Alienation of the quadratic, exponential and d’Alembert functional equations

Abstract

Let $(S,+,0)$ be a commutative monoid, $\sigma :S\rightarrow S$ be an endomorphism with $\sigma ^2=id$ and let K be a field of characteristic different from 2. We study the solutions $f,g,h:S\rightarrow K$ of the Pexider type functional equation

$$\begin{aligned} f(x+y)+f(x+\sigma y)+g(x+y)=2f(x)+2f(y)+g(x)g(y) \end{aligned}$$

resulting from summing up the well known quadratic and exponential functional equations side by side. We show that under some additional assumptions the above equation forces f and g to split back into the system of two equations

$$\begin{aligned} \left\{ \begin{array}{ll}f(x+y)+f(x+\sigma y)=2f(x)+2f(y)\\ g(x+y)=g(x)g(y)\end{array}\right. \end{aligned}$$

for all $x,y\in S$ (alienation phenomenon). We also consider an analogous problem for the quadratic and d’Alembert functional equations as well as for the quadratic, exponential and d’Alembert functional equations.

1 Introduction

The notion of alienation of functional equations was introduced by J. Dhombres in the paper [5]. Consider the functional equation $E(f)=0$ resulting from summing up two functional equations $E_1(f)=0$ and $E_2(f)=0$ side by side. We ask whether or not the equation $E(f)=0$ splits back to the system of these two functional equations. If the answer is positive, then we say that the equations $E_1(f)=0$ and $E_2(f)=0$ are alien.

We may consider functional equations $E_1(f)=0$ and $E_2(f)=0$ with the same function f, but we are also able to consider these equations with two different functions f and g, i.e. $E_1(f)=0$ and $E_2(g)=0$. Adding up these equations side by side gives a new functional equation of Pexider type $E(f,g)=0$. If in this case the alienation phenomenon holds, we say that the functional equations $E_1(f)=0$ and $E_2(g)=0$ are strongly alien.

Throughout this paper, unless otherwise stated, $(S,+,0)$ is a commutative monoid, $\sigma$ is an involutive endomorphism of S, that is $\sigma (\sigma (x))=x$ for all $x\in S$, and K is a field of characteristic different from 2. By the symbols ${\mathbb {R}}$ and ${\mathbb {C}}$ we denote the set of all real and complex numbers, respectively. To simplify our writing, we will write $\sigma x$ instead of $\sigma (x)$.

In the present paper we consider the classical exponential Cauchy functional equation

$$\begin{aligned} g(x+y)=g(x)g(y),\ \ \ x,y\in S, \end{aligned}$$

(E)

and the generalized versions of the classical quadratic and d’Alembert functional equations, namely

$$\begin{aligned} f(x+y)+f(x+\sigma y)=2f(x)+2f(y),\quad x,y\in S, \end{aligned}$$

(Q)

$$\begin{aligned} h(x+y)+h(x+\sigma y)=2h(x)h(y),\quad x,y\in S, \end{aligned}$$

(A)

respectively. The general solutions of equation (Q) for a map $f:S\rightarrow H$, where S is a commutative semigroup, H is an abelian group uniquely divisible by 2, and equation (A) for map $h:S\rightarrow K$, where additionally the field K is quadratically closed, were described in [20]. These equations in the classical forms (i.e. when $\sigma y=-y$) are well known and have been studied by numerous authors. Every function satisfying equation (E) and equation (Q) in the classical forms, is called exponential and quadratic, respectively.

Let $f,g,h:S\rightarrow K$. In order to simplify reading, we define the following operators:

$$\begin{aligned}{} & {} E_g(x,y)=g(x+y)-g(x)g(y),\\{} & {} Q_f(x,y)=f(x+y)+f(x+\sigma y)-2f(x)-2f(y),\\{} & {} A_h(x,y)=h(x+y)+h(x+\sigma y)-2h(x)h(y) \end{aligned}$$

for all $x,y\in S$.

The problems of alienation of functional equations have been discussed (among others) in the following papers [1,2,3, 6,7,8,9,10,11,12,13,14, 16,17,19, 22, 23]. The survey paper [15] constitutes a valuable source of information on the alienation of various functional equations. Inspired by these results, especially by the paper [21], we study the solutions $f,g,h:S\rightarrow K$ of the Pexider type functional equations

$$\begin{aligned} Q_f(x,y)+E_g(x,y)=0, \end{aligned}$$

(QE)

$$\begin{aligned} Q_f(x,y)+A_h(x,y)=0, \end{aligned}$$

(QA)

$$\begin{aligned} Q_f(x,y)+E_g(x,y)+A_h(x,y)=0, \end{aligned}$$

(QEA)

resulting from summing up equations (Q) and (E), equations (Q) and (A), and finally equations (Q), (E) and (A) side by side, respectively.

2 Alienation of the quadratic and exponential equations

In this section we deal with the alienation problem of equation (QE). The following simple example shows that, in general, the exponential and quadratic functional equations are not strongly alien to each other.

Example

Let $f,g:S\rightarrow {\mathbb {C}}$ be constant functions, i.e. $f(x)=c$ and $g(x)=d$ for all $x\in G$, where $d\in {\mathbb {C}}{\setminus }\{0,1\}$ such that $2c+d^2=d$. Then obviously these functions satisfy equation (QE), but neither f fulfills (Q), nor g satisfies (E).

In the next theorem we will prove that under an additional assumption, the exponential and quadratic functional equations are strongly alien to each other. We start with the following basic lemma.

Lemma 2.1

Assume that functions $f,g:S\rightarrow K$ satisfy equation (QE) for all $x,y\in S$. Then the function f is even with respect to $\sigma$, i.e. $f(x)=f(\sigma x)$ for all $x\in S$, and the function g satisfies the equation

$$\begin{aligned} g(x+y)-g(x+\sigma y)=g(x)[g(y)-g(\sigma y)],\quad x,y\in S. \end{aligned}$$

(2.1)

Proof

Interchanging the role of x and y in (QE), and next comparing the resulting equation with (QE), we easily obtain

$$\begin{aligned} f(x+\sigma y)=f(\sigma x+y),\quad x,y\in S. \end{aligned}$$

(2.2)

Setting $y=0$ we see that a function f is even with respect to $\sigma$. Furthermore, replacing y by $\sigma y$ in (QE) and making use of the evenness of f, we get

$$\begin{aligned} Q_f(x,y)+E_g(x,\sigma y)=0,\quad x,y\in S. \end{aligned}$$

(2.3)

Subtracting (2.3) from (QE) side by side, we conclude that the function g satisfies (2.1). $\square$

Theorem 2.2

Assume that functions $f,g:S\rightarrow K$ satisfy the equation

$$\begin{aligned} Q_f(x,y)+E_g(x,y)=0,\quad x,y\in S, \end{aligned}$$

(QE)

and $g(z_0)\ne g(\sigma z_0)$ for some $z_0\in S$. Then the functions f and g satisfy equations (Q) and (E), respectively.

