Stability Analysis of a New Class of Series Type Additive Functional Equation in Banach Spaces: Direct and Fixed Point Techniques

Abstract

In this paper, the authors introduce two new classes of series type additive functional Equations (FEs). The first class of equations is derived from the sum of the squares of the alternative series and the second one is obtained from the sum of the cubes of the series. The solution of the FE is investigated using the principle of mathematical induction. The beauty of this method lies in the fact that it satisfies the property of the additive FE as well as the series. Banach spaces are one of the widely-used spaces that are very helpful to analyse the stability results of various FEs. The Banach space conditions have been applied and the stability results are established for both of the equations. Furthermore, the Banach Contraction principle and alternative of fixed point theorem are used to derive the stability results in a fixed point technique (FPT). The relationship between the FEs and both the series is established through the principle of mathematical induction in the Application section, which adds novelty to the derived results.

1. Introduction

Functional Equations (FEs) are one of the fascinating areas of research in modern-day mathematics. There are a variety of intriguing FEs introduced by notable mathematicians, such as Cauchy FE, Abel’s FE, Schröder’s FE, Jensen’s FE, D’ Alembert FE or Poisson’s FE, Pythagorean FE, and Davison FE. Ulam’s question [1] in 1940 kickstarted the journey of the research in the stability theory of FEs. Many mathematicians have studied and published several novel results in the field of stability theory, such as, Donald H. Hyers Theorem (1941) [2], Tosio Aoki Theorem (1950) [3], Th.M. Rassias (1978) [4], John M. Rassias Theorem (1982) [5], Z. Gadja Theorem (1991) [6], P. Gǎvrutǎ Theorem (1994) [7], and K. Ravi, M. Arunkumar, and John M. Rassias Theorem (2008) [8].

The first result concerning the stability of functional equations was presented by D.H.Hyers in 1941. He comprehensively answered Ulam’s question by assuming $\mathcal{M}$ to be a normed space and $\mathcal{N}$ to be a Banach space. He proved the following celebrated theorem.

Theorem 1.

Let $\mathcal{M}$ be a normed space, $\mathcal{N}$ be a Banach space, and let $f:\mathcal{M}\to \mathcal{N}$ be a mapping satisfying

$$∥f(\mathfrak{x}+\mathfrak{y})-f\left(\mathfrak{x}\right)-f\left(\mathfrak{y}\right)∥\le ϵ$$

(1)

for all $\mathfrak{x},\mathfrak{y}\in \mathcal{M}$. Then the limit

$$A\left(\mathfrak{x}\right)=\underset{n\to \infty }{lim}\frac{f\left(2^n\mathfrak{x}\right)}{2^n}$$

(2)

exists for all $\mathfrak{x}\in \mathcal{M}$ and $A:\mathcal{M}\to \mathcal{N}$ is the unique additive mapping satisfying

$$∥f\left(\mathfrak{x}\right)-A\left(\mathfrak{x}\right)∥\le ϵ$$

(3)

for all $\mathfrak{x}\in \mathcal{M}$. Moreover, if $f\left(t\mathfrak{x}\right)$ is continuous in t for each fixed $\mathfrak{x}\in \mathcal{M}$, then the function A is linear.

Proof. 

In order to prove the stability results, the following have to be proved.

$(\ast )$

The sequence ${\displaystyle \left\{\frac{f\left(2^n\mathfrak{x}\right)}{2^n}\right\}}$ is a Cauchy sequence.

$(\ast )$

If $A\left(\mathfrak{x}\right)=\underset{n\to \infty }{lim}\frac{f\left(2^n\mathfrak{x}\right)}{2^n}$ then A is additive in R.

$(\ast )$

Further, A satisfies

$$∥f\left(\mathfrak{x}\right)-A\left(\mathfrak{x}\right)∥\le ϵ$$

for $\mathfrak{x}\in R$.

$(\ast )$

A is unique.

The stability results of the newly proposed FEs have been derived using the above listed concepts. □

Recently, Hoc et al. [9] studied the existence of a minima of functions in partial metric spaces and its applications to fixed point theory. Mureşan et al. [10] presented some applications of Perov’s fixed point (FP) theorem. Lingxiao et al. [11] derived the Ulam–Hyers stability (UHS) of cubic FEs in fuzzy normed spaces. Tudor Bînzar et al. [12] studied the FP theorems in fuzzy normed linear spaces for contractive mappings with applications to dynamic programming. Shahram Rezapour et al. [13] analysed the FP theory and the Liouville–Caputo integro-differential functional boundary value problem using multiple nonlinear terms. Romaguera [14] studied the basic contractions of Suzuki-type on quasi-metric spaces and FP results. Bodaghi et al. [15] studied the structure of multimixed quadratic-cubic mappings and the application of FP theory. Pathak et al. [16] analysed the application of FP theorem to solvability for non-linear fractional Hadamard functional integral equations. Monairah et al. [17] studied the analysis of fractional differential inclusion models for COVID-19 via (FP) results in metric spaces. Kanwal et al. [18] investigated an FP approach to lattice fuzzy sets via F-contraction. Agilan et al. studied the generalised UHS of additive FEs [19,20,21,22,23,24,25,26] in various normed spaces.

Motivated by the above fact, this is the first time in the literature that the generalised UHS for new classes of series type additive functional equations are analysed using two distinct techniques in Banach Spaces (BSs). Consequently, the findings that will be discussed in the subsequent sections are both novel and essential to the study of FEs. In Section 2, the general solution of the new class of series type additive type FE is derived using the principle of mathematical induction. In Section 3, preliminaries and basic definitions of Banach spaces are provided. Section 4 and Section 5 cover the Ulam stability analysis of the new classes of equations using the direct and FPT, respectively, in Banach spaces. Finally, in Section 6, the application part is discussed.

In this paper, the following two new classes of series type additive FEs are introduced and their general solutions and generalised UHS are investigated in BSs using direct and FPT.

$${\yen }_1\left(\left(\sum _{\vartheta =1}^n{(-1)}^{\vartheta -1}{\vartheta }^2\right)\mathfrak{x}\right)={(-1)}^{n-1}\frac{n(n+1)}{2}{\yen }_1\left(\mathfrak{x}\right)$$

(4)

$${\yen }_2\left(\left(\sum _{\vartheta =1}^{n}n(n-1)+2\vartheta -1\right)\mathfrak{x}\right)=n^3{\yen }_2\left(\mathfrak{x}\right)$$

(5)

2. General Solutions

2.1. Solution of (4) Using the Principle of Mathematical Induction (PMI)

Theorem 2.

Assume $\mathcal{M}$ and $\mathcal{N}$ to be vector spaces. If ${\yen }_1:\mathcal{M}\to \mathcal{N}$ satisfies the FE

$$\begin{array}{cc}\hfill & {\yen }_1\left(\left(\sum _{\vartheta =1}^n{(-1)}^{\vartheta -1}{\vartheta }^2\right)\mathfrak{x}\right)={(-1)}^{n-1}\frac{n(n+1)}{2}{\yen }_1\left(\mathfrak{x}\right)\hfill \end{array}$$

(6)

∀$\mathfrak{x}\in \mathcal{M}$ then ${\yen }_1$ is additive.

Proof. 

$$Let\mathbb{P}\left(n\right):{\yen }_1\left(\left(\sum _{\vartheta =1}^n{(-1)}^{\vartheta -1}{\vartheta }^2\right)\mathfrak{x}\right)={(-1)}^{n-1}\frac{n(n+1)}{2}{\yen }_1\left(\mathfrak{x}\right)$$

(7)

Step 1: To prove $\mathbb{P}\left(1\right)$ is true

For $n=1$, we have

$$\begin{array}{cc}\hfill & {\yen }_1\left(\left(\sum _{\vartheta =1}^1{(-1)}^{\vartheta -1}{\vartheta }^2\right)\mathfrak{x}\right)={(-1)}^{1-1}\frac{1(1+1)}{2}{\yen }_1\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_1\left(\left({(-1)}^{1-1}{1}^2\right)\mathfrak{x}\right)={(-1)}^{1-1}\frac{1(1+1)}{2}{\yen }_1\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_1\left(\mathfrak{x}\right)={\yen }_1\left(\mathfrak{x}\right)\hfill \end{array}$$

so $\mathbb{P}\left(1\right)$ is True.

Step 2: To prove $\mathbb{P}\left(\mathcal{L}\right)$ is true

For $n=\mathcal{L}$, we have

$$\begin{array}{cc}\hfill & {\yen }_1\left(\left(\sum _{\vartheta =1}^{\mathcal{L}}{(-1)}^{\vartheta -1}{\vartheta }^2\right)\mathfrak{x}\right)={(-1)}^{\mathcal{L}-1}\frac{\mathcal{L}(\mathcal{L}+1)}{2}{\yen }_1\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_1\left(\left({(-1)}^{1-1}{1}^1+{(-1)}^{2-1}{2}^2+{(-1)}^{3-1}{3}^2+{(-1)}^{4-1}{4}^2+...+{(-1)}^{\mathcal{L}-1}{\mathcal{L}}^2\right)\mathfrak{x}\right)\hfill \\ \hfill & ={(-1)}^{\mathcal{L}-1}\frac{\mathcal{L}(\mathcal{L}+1)}{2}{\yen }_1\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_1\left(\left(1^2-2^2+3^2-4^2+...+{(-1)}^{\mathcal{L}-1}{\mathcal{L}}^2\right)\mathfrak{x}\right)={(-1)}^{\mathcal{L}-1}\frac{\mathcal{L}(\mathcal{L}+1)}{2}{\yen }_1\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_1\left({(-1)}^{\mathcal{L}-1}\frac{\mathcal{L}(\mathcal{L}+1)}{2}\mathfrak{x}\right)={(-1)}^{\mathcal{L}-1}\frac{\mathcal{L}(\mathcal{L}+1)}{2}{\yen }_1\left(\mathfrak{x}\right),\mathcal{L}=1,2,3...\hfill \end{array}$$

so $\mathbb{P}\left(\mathcal{L}\right)$ is true.

