Abstract
In this paper we give some hyperstability and stability results for the Cauchy and Jensen functional equations on restricted domains. We provide a simple and short proof for Brzdȩk’s result concerning a hyperstability result for the Cauchy equation.
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1 Introduction
Let V and W be linear spaces. A function \(f:V\to W\) is called
-
additive if \(f(x+y)=f(x)+f(y)\) for all \(x,y\in V\);
-
Jensen if \(2f (\frac{x+y}{2} )=f(x)+f(y)\) for all \(x,y\in V\).
The main motivation for the investigation of the stability of functional equations originated from a question of Ulam [21] concerning the stability of group homomorphisms. Hyers [9] gave an affirmative answer to the question of Ulam. The stability and hyperstability problems for various functional equations have been investigated by numerous mathematicians. For more information on this area of research and further references, see [1, 2, 4, 7, 10, 11, 13–15, 20].
Let us state the following theorem that is one of the classical results concerning the stability problem for the Cauchy functional equation \(f(x+y)=f(x)+f(y)\).
Theorem 1.1
Let \(\varepsilon \geqslant 0\) and \(f:X\to Y\), where X is a normed space and Y is a Banach space. Let \(p\neq 1\) be a real number and
Then, there exists a unique additive function \(A:X\to Y\) such that
Rassias [16, 17] considered the case \(\|f(x+y)-f(x)-f(y)\|\leqslant \varepsilon \|x\|^{p}\|y\|^{q}\), where p, q are real numbers with \(p+q\in [0,1)\). Brzdȩk [6, Theprem 1.3] provided a complement for this result in the case \(p+q<0\). His proof was based on a fixed-point theorem. We provide a simple and short proof for Brzdȩk’s result. In addition, some further results on the hyperstability of the Cauchy and Jensen functional equations are investigated.
2 Superstability
Denote by the set of positive integers. A version of the following theorem is introduced by Brzdȩk [6, Theorem 1.3] and its proof is based on a fixed-point theorem. A simple and brief proof is given here.
Theorem 2.1
Let \(\mathcal{X}\) and \(\mathcal{Y}\) be normed spaces, and \(E\subseteq \mathcal{X}\setminus \{0\}\) be a nonempty set. Take \(\varepsilon \geqslant 0\) and let p, q be real numbers with \(p+q<0\). Assume that for each \(x\in E\) there exists a positive integer \(m_{x}\) such that \(nx\in E\) for all with \(n\geqslant m_{x}\). Then, every function \(f:\mathcal{X}\to \mathcal{Y}\) satisfying the inequality
is additive on E, that is
Proof
Without loss of generality, we may assume that \(q<0\). Let \(x, y\in E\) with \(x+y\in E\). By assumption, there exists a positive integer m such that \(nx, ny, n(x+y)\in E\) for all \(n\geqslant m\). Then, (2.1) yields
Letting \(n\to \infty \) in the above inequalities, we obtain
Then,
Hence, \(f(x+y)=f(x)+f(y)\) for all \(x,y\in E\) with \(x+y\in E\). This completes the proof. □
Remark 2.2
The assumption \(p+q<0\) is necessary in Theorem 2.1. For example, the function given by \(f(x)=x^{2}\) fulfilling \(|f(x+y)-f(x)-f(y)|=2|x||y|\) for all . However, f is not additive.
The following theorem states a hyperstability result for the Jensen functional equation on a restricted domain.
