On a General Functional Equation

Abstract

In this paper, we deal with a general functional equation in several variables. We prove the hyperstability of this equation in (m + 1)-normed spaces and describe its general solution in some special cases. In this way, we solve the problems posed by Ciepliński. The considered equation was introduced as a generalization of the equation characterizing n-quadratic functions and has symmetric coefficients (up to sign), and it also generalizes many other known functional equations with symmetric coefficients, such as the multi-Cauchy equation, the multi-Jensen equation, and the multi-Cauchy–Jensen equation. Our results generalize several known results.

1. Introduction

The characterization of inner product spaces among normed spaces is given by the parallelogram law

$${\parallel x+y\parallel }^2+{\parallel x-y\parallel }^2={2\parallel x\parallel }^2+2{\parallel y\parallel }^2,$$

which leads naturally to the following functional equation

$$f(x+y)+f(x-y)=2f\left(x\right)+2f\left(y\right).$$

(1)

It is known, e.g., roughly speaking, that f is a solution of the above equation if and only if there exists a symmetric biadditive mapping B such that $f\left(x\right)=B(x,x)$ for all x. In the case when $f:\mathbb{R}\to \mathbb{R}$, its regular solutions have a form $f\left(x\right)=\alpha x^2$ and for this reason, Equation (1) is called a quadratic equation, and its solutions are called quadratic mappings (see, e.g., [1,2]).

In [3], the authors introduced a counterpart of (1) for multivariable functions:

$$\begin{array}{cc}\hfill \sum _{i_1,i_2,\dots ,i_n\in \{-1,1\}}f\left(x_{11}+i_1{x}_{12},\dots ,x_{n1}+i_n{x}_{n2}\right)& \hfill \\ \hfill =\sum _{j_1,j_2,\dots ,j_n\in \{1,2\}}f(x_{1j_1},\dots ,x_{n{j}_n}),& x_{11},x_{12},\dots ,x_{n1},x_{n2}\in X,\hfill \end{array}$$

(2)

which characterizes the so-called n-quadratic functions (quadratic in each variable). As a natural generalization of the Equation (2) was introduced and investigated in [4,5,6] the following equation

$$\begin{array}{cc}\hfill \sum _{i_1,i_2,\dots ,i_n\in \{-1,1\}}f(a_{1,i_1,i_2,\dots ,i_n}(x_{11}+i_1{x}_{12}),\dots ,a_{n,i_1,i_2,\dots ,i_n}& (x_{n1}+i_n{x}_{n2}))\hfill \\ \hfill =\sum _{j_1,j_2,\dots ,j_n\in \{1,2\}}{A}_{j_1,\dots ,j_n}f(x_{1j_1},\dots ,x_{n{j}_n}),& x_{11},x_{12},\dots ,x_{n1},x_{n2}\in X\hfill \end{array}$$

(3)

for some $a_{1,i_1,i_2,\dots ,i_n},\dots ,a_{n,i_1,i_2,\dots ,i_n}\in \mathbb{F},$ $A_{j_1,\dots ,j_n}\in \mathbb{K}$, where $f:X^n\to Y$, X is a linear space over the field $\mathbb{F}$, and Y is a linear space over the field $\mathbb{K}$.

Let us also mention that some special cases of (3) (one of them is clearly (2)) have been considered for years, because putting

$$g(x_1,x_2,\dots ,x_n):=f(x_1,x_2,\dots ,x_n)-f(0,\dots ,0),x_1,x_2,\dots ,x_n\in X,$$

from (3), we obtain, for example, multi-Cauchy (4), multi-Jensen (5) and multi-Cauchy–Jensen (6) equations:

$$g(x_{11}+x_{12},\dots ,x_{n1}+x_{n2})=\sum _{j_1,j_2,\dots ,j_n\in \{1,2\}}g(x_{1j_1},\dots ,x_{n{j}_n}),$$

(4)

with $a_{l,1,\dots ,1}=1$, $a_{l,i_1,i_2,\dots ,i_n}=0$ for others $i_1,i_2,\dots ,i_n\in \{-1,1\}$ and all $l\in \{1,\dots ,n\},$ $A_{j_1,\dots ,j_n}=1$ for all $j_1,j_2,\dots ,j_n\in \{1,2\};$

$$g\left({\displaystyle \frac{1}{2}}{x}_{11}+{\displaystyle \frac{1}{2}}{x}_{12},\dots ,{\displaystyle \frac{1}{2}}{x}_{n1}+{\displaystyle \frac{1}{2}}{x}_{n2}\right)=\sum _{j_1,j_2,\dots ,j_n\in \{1,2\}}{\displaystyle \frac{1}{2^n}}g(x_{1j_1},\dots ,x_{n{j}_n}),$$

(5)

with $a_{l,1,\dots ,1}={\displaystyle \frac{1}{2}}$, $a_{l,i_1,i_2,\dots ,i_n}=0$ for others $i_1,i_2,\dots ,i_n\in \{-1,1\}$ and all $l\in \{1,\dots ,n\},$ $A_{j_1,\dots ,j_n}={\displaystyle \frac{1}{2^n}}$ for all $j_1,j_2,\dots ,j_n\in \{1,2\};$

$$\begin{array}{cc}\hfill g(x_{11}+x_{12},\dots ,x_{k1}+x_{k2},{\displaystyle \frac{1}{2}}{x}_{k+1,1}+& {\displaystyle \frac{1}{2}}{x}_{k+1,2},\dots ,{\displaystyle \frac{1}{2}}{x}_{n1}+{\displaystyle \frac{1}{2}}{x}_{n2})\hfill \\ \hfill & =\sum _{j_1,j_2,\dots ,j_n\in \{1,2\}}{\displaystyle \frac{1}{2^{n-k}}}g(x_{1j_1},\dots ,x_{n{j}_n}),\hfill \end{array}$$

(6)

where $k\in \{1,\dots ,n-1\}$ is fixed, with $a_{l,1,\dots ,1}=1$ for $l\in \{1,\dots ,k\}$, $a_{l,1,1\dots ,1}={\displaystyle \frac{1}{2}}$ for $l\in \{k+1,\dots ,n\}$, $a_{l,i_1,i_2,\dots ,i_n}=0$ for others $i_1,i_2,\dots ,i_n\in \{-1,1\}$ and $A_{j_1,\dots ,j_n}={\displaystyle \frac{1}{2^{n-k}}}$ for all $j_1,j_2,\dots ,j_n\in \{1,2\}.$

Many physical processes are described by functional equations, and while modeling such processes, various deviations and errors occur. Therefore, it is natural to investigate stability problems in such situations because, considering such problems, we ask how much a slight disturbance of a state affects that state. More classical results on the Ulam stability of functional equations can be found, for example, in the monograph [7].

Over the last few decades, the Ulam stability of various objects (for example, functional equations and inequalities, as well as differential, difference, and integral equations) has been studied by many researchers (see [8,9,10,11,12,13,14,15,16,17,18] for more information on this concept as well as its applications).

Stability problems are studied in various spaces, including m-Banach spaces. Let us mention that such spaces were defined in [19] by A. Misiak.

Let $m\in \mathbb{N}\setminus \left\{1\right\}$ and Y be an at least m-dimensional real linear space. If a mapping $\parallel ·,\dots ,·\parallel :Y^m\to \mathbb{R}$ fulfills the following four conditions:

(i)

$\parallel x_1,\dots ,x_m\parallel =0$ if and only if $x_1,\dots ,x_m$ are linearly dependent,

(ii)

$\parallel x_1,\dots ,x_m\parallel$ is invariant under the permutation of $x_1,\dots ,x_m$,

(iii)

$\parallel \gamma x_1,\dots ,x_m\parallel =\left|\gamma \right|\parallel x_1,\dots ,x_m\parallel$,

(iv)

$\parallel x_{11}+x_{12},x_2,\dots ,x_m\parallel \le \parallel x_{11},x_2,\dots ,x_m\parallel +\parallel x_{12},x_2,\dots ,x_m\parallel$,

for every $\gamma \in \mathbb{R}$ and $x_{11},x_{12},x_1,\dots ,x_m\in Y$, then $\parallel ·,\dots ,·\parallel$ is called an m-norm on Y, and the pair $(Y,\parallel ·,\dots ,·\parallel )$ is said to be an m-normed space.

A sequence ${\left(y_k\right)}_{k\in \mathbb{N}}$ of elements of an m-normed space $(Y,\parallel ·,\dots ,·\parallel )$ is called a Cauchy sequence if

$$\underset{n,k\to \infty }{lim}\parallel y_n-y_k,x_2,\dots ,x_m\parallel =0,x_2,\dots ,x_m\in Y,$$

whereas ${\left(y_k\right)}_{k\in \mathbb{N}}$ is said to be convergent if there exists an element $y\in Y$ (called the limit of this sequence and denoted by ${lim}_{k\to \infty }{y}_k$) with

$$\underset{k\to \infty }{lim}\parallel y_k-y,x_2,\dots ,x_m\parallel =0,x_2,\dots ,x_m\in Y.$$

An m-normed space in which every Cauchy sequence is convergent is called an m-Banach space. From the definition of the m-normed space, we immediately obtain the following well-known property

$$\mathrm{if}{x}_1\in Y\mathrm{and}\parallel x_1,x_2,\dots ,x_m\parallel =0,\mathrm{for}\mathrm{all}{x}_2,\dots ,x_m\in Y,\mathrm{then}{x}_1=0.$$

(7)

For more details, we refer the reader to [19,20,21].

In [6], the author studied the Ulam stability of a general functional equation in several variables and determined, among others, that Equation (3) is Ulam stable in m-Banach spaces. Namely, he proved (Theorem 2, [6]) the following theorem.

Theorem 1. 

Let $m\in \mathbb{N},\epsilon >0,X$ be a linear space, Y be an $(m+1)$-Banach space, and

$$\left|\sum _{j_1,j_2,\dots ,j_n\in \{1,2\}}{A}_{j_1,\dots ,j_n}\right|>1.$$

If $f:X^n\to Y$ is a mapping such that $f(x_{11},\dots ,x_{n1})=0$ for any $(x_{11},\dots ,x_{n1})\in X^n$ with at least one component which is equal to zero and

$$\begin{array}{cc}\hfill \parallel \sum _{i_1,i_2,\dots ,i_n\in \{-1,1\}}& f\left(a_{1,i_1,i_2,\dots ,i_n}(x_{11}+i_1{x}_{12}),\dots ,a_{n,i_1,i_2,\dots ,i_n}(x_{n1}+i_n{x}_{n2})\right)\hfill \\ \hfill & -\sum _{j_1,j_2,\dots ,j_n\in \{1,2\}}{A}_{j_1,\dots ,j_n}f(x_{1j_1},\dots ,x_{n{j}_n}),z\parallel \le \epsilon \hfill \end{array}$$

(8)

for $x_{11},x_{12},\dots ,x_{n1},x_{n2}\in X$ and $z\in Y^m$, then there exists a function $F:X^n\to Y$ satisfying (3) and the condition

$$\parallel f(x_1,\dots ,x_n)-F(x_1,\dots ,x_n),z\parallel \le {\displaystyle \frac{\epsilon }{|{\sum }_{j_1,j_2,\dots ,j_n\in \{1,2\}}{A}_{j_1,\dots ,j_n}|-1}},$$

for all $x_1,\dots ,x_n\in X$ and $z\in Y^m.$

Moreover, he posed three problems:

Problem 1. 

Does the assertion of Theorem 1 hold without the assumption that $f(x_{11},\dots ,x_{n1})=0$ for any $(x_{11},\dots ,x_{n1})\in X^n$ with at least one component which is equal to zero?

Problem 2. 

Is Equation (3) stable for

$$\left|\sum _{j_1,j_2,\dots ,j_n\in \{1,2\}}{A}_{j_1,\dots ,j_n}\right|\le 1?$$

Problem 3. 

Find a general solution of Equation (3).

In this paper, we provide positive answers to Problems 1 and 2 and describe the general solution of the Equation (3) for $n\in \{1,2\}.$ To answer the questions posed as Problems 1 and 2, we present a rather short proof of hyperstability Equation (3) in $(m+1)$-normed spaces. We speak about the hyperstability phenomenon when no deviation of a state affects that state. Our result also improves a result from Theorem 1, where the stability of the Equation (3) was shown. Let us mention that the problem that was put forth in [4] about the Ulam stability of Equation (3) in 2-Banach spaces was considered in [22] (see also [20]).

By $\mathbb{N}$, we will understand the sets of positive integers.

2. Hyperstability of (3) in $\mathbf{(}\mathit{m}\mathbf{+}\mathbf{1}\mathbf{)}$-Banach Spaces

In this section, we prove that functional Equation (3) is Ulam hyperstable in $(m+1)$-normed spaces. In [6], the author proved Theorem 1 and showed that Equation (3) is Ulam stable in m-Banach spaces. In fact, we can obtain more, namely the hyperstability of (3) in $(m+1)$-normed spaces, that is, we do not only obtain an approximation of f by a function satisfying the equation, but f itself has to satisfy already the equation.

