Representation and Stability of General Nonic Functional Equation

Abstract

In this paper, we introduce a way of representing a given mapping as the sum of odd and even mappings. Then, using this representation, we investigate the stability of various forms of the following general nonic functional equation: ${\displaystyle \sum _{i=0}^{10}}{}_{10}{C}_i{(-1)}^{10-i}f(x+iy)=0.$

1. Introduction

The various types of stability for functional equations are a very interesting subject in the field of mathematical analysis. The stability problems of functional equations have been studied by several authors (see [1,2,3,4]). In paricular, Gilányi [5] investigated the stability of a monomial functional equation in real normed spaces. Subsequent studies improved upon the results of Gilányi (for example, [6,7,8]). Moreover, hyperstability results for monomial functional equations can be found in [8,9,10]. The hyperstability of a functional equation requires that any mapping satisfying the equation approximately (in some sense) must be a real solution to it (refer to [11]).

Let X be a real normed space and Y a real Banach space. We also assume that $f:X\to Y$ is a mapping.

Now, we consider the following functional equation:

$$\sum _{i=0}^{n+1}{}_{n+1}{C}_i{(-1)}^{n+1-i}f(x+iy)=0$$

(1)

and the n-monomial functional equation

$$\begin{array}{c}\hfill \sum _{i=0}^n{}_n{C}_i{(-1)}^{n-i}f(x+iy)-n!f\left(y\right)=0.\end{array}$$

The n-monomial functional equation is called a Cauchy, quadratic, cubic, quartic, quintic, sextic, septic, octic, or nonic functional equation for $n=1,2,3,4,5,6,7,8,9$, respectively. In this paper, we discuss the general nonic functional equation:

$$\sum _{i=0}^{10}{}_{10}{C}_i{(-1)}^{10-i}f(x+iy)=0.$$

(2)

If we let

$$C_{n}f(x,y):=\sum _{i=0}^n{}_n{C}_i{(-1)}^{n-i}f(x+iy)-n!f\left(y\right)$$

and

$$D_{n}f(x,y):=\sum _{i=0}^n{}_n{C}_i{(-1)}^{n-i}f(x+iy),$$

then $D_{n}f(x,y)=C_{n-1}f(x+y,y)-C_{n-1}f(x,y)$ and $D_{n}f(x,y)=D_{n-1}f(x+y,x)-$$D_{n-1}f(x,y)$ hold. Moreover, a mapping f satisfying one of the functional equations $C_{1}f(x,y)=0$, $C_{2}f(x,y)=0$, …, $C_{9}f(x,y)=0$, $D_{2}f(x,y)=0$, $D_{3}f(x,y)=0$, …, $D_{8}f(x,y)=0$, or $D_{9}f(x,y)=0$ satisfies the general nonic function equation $D_{10}f(x,y)=0$. Hence, we refer to the functional equation described in Equation (2) as the general nonic functional equation.

In other words, the function $f:\mathbb{R}\to \mathbb{R}$ given by $f\left(x\right):=a{x}^n$ satisfies the n-monomial functional equation for $n\le 9$ (Lemma 1 in [12]), and the function f given by $f\left(x\right):=a{x}^n$ for $n\le 9$ satisfies the general nonic functional equation. Therefore, the function $g:\mathbb{R}\to \mathbb{R}$ given by $g\left(x\right):={\sum }_{i=0}^9{a}_i{x}^i$ is a particular solution of the functional equation $D_{10}f(x,y)=0$. More information on the functional equation $D_{n}f(x,y)=0$ can be found in Baker’s paper [13], especially in Theorem A.

The stability results for n-monomial functional equations with $n<9$ can be found in [6,14,15,16,17,18,19,20,21,22,23,24]. In the papers [16,25,26,27,28,29,30,31,32,33], one can find hyperstability results for various types of functional equation. Recent results regarding the stability of the functional equation described in Equation (1) with $n<9$ can be found in [34,35,36,37,38,39], and hyperstability results for this functional equation with $n<9$ can be found in [38,40]. Note that the superstability of a functional equation requires that any mapping satisfying the equation approximately (in some sense) must be either a real solution to it or a bounded mapping, while hyperstability requires that any mapping satisfying the equation approximately (in some sense) must be a real solution to it. Superstability results for several functional equations can be found in [28,41,42].

However, no study has yet been conducted on the stability of the general nonic functional equation. The significant advantage of this paper is the uniqueness of the solution for the stability of the general nonic functional equation. The uniqueness of the solution for the stability of the monomial functional equation has been discussed in many research papers. However, the uniqueness of the solution for the stability of general nonic functional equations is a more complicated problem. Using a special representation of a given mapping, we solved this problem.

The  rest of this paper is organized as follows. In Section 2, we study a way of representing a given mapping as the sum of odd and even mappings. In Section 3, we present hyperstability results for the general nonic functional equation. Precisely, let $p<0$ be a real number. If f: $X\to Y$ satisfies the inequality

$$\frac{\parallel {\sum }_{i=0}^{10}{}_{10}{C}_i{(-1)}^{10-i}f(x+iy)\parallel }{{\parallel x\parallel }^p+{\parallel y\parallel }^p}\le \theta \mathrm{for}\mathrm{all}x,y\in X\setminus \left\{0\right\},$$

then we prove that f must be a solution mapping of the general nonic functional equation. Finally, in Section 4, we discuss the stability problem of the general nonic functional equation.

The functional equation described in Equation (1) is a particular case of a general linear functional equation, which was labeled as Equation (4) in [25]. Stability results for general linear functional equations can be found in [43,44,45,46]. Moreover, hyperstability results for general linear functional equations can be found in [25,47,48].

Lastly, readers should recall that X is a normed space and Y is a Banach space throughout this paper.

2. Representation of a Given Mapping

In this section, we will introduce a way of representing a given mapping as the sum of odd and even mappings.

For a given mapping f: $X\to Y,$ we denote

$$f_o\left(x\right):=\frac{f\left(x\right)-f(-x)}{2},f_e\left(x\right):=\frac{f\left(x\right)+f(-x)}{2}.$$

Let us consider the following system of nonhomogeneous linear equations:

$$\begin{array}{c}\hfill \left\{\begin{array}{cccccc}{f}_1\left(x\right)& +f_3\left(x\right)& +f_5\left(x\right)& +f_7\left(x\right)& +f_9\left(x\right)& =f_o\left(x\right),\\ 2^1{f}_1\left(x\right)& +8^1{f}_3\left(x\right)& +{32}^1{f}_5\left(x\right)& +{128}^1{f}_7\left(x\right)& +512f_9\left(x\right)& =f_o\left(2x\right),\\ 2^2{f}_1\left(x\right)& +8^2{f}_3\left(x\right)& +{32}^2{f}_5\left(x\right)& +{128}^2{f}_7\left(x\right)& +{512}^2{f}_9\left(x\right)& =f_o\left(4x\right),\\ 2^3{f}_1\left(x\right)& +8^3{f}_3\left(x\right)& +{32}^3{f}_5\left(x\right)& +{128}^3{f}_7\left(x\right)& +{512}^3{f}_9\left(x\right)& =f_o\left(8x\right)),\\ 2^4{f}_1\left(x\right)& +8^4{f}_3\left(x\right)& +{32}^4{f}_5\left(x\right)& +{128}^4{f}_7\left(x\right)& +{512}^4{f}_9\left(x\right)& =f_o\left(16x\right)\end{array}\right.\end{array}$$

and

$$\begin{array}{c}\hfill \left\{\begin{array}{ccccc}{f}_2\left(x\right)& +f_4\left(x\right)& +f_6\left(x\right)& +f_8\left(x\right)& =f_e\left(x\right),\\ 4^1{f}_2\left(x\right)& +{16}^1{f}_4\left(x\right)& +{64}^1{f}_6\left(x\right)& +{256}^1{f}_8\left(x\right)& =f_e\left(2x\right),\\ 4^2{f}_2\left(x\right)& +{16}^2{f}_4\left(x\right)& +{64}^2{f}_6\left(x\right)& +{256}^2{f}_8\left(x\right)& =f_e\left(4x\right),\\ 4^3{f}_2\left(x\right)& +{16}^3{f}_4\left(x\right)& +{64}^3{f}_6\left(x\right)& +{256}^3{f}_8\left(x\right)& =f_e\left(8x\right)\end{array}\right.\end{array}$$

for all $x\in X.$ Then, we obtain the following lemmas by the uniqueness of the solution as stated in Cramer’s rule.

Lemma 1. 

Let f: $X\to Y$ be a given mapping and

$$M:=\left|\begin{array}{ccccc}1& 1& 1& 1& 1\\ 2& 8& 32& 128& 512\\ 2^2& 8^2& {32}^2& {128}^2& {512}^2\\ 2^3& 8^3& {32}^3& {128}^3& {512}^3\\ 2^4& 8^4& {32}^4& {128}^4& {512}^4\end{array}\right|.$$

Then, we have the mappings $f_1,f_3,f_5,f_7,f_9$: $X\to Y$ defined by the formulas

$$\begin{array}{ccc}\hfill f_1\left(x\right)& =\frac{\left|\begin{array}{ccccc}{f}_o\left(x\right)& 1& 1& 1& 1\\ f_o\left(2x\right)& 8& 32& 128& 512\\ f_o\left(4x\right)& 8^2& {32}^2& {128}^2& {512}^2\\ f_o\left(8x\right)& 8^3& {32}^3& {128}^3& {512}^3\\ f_o\left(16x\right)& 8^4& {32}^4& {128}^4& {512}^4\end{array}\right|}{M},f_3\left(x\right)& =\frac{\left|\begin{array}{ccccc}1& f_o\left(x\right)& 1& 1& 1\\ 2& f_o\left(2x\right)& 32& 128& 512\\ 2^2& f_o\left(4x\right)& {32}^2& {128}^2& {512}^2\\ 2^3& f_o\left(8x\right)& {32}^3& {128}^3& {512}^3\\ 2^4& f_o\left(16x\right)& {32}^4& {128}^4& {512}^4\end{array}\right|}{M},\hfill \\ \hfill f_5\left(x\right)& =\frac{\left|\begin{array}{ccccc}1& 1& f_o\left(x\right)& 1& 1\\ 2& 8& f_o\left(2x\right)& 128& 512\\ 2^2& 8^2& f_o\left(4x\right)& {128}^2& {512}^2\\ 2^3& 8^3& f_o\left(8x\right)& {128}^3& {512}^3\\ 2^4& 8^4& f_o\left(16x\right)& {128}^4& {512}^4\end{array}\right|}{M},f_7\left(x\right)& =\frac{\left|\begin{array}{ccccc}1& 1& 1& f_o\left(x\right)& 1\\ 2& 8& 32& f_o\left(2x\right)& 512\\ 2^2& 8^2& {32}^2& f_o\left(4x\right)& {512}^2\\ 2^3& 8^3& {32}^3& f_o\left(8x\right)& {512}^3\\ 2^4& 8^4& {32}^4& f_o\left(16x\right)& {512}^4\end{array}\right|}{M},\hfill \\ \hfill f_9\left(x\right)& =\frac{\left|\begin{array}{ccccc}1& 1& 1& 1& f_o\left(x\right)\\ 2& 8& 32& 128& f_o\left(2x\right)\\ 2^2& 8^2& {32}^2& {128}^2& f_o\left(4x\right)\\ 2^3& 8^3& {32}^3& {128}^3& f_o\left(8x\right)\\ 2^4& 8^4& {32}^4& {128}^4& f_o\left(16x\right)\end{array}\right|}{M}\end{array}$$

for all $x\in X.$ Furthermore, $f_o\left(x\right)=f_1\left(x\right)+f_3\left(x\right)+f_5\left(x\right)+f_7\left(x\right)+f_9\left(x\right)$ for all $x\in X$.

