Some improvements on the stability results for additive functional inequalities in Banach spaces
- Open Access
- 08-12-2025
- Research
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Abstract
1 Introduction
The study of functional equations has been an important area of mathematics for many years. A functional equation is a mathematical equation where the unknown is a function, not just a number. These equations appear in many different fields of mathematics and science, including probability theory, physics, and engineering.
Suppose that X and Y are two vector spaces. One of the most famous examples is Cauchy’s functional equation, that is, \(f:X\to Y\) satisfies for all \(x,y\in X\). In this paper, we call such a mapping an additive mapping. However, in real-world applications, we often deal with measurements that are not completely exact. This leads us to an important question: what happens when a function almost satisfies a functional equation, but not exactly? This question brings us to the concept of stability of functional equations. This idea was first introduced by Ulam (see [15]) in 1940 during a mathematics conference. He asked whether every function that approximately satisfies the additivity property must be close to an additive function. This question started a new and exciting area of research in mathematics.
$$ f(x + y) = f(x) + f(y) $$
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Hyers [6] gave the first positive answer to Ulam’s question in 1941. He proved that if f is a mapping from a Banach space \(X:=(X,\|\cdot \|_{X})\) into a Banach space \(Y:=(Y,\|\cdot \|_{Y})\) satisfying for some small positive number ε, then there exists an additive mapping \(F:X\to Y\) such that \(\|F(x)-f(x)\|_{Y}\le \varepsilon \). Such a result is always regarded as the Hyers–Ulam stability.
$$ \|f(x + y) - f(x) - f(y)\|_{Y} \leq \varepsilon $$
Gilányi [5] discussed the following result: if \(f:X\to Y\) where Y is a Hilbert space satisfies for some small positive number ε, then there exists a quadratic mapping \(Q:X\to Y\) such that \(\|Q(x)-f(x)\|_{Y}\le \frac{5}{2}\varepsilon \). Recall that Q is a quadratic mapping if \(2Q(x)+2Q(y)- Q(x-y)-Q(x+y)=0\).
$$ \|2f(x)+2f(y)- f(x-y)\|_{Y} \leq \|f(x+y)\|_{Y}+\varepsilon $$
Inspired by this result, the following two functional inequalities [9] are studied: Suppose that X and Y are two normed spaces over the same scalar field and \(f:X\to Y\) satisfies one of the following and where \(\omega :X^{3}\to [0,\infty )\) and ϱ is a nonzero scalar.
$$ \|f(x)+f(y)+f(z)\|_{Y}\le \Big\| \varrho f\Big(\frac{x+y+z}{\varrho} \Big)\Big\| _{Y}+\omega (x,y,z) $$
(A)
$$ \|f(x)+f(y)+\varrho f(z)\|_{Y}\le \Big\| \varrho f\Big( \frac{x+y}{\varrho}+z\Big)\Big\| _{Y}+\omega (x,y,z) $$
(B)
Park et al. [10] proved the Hyers–Ulam stability of the functional inequalities (A) with \(\varrho :=1,2\) and (B) with \(\varrho :=2\) where \(\omega (x,y,z):=\theta (\|x\|_{X}^{p}+\|y\|_{X}^{p}+\|z\|_{X}^{p})\) or \(\omega (x,y,z):=\theta (\|x\|_{X}^{p}\cdot \|y\|_{X}^{p}\cdot \|z\|_{X}^{p})\). The first formulation of \(\omega (x,y,z)\) follows the ideas of Aoki [1], Rassias [13], and Gajda [3], whereas the second formulation is based on the work of Rassias [12]. However, there is an error in their results. In [11], Park and Lee revised the statements by adding the assumption that f is an odd mapping, that is, \(f(-x)=-f(x)\) for all \(x\in X\). Jun and Roh [7] proved that in some particular cases of the results in [10] yields the hyperstability, that is, any mapping f approximately satisfies (B) where \(\varrho :=2\) is exactly an additive mapping.
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The following two stability results are due to Lu and Park [9] with a more general choice of ϱ and a more general form of \(\omega (x,y,z)\) ([4]).
Theorem LuP1
([9, Theorem 2.2])
Suppose that X is a vector space and \(Y:=(Y,\|\cdot \|_{Y})\) is a Banach space. Let ϱ be a real number such that \(0<|\varrho |<3\). Suppose that a mapping \(f:X\to Y\) satisfies (A) where Then there exists a unique additive mapping \(F:X\to Y\) such that
$$ \Omega _{1}(x,y,z):=\sum _{n=1}^{\infty }2^{n}\omega \left ( \frac{x}{2^{n}},\frac{y}{2^{n}},\frac{z}{2^{n}}\right )< \infty \quad \textit{for all $x,y,z\in X$.} $$
$$ \|F(x)-f(x)\|_{Y}\leq \frac{1}{2}\Omega _{1}(-x,-x,2x)+\Omega _{1}(x,-x,0) \quad \textit{for all $x\in X$.} $$
Theorem LuP2
([9, Theorems 3.2 and 3.4])
Suppose that X is a vector space and \(Y:=(Y,\|\cdot \|_{Y})\) is a Banach space. Let ϱ be a positive real number. Suppose that a mapping \(f:X\to Y\) satisfies (B). Then the following statements are true.
1.
If \(\varrho <2\) and then there exists a unique additive mapping \(F:X\to Y\) such that
$$ \Omega _{2}(x,y,z):=\sum _{n=1}^{\infty}\Big(\frac{2}{\varrho}\Big)^{n} \omega \bigg(\frac{x}{\big(\frac{2}{\varrho}\big)^{n}}, \frac{y}{\big(\frac{2}{\varrho}\big)^{n}}, \frac{z}{\big(\frac{2}{\varrho}\big)^{n}}\bigg)< \infty \quad \textit{for all $x,y,z\in X$}, $$
$$ \|F(x)-f(x)\|_{Y}\leq \frac{1}{2}\Omega _{2}\Big(-x,-x, \frac{2}{\varrho}x\Big)+\Omega _{2}(x,-x,0)\quad \textit{for all $x\in X$.} $$
2.
If \(\varrho >2\) and then there exists a unique additive mapping \(F:X\to Y\) such that for all \(x\in X\).
$$ \Omega _{3}(x,y,z):=\sum _{n=0}^{\infty } \frac{1}{\big(\frac{2}{\varrho}\big)^{n}}\omega \Big({\Big( \frac{2}{\varrho}\Big)^{n}}x,{\Big(\frac{2}{\varrho}\Big)^{n}}y,{ \Big(\frac{2}{\varrho}\Big)^{n}}z\Big)< \infty \quad \textit{for all $x,y,z\in X$}, $$
$$ \|F(x)-f(x)\|_{Y}\leq \frac{1}{2}\Omega _{3}\Big(x,x,- \frac{2}{\varrho}x\Big)+\Omega _{3}\Big(\frac{2}{\varrho}x,- \frac{2}{\varrho}x,0\Big) $$
Khodaei [8] proved some stability results of the functional inequality (A) with \(\varrho :=1\) as follows.
Theorem K1
([8, Corollary 2.4 with \(l:=-1\)])
Suppose that X is a vector space and \(Y:=(Y,\|\cdot \|_{Y})\) is a Banach space. Suppose that \(f:X\to Y\) satisfies (A) with \(\varrho :=1\) where Then there exists a unique additive mapping \(F:X\to Y\) such that
1.
\(\overline{\Omega}(x):=\sum _{n=1}^{\infty }2^{n}\omega \big( \frac{x}{(-2)^{n}},\frac{x}{(-2)^{n-1}},\frac{x}{(-2)^{n}}\big)< \infty \) for all \(x\in X\);
2.
\(\lim _{n\to \infty}(-2)^{n}\omega \big(\frac{x}{2^{n}}, \frac{y}{2^{n}},\frac{-x-y}{2^{n}}\big)=0\) for all \(x,y\in X\).
$$ \|F(x)-f(x)\|_{Y}\leq \frac{1}{2}\overline{\Omega}(x)\quad \textit{for all $x\in X$.} $$
Theorem K2
([8, Corollary 2.4 with \(l:=1\)])
Suppose that X is a vector space and \(Y:=(Y,\|\cdot \|_{Y})\) is a Banach space. Suppose that \(f:X\to Y\) satisfies (A) with \(\varrho :=1\) where Then there exists a unique additive mapping \(F:X\to Y\) such that
1.
\(\underline{\Omega}(x):=\sum _{n=0}^{\infty }\frac{1}{2^{n}}\omega \left ((-2)^{n}x,(-2)^{n+1}x,(-2)^{n}x\right )<\infty \) for all \(x\in X\);
2.