Step 3: To prove $\mathbb{P}(\mathcal{L}+1)$ is true

For $n=\mathcal{L}+1$, we have

$$\begin{array}{cc}\hfill & {\yen }_1\left(\left(\sum _{\vartheta =1}^{\mathcal{L}+1}{(-1)}^{\vartheta -1}{\vartheta }^2\right)\mathfrak{x}\right)={(-1)}^{\mathcal{L}}\frac{(\mathcal{L}+1)(\mathcal{L}+2)}{2}{\yen }_1\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_1\left(\left({(-1)}^{1-1}{1}^1+{(-1)}^{2-1}{2}^2+...+{(-1)}^{\mathcal{L}-1}{\mathcal{L}}^2+{(-1)}^{\mathcal{L}}{(\mathcal{L}+1)}^2\right)\mathfrak{x}\right)\hfill \\ \hfill & ={(-1)}^{\mathcal{L}}\frac{(\mathcal{L}+1)(\mathcal{L}+2)}{2}{\yen }_1\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_1\left(\left(1^2-2^2+3^2-4^2+...+{(-1)}^{\mathcal{L}-1}{\mathcal{L}}^2+{(-1)}^{\mathcal{L}}{(\mathcal{L}+1)}^2\right)\mathfrak{x}\right)\hfill \\ \hfill & ={(-1)}^{\mathcal{L}}\frac{(\mathcal{L}+1)(\mathcal{L}+2)}{2}{\yen }_1\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_1\left({(-1)}^{\mathcal{L}}\frac{(\mathcal{L}+1)(\mathcal{L}+2)}{2}\mathfrak{x}\right)={(-1)}^{\mathcal{L}-1}\frac{(\mathcal{L}+1)(\mathcal{L}+2)}{2}{\yen }_1\left(\mathfrak{x}\right),\mathcal{L}=0,1,2,3...\hfill \end{array}$$

so $\mathbb{P}(\mathcal{L}+1)$ is true.

Using mathematical induction, $\mathbb{P}\left(n\right)$ is true ∀ positive integer n. □

2.2. Solution of (5) Using PMI

Theorem 3.

If ${\yen }_2:\mathcal{M}\to \mathcal{N}$ satisfies the FE

$$\begin{array}{cc}\hfill & {\yen }_2\left(\left(\sum _{\vartheta =1}^{n}n(n-1)+2\vartheta -1\right)\mathfrak{x}\right)=n^3{\yen }_2\left(\mathfrak{x}\right)\hfill \end{array}$$

(8)

∀$\mathfrak{x}\in \mathcal{M}$ then ${\yen }_2$ is additive.

Proof. 

$$Let{\yen }_2\left(\left(\sum _{\vartheta =1}^{n}n(n-1)+2\vartheta -1\right)\mathfrak{x}\right)=n^3{\yen }_2\left(\mathfrak{x}\right)$$

(9)

Step 1: To Prove $\mathbb{P}\left(1\right)$ is true

For $n=1$, we have

$$\begin{array}{cc}\hfill & {\yen }_2\left(\left(\sum _{\vartheta =1}^{1}1(1-1)+2\vartheta -1\right)\mathfrak{x}\right)=1^3{\yen }_2\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_2\left(\left(1(1-1)+2.1-1\right)\mathfrak{x}\right)=1^3{\yen }_2\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_2\left(\mathfrak{x}\right)={\yen }_2\left(\mathfrak{x}\right)\hfill \end{array}$$

so $\mathbb{P}\left(1\right)$ is True.

Step 2: Assume that $\mathbb{P}\left(\mathcal{L}\right)$ is true

For $n=\mathcal{L}$, we have

$$\begin{array}{cc}\hfill & {\yen }_2\left(\left(\sum _{\vartheta =1}^{\mathcal{L}}\mathcal{L}(\mathcal{L}-1)+2\vartheta -1\right)\mathfrak{x}\right)={\mathcal{L}}^3{\yen }_2\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_2\left(\left(\mathcal{L}(\mathcal{L}-1)+2.1-1+\mathcal{L}(\mathcal{L}-1)+2.2-1+...+\mathcal{L}(\mathcal{L}-1)+2.\mathcal{L}-1\right)\mathfrak{x}\right)\hfill \\ \hfill & ={\mathcal{L}}^3{\yen }_2\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_2\left(\left(\mathcal{L}(\mathcal{L}-1)is\mathcal{L}times+(1+3+5+...+(2\mathcal{L}-1\left)\right)\right)\mathfrak{x}\right)\hfill \\ \hfill & ={\mathcal{L}}^3{\yen }_2\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_2\left(\left(\mathcal{L}(\mathcal{L}-1)is\mathcal{L}times+\left(sumof\mathcal{L}oddtermsis{\mathcal{L}}^2\right)\right)\mathfrak{x}\right)\hfill \\ \hfill & ={\mathcal{L}}^3{\yen }_2\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_2\left(\left({\mathcal{L}}^2(\mathcal{L}-1)+{\mathcal{L}}^2\right)\mathfrak{x}\right)={\mathcal{L}}^3{\yen }_2\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_2\left({\mathcal{L}}^3\mathfrak{x}\right)={\mathcal{L}}^3{\yen }_2\left(\mathfrak{x}\right)\hfill \end{array}$$

so $\mathbb{P}\left(\mathcal{L}\right)$ is True.

Step 3: To prove $\mathbb{P}(\mathcal{L}+1)$ is true

For $n=\mathcal{L}+1$, we have

$$\begin{array}{cc}\hfill & {\yen }_2\left(\left(\sum _{\vartheta =1}^{\mathcal{L}+1}\mathcal{L}(\mathcal{L}+1)+2\vartheta -1\right)\mathfrak{x}\right)\hfill \\ \hfill & ={(\mathcal{L}+1)}^3{\yen }_2\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_2\left(\left(\mathcal{L}(\mathcal{L}+1)+2.1-1+\mathcal{L}(\mathcal{L}+1)+2.2-1+...+\mathcal{L}(\mathcal{L}+1)+2(\mathcal{L}+1)-1\right)\mathfrak{x}\right)\hfill \\ \hfill & ={(\mathcal{L}+1)}^3{\yen }_2\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_2\left(\left(\mathcal{L}(\mathcal{L}+1)is\mathcal{L}times+(1+3+5+...+(2\mathcal{L}-1\left)\right)\right)\mathfrak{x}\right)={(\mathcal{L}+1)}^3{\yen }_2\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_2\left(\left(\mathcal{L}(\mathcal{L}+1)is\mathcal{L}times+(sumof\mathcal{L}+1oddtermsis{(\mathcal{L}+1)}^2)\right)\mathfrak{x}\right)\hfill \\ \hfill & ={(\mathcal{L}+1)}^3{\yen }_2\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_2\left(\left(\mathcal{L}{(\mathcal{L}+1)}^2+{(\mathcal{L}+1)}^2\right)\mathfrak{x}\right)={(\mathcal{L}+1)}^3{\yen }_2\left(\mathfrak{x}\right)\hfill \\ \hfill & {\yen }_2\left({(\mathcal{L}+1)}^3\mathfrak{x}\right)={(\mathcal{L}+1)}^3{\yen }_2\left(\mathfrak{x}\right)\hfill \end{array}$$

so $\mathbb{P}(\mathcal{L}+1)$ is true.

Using mathematical induction, $\mathbb{P}\left(n\right)$ is true ∀ positive integer n. □

3. Preliminaries and Basic Definitions of Banach Spaces

Definition 1.

A norm on a linear space $\mathcal{M}$ is a function $∥.∥:\mathcal{M}\to \mathbb{R}$ with the following properties

$\left(B1\right)∥\mathfrak{x}∥\ge 0$, for all $\mathfrak{x}\in \mathcal{M}$,

$\left(B2\right)∥\mathfrak{x}∥=0$ iff $\mathfrak{x}=0$, for all $\mathfrak{x}\in \mathcal{M}$,

$\left(B3\right)∥\lambda \mathfrak{x}∥=\left|\lambda \right|∥\mathfrak{x}∥$, for all $\mathfrak{x}\in \mathcal{M}$ and $\lambda \in \mathbb{R}\left(or\right)\mathbb{C},$

$\left(B4\right)∥\mathfrak{x}+\mathfrak{y}∥\le ∥\mathfrak{x}∥+∥\mathfrak{y}∥.$

A norm on a linear space $\left(\mathcal{M},∥.∥\right)$ is a linear space $\mathcal{M}$ equipped with norm $∥.∥$.

A Banach space is a complete normed vector space in a mathematical analysis. That is, the distance between the vectors converges as the sequence goes on. In functional analysis, a Banach space is a normed vector space that allows for the vector length to be computed. When the vector space is normed, it means that each vector other than the zero vector has a length greater than zero. The length and distance between two vectors can thus be computed. A vector space is complete if a Cauchy sequence of vectors in the space will converge toward a limit. As the sequence goes on, the vectors arbitrarily become closer together.

A normed linear space is complete if all Cauchy convergent sequences are convergent. A complete normed linear space is called a Banach space.

A metric space $\mathcal{M}$ is said to be complete if every Cauchy sequence in $\mathcal{M}$ converges to a point in $\mathcal{M}$.

Example 1.

The set of all real numbers $\mathbb{R}$ with absolute value norm $∥\mathfrak{x}∥=\left|\mathfrak{x}\right|$ is a one-dimensional real normed linear space.

Our proposed functional Equations (4) and (5) are also one-dimensional series type additive FEs.

4. Stability Analysis Using Direct Method

In this section, the stability of the new series type additive FEs (4) and (5) are investigated. Let $\mathcal{M}$ be a normed space and $\mathcal{N}$ be a Banach space.

Theorem 4.

Assume that the function ${\mathcal{Z}}_1,{\mathcal{Z}}_2:\mathcal{M}\to [0,\infty )$, then

$$\underset{\lambda \to \infty }{lim}\frac{{\mathcal{Z}}_1\left({\mathcal{H}}_1^{\lambda \varpi }\mathfrak{x}\right)}{{\mathcal{H}}_1^{\lambda j}}=0$$

(10)

$$\underset{\lambda \to \infty }{lim}\frac{{\mathcal{Z}}_2\left({\mathcal{H}}_2^{\lambda \varpi }\mathfrak{x}\right)}{{\mathcal{H}}_2^{\lambda j}}=0$$

(11)

∀$\mathfrak{x}\in \mathcal{M}$. Let the function ${\yen }_1,{\yen }_2:\mathcal{M}\to \mathcal{N}$ satisfying

$$∥{\yen }_1\left(\left(\sum _{\vartheta =1}^n{(-1)}^{\vartheta -1}{\vartheta }^2\right)\mathfrak{x}\right)-{(-1)}^{n-1}\frac{n(n+1)}{2}{\yen }_1\left(\mathfrak{x}\right)∥\le {\mathcal{Z}}_1\left(\mathfrak{x}\right)$$

(12)

$$∥{\yen }_2\left(\left(\sum _{\vartheta =1}^{n}n(n-1)+2\vartheta -1\right)\mathfrak{x}\right)-n^3{\yen }_2\left(\mathfrak{x}\right)∥\le {\mathcal{Z}}_2\left(\mathfrak{x}\right)$$

(13)