Lemma 2. 

Let f: $X\to Y$ be a given mapping and

$$M^{\prime }:=\left|\begin{array}{cccc}1& 1& 1& 1\\ 4& 16& 64& 256\\ 4^2& {16}^2& {64}^2& {256}^2\\ 4^3& {16}^3& {64}^3& {256}^3\end{array}\right|.$$

Then, we have the mappings $f_2,f_4,f_6,f_8$: $X\to Y$ defined by the formulas

$$\begin{array}{ccc}\hfill f_2\left(x\right)& =\frac{\left|\begin{array}{cccc}{f}_e\left(x\right)& 1& 1& 1\\ f_e\left(2x\right)& 16& 64& 256\\ f_e\left(4x\right)& {16}^2& {64}^2& {256}^2\\ f_e\left(8x\right)& {16}^3& {64}^3& {256}^3\end{array}\right|}{M^{\prime }},f_4\left(x\right)& =\frac{\left|\begin{array}{cccc}1& f_e\left(x\right)& 1& 1\\ 4& f_e\left(2x\right)& 64& 256\\ 4^2& f_e\left(4x\right)& {64}^2& {256}^2\\ 4^3& f_e\left(8x\right)& {64}^3& {256}^3\end{array}\right|}{M^{\prime }},\hfill \\ \hfill f_6\left(x\right)& =\frac{\left|\begin{array}{cccc}1& 1& f_e\left(x\right)& 1\\ 4& 16& f_e\left(2x\right)& 256\\ 4^2& {16}^2& f_e\left(4x\right)& {256}^2\\ 4^3& {16}^3& f_e\left(8x\right)& {256}^3\end{array}\right|}{M^{\prime }},f_8\left(x\right)& =\frac{\left|\begin{array}{cccc}1& 1& 1& f_e\left(x\right)\\ 4& 16& 64& f_e\left(2x\right)\\ 4^2& {16}^2& {64}^2& f_e\left(4x\right)\\ 4^3& {16}^3& {64}^3& f_e\left(8x\right)\end{array}\right|}{M^{\prime }}\hfill \end{array}$$

for all $x\in X$ and $f_e\left(x\right)=f_2\left(x\right)+f_4\left(x\right)+f_6\left(x\right)+f_8\left(x\right)$ for all $x\in X$.

Remark 1. 

By Lemmas 1 and 2, we have the following results. For all $x\in X,$

$$\begin{array}{cc}\hfill f_1\left(x\right):=& \frac{f_o\left(16x\right)-680f_o\left(8x\right)+91392f_o\left(4x\right)-2785280f_o\left(2x\right)+16777216f_o\left(x\right)}{722925·16},\hfill \\ \hfill f_2\left(x\right):=& \frac{f_e\left(8x\right)-336f_e\left(4x\right)-21504f_e\left(2x\right)-262144f_e\left(x\right)}{2835·64},\hfill \\ \hfill f_3\left(x\right):=& -\frac{5440(f_o\left(16x\right)-674f_o\left(8x\right)+87360f_o\left(4x\right)-2269184f_o\left(2x\right)+4194304f_o\left(x\right))}{722925·65536},\hfill \\ \hfill f_4\left(x\right):=& -\frac{f_e\left(8x\right)-324f_e\left(4x\right)-17664f_e\left(2x\right)-65536f_e\left(x\right))}{135·1024},\hfill \\ \hfill f_5\left(x\right):=& \frac{1428(f_o\left(16x\right)-650f_o\left(8x\right)+71952f_o\left(4x\right)-665600f_o\left(2x\right)+1048576f_o\left(x\right)}{722925·65536},\hfill \\ \hfill f_6\left(x\right):=& \frac{f_e\left(8x\right)-276f_e\left(4x\right)-5184f_e\left(2x\right)-16384f_e\left(x\right))}{135·4096},\hfill \\ \hfill f_7\left(x\right):=& -\frac{85(f_o\left(16x\right)-554f_o\left(8x\right)+21840f_o\left(4x\right)-172544f_o\left(2x\right)+262144f_o\left(x\right))}{722925·65536},\hfill \\ \hfill f_8\left(x\right):=& -\frac{f_e\left(8x\right)-84f_e\left(4x\right)-1344f_e\left(2x\right)-4096f_e\left(x\right))}{2835·4096},\hfill \\ \hfill f_9\left(x\right):=& \frac{f_o\left(16x\right)-170f_o\left(8x\right)+5712f_o\left(4x\right)-43520f_o\left(2x\right)+65536f_o\left(x\right)}{722925·65536}.\hfill \end{array}$$

Moreover,

$$f\left(x\right)=\sum _{i=1}^9{f}_i\left(x\right)\mathit{for}\mathit{all}x\in X.$$

Below, we define the mappings needed to prove the main theorems.

Definition 1. 

For a given mapping f: $X\to Y,$ we define

$$\begin{array}{cc}\hfill \tilde{f}\left(x\right)& :=f\left(x\right)-f\left(0\right),\hfill \\ \hfill Df(x,y)& :=\sum _{i=0}^{10}{}_{10}{C}_i{(-1)}^{10-i}f(x+iy),\hfill \\ \hfill \Gamma f\left(x\right):=& D{f}_o(12x,4x)+10D{f}_o(8x,4x)+55D{f}_o(4x,4x)+220D{f}_o(10x,2x)\hfill \\ \hfill & +2200D{f}_o(8x,2x)+9988D{f}_o(6x,2x)+27280D{f}_o(4x,2x)\hfill \\ \hfill & +45352D{f}_o(2x,2x)+17920D{f}_o(5x,x)+179200D{f}_o(4x,x)\hfill \\ \hfill & +760320D{f}_o(3x,x)+1689600D{f}_o(2x,x)+1790976D{f}_o(x,x),\hfill \\ \hfill \Delta f\left(x\right):=& D{f}_e(6x,2x)+10D{f}_e(4x,2x)+55D{f}_e(2x,2x)+110D{f}_e(0,2x)\hfill \\ \hfill & +320D{f}_e(3x,x)+3200D{f}_e(2x,x)+12992D{f}_e(x,x)+12160D{f}_e(0,x)\hfill \end{array}$$

for all$x,y\in X.$

As the results of tedious calculations, we obtain the following lemmas.

Lemma 3. 

Let f: $X\to Y$ be an arbitrary mapping. Then, the equalities

$$\begin{array}{cc}\hfill & \tilde{f}\left(x\right)=\sum _{i=1}^9{\tilde{f}}_i\left(x\right),\hfill \\ \hfill & D\tilde{f}(x,y)=Df(x,y),\hfill \\ \hfill & \Gamma \tilde{f}\left(x\right)=f_o\left(32x\right)-682f_o\left(16x\right)+92752f_o\left(8x\right)-2968064f_o\left(4x\right),\hfill \\ \hfill & +22347776f_o\left(2x\right)-33554432f_o\left(x\right),\hfill \\ \hfill & \Delta \tilde{f}\left(x\right)=f_e\left(16x\right)-340f_e\left(8x\right)+22848f_e\left(4x\right)-348160f_e\left(2x\right)+1048576f_e\left(x\right)\hfill \end{array}$$

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

Lemma 4. 

Let f: $X\to Y$ be an arbitrary mapping. Then,

$$\begin{array}{cc}\hfill & {\tilde{f}}_1\left(x\right)-\frac{{\tilde{f}}_1\left(2x\right)}{2}=\frac{\Gamma {\tilde{f}}_o\left(x\right)}{722925·32},{\tilde{f}}_2\left(x\right)-\frac{{\tilde{f}}_2\left(2x\right)}{4}=\frac{\Delta {\tilde{f}}_e\left(x\right)}{2835·256},\hfill \\ \hfill & {\tilde{f}}_3\left(x\right)-\frac{{\tilde{f}}_3\left(2x\right)}{8}=-\frac{5440\Gamma {\tilde{f}}_o\left(x\right)}{722925·65536·8},{\tilde{f}}_4\left(x\right)-\frac{{\tilde{f}}_4\left(2x\right)}{16}=-\frac{21\Delta {\tilde{f}}_e\left(x\right)}{2835·256·64},\hfill \\ \hfill & {\tilde{f}}_5\left(x\right)-\frac{{\tilde{f}}_5\left(2x\right)}{32}=\frac{1428\Gamma {\tilde{f}}_o\left(x\right)}{722925·32·65536},{\tilde{f}}_6\left(x\right)-\frac{{\tilde{f}}_6\left(2x\right)}{64}=\frac{21\Delta {\tilde{f}}_e\left(x\right)}{2835·256·1024},\hfill \\ \hfill & {\tilde{f}}_7\left(x\right)-\frac{{\tilde{f}}_7\left(2x\right)}{128}=-\frac{85\Gamma {\tilde{f}}_o\left(x\right)}{722925·65536·128},{\tilde{f}}_8\left(x\right)-\frac{{\tilde{f}}_8\left(2x\right)}{256}=-\frac{\Delta {\tilde{f}}_e\left(x\right)}{2835·256·4096},\hfill \\ \hfill & {\tilde{f}}_9\left(x\right)-\frac{{\tilde{f}}_9\left(2x\right)}{512}=\frac{\Gamma {\tilde{f}}_o\left(x\right)}{722925·65536·512}\hfill \end{array}$$

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

Lemma 5. 