\(\lim _{n\to \infty}\frac{1}{2^{n}}\omega ((-2)^{n}x,(-2)^{n}y,(-2)^{n}(-x-y))=0\) for all \(x,y\in X\);
3.
\(f(0)=0\) if \(\omega (0,0,0)\ne 0\).
$$ \|F(x)-f(x)\|_{Y}\leq \frac{1}{2}\underline{\Omega}(x)\quad \textit{for all $x\in X$.} $$
The Tables 1 and 2 summarize the stability bounds of the functional inequalities (A) and (B), respectively, obtained in Theorems PLe, LuP, and K where \(\omega (x,y,z):=\theta (\|x\|_{X}^{p}+\|y\|_{X}^{p}+\|z\|_{X}^{p})\).
Table 1
Stability bounds of the functional inequality (A)
p | p>1 | 0<p<1 |
|---|---|---|
Theorem PLe (ϱ: = 1,2 and f is odd) | \(\frac{2^{p}+2}{2^{p}-2}\theta \|x\|_{X}^{p}\) | \(\frac{2^{p}+2}{2-2^{p}}\theta \|x\|_{X}^{p}\) |
Theorem LuP (0<|ϱ|<3) | \(\frac{2^{p}+6}{2^{p}-2}\theta \|x\|_{X}^{p}\) | NA |
Theorem K (ϱ: = 1) | \(\frac{2^{p}+2}{2^{p}-2}\theta \|x\|_{X}^{p}\) | \(\frac{2^{p}+2}{2-2^{p}}\theta \|x\|_{X}^{p}\) |
Table 2
Stability bounds of the functional inequality (B)
ϱ | ϱ>0 and ϱ ≠ 2 | ϱ = 2 |
|---|---|---|
Theorem PLe (0<p ≠ 1 and f is odd) | NA | \(\frac{2^{p}+1}{|2^{p}-2|}\theta \|x\|_{X}^{p}\) |
Theorem LuP (p>1) | \(\frac{2^{p}+6\varrho ^{p}}{2^{p}\varrho -2\varrho ^{p}}\theta \|x\|_{X}^{p}\) | NA |
The purpose of this paper is to unify and generalize the stability results of functional inequalities (A) and (B). Moreover, we give a sufficient condition in terms of \(\omega (x,y,z)\) to guarantee the hyperstability.
In this paper, we collect together some preliminaries in Sect. 2. Section 3 is organized as follows. In Sect. 3.1, we introduce the notion of γ-additive mappings. In Sect. 3.2, we employ the core idea of the direct method studied by Forti [2] to obtain the stability result of the γ-additive mappings presented in Sect. 3.1. Section 3.3 is devoted to explaining the stability of functional inequality (A) by applying the results established in Sect. 3.2. Similarly, in Sect. 3.4, we use the results from Sect. 3.2 to study the stability of functional inequality (B). Moreover, we show that the oddness assumption as was the case in Park–Lee’s results is superfluous. An example is given to show that some of our estimate is sharp. Finally, in Sect. 3.5, we discuss the hyperstability of the functional inequalities (A) and (B).
2 Preliminaries
Let \((Y,d)\) be a complete metric space and X a non-empty set. Let \(G:X\to X\) and \(H:Y\to Y\) be two given mappings. We recall the following two interesting results of Forti [2].
Theorem F1
([2, Theorem 1])
Suppose that \(\varphi :[0,\infty )\to [0,\infty )\), \(\delta :X\to [0,\infty )\), and \(f:X\to Y\) are such that If H is continuous, then there exists a unique mapping \(F:X\to Y\) such that the following properties hold: For all \(x\in X\), Moreover, \(F(x)=\lim _{n\to \infty}(H^{n}\circ f\circ G^{n})(x)\) for all \(x\in X\).
1.
\(\varphi (a+b)\leq \varphi (a)+\varphi (b)\) for all \(a,b\in [0,\infty )\);
2.
\(d(Hu,Hv)\leq \varphi (d(u,v))\) for all \(u,v\in Y\);
3.
\(\Phi (x):=\sum _{n=0}^{\infty}\varphi ^{n}(\delta [G^{n}(x)])< \infty \) for all \(x\in X\);
4.
\(d(f(x),(H\circ f\circ G)(x))\leq \delta (x)\) for all \(x\in X\).
-
\((H\circ F\circ G)(x)=F(x)\);
-
\(d(F(x),f(x))\leq \Phi (x)\).
Theorem F2
([2, Theorem 2])
Suppose that G and H are invertible, that is, \(G^{-1}\) and \(H^{-1}\) exist. Suppose that \(\psi :[0,\infty )\to [0,\infty )\), \(\delta :X\to [0,\infty )\) and \(f:X\to Y\) are such that If \(H^{-1}\) is continuous, then there exists a unique mapping \(F:X\to Y\) such that the following properties hold: For all \(x\in X\), Moreover, \(F(x)=\lim _{n\to \infty}(H^{-n}\circ f\circ G^{-n})(x)\) for all \(x\in X\).
1.
\(\psi (a+b)\leq \psi (a)+\psi (b)\) for all \(a,b\in [0,\infty )\);
2.
\(d(H^{-1}u,H^{-1}v)\leq \psi (d(u,v))\) for all \(u,v\in Y\);
3.
\(\Psi (x):=\sum _{n=1}^{\infty}\psi ^{n}(\delta [G^{-n}(x)])<\infty \) for all \(x\in X\);
4.
\(d(f(x),(H\circ f\circ G)(x))\leq \delta (x)\) for all \(x\in X\).
-
\((H\circ F\circ G)(x)=F(x)\);
-
\(d(F(x),f(x))\leq \Psi (x)\).
3 Main results
3.1 γ-Additive mappings
Suppose that X and Y are two vector spaces over the same scalar field where . In this paper, we are interested in the following γ-additive mapping \(F:X\to Y\) such that where . It is obvious that \((-1)\)-additive mapping is nothing but the classical additive mapping.
$$ F(x)+F(y)+\gamma F\Big(\frac{-x-y}{\gamma}\Big)=0\quad (x,y\in X) $$
Lemma 1
Suppose that X and Y are vector spaces over the same scalar field such that \(Y\neq \{0\}\) and . Then the following two statements are equivalent:
1.
\(\gamma \neq -2\).
2.
If \(F:X\to Y\) is γ-additive, then F is additive and \(F(x)=\gamma F(x/\gamma )\) for all \(x\in X\).
Proof
(a) ⇒ (b). Assume that \(\gamma \neq -2\) and \(F:X\to Y\) is γ-additive. Note that \((2+\gamma )F(0)=F(0)+F(0)+\gamma F\big(\frac{-0-0}{\gamma}\big)=0\) and hence \(F(0)=0\) because \(\gamma +2\neq 0\). Moreover, F is odd. To see this, let \(x\in X\). It follows that \(F(x)+F(-x)=F(x)+F(-x)+\gamma F\big(\frac{-x-(-x)}{\gamma}\big)=0\). In particular, since F is odd, we also have \(F(x)-\gamma F(x/\gamma )=F(x)+F(0)-\gamma F(x/\gamma )=0\), that is, \(F(x)=\gamma F(x/\gamma )\). This implies that \(F(x)+F(y)-F(x+y)=F(x)+F(y)+\gamma F\big(\frac{-x-y}{\gamma}\big)=0\) for all \(x,y\in X\), that is, F is additive.
(b) ⇒ (a). Assume that \(\gamma =-2\) and let \(y_{0}\in Y\setminus \{0\}\). We define \(F:X\to Y\) by \(F(x):=y_{0}\neq 0\) for all \(x\in X\). It follows that \(F(x)+F(y)+\gamma F\big(\frac{-x-y}{\gamma}\big)=y_{0}+y_{0}-2y_{0}=0\) for all \(x,y\in X\), that is, F is γ-additive. But \(F(0)\neq 0\) and hence F is not additive. □
Note that the property \(F(x)=\gamma F(x/\gamma )\) for all \(x\in X\) where γ is not a rational number and the additivity of F does not imply the linearity as shown in the following example.
Example 1
There exists a mapping such that
but F is not linear.
