$C^{\ast }$-ternary 3-derivations on $C^{\ast }$-ternary algebras

Abstract

In this paper, we prove the Hyers-Ulam stability of $C^{\ast }$-ternary 3-derivations and of $C^{\ast }$-ternary 3-homomorphisms for the functional equation

$$f(x_1+x_2,y_1+y_2,z_1+z_2)=\sum _{1\le i,j,k\le 2}f(x_i,y_j,z_k)$$

in $C^{\ast }$-ternary algebras.

MSC:17A40, 39B52, 46Lxx, 46K70, 46L05, 46B99.

1 Introduction and preliminaries

Ternary algebraic operations were considered in the nineteenth century by several mathematicians such as Cayley [1] who introduced the notion of a cubic matrix, which in turn was generalized by Kapranov, Gelfand and Zelevinskii [2]. The simplest example of such non-trivial ternary operation is given by the following composition rule:

$${\{a,b,c\}}_{ijk}=\sum _{1\le l,m,n\le N}{a}_{nil}{b}_{ljm}{c}_{mkn}(i,j,k=1,2,\dots ,N).$$

Ternary structures and their generalization, the so-called n-ary structures, raise certain hopes in view of their applications in physics. Some significant physical applications are as follows (see [3, 4]).

• (1) The algebra of nonions generated by two matrices

  $$\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right)\text{and}\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& \omega \\ {\omega }^2& 0& 0\end{array}\right)(\omega =e^{\frac{2\pi i}{3}})$$

was introduced by Sylvester as a ternary analog of Hamiltons quaternions (cf. [5]).

• (2) The quark model inspired a particular brand of ternary algebraic systems. The so-called Nambu mechanics is based on such structures (see [6]).

There are also some applications, although still hypothetical, in the fractional quantum Hall effect, the nonstandard statistics, supersymmetric theory, and Yang-Baxter equation (cf. [3, 5, 7]).

A $C^{\ast }$-ternary algebra is a complex Banach space A, equipped with a ternary product $(x,y,z)\to [x,y,z]$ of $A^3$ into A, which is ℂ-linear in the outer variables, conjugate ℂ-linear in the middle variable, and associative in the sense that $[x,y,[z,w,v]]=[x,[w,z,y],v]=[[x,y,z],w,v]$, and satisfies $\parallel [x,y,z]\parallel \le \parallel x\parallel \cdot \parallel y\parallel \cdot \parallel z\parallel$ and $\parallel [x,x,x]\parallel ={\parallel x\parallel }^3$ (see [8]). Every left Hilbert $C^{\ast }$-module is a $C^{\ast }$-ternary algebra via the ternary product $[x,y,z]:=〈x,y〉z$.

If a $C^{\ast }$-ternary algebra $(A,[\cdot ,\cdot ,\cdot ])$ has an identity, i.e., an element $e\in A$ such that $x=[x,e,e]=[e,e,x]$ for all $x\in A$, then it is routine to verify that A, endowed with $x\circ y:=[x,e,y]$ and $x^{\ast }:=[e,x,e]$, is a unital $C^{\ast }$-algebra. Conversely, if $(A,\circ )$ is a unital $C^{\ast }$-algebra, then $[x,y,z]:=x\circ y^{\ast }\circ z$ makes A into a $C^{\ast }$-ternary algebra.

Throughout this paper, assume that $C^{\ast }$-ternary algebras A and B are induced by unital $C^{\ast }$-algebras with units e and $e^{\prime }$, respectively.

A ℂ-linear mapping $H:A\to B$ is called a $C^{\ast }$-ternary homomorphism if $H([x,y,z])=[H(x),H(y),H(z)]$ for all $x,y,z\in A$. If, in addition, the mapping H is bijective, then the mapping $H:A\to B$ is called a $C^{\ast }$-ternary algebra isomorphism. A ℂ-linear mapping $\delta :A\to A$ is called a $C^{\ast }$-ternary derivation if

$$\delta ([x,y,z])=[\delta (x),y,z]+[x,\delta (y),z]+[x,y,\delta (z)]$$

for all $x,y,z\in A$ (see [9]).