If f: $X\to Y$ is a mapping such that $Df(x,y)=0$ for all $x,y\in X$ with $f\left(0\right)=0,$ then for each $i\in \{1,2,\dots ,9\},$ the mappings $f_i\left(x\right)$ in Remark 1 satisfy the equalities $D{f}_i(x,y)=0$ for all $x\in X$ and $f_i\left(2x\right)=2^i{f}_i\left(x\right)$ for all $x,y\in X.$

Proof. 

Since $Df(x,y)=0$ for all $x,y\in X,$ by the definition of $f_i,\Gamma f$ and $\Delta f,$ we have

$$D{f}_i(x,y)=0,\Delta f\left(x\right)=0,\Gamma f\left(x\right)=0.$$

Applying Lemma 4, we arrive at $f_i\left(2x\right)=2^i{f}_i\left(x\right).$    □

3. Hyperstability of the General Nonic Functional Equation

In this section, we will prove the hyperstability of the general nonic functional equation. To prove the main theorem, we will use the functions introduced in the previous section.

Theorem 1. 

Let $p<0$ be a real number. Suppose that f: $X\to Y$ is a mapping such that

$$\begin{array}{c}\hfill \frac{\parallel Df(x,y)\parallel }{{\parallel x\parallel }^p+{\parallel y\parallel }^p}\le \theta \mathit{for}\mathit{all}x,y\in X\setminus \left\{0\right\}.\end{array}$$

(3)

Then, $Df(x,y)=0$ is fulfilled for all $x,y\in X.$

Proof. 

STEP 1: Using the definition of $\tilde{f}$ together with

$$\sum _{i=0}^{10}{}_{10}{C}_i{(-1)}^{10-i}=0,$$

we have

$$D\tilde{f}(x,y)=Df(x,y),\tilde{f}\left(0\right)=0.$$

From the expression (3) and the definitions of $\Gamma f$ and $\Delta f,$ we obtain

$$\parallel \Gamma \tilde{f}\left(x\right)\parallel \le 47377612800K^{\prime }{\theta \parallel x\parallel }^p,\parallel \Delta \tilde{f}{\left(x\right)\parallel \le 11612160K\theta \parallel x\parallel }^p$$

for all $x\in X\setminus \left\{0\right\},$ where

$$\begin{array}{cc}\hfill K^{\prime }:=& \frac{220·{20}^p+55·{16}^p+10·{12}^p+17920·{10}^p+45353·8^p+27280·6^p}{47377612800}\hfill \\ \hfill & +\frac{1801030·4^p+1689600·3^p+847560·2^p+4617216}{47377612800},\hfill \\ \hfill K:=& \frac{110·{10}^p+55·8^p+10·6^p+12160·5^p+12993·4^p+3200·3^p+496·2^p+28672}{11612160}.\hfill \end{array}$$

Then, Lemma 3 and the definitions of $f_k$ ($k\in \{1,2,\dots ,9\}$) ensure the inequalities

$$\begin{array}{cc}\hfill & ∥\frac{{\tilde{f}}_1\left(2^{i}x\right)}{2^i}-\frac{{\tilde{f}}_1\left(2^{i+1}x\right)}{2^{i+1}}∥=∥-\frac{4096\Gamma {\tilde{f}}_o\left(x\right)}{47377612800·2^{i+1}}∥\le \frac{4096K^{\prime }\theta {\parallel x\parallel }^p}{2^{i+1}},\hfill \\ \hfill & ∥\frac{{\tilde{f}}_2\left(2^{i}x\right)}{4^i}-\frac{{\tilde{f}}_2\left(2^{i+1}x\right)}{4^{i+1}}∥=∥-\frac{64\Delta {\tilde{f}}_e\left(x\right)}{11612160·4^{i+1}}∥\le \frac{{64K\theta \parallel x\parallel }^p}{4^{i+1}},\hfill \\ \hfill & ∥\frac{{\tilde{f}}_3\left(2^{i}x\right)}{8^i}-\frac{{\tilde{f}}_3\left(2^{i+1}x\right)}{8^{i+1}}∥=∥\frac{5440\Gamma {\tilde{f}}_o\left(x\right)}{47377612800·8^{i+1}}∥\le \frac{5440K^{\prime }\theta {\parallel x\parallel }^p}{8^{i+1}},\hfill \\ \hfill & ∥\frac{{\tilde{f}}_4\left(2^{i}x\right)}{{16}^i}-\frac{{\tilde{f}}_4\left(2^{i+1}x\right)}{{16}^{i+1}}∥=∥\frac{84\Delta {\tilde{f}}_e\left(x\right)}{11612160·{16}^{i+1}}∥\le \frac{{84K\theta \parallel x\parallel }^p}{{16}^{i+1}}\hfill \\ \hfill & ∥\frac{{\tilde{f}}_5\left(2^{i}x\right)}{{32}^i}-\frac{{\tilde{f}}_5\left(2^{i+1}x\right)}{{32}^{i+1}}∥=∥-\frac{1428\Gamma {\tilde{f}}_o\left(x\right)}{47377612800·{32}^{i+1}}∥\le \frac{1428K^{\prime }\theta {\parallel x\parallel }^p}{{32}^{i+1}},\hfill \\ \hfill & ∥\frac{{\tilde{f}}_6\left(2^{i}x\right)}{{64}^i}-\frac{{\tilde{f}}_6\left(2^{i+1}x\right)}{{64}^{i+1}}∥=∥-\frac{21\Delta {\tilde{f}}_e\left(x\right)}{11612160·{64}^{i+1}}∥\le \frac{{21K\theta \parallel x\parallel }^p}{{64}^{i+1}},\hfill \\ \hfill & ∥\frac{{\tilde{f}}_7\left(2^{i}x\right)}{{128}^i}-\frac{{\tilde{f}}_7\left(2^{i+1}x\right)}{{128}^{i+1}}∥=∥\frac{85\Gamma {\tilde{f}}_o\left(x\right)}{47377612800·{128}^{i+1}}∥\le \frac{85K^{\prime }\theta {\parallel x\parallel }^p}{{128}^{i+1}},\hfill \\ \hfill & ∥\frac{{\tilde{f}}_8\left(2^{i}x\right)}{{256}^i}-\frac{{\tilde{f}}_8\left(2^{i+1}x\right)}{{256}^{i+1}}∥=∥\frac{\Delta {\tilde{f}}_e\left(x\right)}{11612160·{256}^{i+1}}∥\le \frac{{K\theta \parallel x\parallel }^p}{{256}^{i+1}},\hfill \\ \hfill & ∥\frac{{\tilde{f}}_9\left(2^{i}x\right)}{{512}^i}-\frac{{\tilde{f}}_9\left(2^{i+1}x\right)}{{512}^{i+1}}∥=∥\frac{\Gamma {\tilde{f}}_o\left(x\right)}{47377612800·{512}^{i+1}}∥\le \frac{K^{\prime }\theta {\parallel x\parallel }^p}{{512}^{i+1}}\hfill \end{array}$$

for all $x\in X\setminus \left\{0\right\}$. Note that

$$\sum _{k=1}^9\frac{{\tilde{f}}_k\left(2^{n}x\right)}{2^{kn}}-\sum _{k=1}^9\frac{{\tilde{f}}_k\left(2^{n+m}x\right)}{2^{k(n+m)}}=\sum _{i=n}^{n+m-1}\left(\sum _{k=1}^9\frac{{\tilde{f}}_k\left(2^{i}x\right)}{2^{ki}}-\sum _{k=1}^9\frac{{\tilde{f}}_k\left(2^{i+1}x\right)}{2^{k(i+1)}}\right).$$

This implies that

$$\begin{array}{cc}\hfill & ∥\sum _{k=1}^9\frac{{\tilde{f}}_k\left(2^{n}x\right)}{2^{kn}}-\sum _{k=1}^9\frac{{\tilde{f}}_k\left(2^{n+m}x\right)}{2^{k(n+m)}}∥\le \sum _{i=n}^{n+m-1}(\frac{4096K^{\prime }}{2^{i+1}}+\frac{64K}{4^{i+1}}+\frac{5440K^{\prime }}{8^{i+1}}\hfill \\ \hfill & +\frac{84K}{{16}^{i+1}}+\frac{1428K^{\prime }}{{32}^{i+1}}+\frac{21K}{{64}^{i+1}}+\frac{85K^{\prime }}{{128}^{i+1}}+\frac{K}{{256}^{i+1}}+\frac{K^{\prime }}{{512}^{i+1}})·\theta {\parallel x\parallel }^p\hfill \end{array}$$

(4)

for all $x\in X\setminus \left\{0\right\}$ and $n,m\in \mathbb{N}\cup \left\{0\right\}.$ If $p<0,$ the sequences $\{\frac{{\tilde{f}}_1\left(2^{n}x\right)}{2^n}\},\dots ,\{\frac{{\tilde{f}}_9\left(2^{n}x\right)}{2^{9n}}\}$ and $\{{\sum }_{k=1}^9\frac{{\tilde{f}}_k\left(2^{n}x\right)}{2^{kn}}\}$ are Cauchy for all $x\in X\setminus \left\{0\right\}.$ Since Y is complete and $\tilde{f}\left(0\right)=0,$ the sequences $\{\frac{{\tilde{f}}_1\left(2^{n}x\right)}{2^n}\},\dots ,\{\frac{{\tilde{f}}_9\left(2^{n}x\right)}{2^{9n}}\}$ and $\{{\sum }_{k=1}^9\frac{{\tilde{f}}_k\left(2^{n}x\right)}{2^{kn}}\}$ converge for all $x\in X.$ Hence, for each $k\in \{1,2,\dots ,9\},$ we can define mappings $F_k,F$: $X\to Y$ by

$$\begin{array}{c}\hfill F_k\left(x\right):=\underset{n\to \infty }{lim}\frac{{\tilde{f}}_k\left(2^{n}x\right)}{2^{kn}},F\left(x\right):=\underset{n\to \infty }{lim}\sum _{k=1}^9\frac{{\tilde{f}}_k\left(2^{n}x\right)}{2^{kn}}(x\in X).\end{array}$$

