Almost Generalized Derivation on Banach Algebras

Abstract

We take into consideration generalized derivations. First, we study the stability of generalized derivations on Banach algebras under consideration. Then we prove some theorems involving approximate generalized derivations on Banach algebras. These results can be applied to $C^{*}$-algebras.

1. Introduction

Let $\mathcal{A}$ be an algebra. An additive mapping $\mathcal{D}:\mathcal{A}\to \mathcal{A}$ is said to be a derivation if $\mathcal{D}\left(xy\right)=\mathcal{D}\left(x\right)y+x\mathcal{D}\left(y\right)$ for all $x,y\in \mathcal{A}.$ Furthermore, if $\delta \left(tx\right)=t\delta \left(x\right)$ for all $x\in \mathcal{A}$ and all $t\in \mathbb{C},$ then $\delta$ is called a linear derivation. An additive mapping $\delta :\mathcal{A}\to \mathcal{A}$ is said to be a generalized derivation if there exists a derivation $\mathcal{D}$ of $\mathcal{A}$ such that $\delta \left(xy\right)=\delta \left(x\right)y+x\mathcal{D}\left(y\right)$ for all $x,y\in \mathcal{A}.$ In addition, if $\delta \left(tx\right)=t\delta \left(x\right),\mathcal{D}\left(tx\right)=t\mathcal{D}\left(x\right)$ for all $x\in \mathcal{A}$ and all $t\in \mathbb{C},$ then we say that $\delta$ is a linear generalized derivation.

The concept of stability for functional equations arises when we replace the functional equation with an inequality that acts as a perturbation of the equation. Ulam [1] brought up the question concerning the stability of group homomorphisms. Hyers [2] first proved this problem for the case of approximately additive mappings $f:\mathcal{X}\to \mathcal{Y},$ where $\mathcal{X}$ and $\mathcal{Y}$ are Banach spaces. Since then, a number of mathematicians have generalized the result of Hyers; see, e.g., [3,4,5]. The stability result, i.e., superstability concerning derivations between operator algebras, was first obtained by Šemrl [6]. Badora obtained the results about the stability and superstability of the Bourgin-type for derivations in reference [7]. In addition, many interesting studies on the stability problems of various functional equations (or involving derivations) are still being conducted. The reader may refer to books and papers for more information on the stability problem with a large variety of applications (for example, [8,9,10,11,12,13]).

Recently, in [14], Chang et al. obtained the results concerning the stability of the following functional inequality on quasi-$\beta$-normed spaces

$$\begin{array}{c}\hfill \parallel \sum _{j=1}^l{a}_j\delta \left(x_j\right)\parallel \le \parallel \delta \left(\sum _{j=1}^l{a}_j{x}_j\right)\parallel +\varphi (x_1,\cdots ,x_l),\end{array}$$

(1)

where ${\prod }_{j=1}^l{a}_j\ne 0,l\ge 3,a_j\in \mathbb{R}.$

Our principal purpose is to investigate some theorems regarding functional inequality (1) on Banach algebras together with approximate generalized derivations. More precisely, we establish the stability problems of the functional inequality (1) with approximate generalized derivations and then obtained some theorems and properties for approximate generalized derivations on Banach algebras. Moreover, we deal with some results concerning approximate multiplicative (generalized) derivations.

2. Results and Proofs

First, we introduce an example of a generalized derivation of a Banach algebra, as follows.

Example 1.

Let $\mathcal{A}=M_2\left(\mathbb{C}\right)$ be the Banach algebra of all $2\times 2$ upper triangle matrices over the complex field $\mathbb{C}.$ We define a map $\delta :\mathcal{A}\to \mathcal{A}$ by

$$\delta \left(\left[\begin{array}{cc}a& b\\ 0& c\end{array}\right]\right)=\left[\begin{array}{cc}a& 2b\\ 0& 0\end{array}\right].$$

Then, we see that δ is a generalized derivation associated with derivation $\mathcal{D},$ where $\mathcal{D}$ is a map defined by

$$\mathcal{D}\left(\left[\begin{array}{cc}a& b\\ 0& c\end{array}\right]\right)=\left[\begin{array}{cc}0& b\\ 0& 0\end{array}\right].$$

We now address the problem of stability for the generalized derivation of a Banach algebra. Before dealing with the main topic, we can obtain the following lemmas on a Banach space through calculations that are not difficult (refer to [14]). For that reason, the proof will be omitted here.

