In this section, we introduce the main result in this article that concerns the stability of Eq. (1); it corresponds in particular to some results in [12].

###
**Theorem 2**

Let (A1) be valid, \(p\in \mathbb {K}, A,k\in (0,\infty), |p|^{\alpha k}+|1-p|^{\alpha k}<1\), and

satisfy

$$\begin{array}{*{20}l} \|g(px_{1}+(1-p)x_{2})+g((1-p)x_{1}+px_{2})&-g(x_{1})-g(x_{2}),y \|_{\alpha} \\ \leq A\left(\|x_{1},y\|_{\alpha}^{k}+\|x_{2},y\|_{\alpha}^{k}\right)&,\,\, x_{1},x_{2} \in E, y\in Y_{0}. \end{array} $$

(8)

Then there exists a unique solution *G*:*X*→*Y* of Eq. (1) such that

$$ \|g(x)-G(x),y \|_{\alpha} \leq \frac{A\|x,y\|_{\alpha}^{k}}{1-|p|^{\alpha k}-|1-p|^{\alpha k}}, x\in E $$

(9)

and *G* is given by

$$ G(x):= g(0)+ {\lim}_{n\rightarrow \infty}\left(\mathrm{T}^{n}g_{0}\right)(x),\,\, x\in E, $$

(10)

where *g*_{0} and T are defined by (13) and (14). Moreover, *G* is the unique solution of Eq. (1) such that there exists a constant *M*∈(0,*∞*) with

$$ \|g(x)-G(x),y\|_{\alpha}\leq M \|x,y\|_{\alpha}^{k},\,\, x\in E, y\in Y_{0} $$

(11)

###
*Proof*

Note that (8) with *x*_{2}=0 gives

$$\begin{array}{*{20}l} \|g(px_{1})+g((1-p)x_{1})-g(x_{1})-g(0),y \|_{\alpha} & \\ \leq A\left(\|x_{1},y\|_{\alpha}^{k}+\|y\|_{\alpha}^{k}\right),\,\, x_{1} \in E, y\in Y_{0} &\end{array} $$

(12)

Write

$$ g_{0}(x_{1})=g(x_{1})-g(0),\,\, x_{1} \in E $$

(13)

and

$$ \mathrm{T} \xi(x_{1})=\xi(px_{1})+\xi((1-p)x_{1}),\,\, x_{1}\in E, \xi\in Y^{E}. $$

(14)

Then (12) implies the inequality

$$\begin{array}{*{20}l} \|g_{0}(px_{1})+g_{0}((1-p)x_{1})&\;-g(x_{1})-g(0),y \|_{\alpha} \\ &\;\leq A (\|x_{1},y\|_{\alpha}^{k}),\,\, x_{1}\in E \end{array} $$

(15)

which means that

$$ \|\mathrm{T} g_{0}(x_{1})-g_{0}(x_{1}),y \|_{\alpha} \leq A \left(\|x_{1},y\|_{\alpha}^{k}\right),\,\, x_{1}\in E. $$

(16)

Further, note that (A3) holds with *k*=2, *f*_{1}(*x*)=*p**x*,*f*_{2}(*x*)=(1−*p*)*x*,*L*_{i}(*x*)=1 for *i*=1,2,*x*∈*E*. Define *Λ* as in (A3). Clearly, with \(\varepsilon (x):= A(\|x_{1},y\|_{\alpha }^{k})\) for *x*∈*E*, we have

$$\begin{array}{*{20}l} \varepsilon^{*}(x_{1}):=\sum\limits_{n=0}^{\infty} (\Lambda^{n} \varepsilon)(x_{1}) &\; \\ \leq A (\|x_{1},y\|_{\alpha}^{k})\sum\limits_{n=0}^{\infty} &\;(|p|^{\alpha k}+|1-p|^{\alpha k})^{n} \\ =\frac{A(\|x_{1},y\|_{\alpha}^{k})}{1-|p|^{\alpha k}-|1-p|^{\alpha k}}&\;,\,\, x_{1}\in E. \end{array} $$

(17)

