Stability of Some Inequalities in Banach ∗-Algebras

Abstract

In this paper, we investigate the stability and superstability of a specific class of functional inequalities associated with centrally extended ∗-derivations on Banach ∗-algebras. A CE ∗-derivation $\delta :\mathcal{R}\to \mathcal{R}$ is defined as an additive mapping satisfying $\delta (x+y)-\delta \left(x\right)-\delta \left(y\right)\in Z\left(\mathcal{R}\right)$ and $\delta \left(xy\right)-\delta \left(x\right)y^{\ast }-x\delta \left(y\right)\in Z\left(\mathcal{R}\right)$ for all $x,y\in \mathcal{R}$, where $Z\left(\mathcal{R}\right)$ denotes the center of the ring. We consider the functional inequality $\parallel [a_1\delta \left(x_1\right)+a_2\delta \left(x_2\right)+a_3\delta \left(x_3\right),w]\parallel \le \parallel [\delta (a_1{x}_1+a_2{x}_2+a_3{x}_3),w]\parallel + \mathrm{\Phi }(x_1,x_2,x_3,w),$ where $\mathrm{\Phi }$ is a perturbing term. By employing the direct method, we establish several theorems concerning the Hyers–Ulam stability of this inequality in the context of unital Banach ∗-algebras. Furthermore, we provide sufficient conditions under which these functional inequalities exhibit superstability. We also explore the implications of our results for linear ∗-derivations in semiprime Banach ∗-algebras with no nonzero central ideals.

1. Introduction

In this paper, let $\mathcal{R}$ be an associative ring with center $Z\left(\mathcal{R}\right).$ For all $x,y\in \mathcal{R},$ the symbol $[x,y]$ will denote the commutator $xy-yx.$ The set $\mathtt{a}\mathtt{n}\mathtt{n}\left(\mathcal{R}\right)=\{x\in \mathcal{R}:\mathcal{R}x=0\}$ denotes the two-sided annihilator of $\mathcal{R}.$ For completeness, we recall some well-known maps. An additive mapping $x\mapsto x^{\ast }$ of $\mathcal{R}$ into itself is called an involution if ${\left(xy\right)}^{\ast }=y^{\ast }{x}^{\ast }$ and ${\left(x^{\ast }\right)}^{\ast }=x$ are fulfilled for all $x,y\in \mathcal{R}.$ Let $\mathcal{R}$ be a ring with an involution. An additive mapping $\delta :\mathcal{R}\to \mathcal{R}$ is said to be a ∗-derivation if $\delta \left(xy\right)=\delta \left(x\right)y^{\ast }+x\delta \left(y\right)$ holds for all $x,y\in \mathcal{R}.$ Bell and Daif [1] introduced centrally extended derivations and discussed their existence. Motivated by this concept, El-Deken and Nabiel [2] introduced the notion that a centrally extended ∗-derivation $\delta :\mathcal{R}\to \mathcal{R}$ is a mapping that satisfies

$$\begin{array}{cc}\hfill & \delta (x+y)-\delta \left(x\right)-\delta \left(y\right)\in Z\left(\mathcal{R}\right)\mathrm{for}\mathrm{all}x,y\in \mathcal{R},\hfill \\ \hfill & \delta \left(xy\right)-\delta \left(x\right)y^{\ast }-x\delta \left(y\right)\in Z\left(\mathcal{R}\right)\mathrm{for}\mathrm{all}x,y\in \mathcal{R}.\hfill \end{array}$$

We shall abbreviate this map as CE ∗-derivation. In addition, let $\mathcal{A}$ be an algebra. A linear mapping $\delta :\mathcal{A}\to \mathcal{A}$ is an additive mapping such that $\delta \left(tx\right)=t\delta \left(x\right)$ for all $x\in \mathcal{A}$ and all $t\in \mathbb{C}.$ Quite recently, centrally extended maps under various conditions have been investigated (for example, see [3,4,5]). CE ∗-derivations are crucial in both Banach algebras and ring theory because they generalize the traditional concept of derivations, providing a broader perspective on algebraic structures and their properties. On the other hand, the study of such characterizations is not limited to associative structures but also extends to nonassociative structures such as alternative rings and Jordan algebras (cf. [6,7,8]).

A particular type of stability was studied by Baker, Lawrence and Zorzitto [9]. Indeed, they proved that if a function is approximately exponential, then it is either a true exponential function or bounded. Then the exponential functional equation is said to be superstable. It was the first result concerning the superstability phenomenon of functional equations. The following year, this famous result was generalized with a simplified proof by Baker (cf. [10]). Later, the superstability for derivations between operator algebras was investigated by Šemrl [11]. Badora [12] studied the stability of derivations in Banach algebras. The study of stability has its origin in the famous talk of Ulam [13]. Hyers [14] answered Ulam’s question affirmatively for Banach spaces. Since then, many authors have generalized Hyers’ results in [15,16,17,18]. Numerous subsequent studies on the stability of various functional equations involving derivations and mappings of derivation type are still being done.

The study of the stability of CE ∗-derivations on Banach ∗-algebras helps us to understand the behavior of almost CE ∗-derivations and their relation to actual CE ∗-derivations. In particular, CE ∗-derivations and their stability are relevant to the study of approximate symmetries in quantum mechanics, where derivations on operator algebras model infinitesimal symmetries, and small perturbations of these symmetries arise naturally in physical models. We also note connections to the theory of automatic continuity and the structure of Banach ∗-algebras.

Therefore, it is worthwhile to investigate the stability of CE ∗-derivations. On the other hand, extending the stability of ∗-derivations to the stability problem concerning CE ∗-derivations contributes to the broader understanding of ∗-derivations and mappings of derivation type in Banach ∗-algebras. In this work, we consider the following functional inequality

$$\begin{array}{cc}\hfill & \parallel \left[a_1\delta \left(x_1\right)+a_2\delta \left(x_2\right)+a_3\delta \left(x_3\right),w\right]\parallel \hfill \\ \hfill & \le \parallel \left[\delta (a_1{x}_1+a_2{x}_2+a_3{x}_3),w\right]\parallel +\mathrm{\Phi }(x_1,x_2,x_3,w),\hfill \end{array}$$

(1)

where the function $\mathrm{\Phi }$ is a perturbing term of the inequality

$$\begin{array}{c}\hfill \parallel \left[a_1\delta \left(x_1\right)+a_2\delta \left(x_2\right)+a_3\delta \left(x_3\right),w\right]\parallel \le \parallel \left[\delta (a_1{x}_1+a_2{x}_2+a_3{x}_3),w\right]\parallel \mathrm{and}\prod _{j=1}^3{a}_j\ne 0,\end{array}$$

which is associated with CE ∗-derivation. Specifically, the inequality (1) arises naturally as a perturbed version of the exact centrally additive condition characterizing CE ∗-derivations. The commutator formulation via $[·,w]$ is motivated by the fact that CE ∗-derivations are defined modulo the center $Z\left(A\right)$, and measuring proximity to the center through commutators with an arbitrary element $w\in A$ is the natural norm-based approach in this setting.

In this paper, we first prove some theorems related to the stability of a functional inequality (1) associated with CE ∗-derivation on a Banach ∗-algebra. Furthermore, under certain conditions, we further prove the superstability of functional inequality (1) associated with a CE ∗-derivation on a Banach ∗-algebra.

