Approximate lie brackets: a fixed point approach

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 $\mathbb{F}$ 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) 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,

$${\mathbb{T}}^1:=\left\{z\in ℂ:ǀzǀ=1\right\},{\mathbb{T}}_{\frac{1}{n_o}}^1:=\left\{e^{i\theta }:0\le \theta \le \frac{2\pi }{n_0}\right\}.$$

It is easy to see that ${\mathbb{T}}^1={\mathbb{T}}_{\frac{1}{1}}^1$. Moreover, we suppose that A is a complex Banach space. For any mapping f : A × A → A, we define

$$\begin{array}{cc}\hfill D_{\mu }f\left(x,y,z,t\right):& =4\mu f\left(\frac{x+y}{2},\frac{z+t}{2}\right)+4\mu f\left(\frac{x-y}{2},\frac{z+t}{2}\right)\hfill \\ +4\mu f\left(\frac{x+y}{2}+\frac{z-t}{2}\right)+4\mu f\left(\frac{x-y}{2},\frac{z-t}{2}\right)\hfill \\ -4f\left(\mu x,z\right)\hfill \end{array}$$

for all $\mu \in {\mathbb{T}}_{\frac{1}{n_o}}^1$ 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^mx, T^m+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 {T^mx} is convergent to a fixed point y* of T;

• (3) y* is the unique fixed point of T in the set $\Lambda =\{y\in \Omega :d\left(T^{m_0}x,y\right)<\infty$;

• (4) $d\left(y,y^{*}\right)\le \frac{1}{1-L}d\left(y,Ty\right)$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

$$ǁD_{\mu }f\left(x,y,z,t\right)ǁ\le \varphi \left(x,y,z,t\right),$$

(2.1)

$$\begin{array}{c}\underset{n\to \infty }{\text{lim}}ǁ4^{-n}f\left(2^{n}z,2^{n}f\left(x,y\right)\right)-f\left(4^{-n}f\left(2^{n}z,2^{n}x\right),y\right)-f\left(x,4^{-n}f\left(2^{n}z,2^{n}y\right)\right)ǁ\hfill \\ \le \varphi \left(x,y,0,0\right),\hfill \end{array}$$

(2.2)

$$\underset{n\to \infty }{\text{lim}}{4}^{-n}f\left(2^{n}x,2^{n}x\right)=0$$

(2.3)

for all $\mu \in {\mathbb{T}}_{\frac{1}{n_o}}^1$ and x, y, z, t ∈ A. If there exists L < 1 such that $\varphi \left(x,y,z,t\right)\le 4L\varphi \left(\frac{x}{2},\frac{y}{2},\frac{z}{2},\frac{t}{2}\right)$for all x, y, z, t ∈ A, then there exists a unique bilinear mapping T : A × A → A such that

$$ǁf\left(x,z\right)-T\left(x,z\right)ǁ\le \frac{L}{1-L}\varphi \left(x,0,z,0\right)$$

(2.4)

for all x, z ∈ M. Moreover, for any sequence {a _m } in A, if

$$\underset{m\to \infty }{\text{lim}}\underset{n\to \infty }{\text{lim}}{4}^{-n}f\left(2^{n}x,2^n{a}_m\right)=\underset{n\to \infty }{\text{lim}}\underset{m\to \infty }{\text{lim}}{4}^{-n}f\left(2^{n}x,2^n{a}_m\right)$$

(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

$$ǁ4f\left(\frac{x}{2},\frac{z}{2}\right)-f\left(x,z\right)ǁ\le \varphi \left(x,0,z,0\right)$$

for all x, z ∈ A and so

$$ǁ\frac{1}{4}f\left(2x,2z\right)-f\left(x,z\right)ǁ\le \frac{1}{4}\varphi \left(2x,0,2z,0\right)\le L\varphi \left(x,0,z,0\right)$$