For convenience, we will write

$$\begin{array}{c}\hfill {\varphi }_{\alpha \beta \gamma }(x,y,z):=\varphi (0,\cdots ,0,\underset{\alpha -th}{\underbrace{x}},0,\cdots ,0,\underset{\beta -th}{\underbrace{y}},0,\cdots ,0,\underset{\gamma -th}{\underbrace{z}},0,\cdots ,0)\end{array}$$

in the following lemmas.

Lemma 1.

Let $\mathcal{A}$ be a Banach space. Assume that a mapping $\varphi :{\mathcal{A}}^l\to [0,\infty )$ satisfies the series

$$\begin{array}{c}\hfill \sum _{j=0}^{\infty }{2}^j\varphi \left(\frac{x_1}{2^j},\cdots ,\frac{x_l}{2^j}\right)<\infty \end{array}$$

(2)

for all $x_1,\cdots ,x_l\in \mathcal{A}.$ Suppose that $\delta :\mathcal{A}\to \mathcal{A}$ is a mapping with $\delta \left(0\right)=0$ subjected to inequality (1), where ${\prod }_{j=1}^l{a}_j\ne 0,l\ge 3,a_j\in \mathbb{R}.$ Then, there exists a unique additive mapping $\mathcal{L}:\mathcal{A}\to \mathcal{A}$ defined by

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

(3)

such that

$$\begin{array}{cc}\hfill \parallel \delta \left(x\right)-\mathcal{L}\left(x\right)\parallel & \le \frac{1}{|a_3|}\sum _{j=0}^{\infty }{2}^j[{\varphi }_{123}\left(-\frac{a_{3}x}{2^{j+1}{a}_1},-\frac{a_{3}x}{2^{j+1}{a}_2},\frac{x}{2^j}\right)\hfill \\ \hfill & +{\varphi }_{123}\left(-\frac{a_{3}x}{2^{j+1}{a}_1},0,\frac{x}{2^{j+1}}\right)+{\varphi }_{123}\left(0,-\frac{a_{3}x}{2^{j+1}{a}_2},\frac{x}{2^{j+1}}\right)]\hfill \end{array}$$

(4)

for all $x\in \mathcal{A}.$

Lemma 2.

Let $\mathcal{A}$ be a Banach space. Assume that mappings $\varphi :{\mathcal{A}}^l\to [0,\infty )$ satisfy the series

$$\begin{array}{c}\hfill \sum _{j=0}^{\infty }\frac{1}{2^j}\varphi (2^j{x}_1,\cdots ,2^j{x}_l)<\infty \end{array}$$

(5)

for all $x_1,\cdots ,x_l\in \mathcal{A}.$ Suppose that $\delta :\mathcal{A}\to \mathcal{A}$ is a mapping with $\delta \left(0\right)=0$ subjected to the inequality (1), where ${\prod }_{j=1}^l{a}_j\ne 0,l\ge 3,a_j\in \mathbb{R}.$ Then, there exists a unique additive mapping $\mathcal{L}:\mathcal{A}\to \mathcal{A}$ defined by

$$\begin{array}{c}\hfill \mathcal{L}\left(x\right):=\underset{n\to \infty }{lim}\frac{1}{2^n}\delta \left(2^{n}x\right)\mathit{for}\mathit{all}x\in \mathcal{A}\end{array}$$

(6)

such that

$$\begin{array}{cc}\hfill \parallel \delta \left(x\right)-\mathcal{L}\left(x\right)\parallel & \le \frac{1}{|a_3|}\sum _{j=0}^{\infty }\frac{1}{2^{j+1}}[{\varphi }_{123}\left(-\frac{2^j{a}_{3}x}{a_1},-\frac{2^j{a}_{3}x}{a_2},2^{j+1}x\right)\hfill \\ \hfill & +{\varphi }_{123}\left(-\frac{2^j{a}_{3}x}{a_1},0,2^{j}x\right)+{\varphi }_{123}\left(0,-\frac{2^j{a}_{3}x}{a_2},2^{j}x\right)]\hfill \end{array}$$

(7)

for all $x\in \mathcal{A}.$

We should note that e stands for the unit element of any unital algebra, and for a given $\epsilon >0,$ we set ${\mathbb{T}}_{\epsilon }:=\{e^{i\theta }:0\le \theta \le \epsilon \}.$ We now prove our first theorem.

