Hyers-Ulam stability of derivations on proper Jordan CQ*-algebras

Abstract

Eskandani and Vaezi proved the Hyers-Ulam stability of derivations on proper Jordan CQ*-algebras associated with the following Pexiderized Jensen type functional equation

$$kf\left(\frac{x+y}{k}\right)=f_0\left(x\right)+f_1\left(y\right)$$

by using direct method. Using fixed point method, we prove the Hyers-Ulam stability of derivations on proper Jordan CQ*-algebras. Moreover, we investigate the Pexiderized Jensen type functional inequality in proper Jordan CQ*-algebras.

Mathematics Subject Classification 2010: Primary, 17B40; 39B52; 47N50; 47L60; 46B03; 47H10.

1. Introduction and preliminaries

In 1940, Ulam [1] asked the first question on the stability problem. In 1941, Hyers [2] solved the problem of Ulam. This result was generalized by Aoki [3] for additive mappings and by Rassias [4] for linear mappings by considering an unbounded Cauchy difference. In 1994, a generalization of Rassias' Theorem was obtained by Găvruta [5]. Since then, several stability problems for various functional equations have been investigated by numerous mathematicians (see [6–25], M Eshaghi Gordji, unpublished work).

The Jensen equation is $2f\left(\frac{x+y}{2}\right)=f\left(x\right)+f\left(y\right)$, where f is a mapping between linear spaces. It is easy to see that a mapping f : X → Y between linear spaces with f(0) = 0 satisfies the Jensen equation if and only if it is additive [26]. Stability of the Jensen equation has been studied at first by Kominek [27].

We recall some basic facts concerning quasi *-algebras.

Definition 1.1. Let A be a linear space and let A₀ be a *-algebra contained in A as a subspace. We say that A is a quasi *-algebra over A₀ if

(i) the right and left multiplications of an element of A and an element of A₀ are defined and linear;

(ii) x₁(x₂a) = (x₁x₂)a, (ax 1)x 2 = a(x₁x₂) and x₁(ax₂) = (x₁a)x₂ for all x₁, x₂ ∈ A₀ and all a ∈ A;

(iii) an involution *, which extends the involution of A₀, is defined in A with the property (ab)* = b*a* whenever the multiplication is defined.

Quasi *-algebras [28, 29] arise in natural way as completions of locally convex *-algebras whose multiplication is not jointly continuous; in this case one has to deal with topological quasi *-algebras.

A quasi *-algebra (A, A _o ) is said to be a locally convex quasi *-algebra if in A a locally convex topology τ is defined such that

(i) the involution is continuous and the multiplications are separately continuous;

(ii) A _o is dense in A[τ].

Throughout this article, we suppose that a locally convex quasi *-algebra (A, A₀) is complete. For an overview on partial *-algebra and related topics we refer to [30].

In a series of articles [31–35] many authors have considered a special class of quasi * algebras, called proper CQ*-algebras, which arise as completions of C*-algebras. They can be introduced in the following way:

Let A be a Banach module over the C*-algebra A₀ with involution * and C*-norm || . ||₀ such that A₀ ⊂ A. We say that (A, A₀) is a proper CQ*-algebra if

(i) A₀ is dense in A with respect to its norm || · ||;

(ii) (ab)* = b*a* whenever the multiplication is defined;

(iii) || y ||₀= sup_{a ∈ A, || a ||≤1}|| ay || for all y ∈ A₀.

Definition 1.2. A proper CQ*-algebra (A, A₀), endowed with the Jordan product

$$z\circ x=\frac{zx+xz}{2}$$

for all x ∈ A and all z ∈ A₀, is called a proper Jordan CQ*-algebra.

Definition 1.3. Let (A, A₀) be proper Jordan CQ*-algebras.

A ℂ-linear mapping δ: A₀ → A is called a Jordan derivation if

$$\delta \left(x\circ y\right)=x\circ \delta \left(y\right)+\delta \left(x\right)\circ y$$

for all x, y ∈ A₀.