(2.6)

for all x, z ∈ A. Consider the set X : = {g : g : A × A → A} and introduce the generalized metric on X by:

$$d\left(h,g\right):=\text{inf}\left\{C\in {ℝ}^{+}:ǁg\left(x,z\right)-h\left(x,z\right)ǁ\le C\varphi \left(x,0,z,0\right)\text{for all }x,z\in A\right\}.$$

It is easy to show that (X, d) is a complete generalized metric space. Now, we define the mapping J : X → X by

$$J\left(h\right)\left(x,z\right)=\frac{1}{4}h\left(2x,2z\right)$$

for all x, z ∈ A. For any g, h ∈ X, we have

$$\begin{array}{cc}\hfill d\left(g,h\right)<C& \Rightarrow ǁg\left(x,z\right)-h\left(x,z\right)ǁ\le C\varphi \left(x,0,z,0\right)\hfill \\ \Rightarrow ǁ\frac{1}{4}g\left(2x,2z\right)-\frac{1}{4}h\left(2x,2z\right)ǁ\le \frac{1}{4}C\varphi \left(2x,0,2z,0\right)\hfill \\ \Rightarrow ǁ\frac{1}{4}g\left(2x,2z\right)-\frac{1}{4}h\left(2x,2z\right)ǁ\le LC\varphi \left(x,0,z,0\right)\hfill \\ \Rightarrow d\left(J\left(g\right),J\left(h\right)\right)\le LC\hfill \end{array}$$

for all x, z ∈ A, which means that

$$d\left(J\left(g\right),J\left(h\right)\right)\le Ld\left(g,h\right)$$

for all g, h ∈ X. It follows from (2.6) that

$$d\left(f,J\left(f\right)\right)\le L.$$

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 lim_n→∞d(Jⁿ (f), T) = 0 and

$$\underset{n\to \infty }{\text{lim}}\frac{1}{4^n}f\left(2^{n}x,2^{n}z\right)=T\left(x,z\right)$$

(2.7)

for all x, z ∈ A. By the inequality d(f, J(f)) ≤ L and J(T) = T, we have

$$d\left(f,T\right)\le d\left(f,J\left(f\right)\right)+d\left(J\left(f\right),J\left(T\right)\right)\le L+Ld\left(f,T\right)$$

and so

$$d\left(f,T\right)\le \frac{L}{1-L}.$$

This implies the inequality (2.4). From $\varphi \left(x,y,z,t\right)\le 4L\varphi \left(\frac{x}{2},\frac{y}{2},\frac{z}{2},\frac{t}{2}\right)$, we have

$$\underset{j\to \infty }{\text{lim}}{4}^{-j}\varphi \left(2^{j}x,2^{j}y,2^{j}z,2^{j}t\right)=0$$

(2.8)

for all x, y, z ∈ A. Thus it follows from (2.1), (2.7) and (2.8) that

$$\begin{array}{c}ǁ4T\left(\frac{x+y}{2},\frac{z+t}{2}\right)+4T\left(\frac{x-y}{2},\frac{z+t}{2}\right)\hfill \\ -4T\left(\frac{x+y}{2},\frac{z-t}{2}\right)+4T\left(\frac{x-y}{2},\frac{z-t}{2}\right)-4T\left(x,z\right)ǁ\hfill \\ =\underset{n\to \infty }{\text{lim}}\frac{1}{4^n}ǁ4f\left(\frac{2^{n}x+2^{n}y}{2},\frac{2^{n}z+2^{n}t}{2}\right)+4f\left(\frac{2^{n}x-2^{n}y}{2}-\frac{2^{n}z+2^{n}t}{2}\right)\hfill \\ -4f\left(\frac{2^{n}x+2^{n}y}{2},\frac{2^{n}z-2^{n}t}{2}\right)-4f\left(\frac{2^{n}x-2^{n}y}{2},\frac{2^{n}z-2^{n}t}{2}\right)-4f\left(2^{n}x,2^{n}z\right)ǁ\hfill \\ \le \underset{n\to \infty }{\text{lim}}\frac{1}{4^n}\varphi \left(2^{n}x,2^{n}y,2^{n}z,2^{n}t\right)=0\hfill \end{array}$$

