On the stability of a mixed type quadratic-additive functional equation

Abstract

In this paper we establish the general solutions of the following mixed type quadratic-additive functional equation:

$$9f\left(\frac{x+y+z}{3}\right)+4[f\left(\frac{x-y}{2}\right)+f\left(\frac{y-z}{2}\right)+f\left(\frac{z-x}{2}\right)]=3[f(x)+f(y)+f(z)]$$

in the class of functions between real vector spaces. Moreover, we prove the generalized Hyers-Ulam-Rassias stability of this equation in Banach spaces.

MSC:39B82, 39B52.

1 Introduction

The stability problems of functional equations go back to 1940 when Ulam [1] proposed the following question: This question was solved affirmatively by Hyers [2] under the assumption that $G_2$ is a Banach space. He proved that if f is a mapping between Banach spaces satisfying $\parallel f(x+y)-f(x)-f(y)\parallel \le ϵ$ for some fixed $ϵ\ge 0$, then there exists a unique additive mapping A such that $\parallel f(x)-A(x)\parallel \le ϵ$. In 1978, Rassias [3] generalized Hyers’ result to the unbounded Cauchy difference. Since then, the stability problems of various functional equations have been extensively studied and generalized by a number of authors (see [4–9]).

Let f be a mapping from a group $G_1$ to a metric group $G_2$ with the metric $d(\cdot ,\cdot )$ such that

$$d(f(xy),f(x)f(y))\le ϵ.$$

Then does there exist a group homomorphism $L:G_1\to G_2$ and ${\delta }_{ϵ}>0$ such that

$$d(f(x),L(x))\le {\delta }_{ϵ}$$

for all $x\in G_1$?

In particular, Kannappan [10] introduced the following mixed type quadratic-additive functional equation:

$$f(x+y+z)+f(x)+f(y)+f(z)=f(x+y)+f(y+z)+f(z+x)$$

(1.1)

and proved that a function on a real vector space is a solution of (1.1) if and only if there exist a symmetric biadditive function B and an additive function A such that $f(x)=B(x,x)+A(x)$. In addition, Jung [11] investigated the Hyers-Ulam stability of (1.1) on restricted domains and applied the result to the study of an interesting asymptotic behavior of the quadratic functions. More generally, Jun and Kim [12] solved the general solutions and proved the stability of the following functional equation, which is a generalization of (1.1):

$$f\left(\sum _{i=1}^n{x}_i\right)+(n-2)\sum _{i=1}^{n}f(x_i)=\sum _{1\le i<j\le n}f(x_i+x_j)(n>2).$$

Najati and Moghimi [13] introduced another mixed type quadratic-additive functional equation

$$f(2x+y)+f(2x-y)=f(x+y)+f(x-y)+2f(2x)-2f(x)$$

and investigated the generalized Hyers-Ulam-Rassias stability of this equation in quasi-Banach spaces.

In this paper, we introduce the following quadratic-additive functional equation:

$$\begin{array}{r}9f\left(\frac{x+y+z}{3}\right)+4[f\left(\frac{x-y}{2}\right)+f\left(\frac{y-z}{2}\right)+f\left(\frac{z-x}{2}\right)]\\ =3[f(x)+f(y)+f(z)]\end{array}$$

(1.2)

to establish the general solutions and stability problems of this equation. For real vector spaces X and Y, we prove in Section 2 that a mapping $f:X\to Y$ satisfies (1.2) if and only if there exist a quadratic mapping $Q:X\to Y$ satisfying

$$Q(x+y)+Q(x-y)=2Q(x)+2Q(y)$$

(1.3)

and an additive mapping $A:X\to Y$ satisfying

$$A(x+y)=A(x)+A(y)$$

(1.4)

such that

$$f(x)=Q(x)+A(x)$$

for all $x\in X$. We refer to [14–24] for the stability results of other mixed type functional equations. In Section 3, we prove the generalized Hyers-Ulam-Rassias stability of (1.2) in Banach spaces.

2 General solutions of (1.2)

Throughout this section, X and Y will be real vector spaces. In order to solve the general solutions of (1.2), we need the following two lemmas.

Lemma 2.1 If an even mapping $f:X\to Y$ satisfies (1.2) for all $x,y\in X$, then f is quadratic.