Proof
Note that is a field. We now treat as a vector space over the scalar field and let \(\mathcal{B}:=\{b_{\alpha}\}_{\alpha \in \Lambda}\) be a Hamel basis for . Let \(b_{0}\in \mathcal{B}\) be fixed. In particular, for each , we can write \(x:=q_{0}b_{0}+\sum _{k=1}^{n}q_{k}b_{\alpha _{k}}\) for some and \(b_{\alpha _{1}},\ldots ,b_{\alpha _{n}}\in \mathcal{B}\setminus \{b_{0} \}\). Note that such a representation is unique and we can define We show that is our candidate. First, we show that F is additive. To see this, let . Without loss of generality, we assume that \(x:=q_{0}b_{0}+\sum _{k=1}^{n}q_{k}b_{\alpha _{k}}\) and \(y:=r_{0}b_{0}+\sum _{k=1}^{n}r_{k}b_{\alpha _{k}}\) for some and \(b_{\alpha _{1}},\ldots ,b_{\alpha _{n}}\in \mathcal{B}\setminus \{b_{0} \}\). This implies that \(x+y=(q_{0}+r_{0})b_{0}+\sum _{k=1}^{n}(q_{k}+r_{k})b_{\alpha _{k}}\) and hence For the representation \(x:=q_{0}b_{0}+\sum _{k=1}^{n}q_{k}b_{\alpha _{k}}\), we write \(q_{k}:=q^{(1)}_{k}+q^{(2)}_{k}\sqrt{2}\) where for all \(k=0,1,\dots ,n\). It follows that \(\sqrt{2}x=\big(q^{(1)}_{0}\sqrt{2}+2q^{(2)}_{0}\big)b_{0}+\sum _{k=1}^{n} \big(q^{(1)}_{k}\sqrt{2}+2q^{(2)}_{k}\big)b_{\alpha _{k}}\) and hence Finally, we show that F is not linear. To see this, let \(x\in \mathcal{B}\setminus \{b_{0}\}\) be a nonzero. Then □
$$ F(x):=q_{0}. $$
$$ F(x+y)=q_{0}+r_{0}=F(x)+F(y). $$
$$ F(\sqrt{2}x)=q^{(1)}_{0}\sqrt{2}+2q^{(2)}_{0}=\sqrt{2}\big(q^{(1)}_{0}+q^{(2)}_{0} \sqrt{2}\big)=\sqrt{2}F(x). $$
$$ \frac{F(x)}{x}=0\neq \frac{1}{b_{0}}=\frac{F(b_{0})}{b_{0}}. $$
3.2 Stability results via Theorems F1 and F2
3.2.1 Stability of γ-additive mappings via Theorem F1
Suppose that X and \(Y:=(Y,\|\cdot \|_{Y})\) are a vector space and a Banach space, respectively. Suppose that \(\gamma \notin \{-2,0\}\). Suppose that \(\lambda :X^{2}\to [0,\infty )\) and \(f:X\to Y\) satisfy the following properties:
$$ \Big\| f(x)+f(y)+\gamma f\Big(\frac{-x-y}{\gamma}\Big)\Big\| _{Y}\le \lambda (x,y)\quad \text{for all $x,y\in X$}. $$
Theorem 1
Suppose that Then there exists a unique γ-additive mapping \(F:X\to Y\) such that Moreover, \(F(x)=\lim _{n\to \infty}\frac{1}{(\frac{2}{-\gamma})^{n}} f\big( \big(\frac{2}{-\gamma}\big)^{n}x\big)\) for all \(x\in X\).
1.
\(\Lambda (x):=\sum _{n=0}^{\infty}\frac{1}{(\frac{2}{|\gamma |})^{n}} \lambda \big(\big(\frac{2}{-\gamma}\big)^{n} x,\big(\frac{2}{-\gamma} \big)^{n} x\big)<\infty \) for all \(x\in X\);
2.
\(\lim _{n\to \infty}\frac{1}{(\frac{2}{|\gamma |})^{n}}\lambda \big( \big(\frac{2}{-\gamma}\big)^{n} x,\big(\frac{2}{-\gamma}\big)^{n} y \big)=0\) for all \(x,y\in X\).
$$ \|F(x)-f(x)\|_{Y}\le \frac{1}{2}\Lambda (x)\quad \textit{for all $x\in X$}. $$
Proof
Existence: Let \(x\in X\). Then Letting \(\alpha :=\frac{2}{-\gamma}\), we get that To apply Theorem F1, we define \(G:X\to X\), \(H:Y\to Y\), \(\delta :X\to [0,\infty )\), and \(\varphi :[0,\infty )\to [0,\infty )\) by By Theorem F1, there exists a unique mapping \(F:X\to Y\) satisfying the following properties: for all \(x\in X\). Moreover, \(F(x)=\lim _{n\to \infty}(H^{n}\circ f\circ G^{n})(x)=\lim _{n\to \infty}\frac{1}{\alpha ^{n}} f\left (\alpha ^{n}x\right )\) for all \(x\in X\).
$$ \Big\| 2f(x)+\gamma f\Big(\Big(\frac{2}{-\gamma}\Big)x\Big)\Big\| _{Y} \leq \lambda (x,x). $$
$$ \left \|f(x)-\frac{1}{\alpha}f(\alpha x)\right \|_{Y}\leq \frac{1}{2} \lambda (x,x). $$
Note that H is continuous and \((H\circ f\circ G)(x)=\frac{1}{\alpha}f\left (\alpha x\right )\) for all \(x\in X\). Moreover, we also have the following assertions:
1.
\(\varphi (a+b)=\frac{1}{|\alpha |}(a+b)=\frac{1}{|\alpha |}a+ \frac{1}{|\alpha |}b=\varphi (a)+\varphi (b)\) for all \(a,b\in [0,\infty )\);
2.
\(\|H(u)-H(v)\|_{Y}=\left \|\frac{1}{\alpha}u-\frac{1}{\alpha}v\right \|_{Y}=\frac{1}{|\alpha |}\|u-v\|_{Y}=\varphi (\|u-v\|_{Y})\) for all \(u,v\in Y\);
3.
\(\Phi (x):=\sum _{n=0}^{\infty}\varphi ^{n}(\delta (G^{n}(x)))=\sum _{n=0}^{ \infty}\frac{1}{2}\cdot \frac{1}{|\alpha |^{n}}\lambda \left (\alpha ^{n}x, \alpha ^{n}x\right )=\frac{1}{2}\Lambda (x)<\infty \) for all \(x\in X\);
4.
\(\|f(x)-(H\circ f\circ G)(x)\|_{Y}=\left \|f(x)-\frac{1}{\alpha}f( \alpha x)\right \|_{Y}\leq \delta (x)\) for all \(x\in X\).
-
\(\frac{1}{\alpha}F\left (\alpha x\right )=(H\circ F\circ G)(x)=F(x)\)
-
\(\|F(x)-f(x)\|_{Y}\leq \Phi (x)=\frac{1}{2}\Lambda (x)\)
We first claim that F is γ-additive. To see this, let \(x,y\in X\). It follows that This implies that It follows from (b) that Hence the claim is proved.
$$ \Big\| f(\alpha ^{n}x) +f(\alpha ^{n}y) +\gamma f\Big( \frac{-\alpha ^{n}x-\alpha ^{n}y}{\gamma}\Big)\Big\| _{Y} \le \lambda ( \alpha ^{n}x, \alpha ^{n}y). $$
$$ \left \|\frac{1}{\alpha ^{n}}f\left (\alpha ^{n}x\right ) + \frac{1}{\alpha ^{n}}f\left (\alpha ^{n}y\right ) + \frac{\gamma}{\alpha ^{n}}f\Big(\alpha ^{n}\frac{-x-y}{\gamma}\Big) \right \|_{Y}\le \frac{1}{|\alpha |^{n}}\lambda \left (\alpha ^{n}x, \alpha ^{n}y\right ). $$
$$ \Big\| F(x)+F(y)+\gamma F\Big(\frac{-x-y}{\gamma}\Big)\Big\| _{Y}=\lim _{n \to \infty}\Big\| \frac{1}{\alpha ^{n}}(\alpha ^{n}x) + \frac{1}{\alpha ^{n}}f(\alpha ^{n}y) +\frac{\gamma}{\alpha ^{n}}f \Big(\alpha ^{n}\frac{-x-y}{\gamma}\Big)\Big\| _{Y}=0. $$
Uniqueness: To see this, suppose that there exists a γ-additive mapping \(G:X\to Y\) such that To see that \(G=F\), it suffices to show that \(\frac{1}{\alpha }G(\alpha x)=G(x)\) for all \(x\in X\) and hence the conclusion follows from the uniqueness of F. In fact, it follows from the γ-additivity and the oddness of G that □
$$ \|G(x)-f(x)\|_{Y}\le \frac{1}{2}\Lambda (x)\quad \text{for all $x\in X$.} $$
$$ \frac{1}{\alpha }G(\alpha x)=\frac{-\gamma}{2}G\Big(\frac{2}{-\gamma}x \Big)=\frac{\gamma}{2}G\Big(\frac{2}{\gamma}x\Big)=\frac{1}{2}G(2x)=G(x) \quad \text{for all $x\in X$}. $$
Theorem 2
Suppose that Then there exists a unique γ-additive mapping \(F:X\to Y\) such that Moreover, \(F(x)=\lim _{n\to \infty}(-\gamma )^{n}f\big(\frac{x}{(-\gamma )^{n}} \big)\) for all \(x\in X\).