∀$\mathfrak{x}\in \mathcal{M}$ with $\varpi \in \{-1,1\}$. Then, ∃ a function $A_1,A_2:\mathcal{M}\to \mathcal{N}$ and satisfying the FEs (4) and (5),

$$∥{\yen }_1\left(\mathfrak{x}\right)-A_1\left(\mathfrak{x}\right)∥\le \frac{1}{{\mathcal{H}}_1}\sum _{\chi =\frac{1-\varpi }{2}}^{\infty }\frac{{\mathcal{Z}}_1\left({\mathcal{H}}_1^{\chi \varpi }\mathfrak{x}\right)}{{\mathcal{H}}_1^{\chi \varpi }}$$

(14)

$$∥{\yen }_2\left(\mathfrak{x}\right)-A_2\left(\mathfrak{x}\right)∥\le \frac{1}{{\mathcal{H}}_1}\sum _{\chi =\frac{1-\varpi }{2}}^{\infty }\frac{{\mathcal{Z}}_2\left({\mathcal{H}}_2^{\chi \varpi }\mathfrak{x}\right)}{{\mathcal{H}}_2^{\chi \varpi }}$$

(15)

∀$\mathfrak{x}\in \mathcal{M}$. The functions $A_1\left(\mathfrak{x}\right)$ and $A_2\left(\mathfrak{x}\right)$ are defined as

$$A_1\left(\mathfrak{x}\right)=\underset{\lambda \to \infty }{lim}\frac{{\mathcal{Z}}_1\left({\mathcal{H}}_1^{\lambda \varpi }\mathfrak{x}\right)}{{\mathcal{H}}_1^{\lambda j}}$$

(16)

$$A_2\left(\mathfrak{x}\right)=\underset{\lambda \to \infty }{lim}\frac{{\mathcal{Z}}_2\left({\mathcal{H}}_2^{\lambda \varpi }\mathfrak{x}\right)}{{\mathcal{H}}_2^{\lambda j}}$$

(17)

∀$\mathfrak{x}\in \mathcal{M}$ with ${\mathcal{H}}_1={(-1)}^{n-1}\frac{n(n+1)}{2}$ and ${\mathcal{H}}_2=n^3$.

Proof. 

Let $\varpi =1$. Using ${\yen }_1$ and ${\yen }_2$ in (12) and (13), we obtain

$$∥{\yen }_1\left({(-1)}^{n-1}\frac{n(n+1)}{2}\mathfrak{x}\right)-{(-1)}^{n-1}\frac{n(n+1)}{2}{\yen }_1\left(\mathfrak{x}\right)∥\le {\mathcal{Z}}_1\left(\mathfrak{x}\right)$$

(18)

$$∥{\yen }_2\left(n^3\mathfrak{x}\right)-n^3{\yen }_2\left(\mathfrak{x}\right)∥\le {\mathcal{Z}}_2\left(\mathfrak{x}\right)$$

(19)

∀$\mathfrak{x}\in \mathcal{M}$. If ${\mathcal{H}}_1={(-1)}^{n-1}\frac{n(n+1)}{2}$ and ${\mathcal{H}}_2=n^3$, then

$$∥{\yen }_1\left({\mathcal{H}}_1\mathfrak{x}\right)-{\mathcal{H}}_1{\yen }_1\left(\mathfrak{x}\right)∥\le {\mathcal{Z}}_1\left(\mathfrak{x}\right)$$

(20)

$$∥{\yen }_2\left({\mathcal{H}}_2\mathfrak{x}\right)-{\mathcal{H}}_2{\yen }_2\left(\mathfrak{x}\right)∥\le {\mathcal{Z}}_2\left(\mathfrak{x}\right)$$

(21)

∀$\mathfrak{x}\in \mathcal{M}$. From (20) and (21),

$$∥{\yen }_1\left(\mathfrak{x}\right)-\frac{{\yen }_1\left(\mathfrak{x}\right)}{{\mathcal{H}}_1}∥\le \frac{{\mathcal{Z}}_1\left(\mathfrak{x}\right)}{{\mathcal{H}}_1}$$

(22)

$$∥{\yen }_2\left(\mathfrak{x}\right)-\frac{{\yen }_2\left(\mathfrak{x}\right)}{{\mathcal{H}}_2}∥\le \frac{{\mathcal{Z}}_2\left(\mathfrak{x}\right)}{{\mathcal{H}}_2}$$

(23)

∀$\mathfrak{x}\in \mathcal{M}$. Replacing $\mathfrak{x}$ by ${\mathcal{H}}_1\mathfrak{x}$ and divide by ${\mathcal{H}}_1$ in (22), $\mathfrak{x}$ by ${\mathcal{H}}_2\mathfrak{x}$ and by ${\mathcal{H}}_2$ in (23), the following result is obtained.

$$∥\frac{{\yen }_1\left(\mathfrak{x}\right)}{{\mathcal{H}}_1}-\frac{{\yen }_1\left(\mathfrak{x}\right)}{{\mathcal{H}}_1^2}∥\le \frac{{\mathcal{Z}}_1\left({\mathcal{H}}_1\mathfrak{x}\right)}{{\mathcal{H}}_1^2}$$

(24)

$$∥\frac{{\yen }_2\left(\mathfrak{x}\right)}{{\mathcal{H}}_2}-\frac{{\yen }_2\left(\mathfrak{x}\right)}{{\mathcal{H}}_2^2}∥\le \frac{{\mathcal{Z}}_2\left({\mathcal{H}}_1\mathfrak{x}\right)}{{\mathcal{H}}_2^2}$$

(25)

∀$\mathfrak{x}\in \mathcal{M}$. From (22)–(25), we achieve the subsequent inequalities

$$\begin{array}{cc}\hfill & ∥{\yen }_1\left(\mathfrak{x}\right)-\frac{{\yen }_1\left(\mathfrak{x}\right)}{{\mathcal{H}}_1^2}∥\le ∥{\yen }_1\left(\mathfrak{x}\right)-\frac{{\yen }_1\left(\mathfrak{x}\right)}{{\mathcal{H}}_1}∥+∥\frac{{\yen }_1\left(\mathfrak{x}\right)}{{\mathcal{H}}_1}-\frac{{\yen }_1\left(\mathfrak{x}\right)}{{\mathcal{H}}_1^2}∥\le \frac{1}{{\mathcal{H}}_1}\left[{\mathcal{Z}}_1\left(\mathfrak{x}\right)+\frac{{\mathcal{Z}}_1\left({\mathcal{H}}_1\mathfrak{x}\right)}{{\mathcal{H}}_1}\right]\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill & ∥{\yen }_2\left(\mathfrak{x}\right)-\frac{{\yen }_2\left(\mathfrak{x}\right)}{{\mathcal{H}}_1^2}∥\le ∥{\yen }_2\left(\mathfrak{x}\right)-\frac{{\yen }_2\left(\mathfrak{x}\right)}{{\mathcal{H}}_2}∥+∥\frac{{\yen }_2\left(\mathfrak{x}\right)}{{\mathcal{H}}_2}-\frac{{\yen }_2\left(\mathfrak{x}\right)}{{\mathcal{H}}_2^2}∥\le \frac{1}{{\mathcal{H}}_2}\left[{\mathcal{Z}}_2\left(\mathfrak{x}\right)+\frac{{\mathcal{Z}}_2\left({\mathcal{H}}_2\mathfrak{x}\right)}{{\mathcal{H}}_2}\right]\hfill \end{array}$$

(27)

∀$\mathfrak{x}\in \mathcal{M}$.

For any positive integer $\eta$, we obtain

$$\begin{array}{cc}\hfill ∥{\yen }_1\left(\mathfrak{x}\right)-\frac{{\yen }_1\left({\mathcal{H}}_1\mathfrak{x}\right)}{{\mathcal{H}}_1^{\eta }}∥& \le \frac{1}{{\mathcal{H}}_1}\sum _{\lambda =0}^{\eta -1}\frac{{\mathcal{Z}}_1\left({\mathcal{H}}_1^{\lambda }\mathfrak{x}\right)}{{\mathcal{H}}_1^{\lambda }}\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill ∥{\yen }_1\left(\mathfrak{x}\right)-\frac{{\yen }_2\left({\mathcal{H}}_2\mathfrak{x}\right)}{{\mathcal{H}}_2^{\eta }}∥& \le \frac{1}{{\mathcal{H}}_2}\sum _{\lambda =0}^{\eta -1}\frac{{\mathcal{Z}}_2\left({\mathcal{H}}_2^{\lambda }\mathfrak{x}\right)}{{\mathcal{H}}_2^{\lambda }}\hfill \end{array}$$

(29)

∀$\mathfrak{x}\in \mathcal{M}$.

To prove the convergence of the sequence

$${\displaystyle \left\{\frac{{\mathcal{Z}}_1\left({\mathcal{H}}_1^{\lambda }\mathfrak{x}\right)}{{\mathcal{H}}_1^{\lambda }}\right\},\left\{\frac{{\mathcal{Z}}_2\left({\mathcal{H}}_2^{\lambda }\mathfrak{x}\right)}{{\mathcal{H}}_2^{\lambda }}\right\},}$$

replacing $\mathfrak{x}$ by ${\mathcal{H}}_1^{\tau }\mathcal{M}$ and dividing by ${\mathcal{H}}_1^{\tau }$ in (28) and $\mathfrak{x}$ by ${\mathcal{H}}_2^{\tau }\mathfrak{x}$ and by ${\mathcal{H}}_2^{\tau }$ in (29), for any $\tau ,\eta >0$, we deduce

$$\begin{array}{cc}\hfill ∥\frac{{\yen }_1\left({\mathcal{H}}_1^{\tau }\mathfrak{x}\right)}{{\mathcal{H}}_1^{\tau }}-\frac{{\yen }_1\left({\mathcal{H}}_1^{\eta +\tau }\mathfrak{x}\right)}{{\mathcal{H}}_1^{(\eta +\tau )}}∥& =\frac{1}{{\mathcal{H}}_1^{\tau }}∥{\yen }_1\left({\mathcal{H}}_1^{\tau }\mathfrak{x}\right)-\frac{{\yen }_1({\mathcal{H}}_1^{\eta }·{\mathcal{H}}_1^{\tau }\mathfrak{x})}{{\mathcal{H}}_1^{\eta }}∥\hfill \\ \hfill & \le \frac{1}{{\mathcal{H}}_1}\sum _{\lambda =0}^{\eta -1}\frac{{\mathcal{Z}}_1\left({\mathcal{H}}_1^{\lambda +\tau }\mathfrak{x}\right)}{{\mathcal{H}}_1^{(\lambda +\tau )}}\hfill \\ \hfill & \le \frac{1}{{\mathcal{H}}_1}\sum _{\lambda =0}^{\infty }\frac{{\mathcal{Z}}_1\left({\mathcal{H}}_1^{\lambda +\tau }\mathfrak{x}\right)}{{\mathcal{H}}_1^{(\lambda +\tau )}}\hfill \\ \hfill & \to 0as\tau \to \infty \hfill \end{array}$$