Theorem 1 can be easily generalized in the following way.

Theorem 2. 

Let $m\in \mathbb{N},$ X be a linear space, Y be an $(m+1)$-normed space, and $\phi :X^{2n}\to [0,\infty ).$ Assume also that $f:X^n\to Y$ is a mapping satisfying

$$\begin{array}{cc}\hfill \parallel \sum _{i_1,i_2,\dots ,i_n\in \{-1,1\}}f(a_{1,i_1,i_2,\dots ,i_n}(x_{11}+i_1{x}_{12}),\dots ,& a_{n,i_1,i_2,\dots ,i_n}(x_{n1}+i_n{x}_{n2}))\hfill \\ \hfill -\sum _{j_1,j_2,\dots ,j_n\in \{1,2\}}{A}_{j_1,\dots ,j_n}f(x_{1j_1},\dots ,x_{n{j}_n}),z_1,\dots ,& z_m\parallel \le \phi (x_{11},x_{12},\dots ,x_{n1},x_{n2}),\hfill \end{array}$$

(9)

for $x_{11},x_{12},\dots ,x_{n1},x_{n2}\in X$ and $z_1,\dots ,z_m\in Y.$ Then, function f satisfies the Equation (3).

Proof. 

Denote $(\Phi f)(x_{11},x_{12},\dots ,x_{n1},x_{n2}):=$

$$\sum _{i_1,i_2,\dots ,i_n\in \{-1,1\}}f\left(a_{1,i_1,i_2,\dots ,i_n}(x_{11}+i_1{x}_{12}),\dots ,a_{n,i_1,i_2,\dots ,i_n}(x_{n1}+i_n{x}_{n2})\right)$$

$$-\sum _{j_1,j_2,\dots ,j_n\in \{1,2\}}{A}_{j_1,\dots ,j_n}f(x_{1j_1},\dots ,x_{n{j}_n})$$

for $x_{11},x_{12},\dots ,x_{n1},x_{n2}\in X$.

Fix $x_{11},x_{12},\dots ,x_{n1},x_{n2}\in X$ and $z_1,\dots ,z_m\in Y$. Then,

$$\parallel (\Phi f)(x_{11},x_{12},\dots ,x_{n1},x_{n2}),l{z}_1,\dots ,l{z}_m\parallel \le \phi (x_{11},x_{12},\dots ,x_{n1},x_{n2}),l\in \mathbb{N},$$

and consequently

$$\parallel (\Phi f)(x_{11},x_{12},\dots ,x_{n1},x_{n2}),z_1,\dots ,z_m\parallel \le {\displaystyle \frac{1}{l^m}}\phi (x_{11},x_{12},\dots ,x_{n1},x_{n2}),l\in \mathbb{N}.$$

Hence, letting $l\to \infty$, we obtain

$$\parallel (\Phi f)(x_{11},x_{12},\dots ,x_{n1},x_{n2}),z_1,\dots ,z_m\parallel =0,$$

(10)

as a consequence, we indicate that for every $x_{11},x_{12},\dots ,x_{n1},x_{n2}\in X$ and $z_1,\dots ,z_m\in Y$ (10) holds, which means, using (7), that f satisfies Equation (3). □

We notice that Theorem 2 gives a positive answer to both problems (for the second in $(m+1)$-normed spaces), because if function f satisfies (9), then it must satisfy (3).

From the proof of Theorem 2, we determine that it is sufficient to assume that Y is an $(m+1)$-normed space (in Theorem 1, Y is an $(m+1)$-Banach space), since in the proof, we only use the property (7) that holds in $(m+1)$-normed spaces.

Moreover, the proof of Theorem 2 shows that it can be repeated for any functional equation, and for this reason, the study of Ulam stability of functional equations (with a control function $\epsilon$ or another one independent of z) in $(m+1)$-Banach spaces or $(m+1)$-normed spaces seems pointless.

3. General Solution of (3) for $\mathit{n}\mathbf{\in }\mathbf{\{}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{\}}$

In this section, X and Y will denote linear spaces over the field of real or complex numbers, and unless otherwise noted, equations containing some of the variables $x,y,x_1,x_2,y_1,y_2$ will be considered for all variables in them.

We start with the following observation. If

$$\sum _{j_1,j_2,\dots ,j_n\in \{1,2\}}{A}_{j_1,\dots ,j_n}=2^n,$$

then every constant function $f:X^n\to Y$ satisfies (3). Otherwise, $f=0$ is the only constant function satisfying (3).

3.1. General Solution of (3) for $n=1$

For $n=1$, we can write the Equation (3) in a very simple form

$$f\left(a(x+y)\right)+f\left(b(x-y)\right)=Af\left(x\right)+Bf\left(y\right),x,y\in X,$$

(11)

where $a_{1,1}=a,a_{1,-1}=b,A_1=A,A_2=B,x_{11}=x,x_{12}=y.$

Now, we compare the following examples.

Example 1. 

(a) Function $f\left(x\right)=x+1,x\in \mathbb{R}$ satisfies (11) with $a=1,b=2,A=3,$ $B=-1.$

(b) Function $f\left(x\right)=x^2,x\in \mathbb{R}$ satisfies (11) with $a=-2,b=2,A=8,B=8.$

(c) Function $f\left(x\right)=x^2+x+1,x\in \mathbb{R}$ does not satisfy (11) with any coefficients $a,b,A,B$.

Theorem 3. 

If a function $f:X\to Y$ satisfies (11), then there exist an additive function $\theta :X\to Y$, a quadratic function $q:X\to Y$, and a constant $\delta \in Y$ such that

$$f\left(x\right)=q\left(x\right)+\theta \left(x\right)+\delta ,$$

(12)

and

$$\begin{array}{c}2\theta \left(ax\right)=(A+B)\theta \left(x\right),\hfill \\ 2\theta \left(bx\right)=(A-B)\theta \left(x\right),\hfill \end{array}$$

(13)

$$\begin{array}{c}2q\left(ax\right)=2q\left(bx\right)=Bq\left(x\right)=Aq\left(x\right),\hfill \end{array}$$

(14)

for all $x\in X$ and, moreover,

$$\begin{array}{c}\delta =0,\mathit{whenever}A+B\ne 2;\hfill \\ \theta =0,\mathit{whenever}A=B\mathit{and}f\left(ax\right)=f\left(bx\right)\mathit{for}\mathit{all}x\in X;\hfill \\ q=0,\mathit{whenever}A\ne B\mathit{or}f\left(ax\right)\ne f\left(bx\right)\mathit{for}\mathit{some}x\in X.\hfill \end{array}$$

(15)

Conversely, for every additive function $\theta :X\to Y$, quadratic function $q:X\to Y$, such that Conditions (13) and (14) hold for all $x\in X$ and for every $\delta \in Y$ such that

$$\delta (A+B-2)=0,$$

the function $f:X\to Y$ of the form (12) is a solution to (11).

Proof. 

Assume f satisfies (11) and define the function $g\left(x\right):=f\left(x\right)-f\left(0\right),$ for $x\in X.$ Then, the function g satisfies the equation

$$g\left(a(x+y)\right)+g\left(b(x-y)\right)=Ag\left(x\right)+Bg\left(y\right),x,y\in X,$$

(16)

$g\left(0\right)=0$ and $f\left(x\right)=g\left(x\right)+f\left(0\right).$

Putting in (16) $y=0$, and then $x=0$, we obtain

$$g\left(ax\right)+g\left(bx\right)=Ag\left(x\right),$$

(17)

$$g\left(ax\right)+g(-bx)=Bg\left(x\right),$$

(18)

respectively. Substituting $-x$ for x in (17) and (18), we obtain

$$g(-ax)+g(-bx)=Ag(-x),$$

(19)

$$g(-ax)+g\left(bx\right)=Bg(-x),$$

(20)

respectively. Subtracting Equation (20) from Equation (17), and Equation (19) from Equation (18), we have

$$g\left(ax\right)-g(-ax)=Ag\left(x\right)-Bg(-x),$$

$$g\left(ax\right)-g(-ax)=Bg\left(x\right)-Ag(-x),$$

hence,

$$Ag\left(x\right)-Bg(-x)=Bg\left(x\right)-Ag(-x),$$

and consequently

$$A\left(g\left(x\right)+g(-x)\right)=B\left(g\left(x\right)+g(-x)\right),$$

which means that $A=B$ or g is an odd function.

First, we consider the case when $A=B$. Then, subtracting Equation (18) from Equation (17), we have

$$g\left(bx\right)-g(-bx)=0,$$

which means that $b=0$ or g is an even function.

If $b=0$ from the Equation (17), we have $g\left(ax\right)=Ag\left(x\right)$, and then Equation (16) has the form

$$g\left(a(x+y)\right)=A\left(g\left(x\right)+g\left(y\right)\right),$$

hence,

$$Ag(x+y)=A\left(g\left(x\right)+g\left(y\right)\right),$$

which means that g is an additive function such that $g\left(ax\right)=Ag\left(x\right).$

If $A=B$ and g is an even function, putting in (16) $y=x$, then $y=-x$, we obtain

$$g\left(2ax\right)=2Ag\left(x\right),$$

$$g\left(2bx\right)=2Ag\left(x\right),$$

respectively, and hence

$$g\left(2ax\right)=g\left(2bx\right)=2Ag\left(x\right).$$

Therefore the Equations (16) and (17) have the following forms

$$g\left(a(x+y)\right)+g\left(a(x-y)\right)=A\left(g\left(x\right)+g\left(y\right)\right),$$

and

$$2g\left(ax\right)=Ag\left(x\right),$$

and hence

$$A\left(g(x+y)+g(x-y)\right)=2A\left(g\left(x\right)+g\left(y\right)\right),$$

which means that g is a quadratic function such that $2g\left(ax\right)=2g\left(bx\right)=Ag\left(x\right).$

Now, we consider the case when g is an odd function. Adding Equations (17) and (18) and then Equations (17) and (20), we have

$$2g\left(ax\right)=(A+B)g\left(x\right)\mathrm{and}2g\left(bx\right)=(A-B)g\left(x\right).$$

Putting the above equations into the Equation (16), we obtain

$$(A+B)g(x+y)+(A-B)g(x-y)=2Ag\left(x\right)+2Bg\left(y\right),$$

(21)

then, replacing x with y in the above equation, we get

$$(A+B)g(x+y)+(A-B)g(y-x)=2Ag\left(y\right)+2Bg\left(x\right).$$

(22)

Adding Equations (21) and (22), we have

$$2(A+B)g(x+y)=2(A+B)\left(g\left(x\right)+g\left(y\right)\right),$$

which means that $A+B=0$ or g is an additive function such that

$$2g\left(ax\right)=(A+B)g\left(x\right)\mathrm{and}2g\left(bx\right)=(A-B)g\left(x\right).$$

If $A+B=0$ and g is a non-zero function, then $a=0,$ and using (17), we obtain $g\left(bx\right)=Ag\left(x\right)$, and then the Equation (16) has the form

$$Ag(x-y)=A\left(g\left(x\right)-g\left(y\right)\right).$$

Inserting in the above equation $-y$ in place of y and using the oddness of the function g, we obtain

$$Ag(x+y)=A\left(g\left(x\right)+g\left(y\right)\right),$$

which means that g is an additive function such that $g\left(bx\right)=Ag\left(x\right).$

Putting $x=y=0$ in (11), we have

$$2f\left(0\right)=(A+B)f\left(0\right),$$

which means that $f\left(0\right)=0$ or $A+B=2.$

Putting $f\left(0\right)=\delta$ and $f\left(x\right)=g\left(x\right)+\delta$, where

$$\begin{array}{ccc}g\left(x\right)=q\left(x\right)\hfill & (\theta =0)\hfill & \mathrm{for}A=B\wedge g\left(ax\right)=g\left(bx\right),x\in X,\hfill \\ g\left(x\right)=\theta \left(x\right)\hfill & (q=0)\hfill & \mathrm{for} \mathrm{others},\hfill \end{array}$$

we obtain as a consequence (15).

The proof of the converse is a direct computation. □

Corollary 1. 

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous solution of (11). Then, there exist $\alpha ,\beta ,\delta \in \mathbb{R}$ such that

$$f\left(x\right)=\alpha x^2+\beta x+\delta ,x\in \mathbb{R},$$

(23)

and moreover,

$$\begin{array}{ccc}\alpha =0,\hfill & \mathit{whenever}\hfill & A\ne B\vee \left|a\right|\ne \left|b\right|\vee 2a^2\ne A;\hfill \\ \beta =0,\hfill & \mathit{whenever}\hfill & 2a\ne A+B\vee 2b\ne A-B;\hfill \\ \delta =0,\hfill & \mathit{whenever}\hfill & A+B\ne 2.\hfill \end{array}$$

Proof. 

