Open Access

Stability of a Jensen type quadratic-additive functional equation under the approximately conditions

Advances in Difference Equations20152015:61

https://doi.org/10.1186/s13662-015-0392-z

Received: 26 August 2014

Accepted: 28 January 2015

Published: 25 February 2015

Abstract

We prove the Hyers-Ulam-Rassias stability of the Jensen type quadratic and additive functional equation \(9f ( \frac{x+y+z}{3} ) + 4 [ f (\frac{x-y}{2} ) + f (\frac{y-z}{2} ) + f (\frac{z-x}{2} ) ] = 3 [ f(x)+f(y)+f(z) ]\) under the approximately conditions such as even, odd, quadratic, and additive in Banach spaces.

Keywords

quadratic functional equation additive functional equation mixed type functional equation stability

MSC

39B82 39B52

1 Introduction

The study of stability problems for functional equations has originally been raised by Ulam [1]: under what condition does there exist a homomorphism near an approximate homomorphism? In 1941, Hyers [2] had answered affirmatively the question of Ulam for Banach spaces. The theorem of Hyers was generalized by Aoki [3] for additive mappings and by Rassias [4] for linear mappings by considering an unbounded Cauchy difference. The work of Rassias [4] has had a lot of influence in the development of a generalization of the Hyers-Ulam stability concept. The terminology Hyers-Ulam-Rassias stability originates from these historical backgrounds and this terminology is also applied to the case of other functional equations (see [510]).

In particular, Kannappan [11] introduced the following functional equation:
$$\begin{aligned} f(x+y+z) + f(x) + f(y) + f(z) = f(x+y) + f(y+z) + f(z+x) \end{aligned}$$
(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 \(f(x)=B(x,x)+A(x)\). For this reason, we call (1.1) as the mixed type quadratic and additive functional equation. In addition, Jung [12] investigated 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 [13] solved the general solutions and proved the stability of the following functional equation, which is a generalization of (1.1):
$$\begin{aligned} f \Biggl( \sum_{i=1}^{n} x_{i} \Biggr) + (n-2) \sum_{i=1}^{n} f(x_{i}) = \sum_{1\leq i< j\leq n} f(x_{i} + x_{j}) \quad(n>2). \end{aligned}$$
Najati and Moghimi [14] introduced another type quadratic and additive functional equation
$$\begin{aligned} f(2x+y)+f(2x-y) = f(x+y)+f(x-y)+2f(2x)-2f(x) \end{aligned}$$
and investigated the stability of this equation in quasi-Banach spaces.
Recently, Lee et al. [15] introduced the following Jensen type quadratic and additive functional equation:
$$\begin{aligned} 9f \biggl( \frac{x+y+z}{3} \biggr) + 4 \biggl[ f \biggl( \frac{x-y}{2} \biggr) + f \biggl(\frac{y-z}{2} \biggr) + f \biggl( \frac{z-x}{2} \biggr) \biggr] = 3 \bigl[ f(x)+f(y)+f(z) \bigr] \end{aligned}$$
(1.2)
and proved the stability of this equation. Also they established the general solutions of (1.2) as follows:
Let X and Y be real vector spaces. A mapping \(f:X\to Y\) satisfies (1.2) if and only if there exists a quadratic mapping \(Q:X\to Y\) satisfying
$$\begin{aligned} Q(x+y) + Q(x-y) = 2Q(x) + 2Q(y) \end{aligned}$$
(1.3)
and an additive mapping \(A:X\to Y\) satisfying
$$\begin{aligned} A(x+y) = A(x) + A(y) \end{aligned}$$
(1.4)
such that
$$\begin{aligned} f(x) = Q(x) + A(x) \end{aligned}$$
for all \(x\in X\).

We refer to [1626] for stability results of the mixed type and Jensen type functional equations. Also we refer to [2730] for more stability results of another functional equations.

In this paper, we prove the stability of (1.2) in addition to a suitable condition of the function. In Section 2, we first induce two lemmas which plays a crucial role in the proof of stability theorems. Based on the two lemmas, we evaluate the stability of (1.2) under the approximately even condition and the approximately odd condition in Section 3 and Section 4, respectively. Using similar approaches, we prove the stability of (1.2) under the approximately quadratic condition and the approximately additive condition in Section 5 and Section 6, respectively.

2 Preliminaries

Throughout this paper, we assume that X is a real vector space and Y is a Banach space. For simplicity, given a mapping \(f:X\to Y\), we use the abbreviation
$$\begin{aligned} Df(x,y,z) :={}& 9f \biggl( \frac{x+y+z}{3} \biggr) + 4 \biggl[ f \biggl( \frac{x-y}{2} \biggr) + f \biggl(\frac{y-z}{2} \biggr) + f \biggl( \frac{z-x}{2} \biggr) \biggr]\\ &{} - 3 \bigl[ f(x)+f(y)+f(z) \bigr] \end{aligned}$$
for all \(x,y,z\in X\). Here we need the following lemmas.

Lemma 2.1

Let \(\varphi: X\times X\times X \to[0,\infty)\) be a given mapping. Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\| Df(x,y,z) \bigr\| \leq\varphi(x,y,z) \end{aligned}$$
(2.1)
for all \(x,y,z\in X\). We then have
$$\begin{aligned} & \biggl\Vert f(x) - \frac{3^{n} + 1}{2\cdot9^{n}} f \bigl(3^{n} x\bigr) + \frac {3^{n}-1}{2 \cdot9^{n}} f\bigl(-3^{n} x\bigr) + \frac{9^{n}-1}{2\cdot9^{n}}f(0) \biggr\Vert \\ &\quad \leq \sum_{k=1}^{n} \biggl[ \frac{3^{k-1} + 1}{2\cdot9^{k}} \varphi\bigl(3^{k} x, 3^{k} x, -3^{k} x\bigr) + \frac{3^{k-1}-1}{2 \cdot9^{k}} \varphi\bigl(-3^{k} x, -3^{k} x, 3^{k} x\bigr) \biggr] \end{aligned}$$
(2.2)
for all \(x\in X\) and \(n\in\mathbb{N}\).

Proof

Putting \(y=x\), \(z=-x\) in (2.1) yields
$$\begin{aligned} \biggl\Vert 9 g \biggl( \frac{x}{3} \biggr) - 2g(x) + g(-x) \biggr\Vert \leq \varphi(x,x,-x) \end{aligned}$$
(2.3)
for all \(x\in X\), where \(g(x):=f(x)+\frac{1}{2}f(0)\). Replacing x by 3x in (2.3) and dividing by 9 gives
$$\begin{aligned} \biggl\Vert g(x) - \frac{2}{9}g(3x) + \frac{1}{9}g(-3x) \biggr\Vert \leq\frac{1}{9}\varphi(3x,3x,-3x) \end{aligned}$$
(2.4)
for all \(x\in X\). We use mathematical induction on n to prove lemma. Note that (2.4) proves the validity of inequality (2.2) for the case \(n=1\). Assume that inequality (2.2) holds for some \(n\in\mathbb{N}\). Using (2.4) we have the following relation:
$$\begin{aligned} & \biggl\Vert g(x) - \frac{3^{n+1} + 1}{2\cdot9^{n+1}} g\bigl(3^{n+1} x\bigr) + \frac{3^{n+1}-1}{2 \cdot9^{n+1}} g\bigl(-3^{n+1} x\bigr) \biggr\Vert \\ &\quad \leq\biggl\Vert g(x) - \frac{3^{n} + 1}{2\cdot9^{n}} g\bigl(3^{n} x\bigr) + \frac{3^{n}-1}{2 \cdot9^{n}} g\bigl(-3^{n} x\bigr) \biggr\Vert \\ &\qquad{} + \frac{3^{n}+1}{2\cdot9^{n}} \biggl\Vert g\bigl(3^{n} x\bigr) - \frac{2}{9} g\bigl(3^{n+1} x\bigr) + \frac{1}{9} g \bigl(-3^{n+1} x\bigr) \biggr\Vert \\ &\qquad{} + \frac{3^{n}-1}{2\cdot9^{n}} \biggl\Vert - g\bigl(-3^{n} x\bigr) + \frac{2}{9} g\bigl(-3^{n+1} x\bigr) - \frac{1}{9} g \bigl(3^{n+1} x\bigr) \biggr\Vert \\ &\quad \leq \sum_{k=1}^{n+1} \biggl( \frac{3^{k-1} + 1}{2\cdot9^{k}} \varphi\bigl(3^{k} x, 3^{k} x, -3^{k} x\bigr) + \frac{3^{k-1}-1}{2 \cdot9^{k}} \varphi\bigl(-3^{k} x, -3^{k} x, 3^{k} x\bigr) \biggr) \end{aligned}$$
for all \(x\in X\). This proves the validity of inequality (2.2) for the case \(n+1\). □

