Approximation of involution in multi-Banach algebras: Fixed point technique

1.   Introduction

Ulam ^[25] suggested a problem of stability on group homomorphisms in metric groups in 1940. Hyers is the first mathematician who answered the question of Ulam in 1941. He demonstrated the following theorem in ^[12].

Theorem 1.1. Let $M$ and $N$ be two Banach spaces and $f:M\to N$ be a mapping such that

for some $\delta > 0$ and all $m, n\in M$. Then there exists a unique additive mapping $A:M\to N$ such that

for all $m\in M$.

The results of Hyers' stability theorem were developed by other mathematicians. Recently, numerous consequences concerning the stability of various functional equations in different normed spaces and various control functions have been obtained.

The problem of stability of some functional equations have been widely explored by direct methods and there are numerous exciting outcomes regarding this problem ^{[1,2,9,13,14,15,16,19,21,23,26]}. The fixed point approach has been used to study the Hyers-Ulam stability investigations. The relationship between Hyers-Ulam stability and fixed point theory has been defined in ^{[3,4,20,22,24]}.

Definition 1.2. ^[22] A function $d:M\times M \to [0, \infty]$ is named a generalized metric on the set $M$ if $d$ satisfies the following conditions

$(a)$ for each $m, n\in M$, $d(m, n) = 0$ if and only if $m = n$;

$(b)$ for all $m, n\in M$, $d(m, n) = d(n, m)$;

$(c)$ for all $m, n, l\in M$, $d(m, l)\leq d(m, n)+d(n, l)$.

Notice that the just generalized metric significant difference from the metric is that the generalized metric range contains the infinity.

Theorem 1.3. ^[22] et $(M, d)$ be a complete generalized metric space and $J:M\to M$ be a contractive mapping with Lipschitz constant $L < 1$. Then for every $m\in M$, either

for all nonnegative integers $s$ or there exists an integer $s_0\ge 0$ so that

$(a)$ For all $s\geq s_0$, $d(J^s m, J^{s+1}m) < \infty$;

$(b) \; J^s (m)\to n^*$, where $n^*$ is a fixed point of $J$;

$(c) \; n^*$ is the unique fixed point of $J$ in the set $N = \{n\in M\; :\; d(J^{s_0}(m), n) < \infty\}$;

$(d)$ For each $n\in N$, $d(n^*, n)\leq \frac{1}{1-L}d(Jn, n)$.

Dales and Polyakov ^[8] introduced the concept of multi-normed spaces. We have collected some properties of multi-normed spaces which will be used in this article. We refer readers to ^[8,17,18] for more details.

2.   Multi-Banach algebras

Suppose that $(A, \|.\|)$ is a complex normed space and $k\in\mathbb{N}$. The linear space $A\oplus\cdots\oplus A$ consists of $k$-tuples $(x_{1}, \cdots, x_{k})$ denote by $A^k$, where $x_1, \cdots, x_k\in A$. The linear operations on $A^{k}$ are defined coordinate-wise. The zero element of either $A$ or $A^{k}$ is denoted by 0. The set $\{1, 2, \cdots, k\}$ is indicated by $\mathbb{N}_k$ and we denote by $\Sigma_{k}$ the group of permutations on $k$ symbols.

Definition 2.1. ^[8] A multi-norm on $\{ {A}^{k}: k\in\mathbb{N}\}$ is a sequence

such that $\|\cdot\|_{k}$ is a norm on $A^k$ for every $k\in\mathbb{N}$ with $\| \cdot\|_1 = \| \cdot\|$:

$(M1) \; \|(x_{\sigma(1)}, \ldots, x_{\sigma(k)})\|_{k} = \|(x_{1}, \ldots, x_{k})\|_{k}$ for every $\sigma\in\Sigma_{k}$ and $x_1, \cdots, x_k\in A$.

$(M2) \; \|(\alpha_{1}x_{1}, \ldots, \alpha_{k}x_{k})\|_{k}\leq (\max_{i\in{\mathbb{N}}_{k}}|\alpha_{i}|) \|(x_{1}, \cdots, x_{k})\| _{k}$ for all $\alpha_{1}, \cdots, \alpha_{k} \in\mathbb{C}$ and $x_{1}, \cdots, x_{k}\in A.$

$(M3)$ $\|(x_{1}, \cdots, x_{k-1}, 0)\|_{k} = \|(x_{1}, \ldots, x_{k-1})\|_{k-1}$ for all $x_1, \cdots, x_{k-1}\in A.$

$(M4)$ $\|(x_{1}, \cdots, x_{k-1}, x_{k-1})\|_{k} = \|(x_{1}, \cdots, x_{k-1})\| _{k-1}$ for all $x_1, \ldots, x_{k-1}\in A.$

With these properties, we say that $((A^{k}, \|\cdot\|_{k}): k\in\mathbb{N})$ is a multi-normed space.

