Approximation of linear mappings in Banach modules over $C^{\ast }$-algebras

Abstract

Let X, Y be Banach modules over a $C^{\ast }$-algebra and let $r_1,\dots ,r_n\in \mathbb{R}$ be given. Using fixed-point methods, we prove the stability of the following functional equation in Banach modules over a unital $C^{\ast }$-algebra:

$$\sum _{j=1}^{n}f(\frac{1}{2}\sum _{1\le i\le n,i\ne j}{r}_i{x}_i-\frac{1}{2}{r}_j{x}_j)+\sum _{i=1}^n{r}_{i}f(x_i)=nf\left(\frac{1}{2}\sum _{i=1}^n{r}_i{x}_i\right).$$

As an application, we investigate homomorphisms in unital $C^{\ast }$-algebras.

MSC:39B72, 46L05, 47H10, 46B03, 47B48.

1 Introduction and preliminaries

We say a functional equation $(\zeta )$ is stable if any function g satisfying the equation $(\zeta )$ approximately is near to the true solution of $(\zeta )$. We say that a functional equation is superstable if every approximate solution is an exact solution of it (see [1]). The stability problem of functional equations was originated from a question of Ulam [2] concerning the stability of group homomorphisms. Hyers [3] gave a first affirmative partial answer to the question of Ulam in Banach spaces. Hyers’ theorem was generalized by Aoki [4] for additive mappings and by T.M. Rassias [5] for linear mappings by considering an unbounded Cauchy difference. A generalization of the T.M. Rassias theorem was obtained by Găvruta [6] by replacing the unbounded Cauchy difference by a general control function in the spirit of T.M. Rassias’ approach.

The functional equation

$$f(x+y)+f(x-y)=2f(x)+2f(y)$$

is called a quadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic mapping. A Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof [7] for mappings $f:X\to Y$, where X is a normed space and Y is a Banach space. Cholewa [8] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group. Czerwik [9] proved the Hyers-Ulam stability of the quadratic functional equation. J.M. Rassias [10, 11] introduced and investigated the stability problem of Ulam for the Euler-Lagrange quadratic functional equation

$$f(a_1{x}_1+a_2{x}_2)+f(a_2{x}_1-a_1{x}_2)=(a_1^2+a_2^2)[f(x_1)+f(x_2)].$$

(1.1)

Grabiec [12] has generalized these results mentioned above.

The stability problems of several functional equations have been extensively investigated by a number of authors and there are many interesting results concerning this problem (see [13–43]).

Let X be a set. A function $d:X\times X\to [0,\mathrm{\infty }]$ 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 $x,y,z\in X$.

We recall a fundamental result in fixed-point theory.

Theorem 1.1 [44, 45]

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

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

for all nonnegative integers n or there exists a positive 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 $Y=\{y\in X\mid d(J^{n_0}x,y)<\mathrm{\infty }\}$;

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

In 1996, Isac and T.M. Rassias [46] were the first to provide applications of stability theory of functional equations for the proof of new fixed-point theorems with applications. By using fixed-point methods, the stability problems of several functional equations have been extensively investigated by a number of authors (see [47–58]).

Recently, Park and Park [59] introduced and investigated the following additive functional equation of Euler-Lagrange type:

(1.2)

whose solution is said to be a generalized additive mapping of Euler-Lagrange type.

In this paper, we introduce the following additive functional equation of Euler-Lagrange type which is somewhat different from (1.2):

$$\sum _{j=1}^{n}f(\frac{1}{2}\sum _{1\le i\le n,i\ne j}{r}_i{x}_i-\frac{1}{2}{r}_j{x}_j)+\sum _{i=1}^n{r}_{i}f(x_i)=nf\left(\frac{1}{2}\sum _{i=1}^n{r}_i{x}_i\right),$$

(1.3)

where $r_1,\dots ,r_n\in \mathbb{R}$. Every solution of the functional equation (1.3) is said to be a generalized Euler-Lagrange type additive mapping.

Using fixed-point methods, we investigate the Hyers-Ulam stability of the functional equation (1.3) in Banach modules over a $C^{\ast }$-algebra. These results are applied to investigate $C^{\ast }$-algebra homomorphisms in unital $C^{\ast }$-algebras. Also, ones can get the super-stability results after all theorems by putting the product of powers of norms as the control functions (see for more details [60, 61]).

Throughout this paper, assume that A is a unital $C^{\ast }$-algebra with the norm ${\parallel \cdot \parallel }_A$ and the unit e, B is a unital $C^{\ast }$-algebra with the norm ${\parallel \cdot \parallel }_B$, and X, Y are left Banach modules over a unital $C^{\ast }$-algebra A with the norms ${\parallel \cdot \parallel }_X$ and ${\parallel \cdot \parallel }_Y$, respectively. Let $U(A)$ be the group of unitary elements in A and let $r_1,\dots ,r_n\in \mathbb{R}$.