for all x, y, z ∈ M and so

$$\begin{array}{c}T\left(\frac{x+y}{2},\frac{z+t}{2}\right)+T\left(\frac{x-y}{2},\frac{z+t}{2}\right)-T\left(\frac{x+y}{2},\frac{z-t}{2}\right)+T\left(\frac{x-y}{2},\frac{z-t}{2}\right)\hfill \\ =T\left(x,z\right)\hfill \end{array}$$

for all x, y, z, t ∈ A. This shows that

$$T\left(x+y,z+t\right)=T\left(x,z\right)+T\left(y,z\right)+T\left(x,t\right)+T\left(y,t\right)$$

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

$$ǁ4\mu f\left(x,z\right)-4f\left(\mu x,z\right)ǁ\le \varphi \left(x,x,z,z\right)ǁ$$

(2.9)

for all x, z ∈ A. and $\mu \in {\mathbb{T}}_{\frac{1}{n_o}}^1$ and so

$$\begin{array}{cc}\hfill ǁ4\mu T\left(x,z\right)-4T\left(\mu x,z\right)ǁ& =\underset{n\to \infty }{\text{lim}}\frac{1}{4^n}ǁ4\mu f\left(2^{n}x,2^{n}z\right)-4f\left(2^n\mu x,2^{n}z\right)\hfill \\ \le \underset{n\to \infty }{\text{lim}}\frac{1}{4^n}\varphi \left(2^{n}x,2^{n}x,2^{n}z,2^{n}z\right)=0\hfill \end{array}$$

for all x, z ∈ A and $\mu \in {\mathbb{T}}_{\frac{1}{n_o}}^1$, that is,

$$T\left(\mu x,z\right)=\mu T\left(x,z\right)$$

(2.10)

for all x, z ∈ A.

If λ belongs to ${\mathbb{T}}^{\text{1}}$, then there exists θ ∈ [0, 2π] such that λ = e^iθ. If we set ${\lambda }_1=e^{\frac{i\theta }{n_o}}$, then λ₁ belongs to ${\mathbb{T}}_{\frac{1}{n_o}}^1$. By using (2.10), we have

$$T\left(\lambda x,z\right)=T\left({\lambda }_1^{n_0}x,z\right)={\lambda }_1^{n_0}T\left(x,z\right)=\lambda T\left(x,z\right)$$

for all x, z ∈ M.

If λ belongs to $n{\mathbb{T}}^1=\left\{nz:z\in {\mathbb{T}}^1\right\}$ for some n ∈ ℕ, then, by (2.9), we have

$$\begin{array}{cc}\hfill T\left(\lambda x,z\right)=T\left(n{\lambda }_{1}x,z\right)& =T\left({\lambda }_1\left(nx\right),z\right)\hfill \\ ={\lambda }_{1}T\left(nx,z\right)\hfill \\ ={\lambda }_{1}nT\left(x,z\right)\hfill \\ =\lambda T\left(x,z\right)\hfill \end{array}$$

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 $s_1:=s+\sqrt{n^2-s^2}i$ and $s_2:=t-\sqrt{n^2-s^2}i$, we have $s=\frac{s_1+s_2}{2}$ and $s_1,s_2\in n{\mathbb{T}}^1$. Thus, by (2.9), we have

$$\begin{array}{cc}\hfill T\left(sx,z\right)=T\left(\frac{s_1+s_2}{2}x,z\right)& =T\left(s_1\frac{x}{2},z\right)+T\left(s_2\frac{x}{2},z\right)\hfill \\ =s_{1}T\left(\frac{x}{2},z\right)+s_{2}T\left(\frac{x}{2},z\right)\hfill \\ =4\left(\frac{s_1+s_2}{2}\right)T\left(\frac{x}{2},\frac{z}{2}\right)\hfill \\ =sT\left(x,z\right)\hfill \end{array}$$