In 1940, Ulam [10] gave a talk before the Mathematics Club of the University of Wisconsin in which he discussed a number of unsolved problems. Among these was the following question concerning the stability of homomorphisms:

We are given a group G and a metric group $G^{\mathrm{\prime }}$ with metric $\rho (\cdot ,\cdot )$ . Given $ϵ>0$ , does there exist a $\delta >0$ such that if $f:G\to G^{\mathrm{\prime }}$ satisfies $\rho (f(xy),f(x)f(y))<\delta$ for all $x,y\in G$ , then a homomorphism $h:G\to G$ exists with $\rho (f(x),h(x))<ϵ$ for all $x\in G$ ?

In 1941, Hyers [11] gave the first partial solution to Ulam’s question for the case of approximate additive mappings under the assumption that G and $G^{\prime }$ are Banach spaces. Then, Aoki [12] and Bourgin [13] considered the stability problem with unbounded Cauchy differences. In 1978, Rassias [14] generalized the theorem of Hyers [11] by considering the stability problem with unbounded Cauchy differences. In 1991, Gajda [15], following the same approach as that by Rassias [14], gave an affirmative solution to this question for $p>1$. It was shown by Gajda [15] as well as by Rassias and Šemrl [16], that one cannot prove a Rassias-type theorem when $p=1$. Gǎvruta [17] obtained the generalized result of the Rassias theorem which allows the Cauchy difference to be controlled by a general unbounded function. During the last two decades, a number of articles and research monographs have been published on various generalizations and applications of the Hyers-Ulam stability to a number of functional equations and mappings, for example, Cauchy-Jensen mappings, k-additive mappings, invariant means, multiplicative mappings, bounded n th differences, convex functions, generalized orthogonality mappings, Euler-Lagrange functional equations, differential equations, and Navier-Stokes equations. The stability problems of several functional equations have been extensively investigated by a number of authors and there are many interesting results, containing ternary homomorphisms and ternary derivations, concerning this problem (see [18–31]).

Let X and Y be complex vector spaces. A mapping $f:X\times X\times X\to Y$ is called a 3-additive mapping if f is additive for each variable, and a mapping $f:X\times X\times X\to Y$ is called a 3-ℂ-linear mapping if f is ℂ-linear for each variable.

A 3-ℂ-linear mapping $H:A\times A\times A\to B$ is called a $C^{\ast }$-ternary 3-homomorphism if it satisfies

$$H([x_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])=[H(x_1,x_2,x_3),H(y_1,y_2,y_3),H(z_1,z_2,z_3)]$$

for all $x_1,y_1,z_1,x_2,y_2,z_2,x_3,y_3,z_3\in A$.

For a given mapping $f:A^3\to B$, we define

$$\begin{array}{c}{D}_{\lambda ,\mu ,\nu }f(x_1,x_2,y_1,y_2,z_1,z_2)\hfill \\ :=f(\lambda x_1+\lambda x_2,\mu y_1+\mu y_2,\nu z_1+\nu z_2)-\lambda \mu \nu \sum _{1\le i,j,k\le 2}f(x_i,y_j,z_k)\hfill \end{array}$$

for all $\lambda ,\mu ,\nu \in S^1:=\{\lambda \in \mathbb{C}:|\lambda |=1\}$ and all $x_1,x_2,y_1,y_2,z_1,z_2\in A$.

Bae and Park [32] proved the Hyers-Ulam stability of 3-homomorphisms in $C^{\ast }$-ternary algebras for the functional equation

$$D_{\lambda ,\mu ,\nu }f(x_1,x_2,y_1,y_2,z_1,z_2)=0.$$

Lemma 1.1 [32]

Let X and Y be complex vector spaces, and let $f:X\times X\times X\to Y$ be a 3-additive mapping such that $f(\lambda x,\mu y,\nu z)=\lambda \mu \nu f(x,y,z)$ for all $\lambda ,\mu ,\nu \in S^1$ and all $x,y,z\in X$. Then f is 3-ℂ-linear.

