c}}}\left)\right].\hfill \end{array}$$

(18)

Using (18) and assuming that $\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})>\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)$, we obtain

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \theta (\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))& \le \varphi [\theta ([d_1{(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n))}^c+d_3{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c+\hfill \\ \hfill & d_4{\left[{\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)+\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))\right]}^c{]}^{{\displaystyle \frac{1}{c}}})]\hfill \\ \hfill & <\varphi [\theta ([d_1{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c+d_2{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c+d_3{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c+\hfill \\ \hfill & d_4{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c{]}^{{\displaystyle \frac{1}{c}}})]\hfill \\ \hfill & =\varphi [\theta ({[(d_1+d_2+d_3+d_4){(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c]}^{{\displaystyle \frac{1}{c}}})]\hfill \\ \hfill & =\varphi [\theta ({[{(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))}^c]}^{{\displaystyle \frac{1}{c}}})]\hfill \\ \hfill & =\varphi [\theta ((\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})))]\hfill \\ \hfill & <\theta ((\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))),\hfill \end{array}\end{array}$$

which is a contradiction. Hence, $\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\le \mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n).$

Going back to inequality (18), it is clear that

$$\theta \left(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\right)\le \varphi \left[\theta \left(\left(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)\right)\right)\right].$$

(19)

This brings us to the inductive conclusion that

$$\theta \left(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\right)\le {\varphi }^n\left[\theta \left(\left(\mathfrak{h}({\mathfrak{u}}_0,{\mathfrak{u}}_1)\right)\right)\right].$$

(20)

The definition of $\mathrm{\Theta }$ and $\mathrm{\Phi }$ gives us

$${\displaystyle \underset{n\to \infty }{lim}{\varphi }^n\left[\theta \left(\mathfrak{h}({\mathfrak{u}}_0,{\mathfrak{u}}_1)\right)\right]=1,}$$

which implies

$${\displaystyle \underset{n\to \infty }{lim}\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})=0.}$$

Now, our aim is to prove that $\left\{{\mathfrak{u}}_n\right\}$ is a Cauchy sequence. On the contrary, assume that there exist $ϵ>0$ and the subsequences $n_k$ and $m_k$ of $\mathbb{N}$ such that $n_k>m_k>k$ and

$$\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k})\ge ϵ,$$

(21)

where $n_k$ and $m_k$ are chosen as the smallest integers that fulfill Equation (21), that is

$$\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k-1})<ϵ.$$

(22)

By using (21) and (22), and the triangular inequality, it is easily observed that

$$\begin{array}{cc}ϵ\le \mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k})\hfill & \le \mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k-1})+\mathfrak{h}({\mathfrak{u}}_{n_k-1},{\mathfrak{u}}_{n_k})\hfill \\ \hfill & <ϵ+\mathfrak{h}({\mathfrak{u}}_{n_k-1},{\mathfrak{u}}_{n_k}).\hfill \end{array}$$

(23)

Using (23),

$${\displaystyle \underset{k\to \infty }{lim}\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k})=ϵ.}$$

(24)

