Homomorphisms and derivations in $C^{\ast }$-ternary algebras via fixed point method

Abstract

Park (J. Math. Phys. 47:103512, 2006) proved the Hyers-Ulam stability of homomorphisms in $C^{\ast }$-ternary algebras and of derivations on $C^{\ast }$-ternary algebras for the following generalized Cauchy-Jensen additive mapping:

$$2f(\frac{{\sum }_{j=1}^p{x}_j}{2}+\sum _{j=1}^d{y}_j)=\sum _{j=1}^{p}f(x_j)+2\sum _{j=1}^{d}f(y_j).$$

In this paper, we improve and generalize some results concerning this functional equation via the fixed-point method.

MSC:39B52, 17A40, 46B03, 47Jxx.

1 Introduction and preliminaries

The stability problem of functional equations originated from a question of Ulam [1] concerning the stability of group homomorphisms. Hyers [2] gave a first affirmative partial answer to the question of Ulam for Banach spaces. Let X and Y be Banach spaces. Hyers’ theorem was generalized by Aoki [3] for additive mappings and by Rassias [4] for linear mappings by considering an unbounded Cauchy difference.

Theorem 1.1 (Th.M. Rassias)

Let $f:E\to E^{\prime }$ be a mapping from a normed vector space E into a Banach space $E^{\prime }$ subject to the inequality

$$\parallel f(x+y)-f(x)-f(y)\parallel \le ϵ({\parallel x\parallel }^p+{\parallel y\parallel }^p)$$

(1.1)

for all $x,y\in E$, where ϵ and p are constants with $ϵ>0$ and $p<1$. Then the limit

$$L(x)=\underset{n\to \mathrm{\infty }}{lim}\frac{f(2^{n}x)}{2^n}$$

exists for all $x\in E$ and $L:E\to E^{\prime }$ is the unique additive mapping which satisfies

$$\parallel f(x)-L(x)\parallel \le \frac{2ϵ}{2-2^p}{\parallel x\parallel }^p$$

(1.2)

for all $x\in E$. If $p<0$ then inequality (1.1) holds for $x,y\ne 0$ and (1.2) for $x\ne 0$. Also, if for each $x\in E$ the mapping $f(tx)$ is continuous in $t\in \mathbb{R}$, then L is linear.

Rassias [5] during the 27th International Symposium on Functional Equations asked the question whether such a theorem can also be proved for $p\ge 1$. Gajda [6] following the same approach as in Rassias [4], gave an affirmative solution to this question for $p>1$. It was shown by Gajda [6], as well as by Rassias and Šemrl [7] that one cannot prove a Rassias’ type theorem when $p=1$. The counterexamples of Gajda [6], as well as of Rassias and Šemrl [7] have stimulated several mathematicians to invent new definitions of approximately additive or approximately linear mappings, cf. Găvruta [8], Jung [9], who among others studied the Hyers-Ulam stability of functional equations. The inequality (1.1) that was introduced for the first time by Rassias [4] provided a lot of influence in the development of a generalization of the Hyers-Ulam stability concept (cf. the books of Czerwik [10], Hyers, Isac, and Rassias [11]).

Following the terminology of [12], a nonempty set G with a ternary operation $[\cdot ,\cdot ,\cdot ]:G\times G\times G\to G$ is called a ternary groupoid and is denoted by $(G,[\cdot ,\cdot ,\cdot ])$. The ternary groupoid $(G,[\cdot ,\cdot ,\cdot ])$ is called commutative if $[x_1,x_2,x_3]=[x_{\sigma (1)},x_{\sigma (2)},x_{\sigma (3)}]$ for all $x_1,x_2,x_3\in G$ and all permutations σ of $\{1,2,3\}$.

If a binary operation ∘ is defined on G such that $[x,y,z]=(x\circ y)\circ z$ for all $x,y,z\in G$, then we say that $[\cdot ,\cdot ,\cdot ]$ is derived from ∘. We say that $(G,[\cdot ,\cdot ,\cdot ])$ is a ternary semigroup if the operation $[\cdot ,\cdot ,\cdot ]$ is associative, i.e., if $[[x,y,z],u,v]=[x,[y,z,u],v]=[x,y,[z,u,v]]$ holds for all $x,y,z,u,v\in G$ (see [13]).

A $C^{\ast }$-ternary algebra is a complex Banach space A, equipped with a ternary product $(x,y,z)\mapsto [x,y,z]$ of $A^3$ into A, which are $\mathbb{C}$-linear in the outer variables, conjugate $\mathbb{C}$-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 [12, 14]). 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.