1.
\(\Lambda (x):=\sum _{n=0}^{\infty}|\gamma |^{n}\lambda \big( \frac{x}{(-\gamma )^{n}},0\big)<\infty \) for all \(x\in X\);
2.
\(\lim _{n\to \infty}|\gamma |^{n}\lambda \big( \frac{x}{(-\gamma )^{n}},\frac{y}{(-\gamma )^{n}}\big)=0\) for all \(x,y\in X\);
3.
\(\lambda (0,0)=0\) if \(|\gamma |<1\).
$$ \|F(x)-f(x)\|_{Y}\le \Lambda (x)\quad \textit{for all $x\in X$}. $$
Proof
If \(|\gamma |<1\), then \(\|(2+\gamma )f(0)\|_{Y}\le \lambda (0,0)=0\) and hence \(f(0)=0\). On the other hand, if \(|\gamma |\ge 1\), then it follows from \(\lim _{n\to \infty}|\gamma |^{n}\lambda (0,0)=0\) that \(\lambda (0,0)=0\) and hence \(f(0)=0\). Let \(x\in X\). Replace y in the first inequality by 0. We obtain To apply Theorem F1, we define \(G:X\to X\), \(H:Y\to Y\), \(\delta :X\to [0,\infty )\), and \(\varphi :[0,\infty )\to [0,\infty )\) by
$$ \Big\| f(x)-(-\gamma ) f\Big(\frac{x}{-\gamma}\Big)\Big\| _{Y}\le \lambda (x,0). $$
By using the same technique as in the preceding theorem, the conclusion follows. □
3.2.2 Stability of γ-additive mappings via Theorem F2
Theorem 3
Suppose that Then there exists a unique γ-additive mapping \(F:X\to Y\) such that Moreover, \(F(x)=\lim _{n\to \infty}\big(\frac{2}{-\gamma}\big)^{n} f\Big( \frac{1}{(\frac{2}{-\gamma})^{n}}x\Big)\) for all \(x\in X\).
1.
\(\Lambda (x):=\sum _{n=1}^{\infty}\big(\frac{2}{|\gamma |}\big)^{n} \lambda \Big(\frac{x}{(\frac{2}{-\gamma})^{n}}, \frac{x}{(\frac{2}{-\gamma})^{n}}\Big)<\infty \) for all \(x\in X\);
2.
\(\lim _{n\to \infty}\big(\frac{2}{|\gamma |}\big)^{n}\lambda \Big( \frac{x}{(\frac{2}{-\gamma})^{n}},\frac{y}{(\frac{2}{-\gamma})^{n}} \Big)=0\) for all \(x,y\in X\).
$$ \|F(x)-f(x)\|_{Y}\leq \frac{1}{2}\Lambda (x)\quad \textit{for all $x\in X$.} $$
Proof
Existence: Let \(x\in X\). Then Letting \(\alpha :=\frac{2}{-\gamma}\), we get that To apply Theorem F2, we define \(G:X\to X\), \(H:Y\to Y\), \(\delta :X\to [0,\infty )\), and \(\psi :[0,\infty )\to [0,\infty )\) by By Theorem F2, there exists a unique mapping \(F:X\to Y\) satisfying the following properties: for all \(x\in X\). Moreover, \(F(x)=\lim _{n\to \infty}(H^{-n}\circ f\circ G^{-n})(x)=\lim _{n\to \infty}\alpha ^{n} f\left (\frac{1}{\alpha ^{n}}x\right )\) for all \(x\in X\).
$$ \Big\| 2f(x)+\gamma f\Big(\Big(\frac{2}{-\gamma}\Big)x\Big)\Big\| _{Y} \leq \lambda (x,x). $$
$$ \Big\| f(x)-\frac{1}{\alpha}f(\alpha x)\Big\| _{Y}\leq \frac{1}{2} \lambda (x,x). $$
This implies that
Note that \(H^{-1}\) is continuous. Moreover, we also have the following assertions:
1.
\(\psi (a+b)=|\alpha |(a+b)=|\alpha |a+|\alpha |b=\psi (a)+\psi (b)\) for all \(a,b\in [0,\infty )\);
2.
\(\|H^{-1}(u)-H^{-1}(v)\|_{Y}=\left \|\alpha u-\alpha v\right \|_{Y}=| \alpha |\|u-v\|_{Y}=\psi (\|u-v\|_{Y})\) for all \(u,v\in Y\);
3.
\(\Psi (x):=\sum _{n=1}^{\infty}\psi ^{n}(\delta (G^{-n}(x)))=\sum _{n=1}^{ \infty}\frac{1}{2}|\alpha |^{n}\lambda \left (\frac{x}{\alpha ^{n}}, \frac{x}{\alpha ^{n}}\right )=\frac{1}{2}\Lambda (x)<\infty \) for all \(x\in X\);
4.
\(\|f(x)-(H\circ f\circ G)(x)\|_{Y}=\|f(x)-\frac{1}{\alpha}f(\alpha x) \|_{Y}\leq \delta (x)\) for all \(x\in X\).
-
\(\frac{1}{\alpha}F\left (\alpha x\right )=(H\circ F\circ G)(x)=F(x)\)
-
\(\|F(x)-f(x)\|_{Y}\leq \Psi (x)=\frac{1}{2}\Lambda (x)\)
We first claim that F is γ-additive. To see this, let \(x,y\in X\). It follows that This implies that It follows from (b) that Hence the claim is proved.
$$ \Big\| f\Big(\frac{x}{\alpha ^{n}}\Big)+f\Big(\frac{y}{\alpha ^{n}} \Big)+\gamma f\bigg( \frac{-\frac{x}{\alpha ^{n}}-\frac{y}{\alpha ^{n}}}{\gamma}\bigg) \Big\| _{Y}\leq \lambda \Big(\frac{x}{\alpha ^{n}}, \frac{y}{\alpha ^{n}}\Big). $$
$$ \Big\| \alpha ^{n}f\Big(\frac{x}{\alpha ^{n}}\Big)+\alpha ^{n}f\Big( \frac{y}{\alpha ^{n}}\Big)+\alpha ^{n}\gamma f\bigg( \frac{-(x+y)}{\alpha ^{n}\gamma}\bigg)\Big\| _{Y} \le \alpha ^{n} \lambda \Big(\frac{x}{\alpha ^{n}},\frac{y}{\alpha ^{n}}\Big). $$
$$ \Big\| F(x)+F(y)+\gamma F\Big(\frac{-x-y}{\gamma}\Big)\Big\| _{Y}=\lim _{n \to \infty}\Big\| \alpha ^{n}f\Big(\frac{x}{\alpha ^{n}}\Big)+\alpha ^{n}f \Big(\frac{y}{\alpha ^{n}}\Big)+\alpha ^{n}\gamma f\bigg( \frac{-(x+y)}{\alpha ^{n}\gamma}\bigg)\Big\| _{Y}=0. $$
Uniqueness: To see this, suppose that there exists a γ-additive mapping \(G:X\to Y\) such that To see that \(G=F\), it suffices to show that \(\alpha G\big(\frac{x}{\alpha}\big)=G(x)\) for all \(x\in X\). In fact, it follows from the γ-additivity and the oddness of G that □
$$ \|G(x)-f(x)\|_{Y}\le \frac{1}{2} \Lambda (x)\quad \text{for all $x\in X$.} $$
$$ \alpha G\Big(\frac{x}{\alpha}\Big)=\frac{2}{-\gamma}G\Big( \frac{-\gamma}{2}x\Big)=\frac{2}{\gamma}G\Big(\frac{\gamma}{2}x\Big)=2G \Big(\frac{x}{2}\Big)=G(x)\quad \text{for all $x\in X$.} $$
Theorem 4
Suppose that Then there exists a unique γ-additive mapping \(F:X\to Y\) such that Moreover, \(F(x)=\lim _{n\to \infty}\big(\frac{1}{-\gamma}\big)^{n}f((-\gamma )^{n}x)\) for all \(x\in X\).
1.
\(\Lambda (x):=\sum _{n=1}^{\infty}\big(\frac{1}{|\gamma |}\big)^{n} \lambda \big((-\gamma )^{n}x,0\big)<\infty \) for all \(x\in X\);
2.