The function f is continuous and so, by Theorem 3, any solution of (11) is of the form $f\left(x\right)=\alpha x^2+\beta x+\delta ,x\in \mathbb{R}$ with some $\alpha ,\beta ,\delta \in \mathbb{R}$. Whence, we derive the following conditions

$$\begin{array}{c}2\alpha a^2=2\alpha b^2=\alpha B=\alpha A\hfill \\ 2\beta a=\beta (A+B)\hfill \\ 2\beta b=\beta (A-B)\hfill \\ 2\delta =(A+B)\delta ,\hfill \end{array}$$

which leads to our assertion. □

Remark 1. 

We notice that for $a,b,A,B\in \mathbb{R}$, such that at least one of them is not zero, we have:

• if $A=B$ and $\left|a\right|=\left|b\right|$ and $2a^2=A$, then $2a\ne A+B$ or $2b\ne A-B$;
• if $2a=A+B$ and $2b=A-B$, then $A\ne B$ or $\left|a\right|\ne \left|b\right|$ or $2a^2\ne A$,

which means that $\alpha =0$ or $\beta =0$ in (23).

3.2. General Solution of (3) for $n=2$

For $n=2$, Equation (3) can be written in the following form

$$\begin{array}{cc}\hfill & f\left(a_{11}(x_1+x_2),b_{11}(y_1+y_2)\right)+f\left(a_{12}(x_1+x_2),b_{12}(y_1-y_2)\right)\hfill \\ \hfill & +f\left(a_{21}(x_1-x_2),b_{21}(y_1+y_2)\right)+f\left(a_{22}(x_1-x_2),b_{22}(y_1-y_2)\right)\hfill \\ =\hfill & A_{11}f(x_1,y_1)+A_{12}f(x_1,y_2)+A_{21}f(x_2,y_1)+A_{22}f(x_2,y_2),x_1,x_2,y_1,y_2\in X,\hfill \end{array}$$

(24)

where $a_{1,1,1}=a_{11},a_{2,1,1}=b_{11},a_{1,1,-1}=a_{12},a_{2,1,-1}=b_{12},a_{1,-1,1}=a_{21},a_{2,-1,1}=b_{21},$ $a_{1,-1,-1}=a_{22},a_{2,-1,-1}=b_{22},x_{1i}=x_i,x_{2i}=y_i$ for $i\in \{1,2\}.$

Now, we give examples of functions that satisfy the Equation (24) with certain coefficients $a_{ij},b_{ij},A_{ij}$ for $i,j\in \{1,2\}.$

Example 2. 

(a) Function $f(x,y)=x{y}^2+x+y,x,y\in \mathbb{R}$ satisfies the equation

$$\begin{array}{cc}\hfill f\left(2(x_1+x_2),{\displaystyle \frac{\sqrt{2}}{2}}(y_1+y_2)\right)& +f\left(2(x_1+x_2),{\displaystyle \frac{\sqrt{2}}{2}}(y_1-y_2)\right)+f\left(0,\left(4-{\displaystyle \frac{\sqrt{2}}{2}}\right)(y_1+y_2)\right)\hfill \\ \hfill +f\left(0,-{\displaystyle \frac{\sqrt{2}}{2}}(y_1-y_2)\right)& =2f(x_1,y_1)+2f(x_1,y_2)+2f(x_2,y_1)+2f(x_2,y_2).\hfill \end{array}$$

(b) 

Function $f(x,y)=x^{2}y+x^2+y^2+y,x,y\in \mathbb{R}$ satisfies the equation

$$\begin{array}{cc}\hfill f\left({\displaystyle \frac{\sqrt{2}}{2}}(x_1+x_2),{\displaystyle \frac{1}{2}}(y_1+y_2)\right)& +f\left(0,-{\displaystyle \frac{1}{2}}(y_1-y_2)\right)+f\left({\displaystyle \frac{\sqrt{2}}{2}}(x_1-x_2),{\displaystyle \frac{1}{2}}(y_1+y_2)\right)\hfill \\ \hfill +f\left(0,{\displaystyle \frac{1}{2}}(y_1-y_2)\right)& ={\displaystyle \frac{1}{2}}f(x_1,y_1)+{\displaystyle \frac{1}{2}}f(x_1,y_2)+{\displaystyle \frac{1}{2}}f(x_2,y_1)+{\displaystyle \frac{1}{2}}f(x_2,y_2).\hfill \end{array}$$

(c) 

Function $f(x,y)=x^2{y}^2+x+y^2,x,y\in \mathbb{R}$ satisfies the equation

$$\begin{array}{cc}\hfill & f\left({\displaystyle \frac{\sqrt{2}}{2}}(x_1+x_2),{\displaystyle \frac{\sqrt[4]{2}}{2}}(y_1+y_2)\right)+f\left({\displaystyle \frac{\sqrt{2}}{2}}(x_1+x_2),{\displaystyle \frac{\sqrt[4]{2}}{2}}(y_1-y_2)\right)\hfill \\ & +f\left({\displaystyle \frac{\sqrt{2}}{2}}(x_1-x_2),{\displaystyle \frac{\sqrt[4]{2}}{2}}(y_1+y_2)\right)+f\left(-{\displaystyle \frac{\sqrt{2}}{2}}(x_1-x_2),{\displaystyle \frac{\sqrt[4]{2}}{2}}(y_1-y_2)\right)\hfill \\ \hfill & ={\displaystyle \frac{\sqrt{2}}{2}}f(x_1,y_1)+{\displaystyle \frac{\sqrt{2}}{2}}f(x_1,y_2)+{\displaystyle \frac{\sqrt{2}}{2}}f(x_2,y_1)+{\displaystyle \frac{\sqrt{2}}{2}}f(x_2,y_2).\hfill \end{array}$$

(d) 

Function $f(x,y)=xy+x^2+x+y,x,y\in \mathbb{R}$ satisfies the equation

$$\begin{array}{cc}\hfill f\left({\displaystyle \frac{1}{2}}(x_1+x_2),y_1+y_2\right)& +f\left({\displaystyle \frac{1}{2}}(x_1+x_2),0\right)+f\left(-{\displaystyle \frac{1}{2}}(x_1-x_2),0\right)+f\left({\displaystyle \frac{1}{2}}(x_1-x_2),0\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}f(x_1,y_1)+{\displaystyle \frac{1}{2}}f(x_1,y_2)+{\displaystyle \frac{1}{2}}f(x_2,y_1)+{\displaystyle \frac{1}{2}}f(x_2,y_2).\hfill \end{array}$$

(e) 

Function $f(x,y)=sinx+y,x,y\in \mathbb{R}$ satisfies the equation

$$\begin{array}{cc}\hfill & f\left(2(x_1+x_2),{\displaystyle \frac{\sqrt{2}}{2}}(y_1+y_2)\right)+f\left(-2(x_1+x_2),2(y_1-y_2)\right)+f\left(3(x_1-x_2),-{\displaystyle \frac{\sqrt{2}}{2}}(y_1+y_2)\right)\hfill \\ \hfill & +f\left(-3(x_1-x_2),2(y_1-y_2)\right)=2f(x_1,y_1)-2f(x_1,y_2)+2f(x_2,y_1)-2f(x_2,y_2).\hfill \end{array}$$

(f) 

Function $f(x,y)=2xy+x^3+3sinx+y^7-4siny,x,y\in \mathbb{R}$ satisfies the equation

$$\begin{array}{cc}\hfill f\left(0,-(y_1-y_2)\right)+& f\left(-(x_1-x_2),0)\right)+f\left(x_1-x_2,y_1-y_2\right)\hfill \\ \hfill & =f(x_1,y_1)-f(x_1,y_2)-f(x_2,y_1)+f(x_2,y_2).\hfill \end{array}$$

Theorem 4. 

If a function $f:X^2\to Y$ satisfies (24), then there exist the functions $h,h_{k_1{k}_2}:X^2\to Y$, $\phi ,\psi ,{\phi }_{l_1},{\psi }_{l_2}:X\to Y,$ $k_1,k_2,l_1,l_2\in \{a,q\}$ and a constant $\delta \in Y$, such that

$$\begin{array}{ccc}\hfill & \hfill f(x,y)& =h(x,y)+\phi \left(x\right)+\psi \left(y\right)+\delta ,\hfill \\ \mathit{where}\hfill & \hfill h(x,y)& =h_{aa}(x,y)+h_{qq}(x,y)+h_{aq}(x,y)+h_{qa}(x,y),\hfill \\ \hfill & \hfill \phi \left(x\right)& ={\phi }_a\left(x\right)+{\phi }_q\left(x\right),\hfill \\ \hfill & \hfill \psi \left(y\right)& ={\psi }_a\left(y\right)+{\psi }_q\left(y\right),\hfill \end{array}$$

(25)

for all $x,y\in X$ and $h_{aa}$ are biadditive, $h_{qq}$ is biquadratic, $h_{aq}$ is additive with respect to the first variable and quadratic with respect to the second variable, $h_{qa}$ is quadratic with respect to the first variable and additive with respect to the second variable, ${\phi }_q,{\psi }_q$ are quadratic, ${\phi }_a$ is odd when $A_{11}+A_{12}+A_{21}+A_{22}=A_{11}+A_{12}=0$ and additive in others, ${\psi }_a$ is odd when $A_{11}+A_{12}+A_{21}+A_{22}=A_{11}+A_{21}=0$ and additive in others, and such that

$$\begin{array}{cc}4h_{aa}(a_{11}x,b_{11}y)\hfill & =\left(A_{11}+A_{12}+A_{21}+A_{22}\right)h_{aa}(x,y),\hfill \\ 4h_{aa}(a_{12}x,b_{12}y)\hfill & =\left((A_{11}+A_{21})-(A_{12}+A_{22})\right)h_{aa}(x,y),\hfill \\ 4h_{aa}(a_{21}x,b_{21}y)\hfill & =\left((A_{11}+A_{12})-(A_{21}+A_{22})\right)h_{aa}(x,y),\hfill \\ 4h_{aa}(a_{22}x,b_{22}y)\hfill & =\left((A_{11}+A_{22})-(A_{12}+A_{21})\right)h_{aa}(x,y),\hfill \end{array}$$

(h_aa)

$$\begin{array}{cc}4h_{qq}(a_{ij}x,b_{ij}y)=A_{11}{h}_{qq}(x,y)\hfill & \wedge A_{ij}=A_{11}foralli,j\in \{1,2\},\hfill \end{array}$$

(h_qq)

$$\begin{array}{c}2h_{aq}(a_{i1}x,b_{i1}y)=2h_{aq}(a_{i2}x,b_{i2}y)=A_{11}{h}_{aq}(x,y)=A_{12}{h}_{aq}(x,y)\wedge \hfill \\ a_{3-i1}{b}_{3-i1}=a_{3-i2}{b}_{3-i2}=0\wedge A_{21}=A_{22}={(-1)}^{3-i}{A}_{11}forsomei\in \{1,2\},\hfill \end{array}$$

(h_aq)

$$\begin{array}{c}2h_{qa}(a_{1j}x,b_{1j}y)=2h_{qa}(a_{2j}x,b_{2j}y)=A_{11}{h}_{qa}(x,y)=A_{21}{h}_{qa}(x,y)\wedge \hfill \\ a_{13-j}{b}_{13-j}=a_{23-j}{b}_{23-j}=0\wedge A_{12}=A_{22}={(-1)}^{3-j}{A}_{11}forsomej\in \{1,2\},\hfill \end{array}$$

(h_qa)

$$\begin{array}{cc}\hfill 2\left({\phi }_a\left(a_{11}x\right)+{\phi }_a\left(a_{12}x\right)\right)& =\left(A_{11}+A_{12}+A_{21}+A_{22}\right){\phi }_a\left(x\right),\hfill \\ \hfill 2\left({\phi }_a\left(a_{21}x\right)+{\phi }_a\left(a_{22}x\right)\right)& =\left((A_{11}+A_{12})-(A_{21}+A_{22})\right){\phi }_a\left(x\right),\hfill \end{array}$$

(φ_a)

$$\begin{array}{cc}\hfill 2\left({\psi }_a\left(b_{11}y\right)+{\psi }_a\left(b_{21}y\right)\right)& =\left(A_{11}+A_{12}+A_{21}+A_{22}\right){\psi }_a\left(y\right),\hfill \\ \hfill 2\left({\psi }_a\left(b_{12}y\right)+{\psi }_a\left(b_{22}y\right)\right)& =\left((A_{11}+A_{21})-(A_{12}+A_{22})\right){\psi }_a\left(y\right),\hfill \end{array}$$