A $\mathbb{C}$-linear mapping $H:A\to B$ is called a $C^{\ast }$-ternary algebra 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 $\mathbb{C}$-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 [12, 15]).

There are some applications, although still hypothetical, in the fractional quantum Hall effect, the nonstandard statistics, supersymmetric theory, and Yang-Baxter equation (cf. [16–18]).

Throughout this paper, assume that p, d are nonnegative integers with $p+d\ge 3$, and that A and B are $C^{\ast }$-ternary algebras.

The aim of the present paper is to establish the stability problem of homomorphisms and derivations in $C^{\ast }$-ternary algebras by using the fixed-point method.

Let E be a set. A function $d:E\times E\to [0,1]$ is called a generalized metric on E if d satisfies

• (i) $d(x,y)=0$ if and only if $x=y$;

• (ii) $d(x,y)=d(y,x)$ for all $x,y\in E$;

• (iii) $d(x,z)\le d(x,y)+d(y,z)$ for all $x,y,z\in E$.

Theorem 1.2 Let $(E,d)$ be a complete generalized metric space and let $J:E\to E$ be a strictly contractive mapping with constant $L<1$. Then for each given element $x\in E$, either

$$d(J^{n}x,J^{n+1}x)=\mathrm{\infty }$$

for all nonnegative integers n or there exists a nonnegative integer $n_0$ such that

• (1) $d(J^{n}x,J^{n+1}x)<\mathrm{\infty }$ for all $n\ge n_0$;

• (2) the sequence $J^{n}x$ converges to a fixed point $y^{\ast }$ of J;

• (3) $y^{\ast }$ is the unique fixed point of J in the set $Y=y\in E:d(J^{n_0},y)<\mathrm{\infty }$;

• (4) $d(y,y^{\ast })\le \frac{1}{1-L}d(y,Jy)$ for all $y\in Y$.

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

Throughout this section, assume that A is a unital $C^{\ast }$-ternary algebra with norm $\parallel \cdot \parallel$ and unit e, and that B is a unital $C^{\ast }$-ternary algebra with norm $\parallel \cdot \parallel$ and unit $e^{\prime }$.

The stability of homomorphisms in $C^{\ast }$-ternary algebras has been investigated in [19]via direct method. In this note, we improve some results in [19]via the fixed-point method. For a given mapping $f:A\to B$, we define

$$C_{\mu }f(x_1,\dots ,x_p,y_1,\dots ,y_d):=2f(\frac{{\sum }_{j=1}^p\mu x_j}{2}+\sum _{j=1}^d\mu y_j)-\sum _{j=1}^p\mu f(x_j)-2\sum _{j=1}^d\mu f(y_j)$$

for all $\mu \in {\mathbb{T}}^1:=\{\lambda \in \mathbb{C}:|\lambda |=1\}$ and all $x_1,\dots ,x_p,y_1,\dots ,y_d\in A$.

One can easily show that a mapping $f:A\to B$ satisfies

$$C_{\mu }f(x_1,\dots ,x_p,y_1,\dots ,y_d)=0$$

for all $\mu \in {\mathbb{T}}^1$ and all $x_1,\dots ,x_p,y_1,\dots ,y_d\in A$ if and only if

$$f(\mu x+\lambda y)=\mu f(x)+\lambda f(y)$$

for all $\mu ,\lambda \in {\mathbb{T}}^1$ and all $x,y\in A$.

We will use the following lemma in this paper.

Lemma 2.1 ([20])

Let $f:A\to B$ be an additive mapping such that $f(\mu x)=\mu f(x)$ for all $x\in A$ and all $\mu \in {\mathbb{T}}^1$. Then the mapping f is $\mathbb{C}$-linear.

Lemma 2.2 Let ${\{x_n\}}_n$, ${\{y_n\}}_n$ and ${\{z_n\}}_n$ be convergent sequences in A. Then the sequence $\{[x_n,y_n,z_n]\}$ is convergent in A.