Proof

According to Lemma 2.1, the function g satisfies (2.1). Putting $x=0$ and $y=z_0$ in (2.1) we get $g(z_0)-g(\sigma z_0)=g(0)[g(z_0)-g(\sigma z_0)]$. Hence $g(0)=1$. In view of (QE) and (2.2) we have

$$\begin{aligned}{} & {} Q_f(x+y,z)+E_g(x+y,z)=0, \end{aligned}$$

(2.4)

$$\begin{aligned}{} & {} Q_f(x+\sigma y,z)+E_g(x+\sigma y,z)=0,\end{aligned}$$

(2.5)

$$\begin{aligned}{} & {} Q_f(x,y+z)+E_g(x,y+z)=0,\end{aligned}$$

(2.6)

$$\begin{aligned}{} & {} Q_f(x,\sigma y+z)+E_g(x,\sigma y+z)=0 \end{aligned}$$

(2.7)

for all $x,y,z\in S$. Summing up equations (2.4) and (2.5) side by side, and subtracting the sum of equalities (2.6) and (2.7) from the resulting equation, we obtain

$$\begin{aligned}&2f(x+y)+2f(x+\sigma y)+4f(z)+g(z)[g(x+y)+g(x+\sigma y)]\\&\quad =2f(y+z)+2f(y+\sigma z)+4f(x)+g(x)[g(y+z)+g(\sigma y+z)] \end{aligned}$$

for all $x,y,z\in S$. Thus, applying (QE) again, we get

$$\begin{aligned}&g(x)[g(y+z)+g(\sigma y+z)-2g(y)]+2g(x+y)\\&\quad =g(z)[g(x+y)+g(x+\sigma y)-2g(y)]+2g(y+z),\quad x,y,z\in S. \end{aligned}$$

Replacing in the above equality y by $\sigma y$ and adding the resulting equality to the original one, we have

$$\begin{aligned}&g(x)[g(y+z)+g(\sigma y+z)-g(y)-g(\sigma y)]+g(x+y)+g(x+\sigma y)\\&\quad =g(z)[g(x+y)+g(x+\sigma y)-g(y)-g(\sigma y)]+g(y+z)+g(\sigma y+z), \end{aligned}$$

i.e.

$$\begin{aligned}&[g(x)-1][g(y+z)+g(\sigma y+z)-g(y)-g(\sigma y)]\nonumber \\&\quad =[g(z)-1][g(x+y)+g(x+\sigma y)-g(y)-g(\sigma y)] \end{aligned}$$

(2.8)

for all $x,y,z\in S$. Let us observe that the assumption $g(z_0)\ne g(\sigma z_0)$ implies that at least one of the values $g(z_0)$ or $g(\sigma z_0)$ is not equal to 1. Therefore, without loss of generality, we may assume that $g(z_0)\ne 1$ and define a function $H:S\rightarrow K$ by the formula

$$\begin{aligned} H(y):=\frac{g(y+z_0)+g(\sigma y+z_0)-g(y)-g(\sigma y)}{g(z_0)-1},\quad y\in S. \end{aligned}$$

Otherwise, we define a function $H_1$ by

$$\begin{aligned} H_1(y):=\frac{g(y+\sigma z_0)+g(\sigma y+\sigma z_0)-g(y)-g(\sigma y)}{g(\sigma z_0)-1},\quad y\in S, \end{aligned}$$

and the rest of the proof will proceed in the same way as for the function H. In view of (2.8), we have

$$\begin{aligned} g(x+y)+g(x+\sigma y)-g(y)-g(\sigma y)=[g(x)-1]H(y),\quad x,y\in S. \end{aligned}$$

(2.9)

From the definition of the function H and on account of (2.9) we obtain the following equalities

$$\begin{aligned}{} & {} [g(z_0)-1]H(y)=g(y+z_0)+g(\sigma y+z_0)-g(y)-g(\sigma y),\quad y\in S, \quad \end{aligned}$$

(2.10)

$$\begin{aligned}{} & {} [g(\sigma z_0)-1]H(y)=g(y+\sigma z_0)+g(\sigma y+\sigma z_0)-g(y)-g(\sigma y),\quad y\in S.\qquad \qquad \end{aligned}$$

(2.11)

Moreover, from (2.1) we also get

$$\begin{aligned}{} & {} g(y+z_0)-g(y+\sigma z_0)=g(y)[g(z_0)-g(\sigma z_0)],\quad y\in S, \end{aligned}$$

(2.12)

$$\begin{aligned}{} & {} g(\sigma y+z_0)-g(\sigma y+\sigma z_0)=g(\sigma y)[g(z_0)-g(\sigma z_0)],\quad y\in S. \end{aligned}$$

(2.13)

Subtracting equation (2.11) from (2.10) side by side, and applying equalities (2.12) and (2.13), we arrive at

$$\begin{aligned}&[g(z_0)-g(\sigma z_0)]H(y)=g(y+z_0)+g(\sigma y+z_0)-g(y+\sigma z_0)-g(\sigma y+\sigma z_0)\\&\quad =[g(z_0)-g(\sigma z_0)][g(y)+g(\sigma y)],\quad y\in S. \end{aligned}$$

Therefore, taking into account the assumption that $g(z_0)-g(\sigma z_0)\ne 0$, we infer that

$$\begin{aligned} H(y)=g(y)+g(\sigma y),\quad y\in S. \end{aligned}$$

Consequently, making use of (2.1) and (2.9), we get

$$\begin{aligned}&2g(x+y)=[g(x)-1]H(y)+g(x)[g(y)-g(\sigma y)]+g(y)+g(\sigma y)\\&\quad =[g(x)-1][g(y)+g(\sigma y)]+g(x)[g(y)-g(\sigma y)]+g(y)+g(\sigma y)\\&\quad =g(x)[g(y)+g(\sigma y)]+g(x)[g(y)-g(\sigma y)]\\&\quad =2g(x)g(y),\quad x,y\in S. \end{aligned}$$

Therefore, the function g satisfies equation (E), and hence the function f satisfies equation (Q), which finishes the proof. $\square$

Let us observe that if the function f occurring in Theorem 2.2 is not identically equal to 0, then it is in fact nonconstant.

As we have seen, equation (2.1) plays an important role in the proof of Theorem 2.2. The same functional equation was investigated in the paper [21], where the problems of alienation of the Jensen, exponential and d’Alembert functional equations were considered. We show below that (2.1) is closely related to equation (E).

Proposition 2.3

Let $(G,+)$ be an abelian group uniquely divisible by 2. Assume that a function $g:G\rightarrow K$ satisfies the equation

$$\begin{aligned} g(x+y)-g(x-y)=g(x)[g(y)-g(-y)],\quad x,y\in G. \end{aligned}$$

(2.14)

Then the function g is an arbitrary constant or satisfies equation (E).

Proof

It is easy to observe that an arbitrary constant function g satisfies equation (2.14). So, from now on, we assume that g is nonconstant. Suppose on the contrary that $g(y)=g(-y)$ for every $y\in G$. Then, in view of Lemma 2.1, we obtain $g(x+y)-g(x-y)=0$ for all $x,y\in G$, i.e. $g(2x)=g(0)$ for all $x\in G$. Since the group G is uniquely divisible by 2, $g(x)=g(0)$ for all $x\in G$, which yields a contradiction. Therefore, there exists a $z_0\in G$ such that $g(z_0)\ne g(-z_0)$, and consequently on account of the proof of Theorem 2.2 we have $g(0)=1$. Interchanging the role of x and y, and next putting $-x$ instead of x in (2.14), we get

$$\begin{aligned}{} & {} g(x+y)-g(-x+y)=g(y)[g(x)-g(-x)],\quad x,y\in G, \end{aligned}$$

(2.15)

$$\begin{aligned}{} & {} g(-x+y)-g(-x-y)=g(-x)[g(y)-g(-y)],\quad x,y\in G, \end{aligned}$$

(2.16)

respectively. Therefore, from the equalities (2.15) and (2.16) we deduce that

$$\begin{aligned} g(x+y)-g(-x-y)=g(x)g(y)-g(-x)g(-y),\quad x,y\in G. \end{aligned}$$

(2.17)

Substituting $x-y$ instead of x in (2.17), we obtain

$$\begin{aligned} g(x)-g(-x)=g(x-y)g(y)-g(-x+y)g(-y),\quad x,y\in G. \end{aligned}$$

(2.18)

Replace y by $-y$ in (2.18) to get

$$\begin{aligned} g(x)-g(-x)=g(x+y)g(-y)-g(-x-y)g(y),\quad x,y\in G. \end{aligned}$$

(2.19)

Comparing equations (2.18) and (2.19) we see that

$$\begin{aligned} g(y)[g(x-y)+g(-x-y)]=g(-y)[g(x+y)+g(-x+y)],\quad x,y\in G. \end{aligned}$$

