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Superstability of generalized cauchy functional equations
Advances in Difference Equations volume 2011, Article number: 23 (2011)
Abstract
In this paper, we consider the stability of generalized Cauchy functional equations such as
Especially interesting is that such equations have the HyersUlam stability or superstability whether g is identically one or not.
2000 Mathematics Subject Classification: 39B52, 39B82.
1. Introduction
The most famous functional equations are the following Cauchy functional equations:
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:
Let f be a mapping from a group G _{1} to a metric group G _{2} with metric d(·,·) such that
Then does there exist a group homomorphism L : G _{1} → G _{2} and δ _{ ε } > 0 such that
for all × ∈ G _{1} ?
The case of (1.1) was solved by Hyers [7]. He proved that if 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
is said to have the HyersUlam stability if for an approximate solution φ _{ s } such that
for some fixed constant ε > 0 there exists a solution φ of (1.5) such that
for some positive constant ≤ δ _{ ε }.
During the last decades, HyersUlam stability of various functional equations has been extensively studied by a number of authors (see [3–5, 8–10]). Especially, Forti [11] proved the HyersUlam 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 HyersUlam stability, we call this phenomenon superstability. Generalizing results as in [12], Baker [13] proved that the superstability for (1.4) does also hold.
In this paper, we consider the stability of generalized Cauchy functional equations such as
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 HyersUlam stability or superstability due to the conditions of g. Actually, if g is identically one in (1.6), then HyersUlam 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 HyersUlam 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)
We first consider the stability of (1.6). The general solutions of (1.6) are given by
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 x, y in E _{1}, where ε > 0 is a constant. Then the limit
exists for all × in E _{1} and A : E _{1} → E _{2} is a unique additive mapping satisfying
for all × in E _{1}.
According to the above result, we know that HyersUlam 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)
Suppose that f is bounded and f ≢ 0. Then, there exists a constant M > 0 such that f(x) ≤ M for all x ∈ V. From (2.1), it follows that
(2.2)
for all 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.
Now assume that f(0) ≠ 0. Putting x = 0 in (2.1) yields
for all y ∈ V. We see that g is bounded, since f(0) ≠ 0.

(iii)
Finally, we are going to prove the case that f is unbounded. Since f is unbounded, we can take a sequence {x _{ n }} such that f(x _{ n }) → ∞. Putting x = x _{ n } in (2.1) and dividing both sides by f(x _{ n }) we have
Letting n → ∞ we obtain
Substituting x = x + x _{ n } in (2.1) gives
Dividing both sides by f(x _{ n }) and then letting n → ∞ we have
for all x, y ∈ V. We observe that g is also unbounded. If g ≡ 0, then from (2.1) we have
for all 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
Letting n → ∞ yields
We note that f(0) = 0. Substituting y = y + y _{ n } in (2.1) and using (2.3) we obtain
Dividing both sides in the above inequality by g(y _{ n }) and then letting n → ∞ we have
This completes the proof. □
Analogously, we are going to consider the stability of (1.7). The general solutions of (1.7) are given by
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 x, y in S. Then the limit
exists for all × in S and L : S → Y is a unique mapping satisfying
and
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)
Suppose that f is bounded and f ≢ 0. Then, there exists a constant N > 0 such that f(x) ≤ N for all x ∈ V. It follows from (2.4) that we calculate
for all x, y ∈ V. Since f ≢0, we see that g is bounded.
Assume that f(1) ≠ 0. Putting x = 1 in (2.4) we have g is bounded.

(iii)
Now we prove the case that f is unbounded. Since f is unbounded, we can take a sequence {x _{ n }} such that f(x _{ n }) → ∞. Putting x = x _{ n } in (2.4) and dividing both sides by f(x _{ n }) we have
Letting n → ∞ we obtain
Replacing x by xx _{ n } in (2.4) yields
Dividing both sides by f(x _{ n }) and then letting n → ∞ we have
for all x, y ∈ V. If g ≡ 0, then from (2.4) we have
for all 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
Letting n → ∞ gives
Putting x = 1 yields f(1) = 0. Replacing y by yy _{ n } in (2.4) and using (2.5) we have
Dividing both sides by g(y _{ n }) and letting n → ∞ we obtain
This completes the proof. □
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Acknowledgements
This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MEST) (No.20110000092).
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YL carried out the main part of this manuscript. SC participated discussion and corrected the main theorem. All authors read and approved the final manuscript.
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Keywords
 Cauchy functional equation
 stability; superstability