Proof Let $x,y,z\in A$ such that

$$\underset{n\to \mathrm{\infty }}{lim}{x}_n=x,\underset{n\to \mathrm{\infty }}{lim}{v}_n=y,\underset{n\to \mathrm{\infty }}{lim}{z}_n=z.$$

Since

$$\begin{array}{rcl}[x_n,y_n,z_n]-[x,y,z]& =& [x_n-x,y_n-y,z_n,z]+[x_n,y_n,z]\\ +[x,y_n-y,z_n]+[x_n,y,z_n-z]\end{array}$$

for all n, we get

$$\begin{array}{rcl}\parallel [x_n,y_n,z_n]-[x,y,z]\parallel & =& \parallel x_n-x\parallel \parallel y_n-y\parallel \parallel z_n-z\parallel +\parallel x_n-x\parallel \parallel y_n\parallel \parallel z\parallel \\ +\parallel x\parallel \parallel y_n-y\parallel \parallel z_n\parallel +\parallel x_n\parallel \parallel y\parallel \parallel z_n-z\parallel \end{array}$$

for all n. So

$$\underset{n\to \mathrm{\infty }}{lim}[x_n,y_n,z_n]=[x,y,z].$$

This completes the proof. □

Theorem 2.3 Let $f:A\to B$ be a mapping for which there exist functions $\phi :A^{p+d}\to [0,\mathrm{\infty })$ and $\psi :A^3\to [0,\mathrm{\infty })$ such that

(2.1)

(2.2)

for all $\mu \in {\mathbb{T}}^1$ and all $x,y,z,x_1,\dots ,x_p,y_1,\dots ,y_d\in A$, where $\gamma =\frac{p+2d}{2}$. If there exists constant $L<1$ such that

$$\phi (\gamma x,\dots ,\gamma x)\le \gamma L\phi (x,\dots ,x)$$

for all $x\in A$, then there exists a unique $C^{\ast }$-ternary algebras homomorphism $H:A\to B$ satisfying

$$\parallel f(x)-H(x)\parallel \le \frac{1}{(1-L)2\gamma }\phi (x,\dots ,x)$$

(2.3)

for all $x\in A$.

Proof Let us assume $\mu =1$ and $x_1=\cdots =x_p=y_1=\cdots =y_d=x$ in (2.1). Then we get

$$\parallel f(\gamma x)-\gamma f(x)\parallel \le \frac{1}{2}\phi (x,\dots ,x)$$

(2.4)

for all $x\in A$. Let $E:=\{g:A\to B\}$. We introduce a generalized metric on E as follows:

$$d(g,h):=inf\{C\in [0,\mathrm{\infty }]:\parallel g(x)-h(x)\parallel \le C\phi (x,\dots ,x)\text{ for all }x\in A\}.$$

It is easy to show that $(E,d)$ is a generalized complete metric space.

Now, we consider the mapping $\mathrm{\Lambda }:E\to E$ defined by

$$(\mathrm{\Lambda }g)(x)=\frac{1}{\gamma }g(\gamma x),\text{for all }g\in E\text{ and }x\in A.$$

Let $g,h\in E$ and let $C\in [0,\mathrm{\infty }]$ be an arbitrary constant with $d(g,h)\le C$. From the definition of d, we have

$$\parallel g(x)-h(x)\parallel \le C\phi (x,\dots ,x)$$

for all $x\in A$. By the assumption and the last inequality, we have

$$\parallel (\mathrm{\Lambda }g)(x)-(\mathrm{\Lambda }h)(x)\parallel =\frac{1}{\gamma }\parallel g(\gamma x)-h(\gamma x)\parallel \le \frac{C}{\gamma }\phi (\gamma x,\dots ,\gamma x)\le CL\phi (x,\dots ,x)$$

for all $x\in A$. So $d(\mathrm{\Lambda }g,\mathrm{\Lambda }h)\le Ld(g,h)$ for any $g,h\in E$. It follows from (2.4) that $d(\mathrm{\Lambda }f,f)\le \frac{1}{2\gamma }$. Therefore according to Theorem 1.2, the sequence $\{{\mathrm{\Lambda }}^{n}f\}$ converges to a fixed point H of Λ, i.e.,

$$H:A\to B,H(x)=\underset{n\to \mathrm{\infty }}{lim}\left({\mathrm{\Lambda }}^{n}f\right)(x)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{\gamma }^n}f\left({\gamma }^{n}x\right)$$

(2.5)

and $H(\gamma x)=\gamma H(x)$ for all $x\in A$. Also H is the unique fixed point of Λ in the set $E=\{g\in E:d(f,g)<\mathrm{\infty }\}$ and

$$d(H,f)\le \frac{1}{1-L}d(\mathrm{\Lambda }f,f)\le \frac{1}{(1-L)2\gamma }$$

i.e., the inequality (2.3) holds true for all $x\in A$. It follows from the definition of H that

for all $\mu \in {\mathbb{T}}^1$ and all $x_1,\dots ,x_p,y_1,\dots ,y_d\in A$. Hence

$$2H(\frac{{\sum }_{j=1}^p\mu x_j}{2}+\sum _{j=1}^d\mu y_j)=\sum _{j=1}^p\mu H(x_j)+2\sum _{j=1}^d\mu H(y_j)$$

for all $\mu \in {\mathbb{T}}^1$ and all $x_1,\dots ,x_p,y_1,\dots ,y_d\in A$. So $H(\lambda x+\mu y)=\lambda H(x)+\mu H(y)$ for all $\lambda ,\mu \in {\mathbb{T}}^1$ and all $x,y\in A$.