By using (17),

$$\begin{array}{cc}\hfill \theta \left(\mathfrak{h}({\mathfrak{u}}_{m_k+1},{\mathfrak{u}}_{n_k+1})\right)& =\theta \left(\mathfrak{h}(\widehat{R}{\mathfrak{u}}_{m_k},\widehat{R}{\mathfrak{u}}_{n_k})\right)\hfill \\ \hfill & \le \varphi \left[\theta \left(\left(M({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k})\right)\right)\right]\hfill \\ \hfill & =\varphi \left[\theta \right([d_1{\left(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k})\right)}^c+d_2{\left(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{m_k+1})\right)}^c+d_3{\left(\mathfrak{h}({\mathfrak{u}}_{n_k},{\mathfrak{u}}_{n_k+1})\right)}^c+\hfill \\ \hfill & d_4{\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k+1})+\mathfrak{h}({\mathfrak{u}}_{n_k},{\mathfrak{u}}_{m_k+1}))\right)}^c{]}^{{\displaystyle \frac{1}{c}}}\left)\right]\hfill \\ \hfill & <\theta \left(\right[d_1{\left(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k})\right)}^c+d_2{\left(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{m_k+1})\right)}^c+d_3{\left(\mathfrak{h}({\mathfrak{u}}_{n_k},{\mathfrak{u}}_{n_k+1})\right)}^c+\hfill \\ \hfill & d_4{\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k})+\mathfrak{h}({\mathfrak{u}}_{n_k},{\mathfrak{u}}_{n_k+1})+\mathfrak{h}({\mathfrak{u}}_{n_k},{\mathfrak{u}}_{m_k})+\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{m_k+1}))\right)}^c{]}^{{\displaystyle \frac{1}{c}}})\hfill \\ \hfill \mathfrak{h}({\mathfrak{u}}_{m_k+1},{\mathfrak{u}}_{n_k+1})& <[d_1{\left(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k})\right)}^c+d_2{\left(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{m_k+1})\right)}^c+d_3{\left(\mathfrak{h}({\mathfrak{u}}_{n_k},{\mathfrak{u}}_{n_k+1})\right)}^c+\hfill \\ \hfill & d_4{\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{n_k})+\mathfrak{h}({\mathfrak{u}}_{n_k},{\mathfrak{u}}_{n_k+1})+\mathfrak{h}({\mathfrak{u}}_{n_k},{\mathfrak{u}}_{m_k})+\mathfrak{h}({\mathfrak{u}}_{m_k},{\mathfrak{u}}_{m_k+1}))\right)}^c{]}^{{\displaystyle \frac{1}{c}}}.\hfill \end{array}$$

(25)

Let $k\to \infty$ in Equation (25),

$$ϵ<{(d_1+d_4)}^{{\displaystyle \frac{1}{c}}}.ϵ$$

which is a contradiction, as ${(d_1+d_4)}^{{\displaystyle \frac{1}{c}}}<1.$ Hence, $ϵ=0.$ Therefore, sequence $\left\{{\mathfrak{u}}_n\right\}$ is a Cauchy sequence.

Since $(H,\mathfrak{h},W)$ is a complete convex metric space, there exists $\mathfrak{u}\in H$ such that the sequence $\left\{{\mathfrak{u}}_n\right\}$ converges to $\mathfrak{u}$ and

$${\displaystyle \underset{n\to \infty }{lim}\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u})=0.}$$

(26)

Using Equation (17) with ${\mathfrak{u}}_1={\mathfrak{u}}_n$ and ${\mathfrak{u}}_2=\mathfrak{u}$.

$$\begin{array}{cc}\hfill \theta \left(\mathfrak{h}({\mathfrak{u}}_{n+1},\widehat{R}\mathfrak{u})\right)& =\theta \left(\mathfrak{h}(\widehat{R}{\mathfrak{u}}_n,\widehat{R}\mathfrak{u})\right)\hfill \\ \hfill & =\varphi \left[\theta \right([d_1{\left(\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u})\right)}^c+d_2{\left(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\right)}^c+d_3{\left(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u})\right)}^c+\hfill \\ \hfill & d_4{\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}\mathfrak{u})+\mathfrak{h}(\mathfrak{u},{\mathfrak{u}}_{n+1}))\right)}^c{]}^{{\displaystyle \frac{1}{c}}}\left)\right]\hfill \\ \hfill & <\theta \left(\right[d_1{\left(\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u})\right)}^c+d_2{\left(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\right)}^c+d_3{\left(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u})\right)}^c+\hfill \\ \hfill & d_4{\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}\mathfrak{u})+\mathfrak{h}(\mathfrak{u},{\mathfrak{u}}_{n+1}))\right)}^c{]}^{{\displaystyle \frac{1}{c}}})\hfill \\ \hfill \mathfrak{h}({\mathfrak{u}}_{n+1},\widehat{R}\mathfrak{u})& <[d_1{\left(\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u})\right)}^c+d_2{\left(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\right)}^c+d_3{\left(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u})\right)}^c+\hfill \\ \hfill & d_4{\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}\mathfrak{u})+\mathfrak{h}(\mathfrak{u},{\mathfrak{u}}_{n+1}))\right)}^c{]}^{{\displaystyle \frac{1}{c}}}.\hfill \end{array}$$

(27)

Assume that $\mathfrak{u}\ne \widehat{R}\mathfrak{u}$.

