RETRACTED ARTICLE: Approximation of Jordan homomorphisms in Jordan-Banach algebras

Abstract

Using the direct method based on the Hyers-Ulam-Rassias stability, we investigate and prove the Hyers-Ulam stability of Jordan homomorphisms in Jordan-Banach algebras for the functional equation

$$\begin{array}{l}\sum _{k=2}^n\sum _{i_1=2}^k\sum _{i_2=i_1+1}^{k+1}\cdots \sum _{i_n-k+1=i_{n-k}+1}^{n}f\left(\sum _{i=1,i\ne i_1,\cdots ,i_{n-k+1}}^n{x}_i-\sum _{r=1}^{n-k+1}{x}_{i_r}\right)+f\left(\sum _{i=1}^n{x}_i\right)-2^{n-1}f\left(x_1\right)=0,\end{array}$$

where n is an integer greater than 1.

We have proved the Hyers-Ulam stability of Jordan homomorphisms in Jordan-Banach algebras for the above functional equation.

Introduction

A classical question in the theory of functional equations is that ‘when is it true that a function which approximately satisfies a functional equation $\mathcal{E}$ must be somehow close to an exact solution of $\mathcal{E}$.’ Such a problem was formulated by Ulam [1] in 1940 and solved in the next year for the Cauchy functional equation by Hyers [2]. It gave rise to the stability theory for functional equations. The result of Hyers was generalized by Aoki [3] for approximate additive functions and by Rassias [4] for approximate linear functions. The stability phenomenon that was proved by Rassias is called the Hyers-Ulam-Rassias stability or the generalized Hyers-Ulam stability of functional equations. In 1994, a generalization of the Th.M. Rassias’ theorem was obtained by Gǎvruta [5] as follows: Suppose that (G, +)is an abelian group and E is a Banach space and that the so-called admissible control function $\phi :G\times G\to \mathbb{R}$ satisfies

$$\tilde{\phi }(x,y):=\sum _{n=0}^{\infty }{2}^{-n}\phi (2^{n}x,2^{n}y)<\infty$$

for all x,y ∈ G. If f : G → E is a mapping with

$$\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel \le \phi (x,y)$$

for all x,y ∈ G, then there exists a unique mapping T : G → E such that T(x + y) = T(x) + T(y) and $\parallel f\left(x\right)-T\left(x\right)\parallel \le \tilde{\phi }(x,x)$ for all x,y ∈ G. If, moreover, G is a real normed space and f(tx) is continuous in t for each fixed x in G, then T is a linear function.

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 [6–27]).

Recently, Eshaghi Gordji et al. (unpublished work) defined the following n-dimensional additive functional equation

$$\begin{array}{l}{D}_f(x_1,\cdots ,x_n):=\sum _{k=2}^n\sum _{i_1=2}^k\sum _{i_2=i_1+1}^{k+1}\cdots \sum _{i_n-k+1=i_{n-k}+1}^n\\ \times f\left(\sum _{i=1,i\ne i_1^n,\cdots ,i_{n-k+1}}^n{x}_i-\sum _{r=1}^{n-k+1}{x}_{i_r}\right)\\ +f\left(\sum _{i=1}^n{x}_i\right)-2^{n-1}f\left(x_1\right)=0,\end{array}$$

(1.1)

where n is an integer greater than 1, and investigated the functional equation (1.1) in random normed spaces the via the fixed point method.

Note that a unital algebra A, endowed with the Jordan product $x\circ y=\frac{1}{2}(\mathit{xy}+\mathit{yx})$ on A, is called a Jordan algebra. A $\mathbb{C}$-linear mapping L of a Jordan algebra A into a Jordan algebra B is called a Jordan homomorphism if L(x ∘ y) = (L(x) ∘ L(y)) holds for all x, y ∈ A.

Throughout this paper, let A be a Jordan-Banach algebra with norm ∥ · ∥ and unit e, and let B be a Jordan-Banach algebra with norm ∥ · ∥.

Methods

Using the direct method based on the Hyers-Ulam-Rassias stability, we prove the Hyers-Ulam stability of Jordan homomorphisms in Jordan-Banach algebras for the functional equation (1.1).

Results and discussion

We need the following lemma in the proof of our main theorem.