Theorem 1.2 [32]

Let $p,q,r\in (0,\mathrm{\infty })$ with $p+q+r<3$ and $\theta \in (0,\mathrm{\infty })$, and let $f:A^3\to B$ be a mapping such that

$$\begin{array}{c}\parallel D_{\lambda ,\mu ,\nu }f(x_1,x_2,y_1,y_2,z_1,z_2)\parallel \hfill \\ \le \theta \cdot max{\{\parallel x_1\parallel ,\parallel x_2\parallel \}}^p\cdot max{\{\parallel y_1\parallel ,\parallel y_2\parallel \}}^q\cdot max{\{\parallel z_1\parallel ,\parallel z_2\parallel \}}^r\hfill \end{array}$$

(1)

and

$$\begin{array}{c}\parallel f([x_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])-[f(x_1,x_2,x_3),f(y_1,y_2,y_3),f(z_1,z_2,z_3)]\parallel \hfill \\ \le \theta \sum _{i=1}^3{\parallel x_i\parallel }^p\cdot {\parallel y_i\parallel }^q\cdot {\parallel z_i\parallel }^r\hfill \end{array}$$

(2)

for all $\lambda ,\mu ,\nu \in S^1$ and all $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3\in A$. Then there exists a unique $C^{\ast }$-ternary 3-homomorphism $H:A^3\to B$ such that

$$\parallel f(x,y,z)-H(x,y,z)\parallel \le \frac{\theta }{2^3-2^{p+q+r}}{\parallel x\parallel }^p\cdot {\parallel y\parallel }^q\cdot {\parallel z\parallel }^r$$

(3)

for all $x,y,z\in A$.

2 $C^{\ast }$-ternary 3-homomorphisms in $C^{\ast }$-ternary algebras

Theorem 2.1 Let $p,q,r\in (0,\mathrm{\infty })$ with $p+q+r<3$ and $\theta \in (0,\mathrm{\infty })$, and let $f:A^3\to B$ be a mapping satisfying (1) and (2). If there exists an $(x_0,y_0,z_0)\in A^3$ such that ${lim}_{n\to \mathrm{\infty }}\frac{1}{8^n}f(2^n{x}_0,2^n{y}_0,2^n{z}_0)=e^{\mathrm{\prime }}$, then the mapping f is a $C^{\ast }$-ternary 3-homomorphism.

Proof By Theorem 1.2, there exists a unique $C^{\ast }$-ternary 3-homomorphism $H:A^3\to B$ satisfying (3). Note that

$$H(x,y,z):=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{8^n}f(2^{n}x,2^{n}y,2^{n}z)$$

for all $x,y,z\in A$. By the assumption, we get that

$$H(x_0,y_0,z_0)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{8^n}f(2^n{x}_0,2^n{y}_0,2^n{z}_0)=e^{\mathrm{\prime }}.$$

It follows from (2) that

$$\begin{array}{c}\parallel [H(x_1,x_2,x_3),H(y_1,y_2,y_3),H(z_1,z_2,z_3)]-[H(x_1,x_2,x_3),H(y_1,y_2,y_3),f(z_1,z_2,z_3)]\parallel \hfill \\ =\parallel H([x_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])\hfill \\ -[H(x_1,x_2,x_3),H(y_1,y_2,y_3),f(z_1,z_2,z_3)]\parallel \hfill \\ =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{8^{2n}}\parallel f([2^n{x}_1,2^n{y}_1,z_1],[2^n{x}_2,2^n{y}_2,z_2],[2^n{x}_3,2^n{y}_3,z_3])\hfill \\ -[f(2^n{x}_1,2^n{x}_2,2^n{x}_3),f(2^n{y}_1,2^n{y}_2,2^n{y}_3),f(z_1,z_2,z_3)]\parallel \hfill \\ \le \underset{n\to \mathrm{\infty }}{lim}\frac{\theta 2^{n(p+q)}}{8^{2n}}\sum _{i=1}^3{\parallel x_i\parallel }^p\cdot {\parallel y_i\parallel }^q\cdot {\parallel z_i\parallel }^r=0\hfill \end{array}$$

for all $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3\in A$. So,

$$[H(x_1,x_2,x_3),H(y_1,y_2,y_3),H(z_1,z_2,z_3)]=[H(x_1,x_2,x_3),H(y_1,y_2,y_3),f(z_1,z_2,z_3)]$$

for all $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3\in A$. Letting $x_1=y_1=x_0$, $x_2=y_2=y_0$ and $x_3=y_3=z_0$ in the last equality, we get $f(z_1,z_2,z_3)=H(z_1,z_2,z_3)$ for all $z_1,z_2,z_3\in A$. Therefore, the mapping f is a $C^{\ast }$-ternary 3-homomorphism. □