2 Hyers-Ulam stability of the functional equation (1.3) in Banach modules over a $C^{\ast }$-algebra

For any given mapping $f:X\to Y$, $u\in U(A)$ and $\mu \in \mathbb{C}$, we define $D_{u,r_1,\dots ,r_n}f$ and $D_{\mu ,r_1,\dots ,r_n}f:X^n\to Y$ by

$$\begin{array}{r}{D}_{u,r_1,\dots ,r_n}f(x_1,\dots ,x_n)\\ :=\sum _{j=1}^{n}f(\frac{1}{2}\sum _{1\le i\le n,i\ne j}{r}_{i}u{x}_i-\frac{1}{2}{r}_{j}u{x}_j)+\sum _{i=1}^n{r}_{i}uf(x_i)-nf\left(\frac{1}{2}\sum _{i=1}^n{r}_{i}u{x}_i\right)\end{array}$$

and

$$\begin{array}{r}{D}_{\mu ,r_1,\dots ,r_n}f(x_1,\dots ,x_n)\\ :=\sum _{j=1}^{n}f(\frac{1}{2}\sum _{1\le i\le n,i\ne j}\mu r_i{x}_i-\frac{1}{2}\mu r_j{x}_j)+\sum _{i=1}^n\mu r_{i}f(x_i)-nf\left(\frac{1}{2}\sum _{i=1}^n\mu r_i{x}_i\right)\end{array}$$

for all $x_1,\dots ,x_n\in X$.

Lemma 2.1 Let X and Y be linear spaces and let $r_1,\dots ,r_n$ be real numbers with ${\sum }_{k=1}^n{r}_k\ne 0$ and $r_i\ne 0$, $r_j\ne 0$ for some $1\le i<j\le n$. Assume that a mapping $L:X\to Y$ satisfies the functional equation (1.3) for all $x_1,\dots ,x_n\in X$. Then the mapping L is additive. Moreover, $L(r_{k}x)=r_{k}L(x)$ for all $x\in X$ and $1\le k\le n$.

Proof One can find a complete proof at [62]. □

Lemma 2.2 Let X and Y be linear spaces and let $r_1,\dots ,r_n$ be real numbers with $r_i\ne 0$, $r_j\ne 0$ for some $1\le i<j\le n$. Assume that a mapping $L:X\to Y$ with $L(0)=0$ satisfies the functional equation (1.3) for all $x_1,\dots ,x_n\in X$. Then the mapping L is additive. Moreover, $L(r_{k}x)=r_{k}L(x)$ for all $x\in X$ and $1\le k\le n$.

Proof One can find a complete proof at [62]. □

We investigate the Hyers-Ulam stability of a generalized Euler-Lagrange type additive mapping in Banach modules over a unital $C^{\ast }$-algebra. Throughout this paper, let $r_1,\dots ,r_n$ be real numbers such that $r_i\ne 0$, $r_j\ne 0$ for fixed $1\le i<j\le n$.

Theorem 2.3 Let $f:X\to Y$ be a mapping satisfying $f(0)=0$ for which there is a function $\phi :X^n\to [0,\mathrm{\infty })$ such that

$${\parallel D_{e,r_1,\dots ,r_n}f(x_1,\dots ,x_n)\parallel }_Y\le \phi (x_1,\dots ,x_n)$$

(2.1)

for all $x_1,\dots ,x_n\in X$. Let

$${\phi }_{ij}(x,y):=\phi (0,\dots ,0,\underset{i\mathrm{th}}{\underbrace{x}},0,\dots ,0,\underset{j\mathrm{th}}{\underbrace{y}},0,\dots ,0)$$

for all $x,y\in X$ and $1\le i<j\le n$. If there exists $0<C<1$ such that

$$\phi (2x_1,\dots ,2x_n)\le 2C\phi (x_1,\dots ,x_n)$$

for all $x_1,\dots ,x_n\in X$, then there exists a unique generalized Euler-Lagrange type additive mapping $L:X\to Y$ such that

$$\begin{array}{rl}{\parallel f(x)-L(x)\parallel }_Y\le & \frac{1}{4-4C}\{{\phi }_{ij}(\frac{2x}{r_i},\frac{2x}{r_j})+2{\phi }_{ij}(\frac{x}{r_i},-\frac{x}{r_j})\\ +{\phi }_{ij}(\frac{2x}{r_i},0)+2{\phi }_{ij}(\frac{x}{r_i},0)+{\phi }_{ij}(0,\frac{2x}{r_j})+2{\phi }_{ij}(0,-\frac{x}{r_j})\}\end{array}$$

(2.2)

for all $x\in X$. Moreover, $L(r_{k}x)=r_{k}L(x)$ for all $x\in X$ and $1\le k\le n$.