$$\begin{array}{cc}0<\mathfrak{h}(\widehat{R}\mathfrak{u},\mathfrak{u})\hfill & \le \mathfrak{h}(\widehat{R}\mathfrak{u},{\mathfrak{u}}_{n+1})+\mathfrak{h}({\mathfrak{u}}_{n+1},\mathfrak{u})\hfill \\ \hfill & <[d_1{\left(\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u})\right)}^c+d_2{\left(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\right)}^c+d_3{\left(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u})\right)}^c+\hfill \\ \hfill & d_4{\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}\mathfrak{u})+\mathfrak{h}(\mathfrak{u},{\mathfrak{u}}_{n+1}))\right)}^c{]}^{{\displaystyle \frac{1}{c}}}+\mathfrak{h}({\mathfrak{u}}_{n+1},\mathfrak{u})\hfill \\ \hfill & <[d_1{\left(\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u})\right)}^c+d_2{\left(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\right)}^c+d_3{\left(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u})\right)}^c+\hfill \\ \hfill & d_4{\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u})+\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u})+\mathfrak{h}(\mathfrak{u},{\mathfrak{u}}_{n+1}))\right)}^c{]}^{{\displaystyle \frac{1}{c}}}+\mathfrak{h}({\mathfrak{u}}_{n+1},\mathfrak{u}).\hfill \end{array}$$

(28)

Take $n\to \infty$ in the above equation,

$$\mathfrak{h}(\widehat{R}\mathfrak{u},\mathfrak{u})\le {\left(d_3+{\displaystyle \frac{d_4}{2^c}}\right)}^{{\displaystyle \frac{1}{c}}}\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u}),$$

which is a contradiction as ${\left(d_3+{\displaystyle \frac{d_4}{2^c}}\right)}^{{\displaystyle \frac{1}{c}}}<1.$ Hence, $\mathfrak{u}=\widehat{R}\mathfrak{u}$.

Case II: Let c = 0. By using (17) with ${\mathfrak{u}}_1={\mathfrak{u}}_{n-1}$ and ${\mathfrak{u}}_2={\mathfrak{u}}_n,$ we get

$$\begin{array}{cc}\hfill \theta \left(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\right)& =\theta \left(\mathfrak{h}(\widehat{R}{\mathfrak{u}}_{n-1},\widehat{R}{\mathfrak{u}}_n)\right)\hfill \\ \hfill & \le \varphi \left[\theta \left(M({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)\right)\right]\hfill \\ \hfill & =\varphi \left[\theta \right({\left(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)\right)}^{d_1}.{\left(\mathfrak{h}({\mathfrak{u}}_{n-1},\widehat{R}{\mathfrak{u}}_{n-1})\right)}^{d_2}.{\left(\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}{\mathfrak{u}}_n)\right)}^{d_3}.\hfill \\ \hfill & {\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_{n-1},\widehat{R}{\mathfrak{u}}_n)+\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}{\mathfrak{u}}_{n-1}))\right)}^{d_4}\left)\right]\hfill \\ \hfill & <\theta ({\left(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)\right)}^{d_1}.{\left(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)\right)}^{d_2}.{\left(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\right)}^{d_3}.\hfill \\ \hfill & {\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_{n+1})+\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_n))\right)}^{d_4})\hfill \\ \hfill \mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})& <{\left(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)\right)}^{d_1+d_2}.{\left(\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\right)}^{d_3}.{\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)+\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1}))\right)}^{d_4}.\hfill \end{array}$$

(29)

Assume that $\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})>\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)$, then, using (29), we get

$$\begin{array}{cc}\mathfrak{h}{({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})}^{1-d_3-d_4}\hfill & <\mathfrak{h}{({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)}^{d_1+d_2}\hfill \\ \hfill \mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})& <\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n),\hfill \end{array}$$

(30)

which contradicts our assumption. Hence, $\mathfrak{h}({\mathfrak{u}}_n,{\mathfrak{u}}_{n+1})\le \mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n).$

Using the same methods as those in the case of $c>0,$ we immediately observe that $\left\{{\mathfrak{u}}_n\right\}$ forms a Cauchy sequence in a complete convex metric space. As a result, there exists a $\mathfrak{u}\in H$ such that ${\displaystyle \underset{n\to \infty }{lim}\mathfrak{h}({\mathfrak{u}}_n,\mathfrak{u})=0.}$

Using Equation (17) with ${\mathfrak{u}}_1={\mathfrak{u}}_n$ and ${\mathfrak{u}}_2=\mathfrak{u}$.