Theorem 1.

Let $\mathcal{A}$ be either a semiprime Banach algebra or a unital Banach algebra. Assume that mappings $\varphi :{\mathcal{A}}^l\to [0,\infty )$ and $\phi :{\mathcal{A}}^2\to [0,\infty )$ satisfy (2) and

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{2}^n\phi \left(\frac{x}{2^n},y\right)=\underset{n\to \infty }{lim}{2}^n\phi \left(x,\frac{y}{2^n}\right)=0\mathit{for}\mathit{all}x,y\in \mathcal{A}.\end{array}$$

(8)

Suppose that $\delta :\mathcal{A}\to \mathcal{A}$ and $\mathcal{D}:\mathcal{A}\to \mathcal{A}$ are mappings with $\delta \left(0\right)=0$ subjected to the inequality (1) and

$$\begin{array}{c}\hfill \parallel \delta \left(xy\right)-\delta \left(x\right)y-x\mathcal{D}\left(y\right)\parallel \le \phi (x,y)\mathit{for}\mathit{all}x,y\in \mathcal{A}.\end{array}$$

(9)

Then, δ is a generalized derivation.

Proof. 

According to Lemma 1, there exists a unique additive mapping $\mathcal{L}:\mathcal{A}\to \mathcal{A}$ given by (3) with the estimation (4). In addition, by virtue of (9), we see that

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

for all $x,y\in \mathcal{A}$, which means that

$$\begin{array}{c}\hfill \mathcal{L}\left(xy\right)=\mathcal{L}\left(x\right)y+x\mathcal{D}\left(y\right).\end{array}$$

(10)

Brešar’s result [15] guarantees that if $\mathcal{A}$ is either a semi-prime algebra or an algebra with a unit, then $\mathcal{D}$ is a derivation. On the other hand, by additivity of $\mathcal{D},$ we have that

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

for all $x,y\in \mathcal{A},$ which gives that

$$\begin{array}{c}\hfill \mathcal{L}\left(xy\right)=\delta \left(x\right)y+x\mathcal{D}\left(y\right).\end{array}$$

(11)

Considering (10) and (11), we arrive at

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

(12)

If $\mathcal{A}$ is unital, set $y:=e$ in (12). Then, $\mathcal{L}\left(x\right)=\delta \left(x\right)$ for all $x\in \mathcal{A}.$ If $\mathcal{A}$ is non-unital, then $\mathcal{L}\left(x\right)-\delta \left(x\right)$ lies in the left annihilator $lan\left(\mathcal{A}\right)$ of $\mathcal{A}.$ If $\mathcal{A}$ is semiprime, then $lan\left(\mathcal{A}\right)=\left\{0\right\},$ so that $\mathcal{L}\left(x\right)=\delta \left(x\right)$ for all $x\in \mathcal{A}.$ Thus, Equation (10) (or (11)) implies that

$$\begin{array}{c}\hfill \delta \left(xy\right)=\delta \left(x\right)y+x\mathcal{D}\left(y\right)\mathrm{for}\mathrm{all}x,y\in \mathcal{A},\end{array}$$

(13)

where $\mathcal{D}$ is a derivation. Therefore, $\delta$ is a generalized derivation. □

Based on Lemma 2, the following theorem can be derived by using the same argument as in the proof of Theorem 1.

Theorem 2.