Proof For each $1\le k\le n$ with $k\ne i,j$, let $x_k=0$ in (2.1). Then we get the following inequality:

$$\begin{array}{r}{\parallel f\left(\frac{-r_i{x}_i+r_j{x}_j}{2}\right)+f\left(\frac{r_i{x}_i-r_j{x}_j}{2}\right)-2f\left(\frac{r_i{x}_i+r_j{x}_j}{2}\right)+r_{i}f(x_i)+r_{j}f(x_j)\parallel }_Y\\ \le \phi (0,\dots ,0,\underset{i\mathrm{th}}{\underbrace{x_i}},0,\dots ,0,\underset{j\mathrm{th}}{\underbrace{x_j}},0,\dots ,0)\end{array}$$

(2.3)

for all $x_i,x_j\in X$. Letting $x_i=0$ in (2.3), we get

$${\parallel f(-\frac{r_j{x}_j}{2})-f\left(\frac{r_j{x}_j}{2}\right)+r_{j}f(x_j)\parallel }_Y\le {\phi }_{ij}(0,x_j)$$

(2.4)

for all $x_j\in X$. Similarly, letting $x_j=0$ in (2.3), we get

$${\parallel f(-\frac{r_i{x}_i}{2})-f\left(\frac{r_i{x}_i}{2}\right)+r_{i}f(x_i)\parallel }_Y\le {\phi }_{ij}(x_i,0)$$

(2.5)

for all $x_i\in X$. It follows from (2.3), (2.4) and (2.5) that

$$\begin{array}{r}\parallel f\left(\frac{-r_i{x}_i+r_j{x}_j}{2}\right)+f\left(\frac{r_i{x}_i-r_j{x}_j}{2}\right)-2f\left(\frac{r_i{x}_i+r_j{x}_j}{2}\right)\\ +f\left(\frac{r_i{x}_i}{2}\right)+f\left(\frac{r_j{x}_j}{2}\right)-f(-\frac{r_i{x}_i}{2})-f(-\frac{r_j{x}_j}{2}){\parallel }_Y\\ \le {\phi }_{ij}(x_i,x_j)+{\phi }_{ij}(x_i,0)+{\phi }_{ij}(0,x_j)\end{array}$$

(2.6)

for all $x_i,x_j\in X$. Replacing $x_i$ and $x_j$ by $\frac{2x}{r_i}$ and $\frac{2y}{r_j}$ in (2.6), we get

$$\begin{array}{r}{\parallel f(-x+y)+f(x-y)-2f(x+y)+f(x)+f(y)-f(-x)-f(-y)\parallel }_Y\\ \le {\phi }_{ij}(\frac{2x}{r_i},\frac{2y}{r_j})+{\phi }_{ij}(\frac{2x}{r_i},0)+{\phi }_{ij}(0,\frac{2y}{r_j})\end{array}$$

(2.7)

for all $x,y\in X$. Putting $y=x$ in (2.7), we get

$${\parallel 2f(x)-2f(-x)-2f(2x)\parallel }_Y\le {\phi }_{ij}(\frac{2x}{r_i},\frac{2x}{r_j})+{\phi }_{ij}(\frac{2x}{r_i},0)+{\phi }_{ij}(0,\frac{2x}{r_j})$$

(2.8)

for all $x\in X$. Replacing x and y by $\frac{x}{2}$ and $-\frac{x}{2}$ in (2.7), respectively, we get

$${\parallel f(x)+f(-x)\parallel }_Y\le {\phi }_{ij}(\frac{x}{r_i},-\frac{x}{r_j})+{\phi }_{ij}(\frac{x}{r_i},0)+{\phi }_{ij}(0,-\frac{x}{r_j})$$

(2.9)

for all $x\in X$. It follows from (2.8) and (2.9) that

$${\parallel \frac{1}{2}f(2x)-f(x)\parallel }_Y\le \frac{1}{4}\psi (x)$$

(2.10)

for all $x\in X$, where

$$\begin{array}{rl}\psi (x):=& {\phi }_{ij}(\frac{2x}{r_i},\frac{2x}{r_j})+2{\phi }_{ij}(\frac{x}{r_i},-\frac{x}{r_j})\\ +{\phi }_{ij}(\frac{2x}{r_i},0)+2{\phi }_{ij}(\frac{x}{r_i},0)+{\phi }_{ij}(0,\frac{2x}{r_j})+2{\phi }_{ij}(0,-\frac{x}{r_j}).\end{array}$$

Consider the set $\mathcal{W}:=\{g:X\to Y\}$ and introduce the generalized metric on $\mathcal{W}$:

$$d(g,h)=inf\{C\in {\mathbb{R}}_{+}:{\parallel g(x)-h(x)\parallel }_Y\le C\psi (x),\mathrm{\forall }x\in X\}.$$

It is easy to show that $(\mathcal{W},d)$ is complete.

Now, we consider the linear mapping $J:\mathcal{W}\to \mathcal{W}$ such that

$$Jg(x):=\frac{1}{2}g(2x)$$

(2.11)

for all $x\in X$. By Theorem 3.1 of [44], $d(Jg,Jh)\le Cd(g,h)$ for all $g,h\in \mathcal{W}$. Hence, $d(f,Jf)\le \frac{1}{4}$.