STEP 2: By (3) and the definitions of $Df,F_1$, and $f_1,$ since $D{f}_1(x,y)=D{\tilde{f}}_1(x,y),$ we have

$$\begin{array}{c}\parallel D{F}_1(x,y)\parallel =\underset{n\to \infty }{lim}\parallel \frac{D{\tilde{f}}_1(2^{n}x,2^{n}y)}{2^n}\parallel =\underset{n\to \infty }{lim}\parallel \frac{D{f}_1(2^{n}x,2^{n}y)}{2^n}\parallel \hfill \\ =\underset{n\to \infty }{lim}\parallel \frac{16777216D{f}_o(2^{n}x,2^{n}y)}{11566800·2^n}-\frac{2785280D{f}_o(2^{n+1}x,2^{n+1}y)}{11566800·2^n}\hfill \\ +\frac{91392f_o(2^{n+2}x,2^{n+2}y)}{11566800·2^n}-\frac{680D{f}_o(2^{n+3}x,2^{n+3}y)}{11566800·2^n}+\frac{D{f}_o(2^{n+4}x,2^{n+4}y)}{11566800·2^n}\parallel \hfill \\ \le \underset{n\to \infty }{lim}\left(\frac{16777216+2785280·2^p+91392·2^{2p}+680·2^{3p}+2^{4p}}{11566800}\right)·\frac{2^{np}}{2^n}\theta \left({\parallel x\parallel }^p+{\parallel y\parallel }^p\right)\hfill \\ =0\hfill \end{array}$$

(5)

for all $x,y\in X\setminus \left\{0\right\}.$ On the other hand, from the definition of $D{F}_1,$ we have

$$D{F}_1(x,0)=\sum _{i=0}^{10}{}_{10}{C}_i{(-1)}^{10-i}{F}_1(x+i0)=F_1\left(x\right)\sum _{i=0}^{10}{}_{10}{C}_i{(-1)}^{10-i}=0$$

(6)

for all $x\in X.$ And, in view of (5), we obtain

$$D{F}_1(0,y)=D{F}_1(-10y,y)=0\mathsf{for}\mathsf{all}y\in X\setminus \left\{0\right\}.$$

(7)

Therefore, the relations (5)–(7) yield that $D{F}_1(x,y)=0$ for all $x,y\in X.$ Similarly, we can show that $D{F}_k(x,y)=0$ for each $k\in \{2,3,\dots ,9\}$ and all $x,y\in X.$ Since $DF(x,y)={\sum }_{k=1}^{9}D{F}_k(x,y)$ for all $x,y\in X,$ we obtain $DF(x,y)=0$ for all $x,y\in X.$

STEP 3: Observe that for all $x\in X,$

$$\begin{array}{cc}\hfill & D\tilde{f}(x,y)-DF(x,y)=\sum _{i=0}^{10}{}_{10}{C}_i{(-1)}^{10-i}\left(\tilde{f}(x+iy)-F(x+iy)\right),\hfill \\ \hfill & DF\left(\right(1-n)x,nx)=0.\hfill \end{array}$$

Then, we see that

$$\begin{array}{cc}\hfill D\tilde{f}((1-n)x,nx)& =\tilde{f}\left((1-n)x\right)-F\left((1-n)x\right)-10\tilde{f}\left(x\right)+10F\left(x\right)\hfill \\ \hfill & +\sum _{i=2}^{10}{}_{10}{C}_i{(-1)}^{10-i}\left(\tilde{f}((1-n)x+inx)-F((1-n)x+inx)\right)\hfill \end{array}$$

(8)

for any $n\in N$ and $x\in X\setminus \left\{0\right\}.$ Moreover, by letting $n=0$ and taking the limit $m\to \infty$ in (4), we obtain the inequalities

$$\begin{array}{cc}\hfill \parallel \tilde{f}\left(x\right)-F\left(x\right)\parallel \le & (\frac{4096K^{\prime }}{2-2^p}+\frac{64K}{4-2^p}+\frac{5440K^{\prime }}{8-2^p}+\frac{84K}{16-2^p}\hfill \\ \hfill & +\frac{1428K^{\prime }}{32-2^p}+\frac{21K}{64-2^p}+\frac{85K^{\prime }}{128-2^p}+\frac{K}{256-2^p}+\frac{K^{\prime }}{512-2^p})·\theta {\parallel x\parallel }^p\hfill \end{array}$$

(9)

for all $x\in X.$

Since $p<0,$ by (3), (8), and (9), we are forced to conclude that

$$\begin{array}{cc}\hfill 10·\parallel \tilde{f}\left(x\right)-F\left(x\right)\parallel \le & \underset{n\to \infty }{lim}\parallel D\tilde{f}((1-n)x,nx)\parallel +\underset{n\to \infty }{lim}\parallel \tilde{f}\left((1-n)x\right)-F((1-n)x\parallel \hfill \\ \hfill & +\sum _{i=2}^{10}\underset{n\to \infty }{lim}\parallel {}_{10}{C}_i(\tilde{f}\left(((i-1)n+1)x\right)-F\left(((i-1)n+1)x\right)\parallel \hfill \\ \hfill \le & \underset{n\to \infty }{lim}\left({(n-1)}^p+n^p\right)+M{(n-1)}^p+\sum _{i=2}^{10}{}_{10}{C}_i\left({(i-1)n+1)}^p\right)·\theta {\parallel x\parallel }^p\hfill \\ \hfill =& 0\hfill \end{array}$$

for all $x\in X\setminus \left\{0\right\},$ where

$$M:=\frac{4096K^{\prime }}{2-2^p}+\frac{64K}{4-2^p}+\frac{5440K^{\prime }}{8-2^p}+\frac{84K}{16-2^p}+\frac{1428K^{\prime }}{32-2^p}+\frac{21K}{64-2^p}+\frac{85K^{\prime }}{128-2^p}+\frac{K}{256-2^p}+\frac{K^{\prime }}{512-2^p}.$$

Thus, we have $\parallel \tilde{f}\left(x\right)-F\left(x\right)\parallel =0$ for all $x\in X\setminus \left\{0\right\}.$ Since $\tilde{f}\left(0\right)=0=F\left(0\right),$ we have $\tilde{f}\left(x\right)=F\left(x\right)$ for all $x\in X.$ Therefore, $D\tilde{f}(x,y)=DF(x,y)=0$ for all $x,y\in X.$ From the fact that $D\tilde{f}(x,y)=Df(x,y),$ we finally have

$$Df(x,y)=D\tilde{f}(x,y)=DF(x,y)=0,$$

which completes the proof. □

4. Stability of the General Nonic Functional Equation

In this section, we will consider the stability of the general nonic functional equation

$$\sum _{i=0}^{10}{}_{10}{C}_i{(-1)}^{10-i}f(x+iy)=0.$$

Theorem 2. 

Let $p\ne 1,2,3,4,5,6,7,8,9$ be a non-negative real number. Suppose that $f:X\to Y$ is a mapping such that for all $x,y\in X,$

$$\begin{array}{c}\hfill \parallel Df(x,y)\parallel \le \theta \left({\parallel x\parallel }^p+{\parallel y\parallel }^p\right).\end{array}$$

(10)

Then, there exists a unique mapping F satisfying $DF(x,y)=0$ and

$$\begin{array}{cc}\hfill \parallel \tilde{f}\left(x\right)-F\left(x\right)\parallel \le & \left(\frac{4096}{|2-2^p|}+\frac{5440}{|8-2^p|}+\frac{1428}{|32-2^p|}+\frac{85}{|128-2^p|}+\frac{1}{|512-2^p|}\right)·\frac{K^{\prime }\theta {\parallel x\parallel }^p}{4096}\hfill \\ \hfill & +\left(\frac{64}{|4-2^p|}+\frac{84}{|16-2^p|}+\frac{21}{|64-2^p|}+\frac{1}{|256-2^p|}\right)·\frac{{K\theta \parallel x\parallel }^p}{64},\hfill \end{array}$$

(11)

for all $x\in X,$ where $K:=6^p+10·4^p+320·3^p+3431·2^p+41664$ and

$$\begin{array}{c}\hfill K^{\prime }:={12}^p+220·{10}^p+2210·8^p+9988·6^p+17920·5^p+206601·4^p+760320·3^p+1819992·2^p+6228992.\end{array}$$

Proof. 

From the definition of $\tilde{f},$ we obtain $D\tilde{f}(x,y)=Df(x,y)$ and $\tilde{f}\left(0\right)=0.$ By (10) and the definition of $\Gamma f$ and $\Delta f,$ we have

$$\begin{array}{cc}\hfill \parallel \Gamma \tilde{f}\left(x\right)\parallel =& \parallel D{f}_o(12x,4x)+10D{f}_o(8x,4x)+55D{f}_o(4x,4x)+220D{f}_o(10x,2x)\hfill \\ \hfill & +2200D{f}_o(8x,2x)+9988D{f}_o(6x,2x)+27280D{f}_o(4x,2x)\hfill \\ \hfill & +45352D{f}_o(2x,2x)+17920D{f}_o(5x,x)+179200D{f}_o(4x,x)\hfill \\ \hfill & +760320D{f}_o(3x,x)+1689600D{f}_o(2x,x)+1790976D{f}_o(x,x)\parallel \hfill \\ \hfill \le & ({12}^p+4^p+10·8^p+10·4^p+110·4^p+220·{10}^p+220·2^p\hfill \\ \hfill & +2200·8^p+2200·2^p+9988·6^p+9988·2^p+27280·4^p+27280·2^p\hfill \\ \hfill & +90704·2^p+17920·5^p+17920+179200·4^p+179200\hfill \\ \hfill & +760320·3^p+760320+1689600·2^p+1689600+3581952)·\theta {\parallel x\parallel }^p\hfill \\ \hfill \le & 722925·16K^{\prime }\theta {\parallel x\parallel }^p,\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill \parallel \Delta \tilde{f}\left(x\right)\parallel =& \parallel D{f}_e(6x,2x)+10D{f}_e(4x,2x)+55D{f}_e(2x,2x)+110D{f}_e(0,2x)\hfill \\ \hfill & +320D{f}_e(3x,x)+3200D{f}_e(2x,x)+12992D{f}_e(x,x)+12160D{f}_e(0,x)\parallel \hfill \\ \hfill \le & (6^p+2^p+10·4^p+10·2^p+110·2^p+110·2^p\hfill \\ \hfill & +320·3^p+320+3200·2^p+3200+25984+12160)·\theta {\parallel x\parallel }^p\hfill \\ \hfill \le & 2835·{64K\theta \parallel x\parallel }^p\hfill \end{array}$$

