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Approximate lie brackets: a fixed point approach

Journal of Inequalities and Applications20122012:125

https://doi.org/10.1186/1029-242X-2012-125

Received: 10 March 2012

Accepted: 7 June 2012

Published: 7 June 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.

Keywords

generalized Hyers-Ulam stabilityfixed pointsuperstabilityLie algebraskew-symmetryJacobi identity.

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 [418]).

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 × AA 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 F 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. (1)

    Anti-commutativity: [x, x] = 0 for any x A;

     
  2. (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 [1922]. Throughout this article, we assume that n0 is a positive integer,
T 1 : = { z : ǀ z ǀ = 1 } , T 1 n o 1 : = { e i θ : 0 θ 2 π n 0 } .
It is easy to see that T 1 = T 1 1 1 . Moreover, we suppose that A is a complex Banach space. For any mapping f : A × AA, we define
D μ f ( x , y , z , t ) : = 4 μ f x + y 2 , z + t 2 + 4 μ f x - y 2 , z + t 2 + 4 μ f x + y 2 + z - t 2 + 4 μ f x - y 2 , z - t 2 - 4 f ( μ x , z )

for all μ T 1 n o 1 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(T m x, Tm+1x) = ∞ for all m ≥ 0 or there exists a natural number m0 such that
  1. (1)

    d(T m x, T m+1x) < 1 for all mm0;

     
  2. (2)

    the sequence {T m x} is convergent to a fixed point y* of T;

     
  3. (3)

    y* is the unique fixed point of T in the set Λ = { y Ω : d ( T m 0 x , y ) < ;

     
  4. (4)

    d ( y , y * ) 1 1 - L d ( y , T y ) for all y Λ.

     

Now, we give our main results by using. Theorem 2.1.

Theorem 2.2. Let f : A × AA be a continuous mapping and let ϕ : A4 = A × A × A × A → [0, ∞) be a mapping such that
ǁ D μ f ( x , y , z , t ) ǁ ϕ ( x , y , z , t ) ,
(2.1)
lim n 4 - n f ( 2 n z , 2 n f ( x , y ) ) - f ( 4 - n f ( 2 n z , 2 n x ) , y ) - f ( x , 4 - n f ( 2 n z , 2 n y ) ) ϕ ( x , y , 0 , 0 ) ,
(2.2)
lim n 4 - n f ( 2 n x , 2 n x ) = 0
(2.3)
for all μ T 1 n o 1 and x, y, z, t A. If there exists L < 1 such that ϕ ( x , y , z , t ) 4 L ϕ ( x 2 , y 2 , z 2 , t 2 ) for all x, y, z, t A, then there exists a unique bilinear mapping T : A × AA such that
f ( x , z ) - T ( x , z ) L 1 - L ϕ ( x , 0 , z , 0 )
(2.4)
for all x, z M. Moreover, for any sequence {a m } in A, if
lim m lim n 4 - n f ( 2 n x , 2 n a m ) = lim n lim m 4 - n f ( 2 n x , 2 n a m )
(2.5)

