- Open Access
Superstability of generalized cauchy functional equations
© Lee and Chung; licensee Springer. 2011
- Received: 26 February 2011
- Accepted: 28 July 2011
- Published: 28 July 2011
- Cauchy functional equation
- stability; superstability
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  proposed the following question:
for all × ∈ G 1 ?
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  proved the Hyers-Ulam stability of (1.3). The stability of (1.2) was proved by Baker, Lawrence and Zorzitto . 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 , Baker  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 . 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 . 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).
where A is an additive mapping, E is an exponential mapping and a is an arbitrary nonzero constant. For the proof we refer to [, 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.
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.
If f ≡ 0, then g is arbitrary;
If f(≢ 0) is bounded or f(0) ≠ 0 , then g is also bounded;
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.
for all y ∈ V. This shows that g is bounded.
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:
for all × in S. If S is commutative, then L is logarithmic.
For that reason, we only consider the case g ≢ 1.
If f ≡ 0, then g is arbitrary;
If f(≢ 0) is bounded or f(1) ≠ 0 , then g is also bounded;
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.
for all x, y ∈ V. Since f ≢0, we see that g is bounded.
This completes the proof. □
This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MEST) (No.2011-0000092).
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