Lemma 2.2. ^[18] Assume that $((A^{k}, \|\cdot\|_{k}): k\in\mathbb{N})$ is a multi-normed space and let $k\in\mathbb{N}$. Then

$(i) \; \|(x, \ldots, x)\|_{k} = \|x\|$ for all $x\in A$.

$(ii) \; \max_{i\in\mathbb{N}_{k}}\|x_{i}\|\leq \|(x_{1}, \cdots, x_{k})\|_{k}\leq\sum_{i = 1}^{k}\|x_{i}\|\leq k \max_{i\in {\mathbb{N}}_{k}}\|x_{i}\|$

for all $x_1, \cdots, x_k\in A$.

By the second part of the above lemma, we conclude that, if $(A, \|\cdot\|)$ is a Banach space, then $(A^{k}, \|\cdot\|_{k})$ is a Banach space for every $k\in\mathbb{N}$. In this case, $((A^{k}, \|\cdot\|_{k}): k\in\mathbb{N})$ is a multi-Banach space.

Example 2.3. ^[8] We define the sequence $((A^{k}, \|\cdot\|_{k}): k\in\mathbb{N})$ on $A^k$ as follows:

This is a minimum multi-norm on $A^k$.

Definition 2.4. ^[8,17] Suppose that $(A, \|\cdot\|)$ is a normed algebra such that $((A^{k}, \|\cdot\|_{k}): k\in\mathbb{N})$ is a multi-normed space. Then $((A^{k}, \|\cdot\|_{k}): k\in\mathbb{N})$ is called a multi-normed algebra if

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

It is obvious that every Banach algebra is a multi-Banach algebra with minimum multi-norm.

A mapping $^* : A\to A$, denoted by

is an involution on $A$ if

(1) $x^{**} = x$,

(2) $(\lambda x+\mu y)^* = \overline{\lambda}x^*+\overline{\mu}y^*$, for all $x, y\in A$ and $\lambda, \mu\in\mathbb{C}$,

(3)$(xy)^* = y^* x^*$, for all $x, y\in A$.

Definition 2.5. Assume that $((A^{k}, \|\cdot\|_{k}): k\in\mathbb{N})$ is a multi-Banach algebra. A multi-$C^*$-algebra is a complex multi-Banach algebra $((A^{k}, \|\cdot\|_{k}): k\in\mathbb{N})$ with an involution $^*$ satisfying

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

3.   Main results

In all sections of this article, wherever needed, we assume that

for $n_0\in \mathbb{N}.$ It is obvious that $\mathbb{S}^1 = \mathbb{S}_{\frac{1}{1}}^1$. Furthermore, suppose $\left((A^{k}, \|.\|_k):k\in\mathbb{N}\right)$ is a multi-Banach algebra. Let $I: = \frac{2+\alpha}{\alpha}$ for real $\alpha$ greater than or equal to 2. When $A$ is a Banach algebra, for a mapping $f: A\to A$, we define

for all $x, y, z\in A$ and $\mu, \nu, \eta\in \mathbb{S}_{\frac{1}{n_0}}^1$.

An algebra $A$ is called $C^*$-algebra if it is a Banach algebra with an involution $^*$ such that $\|xx^*\| = \|x\|^2$. We suggest ^[6] as a reference for stability in Banach algebras and $C^*$-algebras.

Lemma 3.1. Let $I$ be a positive real number and $\varphi: A^{3k}\to [0, \infty)$ be a function such that there exists $L < 1$ satisfying

for all $x_1, x_2, \cdots, x_k, y_1, y_2, \cdots, y_k, z_1, z_2, \cdots, z_k\in A$. Then

Proof. Putting $x_i = Ix_i$, $y_i = Iy_i$ and $z_i = Iz_i$ for $1\leq i\leq k$ in (3.1), we have

Replacing $x_i, y_i, z_i$ by $Ix_i, Iy_i, Iz_i$, respectively, in the above inequality, we get

By induction, we obtain

Since $L < 1$, we get

as desired.