We emphasize that the present work is set in the general framework of Banach ∗-algebras, which is strictly broader than the $C^{\ast }$-algebra setting. In the $C^{\ast }$-algebra case, the question addressed in this paper can be handled more directly: the classical Kadison–Sakai theorem guarantees that every bounded derivation on a $C^{\ast }$-algebra is inner, and thus the limit $L\left(a\right)={lim}_{n\to \infty }[x_n,a]$ is automatically of the form $[z,a]$ for some $z\in \mathcal{A}$. However, in the general Banach ∗-algebra setting, neither the $C^{\ast }$-identity nor the Kadison–Sakai theorem is available, and one must impose explicit conditions to recover analogous conclusions.

• Condition: Kadison–Sakai inequality We impose the condition that $\mathcal{A}$ satisfies the norm estimate

  $$\parallel x\parallel \le C\underset{\parallel a\parallel \le 1}{sup}\parallel [x,a]\parallel ,x\in \mathcal{A},$$

  (2)

  for some constant $C>0$. This inequality, which we refer to as the Kadison–Sakai inequality, asserts that the commutator map controls the norm of elements in $\mathcal{A}$. Under this condition, the convergence of $[x_n,a]$ for all $a\in A$ directly implies that $\left(x_n\right)$ is a Cauchy sequence in $\mathcal{A}$, and hence converges to some $z\in \mathcal{A}$ by completeness. Since the commutator is continuous, it follows immediately that $[x_n,a]\to [z,a]$ for all $a\in \mathcal{A}$. We impose this condition because it is the minimal analytic assumption on $\mathcal{A}$ that allows one to pass from commutator convergence to norm convergence of the sequence itself. While this condition is strong—it implies in particular that $Z\left(\mathcal{A}\right)=\left\{0\right\}$ and that the adjoint representation $\mathrm{ad}:\mathcal{A}\to B\left(\mathcal{A}\right)$ has closed range—it is satisfied in natural and important examples, such as $B\left(H\right)$ for an infinite-dimensional Hilbert space H. We note that this condition does not hold in general for $C^{\ast }$-algebras: for instance, any commutative $C^{\ast }$-algebra such as $C\left(\right[0,1\left]\right)$ satisfies $[x,a]=0$ for all $x,a$, and hence the inequality fails trivially.

2. Some Results

Let $\mathcal{A}$ be an algebra. An additive mapping $x\mapsto x^{\ast }$ on $\mathcal{A}$ is called an involution if it has the following properties: (i) ${\left(xy\right)}^{\ast }=y^{\ast }{x}^{\ast },$ (ii) ${\left(x^{\ast }\right)}^{\ast }=x,$ and (iii) ${\left(tx\right)}^{\ast }=\overline{t}{x}^{\ast }$ for all $x,y\in \mathcal{A}$ and all $t\in \mathbb{C},$ where $\overline{t}$ is a complex conjugate of $t.$ An algebra equipped with an involution ∗ is said to be a ∗-algebra. A Banach algebra is a complete normed algebra (i.e., every Cauchy sequence converges) (cf. [19]). A Banach ∗-algebra is a Banach algebra with involution.

As an example of CE ∗-derivation in algebra, we can consider the following:

Example 1.

Let $\mathcal{A}=\left\{\left[\begin{array}{ccc}0& a& b\\ 0& 0& c\\ 0& 0& 0\end{array}\right]:a,b,c\in \mathbb{C}\right\}$ be an algebra. We define maps $\delta ,\ast :\mathcal{A}\to \mathcal{A}$ by

$$\delta \left(\left[\begin{array}{ccc}0& a& b\\ 0& 0& c\\ 0& 0& 0\end{array}\right]\right)=\left[\begin{array}{ccc}0& 0& c-a\\ 0& 0& -c\\ 0& 0& 0\end{array}\right]\mathrm{and}{\left[\begin{array}{ccc}0& a& b\\ 0& 0& c\\ 0& 0& 0\end{array}\right]}^{\ast }=\left[\begin{array}{ccc}0& c& b\\ 0& 0& a\\ 0& 0& 0\end{array}\right]$$

for all $a,b,c\in \mathbb{C}.$ Then $Z\left(\mathcal{A}\right)=\left\{\left[\begin{array}{ccc}0& 0& b\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]:b\in \mathbb{C}\right\}.$ It is easy to show that δ is a CE ∗-derivation, but not a ∗-derivation.

Definition 1.

A Banach ∗-algebra $\mathcal{A}$ is said to satisfy the Kadison–Sakai inequality if there exists a constant $C>0$ such that

$$\parallel x\parallel \le C\underset{\parallel a\parallel \le 1}{sup}\parallel [x,a]\parallel ,$$

for all $x\in \mathcal{A}$, where $[x,a]=xa-ax$ denotes the commutator.

Theorem 1.

Let $\mathcal{A}$ be a Banach ∗-algebra satisfying the Kadison–Sakai inequality. Let $\left\{x_n\right\}$ be a sequence in $\mathcal{A}$ such that for $a\in \mathcal{A}$, the sequence of commutators

$$[x_n,a]=x_{n}a-a{x}_n$$

converges in $\mathcal{A}$. Then $\left\{x_n\right\}$ converges in $\mathcal{A}$ and

$$\underset{n\to \infty }{lim}[x_n,a]=[\underset{n\to \infty }{lim}{x}_n,a].$$

Proof. 

Since $[x_n,a]$ converges for every $a\in \mathcal{A}$, it is Cauchy in $\mathcal{A}$. That is, for every $a\in \mathcal{A}$,

$$\parallel [x_n,a]-[x_m,a]\parallel \to 0(m,n\to \infty ).$$

Observing that $[x_n,a]-[x_m,a]=[x_n-x_m,a]$, we obtain

$$\underset{\parallel a\parallel \le 1}{sup}\parallel [x_n-x_m,a]\parallel \to 0(m,n\to \infty ).$$

Applying the Kadison–Sakai inequality to $x=x_n-x_m$, we get

$$\parallel x_n-x_m\parallel \le C\underset{\parallel a\parallel \le 1}{sup}\parallel [x_n-x_m,a]\parallel \to 0(m,n\to \infty ).$$

Hence $\left\{x_n\right\}$ is a Cauchy sequence in $\mathcal{A}$. Since $\mathcal{A}$ is a Banach space, $\left\{x_n\right\}$ converges in $\mathcal{A}$. □

Now, before getting into the main topic, one obtains the following lemma through some basic calculations.

Theorem 2.

Let $\mathcal{A}$ be a normed algebra. Suppose that a mapping $\delta :\mathcal{A}\to \mathcal{A}$ with $\delta \left(0\right)\in Z\left(\mathcal{A}\right)$ is such that the inequality

$$\begin{array}{c}\hfill \parallel \left[a_1\delta \left(x_1\right)+a_2\delta \left(x_2\right)+a_3\delta \left(x_3\right),w\right]\parallel \le \parallel \left[\delta (a_1{x}_1+a_2{x}_2+a_3{x}_3),w\right]\parallel \end{array}$$

(3)

for all $x_1,x_2,x_3,w\in \mathcal{A},$ where ${\prod }_{j=1}^3{a}_j\ne 0$ and $a_j$ are real numbers. Then δ is centrally additive; i.e.,

$$\begin{array}{c}\hfill \delta (x+y)-\delta \left(x\right)-\delta \left(y\right)\in Z\left(\mathcal{A}\right)\mathit{for}\mathit{all}x,y\in \mathcal{A}.\end{array}$$

Proof. 