Hence, for $y=x$ we have

$$\begin{aligned} g(x)[g(-2x)+1]=g(-x)[g(2x)+1],\quad x\in G. \end{aligned}$$

(2.20)

Putting $y=x$ in (2.14) and using the resulting equality in (2.20), we obtain

$$\begin{aligned} g(x)[g(-x)^2-g(x)g(-x)+2]=g(-x)[g(x)^2-g(x)g(-x)+2],\quad x\in G, \end{aligned}$$

i.e.

$$\begin{aligned}{}[g(x)-g(-x)][g(x)g(-x)-1]=0,\quad x\in G. \end{aligned}$$

(2.21)

Therefore, $g(x)=g(-x)$ or $g(x)g(-x)=1$ for all $x\in G$. If $g(x)=g(-x)$ for all $x\in G$, then from the first part of the proof we obtain that the function g is constant, which is in contradiction with the assumption that g is nonconstant. Therefore, $g(x)g(-x)=1$ for all $x\in G$, and hence $g(x)\ne 0$ for all $x\in G$. Applying this property, from the equalities (2.15) and (2.18) we deduce that

$$\begin{aligned} g(x+y)=g(y)^2g(x-y),\quad x,y\in G. \end{aligned}$$

Multiplying both sides of (2.14) by $g(y)^2$ and using the above equality we get

$$\begin{aligned} g(y)^2g(x+y)-g(x+y)=g(x)g(y)^2[g(y)-g(-y)]\\=g(x)g(y)[g(y)^2-1],\quad x,y\in G, \end{aligned}$$

i.e.

$$\begin{aligned}{}[g(y)^2-1][g(x+y)-g(x)g(y)]=0,\quad x,y\in G. \end{aligned}$$

Since the function g is nonconstant, $g(x+y)=g(x)g(y)$ for all $x,y\in G$. The proof is complete. $\square$

3 Alienation of the quadratic and d’Alembert equations

In this section we deal with the alienation problem of equation (QE). The following simple example shows that, in general, the exponential and quadratic functional equations are not strongly alien to each other.

Example

Let $f,h:S\rightarrow {\mathbb {C}}$ be constant functions, i.e. $f(x)=c$ and $h(x)=d$ for all $x\in G$, where $d\in {\mathbb {C}}{\setminus }\{0,1\}$ such that $c+d^2=d$. Then obviously these functions satisfy equation (QA), but neither f fulfills equation (Q), nor h satisfies equation (A).

In the next theorem we will prove that under an additional assumption, the quadratic and d’Alembert functional equations are strongly alien to each other.

Theorem 3.1

Assume that functions $f,h:S\rightarrow K$ satisfy the equation

$$\begin{aligned} Q_f(x,y)+A_h(x,y)=0 \end{aligned}$$

(QA)

for all $x,y\in S$. Moreover, let $h\ne e$, where $e\in K\setminus \{0,1\}$. Then f satisfies equation (Q) and h satisfies equation (A).

Proof

If $h=0$ or $h=1$, then h satisfies equation (A), which immediately implies that f solves (Q). So, from now on, we assume that $h\ne 0$ and $h\ne 1$, i.e. the function h is, in fact, assumed to be nonconstant. Interchanging the roles of x and y in (QA), and next comparing the resulting equation with (QA), we easily obtain that

$$\begin{aligned} f(x+\sigma y)+h(x+\sigma y)=f(\sigma x+y)+h(\sigma x+y),\quad x,y\in S. \end{aligned}$$

(3.1)

In view of (QA), we have

$$\begin{aligned}{} & {} Q_f(x+y,z)+A_h(x+y,z)=0, \end{aligned}$$

(3.2)

$$\begin{aligned}{} & {} Q_f(x+\sigma y,z)+A_h(x+\sigma y,z)=0,\end{aligned}$$

(3.3)

$$\begin{aligned}{} & {} Q_f(x,y+z)+A_h(x,y+z)=0,\end{aligned}$$

(3.4)

$$\begin{aligned}{} & {} Q_f(x,y+\sigma z)+A_h(x,y+\sigma z)=0 \end{aligned}$$

(3.5)

for all $x,y,z\in S$. Summing up equations (3.2) and (3.3) side by side, and subtracting the sum of equalities (3.4) and (3.5) from the resulting equation, we obtain

$$\begin{aligned}&f(x+y)+f(x+\sigma y)+2f(z)+h(z)[h(x+y)+h(x+\sigma y)]\\&\quad =f(y+z)+f(y+\sigma z)+2f(x)+h(x)[h(y+z)+h(y+\sigma z)] \end{aligned}$$

for all $x,y,z\in S$. Thus, applying (QA) again, we get

$$\begin{aligned}&[h(z)-1][h(x+y)+h(x+\sigma y)]+2h(x)h(y)\\&\quad =[h(x)-1][h(y+z)+h(y+\sigma z)]+2h(y)h(z), \end{aligned}$$

i.e.

$$\begin{aligned}&[h(z)-1][h(x+y)+h(x+\sigma y)-2h(y)]\nonumber \\&\quad =[h(x)-1][h(y+z)+h(y+\sigma z)-2h(y)] \end{aligned}$$

(3.6)

for all $x,y,z\in S$. Setting $z=x$ in (3.6), we obtain

$$\begin{aligned}&[h(x)-1][h(x+y)+h(\sigma x+y)-2h(y)]\\&\quad =[h(x)-1][h(x+y)+h(x+\sigma y)-2h(y)], \end{aligned}$$

i.e.

$$\begin{aligned}{}[h(x)-1][h(x+\sigma y)-h(\sigma x+y)]=0,\quad x,y\in S. \end{aligned}$$

Since $h\ne 1$,

$$\begin{aligned} h(x+\sigma y)=h(\sigma x+y),\quad x,y\in S. \end{aligned}$$

Setting $y=0$ we see that the function h is even with respect to $\sigma$, and in view of (3.1), the function f is also even with respect to $\sigma$. Moreover, putting $y=z=0$ in (3.6), we get

$$\begin{aligned}{}[h(0)-1][h(x)-h(0)]=0,\quad x\in S. \end{aligned}$$

Since the function h is nonconstant, $h(0)=1$.

Furthermore, fix $z_0\in S$ with $h(z_0)\ne 1$ and define a function $H:S\rightarrow K$ by the formula

$$\begin{aligned} H(y):=\frac{h(y+z_0)+h(y+\sigma z_0)-2h(y)}{h(z_0)-1},\quad y\in S. \end{aligned}$$

Then

$$\begin{aligned} h(x+y)+h(x+\sigma y)-2h(y)=[h(x)-1]H(y),\quad x,y\in S. \end{aligned}$$

(3.7)

From (3.7) we have the following equalities:

$$\begin{aligned}{} & {} h(2x+y)+h(x+\sigma x+\sigma y)-2h(x+y)=[h(x)-1]H(x+y), \end{aligned}$$

(3.8)

$$\begin{aligned}{} & {} h(2x+\sigma y)+h(x+\sigma x+y)-2h(x+\sigma y)=[h(x)-1]H(x+\sigma y),\qquad \end{aligned}$$

(3.9)

$$\begin{aligned}{} & {} h(x+\sigma x+y)+h(x+\sigma x+\sigma y)-2h(y)=[h(x+\sigma x)-1]H(y),\qquad \end{aligned}$$

(3.10)

$$\begin{aligned}{} & {} h(2x+y)+h(2x+\sigma y)-2h(y)=[h(2x)-1]H(y), \end{aligned}$$