(ψ_a)

$$\begin{array}{cc}\hfill 2\left({\phi }_q\left(a_{11}x\right)+{\phi }_q\left(a_{12}x\right)\right)& =2\left({\phi }_q\left(a_{21}x\right)+{\phi }_q\left(a_{22}x\right)\right)\hfill \\ \hfill & =\left(A_{11}+A_{12}\right){\phi }_q\left(x\right)=\left(A_{21}+A_{22}\right){\phi }_q\left(x\right),\hfill \end{array}$$

(φ_q)

$$\begin{array}{cc}\hfill 2\left({\psi }_q\left(b_{11}y\right)+{\psi }_q\left(b_{21}y\right)\right)& =2\left({\psi }_q\left(b_{12}y\right)+{\psi }_q\left(b_{22}y\right)\right)\hfill \\ \hfill & =\left(A_{11}+A_{21}\right){\psi }_q\left(y\right)=\left(A_{12}+A_{22}\right){\psi }_q\left(y\right),\hfill \end{array}$$

(ψ_q)

for all $x,y\in X$ and, moreover,

$$\delta =0,\mathit{whenever}{A}_{11}+A_{21}+A_{12}+A_{22}\ne 4.$$

Conversely, for every function $h,h_{k_1{k}_2}:X^2\to Y$, $\phi ,\psi ,{\phi }_{l_1},{\psi }_{l_2}:X\to Y,$ where $k_1,k_2,l_1,l_2\in \{a,q\}$ and $h_{aa}$ are biadditive, $h_{qq}$ is biquadratic, $h_{aq}$ is additive with respect to the first variable and quadratic with respect to the second variable, $h_{qa}$ is quadratic with respect to the first variable and additive with respect to the second variable, ${\phi }_q,{\psi }_q$ are quadratic, ${\phi }_a$ is odd when $A_{11}+A_{12}+A_{21}+A_{22}=A_{11}+A_{12}=0$ and additive in others, ${\psi }_a$ is odd when $A_{11}+A_{12}+A_{21}+A_{22}=A_{11}+A_{21}=0$ and additive in others, and such that conditions $\left(h_{k_1{k}_2}\right)$, $\left({\phi }_{l_1}\right)$, $\left({\psi }_{l_2}\right)$ hold for all $x,y\in X$ and for every constant $\delta \in Y$, such that

$$\delta (A_{11}+A_{12}+A_{21}+A_{22}-4)=0,$$

the function $f:X^2\to Y$ of the form (25) is a solution to (24).

Proof. 

Assume f satisfies (24) and define the function $g(x,y):=f(x,y)-f(0,0)$ for $x,y\in X.$ Then, the function g for all $x_1,x_2,y_1,y_2\in X$ satisfies the equation

$$\begin{array}{cc}\hfill g(a_{11}(x_1+x_2),b_{11}& (y_1+y_2))+g\left(a_{12}(x_1+x_2),b_{12}(y_1-y_2)\right)\hfill \\ \hfill +g(a_{21}(x_1-x_2),& b_{21}(y_1+y_2))+g\left(a_{22}(x_1-x_2),b_{22}(y_1-y_2)\right)\hfill \\ \hfill & =A_{11}g(x_1,y_1)+A_{12}g(x_1,y_2)+A_{21}g(x_2,y_1)+A_{22}g(x_2,y_2),\hfill \end{array}$$

(26)

$g(0,0)=0$ and $f(x,y)=g(x,y)+f(0,0).$

Putting in (26) $y_1=y_2=0,$ $x_1=x,$ $x_2=-x$ and then $x_1=x_2=0,$ $y_1=y,$ $y_2=-y$, we have

$$g\left(2a_{21}x,0\right)+g\left(2a_{22}x,0\right)=\left(A_{11}+A_{12}\right)g(x,0)+\left(A_{21}+A_{22}\right)g(-x,0),$$

(27)

and

$$g\left(0,2b_{12}y\right)+g\left(0,2b_{22}y\right)=\left(A_{11}+A_{21}\right)g(0,y)+\left(A_{12}+A_{22}\right)g(0,-y),$$

(28)

respectively. Putting in (26) $y_1=y_2=y$, we obtain

$$\begin{array}{cc}\hfill g\left(a_{11}(x_1+x_2),2b_{11}y\right)+& g\left(a_{12}(x_1+x_2),0\right)+g\left(a_{21}(x_1-x_2),2b_{21}y\right)\hfill \\ \hfill +g\left(a_{22}(x_1-x_2),0\right)=& (A_{11}+A_{12})g(x_1,y)+(A_{21}+A_{22})g(x_2,y).\hfill \end{array}$$

(29)

Putting in (29) $x_1=x_2=x$, then $x_1=x$ and $x_2=-x$, and finally $x_1=x_2=0$, we get

$$\begin{array}{c}g\left(2a_{11}x,2b_{11}y\right)+g\left(2a_{12}x,0\right)+g\left(0,2b_{21}y\right)=(A_{11}+A_{12}+A_{21}+A_{22})g(x,y),\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill g\left(0,2b_{11}y\right)+& g\left(2a_{21}x,2b_{21}y\right)+g\left(2a_{22}x,0\right)\hfill \\ \hfill & =(A_{11}+A_{12})g(x,y)+(A_{21}+A_{22})g(-x,y)\hfill \end{array}$$

(31)

and

$$\begin{array}{c}g\left(0,2b_{11}y\right)+g\left(0,2b_{21}y\right)=(A_{11}+A_{12}+A_{21}+A_{22})g(0,y),\hfill \end{array}$$

(32)

respectively.

Putting (30) (with $x=x_1+x_2$) and (31) (with $x=x_1-x_2$) into (29), we have

$$\begin{array}{cc}\hfill (A_{11}+A_{12}+A_{21}+A_{22})& g(x_1+x_2,y)-g\left(0,2b_{21}y\right)\hfill \\ \hfill +(A_{11}+A_{12})& g(x_1-x_2,y)+(A_{21}+A_{22})g(x_2-x_1,y)-g(0,2b_{11}y)\hfill \\ \hfill & =(A_{11}+A_{12})g(2x_1,y)+(A_{21}+A_{22})g(2x_2,y),\hfill \end{array}$$

(33)

hence using (32), we obtain

$$\begin{array}{cc}\hfill (A_{11}+A_{12}+A_{21}+A_{22})& g(x_1+x_2,y)-(A_{11}+A_{12}+A_{21}+A_{22})g(0,y)\hfill \\ \hfill +(A_{11}+A_{12})& g(x_1-x_2,y)+(A_{21}+A_{22})g(x_2-x_1,y)\hfill \\ \hfill & =(A_{11}+A_{12})g(2x_1,y)+(A_{21}+A_{22})g(2x_2,y).\hfill \end{array}$$

(34)

Replacing $x_1$ and $x_2$ in the Equation (34), we have

$$\begin{array}{cc}\hfill (A_{11}+A_{12}+A_{21}+A_{22})& g(x_1+x_2,y)-(A_{11}+A_{12}+A_{21}+A_{22})g(0,y)\hfill \\ \hfill +(A_{11}+A_{12})& g(x_2-x_1,y)+(A_{21}+A_{22})g(x_1-x_2,y)\hfill \\ \hfill & =(A_{11}+A_{12})g(2x_2,y)+(A_{21}+A_{22})g(2x_1,y).\hfill \end{array}$$

(35)

We denote $A:=A_{11}+A_{12}+A_{21}+A_{22}.$

Adding Equations (34) and (35), we obtain

$$\begin{array}{cc}A(2g(x_1+x_2,y)-2g(0,y)+\hfill & g(x_1-x_2,y)+g(x_2-x_1,y))\hfill \\ \hfill & =A\left(g(2x_1,y)+g(2x_2,y)\right).\hfill \end{array}$$

(36)

Putting $y=0$ in the above equation, we have

$$\begin{array}{c}A\left(2g(x_1+x_2,0)+g(x_1-x_2,0)+g(x_2-x_1,0)\right)=A\left(g(2x_1,0)+g(2x_2,0)\right).\hfill \end{array}$$

(37)

Putting in (26) $x_1=x_2=x$, we get

$$\begin{array}{cc}\hfill g\left(2a_{11}x,b_{11}(y_1+y_2)\right)+& g\left(2a_{12}x,b_{12}(y_1-y_2)\right)+g\left(0,b_{21}(y_1+y_2)\right)\hfill \\ \hfill +g\left(0,b_{22}(y_1-y_2)\right)=& (A_{11}+A_{21})g(x,y_1)+(A_{12}+A_{22})g(x,y_2).\hfill \end{array}$$

(38)

Putting in (38) $y_1=y_2=y$, then $y_1=y$ and $y_2=-y$, and finally $y_1=y_2=0$, we get (30) and

$$\begin{array}{cc}g\left(2a_{11}x,0\right)+\hfill & g\left(2a_{12}x,2b_{12}y\right)+g\left(0,2b_{22}y\right)\hfill \\ \hfill & =(A_{11}+A_{21})g(x,y)+(A_{12}+A_{22})g(x,-y)\hfill \end{array}$$

(39)

and

$$\begin{array}{c}g\left(2a_{11}x,0\right)+g\left(2a_{12}x,0\right)=(A_{11}+A_{21}+A_{12}+A_{22})g(x,0),\hfill \end{array}$$

(40)

respectively.

Putting (30) (with $y=y_1+y_2$) and (39) (with $y=y_1-y_2$) into (38), we have

$$\begin{array}{cc}\hfill (A_{11}+A_{12}+A_{21}+A_{22})& g(x,y_1+y_2)-g\left(2a_{12}x,0\right)\hfill \\ \hfill +(A_{11}+A_{21})& g(x,y_1-y_2)+(A_{12}+A_{22})g(x,y_2-y_1)-g\left(2a_{11}x,0\right)\hfill \\ \hfill & =(A_{11}+A_{21})g(x,2y_1)+(A_{12}+A_{22})g(x,2y_2),\hfill \end{array}$$

hence, using (40), we obtain

$$\begin{array}{cc}\hfill (A_{11}+A_{12}+A_{21}+A_{22})& g(x,y_1+y_2)-(A_{11}+A_{12}+A_{21}+A_{22})g(x,0)\hfill \\ \hfill +(A_{11}+A_{21})& g(x,y_1-y_2)+(A_{12}+A_{22})g(x,y_2-y_1)\hfill \\ \hfill =& (A_{11}+A_{21})g(x,2y_1)+(A_{12}+A_{22})g(x,2y_2).\hfill \end{array}$$

(41)

Replacing $y_1$ and $y_2$ in the Equation (41), we have

$$\begin{array}{cc}\hfill (A_{11}+A_{12}+A_{21}+A_{22})& g(x,y_1+y_2)-(A_{11}+A_{12}+A_{21}+A_{22})g(x,0)\hfill \\ \hfill +(A_{11}+A_{21})& g(x,y_2-y_1)+(A_{12}+A_{22})g(x,y_1-y_2)\hfill \\ \hfill & =(A_{11}+A_{21})g(x,2y_2)+(A_{12}+A_{22})g(x,2y_1).\hfill \end{array}$$

(42)

Adding Equations (41) and (42), we obtain

$$\begin{array}{cc}A(2g(x,y_1+y_2)-2g(x,0)+\hfill & g(x,y_1-y_2)+g(x,y_2-y_1))\hfill \\ \hfill & =A\left(g(x,2y_1)+g(x,2y_2)\right).\hfill \end{array}$$

(43)

Putting $x=0$ in the above equation, we have

$$\begin{array}{c}A\left(2g(0,y_1+y_2)+g(0,y_1-y_2)+g(0,y_2-y_1)\right)=A\left(g(0,2y_1)+g(0,2y_2)\right).\hfill \end{array}$$

(44)

Now, we consider two cases:

(i)

$A\ne 0$;

(ii)

$A=0$.