Theorem 3.2. Let $((A^n, \|.\|_n):n\in\mathbb{N})$ be a multi-Banach algebra and $f : A\to A$ be a mapping and $\varphi:A^{3k}\to [0, \infty)$ be a function satisfying (3.1) and

for all $x, x_1, y_1, z_1, \cdots, x_k, y_k, z_k\in A$ and $\mu, \eta, \nu\in \mathbb{S}_{\frac{1}{n_0}}^{1}$. Then there exists a unique involution mapping $F:A\to A$ which satisfies

Furthermore, $(a)$ if

for all $x_1, x_2, \ldots, x_n\in A$, then the involution mapping $F:A\to A$ is continuous;

$(b)$ if

for all $x_1, x_2, \ldots, x_k\in A$, then $A$ is a $C^*$-algebra with involution $x^* = F(x)$ for all $x\in A$.

Proof. Let $(\Delta, d)$ be a generalized metric space given by the following definition:

for all $x_1, x_2, \ldots x_k\in A$. We can show that the metric space $(\Delta, d)$ is complete (see [5, Theorem 2.5]). Next, we define an operator

for all $x\in A$. We claim that the operator $\Upsilon$ is strictly contractive on $\Delta$. For this, assume that $d(g, h) = \kappa$ for any $g, h\in \Delta$, for some $\kappa\in \mathbb{R}$, we have

for all $x_1, x_2, \cdots, x_k\in A$. Replacing $x_1, x_2, \cdots, x_k$ by $Ix_1, Ix_2, \cdots, Ix_k$, respectively, in the above inequality, we get

and by (3.1), we obtain

So $d(\Upsilon g, \Upsilon h)\leq \kappa L$. Hence we get

Then $\Upsilon$ is strictly contactive on $\Delta$ by Lipschitz constant $L < 1$.

Letting $\mu = \eta = \nu = 1$ and $y_i = z_i: = x_i$ for $1\leq i\leq k$ in (3.2), we obtain

for all $x_1, x_2, \cdots, x_k\in A$. Note that

By (3.10), we have

for all $x_1, x_2, \ldots, x_k\in A$. Thus we get

for all $x_1, x_2, \ldots, x_k\in A$. It follows from (3.9) that

for all $x_1, x_2, \ldots, x_k\in A$, and by (3.8), we get

The conditions of the fixed point theorem are estabilished, so there exists a mapping $F:A\to A$ such that $F(0) = 0$ and the following conditions hold.

(ⅰ) $F$ is a fixed point of $\Upsilon$. This means that

and thus

for all $x\in A$. Moreover, the mapping $F$ is a unique fixed point in the set

Note that $n_0 = 0$ in Theorem 1.3. From (3.8), there is $c\in[0, \infty)$ satisfying

(ⅱ) The sequence $\{\Upsilon^{n}f\}$ converges to $F$. This implies that

(ⅲ) We obtain

and so (3.5) holds. By Lemma 3.1, (3.1), (3.2) and (3.11), we have

So we obtain

Replacing $x_1, x_2, \ldots, x_k$ and $y_1, y_2, \cdots, y_k$ and $z_1, z_2, \cdots, z_k$ with $x, y, z$, respectively, we have

and thus

and so $F$ is an additive mapping, that is,

for all $x, y\in A$ (see ^[11]).

We will show that $F(xy) = F(y)F(x)$ for all $x, y\in A$. Substituting $x_1, x_2, \cdots., x_k$ with $I^{n}x_1, I^{n}x_2, \ldots, I^{n}x_k$ and $y_1, y_2, \cdots, y_k$ with $I^ny_1, I^ny_2, \cdots, I^ny_k$ in (3.3) and dividing on $I^{-2n}$, we obtain

Letting $n\to\infty$, we have

Replacing $x_1, x_2, \cdots, x_k$ and $y_1, y_2, \cdots, y_k$ with $x, y$, respectively in the above inequality, we get

for all $x, y\in A$. We will show that $F(\mu x) = \overline{\mu}F(x)$ for all $x\in A$ and all $\mu\in\mathbb{S}_{\frac{1}{n_0}}^{1}$.