Proof Putting $x=y=z=0$ in (1.2), we have $f(0)=0$. Putting $z=-x$ in (1.2) yields

$$9f\left(\frac{y}{3}\right)+4[f\left(\frac{x-y}{2}\right)+f\left(\frac{x+y}{2}\right)+f(x)]=6f(x)+3f(y)$$

(2.1)

for all $x,y\in X$. Replacing y with x in (2.1) gives

$$9f\left(\frac{x}{3}\right)=f(x)$$

(2.2)

for all $x\in X$. Putting $y=0$ in (2.1), we have

$$4f\left(\frac{x}{2}\right)=f(x)$$

(2.3)

for all $x\in X$. Using (2.2) and (2.3), we can rewrite (1.2) as

$$f(x+y+z)+f(x-y)+f(y-z)+f(z-x)=3[f(x)+f(y)+f(z)]$$

(2.4)

for all $x,y,z\in X$. Putting $z=0$ in (2.4), we obtain

$$f(x+y)+f(x-y)=2f(x)+2f(y)$$

for all $x,y\in X$. □

Lemma 2.2 If an odd mapping $f:X\to Y$ satisfies (1.2) for all $x,y\in X$, then f is additive.

Proof Putting $x=y=z=0$ in (1.2), we have $f(0)=0$. Putting $z=-x$ in (1.2) yields

$$9f\left(\frac{y}{3}\right)+4[f\left(\frac{x-y}{2}\right)+f\left(\frac{x+y}{2}\right)-f(x)]=3f(y)$$

(2.5)

for all $x,y\in X$. Replacing y by x in (2.5) gives

$$9f\left(\frac{x}{3}\right)=3f(x)$$

(2.6)

for all $x\in X$. Putting $y=0$ in (2.5), we have

$$2f\left(\frac{x}{2}\right)=f(x)$$

(2.7)

for all $x\in X$. It follows from (2.5), (2.6) and (2.7) that

$$f\left(\frac{x+y}{2}\right)+f\left(\frac{x-y}{2}\right)=2f(x)$$

(2.8)

for all $x,y\in X$. Replacing x with $x+y$ and y with $x-y$ in (2.8), we obtain

$$f(x+y)=f(x)+f(y)$$

for all $x,y\in X$. □

Now we are ready to establish the general solutions of (1.2).

Theorem 2.3 A function $f:X\to Y$ satisfies (1.2) for all $x,y,z\in X$ if and only if there exist a symmetric biadditive mapping $B:X\times X\to Y$ and an additive mapping $A:X\to Y$ such that

$$f(x)=B(x,x)+A(x)$$

for all $x\in X$.

Proof (Necessity) We decompose f into the even part and the odd part by putting

$$f_e(x)=\frac{1}{2}(f(x)+f(-x)),f_o(x)=\frac{1}{2}(f(x)-f(-x))$$

for all $x\in X$. By Lemmas 2.1 and 2.2 we have the result.

(Sufficiency) This is obvious. □

3 Stability of (1.2)

In what follows, X and Y will be a real normed linear space and a real Banach space, respectively. For convenience, we define

$$\begin{array}{rcl}Df(x,y,z)& :=& 9f\left(\frac{x+y+z}{3}\right)+4[f\left(\frac{x-y}{2}\right)+f\left(\frac{y-z}{2}\right)+f\left(\frac{z-x}{2}\right)]\\ -3[f(x)+f(y)+f(z)]\end{array}$$

for all $x,y,z\in X$. Let $\phi :X\times X\times X\to [0,\mathrm{\infty })$ be a mapping satisfying one of the conditions (), (ℬ) and one of the conditions (), ():

$$\begin{array}{rl}(\mathcal{A})& {\mathrm{\Phi }}_1(x,y,z):=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{9^k}\phi (3^{k}x,3^{k}y,3^{k}z)<\mathrm{\infty },\\ (\mathcal{B})& {\mathrm{\Phi }}_2(x,y,z):=\sum _{k=0}^{\mathrm{\infty }}{9}^k\phi (\frac{x}{3^k},\frac{y}{3^k},\frac{z}{3^k})<\mathrm{\infty },\\ (\mathcal{C})& {\mathrm{\Psi }}_1(x,y,z):=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{3^k}\phi (3^{k}x,3^{k}y,3^{k}z)<\mathrm{\infty },\\ (\mathcal{D})& {\mathrm{\Psi }}_2(x,y,z):=\sum _{k=0}^{\mathrm{\infty }}{3}^k\phi (\frac{x}{3^k},\frac{y}{3^k},\frac{z}{3^k})<\mathrm{\infty },\end{array}$$

for all $x,y,z\in X$. We note that the condition () implies (). Similarly, the condition (ℬ) implies (). One of the conditions (), (ℬ) will be needed to derive a quadratic mapping, and one of the conditions (), () will be required to derive an additive mapping in the following theorem.