(3.11)

$$\begin{aligned}{} & {} h(2x)+h(x+\sigma x)-2h(x)=[h(x)-1]H(x) \end{aligned}$$

(3.12)

for all $x,y\in S$. Summing up equations (3.8) and (3.9) side by side, and next using (3.7), (3.10), (3.11) and (3.12), we obtain

$$\begin{aligned}&[h(x)-1][H(x+y)+H(x+\sigma y)]\\&\quad =h(2x+y)+h(2x+\sigma y)+h(x+\sigma x+y)\\&\qquad +h(x+\sigma x+\sigma y)-2h(x+y)-2h(x+\sigma y)\\&\quad =[h(2x)-1]H(y)+[h(x+\sigma x)-1]H(y)-2[h(x)-1]H(y)\\&\quad =[h(x)-1]H(x)H(y),\quad x,y\in S. \end{aligned}$$

Since $h\ne 1$,

$$\begin{aligned} H(x+y)+H(x+\sigma y)=H(x)H(y),\quad x,y\in S. \end{aligned}$$

(3.13)

Now, we are going to show that $H=2h$. Interchanging the roles of x and y in (3.7), we get

$$\begin{aligned} h(x+y)+h(\sigma x+y)-2h(x)=[h(y)-1]H(x),\quad x,y\in S. \end{aligned}$$

(3.14)

Since $h(x+\sigma y)=h(\sigma x+y)$ for all $x,y\in S$, comparing (3.7) and (3.14) we obtain

$$\begin{aligned}{} & {} [h(x)-1]H(y)+2h(y)=[h(y)-1]H(x)+2h(x), \end{aligned}$$

i.e.

$$\begin{aligned}{}[h(x)-1][H(y)-2h(y)]=[h(y)-1][H(x)-2h(x)],\quad x,y\in S. \end{aligned}$$

Then

$$\begin{aligned} H(x)=(\alpha +2)h(x)-\alpha , \end{aligned}$$

where $\alpha :=\frac{H(z_0)-2h(z_0)}{h(z_0)-1}$. Therefore, we can rewrite equation (3.7) in the form

$$\begin{aligned} h(x+y)+h(x+\sigma y)=2h(x)h(y)+\alpha [h(x)-1][h(y)-1],\quad x,y\in S. \end{aligned}$$

We show that $\alpha =0$. Suppose that $\alpha \ne 0$. Then from the definition of the constant $\alpha$ we deduce $H(z_0)\ne 2h(z_0)$. Hence, from (3.7) we obtain

$$\begin{aligned}{} & {} h(x+z_0)+h(x+\sigma z_0)-2h(z_0)=[h(x)-1]H(z_0)\\{} & {} \quad \ne 2h(x)h(z_0)-2h(z_0),\quad x\in S, \end{aligned}$$

i.e.

$$\begin{aligned} h(x+z_0)+h(x+\sigma z_0)\ne 2h(x)h(z_0),\quad x\in S. \end{aligned}$$

Setting $x=0$ and using the evenness of the function h with respect to $\sigma$ we get a contradiction. Therefore $\alpha =0$, and consequently $H=2h$. Moreover, the function h satisfies equation (A) and so, in view of (QA), f solves (Q). The proof is complete. $\square$

Applying [20, Theorems 1, 3], from Theorem 3.1 we deduce the following result.

Corollary 3.2

Let K be a quadratically closed field of characteristic different from 2. Assume that functions $f,h:S\rightarrow K$ satisfy the equation (QA) for all $x,y\in S$ and $h\ne e$, where $e\in K\setminus \{0,1\}$. Then there exist a function $\chi :S\rightarrow K$ satisfying equation (E), a symmetric biadditive function $B:S\times S\rightarrow K$ and an additive function $A:S\rightarrow K$ such that $B(\sigma x,y)=-B(x,y)$, $A(\sigma x)=A(x)$ for all $x\in S$ and

$$\begin{aligned} f(x)=B(x,x)+A(x),\quad h(x)=\frac{\chi (x)+\chi (\sigma x)}{2},\quad x\in S. \end{aligned}$$

As a corollary of Theorem 3.1 we obtain a result which states that in the class of nonconstant functions mapping an abelian group into a field of characteristic different from 2, the quadratic and d’Alembert functional equations are strongly alien to each other. In what follows of this section, let $\sigma y=-y$.

Corollary 3.3

Let $(G,+)$ be an abelian group. Assume that functions $f,h:G\rightarrow K$ satisfy equation

$$\begin{aligned} Q_f(x,y)+A_h(x,y)=0 \end{aligned}$$

(3.15)

for all $x,y\in G$. Then:

• (i) the functions f, h satisfy equation (Q) and equation (A), respectively,

  or

• (ii) there exist constants $c,e\in K$ with $c+e^2=e$ such that $f(x)=Q(x)+c$ and $h(x)=e$ for all $x\in G$, where Q is an arbitrary quadratic function.

Proof

Consider the case where the functions f, h are nonconstant. Then, in view of Theorem 3.1, we get (i).

Now, consider the case where $h(x)=e$ for all $x\in G$ and for some $e\in K$. If $e\in \{0,1\}$, then obviously the function h satisfies the d’Alembert functional equation, which immediately implies that f is a quadratic function. Hence, (i) holds. Now, suppose that $e\notin \{0,1\}$. Let $\alpha :=2e(e-1)$. Then $\alpha \ne 0$ and from (3.15) we obtain

$$\begin{aligned} Q_f(x,y)=\alpha ,\quad x,y\in G. \end{aligned}$$

Defining a new function $F:S\rightarrow K$ by the formula $F(x):=f(x)+\frac{1}{2}\alpha$ we can easily see that F satisfies the quadratic functional equation. Finally, $f(x)=Q(x)+c$ for all $x\in G$, where $c:=e(1-e)$ and $Q:S\rightarrow K$ is an arbitrary quadratic function. Hence, (ii) holds.

Now, we assume that the function f is constant, i.e. $f(x)=c$ for all $x\in G$ and for some $c\in K$. If $c=0$, then f is a quadratic function, and consequently the function h satisfies the d’Alembert functional equation. Hence, (i) holds. Now, suppose that $c\ne 0$. Then from (3.15) we have

$$\begin{aligned} A_h(x,y)=2c,\quad x,y\in G. \end{aligned}$$

We are going to show that the function h is constant. Indeed, from the above equation we obtain

$$\begin{aligned}{} & {} A_h(x+y,z)=2c,\\{} & {} A_h(x,y+z)=2c,\\{} & {} A_h(x-y,z)=2c,\\{} & {} A_h(x,y-z)=2c \end{aligned}$$

for all $x,y,z\in G$. Combining these equalities, we get

$$\begin{aligned}{} & {} h(x)[h(y+z)+h(y-z)]=h(z)[h(x+y)+h(x-y)],\quad x,y,z\in G, \end{aligned}$$

i.e.

$$\begin{aligned} h(x)[h(y)h(z)+c]=h(z)[h(x)h(y)+c],\quad x,y,z\in G. \end{aligned}$$

Hence, $ch(x)=ch(z)$ for all $x,z\in G$. Since $c\ne 0$, h is a constant function, i.e. $h(x)=e$ for all $x\in G$ and for some $e\in K$. Therefore, $c+e^2=e$ and (ii) holds (with $Q=0$). This finishes the proof. $\square$

Let us observe that if the function f occurring in Theorem 3.1 is not identically equal to 0, then it is in fact nonconstant.

4 Alienation of the quadratic, exponential and d’Alembert equations

In this section, we deal with the alienation problem of equation (QEA). The following simple examples show that, in general, the quadratic, exponential and d’Alembert functional equations are not strongly alien to each other.

Example

Let $f,g,h:S\rightarrow {\mathbb {C}}$ be constant functions, i.e. $f(x)=c$, $g(x)=d$ and $h(x)=e$ for all $x\in S$, where $d,e\in {\mathbb {C}}{\setminus }\{0,1\}$ such that $2c+d^2+2e^2=d+2e$. Then, as one can easily check, these functions satisfy equation (QEA), but neither f fulfills (Q), nor g satisfies (E), nor h satisfies (A).

Moreover, in the next example we can see that the function h satisfying equation (QEA) does not have to be even.