Park and Rassias [36] investigated homomorphisms and derivations on proper JCQ*- triples.

Throughout this article, assume that k is a fixed positive integer.

Eskandani and Vaezi [37] proved the Hyers-Ulam stability of derivations on proper Jordan CQ*-algebras associated with the following Pexiderized Jensen type functional equation

$$kf\left(\frac{x+y}{k}\right)=f_0\left(x\right)+f_1\left(y\right)$$

by using direct method.

In this article, using fixed point method, we prove the Hyers-Ulam stability of derivations on proper Jordan CQ*-algebras.

Moreover, we investigate the Pexiderized Jensen type functional inequality in proper Jordan CQ*-algebras.

2. Derivations on proper Jordan CQ*-algebras

Throughout this section, assume that (A, A₀) is a proper Jordan CQ*-algebra with C*-norm || · ||_A0 and norm || · || _A .

Theorem 2.1. Let φ : A₀ × A₀ → [0, + ∞) be a function such that

$$\underset{n\to \infty }{\mathsf{\text{lim}}}\frac{1}{4^n}\phi \left(2^{n}x,2^{n}y\right)=0$$

(2.1)

for all x, y ∈ A₀ . Suppose that f, f₀, f₁ : A₀ → A are mappings with f(0) = 0 and

$$\left|\right|\mu f\left(x\right)-f_0\left(y\right)-f_1\left(z\right)|{|}_A\le {∥kf\left(\frac{\mu x+y+z}{k}\right)∥}_A$$

(2.2)

$$\left|\right|f\left(x\circ y\right)+x\circ f_1\left(y\right)+f_0\left(x\right)\circ y|{|}_A\le \phi \left(x,y\right)$$

(2.3)

for all $\mu \in T^1:\left\{\mu \in ℂ:\left|\mu \right|=1\right\}$ and all x, y, z ∈ A₀. Then the mapping f : A₀ → A is a Jordan derivation. Moreover,

$$f\left(x\right)=f_0\left(0\right)-f_0\left(x\right)=f_1\left(0\right)-f_1\left(x\right)$$

for all x ∈ A₀.

Proof. Letting x = yz = 0 in (2.2), we get f₀(0) + f₁(0) = 0.

Letting µ = 1, y = -x and z = 0 in (2.2), we get

$$f\mathsf{\text{(}}x\mathsf{\text{)}}=f_{\mathsf{\text{0}}}\left(-x\right)+f_{\mathsf{\text{1}}}\mathsf{\text{(0)}}=f_{\mathsf{\text{0}}}\left(-x\right)-f_{\mathsf{\text{0}}}\mathsf{\text{(0)}}$$

(2.4)

for all x ∈ A₀. Similarly, we have

$$f\mathsf{\text{(}}x\mathsf{\text{)}}=f_{\mathsf{\text{1}}}\mathsf{\text{(-}}x\mathsf{\text{)}}+f_{\mathsf{\text{0}}}\mathsf{\text{(0)}}=f_{\mathsf{\text{1}}}\mathsf{\text{(-}}x\mathsf{\text{)}}-f_{\mathsf{\text{1}}}\mathsf{\text{(0)}}$$

(2.5)

for all x ∈ A₀. By (2.2), we have

$$\begin{array}{ll}\hfill \left|\right|f\left(x+y\right)-f\left(x\right)-f\left(y\right)|{|}_A& =\left|\right|f\left(x+y\right)-\left(f_0\left(-x\right)+f_1\left(0\right)\right)-\left(f_1\left(-y\right)+f_0\left(0\right)\right)|{|}_A\\ =\left|\right|f\left(x+y\right)-f_0\left(-x\right)-f_1\left(-y\right)|{|}_A=0\end{array}$$

for all x, y ∈ A₀. So the mapping f : A₀ → A is additive. Letting y = -µx and z = 0 in (2.2), we get

$$\mu f\left(x\right)=f_0\left(-\mu x\right)+f_1\left(0\right)=f\left(\mu x\right)$$