Lemma 2.2

Let \(\varphi:X\times X\times X \to[0,\infty)\) be a given mapping. Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\| Df(x,y,z) \bigr\| \leq\varphi(x,y,z) \end{aligned}$$
(2.5)
for all \(x,y,z\in X\). We then have
$$\begin{aligned} & \biggl\Vert f(x) - \frac{9^{n} + 3^{n}}{2} f \biggl(\frac{x}{3^{n}} \biggr) - \frac{9^{n} - 3^{n}}{2} f \biggl(- \frac{x}{3^{n}} \biggr) - \frac {9^{n}-1}{2}f(0) \biggr\Vert \\ &\quad \leq \sum_{k=0}^{n-1} \biggl[ \frac{3\cdot9^{k} + 3^{k}}{6} \varphi \biggl(\frac {x}{3^{k}}, \frac{x}{3^{k}}, - \frac{x}{3^{k}} \biggr) + \frac{3\cdot9^{k} - 3^{k}}{6} \varphi \biggl(- \frac{x}{3^{k}}, -\frac {x}{3^{k}}, \frac{x}{3^{k}} \biggr) \biggr] \end{aligned}$$
(2.6)
for all \(x\in X\) and \(n\in\mathbb{N}\).

Proof

Putting \(y=x\), \(z=-x\) in (2.5) gives
$$\begin{aligned} \biggl\Vert 9 g \biggl( \frac{x}{3} \biggr) - 2g(x) + g(-x) \biggr\Vert \leq \varphi(x,x,-x) \end{aligned}$$
(2.7)
for all \(x\in X\), where \(g(x):=f(x)+\frac{1}{2}f(0)\). Interchanging x with −x in (2.7) we have
$$\begin{aligned} \biggl\Vert 9 g \biggl( -\frac{x}{3} \biggr) - 2g(-x) + g(x) \biggr\Vert \leq \varphi(-x,-x,x) \end{aligned}$$
(2.8)
for all \(x\in X\). It follows from (2.7) and (2.8) that
$$\begin{aligned} \biggl\Vert g(x) - 6 g \biggl( \frac{x}{3} \biggr) - 3 g \biggl( -\frac{x}{3} \biggr) \biggr\Vert \leq\frac{2}{3} \varphi(x,x,-x) + \frac{1}{3}\varphi(-x,-x,x) \end{aligned}$$
(2.9)
which proves the validity of inequality (2.6) for the case \(n=1\). Applying (2.9) we obtain
$$\begin{aligned} & \biggl\Vert g(x) - \frac{9^{n+1} + 3^{n+1}}{2} g \biggl(\frac {x}{3^{n+1}} \biggr) - \frac{9^{n+1} - 3^{n+1}}{2} g \biggl(-\frac{x}{3^{n+1}} \biggr) \biggr\Vert \\ &\quad \leq \biggl\Vert g(x) - \frac{9^{n} + 3^{n}}{2} g \biggl(\frac{x}{3^{n}} \biggr) - \frac{9^{n} - 3^{n}}{2} g \biggl(-\frac{x}{3^{n}} \biggr) \biggr\Vert \\ &\qquad{} + \frac{9^{n} + 3^{n}}{2} \biggl\Vert g \biggl(\frac {x}{3^{n}} \biggr) - 6 g \biggl( \frac{x}{3^{n+1}} \biggr) - 3 g \biggl( -\frac{x}{3^{n+1}} \biggr) \biggr\Vert \\ &\qquad{} + \frac{9^{n} - 3^{n}}{2} \biggl\Vert g \biggl(-\frac {x}{3^{n}} \biggr) - 6 g \biggl(-\frac{x}{3^{n+1}} \biggr) - 3 g \biggl( \frac{x}{3^{n+1}} \biggr) \biggr\Vert \\ & \quad\leq\sum_{k=0}^{n} \biggl[ \frac{3\cdot9^{k} + 3^{k}}{6} \varphi \biggl(\frac{x}{3^{k}}, \frac{x}{3^{k}}, - \frac{x}{3^{k}} \biggr) + \frac{3\cdot9^{k} - 3^{k}}{6} \varphi \biggl(- \frac{x}{3^{k}}, -\frac {x}{3^{k}}, \frac{x}{3^{k}} \biggr) \biggr] \end{aligned}$$
for all \(x\in X\). This proves the validity of inequality (2.6) for the case \(n+1\). □
In order to prove the stability of (1.2), we suppose that \(\varphi:X\times X\times X\to[0,\infty)\) is a mapping satisfying one of the conditions (\(\mathcal{A}_{1}\)) and (\(\mathcal{B}_{1}\)),
$$\begin{aligned} &(\mathcal{A}_{1})\quad \sum_{k=1}^{\infty}\frac{1}{3^{k}} \varphi \bigl(3^{k} x, 3^{k} y, 3^{k} z\bigr)< \infty, \\ &(\mathcal{B}_{1}) \quad\sum_{k=0}^{\infty}9^{k} \varphi \biggl( \frac{x}{3^{k}}, \frac{y}{3^{k}}, \frac{z}{3^{k}} \biggr) < \infty \end{aligned}$$
for all \(x,y,z\in X\). Also we assume that \(\psi:X\to[0,\infty)\) is a mapping satisfying one of the conditions (\(\mathcal{A}_{2}\)) and (\(\mathcal{B}_{2}\)),
$$\begin{aligned} &(\mathcal{A}_{2})\quad \lim_{n\to\infty} \frac{\psi(3^{n} x)}{3^{n}} = 0, \\ &(\mathcal{B}_{2})\quad \lim_{n\to\infty} 9^{n} \psi \biggl( \frac{x}{3^{n}} \biggr) = 0 \end{aligned}$$
for all \(x\in X\).

3 Approximately even case

Now we are going to state and prove the stability of (1.2) under the approximately even condition.

Theorem 3.1

Let \(\varphi:X\times X\times X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{1}\)) and \(\psi:X\to[0,\infty)\) satisfies the condition (\(\mathcal{A}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\| Df(x,y,z) \bigr\| \leq\varphi(x,y,z) \end{aligned}$$
(3.1)
for all \(x,y,z\in X\) and
$$\begin{aligned} \bigl\| f(x) - f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
(3.2)
for all \(x\in X\). Then there exists a unique quadratic mapping \(Q:X\to Y\) satisfying (1.3) such that
$$\begin{aligned} &\biggl\Vert f(x) - Q(x) +\frac{1}{2}f(0) \biggr\Vert \\ &\quad\leq \sum_{k=1}^{\infty} \biggl[ \frac{3^{k-1} + 1}{2\cdot9^{k}} \varphi \bigl(3^{k} x, 3^{k} x, -3^{k} x\bigr) + \frac{3^{k-1}-1}{2 \cdot9^{k}} \varphi\bigl(-3^{k} x, -3^{k} x, 3^{k} x\bigr) \biggr] \end{aligned}$$
(3.3)
for all \(x\in X\).