$$\begin{array}{cc}\hfill ∥\frac{{\yen }_2\left({\mathcal{H}}_2^{\tau }\mathfrak{x}\right)}{{\mathcal{H}}_2^{\tau }}-\frac{{\yen }_2\left({\mathcal{H}}_2^{\eta +\tau }\mathfrak{x}\right)}{{\mathcal{H}}_2^{(\eta +\tau )}}∥& =\frac{1}{{\mathcal{H}}_2^{\tau }}∥{\yen }_2\left({\mathcal{H}}_2^{\tau }\mathfrak{x}\right)-\frac{{\yen }_2({\mathcal{H}}_2^{\eta }·{\mathcal{H}}_2^{\tau }\mathfrak{x})}{{\mathcal{H}}_2^{\eta }}∥\hfill \\ \hfill & \le \frac{1}{{\mathcal{H}}_2}\sum _{\lambda =0}^{\eta -1}\frac{{\mathcal{Z}}_2\left({\mathcal{H}}_2^{\lambda +\tau }\mathfrak{x}\right)}{{\mathcal{H}}_2^{(\lambda +\tau )}}\hfill \\ \hfill & \le \frac{1}{{\mathcal{H}}_2}\sum _{\lambda =0}^{\infty }\frac{{\mathcal{Z}}_2\left({\mathcal{H}}_2^{\lambda +\tau }\mathfrak{x}\right)}{{\mathcal{H}}_2^{(\lambda +\tau )}}\hfill \\ \hfill & \to 0as\tau \to \infty \hfill \end{array}$$

∀$\mathfrak{x}\in \mathcal{M}.$ Hence the sequence

$${\displaystyle \left\{\frac{{\yen }_1\left({\mathcal{H}}_1^{\eta }\mathfrak{x}\right)}{{\mathcal{H}}_1^{\eta }}\right\},\left\{\frac{{\yen }_2\left({\mathcal{H}}_2^{\eta }\mathfrak{x}\right)}{{\mathcal{H}}_2^{\eta }}\right\}}$$

is a Cauchy sequence.

Since $\mathcal{N}$ is complete, there exists a mapping $A_1,A_2:\mathcal{M}\to \mathcal{N}$ such that

$$A_1\left(\mathfrak{x}\right)=\underset{\eta \to \infty }{lim}\frac{{\yen }_1\left({\mathcal{H}}_1^{\eta }\mathfrak{x}\right)}{{\mathcal{H}}_1^{\eta }}\forall \mathfrak{x}\in \mathcal{M}.$$

$$A_2\left(\mathfrak{x}\right)=\underset{\eta \to \infty }{lim}\frac{{\yen }_2\left({\mathcal{H}}_2^{\eta }\mathfrak{x}\right)}{{\mathcal{H}}_1^{\eta }}\forall \mathfrak{x}\in \mathcal{M}.$$

Letting $\eta \to \infty$ in (28) and (29), we see that (14) and (15) holds ∀$\mathfrak{x}\in \mathcal{M}$.

To prove that $A_1,A_2$ satisfies (4) and (5) replacing $\mathfrak{x}$ by ${\mathcal{H}}_1^{\eta }\mathfrak{x}$ and dividing by ${\mathcal{H}}_1^{\eta }$ in (12) and $\mathfrak{x}$ by ${\mathcal{H}}_1^{\eta }\mathfrak{x}$ and by ${\mathcal{H}}_1^{\eta }$ in (13), we obtain

$$\begin{array}{cc}\hfill & \frac{1}{{\mathcal{H}}_1^{\eta }}\parallel {\yen }_1\left(\left(\sum _{\vartheta =1}^n{(-1)}^{\vartheta -1}{\vartheta }^2\right){\mathcal{H}}_1^{\eta }\mathfrak{x}\right)-{(-1)}^{n-1}\frac{n(n+1)}{2}{\yen }_1\left({\mathcal{H}}_1^{\eta }\mathfrak{x}\right)\parallel \le \frac{1}{{\mathcal{H}}_1^{\eta }}{\mathcal{Z}}_1\left({\mathcal{H}}_1^{\eta }\mathfrak{x}\right)\hfill \end{array}$$

$$\begin{array}{cc}\hfill & \frac{1}{{\mathcal{H}}_2^{\eta }}\parallel {\yen }_2\left(\left(\sum _{\vartheta =1}^{n}n(n-1)+2\vartheta -1\right){\mathcal{H}}_2^{\eta }\mathfrak{x}\right)-n^3{\yen }_2\left({\mathcal{H}}_2^{\eta }\mathfrak{x}\right)\parallel \le \frac{1}{{\mathcal{H}}_2^{\eta }}{\mathcal{Z}}_2\left({\mathcal{H}}_2^{\eta }\mathfrak{x}\right)\hfill \end{array}$$

∀$\mathfrak{x}\in \mathcal{M}$.

Assuming $\eta \to \infty$ in the above inequality and using the definition of $A_1\left(\mathfrak{x}\right)$ and $A_2\left(\mathfrak{x}\right)$, we see that

$$A_1\left(\left(\sum _{\vartheta =1}^n{(-1)}^{\vartheta -1}{\vartheta }^2\right)\mathfrak{x}\right)={(-1)}^{n-1}\frac{n(n+1)}{2}{A}_1\left(\mathfrak{x}\right)$$

$$A_2\left(\left(\sum _{\vartheta =1}^{n}n(n-1)+2\vartheta -1\right)\mathfrak{x}\right)=n^3{A}_2\left(\mathfrak{x}\right)$$

Hence, $A_1\left(\mathfrak{x}\right)$ and $A_2\left(\mathfrak{x}\right)$ satisfy (4) and (5) ∀$\mathfrak{x}\in \mathcal{M}$.

To prove $A_1\left(\mathfrak{x}\right)$ and $A_2\left(\mathfrak{x}\right)$ is unique, consider $\check{A_1}\left(\mathfrak{x}\right)$ and $\check{A_2}\left(\mathfrak{x}\right)$ to be another mapping satisfying (4), (5), (14), and (15). Then,

$$\begin{array}{cc}\hfill & ∥A_1\left(\mathfrak{x}\right)-\check{A_1}\left(\mathfrak{x}\right)∥=\frac{1}{{\mathcal{H}}_1^{\eta }}∥A_1\left({\mathcal{H}}_1^{\eta }\mathfrak{x}\right)-\check{A_1}\left({\mathcal{H}}_1^{\eta }\mathfrak{x}\right)∥\hfill \\ \hfill & \le \frac{1}{{\mathcal{H}}_1^{\eta }}\left\{∥A_1\left({\mathcal{H}}_1^{\eta }\mathfrak{x}\right)-{\yen }_1\left({\mathcal{H}}_1^{\eta }\mathfrak{x}\right)∥+∥{\yen }_1\left({\mathcal{H}}_1^n\mathfrak{x}\right)-\check{A_1}\left({\mathcal{H}}_1^{\eta }\mathfrak{x}\right)∥\right\}\hfill \\ \hfill & \le \sum _{\lambda =0}^{\infty }\frac{2{\mathcal{Z}}_1\left({\mathcal{H}}_1^{\lambda +\eta }\mathfrak{x}\right)}{{\mathcal{H}}_1^{(\lambda +\eta )}}\hfill \\ \hfill & \to 0as\eta \to \infty \hfill \end{array}$$

$$\begin{array}{cc}\hfill & ∥A_2\left(\mathfrak{x}\right)-\check{A_2}\left(\mathfrak{x}\right)∥=\frac{1}{{\mathcal{H}}_2^{\eta }}∥A_2\left({\mathcal{H}}_2^{\eta }\mathfrak{x}\right)-\check{A_2}\left({\mathcal{H}}_2^{\eta }\mathfrak{x}\right)∥\hfill \\ \hfill & \le \frac{1}{{\mathcal{H}}_2^{\eta }}\left\{∥A_2\left({\mathcal{H}}_2^{\eta }\mathfrak{x}\right)-{\yen }_2\left({\mathcal{H}}_2^{\eta }\mathfrak{x}\right)∥+∥{\yen }_2\left({\mathcal{H}}_2^n\mathfrak{x}\right)-\check{A_2}\left({\mathcal{H}}_1^{\eta }\mathfrak{x}\right)∥\right\}\hfill \\ \hfill & \le \sum _{\lambda =0}^{\infty }\frac{2{\mathcal{Z}}_2\left({\mathcal{H}}_2^{\lambda +\eta }\mathfrak{x}\right)}{{\mathcal{H}}_2^{(\lambda +\eta )}}\hfill \\ \hfill & \to 0as\eta \to \infty \hfill \end{array}$$

∀$\mathfrak{x}\in \mathcal{M}$.

Hence, $A_1$ and $A_2$ are unique. □

Corollary 1.

Let a function ${\yen }_1,{\yen }_2:\mathcal{M}\to \mathcal{N}$ satisfy

$$\begin{array}{c}\hfill ∥{\yen }_1\left(\left(\sum _{\vartheta =1}^n{(-1)}^{\vartheta -1}{\vartheta }^2\right)\mathfrak{x}\right)-{(-1)}^{n-1}\frac{n(n+1)}{2}{\yen }_1\left(\mathfrak{x}\right)∥\le \left\{\begin{array}{cc}T,\hfill & \\ {T\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }},\hfill & \mathsf{\Phi }\ne 1;\hfill \end{array}\right.\end{array}$$

(30)

$$\begin{array}{c}\hfill ∥{\yen }_1\left(\left(\sum _{\vartheta =1}^{n}n(n-1)+2\vartheta -1\right)\mathfrak{x}\right)-n^3{\yen }_1\left(\mathfrak{x}\right)∥\le \left\{\begin{array}{cc}T,\hfill & \\ {T\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }},\hfill & \mathsf{\Phi }\ne 1;\hfill \end{array}\right.\end{array}$$

(31)

∀$\mathfrak{x}\in \mathcal{M}$ with T and Φ be non-negative real numbers. Then, ∃ a function $A_1,A_2:\mathcal{M}\to \mathcal{N}$ such that

$$∥{\yen }_1\left(\mathfrak{x}\right)-A_1\left(\mathfrak{x}\right)∥\le \left\{\begin{array}{c}{\displaystyle \frac{T}{{\mathcal{H}}_1-1},}\hfill \\ {\displaystyle \frac{{T\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }}}{|{\mathcal{H}}_1-{\mathcal{H}}_1^{\mathsf{\Phi }}|},}\hfill \end{array}\right.$$

(32)

$$∥{\yen }_2\left(\mathfrak{x}\right)-A_2\left(\mathfrak{x}\right)∥\le \left\{\begin{array}{c}{\displaystyle \frac{T}{{\mathcal{H}}_2-1},}\hfill \\ {\displaystyle \frac{{T\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }}}{|{\mathcal{H}}_2-{\mathcal{H}}_2^{\mathsf{\Phi }}|},}\hfill \end{array}\right.$$

(33)

∀$\mathfrak{x}\in \mathcal{M}.$

5. Stability Analysis Using FPT

In this section, the stability of the new series type additive FEs (4) and (5) are derived using a fixed point technique. Let $\mathcal{M}$ be a normed space and $\mathcal{N}$ be a Banach space.