Theorem 2.3
Let \(\mathcal{X}\) and \(\mathcal{Y}\) be normed spaces, and \(E\subseteq \mathcal{X}\setminus \{0\}\) be a nonempty set. Take \(\varepsilon \geqslant 0\) and let p, q be real numbers with \(p+q<0\). Assume that for each \(x\in E\) there exists a positive integer \(m_{x}\) such that \(\frac{nx}{2}\in E\) for all with \(n\geqslant m_{x}\). Then, every function \(f:\mathcal{X}\to \mathcal{Y}\) satisfying the inequality
is Jensen on E, that is
Proof
Without loss of generality, we may assume that \(q<0\). Let \(x, y\in E\) with \(\frac{x+y}{2}\in E\). By assumption, there exists a positive integer m such that \(\{\frac{nx}{2}, \frac{nx}{2}, \frac{n(x+y)}{4}\}\subseteq E\) for all \(n\geqslant m\). Then, (2.2) yields
Letting \(n\to \infty \) in the above inequalities, we obtain
Then,
Therefore, \(2f (\frac{x+y}{2} )=f(x)+f(y)\) for all \(x,y\in E\) with \(\frac{x+y}{2}\in E\). This ends the proof. □
Example 2.4
Let \(E=[1,+\infty )\) and define by \(f(x)=x^{2}\). It is easy to see that
Then, f satisfies (2.2) with \(p+q>0\). However, f is not Jensen on E.
In the following, we obtain other hyperstability results for the Cauchy and Jensen functional equations.
Theorem 2.5
Let \(\mathcal{X}\) and \(\mathcal{Y}\) be normed spaces, and \(E\subseteq \mathcal{X}\setminus \{0\}\) be a nonempty set. Take \(\theta , \varepsilon \geqslant 0\) and let p, q, r be real numbers with \(p+q+r<0\) and \(p+q+2r<0\). Assume that for each \(x\in E\) there exists a positive integer \(m_{x}\) such that \(nx\in E\) for all with \(n\geqslant m_{x}\). Then, every function \(f:\mathcal{X}\to \mathcal{Y}\) satisfying the inequality
is additive on E.
Proof
Put
Since \(p+q+2r<0\), we may assume that \(q+r<0\) without loss of generality. Let \(x, y\in E\) with \(x+y\in E\). By assumption, there exists a positive integer m such that \(nx, ny, n(x+y)\in E\) for all \(n\geqslant m\). By a similar argument as in the proof of Theorem 2.1, we obtain
Then,
Hence, \(f(x+y)=f(x)+f(y)\) for all \(x,y\in E\) with \(x+y\in E\). This completes the proof. □
Example 2.6
Let \(E=[1,+\infty )\) and f be a function defined by \(f(x)=x^{3}\). It is clear that
Then, f satisfies (2.3) with \(p=q=r=1\). However, f is not additive on E.
Theorem 2.7
Let \(\mathcal{X}\) and \(\mathcal{Y}\) be normed spaces, and \(E\subseteq \mathcal{X}\setminus \{0\}\) be a nonempty set. Take \(\theta , \varepsilon \geqslant 0\) and let p, q, r be real numbers with \(p+q+r<0\) and \(p+q+2r<0\). Assume that for each \(x\in E\) there exists a positive integer \(m_{x}\) such that \(\frac{nx}{2}\in E\) for all with \(n\geqslant m_{x}\). Suppose that a function \(f:\mathcal{X}\to \mathcal{Y}\) satisfies the inequality
for all \(x,y\in E\) with \(\frac{x+y}{2}\in E\). Then, f is Jensen on E.
Proof
Put
Without loss of generality we may assume that \(q+r<0\). Let \(x, y\in E\) with \(\frac{x+y}{2}\in E\). By assumption, there exists a positive integer m such that \(\{\frac{nx}{2}, \frac{ny}{2}, \frac{n(x+y)}{4}\}\subseteq E\) for all \(n\geqslant m\). By a similar argument as in the proof of Theorem 2.3, we obtain
Then,
Therefore, \(2f (\frac{x+y}{2} )=f(x)+f(y)\) for all \(x,y\in E\) with \(\frac{x+y}{2}\in E\). This completes the proof. □
Example 2.8
Let \(E=[1,+\infty )\) and f be a function defined by \(f(x)=x^{2}\). It is clear that
Then, f satisfies (2.4) with \(p=q=2\) and \(r=1\). However, f is not Jensen on E.
