# Superstability of generalized cauchy functional equations

- Young-Su Lee
^{1}Email author and - Soon-Yeong Chung
^{2}

**2011**:23

https://doi.org/10.1186/1687-1847-2011-23

© Lee and Chung; licensee Springer. 2011

**Received: **26 February 2011

**Accepted: **28 July 2011

**Published: **28 July 2011

## Abstract

### Keywords

Cauchy functional equation stability; superstability## 1. Introduction

Usually, the solutions of (1.1)-(1.4) are called additive, exponential, logarithmic and multiplicative, respectively. Many authors have been interested in the general solutions and the stability problems of (1.1)-(1.4) (see [1–5]).

The stability problems of functional equations go back to 1940 when Ulam [6] proposed the following question:

*for all ×* ∈ *G*
_{1}
*?*

*f*is a function between Banach spaces satisfying ||

*f*(

*x*+

*y*) -

*f*(

*x*) -

*f*(

*y*)|| ≤

*ε*for some fixed

*ε >*0, then there exists a unique additive mapping

*A*such that ||

*f*(

*x*) -

*A*(

*x*)|| ≤

*ε*. From these historical backgrounds, the functional equation

for some positive constant ≤ *δ*
_{
ε
}.

During the last decades, Hyers-Ulam stability of various functional equations has been extensively studied by a number of authors (see [3–5, 8–10]). Especially, Forti [11] proved the Hyers-Ulam stability of (1.3). The stability of (1.2) was proved by Baker, Lawrence and Zorzitto [12]. They proved that if *f* is a function satisfying |*f*(*x* + *y*) - *f*(*x*)*f*(*y*)| ≤ *ε* for some fixed *ε >* 0 then *f* is either bounded or else *f*(*x*+*y*) = *f*(*x*)*f*(*y*). In order to distinguish this phenomenon from the Hyers-Ulam stability, we call this phenomenon superstability. Generalizing results as in [12], Baker [13] proved that the superstability for (1.4) does also hold.

We say that (1.6) and (1.7) are generalized Cauchy functional equations because these are reduced the Cauchy functional equations if *g* is identically one. It is easily checked that the general solutions of (1.6) are additive or exponential whether *g* is identically one or not. From this point of view, we can expect that (1.6) has the Hyers-Ulam stability or superstability due to the conditions of *g*. Actually, if *g* is identically one in (1.6), then Hyers-Ulam stability holds [7]. On the other hand, if *g* is not identically one in (1.6), then we shall see in Section 2 that superstability holds in this case. That is, *f* and *g* are either bounded or else *f*(*x* + *y*) = *f*(*x*)*g*(*y*) + *f*(*y*).

Analogously, it is easy to see that the general solutions of (1.7) are logarithmic or multiplicative whether *g* is identically one or not. If *g* is identically one in (1.7), then this case is exactly the same as in [11]. And hence Hyers-Ulam stability holds in this case. We shall prove that if *g* is not identically one in (1.7), then *f* and *g* are either bounded or else *f*(*xy*) = *f*(*x*)*g*(*y*)+*f*(*y*).

## 2. Stability of (1.6) and (1.7)

where *A* is an additive mapping, *E* is an exponential mapping and *a* is an arbitrary nonzero constant. For the proof we refer to [[14], Lemma 1]. Although (1.6) is slightly different from (1.1), the general solutions of (1.6) are related to (1.2) rather than (1.1) if *g* is not identically one. The stability result in the case of *g* ≡ 1 in (1.6) is well known as follows.

**Theorem 2.1**. [4, 7]

*Let E*

_{1}

*be a normed vector space and E*

_{2}

*a Banach space. Suppose that f*:

*E*

_{1}→

*E*

_{2}

*satisfies the inequality*

*for all × in E*
_{1}.

According to the above result, we know that Hyers-Ulam stability holds if *g* is identically one. Thus, it suffices to show the case *g* ≢ 1. Especially interesting is that superstability holds if *g* is not identically one as follows.

**Theorem 2.2**.

*Let V be a vector space and let f*,

*g*:

*V*→≤

*be complex valued functions with g*≢ 1.

*Suppose that f and g satisfy the inequality*

*Then, one of the following conditions holds:*

- (i)
*If f*≡ 0,*then g is arbitrary*; - (ii)
*If f*(≢ 0)*is bounded or f*(0) ≠ 0 ,*then g is also bounded;* - (iii)
*If f is unbounded, then f*(0) = 0,*g is also unbounded and f*(*x*+*y*) =*f*(*x*)*g*(*y*) +*f*(*y*)*for all x*,*y*∈*V*.

*Proof*. (i) If

*f*≡ 0, then we easily see that

*g*is arbitrary.

- (ii)

*x*,

*y*∈

*V*. Since

*f*≢ ≡ 0, there exists a point

*x*

_{0}such that

*f*(

*x*

_{0})

*≠*0. Putting

*x*=

*x*

_{0}in (2.2) and dividing the result by |

*f*(

*x*

_{0})| we have

for all *y* ∈ *V*. This shows that *g* is bounded.

*y*∈

*V*. We see that

*g*is bounded, since

*f*(0) ≠ 0.

- (iii)

*x*,

*y*∈

*V*. This shows that

*f*is bounded and hence this reduces a contradiction. Since

*g*satisfies (2.3) with

*g*≢ 0 and

*g*≢ 1, we conclude that

*g*is unbounded. Choose a sequence {

*y*

_{ n }} such that |

*g*(

*y*

_{ n })| → ∞. Putting

*y*=

*y*

_{ n }in (2.1) and dividing both sides by |

*g*(

*y*

_{ n })| we have

This completes the proof. □

where *L* is a logarithmic mapping, *M* is a multiplicative mapping and *b* is an arbitrary nonzero constant. In case of *g* ≡ 1, the stability result is well known as follows:

**Theorem 2.3**. [5, 11]

*Let S be a semigroup and Y a Banach space. Further, let f*:

*S*→

*Y be a mapping satisfying*

*for all × in S. If S is commutative, then L is logarithmic*.

For that reason, we only consider the case *g* ≢ 1.

**Theorem 2.4**.

*Let V be a vector space and let f*,

*g*:

*V*→ ≤

*be complex valued functions with g*≢ 1.

*Suppose that f and g satisfy the inequality*

*Then, one of the following conditions holds:*

- (i)
*If f*≡ 0,*then g is arbitrary*; - (ii)
*If f*(≢ 0)*is bounded or f*(1) ≠ 0 ,*then g is also bounded*; - (iii)
*If f is unbounded, then f*(1) = 0,*g is also unbounded and f*(*xy*) =*f*(*x*)*g*(*y*) +*f*(*y*)*for all x*,*y*∈*V*.

*Proof*. (i) If

*f*≡ 0, then from (2.4) we see that

*g*is arbitrary.

- (ii)

for all *x*, *y* ∈ *V*. Since *f* ≢0, we see that *g* is bounded.

*f*(1) ≠ 0. Putting

*x*= 1 in (2.4) we have

*g*is bounded.

- (iii)

*x*,

*y*∈

*V*. Putting

*y*= 1 in (2.6) we see that

*f*is bounded. This reduces a contradiction. Since

*g*satisfies (2.5) with

*g*≢ 0 and

*g*≢ 1, we can choose a sequence {

*y*

_{ n }} such that |

*g*(

*y*

_{ n })| → ∞. Putting

*y*=

*y*

_{ n }in (2.4) and dividing the result by |

*g*(

*y*

_{ n })| we have

This completes the proof. □

## Declarations

### Acknowledgements

This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MEST) (No.2011-0000092).

## Authors’ Affiliations

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