The following are some fundamental concepts of fixed point theory.

Theorem 5.

(Banach’s contraction principle) Let $(\mathfrak{x},d)$ be a complete metric space and consider a mapping $T:\mathcal{M}\to \mathcal{M}$, which is strictly a contractive mapping, that is

(A1) 

$d(T\mathfrak{x},T\mathfrak{y})\le Ld(\mathfrak{x},\mathfrak{y})$ for some (Lipschitz constant) $L<1$. Then,

(i) The mapping T has only one fixed point ${\mathfrak{x}}^{\ast }=T\left({\mathfrak{x}}^{\ast }\right);$

(ii) The fixed point for each given element ${\mathfrak{x}}^{\ast }$ is globally attractive, that is

(A2) 

$li{m}_{n\to \infty }{T}^n\mathfrak{x}={\mathfrak{x}}^{\ast },$ for any starting point $\mathfrak{x}\in \mathcal{M}$;

(iii) One has the following estimation inequalities:

(A3) 

$d(T^n\mathfrak{x},{\mathfrak{x}}^{\ast })\le \frac{1}{1-L}d(T^n\mathfrak{x},T^{n+1}\mathfrak{x},\forall n\ge 0,\forall \mathfrak{x}\in \mathcal{M};$

(A4) 

$d(\mathfrak{x},{\mathfrak{x}}^{\ast })\le \frac{1}{1-L}d(\mathfrak{x},{\mathfrak{x}}^{\ast }),\forall \mathfrak{x}\in \mathcal{M}.$

Theorem 6.

(The alternative of fixed point) Suppose that for a complete generalized metric space $(\mathcal{M},d)$ and a strictly contractive mapping $T:\mathcal{M}\to \mathcal{M}$ with Lipschitz constant L. Then, for each given element $\mathfrak{x}\in \mathcal{M},$ either

$\left({\mathcal{C}}_1\right)d(T^n\mathfrak{x},T^{n+1}\mathfrak{x})=\infty \forall n\ge 0,$

or

$\left({\mathcal{C}}_2\right)$ there exists a natural number $n_0$ such that:

$\left(1\right)$$d(T^n\mathfrak{x},T^{n+1}\mathfrak{x})<\infty$ for all $n\ge n_0$;

$\left(2\right)$ The sequence $\left(T^n\mathfrak{x}\right)$ is convergent to a fixed point ${\mathfrak{y}}^{\ast }$ of T;

$\left(3\right)$${\mathfrak{y}}^{\ast }$ is the unique fixed point of T in the set $\mathcal{N}=\{\mathfrak{y}\in \mathcal{M}:d(T^{n_0}\mathfrak{x},\mathfrak{y})<\infty \};$

$\left(4\right)$$d({\mathfrak{y}}^{\ast },\mathfrak{y})\le \frac{1}{1-L}d(\mathfrak{y},T\mathfrak{y})$ for all $\mathfrak{y}\in \mathcal{M}.$

Some fundamentals and derived results for FPT can be referenced in [27,28,29,30,31,32,33].

Theorem 7.

Consider the mapping ${\yen }_1,{\yen }_2:\mathcal{M}\to \mathcal{N}$ for which ∃ a function ${\mathcal{Z}}_1,{\mathcal{Z}}_2:\mathcal{M}\to [0,\infty )$ with

$$\underset{\lambda \to \infty }{lim}\frac{{\mathcal{Z}}_1\left({\mathcal{J}}_{\nu }^{\lambda }\mathfrak{x}\right)}{{\mathcal{J}}_{\nu }^{\lambda }}=0$$

(34)

$$\underset{\lambda \to \infty }{lim}\frac{{\mathcal{Z}}_2\left({\mathcal{J}}_{\nu }^{\lambda }\mathfrak{x}\right)}{{\mathcal{J}}_{\nu }^{\lambda }}=0$$

(35)

where ${\mathcal{J}}_{\nu }={\mathcal{H}}_1,{\mathcal{H}}_2$ if $\nu =0$ and ${\mathcal{J}}_{\nu }=\frac{1}{{\mathcal{H}}_1},\frac{1}{{\mathcal{H}}_2}$ if $\nu =1$, then

$$∥{\yen }_1\left(\left(\sum _{\vartheta =1}^n{(-1)}^{\vartheta -1}{\vartheta }^2\right)\mathfrak{x}\right)-{(-1)}^{n-1}\frac{n(n+1)}{2}{\yen }_1\left(\mathfrak{x}\right)∥\le {\mathcal{Z}}_1\left(\mathfrak{x}\right)$$

(36)

$$∥{\yen }_2\left(\left(\sum _{\vartheta =1}^{n}n(n-1)+2\vartheta -1\right)\mathfrak{x}\right)-n^3{\yen }_2\left(\mathfrak{x}\right)∥\le {\mathcal{Z}}_2\left(\mathfrak{x}\right)$$

(37)

If ∃$L=L\left(\nu \right)<1$ such that the function

$$\mathcal{B}\left(\mathfrak{x}\right)={\mathcal{Z}}_1\left(\frac{\mathfrak{x}}{{\mathcal{H}}_1}\right),$$

$$\mathcal{B}\left(\mathfrak{x}\right)={\mathcal{Z}}_2\left(\frac{\mathfrak{x}}{{\mathcal{H}}_2}\right),$$

has the property

$$L\mathcal{B}\left(\mathfrak{x}\right)\le \frac{1}{{\mathcal{J}}_{\nu }}\mathcal{B}\left(\mathfrak{x}{\mathcal{J}}_{\nu }\right)$$

(38)

$$L\mathcal{B}\left(\mathfrak{x}\right)\le \frac{1}{{\mathcal{J}}_{\nu }}\mathcal{B}\left(\mathfrak{x}{\mathcal{J}}_{\nu }\right)$$

(39)

∀$\mathfrak{x}\in \mathcal{M}$. Then, ∃ a unique additive mapping $A_1,A_2:\mathcal{M}\to \mathcal{N}$ satisfying the FE (4) and (5) such that

$$∥{\yen }_1\left(\mathfrak{x}\right)-A_1\left(\mathfrak{x}\right)∥\le \frac{L^{1-\nu }}{1-L}\mathcal{B}\left(\mathfrak{x}\right)$$

(40)

$$∥{\yen }_2\left(\mathfrak{x}\right)-A_2\left(\mathfrak{x}\right)∥\le \frac{L^{1-\nu }}{1-L}\mathcal{B}\left(\mathfrak{x}\right)$$

(41)

holds ∀$\mathfrak{x}\in \mathcal{M}$.

Proof. 

Let us assume that the set $\mathsf{\Psi }=\{{\eta }_1/{\eta }_1:\mathcal{M}\to \mathcal{N},{\eta }_1\left(0\right)=0\}$, $\mathsf{\Psi }=\{{\eta }_2/{\eta }_2:\mathcal{M}\to \mathcal{N},{\eta }_2\left(0\right)=0\}$.

Introduce the generalized metric on $\mathsf{\Psi }$,

$$d({\eta }_1,{\eta }_2)=inf\{K\in (0,\infty ):\Vert {\eta }_1\left(\mathfrak{x}\right)-{\eta }_2\left(\mathfrak{x}\right)\Vert \le K\mathcal{B}\left(\mathfrak{x}\right),\mathfrak{x}\in \mathcal{M}\}.$$

i.e., $(\mathsf{\Psi },d)$ is complete.

Define the mapping $T:\mathsf{\Psi }\to \mathsf{\Psi }$, then

$$T{\yen }_1\left(\mathfrak{x}\right)=\frac{1}{{\mathcal{J}}_{\nu }}{\yen }_1\left({\mathcal{J}}_{\nu }\mathfrak{x}\right)$$

$$T{\yen }_2\left(\mathfrak{x}\right)=\frac{1}{{\mathcal{J}}_{\nu }}{\yen }_2\left({\mathcal{J}}_{\nu }\mathfrak{x}\right)$$

∀$\mathfrak{x}\in \mathsf{\Psi }$.

Let us use ${\eta }_1,{\eta }_2\in \mathsf{\Psi }$,

$$\begin{array}{cc}\hfill & d({\eta }_1,{\eta }_2)\le K\hfill \\ \hfill \Rightarrow & \Vert {\eta }_1\left(\mathfrak{x}\right)-{\eta }_2\left(\mathfrak{x}\right)\Vert \le K\mathcal{B}\left(\mathfrak{x}\right),\mathfrak{x}\in \mathcal{M}.\hfill \\ \hfill \Rightarrow & ∥\frac{1}{{\mathcal{J}}_{\nu }}{\eta }_1\left({\mathcal{J}}_{\nu }\mathfrak{x}\right)-\frac{1}{{\mathcal{J}}_{\nu }}{\eta }_2\left({\mathcal{J}}_{\nu }\mathfrak{x}\right)∥\le \frac{1}{{\mathcal{J}}_{\nu }}K\mathcal{B}\left({\mathcal{J}}_{\nu }\mathfrak{x}\right),\mathfrak{x}\in \mathcal{M},\hfill \\ \hfill \Rightarrow & ∥\frac{1}{{\mathcal{J}}_{\nu }}{\eta }_1\left({\mathcal{J}}_{\nu }\mathfrak{x}\right)-\frac{1}{{\mathcal{J}}_{\nu }}{\eta }_2\left({\mathcal{J}}_{\nu }\mathfrak{x}\right)∥\le LK\mathcal{B}\left(\mathfrak{x}\right),\mathfrak{x}\in \mathcal{M},\hfill \\ \hfill \Rightarrow & \Vert T{\eta }_1\left(\mathfrak{x}\right)-T{\eta }_2\left(\mathfrak{x}\right)\Vert \le LK\mathcal{B}\left(\mathfrak{x}\right),\mathfrak{x}\in \mathcal{M},\hfill \\ \hfill \Rightarrow & d(T{\eta }_1,T{\eta }_2)\le LK.\hfill \end{array}$$