Example

Let $f,g,h:{\mathbb {R}}\rightarrow {\mathbb {R}}$ be nonconstant functions given by the formulae $f(x)=-\frac{\sqrt{2}}{4}E(x)$, $g(x)=\frac{1}{2}E(x)$ and $h(x)=1+\frac{\sqrt{2}}{4}E(x)$ for all $x\in {\mathbb {R}}$, where E is an arbitrary nonconstant exponential function. Assume that $\sigma y=-y$. Then it is not hard to check that these functions satisfy equation (QEA), but the function h is not even.

We prove the following basic lemma.

Lemma 4.1

Assume that functions $f,g,h:S\rightarrow K$ satisfy equation (QEA) for all $x,y\in S$. Then

$$\begin{aligned} f(x)-f(\sigma x)=h(\sigma x)-h(x),\quad x\in S, \end{aligned}$$

(4.1)

and the functions g, h satisfy the equation

$$\begin{aligned} g(x+y)-g(x+\sigma y)=g(x)[g(y)-g(\sigma y)]+2[h(x)-1][h(y)-h(\sigma y)]\qquad \nonumber \\ \end{aligned}$$

(4.2)

for all $x,y\in S$.

Proof

Interchanging the roles of x and y in (QEA), and next comparing the resulting equation with (QEA) we easily obtain that

$$\begin{aligned} f(x+\sigma y)-f(\sigma x+y)=h(\sigma x+y)-h(x+\sigma y),\quad x,y\in S. \end{aligned}$$

(4.3)

Setting $y=0$ we see that the condition (4.1) holds. Furthermore, replacing y by $\sigma y$ in (QEA), and next comparing the resulting equation with (QEA) and making use of (4.1), we get

$$\begin{aligned}&g(x+y)-g(x+\sigma y)\\&\quad =2[f(y)-f(\sigma y)]+g(x)[g(y)-g(\sigma y)]+2h(x)[h(y)-h(\sigma y)]\\&\quad =g(x)[g(y)-g(\sigma y)]+2[h(x)-1][h(y)-h(\sigma y)],\quad x,y\in S. \end{aligned}$$

The proof is complete. $\square$

As we could see in Proposition 2.3, the function g satisfying (2.14) is constant or exponential. In the next example, one can observe that the functions g and h satisfying equation (4.2) may be far from the solutions of equations (E) and (A), respectively.

Example

Let $g,h:{\mathbb {R}}\rightarrow {\mathbb {R}}$ be functions given by the formulae $g(x)=x^2+1$ and $h(x)=x+1$ for all $x\in {\mathbb {R}}$ and let $\sigma y=-y$. Then it is not difficult to verify that these functions satisfy equation (4.2), but neither g satisfies equation (E), nor h satisfies equation (A).

Now, we prove the following statement which under some assumptions on the functions g and h, gives us general solutions of equation (QEA).

Proposition 4.2

Let K be a quadratically closed field of characteristic different from 2. Assume that functions $f,g,h:S\rightarrow K$ satisfy the equation

$$\begin{aligned} Q_f(x,y)+E_g(x,y)+A_h(x,y)=0 \end{aligned}$$

(QEA)

for all $x,y\in S$. Moreover, let $g(z_0)\ne g(\sigma z_0)$, $h(z_0)=h(\sigma z_0)$ for some $z_0\in S$ and $h\ne e$, where $e\in K\setminus \{0,1\}$. Then:

• (i) the functions f, g, h satisfy equations (Q), (E) and (A), respectively,

  or

• (ii) there exist constants $a_i,b_i\in K$, $i\in \{1,2,3\}$ with

  $$\begin{aligned}\left[ \begin{array}{cc} a_1 &{} \sqrt{2}b_1 \\ a_2 &{} \sqrt{2}b_2 \\ a_3 &{} \sqrt{2}b_3 \\ \end{array}\right] \cdot \left[ \begin{array}{ccc} a_1 &{} a_2 &{} a_3 \\ \sqrt{2}b_1 &{} \sqrt{2}b_2 &{} \sqrt{2}b_3 \\ \end{array}\right] =\left[ \begin{array}{ccc} a_1 &{} 0 &{} 0 \\ 0 &{} a_2 &{} a_3 \\ 0 &{} a_3 &{} 0 \\ \end{array}\right] \end{aligned}$$

  such that

  $$\begin{aligned} \left\{ \begin{array}{l} f(x)=B(x,x)+A_1(x)-b_1E_1(x)-b_2E(x)-b_3E(x)A(x)\\ g(x)=a_1E_1(x)+a_2E(x)+a_3E(x)A(x)\\ h(x)=1+b_1E_1(x)+b_2E(x)+b_3E(x)A(x) \end{array}\right. \end{aligned}$$

  for all $x\in S$, where $A:S\rightarrow K$ is an arbitrary additive function, $E,E_1:S\rightarrow K$ satisfy equation (E), $B:S\times S\rightarrow K$ is an arbitrary symmetric biadditive function with $B(\sigma x, y)=-B(x,y)$, and $A_1:S\rightarrow K$ is an arbitrary additive function with $A_1(\sigma x)=A_1(x)$ for all $x\in S$.

Proof

If $h=0$ or $h=1$, then h satisfies equation (A), which, in view of Theorem 2.2, immediately implies that the functions f and g solve (Q) and (E), respectively. So, from now on we assume that $h\ne 0$ and $h\ne 1$, i.e. the function h is nonconstant. In view of (QEA), we have

$$\begin{aligned}{} & {} Q_f(x+y,z)+E_g(x+y,z)+A_h(x+y,z)=0, \end{aligned}$$

(4.4)

$$\begin{aligned}{} & {} Q_f(x+\sigma y,z)+E_g(x+\sigma y,z)+A_h(x+\sigma y,z)=0, \end{aligned}$$

(4.5)

$$\begin{aligned}{} & {} Q_f(x,y+z)+E_g(x,y+z)+A_h(x,y+z)=0, \end{aligned}$$

(4.6)

$$\begin{aligned}{} & {} Q_f(x,\sigma y+z)+E_g(x,\sigma y+z)+A_h(x,\sigma y+z)=0 \end{aligned}$$

(4.7)

for all $x,y,z\in S$. Summing up equations (4.4) and (4.5) side by side, and subtracting the sum of equalities (4.6) and (4.7) from the resulting equation, we obtain

$$\begin{aligned}&2f(x+y)+2f(x+\sigma y)+4f(z)+g(z)[g(x+y)+g(x+\sigma y)]\\&\qquad +2h(z)[h(x+y)+h(x+\sigma y)]\\&\quad =2f(y+z)+2f(\sigma y+z)+4f(x)+g(x)[g(y+z)+g(\sigma y+z)]\\&\qquad +2h(x)[h(y+z)+h(\sigma y+z)] \end{aligned}$$

for all $x,y,z\in S$. Thus, applying (QEA) again, we get

$$\begin{aligned}&g(x)[g(y+z)+g(\sigma y+z)-2g(y)]\\&\qquad +2[h(x)-1][h(y+z)+h(\sigma y+z)-2h(y)]+2g(x+y)\\&\quad =g(z)[g(x+y)+g(x+\sigma y)-2g(y)]\\&\qquad +2[h(z)-1][h(x+y)+h(x+\sigma y)-2h(y)]+2g(y+z) \end{aligned}$$

for all $x,y,z\in S$. Replacing y by $\sigma y$ in the above equality and adding the resulting equality to the original one, we have

$$\begin{aligned}&[g(x)-1][g(y+z)+g(\sigma y+z)-g(y)-g(\sigma y)]\\&\quad +2[h(x)-1][h(y+z)+h(\sigma y+z)-2h(y)]\\&\quad =[g(z)-1][g(x+y)+g(x+\sigma y)-g(y)-g(\sigma y)]\\&\qquad +2[h(z)-1][h(x+y)+h(x+\sigma y)-2h(y)] \end{aligned}$$

for all $x,y,z\in S$. Furthermore, replacing in the above equality z by $\sigma z$ and subtracting the resulting equality from the original one, we obtain

$$\begin{aligned}&[g(x)-1][g(y+z)+g(\sigma y+z)-g(y+\sigma z)-g(\sigma y+\sigma z)]\nonumber \\&\qquad +2[h(x)-1][h(y+z)+h(\sigma y+z)-h(y+\sigma z)-h(\sigma y+\sigma z)]\nonumber \\&\quad =[g(z)-g(\sigma z)][g(x+y)+g(x+\sigma y)-g(y)-g(\sigma y)]\nonumber \\&\qquad +2[h(z)-h(\sigma z)][h(x+y)+h(x+\sigma y)-h(y)-h(\sigma y)] \end{aligned}$$

(4.8)

for all $x,y,z\in S$.

