Multi-$C^{*}$-ternary algebras and applications

Abstract

In this paper, we look at the concept of multi-$C^{*}$-ternary algebras and consider some properties. As an application we approximate multi-$C^{*}$-ternary algebra homomorphisms and derivations in these spaces.

1 Introduction and preliminaries

Ternary algebraic structures arise naturally in theoretical and mathematical physics, for example, the quark model inspired a particular brand of ternary algebraic system. We also refer the reader to ‘Nambu mechanics’ [1] (see also [2, 3] and [4]).

A $C^{*}$-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 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 $\|[x, y, z]\| \le\|x\| \cdot\|y\| \cdot \|z\|$ and $\|[x, x, x]\| = \|x\|^{3}$ (see [4]).

If a $C^{*}$-ternary algebra $(A, [\cdot, \cdot, \cdot] )$ has the identity, i.e., the 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 $x^{*}:=[e, x, e]$, is a unital $C^{*}$-algebra. Conversely, if $(A, \circ)$ is a unital $C^{*}$-algebra, then $[x, y, z] : = x \circ y^{*} \circ z$ makes A into a $C^{*}$-ternary algebra.

A C-linear mapping $H: A \rightarrow B$ is called a $C^{*}$-ternary algebra homomorphism if

$$H\bigl([x, y, z]\bigr) = \bigl[H(x), H(y), H(z)\bigr] $$

for all $x, y, z \in A$. A C-linear mapping $\delta: A \rightarrow A$ is called a $C^{*}$-ternary derivation if

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

for all $x, y, z \in A$ (see [5]).

Ternary structures and their generalization, the so-called n-ary structures, are important in view of their applications in physics (see [6]).

Let X be a set. A function $d: X \times X \rightarrow[0, \infty]$ is called a generalized metric on X if d satisfies the following conditions:

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

• (2) $d(x, y) = d(y, x)$ for all $x, y \in X$;

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

Theorem 1.1

([7])

Let $(X, d)$ be a complete generalized metric space and let $J: X \rightarrow X$ be a strictly contractive mapping with Lipschitz constant $L<1$. Then, for each $x\in X$, either

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

for all non-negative integers n or there exists a positive integer $n_{0}$ such that

• (1) $d(J^{n} x, J^{n+1}x) <\infty$ for all $n\ge n_{0}$;

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

• (3) $y^{*}$ is the unique fixed point of J in the set $Y = \{y\in X \mid d(J^{n_{0}} x, y) <\infty\}$;

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

2 Multi-normed spaces

The notion of a multi-normed space was introduced by Dales and Polyakov in [8] and many examples are given in [8–10].

Let $( {\mathcal{E}},\|\cdot\|)$ be a complex normed space and let $k\in\mathbf{N}$. We denote by $\mathcal{E}^{k}$ the linear space $\mathcal{E}\oplus\cdots\oplus\mathcal{E}$ consisting of k-tuples $(x_{1}, \ldots, x_{k})$, where $x_{1}, \ldots, x_{k}\in\mathcal{E}$. The linear operations on $\mathcal{E}^{k}$ are defined coordinate-wise. The zero element of either $\mathcal{E}$ or $\mathcal{E}^{k}$ is denoted by 0. We denote by $\mathbf{N}_{k}$ the set $\{1, 2, \ldots ,k\}$ and by $\Sigma_{k}$ the group of permutations on k symbols.

Definition 2.1

A multi-norm on $\{ {\mathcal{E}}^{k}: k\in\mathbf{N}\}$ is a sequence

$$\bigl(\Vert \cdot \Vert _{k}\bigr)=\bigl(\Vert \cdot \Vert _{k}:k\in\mathbf{N}\bigr) $$

such that $\|\cdot\|_{k}$ is a norm on ${\mathcal{E}}^{k}$ for each $k\in\mathbf{N}$ with $k\geq2$:

• (A1) $\|(x_{\sigma(1)},\ldots,x_{\sigma(k)})\|_{k}=\|(x_{1},\ldots,x_{k})\|_{k}$ for any $\sigma\in\Sigma_{k}$ and $x_{1},\ldots,x_{k}\in\mathcal{E}$;

• (A2) $\|(\alpha_{1}x_{1},\ldots,\alpha_{k}x_{k})\|_{k}\leq (\max_{i\in{\mathbf{N}}_{k}}|\alpha_{i}| ) \|(x_{1},\ldots,x_{k})\| _{k}$ for any $\alpha_{1},\ldots,\alpha_{k} \in\mathbf{C}$ and $x_{1}, \ldots, x_{k}\in\mathcal{E}$;

• (A3) $\|(x_{1},\ldots,x_{k-1},0)\|_{k}=\|(x_{1},\ldots,x_{k-1})\|_{k-1}$ for any $x_{1}, \ldots, x_{k-1}\in\mathcal{E}$;

• (A4) $\|(x_{1},\ldots,x_{k-1},x_{k-1})\|_{k}=\|(x_{1},\ldots,x_{k-1})\| _{k-1}$ for any $x_{1},\ldots, x_{k-1}\in\mathcal{E}$.

In this case, we say that $((\mathcal{E}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ is a multi-normed space.

Lemma 2.2

([10])

Suppose that $((\mathcal{E}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ is a multi-normed space and let $k\in\mathbf{N}$. Then

• (1) $\|(x,\ldots,x)\|_{k}=\|x\|$ for any $x\in\mathcal{E}$;

• (2) $\max_{i\in\mathbf{N}_{k}}\|x_{i}\|\leq \|x_{1},\ldots,x_{k}\|_{k}\leq\sum_{i=1}^{k}\|x_{i}\|\leq k \max_{i\in {\mathbf{N}}_{k}}\|x_{i}\|$ for any $x_{1},\ldots, x_{k}\in\mathcal{E}$.

It follows from (2) that, if $( \mathcal{E},\|\cdot\|)$ is a Banach space, then $( \mathcal{E}^{k},\|\cdot\|_{k})$ is a Banach space for each $k\in\mathbf{N}$. In this case, $((\mathcal{E}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ is a multi-Banach space.

Now, we present two examples (see [8]).

Example 2.3

The sequence $(\|\cdot\|_{k}: k\in\mathbf{N})$ on $\{\mathcal{E}^{k}: k\in\mathbf{N}\}$ defined by

$$\bigl\Vert (x_{1},\ldots,x_{k})\bigr\Vert _{k}:=\max_{i\in\mathbf{N}_{k}}\|x_{i}\| $$

for any $x_{1}, \ldots, x_{k}\in\mathcal{E}$ is a multi-norm, which is called the minimum multi-norm.

Example 2.4

Let $\{(\|\cdot\|_{k}^{\alpha}: k\in\mathbf{N}):\alpha\in A\}$ be the (non-empty) family of all multi-norms on $\{\mathcal{E}^{k}:k\in\mathbf{N}\}$. For each $k\in\mathbf{N}$, set

$$\bigl\Vert (x_{1},\ldots,x_{k})\bigr\Vert _{k}:=\sup_{\alpha\in A}\bigl\Vert (x_{1}, \ldots,x_{k})\bigr\Vert _{k}^{\alpha}$$

for any $x_{1}, \ldots, x_{k}\in \mathcal{E}$. Then $( \|\cdot\|_{k} : k\in\mathbf{N})$ is a multi-norm on $\{\mathcal{E}^{k}: k\in\mathbf{N}\}$, which is called the maximum multi-norm.

Now, we need the following observation which can easily be deduced from Lemma 2.2(2) of multi-norms.

Lemma 2.5

Suppose that $k\in\mathbf{N}$ and $(x_{1},\ldots, x_{k})\in \mathcal{E}^{k}$. For each $j\in\{1,\ldots,k\}$, let $(x_{n}^{j})$ be a sequence in $\mathcal{E}$ such that $\lim_{n\to\infty}x_{n}^{j}=x_{j}$. Then, for each $(y_{1},\ldots,y_{k})\in\mathcal{E}^{k}$,

$$\lim_{n\to \infty}\bigl(x_{n}^{1}-y_{1}, \ldots,x_{n}^{k}-y_{k}\bigr)=(x_{1}-y_{1}, \ldots,x_{k}-y_{k}). $$

Definition 2.6

Let $((\mathcal{E}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ be a multi-normed space. A sequence $(x_{n})$ in $\mathcal{E}$ is a multi-null sequence if, for each $\epsilon>0$, there exists $n_{0}\in\mathbf{N}$ such that

$$\sup_{k\in\mathbf{N}}\bigl\Vert (x_{n},\ldots,x_{n+k-1}) \bigr\Vert _{k}< \epsilon $$

for any $n\geq n_{0}$. Let $x\in\mathcal{E}$. We say that the sequence $(x_{n})$ is multi-convergent to $x\in\mathcal{E}$ and write

$$\lim_{n\to\infty}x_{n}=x $$

if $(x_{n}-x)$ is a multi-null sequence.

Definition 2.7

([8, 11])

Let $({A},\|\cdot\|)$ be a normed algebra such that $(({A}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ is a multi-normed space. Then $(({A}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ is called a multi-normed algebra if

$$\bigl\Vert (a_{1}b_{1},\ldots, a_{k}b_{k}) \bigr\Vert _{k}\leq\bigl\Vert (a_{1},\ldots, a_{k})\bigr\Vert _{k} \cdot \bigl\Vert (b_{1}, \ldots,b_{k})\bigr\Vert _{k} $$

for all $k\in\mathbf{N}$ and $a_{1},\ldots,a_{k},b_{1},\ldots,b_{k}\in {A}$. Further, the multi-normed algebra $(({A}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ is a multi-Banach algebra if $(({A}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ is a multi-Banach space.

Example 2.8

([8, 11])

Let p, q with $1 \leq p \leq q < \infty$ and let ${A}=\ell^{p}$. The algebra A is a Banach sequence algebra with respect to a coordinate-wise multiplication of sequences (see [12]). Let $(\|\cdot\|_{k}: k\in\mathbf{N})$ be the standard $(p, q)$-multi-norm on $\{{A}^{k}: k\in\mathbf{N}\}$. Then $(({A}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ is a multi-Banach algebra.

Definition 2.9

Let $(({ A}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ be a multi-Banach algebra. A multi-$C^{*}$-algebra is a complex multi-Banach algebra $(({ A}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ with an involution ∗ satisfying

$$\bigl\Vert \bigl(a_{1}^{*}a_{1},\ldots, a_{k}^{*}a_{k} \bigr)\bigr\Vert _{k}= \bigl\Vert (a_{1},\ldots, a_{k})\bigr\Vert _{k} ^{2} $$

for all $k\in\mathbf{N}$ and $a_{1},\ldots,a_{k}\in {A}$.

Definition 2.10

Let $(({ A}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ be a multi-Banach space. A multi-$C^{*}$-ternary algebra is a complex multi-Banach space $(({ A}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ equipped with a ternary product.

3 Approximation of homomorphisms in multi-Banach algebras

Throughout this paper, assume that A, B are $C^{*}$-ternary algebras.