Lemma 2.1. (Eshaghi Gordji et al., unpublished work) A mapping $f:\mathcal{A}\to \mathcal{B}$ with f(0) = 0 satisfies (1.1) if and only if $f:\mathcal{A}\to \mathcal{B}$ is additive.

We are going to prove the main result.

Theorem 2.2. Let $h:\mathcal{A}\to \mathcal{B}$ be a mapping with h(0) = 0 for which there exists a function $\phi :{\mathcal{A}}^{n+2}\to [0,\infty )$ such that

$$\tilde{\phi }(x_1,\cdots x_n,z,w):=\sum _{j=0}^{\infty }{2}^{-j}\phi (2^j{x}_1,\cdots 2^j{x}_n,2^{j}z,2^{j}w)<\infty ,$$

(2.1)

$$\begin{array}{l}∥\sum _{k=2}^n\sum _{i_1=2}^k\sum _{i_2=i_1+1}^{k+1}\cdots \sum _{i_n-k+1=i_{n-k}+1}^n\\ \times h\left(\sum _{i=1,i\ne i_1,\cdots ,i_{n-k+1}}^n\mu x_i-\sum _{r=1}^{n-k+1}\mu x_{i_r}\right)\\ +\mathrm{\mu h}\left(\sum _{i=1}^n{x}_i\right)-\mu 2^{n-1}h\left(x_1\right)+h\left(z,\circ ,w\right)-h\left(z\right)\circ h\left(w\right)∥\\ \le \phi (x_1,\dots x_n,z,w)\end{array}$$

(2.2)

for all $\mu \in T^1:=\{\lambda \in \mathbb{C}|\left|\lambda \right|=1\}$ and $x_1,\dots x_n,z,w\in \mathcal{A.}$ Then, there exists a unique Jordan homomorphism $L:\mathcal{A}\to \mathcal{B}$ such that

$$\begin{array}{l}\parallel h\left(x\right)-L\left(x\right)\parallel \le \frac{1}{2^{n-1}}\tilde{\phi }(x,x,\underset{n-\mathit{times}}{\underbrace{0\dots 0}})\end{array}$$

(2.3)

for all $x\in \mathcal{A}$.

Proof. Let μ = 1. Using the relation

$$\begin{array}{l}1+\sum _{k=1}^{n-k}\left(\begin{array}{c}n-k\\ k\end{array}\right)=\sum _{k=0}^{n-k}\left(\begin{array}{c}n-k\\ k\end{array}\right)=2^{n-k}\end{array}$$

(2.4)

for all n > k and putting x ₁ = x ₂ = x and x _i  = z = w = 0(i = 3, …, n) in (2.2), we obtain

$$\begin{array}{l}∥2^{n-2}h\left(2,x\right)-2^{n-1}h\left(x\right)∥\le \phi \left(x,,,x,,,\underset{n-\mathit{times}}{\underbrace{0,\dots ,0}}\right)\end{array}$$

(2.5)

for all $x\in \mathcal{A}$. So,

$$\begin{array}{l}∥\frac{h\left(2x\right)}{2}-h\left(x\right)∥\le \frac{1}{2^{n-1}}\phi (x,x,\underset{n-\mathit{times}}{\underbrace{0,\dots ,0}})\end{array}$$

(2.6)

for all $x\in \mathcal{A}$. By induction on m, we can show that

$$\begin{array}{l}∥\frac{h\left(2^{m}x\right)}{2^m}-h\left(x\right)∥\le \frac{1}{2^{n-1}}\sum _{j=0}^{m-1}\frac{1}{2^j}\phi (2^{j}x,2^{j}x,\underset{n-\mathit{times}}{\underbrace{0,\dots ,0}})\end{array}$$