for all x ∈ A₀. By the same reasoning as in the proof of [[38], Theorem 2.1], the mapping f : A₀ → A is ℂ-linear. By (2.1) and (2.3), we have

$$\begin{array}{l}\left|\right|f\left(x\circ y\right)-x\circ f\left(y\right)-f\left(x\right)\circ y|{|}_A\\ =\underset{n\to \infty }{\mathsf{\text{lim}}}\frac{1}{4^n}\left|\right|f\left(2^{n}x\circ 2^{n}y\right)-2^{n}x\circ \left(f_1\left(-2^{n}y\right)-f_1\left(0\right)\right)-\left(f_0\left(-2^{n}x\right)-f_0\left(0\right)\right)\circ 2^{n}y|{|}_A\\ =\underset{n\to \infty }{\mathsf{\text{lim}}}\frac{1}{4^n}\left|\right|f\left(2^{n}x\circ 2^{n}y\right)-2^{n}x\circ f_1\left(-2^{n}y\right)-f_0\left(-2^{n}x\right)\circ 2^{n}y|{|}_A\\ \le \underset{n\to \infty }{\mathsf{\text{lim}}}\frac{\phi \left(-2^{n}x,-2^{n}y\right)}{4^n}=0\end{array}$$

for all x, y ∈ A₀. So

$$f\left(x\circ y\right)=x\circ f\left(y\right)+f\left(x\right)\circ y$$

for all x, y ∈ A₀. Therefore, the mapping f : A₀ → A is a Jordan derivation.

Since f(-x) = -f(x) for all x ∈ A₀, it follows from (2.4) that

$$f\mathsf{\text{(}}x\mathsf{\text{)}}=-f\left(-x\right)=-\mathsf{\text{(}}{f}_{\mathsf{\text{0}}}\mathsf{\text{(}}x\mathsf{\text{)}}-f_{\mathsf{\text{0}}}\mathsf{\text{(0))}}=f_{\mathsf{\text{0}}}\mathsf{\text{(0)}}-f_{\mathsf{\text{0}}}\mathsf{\text{(}}x\mathsf{\text{)}}$$

for all x ∈ A₀. It follows from (2.5) that

$$f\mathsf{\text{(}}x\mathsf{\text{)}}=-f\mathsf{\text{(}}-x\mathsf{\text{)}}=-\mathsf{\text{(}}{f}_{\mathsf{\text{1}}}\mathsf{\text{(}}x\mathsf{\text{)}}-f_{\mathsf{\text{1}}}\mathsf{\text{(0))}}=f_{\mathsf{\text{1}}}\mathsf{\text{(0)}}-f_{\mathsf{\text{1}}}\mathsf{\text{(}}x\mathsf{\text{)}}$$

for all x ∈ A₀. This completes the proof.

   □

Corollary 2.2. Let θ, r₀, r₁ be nonnegative real numbers with r₀ + r₁ < 2, and let f, f₀, f₁ : A₀ → A be mappings satisfying f (0) = 0, (2.2) and

$$\left|\right|f\left(x\circ y\right)+x\circ f_1\left(y\right)+f_0\left(x\right)\circ y|{|}_A\le \theta |\left|x\right|{|}_{A_0}^{r_0}\left|\right|y|{|}_{A_0}^{r_1}$$

for all x, y ∈ A₀. Then the mapping f : A₀ → A is a Jordan derivation. Moreover,

$$f\left(x\right)=f_0\left(0\right)-f_0\left(x\right)=f_1\left(0\right)-f_1\left(x\right)$$

for all x ∈ A₀.

Proof. The proof follows from Theorem 2.1.

   □

Corollary 2.3. Let θ, r₀, r₁ be nonnegative real numbers with r < 2 and let f, f₀, f₁ : A₀ → A be mappings satisfying f(0) = 0, (2.2) and

$$\left|\right|f\left(x\circ y\right)+x\circ f_1\left(y\right)+f_0\left(x\right)\circ y|{|}_A\le \theta \left(\left|\right|x|{|}_{A_0}^r+|\left|y\right|{|}_{A_0}^r\right)$$

for all x, y ∈ A₀. Then the mapping f : A₀ → A is a Jordan derivation. Moreover,

$$f\left(x\right)=f_0\left(0\right)-f_0\left(x\right)=f_1\left(0\right)-f_1\left(x\right)$$

for all x ∈ A.