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

$$C_{\mu}f(x_{1},\ldots,x_{p},y_{1}, \ldots,y_{d}):= 2f \Biggl(\frac{\sum_{j=1}^{p}\mu x_{j}}{2}+\sum _{j=1}^{d}\mu y_{j} \Biggr)-\sum _{j=1}^{p}\mu f(x_{j})-2\sum _{j=1}^{d}\mu f(y_{j}) $$

for all $\mu\in{\mathbf{T}}^{1}:=\{ \lambda\in\mathbf{C}: |\lambda|=1\}$ and $x_{1},\ldots,x_{p},y_{1},\ldots,y_{d}\in A$.

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

$$C_{\mu}f(x_{1}, \ldots, x_{p}, y_{1}, \ldots, y_{d}) =0 $$

for all $\mu\in {\mathbf{T}}^{1}$ and all $x_{1},\ldots,x_{p},y_{1},\ldots,y_{d}\in A$ if and only if

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

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

Lemma 3.1

([13])

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

Lemma 3.2

Let $\{x_{n}\}$, $\{y_{n}\}$, and $\{z_{n}\}$ be the convergent sequences in A. Then the sequence $\{[x_{n},y_{n},z_{n}]\}$ is convergent in A.

Proof

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

$$ \lim_{n\to\infty}x_{n}=x,\qquad \lim_{n\to\infty}y_{n}=y, \qquad \lim_{n\to\infty}z_{n}=z. $$

Since

$$\begin{aligned}& [x_{n},y_{n},z_{n}]-[x,y,z] \\& \quad = [x_{n}-x,y_{n}-y,z_{n},z]+[x_{n},y_{n},z] +[x,y_{n}-y,z_{n}]+[x_{n},y,z_{n}-z] \end{aligned}$$

for all $n\geq1$, we get

$$\begin{aligned} \bigl\Vert [x_{n},y_{n},z_{n}]-[x,y,z]\bigr\Vert =&\|x_{n}-x\|\|y_{n}-y\|\|z_{n}-z\|+ \|x_{n}-x\|\|y_{n}\| \|z\| \\ &{}+\|x\|\|y_{n}-y\|\|z_{n}\|+\|x_{n}\|\|y\| \|z_{n}-z\| \end{aligned}$$

for all $n\geq1$, and so

$$ \lim_{n\to\infty}[x_{n},y_{n},z_{n}]=[x,y,z]. $$

This completes the proof. □

Using Theorem 1.1, we approximate homomorphisms in multi-$C^{*}$-ternary algebras for the functional equation $C_{\mu}f(x_{1},\ldots,x_{m}) =0$.

Theorem 3.3

Let $(( {B}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ be a multi-$C^{*}$-ternary algebra. Let $f: A \rightarrow B$ be a mapping for which there are functions $\varphi: A^{(p+d)k} \rightarrow[0, \infty)$ and $\psi: A^{3k} \rightarrow[0, \infty)$ such that

$$\begin{aligned}& \lim_{n\to\infty} {\gamma}^{-n} \varphi\bigl( \gamma^{n} x_{11},\ldots,\gamma^{n} x_{1p},\gamma^{n} y_{11},\ldots, \gamma^{n} y_{1p}, \\& \quad \ldots, \gamma^{n} x_{k1},\ldots,\gamma^{n} x_{kp},\ldots,\gamma^{n} y_{k1},\ldots, \gamma^{n} y_{kd}\bigr) = 0, \end{aligned}$$

(1)

$$\begin{aligned}& \bigl\Vert \bigl( c_{\mu}f(x_{11},\ldots,x_{1p},y_{11}, \ldots ,y_{1d}),\ldots,c_{\mu}f(x_{k1}, \ldots,x_{kp},y_{k1},\ldots,y_{kd} ) \bigr)\bigr\Vert _{k} \\& \quad \leq \varphi(x_{11},\ldots,x_{1p},y_{11}, \ldots ,y_{1d},\ldots,x_{k1},\ldots,x_{kp},y_{k1}, \ldots,y_{kd}) , \end{aligned}$$

(2)

$$\begin{aligned}& \bigl\Vert \bigl(f\bigl([x_{1}, y_{1}, z_{1}] \bigr) - \bigl[f(x_{1}), f(y_{1}), f(z_{1})\bigr], \\& \qquad \ldots,f\bigl([x_{k}, y_{k}, z_{k}]\bigr) - \bigl[f(x_{k}), f(y_{k}), f(z_{k})\bigr] \bigr) \bigr\Vert _{k} \\& \quad \le\psi(x_{1}, y_{1},z_{1}, \ldots,x_{k},y_{k},z_{k}) , \end{aligned}$$

(3)

$$\begin{aligned}& \lim_{n\to\infty} \gamma^{-3n} \psi\bigl( \gamma^{n} x_{1},\gamma^{n} y_{1}, \gamma^{n} z_{1},\ldots,\gamma^{n} x_{k}, \gamma^{n} y_{k},\gamma^{n} z_{k}\bigr) = 0, \end{aligned}$$

(4)

$$\begin{aligned}& \lim_{n\to\infty} \gamma^{-2n} \psi\bigl( \gamma^{n} x_{1},\gamma^{n} y_{1}, z_{1},\ldots,\gamma^{n} x_{k},\gamma^{n} y_{k},z_{k}\bigr) = 0 \end{aligned}$$

(5)

for all $\mu\in\mathbf{T}^{1}$ and $x_{11},\ldots,x_{1p},y_{11},\ldots ,y_{1d},\ldots,x_{k1},\ldots,x_{kp},y_{k1},\ldots, y_{kd}, x_{1},\ldots,x_{k},y_{1}, \ldots,y_{k},z_{1}, \ldots,z_{k}\in A$, where $\gamma=\frac{p+2d}{2}$. If there exists a constant $L<1$ such that

$$\begin{aligned}& \varphi\bigl(\overbrace{\gamma x_{1},\ldots,\gamma x_{1}}^{p+d},\overbrace {\gamma x_{2},\ldots,\gamma x_{2} }^{p+d},\ldots,\overbrace{\gamma x_{k}, \ldots,\gamma x_{k}}^{p+d}\bigr) \\& \quad \le\gamma L \varphi\bigl(\overbrace{x_{1},\ldots ,x_{1}}^{p+d},\overbrace{x_{2}, \ldots,x_{2}}^{p+d},\ldots,\overbrace {x_{k}, \ldots,x_{k}}^{p+d}\bigr) \end{aligned}$$

(6)

for all $x_{1},x_{2},\ldots,x_{k} \in A$, then there exists a unique homomorphism $H : A \rightarrow B$ such that

$$\begin{aligned}& \bigl\Vert \bigl(f(x_{1}) - H(x_{1}), \ldots,f(x_{k}) - H(x_{k}) \bigr) \bigr\Vert _{k} \\& \quad \le\frac{1}{(1-L)2\gamma} \varphi\bigl(\overbrace{x_{1}, \ldots,x_{1}}^{p+d},\overbrace{x_{2},\ldots ,x_{2}}^{p+d},\ldots,\overbrace{x_{k}, \ldots,x_{k}}^{p+d}\bigr) \end{aligned}$$

(7)

for all $x_{1},\ldots,x_{k} \in A$.

Proof

Let $\mu= 1$ and $x_{ij} = y_{ij} = x_{i}$ for $1\leq i\leq k$ in (2). Then we get

$$\begin{aligned}& \bigl\Vert \bigl(f(\gamma x_{1})-\gamma f(x_{1}),\ldots,f(\gamma x_{k})-\gamma f(x_{k}) \bigr)\bigr\Vert _{k} \\& \quad \leq\frac{1}{2} \varphi\bigl(\overbrace{ x_{1}, \ldots,x_{1}}^{p+d},\overbrace{x_{2}, \ldots,x_{2}}^{p+d},\ldots,\overbrace {x_{k}, \ldots,x_{k}}^{p+d}\bigr) \end{aligned}$$

(8)

for all $x_{1},\ldots,x_{k} \in A$. Consider the set

$$E: = \{ g : A \rightarrow B\} $$

and introduce the generalized metric on E:

$$\begin{aligned} d(g, h) = &\inf \bigl\{ C\in{\mathbf{R}}_{+} : \bigl\Vert \bigl(g(x_{1})-h(x_{1}), \ldots,g(x_{k})-h(x_{k}) \bigr)\bigr\Vert _{k} \\ &\le C\varphi\bigl(\overbrace{x_{1},\ldots,x_{1}}^{p+d}, \overbrace{x_{2},\ldots ,x_{2}}^{p+d},\ldots, \overbrace{x_{k},\ldots,x_{k}}^{p+d}\bigr), \forall x_{1},\ldots,x_{k} \in A \bigr\} . \end{aligned}$$

It is easy to see that $(E, d)$ is complete (see also [9]).

First we show that d is metric on E. It is obvious $d(g,g)=0$ for all $g\in E$. If $d(g,h)=0$, then, for every fixed $x_{1},\ldots,x_{k} \in A$,

$$\bigl\Vert \bigl(g(x_{1}) -h(x_{1}),\ldots,g(x_{k})-h(x_{k}) \bigr)\bigr\Vert _{k}=0 $$

and therefore $g=h$. If $d(g,h)=a<\infty$ and $d(h,l)=b<\infty$ for all $g,h,l\in E$, then

$$\begin{aligned}& \bigl\Vert \bigl(g(x_{1})-l(x_{1}),\ldots,g(x_{k})-l(x_{k}) \bigr)\bigr\Vert _{k} \\& \quad = \bigl\Vert \bigl(g(x_{1})-h(x_{1})+h(x_{1})-l(x_{1}), \ldots,g(x_{k})-h(x_{k})+h(x_{k})-l(x_{k}) \bigr)\bigr\Vert _{k} \\& \quad \leq\bigl\Vert \bigl(g(x_{1})-h(x_{1}), \ldots,g(x_{k})-h(x_{k})\bigr)\bigr\Vert _{k} + \bigl\Vert \bigl(h(x_{1})-l(x_{1}),\ldots,h(x_{k})-l(x_{k}) \bigr)\bigr\Vert _{k} \\& \quad \leq a \varphi\bigl(\overbrace{x_{1},\ldots,x_{1}}^{p+d}, \overbrace {x_{2},\ldots,x_{2}}^{p+d},\ldots, \overbrace{x_{k},\ldots,x_{k}}^{p+d}\bigr) + b \varphi \bigl(\overbrace{x_{1},\ldots,x_{1}}^{p+d}, \overbrace {x_{2},\ldots,x_{2}}^{p+d},\ldots, \overbrace{x_{k},\ldots,x_{k}}^{p+d}\bigr) \\& \quad = (a+b) \varphi\bigl(\overbrace{x_{1},\ldots,x_{1}}^{p+d}, \overbrace {x_{2},\ldots,x_{2}}^{p+d},\ldots, \overbrace{x_{k},\ldots,x_{k}}^{p+d}\bigr). \end{aligned}$$

So we have $d(g,l)\leq d(g,h) + d(h,l)$.