\(\lim _{n\to \infty}\big(\frac{1}{|\gamma |}\big)^{n}\lambda \left ((- \gamma )^{n}x,(-\gamma )^{n}y\right )=0\) for all \(x,y\in X\);
3.
\(\lambda (0,0)=0\) if \(|\gamma |>1\).
$$ \|F(x)-f(x)\|_{Y}\le \Lambda (x)\quad \textit{for all $x\in X$}. $$
Proof
If \(|\gamma |>1\), then \(\|(2+\gamma )f(0)\|_{Y}\le \lambda (0,0)=0\) and hence \(f(0)=0\). On the other hand, if \(|\gamma |\le 1\), then it follows from \(\lim _{n\to \infty}\big(\frac{1}{|\gamma |}\big)^{n}\lambda (0,0)=0\) that \(\lambda (0,0)=0\) and hence \(f(0)=0\).
Let \(x\in X\). Replace y in the first inequality by 0. We obtain To apply Theorem F2, we define \(G:X\to X\), \(H:Y\to Y\), \(\delta :X\to [0,\infty )\), and \(\varphi :[0,\infty )\to [0,\infty )\) by
$$ \Big\| f(x)-(-\gamma ) f\Big(\frac{x}{-\gamma}\Big)\Big\| _{Y}\le \lambda \left (x,0\right ). $$
We can follow the technique as in the preceding theorem and obtain the conclusion of the theorem. □
3.3 Stability results for the functional inequality (A)
By using Theorem 1, we obtain the following result.
Theorem 5
Suppose that \(f:X\to Y\) satisfies (A) with \(\varrho \neq 0\) where Then there exists a unique additive mapping \(F:X\to Y\) such that Moreover, \(F(x)=\lim _{n\to \infty}\frac{1}{4^{n}} f\left (4^{n}x\right )\) for all \(x\in X\).
1.
\(\Omega (x):=\sum _{n=0}^{\infty}\frac{1}{4^{n}}\omega \big(4^{n} x,4^{n} x,4^{n}(-2x)\big)<\infty \) for all \(x\in X\);
2.
\(\lim _{n\to \infty}\frac{1}{4^{n}}\omega \big(4^{n}x,4^{n}y,4^{n}(-x-y) \big)=0\) for all \(x,y\in X\);
3.
\(f(0)=0\).
$$ \|F(x)-f(x)\|_{Y}\le \frac{1}{2} \Omega (x)+\frac{1}{4}\Omega (-2x) \quad \textit{for all $x\in X$}. $$
Proof
Note that \(f(0)=0\). Let \(x,y\in X\). It follows from the inequality (A) that To apply Theorem 1 where \(\gamma :=1\), we define \(\lambda (x,y):=\omega (x,y,-x-y)\). It follows from (a) that To see that \(\lim _{n\to \infty}\frac{1}{2^{n}}\lambda \big((-2)^{n}x,(-2)^{n}y \big)=0\), it follows from (b) that and It follows from Theorem 1 that there exists a unique additive mapping \(F:X\to Y\) such that Moreover, \(F(x)=\lim _{n\to \infty}\frac{1}{(-2)^{n}} f\left ((-2)^{n}x\right )= \lim _{n\to \infty}\frac{1}{4^{n}} f\left (4^{n}x\right )\) for all \(x\in X\). □
$$ \|f(x)+f(y)+f(-x-y)\|_{Y}\leq \bigg\| \varrho f\bigg( \frac{x+y+(-x-y)}{\varrho}\bigg)\bigg\| _{Y}+\omega (x,y,-x-y)=\omega (x,y,-x-y). $$
$$\begin{aligned} \Lambda (x)&:=\sum _{n=0}^{\infty}\frac{1}{2^{n}}\lambda \big((-2)^{n}x,(-2)^{n}x \big) \\ &=\left (\sum _{n\text{-even}}+\sum _{n\text{-odd}}\right ) \frac{1}{2^{n}}\lambda \big((-2)^{n}x,(-2)^{n}x\big) \\ &=\Omega (x)+\frac{1}{2}\Omega (-2x)< \infty \quad \text{for all $x\in X$.} \end{aligned}$$
$$\begin{aligned} \lim _{n\to \infty}\frac{1}{2^{2n}}\lambda \big((-2)^{2n}x,(-2)^{2n}y \big) &=\lim _{n\to \infty}\frac{1}{4^{n}}\lambda \big(4^{n}x,4^{n}y \big) \\ &=\lim _{n\to \infty}\frac{1}{4^{n}}\omega \big(4^{n}x,4^{n}y,4^{n}(-x-y) \big)=0 \end{aligned}$$
$$\begin{aligned} \lim _{n\to \infty}\frac{1}{2^{2n+1}}\lambda \big((-2)^{2n+1}x,(-2)^{2n+1}y \big) &=\frac{1}{2}\lim _{n\to \infty}\frac{1}{4^{n}}\lambda \big(4^{n}(-2x),4^{n}(-2y) \big) \\ &=\frac{1}{2}\lim _{n\to \infty}\frac{1}{4^{n}}\omega \big(4^{n}(-2x),4^{n}(-2y),4^{n}(-(-2x)-(-2y)) \big)\\ &=0. \end{aligned}$$
$$ \|F(x)-f(x)\|_{Y}\le \frac{1}{2}\Lambda (x)=\frac{1}{2}\Omega (x)+ \frac{1}{4}\Omega (-2x)\quad \text{for all $x\in X$}. $$
Set \(\omega (x,y,z):=\theta \ge 0\) for all \(x,y,z\in X\) and the following result is obtained.
Corollary 1
Suppose that \(\theta \ge 0\) and \(f:X\to Y\) satisfies \(f(0)=0\) and Then there exists a unique additive mapping \(F:X\to Y\) such that Moreover, \(F(x)=\lim _{n\to \infty}\frac{1}{4^{n}}f(4^{n}x)\) for all \(x\in X\).
$$ \|f(x)+f(y)+f(z)\|_{Y} \le \Big\| 2f\Big(\frac{x+y+z}{2}\Big)\Big\| _{Y}+ \theta \quad \textit{for all $x,y,z\in X$.} $$
$$ \|F(x)-f(x)\|_{Y}\le \theta \quad \textit{for all $x\in X$}. $$
We now discuss a supplementary of Theorem 5.
Theorem 6
Suppose that \(f:X\to Y\) satisfies (A) with \(0<|\varrho |<3\) where Then there exists a unique additive mapping \(F:X\to Y\) such that Moreover, \(F(x)=\lim _{n\to \infty}4^{n}f\left (\frac{x}{4^{n}}\right )\) for all \(x\in X\).
1.
\(\Omega (x):=\sum _{n=1}^{\infty }4^{n}\omega \Big(\frac{x}{4^{n}}, \frac{x}{4^{n}},\frac{-2x}{4^{n}}\Big)<\infty \) for all \(x\in X\);
2.
\(\lim _{n\to \infty}4^{n}\omega \Big(\frac{x}{4^{n}},\frac{y}{4^{n}}, \frac{-x-y}{4^{n}}\Big)=0\) for all \(x,y\in X\).