Theorem 3.1 Suppose that a mapping $f:X\to Y$ satisfies

$$\parallel Df(x,y,z)\parallel \le \phi (x,y,z)$$

(3.1)

for all $x,y,z\in X$. Then there exist a unique quadratic mapping $Q:X\to Y$ satisfying (1.3) and an additive mapping $A:X\to Y$ satisfying (1.4) such that

$$\begin{array}{c}\parallel f(x)-Q(x)-A(x)+\frac{1}{2}f(0)\parallel \hfill \\ \le \frac{1}{2}[{\mathrm{\Phi }}_i(x,x,-x)+{\mathrm{\Phi }}_i(-x,-x,x)]+\frac{1}{6}[{\mathrm{\Psi }}_j(x,x,-x)+{\mathrm{\Psi }}_j(-x,-x,x)],\hfill \\ \parallel \frac{f(x)+f(-x)}{2}-Q(x)+\frac{1}{2}f(0)\parallel \le \frac{1}{2}[{\mathrm{\Phi }}_i(x,x,-x)+{\mathrm{\Phi }}_i(-x,-x,x)],\hfill \end{array}$$

and

$$\parallel \frac{f(x)-f(-x)}{2}-A(x)\parallel \le \frac{1}{6}[{\mathrm{\Psi }}_j(x,x,-x)+{\mathrm{\Psi }}_j(-x,-x,x)]$$

for all $x\in X$ and for $i=1$ or 2, $j=1$ or 2. The mappings Q and A are given by

$$\{\begin{array}{cc}Q(x)={lim}_{n\to \mathrm{\infty }}\frac{1}{2\cdot 9^n}[f(3^{n}x)+f(-3^{n}x)]\hfill & \mathit{\text{if condition}}\text{ (}\mathcal{A}\text{) }\mathit{\text{holds}},\hfill \\ Q(x)={lim}_{n\to \mathrm{\infty }}\frac{9^n}{2}[f(\frac{x}{3^n})+f(-\frac{x}{3^n})],f(0)=0\hfill & \mathit{\text{if condition}}\text{ (}\mathcal{B}\text{) }\mathit{\text{holds}},\hfill \\ A(x)={lim}_{n\to \mathrm{\infty }}\frac{1}{2\cdot 3^n}[f(3^{n}x)-f(-3^{n}x)]\hfill & \mathit{\text{if condition}}\text{ (}\mathcal{C}\text{) }\mathit{\text{holds}},\hfill \\ A(x)={lim}_{n\to \mathrm{\infty }}\frac{3^n}{2}[f(\frac{x}{3^n})-f(-\frac{x}{3^n})],f(0)=0\hfill & \mathit{\text{if condition}}\text{ (}\mathcal{D}\text{) }\mathit{\text{holds}}\hfill \end{array}$$

for all $x\in X$.

Proof We first consider the even part of f. Let $f_e:X\to Y$ be a function defined by $f_e(x):=\frac{f(x)+f(-x)}{2}$ for all $x\in X$. Then $f_e(-x)=f_e(x)$ and

$$\parallel D{f}_e(x,y,z)\parallel \le \frac{1}{2}[\phi (x,y,z)+\phi (-x,-y,-z)]$$

(3.2)

for all $x,y,z\in X$. Putting $y=x$, $z=-x$ in (3.2), we have

$$\parallel 9g\left(\frac{x}{3}\right)-g(x)\parallel \le \frac{1}{2}[\phi (x,x,-x)+\phi (-x,-x,x)]$$

(3.3)

for all $x\in X$, where $g(x):=f_e(x)+\frac{1}{2}f(0)$.