Hence, according to Theorem 2, there exists a unique solution *G*_{0}:*X*→*Y* of the equation

$$ G_{0}(x_{1})=G_{0}(px_{1})+G_{0}((1-p)x_{1}), \,\, x_{1} \in E $$

(18)

such that

$$ \|g_{0}(x_{1})-G_{0}(x_{1}),y\|_{\alpha} \leq \frac{A\left(\|x_{1},y\|_{\alpha}^{k}\right)}{1-|p|^{\alpha k}-|1-p|^{\alpha k}}, \,\, x_{1}\in E; $$

(19)

moreover,

$$ G_{0}(x_{1}):={\lim}_{n\rightarrow \infty}(\mathrm{T}^{n}g_{0})(x_{1}), \,\, x_{1}\in E. $$

(20)

Now we show that, for every \(x_{1}, x_{2} \in E, n \in \mathbb {N}_{0}\) (nonnegative integers),

$$\begin{array}{*{20}l} \|\mathrm{T}^{n} g_{0}(px_{1}+(1-p)x_{2})+\mathrm{T}^{n} g_{0}((1-p)x_{1}+px_{2})& -\mathrm{T}^{n} g(x_{1})-\mathrm{T}^{n} g(x_{2}),y \|_{\alpha} \\ \leq A(|p|^{\alpha k}+|1-p|^{\alpha k})^{n} \left(\|x_{1},y\|_{\alpha}^{k}+\|x_{2},y\|_{\alpha}^{k}\right), & \,\, x_{1},x_{2} \in E, y\in Y_{0} \end{array} $$

(21)

It is easy to see that the case *n*=0 is just (8). Next, fix \(m \in \mathbb {N}_{0}\) and assume that (21) holds for every *x*_{1},*x*_{2}∈*E* with *n*=*m*. Then

$$\begin{array}{*{20}l} \|\mathrm{T}^{m+1} g_{0}(px_{1}+(1-p)x_{2})&\;+\mathrm{T}^{m+1} g_{0}((1-p)x_{1}+px_{2}) \\ -\mathrm{T}^{m+1} g(x_{1})-\mathrm{T}^{m+1}&\; g(x_{2}),y \|_{\alpha}=\|\mathrm{T}^{m} g_{0}(p(px_{1}+(1-p)x_{2})) \\ +\mathrm{T}^{m} g_{0}((1-p)(px_{1}&\;+(1-p)x_{2}))+\mathrm{T}^{m}g_{0}(p((1-p)x_{1}+px_{2})) \\ +\mathrm{T}^{m} &\;g_{0}((1-p)(1-p)x_{1}+px_{2}))-\mathrm{T}^{m} g_{0}(px_{1}) \\ -\mathrm{T}^{m} &\;g_{0}((1-p)x_{1})-\mathrm{T}^{m} g_{0}(px_{2})-\mathrm{T}^{m} g_{0}((1-p)x_{2}),y \|_{\alpha} \end{array} $$

(22)

which is clearly

$$\begin{array}{*{20}l} \leq \|\mathrm{T}^{m} g_{0}(ppx_{1}+(1-p)px_{2})+&\;\mathrm{T}^{m} g_{0}((1-p)px_{1}+ppx_{2})-\mathrm{T}^{m} g_{0}(px_{1}) \\ -\mathrm{T}^{m} g_{0}(px_{2}),y\|_{\alpha}+\|\mathrm{T}&\;^{m} g_{0}(p(1-p)x_{1}+(1-p)(1-p)x_{2}) \\ +\mathrm{T}^{m}g_{0}((1-p)(1-p)&\;x_{1}+p(1-p)x_{2}) \\ -\mathrm{T}^{m} g_{0}((1-p)x_{1})&\;-\mathrm{T}^{m} g_{0}(p(1-p)x_{2}),y\|_{\alpha} \\ \leq A (|p|^{\alpha k}+&\;|1-p|^{\alpha k})^{m} \left((p\|x_{1},y\|_{\alpha})^{k}+(p\|x_{2},y\|_{\alpha})^{k}\right) \\ +(|p|^{\alpha k}+|1-p|^{\alpha k})&\;^{m} \left(((1-p)\|x_{1},y\|_{\alpha})^{k}+((1-p)\|x_{2},y\|_{\alpha})^{k}\right) \\ =(|p|^{\alpha k}+|1-p|^{\alpha k})&\;^{m} \left((\|x_{1},y\|_{\alpha})^{k}+(\|x_{2},y\|_{\alpha})^{k}\right), \\ &\; x_{1},x_{2} \in E, y\in Y_{0}. \end{array} $$