$$\begin{array}{cc}\hfill \theta \left(\mathfrak{h}({\mathfrak{u}}_{n+1},\widehat{R}\mathfrak{u})\right)& =\theta \left(\mathfrak{h}(\widehat{R}{\mathfrak{u}}_n,\widehat{R}\mathfrak{u})\right)\hfill \\ \hfill & =\varphi \left[\theta \right([{\left(\mathfrak{h}({\mathfrak{u}}_{n-1},\mathfrak{u})\right)}^{d_1}.{\left(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)\right)}^{d_2}.{\left(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u})\right)}^{d_3}.\hfill \\ \hfill & {\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}\mathfrak{u})+\mathfrak{h}(\mathfrak{u},\widehat{R}{\mathfrak{u}}_n))\right)}^{d_4}\left]\right)]\hfill \\ \hfill & <\theta \left(\right[{\left(\mathfrak{h}({\mathfrak{u}}_{n-1},\mathfrak{u})\right)}^{d_1}.{\left(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)\right)}^{d_2}.{\left(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u})\right)}^{d_3}.\hfill \\ \hfill & {\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}\mathfrak{u})+\mathfrak{h}(\mathfrak{u},\widehat{R}{\mathfrak{u}}_n))\right)}^{d_4}\left]\right)\hfill \\ \hfill \mathfrak{h}({\mathfrak{u}}_{n+1},\widehat{R}\mathfrak{u})& <{\left(\mathfrak{h}({\mathfrak{u}}_{n-1},\mathfrak{u})\right)}^{d_1}.{\left(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)\right)}^{d_2}.{\left(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u})\right)}^{d_3}.\hfill \\ \hfill & {\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}\mathfrak{u})+\mathfrak{h}(\mathfrak{u},\widehat{R}{\mathfrak{u}}_n))\right)}^{d_4}.\hfill \end{array}$$

(31)

Assume that $\mathfrak{u}\ne \widehat{R}\mathfrak{u}$.

$$\begin{array}{cc}0<\mathfrak{h}(\widehat{R}\mathfrak{u},\mathfrak{u})\hfill & \le \mathfrak{h}(\widehat{R}\mathfrak{u},{\mathfrak{u}}_{n+1})+\mathfrak{h}({\mathfrak{u}}_{n+1},\mathfrak{u})\hfill \\ \hfill & \le {\left(\mathfrak{h}({\mathfrak{u}}_{n-1},\mathfrak{u})\right)}^{d_1}.{\left(\mathfrak{h}({\mathfrak{u}}_{n-1},{\mathfrak{u}}_n)\right)}^{d_2}.{\left(\mathfrak{h}(\mathfrak{u},\widehat{R}\mathfrak{u})\right)}^{d_3}.\hfill \\ \hfill & {\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_n,\widehat{R}\mathfrak{u})+\mathfrak{h}(\mathfrak{u},\widehat{R}{\mathfrak{u}}_n))\right)}^{d_4}+\mathfrak{h}({\mathfrak{u}}_{n+1},\mathfrak{u}).\hfill \end{array}$$

(32)

Take $n\to \infty$ in the above equation

$$\mathfrak{h}(\widehat{R}\mathfrak{u},\mathfrak{u})\le 0,$$

Hence, $\mathfrak{u}=\widehat{R}\mathfrak{u}$.

By Lemma 4 of [24], we have $R\mathfrak{u}=\mathfrak{u}.$ □

To demonstrate the validity of this result, we now present an example.

Example 2. 

Let H = $[0,1]$ with the metric $\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)$ = $|{\mathfrak{u}}_1-{\mathfrak{u}}_2|$. Let $W({\mathfrak{u}}_1,{\mathfrak{u}}_2,\mu )=\mu {\mathfrak{u}}_1+(1-\mu ){\mathfrak{u}}_2.$ Assume that the mapping R is defined as

$$R\left(\mathfrak{u}\right)=1-{\displaystyle \frac{3}{4}}\mathfrak{u}.$$

The mapping $R:H\to H$ satisfies the conditions of Theorem 3 with $\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)=|{\mathfrak{u}}_1-{\mathfrak{u}}_2|$,$d_1=d_2=d_3=d_4={\displaystyle \frac{1}{4}}$, $c=0$, $\theta \left(\mathfrak{u}\right)=2^{\mathfrak{u}}$$\forall \mathfrak{u}\in [0,+\infty ),$ and $\varphi \left(\mathfrak{u}\right)={\mathfrak{u}}^{0.9}$ for all $\mathfrak{u}\in [1,+\infty ).$