Theorem 2.2 Let $p_i,q_i,r_i\in (0,\mathrm{\infty })$ ($i=1,2,3$) such that $p_i\ne 1$ or $q_i\ne 1$ or $r_i\ne 1$ for some $1\le i\le 3$ and $\theta ,\eta \in (0,\mathrm{\infty })$, and let $f:A^3\to B$ be a mapping such that

$$\begin{array}{c}\parallel D_{\lambda ,\mu ,\nu }f(x_1,x_2,y_1,y_2,z_1,z_2)\parallel \hfill \\ \le \theta ({\parallel x_1\parallel }^{p_1}\cdot {\parallel x_2\parallel }^{p_2}\cdot {\parallel y_1\parallel }^{q_1}\cdot {\parallel y_2\parallel }^{q_2}\hfill \\ +{\parallel y_1\parallel }^{q_1}\cdot {\parallel y_2\parallel }^{q_2}\cdot {\parallel z_1\parallel }^{r_1}\cdot {\parallel z_2\parallel }^{r_2}+{\parallel x_1\parallel }^{p_1}\cdot {\parallel x_2\parallel }^{p_2}\cdot {\parallel z_1\parallel }^{r_1}\cdot {\parallel z_2\parallel }^{r_2})\hfill \end{array}$$

(4)

and

$$\begin{array}{c}\parallel f([x_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])-[f(x_1,x_2,x_3),f(y_1,y_2,y_3),f(z_1,z_2,z_3)]\parallel \hfill \\ \le \eta {\parallel x_1\parallel }^{p_1}\cdot {\parallel x_2\parallel }^{p_2}\cdot {\parallel x_3\parallel }^{p_3}\cdot {\parallel y_1\parallel }^{q_1}\cdot {\parallel y_2\parallel }^{q_2}\cdot {\parallel y_3\parallel }^{q_3}\cdot {\parallel z_1\parallel }^{r_1}\cdot {\parallel z_2\parallel }^{r_2}\cdot {\parallel z_3\parallel }^{r_3}\hfill \end{array}$$

(5)

for all $\lambda ,\mu ,\nu \in S^1$ and all $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3\in A$. Then the mapping $f:A^3\to B$ is a $C^{\ast }$-ternary 3-homomorphism.

Proof Letting $x_i=y_j=z_k=0$ ($i,j,k=1,2$) in (4), we get $f(0,0,0)=0$. Putting $\lambda =\mu =\nu =1$, $x_2=0$ and $y_j=z_k=0$ ($j,k=1,2$) in (4), we have $f(x_1,0,0)=0$ for all $x_1\in A$. Similarly, we get $f(0,y_1,0)=f(0,0,z_1)=0$ for all $y_1,z_1\in A$. Setting $\lambda =\mu =\nu =1$, $x_2=0$, $y_2=0$ and $z_1=z_2=0$, we have $f(x_1,y_1,0)=0$ for all $x_1,y_1\in A$. Similarly, we get $f(x_1,0,z_1)=f(0,y_1,z_1)=0$ for all $x_1,y_1,z_1\in A$. Now letting $\lambda =\mu =\nu =1$ and $y_2=z_2=0$ in (4), we have

$$f(x_1+x_2,y_1,z_1)=f(x_1,y_1,z_1)+f(x_2,y_1,z_1)$$

for all $x_1,x_2,y_1,z_1\in A$.

Similarly, one can show that the other equations hold. So, f is 3-additive.

Letting $x_2=y_2=z_2=0$ in (4), we get $f(\lambda x_1,\mu y_1,\nu z_1)=\lambda \mu \nu f(x_1,y_1,z_1)$ for all $\lambda ,\mu ,\nu \in S^1$ and all $x_1,y_1,z_1\in A$. So, by Lemma 1.1, the mapping f is 3-ℂ-linear.

Without any loss of generality, we may suppose that $p_1\ne 1$.