Case 1. Assume that φ satisfies the condition (). Replacing x by 3x in (3.3) and dividing by 9 yield

$$\parallel g(x)-\frac{g(3x)}{9}\parallel \le \frac{1}{2\cdot 9}[\phi (3x,3x,-3x)+\phi (-3x,-3x,3x)]$$

(3.4)

for all $x\in X$. Making use of an induction argument in (3.4) implies

$$\parallel g(x)-\frac{g(3^{n}x)}{9^n}\parallel \le \frac{1}{2}\sum _{k=1}^n\frac{1}{9^k}[\phi (3^{k}x,3^{k}x,-3^{k}x)+\phi (-3^{k}x,-3^{k}x,3^{k}x)]$$

(3.5)

for all $n\in \mathbb{N}$ and $x\in X$. From (3.5) we figure out

$$\begin{array}{rcl}\parallel \frac{g(3^{m}x)}{9^m}-\frac{g(3^{n}x)}{9^n}\parallel & =& \frac{1}{9^m}\parallel g\left(3^{m}x\right)-\frac{g(3^{n-m}\cdot 3^{m}x)}{9^{n-m}}\parallel \\ \le & \frac{1}{2}\sum _{k=m+1}^n\frac{1}{9^k}[\phi (3^{k}x,3^{k}x,-3^{k}x)+\phi (-3^{k}x,-3^{k}x,3^{k}x)]\end{array}$$

for all $n,m\in \mathbb{N}$ with $n>m$ and $x\in X$. The right-hand side of the inequality above tends to 0 as $m\to \mathrm{\infty }$, the sequence $\{g(3^{n}x)/9^n\}$ is a Cauchy sequence for all $x\in X$ and thus converges by the completeness of Y. Therefore, we can define a mapping $Q:X\to Y$ by

$$Q(x):=\underset{n\to \mathrm{\infty }}{lim}\frac{g(3^{n}x)}{9^n}=\underset{n\to \mathrm{\infty }}{lim}\frac{f(3^{n}x)+f(-3^{n}x)}{2\cdot 9^n}$$

for all $x\in X$. Note that $Q(0)=0$, $Q(-x)=Q(x)$ for all $x\in X$. It follows from the condition () and (3.2) that Q satisfies

$$9Q\left(\frac{x+y+z}{3}\right)+4[Q\left(\frac{x-y}{2}\right)+Q\left(\frac{y-z}{2}\right)+Q\left(\frac{z-x}{2}\right)]=3[Q(x)+Q(y)+Q(z)]$$

for all $x,y,z\in X$. According to Lemma 2.1, the mapping Q satisfies (1.3). Letting $n\to \mathrm{\infty }$ in (3.5), we have

$$\parallel g(x)-Q(x)\parallel \le \frac{1}{2}[{\mathrm{\Phi }}_1(x,x,-x)+{\mathrm{\Phi }}_1(-x,-x,x)]$$

(3.6)

for all $x\in X$. Now we are going to prove the uniqueness of Q. Assume that $Q^{\prime }$ is another quadratic function satisfying (1.3) and (3.6). Obviously, we have $Q(3^{n}x)=9^{n}Q(x)$ and $Q^{\prime }(3^{n}x)=9^n{Q}^{\prime }(x)$ for all $x\in X$. Then we figure out

$$\begin{array}{rcl}\parallel Q(x)-Q^{\prime }(x)\parallel & =& 9^{-n}\parallel Q\left(3^{n}x\right)-Q^{\prime }\left(3^{n}x\right)\parallel \\ \le & 9^{-n}\parallel Q\left(3^{n}x\right)-g\left(3^{n}x\right)\parallel +9^{-n}\parallel g\left(3^{n}x\right)-Q^{\prime }\left(3^{n}x\right)\parallel \\ \le & \sum _{k=n+1}^{\mathrm{\infty }}\frac{1}{9^k}[\phi (3^{k}x,3^{k}x,-3^{k}x)+\phi (-3^{k}x,-3^{k}x,3^{k}x)]\end{array}$$

for all $n\in \mathbb{N}$ and $x\in X$. Taking the limit as $n\to \mathrm{\infty }$, we conclude that $Q(x)=Q^{\prime }(x)$ for all $x\in X$.