Let $\mathcal{A}$ be either a semiprime Banach algebra or a unital Banach algebra. Assume that mappings $\varphi :{\mathcal{A}}^l\to [0,\infty )$ and $\phi :{\mathcal{A}}^2\to [0,\infty )$ satisfy (5) and

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\frac{1}{2^n}\phi (2^{n}x,y)=\underset{n\to \infty }{lim}\frac{1}{2^n}\phi (x,2^{n}y)=0\mathit{for}\mathit{all}x,y\in \mathcal{A}.\end{array}$$

(14)

Suppose that $\delta :\mathcal{A}\to \mathcal{A}$ and $\mathcal{D}:\mathcal{A}\to \mathcal{A}$ are mappings with $\delta \left(0\right)=0$ subjected to inequalities (1) and (9). Then, δ is a generalized derivation.

Theorem 3.

Let $\mathcal{A}$ be either a semiprime Banach algebra or a unital Banach algebra. Assume that mappings $\varphi :{\mathcal{A}}^l\to [0,\infty )$ and $\phi :{\mathcal{A}}^2\to [0,\infty )$ satisfy (2) and (8) (resp., (5) and (14)). Suppose that $\delta :\mathcal{A}\to \mathcal{A}$ and $\mathcal{D}:\mathcal{A}\to \mathcal{A}$ are mappings with $\delta \left(0\right)=0$ subjected to the inequalities (1) and (9). If $\mathcal{D}\left(tx\right)=t\mathcal{D}\left(x\right)$ for all $t\in {\mathbb{T}}_{\epsilon }$ and all $x\in \mathcal{A},$ then δ is a linear generalized derivation.

Proof. 

We have by Theorem 1 (resp., Theorem 2) that $\delta$ satisfies (13). In this case, $\mathcal{D}$ is a derivation. However, from assumption, we see that $\mathcal{D}$ is linear; cf. [16]. Now, we are in a position to show that $\delta$ is a linear mapping. With the help of (13), we figure out

$$\begin{array}{c}\hfill \delta \left(tx\right)y+tx\mathcal{D}\left(y\right)=\delta (tx·y)=\delta (x·ty)=t\delta \left(x\right)y+tx\mathcal{D}\left(y\right).\end{array}$$

for all $x,y\in \mathcal{A}$ and all $t\in \mathbb{C}.$ So, we are forced to conclude that

$$\begin{array}{c}\hfill \left(\delta \left(tx\right)-t\delta \left(x\right)\right)y=0\mathrm{for}\mathrm{all}x,y\in \mathcal{A}\mathrm{and}\mathrm{all}t\in \mathbb{C}.\end{array}$$

(15)

If $\mathcal{A}$ contains a unit element, set $y:=e$ in (15). Then, $\delta \left(tx\right)=t\delta \left(x\right)$ for all $x\in \mathcal{A}$ and all $t\in \mathbb{C}.$ If $\mathcal{A}$ has no unit element, then $\delta \left(tx\right)-t\delta \left(x\right)$ lies in the left annihilator $lan\left(\mathcal{A}\right)$ of $\mathcal{A}.$ If $\mathcal{A}$ is semiprime, then $lan\left(\mathcal{A}\right)=\left\{0\right\},$ so that $\delta \left(tx\right)=t\delta \left(x\right)$ for all $x\in \mathcal{A}$ and all $t\in \mathbb{C}.$ Therefore, $\delta$ is a linear mapping. □

Here, we take note of the following: use the fact that any $C^{*}$-algebra is semiprime (viz., [17] for details) and then we can conclude that Theorems 2 and 3 are also valid in a $C^{*}$-algebra.

Theorem 4.

Let $\mathcal{A}$ be a unital Banach algebra. Assume that mappings $\varphi :{\mathcal{A}}^l\to [0,\infty )$ and $\phi :{\mathcal{A}}^2\to [0,\infty )$ satisfy (2) and (8) (resp., (5) and (14)). Suppose that $\delta :\mathcal{A}\to \mathcal{A}$ and $\mathcal{D}:\mathcal{A}\to \mathcal{A}$ with $\delta \left(0\right)=0$ are mappings subjected to inequalities (1) and (9). If $\delta \left(te\right)=t\delta \left(e\right)$ for all $t\in \mathbb{C}.$ then δ is a linear generalized derivation.

Proof. 