Let us define functions $U,V:S\rightarrow K$ by the formulae

$$\begin{aligned}{} & {} U(y):=\frac{g(y+z_0)+g(\sigma y+z_0)-g(y+\sigma z_0)-g(\sigma y+\sigma z_0)}{g(z_0)-g(\sigma z_0)},\quad y\in S,\\{} & {} V(y):=\frac{h(y+z_0)+h(\sigma y+z_0)-h(y+\sigma z_0)-h(\sigma y+\sigma z_0)}{g(z_0)-g(\sigma z_0)},\quad y\in S. \end{aligned}$$

As we can see, both functions U and V are even with respect to $\sigma$. In view of (4.8) and the condition $h(z_0)=h(\sigma z_0)$, we have

$$\begin{aligned} g(x+y)+g(x+\sigma y)-g(y)-g(\sigma y)=[g(x)-1]U(y)+2[h(x)-1]V(y) \qquad \end{aligned}$$

(4.9)

for all $x,y\in S$. From (4.9), we obtain

$$\begin{aligned}{} & {} [g(z_0)-1]U(y)+2[h(z_0)-1]V(y)\nonumber \\{} & {} \qquad =g(y+z_0)+g(\sigma y+z_0)-g(y)-g(\sigma y), \end{aligned}$$

(4.10)

$$\begin{aligned}{} & {} [g(\sigma z_0)-1]U(y)+2[h(\sigma z_0)-1]V(y)\nonumber \\{} & {} \qquad =g(y+\sigma z_0)+g(\sigma y+\sigma z_0)-g(y)-g(\sigma y) \end{aligned}$$

(4.11)

for all $y\in S$. Moreover, from (4.2) we also get

$$\begin{aligned}{} & {} g(y+z_0)-g(y+\sigma z_0)=g(y)[g(z_0)-g(\sigma z_0)],\quad y\in S, \end{aligned}$$

(4.12)

$$\begin{aligned}{} & {} g(\sigma y+z_0)-g(\sigma y+\sigma z_0)=g(\sigma y)[g(z_0)-g(\sigma z_0)],\quad y\in S. \end{aligned}$$

(4.13)

Subtracting equation (4.11) from (4.10) side by side, and applying equalities (4.12) and (4.13), we arrive at

$$\begin{aligned}&[g(z_0)-g(\sigma z_0)]U(y)=g(y+z_0)+g(\sigma y+z_0)-g(y+\sigma z_0)-g(\sigma y+\sigma z_0)\\&\quad =[g(z_0)-g(\sigma z_0)][g(y)+g(\sigma y)],\quad y\in S.\end{aligned}$$

Therefore, taking into account the assumption $g(z_0)-g(\sigma z_0)\ne 0$, we infer that

$$\begin{aligned} U(y)=g(y)+g(\sigma y),\quad y\in S. \end{aligned}$$

Consequently, making use of (4.2) and (4.9), we get

$$\begin{aligned}&2g(x+y)=[g(x)-1]U(y)+2[h(x)-1]V(y)+g(x)[g(y)-g(\sigma y)]\\&\qquad +g(y)+g(\sigma y)+2[h(x)-1][h(y)-h(\sigma y)]\\&\quad =2g(x)g(y)+2[h(x)-1][V(y)+h(y)-h(\sigma y)],\quad x,y\in S, \end{aligned}$$

i.e.

$$\begin{aligned} g(x+y)=g(x)g(y)+[h(x)-1][V(y)+h(y)-h(\sigma y)],\quad x,y\in S.\qquad \end{aligned}$$

(4.14)

Now, we determine the function V. Interchanging the roles of x and y in (4.14), and next comparing the resulting equation with the original one, we obtain

$$\begin{aligned}{}[h(x)-1][V(y)+h(y)-h(\sigma y)]=[h(y)-1][V(x)+h(x)-h(\sigma x)],\quad x,y\in S. \end{aligned}$$

Furthermore, replacing y by $\sigma y$ in the above equality, using the evenness of the function V, and next subtracting the resulting equation from the last one, we get

$$\begin{aligned}{}[h(y)-h(\sigma y)][V(x)-h(x)-h(\sigma x)+2]=0,\quad x,y\in S. \end{aligned}$$

Hence, $h(y)=h(\sigma y)$ for all $y\in S$ or $V(x)=h(x)+h(\sigma x)-2$ for all $x\in S$.

Let us consider the first case, where the function h is even with respect to $\sigma$. Then, on account of the definition of the function V, we have $V=0$. Hence, from (4.14), we see that the function g satisfies equation (E). Consequently, equation (QEA) reduces to equation (QA). Therefore, in view of Theorem 3.1, a function f satisfies equation (Q), and the function h satisfies equation (A). Hence, the condition (i) holds.

Now, we consider the second case, where $V(x)=h(x)+h(\sigma x)-2$ for all $x\in S$. Then we can rewrite equality (4.14) in the form

$$\begin{aligned} g(x+y)=g(x)g(y)+2[h(x)-1][h(y)-1],\quad x,y\in S. \end{aligned}$$

(4.15)

Applying this equality to (QEA), we get

$$\begin{aligned}{} & {} f(x+y)+f(x+\sigma y)+h(x+y)+h(x+\sigma y)\\{} & {} \qquad =2f(x)+2f(y)+2h(x)+2h(y)-2 \end{aligned}$$

for all $x,y\in S$, i.e. the function $f+h-1$ satisfies (Q). Therefore, in view of [20, Theorem 3], we obtain $f(x)=B(x,x)+A_1(x)-h(x)+1$, where $B:S\times S\rightarrow K$ is an arbitrary symmetric biadditive function with $B(\sigma x, y)=-B(x,y)$ and $A_1:S\rightarrow K$ is an arbitrary additive function with $A_1(\sigma x)=A_1(x)$ for all $x\in S$. Let us define a new function $\varphi :S\rightarrow K$ by the formula $\varphi (x):=\sqrt{2}[h(x)-1]$ for all $x\in S$. Then equation (4.15) reduces to

$$\begin{aligned} g(x+y)=g(x)g(y)+\varphi (x)\varphi (y),\quad x,y\in S. \end{aligned}$$

(4.16)

The above functional equation was considered in the paper [4, Lemma 4], and its general solutions are of the form

$$\begin{aligned}{} & {} g(x)=a_1E_1(x)+a_2E(x)+a_3E(x)A(x),\\{} & {} \varphi (x)=c_1E_1(x)+c_2E(x)+c_3E(x)A(x) \end{aligned}$$

for all $x\in S$, where A is an arbitrary additive function, $E,E_1$ are arbitrary functions satisfying equation (E), and $a_i,c_i\in K$, $i\in \{1,2,3\}$ with

$$\begin{aligned}\left[ \begin{array}{cc} a_1 &{} c_1 \\ a_2 &{} c_2 \\ a_3 &{} c_3 \\ \end{array}\right] \cdot \left[ \begin{array}{ccc} a_1 &{} a_2 &{} a_3 \\ c_1 &{} c_2 &{} c_3 \\ \end{array}\right] =\left[ \begin{array}{ccc} a_1 &{} 0 &{} 0 \\ 0 &{} a_2 &{} a_3 \\ 0 &{} a_3 &{} 0 \\ \end{array}\right] . \end{aligned}$$

Finally, since $h(x)=1+\frac{\sqrt{2}}{2}\varphi (x)$, setting $b_i:=\frac{\sqrt{2}}{2}c_i$ and applying the above solutions, we obtain (ii). The proof is complete. $\square$

It is an open problem to prove Proposition 4.2 without assuming that $h(z_0)=h(\sigma z_0)$ for some $z_0\in S$.