By Theorem 1.1, there exists a mapping $L:X\to Y$ such that

• (1) L is a fixed point of J, i.e.,

  $$L(2x)=2L(x)$$

  (2.12)

for all $x\in X$. The mapping L is a unique fixed point of J in the set

$$Z=\{g\in \mathcal{W}:d(f,g)<\mathrm{\infty }\}.$$

This implies that L is a unique mapping satisfying (2.12) such that there exists $C\in (0,\mathrm{\infty })$ satisfying

$${\parallel L(x)-f(x)\parallel }_Y\le C\psi (x)$$

for all $x\in X$.

• (2) $d(J^{n}f,L)\to 0$ as $n\to \mathrm{\infty }$. This implies the equality

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

for all $x\in X$.

• (3) $d(f,L)\le \frac{1}{1-C}d(f,Jf)$, which implies the inequality $d(f,L)\le \frac{1}{4-4C}$. This implies that the inequality (2.2) holds.

Since $\phi (2x_1,\dots ,2x_n)\le 2C\phi (x_1,\dots ,x_n)$, it follows that

$$\begin{array}{rl}{\parallel D_{e,r_1,\dots ,r_n}L(x_1,\dots ,x_n)\parallel }_Y& =\underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}{\parallel D_{e,r_1,\dots ,r_n}f(2^k{x}_1,\dots ,2^k{x}_n)\parallel }_Y\\ \le \underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}\phi (2^k{x}_1,\dots ,2^k{x}_n)\\ \le \underset{k\to \mathrm{\infty }}{lim}{C}^k\phi (x_1,\dots ,x_n)=0\end{array}$$

for all $x_1,\dots ,x_n\in X$. Therefore, the mapping $L:X\to Y$ satisfies the equation (1.3) and $L(0)=0$. Hence, by Lemma 2.2, L is a generalized Euler-Lagrange type additive mapping and $L(r_{k}x)=r_{k}L(x)$ for all $x\in X$ and $1\le k\le n$. This completes the proof. □

Theorem 2.4 Let $f:X\to Y$ be a mapping satisfying $f(0)=0$ for which there is a function $\phi :X^n\to [0,\mathrm{\infty })$ satisfying

$$\parallel D_{u,r_1,\dots ,r_n}f(x_1,\dots ,x_n)\parallel \le \phi (x_1,\dots ,x_n)$$

(2.13)

for all $x_1,\dots ,x_n\in X$ and $u\in U(A)$. If there exists $0<C<1$ such that

$$\phi (2x_1,\dots ,2x_n)\le 2C\phi (x_1,\dots ,x_n)$$

for all $x_1,\dots ,x_n\in X$, then there exists a unique A-linear generalized Euler-Lagrange type additive mapping $L:X\to Y$ satisfying (2.2) for all $x\in X$. Moreover, $L(r_{k}x)=r_{k}L(x)$ for all $x\in X$ and $1\le k\le n$.

Proof By Theorem 2.3, there exists a unique generalized Euler-Lagrange type additive mapping $L:X\to Y$ satisfying (2.2), and moreover $L(r_{k}x)=r_{k}L(x)$ for all $x\in X$ and $1\le k\le n$. By the assumption, for each $u\in U(A)$, we get

$$\begin{array}{r}{\parallel D_{u,r_1,\dots ,r_n}L(0,\dots ,0,\underset{i\mathrm{th}}{\underbrace{x}},0,\dots ,0)\parallel }_Y\\ =\underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}{\parallel D_{u,r_1,\dots ,r_n}f(0,\dots ,0,\underset{i\mathrm{th}}{\underbrace{2^{k}x}},0,\dots ,0)\parallel }_Y\\ \le \underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}\phi (0,\dots ,0,\underset{i\mathrm{th}}{\underbrace{2^{k}x}},0,\dots ,0)\\ \le \underset{k\to \mathrm{\infty }}{lim}{C}^k\phi (0,\dots ,0,\underset{i\mathrm{th}}{\underbrace{x}},0,\dots ,0)=0\end{array}$$

for all $x\in X$. So, we have

$$r_{i}uL(x)=L(r_{i}ux)$$

for all $u\in U(A)$ and $x\in X$. Since $L(r_{i}x)=r_{i}L(x)$ for all $x\in X$ and $r_i\ne 0$,

$$L(ux)=uL(x)$$

for all $u\in U(A)$ and $x\in X$. By the same reasoning as in the proofs of [63] and [64],

$$L(ax+by)=L(ax)+L(by)=aL(x)+bL(y)$$

for all $a,b\in A$ ($a,b\ne 0$) and $x,y\in X$. Since $L(0x)=0=0L(x)$ for all $x\in X$, the unique generalized Euler-Lagrange type additive mapping $L:X\to Y$ is an A-linear mapping. This completes the proof. □