$$ \|F(x)-f(x)\|_{Y}\le \frac{1}{2}\Omega (x)+\Omega \Big(-\frac{1}{2}x \Big)\quad \textit{for all $x\in X$}. $$
Proof
Since \(\lim _{n\to \infty}4^{n}\omega (0,0,0)=0\), we have \(\omega (0,0,0)=0\). It follows from the inequality (A) that \(3\|f(0)\|_{Y}\leq \|\varrho f(0)\|_{Y}+\omega (0,0,0)=|\varrho |\|f(0) \|_{Y}\), that is, \(f(0)=0\). Let \(x,y\in X\). We get that To apply Theorem 3 where \(\gamma :=1\), we define \(\lambda (x,y):=\omega (x,y,-x-y)\). It follows from (a) that To see that \(\lim _{n\to \infty}2^{n}\lambda \Big(\frac{x}{(-2)^{n}}, \frac{y}{(-2)^{n}}\Big)=0\), we consider and It follows from Theorem 3 that there exists a unique additive mapping \(F:X\to Y\) such that Moreover, \(F(x)=\lim _{n\to \infty}(-2)^{n}f\left (\frac{1}{(-2)^{n}}x\right )= \lim _{n\to \infty}4^{n}f\left (\frac{1}{4^{n}}x\right )\) for all \(x\in X\). □
$$ \|f(x)+f(y)+f(-x-y)\|_{Y}\leq \Big\| \varrho f\bigg( \frac{x+y+(-x-y)}{\varrho}\bigg)\Big\| _{Y}+\omega (x,y,-x-y)=\omega (x,y,-x-y). $$
$$ \Lambda (x):=\sum _{n=1}^{\infty}2^{n}\lambda \Big(\frac{x}{(-2)^{n}}, \frac{x}{(-2)^{n}}\Big)=\Omega (x)+2\Omega \Big(-\frac{1}{2}x\Big)< \infty . $$
$$\begin{aligned} \lim _{n\to \infty}2^{2n}\lambda \Big(\frac{x}{(-2)^{2n}}, \frac{y}{(-2)^{2n}}\Big) &=\lim _{n\to \infty}4^{n}\omega \Big( \frac{x}{4^{n}},\frac{y}{4^{n}},\frac{-x-y}{4^{n}}\Big)=0 \end{aligned}$$
$$\begin{aligned} &\lim _{n\to \infty}2^{2n+1}\lambda \Big(\frac{x}{(-2)^{2n+1}}, \frac{y}{(-2)^{2n+1}}\Big) \\ &\quad =2\lim _{n\to \infty}4^{n}\omega \Big( \frac{1}{4^{n}}\Big(-\frac{1}{2}x\Big),\frac{1}{4^{n}}\Big(- \frac{1}{2}y\Big),\frac{1}{4^{n}}\Big(\frac{1}{2}x+\frac{1}{2}y\Big) \Big) =0. \end{aligned}$$
$$ \|F(x)-f(x)\|_{Y}\le \frac{1}{2} \Lambda (x)=\frac{1}{2}\Omega (x)+ \Omega \Big(-\frac{1}{2}x\Big)\quad \text{for all $x\in X$}. $$
Remark 1
Theorems 5 and 6 improve several results in the literature as follows.
1.
2.
3.
Our Theorems 5 and 6 with \(\omega (x,y,z):=\theta (\|x\|_{X}^{p}+\|y\|_{X}^{p}+\|z\|_{X}^{p})\) where \(\theta \ge 0\) and \(0< p\neq 1\) provide the same bounds obtained in [11, Theorems 2.2, 2.3, 3.2, and 3.3] without assuming that f is odd. Example 3 (in the Appendix) provides two functions such that \(f_{1}\) (\(f_{2}\), respectively) satisfies the inequality (A) with \(p:=2\) (\(p:=1/2\), respectively) but they are not odd. Hence our result gives a significant improvement of [11]. Example 4 (in the Appendix) shows the failure of the stability where \(p:=1\).
4.
Our bound in Theorem 6 gives a better approximation for \(\|F(x)-f(x)\|_{Y}+\|F(-x)-f(-x)\|_{Y}\) than the one in Theorem LuP1 as shown in Table 3. Moreover, our assumption in Theorem 6 is weaker than the one in Theorem LuP1.
Bound of \(\|F(x)-f(x)\|_{Y}+\|F(-x)-f(-x)\|_{Y}\) | |
|---|---|
Theorem LuP1 | \(\begin{array}{@{}l@{}} \displaystyle \frac{1}{2}\sum _{n=1}^{\infty }2^{n} \omega \Big(\frac{-x}{2^{n}},\frac{-x}{2^{n}},\frac{2x}{2^{n}}\Big)+\sum _{n=1}^{\infty }2^{n} \omega \Big(\frac{x}{2^{n}},\frac{-x}{2^{n}},\frac{0}{2^{n}}\Big)\\ \displaystyle +\frac{1}{2}\sum _{n=1}^{\infty }2^{n} \omega \Big(\frac{x}{2^{n}},\frac{x}{2^{n}},\frac{-2x}{2^{n}}\Big)+\sum _{n=1}^{\infty }2^{n} \omega \Big(\frac{-x}{2^{n}},\frac{x}{2^{n}},\frac{0}{2^{n}}\Big) \end{array}\) |
Our Theorem 6 | \(\displaystyle \frac{1}{2}\sum _{n=1}^{\infty }2^{n}\omega \Big(\frac{-x}{2^{n}},\frac{-x}{2^{n}},\frac{2x}{2^{n}}\Big)+\frac{1}{2}\sum _{n=1}^{\infty }2^{n}\omega \Big(\frac{x}{2^{n}},\frac{x}{2^{n}},\frac{-2x}{2^{n}}\Big)\) |
3.4 Stability results for the functional inequality (B)
Theorem 7
Suppose that \(f:X\to Y\) satisfies (B) with \(\varrho \notin \{-2,0\}\) where Then there exists a unique ϱ-additive mapping \(F:X\to Y\) such that Moreover, \(F(x)=\lim _{n\to \infty}\frac{1}{(\frac{-2}{\varrho})^{n}} f\big( \big(\frac{-2}{\varrho}\big)^{n}x\big)\) for all \(x\in X\).
1.
\(\Omega (x):=\sum _{n=0}^{\infty}\frac{1}{(\frac{2}{|\varrho |})^{n}} \omega \big(\big(\frac{2}{-\varrho}\big)^{n} x,\big( \frac{2}{-\varrho}\big)^{n} x,\big(\frac{2}{-\varrho}\big)^{n+1}x \big)<\infty \) for all \(x\in X\);
2.
\(\lim _{n\to \infty}\frac{1}{(\frac{2}{|\varrho |})^{n}}\omega \big( \big(\frac{2}{-\varrho}\big)^{n} x,\big(\frac{2}{-\varrho}\big)^{n} y, \big(\frac{2}{-\varrho}\big)^{n}\big(\frac{-x-y}{\varrho}\big)\big)=0\) for all \(x,y\in X\);
3.
\(\omega (0,0,0)=0\) if \(|\varrho |<2\).
$$ \|F(x)-f(x)\|_{Y}\le \frac{1}{2} \Omega (x)\quad \textit{for all $x\in X$}. $$
Proof
First, we show that \(f(0)=0\). If \(|\varrho |<2\), then \(\|(2+\varrho )f(0)\|_{Y}\le \|\varrho f(0)\|_{Y}+\omega (0,0,0)=\| \varrho f(0)\|_{Y}\) and hence \(f(0)=0\). On the other hand, if \(|\varrho |\ge 2\), then it follows from \(\lim _{n\to \infty}\frac{1}{(\frac{2}{|\varrho |})^{n}}\omega (0,0,0)=0\) that \(\omega (0,0,0)=0\) and hence \(f(0)=0\). Let \(x,y\in X\). We get that To apply Theorem 1 where \(\gamma :=\varrho \), we define \(\lambda (x,y):=\omega \left (x,y,\frac{-x-y}{\varrho}\right )\). This implies that Then there exists a unique ϱ-additive mapping \(F:X\to Y\) such that \(\|F(x)-f(x)\|_{Y}\le \frac{1}{2} \Omega (x)\) for all \(x\in X\). Moreover, \(F(x)=\lim _{n\to \infty}\frac{1}{(\frac{-2}{\varrho})^{n}} f\Big( \big(\frac{-2}{\varrho}\big)^{n}x\Big)\) for all \(x\in X\). □
$$\begin{aligned} \Big\| f(x)+f(y)+\varrho f\Big(\frac{-x-y}{\varrho}\Big)\Big\| _{Y} &\leq \Big\| \varrho f\Big(\frac{x+y}{\varrho}+\frac{-x-y}{\varrho} \Big)\Big\| _{Y}+\omega \Big(x,y,\frac{-x-y}{\varrho}\Big)\\ &=\omega \Big(x,y,\frac{-x-y}{\varrho}\Big). \end{aligned}$$
-
\(\Lambda (x):=\sum _{n=0}^{\infty} \frac{1}{(\frac{2}{|\varrho |})^{n}}\lambda \Big(\big( \frac{2}{-\varrho}\big)^{n} x,\big(\frac{2}{-\varrho}\big)^{n} x\Big)= \Omega (x)<\infty \) for all \(x\in X\);
-
\(\lim _{n\to \infty}\frac{1}{(\frac{2}{|\varrho |})^{n}}\lambda \Big( \big(\frac{2}{-\varrho}\big)^{n} x,\big(\frac{2}{-\varrho}\big)^{n} y \Big)=\lim _{n\to \infty}\frac{1}{(\frac{2}{|\varrho |})^{n}}\omega \Big(\big(\frac{2}{-\varrho}\big)^{n} x,\big(\frac{2}{-\varrho}\big)^{n} y,\big(\frac{2}{-\varrho}\big)^{n}\big(\frac{-x-y}{\varrho}\big)\Big)=0\) for all \(x,y\in X\).