⟹$d(T{\eta }_1,T{\eta }_2)\le Ld({\eta }_1,{\eta }_2)$, ∀${\eta }_1,{\eta }_2\in \mathsf{\Psi }$.

i.e., T is a strictly contractive mapping on $\mathsf{\Psi }$ with Lipschitz constant L. Using ${\yen }_1$ and ${\yen }_2$ in (36) and (37), we obtain

$$∥{\yen }_1\left(\mathfrak{x}\right)-\frac{{\yen }_2\left({\mathcal{H}}_1\mathfrak{x}\right))}{{\mathcal{H}}_1}∥\le \frac{{\mathcal{Z}}_1\left(\mathfrak{x}\right)}{{\mathcal{H}}_1}$$

(42)

$$∥{\yen }_2\left(\mathfrak{x}\right)-\frac{{\yen }_2\left({\mathcal{H}}_2\mathfrak{x}\right)}{{\mathcal{H}}_2}∥\le \frac{{\mathcal{Z}}_2\left(\mathfrak{x}\right)}{{\mathcal{H}}_2}$$

(43)

∀$\mathfrak{x}\in \mathcal{M}$. For the case $\nu =0$, (38) and (39) reduce to

$$∥{\yen }_1\left(\mathfrak{x}\right)-\frac{{\yen }_1\left({\mathcal{H}}_1\mathfrak{x}\right)}{{\mathcal{H}}_1}∥\le \frac{1}{{\mathcal{H}}_1}\mathcal{B}\left(\mathfrak{x}\right)$$

$$∥{\yen }_2\left(\mathfrak{x}\right)-\frac{{\yen }_2\left({\mathcal{H}}_2\mathfrak{x}\right)}{{\mathcal{H}}_2}∥\le \frac{1}{{\mathcal{H}}_2}\mathcal{B}\left(\mathfrak{x}\right)$$

∀$\mathfrak{x}\in \mathcal{M}$.

$$\mathrm{i}.\mathrm{e}.,d({\yen }_1,T{\yen }_1)\le \frac{1}{{\mathcal{H}}_1}\Rightarrow d({\yen }_1,T{\yen }_1)\le \frac{1}{{\mathcal{H}}_1}=L=L^1<\infty .$$

$$\mathrm{i}.\mathrm{e}.,d({\yen }_2,T{\yen }_2)\le \frac{1}{{\mathcal{H}}_2}\Rightarrow d({\yen }_2,T{\yen }_2)\le \frac{1}{{\mathcal{H}}_2}=L=L^1<\infty .$$

Again, replacing $\mathfrak{x}=\frac{\mathfrak{x}}{{\mathcal{H}}_1}$ in (42) and $\mathfrak{x}=\frac{\mathfrak{x}}{{\mathcal{H}}_2}$ in (43), we obtain

$$∥{\yen }_1\left(\mathfrak{x}\right)-{\mathcal{H}}_1{\yen }_1\left(\frac{\mathfrak{x}}{{\mathcal{H}}_1}\right)∥\le {\mathcal{Z}}_1\left(\frac{\mathfrak{x}}{{\mathcal{H}}_1}\right).$$

$$∥{\yen }_2\left(\mathfrak{x}\right)-{\mathcal{H}}_2{\yen }_2\left(\frac{\mathfrak{x}}{{\mathcal{H}}_2}\right)∥\le {\mathcal{Z}}_2\left(\frac{\mathfrak{x}}{{\mathcal{H}}_2}\right).$$

∀$\mathfrak{x}\in \mathcal{M}$. For the case $\nu =1$, (38) and (39) reduce to

$$∥{\yen }_1\left(\mathfrak{x}\right)-{\mathcal{H}}_1{\yen }_1\left(\frac{\mathfrak{x}}{{\mathcal{H}}_1}\right)∥\le \mathcal{B}\left(\mathfrak{x}\right)$$

$$∥{\yen }_2\left(\mathfrak{x}\right)-{\mathcal{H}}_2{\yen }_2\left(\frac{\mathfrak{x}}{{\mathcal{H}}_2}\right)∥\le \mathcal{B}\left(\mathfrak{x}\right)$$

∀$\mathfrak{x}\in \mathcal{M}$.

$$\mathrm{i}.\mathrm{e}.,d({\yen }_1,T{\yen }_1)\le 1\Rightarrow d({\yen }_1,T{\yen }_1\le 1=L^0<\infty .$$

$$\mathrm{i}.\mathrm{e}.,d({\yen }_2,T{\yen }_2)\le 1\Rightarrow d({\yen }_2,T{\yen }_2)\le 1=L^0<\infty .$$

In both the cases, the following inequality is obtained.

$$d({\yen }_1,T{\yen }_1)\le L^{1-\nu }$$

$$d({\yen }_2,T_2)\le L^{1-\nu }$$

Therefore, $\left({\mathcal{C}}_2\left(1\right)\right)$ holds.

By $\left({\mathcal{C}}_2\left(2\right)\right)$, it follows that ∃ fixed points $A_1$ and $A_2$ of T in $\mathcal{M}$ such that

$$A_1\left(\mathfrak{x}\right)=\underset{\lambda \to \infty }{lim}\frac{{\yen }_1\left({\mathcal{J}}_{\nu }^{\lambda }\mathfrak{x}\right)}{{\mathcal{J}}_{\nu }^{\lambda }}\forall \mathfrak{x}\in \mathcal{M}.$$

(44)

$$A_2\left(\mathfrak{x}\right)=\underset{\lambda \to \infty }{lim}\frac{{\yen }_2\left({\mathcal{J}}_{\nu }^{\lambda }\mathfrak{x}\right)}{{\mathcal{J}}_{\nu }^{\lambda }}\forall \mathfrak{x}\in \mathcal{M}.$$

(45)

To prove $A_1,A_2:\mathcal{M}\to \mathcal{N}$ is additive.

Replacing $\mathfrak{x}$ by $\left({\mathcal{J}}_{\nu }^{\lambda }\mathfrak{x}\right)$ in (36) and (37) and dividing by ${\mathcal{J}}_{\nu }^{\lambda },$ it follows from (34), (35), (44), and (45), that $A_1$ and $A_2$ satisfy (4) and (5) ∀$\mathfrak{x}\in \mathcal{M}$, i.e., $A_1$ and $A_2$ satisfy the FEs (4) and (5).

By $\left({\mathcal{C}}_2\left(3\right)\right)$, the functions $A_1$ and $A_2$ are the unique fixed points of T in the set

$$Y=\{A_1\left(\mathfrak{x}\right)\in \mathcal{M}:d(T{\yen }_1\left(\mathfrak{x}\right),A_1\left(\mathfrak{x}\right))<\infty \},$$

$$Y=\{A_2\left(\mathfrak{x}\right)\in \mathcal{M}:d(T{\yen }_1\left(\mathfrak{x}\right),A_2\left(\mathfrak{x}\right))<\infty \},$$

The functions $A_1$ and $A_2$ are unique. Using the fixed point alternative result,

$$∥{\yen }_1\left(\mathfrak{x}\right)-A_1\left(\mathfrak{x}\right)∥\le K\mathcal{B}\left(\mathfrak{x}\right)$$

$$∥{\yen }_2\left(\mathfrak{x}\right)-A_2\left(\mathfrak{x}\right)∥\le K\mathcal{B}\left(\mathfrak{x}\right)$$

Finally, using $\left({\mathcal{C}}_2\left(4\right)\right)$, the following result is obtained.

$$d({\yen }_1,A_1)\le \frac{1}{1-L}d({\yen }_1,T{\yen }_1)$$

$$d({\yen }_2,A_2)\le \frac{1}{1-L}d({\yen }_2,T{\yen }_2)$$

This implies that

$$d({\yen }_1,A_1)\le \frac{L^{1-\nu }}{1-L}.$$

$$d({\yen }_2,A_2)\le \frac{L^{1-\nu }}{1-L}$$

Thus, it can be concluded that

$$\Vert {\yen }_1\left(\mathfrak{x}\right)-A_1\left(\mathfrak{x}\right)\Vert \le \frac{L^{1-\nu }}{1-L}\mathcal{B}\left(\mathfrak{x}\right)$$

$$\Vert {\yen }_1\left(\mathfrak{x}\right)-A_1\left(\mathfrak{x}\right)\Vert \le \frac{L^{1-\nu }}{1-L}\mathcal{B}\left(\mathfrak{x}\right),\forall \mathfrak{x}\in \mathcal{M}$$

□

Corollary 2.

Consider the function ${\yen }_1,{\yen }_2:\mathcal{M}\to \mathcal{N}$ satisfies

$$\begin{array}{c}\hfill ∥{\yen }_1\left(\left(\sum _{\vartheta =1}^n{(-1)}^{\vartheta -1}{\vartheta }^2\right)\mathfrak{x}\right)-{(-1)}^{n-1}\frac{n(n+1)}{2}{\yen }_1\left(\mathfrak{x}\right)∥\le \left\{\begin{array}{cc}T,\hfill & \\ {T\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }},\hfill & \mathsf{\Phi }\ne 1;\hfill \end{array}\right.\end{array}$$

(46)

$$\begin{array}{c}\hfill ∥{\yen }_2\left(\left(\sum _{\vartheta =1}^{n}n(n-1)+2\vartheta -1\right)\mathfrak{x}\right)-n^3{\yen }_2\left(\mathfrak{x}\right)∥\le \left\{\begin{array}{cc}T,\hfill & \\ {T\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }},\hfill & \mathsf{\Phi }\ne 1;\hfill \end{array}\right.\end{array}$$

(47)

∀$\mathfrak{x}\in \mathcal{M}$ with T and Φ be real numbers. Then, ∃ additive mapping $A_1,A_2:\mathcal{M}\to \mathcal{N}$ that

$$∥{\yen }_1\left(\mathfrak{x}\right)-A_1\left(\mathfrak{x}\right)∥\le \left\{\begin{array}{c}{\displaystyle \frac{T}{{\mathcal{H}}_1-1},}\hfill \\ {\displaystyle \frac{{T\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }}}{|{\mathcal{H}}_1-{\mathcal{H}}_1^{\mathsf{\Phi }}|},}\hfill \end{array}\right.$$

(48)

$$∥{\yen }_2\left(\mathfrak{x}\right)-A_2\left(\mathfrak{x}\right)∥\le \left\{\begin{array}{c}{\displaystyle \frac{T}{{\mathcal{H}}_2-1},}\hfill \\ {\displaystyle \frac{{T\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }}}{|{\mathcal{H}}_2-{\mathcal{H}}_2^{\mathsf{\Phi }}|},}\hfill \end{array}\right.$$

(49)

Proof. 