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
4 f ( x 2 , z 2 ) - f ( x , z ) ϕ ( x , 0 , z , 0 )
for all x, z A and so
1 4 f ( 2 x , 2 z ) - f ( x , z ) 1 4 ϕ ( 2 x , 0 , 2 z , 0 ) L ϕ ( x , 0 , z , 0 )
(2.6)
for all x, z A. Consider the set X : = {g : g : A × AA} and introduce the generalized metric on X by:
d ( h , g ) : = inf { C + : ǁ g ( x , z ) - h ( x , z ) ǁ C ϕ ( x , 0 , z , 0 ) for all  x , z A } .
It is easy to show that (X, d) is a complete generalized metric space. Now, we define the mapping J : XX by
J ( h ) ( x , z ) = 1 4 h ( 2 x , 2 z )
for all x, z A. For any g, h X, we have
d ( g , h ) < C g ( x , z ) - h ( x , z ) C ϕ ( x , 0 , z , 0 ) 1 4 g ( 2 x , 2 z ) - 1 4 h ( 2 x , 2 z ) 1 4 C ϕ ( 2 x , 0 , 2 z , 0 ) 1 4 g ( 2 x , 2 z ) - 1 4 h ( 2 x , 2 z ) L C ϕ ( x , 0 , z , 0 ) d ( J ( g ) , J ( h ) ) L C
for all x, z A, which means that
d ( J ( g ) , J ( h ) ) L d ( g , h )
for all g, h X. It follows from (2.6) that
d ( f , J ( f ) ) L .
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(J n (f), T) = 0 and
lim n 1 4 n f ( 2 n x , 2 n z ) = T ( x , z )
(2.7)
for all x, z A. By the inequality d(f, J(f)) ≤ L and J(T) = T, we have
d ( f , T ) d ( f , J ( f ) ) + d ( J ( f ) , J ( T ) ) L + L d ( f , T )
and so
d ( f , T ) L 1 - L .
This implies the inequality (2.4). From ϕ ( x , y , z , t ) 4 L ϕ x 2 , y 2 , z 2 , t 2 , we have
lim j 4 - j ϕ ( 2 j x , 2 j y , 2 j z , 2 j t ) = 0
(2.8)
for all x, y, z A. Thus it follows from (2.1), (2.7) and (2.8) that
4 T x + y 2 , z + t 2 + 4 T x - y 2 , z + t 2 - 4 T x + y 2 , z - t 2 + 4 T x - y 2 , z - t 2 - 4 T ( x , z ) = lim n 1 4 n 4 f 2 n x + 2 n y 2 , 2 n z + 2 n t 2 + 4 f 2 n x - 2 n y 2 - 2 n z + 2 n t 2 - 4 f 2 n x + 2 n y 2 , 2 n z - 2 n t 2 - 4 f 2 n x - 2 n y 2 , 2 n z - 2 n t 2 - 4 f ( 2 n x , 2 n z ) lim n 1 4 n ϕ ( 2 n x , 2 n y , 2 n z , 2 n t ) = 0
for all x, y, z M and so
T x + y 2 , z + t 2 + T x - y 2 , z + t 2 - T x + y 2 , z - t 2 + T x - y 2 , z - t 2 = T ( x , z )
for all x, y, z, t A. This shows that
T ( x + y , z + t ) = T ( x , z ) + T ( y , z ) + T ( x , t ) + T ( y , t )
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
4 μ f ( x , z ) - 4 f ( μ x , z ) ϕ ( x , x , z , z )
(2.9)
for all x, z A. and μ T 1 n o 1 and so
4 μ T ( x , z ) - 4 T ( μ x , z ) = lim n 1 4 n 4 μ f ( 2 n x , 2 n z ) - 4 f ( 2 n μ x , 2 n z ) lim n 1 4 n ϕ ( 2 n x , 2 n x , 2 n z , 2 n z ) = 0
for all x, z A and μ T 1 n o 1 , that is,
T ( μ x , z ) = μ T ( x , z )
(2.10)

for all x, z A.

If λ belongs to T 1 , then there exists θ [0, 2π] such that λ = e . If we set λ 1 = e i θ n o , then λ1 belongs to T 1 n o 1 . By using (2.10), we have
T ( λ x , z ) = T ( λ 1 n 0 x , z ) = λ 1 n 0 T ( x , z ) = λ T ( x , z )

for all x, z M.

If λ belongs to n T 1 = { n z : z T 1 } for some n , then, by (2.9), we have
T ( λ x , z ) = T ( n λ 1 x , z ) = T ( λ 1 ( n x ) , z ) = λ 1 T ( n x , z ) = λ 1 n T ( x , z ) = λ T ( x , z )
for all x, z A. Let s (0, ∞). Then, by Archimedean property of , there exists a positive real number n such that the point (s, 0) 2 lies in the interior of circle with center at origin and radius n in 2. Putting s 1 : = s + n 2 - s 2 i and s 2 : = t - n 2 - s 2 i , we have s = s 1 + s 2 2 and s 1 , s 2 n T 1 . Thus, by (2.9), we have
T ( s x , z ) = T s 1 + s 2 2 x , z = T s 1 x 2 , z + T s 2 x 2 , z = s 1 T x 2 , z + s 2 T ( x 2 , z ) = 4 s 1 + s 2 2 T x 2 , z 2 = s T ( x , z )

for all x, z s. Moreover, there exists θ [0, 2π] such that λ = ǀλ ǀ e .

Therefore, we have
T ( λ x , z ) = T ( λ e i θ x , z ) = λ T ( e i θ x , z ) = λ e i θ T ( x , z ) = λ T ( x , z )
(2.11)

for all x, z A and so T : A × AA is homogeneous with respect to the first variable. It follows from (2.9) and (2.11) that T is -Linear with respect to the first variable.