(13)

for all $x\in X.$

Next, for $i\in \{1,2,\dots ,9\},$ we will find $F_k$ to reach $F\left(x\right)$ = ${\sum }_{k=1}^9{F}_k\left(x\right).$ For each given $p\ne 1,2,3,4,5,6,7,8,9,$ we will use a different approach to find the functions $F_k.$

• Setting ${\mathit{F}}_{\mathbf{1}}$: Let $0\le p<1$. It follows from Lemma 4 and (12) that

$$\begin{array}{cc}\hfill ∥\frac{{\tilde{f}}_1\left(2^{i}x\right)}{2^i}-\frac{{\tilde{f}}_1\left(2^{i+1}x\right)}{2^{i+1}}∥=∥\frac{\Gamma \tilde{f}\left(2^{i}x\right)}{722925·32·2^i}∥\le & \frac{K^{\prime }\theta {\parallel x\parallel }^p}{2·2^i},\hfill \end{array}$$

for all $x\in X.$ Notice that for all $x\in X,$

$$\frac{{\tilde{f}}_1\left(2^{n}x\right)}{2^n}-\frac{{\tilde{f}}_1\left(2^{n+m}x\right)}{2^{n+m}}=\sum _{i=n}^{n+m-1}\left(\frac{{\tilde{f}}_1\left(2^{i}x\right)}{2^i}-\frac{{\tilde{f}}_1\left(2^{i+1}x\right)}{2^{i+1}}\right),$$

which implies that

$$\begin{array}{c}\hfill ∥\frac{{\tilde{f}}_1\left(2^{n}x\right)}{2^n}-\frac{{\tilde{f}}_1\left(2^{n+m}x\right)}{2^{n+m}}∥\le \sum _{i=n}^{n+m-1}\frac{K^{\prime }{2}^{ip}\theta {\parallel x\parallel }^p}{2·2^i}\end{array}$$

(14)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}$. By (14), the sequence $\left\{\frac{{\tilde{f}}_1\left(2^{n}x\right)}{2^n}\right\}$ is a Cauchy sequence for all $x\in X,$ because of the fact that $0\le p<1.$ Since Y is complete, the sequence $\left\{\frac{{\tilde{f}}_1\left(2^{n}x\right)}{2^n}\right\}$ converges. Hence, we may define a mapping $F_1:X\to Y$ by

$$F_1\left(x\right):=\underset{n\to \infty }{lim}\frac{{\tilde{f}}_1\left(2^{n}x\right)}{2^n}\mathrm{for}\mathrm{all}x\in X.$$

Moreover, letting $n=0$ and taking the limit $m\to \infty$ in (14), we obtain the inequality

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_1\left(x\right)-F_1\left(x\right)\parallel \le \frac{K^{\prime }\theta {\parallel x\parallel }^p}{2-2^p}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

By the definition of $F_1,$ we easily obtain $F_1\left(2x\right)=2F_1\left(x\right)$ for all $x\in X$ and

$$\begin{array}{cc}\hfill \parallel D{F}_1(x,y)\parallel =& \underset{n\to \infty }{lim}\parallel \frac{D{f}_1(2^{n}x,2^{n}y)}{2^n}\parallel \hfill \\ \hfill =& \underset{n\to \infty }{lim}\parallel \frac{16777216D{f}_o(2^{n}x,2^{n}y)}{11566800·2^n}-\frac{2785280D{f}_o(2^{n+1}x,2^{n+1}y)}{11566800·2^n}\hfill \\ \hfill & +\frac{91392D{f}_o(2^{n+2}x,2^{n+2}y)}{11566800·2^n}-\frac{680D{f}_o(2^{n+3}x,2^{n+3}y)}{11566800·2^n}-\frac{D{f}_o(2^{n+4}x,2^{n+4}y)}{11566800·2^n}\parallel \hfill \\ \hfill \le & \underset{n\to \infty }{lim}\frac{16777216·2^{np}{\theta (\parallel x\parallel }^p+{\parallel y\parallel }^p)}{11566800·2^n}+\underset{n\to \infty }{lim}\frac{2785280·2^{(n+1)p}{\theta (\parallel x\parallel }^p+{\parallel y\parallel }^p)}{11566800·2^n}\hfill \\ \hfill & +\underset{n\to \infty }{lim}\frac{91392·2^{(n+2)p}{\theta (\parallel x\parallel }^p+{\parallel y\parallel }^p\left)\right)}{11566800·2^n}+\underset{n\to \infty }{lim}\frac{680·2^{(n+3)p}{\theta (\parallel x\parallel }^p+{\parallel y\parallel }^p\left)\right)}{11566800·2^n}\hfill \\ \hfill & +\underset{n\to \infty }{lim}\frac{2^{(n+4)p}{\theta (\parallel x\parallel }^p+{\parallel y\parallel }^p\left)\right)}{11566800·2^n}=0\hfill \end{array}$$

for all $x,y\in X.$

Let $p>1.$ It follows from Lemma 4 and (12) that

$$\begin{array}{cc}\hfill ∥2^i{\tilde{f}}_1\left(2^{-i}x\right)-2^{i+1}{\tilde{f}}_1\left(2^{-i-1}x\right)∥\le ∥\frac{2^i\Gamma \tilde{f}\left(2^{-i-1}x\right)}{722925·16}∥\le & \frac{K^{\prime }{2}^i\theta {\parallel x\parallel }^p}{2^{(i+1)p}}\hfill \end{array}$$

for all $x\in X.$ Because of the fact that

$$2^n{\tilde{f}}_1\left(2^{-n}x\right)-2^{n+m}{\tilde{f}}_1\left(2^{-n-m}x\right)=\sum _{i=n}^{n+m-1}\left(2^i{\tilde{f}}_1\left(2^{-i}x\right)-(2^{i+1}{\tilde{f}}_1(2^{-i-1}x))\right)$$

for all $x\in X,$ we have

$$\begin{array}{c}\hfill \parallel 2^n{\tilde{f}}_1\left(2^{-n}x\right)-2^{n+m}{\tilde{f}}_1\left(2^{-n-m}x\right)\parallel \le \sum _{i=n}^{n+m-1}\frac{K^{\prime }{2}^i\theta {\parallel x\parallel }^p}{2^{(i+1)p}}\end{array}$$

(15)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}$. Since $p>1$, by (15), the sequence $\left\{2^n{\tilde{f}}_1\left(2^{-n}x\right)\right\}$ is a Cauchy sequence for all $x\in X$. By the completeness of $Y,$ we know that the sequence $\left\{2^n{\tilde{f}}_1\left(2^{-n}x\right)\right\}$ converges. Hence, we can define a mapping $F_1$: $X\to Y$ by

$$F_1\left(x\right):=\underset{n\to \infty }{lim}{2}^n{\tilde{f}}_1\left(2^{-n}x\right)\mathrm{for}\mathrm{all}x\in X.$$

However, letting $n=0$ and passing the limit $m\to \infty$ in (15), we obtain the inequality

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_1\left(x\right)-F_1\left(x\right)\parallel \le \frac{K^{\prime }\theta {\parallel x\parallel }^p}{2^p-2}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

From the definition of $F_1$, we easily obtain $F_1\left(2x\right)=2F_1\left(x\right)$ for all $x\in X$ and $D{F}_1(x,y)=0$ for all $x,y\in X.$

• Setting ${\mathit{F}}_{\mathbf{2}}$: Let $p<2.$ It follows from Lemma 4 and (13) that

$$\begin{array}{c}\hfill ∥\frac{{\tilde{f}}_2\left(2^{n}x\right)}{4^n}-\frac{{\tilde{f}}_2\left(2^{n+m}x\right)}{4^{n+m}}∥=∥\frac{\Delta {\tilde{f}}_e\left(2^{i}x\right)}{2835·256·4^i}∥\le \sum _{i=n}^{n+m-1}\frac{K{2}^{ip}\theta {\parallel x\parallel }^p}{4·4^i}\end{array}$$

(16)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}.$ Since $p<2,$ we have from (16) that the sequence $\left\{\frac{{\tilde{f}}_2\left(2^{n}x\right)}{4^n}\right\}$ is a Cauchy sequence for all $x\in X$. By the completeness of $Y,$ the sequence $\left\{\frac{{\tilde{f}}_2\left(2^{n}x\right)}{4^n}\right\}$ converges. Hence, we can define a mapping $F_2$: $X\to Y$ by

$$F_2\left(x\right):=\underset{n\to \infty }{lim}\frac{{\tilde{f}}_2\left(2^{n}x\right)}{4^n}\mathrm{for}\mathrm{all}x\in X.$$

Now, letting $n=0$ and passing the limit $m\to \infty$ in (16), we obtain

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_2\left(x\right)-F_2\left(x\right)\parallel \le \frac{{K\theta \parallel x\parallel }^p}{4-2^p}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

(17)

From the definition of $F_2$, we then have $F_2\left(2x\right)=4F_2\left(x\right)$ for all $x\in X$ and $D{F}_2(x,y)=0$ for all $x,y\in X.$

Let $p>2.$ It follows from Lemma 4 and (13) that

$$\begin{array}{c}\hfill ∥4^n{\tilde{f}}_2\left(2^{-n}x\right)-4^{n+m}{\tilde{f}}_2\left(2^{-n-m}x\right)∥\le \sum _{i=n}^{n+m-1}\frac{K{4}^i\theta {\parallel x\parallel }^p}{2^{(i+1)p}}\end{array}$$

(18)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}.$ Since $p>2,$ we have by (18) that the sequence $4^n{\tilde{f}}_2\left(2^{-n}x\right)$ is a Cauchy sequence for all $x\in X.$ Since Y is complete, the sequence $\left\{4^n{\tilde{f}}_2\left(2^{-n}x\right)\right\}$ converges. Then, we can define a mapping $F_2$: $X\to Y$ by

$$F_2\left(x\right):=\underset{n\to \infty }{lim}{4}^n{\tilde{f}}_2\left(2^{-n}x\right)\mathrm{for}\mathrm{all}x\in X.$$

In (18), letting $n=0$ and passing the limit $m\to \infty ,$ we obtain

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_2\left(x\right)-F_2\left(x\right)\parallel \le \frac{{K\theta \parallel x\parallel }^p}{2^p-4}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