Let $\{g_{n}\}$ be a Cauchy sequence in $(E,d)$. Then for all $\epsilon >0$ there exists N such that $d(g_{n},g_{i}) < \epsilon$, if $n,i \geq N$, Let $n,i\geq N$. Since $d(g_{n},g_{i}) < \epsilon$ there exists $C\in [0,\epsilon)$ such that

$$\begin{aligned}& \bigl\Vert \bigl(g_{n}(x_{1}) -g_{i}(x_{1}),\ldots,g_{n}(x_{k})-g_{i}(x_{k}) \bigr)\bigr\Vert _{k} \\& \quad \leq C\varphi\bigl(\overbrace{x_{1},\ldots,x_{1}}^{p+d}, \overbrace {x_{2},\ldots,x_{2}}^{p+d},\ldots, \overbrace{x_{k},\ldots,x_{k}}^{p+d}\bigr) \\& \quad \leq\epsilon\varphi\bigl(\overbrace{x_{1},\ldots,x_{1}}^{p+d}, \overbrace {x_{2},\ldots,x_{2}}^{p+d},\ldots, \overbrace{x_{k},\ldots,x_{k}}^{p+d}\bigr) \end{aligned}$$

(9)

for all $x_{1},\ldots,x_{k} \in A$, so for each $x_{1},\ldots,x_{k}\in A$, $\{g_{n}(x_{1},\ldots,x_{k})\}$ is a Cauchy sequence in B. Since B is complete, there exists $g(x_{1},\ldots,x_{k})\in B$ such that $g_{n}(x_{1},\ldots,x_{k})\rightarrow g(x_{1},\ldots,x_{k})$ as $n\rightarrow\infty$. Thus, we have $g\in E$. Taking the limit as $i\rightarrow\infty$ in (9) we obtain, for $n\geq N$,

$$ \bigl\Vert \bigl(g_{n}(x_{1}) -g(x_{1}), \ldots,g_{n}(x_{k})-g(x_{k})\bigr)\bigr\Vert _{k} \leq\epsilon\varphi\bigl(\overbrace{x_{1},\ldots,x_{1}}^{p+d}, \overbrace {x_{2},\ldots,x_{2}}^{p+d},\ldots, \overbrace{x_{k},\ldots,x_{k}}^{p+d}\bigr). $$

Therefore $d(g_{n},g)\leq\epsilon$. Hence $g_{n}\rightarrow g$ as $n\rightarrow\infty$, so $(E,d)$ is complete. Now, we consider the linear mapping $\Lambda: E \rightarrow E$ such that

$$\Lambda g(x): = \frac{1}{\gamma} g(\gamma x) $$

for all $x \in A$. From Theorem 3.1 of [14] (also see Lemma 3.2 of [9]),

$$d(\Lambda g, \Lambda h) \le L d(g, h) $$

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

$$ \bigl\Vert \bigl(g(x_{1}) - h(x_{1}) , \ldots,g(x_{k}) - h(x_{k}) \bigr)\bigr\Vert _{k} \leq C \varphi\bigl(\overbrace{x_{1},\ldots,x_{1}}^{p+d}, \ldots,\overbrace {x_{k},\ldots,x_{k}}^{p+d}\bigr) $$

for all $x_{1},\ldots,x_{k}\in A$. From our assumption and the last inequality, we have

$$\begin{aligned}& \bigl\Vert \bigl(\Lambda g(x_{1}) - \Lambda h(x_{1}), \ldots,\Lambda g(x_{k})- \Lambda h(x_{k}) \bigr)\bigr\Vert _{k} \\& \quad =\frac{1}{\gamma} \bigl\Vert \bigl(g(\gamma x_{1})- h (\gamma x_{1}) ,\ldots,g(\gamma x_{k})- h (\gamma x_{k}) \bigr)\bigr\Vert _{k} \\& \quad \leq\frac{C}{\gamma}\varphi\bigl(\overbrace{\gamma x_{1}, \ldots,\gamma x_{1} }^{p+d},\ldots,\overbrace{\gamma x_{k},\ldots,\gamma x_{k} }^{p+d}\bigr) \\& \quad \leq C L \varphi\bigl(\overbrace {x_{1},\ldots,x_{1}}^{p+d}, \ldots,\overbrace{x_{k},\ldots,x_{k}}^{p+d}\bigr) \end{aligned}$$

for all $x_{1},\ldots,x_{k} \in A$ and so

$$\begin{aligned}& \bigl\Vert \bigl(\Lambda f(x_{1}) - f(x_{1}),\ldots, \Lambda f(x_{k}) - f(x_{k}) \bigr)\bigr\Vert _{k} \\& \quad = \biggl\Vert \biggl(\frac{1}{\gamma}f(\gamma x_{1}) - f(x_{1}),\ldots,\frac{1}{\gamma}f(\gamma x_{k}) - f(x_{k}) \biggr)\biggr\Vert _{k} \\& \quad = \frac{1}{\gamma}\bigl\Vert \bigl(f(\gamma x_{1}) - \gamma f(x_{1}),\ldots,f(\gamma x_{k}) - \gamma f(x_{k}) \bigr)\bigr\Vert _{k} \\& \quad \leq \frac{1}{2\gamma}\varphi\bigl(\overbrace{x_{1}, \ldots,x_{1}}^{p+d},\ldots ,\overbrace{x_{k}, \ldots,x_{k}}^{p+d}\bigr) \end{aligned}$$

for all $x_{1},\ldots,x_{k} \in A$. Hence $d(\Lambda f, f) \le \frac{1}{2\gamma}$. From Theorem 1.1, the sequence $\{\Lambda^{n} f\}$ converges to a fixed point H of Λ, i.e., $H:A\rightarrow B$ is a mapping defined by

$$ H(x) = \lim_{n\to\infty}\bigl(\Lambda^{n} f\bigr) (x) = \lim_{n\to\infty}\frac{1}{\gamma^{n}}f\bigl(\gamma^{n} x\bigr) $$

(10)

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

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

i.e., the inequality (7) hold for all $x_{1},\ldots ,x_{k} \in A$. Thus it follows from the definition of H, (1), and (2) that

$$\begin{aligned}& \Biggl\Vert \Biggl(2H \Biggl(\frac{\sum_{j=1}^{p} \mu x_{1j}}{2}+\sum _{j=1}^{d} \mu y_{1j} \Biggr) - \sum _{j=1}^{p} \mu H(x_{1j})-2 \sum _{j=1}^{d} \mu H(y_{1j}), \\& \qquad \ldots, 2H \Biggl(\frac{\sum_{j=1}^{p} \mu x_{kj}}{2}+\sum_{j=1}^{d} \mu y_{kj} \Biggr) - \sum_{j=1}^{p} \mu H(x_{kj})-2 \sum_{j=1}^{d} \mu H(y_{kj}) \Biggr) \Biggr\Vert _{k} \\& \quad = \lim_{n\to\infty} \frac{1}{\gamma^{n}} \Biggl\Vert \Biggl(2 f \Biggl(\gamma^{n}\frac{\sum_{j=1}^{p} \mu x_{1j}}{2}+\gamma^{n}\sum _{j=1}^{d} \mu y_{1j} \Biggr) - \sum _{j=1}^{p} \mu f\bigl(\gamma^{n} x_{1j}\bigr)-2 \sum_{j=1}^{d} \mu f\bigl(\gamma^{n}y_{1j}\bigr), \\& \qquad \ldots, 2 f \Biggl(\gamma^{n}\frac{\sum_{j=1}^{p} \mu x_{kj}}{2}+ \gamma^{n}\sum_{j=1}^{d} \mu y_{kj} \Biggr) - \sum_{j=1}^{p} \mu f\bigl(\gamma^{n} x_{kj}\bigr)-2 \sum _{j=1}^{d} \mu f\bigl(\gamma ^{n}y_{kj} \bigr) \Biggr) \Biggr\Vert _{k} \\& \quad \leq\lim_{n\to\infty} \frac{1}{\gamma^{n}}\bigl\Vert \bigl(C_{\mu}f\bigl(\gamma^{n} x_{11},\ldots, \gamma^{n} x_{1p},\gamma^{n} y_{11}, \ldots,\gamma^{n}y_{1d}\bigr), \\& \qquad \ldots, C_{\mu}f\bigl(\gamma^{n} x_{k1}, \ldots,\gamma^{n} x_{kp},\gamma^{n} y_{k1},\ldots,\gamma^{n}y_{kd}\bigr) \bigr)\bigr\Vert _{k} \\& \quad \leq\lim_{n\to\infty}\frac{1}{\gamma^{n}}\varphi\bigl( \gamma^{n} x_{11},\ldots,\gamma^{n} x_{1p},\gamma^{n} y_{11},\ldots, \gamma^{n} y_{1d}, \\& \qquad \ldots, \gamma^{n} x_{k1},\ldots,\gamma^{n} x_{kp},\gamma^{n} y_{k1},\ldots, \gamma^{n} y_{kd}\bigr)=0 \end{aligned}$$

for all $\mu\in\mathbf{T}^{1}$ and $x_{11},\ldots,x_{1p},y_{11},\ldots ,y_{1d},\ldots,x_{k1},\ldots,x_{kp},y_{k1},\ldots,y_{kd} \in A$. Hence we have

$$ 2H \Biggl(\frac{\sum_{j=1}^{p} \mu x_{ij}}{2}+\sum_{j=1}^{d} \mu y_{ij} \Biggr) = \sum_{j=1}^{p} \mu H(x_{ij})+2 \sum_{j=1}^{d} \mu H(y_{ij}) $$

for all $\mu\in\mathbf{T}^{1}$, $x_{11},\ldots,x_{1p},y_{11},\ldots,y_{1d},\ldots,x_{k1},\ldots ,x_{kp},y_{k1},\ldots,y_{kd} \in A$ and $1\leq i \leq k$ and so $H(\lambda x + \mu y)=\lambda H(x) + \mu H(y)$ for all $\lambda, \mu\in\mathbf{T}^{1}$ and $x,y \in A$. Therefore, by Lemma 3.1, the mapping $H : A \rightarrow B$ is C-linear.

Also it follows from (3) and (4) that

$$\begin{aligned}& \bigl\Vert \bigl(H\bigl([x_{1}, y_{1}, z_{1}] \bigr)- \bigl[H(x_{1}), H(y_{1}), H(z_{1})\bigr], \ldots ,H\bigl([x_{k}, y_{k}, z_{k}]\bigr)- \bigl[H(x_{k}), H(y_{k}), H(z_{k})\bigr] \bigr)\bigr\Vert _{k} \\& \quad =\lim_{n\to\infty}\frac{1}{\gamma^{3n}} \bigl\Vert \bigl(f \bigl(\bigl[\gamma^{n} x_{1}, \gamma^{n} y_{1}, \gamma^{n} z_{1}\bigr] \bigr) - \bigl[f \bigl(\gamma^{n} x_{1}\bigr), f\bigl(\gamma^{n} y_{1}\bigr), f\bigl(\gamma^{n} z_{1}\bigr) \bigr], \\& \qquad \ldots, f \bigl(\bigl[\gamma^{n} x_{k}, \gamma^{n} y_{k}, \gamma^{n} z_{k}\bigr] \bigr) - \bigl[f\bigl(\gamma^{n} x_{k}\bigr), f\bigl( \gamma^{n} y_{k}\bigr), f\bigl(\gamma^{n} z_{k}\bigr) \bigr] \bigr) \bigr\Vert _{k} \\& \quad \leq\lim_{n\to\infty} \frac{1}{\gamma^{3n}}\psi\bigl( \gamma^{n}x_{1},\gamma^{n}y_{1},\gamma ^{n}z_{1},\ldots,\gamma^{n}x_{k}, \gamma^{n}y_{k},\gamma^{n}z_{k}\bigr)=0 \end{aligned}$$

for all $x_{1}, y_{1}, z_{1},\ldots,x_{k},y_{k},z_{k} \in A$. Thus we have

$$H\bigl([x, y,z]\bigr) = \bigl[H(x), H(y), H(z)\bigr] $$

for all $x, y, z \in A$. Thus $H : A \rightarrow B$ is a homomorphism satisfying (7).