Remark 4.3

On account of [4, Remark 3], we can rewrite the solutions of equation (QEA) contained in Proposition 4.2 (ii) as follows:

$$\begin{aligned} g(x)=E(x)[1+A(x)],\quad \varphi (x)=\pm iE(x)A(x) \end{aligned}$$

or

$$\begin{aligned} g(x)=aE(x)+(1-a)E_1(x),\quad \varphi (x)=cE(x)-cE_1(x) \end{aligned}$$

for all $x\in S$, where $a,c\in K$, $a^2+c^2=a$ and $i^2=-1$. Since $h(x)=1+\frac{\sqrt{2}}{2}\varphi (x)$, setting $b:=\frac{\sqrt{2}}{2}c$ and applying the above solutions, we obtain

$$\begin{aligned} \left\{ \begin{array}{l} f(x)=B(x,x)+A_1(x)\mp \frac{\sqrt{2}}{2}iE(x)A(x)\\ g(x)=E(x)[1+A(x)]\\ h(x)=1\pm \frac{\sqrt{2}}{2}iE(x)A(x) \end{array}\right. \end{aligned}$$

or

$$\begin{aligned} \left\{ \begin{array}{l} f(x)=B(x,x)+A_1(x)-bE(x)+bE_1(x)\\ g(x)=aE(x)+(1-a)E_1(x)\\ h(x)=1+bE(x)-bE_1(x) \end{array}\right. \end{aligned}$$

for all $x\in S$, where $a,b\in K$ with $a^2+2b^2=a$, $A:S\rightarrow K$ is an arbitrary additive function, $E:S\rightarrow K$ satisfies equation (E), $B:S\times S\rightarrow K$ is an arbitrary symmetric biadditive function with $B(\sigma x, y)=-B(x,y)$, and $A_1:S\rightarrow K$ is an arbitrary additive function with $A_1(\sigma x)=A_1(x)$ for all $x\in S$.

As an immediate consequence of Proposition 4.2, under the assumption that the function h is even with respect to $\sigma$, we obtain the following theorem, which states that the quadratic, exponential and d’Alembert functional equations are strongly alien to each other. Note that we do not have to assume that the field K is quadratically closed (this assumption is only used in the second case of the proof of Proposition 4.2).

Theorem 4.4

Assume that functions $f,g,h:S\rightarrow K$ satisfy equation (QEA) for all $x,y\in S$, and $g(z_0)\ne g(\sigma z_0)$ for some $z_0\in S$. Moreover, let the function h be even with respect to $\sigma$ and $h\ne e$, where $e\in K\setminus \{0,1\}$. Then the functions f, g, h satisfy equations (Q), (E) and (A), respectively.

Remark 4.5

The following functional equation

$$\begin{aligned} E_g(x,y)+A_h(x,y)=0,\quad x,y\in S, \end{aligned}$$

(EA)

which is a particular case of equation (QEA), was investigated in the paper [21]. Let us notice that Theorem 4.4 is a generalization of Theorem 4.2 from that paper. Indeed, if we assume that $f=0$ in (QEA), it reduces to equation (EA). Moreover, the assumption of Theorem 4.4 that the function h is even with respect to $\sigma$ follows now immediately from (4.1). Furthermore, the only constant functions h that satisfy equation (EA) are $h=0$ and $h=1$. If $h=e$, where $e\in K\setminus \{0,1\}$, then from (EA) we get

$$\begin{aligned}{} & {} g(x+y)=g(x)g(y)+\alpha ,\quad x,y\in S, \end{aligned}$$

where $\alpha :=2e^2-2e$. Then $\alpha \ne 0$ and

$$\begin{aligned}{} & {} g(x+y+z)=g(x+y)g(z)+\alpha =g(x)g(y)g(z)+\alpha g(z)+\alpha ,\quad x,y\in S,\\{} & {} g(x+y+z)=g(x)g(y+z)+\alpha =g(x)g(y)g(z)+\alpha g(x)+\alpha ,\quad x,y\in S. \end{aligned}$$

Comparing the above equalities, we obtain $\alpha g(x)=\alpha g(z)$ for all $x,z\in S$. Since $\alpha \ne 0$, the function g is constant. Thus, we get a contradiction with the assumption that $g(z_0)\ne g(\sigma z_0)$ for some $z_0\in S$. So, in the case when $f=0$ in Theorem 4.4, the assumptions that the function h is even with respect to $\sigma$ and $h\ne e$, where $e\in K\setminus \{0,1\}$, are automatically satisfied. Therefore, from Theorem 4.4 we obtain the following result.

Corollary 4.6

(cf. [21], Theorem 4.2) Assume that functions $g,h:S\rightarrow K$ satisfy equation (EA) for all $x,y\in S$, and $g(z_0)\ne g(\sigma z_0)$ for some $z_0\in S$. Then the functions g, h satisfy equations (E) and (A), respectively.

Using [20, Theorems 1, 3], from Theorem 4.4, under the assumption that the function h is even with respect to $\sigma$, we get the following result.

Corollary 4.7

Let K be a quadratically closed field of characteristic different from 2. Assume that functions $f,g,h:S\rightarrow K$ satisfy equation (QEA) for all $x,y\in S$, and $g(z_0)\ne g(\sigma z_0)$ for some $z_0\in S$. Moreover, let the function h be even with respect to $\sigma$ and $h\ne e$, where $e\in K\setminus \{0,1\}$. Then the function g satisfies equation (E), and there exist a function $\chi :S\rightarrow K$ satisfying equation (E), a symmetric biadditive function $B:S\times S\rightarrow K$ and an additive function $A:S\rightarrow K$ such that $B(\sigma x,y)=-B(x,y)$, $A(\sigma x)=A(x)$ for all $x\in S$ and

$$\begin{aligned} f(x)=B(x,x)+A(x),\quad h(x)=\frac{\chi (x)+\chi (\sigma x)}{2},\quad x\in S. \end{aligned}$$

As a special case of the above corollary ($h=0$), we obtain a similar result for equation (QE).

As a corollary of Theorem 4.4, we obtain a result which states that in the class of nonconstant functions mapping a 2-divisible abelian group into a field of characteristic different from 2, the quadratic, exponential and d’Alembert functional equations are strongly alien to each other. In what follows in this section, let $\sigma y=-y$.

Corollary 4.8

Let $(G,+)$ be an abelian group uniquely divisible by 2. Assume that functions $f,g,h:G\rightarrow K$ satisfy the equation

$$\begin{aligned} Q_f(x,y)+E_g(x,y)+A_h(x,y)=0 \end{aligned}$$

(4.17)

for all $x,y\in G$. Moreover, let the function h be even. Then:

• (i) the functions f, g, h satisfy equations (Q), (E), and (A), respectively,

  or

• (ii) there exist constants $c,e\in K$ with $c+e^2=e$ such that

  $$\begin{aligned} f(x)=Q(x)+c,\quad g(x)=E(x),\quad h(x)=e, \end{aligned}$$

  or

• (iii) there exist constants $c,d\in K$ with $2c+d^2=d$ such that

  $$\begin{aligned} f(x)=Q(x)+c,\quad g(x)=d,\quad h(x)=H(x), \end{aligned}$$

  or

• (iv) there exist constants $c,d,e\in K$ with $2c+d^2+2e^2=d+2e$ such that

  $$\begin{aligned} f(x)=Q(x)+c,\quad g(x)=d,\quad h(x)=e \end{aligned}$$

for all $x\in G$, where the functions Q, E, H satisfy the quadratic, exponential and d’Alembert functional equations, respectively.