We start with case (i), so we assume that $A\ne 0$. Dividing the Equation (36) by A and subtracting $2g(0,y)$ from both its sides, we obtain

$$\begin{array}{cc}\hfill 2g(x_1+x_2,y)& -2g(0,y)+g(x_1-x_2,y)-g(0,y)+g(x_2-x_1,y)-g(0,y)\hfill \\ \hfill & =g(2x_1,y)-g(0,y)+g(2x_2,y)-g(0,y).\hfill \end{array}$$

Subtracting Equation (37) and dividing by A from the above equation, we obtain

$$\begin{array}{cc}\hfill 2g(& x_1+x_2,y)-2g(x_1+x_2,0)-2g(0,y)+g(x_1-x_2,y)-g(x_1-x_2,0)\hfill \\ \hfill -& g(0,y)+g(x_2-x_1,y)-g(x_2-x_1,0)-g(0,y)\hfill \\ \hfill & =g(2x_1,y)-g(2x_1,0)-g(0,y)+g(2x_2,y)-g(2x_2,0)-g(0,y).\hfill \end{array}$$

We fix $y\in X$ and denote $h_y\left(x\right):=g(x,y)-g(x,0)-g(0,y),$ $x\in X.$ Then, we can write the above equation in the form

$$\begin{array}{c}2h_y(x_1+x_2)+h_y(x_1-x_2)+h_y(x_2-x_1)=h_y\left(2x_1\right)+h_y\left(2x_2\right).\hfill \end{array}$$

(45)

We decompose the function $h_y$ into even and odd parts, namely

$$h_y\left(x\right)=h_{ye}\left(x\right)+h_{yo}\left(x\right),x\in X,$$

$$\mathrm{where}{h}_{ye}\left(x\right)={\displaystyle \frac{h_y\left(x\right)+h_y(-x)}{2}}\mathrm{and}{h}_{yo}\left(x\right)={\displaystyle \frac{h_y\left(x\right)-h_y(-x)}{2}}.$$

By inserting in (45) $-x_1$ in place of $x_1$ and $-x_2$ in place of $x_2$, we obtain

$$\begin{array}{c}2h_y(-x_1-x_2)+h_y(x_2-x_1)+h_y(x_1-x_2)=h_y(-2x_1)+h_y(-2x_2).\hfill \end{array}$$

(46)

Adding Equations (45) and (46), we obtain

$$h_{ye}(x_1+x_2)+h_{ye}(x_1-x_2)={\displaystyle \frac{h_{ye}\left(2x_1\right)+h_{ye}\left(2x_2\right)}{2}}.$$

Putting in the above equation $x_2=0$, we have

$$2h_{ye}\left(x_1\right)={\displaystyle \frac{h_{ye}\left(2x_1\right)}{2}},$$

therefore

$$h_{ye}(x_1+x_2)+h_{ye}(x_1-x_2)=2h_{ye}\left(x_1\right)+2h_{ye}\left(x_2\right),$$

which means that $h_{ye}$ is a quadratic mapping.

Subtracting Equation (46) from Equation (45) we obtain

$$h_{yo}(x_1+x_2)={\displaystyle \frac{h_{yo}\left(2x_1\right)+h_{yo}\left(2x_2\right)}{2}}.$$

Putting in the above equation $x_2=0$, we have

$$h_{yo}\left(x_1\right)={\displaystyle \frac{h_{yo}\left(2x_1\right)}{2}},$$

therefore

$$h_{yo}(x_1+x_2)=h_{yo}\left(x_1\right)+h_{yo}\left(x_2\right),$$

which means that $h_{yo}$ is an additive mapping.

Dividing the Equation (37) by A, we obtain the equation

$$\begin{array}{c}2g(x_1+x_2,0)+g(x_1-x_2,0)+g(x_2-x_1,0)=g(2x_1,0)+g(2x_2,0),\hfill \end{array}$$

which means that function $\phi \left(x\right):=g(x,0),$ $x\in X$ satisfies functional Equation (45), and hence

$$\phi \left(x\right)={\phi }_a\left(x\right)+{\phi }_q\left(x\right),x\in X,$$

where ${\phi }_a$ is an additive mapping and ${\phi }_q$ is a quadratic mapping.

Dividing the Equation (43) by A and subtracting $2g(x,0)$ from both its sides, we obtain

$$\begin{array}{cc}\hfill 2g(x,y_1+y_2)& -2g(x,0)+g(x,y_1-y_2)-g(x,0)+g(x,y_2-y_1)-g(x,0)\hfill \\ \hfill & =g(x,2y_1)-g(x,0)+g(x,2y_2)-g(x,0).\hfill \end{array}$$

Subtracting Equation (44) divided by A from the above equation, we obtain

$$\begin{array}{cc}\hfill 2g(& x,y_1+y_2)-2g(x,0)-2g(0,y_1+y_2)+g(x,y_1-y_2)-g(x,0)\hfill \\ \hfill -& g(0,y_1-y_2)+g(x,y_2-y_1)-g(x,0)-g(0,y_2-y_1)\hfill \\ \hfill & =g(x,2y_1)-g(x,0)-g(0,2y_1)+g(x,2y_2)-g(x,0)-g(0,2y_2).\hfill \end{array}$$

We denote $h_x\left(y\right):=g(x,y)-g(x,0)-g(0,y)$ for fixed $x\in X$ and all $y\in X.$ Then, we can write the above equation in the form

$$\begin{array}{c}2h_x(y_1+y_2)+h_x(y_1-y_2)+h_x(y_2-y_1)=h_x\left(2y_1\right)+h_x\left(2y_2\right),\hfill \end{array}$$

(47)

which means that the function $h_x\left(y\right)$ satisfies functional Equation (45), and hence

$$h_x\left(y\right)=h_{xe}\left(y\right)+h_{xo}\left(y\right),y\in X,$$

where $h_{xo}$ is an additive mapping and $h_{xe}$ is a quadratic mapping.

Dividing the Equation (37) by A, we obtain

$$\begin{array}{c}2g(0,y_1+y_2)+g(0,y_1-y_2)+g(0,y_2-y_1)=g(0,2y_1)+g(0,2y_2),\hfill \end{array}$$

which means that function $\psi \left(y\right):=g(0,y),$ $y\in X$ satisfies functional Equation (45), and hence

$$\psi \left(y\right)={\psi }_a\left(y\right)+{\psi }_q\left(y\right),y\in X,$$

where ${\psi }_a$ is an additive mapping and ${\psi }_q$ is a quadratic mapping.

Putting in (26) $x_1=x,x_2=-x,y_1=y$ and $y_2=-y$, we obtain

$$\begin{array}{cc}\hfill g\left(0,2b_{12}y\right)+& g\left(2a_{21}x,0\right)+g\left(2a_{22}x,2b_{22}y\right)\hfill \\ \hfill & =A_{11}g(x,y)+A_{12}g(x,-y)+A_{21}g(-x,y)+A_{22}g(-x,-y).\hfill \end{array}$$

(48)

It follows from our considerations that

$$g(x,y)=h(x,y)+g(x,0)+g(0,y),$$

where

$$\begin{array}{cc}h(x,y)\hfill & =h_{aa}(x,y)+h_{qq}(x,y)+h_{aq}(x,y)+h_{qa}(x,y),\hfill \\ g(x,0)\hfill & ={\phi }_a\left(x\right)+{\phi }_q\left(x\right),g(0,y)={\psi }_a\left(y\right)+{\psi }_q\left(y\right),\hfill \\ f(x,y)\hfill & =g(x,y)+f(0,0),\hfill \end{array}$$

for all $x,y\in X,$ and $h_{aa}$ are biadditive, $h_{qq}$ is biquadratic, $h_{aq}$ is additive with respect to the first variable and quadratic with respect to the second variable, $h_{qa}$ is quadratic with respect to the first variable and additive with respect to the second variable, ${\phi }_a,{\psi }_a$ are additive, ${\phi }_q,{\psi }_q$ are quadratic, and such that the conditions $\left(h_{aa}\right),\left(h_{qq}\right),$ and $\left(h_{aq}\right)$ with $i=1,\left(h_{qa}\right)$ with $j=1$ (they result from Conditions (30), (39), (31) and (48)),$\left({\phi }_a\right),\left({\psi }_a\right),\left({\phi }_q\right)$ and $\left({\psi }_q\right)$ (they result from Conditions (27), (28), (32) and (40)) hold.

From the conditions $\left(h_{aa}\right),\left(h_{qq}\right),$ $\left(h_{aq}\right)$ with $i=1$, and $\left(h_{qa}\right)$ with $j=1$, we determine that at most one of the functions $h_{aa},h_{qq},h_{aq},h_{qa}$ is non-zero, so the function f has the form (25), where $\delta =f(0,0).$

Now, we consider case (ii), when $A=0.$ From (40) and (32), we obtain

$$\begin{array}{c}g\left(2a_{11}x,0\right)+g\left(2a_{12}x,0\right)=0\hfill \end{array}$$

(49)

and

$$g\left(0,2b_{11}y\right)+g\left(0,2b_{21}y\right)=0,$$

(50)

respectively. Putting in (49) $x=a_{11}x,$ and then $x=a_{12}x$, we have

$$g\left(2a_{11}^2,0\right)+g\left(2a_{12}{a}_{11}x,0\right)=0$$

and

$$g\left(2a_{11}{a}_{12}x,0\right)+g\left(2a_{12}^{2}x,0\right)=0,$$

which means that

$$g\left(2a_{11}^{2}x,0\right)=g\left(2a_{12}^{2}x,0\right).$$

Whence and from (49), using the function $\phi \left(x\right)=g(x,0),x\in X$, we obtain

$$\phi \left(x\right)=0\mathrm{or}{a}_{12}=-a_{11}\ne 0\mathrm{and}\mathrm{the}\mathrm{function}\phi \mathrm{is}\mathrm{odd}.$$

When $\phi$ is non-zero, using (27), we get

$$\phi \left(2a_{21}x\right)+\phi \left(2a_{22}x\right)=2(A_{11}+A_{12})\phi \left(x\right).$$

(51)

From (26) with $y_1=y_2=0$, we obtain

$$\begin{array}{cc}\phi \left(a_{11}(x_1+x_2)\right)+\hfill & \phi \left(a_{12}(x_1+x_2)\right)+\phi \left(a_{21}(x_1-x_2)\right)+\phi \left(a_{22}(x_1-x_2)\right)\hfill \\ \hfill & =\left(A_{11}+A_{12}\right)\phi \left(x_1\right)+\left(A_{21}+A_{22}\right)\phi \left(x_2\right).\hfill \end{array}$$

Using our assumption from the above equation, we have

$$\begin{array}{c}\phi \left(a_{21}(x_1-x_2)\right)+\phi \left(a_{22}(x_1-x_2)\right)=\left(A_{11}+A_{12}\right)\left(\phi \left(x_1\right)-\phi \left(x_2\right)\right).\hfill \end{array}$$

(52)

Inserting $x_1-x_2$ in place of $x_1$ and 0 in place of $x_2$ in (52), we obtain

$$\phi \left(a_{21}(x_1-x_2)\right)+\phi \left(a_{22}(x_1-x_2)\right)=\left(A_{11}+A_{12}\right)\phi (x_1-x_2).$$

(53)

From (52) and (53), we get

$$\left(A_{11}+A_{12}\right)\left(\phi \left(x_1\right)-\phi \left(x_2\right)\right)=\left(A_{11}+A_{12}\right)\phi (x_1-x_2).$$

Inserting $-x_2$ in the above equation in place of $x_2$ and using the oddness of the function $\phi$, we have

$$\left(A_{11}+A_{12}\right)\left(\phi \left(x_1\right)+\phi \left(x_2\right)\right)=\left(A_{11}+A_{12}\right)\phi (x_1+x_2).$$

(54)

Using (51) and (54), we obtain that

• if $A_{11}+A_{12}=0$, then $a_{22}=-a_{21}$ and $\phi$ is an odd;
• if $A_{11}+A_{12}\ne 0$, then the function $\phi$ is additive, such that

  $$\phi \left(a_{21}x\right)+\phi \left(a_{22}x\right)=\left(A_{11}+A_{12}\right)\phi \left(x\right).$$

In an analogous manner to (50), using the function $\psi \left(y\right)=g(0,y),y\in X$, we obtain

$$\psi \left(y\right)=0\mathrm{or}{b}_{21}=-b_{11}\ne 0\mathrm{and}\mathrm{the}\mathrm{function}\psi \mathrm{is}\mathrm{odd}.$$

Moreover, when $\psi$ is non-zero, we have

• if $A_{11}+A_{21}=0$, then $b_{22}=-b_{12}$ and $\psi$ is an odd;
• if $A_{11}+A_{21}\ne 0$, then the function $\psi$ is additive, such that

  $$\psi \left(b_{12}y\right)+\psi \left(b_{22}y\right)=\left(A_{11}+A_{21}\right)\psi \left(y\right).$$

Therefore, in the case when $A=0$, (then $f(x,y)=g(x,y)$) the functions $\phi \left(x\right)={\phi }_a\left(x\right),$ $\psi \left(y\right)={\psi }_a\left(y\right)$ are such that

$$\begin{array}{c}{\phi }_a\left(a_{12}x\right)=-{\phi }_a\left(a_{11}x\right)\wedge {\phi }_a\left(a_{21}x\right)+{\phi }_a\left(a_{22}x\right)=\left(A_{11}+A_{12}\right){\phi }_a\left(x\right),\hfill \end{array}$$

(55)

$$\begin{array}{c}{\psi }_a\left(b_{21}y\right)=-{\psi }_a\left(b_{11}y\right)\wedge {\psi }_a\left(b_{12}y\right)+{\psi }_a\left(b_{22}y\right)=\left(A_{11}+A_{21}\right){\psi }_a\left(y\right),\hfill \end{array}$$

(56)

and

$$\begin{array}{cc}{\phi }_a\hfill & \mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{when}{A}_{11}+A_{12}=0\mathrm{and}\mathrm{additive}\mathrm{in}\mathrm{others},\hfill \end{array}$$

(57)

$$\begin{array}{cc}{\psi }_a\hfill & \mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{when}{A}_{11}+A_{21}=0\mathrm{and}\mathrm{additive}\mathrm{in}\mathrm{others}.\hfill \end{array}$$

(58)

From (30), we obtain

$$g\left(2a_{11}x,2b_{11}y\right)+g\left(2a_{12}x,0\right)+g\left(0,2b_{21}y\right)=0,$$

(59)

hence, and from (49) and (50), we obtain

$$g\left(a_{11}x,b_{11}y\right)=g\left(a_{11}x,0\right)+g\left(0,b_{11}y\right).$$

Therefore, in the case when $A=0$ and $a_{11}{b}_{11}\ne 0$, the function

$$f(x,y)={\phi }_a\left(x\right)+{\psi }_a\left(y\right),x,y\in X,$$

(60)

and the functions ${\phi }_a$ and ${\psi }_a$ satisfy Conditions (55)–(58).