Now, let $T : A \rightarrow B$ be another $C^{*}$-ternary-algebras homomorphism satisfying (7). Since $d(f,T)\leq \frac{1}{(1-L)2\gamma}$ and T is C-linear, we get $T\in E'$ and $(\Lambda T)(x)=\frac{1}{\gamma}(T\gamma x)=T(x)$ for all $x\in A$, i.e., T is a fixed point of Λ. Since H is the unique fixed point of $\Lambda\in E'$, we get $H=T$. This completes the proof. □

Theorem 3.4

Let $(( {B}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ be a multi-$C^{*}$-ternary algebra. Let $f: A \rightarrow B$ be a mapping for which there are the functions $\varphi: A^{(p+d)k} \rightarrow[0, \infty)$ and $\psi: A^{3k} \rightarrow[0, \infty)$ satisfying the inequalities (2) and (3) such that

$$\begin{aligned}& \lim_{n\to\infty} {\gamma}^{n} \varphi \biggl( \frac{x_{11}}{\gamma^{n}} ,\ldots,\frac{x_{1p}}{\gamma^{n}} ,\frac{y_{11}}{\gamma^{n}} ,\ldots, \frac{y_{1p}}{\gamma^{n}}, \ldots,\frac{x_{k1}}{\gamma^{n}} ,\ldots,\frac{x_{kp}}{\gamma ^{n}} ,\ldots, \frac{y_{k1}}{\gamma^{n}} ,\ldots,\frac{y_{kd}}{\gamma^{n}} \biggr) = 0, \end{aligned}$$

(11)

$$\begin{aligned}& \lim_{n\to\infty} \gamma^{3n} \psi \biggl( \frac{x_{1}}{\gamma^{n}} ,\frac{y_{1}}{\gamma^{n}} ,\frac{z_{1}}{\gamma^{n}} ,\ldots, \frac{x_{k}}{\gamma^{n}} ,\frac{y_{k}}{\gamma^{n}} ,\frac{z_{k}}{\gamma^{n}} \biggr) = 0, \end{aligned}$$

(12)

$$\begin{aligned}& \lim_{n\to\infty} \gamma^{2n} \psi \biggl( \frac{x_{1}}{\gamma^{n}} ,\frac{y_{1}}{\gamma^{n}} ,z_{1} ,\ldots,\frac{x_{k}}{\gamma^{n}} ,\frac{y_{k}}{\gamma^{n}} ,z_{k} \biggr) = 0 \end{aligned}$$

(13)

for all $\mu\in\mathbf{T}^{1}$ and $x_{11},\ldots,x_{1p},y_{11},\ldots ,y_{1d},\ldots,x_{k1},\ldots,x_{kp},y_{k1},\ldots, y_{kd}, x_{1},\ldots,x_{k}, y_{1},\ldots,y_{k},z_{1}, \ldots,z_{k} \in A$, where $\gamma=\frac{p+2d}{2}$. If the constant $L<1$ exists such that

$$\begin{aligned}& \varphi\biggl(\overbrace{\frac{x_{1}}{\gamma} ,\ldots, \frac{x_{1}}{\gamma} }^{p+d},\overbrace{\frac{x_{2}}{\gamma} ,\ldots, \frac{x_{2}}{\gamma} }^{p+d},\ldots,\overbrace{\frac{x_{k}}{\gamma} ,\ldots, \frac{x_{k}}{\gamma } }^{p+d}\biggr) \\& \quad \le\frac{L}{\gamma} \varphi\bigl(\overbrace{x_{1},\ldots ,x_{1}}^{p+d},\overbrace{x_{2}, \ldots,x_{2}}^{p+d},\ldots,\overbrace {x_{k}, \ldots,x_{k}}^{p+d}\bigr) \end{aligned}$$

(14)

for all $x_{1},x_{2},\ldots,x_{k} \in A$, then there exists a unique homomorphism $H : A \rightarrow B$ such that

$$\begin{aligned}& \bigl\Vert \bigl(f(x_{1}) - H(x_{1}), \ldots,f(x_{k}) - H(x_{k}) \bigr) \bigr\Vert _{k} \\& \quad \le\frac{1}{(1-L)2\gamma} \varphi\bigl(\overbrace{x_{1}, \ldots,x_{1}}^{p+d},\overbrace{x_{2},\ldots ,x_{2}}^{p+d},\ldots,\overbrace{x_{k}, \ldots,x_{k}}^{p+d}\bigr) \end{aligned}$$

(15)

for all $x_{1},\ldots,x_{k} \in A$.

Proof

If we replace $x_{i}$ in (8) by $\frac{x_{i}}{\gamma}$ for $1\leq i\leq k$, then we get

$$\begin{aligned}& \biggl\Vert \biggl(f(x_{1})-\gamma f\biggl( \frac{1}{x_{1}}\biggr),\ldots,f(x_{k})-\gamma f\biggl( \frac{1}{x_{k}}\biggr) \biggr)\biggr\Vert _{k} \\& \quad \leq\frac{1}{2} \varphi\biggl(\overbrace{ \frac{1}{x_{1}},\ldots, \frac{1}{x_{1}}}^{p+d},\overbrace{\frac {1}{x_{2}},\ldots, \frac{1}{x_{2}}}^{p+d},\ldots,\overbrace{\frac {1}{x_{k}},\ldots, \frac{1}{x_{k}}}^{p+d}\biggr) \end{aligned}$$

(16)

for all $x_{1},\ldots,x_{k} \in A$. Consider the set

$$E: = \{ g : A \rightarrow B\} $$

and introduce the generalized metric on E:

$$\begin{aligned} d(g, h) =& \inf \bigl\{ C\in{\mathbf{R}}_{+} : \bigl\Vert \bigl(g(x_{1})-h(x_{1}), \ldots,g(x_{k})-h(x_{k}) \bigr)\bigr\Vert _{k} \\ &\le C\varphi\bigl(\overbrace{x_{1},\ldots,x_{1}}^{p+d}, \overbrace{x_{2},\ldots ,x_{2}}^{p+d},\ldots, \overbrace{x_{k},\ldots,x_{k}}^{p+d}\bigr), \forall x_{1},\ldots,x_{k} \in A \bigr\} . \end{aligned}$$

It is easy to see that $(E, d)$ is complete (see [9]).

Now, we consider the linear mapping $\Lambda: E \rightarrow E$ such that

$$\Lambda g(x): = \gamma g\biggl(\frac{x}{\gamma} \biggr) $$

for all $x \in A$. From Theorem 3.1 of [14] (also see Lemma 3.2 of [9]),

$$d(\Lambda g, \Lambda h) \le L d(g, h) $$

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

$$ \bigl\Vert \bigl(g(x_{1}) - h(x_{1}) , \ldots,g(x_{k}) - h(x_{k}) \bigr)\bigr\Vert _{k} \leq C \varphi\bigl(\overbrace{x_{1},\ldots,x_{1}}^{p+d}, \ldots,\overbrace {x_{k},\ldots,x_{k}}^{p+d}\bigr) $$

for all $x_{1},\ldots,x_{k}\in A$. From our assumption and the last inequality, we have

$$\begin{aligned}& \bigl\Vert \bigl(\Lambda g(x_{1}) - \Lambda h(x_{1}), \ldots,\Lambda g(x_{k})- \Lambda h(x_{k}) \bigr)\bigr\Vert _{k} \\& \quad =\gamma\biggl\Vert \biggl(g\biggl(\frac{x_{1}}{\gamma} \biggr)- h \biggl( \frac{x_{1}}{\gamma} \biggr) ,\ldots,g\biggl(\frac{x_{k}}{\gamma}\biggr)- h \biggl( \frac{x_{k}}{\gamma}\biggr) \biggr)\biggr\Vert _{k} \\& \quad \leq{C} {\gamma}\varphi\biggl(\overbrace{\frac{x_{1}}{\gamma} ,\ldots, \frac{x_{1}}{\gamma} }^{p+d},\ldots,\overbrace{\frac{x_{k}}{\gamma} ,\ldots, \frac{x_{k} }{\gamma} }^{p+d}\biggr) \\& \quad \leq C L \varphi\bigl(\overbrace {x_{1},\ldots,x_{1}}^{p+d}, \ldots,\overbrace{x_{k},\ldots,x_{k}}^{p+d}\bigr) \end{aligned}$$

for all $x_{1},\ldots,x_{k} \in A$ and so $d(\Lambda g ,\Lambda h)\leq Ld(g,h)$ for any $g,h \in E$. It follows from (16) that $d(\Lambda f,f)\leq\frac{1}{2\gamma}$. Therefore, according to Theorem 1.1, the sequence $\{\Lambda^{n} f\}$ converges to a fixed point H of Λ, i.e., $H:A\rightarrow B$ is a mapping defined by

$$ H(x) = \lim_{n\to\infty}\bigl(\Lambda^{n} f\bigr) (x) = \lim_{n\to\infty}{\gamma^{n}}f\biggl(\frac{ x}{\gamma^{n}} \biggr) $$

(17)

for all $x \in A$.

The rest of the proof is similar to the proof of Theorem 3.3 and so we omit it. This completes the proof. □

Theorem 3.5

Let r and θ be non-negative real numbers such that $r\notin[1,3]$ and let $(( {B}^{k},{\|\cdot\|_{k}}): k\in\mathbf{N})$ be a multi-$C^{*}$-ternary algebra. Let $f : A \rightarrow B$ be a mapping such that

$$\begin{aligned}& \bigl\Vert \bigl(C_{\mu}f(x_{11}, \ldots,x_{1p},y_{11},\ldots ,y_{1d}),\ldots, C_{\mu}f(x_{k1},\ldots,x_{kp},y_{k1}, \ldots,y_{kd}) \bigr)\bigr\Vert _{k} \\& \quad \le\theta \Biggl( \sum_{j=1}^{p} \|x_{1j}\|^{r}_{A} + \sum _{j=1}^{d}\|y_{1j}\| ^{r}_{A} + \cdots+\sum_{j=1}^{p}\|x_{kj} \|^{r}_{A} + \sum_{j=1}^{d} \|y_{kj}\| ^{r}_{A} \Biggr) \end{aligned}$$

(18)

and

$$\begin{aligned}& \bigl\Vert \bigl(f\bigl([x_{1},y_{1},z_{1}] \bigr)-\bigl[f(x_{1}),f(y_{1}),f(z_{1})\bigr],\ldots ,f\bigl([x_{k},y_{k},z_{k}]\bigr)- \bigl[f(x_{k}),f(y_{k}),f(z_{k})\bigr] \bigr)\bigr\Vert _{k} \\& \quad \le\theta\bigl(\|x_{1}\|^{r}_{A}\cdot \|y_{1}\|^{r}_{A} . \|z_{1} \|^{r}_{A} + \cdots+ \| x_{k}\|^{r}_{A} \cdot\|y_{k}\|^{r}_{A} \cdot\|z_{k} \|^{r}_{A}\bigr) \end{aligned}$$

(19)

for all $\mu\in\mathbf{T}^{1}$ and $x_{11},\ldots,x_{1p},y_{11},\ldots,y_{1d},\ldots,x_{k1},\ldots ,x_{kp},y_{k1},\ldots,y_{kd},x_{1},\ldots,x_{k}, y_{1},\ldots,y_{k},z_{1}, \ldots,z_{k} \in A$. Then there exists a unique $C^{*}$-ternary algebra homomorphism $H : A \rightarrow B$ such that

$$\begin{aligned} \begin{aligned}[b] &\bigl\Vert \bigl(f(x_{1}) - H(x_{1}), \ldots,f(x_{k}) - H(x_{k}) \bigr)\bigr\Vert _{B} \\ &\quad \le\frac{2^{r}(p+d)\theta}{|2(p+2d)^{r} -(p+2d)2^{r}|}\bigl(\|x_{1}\|^{r}_{A} + \cdots + \|x_{k}\|^{r}_{A}\bigr) \end{aligned} \end{aligned}$$

(20)

for all $x_{1},\ldots,x_{k} \in A$.