Case 2. If φ satisfies the condition (ℬ) (and hence implies ()), the proof is analogous to that of Case 1. By virtue of the condition (ℬ) and (3.1), we have $\phi (0,0,0)=0$ and $f(0)=0$. An induction argument on (3.3) implies

$$\parallel 9^n{f}_e\left(\frac{x}{3^n}\right)-f_e(x)\parallel \le \frac{1}{2}\sum _{k=0}^{n-1}{9}^k[\phi (\frac{x}{3^k},\frac{x}{3^k},-\frac{x}{3^k})+\phi (-\frac{x}{3^k},-\frac{x}{3^k},\frac{x}{3^k})]$$

(3.7)

for all $n\in \mathbb{N}$ and $x\in X$. Using a similar argument to that of Case 1, we see that the sequence $\{9^n{f}_e(\frac{x}{3^n})\}$ is a Cauchy sequence for all $x\in X$. Thus we can define a mapping $Q:X\to Y$ by

$$Q(x):=\underset{n\to \mathrm{\infty }}{lim}{9}^n{f}_e\left(\frac{x}{3^n}\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{9^n}{2}[f\left(\frac{x}{3^n}\right)+f(-\frac{x}{3^n})]$$

for all $x\in X$. Note that $Q(0)=0$ and $Q(-x)=Q(x)$ for all $x\in X$. From the condition (ℬ) and (3.2), we see that Q satisfies

$$9Q\left(\frac{x+y+z}{3}\right)+4[Q\left(\frac{x-y}{2}\right)+Q\left(\frac{y-z}{2}\right)+Q\left(\frac{z-x}{2}\right)]=3[Q(x)+Q(y)+Q(z)]$$

for all $x,y,z\in X$. By Lemma 2.1 the mapping Q satisfies (1.3). Taking the limit as $n\to \mathrm{\infty }$ in (3.7), we obtain

$$\parallel f_e(x)-Q(x)\parallel \le \frac{1}{2}[{\mathrm{\Phi }}_2(x,x,-x)+{\mathrm{\Phi }}_2(-x,-x,x)]$$

for all $x\in X$. The rest of the proof is similar to that of Case 1.

Next, we consider the odd part of f. Now, let $f_o:X\to Y$ be a function defined by $f_o(x):=\frac{1}{2}[f(x)-f(-x)]$ for all $x\in X$. Then $f_o(0)=0$, $f_o(-x)=-f_o(x)$ and

$$\parallel D{f}_o(x,y,z)\parallel \le \frac{1}{2}[\phi (x,y,z)+\phi (-x,-y,-z)]$$

(3.8)

for all $x,y,z\in X$. Putting $y=x$, $z=-x$ in (3.8) and dividing by 3 yield

$$\parallel 3f_o\left(\frac{x}{3}\right)-f_o(x)\parallel \le \frac{1}{6}[\phi (x,x,-x)+\phi (-x,-x,x)]$$

(3.9)

for all $x\in X$.

Case 3. Assume that φ satisfies the condition () (and hence implies ()). Replacing x by 3x in (3.9) and dividing by 3, we have

$$\parallel f_o(x)-\frac{f_o(3x)}{3}\parallel \le \frac{1}{6\cdot 3}[\phi (3x,3x,-3x)+\phi (-3x,-3x,3x)]$$

(3.10)

for all $x\in X$. Making use of an induction argument in (3.10) implies

$$\parallel f_o(x)-\frac{f_o(3^{n}x)}{3^n}\parallel \le \frac{1}{6}\sum _{k=1}^n\frac{1}{3^k}[\phi (3^{k}x,3^{k}x,-3^{k}x)+\phi (-3^{k}x,-3^{k}x,3^{k}x)]$$