For $\mu ={\displaystyle \frac{1}{4}}$, $\widehat{R}$ is defined as

$$\widehat{R}\left(\mathfrak{u}\right)={\displaystyle \frac{3}{4}}.$$

$\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)=0$, for all ${\mathfrak{u}}_1,{\mathfrak{u}}_2\in H$. Since, R is a hybrid enriched contraction of a Hardy–Rogers type, $\widehat{R}$ therefore has a fixed point and ${\displaystyle \frac{3}{4}}$ is the fixed point of $\widehat{R}$ (as shown in Figure 3). By using Lemma 4 of [24], we have, $R\mathfrak{u}=\widehat{R}\mathfrak{u}={\displaystyle \frac{3}{4}}$ (as shown in Figure 4).

We now present the following results which are direct consequences of Theorem 2.

Corollary 5. 

Let $(R,\mathfrak{h},W)$ be a complete convex metric space. Consider a function $R:H\to H$ such that $\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)\le aM({\mathfrak{u}}_1,{\mathfrak{u}}_2)$ holds for $a\in (0,1)$ and

$$M({\mathfrak{u}}_1,{\mathfrak{u}}_2)=\left\{\begin{array}{c}[d_1{\left(\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)\right)}^c+d_2{\left(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1)\right)}^c+d_3{\left(\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2)\right)}^c+\hfill \\ d_4{\left[{\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_1))\right]}^c{]}^{{\displaystyle \frac{1}{c}}},\hfill & forc>0\hfill \\ \\ {\left(\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)\right)}^{d_1}.{\left(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1)\right)}^{d_2}.{\left(\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2)\right)}^{d_3}.\hfill \\ {\left({\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_1))\right)}^{d_4},\hfill & forc=0,{\mathfrak{u}}_1,{\mathfrak{u}}_2\in R\setminus F\left(\widehat{R}\right)\hfill \end{array}\right.$$

where $c\ge 0$, $d_i>0$, $i=1,2,3,4$ with $d_1+d_2+d_3+d_4=1$ and $F\left(\widehat{R}\right)$ = $\{\mathfrak{u}\in H:\widehat{R}\mathfrak{u}=\mathfrak{u}\}$. Then, R has a fixed point.

Proof. 

Assume that $\theta \left(\mathfrak{u}\right)=exp\left(\mathfrak{u}\right)$$\forall \mathfrak{u}\in [0,+\infty ),$ and $\varphi \left(\mathfrak{u}\right)={\mathfrak{u}}^a$ for all $\mathfrak{u}\in [1,+\infty )$ and $a\in (0,1).$ It is clear that $\theta \in \mathrm{\Theta }$, $\varphi \in \mathrm{\Phi }.$

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \theta \left(\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)\right)& =e^{\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)}\hfill \\ \hfill & \le e^{aM({\mathfrak{u}}_1,{\mathfrak{u}}_2)}\hfill \\ \hfill & ={\left(e^{\left(M({\mathfrak{u}}_1,{\mathfrak{u}}_2)\right)}\right)}^a\hfill \\ \hfill & =\varphi \left(\theta \left(M({\mathfrak{u}}_1,{\mathfrak{u}}_2)\right)\right).\hfill \end{array}\end{array}$$

As a result, all requirements of Theorem 2 are met. Therefore, R has a fixed point. □

Corollary 6. 

Let $(R,\mathfrak{h},W)$ be a complete convex metric space. Let $R:H\to H$ be a mapping such that

$$\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)\le {\displaystyle \frac{a}{4}}\left(\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1)+\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2)+{\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_1))\right).$$

Then, R has a unique fixed point.

Proof. 