Proof

The proof follows from Theorem 3.3 by taking

$$\begin{aligned}& \varphi(x_{11},\ldots,x_{1p},y_{11}, \ldots,y_{1d},\ldots,x_{k1},\ldots ,x_{kp},y_{k1}, \ldots,y_{kd}) \\& \quad := \theta \Biggl( \sum_{j=1}^{p} \|x_{ij}\|^{r}_{A} + \sum _{j=1}^{d}\|y_{ij}\|^{r}_{A} + \cdots+ \sum_{j=1}^{p}\|x_{kj} \|^{r}_{A} + \sum_{j=1}^{d} \|y_{kj}\|^{r}_{A} \Biggr), \\& \psi(x_{1},y_{1},z_{1},\ldots,x_{k},y_{k},z_{k}) \\& \quad :=\theta\bigl(\|x_{1}\|^{r}_{A} \cdot \|y_{1}\|^{r}_{A} \cdot\|z_{1} \|^{r}_{A} + \cdots+\| x_{k}\|^{r}_{A} \cdot\|y_{k}\|^{r}_{A} \cdot\|z_{k} \|^{r}_{A} \bigr) \end{aligned}$$

for all $\mu\in\mathbf{T}^{1}$ and $x_{11},\ldots,x_{1p},y_{11},\ldots ,y_{1d},\ldots,x_{k1},\ldots,x_{kp},y_{k1},\ldots, y_{kd},x_{1},\ldots,x_{k}, y_{1},\ldots,y_{k},z_{1}, \ldots,z_{k}\in A$. 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 so we get the desired result. This completes the proof. □

Theorem 3.6

Let $(( {B}^{k},\|\cdot\|_{k}): k\mathbf{N})$ be a multi-$C^{*}$-ternary algebra. Let $f: A \rightarrow B$ be a mapping for which there are functions $\varphi: A^{(p+d)k} \rightarrow[0, \infty)$ and $\psi: A^{3k} \rightarrow[0, \infty)$ such that

$$\begin{aligned}& \lim_{n\to\infty} {d}^{-n} \varphi \bigl(d^{n} x_{11},\ldots,d^{n} x_{1p},d^{n} y_{11},\ldots,d^{n} y_{1p}, \\& \quad \ldots, d^{n} x_{k1},\ldots,d^{n} x_{kp},\ldots,d^{n} y_{k1},\ldots,d^{n} y_{kd}\bigr) = 0, \end{aligned}$$

(21)

$$\begin{aligned}& \bigl\Vert \bigl( c_{\mu}f(x_{11},\ldots,x_{1p},y_{11}, \ldots,y_{1d}),\ldots,c_{\mu }f(x_{k1}, \ldots,x_{kp},y_{k1},\ldots,y_{kd} ) \bigr)\bigr\Vert _{k} \\& \quad \leq \varphi(x_{11},\ldots,x_{1p},y_{11}, \ldots ,y_{1d},\ldots,x_{k1},\ldots,x_{kp},y_{k1}, \ldots,y_{kd}) , \end{aligned}$$

(22)

$$\begin{aligned}& \bigl\Vert \bigl(f\bigl([x_{1}, y_{1}, z_{1}] \bigr) - \bigl[f(x_{1}), f(y_{1}), f(z_{1})\bigr], \\& \qquad \ldots,f\bigl([x_{k}, y_{k}, z_{k}]\bigr) - \bigl[f(x_{k}), f(y_{k}), f(z_{k})\bigr]\bigr) \bigr\Vert _{k} \\& \quad \le\psi(x_{1}, y_{1},z_{1}, \ldots,x_{k},y_{k},z_{k}) , \end{aligned}$$

(23)

$$\begin{aligned}& \lim_{n\to\infty} d^{-3n} \psi \bigl(d^{n} x_{1},d^{n} y_{1},d^{n} z_{1},\ldots,d^{n} x_{k},d^{n} y_{k},d^{n} z_{k}\bigr)= 0, \end{aligned}$$

(24)

$$\begin{aligned}& \lim_{n\to\infty} d^{-2n} \psi \bigl(d^{n} x_{1},d^{n} y_{1},z_{1}, \ldots,d^{n} x_{k},d^{n} y_{k}, z_{k}\bigr) = 0 \end{aligned}$$

(25)

for all $\mu\in\mathbf{T}^{1}$ and $x_{11},\ldots,x_{1p},y_{11},\ldots ,y_{1d},\ldots,x_{k1},\ldots,x_{kp},y_{k1},\ldots,y_{kd}, x_{1},\ldots,x_{k}, y_{1},\ldots,y_{k},z_{1}, \ldots,z_{k}\in A$, where $\gamma=\frac{p+2d}{2}$. If there exists the constant $L<1$ such that

$$\begin{aligned}& \varphi\bigl(\overbrace{d x_{1},\ldots,d x_{1}}^{p+d},\overbrace{d x_{2},\ldots,d x_{2} }^{p+d},\ldots,\overbrace{d x_{k},\ldots,d x_{k}}^{p+d}\bigr) \\& \quad \le d L \varphi\bigl( \overbrace{0,\ldots,0}^{p},\overbrace {x_{1},\ldots,x_{1}}^{d},\overbrace{0, \ldots,0}^{p}, \overbrace{x_{2},\ldots,x_{2}}^{d}, \ldots,\overbrace{0,\ldots ,0}^{p},\overbrace{x_{k}, \ldots,x_{k}}^{d}\bigr) \end{aligned}$$

(26)

for all $x_{1},x_{2},\ldots,x_{k} \in A$, then there exists a unique homomorphism $H : A \rightarrow B$ such that

$$\begin{aligned}& \bigl\Vert \bigl(f(x_{1}) - H(x_{1}), \ldots,f(x_{k}) - H(x_{k})\bigr) \bigr\Vert _{k} \\& \quad \le\frac{1}{(1-L)2d} \varphi\bigl(\overbrace{0,\ldots,0}^{p}, \overbrace{x_{1},\ldots ,x_{1}}^{d},\overbrace{0, \ldots,0}^{p},\overbrace{x_{2},\ldots ,x_{2}}^{d}, \ldots, \overbrace{0,\ldots,0}^{p},\overbrace{x_{k}, \ldots,x_{k}}^{d}\bigr) \end{aligned}$$

(27)

for all $x_{1},\ldots,x_{k} \in A$.

Proof

Let $\mu= 1$ and $x_{ij} = 0$, $y_{ij} = x_{i}$ for $1\leq i\leq k$ in (22). Then we get

$$\begin{aligned}& \bigl\Vert \bigl(f(d x_{1})- d f(x_{1}), \ldots,f(d x_{k})-d f(x_{k})\bigr)\bigr\Vert _{k} \\ & \quad \leq\frac{1}{2} \varphi\bigl(\overbrace{0,\ldots,0}^{p}, \overbrace{x_{1},\ldots ,x_{1}}^{d},\overbrace{0, \ldots,0}^{p},\overbrace{x_{2},\ldots ,x_{2}}^{d}, \ldots, \overbrace{0,\ldots,0}^{p},\overbrace{x_{k}, \ldots,x_{k}}^{d}\bigr) \end{aligned}$$

(28)

for all $x_{1},\ldots,x_{k} \in A$. Consider the set

$$E: = \{ g : A \rightarrow B\} $$

and introduce the generalized metric on E:

$$\begin{aligned} d(g, h) =& \inf \bigl\{ C\in{\mathbf{R}}_{+} : \bigl\Vert \bigl(g(x_{1})-h(x_{1}), \ldots,g(x_{k})-h(x_{k})\bigr)\bigr\Vert _{k} \\ & \le C\varphi\bigl(\overbrace{0,\ldots,0}^{p},\overbrace{x_{1}, \ldots ,x_{1}}^{d},\overbrace{0,\ldots,0}^{p}, \overbrace{x_{2},\ldots ,x_{2}}^{d},\ldots, \overbrace{0,\ldots,0}^{p},\overbrace{x_{k}, \ldots,x_{k}}^{d}\bigr), \\ &\forall x_{1}, \ldots,x_{k} \in A \bigr\} . \end{aligned}$$

It is easy to see that $(E, d)$ is complete (see [9]).

Now, we consider the linear mapping $\Lambda: E \rightarrow E$ such that

$$\Lambda g(x): = \frac{1}{d} g(d x) $$

for all $x \in A$. From Theorem 3.1 of [14] (also see Lemma 3.2 of [9]),

$$d(\Lambda g, \Lambda h) \le L d(g, h) $$

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

$$ \bigl\Vert \bigl(g(x_{1}) - h(x_{1}) , \ldots,g(x_{k}) - h(x_{k}) \bigr)\bigr\Vert _{k} \leq C \varphi\bigl(\overbrace{0,\ldots,0}^{p},\overbrace {x_{1},\ldots,x_{1}}^{d},\ldots,\overbrace{0, \ldots,0}^{p},\overbrace {x_{k},\ldots,x_{k}}^{d} \bigr) $$

for all $x_{1},\ldots,x_{k}\in A$. From our assumption and the last inequality, we have

$$\begin{aligned}& \bigl\Vert \bigl(\Lambda g(x_{1}) - \Lambda h(x_{1}), \ldots,\Lambda g(x_{k})- \Lambda h(x_{k})\bigr)\bigr\Vert _{k} \\& \quad =\frac{1}{d} \bigl\Vert \bigl(g(d x_{1})- h (d x_{1}) ,\ldots,g(d x_{k})- h (d x_{k})\bigr)\bigr\Vert _{k} \\& \quad \leq\frac{C}{d}\varphi \bigl(\overbrace{0,\ldots,0}^{p}, \overbrace{d x_{1},\ldots,d x_{1}}^{d},\ldots, \overbrace{0,\ldots,0}^{p},\overbrace{d x_{k},\ldots,d x_{k} }^{d}\bigr) \\& \quad \leq C L \varphi\bigl(\overbrace{0,\ldots,0}^{p},\overbrace {x_{1},\ldots,x_{1}}^{d},\ldots,\overbrace{0, \ldots,0}^{p},\overbrace {x_{k},\ldots,x_{k}}^{d} \bigr) \end{aligned}$$

for all $x_{1},\ldots,x_{k} \in A$. Thus we have

$$\begin{aligned}& \bigl\Vert \bigl(\Lambda f(x_{1}) - f(x_{1}),\ldots, \Lambda f(x_{k}) - f(x_{k}) \bigr) \bigr\Vert _{k} \\& \quad = \biggl\Vert \biggl(\frac{1}{d}f(d x_{1}) - f(x_{1}),\ldots,\frac{1}{d}f(d x_{k}) - f(x_{k})\biggr)\biggr\Vert _{k} \\& \quad = \frac{1}{d }\bigl\Vert \bigl(f(d x_{1}) - d f(x_{1}),\ldots,f(d x_{k}) - d f(x_{k})\bigr)\bigr\Vert _{k} \\& \quad \leq \frac{1}{2d}\varphi\bigl(\overbrace{0,\ldots,0}^{p}, \overbrace{x_{1},\ldots ,x_{1}}^{d},\ldots, \overbrace{0,\ldots,0}^{p},\overbrace{x_{k}, \ldots,x_{k}}^{d}\bigr) \end{aligned}$$

for all $x_{1},\ldots,x_{k} \in A$. Hence $d(\Lambda f, f) \le \frac{1}{2d}$. From Theorem 1.1, the sequence $\{\Lambda^{n} f\}$ converges to a fixed point H of Λ, i.e., $H:A\rightarrow B$ is a mapping defined by

$$ H(x) = \lim_{n\to\infty}\bigl(\Lambda^{n} f\bigr) (x) = \lim_{n\to\infty}\frac{1}{d^{n}}f\bigl(d^{n} x\bigr) $$