Letting $x_1:=x,x_2:=0$ and $x_3:=-{\displaystyle \frac{a_{1}x}{a_3}}$ in (3), one obtains that

$$\begin{array}{c}\hfill \parallel \left[a_1\delta \left(x\right)+a_3\delta \left({\displaystyle \frac{-a_{1}x}{a_3}}\right),w\right]\parallel \le \parallel \left[\delta \left(0\right),w\right]\parallel =0.\end{array}$$

Hence we arrive at

$$\begin{array}{c}\hfill \left[a_1\delta \left(x\right)+a_3\delta \left({\displaystyle \frac{-a_{1}x}{a_3}}\right),w\right]=0\mathrm{for}\mathrm{all}x,w\in \mathcal{A}.\end{array}$$

(4)

Set $x_1:=0,x_2:=y$ and $x_3:={\displaystyle \frac{-a_{2}y}{a_3}}$ in (3). Then we get

$$\begin{array}{c}\hfill \parallel \left[a_2\delta \left(y\right)+a_3\delta \left({\displaystyle \frac{-a_{2}y}{a_3}}\right),w\right]\parallel \le \parallel \left[\delta \left(0\right),w\right]\parallel =0.\end{array}$$

Then we have that

$$\begin{array}{c}\hfill \left[a_2\delta \left(y\right)+a_3\delta \left({\displaystyle \frac{-a_{2}y}{a_3}}\right),w\right]=0\mathrm{for}\mathrm{all}y,w\in \mathcal{A}.\end{array}$$

(5)

By putting $x_1:=x,x_2:=y$ and $x_3:={\displaystyle \frac{-a_{1}x-a_{2}y}{a_3}}$ in (3), we see that

$$\begin{array}{c}\hfill \parallel \left[a_1\delta \left(x\right)+a_2\delta \left(y\right)+a_3\delta \left({\displaystyle \frac{-a_{1}x-a_{2}y}{a_3}}\right),w\right]\parallel \le \parallel \left[\delta \left(0\right),w\right]\parallel =0.\end{array}$$

This means that

$$\begin{array}{c}\hfill \left[a_1\delta \left(x\right)+a_2\delta \left(y\right)+a_3\delta \left({\displaystyle \frac{-a_{1}x-a_{2}y}{a_3}}\right),w\right]=0\mathrm{for}\mathrm{all}x,y,w\in \mathcal{A}.\end{array}$$

(6)

It follows from (4)–(6) that

$$\begin{array}{cc}\hfill & \left[a_3\delta \left({\displaystyle \frac{-a_{1}x-a_{2}y}{a_3}}\right)-a_3\delta \left({\displaystyle \frac{-a_{1}x}{a_3}}\right)-a_3\delta \left({\displaystyle \frac{-a_{2}y}{a_3}}\right),w\right]\hfill \\ \hfill & =\left[a_1\delta \left(x\right)+a_2\delta \left(y\right)+a_3\delta \left({\displaystyle \frac{-a_{1}x-a_{2}y}{a_3}}\right),w\right]-\left[a_1\delta \left(x\right)+a_3\delta \left({\displaystyle \frac{-a_{1}x}{a_3}}\right),w\right]\hfill \\ \hfill & -\left[a_2\delta \left(y\right)+a_3\delta \left({\displaystyle \frac{-a_{2}y}{a_3}}\right),w\right]=0.\hfill \end{array}$$

(7)

for all $x,y,w\in \mathcal{A}.$

Letting $x:={\displaystyle \frac{-a_{3}x}{a_1}}$ and $y:={\displaystyle \frac{-a_{3}y}{a_2}}$ in (7), we are forced to

$$\begin{array}{c}\hfill \left[\delta (x+y)-\delta \left(x\right)-\delta \left(y\right),w\right]=0\mathrm{for}\mathrm{all}x,y,w\in \mathcal{A}.\end{array}$$

Therefore, the mapping $\delta$ is centrally additive. □

Theorem 3.

Let $\mathcal{A}$ be a Banach ∗-algebra. Assume that mappings $\mathrm{\Phi }:{\mathcal{A}}^4\to [0,\infty )$ and $\phi :{\mathcal{A}}^3\to [0,\infty )$ satisfy

$$\begin{array}{cc}\hfill & {\mathrm{\Sigma }}_{j=0}^{\infty }{2}^j\mathrm{\Phi }\left({\displaystyle \frac{x_1}{2^j}},{\displaystyle \frac{x_2}{2^j}},{\displaystyle \frac{x_3}{2^j}},w\right)<\infty (x_1,x_2,x_3,w\in \mathcal{A}),\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill & \underset{n\to \infty }{lim}{2}^{2n}\phi \left({\displaystyle \frac{x}{2^n}},{\displaystyle \frac{y}{2^n}},z\right)=0(x,y,z\in \mathcal{A}).\hfill \end{array}$$

(9)

Suppose that a mapping $\delta :\mathcal{A}\to \mathcal{A}$ with $\delta \left(0\right)\in Z\left(\mathcal{A}\right)$ is such that

$$\begin{array}{cc}\hfill & \parallel \left[a_1\delta \left(x_1\right)+a_2\delta \left(x_2\right)+a_3\delta \left(x_3\right),w\right]\parallel \hfill \\ \hfill & \le \parallel \left[\delta (a_1{x}_1+a_2{x}_2+a_3{x}_3),w\right]\parallel +\mathrm{\Phi }(x_1,x_2,x_3,w)(x_1,x_2,x_3,w\in \mathcal{A}),\hfill \end{array}$$

(10)

where ${\prod }_{j=1}^3{a}_j\ne 0,$ and $a_j$ are real numbers.

$$\begin{array}{c}\hfill \parallel \left[\delta \left(xy\right)-\delta \left(x\right)y^{\ast }-x\delta \left(y\right),z\right]\parallel \le \phi (x,y,z)(x,y,z\in \mathcal{A}).\end{array}$$

(11)

If $\mathcal{A}$ satisfies the Kadison–Sakai inequality, then there exists a CE ∗-derivation $\mathcal{L}:\mathcal{A}\to \mathcal{A}$ satisfying

$$\begin{array}{c}\hfill \parallel \left[\mathcal{L}\left(x\right)-\delta \left(x\right),w\right]\parallel \le \eta (x,w)\end{array}$$

(12)

that holds for all $x,w\in \mathcal{A},$ where

$$\begin{array}{c}\hfill \eta (x,w)={\displaystyle \frac{1}{|a_3|}}\sum _{j=0}^{\infty }{2}^j\left\{\mathrm{\Phi }\left({\displaystyle \frac{-a_{3}x}{2^{j+1}{a}_1}},{\displaystyle \frac{-a_{3}x}{2^{j+1}{a}_2}},{\displaystyle \frac{x}{2^j}},w\right)+\mathrm{\Phi }\left({\displaystyle \frac{-a_{3}x}{2^{j+1}{a}_2}},0,{\displaystyle \frac{x}{2^{j+1}}},w\right)+\mathrm{\Phi }\left(0,{\displaystyle \frac{-a_{3}x}{2^{j+1}{a}_2}},{\displaystyle \frac{x}{2^{j+1}}},w\right)\right\}.\end{array}$$

Proof. 