Theorem 2.5 Let $f:X\to Y$ be a mapping satisfying $f(0)=0$ for which there is a function $\phi :X^n\to [0,\mathrm{\infty })$ such that

$${\parallel D_{e,r_1,\dots ,r_n}f(x_1,\dots ,x_n)\parallel }_Y\le \phi (x_1,\dots ,x_n)$$

(2.14)

for all $x_1,\dots ,x_n\in X$. If there exists $0<C<1$ such that

$$\phi (x_1,\dots ,2_n)\le \frac{C}{2}\phi (2x_1,\dots ,2x_n)$$

for all $x_1,\dots ,x_n\in X$, then there exists a unique generalized Euler-Lagrange type additive mapping $L:X\to Y$ such that

$$\begin{array}{rl}{\parallel f(x)-L(x)\parallel }_Y\le & \frac{C}{4-4C}\{{\phi }_{ij}(\frac{2x}{r_i},\frac{2x}{r_j})+2{\phi }_{ij}(\frac{x}{r_i},-\frac{x}{r_j})\\ +{\phi }_{ij}(\frac{2x}{r_i},0)+2{\phi }_{ij}(\frac{x}{r_i},0)+{\phi }_{ij}(0,\frac{2x}{r_j})+2{\phi }_{ij}(0,-\frac{x}{r_j})\}\end{array}$$

(2.15)

for all $x\in X$, where ${\phi }_{ij}$ is defined in the statement of Theorem  2.3. Moreover, $L(r_{k}x)=r_{k}L(x)$ for all $x\in X$ and $1\le k\le n$.

Proof It follows from (2.10) that

$${\parallel f(x)-f\left(\frac{x}{2}\right)\parallel }_Y\le \frac{1}{2}\psi \left(\frac{x}{2}\right)\le \frac{C}{4}\psi (x)$$

for all $x\in X$, where ψ is defined in the proof of Theorem 2.3. The rest of the proof is similar to the proof of Theorem 2.3. □

Theorem 2.6 Let $f:X\to Y$ be a mapping with $f(0)=0$ for which there is a function $\phi :X^n\to [0,\mathrm{\infty })$ satisfying

$$\parallel D_{u,r_1,\dots ,r_n}f(x_1,\dots ,x_n)\parallel \le \phi (x_1,\dots ,x_n)$$

(2.16)

for all $x_1,\dots ,x_n\in X$ and $u\in U(A)$. If there exists $0<C<1$ such that

$$\phi (x_1,\dots ,2_n)\le \frac{C}{2}\phi (2x_1,\dots ,2x_n)$$

for all $x_1,\dots ,x_n\in X$, then there exists a unique A-linear generalized Euler-Lagrange type additive mapping $L:X\to Y$ satisfying (2.15) for all $x\in X$. Moreover, $L(r_{k}x)=r_{k}L(x)$ for all $x\in X$ and all $1\le k\le n$.

Proof The proof is similar to the proof of Theorem 2.4. □

Remark 2.7 In Theorems 2.5 and 2.6, one can assume that ${\sum }_{k=1}^n{r}_k\ne 0$ instead of $f(0)=0$.

3 Homomorphisms in unital $C^{\ast }$-algebras

In this section, we investigate $C^{\ast }$-algebra homomorphisms in unital $C^{\ast }$-algebras. We use the following lemma in the proof of the next theorem.

Lemma 3.1 [64]

Let $f:A\to B$ be an additive mapping such that $f(\mu x)=\mu f(x)$ for all $x\in A$ and $\mu \in {\mathbb{S}}_{\frac{1}{n_o}}^1:=\{e^{i\theta };0\le \theta \le 2\pi n_o\}$. Then the mapping $f:A\to B$ is ℂ-linear.

Note that a ℂ-linear mapping $H:A\to B$ is called a homomorphism in $C^{\ast }$-algebras if H satisfies $H(xy)=H(x)H(y)$ and $H(x^{\ast })=H{(x)}^{\ast }$ for all $x,y\in A$.

Theorem 3.2 Let $f:A\to B$ be a mapping with $f(0)=0$ for which there is a function $\phi :A^n\to [0,\mathrm{\infty })$ satisfying

(3.1)

(3.2)

(3.3)

for all $x,x_1,\dots ,x_n\in A$, $u\in U(A)$, $k\in \mathbb{N}$ and $\mu \in {\mathbb{S}}^1$. If there exists $0<C<1$ such that

$$\phi (2x_1,\dots ,2x_n)\le 2C\phi (x_1,\dots ,x_n)$$

for all $x_1,\dots ,x_n\in A$, then the mapping $f:A\to B$ is a $C^{\ast }$-algebra homomorphism.