3. Hyers-Ulam stability of derivations on proper Jordan CQ*-algebras

We now introduce one of fundamental results of fixed point theory. For the proof, refer to [39, 40]. For an extensive theory of fixed point theorems and other nonlinear methods, the reader is referred to the book of Hyers et al. [8].

Let X be a set. A function d : X × X → [0, ∞] is called a generalized metric on X if and only if d satisfies:

(GM₁) d(x, y) = 0 if and only if x = y;

(GM₂) d(x, y) = d(y, x) for all x, y ∈ X;

(GM₃) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X.

Note that the distinction between the generalized metric and the usual metric is that the range of the former is permitted to include the infinity.

Let (X, d) be a generalized metric space. An operator T : X → X satisfies a Lipschitz condition with Lipschitz constant L if there exists a constant L ≥ 0 such that

$$d\left(Tx,Ty\right)\le Ld\left(x,y\right)$$

for all x, y ∈ X. If the Lipschitz constant L is less than 1, then the operator T is called a strictly contractive operator.

We recall the following theorem by Diaz and Margolis [39].

Theorem 3.1. Suppose that we are given a complete generalized metric space (Ω, d) and a strictly contractive function T : Ω → Ω with Lipschitz constant L. Then for each given x ∈ Ω, either

$$d\left(T^{m}x,T^{m+1}x\right)=\infty forallm\ge 0,$$

or other exists a natural number m ₀ such that

★ d(T^mx, T^m+1x) < ∞ for all m ≥ m₀;

★ the sequence {T^mx} is convergent to a fixed point y* of T;

★ y* is the unique fixed point of T in

$$\mathrm{\Lambda }=\left\{y\in \Omega :d\left(T^{m_0}x,y\right)<\infty \right\};$$

★ $d\left(y,y^{*}\right)\le \frac{1}{1-L}d\left(y,Ty\right)$ for all y ∈ Λ.

Now we prove the Hyers-Ulam stability of derivations on proper Jordan CQ*-algebras by using fixed point method.

Theorem 3.2. Let f, f₀, f₁ : A₀ → A be mappings with f(0) = 0 for which there exists a function $\phi :A_0^2\to \left[0,\infty \right)$ with φ(0, 0) = 0 such that

$${∥kf\left(\frac{\mu x+\mu y}{k}\right)-\mu f_0\left(x\right)-\mu f_1\left(y\right)∥}_A\le \phi \left(x,y\right),$$

(3.1)

$${∥kf\left(\frac{x\circ y}{k}\right)-x\circ f_1\left(y\right)-f_0\left(x\right)\circ y∥}_A\le \phi \left(x,y\right)$$

(3.2)

for all $\mu \in T^1$ and all x, y ∈ A₀. If there exists an L < 1 such that $\phi \left(x,y\right)\le 2L\phi \left(\frac{x}{2},\frac{y}{2}\right)$ for all x, y ∈ A₀, then there exists a unique Jordan derivation δ : A₀ → A such that

$$\begin{array}{c}\left|\right|f\left(x\right)-\delta \left(x\right)|{|}_A\le \frac{1}{2k-2kL}\phi \left(kx,kx\right),\\ \left|\right|f_0\left(x\right)-f_0\left(0\right)-\delta \left(x\right)|{|}_A\le \frac{1}{2-2L}\phi \left(x,x\right)\end{array}$$

(3.3)

for all x ∈ A₀. Moreover, f₀(x) - f₀(0) = f₁(x) - f₁(0) for all x ∈ A₀.

Proof. Letting x = y = 0 and µ = 1 in (3.1), we get f₀(0) + f₁(0) = 0.