According to the definition of $F_2$, we arrive at $F_2\left(2x\right)=4F_2\left(x\right)$ for all $x\in X$ and $D{F}_2(x,y)=0$ for all $x,y\in X.$

• Setting ${\mathit{F}}_{\mathbf{3}}$: Let $p<3.$ It follows by Lemma 4 and (12) that

$$\begin{array}{c}\hfill ∥\frac{{\tilde{f}}_3\left(2^{n}x\right)}{8^n}-\frac{{\tilde{f}}_3\left(2^{n+m}x\right)}{8^{n+m}}∥=∥\frac{5440\Gamma \tilde{f}\left(2^{i}x\right)}{722925·524288·8^i}∥\le \sum _{i=n}^{n+m-1}\frac{5440K^{\prime }{2}^{ip}\theta {\parallel x\parallel }^p}{32768·8^i}\end{array}$$

(19)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}$. Since $p<3,$ we see by (19) that the sequence $\left\{\frac{{\tilde{f}}_3\left(2^{n}x\right)}{8^n}\right\}$ is a Cauchy sequence for all $x\in X$. From the completeness of $Y,$ the sequence $\left\{\frac{{\tilde{f}}_3\left(2^{n}x\right)}{8^n}\right\}$ converges. Thus, we may define a mapping $F_3$: $X\to Y$ by

$$F_3\left(x\right):=\underset{n\to \infty }{lim}\frac{{\tilde{f}}_3\left(2^{n}x\right)}{8^n}\mathrm{for}\mathrm{all}x\in X.$$

Putting $n=0$ and passing to the limit as $m\to \infty$ in (19), we find that

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_3\left(x\right)-F_3\left(x\right)\parallel \le \frac{5440K^{\prime }\theta {\parallel x\parallel }^p}{4096(8-2^p)}\end{array}$$

for all $x\in X.$ Based on the definition of $F_3$, we yield $F_3\left(2x\right)=8F_3\left(x\right)$ for all $x\in X$ and $D{F}_3(x,y)=0$ for all $x,y\in X.$

Let $p>3.$ Lemma 4 and (12) guarantee that

$$\begin{array}{c}\hfill ∥8^n{\tilde{f}}_3\left(2^{-n}x\right)-8^{n+m}{\tilde{f}}_3\left(2^{-n-m}x\right)∥\le \sum _{i=n}^{n+m-1}\frac{5440K^{\prime }{8}^i\theta {\parallel x\parallel }^p}{4096·2^{(i+1)p}}\end{array}$$

(20)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}.$ Since $p>3,$ we have from (20) that $\left\{8^n{\tilde{f}}_3\left(2^{-n}x\right)\right\}$ is a Cauchy sequence for all $x\in X$. Thus, because Y is complete, the sequence $\left\{8^n{\tilde{f}}_3\left(2^{-n}x\right)\right\}$ converges. Then, we may define a mapping $F_3$: $X\to Y$ by

$$F_3\left(x\right):=\underset{n\to \infty }{lim}{8}^n{\tilde{f}}_3\left(2^{-n}x\right)\mathrm{for}\mathrm{all}x\in X.$$

Let $n=0$ and take the limit $m\to \infty$ in (20); then,

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_3\left(x\right)-F_3\left(x\right)\parallel \le \frac{5440K^{\prime }\theta {\parallel x\parallel }^p}{4096(2^p-8)}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

The definition of $F_3$ shows that $F_3\left(2x\right)=8F_3\left(x\right)$ for all $x\in X$ and $D{F}_3(x,y)=0$ for all $x,y\in X.$

• Setting ${\mathit{F}}_{\mathbf{4}}$: Let $p<4.$ It follows from Lemma 4 and (13) that

$$\begin{array}{c}\hfill ∥\frac{{\tilde{f}}_4\left(2^{n}x\right)}{{16}^n}-\frac{{\tilde{f}}_4\left(2^{n+m}x\right)}{{16}^{n+m}}∥=∥\frac{21\Delta {\tilde{f}}_e\left(2^{i}x\right)}{2835·16384·{16}^i}∥\le \sum _{i=n}^{n+m-1}\frac{21K{2}^{ip}\theta {\parallel x\parallel }^p}{256·{16}^i}\end{array}$$

(21)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}.$ Considering $p<4$ and (21), we know that the sequence $\left\{\frac{{\tilde{f}}_4\left(2^{n}x\right)}{{16}^n}\right\}$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\left\{\frac{{\tilde{f}}_4\left(2^{n}x\right)}{{16}^n}\right\}$ converges. Thereby, we can define a mapping $F_4$: $X\to Y$ by

$$F_4\left(x\right):=\underset{n\to \infty }{lim}\frac{{\tilde{f}}_4\left(2^{n}x\right)}{{16}^n}\mathrm{for}\mathrm{all}x\in X.$$

Now, by letting $n=0$ and passing the limit $m\to \infty$ in (21), we obtain the inequality

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_4\left(x\right)-F_4\left(x\right)\parallel \le \frac{{21K\theta \parallel x\parallel }^p}{16(16-2^p)}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

We know from the definition of $F_4$ that $F_4\left(2x\right)=16F_4\left(x\right)$ for all $x\in X$ and $D{F}_4(x,y)=0$ for all $x,y\in X.$

Let $p>4.$ Lemma 4 and (13) imply that

$$\begin{array}{c}\hfill ∥{16}^n{\tilde{f}}_4\left(2^{-n}x\right)-{16}^{n+m}{\tilde{f}}_4\left(2^{-n-m}x\right)∥\le \sum _{i=n}^{n+m-1}\frac{21K{16}^i\theta {\parallel x\parallel }^p}{16·2^{(i+1)p}}\end{array}$$

(22)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}$. Since $p>4,$ we have by (22) that ${16}^n{\tilde{f}}_4\left(2^{-n}x\right)$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\left\{{16}^n{\tilde{f}}_4\left(2^{-n}x\right)\right\}$ converges. Thus, we can define a mapping $F_4$: $X\to Y$ by

$$F_4\left(x\right):=\underset{n\to \infty }{lim}{16}^n{\tilde{f}}_4\left(2^{-n}x\right)\mathrm{for}\mathrm{all}x\in X.$$

Meanwhile, in (22), letting $n=0$ and passing the limit $m\to \infty$, we then have the inequality

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_4\left(x\right)-F_4\left(x\right)\parallel \le \frac{{21K\theta \parallel x\parallel }^p}{16(2^p-16)}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

From the definition of $F_4$, we see that $F_4\left(2x\right)=16F_4\left(x\right)$ for all $x\in X$ and $D{F}_4(x,y)=0$ for all $x,y\in X.$

• Setting ${\mathit{F}}_{\mathbf{5}}$: Let $p<5.$ It follows from Lemma 4 and (12) that

$$\begin{array}{c}\hfill ∥\frac{{\tilde{f}}_5\left(2^{n}x\right)}{{32}^n}-\frac{{\tilde{f}}_5\left(2^{n+m}x\right)}{{32}^{n+m}}∥=∥\frac{1428\Gamma \tilde{f}\left(2^{i}x\right)}{722925·2097152·{32}^i}∥\le \sum _{i=n}^{n+m-1}\frac{1428K^{\prime }{2}^{ip}\theta {\parallel x\parallel }^p}{131072·{32}^i}\end{array}$$

(23)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}.$ We assume that $p<5.$ Thus, we have by (23) that $\left\{\frac{{\tilde{f}}_5\left(2^{n}x\right)}{{32}^n}\right\}$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\left\{\frac{{\tilde{f}}_5\left(2^{n}x\right)}{{32}^n}\right\}$ converges. Hence, we can define a mapping $F_5:X\to Y$ by

$$F_5\left(x\right):=\underset{n\to \infty }{lim}\frac{{\tilde{f}}_5\left(2^{n}x\right)}{{32}^n}\mathrm{for}\mathrm{all}x\in X.$$

Moreover, letting $n=0$ and passing the limit $m\to \infty$ in (23), we obtain the inequality

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_5\left(x\right)-F_5\left(x\right)\parallel \le \frac{1428K^{\prime }\theta {\parallel x\parallel }^p}{4096(32-2^p)}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

With the help of the definition of $F_5$, we obtain $F_5\left(2x\right)=32F_5\left(x\right)$ for all $x\in X$ and $D{F}_5(x,y)=0$ for all $x,y\in X.$

Let $p>5.$ In view of Lemma 4 and (12), we have

$$\begin{array}{c}\hfill ∥{32}^n{\tilde{f}}_5\left(2^{-n}x\right)-{32}^{n+m}{\tilde{f}}_5\left(2^{-n-m}x\right)∥\le \sum _{i=n}^{n+m-1}\frac{1428·{32}^i{K}^{\prime }\theta {\parallel x\parallel }^p}{4096·2^{(i+1)p}}\end{array}$$

(24)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}$. It follows from $p>5$ and (24) that the sequence $\left\{{32}^n{\tilde{f}}_5\left(2^{-n}x\right)\right\}$ is a Cauchy sequence for all $x\in X.$ Since Y is complete, we see that the sequence $\left\{{32}^n{\tilde{f}}_5\left(2^{-n}x\right)\right\}$ converges. Thus, one can define a mapping $F_5$: $X\to Y$ by

$$F_5\left(x\right):=\underset{n\to \infty }{lim}{32}^n{\tilde{f}}_5\left(2^{-n}x\right)\mathrm{for}\mathrm{all}x\in X.$$

Furthermore, setting $n=0$ and passing the limit $m\to \infty$ in (24), we reach

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_5\left(x\right)-F_5\left(x\right)\parallel \le \frac{1428K^{\prime }\theta {\parallel x\parallel }^p}{4096(2^p-32)}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

By virtue of the definition of $F_5$, we find that $F_5\left(2x\right)=32F_5\left(x\right)$ for all $x\in X$ and $D{F}_5(x,y)=0$ for all $x,y\in X.$

• Setting ${\mathit{F}}_{\mathbf{6}}$: Let $p<6.$ We combine Lemma 4 and (13) to find that

$$\begin{array}{c}\hfill ∥\frac{{\tilde{f}}_6\left(2^{n}x\right)}{{64}^n}-\frac{{\tilde{f}}_6\left(2^{n+m}x\right)}{{64}^{n+m}}∥=∥\frac{21\Delta {\tilde{f}}_e\left(2^{i}x\right)}{2835·262144·{64}^i}∥\le \sum _{i=n}^{n+m-1}\frac{21K{2}^{ip}\theta {\parallel x\parallel }^p}{4096·{64}^i}\end{array}$$