Theorem 2.9
Assume that X is a linear space over the field , and \(\mathcal{Y}\) is a normed space over the field . Let and \(\varphi : X\times X \rightarrow [0,+\infty )\) be a function such that
for all \(x,y\in X\setminus \{0\}\). Let , \(C\in \mathcal{Y}\) and \(f:X\to \mathcal{Y}\) satisfy
for all \(x,y\in E_{d}=\{z\in X: \|z\|\geqslant d\}\) for some \(d>0\). Then, f satisfies
for all \(x,y\in X\). Moreover,
for all \(x\in X\).
Proof
Replacing x by \(a^{-1}(m+1)x\) and y by \(-b^{-1}mx\) in (2.6), we obtain
for all \(x\in X\setminus \{0\}\) and positive integers \(m\geqslant n\), where \(a^{-1}(n+1)x, b^{-1}nx\in E_{d}\). Letting \(m\to \infty \) in (2.9) and using (2.5), we obtain
Let \(x\in X\setminus \{0\}\), then (2.5) and (2.10) yield
Hence, we obtain
If we replace x by \(bmx\) and y by \(-amx\) in (2.6), we obtain
for all \(x\in X\setminus \{0\}\) and positive integers \(m\geqslant n\), where \(anx, bnx\in E_{d}\). Therefore,
for all \(x\in X\setminus \{0\}\). Replacing x by \(bmx\) in (2.11) and letting \(m\to \infty \), we obtain from (2.13) that
Therefore, (2.7) holds true for \(x=y=0\), and (2.10) holds for all \(x\in X\). To prove (2.7), let \(x,y\in X\) with \((x,y)\neq (0,0)\). Then,
Therefore, f satisfies (2.7) for all \(x,y\in X\). □
In the following corollaries \(\mathcal{X}\) and \(\mathcal{Y}\) are normed spaces.
Corollary 2.10
Let , , \(C\in Y\) and let \(f:X\to Y\) be a function. Take \(\theta , \varepsilon \geqslant 0\) and let p, q, r be real numbers. Then, f satisfies
if one of the following conditions holds:
- \((i)\):
-
\(p+q+r<0\) and
$$ \bigl\Vert f(ax+by)-Af(x)-Bf(y)-C \bigr\Vert \leqslant \Vert x \Vert ^{p} \Vert y \Vert ^{q}\bigl(\varepsilon \Vert x+y \Vert ^{r}+\theta \Vert x-y \Vert ^{r}\bigr);$$ - \((\mathit{ii})\):
-
\(p+r<0\), \(q+r<0\) and
$$ \bigl\Vert f(ax+by)-Af(x)-Bf(y)-C \bigr\Vert \leqslant \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{q}\bigr) \bigl( \varepsilon \Vert x+y \Vert ^{r}+\theta \Vert x-y \Vert ^{r} \bigr);$$ - \((\mathit{iii})\):
-
\(p, q<0\) and
$$ \bigl\Vert f(ax+by)-Af(x)-Bf(y)-C \bigr\Vert \leqslant \varepsilon \Vert x \Vert ^{p}+\theta \Vert y \Vert ^{q};$$
for all \(x,y\in E_{d}=\{z\in X: \|z\|\geqslant d\}\) for some \(d>0\).
Corollary 2.11
Every function \(f:X\to Y\) satisfies one of the following assertions:
- \((i)\):
-
\(f(ax+by)=Af(x)+Bf(y)+C\), \(x,y\in X\).
- \((\mathit{ii})\):
-
\(\limsup_{\min \{\|x\|,\|y\|\}\to \infty}\|f(ax+by)-Af(x)-Bf(y)-C \|\|x\|^{r}\|y\|^{s}=+\infty \) for all real numbers r, s with \(r+s>0\).
Corollary 2.12
Every function \(f:X\to Y\) satisfies one of the following assertions:
- \((i)\):
-
\(f(ax+by)=Af(x)+Bf(y)+C\), \(x,y\in X\).