Let $p_1<1$. It follows from (5) that

$$\begin{array}{c}\parallel f([x_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])-[f(x_1,x_2,x_3),f(y_1,y_2,y_3),f(z_1,z_2,z_3)]\parallel \hfill \\ =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{3^n}\parallel f([3^n{x}_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])\hfill \\ -[f(3^n{x}_1,x_2,x_3),f(y_1,y_2,y_3),f(z_1,z_2,z_3)]\parallel \hfill \\ \le \eta \underset{n\to \mathrm{\infty }}{lim}\frac{3^{n{p}_1}}{3^n}\hfill \\ \times ({\parallel x_1\parallel }^{p_1}\cdot {\parallel x_2\parallel }^{p_2}\cdot {\parallel x_3\parallel }^{p_3}\cdot {\parallel y_1\parallel }^{q_1}\cdot {\parallel y_2\parallel }^{q_2}\cdot {\parallel y_3\parallel }^{q_3}\cdot {\parallel z_1\parallel }^{r_1}\cdot {\parallel z_2\parallel }^{r_2}\cdot {\parallel z_3\parallel }^{r_3})\hfill \\ =0\hfill \end{array}$$

for all $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3\in A$.

Let $p_1>1$. It follows from (5) that

$$\begin{array}{c}\parallel f([x_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])-[f(x_1,x_2,x_3),f(y_1,y_2,y_3),f(z_1,z_2,z_3)]\parallel \hfill \\ =\underset{n\to \mathrm{\infty }}{lim}{3}^n\parallel f([\frac{1}{3^n}{x}_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])\hfill \\ -[f(\frac{1}{3^n}{x}_1,x_2,x_3),f(y_1,y_2,y_3),f(z_1,z_2,z_3)]\parallel \hfill \\ \le \eta \underset{n\to \mathrm{\infty }}{lim}\frac{3^n}{3^{n{p}_1}}\hfill \\ \times ({\parallel x_1\parallel }^{p_1}\cdot {\parallel x_2\parallel }^{p_2}\cdot {\parallel x_3\parallel }^{p_3}\cdot {\parallel y_1\parallel }^{q_1}\cdot {\parallel y_2\parallel }^{q_2}\cdot {\parallel y_3\parallel }^{q_3}\cdot {\parallel z_1\parallel }^{r_1}\cdot {\parallel z_2\parallel }^{r_2}\cdot {\parallel z_3\parallel }^{r_3})\hfill \\ =0\hfill \end{array}$$

for all $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3\in A$. Therefore,

$$f([x_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])=[f(x_1,x_2,x_3),f(y_1,y_2,y_3),f(z_1,z_2,z_3)]$$

for all $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3\in A$. So, the mapping $f:A^3\to B$ is a $C^{\ast }$-ternary 3-homomorphism. □

Theorem 2.3 Let $\phi :A^6\to [0,\mathrm{\infty })$ and $\psi :A^9\to [0,\mathrm{\infty })$ be functions such that

$$\phi (x_1,\dots ,x_6)=0$$

if $x_i=0$ for some $1\le i\le 6$ and

$$\frac{1}{3^n}\psi (x_1,\dots ,3^n{x}_i,\dots ,x_9)=0\mathit{\text{or}}{3}^n\psi (x_1,\dots ,\frac{1}{3^n}{x}_i,\dots ,x_9)=0.$$

Suppose that $f:A^3\to B$ is a mapping satisfying

$$\parallel D_{\lambda ,\mu ,\nu }f(x_1,x_2,y_1,y_2,z_1,z_2)\parallel \le \phi (x_1,x_2,y_1,y_2,z_1,z_2)$$

and

$$\begin{array}{c}\parallel f([x_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])-[f(x_1,x_2,x_3),f(y_1,y_2,y_3),f(z_1,z_2,z_3)]\parallel \hfill \\ \le \psi (x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3)\hfill \end{array}$$

for all $\lambda ,\mu ,\nu \in S^1$ and all $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3\in A$. Then the mapping f is a $C^{\ast }$-ternary 3-homomorphism.

Proof The proof is similar to the proof of Theorem 2.2. □

Corollary 2.4 Let $p_i,q_i,r_i\in (0,\mathrm{\infty })$ ($i=1,2,3$) such that $p_i\ne 1$ or $q_i\ne 1$ or $r_i\ne 1$ for some $1\le i\le 3$ and $\theta ,\eta \in (0,\mathrm{\infty })$, and let $f:A^3\to B$ be a mapping such that

$$\parallel D_{\lambda ,\mu ,\nu }f(x_1,x_2,y_1,y_2,z_1,z_2)\parallel \le \theta {\parallel x_1\parallel }^{p_1}\cdot {\parallel x_2\parallel }^{p_2}\cdot {\parallel y_1\parallel }^{q_1}\cdot {\parallel y_2\parallel }^{q_2}\cdot {\parallel z_1\parallel }^{r_1}\cdot {\parallel z_2\parallel }^{r_2}$$