Therefore, by Lemma 2.1, the mapping $H:A\to B$ is $\mathbb{C}$-linear.

It follows from (2.2) and (2.5) that

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

$$H([x,y,z])=[H(x),H(y),H(z)]$$

for all $x,y,z\in A$. Therefore, the mapping H is a $C^{\ast }$-ternary algebras homomorphism.

Now, let $T:A\to B$ be another $C^{\ast }$-ternary algebras homomorphism satisfying (2.3). Since $d(f,T)\le \frac{1}{(1-L)2\gamma }$ and T is $\mathbb{C}$-linear, we get $T\in E^{\prime }$ and $(\mathrm{\Lambda }T)(x)=\frac{1}{\gamma }(T\gamma x)=T(x)$ for all $x\in A$, i.e., T is a fixed point of Λ. Since H is the unique fixed point of $\mathrm{\Lambda }\in E^{\prime }$, we get $H=T$. □

Theorem 2.4 Let $f:A\to B$ be a mapping for which there exist functions $\phi :A^{p+d}\to [0,\mathrm{\infty })$ and $\psi :A^3\to [0,\mathrm{\infty })$ satisfying (2.1), (2.2),

for all $x,y,z,x_1,\dots ,x_p,y_1,\dots ,y_d\in A$, where $\gamma =\frac{p+2d}{2}$. If there exists constant $L<1$ such that

$$\phi (\frac{1}{\gamma }x,\dots ,\frac{1}{\gamma }x)\le \frac{1}{\gamma }L\phi (x,\dots ,x)$$

for all $x\in A$, then there exists a unique $C^{\ast }$-ternary algebras homomorphism $H:A\to B$ satisfying

$$\parallel f(x)-H(x)\parallel \le \frac{1}{(1-L)2\gamma }\phi (x,\dots ,x)$$

for all $x\in A$.

Proof If we replace x in (2.4) by $\frac{x}{\gamma }$, then we get

$$\parallel f(x)-\gamma f\left(\frac{x}{\gamma }\right)\parallel \le \frac{1}{2}\phi (\frac{x}{\gamma },\dots ,\frac{x}{\gamma })\le \frac{L}{2\gamma }\phi (x,\dots ,x)$$

(2.6)

for all $x\in A$. Let $E:=\{g:A\to A\}$. We introduce a generalized metric on E as follows:

$$d(g,h):=inf\{C\in [0,\mathrm{\infty }]:\parallel g(x)-h(x)\parallel \le C\phi (x,\dots ,x)\text{ for all }x\in A\}.$$

It is easy to show that $(E,d)$ is a generalized complete metric space.

Now, we consider the mapping $\mathrm{\Lambda }:E\to E$ defined by

$$(\mathrm{\Lambda }g)(x)=\gamma g\left(\frac{x}{\gamma }\right),\text{for all }g\in E\text{ and }x\in A.$$

Let $g,h\in E$ and let $C\in [0,\mathrm{\infty }]$ be an arbitrary constant with $d(g,h)\le C$. From the definition of d, we have

$$\parallel g(x)-h(x)\parallel \le C\phi (x,\dots ,x)$$

for all $x\in A$. By the assumption and the last inequality, we have

$$\parallel (\mathrm{\Lambda }g)(x)-(\mathrm{\Lambda }h)(x)\parallel =\parallel \gamma g\left(\frac{x}{\gamma }\right)-\gamma h\left(\frac{x}{\gamma }\right)\parallel \le \gamma C\phi (\frac{x}{\gamma },\dots ,\frac{x}{\gamma })\le CL\phi (x,\dots ,x)$$

for all $x\in A$, and so $d(\mathrm{\Lambda }g,\mathrm{\Lambda }h)\le Ld(g,h)$ for any $g,h\in E$. It follows from (2.6) that $d(\mathrm{\Lambda }f,f)\le \frac{1}{2\gamma }$. Thus, according to Theorem 1.2, the sequence $\{{\mathrm{\Lambda }}^{n}f\}$ converges to a fixed point H of Λ, i.e.,

$$H:A\to B,H(x)=\underset{n\to \mathrm{\infty }}{lim}\left({\mathrm{\Lambda }}^{n}f\right)(x)=\underset{n\to \mathrm{\infty }}{lim}{\gamma }^{n}f\left(\frac{x}{{\gamma }^n}\right)$$

for all $x\in A$.