The method applied in ^[10] is used to continue the proof. Substituting $x_i = y_i = z_i: = I^{n}x_i$ for every $1\leq i\leq k$ and $\eta = \nu: = \mu$, we obtain

In particular, if $\mu = 1$, then

We have

for all $x_1, x_2, \ldots, x_k\in A$. This implies that

for all $x_1, x_2, \cdots, x_k\in A$. As $n\to\infty$, we have

by Lemma 3.1. Therefore,

Letting $x_1 = x_2 = \cdots = x_k = x$, we get

for all $x\in A$ and $\mu\in\mathbb{S}_{\frac{1}{n_0}}^{1}$. Now, we will show that the above equality holds for all $\mu\in \mathbb{C}$. To prove this, we can consider the following statements:

(1) If $\mu\in \mathbb{S}^{1}$ then there exists $\theta\in [0, 2\pi]$ such that $\mu = e^{i\theta}$. Putting $\mu_1 = e^{i\frac{\theta}{n_0}}\in \mathbb{S}_{\frac{1}{n_0}}^{1}$, by (3.15), we get

for all $x\in A$.

(2) If $\mu\in n\mathbb{S}^1 = \{nz; \; \; z\in\mathbb{S}^1\}$ for some $n\in\mathbb{N}$, then by additivity of $F$, we have

(3) Let $t\in (0, \infty)$. Then there exists a positive integer $n$ such that the point $(t, 0)$ lies in the interior of circle with center at origin and radius $n$. Putting $t_1: = t+i\sqrt{n^2 -t^2}$ and $t_2: = t-i\sqrt{n^2 -t^2}$, we have $t = \frac{t_1+t_2}{2}$ and $t_1, t_2\in n\mathbb{S}^{1}$. It follows that

for all $x\in A$. On the other hand, for any $\lambda \in {\mathbb{C}}$, there is $\theta\in[0, 2\pi]$ such that $\lambda = |\lambda|e^{i\theta}$. It follows that

for all $x\in A$. Hence, for any case, we get

for all $\lambda\in \mathbb{C}$. Using the assumption (3.4), we have

and thus

Hence $F(x)$ is an involution for $A$, since

Therefore $A$ is a Banach algebra with an involution $F(x)$. To prove (a) of Theorem 3.2, replacing $x_1, x_2, \cdots, x_k$ by $I^nx_1, I^nx_2, \cdots, I^nx_k$ in (3.6) and dividing both sides on $I^n$, we have

Letting $n\to\infty$, we have

Replacing $x_1, x_2, \cdots, x_k$ by $x$, we get

This implies that the involution $F(x)$ is isometric. Let $\{x_n\}$ be an arbitrary sequence in $A$ such that $x_n\to x$. Then

which indicates that the involution $F(x)$ is continuous. To show that $A$ is a $C^*$-algebra, we must prove that

For this purpose, we use the assumption (3.7) and the equality (3.12). If

for all $x_1, x_2, \cdots, x_k\in A$, then by replacing $x_1, x_2, \cdots, x_k$ by $I^{n}x_1, I^{n}x_1, \cdots, I^{n}x_k$ and dividing both sides of the above inequality by $I^{-2n}$, we obtain

Thus

Letting $n\to\infty$, we get

Therefore,

Replacing $x_1, x_2, \cdots, x_k$ by $x$ in last inequality, we have

for all $x\in A$. Then $A$ is a $C^*$-algebra with $F(x)$ as an involution.

Corollary 3.3. Let $f:A\to A$ be a mapping such that

for all $x, x_1, x_2, \cdots, x_k, y_1, y_2, \cdots, y_k, z_1, z_2, \cdots, z_k\in A$ and $\mu, \eta, \nu\in \mathbb{S}_{\frac{1}{n_0}}^{1}$, $p\in [0, 1)$ and $\epsilon\in [0, \infty)$. Then there is a unique involution mapping $F:A\to A$ such that

Furthermore, $(a)$ if

for all $x_1, x_2, \cdots, x_n\in A$, then $A = A_{sa}$, which means that for every $x\in A$, we have $x = F(x); \; (b)$ if

for all $x_1, x_2, \cdots, x_k\in A$, then $A$ is a $C^*$-algebra with involution $x^* = F(x)$ for all $x\in A$.

Proof. By taking

in Theorem 3.2, this corollary is proved. Note that

Furthermore, putting $I^{n}x_i$ by $x_i$ for all $1\leq i\leq k$ in (3.18), we get

Dividing both sides of the above inequality by $I^n$, we obtain

Letting $n\to\infty$, we have

Replacing $x_1, x_2, \cdots, x_k$ by $x$ in the above equality, we obtain

for all $x\in A$. The rest of the proof is similar to the previous one.