Put $d_1=d_2=d_3=d_4={\displaystyle \frac{1}{4}}$ and $c=1$ in Corollary 5, then the mapping

$$\mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)\le {\displaystyle \frac{a}{4}}\left(\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1)+\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2)+{\displaystyle \frac{1}{2}}(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_1))\right),$$

has a fixed point. Suppose that ${\mathfrak{u}}_1$ and ${\mathfrak{u}}_2$ are two distinct fixed points of $\widehat{R}$; then,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)& \le {\displaystyle \frac{a}{4}}\left[\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1)+\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2)+{\displaystyle \frac{1}{2}}\left(\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_1)\right)\right]\hfill \\ \hfill \mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)& \le {\displaystyle \frac{a}{4}}\left[\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_1)+\mathfrak{h}({\mathfrak{u}}_2,{\mathfrak{u}}_2)+{\displaystyle \frac{1}{2}}\left(\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)+\mathfrak{h}({\mathfrak{u}}_2,{\mathfrak{u}}_1)\right)\right]\hfill \end{array}\end{array}$$

which is a contradiction. □

Corollary 7. 

Let $(R,\mathfrak{h},W)$ be a complete convex metric space. Let $R:H\to H$ be an interpolative enriched Hardy–Rogers-type contraction mapping. Then, R has a fixed point.

Proof. 

Put $c=0$ in Corollary 5 with $d_1=\mathfrak{b}$, $d_2=\mathfrak{c}$, $d_3=\mathfrak{d}$ and $d_4=1-\mathfrak{b}-\mathfrak{c}-\mathfrak{d}$ for $\mathfrak{b},\mathfrak{c},\mathfrak{d}\in$$(0,1)$. □

Corollary 8 

([27]). Let $(R,\mathfrak{h})$ be a complete metric space. Let $R:H\to H$ be an interpolative Hardy–Rogers-type contraction mapping. Then, R has a fixed point.

Proof. 

Put $c=0$, $d_3=1-d_1-d_2$ and $\mu =0$ in Corollary 5. □

3. Applications

This section deals with the use of a hybrid enriched contraction to check the existence of a solution of nonlinear equations.

Example 3. 

Let $H:\mathbb{R}\to \mathbb{R}$ be a mapping defined by $H\mathfrak{u}=\mathfrak{u}-e^{-e^{-\mathfrak{u}}}$. Suppose we want to solve $H\mathfrak{u}=0;$ then, the root of H will be the fixed point of $H_1=e^{-e^{-\mathfrak{u}}}$.

Assume that $\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)=|{\mathfrak{u}}_1-{\mathfrak{u}}_2|$, $\widehat{R}\mathfrak{u}={\displaystyle \frac{1}{1000}}\mathfrak{u}+{\displaystyle \frac{999}{1000}}{e}^{-e^{-\mathfrak{u}}}$. Then,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)& =|{\displaystyle \frac{1}{1000}}({\mathfrak{u}}_1-{\mathfrak{u}}_2)+{\displaystyle \frac{999}{1000}}(e^{-e^{-{\mathfrak{u}}_1}}-e^{-e^{-{\mathfrak{u}}_2}})|\hfill \\ \hfill & \le {\displaystyle \frac{1}{1000}}|{\mathfrak{u}}_1-{\mathfrak{u}}_2|+{\displaystyle \frac{999}{1000}}|e^{-e^{-{\mathfrak{u}}_1}}-e^{-e^{-{\mathfrak{u}}_2}}|\hfill \\ \hfill & \le {\displaystyle \frac{1}{1000}}|{\mathfrak{u}}_1-{\mathfrak{u}}_2|+{\displaystyle \frac{999}{1000}}.{\displaystyle \frac{1}{e}}|{\mathfrak{u}}_1-{\mathfrak{u}}_2|\hfill \\ \hfill & =\left({\displaystyle \frac{1}{1000}}+{\displaystyle \frac{999}{1000}}.{\displaystyle \frac{1}{e}}\right)|{\mathfrak{u}}_1-{\mathfrak{u}}_2|\hfill \\ \hfill & \le {\displaystyle \frac{4}{10}}|{\mathfrak{u}}_1-{\mathfrak{u}}_2|\hfill \\ \hfill & \le {\displaystyle \frac{4}{10}}\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)+{\displaystyle \frac{3}{20}}\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1)+{\displaystyle \frac{3}{20}}\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2)\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{2}{5}}\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)+{\displaystyle \frac{3}{10}}\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1)+{\displaystyle \frac{3}{10}}\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2)\right].\hfill \end{array}\end{array}$$

Thus, the mapping $H_1$ satisfies all the assumptions of Corollary 1 with $d_1={\displaystyle \frac{2}{5}}$, $d_2=d_3={\displaystyle \frac{3}{10}}$ and $a={\displaystyle \frac{1}{2}}$. Therefore, the mapping H has a solution. The iteration ${\mathfrak{u}}_n={\displaystyle \frac{1}{1000}}.{\mathfrak{u}}_{n-1}+{\displaystyle \frac{999}{1000}}.e^{-e^{-{\mathfrak{u}}_{n-1}}}$ will converge to the solution of mapping H (as shown in Figure 5).