Proof Since $|J|\ge 3$, letting $\mu =1$ and $x_k=0$ for all $1\le k\le n$ ($k\ne i,j$) in (3.1), we get

$$f\left(\frac{-r_i{x}_i+r_j{x}_j}{2}\right)+f\left(\frac{r_i{x}_i-r_j{x}_j}{2}\right)+r_{i}f(x_i)+r_{j}f(x_j)=2f\left(\frac{r_i{x}_i+r_j{x}_j}{2}\right)$$

for all $x_i,x_j\in A$. By the same reasoning as in the proof of Lemma 2.1, the mapping f is additive and $f(r_{k}x)=r_{k}f(x)$ for all $x\in A$ and $k=i,j$. So, by letting $x_i=x$ and $x_k=0$ for all $1\le k\le n$, $k\ne i$, in (3.1), we get $f(\mu x)=\mu f(x)$ for all $x\in A$ and $\mu \in {\mathbb{S}}^1$. Therefore, by Lemma 3.1, the mapping f is ℂ-linear. Hence, it follows from (3.2) and (3.3) that

$$\begin{array}{r}\begin{array}{rl}{\parallel f\left(u^{\ast }\right)-f{(u)}^{\ast }\parallel }_B& =\underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}{\parallel f\left(2^k{u}^{\ast }\right)-f{\left(2^{k}u\right)}^{\ast }\parallel }_B\\ \le \underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}\phi \left(\underset{n\mathrm{times}}{\underbrace{2^{k}u,\dots ,2^{k}u}}\right)\le \underset{k\to \mathrm{\infty }}{lim}{C}^k\phi (\underset{n\mathrm{times}}{\underbrace{u,\dots ,u}})\\ =0,\end{array}\\ \begin{array}{rl}{\parallel f(ux)-f(u)f(x)\parallel }_B& =\underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}{\parallel f\left(2^{k}ux\right)-f\left(2^{k}u\right)f(x)\parallel }_B\\ \le \underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}\phi \left(\underset{n\mathrm{times}}{\underbrace{2^{k}ux,\dots ,2^{k}ux}}\right)\le \underset{k\to \mathrm{\infty }}{lim}{C}^k\phi (\underset{n\mathrm{times}}{\underbrace{ux,\dots ,ux}})\\ =0\end{array}\end{array}$$

for all $x\in A$ and $u\in U(A)$. So, we have $f(u^{\ast })=f{(u)}^{\ast }$ and $f(ux)=f(u)f(x)$ for all $x\in A$ and $u\in U(A)$. Since f is ℂ-linear and each $x\in A$ is a finite linear combination of unitary elements (see [65]), i.e., $x={\sum }_{k=1}^m{\lambda }_k{u}_k$, where ${\lambda }_k\in \mathbb{C}$ and $u_k\in U(A)$ for all $1\le k\le n$, we have

$$\begin{array}{r}\begin{array}{rl}f\left(x^{\ast }\right)& =f\left(\sum _{k=1}^m{\overline{\lambda }}_k{u}_k^{\ast }\right)=\sum _{k=1}^m{\overline{\lambda }}_{k}f\left(u_k^{\ast }\right)=\sum _{k=1}^m{\overline{\lambda }}_{k}f{(u_k)}^{\ast }\\ ={(\sum _{k=1}^m{\lambda }_{k}f(u_k))}^{\ast }=f{\left(\sum _{k=1}^m{\lambda }_k{u}_k\right)}^{\ast }=f{(x)}^{\ast },\end{array}\\ \begin{array}{rl}f(xy)& =f\left(\sum _{k=1}^m{\lambda }_k{u}_{k}y\right)=\sum _{k=1}^m{\lambda }_{k}f(u_{k}y)\\ =\sum _{k=1}^m{\lambda }_{k}f(u_k)f(y)=f\left(\sum _{k=1}^m{\lambda }_k{u}_k\right)f(y)=f(x)f(y)\end{array}\end{array}$$

for all $x,y\in A$. Therefore, the mapping $f:A\to B$ is a $C^{\ast }$-algebra homomorphism. This completes the proof. □

The following theorem is an alternative result of Theorem 3.2.

Theorem 3.3 Let $f:A\to B$ be a mapping with $f(0)=0$ for which there is a function $\phi :A^n\to [0,\mathrm{\infty })$ satisfying

(3.4)

(3.5)

for all $x,x_1,\dots ,x_n\in A$, $u\in U(A)$, $k\in \mathbb{N}$ and $\mu \in {\mathbb{S}}^1$. If there exists $0<C<1$ such that

$$\phi (x_1,\dots ,2_n)\le \frac{C}{2}\phi (2x_1,\dots ,2x_n)$$

for all $x_1,\dots ,x_n\in A$, then the mapping $f:A\to B$ is a $C^{\ast }$-algebra homomorphism.

Remark 3.4 In Theorems 3.2 and 3.3, one can assume that ${\sum }_{k=1}^n{r}_k\ne 0$ instead of $f(0)=0$.