Theorem 8
Suppose that \(f:X\to Y\) satisfies (B) with \(\varrho \notin \{-2,0\}\) where Then there exists a unique ϱ-additive mapping \(F:X\to Y\) such that Moreover, \(F(x)=\lim _{n\to \infty}(-\varrho )^{n}f\Big( \frac{x}{(-\varrho )^{n}}\Big)\) for all \(x\in X\).
1.
\(\Omega (x):=\sum _{n=0}^{\infty}|\varrho |^{n}\omega \Big( \frac{x}{(-\varrho )^{n}},0,\frac{x}{(-\varrho )^{n+1}}\Big)<\infty \) for all \(x\in X\);
2.
\(\lim _{n\to \infty}|\varrho |^{n}\omega \Big( \frac{x}{(-\varrho )^{n}},\frac{y}{(-\varrho )^{n}}, \frac{x+y}{(-\varrho )^{n+1}}\Big)=0\) for all \(x,y\in X\);
3.
\(\omega (0,0,0)=0\) if \(|\varrho |<1\).
$$ \|F(x)-f(x)\|_{Y}\le \Omega (x)\quad \textit{for all $x\in X$}. $$
Proof
Note that \(f(0)=0\) and hence the conclusion follows from Theorem 2. □
Theorem 9
Suppose that \(f:X\to Y\) satisfies (B) with \(\varrho \notin \{-2,0\}\) where Then there exists a unique ϱ-additive mapping \(F:X\to Y\) such that Moreover, \(F(x)=\lim _{n\to \infty}\big(\frac{-2}{\varrho}\big)^{n} f\Big( \frac{1}{(\frac{-2}{\varrho})^{n}}x\Big)\) for all \(x\in X\).
1.
\(\Omega (x):=\sum _{n=1}^{\infty}\big(\frac{2}{|\varrho |}\big)^{n} \omega \Big(\frac{x}{(-\frac{2}{\varrho})^{n}}, \frac{x}{(-\frac{2}{\varrho})^{n}}, \frac{x}{(-\frac{2}{\varrho})^{n-1}}\Big)<\infty \) for all \(x\in X\);
2.
\(\lim _{n\to \infty}\big(\frac{2}{|\varrho |}\big)^{n}\omega \Big( \frac{x}{(-\frac{2}{\varrho})^{n}},\frac{y}{(-\frac{2}{\varrho})^{n}}, \frac{\frac{-x-y}{\varrho}}{(-\frac{2}{\varrho})^{n}}\Big)=0\) for all \(x,y\in X\);
3.
\(\omega (0,0,0)=0\) if \(|\varrho |>2\).
$$ \|F(x)-f(x)\|_{Y}\le \frac{1}{2} \Omega (x)\quad \textit{for all $x\in X$}. $$
Proof
Note that \(f(0)=0\) and hence the conclusion follows from Theorem 3. □
Theorem 10
Suppose that \(f:X\to Y\) satisfies (B) where Then there exists a unique ϱ-additive mapping \(F:X\to Y\) such that Moreover, \(F(x)=\lim _{n\to \infty}\big(\frac{1}{-\varrho}\big)^{n}f((-\varrho )^{n}x)\) for all \(x\in X\).
1.
\(\Omega (x):=\sum _{n=1}^{\infty}\big(\frac{1}{|\varrho |}\big)^{n} \omega \left ((-\varrho )^{n}x,0,(-\varrho )^{n-1}x\right )<\infty \) for all \(x\in X\);
2.
\(\lim _{n\to \infty}\big(\frac{1}{|\varrho |}\big)^{n}\omega \left ((- \varrho )^{n}x,(-\varrho )^{n}y,(-\varrho )^{n-1}(x+y)\right )=0\) for all \(x,y\in X\);
3.
\(\omega (0,0,0)=0\) if \(|\varrho |>1\).
$$ \|F(x)-f(x)\|_{Y}\le \Omega (x)\quad \textit{for all $x\in X$}. $$
Proof
Note that \(f(0)=0\) and hence the conclusion follows from Theorem 4. □
Remark 2
Theorems 7, 8, 9, and 10 improve several results in the literature as follows.
1.
2.
3.
Our bound in Theorem 9 gives a better approximation for \(\|F(x)-f(x)\|_{Y}+\|F(-x)-f(-x)\|_{Y}\) than the one in Theorem LuP2(a) as shown in Table 4.
Bound of \(\|F(x)-f(x)\|_{Y}+\|F(-x)-f(-x)\|_{Y}\) | |
|---|---|
Theorem LuP2(a) | \(\begin{array}{@{}l@{}} \displaystyle \frac{1}{2}\sum _{n=1}^{\infty}\bigg(\frac{2}{|\varrho |}\bigg)^{n}\omega \bigg(\frac{-x}{(\frac{2}{\varrho})^{n}},\frac{-x}{(\frac{2}{\varrho})^{n}},\frac{\frac{2}{\varrho}x}{(\frac{2}{\varrho})^{n}}\bigg)+\sum _{n=1}^{\infty}\bigg(\frac{2}{|\varrho |}\bigg)^{n}\omega \bigg(\frac{x}{(\frac{2}{\varrho})^{n}},\frac{-x}{(\frac{2}{\varrho})^{n}},0\bigg)\\ \displaystyle +\frac{1}{2}\sum _{n=1}^{\infty}\bigg(\frac{2}{|\varrho |}\bigg)^{n}\omega \bigg(\frac{x}{(\frac{2}{\varrho})^{n}},\frac{x}{(\frac{2}{\varrho})^{n}},\frac{-\frac{2}{\rho}x}{(\frac{2}{\varrho})^{n}}\bigg)+\sum _{n=1}^{\infty}\bigg(\frac{2}{|\varrho |}\bigg)^{n}\omega \bigg(\frac{-x}{(\frac{2}{\varrho})^{n}},\frac{x}{(\frac{2}{\varrho})^{n}},0\bigg) \end{array}\) |
Our Theorem 9 | \(\displaystyle \frac{1}{2}\sum _{n=1}^{\infty}\bigg(\frac{2}{|\varrho |}\bigg)^{n}\omega \bigg(\frac{-x}{(\frac{2}{\varrho})^{n}},\frac{-x}{(\frac{2}{\varrho})^{n}},\frac{\frac{2}{\varrho}x}{(\frac{2}{\varrho})^{n}}\bigg)+\frac{1}{2}\sum _{n=1}^{\infty}\bigg(\frac{2}{|\varrho |}\bigg)^{n}\omega \bigg(\frac{x}{(\frac{2}{\varrho})^{n}},\frac{x}{(\frac{2}{\varrho})^{n}},\frac{-\frac{2}{\varrho}x}{(\frac{2}{\varrho})^{n}}\bigg)\) |
4.
Our bound in Theorem 7 gives a better approximation for \(\|F(x)-f(x)\|_{Y}+\|F(-x)-f(-x)\|_{Y}\) than the one in Theorem LuP2(b) as shown in Table 5.
Bound of \(\|F(x)-f(x)\|_{Y}+\|F(-x)-f(-x)\|_{Y}\) | |
|---|---|
Theorem LuP2(b) | \(\begin{array}{@{}l@{}} \displaystyle \frac{1}{2}\sum _{n=0}^{\infty}\frac{1}{(\frac{2}{|\varrho |})^{n}}\omega \Big(\Big(\frac{2}{\varrho}\Big)^{n} x,\Big(\frac{2}{\varrho}\Big)^{n} x,\Big(\frac{2}{\varrho}\Big)^{n}\Big(-\frac{2}{\varrho}x\Big)\Big)\\ \displaystyle +\sum _{n=0}^{\infty}\frac{1}{(\frac{2}{|\rho |})^{n}}\omega \Big(\Big(\frac{2}{\varrho}\Big)^{n} \Big(\frac{2}{\varrho}x\Big),\Big(\frac{2}{\varrho}\Big)^{n} \Big(-\frac{2}{\varrho}x\Big),0\Big)\\ \displaystyle +\frac{1}{2}\sum _{n=0}^{\infty}\frac{1}{(\frac{2}{|\varrho |})^{n}}\omega \Big(\Big(\frac{2}{\varrho}\Big)^{n} (-x),\Big(\frac{2}{\varrho}\Big)^{n} (-x),\Big(\frac{2}{\varrho}\Big)^{n}\Big(\frac{2}{\varrho}x\Big)\Big)\\ \displaystyle +\sum _{n=0}^{\infty}\frac{1}{(\frac{2}{|\varrho |})^{n}}\omega \Big(\Big(\frac{2}{\varrho}\Big)^{n} \Big(-\frac{2}{\varrho}x\Big),\Big(\frac{2}{\varrho}\Big)^{n} \Big(\frac{2}{\varrho}x\Big),0\Big) \end{array}\) |
Our Theorem 7 | \(\begin{array}{@{}l@{}} \displaystyle \frac{1}{2}\sum _{n=0}^{\infty}\frac{1}{(\frac{2}{|\varrho |})^{n}}\omega \Big(\Big(\frac{2}{\varrho}\Big)^{n} x,\Big(\frac{2}{\varrho}\Big)^{n} x,\Big(\frac{2}{\varrho}\Big)^{n}\Big(-\frac{2}{\varrho}x\Big)\Big)\\ \displaystyle +\frac{1}{2}\sum _{n=0}^{\infty}\frac{1}{(\frac{2}{|\varrho |})^{n}}\omega \Big(\Big(\frac{2}{\varrho}\Big)^{n} (-x),\Big(\frac{2}{\varrho}\Big)^{n} (-x),\Big(\frac{2}{\varrho}\Big)^{n}\Big(\frac{2}{\varrho}x\Big)\Big) \end{array}\) |
3.5 Hyperstability result of the functional inequalities (A) and (B)
3.5.1 Hyperstability result of the functional inequality (A)
We now discuss the hyperstability result for the functional inequality (A). We start with some observations.