(29)

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

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

i.e., the inequality (27) hold for all $x_{1},\ldots ,x_{k} \in A$. It follows from the definition of H, (21), and (22) that

$$\begin{aligned}& \Biggl\Vert 2H \Biggl(\frac{\sum_{j=1}^{p} \mu x_{1j}}{2}+\sum_{j=1}^{d} \mu y_{1j} \Biggr) - \sum_{j=1}^{p} \mu H(x_{1j})-2 \sum_{j=1}^{d} \mu H(y_{1j}), \\& \qquad \ldots,2H \Biggl(\frac{\sum_{j=1}^{p} \mu x_{kj}}{2}+\sum_{j=1}^{d} \mu y_{kj} \Biggr) - \sum_{j=1}^{p} \mu H(x_{kj})-2 \sum_{j=1}^{d} \mu H(y_{kj}) \Biggr\Vert _{k} \\& \quad = \lim_{n\to\infty} \frac{1}{d^{n}} \Biggl\Vert 2 f \Biggl(d^{n}\frac{\sum_{j=1}^{p} \mu x_{1j}}{2}+d^{n}\sum _{j=1}^{d} \mu y_{1j} \Biggr) - \sum _{j=1}^{p} \mu f\bigl(d^{n} x_{1j}\bigr)-2 \sum_{j=1}^{d} \mu f\bigl(d^{n}y_{1j}\bigr), \\& \qquad \ldots,2 f \Biggl(d^{n}\frac{\sum_{j=1}^{p} \mu x_{kj}}{2}+d^{n}\sum _{j=1}^{d} \mu y_{kj} \Biggr) - \sum_{j=1}^{p} \mu f\bigl(d^{n} x_{kj}\bigr)-2 \sum_{j=1}^{d} \mu f\bigl(d^{n}y_{kj}\bigr) \Biggr\Vert _{k} \\& \quad \leq\lim_{n\to\infty} \frac{1}{d^{n}}\bigl\Vert \bigl(C_{\mu}f\bigl(d^{n} x_{11}, \ldots,d^{n} x_{1p},d^{n} y_{11}, \ldots,d^{n}y_{1d}\bigr), \\& \qquad \ldots, C_{\mu}f\bigl(d^{n} x_{k1}, \ldots,d^{n} x_{kp},d^{n} y_{k1}, \ldots,d^{n}y_{kd}\bigr)\bigr)\bigr\Vert _{k} \\& \qquad {}+ \lim_{n\to\infty}\frac{1}{d^{n}}\varphi \bigl(d^{n} x_{11},\ldots,d^{n} x_{1p},d^{n} y_{11},\ldots,d^{n} y_{1d}, \\& \qquad \ldots,d^{n} x_{k1},\ldots,d^{n} x_{kp},d^{n} y_{k1},\ldots,d^{n} y_{kd}\bigr)=0 \end{aligned}$$

for all $\mu\in\mathbf{T}^{1}$ and $x_{11},\ldots,x_{1p},y_{11},\ldots,y_{1d},\ldots,x_{k1},\ldots ,x_{kp},y_{k1},\ldots,y_{kd} \in A$. Hence we have

$$ 2H \Biggl(\frac{\sum_{j=1}^{p} \mu x_{ij}}{2}+\sum_{j=1}^{d} \mu y_{ij} \Biggr) = \sum_{j=1}^{p} \mu H(x_{ij})+2 \sum_{j=1}^{d} \mu H(y_{ij}) $$

for all $\mu\in\mathbf{T}^{1}$, $x_{11},\ldots,x_{1p},y_{11},\ldots,y_{1d},\ldots,x_{k1},\ldots ,x_{kp},y_{k1},\ldots,y_{kd} \in A$ and $1\leq i \leq k$ and so $H(\lambda x + \mu y)=\lambda H(x) + \mu H(y)$ for all $\lambda, \mu\in\mathbf{T}^{1}$ and all $x,y \in A$. Therefore, by Lemma 3.1, the mapping $H : A \rightarrow B$ is C-linear.

Also it follows from (23) and (24) that

$$\begin{aligned}& \bigl\Vert H\bigl([x_{1}, y_{1}, z_{1}]\bigr)- \bigl[H(x_{1}), H(y_{1}), H(z_{1})\bigr],\ldots,H \bigl([x_{k}, y_{k}, z_{k}]\bigr)- \bigl[H(x_{k}), H(y_{k}), H(z_{k})\bigr]\bigr\Vert _{k} \\& \quad =\lim_{n\to\infty}\frac{1}{d^{3n}} \bigl\Vert f \bigl( \bigl[d^{n} x_{1}, d^{n} y_{1}, d^{n} z_{1}\bigr] \bigr) - \bigl[f\bigl(d^{n} x_{1}\bigr), f\bigl(d^{n} y_{1}\bigr), f \bigl(d^{n} z_{1}\bigr) \bigr], \\& \qquad \ldots, f \bigl(\bigl[d^{n} x_{k}, d^{n} y_{k}, d^{n} z_{k}\bigr] \bigr) - \bigl[f \bigl(d^{n} x_{k}\bigr), f\bigl(d^{n} y_{k}\bigr), f\bigl(d^{n} z_{k}\bigr) \bigr]\bigr\Vert _{k} \\& \quad \leq\lim_{n\to\infty} \frac{1}{d^{3n}}\psi \bigl(d^{n}x_{1},d^{n}y_{1},d^{n}z_{1}, \ldots ,d^{n}x_{k},d^{n}y_{k},d^{n}z_{k} \bigr)=0 \end{aligned}$$

for all $x_{1}, y_{1}, z_{1},\ldots,x_{k},y_{k},z_{k} \in A$. Thus

$$H\bigl([x, y,z]\bigr) = \bigl[H(x), H(y), H(z)\bigr] $$

for all $x, y, z \in A$.Thus $H : A \rightarrow B$ is a homomorphism satisfying (26).

Now, let $T : A \rightarrow B$ be another $C^{*}$-ternary algebras homomorphism satisfying (27). Since $d(f,T)\leq \frac{1}{(1-L)2d}$ and T is C-linear, we get $T\in E'$ and $(\Lambda T)(x)=\frac{1}{d}(T\gamma x)=T(x)$ for all $x\in A$, i.e., T is a fixed point of Λ. Since H is the unique fixed point of $\Lambda\in E'$, we get $H=T$. This completes the proof. □

Theorem 3.7

Let r, s, and θ be non-negative real numbers such that $0< r\neq1$, $0< s\neq3$, and let $d\ge2$. Suppose that $f : A \rightarrow B$ is a mapping with $f(0)=0$ satisfying (18) and

$$\begin{aligned}& \bigl\Vert \bigl( f\bigl([x_{1},y_{1},z_{1}] \bigr)-\bigl[f(x_{1}),f(y_{1}),f(z_{1})\bigr],\ldots ,f\bigl([x_{k},y_{k},z_{k}]\bigr)- \bigl[f(x_{k}),f(y_{k}),f(z_{k})\bigr] \bigr)\bigr\Vert \\& \quad \leq\theta \bigl( \|x_{1}\|^{s}_{A}\cdot \|y_{1}\|^{s}_{A}\cdot\|z_{1} \|^{s}_{A} + \cdots+ \|x_{k}\|^{s}_{A} \cdot\|y_{k}\|^{s}_{A}\cdot\|z_{k} \|^{s}_{A} \bigr) \end{aligned}$$

(30)

for all $\mu\in\mathbf{T}^{1}$ and $x_{1},\ldots,x_{k}, y_{1},\ldots,y_{k}, z_{1},\ldots,z_{k} \in A$. Then there exists a unique $C^{*}$-ternary algebra homomorphism $H : A \rightarrow B$ such that

$$\begin{aligned}& \bigl\Vert \bigl(f(x_{1}) - H(x_{1}), \ldots,f(x_{k}) - H(x_{k}) \bigr)\bigr\Vert _{K} \\& \quad \le\frac{d\theta}{2|d-d^{r}|} \bigl(\|x_{1}\|^{r}_{A} + \cdots+\|x_{k}\| ^{r}_{A} \bigr) \end{aligned}$$

(31)

for all $x_{1},\ldots,x_{k} \in A$.

Proof

We only prove the theorem when $0< r<1$ and $0< s<3$. One can prove the theorem for the other cases in a similar way. The proof follows from Theorem 3.6 by taking

$$\begin{aligned}& \varphi (x_{11},\ldots,x_{1p},y_{11}, \ldots,y_{1d},\ldots,x_{k1},\ldots ,x_{kp},y_{k1}, \ldots,y_{kd}) \\& \quad :=\theta \Biggl(\sum_{j=1}^{p} \|x_{1j}\|^{r}_{A} + \sum _{j=1}^{d}\|y_{1j}\|^{r}_{A} + \cdots+ \sum_{j=1}^{p}\|x_{kj} \|^{r}_{A} + \sum_{j=1}^{d} \|y_{kj}\|^{r}_{A} \Biggr), \\& \psi(x_{1},y_{1},z_{1},\ldots,x_{k},y_{k},z_{k}):= \theta\bigl(\|x_{1}\|^{s}_{A} \cdot \|y_{1}\|^{s}_{A} \cdot \|z_{1} \|^{s}_{A} + \cdots+\|x_{k}\|^{s}_{A} \cdot \|y_{k}\|^{s}_{A} \cdot \|z_{k} \|^{s}_{A}\bigr) \end{aligned}$$

for all $\mu\in\mathbf{T}^{1}$ and $x_{11},\ldots,x_{1p},y_{11},\ldots,y_{1d},\ldots,x_{k1},\ldots ,x_{kp},y_{k1},\ldots,y_{kd},x_{1},\ldots,x_{k},y_{1},\ldots,y_{k}, z_{1}, \ldots,z_{k} \in A$. Then we can choose $L=d^{r-1}$, when $0< r<1$ and $0< s<3$, and $L=2-d^{r-1}$, when $r>1$ and $s>3$, and so we get the desired result. □

Now, assume that A is a unital $C^{*}$-ternary algebra with norm $\| \cdot\|$ and unit e and B is a unital $C^{*}$-ternary algebra with norm $\| \cdot\|$ and unit $e'$.