Proof

It follows from Lemma 2.1 and (3.2) that we have
$$\begin{aligned} & \biggl\Vert f(x) - \frac{f(3^{n}x)}{9^{n}} + \frac{9^{n}-1}{2\cdot9^{n}}f(0) \biggr\Vert \\ &\quad \leq \sum_{k=1}^{n} \biggl[ \frac{3^{k-1} + 1}{2\cdot9^{k}} \varphi\bigl(3^{k} x, 3^{k} x, -3^{k} x\bigr) + \frac{3^{k-1}-1}{2 \cdot9^{k}} \varphi\bigl(-3^{k} x, -3^{k} x, 3^{k} x\bigr) \biggr] \\ &\qquad{}+ \frac{3^{n}-1}{2\cdot9^{n}} \psi \bigl(3^{n} x\bigr) \end{aligned}$$
(3.4)
for all \(x\in X\) and \(n\in\mathbb{N}\). By virtue of (3.4), for \(n, m\in\mathbb{N}\) with \(n>m\), we obtain
$$\begin{aligned} & \biggl\Vert \frac{f(3^{m} x)}{9^{m}} - \frac{f(3^{n} x)}{9^{n}} \biggr\Vert \\ &\quad = \frac{1}{9^{m}} \biggl\Vert f\bigl(3^{m} x\bigr) - \frac{f(3^{n-m} \cdot3^{m} x)}{9^{n-m}} \biggr\Vert \\ &\quad \leq \sum_{k=1}^{n-m} \biggl[ \frac{3^{k-1} + 1}{2\cdot9^{k+m}} \varphi \bigl(3^{k+m} x, 3^{k+m} x, -3^{k+m}x\bigr) + \frac{3^{k-1} - 1}{2\cdot9^{k+m}} \varphi\bigl(-3^{k+m} x, -3^{k+m} x, 3^{k+m}x\bigr) \biggr] \\ &\qquad{} + \frac{3^{n-m}-1}{2\cdot9^{n}}\psi\bigl(3^{n} x\bigr) + \frac{9^{n-m}-1}{2\cdot9^{n}} \bigl\| f(0) \bigr\| \end{aligned}$$
(3.5)
for all \(x\in X\). Since the right-hand side of inequality (3.5) tends to 0 as \(m\to\infty\), the sequence \(\{ 9^{-n} f(3^{n} x) \}\) is a Cauchy sequence. Completeness of Y allows us to assume that there exists a mapping Q so that
$$\begin{aligned} Q(x) := \lim_{n\to\infty} 9^{-n} f\bigl(3^{n} x \bigr) \end{aligned}$$
for all \(x\in X\). Replacing x, y, z by \(3^{n} x\), \(3^{n} y\), \(3^{n} z\) in (3.1) and dividing both sides by \(9^{n}\) yields
$$\begin{aligned} \frac{1}{9^{n}} \bigl\Vert Df\bigl(3^{n} x, 3^{n} y, 3^{n} z\bigr) \bigr\Vert \leq\frac{1}{9^{n}} \varphi \bigl(3^{n} x, 3^{n} y, 3^{n} z\bigr) \end{aligned}$$
for all \(x,y,z\in X\). Taking the limit in the above inequality we have
$$\begin{aligned} &9Q \biggl( \frac{x+y+z}{3} \biggr) + 4 \biggl[ Q \biggl( \frac{x-y}{2} \biggr) + Q \biggl(\frac{y-z}{2} \biggr) + Q \biggl( \frac{z-x}{2} \biggr) \biggr] \\ &\quad= 3 \bigl[ Q(x)+Q(y)+Q(z) \bigr] \end{aligned}$$
(3.6)
for all \(x,y,z\in X\). Similarly, we get \(Q(-x)=Q(x)\) for all \(x\in X\) by (3.2). Putting \(y=x\), \(z=-x\) in (3.6) and using the evenness of Q we obtain \(9Q(\frac{x}{3})=Q(x)\) for all \(x\in X\). Setting \(y=0\), \(z=-x\) in (3.6) gives \(4Q(\frac{x}{2})=Q(x)\) for all \(x\in X\). Thus, we can rewrite (3.6) as
$$\begin{aligned} Q(x+y+z)+Q(x-y)+Q(y-z)+Q(z-x) = 3 \bigl[ Q(x)+Q(y)+Q(z) \bigr] \end{aligned}$$
(3.7)
for all \(x,y,z\in X\). Putting \(z=0\) in (3.7) we see that Q satisfies the quadratic functional equation (1.3). Letting \(n\to\infty\) in (3.4) we finally obtain the result (3.3).
In order to prove the uniqueness of Q we assume that \(Q ':X\to Y\) be another quadratic mapping satisfying (3.3). From the quadratic property of Q and \(Q '\) we calculate
$$\begin{aligned} & \bigl\| Q(x) - Q '(x) \bigr\| \\ &\quad = \frac{1}{9^{n}} \bigl\| Q\bigl(3^{n} x\bigr) - Q ' \bigl(3^{n}x\bigr) \bigr\| \\ &\quad \leq\frac{1}{9^{n}} \biggl\Vert Q\bigl(3^{n} x\bigr) - f \bigl(3^{n} x\bigr) - \frac {1}{2}f(0) \biggr\Vert + \frac{1}{9^{n}} \biggl\Vert f\bigl(3^{n} x\bigr) - Q '\bigl(3^{n} x\bigr) + \frac{1}{2}f(0) \biggr\Vert \\ &\quad \leq\frac{2}{9^{n}} \sum_{k=1}^{\infty} \biggl[ \frac{3^{k-1} + 1}{2\cdot9^{k}} \varphi \bigl(3^{k+n} x, 3^{k+n} x, -3^{k+n} x\bigr) + \frac{3^{k-1}-1}{2 \cdot9^{k}} \varphi\bigl(-3^{k+n} x, -3^{k+n} x, 3^{k+n} x\bigr) \biggr] \end{aligned}$$
for all \(x\in X\) and \(n\in\mathbb{N}\). Taking \(n\to\infty\) in the preceding inequality we immediately find the uniqueness of Q. □

As a consequence of the above theorem we have the following corollaries immediately.

Corollary 3.2

Let \(\epsilon\geq0\), \(p<1\), and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\Vert Df(x,y,z) \bigr\Vert \leq\epsilon\bigl(\|x\|^{p}+\|y \|^{p}+\|z\|^{p}\bigr) \end{aligned}$$
for all \(x,y,z\in X\) (\(x,y,z\in X\backslash\{0\}\) if \(p<0\)) and
$$\begin{aligned} \bigl\| f(x) - f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
for all \(x\in X\). Then there exists a unique quadratic mapping \(Q:X\to Y\) satisfying (1.3) such that
$$\begin{aligned} \biggl\Vert f(x) - Q(x) +\frac{1}{2}f(0) \biggr\Vert \leq \frac{3^{p}}{3-3^{p}} \epsilon\|x\|^{p} \end{aligned}$$
for all \(x\in X\) (\(x\in X\backslash\{0\}\) if \(p<0\)).

Corollary 3.3

Let \(\epsilon\geq0\) and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\Vert Df(x,y,z) \bigr\Vert \leq\epsilon \end{aligned}$$
for all \(x,y,z\in X\) and
$$\begin{aligned} \bigl\| f(x) - f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
for all \(x\in X\). Then there exists a unique quadratic mapping \(Q:X\to Y\) satisfying (1.3) such that
$$\begin{aligned} \biggl\Vert f(x) - Q(x) + \frac{1}{2}f(0) \biggr\Vert \leq \frac{1}{6}\epsilon \end{aligned}$$
for all \(x\in X\).

We have the following result, which is analogous to Theorem 3.1 for the functional equation (1.2).

Theorem 3.4

Let \(\varphi:X\times X\times X\to[0,\infty)\) satisfy the condition (\(\mathcal{B}_{1}\)) and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{B}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\| Df(x,y,z) \bigr\| \leq\varphi(x,y,z) \end{aligned}$$
(3.8)
for all \(x,y,z\in X\) and
$$\begin{aligned} \bigl\| f(x) - f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
(3.9)
for all \(x\in X\). Then there exists a unique quadratic mapping \(Q:X\to Y\) satisfying (1.3) such that
$$ \begin{aligned}[b] &\bigl\Vert f(x) - Q(x) \bigr\Vert \\ &\quad\leq \sum _{k=0}^{\infty} \biggl[ \frac{3\cdot9^{k} + 3^{k}}{6} \varphi \biggl( \frac{x}{3^{k}}, \frac{x}{3^{k}}, -\frac{x}{3^{k}} \biggr) + \frac{3\cdot9^{k} - 3^{k}}{6} \varphi \biggl(-\frac{x}{3^{k}}, -\frac {x}{3^{k}}, \frac{x}{3^{k}} \biggr) \biggr] \end{aligned} $$
(3.10)
for all \(x\in X\).