(2.7)

for all $x\in \mathcal{A}$ It follows from (2.1) and (2.7) that the sequence $\left\{\frac{h\left(2^{m}x\right)}{2^m}\right\}$ is a Cauchy sequence for all $x\in \mathcal{A}$ Since $\mathcal{A}$ is complete, the sequence $\left\{\frac{h\left(2^{m}x\right)}{2^m}\right\}$ converges. Thus, one can define the mapping $L:\mathcal{A}\to \mathcal{B}$ by

$$L\left(x\right):=\underset{m\to \infty }{lim}\frac{h\left(2^m,x\right)}{2^m}$$

for all $x\in \mathcal{A}$. Let z = w = 0 and μ = 1 in (2.2). By (2.1)

$$\begin{array}{ll}\parallel D_f(x_1,\mathrm{..},x_n)\parallel & =\underset{j\to \infty }{lim}\frac{1}{2^j}∥D_f\left(2^j{x}_1,\mathrm{..},2^j{x}_n\right)∥\\ \le \underset{j\to \infty }{lim}\frac{1}{2^j}\phi \left(2^j{x}_1,\mathrm{..},2^j{x}_n,0,0\right)=0\end{array}$$

for all $x_1,\cdots ,x_n\in \mathcal{A}$. So, D _f (x ₁,⋯,x _n ) = 0. By Lemma 2.1, the mapping $L:\mathcal{A}\to \mathcal{B}$ is additive. Moreover, passing the limit m → ∞ in (2.7) we get the inequality (2.3). □

Now, let $L^{\prime }:\mathcal{A}\to \mathcal{B}$ be another additive mapping satisfying (1.1) and (2.3). Then,

$$\begin{array}{ll}\parallel L\left(x\right)-L^{\prime }\left(x\right)\parallel & =\frac{1}{2^n}\parallel L\left(2^{n}x\right)-L^{\prime }\left(2^{n}x\right)\parallel \\ \le \frac{1}{2^m}(\parallel L(2^{n}x)-h(2^{n}x)\parallel +\parallel L^{\prime }(2^{n}x)\\ -h\left(2^{n}x\right)\parallel )\\ \le \frac{2}{2^m{2}^{n-1}}\tilde{\phi }(2^{m}x,2^{m}x,\underset{n-\mathit{times}}{\underbrace{0,\dots ,0}})\end{array}$$

which tends to zero as m → ∞ for all $x\in \mathcal{A}$. So, we can conclude that L(x) = L ^′(x) for all $x\in \mathcal{A}$ This proves the uniqueness of L.

Let $\mu \in {\mathbb{T}}^1$. Set x ₁ = x and z = w = x _i  = 0 (i = 2, .., n) in (2.2). Then, by (2.1), we get

$$\begin{array}{l}\parallel 2^{n-1}h\left(\mathrm{\mu x}\right)-2^{n-1}\mathrm{\mu h}\left(x\right)\parallel \le \phi (x,0,\mathrm{..},0,0,0)\end{array}$$

(2.8)

for all $x\in \mathcal{A}$ So,

$$\parallel 2^{-m}\left(h\left(2^m\mathrm{\mu x}\right)-\mathrm{\mu h}\left(2^{m}x\right)\right)\parallel \le \frac{2^{-m}}{2^{n-1}}\phi \left(2^{m}x,0,\mathrm{..},0,0,0\right)$$

for all $x\in \mathcal{A}$. Since the right hand side of the above inequality tends to zero as m → ∞, we have

$$L\left(\mathrm{\mu x}\right)=\underset{m\to \infty }{lim}\frac{h\left(2^m\mathrm{\mu x}\right)}{2^m}=\underset{m\to \infty }{lim}\frac{\mathrm{\mu h}\left(2^{m}x\right)}{2^m}=\mathrm{\mu L}\left(x\right)$$

(2.9)

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

Now let $\lambda \in \mathbb{C}(\lambda \ne 0)$ and M an integer greater than 4|λ|. Then, |λ/M| < 1/4 < 1 - 2/3 = 1/3. By Theorem 1 of [28], there exist three elements ${\mu }_1,{\mu }_2,{\mu }_3\in {\mathbb{T}}^1$ such that $3\frac{\lambda }{M}={\mu }_1+{\mu }_2+{\mu }_3$, and $L\left(x\right)=L\left(3·\frac{1}{3}x\right)=3L\left(\frac{1}{3}x\right)$ for all $x\in \mathcal{A}$. So, $L\left(\frac{1}{3}x\right)=\frac{1}{3}L\left(x\right)$ for all $x\in \mathcal{A}$. Thus, by (2.9),