Proof

Assume that the functions f, g and h are nonconstant. Suppose on the contrary that $g(y)=g(-y)$ for every $y\in G$. Then, in view of Lemma 4.1, we obtain $g(x+y)-g(x-y)=0$ for all $x,y\in G$, i.e. $g(2x)=g(0)$ for all $x\in G$. Since the group G is uniquely divisible by 2, $g(x)=g(0)$ for all $x\in G$, which yields a contradiction. Therefore, there exists a $z_0\in G$ such that $g(z_0)\ne g(-z_0)$. Hence, on account of Theorem 4.4, we get (i).

Since the function h is even, on account of Lemma 4.1 the function f is even too. Moreover, equality (4.2) reduces to equation (2.1). Therefore, in view of Proposition 2.3, the function g is an arbitrary constant or exponential.

Consider the first case, where $g(x)=d$ for all $x\in G$ and for some $d\in K$. Let the functions f and h be nonconstant. Then

$$\begin{aligned} Q_f(x,y)+A_h(x,y)=d^2-d \end{aligned}$$

for all $x,y\in G$. Let $\alpha :=d(d-1)$. Then the last equation reduces to

$$\begin{aligned} Q_F(x,y)+A_h(x,y)=0,\quad x,y\in G, \end{aligned}$$

where $F(x):=f(x)+\frac{1}{2}\alpha$. Therefore, on account of Corollary 3.3, the functions F and h satisfy the quadratic and d’Alembert functional equations, respectively. Hence, $f(x)=Q(x)+c$ and $h(x)=H(x)$, where Q is an arbitrary quadratic function, H satisfies the d’Alembert functional equation, and $c:=\frac{1}{2}d(1-d)$. If $\alpha =0$ (i.e. $d\in \{0,1\}$), then the condition (i) holds. If $\alpha \ne 0$ (i.e. $d\notin \{0,1\}$), then (iii) is true. Now, assume that $f(x)=c$ for all $x\in G$ and for some $c\in K$. Then equality (4.17) becomes

$$\begin{aligned} A_h(x,y)=2c+d^2-d,\quad x,y\in G. \end{aligned}$$

If $2c+d^2=d$, then the function h satisfies the d’Alembert functional equation and condition (iii) holds (with $Q=0$). If $2c+d^2-d\ne 0$, then from the proof of the Corollary 3.3 we obtain that the function h is constant, i.e. $h(x)=e$ for all $x\in G$ and for some $e\in K$. Hence, the condition (iv) (with $Q=0$) is true. Finally, let $h(x)=e$ for all $x\in G$ and for some $e\in K$. Then

$$\begin{aligned} Q_f(x,y)=d^2-d+2e^2-2e,\quad x,y\in G. \end{aligned}$$

Defining a new function $F:G\rightarrow K$ by the formula $F(x):=f(x)+\frac{1}{2}\alpha$, where $\alpha :=d^2-d+2e^2-2e$, we can easily see that F is a quadratic function. Finally, $f(x)=Q(x)+c$ for all $x\in G$, where $c:=\frac{1}{2}(d-d^2+2e-2e^2)$ and Q is an arbitrary quadratic function. Hence, (iv) holds.

Now, we consider the second case where the function g is exponential. If $f(x)=c$ for all $x\in G$ and for some $c\in K$, then we can rewrite equation (4.17) in the following form

$$\begin{aligned} A_h(x,y)=2c,\quad x,y\in G. \end{aligned}$$

If $c=0$, then the function h satisfies the d’Alembert functional equation and the condition (i) (with $Q=0$) holds. If $c\ne 0$, then from the proof of Corollary 3.3 we obtain that the function h is constant, i.e. $h(x)=e$ for all $x\in G$ and for some $e\in K$. Hence, (ii) (with $Q=0$) is true. Finally, let $h(x)=e$ for all $x\in G$ and for some $e\in K$. Then from (4.17) we have

$$\begin{aligned} Q_f(x,y)=2e^2-2e,\quad x,y\in G. \end{aligned}$$

Let $\alpha :=2e(e-1)$. Then the function $F(x):=f(x)+\frac{1}{2}\alpha$ is quadratic. Hence, $f(x)=Q(x)+c$, where Q is an arbitrary quadratic function and $c:=e(1-e)$. If $\alpha =0$ (i.e. $e\in \{0,1\}$), then the condition (i) holds. If $\alpha \ne 0$ (i.e. $e\notin \{0,1\}$), then the condition (ii) is true. If $f(x)=c$ and $h(x)=e$ for all $x\in G$ and for some $c,e\in K$, then obviously the condition (ii) (with $Q=0$) holds. The proof is complete. $\square$

It is an open problem to prove Corollary 4.8 without assuming that the function h is even.

As a special case of the above corollary ($h=0$), we obtain the following result, which states that in the class of nonconstant functions, the quadratic and exponential functional equations are strongly alien to each other.

Corollary 4.9

Let $(G,+)$ be an abelian group uniquely divisible by 2. Assume that functions $f,g:G\rightarrow K$ satisfy the equation

$$\begin{aligned} Q_f(x,y)+E_g(x,y)=0,\quad x,y\in G. \end{aligned}$$

(4.18)

Then:

• (i) f is a quadratic function and g is an exponential one, or

• (ii) there exist constants $c,d\in K$ with $2c+d^2=d$ such that $f(x)=Q(x)+c$ and $g(x)=d$ for all $x\in G$, where Q is an arbitrary quadratic function.

5 Some generalizations

During the 17th International Conference on Functional Equations and Inequalities, which took place in Bȩdlewo, Poland, prof. H. Stetkær asked about possible generalizations of equations (QA) and (QEA). Indeed, we can consider these equations in more general forms, namely

$$\begin{aligned} {\tilde{Q}}_f(x,y)+{\tilde{A}}_h(x,y)=0, \end{aligned}$$

(QA')

$$\begin{aligned} {\tilde{Q}}_f(x,y)+E_g(x,y)+{\tilde{A}}_h(x,y)=0 \end{aligned}$$

(QEA')

for all $x,y\in S$, where $\sigma _1,\sigma _2:S\rightarrow S$ are endomorphisms with $\sigma _i^2=id$, $i=1,2$, and

$$\begin{aligned}{} & {} {\tilde{Q}}_f(x,y)=f(x+y)+f(x+\sigma _1 y)-2f(x)-2f(y),\\{} & {} {\tilde{A}}_h(x,y)=h(x+y)+h(x+\sigma _2 y)-2h(x)h(y). \end{aligned}$$

Thus, it is an open problem to prove results analogous to Theorems 3.1 and 4.4for equations (QA’) and (QEA’), respectively.

Now, we show that, under an additional assumption on the endomorphisms $\sigma _1,\sigma _2$, we can in fact consider equations (QA) and (QEA) instead of equations (QA’) and (QEA’), respectively. Indeed, assume additionally that $\sigma _1\sigma _2(x)=x$ for all $x\in S$. From (QA’) we have

$$\begin{aligned}{} & {} {\tilde{Q}}_f(x,y+z)+{\tilde{A}}_h(x,y+z)=0,\nonumber \\{} & {} {\tilde{Q}}_f(x,y+\sigma _1 z)+{\tilde{A}}_h(x,y+\sigma _1 z)=0 \end{aligned}$$

(5.1)

for all $x,y,z\in S$. Replacing z by $\sigma _2z$ in the last equation, and next comparing the resulting equation with (5.1) we obtain the following condition

$$\begin{aligned} f(x+\sigma _1y+\sigma _1z)=f(x+\sigma _1y+\sigma _2z),\quad x,y,z\in S. \end{aligned}$$

If we put $y=0$, then we see that equation (QA’) reduces to (QA) with $\sigma =\sigma _2$. The same arguments can be used for equation (QEA’).