The rest of the proof is similar to the proof of Theorem 2.3, and we omit it. □

Corollary 2.5 ([19])

Let r and θ be nonnegative real numbers such that $r\notin [1,3]$, and let $f:A\to B$ be a mapping such that

$$\parallel C_{\mu }f(x_1,\dots ,x_p,y_1,\dots ,y_d)\parallel \le \theta (\sum _{j=1}^p{\parallel x_j\parallel }^r+\sum _{j=1}^d{\parallel y_j\parallel }^r)$$

(2.7)

and

$$\parallel f([x,y,z])-[f(x),f(y),f(z)]\parallel \le \theta ({\parallel x\parallel }^r+{\parallel y\parallel }^r+{\parallel z\parallel }^r)$$

(2.8)

for all $\mu \in {\mathbb{T}}^1$ and all $x,y,z,x_1,\dots ,x_p,y_1,\dots ,y_d\in A$. Then there exists a unique $C^{\ast }$-ternary algebra homomorphism $H:A\to B$ such that

$$\parallel f(x)-H(x)\parallel \le \frac{2^r(p+d)\theta }{|2{(p+2d)}^r-(p+2d)2^r|}{\parallel x\parallel }^r$$

(2.9)

for all $x\in A$.

Proof The proof follows from Theorems 2.3 and 2.4 by taking

for all $\mu \in {\mathbb{T}}^1$ and all $x,y,z,x_1,\dots ,x_p,y_1,\dots ,y_d\in A$. Then we can choose $L=2^{1-r}{(p+2d)}^{r-1}$, when $0<r<1$ and $L=2-2^{1-r}{(p+2d)}^{r-1}$, when $r>3$ and we get the desired results. □

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

Throughout this section, assume that A is a unital $C^{\ast }$-ternary algebra with norm $\parallel \cdot \parallel$ and unit e, and that B is a unital $C^{\ast }$-ternary algebra with norm $\parallel \cdot \parallel$ and unit $e^{\prime }$.

We investigate homomorphisms in $C^{\ast }$-ternary algebras associated with the functional equation $C_{\mu }f(x_1,\dots ,x_p,y_1,\dots ,y_d)=0$.

Theorem 3.1 ([19])

Let $r>1$ (resp., $r<1$) and θ be nonnegative real numbers, and let $f:A\to B$ be a bijective mapping satisfying (2.1) and

$$f([x,y,z])=[f(x),f(y),f(z)]$$

for all $x,y,z\in A$. If ${lim}_{n\to \mathrm{\infty }}\frac{{(p+2d)}^n}{2^n}f(\frac{2^{n}e}{{(p+2d)}^n})=e^{\prime }$ (resp., ${lim}_{n\to \mathrm{\infty }}\frac{2^n}{{(p+2d)}^n}f(\frac{{(p+2d)}^n}{2^n}e)=e^{\prime }$), then the mapping $f:A\to B$ is a $C^{\ast }$-ternary algebra isomorphism.

In the following theorems we have alternative results of Theorem 3.1.

Theorem 3.2 Let $r<1$ and θ be nonnegative real numbers, and let $f:A\to B$ be a mapping satisfying (2.7) and (2.8). If there exist a real number $\lambda >1$ (resp., $0<\lambda <1$) and an element $x_0\in A$ such that ${lim}_{n\to \mathrm{\infty }}\frac{1}{{\lambda }^n}f({\lambda }^n{x}_0)=e^{\prime }$ (resp., ${lim}_{n\to \mathrm{\infty }}{\lambda }^{n}f(\frac{x_0}{{\lambda }^n})=e^{\prime }$), then the mapping $f:A\to B$ is a $C^{\ast }$-ternary algebra homomorphism.

Proof By using the proof of Corollary 2.5, there exists a unique $C^{\ast }$-ternary algebra homomorphism $H:A\to B$ satisfying (2.9). It follows from (2.9) that

$$H(x)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{\lambda }^n}f\left({\lambda }^{n}x\right),(H(x)=\underset{n\to \mathrm{\infty }}{lim}{\lambda }^{n}f\left(\frac{x}{{\lambda }^n}\right))$$

for all $x\in A$ and all real numbers $\lambda >1$ ($0<\lambda <1$). Therefore, by the assumption, we get that $H(x_0)=e^{\prime }$.