It follows from Theorem 1 (resp., Theorem 2) that $\delta$ is a generalized derivation. In other words, it satisfies the identity (13). For all $t\in \mathbb{C},$ put $x:=e$ and $y:=te$ in (13). By assumption, we then have that $\mathcal{D}\left(te\right)=0.$ Again, letting $y:=te$ in (13) and using the last expression, we deduce that $\delta \left(tx\right)=t\delta \left(x\right)$ for all $x\in \mathcal{A}$ and all $t\in \mathbb{C},$ which means that $\delta$ is linear. Meanwhile, it follows from Lemma 2.1 in [18] that $\mathcal{D}$ is also linear. Therefore, $\delta$ is a linear generalized derivation. □

Theorem 5.

Let $\mathcal{A}$ be a noncommutative unital Banach algebra. Assume that mappings $\varphi :{\mathcal{A}}^l\to [0,\infty )$ and $\phi :{\mathcal{A}}^2\to [0,\infty )$ satisfy (2) and

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{2}^n\phi \left(x,\frac{y}{2^n}\right)=0\mathrm{for}\mathrm{all}x,y\in \mathcal{A}.\end{array}$$

(16)

Suppose that $\delta :\mathcal{A}\to \mathcal{A}$ is a mapping with $\delta \left(0\right)=0$ such that for all $x_1,\cdots ,x_l\in \mathcal{A}$ and all $t\in {\mathbb{T}}_{\epsilon },$

$$\begin{array}{c}\hfill \parallel a_1\delta \left(t{x}_1\right)+t{a}_2\delta \left(x_2\right)+\sum _{j=3}^l{a}_j\delta \left(x_j\right)\parallel \le \parallel \delta (\sum _{j=1}^l{a}_j{x}_j)\parallel +\varphi (x_1,\cdots ,x_l),\end{array}$$

(17)

where ${\prod }_{j=1}^l{a}_j\ne 0,a_1=a_2,l\ge 3$ and

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

(18)

Then, δ is a linear generalized derivation. In particular, if there is a nonnegative constant M, such that $r\left(\delta \right(x\left)\right)\le Mr\left(x\right)$ for all $x\in \mathcal{A},$ where $r(·)$ stands for the spectral radius, then δ leaves each primitive ideal invariant.

Proof. 

We first consider $t=1$ in (17). It follows from Lemma 1 that there exists a unique additive mapping $\mathcal{L}:\mathcal{A}\to \mathcal{A}$ defined by (3) with (4). We intend to prove that $\delta$ is linear. It follows from (17) that

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{2}^n\parallel a_1\delta \left(t\frac{x}{2^n}\right)+t{a}_2\delta \left(-\frac{x}{2^n}\right)\parallel \le \underset{n\to \infty }{lim}{2}^n{\varphi }_{123}\left(\frac{x}{2^n},-\frac{x}{2^n},0\right)=0.\end{array}$$

for all $x\in \mathcal{A}$ and all $t\in {\mathbb{T}}_{\epsilon }.$ Since $a_1=a_2,$ by (3), we have $\mathcal{L}\left(tx\right)=t\mathcal{L}\left(x\right)$ and so, $\mathcal{L}$ is linear (see reference [16]). Expressions (16) and (18) guarantee that

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

for all $x,y\in \mathcal{A},$ which implies that

$$\begin{array}{c}\hfill \mathcal{L}\left(xy\right)=\delta \left(x\right)y-x\delta \left(e\right)y+x\mathcal{L}\left(y\right).\end{array}$$

(19)

Let us define a mapping $\mathcal{D}:\mathcal{A}\to \mathcal{A}$ by

$$\begin{array}{c}\hfill \mathcal{D}\left(x\right)=\mathcal{L}\left(x\right)-\delta \left(e\right)x\mathrm{for}\mathrm{all}x\in \mathcal{A}.\end{array}$$

(20)

Then, the relation (19) can be written in the form

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

(21)