Letting y = 0 and µ = 1 in (3.1), we get

$$kf\left(\frac{x}{k}\right)=f_0\left(x\right)+f_1\left(0\right)=f_0\left(x\right)-f_0\left(0\right)$$

(3.4)

for all x ∈ A₀. Similarly, we get

$$kf\left(\frac{y}{k}\right)=f_1\left(y\right)+f_0\left(0\right)=f_1\left(y\right)-f_1\left(0\right)$$

(3.5)

for all y ∈ A₀. Using (3.4) and (3.5), we get

$$f_0\left(x\right)-f_0\left(0\right)=f_1\left(x\right)-f_1\left(0\right)$$

for all x ∈ A₀.

Let H : A 0 → A be a mapping defined by

$$H\left(x\right):=f_0\left(x\right)-f_0\left(0\right)=f_1\left(x\right)-f_1\left(0\right)=kf\left(\frac{x}{k}\right)$$

for all x ∈ A₀. Then we have

$$\left|\right|H\left(\mu x+\mu y\right)-\mu H\left(x\right)-\mu H\left(y\right)|{|}_A\le \phi \left(x,y\right)$$

(3.6)

for all $\mu \in T^1$ and x, y ∈ A₀.

Consider the set

$$X:=\left\{g:A_0\to A\right\}$$

and introduce the generalized metric on X:

$$d\left(g,h\right)=\mathsf{\text{inf}}\left\{C\in {ℝ}_{+}:\left|\right|g\left(x\right)-h\left(x\right)|{|}_A\le C\phi \left(x,x\right),\forall x\in A_0\right\}.$$

It is easy to show that (X, d) is complete (see [[41], Lemma 2.1]).

Now we consider the linear mapping J : X → X such that

$$Jg\left(x\right):=\frac{1}{2}g\left(2x\right)$$

for all x ∈ A.

By [[41], Theorem 3.1],

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

for all g, h ∈ X.

Letting µ = 1 and y = x in (3.6), we get

$$\left|\right|H\left(2x\right)-2H\left(x\right)\left|\right|\le \phi \left(x,x\right)$$

(3.7)

and so

$$\left|\right|H\left(x\right)-\frac{1}{2}H\left(2x\right)\left|\right|\le \frac{1}{2}\phi \left(x,x\right)$$

for all x ∈ A₀. Hence $d\left(H,JH\right)\le \frac{1}{2}$.

By Theorem 3.1, there exists a mapping δ : A₀ → A such that

• (1) δ is a fixed point of J, i.e.,

  $$\delta \left(2x\right)=2\delta \left(x\right)$$

  (3.8)

for all x ∈ A₀. The mapping δ is a unique fixed point of J in the set

$$Y=\left\{g\in X:d\left(f,g\right)<\infty \right\}.$$

This implies that δ is a unique mapping satisfying (3.8) such that there exists C ∈ (0, ∞) satisfying

$$\left|\right|H\left(x\right)-\delta \left(x\right)|{|}_A\le C\phi \left(x,x\right)$$

for all x ∈ A₀.

• (2) d(JⁿH, δ) → 0 as n → ∞. This implies the equality

  $$\underset{n\to \infty }{\mathsf{\text{lim}}}\frac{H\left(2^{n}x\right)}{2^n}=\delta \left(x\right)$$

  (3.9)

for all x ∈ A₀.

• (3) $d\left(H,\delta \right)\le \frac{1}{1-L}d\left(H,JH\right)$, which implies the inequality

  $$d\left(H,\delta \right)\le \frac{1}{2-2L}.$$

This implies that the inequality (3.3) holds.