Moreover, by (2.3), T (x, x) = 0 for all x A, whence
0 = T ( x + y , x + y ) = T ( x , x ) + T ( x , y ) + T ( y , x ) + T ( y , y ) = T ( x , y ) + T ( y , x )
for all x, y A and so
T ( x , y ) = - T ( y , x )
for all x, y A, that is, T is skew symmetric. Let z A and define a mapping ad(z): AA by
a d ( z ) ( x ) = T ( z , x )
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
lim m a d ( z ) ( a m ) = lim m lim n 4 - n f ( 2 n z , 2 n a m ) = lim n lim m 4 - n f ( 2 n z , 2 n a m ) = lim n 4 - n f ( 2 n z , 0 ) = a d ( z ) ( 0 ) = 0 .
Thus, for all z A, ad(z) is continuous at zero and so ad(z) is a continuous and linear mapping. Substituting x with 2 m x and y with 2 m y in (2.2) and multiplying by 4-mboth sides of the inequality, we have
lim n 4 - m 4 - n f ( 2 n z , 2 n f ( 2 m x , 2 m y ) ) - f ( 4 - n f ( 2 n z , 2 n + m x ) , 2 m y ) - f ( 2 m x , 4 - n f ( 2 n z , 2 n + m y ) ) 4 - m ϕ ( 2 m x , 2 m y , 0 , 0 )
for all x, y, z A and m . Since f is continuous, we have
4 - m a d ( z ) ( f ( 2 m x , 2 m y ) ) - f ( a d ( z ) ( 2 m x ) , 2 m y ) - f ( 2 m x , a d ( z ) 2 m y ) 4 - m ϕ ( 2 m x , 2 m y , 0 , 0 )
for all x, y, z A. Since, for all z A, ad(z) is a linear and continuous mapping, we get
a d ( z ) T ( x , y ) - T ( a d ( z ) ( x ) , y ) - T ( x , a d ( z ) ( y ) ) = 0

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 × AA be another bilinear mapping satisfying (2.7). Then we have
T ( x , z ) - Q ( x , z ) = lim n f ( 2 n x , 2 n z ) 4 n - Q ( 2 n x , 2 n z ) 4 n lim n 1 4 n L 1 - L ϕ ( 2 n x , 0 , 2 n z , 0 ) = 0

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 × AA is a mapping such that
D μ f ( x , y , z , t ) θ x p + y p + z p + t p ,
lim n 4 - n f ( 2 n z , 2 n f ( x , y ) ) - f ( 4 - n f ( 2 n z , 2 n x ) , y ) - f ( x , 4 - n f ( 2 n z , 2 n y ) ) θ ( ǁ x ǀ ǀ p + ǁ y ǀ ǀ p ) ,
lim n 4 - n f ( 2 n x , 2 n x ) = 0
for all μ T 1 n o 1 and x, y, z, t A. Then there exists a unique bilinear mapping T : A × AA such that
ǀ f ( x , z ) - T ( x , z ) ǁ 4 p θ 4 - 4 p ( ǁ x ǀ ǀ p + ǁ z ǀ ǀ p )
for all x, z A. Moreover, for any sequence {a m } in A, if
lim m lim n 4 - n f ( 2 n x , 2 n a m ) = lim n lim m 4 - n f ( 2 n x , 2 n a m )

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): = θx ǁ p y ǁ p z ǁ p 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 p 0 , 1 4 and θ [0, ∞) be real numbers. Suppose that f: A × AA is a mapping such that
ǁ D μ f ( x , y , z , t ) ǁ θ ( ǁ x ǀ ǀ p ǁ y ǀ ǀ p ǁ z ǀ ǀ p ǁ t ǀ ǀ p ) ,
lim n ǁ 4 - n f ( 2 n z , 2 n f ( x , y ) ) - f ( 4 - n f ( 2 n z , 2 n x ) , y ) - f ( x , 4 - n f ( 2 n z , 2 n y ) ) ǁ θ ( ǁ x ǀ ǀ p ǁ y ǀ ǀ p ) ,
lim n 4 - n f ( 2 n x , 2 n x ) = 0
for all μ T 1 n o 1 and x, y, z, t A. Moreover, for any sequence {a m } in A, if
lim m lim n 4 - n f ( 2 n x , 2 n a m ) = lim n lim m 4 - n f ( 2 n x , 2 n a m )

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): = θx ǁ p ǁy ǁ p ǁz ǁ p ǁt ǁ p ) for all x, y, z M and L = 1 2 in Theorem 2.2, the conclusion follows. □

Declarations

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).

Authors’ Affiliations

(1)
Center of excellence in nonlinear analysis and applications, Department of mathematics, Semnan university, Semnan, Iran
(2)
Department of mathematics education and the rins, Gyeongsang national university, Chinju, Korea

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Copyright

© Gordji et al; licensee Springer. 2012

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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