Our goal is to show that $\delta$ is a linear generalized derivation. It follows from (20) and the linearity of $\mathcal{L}$ that for all $x,y\in \mathcal{A}$ and all $t_1,t_2\in \mathbb{C},$

$$\begin{array}{cc}\hfill \mathcal{D}(t_{1}x+t_{2}y)& =\mathcal{L}(t_{1}x+t_{2}y)-\delta \left(e\right)(t_{1}x+t_{2}y)\hfill \\ \hfill & =t_1\left[\mathcal{L}\left(x\right)-\delta \left(e\right)x\right]+t_2\left[\mathcal{L}\left(y\right)-\delta \left(e\right)y\right]\hfill \\ \hfill & =t_1\mathcal{D}\left(x\right)+t_2\mathcal{D}\left(y\right),\hfill \end{array}$$

which means that $\mathcal{D}$ is linear. However, considering (21) and ([18], Lemma 2.1), we have that $\delta$ is also linear. From the definition of $\mathcal{L},$ we are forced to conclude that $\delta =\mathcal{L}.$ Hence, the expression (19) can be represented as

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

(22)

Then, in view of (20) and (22), we then obtain equality (13). By applying (13) and (20), we see that for all $x,y\in \mathcal{A},$

$$\begin{array}{c}\hfill \mathcal{D}\left(xy\right)=\delta \left(xy\right)-\delta \left(e\right)xy=\delta \left(x\right)y+x\mathcal{D}\left(y\right)-\delta \left(e\right)xy=\mathcal{D}\left(x\right)y+x\mathcal{D}\left(y\right).\end{array}$$

This means that $\mathcal{D}$ is a derivation. Therefore, we have proved that $\delta$ is a linear generalized derivation. On the other hand, the equalities (13) and (20) yield that

$$\begin{array}{cc}\hfill \delta \left(xyz\right)& =\delta \left(xy\right)z+xy\mathcal{D}\left(z\right)\hfill \\ \hfill & =\delta \left(xy\right)z+x\mathcal{D}\left(yz\right)-x\mathcal{D}\left(y\right)z\hfill \\ \hfill & =\delta \left(xy\right)z-x\delta \left(y\right)z+x\delta \left(yz\right)\hfill \end{array}$$

(23)

for all $x,y,z\in \mathcal{A}.$ Letting $y:=e$ and $z:=y$ in (23), we obtain (22). In other words, we note that Equation (22) is equivalent to Equation (23). Thus, if $r\left(\delta \right(x\left)\right)\le Mr\left(x\right)$ holds for all $x\in \mathcal{A},$ then $\delta$ is a spectrally bounded linear generalized derivation, so that $\delta \left(P\right)\subseteq P,$ where P is a primitive ideal of $\mathcal{A}$; cf. [19]. □

The following theorem shows alternative results of inequalities (17) and (18) under the conditions (5) and (24), which are similarly verified as in the proof of Theorem 5.

Theorem 6.

Let $\mathcal{A}$ be a noncommutative unital Banach algebra. Assume that mappings $\varphi :{\mathcal{A}}^l\to [0,\infty )$ and (5) satisfy $\phi :{\mathcal{A}}^2\to [0,\infty )$ and

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\frac{1}{2^n}\phi (x,2^{n}y)=0\mathit{for}\mathit{all}x,y\in \mathcal{A}.\end{array}$$

(24)

Suppose that $\delta :\mathcal{A}\to \mathcal{A}$ is a mapping with $\delta \left(0\right)=0$ subjected to inequalities (17) and (18). Then, δ is a linear generalized derivation. In particular, if there is a nonnegative constant M, such that $r\left(\delta \right(x\left)\right)\le Mr\left(x\right)$ for all $x\in \mathcal{A},$ then δ leaves each primitive ideal invariant.

It is well-known that any spectrally bound linear derivation on a noncommutative unital Banach algebra maps into its (Jacobson) radical, which was proved by Brešar and Mathieu in reference [19]. Then, we arrive at the following conclusion.

Corollary 1.

Let $\mathcal{A}$ be a noncommutative unital Banach algebra. Assume that mappings $\varphi :{\mathcal{A}}^l\to [0,\infty )$ and $\phi :{\mathcal{A}}^2\to [0,\infty )$ satisfy (2) and (16) (resp., (5) and (24)). Suppose that $\delta :\mathcal{A}\to \mathcal{A}$ is a mapping with $\delta \left(0\right)=0$ subjected to the inequalities (17) and (18). If $\delta \left(e\right)=0$ is fulfilled and there is a nonnegative constant M, such that $r\left(\delta \right(x\left)\right)\le Mr\left(x\right)$ for all $x\in \mathcal{A},$ then δ maps $\mathcal{A}$ into its radical $\mathrm{rad}\left(\mathcal{A}\right).$

Proof. 