We have to consider the case when $a_{11}=0$ or $b_{11}=0.$

From (34) and then (41), we obtain

$$\begin{array}{c}(A_{11}+A_{12})\left(g(x_1-x_2,y)-g(x_2-x_1,y)\right)=(A_{11}+A_{12})\left(g(2x_1,y)-g(2x_2,y)\right)\hfill \end{array}$$

(61)

and

$$(A_{11}+A_{21})\left(g(x,y_1-y_2)-g(x,y_2-y_1)\right)=(A_{11}+A_{21})\left(g(x,2y_1)-g(x,2y_2)\right),$$

(62)

respectively. We consider four cases:

(a)

$A_{11}+A_{12}\ne 0$;

(b)

$A_{11}+A_{21}\ne 0$;

(c)

$A_{11}+A_{12}=0$;

(d)

$A_{11}+A_{21}=0$.

We start with case (a), so we assume that $A_{11}+A_{12}\ne 0$. Then, dividing Equation (61) by $A_{11}+A_{12}$, we obtain

$$\begin{array}{c}g(x_1-x_2,y)-g(x_2-x_1,y)=g(2x_1,y)-g(2x_2,y).\hfill \end{array}$$

(63)

Putting $x_1=x$ and $x_2=0$, and then $x_1=0$ and $x_2=-x$ in (63), we have

$$\begin{array}{cc}g(x,y)-g(-x,y)\hfill & =g(2x,y)-g(0,y),\hfill \\ g(x,y)-g(-x,y)\hfill & =g(0,y)-g(-2x,y)\hfill \end{array}$$

respectively, and hence

$$g(2x,y)-g(0,y)=-\left(g(-2x,y)-g(0,y)\right),$$

which means that for fixed $y\in X$ the function $g_y\left(x\right)=g(x,y)-g(0,y),$ $x\in X$ is odd and $2g_y\left(x\right)=g_y\left(2x\right)$. Using these properties to (63), we have

$$\begin{array}{c}2g_y(x_1-x_2)=2g_y\left(x_1\right)+2g_y(-x_2),\hfill \end{array}$$

and hence

$$\begin{array}{c}{g}_y(x_1+x_2)=g_y\left(x_1\right)+g_y\left(x_2\right),\hfill \end{array}$$

which means that for fixed $y\in X$, the function $g_y\left(x\right),$ $x\in X$ is additive. From the fact that function ${\phi }_a\left(x\right)=\phi \left(x\right)=g(x,0)$ is additive in this case, we determine that for each fixed $y\in X$ the function $h_y\left(x\right)=g(x,y)-g(0,y)-g(x,0)=g_y\left(x\right)-{\phi }_a\left(x\right),x\in X$ is additive.

In case (b), we assume that $A_{11}+A_{21}\ne 0$. Then, dividing the Equation (62) by $A_{11}+A_{21}$, we obtain

$$g(x,y_1-y_2)-g(x,y_2-y_1)=g(x,2y_1)-g(x,2y_2),$$

and in an analogous way, like in case (a), we determine that, for each fixed $x\in X$, the function $h_x\left(y\right)=g(x,y)-g(x,0)-g(0,y),$ $y\in X$ is additive.

In case (c), we assume that $A_{11}+A_{12}=0$. From (31), we obtain

$$g\left(0,2b_{11}y\right)+g\left(2a_{21}x,2b_{21}y\right)+g\left(2a_{22}x,0\right)=0,$$

(64)

which means that if $A_{11}+A_{12}=0$ and $a_{21}{b}_{21}\ne 0$, the function f has the form (60) and the functions ${\phi }_a$ and ${\psi }_a$ satisfy Conditions (55), (56), and (58), and ${\phi }_a$ is odd.

In case (d), we assume that $A_{11}+A_{21}=0$. From (39), we obtain

$$g\left(2a_{11}x,0\right)+g\left(2a_{12}x,2b_{12}y\right)+g\left(0,2b_{22}y\right)=0,$$

(65)

which means that if $A_{11}+A_{21}=0$ and $a_{12}{b}_{12}\ne 0$, the function f has the form (60) and the functions ${\phi }_a$ and ${\psi }_a$ satisfy Conditions (55)–(57) and ${\psi }_a$ is odd.

We notice that in the case when $A=0$ and $a_{ij}{b}_{ij}=0$ for all $i,j\in \{1,2\}$

• if $A_{i_1{j}_1}\ne 0$ for some $i_1,j_1\in \{1,2\}$, then from (26) with $x_{i_1}=x,y_{j_1}=y$ and $x_{3-i_1}=y_{3-j_1}=0$, we have

  $$\begin{array}{cc}\hfill g(x,y)={\displaystyle \frac{1}{A_{i_1{j}_1}}}(& g(a_{11}x,0)+g(a_{12}x,0)+g\left({(-1)}^{3-i_1}{a}_{21}x,0\right)+g\left({(-1)}^{3-i_1}{a}_{22}x,0\right)\hfill \\ & +g\left(0,b_{11}y\right)+g\left(0,{(-1)}^{3-j_1}{b}_{12}y\right)+g\left(0,b_{21}y\right)+g\left(0,{(-1)}^{3-j_1}{b}_{22}y\right)\hfill \\ \hfill & -A_{i_{1}3-j_1}g(x,0)-A_{3-i_1{j}_1}g(0,y));\hfill \end{array}$$
• if $A_{ij}=0$ for all $i,j\in \{1,2\},$ then from (26) with $x_1=x,y_1=y$ and $x_2=y_2=0$, we have

  $$\sum _{i,j\in \{1,2\}}g(a_{ij}x,0)+g(0,b_{ij}y)=0.$$

As a consequence, when $A=0$ and

$$\begin{array}{cc}\hfill (a_{11}{b}_{11}\ne 0)\vee & (A_{11}+A_{12}=0\wedge a_{21}{b}_{21}\ne 0)\vee (A_{11}+A_{21}=0\wedge a_{12}{b}_{12}\ne 0)\vee \hfill \\ \hfill & (a_{ij}{b}_{ij}=0\mathrm{for} \mathrm{all}i,j\in \{1,2\})\hfill \end{array}$$

(66)

the function f has the form (25), where $\delta =f(0,0),$ $h_{k_1{k}_2}=0,$ for all $k_1,k_2\in \{a,q\},$ ${\phi }_q={\psi }_q=0,$ and ${\phi }_a$ is an odd when $A_{11}+A_{12}+A_{21}+A_{22}=A_{11}+A_{12}=0$ and additive in others; ${\psi }_a$ is an odd when $A_{11}+A_{12}+A_{21}+A_{22}=A_{11}+A_{21}=0$ and additive in others, such that the conditions $\left({\phi }_a\right)$ and $\left({\psi }_a\right)$ hold.

In the case when $A=0$ and Condition (66) does not hold, we consider four possibilities:

(1)

$A_{11}+A_{12}\ne 0\wedge A_{11}+A_{21}\ne 0,$

(2)

$A_{11}+A_{12}\ne 0\wedge A_{11}+A_{21}=0,$

(3)

$A_{11}+A_{12}=0\wedge A_{11}+A_{21}\ne 0,$

(4)

$A_{11}+A_{12}=0\wedge A_{11}+A_{21}=0.$

In case (1), from (a) and (b), we determine that

$$f(x,y)=h_{aa}(x,y)+{\phi }_a\left(x\right)+{\psi }_a\left(y\right),$$

(67)

where $h_{aa}$ is a biadditive function such that

$$\begin{array}{cc}2h_{aa}(a_{12}x,b_{12}y)\hfill & =\left(A_{11}+A_{21}\right)h_{aa}(x,y),\hfill \\ 2h_{aa}(a_{21}x,b_{21}y)\hfill & =\left(A_{11}+A_{12}\right)h_{aa}(x,y),\hfill \\ 2h_{aa}(a_{22}x,b_{22}y)\hfill & =\left(A_{11}+A_{22}\right)h_{aa}(x,y),\hfill \end{array}$$

and the functions ${\phi }_a,{\psi }_a$ are additive and satisfy Conditions (55) and (56).

In case (2), from (a) and (d), we determine that for each fixed $y\in X$, the function $h_y\left(x\right)=g(x,y)-g(x,0)-g(0,y),$ $x\in X$ is additive and $a_{11}{b}_{11}=0$, $a_{12}{b}_{12}=0,$ $A_{21}=-A_{11},$ $A_{22}=-A_{12}.$

Putting in (26) $x_1=x,x_2=-x$ and using our assumptions, we obtain

$$\begin{array}{cc}\hfill g\left(0,b_{11}(y_1+y_2)\right)+& g\left(0,b_{12}(y_1-y_2)\right)+g\left(2a_{21}x,b_{21}(y_1+y_2)\right)+g\left(2a_{22}x,b_{22}(y_1-y_2)\right)\hfill \\ \hfill & =2A_{11}\left(g(x,y_1)-g(0,y_1)\right)+2A_{12}\left(g(x,y_2)-g(0,y_2)\right).\hfill \end{array}$$

(68)

The Conditions (31) and (48), with our assumptions, have the form

$$\begin{array}{c}g\left(0,2b_{11}y\right)+g\left(2a_{21}x,2b_{21}y\right)+g\left(2a_{22}x,0\right)=2(A_{11}+A_{12})\left(g(x,y)-g(0,y)\right)\hfill \end{array}$$

(69)

and

$$\begin{array}{cc}\hfill g\left(0,2b_{12}y\right)+& g\left(2a_{21}x,0\right)+g\left(2a_{22}x,2b_{22}y\right)\hfill \\ \hfill & =2A_{11}\left(g(x,y)-g(0,y)\right)+2A_{12}\left(g(x,-y)-g(0,-y)\right),\hfill \end{array}$$

(70)

respectively.

Putting (69) (with $y=y_1+y_2$) and (70) (with $y=y_1-y_2$) into (68), we have

$$\begin{array}{cc}\hfill 2(A_{11}+& A_{12})\left(g(x,y_1+y_2)-g(0,y_1+y_2)\right)-g(2a_{22}x,0)-g(2a_{21}x,0)\hfill \\ \hfill +2A_{11}& \left(g(x,y_1-y_2)-g(0,y_1-y_2)\right)+2A_{12}\left(g(x,y_2-y_1)-g(0,y_2-y_1)\right)\hfill \\ \hfill & =2A_{11}\left(g(x,2y_1)-g(0,2y_1)\right)+2A_{12}\left(g(x,2y_2)-g(0,2y_2)\right),\hfill \end{array}$$

hence, using (27), we obtain

$$\begin{array}{cc}\hfill (A_{11}+& A_{12})\left(g(x,y_1+y_2)-g(0,y_1+y_2)\right)-(A_{11}+A_{12})g(x,0)\hfill \\ \hfill +A_{11}& \left(g(x,y_1-y_2)-g(0,y_1-y_2)\right)+A_{12}\left(g(x,y_2-y_1)-g(0,y_2-y_1)\right)\hfill \\ \hfill & =A_{11}\left(g(x,2y_1)-g(0,2y_1)\right)+A_{12}\left(g(x,2y_2)-g(0,2y_2)\right).\hfill \end{array}$$

(71)

Replacing $y_1$ and $y_2$ in the Equation (71), we obtain

$$\begin{array}{cc}\hfill (A_{11}+& A_{12})\left(g(x,y_1+y_2)-g(0,y_1+y_2)\right)-(A_{11}+A_{12})g(x,0)\hfill \\ \hfill +A_{11}& \left(g(x,y_2-y_1)-g(0,y_2-y_1)\right)+A_{12}\left(g(x,y_1-y_2)-g(0,y_1-y_2)\right)\hfill \\ \hfill & =A_{11}\left(g(x,2y_2)-g(0,2y_2)\right)+A_{12}\left(g(x,2y_1)-g(0,2y_1)\right).\hfill \end{array}$$

(72)