and

$$\begin{array}{c}\parallel f([x_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])-[f(x_1,x_2,x_3),f(y_1,y_2,y_3),f(z_1,z_2,z_3)]\parallel \hfill \\ \le \eta {\parallel x_1\parallel }^{p_1}\cdot {\parallel x_2\parallel }^{p_2}\cdot {\parallel x_3\parallel }^{p_3}\cdot {\parallel y_1\parallel }^{q_1}\cdot {\parallel y_2\parallel }^{q_2}\cdot {\parallel y_3\parallel }^{q_3}\cdot {\parallel z_1\parallel }^{r_1}\cdot {\parallel z_2\parallel }^{r_2}\cdot {\parallel z_3\parallel }^{r_3}\hfill \end{array}$$

for all $\lambda ,\mu ,\nu \in S^1$ and all $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3\in A$. Then the mapping $f:A^3\to B$ is a $C^{\ast }$-ternary 3-homomorphism.

3 $C^{\ast }$-ternary 3-derivations on $C^{\ast }$-ternary algebras

Definition 3.1 A 3-ℂ-linear mapping $D:A^3\to A$ is called a $C^{\ast }$-ternary 3-derivation if it satisfies

$$\begin{array}{rcl}D([x_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])& =& [D(x_1,x_2,x_3),[y_1,y_2^{\ast },y_3],[z_1,z_2^{\ast },z_3]]\\ +[[x_1,x_2^{\ast },x_3],D(y_1,y_2,y_3),[z_1,z_2^{\ast },z_3]]\\ +[[x_1,x_2^{\ast },x_3],[y_1,y_2^{\ast },y_3],D(z_1,z_2,z_3)]\end{array}$$

for all $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3\in A$.

Theorem 3.2 Let $p,q,r\in (0,\mathrm{\infty })$ with $p+q+r<3$ and $\theta \in (0,\mathrm{\infty })$, and let $f:A^3\to A$ be a mapping such that

$$\begin{array}{c}\parallel D_{\lambda ,\mu ,\nu }f(x_1,x_2,y_1,y_2,z_1,z_2)\parallel \hfill \\ \le \theta \cdot max{\{\parallel x_1\parallel ,\parallel x_2\parallel \}}^p\cdot max{\{\parallel y_1\parallel ,\parallel y_2\parallel \}}^q\cdot max{\{\parallel z_1\parallel ,\parallel z_2\parallel \}}^r\hfill \end{array}$$

(6)

and

$$\begin{array}{c}\parallel f([x_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])-[f(x_1,x_2,x_3),[y_1,y_2^{\ast },y_3],[z_1,z_2^{\ast },z_3]]\hfill \\ -[[x_1,x_2^{\ast },x_3],f(y_1,y_2,y_3),[z_1,z_2^{\ast },z_3]]-[[x_1,x_2^{\ast },x_3],[y_1,y_2^{\ast },y_3],f(z_1,z_2,z_3)]\parallel \hfill \\ \le \theta \sum _{i=1}^3{\parallel x_i\parallel }^p\cdot {\parallel y_i\parallel }^q\cdot {\parallel z_i\parallel }^r\hfill \end{array}$$

(7)

for all $\lambda ,\mu ,\nu \in S^1$ and all $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3\in A$. Then there exists a unique $C^{\ast }$-ternary 3-derivation $\delta :A^3\to A$ such that

$$\parallel f(x,y,z)-\delta (x,y,z)\parallel \le \frac{\theta }{2^3-2^{p+q+r}}{\parallel x\parallel }^p\cdot {\parallel y\parallel }^q\cdot {\parallel z\parallel }^r$$

(8)

for all $x,y,z\in A$.

Proof By the same method as in the proof of [[32], Theorem 1.2], we obtain a 3-ℂ-linear mapping $\delta :A^3\to A$ satisfying (8). The mapping $\delta (x,y,z):={lim}_{j\to \mathrm{\infty }}\frac{1}{8^j}f(2^{j}x,2^{j}y,2^{j}z)$ for all $x,y,z\in A$.