(23)

Thus, by induction, we have shown that (21) holds for every *x*_{1},*x*_{2}∈*E* and \(n\in \mathbb {N}_{0}\). Letting *n*→*∞* in (21), we obtain that

$$\begin{array}{*{20}l} G_{0}(px_{1}+(1-p)x_{2})&\;+ G_{0}((1-p)x_{1}+px_{2})=G_{0}(x_{1}) \\ &\; +G_{0}(x_{2}), \,\, x_{1},x_{2} \in E \end{array} $$

(24)

Write *G*(*x*_{1}):=*G*_{0}(*x*_{1})+*g*(0) for *x*_{1}∈*E*. Then it is easily seen that

$$\begin{array}{*{20}l} G(px_{1}+(1-p)x_{2})&\;+ G((1-p)x_{1}+px_{2})=G(x_{1}) \\ &\;+G(x_{2}), \,\, x_{1},x_{2} \in E \end{array} $$

(25)

and (9) holds. It remains to show the uniqueness of *G*. So suppose that *M*_{0}∈(0,*∞*) and *G*_{1}:*X*→*Y* is a solution to (1) with

$$ \|g(x_{1})-G_{1}(x_{1}),y\|_{\alpha}\leq M_{0}\|x_{1},y\|_{\alpha}, \,\, x_{1}\in E, y\in Y_{0} $$

(26)

Note that

$$ G_{1}(px_{1})+G_{1}((1-p)x_{1})=G_{1}(x_{1})+G_{1}(0),\,\, x_{1}\in E, $$

(27)

$$ G(px_{1})+G((1-p)x_{1})=G(x_{1})+G(0), \,\, x_{1}\in E, $$

(28)

and, by (9),

$$\begin{array}{*{20}l} \|G(x_{1})-G_{1}(x_{1}),y\|_{\alpha} \leq \frac{(M+A)\|x_{1},y\|_{\alpha}^{k}}{1-|p|^{\alpha k}-|1-p|^{\alpha k}} & \\ =(M+A)\|x_{1},y\|_{\alpha}^{k} \sum\limits_{n=j}^{\infty} (|p|^{\alpha k}+|1-p|^{\alpha k})^{n}, \,\, x_{1}\in E &\end{array} $$

(29)

The case *j*=0 is exactly (29). So fix \(l\in \mathbb {N}_{0}\) and assume that (29) holds for *j*=*l*. Then, in view of (27) and (28),

$$\begin{array}{*{20}l} \|G(x_{1})-G_{1}(x_{1}),y\|_{\alpha} &\; \\ =\|G(px_{1})+G((1-p)x_{1})&\;-G_{1}(px_{1})-G_{1}((1-p)x_{1}),y\|_{\alpha}, \\ \leq \|G(px_{1})-G_{1}(px_{1}),y\|_{\alpha}&\;+\|G((1-p)x_{1})-G_{1}((1-p)x_{1}),y\|_{\alpha} \\ \leq (M+A)(\|p\|_{\alpha}^{k}\|x_{1},y\|_{\alpha}^{k}&\; +\|(1-p)\|_{\alpha}^{k}\|x_{1},y\|_{\alpha}^{k}) \sum\limits_{n=l}^{\infty}(|p|^{\alpha k}+|1-p|^{\alpha k})^{n}, \\ \leq (M+A)\|x_{1},y\|_{\alpha}^{k} &\;\sum\limits_{n=l+1}^{\infty}(|p|^{\alpha k}+|1-p|^{\alpha k})^{n}, \,\, x_{1}\in E, y\in Y_{0} \end{array} $$

Thus, we have shown (29). Now, letting *j*→*∞* in (29), we get *G*_{1}=*G*. □