Let $\lambda >1$ and ${lim}_{n\to \mathrm{\infty }}\frac{1}{{\lambda }^n}f({\lambda }^n{x}_0)=e^{\prime }$. It follows from (2.8) that

for all $x\in A$. So $[H(x),H(y),H(z)]=[H(x),H(y),f(z)]$ for all $x,y,z\in A$. Letting $x=y=x_0$ in the last equality, we get $f(z)=H(z)$ for all $z\in A$. Similarly, one can show that $H(x)=f(x)$ for all $x\in A$ when $0<\lambda <1$ and ${lim}_{n\to \mathrm{\infty }}{\lambda }^{n}f(\frac{x_0}{{\lambda }^n})=e^{\prime }$.

Similarly, one can show the theorem for the case $\lambda >1$.

Therefore, the mapping $f:A\to B$ is a $C^{\ast }$-ternary algebra homomorphism. □

Theorem 3.3 Let $r>1$ and θ be nonnegative real numbers, and let $f:A\to B$ be a mapping satisfying (2.7) and (2.8). If there exist a real number $0<\lambda <1$ (resp., $\lambda >1$) and an element $x_0\in A$ such that ${lim}_{n\to \mathrm{\infty }}\frac{1}{{\lambda }^n}f({\lambda }^n{x}_0)=e^{\prime }$ (resp., ${lim}_{n\to \mathrm{\infty }}{\lambda }^{n}f(\frac{x_0}{{\lambda }^n})=e^{\prime }$), then the mapping $f:A\to B$ is a $C^{\ast }$-ternary algebra homomorphism.

Proof The proof is similar to the proof of Theorem 3.2 and we omit it. □

4 Stability of derivations on $C^{\ast }$-ternary algebras

Throughout this section, assume that A is a $C^{\ast }$-ternary algebra with norm $\parallel \cdot \parallel$.

Park [19] proved the Hyers-Ulam stability of derivations on $C^{\ast }$-ternary algebras for the functional equation $C_{\mu }f(x_1,\dots ,x_p,y_1,\dots ,y_d)=0$.

For a given mapping $f:A\to A$, let

$$\mathbf{D}f(x,y,z)=f([x,y,z])-[f(x),y,z]-[x,f(y),z]-[x,y,f(z)]$$

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

Theorem 4.1 ([19])

Let r and θ be nonnegative real numbers such that $r\notin [1,3]$, and let $f:A\to A$ be a mapping satisfying (2.7) and

$$\parallel \mathbf{D}f(x,y,z)\parallel \le \theta ({\parallel x\parallel }^r+{\parallel y\parallel }^r+{\parallel z\parallel }^r)$$

for all $x,y,z\in A$. Then there exists a unique $C^{\ast }$-ternary derivation $\delta :A\to A$ such that

$$\parallel f(x)-\delta (x)\parallel \le \frac{2^r(p+d)}{|2{(p+2d)}^r-(p+2d)2^r|}\theta {\parallel x\parallel }^r$$

for all $x\in A$.

In the following theorem, we generalize and improve the result in Theorems 4.1.

Theorem 4.2 Let $\phi :A^{p+d}\to [0,\mathrm{\infty })$ and $\psi :A^3\to [0,\mathrm{\infty })$ be functions such that

(4.1)

(4.2)

for all $x,y,z,x_1,\dots ,x_p,y_1,\dots ,y_d\in A$, where $\gamma =\frac{p+2d}{2}$. Suppose that $f:A\to A$ is a mapping satisfying

(4.3)

(4.4)

for all $\mu \in {\mathbb{T}}^1$ and all $x,y,z,x_1,\dots ,x_p,y_1,\dots ,y_d\in A$. If there exists a constant $L<1$ such that

$$\phi (\gamma x,\dots ,\gamma x)\le \gamma \phi (x,\dots ,x),$$

then the mapping $f:A\to A$ is a $C^{\ast }$-ternary derivation.

Proof Let us assume $\mu =1$ and $x_1=\cdots =x_p=y_1=\cdots =y_d=x$ in (4.3). Then we get

$$\parallel f(\gamma x)-\gamma f(x)\parallel \le \frac{1}{2}\phi (x,\dots ,x)$$

(4.5)

for all $x\in A$. Let $E:=\{g:A\to A\}$. We introduce a generalized metric on E as follows:

$$d(g,h):=inf\{C\in [0,\mathrm{\infty }]:\parallel g(x)-h(x)\parallel \le C\phi (x,\dots ,x)\text{ for all }x\in A\}.$$

It is easy to show that $(E,d)$ is a generalized complete metric space.