Theorem 3.4. Suppose that $f:A\to A$ is a mapping which satisfies (3.2), (3.3) and (3.4). Let $\psi: A^{3k}\to [0, \infty)$ be a function such that there is $L < 1$ satisfying

for all $x_1, x_2, \cdots, x_k, y_1, y_2, \cdots, y_k, z_1, z_2, \cdots, z_k\in A$ and $\mu, \eta, \nu\in \mathbb{S}_{\frac{1}{n_0}}^{1}$. Then there exists a unique involution mapping $F:A\to A$ which satisfies

Furthermore, $(a)$ if

for all $x_1, x_2, \cdots, x_n\in A$, then $A = A_{sa}$, that is, for every $x\in A$, we have $x = F(x)$; $(b)$ if

for all $x_1, x_2, \cdots, x_n\in A$, then $A$ is $C^*$-algebra with $x^* = F(x)$ as an involution.

Proof. The proof of this theorem is the same as in the previous one. We prove some of the parts and refer the rest on to the readers.

The linear mapping $\Upsilon:\Delta\to \Delta$ is defined by

for all $x\in A$. It is easy to show that the operator $\Upsilon$ is strictly contractive. Putting $\mu = \eta = \nu = 1$ and replacing $x_i, y_i, z_i$ by $I^{-i}x_i$ in (3.2) for all $1\leq i\leq k$, we get

Therefore,

for all $x_1, x_2, \ldots, x_k\in A.$ Hence

for all $x_1, x_2, \ldots, x_k\in A.$ We obtain

There exists a mapping $F:A\to A$ with $F(0) = 0$, which is a fixed point of $\Upsilon$ and $\Upsilon^{n}f\to F$, that is,

The remainder is similar to the proof of Theorem 3.2.

Theorem 3.5. Let $A$ be a Banach algebra and $\left(\left(A^{k}\|\cdot\|_{k}\right): k \in \mathbb{N}\right)$ be a multi-Banach algebra. Let $p\in\mathbb{N}$ with $p\geq 2$ and assume that there are $0\leq L \leq 1$ and functions $\Gamma, \varphi: A^{2 k} \longrightarrow[0, \infty)$ satisfying

for all $x_{1}, \cdots, x_{k}, y_{1}, \cdots, y_{k} \in A.$ Assume that $f: A \rightarrow A$ is a mapping satisfying $f(0) = 0$ and

for all $\mu = 1, i$ and $\nu = 1, i$ and for all $x, x_{1}, x_{2}, \cdots, x_{k}$, $y_{1}, y_{2}, \ldots, y_{k} \in A.$ Suppose that for any fixed $x\in A$ the function $t\longmapsto f(tx)$ is continuous on $\mathbb{R}$. Then there is a unique involution $T:A\to A$ such that

where

for all $x_{1}, x_{2}, \cdots, x_{k}\in A$.

Furthermore, if

for all $x_{1}, x_{2}, \ldots, x_{k}\in A$, then $A$ is a $C^*$-algebra with involution $x^* = T(x)$ for all $x\in A$. Also, if

for all $x_{1}, x_{2}, \cdots, x_{k}\in A$, then the involution mapping $T:A\to A$ is continuous.

Proof. Putting $\mu = \nu = 1$ and $y_{1} = x_{1}, y_{2} = x_{2}, \cdots, y_{k} = x_{k}$ in (3.19), we have

for all $x_{1}, x_{2}, \cdots, x_{k}\in A$. By using induction, we get

Replacing $x_j$ by $\frac{x_j}{p}$ in the above inequality for $1\leq j\leq k$, we get

Set $\Delta: = \{g: A \rightarrow A, g(0) = 0\}$ and define

for all $x_{1}, x_{2}, \cdots, x_{k}\in A$. We can easily show that $(\Delta, d)$ is a complete metric space. We assert that the mapping $J: \Delta\to \Delta$ defined by $Jg(x) = pg(\frac{x}{p})$ is a strictly contractive mapping. Let $g, h\in \Delta$ with $d(g, h) < \infty$ and $d(g, h) < c$. Thus we have

for all $x_{1}, x_{2}, \cdots, x_{k}\in A$. Therefore,

Thus $d(J g, J h) \leq c L$ and so $d(J g, J h) \leq L d(g, h)$ for all $g, h\in\Delta$. Hence $J$ is a strictly contractive mapping with Lipschitz constant $L$. By (3.22), we have