(3.11)

for all $n\in \mathbb{N}$ and $x\in X$. From (3.11) we can show that the sequence $\{f(3^{n}x)/3^n\}$ is a Cauchy sequence for all $x\in X$. Define a mapping $A:X\to Y$ by

$$A(x):=\underset{n\to \mathrm{\infty }}{lim}\frac{f_o(3^{n}x)}{3^n}=\underset{n\to \mathrm{\infty }}{lim}\frac{f(3^{n}x)-f(-3^{n}x)}{2\cdot 3^n}$$

for all $x\in X$. By the oddness of f, we see that $A(-x)=-A(x)$ for all $x\in X$. Also, from the condition () and (3.9), we verify that A satisfies

$$9A\left(\frac{x+y+z}{3}\right)+4[A\left(\frac{x-y}{2}\right)+A\left(\frac{y-z}{2}\right)+A\left(\frac{z-x}{2}\right)]=3[A(x)+A(y)+A(z)]$$

for all $x,y,z\in X$. According to Lemma 2.2, the mapping A satisfies (1.4). Letting $n\to \mathrm{\infty }$ in (3.11), we have

$$\parallel f_o(x)-A(x)\parallel \le \frac{1}{6}[{\mathrm{\Psi }}_1(x,x,-x)+{\mathrm{\Psi }}_1(-x,-x,x)]$$

for all $x\in X$. Using a similar argument to that of Case 1, we can easily see the uniqueness of A.

Case 4. If φ satisfies the condition (), the proof is analogous to that of Case 3. An induction argument on (3.9) implies

$$\parallel f_o(x)-3^n{f}_o\left(\frac{x}{3^n}\right)\parallel \le \sum _{k=0}^{n-1}\frac{3^k}{6}[\phi (\frac{x}{3^k},\frac{x}{3^k},-\frac{x}{3^k})+\phi (-\frac{x}{3^k},-\frac{x}{3^k},\frac{x}{3^k})]$$

(3.12)

for all $n\in \mathbb{N}$ and $x\in X$. It follows from (3.12) that $\{3^{n}f(3^{-n}x)\}$ is a Cauchy sequence for all $x\in X$. Thus we can define a mapping $A:X\to Y$ by

$$A(x):=\underset{n\to \mathrm{\infty }}{lim}{3}^n{f}_o\left(\frac{x}{3^n}\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{3^n}{2}[f\left(\frac{x}{3^n}\right)-f(-\frac{x}{3^n})]$$

for all $x\in X$. Note that $A(-x)=-A(x)$ for all $x\in X$. Also, from the condition () and (3.8), we verify that A satisfies

$$9A\left(\frac{x+y+z}{3}\right)+4[A\left(\frac{x-y}{2}\right)+A\left(\frac{y-z}{2}\right)+A\left(\frac{z-x}{2}\right)]=3[A(x)+A(y)+A(z)]$$

for all $x,y,z\in X$. According to Lemma 2.2, the mapping A satisfies (1.4). Taking the limit as $n\to \mathrm{\infty }$ in (3.12), we have

$$\parallel f_o(x)-A(x)\parallel \le \frac{1}{6}[{\mathrm{\Psi }}_2(x,x,-x)+{\mathrm{\Psi }}_2(-x,-x,x)]$$

for all $x\in X$. Similarly, we can show the uniqueness of A. □

From the theorem above, we have the following corollary immediately.

Corollary 3.2 Let $p\ne 1$, $p\ne 2$ and $ϵ\ge 0$ be real numbers. Suppose that a mapping $f:X\to Y$ satisfies

$$\parallel Df(x,y,z)\parallel \le ϵ({\parallel x\parallel }^p+{\parallel y\parallel }^p+{\parallel z\parallel }^p)$$

for all $x,y,z\in X$ ($x,y,z\in X\mathrm{\setminus }\{0\}$ if $p<0$). Then for each three cases $p<1$, $1<p<2$ and $p>2$, there exist a unique quadratic mapping $Q:X\to Y$ satisfying (1.3) and an additive mapping $A:X\to Y$ satisfying (1.4) such that

$$\begin{array}{c}\parallel f(x)-Q(x)-A(x)+\frac{1}{2}f(0)\parallel \le 3^pϵ(\frac{3}{|9-3^p|}+\frac{1}{|3-3^p|}){\parallel x\parallel }^p,\hfill \\ \parallel \frac{f(x)+f(-x)}{2}-Q(x)+\frac{1}{2}f(0)\parallel \le \frac{3^{p+1}ϵ}{|9-3^p|}{\parallel x\parallel }^p,\hfill \end{array}$$

and

$$\parallel \frac{f(x)-f(-x)}{2}-A(x)\parallel \le \frac{3^pϵ}{|3-3^p|}{\parallel x\parallel }^p$$

for all $x\in X$ ($x\in X\mathrm{\setminus }\{0\}$ if $p<0$), where $f(0)=0$ if $p>1$.