Now, we consider the mapping $\mathrm{\Lambda }:E\to E$ defined by

$$(\mathrm{\Lambda }g)(x)=\frac{1}{\gamma }g(\gamma x),\text{for all }g\in E\text{ and }x\in A.$$

Let $g,h\in E$ and let $C\in [0,\mathrm{\infty }]$ be an arbitrary constant with $d(g,h)\le C$. From the definition of d, we have

$$\parallel g(x)-h(x)\parallel \le C\phi (x,\dots ,x)$$

for all $x\in A$. By the assumption and the last inequality, we have

$$\parallel (\mathrm{\Lambda }g)(x)-(\mathrm{\Lambda }h)(x)\parallel =\frac{1}{\gamma }\parallel g(\gamma x)-h(\gamma x)\parallel \le \frac{C}{\gamma }\phi (\gamma x,\dots ,\gamma x)\le CL\phi (x,\dots ,x)$$

for all $x\in A$. Then $d(\mathrm{\Lambda }g,\mathrm{\Lambda }h)\le Ld(g,h)$ for any $g,h\in E$. It follows from (2.4) that $d(\mathrm{\Lambda }f,f)\le \frac{1}{2\gamma }$. Thus according to Theorem 1.2, the sequence $\{{\mathrm{\Lambda }}^{n}f\}$ converges to a fixed point δ of Λ, i.e.,

$$\delta :A\to A,\delta (x)=\underset{n\to \mathrm{\infty }}{lim}\left({\mathrm{\Lambda }}^{n}f\right)(x)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{\gamma }^n}f\left({\gamma }^{n}x\right)$$

(4.6)

and $\delta (\gamma x)=\gamma \delta (x)$ for all $x\in A$. Also δ is the unique fixed point of Λ in the set $E=\{g\in E:d(f,g)<\mathrm{\infty }\}$ and

$$d(\delta ,f)\le \frac{1}{1-L}d(\mathrm{\Lambda }f,f)\le \frac{1}{(1-L)2\gamma }$$

i.e., the inequality (2.3) holds true for all $x\in A$. It follows from the definition of δ, (4.1), (4.3), and (4.6) that

for all $\mu \in {\mathbb{T}}^1$ and all $x,y,z,x_1,\dots ,x_p,y_1,\dots ,y_d\in A$. Hence,

$$2\delta (\frac{{\sum }_{j=1}^p\mu x_j}{2}+\sum _{j=1}^d\mu y_j)=\sum _{j=1}^p\mu \delta (x_j)+2\sum _{j=1}^d\mu \delta (y_j)$$

for all $\mu \in {\mathbb{T}}^1$ and all $x_1,\dots ,x_p,y_1,\dots ,y_d\in A$. So $\delta (\lambda x+\mu y)=\lambda \delta (x)+\mu \delta (y)$ for all $\lambda ,\mu \in {\mathbb{T}}^1$ and all $x,y\in A$.

Therefore, by Lemma 2.1 the mapping $\delta :A\to A$ is $\mathbb{C}$-linear.

It follows from (4.2) and (4.4) that

$$\parallel \mathbf{D}\delta (x,y,z)\parallel =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{\gamma }^{3n}}\parallel \mathbf{D}f({\gamma }^{n}x,{\gamma }^{n}y,{\gamma }^{n}z)\parallel \le \underset{n\to \mathrm{\infty }}{lim}\frac{1}{{\gamma }^{3n}}\psi ({\gamma }^{n}x,{\gamma }^{n}y,{\gamma }^{n}z)=0$$

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

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

(4.7)

for all $x,y,z\in A$. So the mapping $\delta :A\to A$ is a $C^{\ast }$-ternary derivation.