Letting $x_1:=x,x_2:=0$ and $x_3:={\displaystyle \frac{-a_{1}x}{a_3}}$ in (10), we find that

$$\begin{array}{c}\hfill \parallel \left[a_1\delta \left(x\right)+a_3\delta \left({\displaystyle \frac{-a_{1}x}{a_3}}\right),w\right]\parallel \le \mathrm{\Phi }\left(x,0,{\displaystyle \frac{-a_{1}x}{a_3}},w\right)\mathrm{for}\mathrm{all}x,w\in \mathcal{A}.\end{array}$$

(13)

Put $x_1:=0,x_2:=y$ and $x_3:={\displaystyle \frac{-a_{2}y}{a_3}}$ in (10). We then have

$$\begin{array}{c}\hfill \parallel \left[a_2\delta \left(y\right)+a_3\delta \left({\displaystyle \frac{-a_{2}y}{a_3}}\right),w\right]\parallel \le \mathrm{\Phi }\left(0,y,{\displaystyle \frac{-a_{2}y}{a_3}},w\right)\mathrm{for}\mathrm{all}y,w\in \mathcal{A}.\end{array}$$

(14)

Taking $x_1:=x,x_2:=y$ and $x_3:={\displaystyle \frac{-a_{1}x-a_{2}y}{a_3}}$ in (10), we get

$$\begin{array}{c}\hfill \parallel \left[a_1\delta \left(x\right)+a_2\delta \left(y\right)+a_3\delta \left({\displaystyle \frac{-a_{1}x-a_{2}y}{a_3}}\right),w\right]\parallel \le \mathrm{\Phi }\left(x,y,{\displaystyle \frac{-a_{1}x-a_{2}y}{a_3}},w\right)\end{array}$$

(15)

for all $x,y,w\in \mathcal{A}.$ Combining (13)–(15), we figure out that

$$\begin{array}{cc}\hfill & \parallel \left[a_3\delta \left({\displaystyle \frac{-a_{1}x-a_{2}y}{a_3}}\right)-a_3\delta \left({\displaystyle \frac{-a_{1}x}{a_3}}\right)-a_3\delta \left({\displaystyle \frac{-a_{2}y}{a_3}}\right),w\right]\parallel \hfill \\ \hfill & \le \parallel \left[a_1\delta \left(x\right)+a_2\delta \left(y\right)+a_3\delta \left({\displaystyle \frac{-a_{1}x-a_{2}y}{a_3}}\right),w\right]\parallel +\parallel \left[a_1\delta \left(x\right)+a_3\delta \left({\displaystyle \frac{-a_{1}x}{a_3}}\right),w\right]\parallel \hfill \\ \hfill & +\parallel \left[a_2\delta \left(y\right)+a_3\delta \left({\displaystyle \frac{-a_{2}y}{a_3}}\right),w\right]\parallel \hfill \\ \hfill & \le \mathrm{\Phi }\left(x,y,{\displaystyle \frac{-a_{1}x-a_{2}y}{a_3}},w\right)+\mathrm{\Phi }\left(x,0,{\displaystyle \frac{-a_{1}x}{a_3}},w\right)+\mathrm{\Phi }\left(0,y,{\displaystyle \frac{-a_{2}y}{a_3}},w\right)\hfill \end{array}$$

(16)

for all $x,y,w\in \mathcal{A}.$ Set $x:={\displaystyle \frac{-a_{3}x}{a_1}}$ and $y:={\displaystyle \frac{-a_{3}y}{a_2}}$ in (16) and then divide on both sides by $|a_3|.$ Then we see that

$$\begin{array}{cc}\hfill \parallel \left[\delta (x+y)-\delta \left(x\right)-\delta \left(y\right),w\right]\parallel & \le {\displaystyle \frac{1}{|a_3|}}\{\mathrm{\Phi }\left({\displaystyle \frac{-a_{3}x}{a_1}},{\displaystyle \frac{-a_{3}y}{a_2}},x+y,w\right)\hfill \\ \hfill & + \mathrm{\Phi }\left({\displaystyle \frac{-a_{3}x}{a_1}},0,x,w\right)+ \mathrm{\Phi }\left(0,{\displaystyle \frac{-a_{3}y}{a_2}},y,w\right)\}\hfill \end{array}$$

for all $x,y,w\in \mathcal{A}.$ Consider $y:=x$ in the above relation, we then have that

$$\begin{array}{cc}\hfill \parallel \left[\delta \left(2x\right)-2\delta \left(x\right),w\right]\parallel & \le {\displaystyle \frac{1}{|a_3|}}\{\mathrm{\Phi }\left({\displaystyle \frac{-a_{3}x}{a_1}},{\displaystyle \frac{-a_{3}x}{a_2}},2x,w\right)\hfill \\ \hfill & + \mathrm{\Phi }\left({\displaystyle \frac{-a_{3}x}{a_1}},0,x,w\right)+\mathrm{\Phi }\left(0,{\displaystyle \frac{-a_{3}x}{a_2}},x,w\right)\}\hfill \end{array}$$

(17)

for all $x,w\in \mathcal{A}.$ It follows from (17) that

$$\begin{array}{cc}\hfill \parallel \left[2^m\delta \left({\displaystyle \frac{x}{2^m}}\right)-2^n\delta \left({\displaystyle \frac{x}{2^n}}\right),w\right]\parallel & \le {\displaystyle \frac{1}{|a_3|}}\sum _{j=m}^{n-1}{2}^j\{\mathrm{\Phi }\left({\displaystyle \frac{-a_{3}x}{2^{j+1}{a}_1}},{\displaystyle \frac{-a_{3}x}{2^{j+1}{a}_2}},{\displaystyle \frac{x}{2^j}},w\right)\hfill \\ \hfill & + \mathrm{\Phi }\left({\displaystyle \frac{-a_{3}x}{2^{j+1}{a}_1}},0,{\displaystyle \frac{x}{2^{j+1}}},w\right)+ \mathrm{\Phi }\left(0,{\displaystyle \frac{-a_{3}x}{2^{j+1}{a}_2}},{\displaystyle \frac{x}{2^{j+1}}},w\right)\}\hfill \end{array}$$

(18)

for all $x,w\in \mathcal{A}$ and all integers $m,n\ge 0$ with $n>m.$ Since the right hand-side tends to zero as $m\to \infty ,$ a sequence $\left\{[2^n\delta \left({\displaystyle \frac{x}{2^n}}\right),w]\right\}$ is Cauchy in $\mathcal{A}.$ Since $\mathcal{A}$ is complete, the sequence $\left\{[2^n\delta \left({\displaystyle \frac{x}{2^n}}\right),w]\right\}$ converges. That is, the following

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\left[2^n\delta \left({\displaystyle \frac{x}{2^n}}\right),w\right](x,w\in \mathcal{A})\end{array}$$

(19)

exists. Then, by Lemma 1, we have $\mathcal{L}\left(x\right)\in \mathcal{A}$ such that

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\left[2^n\delta \left({\displaystyle \frac{x}{2^n}}\right),w\right]=\left[\underset{n\to \infty }{lim}{2}^n\delta \left({\displaystyle \frac{x}{2^n}}\right),w\right]=\left[\mathcal{L}\left(x\right),w\right],(x,w\in \mathcal{A}).\end{array}$$

(20)

Letting $m=0$ and passing the limit $n\to \infty$ in (18), we get the desired estimation (12). Next, we verify the centrally additive property of $\mathcal{L}$. For all $x_1,x_2,x_3,w\in \mathcal{A}$, we compute as follows.