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

Therefore, we have

$$T\left(\lambda x,z\right)=T\left(ǀ\lambda ǀe^{i\theta }x,z\right)=ǀ\lambda ǀT\left(e^{i\theta }x,z\right)=ǀ\lambda ǀe^{i\theta }T\left(x,z\right)=\lambda T\left(x,z\right)$$

(2.11)

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

$$0=T\left(x+y,x+y\right)=T\left(x,x\right)+T\left(x,y\right)+T\left(y,x\right)+T\left(y,y\right)=T\left(x,y\right)+T\left(y,x\right)$$

for all x, y ∈ A and so

$$T\left(x,y\right)=-T\left(y,x\right)$$

for all x, y ∈ A, that is, T is skew symmetric. Let z ∈ A and define a mapping ad(z): A →A by

$$ad\left(z\right)\left(x\right)=T\left(z,x\right)$$

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 lim_n→∞a _m = 0, then, by (2.5), we have

$$\begin{array}{cc}\hfill \underset{m\to \infty }{\text{lim}}ad\left(z\right)\left(a_m\right)& =\underset{m\to \infty }{\text{lim}}\underset{n\to \infty }{\text{lim}}{4}^{-n}f\left(2^{n}z,2^n{a}_m\right)\hfill \\ =\underset{n\to \infty }{\text{lim}}\underset{m\to \infty }{\text{lim}}{4}^{-n}f\left(2^{n}z,2^n{a}_m\right)\hfill \\ =\underset{n\to \infty }{\text{lim}}{4}^{-n}f\left(2^{n}z,0\right)=ad\left(z\right)\left(0\right)=0.\hfill \end{array}$$

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^mx and y with 2^my in (2.2) and multiplying by 4^-mboth sides of the inequality, we have

$$\begin{array}{cc}\hfill \underset{n\to \infty }{\text{lim}}{4}^{-m}& ǁ4^{-n}f\left(2^{n}z,2^{n}f\left(2^{m}x,2^{m}y\right)\right)-f\left(4^{-n}f\left(2^{n}z,2^{n+m}x\right),2^{m}y\right)\hfill \\ -f\left(2^{m}x,4^{-n}f\left(2^{n}z,2^{n+m}y\right)\right)ǁ\hfill \\ \le 4^{-m}\varphi \left(2^{m}x,2^{m}y,0,0\right)\hfill \end{array}$$

for all x, y, z ∈ A and m ∈ ℕ. Since f is continuous, we have

$$\begin{array}{cc}\hfill 4^{-m}& ǁad\left(z\right)\left(f\left(2^{m}x,2^{m}y\right)\right)-f\left(ad\left(z\right)\left(2^{m}x\right),2^{m}y\right)-f\left(2^{m}x,ad\left(z\right)2^{m}y\right)ǁ\hfill \\ \le 4^{-m}\varphi \left(2^{m}x,2^{m}y,0,0\right)\hfill \end{array}$$

for all x, y, z ∈ A. Since, for all z ∈ A, ad(z) is a linear and continuous mapping, we get

$$ad\left(z\right)T\left(x,y\right)-T\left(ad\left(z\right)\left(x\right),y\right)-T\left(x,ad\left(z\right)\left(y\right)\right)=0$$

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 × A →A be another bilinear mapping satisfying (2.7). Then we have

$$\begin{array}{cc}\hfill ǁT\left(x,z\right)-Q\left(x,z\right)ǁ& =\underset{n\to \infty }{\text{lim}}ǁ\frac{f\left(2^{n}x,2^{n}z\right)}{4^n}-\frac{Q\left(2^{n}x,2^{n}z\right)}{4^n}ǁ\hfill \\ \le \underset{n\to \infty }{\text{lim}}\frac{1}{4^n}\left(\frac{L}{1-L}\right)\varphi \left(2^{n}x,0,2^{n}z,0\right)=0\hfill \end{array}$$