Let

$${\mathcal{Z}}_1\left(\mathfrak{x}\right)=\left\{\begin{array}{c}T\hfill \\ {\displaystyle T\left\{{\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }}\right\},}\hfill \end{array}\right.$$

$${\mathcal{Z}}_2\left(\mathfrak{x}\right)=\left\{\begin{array}{c}T\hfill \\ {\displaystyle T\left\{{\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }}\right\},}\hfill \end{array}\right.$$

∀$\mathfrak{x}\in \mathcal{M}$. Now,

$$\begin{array}{cc}\hfill \frac{{\mathcal{Z}}_1\left({\mathcal{J}}_{\nu }^{\lambda }\mathfrak{x}\right)}{{\mathcal{J}}_{\nu }^{\lambda }}& =\left\{\begin{array}{c}{\displaystyle \frac{T}{{\mathcal{J}}_{\nu }^{\lambda }}}\hfill \\ {\displaystyle \frac{T}{{\mathcal{J}}_{\nu }^{\lambda }}\left\{\left|\right|{\mathcal{J}}_{\nu }^{\lambda }\mathfrak{x}{\left|\right|}^{\mathsf{\Phi }}\right\}}\hfill \end{array}\right.\hfill \\ \hfill & =\left\{\begin{array}{c}\to 0\mathrm{as}\lambda \to \infty ,\hfill \\ \to 0\mathrm{as}\lambda \to \infty \hfill \end{array}\right.\hfill \end{array}$$

$$\begin{array}{cc}\hfill \frac{{\mathcal{Z}}_2\left({\mathcal{J}}_{\nu }^{\lambda }\mathfrak{x}\right)}{{\mathcal{J}}_{\nu }^{\lambda }}& =\left\{\begin{array}{c}{\displaystyle \frac{T}{{\mathcal{J}}_{\nu }^{\lambda }}}\hfill \\ {\displaystyle \frac{T}{{\mathcal{J}}_{\nu }^{\lambda }}\left\{\left|\right|{\mathcal{J}}_{\nu }^{\lambda }\mathfrak{x}{\left|\right|}^{\mathsf{\Phi }}\right\}}\hfill \end{array}\right.\hfill \\ \hfill & =\left\{\begin{array}{c}\to 0\mathrm{as}\lambda \to \infty ,\hfill \\ \to 0\mathrm{as}\lambda \to \infty \hfill \end{array}\right.\hfill \end{array}$$

i.e., (34) and (35) holds.

The function $\mathcal{B}\left(\mathfrak{x}\right)={\mathcal{Z}}_1\left(\frac{\mathfrak{x}}{{\mathcal{H}}_1}\right)$ and $\mathcal{B}\left(\mathfrak{x}\right)={\mathcal{Z}}_2\left(\frac{\mathfrak{x}}{{\mathcal{H}}_2}\right)$ has the property

$$L\mathcal{B}\left(\mathfrak{x}\right)\le \frac{1}{{\mathcal{J}}_{\nu }}\mathcal{B}\left(\mathfrak{x}{\mathcal{J}}_{\nu }\right),$$

$$L\mathcal{B}\left(\mathfrak{x}\right)\le \frac{1}{{\mathcal{J}}_{\nu }}\mathcal{B}\left(\mathfrak{x}{\mathcal{J}}_{\nu }\right)$$

∀$\mathfrak{x}\in \mathcal{M}$. Hence,

$$\begin{array}{c}\hfill \mathcal{B}\left(\mathfrak{x}\right)={\mathcal{Z}}_1\left(\frac{\mathfrak{x}}{{\mathcal{H}}_1}\right)=\left\{\begin{array}{c}{\displaystyle T}\hfill \\ {\displaystyle \frac{T}{{\mathcal{H}}_1^s}{\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }},}\hfill \end{array}\right.\end{array}$$

$$\begin{array}{c}\hfill \mathcal{B}\left(\mathfrak{x}\right)={\mathcal{Z}}_2\left(\frac{\mathfrak{x}}{{\mathcal{H}}_2}\right)=\left\{\begin{array}{c}{\displaystyle T}\hfill \\ {\displaystyle \frac{T}{{\mathcal{H}}_2^s}{\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }},}\hfill \end{array}\right.\end{array}$$

Furthermore,

$$\begin{array}{cc}\hfill \frac{1}{{\mathcal{J}}_{\nu }}\mathcal{B}\left({\mathcal{J}}_{\nu }\mathfrak{x}\right)& =\left\{\begin{array}{c}{\displaystyle \frac{T}{{\mathcal{J}}_{\nu }}}\hfill \\ {\displaystyle \frac{T}{{\mathcal{J}}_{\nu }}\left|\right|{\mathcal{J}}_{\nu }\mathfrak{x}{\left|\right|}^{\mathsf{\Phi }}}\hfill \end{array}\right.\hfill \\ \hfill & =\left\{\begin{array}{c}{\displaystyle {\mathcal{J}}_{\nu }^{-1}\mathcal{B}\left(\mathfrak{x}\right)}\hfill \\ {\displaystyle {\mathcal{J}}_{\nu }^{s-1}\mathcal{B}\left(\mathfrak{x}\right)}\hfill \end{array}\right.\hfill \end{array}$$

$$\begin{array}{cc}\hfill \frac{1}{{\mathcal{J}}_{\nu }}\mathcal{B}\left({\mathcal{J}}_{\nu }\mathfrak{x}\right)& =\left\{\begin{array}{c}{\displaystyle \frac{T}{{\mathcal{J}}_{\nu }}}\hfill \\ {\displaystyle \frac{T}{{\mathcal{J}}_{\nu }}\left|\right|{\mathcal{J}}_{\nu }\mathfrak{x}{\left|\right|}^{\mathsf{\Phi }}}\hfill \end{array}\right.\hfill \\ \hfill & =\left\{\begin{array}{c}{\displaystyle {\mathcal{J}}_{\nu }^{-1}\mathcal{B}\left(\mathfrak{x}\right)}\hfill \\ {\displaystyle {\mathcal{J}}_{\nu }^{s-1}\mathcal{B}\left(\mathfrak{x}\right)}\hfill \end{array}\right.\hfill \end{array}$$

Now, from (40) and (41),

Axiom 1.$L={\mathcal{H}}_1^{-1}$, if $\nu =0$

$$\begin{array}{c}\hfill ∥{\yen }_1\left(\mathfrak{x}\right)-A_1\left(\mathfrak{x}\right)∥\le \frac{L^{1-i}}{1-L}\mathcal{B}\left(\mathfrak{x}\right)=\frac{{\left({\mathcal{H}}_1^{-1}\right)}^{1-0}}{1-{\left({\mathcal{H}}_1\right)}^{-1}}T=\frac{T}{{\mathcal{H}}_1-1}.\end{array}$$

Axiom 2.$L=\frac{1}{{\mathcal{H}}_1^{-1}}$, if $\nu =1$

$$\begin{array}{c}\hfill ∥{\yen }_1\left(\mathfrak{x}\right)-A_1\left(\mathfrak{x}\right)∥\le \frac{L^{1-i}}{1-L}\mathcal{B}\left(\mathfrak{x}\right)=\frac{{\left(\frac{1}{{\mathcal{H}}_1^{-1}}\right)}^{1-1}}{1-\frac{1}{{\mathcal{H}}_1^{-1}}}T=\frac{T}{1-{\mathcal{H}}_1}.\end{array}$$

Axiom 3.$L={\mathcal{H}}_2^{-1}$, if $\nu =0$

$$\begin{array}{c}\hfill ∥{\yen }_2\left(\mathfrak{x}\right)-A_2\left(\mathfrak{x}\right)∥\le \frac{L^{1-i}}{1-L}\mathcal{B}\left(\mathfrak{x}\right)=\frac{{\left({\mathcal{H}}_2^{-1}\right)}^{1-0}}{1-{\left({\mathcal{H}}_2\right)}^{-1}}T=\frac{T}{{\mathcal{H}}_2-1}.\end{array}$$

Axiom 4.$L=\frac{1}{{\mathcal{H}}_2^{-1}}$, if $\nu =1$

$$\begin{array}{c}\hfill ∥{\yen }_2\left(\mathfrak{x}\right)-A_2\left(\mathfrak{x}\right)∥\le \frac{L^{1-i}}{1-L}\mathcal{B}\left(\mathfrak{x}\right)=\frac{{\left(\frac{1}{{\mathcal{H}}_2^{-1}}\right)}^{1-1}}{1-\frac{1}{{\mathcal{H}}_2^{-1}}}T=\frac{T}{1-{\mathcal{H}}_2}.\end{array}$$

Axiom 5.$L={\mathcal{H}}_1^{\mathsf{\Phi }}$, if $\nu =0$

$$\begin{array}{cc}\hfill & ∥{\yen }_1\left(\mathfrak{x}\right)-A_1\left(\mathfrak{x}\right)∥\le \frac{L^{1-i}}{1-L}\mathcal{B}\left(\mathfrak{x}\right)\hfill \\ \hfill & =\frac{{\left({\mathcal{H}}_1^{\mathsf{\Phi }}\right)}^{1-0}}{1-{\left({\mathcal{H}}_1\right)}^{\mathsf{\Phi }}}\frac{T}{{\mathcal{H}}_1^{\mathsf{\Phi }}}{\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }}\hfill \\ \hfill & =\frac{{T\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }}}{{\mathcal{H}}_1-{\mathcal{H}}_1^{\mathsf{\Phi }}}.\hfill \end{array}$$

Axiom 6.$L=\frac{1}{{\mathcal{H}}_1^{\mathsf{\Phi }}}$, if $\nu =1$

$$\begin{array}{cc}\hfill & ∥{\yen }_2\left(\mathfrak{x}\right)-A_2\left(\mathfrak{x}\right)∥\le \frac{L^{1-i}}{1-L}\mathcal{B}\left(\mathfrak{x}\right)\hfill \\ \hfill & =\frac{{\left(\frac{1}{{\mathcal{H}}_1^{\mathsf{\Phi }}}\right)}^{1-1}}{1-\frac{1}{{\mathcal{H}}_1^{\mathsf{\Phi }}}}\frac{T}{{\mathcal{H}}_1^{\mathsf{\Phi }}}{\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }}\hfill \\ \hfill & =\frac{{T\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }}}{{\mathcal{H}}_1^{\mathsf{\Phi }}-{\mathcal{H}}_1}.\hfill \end{array}$$