Since $\mathcal{L}\left(x\right)={lim}_{n\to \infty }{2}^n\delta \left({\displaystyle \frac{x}{2^n}}\right)$ converges in norm, and since the commutator and the norm are continuous, we have

$$\begin{array}{cc}\hfill & ∥\left[a_1\mathcal{L}\left(x_1\right)+a_2\mathcal{L}\left(x_2\right)+a_3\mathcal{L}\left(x_3\right),w\right]∥\hfill \\ \hfill & =∥\left[\underset{n\to \infty }{lim}{2}^n\left(a_1\delta \left(\frac{x_1}{2^n}\right)+a_2\delta \left(\frac{x_2}{2^n}\right)+a_3\delta \left(\frac{x_3}{2^n}\right)\right),w\right]∥\hfill \\ \hfill & =\underset{n\to \infty }{lim}{2}^n∥\left[a_1\delta \left(\frac{x_1}{2^n}\right)+a_2\delta \left(\frac{x_2}{2^n}\right)+a_3\delta \left(\frac{x_3}{2^n}\right),w\right]∥.\hfill \end{array}$$

(21)

Applying (10) with $x_i$ replaced by ${\displaystyle \frac{x_i}{2^n}}$, we obtain

$$\begin{array}{cc}\hfill & ∥\left[a_1\delta \left(\frac{x_1}{2^n}\right)+a_2\delta \left(\frac{x_2}{2^n}\right)+a_3\delta \left(\frac{x_3}{2^n}\right),w\right]∥\hfill \\ \hfill & \le ∥\left[\delta \left(\frac{a_1{x}_1+a_2{x}_2+a_3{x}_3}{2^n}\right),w\right]∥+\mathrm{\Phi }\left(\frac{x_1}{2^n},\frac{x_2}{2^n},\frac{x_3}{2^n},w\right).\hfill \end{array}$$

(22)

Multiplying both sides of (22) by $2^n$ and taking the limit $n\to \infty$,

$$\begin{array}{cc}\hfill & \underset{n\to \infty }{lim}{2}^n∥\left[a_1\delta \left(\frac{x_1}{2^n}\right)+a_2\delta \left(\frac{x_2}{2^n}\right)+a_3\delta \left(\frac{x_3}{2^n}\right),w\right]∥\hfill \\ \hfill & \le \underset{n\to \infty }{lim}{2}^n∥\left[\delta \left(\frac{a_1{x}_1+a_2{x}_2+a_3{x}_3}{2^n}\right),w\right]∥+\underset{n\to \infty }{lim}{2}^n\mathrm{\Phi }\left(\frac{x_1}{2^n},\frac{x_2}{2^n},\frac{x_3}{2^n},w\right).\hfill \end{array}$$

(23)

By condition (8), the second term on the right-hand side of (23) satisfies

$$\underset{n\to \infty }{lim}{2}^n\mathrm{\Phi }\left(\frac{x_1}{2^n},\frac{x_2}{2^n},\frac{x_3}{2^n},w\right)=0.$$

(24)

Again by the norm convergence of $\mathcal{L}$ and the continuity of the commutator,

$$\underset{n\to \infty }{lim}{2}^n∥\left[\delta \left(\frac{a_1{x}_1+a_2{x}_2+a_3{x}_3}{2^n}\right),w\right]∥=∥\left[\mathcal{L}(a_1{x}_1+a_2{x}_2+a_3{x}_3),w\right]∥.$$

(25)

Therefore, we conclude that

$$∥\left[a_1\mathcal{L}\left(x_1\right)+a_2\mathcal{L}\left(x_2\right)+a_3\mathcal{L}\left(x_3\right),w\right]∥\le ∥\left[\mathcal{L}(a_1{x}_1+a_2{x}_2+a_3{x}_3),w\right]∥,$$

(26)

for all $x_1,x_2,x_3,w\in \mathcal{A}.$ Since $\mathcal{L}\left(0\right)\in Z\left(\mathcal{A}\right)$, by Lemma 2, the mapping $\mathcal{L}$ is centrally additive.

On the other hand, one obtains from (11) that

$$\begin{array}{cc}\hfill \parallel \left[\mathcal{L}\left(xy\right)-\mathcal{L}\left(x\right)y^{\ast }-x\mathcal{L}\left(y\right),z\right]\parallel & =\underset{n\to \infty }{lim}{2}^{2n}\parallel \left[\delta \left({\displaystyle \frac{xy}{2^{2n}}}\right)-\delta \left({\displaystyle \frac{x}{2^n}}\right){\displaystyle \frac{y^{\ast }}{2^n}}-{\displaystyle \frac{x}{2^n}}\delta \left({\displaystyle \frac{y}{2^n}}\right),z\right]\parallel \hfill \\ \hfill & \le \underset{n\to \infty }{lim}{2}^{2n}\phi \left({\displaystyle \frac{x}{2^n}},{\displaystyle \frac{y}{2^n}},z\right)=0,\hfill \end{array}$$

which implies that

$$\begin{array}{c}\hfill \left[\mathcal{L}\left(xy\right)-\mathcal{L}\left(x\right)y^{\ast }-x\mathcal{L}\left(y\right),z\right]=0\mathrm{for}\mathrm{all}x,y,z\in \mathcal{A}.\end{array}$$

(27)

Therefore, $\mathcal{L}$ is a CE ∗-derivation. □

Example 2.

Let $\mathcal{A}$ be a Banach ∗-algebra satisfying the Kadison–Sakai inequality, and let $\theta >0$, $p>1$, and $q>0$ with $p+q>2$. Define $\mathrm{\Phi }:{\mathcal{A}}^4\to [0,\infty )$ and $\phi :{\mathcal{A}}^3\to [0,\infty )$ by

$$\begin{array}{cc}\hfill \mathrm{\Phi }(x_1,x_2,x_3,w)& =\theta \sum _{k=1}^3{\parallel x_k\parallel }^p,\hfill \\ \hfill \phi (x,y,z)& ={\theta \parallel x\parallel }^p{\parallel y\parallel }^q.\hfill \end{array}$$

Then conditions (8) and (9) are satisfied. Indeed, for condition (8), we have

$$\begin{array}{cc}\hfill \sum _{j=0}^{\infty }{2}^j\mathrm{\Phi }\left({\displaystyle \frac{x_1}{2^j}},{\displaystyle \frac{x_2}{2^j}},{\displaystyle \frac{x_3}{2^j}},w\right)& = \theta \sum _{k=1}^3{\parallel x_k\parallel }^p\sum _{j=0}^{\infty }{2}^{j(1-p)}<\infty ,\hfill \end{array}$$

since $p>1$ implies $2^{1-p}<1$. For condition (9), we compute

$$\begin{array}{c}\hfill 2^{2n}\phi \left({\displaystyle \frac{x}{2^n}},{\displaystyle \frac{y}{2^n}},z\right)=\theta 2^{2n}·{\displaystyle \frac{{\parallel x\parallel }^p{\parallel y\parallel }^q}{2^{n(p+q)}}}=\theta 2^{n(2-p-q)}{\parallel x\parallel }^p{\parallel y\parallel }^q\to 0(n\to \infty ),\end{array}$$

since $p+q>2$.