We investigate homomorphisms in $C^{*}$-ternary algebras associated with the functional equation $C_{\mu}f(x_{1}, \ldots, x_{p}, y_{1}, \ldots, y_{d})=0$.

Theorem 3.8

([5])

Let $r > 1$ (resp., $r<1$) and θ be non-negative real numbers and let $f : A \rightarrow B$ be a bijective mapping satisfying (18) and

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

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

Theorem 3.9

Let $r<1$ and θ be non-negative real numbers and let $f : A \rightarrow B$ be a mapping satisfying (18) and (19). If there exist a real number $\lambda>1$ (resp., $0<\lambda<1$) and an element $x_{0}\in A$ such that $\lim_{n\rightarrow\infty} \frac{1}{\lambda^{n}} f(\lambda^{n} x_{0}) = e'$ (resp., $\lim_{n\rightarrow\infty} \lambda^{n} f(\frac{x_{0}}{\lambda^{n}}) = e'$), then the mapping $f : A \rightarrow B$ is a multi-$C^{*}$-ternary algebra homomorphism.

Proof

By using the proof of Theorem 3.5, there exists a unique multi-$C^{*}$-ternary algebra homomorphism $H : A \rightarrow B$ satisfying (20). It follows from (20) that

$$ H(x)=\lim_{n\rightarrow\infty} \frac{1}{\lambda^{n}} f\bigl( \lambda^{n} x\bigr)\qquad \biggl(\mbox{resp.}, H(x)=\lim _{n\rightarrow\infty} \lambda^{n} f\biggl(\frac{x}{\lambda^{n}}\biggr) \biggr) $$

for all $x\in A$ and $\lambda>1$ ($0<\lambda<1$). Therefore, from our assumption, we get $H(x_{0})=e'$.

Let $\lambda>1$ and $\lim_{n\rightarrow\infty} \frac{1}{\lambda^{n}} f(\lambda^{n} x_{0}) = e'$. It follows from (19) that

$$\begin{aligned}& \bigl\Vert \bigl( \bigl[H(x_{1}),H(y_{1}),H(z_{1}) \bigr] - \bigl[H(x_{1}),H(y_{1}),f(z_{1})\bigr], \\& \qquad \ldots,\bigl[H(x_{k}),H(y_{k}),H(z_{k}) \bigr] - \bigl[H(x_{k}),H(y_{k}),f(z_{k})\bigr] \bigr)\bigr\Vert \\& \quad = \bigl\Vert \bigl( H[x_{1},y_{1},z_{1}] - \bigl[H(x_{1}),H(y_{1}),f(z_{1})\bigr], \\& \qquad \ldots,H[x_{k},y_{k},z_{k}] - \bigl[H(x_{k}),H(y_{k}),f(z_{k})\bigr] \bigr)\bigr\Vert \\& \quad =\lim_{n\to\infty}\frac{1}{\lambda ^{2n}}\bigl\Vert \bigl(f\bigl( \bigl[\lambda^{n}x_{1},\lambda^{n}y_{1},z_{1} \bigr]\bigr) - \bigl[f\bigl(\lambda^{n}x_{1}\bigr),f\bigl( \lambda^{n}y_{1}\bigr),f(z_{1})\bigr], \\& \qquad \ldots,f\bigl(\bigl[\lambda^{n}x_{k},\lambda ^{n}y_{k},z_{k}\bigr]\bigr) - \bigl[f\bigl( \lambda^{n}x_{k}\bigr),f\bigl(\lambda^{n}y_{k} \bigr),f(z_{k})\bigr] \bigr)\bigr\Vert \\& \quad \leq\lim_{n\to\infty}\frac{\lambda^{rn}}{\lambda^{3n}}\theta\bigl( \|x_{1}\|^{r}_{A} \cdot\|y_{1} \|^{r}_{A} \cdot\|z_{1}\|^{r}_{A} + \cdots+ \|x_{k}\|^{r}_{A} \cdot\|y_{k} \| ^{r}_{A} \cdot\|z_{k}\|^{r}_{A} \bigr)= 0 \end{aligned}$$

for all $x_{1},\ldots,x_{k}\in A$. Thus $[H(x),H(y),H(z)]=[H(x), H(y), f(z)]$ 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 $\lim_{n\rightarrow\infty} \lambda^{n} f(\frac{x_{0}}{\lambda^{n}})=e'$.

Similarly, one can show the theorem for the case $\lambda>1$. Therefore, the mapping $f : A \rightarrow B$ is a multi-$C^{*}$-ternary algebra homomorphism. This completes the proof. □

Theorem 3.10

Let $r>1$ and θ be non-negative real numbers and let $f : A \rightarrow B$ be a mapping satisfying (18) and (19). If there exist a real number $\lambda>1$ (resp., $0<\lambda<1$) and an element $x_{0}\in A$ such that $\lim_{n\rightarrow\infty} \frac{1}{\lambda^{n}} f(\lambda^{n} x_{0}) = e'$ (resp., $\lim_{n\rightarrow\infty} \lambda^{n} f(\frac{x_{0}}{\lambda^{n}}) = e'$), then the mapping $f : A \rightarrow B$ is a multi-$C^{*}$-ternary algebra homomorphism.

Proof

The proof is similar to the proof of Theorem 3.9 and we omit it. □

4 Approximation of derivations on multi-$C^{*}$-ternary algebras

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

Park [5] studied approximation of derivations on $C^{*}$-ternary algebras for the functional equation $C_{\mu}f(x_{1}, \ldots, x_{p}, y_{1}, \ldots, y_{d}) =0$ (see also [5, 13, 15–59] and [60]).

For any mapping $f : A \rightarrow A$, let

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

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

Theorem 4.1

([13])

Let r and θ be non-negative real numbers such that $r\notin[1,3]$ and let $f:A \rightarrow A$ be a mapping satisfying (19) and

$$ \bigl\Vert {\mathbf{D}}f(x,y,z)\bigr\Vert \le\theta\bigl(\|x\|^{r} + \|y\|^{r} + \|z\|^{r}\bigr) $$

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

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

for all $x \in A$.

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

Theorem 4.2

Let $(( {A}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ be a multi-$C^{*}$-ternary algebra. Let $f:A\rightarrow A$ be a mapping for which there are the functions $\varphi: A^{(p+d)k} \rightarrow[0, \infty)$ and $\psi : A^{3k} \rightarrow[0, \infty)$ satisfying the inequalities (1), (2), and (4) such that

$$ \bigl\Vert \bigl({\mathbf{D}} f(x_{1}, y_{1}, z_{1}),\ldots,{\mathbf{D}} f(x_{k}, y_{k}, z_{k}) \bigr)\bigr\Vert \le\psi(x_{1}, y_{1},z_{1},\ldots,x_{k},y_{k},z_{k}) $$

(32)

for all $\mu\in\mathbf{T}^{1}$ and $x_{11},\ldots,x_{1p},y_{11},\ldots ,y_{1d},\ldots,x_{k1},\ldots,x_{kp},y_{k1},\ldots,y_{kd}, x_{1},\ldots,x_{k}, y_{1},\ldots,y_{k},z_{1}, \ldots,z_{k}\in A$, where $\gamma=\frac{p+2d}{2}$. If the constant $L<1$ exists such that

$$\begin{aligned}& \varphi\bigl(\overbrace{\gamma x_{1},\ldots,\gamma x_{1}}^{p+d},\overbrace {\gamma x_{2},\ldots,\gamma x_{2} }^{p+d},\ldots,\overbrace{\gamma x_{k}, \ldots,\gamma x_{k}}^{p+d}\bigr) \\& \quad \le\gamma L \varphi\bigl(\overbrace{x_{1},\ldots,x_{1}}^{p+d}, \overbrace {x_{2},\ldots,x_{2}}^{p+d},\ldots, \overbrace{x_{k},\ldots,x_{k}}^{p+d}\bigr) \end{aligned}$$

(33)

for all $x_{1},x_{2},\ldots,x_{k} \in A$, then there exists a unique $C^{*}$-ternary derivation $\delta: A \rightarrow B$ such that

$$\begin{aligned}& \bigl\Vert \bigl(f(x_{1}) - \delta(x_{1}), \ldots,f(x_{k}) - \delta(x_{k}) \bigr) \bigr\Vert _{k} \\& \quad \le\frac{1}{(1-L)2\gamma} \varphi\bigl(\overbrace{x_{1}, \ldots,x_{1}}^{p+d},\overbrace{x_{2},\ldots ,x_{2}}^{p+d},\ldots,\overbrace{x_{k}, \ldots,x_{k}}^{p+d}\bigr) \end{aligned}$$

(34)

for all $x_{1},\ldots,x_{k} \in A$.

Proof

The same reasoning as in the proof of Theorem 3.3, guarantees there exists a unique C-linear mapping $\delta :A\rightarrow A$ satisfying (32). The mapping $\delta:A\rightarrow A$ is given by

$$ \delta(x) = \lim_{n\to\infty}\bigl( \Lambda^{n} f\bigr) (x) = \lim_{n\to\infty}\frac{1}{\gamma^{n}}f \bigl(\gamma^{n} x\bigr) $$

(35)

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

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

i.e., the inequality (6) holds for all $x_{1},\ldots ,x_{k} \in A$. It follows from the definition of δ, (1) and (2), and (35) that

$$\begin{aligned}& \bigl\Vert \bigl(C_{\mu}\delta(x_{11},\ldots,x_{1p}y_{11}, \ldots, y_{1d}),\ldots,C_{\mu}\delta(x_{k1}, \ldots,x_{kp}y_{k1},\ldots, y_{kd}) \bigr) \bigr\Vert _{k} \\& \quad = \lim_{n\to\infty} \frac{1}{\gamma^{n}} \bigl\Vert \bigl(C_{\mu}f\bigl(\gamma^{n} x_{11},\ldots, \gamma^{n} x_{1p},\gamma^{n} y_{11}, \ldots,\gamma^{n} y_{1d}\bigr), \\& \qquad \ldots,C_{\mu}f \bigl(\gamma^{n} x_{k1}, \ldots,\gamma^{n} x_{kp},\gamma^{n} y_{k1},\ldots,\gamma^{n} y_{kd}\bigr) \bigr) \bigr\Vert _{k} \\& \quad \leq\lim_{n\to\infty}\frac{1}{\gamma^{n}}\varphi\bigl( \gamma^{n} x_{11},\ldots,\gamma^{n} x_{1p}.\gamma^{n} y_{11},\ldots, \gamma^{n} y_{1d}, \\& \qquad \ldots,\gamma^{n} x_{k1},\ldots,\gamma^{n} x_{kp},\gamma^{n} y_{k1},\ldots, \gamma^{n} y_{kd}\bigr)=0 \end{aligned}$$

for all $\mu\in\mathbf{T}^{1}$ and $x_{11},\ldots,x_{1p},y_{11},\ldots,y_{1d},\ldots,x_{k1},\ldots ,x_{kp},y_{k1},\ldots,y_{kd} \in A$. Hence we have

$$ 2\delta \Biggl(\frac{\sum_{j=1}^{p} \mu x_{ij}}{2}+\sum_{j=1}^{d} \mu y_{ij} \Biggr) = \sum_{j=1}^{p} \mu\delta(x_{ij})+2 \sum_{j=1}^{d} \mu\delta(y_{ij}) $$

for all $\mu\in\mathbf{T}^{1}$, $x_{11},\ldots,x_{1p},y_{11},\ldots,y_{1d},\ldots,x_{k1},\ldots ,x_{kp},y_{k1},\ldots,y_{kd} \in A$ and $1\leq i \leq k$ and so $\delta(\lambda x + \mu y)=\lambda \delta(x) + \mu \delta(y)$ for all $\lambda, \mu\in\mathbf{T}^{1}$ and $x,y \in A$. Therefore, by Lemma 3.1, the mapping $\delta: A \rightarrow B$ is C-linear.