It follows from (3.6) and (3.9) that

$$\begin{array}{l}\left|\right|\delta \left(\mu x+\mu y\right)-\mu \delta \left(x\right)-\mu \delta \left(y\right)|{|}_A\\ =\underset{n\to \infty }{\mathsf{\text{lim}}}\frac{1}{2^n}\left|\right|H\left(2^n\mu x+2^n\mu y\right)-\mu H\left(2^{n}x\right)-\mu H\left(2^{n}y\right)|{|}_A\\ \le \underset{n\to \infty }{\mathsf{\text{lim}}}\frac{1}{2^n}\phi \left(2^{n}x,2^{n}y\right)=0\end{array}$$

for $\mu \in T^1$ and all x, y ∈ A₀. So

$$\delta \left(\mu x+\mu y\right)=\mu \delta \left(x\right)+\mu \delta \left(y\right)$$

for $\mu \in T^1$ and all x, y ∈ A₀. By the same reasoning as in the proof of [[38], Theorem 2.1], the mapping δ : A₀ → A is ℂ-linear.

It follows from $\phi \left(x,y\right)\le 2L\phi \left(\frac{x}{2},\frac{y}{2}\right)$ that

$$\underset{n\to \infty }{\mathsf{\text{lim}}}\frac{1}{4^n}\phi \left(2^{n}x,2^{n}y\right)\le \frac{1}{2^n}\phi \left(2^{n}x,2^{n}y\right)=0$$

(3.10)

for all x, y ∈ A₀.

It follows from (3.2) and (3.10) that

$$\begin{array}{l}\left|\right|\delta \left(x\circ y\right)-x\circ \delta \left(y\right)-\delta \left(x\right)\circ y|{|}_A\\ \le \underset{n\to \infty }{\mathsf{\text{lim}}}\frac{1}{4^n}{∥kf\left(4^n\left(\frac{x\circ y}{k}\right)\right)-2^{n}x\circ \left(f_1\left(2^{n}y\right)+f_0\left(0\right)\right)-\left(f_0\left(2^{n}x\right)+f_1\left(0\right)\right)\circ 2^{n}y∥}_A\\ \le \underset{n\to \infty }{\mathsf{\text{lim}}}\frac{1}{4^n}{∥kf\left(4^n\left(\frac{x\circ y}{k}\right)\right)-2^{n}x\circ f_1\left(2^{n}y\right)-f_0\left(2^{n}x\right)\circ 2^{n}y∥}_A\\ \le \underset{n\to \infty }{\mathsf{\text{lim}}}\frac{1}{4^n}\phi \left(2^{n}x,2^{n}y\right)=0\end{array}$$

for all x, y ∈ A₀. Hence

$$\delta \left(x\circ y\right)=x\circ \delta \left(y\right)+\delta \left(x\right)\circ y$$

for all x, y ∈ A₀. So δ : A₀ → A is a Jordan derivation, as desired.

   □

Corollary 3.3. [[37], Theorem 3.1] Let be a nonnegative real number and r₀, r₁ positive real numbers with λ:= r₀ + r₁ < 1 and let f, f₀, f₁ : A₀ → A be mappings with f (0) = 0 such that

$${∥kf\left(\frac{\mu x+\mu y}{k}\right)-\mu f_0\left(x\right)-\mu f_1\left(y\right)∥}_A\le \theta \left|\right|x\left|{|}_{A_0}^{r_0}\right|\left|y\right|{|}_{A_0}^{r_1},$$

(3.11)

$${∥kf\left(\frac{x\circ y}{k}\right)-x\circ f_1\left(y\right)-f_0\left(x\right)\circ y∥}_A\le \theta \left|\right|x\left|{|}_{A_0}^{r_0}\right|\left|y\right|{|}_{A_0}^{r_1}$$

(3.12)

for all $\mu \in T^1$ and all x, y ∈ A₀. Then there exists a unique Jordan derivation δ : A₀ → A such that

$$\begin{array}{ll}\hfill \left|\right|f\left(x\right)-\delta \left(x\right)|{|}_A& \le \frac{k^{\lambda -1}\theta }{2-2^{\lambda }}\left|\right|x|{|}_{A_0}^{\lambda },\\ \hfill \left|\right|f_0\left(x\right)-f_0\left(0\right)-\delta \left(x\right)|{|}_A& \le \frac{\theta }{2-2^{\lambda }}\left|\right|x|{|}_{A_0}^{\lambda }\end{array}$$

for all x ∈ A₀. Moreover, f₀(x) - f₀(0) = f₁(x) - f₁(0) for all x ∈ A₀.