$$\begin{array}{l}L\left(\mathrm{\lambda x}\right)=L\left(\frac{M}{3}·3\frac{\lambda }{M}x\right)=M·L\left(\frac{1}{3}·3\frac{\lambda }{M}x\right)\\ =\frac{M}{3}L\left(3\frac{\lambda }{M}x\right)\\ =\frac{M}{3}L({\mu }_{1}x+{\mu }_{2}x+{\mu }_{3}x)\\ =\frac{M}{3}\left(L\right({\mu }_{1}x)+L({\mu }_{2}x)+L({\mu }_{3}x\left)\right)\\ =\frac{M}{3}({\mu }_1+{\mu }_2+{\mu }_3)L\left(x\right)\\ =\frac{M}{3}·3\frac{\lambda }{M}L\left(x\right)=\mathrm{\lambda L}\left(x\right)\end{array}$$

for all $x\in \mathcal{A}$. Hence,

$$L(\zeta x_1+\eta x_2)=L\left(\zeta x_1\right)+L\left(\eta x_2\right)=\mathrm{\zeta L}\left(x_1\right)+\mathrm{\eta L}\left(x_2\right)$$

for all $\zeta ,\eta \in \mathbb{C}(\zeta ,\eta \ne 0)$ and all $x_1,x_2\in \mathcal{A}$, and L(0x) = 0 = 0L(x) for all $x\in \mathcal{A}$.

So, $L:\mathcal{A}\to \mathcal{B}$ is $\mathbb{C}$-linear.

Let x _i  = 0 (i ≥ 0) in (2.2). Then, we get

$$\parallel h\left(z\circ w\right)-h\left(z\right)\circ h\left(w\right)\parallel \le \phi \left(\underset{n-\mathit{times}}{\underbrace{0,\cdots ,0}},z,w\right)$$

for all $z,w\in \mathcal{A}$. Since

$$\frac{1}{2^{2m}}\phi \left(\underset{n-\mathit{times}}{\underbrace{0,\cdots ,0}},2^{m}z,2^{m}w\right)\le \frac{1}{2^m}\phi \left(\underset{n-\mathit{times}}{\underbrace{0,\cdots ,0}},2^{m}z,2^{m}w\right),$$

$$\begin{array}{ll}\frac{1}{2^{2m}}& ∥h(2^{m}z\circ 2^{m}w)-h\left(2^{m}z\right)\circ h\left(2^{m}w\right)∥\\ \le \frac{1}{2^{2m}}\phi (\underset{n-\mathit{times}}{\underbrace{0,\dots ,0}},z,w)\\ \le \frac{1}{2^m}\phi (\underset{n-\mathit{times}}{\underbrace{0,\dots ,0}},z,w),\end{array}$$

which tends to zero as m → ∞ for all $z,w\in \mathcal{A}$. Hence,

$$\begin{array}{ll}L(z\circ w)& =\underset{m\to \infty }{lim}\frac{h\left(2^{2m}(z\circ w)\right)}{2^{2m}}\\ =\underset{m\to \infty }{lim}\frac{h(2^{m}z\circ 2^{m}w)}{2^{2m}}\\ =\underset{m\to \infty }{lim}\frac{1}{2^{2m}}\left(h\left(2^{m}z\right)\circ h\left(2^{m}w\right)\right)\\ =\underset{m\to \infty }{lim}\left(\frac{h\left(2^{m}z\right)}{2^m}\circ \frac{h\left(2^{m}w\right)}{2^m}\right)\\ =L\left(z\right)\circ L\left(w\right)\end{array}$$

for all $z,w\in \mathcal{A}$ So, the $\mathbb{C}$-linear mapping $L:\mathcal{A}\to \mathcal{B}$ is a Jordan homomorphism satisfying (2.3).

Corollary 2.3. Let $h:\mathcal{A}\to \mathcal{B}$ be a mapping with h(0) = 0 for which there exist constants ε ≥ 0 and p ∈[0, 1)such that

$$\begin{array}{l}∥\sum _{k=2}^n\sum _{i_1=2}^k\sum _{i_2=i_1+1}^{k+1}\cdots \sum _{i_n-k+1=i_{n-k}+1}^n\\ \times h\left(\sum _{i=1,i\ne i_1,\cdots ,i_{n-k+1}}^n\mu x_i-\sum _{r=1}^{n-k+1}\mu x_{i_r}\right)\\ +\mathrm{\mu h}\left(\sum _{i=1}^n{x}_i\right)-\mu 2^{n-1}h\left(x_1\right)+h\left(z\circ w\right)-h\left(z\right)\circ h\left(w\right)∥\\ \le \epsilon (\parallel x_1{\parallel }^p+\cdots +\parallel x_n{\parallel }^p+\parallel z{\parallel }^p+\parallel w{\parallel }^p)\end{array}$$