- \((\mathit{ii})\):
-
\(\limsup_{\min \{\|x\|,\|y\|\}\to \infty} \frac{\|x\|^{r}\|y\|^{s}}{\|x\|^{r}+\|y\|^{s}}\|f(ax+by)-Af(x)-Bf(y)-C \|=+\infty \) for all real nonnegative numbers r, s.
3 Stability on restricted domains
Jung [12] proved the stability of Jensen’s functional equation on a restricted and unbounded domain. In the following theorem, we improve the bound and thus the result of Jung [12] by obtaining sharper estimates.
Theorem 3.1
Let \(\mathcal{X}\) be a normed space and \(\mathcal{Y}\) a Banach space. Take \(\varepsilon \geqslant 0\) and let a function \(f:\mathcal{X}\to \mathcal{Y}\) satisfy the inequality
for all \(x,y\in E_{d}=\{z\in X: \|z\|\geqslant d\}\) for some \(d>0\). Then, there exists a unique additive function \(T:\mathcal{X}\to \mathcal{Y}\) such that
Proof
Letting \(y=-x\) in (3.1), we obtain
Letting \(y=-3x\) in (3.1), we obtain
Now, adding (3.2) and (3.3), we have
Then,
It is easy to see that
This implies that the sequence \(\{\frac{f(3^{n}x)}{3^{n}}\}_{n}\) is Cauchy for all \(x\in \mathcal{X}\). Define \(T:\mathcal{X}\to \mathcal{Y}\) by
It is clear that \(T(0)=0\) and \(T(3x)=3T(x)\) for all \(x\in \mathcal{X}\). In view of the definition of T, (3.1) yields
Putting \(y=3x\) in (3.6) and using \(T(3x)=3T(x)\), we infer that \(T(2x)=2T(x)\) for all \(x\in \mathcal{X}\). Hence, (3.6) implies that T is Jensen (additive) on \(\mathcal{X}\). Letting \(m=0\) and taking the limit as \(n\to \infty \) in (3.5), one obtains
To extend (3.7) to the whole \(\mathcal{X}\), let \(z\in \mathcal{X}\setminus \{0\}\) and choose a positive integer n such that \(\|nz\|\geqslant d\). Take \(x=2(n+1)z\) and \(y=-2nz\). Then, (3.7) yields
Using these inequalities together with (3.1), we obtain
Since T is Jensen and \(z=\frac{x+y}{2}\), we obtain
This inequality is valid for \(z=0\) because of \(T(0)=0\). The uniqueness of T follows easily from the last inequality. □
Remark 3.2
Since \(E_{d}\times E_{d}\subseteq \{(x,y)\in X\times X: \|x\|+\|y\| \geqslant d\}\), Theorems 2.9 and 3.1 are valid when (2.6) and (3.1) hold for all \(x,y\in X\) with \(\|x\|+\|y\|\geqslant d\).