Proof

Putting \(x=y=z=0\) in (3.8) yields \(f(0)=0\). According to Lemma 2.2 and (3.9) we have
$$\begin{aligned} & \biggl\Vert f(x) - 9^{n} f \biggl(\frac{x}{3^{n}} \biggr) \biggr\Vert \\ &\quad \leq \sum_{k=0}^{n-1} \biggl[ \frac{3\cdot9^{k} + 3^{k}}{6} \varphi \biggl(\frac {x}{3^{k}}, \frac{x}{3^{k}}, - \frac{x}{3^{k}} \biggr) + \frac{3\cdot9^{k} - 3^{k}}{6} \varphi \biggl(- \frac{x}{3^{k}}, -\frac {x}{3^{k}}, \frac{x}{3^{k}} \biggr) \biggr] \\ &\qquad{} + \frac{9^{n}-3^{n}}{2} \psi \biggl(\frac{x}{3^{n}} \biggr) \end{aligned}$$
(3.11)
for all \(x\in X\) and \(n\in\mathbb{N}\). From (3.11) we verify
$$\begin{aligned} & \biggl\Vert 9^{m} f \biggl( \frac{x}{3^{m}} \biggr) - 9^{n} f \biggl( \frac {x}{3^{n}} \biggr) \biggr\Vert \\ &\quad = 9^{m} \biggl\Vert f \biggl(\frac{x}{3^{m}} \biggr) - 9^{n-m} f \biggl(\frac {x}{3^{n-m}\cdot3^{m}} \biggr) \biggr\Vert \\ &\quad \leq \sum_{k=0}^{n-m-1} \biggl[ \frac{3\cdot9^{k+m} + 3^{k+2m}}{6} \varphi \biggl(\frac {x}{3^{k+m}}, \frac{x}{3^{k+m}}, - \frac{x}{3^{k+m}} \biggr) \\ &\qquad{} + \frac{3\cdot9^{k+m} - 3^{k+2m}}{6} \varphi \biggl(-\frac{x}{3^{k+m}}, - \frac{x}{3^{k+m}}, \frac{x}{3^{k+m}} \biggr) \biggr] + \frac{9^{n}-3^{n+m}}{2} \psi \biggl(\frac{x}{3^{n}} \biggr) \end{aligned}$$
(3.12)
for all \(x\in X\) and \(m, n\in\mathbb{N}\) with \(n> m\). Since the right-hand side of inequality (3.12) tends to 0 as \(m\to\infty\), we may define a mapping
$$\begin{aligned} Q(x) := \lim_{n\to\infty} 9^{n} f \biggl(\frac{x}{3^{n}} \biggr) \end{aligned}$$
for all \(x\in X\). Replacing x, y, z by \(\frac{x}{3^{n}}\), \(\frac{y}{3^{n}}\), \(\frac{z}{3^{n}}\) in (3.8) and multiplying both sides by \(9^{n}\), and after taking the limit in the resulting inequality, we see that Q satisfies (1.2). Using a similar method to the proof of Theorem 3.1 we see that Q is the unique quadratic mapping satisfying (1.3). Now letting \(n\to\infty\) in (3.11) we arrive at the desired result (3.10). □

Corollary 3.5

Let \(\epsilon\geq0\), \(p>2\), and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{B}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\Vert Df(x,y,z) \bigr\Vert \leq\epsilon\bigl(\|x\|^{p}+\|y \|^{p}+\|z\|^{p}\bigr) \end{aligned}$$
for all \(x,y,z\in X\) and
$$\begin{aligned} \bigl\| f(x) - f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
for all \(x\in X\). Then there exists a unique quadratic mapping \(Q:X\to Y\) satisfying (1.3) such that
$$\begin{aligned} \bigl\Vert f(x) - Q(x) \bigr\Vert \leq \frac{3^{p+1}}{3^{p}-9} \epsilon\|x \|^{p} \end{aligned}$$
for all \(x\in X\).

4 Approximately odd case

In this section, we establish the stability of (1.2) under the approximately odd condition.

Theorem 4.1

Let \(\varphi:X\times X\times X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{1}\)) and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\| Df(x,y,z) \bigr\| \leq\varphi(x,y,z) \end{aligned}$$
(4.1)
for all \(x, y, z\in X\) and
$$\begin{aligned} \bigl\| f(x) + f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
(4.2)
for all \(x\in X\). Then there exists a unique additive mapping \(A:X\to Y\) satisfying (1.4) such that
$$\begin{aligned} &\biggl\Vert f(x) - A(x) + \frac{1}{2}f(0) \biggr\Vert \\ &\quad\leq \sum_{k=1}^{\infty} \biggl[ \frac{3^{k-1} + 1}{2\cdot9^{k}} \varphi \bigl(3^{k} x, 3^{k} x, -3^{k} x\bigr) + \frac{3^{k-1}-1}{2 \cdot9^{k}} \varphi\bigl(-3^{k} x, -3^{k} x, 3^{k} x\bigr) \biggr] \end{aligned}$$
(4.3)
for all \(x\in X\).

Proof

It follows from Lemma 2.1 and (4.2) that
$$\begin{aligned} & \biggl\Vert f(x) - \frac{f(3^{n}x)}{3^{n}} + \frac{9^{n}-1}{2\cdot9^{n}}f(0) \biggr\Vert \\ &\quad \leq \sum_{k=1}^{n} \biggl[ \frac{3^{k-1} + 1}{2\cdot9^{k}} \varphi\bigl(3^{k} x, 3^{k} x, -3^{k} x\bigr) + \frac{3^{k-1}-1}{2 \cdot9^{k}} \varphi\bigl(-3^{k} x, -3^{k} x, 3^{k} x\bigr) \biggr] \\ &\qquad{} + \frac{3^{n}-1}{2\cdot9^{n}} \psi\bigl(3^{n} x\bigr) \end{aligned}$$
(4.4)
for all \(x\in X\) and \(n\in\mathbb{N}\). From (4.4) we figure out
$$\begin{aligned} \biggl\Vert \frac{f(3^{m} x)}{3^{m}} - \frac{f(3^{n} x)}{3^{n}} \biggr\Vert ={}& \frac{1}{3^{m}} \biggl\Vert f \bigl(3^{m} x\bigr) - \frac{f(3^{n-m} \cdot3^{m} x)}{3^{n-m}} \biggr\Vert \\ \leq{}& \sum_{k=1}^{n-m} \biggl[ \frac{3^{k-1} + 1}{2\cdot3^{2k+m}} \varphi \bigl(3^{k+m} x, 3^{k+m} x, -3^{k+m}x\bigr) \\ &{} + \frac{3^{k-1} - 1}{2\cdot3^{2k+m}} \varphi \bigl(-3^{k+m} x, -3^{k+m} x, 3^{k+m}x\bigr) \biggr] \\ &{} +\frac{3^{n-m}-1}{2\cdot3^{2n-m}}\psi\bigl(3^{n} x\bigr) + \frac{9^{n-m}-1}{2\cdot3^{2n-m}} \bigl\| f(0) \bigr\| \end{aligned}$$
(4.5)
for all \(x\in X\) and \(m, n\in\mathbb{N}\) with \(n> m\). Taking the limit as \(m\to\infty\) in (4.5) we verify that the right-hand side of inequality (4.5) tends to 0. Thus, the mentioned sequence is convergent to the mapping A; that is,
$$\begin{aligned} A(x) := \lim_{n\to\infty} 3^{-n} f\bigl(3^{n} x \bigr) \end{aligned}$$
for all \(x\in X\). Replacing x, y, z by \(3^{n} x\), \(3^{n} y\), \(3^{n} z\) in (4.1) and dividing both sides by \(3^{n}\), and after taking the limit in the resulting inequality, we see that
$$\begin{aligned} &9A \biggl( \frac{x+y+z}{3} \biggr) + 4 \biggl[ A \biggl( \frac{x-y}{2} \biggr) + A \biggl(\frac{y-z}{2} \biggr) + A \biggl( \frac{z-x}{2} \biggr) \biggr] \\ &\quad= 3 \bigl[ A(x)+A(y)+A(z) \bigr] \end{aligned}$$
(4.6)
for all \(x,y,z\in X\). By virtue of (4.2) we have \(A(-x)=-A(x)\) for all \(x\in X\). Setting \(y=x\), \(z=-x\) in (4.6) yields \(3A(\frac{x}{3})=A(x)\) for all \(x\in X\). Similarly, putting \(y=0\), \(z=-x\) in (4.6) gives \(2A(\frac {x}{2})=A(x)\) for all \(x\in X\). Thus, we can rewrite (4.6) as
$$\begin{aligned} 3A(x+y+z) + 2 \bigl[ A(x-y)+A(y-z)+A(z-x) \bigr] = 3 \bigl[ A(x)+A(y)+A(z) \bigr] \end{aligned}$$
(4.7)
for all \(x,y,z\in X\). Replacing z by −x in (4.7) we have
$$\begin{aligned} A(x+y)+A(x-y)=2A(x) \end{aligned}$$
(4.8)
for all \(x,y\in X\). Replacing x, y by \(\frac{x+y}{2}\), \(\frac{x-y}{2}\) in (4.8) we see that A satisfies the Cauchy functional equation (1.4). Letting \(n\to\infty\) in (4.4) we finally obtain the result (4.3).
To show the uniqueness of A we assume that \(A':X\to Y\) be another quadratic mapping satisfying (4.3). Obviously, we have \(A(3^{n} x)=3^{n} A(x)\) and \(A'(3^{n} x)=3^{n} A'(x)\) for all \(x\in X\). According to the additive property of A and \(A'\) we figure out
$$\begin{aligned} & \bigl\| A(x) - A'(x) \bigr\| \\ &\quad = \frac{1}{3^{n}} \bigl\| A\bigl(3^{n} x\bigr) - A' \bigl(3^{n}x\bigr) \bigr\| \\ &\quad \leq\frac{1}{3^{n}} \biggl\Vert A\bigl(3^{n} x\bigr) - f \bigl(3^{n} x\bigr) - \frac {1}{2}f(0) \biggr\Vert + \frac{1}{3^{n}} \biggl\Vert f\bigl(3^{n} x\bigr) - A' \bigl(3^{n} x\bigr) + \frac{1}{2}f(0) \biggr\Vert \\ & \quad\leq \frac{2}{3^{n}} \sum_{k=1}^{\infty} \biggl[ \frac{3^{k-1} + 1}{2\cdot9^{k}} \varphi \bigl(3^{k+n} x, 3^{k+n} x, -3^{k+n} x\bigr) + \frac{3^{k-1}-1}{2 \cdot9^{k}} \varphi\bigl(-3^{k+n} x, -3^{k+n} x, 3^{k+n} x\bigr) \biggr] \end{aligned}$$
for all \(n\in\mathbb{N}\) and \(x\in X\), which means the uniqueness of A. □