Proof. The proof follows from Theorem 3.2 by taking

$$\phi \left(x,y\right):=\theta \left|\right|x\left|{|}_{A_0}^{r_0}\right|\left|y\right|{|}_{A_0}^{r_1}$$

for all x, y ∈ A. Letting L = 2^λ-1, we get the desired result.

   □

Corollary 3.4. [[37], Theorem 3.4] Let θ, r be a nonnegative real numbers with 0 < r < 1, and let f, f₀, f₁ : A₀ → A be mappings with f (0) = 0 such that

$${∥kf\left(\frac{\mu x+\mu y}{k}\right)-\mu f_0\left(x\right)-\mu f_1\left(y\right)∥}_A\le \theta \left(\left|\right|x|{|}_{A_0}^r+|\left|y\right|{|}_{A_0}^r\right),$$

(3.13)

$${∥kf\left(\frac{xoy}{k}\right)-xo{f}_1\left(y\right)-f_0\left(x\right)oy∥}_A\le \theta \left(\left|\right|x|{|}_{A_0}^r+|\left|y\right|{|}_{A_0}^r\right)$$

(3.14)

for all $\mu \in T^1$ and all x, y ∈ A₀. Then there exists a unique Jordan derivation δ : A₀ → A such that

$$\begin{array}{c}\left|\right|f\left(x\right)-\delta \left(x\right)|{|}_A\le \frac{2k^{r-1}\theta }{2-2^r}|\left|x\right|{|}_{A_0}^r,\\ \left|\right|f_0\left(x\right)-f_0\left(0\right)-\delta \left(x\right)|{|}_A\le \frac{2\theta }{2-2^r}|\left|x\right|{|}_{A_0}^r\end{array}$$

for all x ∈ A₀. Moreover, f₀(x) - f₀(0) = f₁(x) - f₁(0) for all x ∈ A₀.

Proof. The proof follows from Theorem 3.2 by taking

$$\phi \left(x,y\right):=\theta \left(\left|\right|x|{|}_{A_0}^r+|\left|y\right|{|}_{A_0}^r\right)$$

for all x, y ∈ A. Letting L = 2^{r- 1}, we get the desired result.

   □

Theorem 3.5. Let f, f₀, f₁ : A₀ → A be mappings with f(0) = f₀(0) = f₁(0) = 0 for which there exists a function $\phi :A_0^2\to \left[0,\infty \right)$ satisfying (3.1) and (3.2). If there exists an L < 1 such that $\phi \left(x,y\right)\le \frac{L}{4}\phi \left(2x,2y\right)$ for all x, y ∈ A₀, then there exists a unique Jordan derivationδ: A₀ → A such that

$$\begin{array}{ll}\hfill \left|\right|f\left(x\right)-\delta \left(x\right)|{|}_A& \le \frac{L}{4k-4kL}\phi \left(kx,kx\right),\\ \hfill \left|\right|f_0\left(x\right)-\delta \left(x\right)|{|}_A& \le \frac{L}{4-4L}\phi \left(x,x\right)\end{array}$$

(3.15)

for all x ∈ A₀. Moreover, f₀(x) = f₁(x) for all x ∈ A₀.

Proof. Let (X, d) be the generalized metric space defined in the proof of Theorem 3.2.

Now we consider the linear mapping J : X → X such that

$$Jg\left(x\right):=2g\left(\frac{x}{2}\right)$$

for all x ∈ X.

Let $H\left(x\right):=f_0\left(x\right)=f_1\left(x\right)=kf\left(\frac{x}{k}\right)$ for all x ∈ A₀. It follows from (3.7) that

$$∥H\left(x\right)-2H\left(\frac{x}{2}\right)∥\le \phi \left(\frac{x}{2},\frac{x}{2}\right)\le \frac{L}{4}\phi \left(x,y\right)$$

for all x ∈ A₀. Thus $d\left(H,JH\right)\le \frac{L}{4}$. One can show that there exists a mapping δ : A₀ → A such that

$$d\left(H,\delta \right)\le \frac{L}{4-4L}.$$

Hence we obtain the inequality (3.15).