Also it follows from (4) and (32) that

$$\begin{aligned}& \bigl\Vert \bigl({\mathbf{D}}\delta(x_{1},y_{1},z_{1}), \ldots,{\mathbf{D}}\delta (x_{k},y_{k},z_{k}) \bigr) \bigr\Vert _{k} \\& \quad =\lim_{n\to\infty}\frac{1}{\gamma^{3n}} \bigl\Vert f \bigl({ \mathbf{D}}f\bigl(\gamma^{n} x_{1}, \gamma^{n} y_{1}, \gamma^{n} z_{1}\bigr),\ldots,f\bigl( \gamma^{n} x_{k}, \gamma^{n} y_{k}, \gamma^{n} z_{k}\bigr) \bigr)\bigr\Vert \\& \quad \leq\lim_{n\to\infty} \frac{1}{\gamma^{3n}}\psi\bigl( \gamma^{n}x_{1},\gamma^{n}y_{1},\gamma ^{n}z_{1},\ldots,\gamma^{n}x_{k}, \gamma^{n}y_{k},\gamma^{n}z_{k}\bigr)=0 \end{aligned}$$

for all $x_{1}, y_{1}, z_{1},\ldots,x_{k},y_{k},z_{k} \in A$ and hence

$$\begin{aligned}& \bigl(\delta\bigl([x_{1}, y_{1},z_{1}] \bigr),\ldots,\delta\bigl([x_{k}, y_{k},z_{k}]\bigr) \bigr) \\& \quad {} + \bigl(\bigl[\delta(x_{1}),(y_{1}), (z_{1})\bigr] + \bigl[x_{1},\delta(y_{1}),z_{1} \bigr] + \bigl[x_{1},y_{1},\delta(z_{1})\bigr], \\& \quad \ldots,\bigl[\delta(x_{k}),(y_{k}), (z_{k}) \bigr] + \bigl[x_{k},\delta(y_{k}),z_{k}\bigr] + \bigl[x_{k},y_{k},\delta(z_{k})\bigr] \bigr) \end{aligned}$$

(36)

for all $x, y, z \in A$ and so the mapping $\delta: A \rightarrow A$ is a $C^{*}$-ternary derivation. It follows from (32) and (4) that

$$\begin{aligned}& \bigl\Vert \bigl( \delta[x_{1},y_{1},z_{1}]- \bigl[\delta(x_{1}),y_{1},z_{1}\bigr]- \bigl[x_{1},\delta (y_{1}),z_{1}\bigr]- \bigl[x,y,f(z_{1})\bigr], \\& \qquad \ldots,\delta[x_{k},y_{k},z_{k}]-\bigl[ \delta(x_{k}),y_{k},z_{k}\bigr]- \bigl[x_{k},\delta (y_{k}),z_{k}\bigr]- \bigl[x,y,f(z_{k})\bigr] \bigr)\bigr\Vert \\& \quad =\lim_{n\to\infty}\frac{1}{\gamma^{2n}} \bigl\Vert \bigl(f\bigl[ \gamma^{n} x_{1},\gamma^{n} y_{1},z_{1} \bigr]-\bigl[f\bigl(\gamma^{n} x_{1}\bigr), \gamma^{n} y_{1},z_{1}\bigr] \\& \qquad {}-\bigl[\gamma^{n} x_{1},f\bigl(\gamma^{n} y_{1}\bigr),z_{1}\bigr]-\bigl[\gamma^{n} x_{1},\gamma^{n} y_{1},f(z_{1})\bigr], \\& \qquad \ldots,f\bigl[\gamma^{n} x_{k},\gamma^{n} y_{k},z_{k}\bigr]-\bigl[f\bigl(\gamma^{n} x_{k}\bigr),\gamma^{n} y_{k},z_{k}\bigr] \\& \qquad {}-\bigl[\gamma^{n} x_{k},f\bigl(\gamma^{n} y_{k}\bigr),z_{k}\bigr]-\bigl[\gamma^{n} x_{k},\gamma^{n} y_{k},f(z_{k})\bigr] \bigr)\bigr\Vert \\& \quad \le\lim_{n\rightarrow\infty}\frac{1}{\gamma^{2n}}\psi\bigl(\gamma ^{n}x_{1},\gamma^{n}y_{1},z_{1}, \ldots,\gamma^{n}x_{k},\gamma^{n}y_{k},z_{k} \bigr)=0 \end{aligned}$$

for all $x_{1},y_{1},z_{1},\ldots,x_{k},y_{k},z_{k}\in A$ and so we have

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

(37)

for all $x,y,z\in A$. Hence it follows from (36) and (37) that

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

(38)

for all $x,y,z\in A$. Letting $x=y=f(z)-\delta(z)$ in (38), we get

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

(39)

for all $z_{1},\ldots,z_{k} \in A$ and hence $f(z)=\delta(z)$ for all $z\in A$. Therefore, the mapping $f:A\rightarrow A$ is a $C^{*}$-ternary derivation. This completes the proof. □

Corollary 4.3

Let $r<1$, $s<2$, and θ be non-negative real numbers and let $f:A \rightarrow A$ be a mapping satisfying (18) and

$$\begin{aligned}& \bigl\Vert \bigl({\mathbf{D}} f(x_{1},y_{1},z_{1}), \ldots,{\mathbf {D}}f(x_{k},y_{k},z_{k}) \bigr)\bigr\Vert \\& \quad \le\theta\bigl(\|x_{1}\|^{s}_{A} \cdot \|y_{1}\|^{s}_{A} \cdot\|z_{1} \|^{s}_{A} + \cdots+\|x_{k}\| ^{s}_{A} \cdot \|y_{k}\|^{s}_{A} \cdot\|z_{k} \|^{s}_{A}\bigr) \end{aligned}$$

for all $x_{1}, y_{1}, z_{1},\ldots,x_{k},y_{k},z_{k} \in A$. Then the mapping $f:A \rightarrow A$ is a $C^{*}$-ternary derivation.

Proof

Define

$$\begin{aligned}& \varphi(x_{11},\ldots,x_{1p},y_{11}, \ldots,y_{1d},\ldots,x_{k1},\ldots ,x_{kp},y_{k1}, \ldots,y_{kd}) \\& \quad =\theta \Biggl( \sum_{j=1}^{p} \|x_{1j}\|^{r}_{A} + \sum _{j=1}^{d}\|y_{1j}\|^{r}_{A}, \ldots,\sum_{j=1}^{p}\|x_{kj} \|^{r}_{A} + \sum_{j=1}^{d} \|y_{kj}\|^{r}_{A} \Biggr) \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} &\psi(x_{1},y_{1},z_{1},\ldots,x_{k},y_{k},z_{k}) \\ &\quad =\theta \bigl(\|x_{1}\|^{s}_{A}\cdot \|y_{1}\|^{s}_{A}\cdot\|z_{1} \|^{s}_{A} + \cdots+ \|x_{k}\|^{s}_{A} \cdot\|y_{k}\|^{s}_{A}\cdot\|z_{k} \|^{s}_{A} \bigr) \end{aligned} \end{aligned}$$

for all $x_{1}, y_{1}, z_{1},\ldots,x_{k},y_{k},z_{k} , x_{11},\ldots,x_{1p},y_{11},\ldots,y_{1d},\ldots,x_{k1},\ldots ,x_{kp},y_{k1},\ldots,y_{kd}\in A$ and applying Theorem 4.2, we get the desired result. □

Theorem 4.4

Let $(( {A}^{k},\|\cdot\|_{k}): k\in\mathbf{N})$ be a multi-$C^{*}$-ternary algebra. Let $f: A \rightarrow A$ be a mapping for which there are the functions $\varphi: A^{(p+d)k} \rightarrow[0, \infty)$ and $\psi: A^{3k} \rightarrow[0, \infty)$ satisfying the inequalities (2), (11), (12), and (32) for all $\mu\in\mathbf{T}^{1}$ and $x_{11},\ldots ,x_{1p},y_{11},\ldots, y_{1d},\ldots,x_{k1}, \ldots,x_{kp},y_{k1}, \ldots,y_{kd}, x_{1},\ldots,x_{k},y_{1},\ldots,y_{k},z_{1},\ldots,z_{k}\in A$, where $\gamma=\frac{p+2d}{2}$. If there exists the constant $L<1$ such that

$$\begin{aligned}& \varphi\biggl(\overbrace{\frac{x_{1}}{\gamma} ,\ldots, \frac{x_{1}}{\gamma} }^{p+d},\overbrace{\frac{x_{2}}{\gamma} ,\ldots, \frac{x_{2}}{\gamma} }^{p+d},\ldots,\overbrace{\frac{x_{k}}{\gamma} ,\ldots, \frac{x_{k}}{\gamma } }^{p+d}\biggr) \\& \quad \le\frac{L}{\gamma} \varphi\bigl(\overbrace{x_{1},\ldots ,x_{1}}^{p+d},\overbrace{x_{2}, \ldots,x_{2}}^{p+d},\ldots,\overbrace {x_{k}, \ldots,x_{k}}^{p+d}\bigr) \end{aligned}$$

(40)

for all $x_{1},x_{2},\ldots,x_{k} \in A$, then there exists a unique homomorphism $\delta: A \rightarrow A$ such that

$$\begin{aligned}& \bigl\Vert \bigl(f(x_{1}) - \delta(x_{1}), \ldots,f(x_{k}) - \delta(x_{k})\bigr) \bigr\Vert _{k} \\& \quad \le\frac{1}{(1-L)2\gamma} \varphi\bigl(\overbrace{x_{1}, \ldots,x_{1}}^{p+d},\overbrace{x_{2},\ldots ,x_{2}}^{p+d},\ldots,\overbrace{x_{k}, \ldots,x_{k}}^{p+d}\bigr) \end{aligned}$$

(41)

for all $x_{1},\ldots,x_{k} \in A$.

Proof

The same reasoning as in the proof of Theorem 3.4 guarantees there exists a unique C-linear mapping $\delta :A\rightarrow A$ satisfying (32). The rest of the proof is similar to the proof of Theorem 4.2 and so we omit it. □