In this section, applying some ideas from [18, 21], we deal with the stability problem for the orthogonally Jensen additive functional equation
2f\left(\frac{x+y}{2}\right)=f(x)+f(y)
for all x,y\in X with x\perp y.
Definition 2.1 An additive mapping f:X\to Y is called an Aadditive mapping if f(ax)=af(x) for all a\in A and x\in X.
Theorem 2.2
Let
\phi :{X}^{2}\to [0,\mathrm{\infty})
be a function such that there exists an
\alpha <1
with
\phi (x,y)\le 2\alpha \phi (\frac{x}{2},\frac{y}{2})
(2.1)
for all x,y\in X with x\perp y. Let f:X\to Y be a mapping satisfying f(0)=0 and
{\parallel 2af\left(\frac{x+y}{2}\right)f(ax)f(ay)\parallel}_{Y}\le \phi (x,y)
(2.2)
for all a\in {A}_{1} and all x,y\in X with x\perp y. If, for each x\in X, the mapping f(tx) is continuous in t\in \mathbb{R}, then there exists a unique orthogonally Jensen Aadditive mapping L:X\to Y such that
{\parallel f(x)L(x)\parallel}_{Y}\le \frac{\alpha}{1\alpha}\phi (x,0)
(2.3)
for all x\in X.
Proof Putting y=0 and a=e in (2.2), we get
{\parallel 2f\left(\frac{x}{2}\right)f(x)\parallel}_{Y}\le \phi (x,0)
(2.4)
for all x\in X since x\perp 0. So, we have
{\parallel f(x)\frac{1}{2}f(2x)\parallel}_{Y}\le \frac{1}{2}\phi (2x,0)\le \alpha \cdot \phi (x,0)
(2.5)
for all x\in X. Consider the set
and introduce the generalized metric on S:
d(g,h)=inf\{\mu \in {\mathbb{R}}_{+}:{\parallel g(x)h(x)\parallel}_{Y}\le \mu \phi (x,0),\mathrm{\forall}x\in X\},
where, as usual, inf\varphi =+\mathrm{\infty}. It is easy to show that (S,d) is complete (see [50]).
Now, we consider the linear mapping J:S\to S such that
for all x\in X. Let g,h\in S be given such that d(g,h)=\epsilon. Then we have
{\parallel g(x)h(x)\parallel}_{Y}\le \phi (x,0)
for all x\in X. Hence,
{\parallel Jg(x)Jh(x)\parallel}_{Y}={\parallel \frac{1}{2}g(2x)\frac{1}{2}h(2x)\parallel}_{Y}\le \alpha \phi (x,0)
for all x\in X. So, d(g,h)=\epsilon implies that d(Jg,Jh)\le \alpha \epsilon. This means that
d(Jg,Jh)\le \alpha d(g,h)
for all g,h\in S. It follows from (2.5) that d(f,Jf)\le \alpha. By Theorem 1.1, there exists a mapping L:X\to Y satisfying the following:

(1)
L is a fixed point of J, i.e.,
for all x\in X. The mapping L is a unique fixed point of J in the set
M=\{g\in S:d(h,g)<\mathrm{\infty}\}.
This implies that L is a unique mapping satisfying (2.6) such that there exists a \mu \in (0,\mathrm{\infty}) satisfying
{\parallel f(x)L(x)\parallel}_{Y}\le \mu \phi (x,0)
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{1}{{2}^{n}}f\left({2}^{n}x\right)=L(x)
for all x\in X;

(3)
d(f,L)\le \frac{1}{1\alpha}d(f,Jf), which implies the inequality
d(f,L)\le \frac{\alpha}{1\alpha}.
This implies that inequality (2.3) holds.
Let a=e in (2.2). It follows from (2.1) and (2.2) that
\begin{array}{rcl}{\parallel 2L\left(\frac{x+y}{2}\right)L(x)L(y)\parallel}_{Y}& =& \underset{n\to \mathrm{\infty}}{lim}\frac{1}{{2}^{n}}{\parallel 2f({2}^{n1}(x+y))f\left({2}^{n}x\right)f\left({2}^{n}y\right)\parallel}_{Y}\\ \le & \underset{n\to \mathrm{\infty}}{lim}\frac{1}{{2}^{n}}\phi ({2}^{n}x,{2}^{n}y)\le \underset{n\to \mathrm{\infty}}{lim}\frac{{2}^{n}{\alpha}^{n}}{{2}^{n}}\phi (x,y)=0\end{array}
for all x,y\in X with x\perp y. So,
2L\left(\frac{x+y}{2}\right)L(x)L(y)=0
for all x,y\in X with x\perp y. Hence, L:X\to Y is an orthogonally Jensen additive mapping.