Now define a mapping $\delta :\mathcal{A}\to \mathcal{A}$ by

$$\begin{array}{c}\hfill \delta \left(x\right)=\lambda x(x\in \mathcal{A}),\end{array}$$

where $\lambda \in \mathbb{R}$. One can verify that δ satisfies conditions (10) and (11). For condition (10), since δ is $\mathbb{R}$-linear, we have

$$\begin{array}{cc}\hfill ∥\left[a_1\delta \left(x_1\right)+a_2\delta \left(x_2\right)+a_3\delta \left(x_3\right),w\right]∥& = ∥\left[\delta (a_1{x}_1+a_2{x}_2+a_3{x}_3),w\right]∥\hfill \\ \hfill & \le ∥\left[\delta (a_1{x}_1+a_2{x}_2+a_3{x}_3),w\right]∥+\mathrm{\Phi }(x_1,x_2,x_3,w).\hfill \end{array}$$

For condition (11), a direct computation gives

$$\begin{array}{c}\hfill \delta \left(xy\right)-\delta \left(x\right)y^{\ast }-x\delta \left(y\right)=\lambda xy-\lambda x{y}^{\ast }-\lambda xy=-\lambda x{y}^{\ast },\end{array}$$

so that

$$\begin{array}{c}\hfill ∥\left[\delta \left(xy\right)-\delta \left(x\right)y^{\ast }-x\delta \left(y\right),z\right]∥=\lambda ∥\left[x{y}^{\ast },z\right]∥\le \phi (x,y,z).\end{array}$$

By Theorem 3, the limit

$$\begin{array}{c}\hfill \mathcal{L}\left(x\right)=\underset{n\to \infty }{lim}{2}^n\delta \left({\displaystyle \frac{x}{2^n}}\right)=\underset{n\to \infty }{lim}{2}^n·{\displaystyle \frac{\lambda x}{2^n}}=\lambda x\end{array}$$

defines a CE ∗-derivation $\mathcal{L}:\mathcal{A}\to \mathcal{A}$. Moreover, the bound (12) holds with

$$\begin{array}{c}\hfill \eta (x,w)={\displaystyle \frac{C(\mathbf{a},\theta )}{|a_3|}}·{\displaystyle \frac{{\parallel x\parallel }^p}{1-2^{1-p}}},\end{array}$$

where $C(\mathbf{a},\theta )>0$ is an explicit constant depending only on $a_1,a_2,a_3$, and θ. Indeed, one can directly check that

$$\begin{array}{c}\hfill ∥\left[\mathcal{L}\left(x\right)-\delta \left(x\right),w\right]∥=∥\left[\lambda x-\lambda x,w\right]∥=0 \le \eta (x,w),\end{array}$$

which confirms the conclusion of Theorem 3.

In the next theorem, we discuss the superstability of the equation in comparison with the Hyers–Ulam stability established in Theorem 3.

Theorem 4.

Let $\mathcal{A}$ be a semiprime Banach ∗-algebra. Assume that mappings $\mathrm{\Phi }:{\mathcal{A}}^4\to [0,\infty )$ and $\phi :{\mathcal{A}}^3\to [0,\infty )$ satisfy (8) and (9). Suppose that $\delta :\mathcal{A}\to \mathcal{A}$ is a mapping subjected to $\delta \left(0\right)\in Z\left(\mathcal{A}\right)$ and

$$\begin{array}{cc}\hfill & \parallel \left[\lambda a_1\delta \left(x_1\right)+a_2\delta \left(x_2\right)+a_3\delta \left(\lambda x_3\right),w\right]\parallel \hfill \\ \hfill & \le \parallel \left[\delta (a_1{x}_1+a_2{x}_2+a_3{x}_3),w\right]\parallel + \mathrm{\Phi }(x_1,x_2,x_3,w)(x_1,x_2,x_3,w\in \mathcal{A}),\hfill \end{array}$$

(28)

where ${\prod }_{j=1}^3{a}_j\ne 0$, $a_j\in \mathbb{R}$ and $\lambda \in \mathbb{U},$ together with the inequality (11). If $\mathcal{A}$ satisfies the Kadison–Sakai inequality, then δ is a CE ∗-derivation.

Proof. 

We first consider $\lambda =1$ in (28). It follows from Theorem 3 that there exists a CE ∗-derivation $\mathcal{L}:\mathcal{A}\to \mathcal{A}$ satisfying (12). In this case, $\mathcal{L}$ is defined as (19).

Based on (28), using a similar method to that in the proof of Theorem 3, we find that

$$\begin{array}{c}\hfill \parallel \left[\lambda a_1\mathcal{L}\left(x_1\right)+a_2\mathcal{L}\left(x_2\right)+a_3\mathcal{L}\left(\lambda x_3\right),w\right]\parallel \le \parallel \left[\mathcal{L}(a_1{x}_1+a_2{x}_2+a_3{x}_3),w\right]\parallel \end{array}$$

(29)

for all $x_1,x_2,x_3,w\in \mathcal{A}$ and all $\lambda \in \mathbb{U},$ together with $\mathcal{L}\left(0\right)\in Z\left(\mathcal{A}\right).$ Since $\mathcal{L}$ is centrally additive, we have that $\left[\mathcal{L}\right(-x),w]=[-\mathcal{L}(x),w].$ So, by letting $x_1:=x,x_2:=0$ and $x_3:={\displaystyle \frac{-a_{1}x}{a_3}}$ in (29), we get $\left[{\displaystyle \frac{\lambda a_1}{a_3}}\mathcal{L}\left(x\right)-\mathcal{L}\left({\displaystyle \frac{\lambda a_{1}x}{a_3}}\right),w\right]=0.$ The last relation ensures that $\left[\mathcal{L}\left(\lambda x\right)-\lambda \mathcal{L}\left(x\right),w\right]=0$ for all $x,w\in \mathcal{A}$ and all $\lambda \in \mathbb{U}.$ Let us assume that $t\in \mathbb{C}$ is a nonzero number and that $M>0$ is an integer greater than $\left|t\right|.$ Then, by applying a geometric argument, there exist ${\lambda }_1,{\lambda }_2\in \mathbb{U}$ such that $2(t/M)={\lambda }_1+{\lambda }_2.$ In particular, by the central additivity of $\mathcal{L},$ one obtains that $\left[\mathcal{L}(x/2),w\right]=\left[(1/2)\mathcal{L}\left(x\right),w\right]$ for all $x,w\in \mathcal{A}.$ Thus we see that

$$\begin{array}{c}\hfill \left[\mathcal{L}\left(tx\right),w\right]=\left[\mathcal{L}\left({\displaystyle \frac{M}{2}}·2·{\displaystyle \frac{t}{M}}x\right),w\right]=\left[{\displaystyle \frac{M}{2}}({\lambda }_1+{\lambda }_2)\mathcal{L}\left(x\right),w\right]=\left[t\mathcal{L}\left(x\right),w\right]\end{array}$$

for all $x,w\in \mathcal{A}.$ Clearly, $\left[\mathcal{L}\left(0x\right),w\right]=\left[0\mathcal{L}\left(x\right),w\right].$ Consequently, $\left[\mathcal{L}\left(tx\right)-t\mathcal{L}\left(x\right),w\right]=0.$ In other words, $\mathcal{L}$ satisfies the following:

$$\begin{array}{c}\hfill \mathcal{L}\left(tx\right)-t\mathcal{L}\left(x\right)\in Z\left(\mathcal{A}\right)(x\in \mathcal{A},t\in \mathbb{C}).\end{array}$$