From the theorem above we obtain the following corollaries immediately.

Corollary 4.2

Let \(\epsilon\geq0\), \(p<1\), and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\Vert Df(x,y,z) \bigr\Vert \leq\epsilon\bigl(\|x\|^{p}+\|y \|^{p}+\|z\|^{p}\bigr) \end{aligned}$$
for all \(x,y,z\in X\) (\(x,y,z\in X\backslash\{0\}\) if \(p<0\)) and
$$\begin{aligned} \bigl\| f(x) + f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
for all \(x\in X\). Then there exists a unique additive mapping \(A:X\to Y\) satisfying (1.4) such that
$$\begin{aligned} \biggl\Vert f(x) - A(x) + \frac{1}{2}f(0) \biggr\Vert \leq \frac{3^{p}}{3-3^{p}} \epsilon\|x\|^{p} \end{aligned}$$
for all \(x\in X\) (\(x\in X\backslash\{0\}\) if \(p<0\)).

Corollary 4.3

Let \(\epsilon\geq0\) and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\Vert Df(x,y,z) \bigr\Vert \leq\epsilon \end{aligned}$$
for all \(x,y,z\in X\) and
$$\begin{aligned} \bigl\| f(x) + f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
for all \(x\in X\). Then there exists a unique additive mapping \(A:X\to Y\) satisfying (1.4) such that
$$\begin{aligned} \biggl\Vert f(x) - A(x) + \frac{1}{2}f(0) \biggr\Vert \leq \frac{1}{6}\epsilon \end{aligned}$$
for all \(x\in X\).

We have the following result, which is analogous to Theorem 4.1.

Theorem 4.4

Let \(\varphi:X\times X\times X\to[0,\infty)\) satisfy the condition (\(\mathcal{B}_{1}\)) and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{B}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\| Df(x,y,z) \bigr\| \leq\varphi(x,y,z) \end{aligned}$$
(4.9)
for all \(x, y, z\in X\) and
$$\begin{aligned} \bigl\| f(x) + f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
(4.10)
for all \(x\in X\). Then there exists a unique additive mapping \(A:X\to Y\) satisfying (1.4) such that
$$ \begin{aligned}[b] &\bigl\Vert f(x) - A(x) \bigr\Vert \\ &\quad\leq \sum _{k=0}^{\infty} \biggl[ \frac{3\cdot9^{k} + 3^{k}}{6} \varphi \biggl( \frac{x}{3^{k}}, \frac{x}{3^{k}}, -\frac{x}{3^{k}} \biggr) + \frac{3\cdot9^{k} - 3^{k}}{6} \varphi \biggl(-\frac{x}{3^{k}}, -\frac {x}{3^{k}}, \frac{x}{3^{k}} \biggr) \biggr] \end{aligned} $$
(4.11)
for all \(x\in X\).

Proof

From Lemma 2.2 and the approximately odd condition (4.10) we have
$$\begin{aligned} & \biggl\Vert f(x) - 3^{n} f \biggl(\frac{x}{3^{n}} \biggr) \biggr\Vert \\ & \quad\leq \sum_{k=0}^{n-1} \biggl[ \frac{3\cdot9^{k} + 3^{k}}{6} \varphi \biggl(\frac {x}{3^{k}}, \frac{x}{3^{k}}, - \frac{x}{3^{k}} \biggr) + \frac{3\cdot9^{k} - 3^{k}}{6} \varphi \biggl(- \frac{x}{3^{k}}, -\frac {x}{3^{k}}, \frac{x}{3^{k}} \biggr) \biggr] \\ &\qquad{}+ \frac{9^{n}-3^{n}}{2} \psi \biggl(\frac{x}{3^{n}} \biggr) \end{aligned}$$
(4.12)
for all \(x\in X\) and \(n\in\mathbb{N}\). By (4.12) we obtain
$$\begin{aligned} & \biggl\Vert 3^{m} f \biggl( \frac{x}{3^{m}} \biggr) - 3^{n} f \biggl( \frac {x}{3^{n}} \biggr) \biggr\Vert \\ &\quad = 3^{m} \biggl\Vert f \biggl(\frac{x}{3^{m}} \biggr) - 3^{n-m} f \biggl(\frac {x}{3^{n-m}\cdot3^{m}} \biggr) \biggr\Vert \\ &\quad \leq \sum_{k=0}^{n-m-1} \biggl[ \frac{3\cdot3^{2k+m} + 3^{k+m}}{6} \varphi \biggl(\frac {x}{3^{k+m}}, \frac{x}{3^{k+m}}, - \frac{x}{3^{k+m}} \biggr) \\ &\qquad{} + \frac{3\cdot3^{2k+m} - 3^{k+m}}{6} \varphi \biggl(-\frac{x}{3^{k+m}}, - \frac{x}{3^{k+m}}, \frac{x}{3^{k+m}} \biggr) \biggr] + \frac{3^{2n-m}-3^{n}}{2} \psi \biggl(\frac{x}{3^{n}} \biggr) \end{aligned}$$
(4.13)
for all \(x\in X\) and \(m, n\in\mathbb{N}\) with \(n> m\). Since the right-hand side of inequality (4.13) tends to 0 as \(m\to\infty\), the sequence \(\{ 3^{n} f(3^{-n} x) \}\) is a Cauchy sequence. Thus, we can define a mapping
$$\begin{aligned} A(x) := \lim_{n\to\infty} 3^{n} f \biggl(\frac{x}{3^{n}} \biggr) \end{aligned}$$
for all \(x\in X\). Replacing x, y, z by \(\frac{x}{3^{n}}\), \(\frac{y}{3^{n}}\), \(\frac{z}{3^{n}}\) in (4.9) and multiplying both sides by \(3^{n}\), and after taking the limit in the resulting inequality, we see that A satisfies (1.2). Using a similar method to the proof of Theorem 4.1 we see that A is the unique additive mapping satisfying (1.4). Taking the limit as \(n\to\infty\) in (4.12) we finally obtain the result (4.11). □

Corollary 4.5

Let \(\epsilon\geq0\), \(p>2\), and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{B}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\Vert Df(x,y,z) \bigr\Vert \leq\epsilon\bigl(\|x\|^{p}+\|y \|^{p}+\|z\|^{p}\bigr) \end{aligned}$$
for all \(x,y,z\in X\) and
$$\begin{aligned} \bigl\| f(x) + f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
for all \(x\in X\). Then there exists a unique additive mapping \(A:X\to Y\) satisfying (1.4) such that
$$\begin{aligned} \bigl\Vert f(x) - A(x) \bigr\Vert \leq \frac{3^{p+1}}{3^{p}-9} \epsilon\|x \|^{p} \end{aligned}$$
for all \(x\in X\) (\(x\in X\backslash\{0\}\) if \(p<0\)).