Theorem 3.5 Let $f:A\to B$ be a mapping with $f(0)=0$ for which there is a function $\phi :A^n\to [0,\mathrm{\infty })$ satisfying (3.2), (3.3) and

$${\parallel D_{\mu ,r_1,\dots ,r_n}f(x_1,\dots ,x_n)\parallel }_B\le \phi (x_1,\dots ,x_n)$$

(3.6)

for all $x_1,\dots ,x_n\in A$ and $\mu \in {\mathbb{S}}^1$. Assume that ${lim}_{k\to \mathrm{\infty }}\frac{1}{2^k}f(2^{k}e)$ is invertible. If there exists $0<C<1$ such that

$$\phi (2x_1,\dots ,2x_n)\le 2C\phi (x_1,\dots ,x_n)$$

for all $x_1,\dots ,x_n\in A$, then the mapping $f:A\to B$ is a $C^{\ast }$-algebra homomorphism.

Proof Consider the $C^{\ast }$-algebras A and B as left Banach modules over the unital $C^{\ast }$-algebra ℂ. By Theorem 2.4, there exists a unique ℂ-linear generalized Euler-Lagrange type additive mapping $H:A\to B$ defined by

$$H(x)=\underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}f\left(2^{k}x\right)$$

for all $x\in A$. By (3.2) and (3.3), we get

$$\begin{array}{r}\begin{array}{rl}{\parallel H\left(u^{\ast }\right)-H{(u)}^{\ast }\parallel }_B& =\underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}{\parallel f\left(2^k{u}^{\ast }\right)-f{\left(2^{k}u\right)}^{\ast }\parallel }_B\\ \le \underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}\phi \left(\underset{n\mathrm{times}}{\underbrace{2^{k}u,\dots ,2^{k}u}}\right)\\ =0,\end{array}\\ \begin{array}{rl}{\parallel H(ux)-H(u)f(x)\parallel }_B& =\underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}{\parallel f\left(2^{k}ux\right)-f\left(2^{k}u\right)f(x)\parallel }_B\\ \le \underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}\phi \left(\underset{n\mathrm{times}}{\underbrace{2^{k}ux,\dots ,2^{k}ux}}\right)\\ =0\end{array}\end{array}$$

for all $u\in U(A)$ and $x\in A$. So, we have $H(u^{\ast })=H{(u)}^{\ast }$ and $H(ux)=H(u)f(x)$ for all $u\in U(A)$ and $x\in A$. Therefore, by the additivity of H, we have

$$H(ux)=\underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}H\left(2^{k}ux\right)=H(u)\underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}f\left(2^{k}x\right)=H(u)H(x)$$

(3.7)

for all $u\in U(A)$ and all $x\in A$. Since H is ℂ-linear and each $x\in A$ is a finite linear combination of unitary elements, i.e., $x={\sum }_{k=1}^m{\lambda }_k{u}_k$, where ${\lambda }_k\in \mathbb{C}$ and $u_k\in U(A)$ for all $1\le k\le n$, it follows from (3.7) that

$$\begin{array}{r}\begin{array}{rl}H(xy)& =H\left(\sum _{k=1}^m{\lambda }_k{u}_{k}y\right)=\sum _{k=1}^m{\lambda }_{k}H(u_{k}y)\\ =\sum _{k=1}^m{\lambda }_{k}H(u_k)H(y)=H\left(\sum _{k=1}^m{\lambda }_k{u}_k\right)H(y)\\ =H(x)H(y),\end{array}\\ \begin{array}{rl}H\left(x^{\ast }\right)& =H\left(\sum _{k=1}^m{\overline{\lambda }}_k{u}_k^{\ast }\right)=\sum _{k=1}^m{\overline{\lambda }}_{k}H\left(u_k^{\ast }\right)=\sum _{k=1}^m{\overline{\lambda }}_{k}H{(u_k)}^{\ast }\\ ={(\sum _{k=1}^m{\lambda }_{k}H(u_k))}^{\ast }=H{\left(\sum _{k=1}^m{\lambda }_k{u}_k\right)}^{\ast }\\ =H{(x)}^{\ast }\end{array}\end{array}$$

for all $x,y\in A$. Since $H(e)={lim}_{k\to \mathrm{\infty }}\frac{1}{2^k}f(2^{k}e)$ is invertible and

$$H(e)H(y)=H(ey)=H(e)f(y)$$

for all $y\in A$, it follows that $H(y)=f(y)$ for all $y\in A$. Therefore, the mapping $f:A\to B$ is a $C^{\ast }$-algebra homomorphism. This completes the proof. □

The following theorem is an alternative result of Theorem 3.5.

Theorem 3.6 Let $f:A\to B$ be a mapping with $f(0)=0$ for which there is a function $\phi :A^n\to [0,\mathrm{\infty })$ satisfying (3.4), (3.5) and

$${\parallel D_{\mu ,r_1,\dots ,r_n}f(x_1,\dots ,x_n)\parallel }_B\le \phi (x_1,\dots ,x_n)$$

for all $x_1,\dots ,x_n\in A$ and $\mu \in {\mathbb{S}}^1$. Assume that ${lim}_{k\to \mathrm{\infty }}{2}^{k}f(\frac{e}{2^k})$ is invertible. If there exists $0<C<1$ such that

$$\phi (x_1,\dots ,2_n)\le \frac{C}{2}\phi (2x_1,\dots ,2x_n)$$

for all $x_1,\dots ,x_n\in A$, then the mapping $f:A\to B$ is a $C^{\ast }$-algebra homomorphism.