(30)

On the other hand, as we did in the proof of Theorem 3, we get the relation (27). Also, with the aid of (10), we yield that

$$\begin{array}{cc}\hfill \parallel \left[\mathcal{L}\left(xy\right)-\mathcal{L}\left(x\right)y^{\ast }-x\delta \left(y\right),z\right]\parallel & =\underset{n\to \infty }{lim}{2}^n\parallel \left[\delta \left({\displaystyle \frac{xy}{2^n}}\right)-\delta \left({\displaystyle \frac{x}{2^n}}\right)y^{\ast }-{\displaystyle \frac{x}{2^n}}\delta \left(y\right),z\right]\parallel \hfill \\ \hfill & \le \underset{n\to \infty }{lim}{2}^n\phi \left({\displaystyle \frac{x}{2^n}},y,z\right)=0,\hfill \end{array}$$

which means that

$$\begin{array}{c}\hfill \left[\mathcal{L}\left(xy\right)-\mathcal{L}\left(x\right)y^{\ast }-x\delta \left(y\right),z\right]=0\mathrm{for}\mathrm{all}x,y,z\in \mathcal{A}.\end{array}$$

(31)

Subtract (27) from (31) to get

$$\begin{array}{c}\hfill \left[x\left(\mathcal{L}\left(y\right)-\delta \left(y\right)\right),z\right]=0\mathrm{for}\mathrm{all}x,y,z\in \mathcal{A}.\end{array}$$

(32)

The identity (32) can be represented as

$$\begin{array}{c}\hfill [x,z]·\left(\mathcal{L}\right(y)-\delta (y\left)\right)+x\left[\mathcal{L}\right(y)-\delta (y),z]=0.\end{array}$$

(33)

Replacing z by $\mathcal{L}\left(y\right)-\delta \left(y\right)$ in (33), we have

$$\begin{array}{c}\hfill \left[\mathcal{L}\right(y)-\delta (y),x]·\left(\mathcal{L}\right(y)-\delta (y\left)\right)=0.\end{array}$$

(34)

Substitute $xz$ instead of x in (34) and then use (34) to get

$$\begin{array}{c}\hfill \left[\mathcal{L}\right(y)-\delta (y),x]z\left(\mathcal{L}\right(y)-\delta (y\left)\right)=0.\end{array}$$

(35)

Using the relation obtained by multiplying x in the right-hand side of (35) and the expression attained by substituting $zx$ instead of z in (35), it can be derived as follows:

$$\begin{array}{c}\hfill \left[\mathcal{L}\right(y)-\delta (y),x]z\left[\mathcal{L}\right(y)-\delta (y),x]=0\end{array}$$

for all $x,y,z\in \mathcal{A}.$ So, by semiprimeness of $\mathcal{A},$ we have $\left[\mathcal{L}\right(y)-\delta (y),x]=0.$ This means that

$$\begin{array}{c}\hfill \left[\mathcal{L}\right(y),x]=\left[\delta \right(y),x]\mathrm{for}\mathrm{all}x,y\in \mathcal{A}.\end{array}$$

(36)

The inequality (10) implies that

$$\begin{array}{cc}\hfill \parallel \left[\mathcal{L}\left(xy\right)-\delta \left(x\right)y^{\ast }-x\mathcal{L}\left(y\right),z\right]\parallel & =\underset{n\to \infty }{lim}{2}^n\parallel \left[\delta \left({\displaystyle \frac{xy}{2^n}}\right)-\delta \left(x\right){\displaystyle \frac{y^{\ast }}{2^n}}-x\delta \left({\displaystyle \frac{y}{2^n}}\right),z\right]\parallel \hfill \\ \hfill & \le \underset{n\to \infty }{lim}{2}^n\phi \left(x,{\displaystyle \frac{y}{2^n}},z\right)=0.\hfill \end{array}$$

Then we are forced to conclude that

$$\begin{array}{c}\hfill \left[\mathcal{L}\left(xy\right)-\delta \left(x\right)y^{\ast }-x\mathcal{L}\left(y\right),z\right]=0\mathrm{for}\mathrm{all}x,y,z\in \mathcal{A}.\end{array}$$

(37)

Comparing (32) and (36) in (37), we get $[\delta \left(xy\right)-\delta \left(x\right)y^{\ast }-x\delta \left(y\right),z]=0,$ that is,

$$\delta \left(xy\right)-\delta \left(x\right)y^{\ast }-x\delta \left(y\right)\in Z\left(\mathcal{A}\right)\mathrm{for}\mathrm{all}x,y\in \mathcal{A}.$$

(38)

Now, since $\mathcal{L}$ is centrally additive, one obtains that $\left[\mathcal{L}\right(x+y)-\mathcal{L}(x)-\mathcal{L}(y),z]=0.$ It then follows from (36) that $\left[\delta \right(x+y)-\delta (x)-\delta (y),z]=0.$ Hence we see that

$$\delta (x+y)-\delta \left(x\right)-\delta \left(y\right)\in Z\left(\mathcal{A}\right)\mathrm{for}\mathrm{all}x,y\in \mathcal{A}.$$

(39)

Therefore, by (38) and (39), $\delta$ is a CE ∗-derivation. □

Theorem 5.

Let $\mathcal{A}$ be a Banach ∗-algebra. Assume that mappings $\mathrm{\Phi }:{\mathcal{A}}^4\to [0,\infty )$ and $\phi :{\mathcal{A}}^3\to [0,\infty )$ satisfy

$$\begin{array}{cc}\hfill & {\mathrm{\Sigma }}_{j=0}^{\infty }{\displaystyle \frac{1}{2^j}}\mathrm{\Phi }\left(2^j{x}_1,2^j{x}_2,2^j{x}_3,w\right)<\infty (x_1,x_2,x_3,w\in \mathcal{A}),\hfill \end{array}$$

(40)

$$\begin{array}{cc}\hfill & \underset{n\to \infty }{lim}{\displaystyle \frac{1}{2^n}}\phi \left(2^{n}x,2^{n}y,z\right)=0(x,y,z\in \mathcal{A}).\hfill \end{array}$$

(41)