Considering the proof of Theorem 5 (resp., Theorem 6), we see that $\delta$ is a linear mapping satisfying (22). It follows from the assumptions that the mapping $\delta$ is a spectrally bounded linear derivation. Therefore, due to the aforementioned fact, we find that $\delta \left(\mathcal{A}\right)\subseteq \mathrm{rad}\left(\mathcal{A}\right).$ □

Let us introduce some definitions used in the next discussion: a multiplicative derivation of $\mathcal{A}$ is a mapping $\delta :\mathcal{A}\to \mathcal{A}$, which satisfies $\delta \left(xy\right)=\delta \left(x\right)y+x\delta \left(y\right)$ for all $x,y\in \mathcal{A}.$ So, a multiplicative derivation will be a derivation when it is an additive mapping. We say that a mapping $\delta :\mathcal{A}\to \mathcal{A}$ is a multiplicative generalized derivation if there exists a derivation $\mathcal{D}$ of $\mathcal{A}$, such that $\delta \left(xy\right)=\delta \left(x\right)y+x\mathcal{D}\left(y\right)$ holds for all $x,y\in \mathcal{A}.$ A mapping $\delta :\mathcal{A}\to \mathcal{A}$ (not necessarily additive) is called a multiplicative(generalized)-derivation if there exists a mapping $\mathcal{D}$ of $\mathcal{A}$ (not necessarily additive or no derivation), such that $\delta \left(xy\right)=\delta \left(x\right)y+x\mathcal{D}\left(y\right)$ for all $x,y\in \mathcal{A}.$ These definitions can be found in [20,21,22,23,24].

Theorem 7.

Let $\mathcal{A}$ be a normed algebra with the unit. Suppose that $\delta :\mathcal{A}\to \mathcal{A}$ is a mapping, such that for all $x,y\in \mathcal{A}$ and some $\epsilon \ge 0,$

$$\begin{array}{c}\hfill \parallel \delta \left(xy\right)-\delta \left(x\right)y+x\delta \left(e\right)y-x\delta \left(y\right)\parallel \le \epsilon .\end{array}$$

(25)

Then, δ is fulfilled with (13), where $\mathcal{D}:\mathcal{A}\to \mathcal{A}$ is a mapping defined by

$$\begin{array}{c}\hfill \mathcal{D}\left(x\right)=\delta \left(x\right)-\delta \left(e\right)x\mathit{for}\mathit{all}x\in \mathcal{A}.\end{array}$$

(26)

i.e., δ is a multiplicative (generalized)-derivation. In this case, $\mathcal{D}$ is a multiplicative derivation, which leaves each center $Z\left(\mathcal{A}\right)$ of the $\mathcal{A}$ invariant.

Proof. 

It follows from (25) that for all $x,y\in \mathcal{A}$, and all positive integers $n,$

$$\begin{array}{c}\hfill \parallel \frac{\delta (nx·y)}{n}-\frac{\delta \left(nx\right)}{n}y+x\delta \left(e\right)y-x\delta \left(y\right)\parallel \le \frac{\epsilon }{n}.\end{array}$$

(27)

Moreover, by (25), for all $x,y\in \mathcal{A}$ and all positive integers $n,$ we have

$$\begin{array}{c}\hfill \parallel \frac{\delta (x·ny)}{n}-\delta \left(x\right)y+x\delta \left(e\right)y-\frac{x\delta \left(ny\right)}{n}\parallel \le \frac{\epsilon }{n}.\end{array}$$

(28)