(25)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}$. Then, since $p<6,$ it follows by (25) that $\left\{\frac{{\tilde{f}}_6\left(2^{n}x\right)}{{64}^n}\right\}$ is a Cauchy sequence for all $x\in X.$ The completeness of Y ensures that the sequence $\left\{\frac{{\tilde{f}}_6\left(2^{n}x\right)}{{64}^n}\right\}$ converges, so that we can define a mapping $F_6$: $X\to Y$ by

$$F_6\left(x\right):=\underset{n\to \infty }{lim}\frac{{\tilde{f}}_6\left(2^{n}x\right)}{{64}^n}\mathrm{for}\mathrm{all}x\in X.$$

Then, we let $n=0$ and take the limit as $m\to \infty$ in (25) to obtain

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_6\left(x\right)-F_6\left(x\right)\parallel \le \frac{{21K\theta \parallel x\parallel }^p}{64(64-2^p)}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

According to the definition of $F_6$, we show that $F_6\left(2x\right)=64F_6\left(x\right)$ for all $x\in X$ and $D{F}_6(x,y)=0$ for all $x,y\in X.$

Let $p>6.$ It follows from Lemma 4 and (13) that

$$\begin{array}{c}\hfill ∥{64}^n{\tilde{f}}_6\left(2^{-n}x\right)-{64}^{n+m}{\tilde{f}}_6\left(2^{-n-m}x\right)∥\le \sum _{i=n}^{n+m-1}\frac{21·{64}^{i}K\theta {\parallel x\parallel }^p}{64·2^{(i+1)p}}\end{array}$$

(26)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}.$ Since $p>6,$ we see from (26) that the sequence $\left\{{64}^n{\tilde{f}}_6\left(2^{-n}x\right)\right\}$ is a Cauchy sequence for all $x\in X.$ Thus, by the completeness of Y, we find that the sequence $\left\{{64}^n{\tilde{f}}_6\left(2^{-n}x\right)\right\}$ converges. Hence, we can define a mapping $F_6$: $X\to Y$ by

$$F_6\left(x\right):=\underset{n\to \infty }{lim}{64}^n{\tilde{f}}_6\left(2^{-n}x\right)\mathrm{for}\mathrm{all}x\in X.$$

In addition, we let $n=0$ and $m\to \infty$ in (26) to obtain

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_6\left(x\right)-F_6\left(x\right)\parallel \le \frac{{21K\theta \parallel x\parallel }^p}{64(2^p-64)}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

From the definition of $F_6$, we reach $F_6\left(2x\right)=64F_6\left(x\right)$ for all $x\in X$ and $D{F}_6(x,y)=0$ for all $x,y\in X.$

• Setting ${\mathit{F}}_{\mathbf{7}}$: Let $p<7.$ We know by Lemma 4 and (12) that

$$\begin{array}{c}\hfill ∥\frac{{\tilde{f}}_7\left(2^{n}x\right)}{{128}^n}-\frac{{\tilde{f}}_7\left(2^{n+m}x\right)}{{128}^{n+m}}∥=∥\frac{85\Gamma \tilde{f}\left(2^{i}x\right)}{722925·8388608·{128}^i}∥\le \sum _{i=n}^{n+m-1}\frac{85K^{\prime }{2}^{ip}\theta {\parallel x\parallel }^p}{524504·{128}^i}\end{array}$$

(27)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}.$ Based on the fact that $p<7$ and (27), the sequence $\left\{\frac{{\tilde{f}}_7\left(2^{n}x\right)}{{128}^n}\right\}$ is a Cauchy sequence for all $x\in X.$ Since Y is complete, the sequence $\left\{\frac{{\tilde{f}}_7\left(2^{n}x\right)}{{128}^n}\right\}$ converges. Thus, one can define a mapping $F_7$: $X\to Y$ by

$$F_7\left(x\right):=\underset{n\to \infty }{lim}\frac{{\tilde{f}}_7\left(2^{n}x\right)}{{128}^n}\mathrm{for}\mathrm{all}x\in X.$$

Moreover, letting $n=0$ and passing the limit $m\to \infty$ in (27), we obtain

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_7\left(x\right)-F_7\left(x\right)\parallel \le \frac{85K^{\prime }\theta {\parallel x\parallel }^p}{4096(128-2^p)}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

By the definition of $F_7$, we have $F_7\left(2x\right)=128F_7\left(x\right)$ for all $x\in X$ and $D{F}_7(x,y)=0$ for all $x,y\in X.$

Let $p>7.$ Note that, by Lemma 4 and (12), we have

$$\begin{array}{c}\hfill ∥{128}^n{\tilde{f}}_7\left(2^{-n}x\right)-{128}^{n+m}{\tilde{f}}_7\left(2^{-n-m}x\right)∥\le \frac{85·{128}^i{K}^{\prime }\theta {\parallel x\parallel }^p}{4096·2^{(i+1)p}}\end{array}$$

(28)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}.$ On the basis of the assumption $p>7$ and (28), we see that $\left\{{128}^n{\tilde{f}}_7\left(2^{-n}x\right)\right\}$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\left\{{128}^n{\tilde{f}}_7\left(2^{-n}x\right)\right\}$ converges for all $x\in X.$ Therefore, we can define a mapping $F_7$: $X\to Y$ by

$$F_7\left(x\right):=\underset{n\to \infty }{lim}{128}^n{\tilde{f}}_7\left(2^{-n}x\right)\mathrm{for}\mathrm{all}x\in X.$$

In (28), we set $n=0$ and then let $m\to \infty$ to find

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_7\left(x\right)-F_7\left(x\right)\parallel \le \frac{85K^{\prime }\theta {\parallel x\parallel }^p}{4096(2^p-128)}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

By the definition of $F_7$, it is shown that $F_7\left(2x\right)=128F_7\left(x\right)$ for all $x\in X$ and $D{F}_7(x,y)=0$ for all $x,y\in X.$

• Setting ${\mathit{F}}_{\mathbf{8}}$: Let $p<8.$ It follows from Lemma 4 and (13) that

$$\begin{array}{c}\hfill ∥\frac{{\tilde{f}}_8\left(2^{n}x\right)}{{256}^n}-\frac{{\tilde{f}}_8\left(2^{n+m}x\right)}{{256}^{n+m}}∥=∥\frac{\Delta {\tilde{f}}_e\left(2^{i}x\right)}{2835·256·4096·{256}^i}∥\le \sum _{i=n}^{n+m-1}\frac{K{2}^{ip}\theta {\parallel x\parallel }^p}{16384·{256}^i}\end{array}$$

(29)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}.$ By the assumption $p<8$ and (29), the sequence $\left\{\frac{{\tilde{f}}_8\left(2^{n}x\right)}{{256}^n}\right\}$ is a Cauchy sequence for all $x\in X.$ Since Y is complete, the sequence $\left\{\frac{{\tilde{f}}_8\left(2^{n}x\right)}{{256}^n}\right\}$ converges. Hence, we can define a mapping $F_8$: $X\to Y$ by

$$F_8\left(x\right):=\underset{n\to \infty }{lim}\frac{{\tilde{f}}_8\left(2^{n}x\right)}{{256}^n}\mathrm{for}\mathrm{all}x\in X.$$

Letting $n=0$ and passing the limit $m\to \infty$ in (29), we then have the following inequality:

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_8\left(x\right)-F_8\left(x\right)\parallel \le \frac{{K\theta \parallel x\parallel }^p}{64(256-2^p)}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

From the definition of $F_8,$ we are forced to conclude that $F_8\left(2x\right)=256F_8\left(x\right)$ for all $x\in X$ and $D{F}_8(x,y)=0$ for all $x,y\in X.$

Let $p>8.$ Observe that, by Lemma 4 and (13), we obtain

$$\begin{array}{c}\hfill ∥{256}^n{\tilde{f}}_8\left(2^{-n}x\right)-{256}^{n+m}{\tilde{f}}_8\left(2^{-n-m}x\right)∥\le \sum _{i=n}^{n+m-1}\frac{{256}^{i}K\theta {\parallel x\parallel }^p}{64·2^{(i+1)p}}\end{array}$$

(30)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}.$ It follows from the assumption $p>8$ and (30) that $\left\{{256}^n{\tilde{f}}_8\left(2^{-n}x\right)\right\}$ is a Cauchy sequence for all $x\in X.$ The completeness of Y implies that the sequence $\left\{{256}^n{\tilde{f}}_8\left(2^{-n}x\right)\right\}$ converges, so that we can define a mapping $F_8$: $X\to Y$ by

$$F_8\left(x\right):=\underset{n\to \infty }{lim}{256}^n{\tilde{f}}_8\left(2^{-n}x\right)\mathrm{for}\mathrm{all}x\in X.$$

We let $n=0$ and then take $m\to \infty$ in (30) to obtain

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_8\left(x\right)-F_8\left(x\right)\parallel \le \frac{{K\theta \parallel x\parallel }^p}{64(2^p-256)}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

According to the definition of $F_8$, we find that $F_8\left(2x\right)=256F_8\left(x\right)$ for all $x\in X$ and $D{F}_8(x,y)=0$ for all $x,y\in X.$

• Setting ${\mathit{F}}_{\mathbf{9}}$: Let p < 9. It follows from Lemma 4 and (12) that

$$\begin{array}{c}\hfill ∥\frac{{\tilde{f}}_9\left(2^{n}x\right)}{{512}^n}-\frac{{\tilde{f}}_9\left(2^{n+m}x\right)}{{512}^{n+m}}∥=∥\frac{\Gamma \tilde{f}\left(2^{i}x\right)}{722925·33554432·{512}^i}∥\le \sum _{i=n}^{n+m-1}\frac{K^{\prime }{2}^{ip}\theta {\parallel x\parallel }^p}{2097152·{512}^i}\end{array}$$

(31)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}.$ We then have by the assumption $p<9$ and (31) that $\left\{\frac{{\tilde{f}}_9\left(2^{n}x\right)}{{512}^n}\right\}$ is a Cauchy sequence for all $x\in X.$ Since Y is complete, the sequence $\left\{\frac{{\tilde{f}}_9\left(2^{n}x\right)}{{512}^n}\right\}$ converges. Then, we can define a mapping $F_9$: $X\to Y$ by

$$F_9\left(x\right):=\underset{n\to \infty }{lim}\frac{{\tilde{f}}_9\left(2^{n}x\right)}{{512}^n}\mathrm{for}\mathrm{all}x\in X.$$