Axiom 7.$L={\mathcal{H}}_2^{\mathsf{\Phi }}$, if $\nu =0$

$$\begin{array}{cc}\hfill & ∥{\yen }_2\left(\mathfrak{x}\right)-A_2\left(\mathfrak{x}\right)∥\le \frac{L^{1-i}}{1-L}\mathcal{B}\left(\mathfrak{x}\right)\hfill \\ \hfill & =\frac{{\left({\mathcal{H}}_2^{\mathsf{\Phi }}\right)}^{1-0}}{1-{\left({\mathcal{H}}_2\right)}^{\mathsf{\Phi }}}\frac{T}{{\mathcal{H}}_2^{\mathsf{\Phi }}}{\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }}\hfill \\ \hfill & =\frac{{T\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }}}{{\mathcal{H}}_2-{\mathcal{H}}_2^{\mathsf{\Phi }}}.\hfill \end{array}$$

Axiom 8.$L=\frac{1}{{\mathcal{H}}_2^{\mathsf{\Phi }}}$, if $\nu =1$

$$\begin{array}{cc}\hfill & ∥{\yen }_2\left(\mathfrak{x}\right)-A_2\left(\mathfrak{x}\right)∥\le \frac{L^{1-i}}{1-L}\mathcal{B}\left(\mathfrak{x}\right)\hfill \\ \hfill & =\frac{{\left(\frac{1}{{\mathcal{H}}_2^{\mathsf{\Phi }}}\right)}^{1-1}}{1-\frac{1}{{\mathcal{H}}_2^{\mathsf{\Phi }}}}\frac{T}{{\mathcal{H}}_2^{\mathsf{\Phi }}}{\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }}\hfill \\ \hfill & =\frac{{T\left|\right|\mathfrak{x}\left|\right|}^{\mathsf{\Phi }}}{{\mathcal{H}}_2^{\mathsf{\Phi }}-{\mathcal{H}}_2}.\hfill \end{array}$$

□

6. Applications

In this section, some applications of the newly proposed FEs are explored.

Example 2.

The solution of the FE

$${\yen }_1\left(\left(\sum _{\vartheta =1}^n{(-1)}^{\vartheta -1}{\vartheta }^2\right)\mathfrak{x}\right)={(-1)}^{n-1}\frac{n(n+1)}{2}{\yen }_1\left(\mathfrak{x}\right)$$

(50)

is ${\yen }_1\left(\mathfrak{x}\right)=\mathfrak{x}$. By applying ${\yen }_1\left(\mathfrak{x}\right)=\mathfrak{x}$ in (50) and expanding the summation, it will satisfy the series $1^2-2^2+3^2-4^2+...+{(-1)}^{n-1}{n}^2={(-1)}^{n-1}\frac{n(n+1)}{2}$. This can also be verified using mathematical induction.

Proof. 

$$Let\mathbb{P}\left(n\right):\sum _{\vartheta =1}^n{(-1)}^{\vartheta -1}{\vartheta }^2={(-1)}^{n-1}\frac{n(n+1)}{2}$$

(51)

Step 1: To Prove $\mathbb{P}\left(1\right)$ is true

For $n=1$,

$$\begin{array}{cc}\hfill & \sum _{\vartheta =1}^1{(-1)}^{\vartheta -1}{\vartheta }^2={(-1)}^{1-1}\frac{1(1+1)}{2}\hfill \\ \hfill & {(-1)}^{1-1}{1}^2={(-1)}^{1-1}\frac{1(1+1)}{2}\hfill \\ \hfill & 1=1\hfill \end{array}$$

so $\mathbb{P}\left(1\right)$ is true.

Step 2: Assume that $\mathbb{P}\left(k\right)$ is true

For $n=\mathcal{L}$,

$$\begin{array}{cc}\hfill & \sum _{\vartheta =1}^{\mathcal{L}}{(-1)}^{\vartheta -1}{\vartheta }^2={(-1)}^{\mathcal{L}-1}\frac{\mathcal{L}(\mathcal{L}+1)}{2}\hfill \\ \hfill & {(-1)}^{1-1}{1}^1+{(-1)}^{2-1}{2}^2+{(-1)}^{3-1}{3}^2+{(-1)}^{4-1}{4}^2+...+{(-1)}^{\mathcal{L}-1}{\mathcal{L}}^2\hfill \\ \hfill & ={(-1)}^{\mathcal{L}-1}\frac{\mathcal{L}(\mathcal{L}+1)}{2}\hfill \\ \hfill & 1^2-2^2+3^2-4^2+...+{(-1)}^{\mathcal{L}-1}{\mathcal{L}}^2={(-1)}^{\mathcal{L}-1}\frac{\mathcal{L}(\mathcal{L}+1)}{2}\hfill \end{array}$$

so $\mathbb{P}\left(\mathcal{L}\right)$ is true.

Step 3: To prove $\mathbb{P}(\mathcal{L}+1)$ is true

For $n=\mathcal{L}+1$,

$$\begin{array}{cc}\hfill & \sum _{\vartheta =1}^{\mathcal{L}+1}{(-1)}^{\vartheta -1}{\vartheta }^2={(-1)}^{\mathcal{L}}\frac{(\mathcal{L}+1)(\mathcal{L}+2)}{2}\hfill \\ \hfill & {(-1)}^{1-1}{1}^1+{(-1)}^{2-1}{2}^2+...+{(-1)}^{\mathcal{L}-1}{\mathcal{L}}^2+{(-1)}^{\mathcal{L}}{(\mathcal{L}+1)}^2\hfill \\ \hfill & ={(-1)}^{\mathcal{L}}\frac{(\mathcal{L}+1)(\mathcal{L}+2)}{2}\hfill \\ \hfill & 1^2-2^2+3^2-4^2+...+{(-1)}^{\mathcal{L}-1}{\mathcal{L}}^2+{(-1)}^{\mathcal{L}}{(\mathcal{L}+1)}^2={(-1)}^{\mathcal{L}}\frac{(\mathcal{L}+1)(\mathcal{L}+2)}{2}\hfill \end{array}$$

so $\mathbb{P}(\mathcal{L}+1)$ is true.

Using mathematical induction, $\mathbb{P}\left(n\right)$ is true ∀ positive integer n. □

Example 3.

The solution of the FE

$${\yen }_2\left(\left(\sum _{\vartheta =1}^{n}n(n-1)+2\vartheta -1\right)\mathfrak{x}\right)=n^3{\yen }_2\left(\mathfrak{x}\right)$$

(52)

is ${\yen }_1\left(\mathfrak{x}\right)=\mathfrak{x}$. By applying ${\yen }_1\left(\mathfrak{x}\right)=\mathfrak{x}$ in (52) and expanding the summation, it will satisfy the series $1+(3+5)+(7+9+11)+(13+15+17+19)+...=n^3$. This can also be verified using mathematical induction.

Proof. 

$$Let\mathbb{P}\left(n\right):\sum _{\vartheta =1}^{n}n(n-1)+2\vartheta -1=n^3$$

(53)

Step 1: To prove $\mathbb{P}\left(1\right)$ is true

For $n=1$, we have

$$\begin{array}{cc}\hfill & \sum _{\vartheta =1}^{1}1(1-1)+2\vartheta -1=1^3\hfill \\ \hfill & 1(1-1)+2.1-1=1^3\hfill \\ \hfill & 1^3=1^3\hfill \end{array}$$

so $\mathbb{P}\left(1\right)$ is true.

Step 2: Assume that $\mathbb{P}\left(\mathcal{L}\right)$ is true

For $n=\mathcal{L}$, we have

$$\begin{array}{cc}\hfill & \sum _{\vartheta =1}^{\mathcal{L}}\mathcal{L}(\mathcal{L}-1)+2\vartheta -1={\mathcal{L}}^3\hfill \\ \hfill & \mathcal{L}(\mathcal{L}-1)+2.1-1+\mathcal{L}(\mathcal{L}-1)+2.2-1+...+\mathcal{L}(\mathcal{L}-1)+2.\mathcal{L}-1={\mathcal{L}}^3\hfill \\ \hfill & \mathcal{L}(\mathcal{L}-1)is\mathcal{L}times+(1+3+5+...+(2\mathcal{L}-1))={\mathcal{L}}^3\hfill \\ \hfill & \mathcal{L}(\mathcal{L}-1)is\mathcal{L}times+\left(sumof\mathcal{L}oddtermsis{\mathcal{L}}^2\right)={\mathcal{L}}^3\hfill \\ \hfill & {\mathcal{L}}^2(\mathcal{L}-1)+{\mathcal{L}}^2={\mathcal{L}}^3\hfill \\ \hfill & {\mathcal{L}}^3={\mathcal{L}}^3\hfill \end{array}$$

so $\mathbb{P}\left(\mathcal{L}\right)$ is true.

Step 3: To prove $\mathbb{P}(\mathcal{L}+1)$ is true

For $n=\mathcal{L}+1$, we have

$$\begin{array}{cc}\hfill & \sum _{\vartheta =1}^{\mathcal{L}+1}\mathcal{L}(\mathcal{L}+1)+2\vartheta -1={(\mathcal{L}+1)}^3\hfill \\ \hfill & \mathcal{L}(\mathcal{L}+1)+2.1-1+\mathcal{L}(\mathcal{L}+1)+2.2-1+...+\mathcal{L}(\mathcal{L}+1)+2(\mathcal{L}+1)-1\hfill \\ \hfill & ={(\mathcal{L}+1)}^3\hfill \\ \hfill & \mathcal{L}(\mathcal{L}+1)is\mathcal{L}times+(1+3+5+...+(2\mathcal{L}-1))={(\mathcal{L}+1)}^3\hfill \\ \hfill & \mathcal{L}(\mathcal{L}+1)is\mathcal{L}times+(sumofk+1oddtermsis{(\mathcal{L}+1)}^2)={(\mathcal{L}+1)}^3\hfill \\ \hfill & \mathcal{L}{(\mathcal{L}+1)}^2+{(\mathcal{L}+1)}^2={(\mathcal{L}+1)}^3\hfill \\ \hfill & {(\mathcal{L}+1)}^3={(\mathcal{L}+1)}^3\hfill \end{array}$$

so $\mathbb{P}(\mathcal{L}+1)$ is true.

Using mathematical induction, $\mathbb{P}\left(n\right)$ is true ∀ positive integer n. □

7. Conclusions

In this article, two novel systems of series type additive FEs (4) and (5) have been introduced. The general solution of the equations are derived using the principle of mathematical induction and the Hyers–Ulam stability has been analysed in BSs using direct and FPT. A few potential applications of the newly introduced equations and their stability analyses are also explored to help the readers appreciate and understand the significance of the FEs. In the future, the UHS for the same Equations (4) and (5) can be determined in other normed spaces such as para-normed spaces, two normed spaces, and matrix normed spaces. This is left as an open problem for future researchers.