It follows from $\phi \left(x,y\right)\le \frac{L}{4}\phi \left(2x,2y\right)$ that

$$\underset{n\to \infty }{\mathsf{\text{lim}}}{4}^n\phi \left(\frac{x}{2^n},\frac{y}{2^n}\right)=0$$

for all x, y ∈ A₀. So

$$\begin{array}{l}\left|\right|\delta \left(x\circ y\right)-x\circ \delta \left(y\right)-\delta \left(x\right)\circ y|{|}_A\\ \le \underset{n\to \infty }{\mathsf{\text{lim}}}{4}^n{∥kf\left(4^n\frac{x\circ y}{4^{n}k}\right)-\frac{x}{2^n}\circ f_1\left(\frac{y}{2^n}\right)-f_0\left(\frac{x}{2^n}\right)\circ \frac{y}{2^n}∥}_A\\ \le \underset{n\to \infty }{\mathsf{\text{lim}}}{4}^n\phi \left(\frac{x}{2^n},\frac{y}{2^n}\right)=0\end{array}$$

for all x, y ∈ A₀. Hence

$$\delta \left(x\circ y\right)=x\circ \delta \left(y\right)+\delta \left(x\right)\circ y$$

for all x, y ∈ A₀. So δ : A₀ → A is a Jordan derivation, as desired.

The rest of the proof is similar to the proof of Theorem 3.2.

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Corollary 3.6. [[37], Theorem 3.2] Let θ be a nonnegative real number and r₀, r₁ positive real numbers with λ:= r₀ + r₁ > 2 and let f, f₀, f₁ : A₀ → A be mappings satisfying f(0) = f₀(0) = f₁(0) = 0, (3.11) and (3.12). Then there exists a unique Jordan derivation δ: A₀ → A such that

$$\begin{array}{c}\left|\right|f\left(x\right)-\delta \left(x\right)|{|}_A\le \frac{k^{\lambda -1}\theta }{2^{\lambda }-4}|\left|x\right|{|}_{A_0}^{\lambda }\\ \left|\right|f_i\left(x\right)-\delta \left(x\right)|{|}_A\le \frac{\theta }{2^{\lambda }-4}|\left|x\right|{|}_{A_0}^{\lambda }\end{array}$$

for all x ∈ A₀. Moreover, f₀(x) = f₁(x) for all x ∈ A₀.

Proof. The proof follows from Theorem 3.3 by taking

$$\phi \left(x,y\right):=\theta \left|\right|x\left|{|}_{A_0}^{r_0}\right|\left|y\right|{|}_{A_0}^{r_1}$$

for all x, y ∈ A. Letting L = 2^2-λ, we get the desired result.

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Corollary 3.7. [[37], Theorem 3.3] Let θ, r be nonnegative real numbers with r > 2, and let f, f₀, f₁ : A₀ → A be mappings satisfying f (0) = f₀(0) = f₁(0) = 0, (3.13) and (3.14). Then there exists a unique Jordan derivation δ : A₀ → A such that

$$\begin{array}{ll}\hfill \left|\right|f\left(x\right)-\delta \left(x\right)|{|}_A& \le \frac{2k^{r-1}\theta }{2^r-4}\left|\right|x|{|}_{A_0}^r,\\ \hfill \left|\right|f_0\left(x\right)-\delta \left(x\right)|{|}_A& \le \frac{2\theta }{2^r-4}\left|\right|x|{|}_{A_0}^r\end{array}$$

for all x ∈ A₀. Moreover, f₀(x) = f₁(x) for all x ∈ A₀.

Proof. The proof follows from Theorem 3.3 by taking

$$\phi \left(x,y\right):=\theta \left(\left|\right|x|{|}_{A_0}^r+|\left|y\right|{|}_{A_0}^r\right)$$

for all x, y ∈ A. Letting L = 2^2-r, we get the desired result.

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