5 Approximately quadratic case

We are going to prove the stability of (1.2) under the approximately quadratic condition.

Theorem 5.1

Let \(\varphi:X\times X\times X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{1}\)) and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\| Df(x,y,z) \bigr\| \leq\varphi(x,y,z) \end{aligned}$$
(5.1)
for all \(x,y,z\in X\) and
$$\begin{aligned} \bigl\| f(3x) - 9f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
(5.2)
for all \(x\in X\). Then there exists a unique quadratic mapping \(Q:X\to Y\) satisfying (1.3) such that
$$\begin{aligned} &\biggl\Vert f(x) - Q(x) + \frac{1}{2}f(0) \biggr\Vert \\ &\quad \leq \sum_{k=1}^{\infty} \biggl[ \frac{3^{k-1} + 1}{2\cdot9^{k}} \varphi \bigl(3^{k} x, 3^{k} x, -3^{k} x\bigr) + \frac{3^{k-1}-1}{2 \cdot9^{k}} \varphi\bigl(-3^{k} x, -3^{k} x, 3^{k} x\bigr) \biggr] \end{aligned}$$
(5.3)
for all \(x\in X\).

Proof

Setting \(y=x\), \(z=-x\) in (5.1) gives
$$\begin{aligned} \biggl\Vert 9f \biggl(\frac{x}{3} \biggr) -2f(x) + f(-x) + 4f(0) \biggr\Vert \leq\varphi(x, x, -x) \end{aligned}$$
(5.4)
for all \(x\in X\) and \(n\in\mathbb{N}\). Combining (5.2), (5.4) and Lemma 2.1 we have
$$\begin{aligned} & \biggl\Vert f(x) - \frac{f(3^{n}x)}{9^{n}} + \frac{9^{n} + 2\cdot3^{n} - 3}{2\cdot9^{n}}f(0) \biggr\Vert \\ &\quad \leq \sum_{k=1}^{n} \biggl[ \frac{3^{k-1} + 1}{2\cdot9^{k}} \varphi\bigl(3^{k} x, 3^{k} x, -3^{k} x\bigr) + \frac{3^{k-1}-1}{2 \cdot9^{k}} \varphi\bigl(-3^{k} x, -3^{k} x, 3^{k} x\bigr) \biggr] \\ &\qquad{} + \frac{3^{n}-1}{4\cdot9^{n}}\varphi\bigl(- 3^{n} x, - 3^{n} x, 3^{n} x\bigr) + \frac{3^{n}-1}{4\cdot9^{n}} \psi\bigl(3^{n-1} x\bigr) \end{aligned}$$
(5.5)
for all \(x\in X\) and \(n\in\mathbb{N}\). For \(n, m\in\mathbb{N}\) with \(n>m\), we verify by (5.5) that
$$\begin{aligned} & \biggl\Vert \frac{f(3^{m} x)}{9^{m}} - \frac{f(3^{n} x)}{9^{n}} \biggr\Vert \\ &\quad = \frac{1}{9^{m}} \biggl\Vert f\bigl(3^{m} x\bigr) - \frac{f(3^{n-m} \cdot3^{m} x)}{9^{n-m}} \biggr\Vert \\ &\quad \leq \sum_{k=1}^{n-m} \biggl[ \frac{3^{k-1} + 1}{2\cdot9^{k+m}} \varphi \bigl(3^{k+m} x, 3^{k+m} x, -3^{k+m}x\bigr) + \frac{3^{k-1} - 1}{2\cdot9^{k+m}} \varphi\bigl(-3^{k+m} x, -3^{k+m} x, 3^{k+m}x\bigr) \biggr] \\ &\qquad{} + \frac{3^{n-m}-1}{4\cdot9^{n}}\varphi\bigl(-3^{n} x, -3^{n} x, 3^{n} x\bigr) + \frac{3^{n-m}-1}{4\cdot9^{n}}\psi\bigl(3^{n-1} x\bigr) \\ &\qquad{} + \frac{9^{n-m} + 2\cdot3^{n-m} - 3}{2\cdot9^{n}} \bigl\| f(0) \bigr\| \end{aligned}$$
(5.6)
for all \(x\in X\). Since the right-hand side of inequality (5.6) tends to 0 as \(m\to\infty\), we can define a mapping
$$\begin{aligned} Q(x) := \lim_{n\to\infty} 9^{-n} f\bigl(3^{n} x \bigr) \end{aligned}$$
for all \(x\in X\). Replacing x, y, z by \(3^{n} x\), \(3^{n} y\), \(3^{n} z\) in (5.1) and dividing both sides by \(9^{n}\), and after taking the limit in the resulting inequality we see that Q satisfies (1.2). Letting \(n\to\infty\) in (5.4) we have the result (5.3). The rest of the proof is similar to that of the proof of Theorem 3.1. □

The following corollaries are immediate consequences of the above theorem.

Corollary 5.2

Let \(\epsilon\geq0\), \(p<1\), and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\Vert Df(x,y,z) \bigr\Vert \leq\epsilon\bigl(\|x\|^{p}+\|y \|^{p}+\|z\|^{p}\bigr) \end{aligned}$$
for all \(x,y,z\in X\) (\(x,y,z\in X\backslash\{0\}\) if \(p<0\)) and
$$\begin{aligned} \bigl\| f(3x) - 9f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
for all \(x\in X\). Then there exists a unique quadratic mapping \(Q:X\to Y\) satisfying (1.3) such that
$$\begin{aligned} \biggl\Vert f(x) - Q(x) +\frac{1}{2}f(0) \biggr\Vert \leq \frac{3^{p}}{3-3^{p}} \epsilon\|x\|^{p} \end{aligned}$$
for all \(x\in X\) (\(x\in X\backslash\{0\}\) if \(p<0\)).

Corollary 5.3

Let \(\epsilon\geq0\) and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\Vert Df(x,y,z) \bigr\Vert \leq\epsilon \end{aligned}$$
for all \(x,y,z\in X\) and
$$\begin{aligned} \bigl\| f(3x) - 9f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
for all \(x\in X\). Then there exists a unique quadratic mapping \(Q:X\to Y\) satisfying (1.3) such that
$$\begin{aligned} \biggl\Vert f(x) - Q(x) + \frac{1}{2}f(0) \biggr\Vert \leq \frac{1}{6}\epsilon \end{aligned}$$
for all \(x\in X\).

We have the following result, which is analogous to Theorem 5.1.

Theorem 5.4

Let \(\varphi:X\times X\times X\to[0,\infty)\) satisfy the condition (\(\mathcal{B}_{1}\)) and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{B}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\| Df(x,y,z) \bigr\| \leq\varphi(x,y,z) \end{aligned}$$
(5.7)
for all \(x,y,z\in X\) and
$$\begin{aligned} \bigl\| f(3x) - 9f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
(5.8)
for all \(x\in X\). Then there exists a unique quadratic mapping \(Q:X\to Y\) satisfying (1.3) such that
$$\begin{aligned} &\bigl\Vert f(x) - Q(x) \bigr\Vert \\ &\quad\leq \sum _{k=0}^{\infty} \biggl[ \frac{3\cdot9^{k} + 3^{k}}{6} \varphi \biggl( \frac{x}{3^{k}}, \frac{x}{3^{k}}, -\frac{x}{3^{k}} \biggr) + \frac{3\cdot9^{k} - 3^{k}}{6} \varphi \biggl(-\frac{x}{3^{k}}, -\frac {x}{3^{k}}, \frac{x}{3^{k}} \biggr) \biggr] \end{aligned}$$
(5.9)
for all \(x\in X\).

