Stability of the Pexiderized Lobacevski Equation
Abstract
The aim of this paper is to investigate the solution and the superstability of the Pexiderized Lobacevski equation f((x + y)/2) 2 = g(x)h(y), where f, g, h : G2 → ℂ are unknown functions on an Abelian semigroup (G, +). The obtained result is a generalization of Gǎvruţa′s result in 1994 and Kim′s result in 2010.
1. Introduction
The stability problem of the functional equation was conjectured by Ulam [1] during the conference in the University of Wisconsin in 1940. In the next year, it was solved by Hyers [2] in the case of additive mapping, which is called the Hyers-Ulam stability. Thereafter, this problem was improved by Bourgin [3], Aoki [4], Rassias [5], Ger [6], and Gǎvruţa et al. [7, 8] in which Rassias’ result is called the Hyers-Ulam-Rassias stability.
Kim [15] improved his result under the condition bounded by an unknown function. In there, author conjectured through an example that the Lobacevski equation (L) will have a solution as an exponential function. Namely, for a simple example of this equation, we can find the functional equation .
Furthermore, the range of the function in all results is expanded to the Banach space.
In this paper, let (G, +) be a uniquely 2-divisible Abelian semigroup (i.e., for each x ∈ G, there exists a unique y ∈ G such that y + y = x: such y will be denoted by x/2), ℂ is the field of complex numbers, ℝ the field of real numbers, and ℝ+ the set of positive reals. We assume that f, g, h : G → ℂ are nonzero and nonconstant functions, ε is a nonnegative real constant, and φ : G → ℝ+ is a mapping.
2. Stability of the Pexiderized Lobacevski Equation (PL)
We will investigate the solution and the superstability of the Pexiderized Lobacevski equation (PL).
Theorem 2.1. Suppose that f, g, h : G → ℝ satisfy the inequality
Then, either there exist C1, C2, C3 > 0 such that
Proof. Replacing x with y in (2.1), and then subtracting them and using triangle inequality, we have
It follows from the inequality (2.5) that there exist constants c1, c2, d1, d2 ≥ 0 such that
Now if h(x) is unbounded, then we can choose (yn) ∈ G so that |h(yn)| → ∞ as n → ∞. Letting y = yn in (2.1), dividing by |h(yn)|, and letting n → ∞, we have
It follows from (2.1) and (2.8) that
Letting z = 0 in (2.10), we get
Exchanging the roles of g and h, by the same proceeding, we have
Putting (2.12) and (2.13) in (2.5), it implies
Let x = 0 in (2.14). Since e1 and e2 are exponentials, this implies that |e1(y) − e2(y)| ≤ M for all y ∈ G. Hence, from this and (2.14), we have
Since g is unbounded from (2.2), we can choose (yn) ∈ G so that g(yn) = g(0)e1(yn) → ∞ as n → ∞. Letting y = yn in (2.16), we get that e1(x) = e2(x). Let it be denoted by e(x). Then (2.12) and (2.13) state nothing but (2.3). Putting (2.3) with x = y in (2.1), we get the inequality (2.4).
Finally, it is immediate that g and h in (2.3) satisfy (L), respectively.
Corollary 2.2. Suppose that f, g : G → ℝ satisfy the inequality
Then, either g is bounded or g satisfies (L). In particular, g is represented by
Corollary 2.3. Suppose that f, g : G → ℝ satisfy the inequality
Then either there exist C1, C2 > 0 such that
Corollary 2.4. Suppose that f : G → ℝ satisfy the inequality
Then either f is bounded or f satisfies (L). In particular, f is represented by
In Corollary 2.4, it is founded in papers [14, 15] that f satisfies (L).
Theorem 2.5. Suppose that f, g, h : G → ℝ satisfy the inequality
Then, either h is bounded or g is an exponential by the multiplying of a scalar g(0) and satisfies (L).
Proof. Suppose that h(x) is unbounded. Then we can choose (yn) ∈ G such that |h(yn)| → ∞ as n → ∞. Letting y = yn in (2.24), dividing by |h(yn)|, and letting n → ∞, we have
Thus, it follows from (2.24) and (2.25) that
Letting z = 0 in (2.27), we get
Theorem 2.6. Suppose that f, g, h : G → ℝ satisfy the inequality
Then, either g is bounded or h is an exponential by the multiplying of a scalar h(0) and satisfies (L).
Proof. The proof runs along a slight change in the step-by-step procedure in Theorem 2.5.
Remark 2.7. (i) As Corollaries 2.2–2.4 of Theorem 2.1, by replacing g and h with f in Theorems 2.5 and 2.6, we can obtain more corollaries for the following functional equations:
(ii) For the results obtained from each equation of the above (i), by applying φ(y) = φ(x) = ε, we can obtain the same number of corollaries.
3. Extension to Banach Algebra
All obtained results can be extended to the stability on the Banach algebras. We will illustrate only for the case of Theorem 2.1 among them.
Theorem 3.1. Let (E, ∥·∥) be a semisimple commutative Banach algebra. Assume that f, g, h : G → E satisfy the inequality
Then, for an arbitrary linear multiplicative functional x* ∈ E*, either there exist C1, C2, C3 > 0 such that
Proof. Assume that (3.1) holds, and fix arbitrarily a linear multiplicative functional x* ∈ E. As well known, we have ∥x*∥ = 1, hence, for every x, y ∈ G, we have
It follows from the inequality (3.6) that there exist constants c1, c2, d1, d2 ≥ 0 such that
By the assumption (3.2), an appeal to Theorem 2.1 shows that
Putting (3.13) in (3.6), following the same proceeding as after (2.13) in Theorem 2.1, then we arrive that e1(x) = e2(x). Indeed, we have
Letting x = 0 in (3.15), it implies |(x*∘e2)(y)−(x*∘e1)(y)| ≤ M/x*(1) = M′ for all y ∈ G. Thus, from this and (3.15), we have
Since x*∘g is unbounded from (3.2), we can choose (yn) ∈ G so that |(x*∘g)(yn)| = |g(0)(x*∘e1)(yn)| → ∞ as n → ∞. Letting y = yn in (3.17), which arrive that
Using the same logic as before, that is, bearing the linear multiplicativity of x* in mind, the difference derived from (3.18), D(3.18)(x): = e1(x) − e2(x), falls into the kernel of x*. Then, the semisimplicity of E implies that e1(x) = e2(x). Let it be denoted by e(x), which arrive the claimed (3.3) and (3.4).
Since e(x) : G → ℝ is exponential, it is immediate from (3.3) that each function g and h satisfies (L).
Acknowledgment
This work was supported by Kangnam University Research Grant in 2010.