Remark 3.7 In Theorem 3.6, one can assume that ${\sum }_{k=1}^n{r}_k\ne 0$ instead of $f(0)=0$.

Theorem 3.8 Let $f:A\to B$ be a mapping with $f(0)=0$ for which there is a function $\phi :A^n\to [0,\mathrm{\infty })$ satisfying (3.2), (3.3) and

$${\parallel D_{\mu ,r_1,\dots ,r_n}f(x_1,\dots ,x_n)\parallel }_B\le \phi (x_1,\dots ,x_n)$$

(3.8)

for all $x_1,\dots ,x_n\in A$ and $\mu =i,1$. Assume that ${lim}_{k\to \mathrm{\infty }}\frac{1}{2^k}f(2^{k}e)$ is invertible and for each fixed $x\in A$ the mapping $t\mapsto f(tx)$ is continuous in $t\in \mathbb{R}$. If there exists $0<C<1$ such that

$$\phi (2x_1,\dots ,2x_n)\le 2C\phi (x_1,\dots ,x_n)$$

for all $x_1,\dots ,x_n\in A$, then the mapping $f:A\to B$ is a $C^{\ast }$-algebra homomorphism.

Proof Put $\mu =1$ in (3.8). By the same reasoning as in the proof of Theorem 2.3, there exists a unique generalized Euler-Lagrange type additive mapping $H:A\to B$ defined by

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

for all $x\in A$. By the same reasoning as in the proof of [58], the generalized Euler-Lagrange type additive mapping $H:A\to B$ is ℝ-linear. By the same method as in the proof of Theorem 2.4, we have

$$\begin{array}{r}{\parallel D_{\mu ,r_1,\dots ,r_n}H(0,\dots ,0,\underset{j\mathrm{th}}{\underbrace{x}},0,\dots ,0)\parallel }_Y\\ =\underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}{\parallel D_{\mu ,r_1,\dots ,r_n}f(0,\dots ,0,\underset{j\mathrm{th}}{\underbrace{2^{k}x}},0,\dots ,0)\parallel }_Y\\ \le \underset{k\to \mathrm{\infty }}{lim}\frac{1}{2^k}\phi (0,\dots ,0,\underset{j\mathrm{th}}{\underbrace{2^{k}x}},0,\dots ,0)=0\end{array}$$

for all $x\in A$ and so

$$r_j\mu H(x)=H(r_j\mu x)$$

for all $x\in A$. Since $H(r_{j}x)=r_{j}H(x)$ for all $x\in X$ and $r_j\ne 0$,

$$H(\mu x)=\mu H(x)$$

for all $x\in A$ and $\mu =i,1$. For each $\lambda \in \mathbb{C}$, we have $\lambda =s+it$, where $s,t\in \mathbb{R}$. Thus, it follows that

$$\begin{array}{rl}H(\lambda x)& =H(sx+itx)=sH(x)+tH(ix)\\ =sH(x)+itH(x)=(s+it)H(x)\\ =\lambda H(x)\end{array}$$

for all $\lambda \in \mathbb{C}$ and $x\in A$ and so

$$H(\zeta x+\eta y)=H(\zeta x)+H(\eta y)=\zeta H(x)+\eta H(y)$$

for all $\zeta ,\eta \in \mathbb{C}$ and $x,y\in A$. Hence, the generalized Euler-Lagrange type additive mapping $H:A\to B$ is ℂ-linear.

The rest of the proof is the same as in the proof of Theorem 3.5. This completes the proof. □

The following theorem is an alternative result of Theorem 3.8.

Theorem 3.9 Let $f:A\to B$ be a mapping with $f(0)=0$ for which there is a function $\phi :A^n\to [0,\mathrm{\infty })$ satisfying (3.4), (3.5) and

$${\parallel D_{\mu ,r_1,\dots ,r_n}f(x_1,\dots ,x_n)\parallel }_B\le \phi (x_1,\dots ,x_n),$$

for all $x,x_1,\dots ,x_n\in A$ and $\mu =i,1$. Assume that ${lim}_{k\to \mathrm{\infty }}{2}^{k}f(\frac{e}{2^k})$ is invertible and for each fixed $x\in A$ the mapping $t\mapsto f(tx)$ is continuous in $t\in \mathbb{R}$. If there exists $0<C<1$ such that

$$\phi (x_1,\dots ,2_n)\le \frac{C}{2}\phi (2x_1,\dots ,2x_n)$$

for all $x_1,\dots ,x_n\in A$, then the mapping $f:A\to B$ is a $C^{\ast }$-algebra homomorphism.

Proof We omit the proof because it is very similar to last theorem. □

Remark 3.10 In Theorem 3.9, one can assume that ${\sum }_{k=1}^n{r}_k\ne 0$ instead of $f(0)=0$.