for all $\mu \in {\mathbb{T}}^1$ and all $x_1,x_2,\mathrm{..},x_n,z,w\in \mathcal{A}$. Then, there exists a unique Jordan homomorphism $L:\mathcal{A}\to \mathcal{B}$ such that

$$\parallel h\left(x\right)-L\left(x\right)\parallel \le \frac{\epsilon }{2^n\left(1,-,2^{p-1}\right)}\parallel x{\parallel }^p$$

for all $x\in \mathcal{A}$.

Proof. Define φ(x ₁,⋯x _n ,z,w) = ε(∥x ₁∥^p + ⋯ + ∥x _n ∥^p + ∥z∥^p + ∥w∥^p)and apply Theorem 2.2 Then, we get the desired result. □

Corollary 2.4. Suppose that $h:\mathcal{A}\to \mathcal{B}$ is mapping with h(0) = 0 satisfying (2.2) If there exists a function $\phi :{\mathcal{A}}^{n+2}\to [0,\infty )$ such that

$$\tilde{\phi }(x_1,\cdots x_n,z,w):=\sum _{j=0}^{\infty }{2}^j\phi (2^{-j}{x}_1,\cdots 2^{-j}{x}_n,2^{-j}z,2^{-j}w)<\infty$$

for all $z,w,x_i\in \mathcal{A}(i=1,\mathrm{..},n),$ then there exists a unique Jordan homomorphism $L:\mathcal{A}\to \mathcal{B}$ such that

$$\parallel h\left(x\right)-L\left(x\right)\parallel \le \frac{1}{2^{n-1}}\tilde{\phi }\left(x,x,\underset{n-\mathit{times}}{\underbrace{0\dots 0}}\right)$$

for all $x\in \mathcal{A}$

Proof. By the same method as in the proof of Theorem 2.2 one can obtain that

$$L\left(x\right)=\underset{m\to \infty }{lim}\frac{h\left(2^{m}x\right)}{2^m}$$

for all $x\in \mathcal{A}$. □

The rest of the proof is similar to the proof of Theorem 2. 2.

Theorem 2.5. Let $h:\mathcal{A}\to \mathcal{B}$ be a mapping with h(0) = 0 for which there exists a function $\phi :{\mathcal{A}}^{n+2}\to [0,\infty )$ satisfying (2.1) such that

$$\begin{array}{l}∥\sum _{k=2}^n\sum _{i_1=2}^k\sum _{i_2=i_1+1}^{k+1}\cdots \sum _{i_n-k+1=i_{n-k}+1}^n\\ \times h\left(\sum _{i=1,i\ne i_1,\cdots ,i_{n-k+1}}\mu x_i-\sum _{r=1}^{n-k+1}\mu x_{i_r}\right)\\ +\mathrm{\mu h}\left(\sum _{i=1}^n{x}_i\right)-\mu 2^{n-1}h\left(x_1\right)+h\left(z\circ w\right)-h\left(z\right)\circ h\left(w\right)∥\\ \le \phi (x_1,\cdots x_n,z,w)\end{array}$$

(2.10)

for μ = 1, i and all $x_1,\cdots ,x_n,z,w\in \mathcal{A}$. If h(tx) is continuous in $t\in \mathbb{R}$ for each fixed $x\in \mathcal{A},$ then there exists a unique Jordan homomorphism $L:\mathcal{A}\to \mathcal{B}$ satisfying (2.3).