Proposition 1
Suppose that X and Y are vector spaces. If \(f:X\to Y\) satisfies then f is odd.
$$ f(x)+f(y)+f(-x-y)=0\quad \textit{ for all $x,y\in X\setminus \{0\}$ with $x+y\ne 0$}, $$
Proof
Let \(x\in X\setminus \{0\}\). Note that \(2f(x)+f(-2x)=0\). That is, \(f(-2x)=-2f(x)\). This implies that It follows that and Then \(f(x)+4f(-x)=f(-x)-2f(x)\). We get that \(f(-x)=-f(x)\). □
$$ f(4x)=f(-2(-2x))=-2f(-2x)=4f(x). $$
$$ f(3x)+f(x)+4f(-x)=f(3x)+f(x)+f(-4x)=f(3x)+f(x)+f(-3x-x)=0 $$
$$ f(3x)+f(-x)-2f(x)=f(3x)+f(-x)+f(-2x)=f(3x)+f(-x)+f(-3x-(-x))=0. $$
Proposition 2
Suppose that X and Y are vector spaces and let \(f:X\to Y\). Then the following two statements are equivalent:
1.
\(f(x)+f(y)+f(-x-y)=0\) for all \(x,y\in X\setminus \{0\}\) with \(x+y\ne 0\) and \(f(0)=0\);
2.
f is additive.
Proof
(b) ⇒ (a) is obvious. We prove (a) ⇒ (b). Suppose that (a) holds. To prove (b), let \(x,y\in X\). Note that \(f(2x)=2f(x)\). By induction hypothesis, \(f(nx)=nf(x)\) for all . It follows that This implies that \(f(-x)=-f(x)\). Let \(x,y\in X\). We consider the following two cases.
□
$$ 3f(x)=f(3x)=f(4x)+f(-x)=4f(x)+f(-x). $$
Case 1
\(x=0\) or \(y=0\). Without loss of generality, we assume that \(x=0\). We get that
$$ f(x+y)=f(y)=f(0)+f(y)=f(x)+f(y). $$
Case 2
\(x+y=0\). We get that
$$ f(x+y)=f(0)=0=f(x)+f(-x)=f(x)+f(y). $$
Theorem 11
Suppose that X is a vector space and \(Y:=(Y,\|\cdot \|_{Y})\) is a normed space. If \(f:X\to Y\) satisfies then \(f(x)+f(y)-f(x+y)=0\) for all \(x,y\in X\setminus \{0\}\) with \(x+y\ne 0\).
1.
\(\lim _{n\to \infty}\|f(x)+f(nx)+f(-(n+1)x)\|_{Y}=0\) for all \(x\in X\setminus \{0\}\);
2.
\(\lim _{n\to \infty}\|f(nx)+f(ny)+f(-n(x+y))\|_{Y}=0\) for all \(x,y\in X\setminus \{0\}\) with \(x+y\neq 0\),
Proof
Let \(x,y\in X\setminus \{0\}\) be such that \(x+y\ne 0\). Note that In fact, \(\alpha _{1}\), \(\alpha _{2}\), \(\alpha _{3}\) follows from (a) while \(\beta _{1}\), \(\beta _{2}\) from (b). It follows then that Hence \(f(x)+f(y)+f(-x-y)=0\) for all \(x,y\in X\setminus \{0\}\) with \(x+y\ne 0\). It follows from Proposition 1 that f is odd and hence the assertion holds. □
$$\begin{aligned} \alpha _{1}&:=\lim _{n\to \infty}\|f(x)+f(nx)+f(-(n+1)x)\|_{Y}=0 \\ \alpha _{2}&:=\lim _{n\to \infty}\|f(y)+f(ny)+f(-(n+1)y)\|_{Y}=0 \\ \alpha _{3}&:=\lim _{n\to \infty}\|f(-x-y)+f(n(-x-y))+f(-(n+1)(-x-y)) \|_{Y}=0 \\ \beta _{1}&:=\lim _{n\to \infty}\|-f(nx)-f(ny)-f(n(-x-y))\|_{Y}=0 \\ \beta _{2}&:=\lim _{n\to \infty}\|-f(-(n+1)x)-f(-(n+1)y)-f(-(n+1)(-x-y)) \|_{Y}=0. \end{aligned}$$
$$\begin{aligned} \lim _{n\to \infty}\|f(x)+f(y)+f(-x-y)\|_{Y} &\le \alpha _{1}+\alpha _{2}+ \alpha _{3}+\beta _{1}+\beta _{2}=0. \end{aligned}$$
3.5.2 Hyperstability result of the functional inequality (B)
We now discuss the hyperstability result of the functional inequality (B). Suppose that X is a vector space and \(Y:=(Y,\|\cdot \|_{Y})\) is a normed space.
Theorem 12
Suppose that a mapping \(f:X\to Y\) satisfies (B) with \(\varrho \ne 0\) where Then f is ϱ-additive.
1.
\(\omega \big(x,0,\frac{-x}{\varrho}\big)=0\) for all \(x\in X\);
2.
\(\lim _{n\to \infty}|\varrho |^{n}\omega \big( \frac{x}{(-\varrho )^{n}},\frac{y}{(-\varrho )^{n}}, \frac{x+y}{(-\varrho )^{n+1}}\big)=0\) for all \(x,y\in X\);
3.
\(\omega (0,0,0)=0\) if \(|\varrho |<1\).
Proof
We may assume that Y is complete. Otherwise, we replace Y with its completion. Note that \(f(0)=0\). Let \(x\in X\). It follows from (a) that This implies that \(f(x)=-\varrho f\big(\frac{x}{-\varrho}\big)\). By induction, \(f(x)=(-\varrho )^{n} f\big(\frac{x}{(-\varrho )^{n}}\big)\) for all \(n\ge 0\). By Theorem 8, \(F:X\to Y\) defined by \(F(x):=\lim _{n\to \infty}(-\varrho )^{n}f\big( \frac{x}{(-\varrho )^{n}}\big)=f(x)\) is a ϱ-additive mapping, as desired. □
$$\begin{aligned} \Big\| f(x)+\varrho f\Big(\frac{-x}{\varrho}\Big)\Big\| _{Y} &=\Big\| f(x)+f(0)+ \varrho f\Big(\frac{-x}{\varrho}\Big)\Big\| _{Y}\le \Big\| \varrho f \Big(\frac{x+0}{\varrho}-\frac{x}{\varrho}\Big)\Big\| _{Y}+\omega \Big(x,0,\frac{-x}{\varrho}\Big)\\ &=0. \end{aligned}$$
Theorem 13
Suppose that a mapping \(f:X\to Y\) satisfies (B) with \(\varrho \neq 0\) where Then f is ϱ-additive.
1.
\(\omega \big(x,0,\frac{-x}{\varrho}\big)=0\) for all \(x\in X\);
2.
\(\lim _{n\to \infty}\big(\frac{1}{|\varrho |}\big)^{n}\omega (- \varrho )^{n}x,(-\varrho )^{n}y,(-\varrho )^{n+1}(x+y))=0\) for all \(x,y\in X\);
3.
\(\omega (0,0,0)=0\) if \(|\varrho |>1\).
Proof
We can follow the proof of the preceding theorem to obtain the conclusion. □
Acknowledgements
The authors would like to thank Mr. Phemmatad Tansoontorn for the valuable discussion on Example 1. They are also grateful to the two referees for their helpful suggestions and insightful comments.
Declarations
Competing interests
The authors declare no competing interests.
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