Adding Equations (71) and (72), we have

$$\begin{array}{cc}\hfill & 2(A_{11}+A_{12})\left(g(x,y_1+y_2)-g(0,y_1+y_2)-g(x,0)\right)\hfill \\ \hfill & +(A_{11}+A_{12})\left(g(x,y_1-y_2)-g(0,y_1-y_2)+g(x,y_2-y_1)-g(0,y_2-y_1)\right)\hfill \\ \hfill & =(A_{11}+A_{12})\left(g(x,2y_1)-g(0,2y_1)+g(x,2y_2)-g(0,2y_2)\right).\hfill \end{array}$$

(73)

Dividing the Equation (73) by $A_{11}+A_{12}$ and subtracting $2g(x,0)$ from both sides, we obtain

$$\begin{array}{cc}\hfill & 2\left(g(x,y_1+y_2)-g(0,y_1+y_2)-g(x,0)\right)\hfill \\ \hfill & +g(x,y_1-y_2)-g(0,y_1-y_2)-g(x,0)+g(x,y_2-y_1)-g(0,y_2-y_1)-g(x,0)\hfill \\ \hfill & =g(x,2y_1)-g(0,2y_1)-g(x,0)+g(x,2y_2)-g(0,2y_2)-g(x,0).\hfill \end{array}$$

Then, we can write the above equation, using function $h_x\left(y\right)=g(x,y)-g(x,0)-g(0,y),$ where $y\in X$ and $x\in X$ is fixed, in the form

$$\begin{array}{c}2h_x(y_1+y_2)+h_x(y_1-y_2)+h_x(y_2-y_1)=h_x\left(2y_1\right)+h_x\left(2y_2\right),\hfill \end{array}$$

which means that the function $h_x$ satisfies functional Equation (47), and hence $h_x$ can be decomposed into even and odd parts, namely

$$h_x\left(y\right)=h_{xe}\left(y\right)+h_{xo}\left(y\right),y\in X,$$

where $h_{xe}$ is a quadratic mapping and $h_{xo}$ is an additive mapping.

Finally, in case (2), we obtain

$$f(x,y)=h_{aa}(x,y)+h_{aq}(x,y)+{\phi }_a\left(x\right)+{\psi }_a\left(y\right),x,y\in X,$$

where $h_{aa}$ is biadditive, and $h_{aq}$ is additive with respect to the first variable and quadratic with respect to the second variable function, such that

$$\begin{array}{cc}2h_{aa}(a_{21}x,b_{21}y)\hfill & =\left(A_{11}+A_{12}\right)h_{aa}(x,y),\hfill \\ 2h_{aa}(a_{22}x,b_{22}y)\hfill & =\left(A_{11}+A_{22}\right)h_{aa}(x,y),\hfill \end{array}$$

$$\begin{array}{c}2h_{aq}(a_{21}x,b_{21}y)=2h_{aq}(a_{22}x,b_{22}y)=A_{11}{h}_{aq}(x,y)=A_{12}{h}_{aq}(x,y),\hfill \end{array}$$

and the functions ${\phi }_a$ and ${\psi }_a$ satisfy Conditions (55)–(57) and ${\psi }_a$ is odd. From the above conditions, we determine that at most one of the functions $h_{aa},h_{aq}$ is non-zero.

In case (3), from (b) and (c), we determine that for each fixed $x\in X$, the function $h_x\left(y\right)=g(x,y)-g(x,0)-g(0,y),$ $y\in X,$ is additive and $a_{11}{b}_{11}=0$, $a_{21}{b}_{21}=0,$ $A_{12}=-A_{11},$ $A_{22}=-A_{21}.$

Starting from the equation

$$\begin{array}{cc}\hfill g\left(a_{11}(x_1+x_2),0\right)& +g\left(a_{12}(x_1+x_2),2b_{12}y\right)+g\left(a_{21}(x_1-x_2),0\right)+g\left(a_{22}(x_1-x_2),2b_{22}y\right)\hfill \\ \hfill & =2A_{11}\left(g(x_1,y)-g(x_1,0)\right)+2A_{21}\left(g(x_2,y)-g(x_2,0)\right),\hfill \end{array}$$

obtained from the Equation (26) with $y_1=y,y_2=-y$, and proceeding analogously as in (2), we obtain

$$\begin{array}{cc}f(x,y)\hfill & =h(x,y)+{\phi }_a\left(x\right)+{\psi }_a\left(y\right),\hfill \\ h(x,y)\hfill & =h_{aa}(x,y)+h_{qa}(x,y),\hfill \end{array}$$

for $x,y\in X$, where $h_{aa}$ is biadditive, $h_{qa}$ is quadratic with respect to the first variable and additive with respect to the second variable function, such that

$$\begin{array}{cc}2h_{aa}(a_{12}x,b_{12}y)\hfill & =\left(A_{11}+A_{21}\right)h_{aa}(x,y),\hfill \\ 2h_{aa}(a_{22}x,b_{22}y)\hfill & =\left(A_{11}+A_{22}\right)h_{aa}(x,y),\hfill \end{array}$$

$$\begin{array}{c}2h_{qa}(a_{12}x,b_{12}y)=2h_{qa}(a_{22}x,b_{22}y)=A_{11}{h}_{qa}(x,y)=A_{21}{h}_{qa}(x,y),\hfill \end{array}$$

and the functions ${\phi }_a$ and ${\psi }_a$ satisfy Conditions (55), (56), and (58), and ${\phi }_a$ is odd.

From the above conditions, we determine that at most one of the functions $h_{aa},h_{qa}$ is non-zero.

In case (4), from (c) and (d), we have $a_{11}{b}_{11}=0$, $a_{12}{b}_{12}=0$, $a_{21}{b}_{21}=0,$ $a_{22}{b}_{22}\ne 0$ and $A_{12}=-A_{11}=-A_{22}=A_{21}\ne 0.$

Putting in (26) $x_1=x,y_1=y$ and $x_2=y_2=0$ and using our assumptions, we obtain

$$\begin{array}{cc}\hfill g\left(a_{11}x,b_{11}y\right)& +g\left(a_{12}x,b_{12}y\right)+g\left(a_{21}x,b_{21}y\right)+g\left(a_{22}x,b_{22}y\right)\hfill \\ \hfill & =A_{11}\left(g(x,y)-g(x,0)-g(0,y)\right).\hfill \end{array}$$

(74)

Putting in (26) $x_1=x,y_2=y,x_2=y_1=0$ and then $x_2=x,y_1=y,x_1=y_2=0,$ and using our assumptions, we get

$$\begin{array}{cc}\hfill g\left(a_{11}x,b_{11}y\right)& +g\left(a_{12}x,-b_{12}y\right)+g\left(a_{21}x,b_{21}y\right)+g\left(a_{22}x,-b_{22}y\right)\hfill \\ \hfill & =A_{11}\left(g(x,0)-g(x,y)+g(0,y)\right),\hfill \end{array}$$

(75)

and

$$\begin{array}{cc}\hfill g\left(a_{11}x,b_{11}y\right)& +g\left(a_{12}x,b_{12}y\right)+g\left(-a_{21}x,b_{21}y\right)+g\left(-a_{22}x,b_{22}y\right)\hfill \\ \hfill & =A_{11}\left(g(0,y)-g(x,y)+g(x,0)\right),\hfill \end{array}$$

(76)

respectively. From (74)–(76), using (59), (65), (64), (49) and (50), we have

$$\begin{array}{c}-g\left(0,b_{22}y\right)-g\left(a_{22}x,0\right)+g\left(a_{22}x,b_{22}y\right)=A_{11}\left(g(x,y)-g(x,0)-g(0,y)\right),\hfill \end{array}$$

(77)

$$\begin{array}{c}-g\left(0,-b_{22}y\right)-g\left(a_{22}x,0\right)+g\left(a_{22}x,-b_{22}y\right)=A_{11}\left(g(x,0)-g(x,y)+g(0,y)\right),\hfill \end{array}$$

(78)

$$\begin{array}{c}-g\left(0,b_{22}y\right)-g\left(-a_{22}x,0\right)+g\left(-a_{22}x,b_{22}y\right)=A_{11}\left(g(0,y)-g(x,y)+g(x,0)\right),\hfill \end{array}$$

(79)

respectively. Adding Equations (77) and (78), and then Equations (77) and (79), using our assumptions, we have

$$g(x,-y)=-g(x,y)+2g(x,0)andg(-x,y)=-g(x,y)+2g(0,y).$$

(80)

From (26), using (49), (50), (59), (64) and (65), we obtain

$$\begin{array}{cc}-g\left(0,b_{22}(y_1-y_2)\right)-\hfill & g\left(a_{22}(x_1-x_2),0\right)+g\left(a_{22}(x_1-x_2),b_{22}(y_1-y_2)\right)\hfill \\ \hfill & =A_{11}\left(g(x_1,y_1)-g(x_1,y_2)-g(x_2,y_1)+g(x_2,y_2)\right).\hfill \end{array}$$

(81)

Putting $x=x_1-x_2$ and $y=y_1-y_2$ in (77), we have

$$\begin{array}{cc}-g(0,b_{22}(y_1-\hfill & y_2\left)\right)-g\left(a_{22}(x_1-x_2),0\right)+g\left(a_{22}(x_1-x_2),b_{22}(y_1-y_2)\right)\hfill \\ \hfill & =A_{11}\left(g(x_1-x_2,y_1-y_2)-g(x_1-x_2,0)-g(0,y_1-y_2)\right).\hfill \end{array}$$

(82)

From (81) and (82), we obtain

$$\begin{array}{cc}g(x_1-x_2,y_1-y_2)-\hfill & g(x_1-x_2,0)-g(0,y_1-y_2)\hfill \\ \hfill & =g(x_1,y_1)-g(x_1,y_2)-g(x_2,y_1)+g(x_2,y_2).\hfill \end{array}$$

Inserting in the above equation $-x_2$ in place of $x_2$ and $-y_2$ in place of $y_2$ and using (80), we get

$$\begin{array}{cc}\hfill g(x_1+x_2,y_1+y_2)-& g(x_1+x_2,0)-g(0,y_1+y_2)\hfill \\ \hfill =& g(x_1,y_1)-g(x_1,0)-g(0,y_1)+g(x_1,y_2)-g(x_1,0)-g(0,y_2)\hfill \\ & +g(x_2,y_1)-g(x_2,0)-g(0,y_1)+g(x_2,y_2)-g(x_2,0)-g(0,y_2),\hfill \end{array}$$

therefore, in this case, $h(x,y)=g(x,y)-g(x,0)-g(0,y)$ is biadditive, and f has Form (67), where $h=h_{aa}$ is a biadditive function and ${\phi }_a,{\psi }_a$ are odd functions, such that

$$\begin{array}{cccc}& \hfill h_{aa}(a_{22}x,b_{22}y)& =& A_{11}{h}_{aa}(x,y),\hfill \\ \hfill \mathrm{and}& \hfill {\phi }_a\left(a_{12}x\right)=-{\phi }_a\left(a_{11}x\right)& \wedge & {\phi }_a\left(a_{21}x\right)=-{\phi }_a\left(a_{22}x\right),\hfill \\ & \hfill {\psi }_a\left(b_{21}y\right)=-{\psi }_a\left(b_{11}y\right)& \wedge & {\psi }_a\left(b_{12}y\right)=-{\psi }_a\left(b_{22}y\right).\hfill \end{array}$$

As a consequence, when $A=0$ and Condition (66) is not satisfied, in each of the four possible cases, we determine that the function f has Form (25), where $\delta =f(0,0),$ $h(x,y)=h_{aa}(x,y)+h_{aq}(x,y)+h_{qa}(x,y),$ and $h_{aa}$ is biadditive, $h_{aq}$ is additive with respect to the first variable and quadratic with respect to the second variable, $h_{qa}$ is quadratic with respect to the the first variable and additive with respect to the second variable, and at most one of the functions $h_{aa},h_{aq},h_{qa}$ is non-zero, ${\phi }_q={\psi }_q=0,$ ${\phi }_a$ is an odd when $A_{11}+A_{12}+A_{21}+A_{22}=A_{11}+A_{12}=0$ and additive in others, ${\psi }_a$ is an odd when $A_{11}+A_{12}+A_{21}+A_{22}=A_{11}+A_{21}=0$ and additive in others, and such that the conditions $\left(h_{aa}\right)$ and $\left(h_{aq}\right)$ with $i=2,\left(h_{qa}\right)$ with $j=2,$ and $\left({\phi }_a\right),\left({\psi }_a\right)$ hold.

The proof of the converse is a direct computation. □

Remark 2. 