Proof

It follows from (5.7), (5.8), and Lemma 2.2 that we have
$$\begin{aligned} &\biggl\Vert f(x) - 9^{n} f \biggl(\frac{x}{3^{n}} \biggr) \biggr\Vert \\ &\quad \leq \sum _{k=0}^{n-1} \biggl[ \frac{3\cdot9^{k} + 3^{k}}{6} \varphi \biggl( \frac {x}{3^{k}}, \frac{x}{3^{k}}, -\frac{x}{3^{k}} \biggr) + \frac{3\cdot9^{k} - 3^{k}}{6} \varphi \biggl(-\frac{x}{3^{k}}, -\frac {x}{3^{k}}, \frac{x}{3^{k}} \biggr) \biggr] \\ &\qquad{} + \frac{9^{n}-3^{n}}{4} \varphi \biggl( -\frac{x}{3^{n}}, -\frac{x}{3^{n}}, \frac {x}{3^{n}} \biggr) + \frac{9^{n}-3^{n}}{4} \psi \biggl(\frac{x}{3^{n+1}} \biggr) \end{aligned}$$
(5.10)
for all \(x\in X\) and \(n\in\mathbb{N}\). From (5.10) we obtain
$$\begin{aligned} & \biggl\Vert 9^{m} f \biggl( \frac{x}{3^{m}} \biggr) - 9^{n} f \biggl( \frac {x}{3^{n}} \biggr) \biggr\Vert \\ &\quad =9^{m} \biggl\Vert f \biggl(\frac{x}{3^{m}} \biggr) - 9^{n-m} f \biggl(\frac{x}{3^{n-m}\cdot3^{m}} \biggr) \biggr\Vert \\ &\quad \leq\sum_{k=0}^{n-m-1} \biggl[ \frac{3\cdot9^{k+m} + 3^{k+2m}}{6} \varphi \biggl(\frac {x}{3^{k+m}}, \frac{x}{3^{k+m}}, - \frac{x}{3^{k+m}} \biggr) \\ &\qquad{} + \frac{3\cdot9^{k+m} - 3^{k+2m}}{6} \varphi \biggl(-\frac{x}{3^{k+m}}, - \frac{x}{3^{k+m}}, \frac{x}{3^{k+m}} \biggr) \biggr] \\ &\qquad{} + \frac{9^{n}-3^{n+m}}{4} \varphi \biggl( -\frac{x}{3^{n}}, -\frac{x}{3^{n}}, \frac{x}{3^{n}} \biggr) + \frac{9^{n}-3^{n+m}}{4} \psi \biggl(\frac{x}{3^{n+1}} \biggr) \end{aligned}$$
(5.11)
for all \(x\in X\) and \(m, n\in\mathbb{N}\) with \(n> m\). Note that the right-hand side of inequality (5.11) tends to 0 as \(m\to\infty\). This means the sequence \(\{ 9^{n} f(3^{-n} x) \}\) is a Cauchy sequence. Now we define a mapping
$$\begin{aligned} Q(x) := \lim_{n\to\infty} 9^{n} f \biggl(\frac{x}{3^{n}} \biggr) \end{aligned}$$
for all \(x\in X\). Replacing x, y, z by \(\frac{x}{3^{n}}\), \(\frac{y}{3^{n}}\), \(\frac{z}{3^{n}}\) in (5.7) and multiplying both sides by \(9^{n}\), and after taking the limit in the resulting inequality, we see that Q satisfies (1.2). Letting \(n\to\infty\) in (5.10) we have the result (5.9). Using a similar method to the proof of Theorem 3.1 we see that Q is the unique quadratic mapping satisfying (1.3). □

Corollary 5.5

Let \(\epsilon\geq0\), \(p>2\), and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{B}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\Vert Df(x,y,z) \bigr\Vert \leq\epsilon\bigl(\|x\|^{p}+\|y \|^{p}+\|z\|^{p}\bigr) \end{aligned}$$
for all \(x,y,z\in X\) and
$$\begin{aligned} \bigl\| f(3x) - 9f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
for all \(x\in X\). Then there exists a unique quadratic mapping \(Q:X\to Y\) satisfying (1.3) such that
$$\begin{aligned} \bigl\Vert f(x) - Q(x) \bigr\Vert \leq \frac{3^{p+1}}{3^{p}-9} \epsilon\|x \|^{p} \end{aligned}$$
for all \(x\in X\).

6 Approximately additive case

In this section, we establish the stability of (1.2) under the approximately additive condition.

Theorem 6.1

Let \(\varphi:X\times X\times X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{1}\)) and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\| Df(x,y,z) \bigr\| \leq\varphi(x,y,z) \end{aligned}$$
(6.1)
for all \(x,y,z\in X\) and
$$\begin{aligned} \bigl\| f(3x) + 3f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
(6.2)
for all \(x\in X\). Then there exists a unique additive mapping \(A:X\to Y\) satisfying (1.4) such that
$$\begin{aligned} &\biggl\Vert f(x) - A(x) + \frac{1}{2}f(0) \biggr\Vert \\ &\quad \leq \sum_{k=1}^{\infty} \biggl[ \frac{3^{k-1} + 1}{2\cdot9^{k}} \varphi \bigl(3^{k} x, 3^{k} x, -3^{k} x\bigr) + \frac{3^{k-1}-1}{2 \cdot9^{k}} \varphi\bigl(-3^{k} x, -3^{k} x, 3^{k} x\bigr) \biggr] \end{aligned}$$
(6.3)
for all \(x\in X\).

Proof

It follows from (6.1), (6.2), and Lemma 2.1 that we have
$$\begin{aligned} & \biggl\Vert f(x) - \frac{f(3^{n}x)}{3^{n}} + \frac{9^{n} + 2\cdot3^{n} - 3}{2\cdot9^{n}}f(0) \biggr\Vert \\ & \quad\leq \sum_{k=1}^{n} \biggl[ \frac{3^{k-1} + 1}{2\cdot9^{k}} \varphi\bigl(3^{k} x, 3^{k} x, -3^{k} x\bigr) + \frac{3^{k-1}-1}{2 \cdot9^{k}} \varphi\bigl(-3^{k} x, -3^{k} x, 3^{k} x\bigr) \biggr] \\ &\qquad{} + \frac{3^{n}-1}{4\cdot9^{n}} \varphi\bigl(-3^{n} x, -3^{n} x, 3^{n} x\bigr) + \frac{3^{n+1}-3}{4\cdot9^{n}} \psi\bigl(3^{n-1} x\bigr) \end{aligned}$$
(6.4)
for all \(x\in X\) and \(n\in\mathbb{N}\). From (6.4) we figure out
$$\begin{aligned} & \biggl\Vert \frac{f(3^{m} x)}{3^{m}} - \frac{f(3^{n} x)}{3^{n}} \biggr\Vert \\ &\quad = \frac{1}{3^{m}} \biggl\Vert f\bigl(3^{m} x\bigr) - \frac{f(3^{n-m} \cdot3^{m} x)}{3^{n-m}} \biggr\Vert \\ &\quad \leq \sum_{k=1}^{n-m} \biggl[ \frac{3^{k-1} + 1}{2\cdot3^{2k+m}} \varphi \bigl(3^{k+m} x, 3^{k+m} x, -3^{k+m}x\bigr) + \frac{3^{k-1} - 1}{2\cdot3^{2k+m}} \varphi\bigl(-3^{k+m} x, -3^{k+m} x, 3^{k+m}x\bigr) \biggr] \\ &\qquad{} + \frac{3^{n-m}-1}{4\cdot3^{2n-m}} \varphi\bigl(-3^{n} x, -3^{n} x, 3^{n} x\bigr) +\frac{3^{n-m+1}-3}{4\cdot3^{2n-m}}\psi\bigl(3^{n-m} x\bigr) \\ &\qquad{} + \frac{9^{n-m} + 2\cdot3^{n-m} - 3}{2\cdot3^{2n-m}} \bigl\| f(0) \bigr\| \end{aligned}$$
(6.5)
for all \(x\in X\) and \(m, n\in\mathbb{N}\) with \(n> m\). We remark that the right-hand side of inequality (6.5) tends to 0 as \(m\to\infty\). Define a mapping
$$\begin{aligned} A(x) := \lim_{n\to\infty} 3^{-n} f\bigl(3^{n} x \bigr) \end{aligned}$$
for all \(x\in X\). Replacing x, y, z by \(3^{n} x\), \(3^{n} y\), \(3^{n} z\) in (6.1) and dividing both sides by \(3^{n}\), and after taking the limit in the resulting inequality, we see that A satisfies (1.2). Letting \(n\to\infty\) in (6.4) we obtain the result (6.3). The remains of the proof are similar to that of the proof of Theorem 4.1. □

From the theorem above we have the following corollaries.