On the other hand, letting $n=0$ and passing the limit $m\to \infty$ in (31), we deduce that

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_9\left(x\right)-F_9\left(x\right)\parallel \le \frac{K^{\prime }\theta {\parallel x\parallel }^p}{4096(512-2^p)}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

We have from the definition of $F_9$ that $F_9\left(2x\right)=512F_9\left(x\right)$ for all $x\in X$ and $D{F}_9(x,y)=0$ for all $x,y\in X.$

Let $p>9.$ Using Lemma 4 and (12), we have

$$\begin{array}{c}\hfill ∥{512}^n{\tilde{f}}_9\left(2^{-n}x\right)-{512}^{n+m}{\tilde{f}}_9\left(2^{-n-m}x\right)∥\le \frac{{512}^i{K}^{\prime }\theta {\parallel x\parallel }^p}{4096·2^{(i+1)p}}\end{array}$$

(32)

for all $x\in X$ and $n,m\in \mathbb{N}\cup \left\{0\right\}.$ From $p>9$ and (32), it follows that the sequence $\left\{{512}^n{\tilde{f}}_9\left(2^{-n}x\right)\right\}$ is a Cauchy sequence for all $x\in X.$ By the completeness of $Y,$ the sequence $\left\{{512}^n{\tilde{f}}_9\left(2^{-n}x\right)\right\}$ converges. Thereby, we can define a mapping $F_9$: $X\to Y$ by

$$F_9\left(x\right):=\underset{n\to \infty }{lim}{512}^n{\tilde{f}}_9\left(2^{-n}x\right)\mathrm{for}\mathrm{all}x\in X.$$

In particular, we let $n=0$ and $m\to \infty$ in (32) to obtain

$$\begin{array}{c}\hfill \parallel {\tilde{f}}_9\left(x\right)-F_9\left(x\right)\parallel \le \frac{K^{\prime }\theta {\parallel x\parallel }^p}{4096(2^p-512)}\mathrm{for}\mathrm{all}x\in X.\end{array}$$

With the aid of the definition of $F_9$, we obtain that $F_9\left(2x\right)=512F_9\left(x\right)$ for all $x\in X$ and $D{F}_9(x,y)=0$ for all $x,y\in X.$

Finally, we set a mapping F as

$$F\left(x\right):=\sum _{k=1}^9{F}_k\left(x\right)\mathrm{for}\mathrm{all}x\in X.$$

Since $D{F}_k(x,y)=0$ for all $k\in \{1,2,\dots ,9\},$ we have

$$DF(x,y)=\sum _{k=1}^{9}D{F}_k(x,y)=0\mathrm{for}\mathrm{all}x,y\in X.$$

Next, we are in the position to prove that the mapping F satisfies Inequality (11). Since $\tilde{f}\left(x\right)={\sum }_{k=1}^9{\tilde{f}}_k\left(x\right),$ we have

$$\parallel \tilde{f}\left(x\right)-F\left(x\right)\parallel \le \sum _{k=1}^9\parallel {\tilde{f}}_k\left(x\right)-F_k\left(x\right)\parallel \mathrm{for}\mathrm{all}x\in X,$$

and so we obtain the desired result (11).

It remains to be proven that F is unique. Suppose that $F^{\prime }:V\to Y$ is another mapping with $F^{\prime }\left(0\right)=0$ satisfying the relation $D{F}^{\prime }(x,y)=0$ and the inequality (11). We know by Lemma 5 that for each $k\in \{1,2,\dots ,9\},$ the mappings $F_k^{\prime }$: $X\to Y$ satisfy

$$\begin{array}{c}\hfill F^{\prime }\left(x\right)=\sum _{k=1}^9{F}_k^{\prime }\left(x\right),F_k^{\prime }\left(2x\right)=2^k{F}_k^{\prime }\left(x\right)(x\in X).\end{array}$$

Next, it is sufficient to show the uniqueness of F only for $2<p<3$. This is because other cases of p can be shown in a similar fashion. Therefore, let us assume that $2<p<3.$ Then, we see that for $2<p<3,$

$$\begin{array}{c}\parallel 4^n{\tilde{f}}_2\left(\frac{x}{2^n}\right)-F_2^{\prime }\left(x\right)\parallel = \parallel 4^n{\tilde{f}}_2\left(\frac{x}{2^n}\right)-4^n{F}_2^{\prime }\left(\frac{x}{2^n}\right)\parallel \hfill \\ =\frac{4^n}{181440}\parallel 262144{\tilde{f}}_e\left(\frac{x}{2^n}\right)-21504{\tilde{f}}_e\left(\frac{2x}{2^n}\right)+336{\tilde{f}}_e\left(\frac{4x}{2^n}\right)-{\tilde{f}}_e\left(\frac{8x}{2^n}\right)\hfill \\ -262144F_e^{\prime }\left(\frac{x}{2^n}\right)+21504F_e^{\prime }\left(\frac{2x}{2^n}\right)-336F_e^{\prime }\left(\frac{4x}{2^n}\right)+F_e^{\prime }\left(\frac{8x}{2^n}\right)\parallel \hfill \\ \le \frac{262144·4^n}{181440}\parallel {\tilde{f}}_e\left(\frac{x}{2^n}\right)-F_e^{\prime }\left(\frac{x}{2^n}\right)\parallel +\frac{21504·4^n}{181440}\parallel {\tilde{f}}_e\left(\frac{2x}{2^n}\right)-F_e^{\prime }\left(\frac{2x}{2^n}\right)\parallel \hfill \\ +\frac{336·4^n}{181440}\parallel {\tilde{f}}_e\left(\frac{4x}{2^n}\right)-F_e^{\prime }\left(\frac{4x}{2^n}\right)\parallel +\frac{4^n}{181440}\parallel {\tilde{f}}_e\left(\frac{8x}{2^n}\right)-F_e^{\prime }\left(\frac{8x}{2^n}\right)\parallel \hfill \\ \le \frac{262144+21504·2^p+336·4^p+8^p}{181440}·\frac{4^n}{2^{np}}M\theta {\parallel x\parallel }^p,\hfill \end{array}$$

and

$$\begin{array}{c}\parallel \frac{{\tilde{f}}_3\left(2^{n}x\right)}{8^n}-F_3^{\prime }\left(x\right)\parallel =\parallel \frac{{\tilde{f}}_3\left(2^{n}x\right)}{8^n}-\frac{F_3^{\prime }\left(2^{n}x\right)}{8^n}\parallel \hfill \\ =\frac{5440}{722925·65536}\parallel \frac{4194304\left((F_o^{\prime }-{\tilde{f}}_o)\left(2^{n}x\right)\right)-2269184\left((F_o^{\prime }-{\tilde{f}}_o)\left(2^{n+1}x\right)\right)}{8^n}\hfill \\ +\frac{87360\left((F_o^{\prime }-{\tilde{f}}_o)\left(2^{n+2}x\right)\right)-674\left((F_o^{\prime }-{\tilde{f}}_o)\left(2^{n+3}x\right)\right)+\left((F_o^{\prime }-{\tilde{f}}_o)\left(2^{n+4}x\right)\right)}{8^n}\parallel \hfill \\ \le \left(\frac{4194304+2269184·2^p+87360·2^{2p}+674·2^{3p}+2^{4p}}{722925·65536}\right)·\frac{5440·2^{np}M\theta {\parallel x\parallel }^p}{8^n}\hfill \end{array}$$

for all $x\in X$ and all positive integers $n,$ where

$$\begin{array}{cc}\hfill M:=& \left(\frac{4096}{|2-2^p|}+\frac{5440}{|8-2^p|}+\frac{1428}{|32-2^p|}+\frac{85}{|128-2^p|}+\frac{1}{|512-2^p|}\right)·\frac{K^{\prime }}{4096}\hfill \\ \hfill & +\left(\frac{64}{|4-2^p|}+\frac{84}{|16-2^p|}+\frac{21}{|64-2^p|}+\frac{1}{|256-2^p|}\right)·\frac{K}{64}.\hfill \end{array}$$

Taking the limit in the above relations as $n\to \infty ,$ we obtain the equality

$$F_2^{\prime }\left(x\right)=\underset{n\to \infty }{lim}{\tilde{4}}^n{\tilde{f}}_2\left(\frac{x}{2^n}\right),F_3^{\prime }\left(x\right)=\underset{n\to \infty }{lim}\frac{{\tilde{f}}_3\left(2^{n}x\right)}{8^n}(x\in X),$$

which means that $F_2\left(x\right)=F_2^{\prime }\left(x\right)$ and $F_3\left(x\right)=F_3^{\prime }\left(x\right)$ for all $x\in X.$ In a similar way, it is easily shown that for each $k\in \{1,4,5,6,7,8,9\},$ the equalities $F_k=F_k^{\prime }$ hold. Note that

$$F\left(x\right)=\sum _{k=1}^9{F}_k\left(x\right)=\sum _{k=1}^9{F}_k^{\prime }\left(x\right)=F^{\prime }\left(x\right).$$

This completes the proof of the uniqueness of F for $2<p<3.$

For other cases of p, one can prove the uniqueness of F using the same reasoning as in the uniqueness proof for $2<p<3.$ □

5. Conclusions

Example 1 in [10] proved that a specially defined mapping $f:\mathbb{R}\to \mathbb{R}$ satisfies the inequality $|{\sum }_{i=0}^n{}_n{C}_i{(-1)}^{n-i}{f(x+iy)-n!f\left(y\right)|\le 4(n+1)!(n+1)}^{2n}{\left(\right|x|}^n+{\left|y\right|}^n)$, but there is no n-monomial mapping $F:\mathbb{R}\to \mathbb{R}$ and constant $d>0$ that satisfy the inequality $|f\left(x\right)-F\left(x\right)|\le {d\left|x\right|}^n$ (see also [15,49] and Theorem 2 in [5]). Since a mapping f satisfying any Cauchy functional equation, quadratic functional equation, …, or nonic functional equation satisfies the general nonic functional equation, Theorem 2 fails to hold for $p=1,2,\dots ,9$.

For further study, this theory can be extended to the case of 11 instead of 10 in Equation (2). We will also investigate whether we can extend our theory to the functional equation described in Equation (1) as well.