Example 4. 

Let $R:[0,1]\to [0,1]$ be a mapping defined by $R\mathfrak{u}=e^{\mathfrak{u}}-10\mathfrak{u}$. Suppose we want to solve $R\mathfrak{u}=0;$ then, the root of R will be the fixed point of $R_1={\displaystyle \frac{e^{\mathfrak{u}}}{10}}$.

Assume that $\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)=|{\mathfrak{u}}_1-{\mathfrak{u}}_2|$, $\widehat{R}\mathfrak{u}={\displaystyle \frac{1}{100}}\mathfrak{u}+{\displaystyle \frac{99}{100}}{\displaystyle \frac{e^{\mathfrak{u}}}{10}}$. Then,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \mathfrak{h}(\widehat{R}{\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_2)& =|{\displaystyle \frac{1}{100}}({\mathfrak{u}}_1-{\mathfrak{u}}_2)+{\displaystyle \frac{99}{1000}}(e^{{\mathfrak{u}}_1}-e^{{\mathfrak{u}}_2})|\hfill \\ \hfill & \le {\displaystyle \frac{1}{100}}|{\mathfrak{u}}_1-{\mathfrak{u}}_2|+{\displaystyle \frac{99}{1000}}|e^{{\mathfrak{u}}_1}-e^{{\mathfrak{u}}_2}|\hfill \\ \hfill & \le {\displaystyle \frac{1}{100}}|{\mathfrak{u}}_1-{\mathfrak{u}}_2|+{\displaystyle \frac{99}{1000}}e|{\mathfrak{u}}_1-{\mathfrak{u}}_2|\hfill \\ \hfill & =\left({\displaystyle \frac{1}{100}}+{\displaystyle \frac{99}{1000}}.e\right)|{\mathfrak{u}}_1-{\mathfrak{u}}_2|\hfill \\ \hfill & \le {\displaystyle \frac{3}{10}}|{\mathfrak{u}}_1-{\mathfrak{u}}_2|\hfill \\ \hfill & \le {\displaystyle \frac{3}{10}}\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)+{\displaystyle \frac{2}{20}}\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1)+{\displaystyle \frac{2}{20}}\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2)\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{3}{5}}\mathfrak{h}({\mathfrak{u}}_1,{\mathfrak{u}}_2)+{\displaystyle \frac{2}{10}}\mathfrak{h}({\mathfrak{u}}_1,\widehat{R}{\mathfrak{u}}_1)+{\displaystyle \frac{2}{10}}\mathfrak{h}({\mathfrak{u}}_2,\widehat{R}{\mathfrak{u}}_2)\right].\hfill \end{array}\end{array}$$

Thus, the mapping $R_1$ satisfies all the assumptions of Corollary 1 with $d_1={\displaystyle \frac{3}{5}}$, $d_2=d_3={\displaystyle \frac{1}{5}}$ and $a={\displaystyle \frac{1}{5}}$. Therefore, the mapping R has a solution. The iteration ${\mathfrak{u}}_n={\displaystyle \frac{1}{100}}.{\mathfrak{u}}_{n-1}+{\displaystyle \frac{99}{1000}}.e^{{\mathfrak{u}}_{n-1}}$ will converge to the solution of mapping R (as shown in Figure 6).

4. Conclusions and Future Scope

In this work, we have investigated the presence and approximation of fixed points for hybrid enriched contractions of the Hardy–Rogers type and the Ćirić–Reich–Rus type in the framework of convex metric space. It is demonstrated that each hybrid enriched contraction in a complete convex metric space has at least one fixed point that can be approximated by means of a Kransnoselskij-type iterative process. Our study extends and generalizes various important related results that already exist in the literature.

The study of existence of fixed points for such kinds of contraction mappings in more spaces would be an interesting topic for future work. Similar results for cyclic contractions are another direction of future study.