Theorem 3.3
Let \(\mathcal{X}\) be a normed space and \(\mathcal{Y}\) a Banach space. Take \(\varepsilon \geqslant 0\) and let a function \(f:\mathcal{X}\to \mathcal{Y}\) satisfy the inequality (3.1) for all \(x,y\in \mathcal{X}\) with \(\|x+y\|\geqslant d\) for some \(d>0\). Then, there exists a unique additive function \(T:\mathcal{X}\to \mathcal{Y}\) such that
Proof
Letting \(y=0\) in (3.1), we obtain
It is easy to see that
Then, the sequence \(\{\frac{f(2^{n}x)}{2^{n}}\}_{n}\) is Cauchy for all \(x\in \mathcal{X}\). Define \(T:\mathcal{X}\to \mathcal{Y}\) by
It is clear that \(T(0)=0\) and \(T(2x)=2T(x)\) for all \(x\in \mathcal{X}\). In view of the definition of T, (3.1) yields
Let \(T_{e}\) and \(T_{o}\) be the even part and the odd part of T. Then, \(T_{e}\) and \(T_{o}\) satisfy (3.9) for all \(x,y\in \mathcal{X}\) with \(x+y\neq 0\). Since \(T_{o}\) is odd, (3.9) yields that \(T_{o}\) is additive on \(\mathcal{X}\). It follows from (3.9) that \(2T (\frac{x}{2} )=T(x)\) for all \(x\neq 0\), and then
Putting \(y=3x\) and using \(T_{e}(2x)=2T_{e}(x)\), we infer that \(T_{e}(x)=0\) for all \(x\in \mathcal{X}\). Hence, (3.9) implies that T is Jensen (additive) on \(\mathcal{X}\). Letting \(m=0\) and taking the limit as \(n\to \infty \) in (3.8), one obtains
To extend (3.7) to the whole \(\mathcal{X}\), let \(z\in \mathcal{X}\setminus \{0\}\) and choose a positive integer n such that \(\|nz\|\geqslant d\). Take \(x=2(n+1)z\) and \(y=-2nz\). Then, (3.7) yields
Using these inequalities together with (3.1), we obtain
Since T is Jensen and \(z=\frac{x+y}{2}\), we obtain
This inequality is valid for \(z=0\) because of \(T(0)=0\). The uniqueness of T follows easily from the last inequality. □
Corollary 3.4
Let \(\mathcal{X}\) and \(\mathcal{Y}\) be linear normed spaces. For a function \(f:\mathcal{X}\to \mathcal{Y}\) the following conditions are equivalent:
- \((i)\):
-
\(\lim_{\|x+y\|\to \infty} [2f (\frac{x+y}{2} )-f(x)-f(y) ]=0\);
- \((\mathit{ii})\):
-
\(\lim_{\|x\|+\|y\|\to \infty} [2f ( \frac{x+y}{2} )-f(x)-f(y) ]=0\);
- \((\mathit{iii})\):
-
\(\lim_{\min \{\|x\|,\|y\|\}\to \infty} [2f ( \frac{x+y}{2} )-f(x)-f(y) ]=0\);
- \((\mathit{iv})\):
-
\(\lim_{\max \{\|x\|,\|y\|\}\to \infty} [2f ( \frac{x+y}{2} )-f(x)-f(y) ]=0\);
- \((v)\):
-
\(2f (\frac{x+y}{2} )=f(x)+f(y)\), \(x,y\in X\).
Proof
The implications \((v)\Rightarrow (\mathit{ii})\Rightarrow (\mathit{iii})\) and \((v)\Rightarrow (\mathit{iv})\Rightarrow (i)\) are obvious. It is enough to prove the implications \((i)\Rightarrow (v)\) and \((\mathit{iii})\Rightarrow (v)\).
To prove \((i)\Rightarrow (v)\), let \(\varepsilon >0\) be an arbitrary real number. By \((i)\) there exists \(d_{\varepsilon }>0\) such that
Let \(\widetilde{\mathcal{Y}}\) be the completion of \(\mathcal{Y}\). In view of Theorem 3.3 there exists a unique additive function \(A_{\varepsilon}:\mathcal{X}\to \widetilde{\mathcal{Y}}\) such that
Then,
Since ε is arbitrary, we obtain that f satisfies \((v)\). Using a similar argument, the implication \((\mathit{iii})\Rightarrow (v)\) is obtained by Theorem 3.1. Hence, the proof is complete. □
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MBM and AN contributed to the study conception and design. The first draft of the manuscript was written by AN and all authors commented on previous versions of the manuscript. All authors read and approved the final manuscript. MBM supervised the project.
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Moghimi, M.B., Najati, A. Some hyperstability and stability results for the Cauchy and Jensen equations. J Inequal Appl 2022, 97 (2022). https://doi.org/10.1186/s13660-022-02837-6
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DOI: https://doi.org/10.1186/s13660-022-02837-6