Suppose that $\delta :\mathcal{A}\to \mathcal{A}$ is a mapping with $\delta \left(0\right)\in Z\left(\mathcal{A}\right)$ subjected to (10) and (11). If $\mathcal{A}$ satisfies the Kadison–Sakai inequality, then there exists a CE ∗-derivation $\mathcal{L}:\mathcal{A}\to \mathcal{A}$ satisfying

$$\begin{array}{cc}\hfill \parallel \left[\mathcal{L}\left(x\right)-\delta \left(x\right),w\right]\parallel & \le \eta (x,w)\hfill \end{array}$$

for all $x,z\in \mathcal{A},$ where

$$\begin{array}{cc}\hfill \eta (x,w)& ={\displaystyle \frac{1}{2|a_3|}}\sum _{j=0}^{\infty }{\displaystyle \frac{1}{2^j}}\{\mathrm{\Phi }\left({\displaystyle \frac{-2^j{a}_{3}x}{a_1}},{\displaystyle \frac{-2^j{a}_{3}x}{a_2}},2^{j+1}x,w\right)+ \mathrm{\Phi }\left({\displaystyle \frac{-2^j{a}_{3}x}{a_1}},0,2^{j}x,w\right)\hfill \\ \hfill & + \mathrm{\Phi }\left(0,{\displaystyle \frac{-2^j{a}_{3}x}{a_2}},2^{j}x,w\right)\}.\hfill \end{array}$$

(42)

Proof. 

It follows from (17) that

$$\begin{array}{cc}\hfill \parallel \left[2^m\delta \left({\displaystyle \frac{x}{2^m}}\right)-2^n\delta \left({\displaystyle \frac{x}{2^n}}\right),w\right]\parallel & \le {\displaystyle \frac{1}{2|a_3|}}\sum _{j=m}^{n-1}{\displaystyle \frac{1}{2^j}}\{\mathrm{\Phi }\left({\displaystyle \frac{-2^j{a}_{3}x}{a_1}},{\displaystyle \frac{-2^j{a}_{2}x}{a_3}},2^{j+1}x,w\right)\hfill \\ \hfill & + \mathrm{\Phi }\left({\displaystyle \frac{-2^j{a}_{3}x}{a_1}},0,2^{j}x,w\right)+\mathrm{\Phi }\left(0,{\displaystyle \frac{-2^j{a}_{3}x}{a_2}},2^{j}x,w\right)\}\hfill \end{array}$$

for all $x,z\in \mathcal{A}$ and all integers $m,n\ge 0$ with $n>m,$ which implies that a sequence $\left\{\left[{\displaystyle \frac{\delta \left(2^{n}x\right)}{2^n}},w\right]\right\}$ is Cauchy in $\mathcal{A}.$ Since $\mathcal{A}$ is complete, the sequence $\left\{\left[{\displaystyle \frac{\delta \left(2^{n}x\right)}{2^n}},w\right]\right\}$ converges, that is, ${lim}_{n\to \infty }\left[{\displaystyle \frac{\delta \left(2^{n}x\right)}{2^n}},w\right]$ exists. As we did in the proof of Theorem 3, we can define $\mathcal{L}:\mathcal{A}\to \mathcal{A}$ by

$$\begin{array}{c}\hfill \mathcal{L}\left(x\right):=\underset{n\to \infty }{lim}{\displaystyle \frac{\delta \left(2^{n}x\right)}{2^n}}\mathrm{for}\mathrm{all}x\in \mathcal{A}.\end{array}$$

The remainder of this proof can be carried out similarly to the corresponding part of Theorem 3. □

In the next theorem, we discuss the superstability of the equation in comparison with the Hyers–Ulam stability established in Theorem 5.

Theorem 6.

Let $\mathcal{A}$ be a semiprime Banach ∗-algebra. Assume that mappings $\mathrm{\Phi }:{\mathcal{A}}^4\to [0,\infty )$ and $\phi :{\mathcal{A}}^3\to [0,\infty )$ satisfy (40) and (41). Suppose that $\delta :\mathcal{A}\to \mathcal{A}$ is a mapping with $\delta \left(0\right)\in Z\left(\mathcal{A}\right)$ subjected to (11) and (28). If $\mathcal{A}$ satisfies the Kadison–Sakai inequality, then δ is a CE ∗-derivation.

Proof. 

This theorem can be proved in the same way as Theorem 4. □

Corollary 1.

Let $\mathcal{A}$ be a Banach ∗-algebra. Assume that mappings $\mathrm{\Phi }:{\mathcal{A}}^4\to [0,\infty )$ and $\phi :{\mathcal{A}}^3\to [0,\infty )$ satisfy the conditions (8) and (9) (resp. (40) and (41)). Suppose that $\delta :\mathcal{A}\to \mathcal{A}$ is a mapping with

$$\delta \left(tx\right)-t\delta \left(x\right)\in Z\left(\mathcal{A}\right)(x\in \mathcal{A},t\in \mathbb{C})$$

subjected to (10) and (11). If $\mathcal{A}$ satisfies the Kadison–Sakai inequality, then there exists a CE ∗-derivation $\mathcal{L}:\mathcal{A}\to \mathcal{A}$ satisfying (12).

Proof. 

Observe that the assumption $\delta \left(tx\right)-t\delta \left(x\right)\in Z\left(\mathcal{A}\right)$ guarantees $\delta \left(0\right)\in Z\left(\mathcal{A}\right).$ We then have by Theorem 3 (resp. Theorem 5) that there exists a CE ∗-derivation $\mathcal{L}:\mathcal{A}\to \mathcal{A}$ satisfying (12) (resp. (42)). □

3. Future Research

Recall that a $J{B}^{\ast }$-algebra is a complex Jordan algebra A equipped with an involution ∗ and a complete norm satisfying $\parallel U_a{\left(b\right)\parallel \le \parallel a\parallel }^2\parallel b\parallel$ and $\parallel a^{\ast }\parallel = \parallel a\parallel$, where $U_a\left(b\right)=2a\circ (a\circ b)-a^2\circ b$ denotes the Jordan triple product.

• Conjecture 1 (Stability in ${\mathit{JB}}^{\ast }$-algebras). Let A be a unital $J{B}^{\ast }$-algebra. Define a centrally extended Jordan ∗-derivation (CE Jordan ∗-derivation) as an additive mapping $\delta :A\to A$ satisfying

  $$\delta (a\circ b)-\delta \left(a\right)\circ b^{\ast }-a\circ \delta \left(b\right)\in Z\left(A\right)\mathrm{for}\mathrm{all}a,b\in A,$$

  where $Z\left(A\right)$ denotes the center of A with respect to the Jordan product. We conjecture that analogs of Theorems 3 and 5 hold in this setting: namely, if $\delta :A\to A$ satisfies a perturbed version of the functional inequality (1) (appropriately reformulated using Jordan products and commutators defined via the associator), then there exists a genuine CE Jordan ∗-derivation $L:A\to A$ close to $\delta$ in the sense of (12).
• Conjecture 2 (Superstability in ${\mathit{JB}}^{\ast }$-algebras). Under semiprimeness conditions appropriate for $J{B}^{\ast }$-algebras and the absence of nonzero central ideals, we conjecture that a mapping $\delta$ satisfying the perturbed inequality is itself a Jordan ∗-derivation. The key technical challenge is that the semiprime condition for Jordan algebras takes a different form than in the associative case [20], and new arguments are likely required to replace the commutator-based reasoning in our proofs.