Proof Let $\phi (x,y,z):=ϵ({\parallel x\parallel }^p+{\parallel y\parallel }^p+{\parallel z\parallel }^p)$ for all $x,y,z\in X$. Then $\phi (x,x-x)=\phi (-x,-x,x)=3ϵ{\parallel x\parallel }^p$ for all $x\in X$ ($x\in X\mathrm{\setminus }\{0\}$ if $p<0$). If $p<2$, the mapping φ satisfies (). Thus, we figure out

$$\frac{1}{2}[{\mathrm{\Phi }}_1(x,x,-x)+{\mathrm{\Phi }}_1(-x,-x,x)]=\sum _{k=1}^{\mathrm{\infty }}3ϵ{\parallel x\parallel }^p\cdot 3^{(p-2)k}=ϵ{\parallel x\parallel }^p\frac{3^{p+1}}{9-3^p}$$

for all $x\in X$ ($x\in X\mathrm{\setminus }\{0\}$ if $p<0$). If $p>2$, the mapping φ satisfies (ℬ). Thus, we have

$$\frac{1}{2}[{\mathrm{\Phi }}_2(x,x,-x)+{\mathrm{\Phi }}_2(-x,-x,x)]=\sum _{k=0}^{\mathrm{\infty }}3ϵ{\parallel x\parallel }^p\cdot 3^{(2-p)k}=ϵ{\parallel x\parallel }^p\frac{3^{p+1}}{3^p-9}$$

for all $x\in X$. If $p<1$, the mapping φ satisfies (). Thus, we get

$$\frac{1}{2}[{\mathrm{\Psi }}_1(x,x,-x)+{\mathrm{\Psi }}_1(-x,-x,x)]=\sum _{k=1}^{\mathrm{\infty }}ϵ{\parallel x\parallel }^p\cdot 3^{(p-1)k}=ϵ{\parallel x\parallel }^p\frac{3^p}{3-3^p}$$

for all $x\in X$ ($x\in X\mathrm{\setminus }\{0\}$ if $p<0$). If $p>1$, the mapping φ satisfies (). Thus, we obtain

$$\frac{1}{2}[{\mathrm{\Psi }}_2(x,x,-x)+{\mathrm{\Psi }}_2(-x,-x,x)]=\sum _{k=0}^{\mathrm{\infty }}ϵ{\parallel x\parallel }^p\cdot 3^{(1-p)k}=ϵ{\parallel x\parallel }^p\frac{3^p}{3^p-3}$$

for all $x\in X$. Therefore, we have

$$\parallel f(x)-Q(x)-A(x)+\frac{1}{2}f(0)\parallel \le \{\begin{array}{cc}{3}^pϵ{\parallel x\parallel }^p(\frac{3}{3^p-9}+\frac{1}{3^p-3})\hfill & \text{if }p>2,\hfill \\ 3^pϵ{\parallel x\parallel }^p(\frac{3}{9-3^p}+\frac{1}{3^p-3})\hfill & \text{if }1<p<2,\hfill \\ 3^pϵ{\parallel x\parallel }^p(\frac{3}{9-3^p}+\frac{1}{3-3^p})\hfill & \text{if }p<1\hfill \end{array}$$

for all $x\in X$ ($x\in X\mathrm{\setminus }\{0\}$ if $p<0$). □

Corollary 3.3 Let $ϵ\ge 0$ be a real number. Suppose that a mapping $f:X\to Y$ satisfies

$$\parallel Df(x,y,z)\parallel \le ϵ$$

for all $x,y,z\in X$. Then there exist a unique quadratic mapping $Q:X\to Y$ satisfying (1.3) and an additive mapping $A:X\to Y$ satisfying (1.4) such that

$$\begin{array}{c}\parallel f(x)-Q(x)-A(x)+\frac{1}{2}f(0)\parallel \le \frac{7}{24}ϵ,\hfill \\ \parallel \frac{f(x)+f(-x)}{2}-Q(x)+\frac{1}{2}f(0)\parallel \le \frac{1}{8}ϵ,\hfill \end{array}$$

and

$$\parallel \frac{f(x)-f(-x)}{2}-A(x)\parallel \le \frac{1}{6}ϵ$$

for all $x\in X$.