It follows from (4.2) and (4.4)

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

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

(4.8)

for all $x,y,z\in A$. Hence, we get from (4.7) and (4.8) that

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

(4.9)

for all $x,y,z\in A$. Letting $x=y=f(z)-\delta (z)$ in (4.9), we get

$${\parallel f(z)-\delta (z)\parallel }^3=\parallel [f(z)-\delta (z),f(z)-\delta (z),f(z)-\delta (z)]\parallel =0$$

for all $z\in A$. Hence, $f(z)=\delta (z)$ for all $z\in A$. So the mapping $f:A\to A$ is a $C^{\ast }$-ternary derivation, as desired. □

Corollary 4.3 Let $r<1$, $s<2$ and θ be nonnegative real numbers, and let $f:A\to A$ be a mapping satisfying (2.7) and

$${\parallel \mathbf{D}f(x,y,z)\parallel }_A\le \theta ({\parallel x\parallel }_A^s+{\parallel y\parallel }_A^s+{\parallel z\parallel }_A^s)$$

for all $x,y,z\in A$. Then the mapping $f:A\to A$ is a $C^{\ast }$-ternary derivation.

Proof Defining

$$\phi (x_1,\dots ,x_p,y_1,\dots ,y_d)=\theta (\sum _{j=1}^p{\parallel x_j\parallel }_A^r+\sum _{j=1}^d{\parallel y_j\parallel }_A^r)$$

and

$$\psi (x,y,z)=\theta ({\parallel x\parallel }_A^s+{\parallel y\parallel }_A^s+{\parallel z\parallel }_A^s)$$

for all $x,y,z,x_1,\dots ,x_p,y_1,\dots ,y_d\in A$, and applying Theorem 4.2, we get the desired result. □

Theorem 4.4 Let $\phi :A^{p+d}\to [0,\mathrm{\infty })$ and $\psi :A^3\to [0,\mathrm{\infty })$ be functions such that

for all $x,y,z,x_1,\dots ,x_p,y_1,\dots ,y_d\in A$ where $\gamma =\frac{p+2d}{2}$. Suppose that $f:A\to A$ is a mapping satisfying (4.3) and (4.4). If there exists a constant $L<1$ such that

$$\phi (\frac{x}{\gamma },\dots ,\frac{x}{\gamma })\le \frac{L}{\gamma }\phi (x,\dots ,x),$$

then the mapping $f:A\to A$ is a $C^{\ast }$-ternary derivation.

Proof If we replace x in (4.5) by $\frac{x}{\gamma }$, then we get

$${\parallel f(x)-\gamma f\left(\frac{x}{\gamma }\right)\parallel }_A\le \frac{1}{2}\phi (\frac{x}{\gamma },\dots ,\frac{x}{\gamma })$$

for all $x\in A$. Let $E:=\{g:A\to A\}$. We introduce a generalized metric on E as follows:

$$d(g,h):=inf\{C\in [0,\mathrm{\infty }]:\parallel g(x)-h(x)\parallel \le C\phi (x,\dots ,x)\text{ for all }x\in A\}$$

It is easy to show that $(E,d)$ is a generalized complete metric space.

Now, we consider the mapping $\mathrm{\Lambda }:E\to E$ defined by

$$(\mathrm{\Lambda }g)(x)=\gamma g\left(\frac{x}{\gamma }\right),\text{for all }g\in E\text{ and }x\in A.$$

Let $g,h\in E$ and let $C\in [0,\mathrm{\infty }]$ be an arbitrary constant with $d(g,h)\le C$. From the definition of d, we have

$$\parallel g(x)-h(x)\parallel \le C\phi (x,\dots ,x)$$

for all $x\in A$. By the assumption and last inequality, we have

$$\parallel (\mathrm{\Lambda }g)(x)-(\mathrm{\Lambda }h)(x)\parallel =\parallel \gamma g\left(\frac{x}{\gamma }\right)-\gamma h\left(\frac{x}{\gamma }\right)\parallel \le \gamma C\phi (\frac{x}{\gamma },\dots ,\frac{x}{\gamma })\le CL\phi (x,\dots ,x)$$

for all $x\in A$. Then $d(\mathrm{\Lambda }g,\mathrm{\Lambda }h)\le Ld(g,h)$ for any $g,h\in E$. It follows from (4.5) that $d(\mathrm{\Lambda }f,f)\le \frac{1}{2\gamma }$. Therefore according to Theorem 1.2, the sequence $\{{\mathrm{\Lambda }}^{n}f\}$ converges to a fixed point δ of Λ, i.e.,

$$\delta :A\to A,\delta (x)=\underset{n\to \mathrm{\infty }}{lim}\left({\mathrm{\Lambda }}^{n}f\right)(x)=\underset{n\to \mathrm{\infty }}{lim}{\gamma }^{n}f\left(\frac{x}{{\gamma }^n}\right)$$

and $\delta (\gamma x)=\gamma \delta (x)$ for all $x\in A$.

The rest of the proof is similar to the proof of Theorem 4.2, and we omit it. □