It follows from (7) that

$$\begin{array}{c}\parallel \delta ([x_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])-[\delta (x_1,x_2,x_3),[y_1,y_2^{\ast },y_3],[z_1,z_2^{\ast },z_3]]\hfill \\ -[[x_1,x_2^{\ast },x_3],\delta (y_1,y_2,y_3),[z_1,z_2^{\ast },z_3]]-[[x_1,x_2^{\ast },x_3],[y_1,y_2^{\ast },y_3],\delta (z_1,z_2,z_3)]\parallel \hfill \\ =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{8^{3n}}\parallel f(2^{3n}[x_1,y_1,z_1],2^{3n}[x_2,y_2,z_2],2^{3n}[x_3,y_3,z_3])\hfill \\ -[f(2^n{x}_1,2^n{x}_2,2^n{x}_3),[2^n{y}_1,2^n{y}_2^{\ast },2^n{y}_3],[2^n{z}_1,2^n{z}_2^{\ast },2^n{z}_3]]\hfill \\ -[[2^n{x}_1,2^n{x}_2^{\ast },2^n{x}_3],f(2^n{y}_1,2^n{y}_2,2^n{y}_3),[2^n{z}_1,2^n{z}_2^{\ast },2^n{z}_3]]\hfill \\ -[[2^n{x}_1,2^n{x}_2^{\ast },2^n{x}_3],[2^n{y}_1,2^n{y}_2^{\ast },2^n{y}_3],f(2^n{z}_1,2^n{z}_2,2^n{z}_3)]\parallel \hfill \\ \le \underset{n\to \mathrm{\infty }}{lim}\frac{\theta 2^{n(p+q+r)}}{8^{3n}}\sum _{i=1}^3{\parallel x_i\parallel }^p\cdot {\parallel y_i\parallel }^q\cdot {\parallel z_i\parallel }^r=0\hfill \end{array}$$

for all $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3\in A$.

Now, let $T:A^3\to A$ be another 3-derivation satisfying (8). Then we have

$$\begin{array}{rcl}\parallel \delta (x,y,z)-T(x,y,z)\parallel & =& \frac{1}{8^n}\parallel \delta (2^{n}x,2^{n}y,2^{n}z)-T(2^{n}x,2^{n}y,2^{n}z)\parallel \\ \le & \frac{1}{8^n}\parallel \delta (2^{n}x,2^{n}y,2^{n}z)-f(2^{n}x,2^{n}y,2^{n}z)\parallel \\ +\frac{1}{8^n}\parallel f(2^{n}x,2^{n}y,2^{n}z)-T(2^{n}x,2^{n}y,2^{n}z)\parallel \\ \le & \frac{\theta 2^{(p+q+r-3)n+1}}{2^3-2^{p+q+r}}{\parallel x\parallel }^p\cdot {\parallel y\parallel }^q\cdot {\parallel z\parallel }^r,\end{array}$$

which tends to zero as $n\to \mathrm{\infty }$ for all $x,y,z\in A$. So, we can conclude that $\delta (x,y,z)=T(x,y,z)$ for all $x,y,z\in A$. This proves the uniqueness of δ.

Therefore, the mapping $\delta :A^3\to A$ is a unique $C^{\ast }$-ternary 3-derivation satisfying (8). □

Corollary 3.3 Let $ϵ\in (0,\mathrm{\infty })$, and let $f:A^3\to A$ be a mapping satisfying

$$\parallel D_{\lambda ,\mu ,\nu }f(x_1,x_2,y_1,y_2,z_1,z_2)\parallel \le ϵ$$

and

$$\begin{array}{c}\parallel f([x_1,y_1,z_1],[x_2,y_2,z_2],[x_3,y_3,z_3])-[f(x_1,x_2,x_3),[y_1,y_2^{\ast },y_3],[z_1,z_2^{\ast },z_3]]\hfill \\ -[[x_1,x_2^{\ast },x_3],f(y_1,y_2,y_3),[z_1,z_2^{\ast },z_3]]-[[x_1,x_2^{\ast },x_3],[y_1,y_2^{\ast },y_3],f(z_1,z_2,z_3)]\parallel \le 3ϵ\hfill \end{array}$$

for all $\lambda ,\mu ,\nu \in S^1$ and all $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3\in A$. Then there exists a unique $C^{\ast }$-ternary 3-derivation $\delta :A^3\to A$ such that

$$\parallel f(x,y,z)-\delta (x,y,z)\parallel \le \frac{ϵ}{7}$$

for all $x,y,z\in A$.