Proof. Put z = w = 0 in (2.10). By the same reasoning as in the proof of Theorem 2. 2, there exists a unique additive mapping $L:\mathcal{A}\to \mathcal{B}$ satisfying (2.3). The additive mapping $L:\mathcal{A}\to \mathcal{B}$ is given by

$$L\left(x\right)=\underset{m\to \infty }{lim}\frac{h\left(2^{m}x\right)}{2^m}$$

for all $x\in \mathcal{A}$. By the same reasoning as in the proof of Theorem 2. 2 the additive mapping $L:\mathcal{A}\to \mathcal{B}$ is $\mathbb{R}$-linear. □

Putting x _i  = z = w = 0 (i = 2, ⋯, n)and μ = i in (2.10), we get

$$∥h\left(\mathit{\text{ix}}\right)-\mathit{\text{ih}}\left(x\right)∥\le \phi \left(x,\underset{\left(n,+,1\right)-\mathit{times}}{\underbrace{0,\cdots ,0}}\right)$$

for all $x\in \mathcal{A}$. So,

$$\frac{1}{2^n}∥h\left(2^m\mathit{ix}\right)-\mathit{ih}\left(2^{m}x\right)∥\le \frac{1}{2^n}\phi (2^{n}x,\underset{(n+1)-\mathit{times}}{\underbrace{0,\dots ,0}}),$$

which tends to zero as m → ∞. Hence,

$$L\left(\mathit{ix}\right)=\underset{m\to \infty }{lim}\frac{h\left(2^m\mathit{ix}\right)}{2^m}=\underset{m\to \infty }{lim}\frac{\mathit{ih}\left(2^{m}x\right)}{2^m}=\mathit{iL}\left(x\right)$$

for all $x\in \mathcal{A}$

For each element $\lambda \in \mathbb{C},$ λ = s + it, where $s,t\in \mathbb{R}\text{.}$ So,

$$\begin{array}{ll}L\left(\mathrm{\lambda x}\right)& =L(\mathit{sx}+\mathit{itx})=\mathit{sL}\left(x\right)+\mathit{tL}\left(\mathit{ix}\right)=\mathit{sL}\left(x\right)+\mathit{itL}\left(x\right)\\ =(s+\mathit{it})L\left(x\right)=\mathrm{\lambda L}\left(x\right)\end{array}$$

for all $x\in \mathcal{A}$. So,

$$L(\zeta x_1+\eta x_2)=L\left(\zeta x_1\right)+L\left(\eta x_2\right)=\mathrm{\zeta L}\left(x_1\right)+\mathrm{\eta L}\left(x_2\right)$$

for all $\zeta ,\eta \in \mathbb{C},$ and all $x_1,x_2\in \mathcal{A}$. Hence, the additive mapping $L:\mathcal{A}\to \mathcal{B}$ is $\mathbb{C}$-linear.

The rest of the proof is the same as in the proof of Theorem 2.2

Corollary 2.6. Let $h:\mathcal{A}\to \mathcal{B}$ be a mapping with h(0) = 0 for which there exist constants ε ≥ 0 and p > 1 such that

$$\begin{array}{l}∥\sum _{k=2}^n\sum _{i_1=2}^k\sum _{i_2=i_1+1}^{k+1}\cdots \sum _{i_n-k+1=i_{n-k}+1}^n\\ \times h\left(\sum _{i=1,i\ne i_1,\cdots ,i_{n-k+1}}^n\mu x_i-\sum _{r=1}^{n-k+1}\mu x_{i_r}\right)\\ +\mathrm{\mu h}\left(\sum _{i=1}^n{x}_i\right)-\mu 2^{n-1}h\left(x_1\right)+h\left(z\circ w\right)-\left(h\left(z\right)\circ h\left(w\right)\right)∥\\ \le \epsilon (\parallel x_1{\parallel }^p+\dots +\parallel x_n{\parallel }^p+\parallel z{\parallel }^p+\parallel w{\parallel }^p)\end{array}$$

for all $z,w,x_i\in \mathcal{A}(i=1,2,\cdots ,n)$ and all $\mu \in {\mathbb{T}}^1.$ Then, there exists a unique Jordan homomorphism $L:\mathcal{A}\to \mathcal{B}$ such that

$$\parallel h\left(x\right)-L\left(x\right)\parallel \le \frac{\epsilon }{2^n(2^{1-p}-1)}\parallel x{\parallel }^p$$

for all $x\in \mathcal{A}$

Proof. Define φ(x ₁,⋯x _n ,z,w) = ε(∥x ₁∥^p + ⋯ + ∥x _n ∥^p + ∥z∥^p + ∥w∥^p)and apply Theorem 2.2 Then, we get the desired result. □

Conclusions

We have proved the Hyers-Ulam stability of Jordan homomorphisms in Jordan-Banach algebras for the functional equation (1.1).