The conditions of the fixed point theorem are satisfied. There exists $n_0\in \mathbb{N}$ such that the sequence $\{J^n f\}$ converges to a fixed point $T$ of $J$ and therefore $T\left(\frac{x}{p}\right) = \frac{1}{p}T(x)$. Also $T$ is a unique fixed point of $J$ in the set $\chi = \left\{g \in \Delta: d\left(J^{n_0} f, g\right) < \infty\right\}$ and

Since $\lim_{n \rightarrow \infty}d\left(J^{n} f, T\right) = 0$, we have

From (3.23), we have $d\left(f, T\right)\leq \frac{1}{1-L}$. Hence

Replacing $x_{1}, x_{2}, \cdots, x_{k}$ by $\frac{x}{p^{n}}$ and $y_{1}, y_{2}, \cdots, y_{k}$ by $\frac{y}{p^{n}}$ and multiplying both sides by $p^{n}$, we get

Letting $n\to\infty$, we obtain

Therefore,

Replacing $x_{1}, x_{2}, \cdots, x_{k}$ by $x$ and $y_{1}, y_{2}, \cdots, y_k$ by $y$, we have

for all $x, y\in A$ and $\mu, \nu = 1, i$.

Next, putting $x_{1}: = \frac{x_{1}}{p^{n}}, \cdots, x_{k}: = \frac{x_{k}}{p^{n}}$ and $y_{1}: = \frac{y_{1}}{p^{n}}, \cdots, y_{k}: = \frac{y_{k}}{p^{n}}$ in (3.20), we obtain

Multiplying both sides in $p^{2n}$, we get

Letting $n\to\infty$ in the above inequality, we have

Therefore, we get

for all $x, y\in A$. We must show that $T$ is linear. For this, we can use the method of ^[7]. Fix $x_0\in A$ and $\rho\in A^{\prime}$(dual space of A). We define the mapping $\Phi; \mathbb{R}\to\mathbb{R}$ as follows:

So

$(i) \; \Phi$ is an additive mapping, i.e., for any $a, b\in\mathbb{R}$, we have

It is easy to show that $\Phi$ is additive by (3.24).

$(ii) \; \Phi$ is a Borel function. Put $\Phi_{n}(t) = p^{n}\rho(f(p^{-n}tx_0))$. Then $\Phi_{n}(t)$ are continuous, and $\Phi(t)$ is the pointwise limit of $\Phi_{n}(t)$ and thus $\Phi(t)$ is a Borel function.

Also, we know that if $\Phi:\mathbb{R}^{n}\to \mathbb{R}^{n}$ is a function such that $\Phi$ is additive and measurable, then $\Phi$ is continuous. Note that, if we replace $\mathbb{R}^{n}$ by any separable, locally compact abelian group and this statement is true. Therefore $\Phi(t)$ is a continuous function. Let $a\in\mathbb{R}$. Then $a = \lim_{n\to \infty}r_{n}$, where $\{r_n\}$ is a sequence of rational numbers. Hence

Thus $\Phi$ is $\mathbb{R}$-linear and therefore $T(ax) = aT(x)$ for all $a\in\mathbb{R}$. For $\varrho = a_1+ia_2\in\mathbb{C}$, where $a_1, a_2\in\mathbb{R}$, we have

Then $T$ is a $\mathbb{C}$-linear mapping. Using the assumption (3.21), we have

Now, we can say that $T(x)$ is an involution for $A$ (see (3.24), (3.25), ($\ref{123}$) and (3.26)).

4.   Conclusion

In this research work, we demonstrated the stability for Cauchy-Jensen functional equation in multi-Banach algebra by using the fixed point technique. In fact, we proved that for a function which is approximately Cauchy-Jensen in multi-Banach algebra, there is a unique involution near it. Next, we showed that under some conditions the involution is continuous, the multi-Banach algebra becomes multi-$C^*$-algebra and the Banach algebra is self-adjoint.

Acknowledgments

We would like to express our sincere gratitude to the anonymous referee for his/her helpful comments that will help to improve the quality of the manuscript.

Conflict of interest

The authors equally conceived of the study, participated in its design and coordination, drafted the manuscript, participated in the sequence alignment, and read and approved the final manuscript.

All authors declare no conflicts of interest in this paper.