On the other hand, we note that for all $x,y\in \mathcal{A}$ and all positive integers $n,$

$$\begin{array}{cc}\hfill & \parallel \delta \left(xy\right)-\delta \left(x\right)y+x\delta \left(e\right)y-x\delta \left(y\right)\parallel \hfill \\ \hfill & =\parallel \delta \left(xy\right)-\delta \left(e\right)xy-\frac{\delta (ne·xy)}{n}+\frac{\delta \left(ne\right)}{n}xy-\delta \left(x\right)y+\frac{\delta (x·ny)}{n}+x\delta \left(e\right)y-x\frac{\delta \left(ny\right)}{n}\hfill \\ \hfill & +x\delta \left(e\right)y-x\delta \left(y\right)+\frac{\delta (nx·y)}{n}-\frac{\delta \left(nx\right)}{n}y+\delta \left(e\right)xy-\frac{\delta \left(ne\right)}{n}xy+x\frac{\delta \left(ny\right)}{n}-x\delta \left(e\right)y\hfill \\ \hfill & -\frac{\delta (nx·y)}{n}+\frac{\delta \left(nx\right)}{n}y-\delta \left(x\right)y+\delta \left(x\right)y\parallel \hfill \\ \hfill & \le \parallel \delta \left(xy\right)-\delta \left(e\right)xy-\frac{\delta (ne·xy)}{n}+\frac{\delta \left(ne\right)}{n}xy\parallel +\parallel -\delta \left(x\right)y+\frac{\delta (x·ny)}{n}+x\delta \left(e\right)y-x\frac{\delta \left(ny\right)}{n}\parallel \hfill \\ \hfill & +\parallel x\delta \left(e\right)y-x\delta \left(y\right)+\frac{\delta (nx·y)}{n}-\frac{\delta \left(nx\right)}{n}y\parallel +\parallel \frac{\delta (ne·x)}{n}-\frac{\delta \left(ne\right)}{n}x+\delta \left(e\right)x-\delta \left(x\right)\parallel \parallel y\parallel \hfill \\ \hfill & +\parallel \delta \left(x\right)y-x\delta \left(e\right)y-\frac{\delta (x·ny)}{n}+x\frac{\delta \left(ny\right)}{n}\parallel \hfill \\ \hfill & \le \frac{\epsilon }{n}\left[4+\parallel y\parallel \right].\hfill \end{array}$$

Applying (25)–(28), we observe that the right-hand side of the last inequality tends to zero when $n\to \infty .$ Hence, we obtain (22). Let $\mathcal{D}:\mathcal{A}\to \mathcal{A}$ be a mapping defined by (26). Then, by virtue of (22), we see that $\delta$ satisfies (13). Thereby, $\delta$ is a multiplicative (generalized)-derivation. Meanwhile, as we did in the proof of Theorem 5 together with (22) and (26), we arrive at $\mathcal{D}\left(xy\right)=\mathcal{D}\left(x\right)y+x\mathcal{D}\left(y\right),$ so that $\mathcal{D}$ is a multiplicative derivation and then, $\mathcal{D}\left(Z\right(\mathcal{A}\left)\right)\subseteq Z\left(\mathcal{A}\right)$ (cf. [25]). □

The following result is a special case of Theorem 7.

Corollary 2.

Let $\mathcal{A}$ be a normed algebra with the unit. Suppose that a mapping $\delta :\mathcal{A}\to \mathcal{A}$ satisfies (25). If $\delta \left(e\right)=0$ holds, then δ is a multiplicative derivation that leaves each center $Z\left(\mathcal{A}\right)$ of the $\mathcal{A}$ invariant.

Proof. 

As we see in the proof of Theorem 7, we know that $\delta$ is a mapping with (22). By assumption, we then have that $\delta$ is a multiplicative derivation. Therefore, we obtain $\delta \left(Z\right(\mathcal{A}\left)\right)\subseteq Z\left(\mathcal{A}\right),$ see [25]. □

3. Conclusions

We presented our results regarding the stability of an approximate generalized derivation in a unital Banach algebra, and a semiprime Banach algebra. We established the conditions in which this approximate generalized derivation can be a linear map. In particular, some of these results can be applied to $C^{*}$-algebra.

On the other hand, an approximate generalized derivation of a noncommutative unital Banach algebra leaves each primitive ideal invariant. Additionally, we designed some theorems concerning the stability of multiplicative (generalized) derivation in a normed algebra with a unit.