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

$$ǁD_{\mu }f\left(x,y,z,t\right)ǁ\le \theta \left({ǁxǁ}^p+{ǁyǁ}^p+{ǁzǁ}^p+{ǁtǁ}^p\right),$$

$$\begin{array}{c}\underset{n\to \infty }{\text{lim}}ǁ4^{-n}f\left(2^{n}z,2^{n}f\left(x,y\right)\right)-f\left(4^{-n}f\left(2^{n}z,2^{n}x\right),y\right)-f\left(x,4^{-n}f\left(2^{n}z,2^{n}y\right)\right)ǁ\hfill \\ \le \theta \left(ǁxǀ{ǀ}^p+ǁyǀ{ǀ}^p\right),\hfill \end{array}$$

$$\underset{n\to \infty }{\text{lim}}{4}^{-n}f\left(2^{n}x,2^{n}x\right)=0$$

for all $\mu \in {\mathbb{T}}_{\frac{1}{n_o}}^1$ and x, y, z, t ∈ A. Then there exists a unique bilinear mapping T : A × A →A such that

$$ǀf\left(x,z\right)-T\left(x,z\right)ǁ\le \frac{4^p\theta }{4-4^p}\left(ǁxǀ{ǀ}^p+ǁzǀ{ǀ}^p\right)$$

for all x, z ∈ A. Moreover, for any sequence {a _m } in A, if

$$\underset{m\to \infty }{\text{lim}}\underset{n\to \infty }{\text{lim}}{4}^{-n}f\left(2^{n}x,2^n{a}_m\right)=\underset{n\to \infty }{\text{lim}}\underset{m\to \infty }{\text{lim}}{4}^{-n}f\left(2^{n}x,2^n{a}_m\right)$$

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 = 4^{p -1}. □

Finally, we prove the superstability of Lie brackets as follows:

Corollary 2.4. Let $p\in \left(0,\frac{1}{4}\right)$ and θ ∈ [0, ∞) be real numbers. Suppose that f: A × A →A is a mapping such that

$$ǁD_{\mu }f\left(x,y,z,t\right)ǁ\le \theta \left(ǁxǀ{ǀ}^pǁyǀ{ǀ}^pǁzǀ{ǀ}^pǁtǀ{ǀ}^p\right),$$

$$\begin{array}{c}\underset{n\to \infty }{\text{lim}}ǁ4^{-n}f\left(2^{n}z,2^{n}f\left(x,y\right)\right)-f\left(4^{-n}f\left(2^{n}z,2^{n}x\right),y\right)-f\left(x,4^{-n}f\left(2^{n}z,2^{n}y\right)\right)ǁ\hfill \\ \le \theta \left(ǁxǀ{ǀ}^pǁyǀ{ǀ}^p\right),\hfill \end{array}$$

$$\underset{n\to \infty }{\text{lim}}{4}^{-n}f\left(2^{n}x,2^{n}x\right)=0$$

for all $\mu \in {\mathbb{T}}_{\frac{1}{n_o}}^1$ and x, y, z, t ∈ A. Moreover, for any sequence {a _m } in A, if

$$\underset{m\to \infty }{\text{lim}}\underset{n\to \infty }{\text{lim}}{4}^{-n}f\left(2^{n}x,2^n{a}_m\right)=\underset{n\to \infty }{\text{lim}}\underset{m\to \infty }{\text{lim}}{4}^{-n}f\left(2^{n}x,2^n{a}_m\right)$$

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 ǁ^p ǁy ǁ^p ǁz ǁ^p ǁt ǁ^p) for all x, y, z ∈ M and $L=\frac{1}{2}$ in Theorem 2.2, the conclusion follows. □