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Approximate lie brackets: a fixed point approach
Journal of Inequalities and Applications volume 2012, Article number: 125 (2012)
Abstract
The aim of this article is to investigate the stability and superstability of Lie brackets on Banach spaces by using fixed point methods.
2010 Mathematics Subject Classification: 46L06; 39B82; 39B52.
1. Introduction
The stability problem of functional equations originated from a question of Ulam [1] concerning the stability of group homomorphisms. Hyers [2] gave a first affirmative answer to the question of Ulam for Banach spaces. Rassias [3] considered the stability problem with unbounded Cauchy differences. The stability problems of several functional equations have extensively been investigated by a number of authors and there are many interesting results concerning this problem (see [4–18]).
In 2003, Cǎdariu and Radu applied the fixed point method and they could present a short and simple proof (different from the "direct method", initiated by Hyers in 1941) for the generalized Hyers-Ulam stability of Jensen functional equation.
In this article, by using the fixed point method, we prove that, if there exists an approximately Lie bracket f : A × A → A on Banach spaces A, then there exists a Lie bracket T : A × A → A which is near to f. Moreover, under some conditions on f, the Banach space A has a Lie algebra structure with Lie bracket T.
We recall a Lie algebra consists of a (finite dimensional) vector space A over a field and a multiplication in A (usually, the product of x, y ∈ A is denoted by [x, y] and called a Lie bracket or commutator) with the following two properties:
-
(1)
Anti-commutativity: [x, x] = 0 for any x ∈ A;
-
(2)
Jacobi identity: [z, [x, y]] = [[z, x], y] + [x, [z, y]] for any x, y, z ∈ A.
For more details about Lie algebras, the readers are referred to [19–22]. Throughout this article, we assume that n0 ∈ ℕ is a positive integer,
It is easy to see that . Moreover, we suppose that A is a complex Banach space. For any mapping f : A × A → A, we define
for all and x, y, z, t ∈ A.
2. Main results
We need the following theorem to prove the main result of this article.
Theorem 2.1. (The alternative of fixed point theorem [23, 24]) Suppose that (Ω, d) is a complete generalized metric space and T : Ω → Ω is a strictly contractive mapping with Lipschitz constant L. Then, for any x ∈ Ω, either d(Tmx, Tm+1x) = ∞ for all m ≥ 0 or there exists a natural number m0 such that
-
(1)
d(T mx, T m+1x) < 1 for all m ≥ m0;
-
(2)
the sequence {Tmx} is convergent to a fixed point y* of T;
-
(3)
y* is the unique fixed point of T in the set ;
-
(4)
for all y ∈ Λ.
Now, we give our main results by using. Theorem 2.1.
Theorem 2.2. Let f : A × A → A be a continuous mapping and let ϕ : A4 = A × A × A × A → [0, ∞) be a mapping such that
for all and x, y, z, t ∈ A. If there exists L < 1 such that for all x, y, z, t ∈ A, then there exists a unique bilinear mapping T : A × A → A such that
for all x, z ∈ M. Moreover, for any sequence {a m } in A, if
for all x ∈ A, then A is a Lie algebra with Lie bracket [x, y] = T (x, y) for all x, y ∈ A.
Proof. Putting μ = 1 and y = t = 0 in (2.1), we get
for all x, z ∈ A and so
for all x, z ∈ A. Consider the set X : = {g : g : A × A → A} and introduce the generalized metric on X by:
It is easy to show that (X, d) is a complete generalized metric space. Now, we define the mapping J : X → X by
for all x, z ∈ A. For any g, h ∈ X, we have
for all x, z ∈ A, which means that
for all g, h ∈ X. It follows from (2.6) that
From Theorem 2.1, it follows that J has a unique fixed point in the set X1:= {I ∈ X: d(f, T) < ∞}. Let T be the fixed point of J. Then we have limn→∞d(Jn (f), T) = 0 and
for all x, z ∈ A. By the inequality d(f, J(f)) ≤ L and J(T) = T, we have
and so
This implies the inequality (2.4). From , we have
for all x, y, z ∈ A. Thus it follows from (2.1), (2.7) and (2.8) that
for all x, y, z ∈ M and so
for all x, y, z, t ∈ A. This shows that
for all x, y, z, t ∈ A. Hence, T is Cauchy additive with respect to the first and second variables. By putting y : = x and t : = z in (2.1), we have
for all x, z ∈ A. and and so
for all x, z ∈ A and , that is,
for all x, z ∈ A.