Let us notice the following:

• From the conditions $\left(h_{aa}\right),\left(h_{qq}\right),\left(h_{aq}\right),\left(h_{qa}\right)$, we determine that at most one of the functions $h_{aa},h_{qq},h_{aq},h_{qa}$ is non-zero;
• The conditions $\left({\phi }_a\right),\left({\phi }_q\right)$ do not imply that at most one of the functions ${\phi }_a,{\phi }_q$ is a non-zero (see Example 2(d));
• The conditions $\left({\psi }_a\right),\left({\psi }_q\right)$ do not imply that at most one of the functions ${\psi }_a,{\psi }_q$ is a non-zero.

Corollary 2. 

Let $f:{\mathbb{R}}^2\to \mathbb{R}$ be a continuous solution of (24). Then, there exist the odd continuous functions ${\phi }_a,{\psi }_a:\mathbb{R}\to \mathbb{R}$ and ${\alpha }_{k_1{k}_2},{\beta }_a,{\beta }_q,{\gamma }_a,{\gamma }_q,\delta \in \mathbb{R},$ where $k_1,k_2\in \{a,q\}$ such that

$$f(x,y)={\alpha }_{k_1{k}_2}{x}^{s_{k_1}}{y}^{s_{k_2}}+{\beta }_a{\phi }_a\left(x\right)+{\beta }_q{x}^2+{\gamma }_a{\psi }_a\left(y\right)+{\gamma }_a{y}^2+\delta ,x,y\in \mathbb{R},$$

$$\begin{array}{cc}\hfill \mathit{where}& s_{k_i}=\left\{\begin{array}{cc}1,\hfill & \mathit{if}{k}_i=a\hfill \\ 2,\hfill & \mathit{if}{k}_i=q\hfill \end{array}\right.,i\in \{1,2\},\hfill \\ \hfill & {\phi }_a\left(x\right)=x,\mathit{when}{A}_{11}+A_{12}+A_{21}+A_{22}\ne 0\vee A_{11}+A_{12}\ne 0,\hfill \\ \hfill & {\psi }_a\left(y\right)=y,\mathit{when}{A}_{11}+A_{12}+A_{21}+A_{22}\ne 0\vee A_{11}+A_{21}\ne 0,\hfill \end{array}$$

and moreover,

$\begin{array}{ccc}{\alpha }_{aa}=0,\hfill & \mathit{whenever}\hfill & \begin{array}{c}4a_{11}{b}_{11}\ne A_{11}+A_{12}+A_{21}+A_{22}\vee \hfill \\ 4a_{12}{b}_{12}\ne (A_{11}+A_{21})-(A_{12}+A_{22})\vee \hfill \\ 4a_{21}{b}_{21}\ne (A_{11}+A_{12})-(A_{21}+A_{22})\vee \hfill \\ 4a_{22}{b}_{22}\ne (A_{11}+A_{22})-(A_{12}+A_{21});\hfill \end{array}\hfill \\ & & \\ {\alpha }_{qq}=0,\hfill & \mathit{whenever}\hfill & \begin{array}{cc}4a_{ij}^2{b}_{ij}^2\ne A_{11}\vee A_{ij}\ne A_{11}\hfill & \mathit{for}\mathit{some}i,j\in \{1,2\};\hfill \end{array}\hfill \end{array}$

$\begin{array}{ccc}{\alpha }_{aq}=0,\hfill & \mathit{whenever}\hfill & \begin{array}{c}{A}_{11}\ne A_{12}\vee A_{21}\ne A_{22}\vee \left(\right(a_{21}{b}_{21}\ne 0\vee a_{22}{b}_{22}\ne 0\vee \hfill \\ 2a_{11}{b}_{11}^2\ne A_{11}\vee 2a_{12}{b}_{12}^2\ne A_{11}\vee A_{11}\ne A_{22})\wedge (a_{11}{b}_{11}\ne 0\vee \hfill \\ a_{12}{b}_{12}\ne 0\vee 2a_{21}{b}_{21}^2\ne A_{11}\vee 2a_{22}{b}_{22}^2\ne A_{11}\vee A_{11}\ne -A_{22}\left)\right);\hfill \end{array}\hfill \end{array}$

$\begin{array}{ccc}{\alpha }_{qa}=0,\hfill & \mathit{whenever}\hfill & \begin{array}{c}{A}_{11}\ne A_{21}\vee A_{12}\ne A_{22}\vee \left(\right(a_{12}{b}_{12}\ne 0\vee a_{22}{b}_{22}\ne 0\vee \hfill \\ 2a_{11}^2{b}_{11}\ne A_{11}\vee 2a_{21}^2{b}_{21}\ne A_{11}\vee A_{11}\ne A_{22})\wedge (a_{11}{b}_{11}\ne 0\vee \hfill \\ a_{21}{b}_{21}\ne 0\vee 2a_{12}^2{b}_{12}\ne A_{11}\vee 2a_{22}^2{b}_{22}\ne A_{11}\vee A_{11}\ne -A_{22}\left)\right);\hfill \end{array}\hfill \end{array}$

$\begin{array}{ccc}{\beta }_a=0,\hfill & \mathit{whenever}\hfill & \begin{array}{c}2(a_{11}+a_{12})\ne A_{11}+A_{12}+A_{21}+A_{22}\vee \hfill \\ 2(a_{21}+a_{22})\ne (A_{11}+A_{12})-(A_{21}+A_{22});\hfill \end{array}\hfill \\ & & \\ {\gamma }_a=0,\hfill & \mathit{whenever}\hfill & \begin{array}{c}2(b_{11}+b_{21})\ne A_{11}+A_{12}+A_{21}+A_{22}\vee \hfill \\ 2(b_{12}+b_{22})\ne (A_{11}+A_{21})-(A_{12}+A_{22});\hfill \end{array}\hfill \end{array}$

$\begin{array}{ccc}{\beta }_q=0,\hfill & \mathrm{whenever}\hfill & \begin{array}{c}2(a_{21}^2+a_{22}^2)\ne A_{21}+A_{22}\vee 2(a_{11}^2+a_{12}^2)\ne A_{11}+A_{12}\vee \hfill \\ A_{11}+A_{12}\ne A_{21}+A_{22};\hfill \end{array}\hfill \end{array}$

$\begin{array}{ccc}{\gamma }_q=0,\hfill & \mathit{whenever}\hfill & \begin{array}{c}2(b_{12}^2+b_{22}^2)\ne A_{12}+A_{22}\vee 2(b_{11}^2+b_{21}^2)\ne A_{11}+A_{21}\vee \hfill \\ A_{11}+A_{21}\ne A_{12}+A_{22};\hfill \end{array}\hfill \\ & & \\ \delta =0,\hfill & \mathit{whenever}\hfill & A_{11}+A_{12}+A_{21}+A_{22}\ne 4.\hfill \end{array}$

Proof. 

The function f is continuous and so, by Theorem 3, any solution of (24) is of the form

$$f(x,y)=\alpha h(x,y)+{\beta }_a{\phi }_a\left(x\right)+{\beta }_q{x}^2+{\gamma }_a{\psi }_a\left(y\right)+{\gamma }_a{y}^2+\delta ,x,y\in \mathbb{R},$$

where $\alpha h(x,y)={\alpha }_{aa}xy+{\alpha }_{qq}{x}^2{y}^2+{\alpha }_{aq}x{y}^2+{\alpha }_{qq}{x}^{2}y,$ ${\alpha }_{k_1{k}_2},{\beta }_{l_1},{\gamma }_{l_2},\delta \in \mathbb{R},$ $k_1,k_2\in \{a,q\},$ ${\phi }_a,{\psi }_a:\mathbb{R}\to \mathbb{R}$ and such that ${\phi }_a$ is an odd continuous function when $A_{11}+A_{12}+A_{21}+A_{22}$ $=A_{11}+A_{12}=0$ and ${\phi }_a\left(x\right)=x$ in others, ${\psi }_a$ is an odd continuous function when $A_{11}+A_{12}+A_{21}+A_{22}$ $=A_{11}+A_{21}=0$ and ${\psi }_a\left(y\right)=y$ in others. From $\left(h_{aa}\right),\left(h_{qq}\right),\left(h_{aq}\right)$ and $\left(h_{qa}\right)$, we derive the following conditions

$$\begin{array}{c}4{\alpha }_{aa}{a}_{11}{b}_{11}={\alpha }_{aa}\left(A_{11}+A_{12}+A_{21}+A_{22}\right)\hfill \\ 4{\alpha }_{aa}{a}_{12}{b}_{12}={\alpha }_{aa}\left((A_{11}+A_{21})-(A_{12}+A_{22})\right)\hfill \\ 4{\alpha }_{aa}{a}_{21}{b}_{21}={\alpha }_{aa}\left((A_{11}+A_{12})-(A_{21}+A_{22})\right)\hfill \\ 4{\alpha }_{aa}{a}_{22}{b}_{22}={\alpha }_{aa}\left((A_{11}+A_{22})-(A_{12}+A_{21})\right)\hfill \\ \\ 4{\alpha }_{qq}{a}_{ij}^2{b}_{ij}^2={\alpha }_{qq}{A}_{11}\wedge A_{ij}=A_{11}\mathrm{for}\mathrm{all}i,j\in \{1,2\}\hfill \\ \\ 2{\alpha }_{aq}{a}_{i1}{b}_{i1}^2=2{\alpha }_{aq}{a}_{i2}{b}_{i2}^2={\alpha }_{aq}{A}_{11}={\alpha }_{aq}{A}_{12}\wedge \hfill \\ a_{3-i1}{b}_{3-i1}=a_{3-i2}{b}_{3-i2}=0\wedge A_{21}=A_{22}={(-1)}^{3-i}{A}_{11}\mathrm{for}\mathrm{some}i\in \{1,2\}\hfill \\ \\ 2{\alpha }_{qa}{a}_{1j}^2{b}_{1j}=2{\alpha }_{qa}{a}_{2j}^2{b}_{2j}={\alpha }_{qa}{A}_{11}={\alpha }_{qa}{A}_{21}\wedge \hfill \\ a_{13-j}{b}_{13-j}=a_{23-j}{b}_{23-j}=0\wedge A_{12}=A_{22}={(-1)}^{3-j}{A}_{11}\mathrm{for}\mathrm{some}j\in \{1,2\},\hfill \end{array}$$

which means that at most one of the numbers ${\alpha }_{aa},{\alpha }_{qq},{\alpha }_{aq},{\alpha }_{qa}$ is non-zero. Therefore,

$$\alpha h(x,y)={\alpha }_{k_1{k}_2}{x}^{s_{k_1}}{y}^{s_{k_2}}x,y\in X,$$

with some $k_1,k_2\in \{a,q\}$ and $s_a=1,s_q=2.$

From conditions $\left({\phi }_a\right),\left({\phi }_q\right),\left({\psi }_a\right)$ and $\left({\psi }_q\right)$, we have

$$\begin{array}{c}2{\beta }_a(a_{11}+a_{12})={\beta }_a(A_{11}+A_{12}+A_{21}+A_{22})\hfill \\ 2{\beta }_a(a_{21}+a_{22})={\beta }_a\left((A_{11}+A_{12})-(A_{21}+A_{22})\right)\hfill \\ \\ 2{\beta }_q(a_{11}^2+a_{12}^2)=2{\beta }_q(a_{21}^2+a_{22}^2)={\beta }_q(A_{11}+A_{12})={\beta }_q(A_{21}+A_{22})\hfill \end{array}$$

$$\begin{array}{c}2{\gamma }_a(b_{11}+b_{21})={\gamma }_a(A_{11}+A_{12}+A_{21}+A_{22})\hfill \\ 2{\gamma }_a(b_{12}+b_{22})={\gamma }_a\left((A_{11}+A_{21})-(A_{12}+A_{22})\right)\hfill \\ \\ 2{\gamma }_q(b_{11}^2+b_{21}^2)=2{\gamma }_q(b_{12}^2+b_{22}^2)={\gamma }_q(A_{11}+A_{21})={\gamma }_q(A_{12}+A_{22})\hfill \end{array}$$

and

$$4\delta =\left(A_{11}+A_{12}+A_{21}+A_{22}\right)\delta ,$$

which leads to our assertion. □

Remark 3. 

We notice that from Theorem 4 and Corollary 2, we obtain results for well-known functional equations, namely bi-Cauchy, bi-Jensen, Cauchy–Jensen, Jensen–Cauchy, and quadratic equations.

4. Conclusions

In this paper, we have presented solutions to the problems posed in [6]. We have given a positive answer to the questions posed in Problems 1 and 2, showing that the Equation (3) is hyperstable in $(m+1)$-normed spaces (Theorem 2). Our result also improves a result from [6,22]. We have also given a partial solution to Problem 3, describing a general solution of the Equation (3) for $n=1$ (Theorem 3) and $n=2$ (Theorem 4), which has a very complicated form.

The functional Equation (3) generalizes not only Equation (2) but also many other well-known functional equations, and for this reason, its study seems interesting.

We end the article with the following problem, which remains open.

Problem 4. 

Find a general solution of Equation (3) for $n>2.$