Corollary 6.2

Let \(\epsilon\geq0\), \(p<1\), and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\Vert Df(x,y,z) \bigr\Vert \leq\epsilon\bigl(\|x\|^{p}+\|y \|^{p}+\|z\|^{p}\bigr) \end{aligned}$$
for all \(x,y,z\in X\) (\(x,y,z\in X\backslash\{0\}\) if \(p<0\)) and
$$\begin{aligned} \bigl\| f(3x) + 3f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
for all \(x\in X\). Then there exists a unique additive mapping \(A:X\to Y\) satisfying (1.4) such that
$$\begin{aligned} \biggl\Vert f(x) - A(x) + \frac{1}{2}f(0) \biggr\Vert \leq \frac{3^{p}}{3-3^{p}} \epsilon\|x\|^{p} \end{aligned}$$
for all \(x\in X\) (\(x\in X\backslash\{0\}\) if \(p<0\)).

Corollary 6.3

Let \(\epsilon\geq0\) and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{A}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\Vert Df(x,y,z) \bigr\Vert \leq\epsilon \end{aligned}$$
for all \(x,y,z\in X\) and
$$\begin{aligned} \bigl\| f(3x) + 3f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
for all \(x\in X\). Then there exists a unique additive mapping \(A:X\to Y\) satisfying (1.4) such that
$$\begin{aligned} \biggl\Vert f(x) - A(x) + \frac{1}{2}f(0) \biggr\Vert \leq \frac{1}{6}\epsilon \end{aligned}$$
for all \(x\in X\).

We have the following result, which is analogous to Theorem 6.1.

Theorem 6.4

Let \(\varphi:X\times X\times X\to[0,\infty)\) satisfy the condition (\(\mathcal{B}_{1}\)) and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{B}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\| Df(x,y,z) \bigr\| \leq\varphi(x,y,z) \end{aligned}$$
(6.6)
for all \(x,y,z\in X\) and
$$\begin{aligned} \bigl\| f(3x) + 3f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
(6.7)
for all \(x\in X\). Then there exists a unique additive mapping \(A:X\to Y\) satisfying (1.4) such that
$$\begin{aligned} &\bigl\Vert f(x) - A(x) \bigr\Vert \\ &\quad\leq \sum _{k=0}^{\infty} \biggl[ \frac{3\cdot9^{k} + 3^{k}}{6} \varphi \biggl( \frac{x}{3^{k}}, \frac{x}{3^{k}}, -\frac{x}{3^{k}} \biggr) + \frac{3\cdot9^{k} - 3^{k}}{6} \varphi \biggl(-\frac{x}{3^{k}}, -\frac {x}{3^{k}}, \frac{x}{3^{k}} \biggr) \biggr] \end{aligned}$$
(6.8)
for all \(x\in X\).

Proof

Combining (6.6), (6.7), and Lemma 2.2 we have
$$\begin{aligned} & \biggl\Vert f(x) - 3^{n} f \biggl(\frac{x}{3^{n}} \biggr) \biggr\Vert \\ &\quad \leq \sum _{k=0}^{n-1} \biggl[ \frac{3\cdot9^{k} + 3^{k}}{6} \varphi \biggl( \frac {x}{3^{k}}, \frac{x}{3^{k}}, -\frac{x}{3^{k}} \biggr) + \frac{3\cdot9^{k} - 3^{k}}{6} \varphi \biggl(-\frac{x}{3^{k}}, -\frac {x}{3^{k}}, \frac{x}{3^{k}} \biggr) \biggr] \\ &\qquad{} + \frac{9^{n}-3^{n}}{4} \varphi \biggl( -\frac{x}{3^{n}}, -\frac{x}{3^{n}}, \frac{x}{3^{n}} \biggr) + \frac{3(9^{n}-3^{n})}{4} \psi \biggl(\frac{x}{3^{n+1}} \biggr) \end{aligned}$$
(6.9)
for all \(x\in X\) and \(n\in\mathbb{N}\). We claim that \(\{9^{n} f(3^{-n} x) \}\) is a Cauchy sequence. From (6.9) we obtain
$$\begin{aligned} & \biggl\Vert 3^{m} f \biggl( \frac{x}{3^{m}} \biggr) - 3^{n} f \biggl( \frac {x}{3^{n}} \biggr) \biggr\Vert \\ &\quad = 3^{m} \biggl\Vert f \biggl(\frac{x}{3^{m}} \biggr) - 3^{n-m} f \biggl(\frac {x}{3^{n-m}\cdot3^{m}} \biggr) \biggr\Vert \\ &\quad \leq \sum_{k=0}^{n-m-1} \biggl[ \frac{3\cdot3^{2k+m} + 3^{k+m}}{6} \varphi \biggl(\frac {x}{3^{k+m}}, \frac{x}{3^{k+m}}, - \frac{x}{3^{k+m}} \biggr) \\ &\qquad{} + \frac{3\cdot3^{2k+m} - 3^{k+m}}{6} \varphi \biggl(-\frac{x}{3^{k+m}}, - \frac{x}{3^{k+m}}, \frac{x}{3^{k+m}} \biggr) \biggr] \\ &\qquad{} + \frac{3^{2n-m}-3^{n}}{4} \varphi \biggl( -\frac{x}{3^{n}}, - \frac{x}{3^{n}}, \frac{x}{3^{n}} \biggr) + \frac{3^{2n-m+1}-3^{n+1}}{4} \psi \biggl( \frac{x}{3^{n}} \biggr) \end{aligned}$$
(6.10)
for all \(x\in X\) and \(m, n\in\mathbb{N}\) with \(n> m\). Since the right-hand side of inequality (6.10) tends to 0 as \(m\to\infty\), the sequence \(\{ 3^{n} f(3^{-n} x) \}\) is a Cauchy sequence. Define a mapping
$$\begin{aligned} A(x) := \lim_{n\to\infty} 3^{n} f \biggl(\frac{x}{3^{n}} \biggr) \end{aligned}$$
for all \(x\in X\). Replacing x, y, z by \(\frac{x}{3^{n}}\), \(\frac{y}{3^{n}}\), \(\frac{z}{3^{n}}\) in (6.6) and multiplying both sides by \(3^{n}\), and after taking the limit in the resulting inequality, we see that A satisfies (1.2). Using a similar method to the proof of Theorem 4.1 we see that A is the unique additive mapping satisfying (1.4). Finally taking the limit as \(n\to\infty\) in (6.9) we have the result (6.8). □

Corollary 6.5

Let \(\epsilon\geq0\), \(p>2\), and \(\psi:X\to[0,\infty)\) satisfy the condition (\(\mathcal{B}_{2}\)). Suppose that a mapping \(f:X\to Y\) satisfies
$$\begin{aligned} \bigl\Vert Df(x,y,z) \bigr\Vert \leq\epsilon\bigl(\|x\|^{p}+\|y \|^{p}+\|z\|^{p}\bigr) \end{aligned}$$
for all \(x,y,z\in X\) and
$$\begin{aligned} \bigl\| f(3x) + 3f(-x) \bigr\| \leq\psi(x) \end{aligned}$$
for all \(x\in X\). Then there exists a unique additive mapping \(A:X\to Y\) satisfying (1.4) such that
$$\begin{aligned} \bigl\Vert f(x) - A(x) \bigr\Vert \leq \frac{3^{p+1}}{3^{p}-9} \epsilon\|x \|^{p} \end{aligned}$$
for all \(x\in X\).

Declarations

Acknowledgements

The authors would like to thank the editor and the referees for their valuable comments and suggestions to improve the quality of this paper.

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
Hana Academy Seoul

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