Let y=0 in (2.2). It follows from (2.1) and (2.2) that
\begin{array}{rcl}{\parallel 2aL\left(\frac{x}{2}\right)L(ax)\parallel}_{Y}& =& \underset{n\to \mathrm{\infty}}{lim}\frac{1}{{2}^{n}}{\parallel 2af\left({2}^{n1}x\right)f\left({2}^{n}ax\right)\parallel}_{Y}\\ \le & \underset{n\to \mathrm{\infty}}{lim}\frac{1}{{2}^{n}}\phi ({2}^{n}x,0)\le \underset{n\to \mathrm{\infty}}{lim}\frac{{2}^{n}{\alpha}^{n}}{{2}^{n}}\phi (x,0)=0\end{array}
for all x\in X. So, we have
2aL\left(\frac{x}{2}\right)L(ax)=0
for all x\in X, and hence
L(ax)=2aL\left(\frac{x}{2}\right)=aL(x)
for all a\in {A}_{1} and x\in X. By the same reasoning as in the proof of Theorem in [15], we can show that L:X\to Y is \mathbb{R}linear since the mapping f(tx) is continuous in t\in \mathbb{R} for each x\in X. For each a\in A with a\ne 0, we have
L(ax)=L(\parallel a\parallel \frac{a}{\parallel a\parallel}x)=\parallel a\parallel L\left(\frac{a}{\parallel a\parallel}x\right)=\parallel a\parallel \frac{a}{\parallel a\parallel}L(x)=aL(x)
for all x\in X. Thus L:X\to Y is a unique orthogonally Jensen Aadditive mapping satisfying (2.3). This completes the proof. □
From now on, in corollaries, assume that (X,\perp ) is an orthogonality normed module over a unital Banach algebra A.
Corollary 2.3 Let θ be a positive real number and p be a real number with 0<p<1. Let f:X\to Y be a mapping satisfying f(0)=0 and
{\parallel 2af\left(\frac{x+y}{2}\right)f(ax)f(ay)\parallel}_{Y}\le \theta ({\parallel x\parallel}^{p}+{\parallel y\parallel}^{p})
(2.7)
for all a\in {A}_{1} and x,y\in X with x\perp y. If, for each x\in X, the mapping f(tx) is continuous in t\in \mathbb{R}, then there exists a unique orthogonally Jensen Aadditive mapping L:X\to Y such that
{\parallel f(x)L(x)\parallel}_{Y}\le \frac{{2}^{p}\theta}{2{2}^{p}}{\parallel x\parallel}^{p}
for all x\in X.
Proof The proof follows from Theorem 2.2 by taking \phi (x,y)=\theta ({\parallel x\parallel}^{p}+{\parallel y\parallel}^{p}) for all x,y\in X with x\perp y. Then we can choose \alpha ={2}^{p1} and we get the desired result. □
Theorem 2.4 Let f:X\to Y be a mapping satisfying (2.2) and f(0)=0 for which there exists a function \phi :{X}^{2}\to [0,\mathrm{\infty}) such that
\phi (x,y)\le \frac{\alpha}{2}\phi (2x,2y)
for all x,y\in X with x\perp y. If, for each x\in X, the mapping f(tx) is continuous in t\in \mathbb{R}, then there exists a unique orthogonally Jensen Aadditive mapping L:X\to Y such that
{\parallel f(x)L(x)\parallel}_{Y}\le \frac{1}{1\alpha}\phi (x,0)
(2.8)
for all x\in X.
Proof Let (S,d) be the generalized metric space defined in the proof of Theorem 2.2.
Now, we consider the linear mapping J:S\to S such that
Jg(x):=2g\left(\frac{x}{2}\right)
for all x\in X. It follows from (2.4) that d(f,Jf)\le 1. So,
d(f,L)\le \frac{1}{1\alpha}.
Thus we obtain inequality (2.8). The rest of the proof is similar to the proof of Theorem 2.2. This completes the proof. □
Corollary 2.5 Let θ be a positive real number and p be a real number with p>1. Let f:X\to Y be a mapping satisfying f(0)=0 and (2.7). If, for each x\in X, the mapping f(tx) is continuous in t\in \mathbb{R}, then there exists a unique orthogonally Jensen Aadditive mapping L:X\to Y such that
{\parallel f(x)L(x)\parallel}_{Y}\le \frac{{2}^{p}\theta}{{2}^{p}2}{\parallel x\parallel}^{p}
for all x\in X.
Proof The proof follows from Theorem 2.4 by taking \phi (x,y)=\theta ({\parallel x\parallel}^{p}+{\parallel y\parallel}^{p}) for all x,y\in X with x\perp y. Then we can choose \alpha ={2}^{1p} and we get the desired result. □