If λ belongs to , then there exists θ ∈ [0, 2π] such that λ = eiθ. If we set , then λ1 belongs to . By using (2.10), we have
for all x, z ∈ M.
If λ belongs to for some n ∈ ℕ, then, by (2.9), we have
for all x, z ∈ A. Let s ∈ (0, ∞). Then, by Archimedean property of ℂ, there exists a positive real number n such that the point (s, 0) ∈ ℝ2 lies in the interior of circle with center at origin and radius n in ℝ2. Putting and , we have and . Thus, by (2.9), we have
for all x, z ∈ s. Moreover, there exists θ ∈ [0, 2π] such that λ = ǀλ ǀ eiθ.
Therefore, we have
for all x, z ∈ A and so T : A × A →A is homogeneous with respect to the first variable. It follows from (2.9) and (2.11) that T is ℂ-Linear with respect to the first variable.
Moreover, by (2.3), T (x, x) = 0 for all x ∈ A, whence
for all x, y ∈ A and so
for all x, y ∈ A, that is, T is skew symmetric. Let z ∈ A and define a mapping ad(z): A →A by
for all x ∈ A. It is clear that ad(z) is a linear and continuous mapping at zero. In fact, if {a m } is a sequence in A such that limn→∞a m = 0, then, by (2.5), we have
Thus, for all z ∈ A, ad(z) is continuous at zero and so ad(z) is a continuous and linear mapping. Substituting x with 2mx and y with 2my in (2.2) and multiplying by 4-mboth sides of the inequality, we have
for all x, y, z ∈ A and m ∈ ℕ. Since f is continuous, we have
for all x, y, z ∈ A. Since, for all z ∈ A, ad(z) is a linear and continuous mapping, we get
for all x, y, z ∈ A. Since T is skew symmetric, it is easy to show that T is satisfies in the Jacobi identity condition. Thus T is a Lie bracket satisfies in (2.4) and (A, T) is a Lie algebra.
To prove the uniqueness property of T, let Q : A × A →A be another bilinear mapping satisfying (2.7). Then we have
for all x, z ∈ A. This means that T = Q. This completes the proof. □
Corollary 2.3. Let p ∈ (0, 1) and θ ∈ [0, ∞) be real numbers. Suppose that f : A × A →A is a mapping such that
for all and x, y, z, t ∈ A. Then there exists a unique bilinear mapping T : A × A →A such that
for all x, z ∈ A. Moreover, for any sequence {a m } in A, if
for all x ∈ A, then A is a Lie algebra with Lie bracket [x, y] = T(x, y) for all x, y ∈ A.
Proof. It follows from Theorem 2.2 by putting ϕ(x, y, z): = θ(ǁ x ǁ p +ǁ y ǁ p +ǁ z ǁ p +ǁ t ǁ p ) for all x, y, z , ∈ M and L = 4p -1. □
Finally, we prove the superstability of Lie brackets as follows:
Corollary 2.4. Let and θ ∈ [0, ∞) be real numbers. Suppose that f: A × A →A is a mapping such that
for all and x, y, z, t ∈ A. Moreover, for any sequence {a m } in A, if
for all x ∈ A, then A is a Lie algebra with Lie bracket [x, y] = f(x, y) for all x, y ∈ A.
Proof. Putting ϕ(x, y, z, t): = θ(ǁx ǁp ǁy ǁp ǁz ǁp ǁt ǁp) for all x, y, z ∈ M and in Theorem 2.2, the conclusion follows. □
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Acknowledgements
This study was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant Number: 2011-0021821).
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Gordji, M.E., Ramezani, M., Cho, Y.J. et al. Approximate lie brackets: a fixed point approach. J Inequal Appl 2012, 125 (2012). https://doi.org/10.1186/1029-242X-2012-125
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DOI: https://doi.org/10.1186/1029-242X-2012-125