ALIENATION OF DRYGAS’ AND CAUCHY’S FUNCTIONAL EQUATIONS

. Inspired by the papers [2, 10] we will study, on 2-divisible groups that need not be abelian, the alienation problem between Drygas’ and the exponential Cauchy functional equations, which is expressed by the equation f ( x + y ) + g ( x + y ) + g ( x − y ) = f ( x ) f ( y ) + 2 g ( x ) + g ( y ) + g ( − y ) . We also consider an analogous problem for Drygas’ and the additive Cauchy functional equations as well as for Drygas’ and the logarithmic Cauchy functional equations. Interesting consequences of these results are presented.


Introduction
The alienation and strong alienation problems are introduced by Dhombres ( [3]), who gave the following definitions. Definition 1.1. Let E 1 (f ) = 0 and E 2 (f ) = 0 be two functional equations for a function f : X → Y , where X and Y are non-empty sets. The equations E 1 and E 2 are alien with respect to X and Y , if any solution f : X → Y of is a solution of the system Definition 1.2. Let E 1 (f ) = 0 and E 2 (g) = 0 be two functional equations for functions f, g : X → Y , where X and Y are non-empty sets. The equations E 1 and E 2 are strongly alien, if any solution f, g : X → Y of is a solution of the system Later on, several papers and lectures have appeared on this subject (see [4-11, 13-16, 18, 19]). For more details concerning the alienation phenomenon in the theory of functional equations we refer to the survey article [12], which was authored by Ger and Sablik.
The aim of the present paper is to study the alienation phenomenon between Drygas' and Cauchy's functional equations. Firstly, we will solve the functional equation (1.1) f (x + y) + g(x + y) + g(x − y) = f (x)f (y) + 2g(x) + g(y) + g(−y), x, y ∈ X, which is strictly connected with the problem of alienation of Drygas' functional equation, that is (1.2) g(x + y) + g(x − y) = 2g(x) + g(y) + g(−y), x, y ∈ X, and the exponential Cauchy functional equation where (X, +) is 2-divisible group that need not be abelian, and f and g are the unknown functions which take their values in an unital commutative ring Y of characteristic different from 2.
As consequences, we will introduce and discuss the solutions of the following functional equations The last equation results from summing up side by side Drygas' functional equation and Lobachevsky's functional equation, that is Secondly, we will describe the solutions of the equation which is derived by summing up side by side the equations (1.2) and the additive Cauchy functional equation, that is We show that modulo a constant, equations (1.2) and (1.7) are strongly alien on a 2-divisible group. As applications, we will examine, on 2-divisible abelian groups, the alienation phenomenon between the equation (1.2) and the Jensen-additive functional equation, that is 2f x and we get the result [9, Theorem 2.1], which was established by Ger about the alienation phenomenon of additivity and quadraticity up to a constant. Finally, we will study the alienation problem, on a ring, of Drygas' and the logarithmic Cauchy functional equations, which is expressed as follows (1.8) f (xy)+g(x+y)+g(x−y) = f (x)+f (y)+2g(x)+g(y)+g(−y), x, y ∈ X.
Furthermore, we will describe the solutions of the equation The monographs by Aczél and Dhombres ([1]) and by Stetkaer ([17]) contain many references about Drygas' and Cauchy's functional equations.
Notation. The following notation will be used throughout the paper unless explicitly stated otherwise. Let (X, +) and (Y, +) be groups. We deal with the very classical functional equations defining additivity and quadraticity functions, i.e.
A(x + y) = A(x) + A(y), x, y ∈ X, and Q(x + y) + Q(x − y) = 2Q(x) + 2Q(y), x, y ∈ X respectively, where A and Q are functions mapping X into Y . A map D : X → Y is called Drygas' map provided that it satisfies the following functional equation D(x + y) + D(x − y) = 2D(x) + D(y) + D(−y), x, y ∈ X.

Drygas' and the exponential Cauchy functional equations
Our first main result describes the solutions f, g : X → Y of Pexider type functional equation resulting from summing up Drygas' and the exponential Cauchy functional equations side by side. (1) f (0) = 1 and g(x + y) + g(x − y) = 2g(x) + g(y) + g(−y), x, y ∈ X.
Proof. Let f, g : X → Y be a solution of (1.1). For x = y = 0 in (1.1) we have If we put y = 0 in (1.1) we get for all x ∈ X. Next, we distinguish between two cases: Case 1: Suppose that f (0) = 1, then we get from (2.1) that 2g(0) = 0. Let where x, y ∈ X. Since Γ satisfies the following equation for all x, y, z ∈ X, i.e., for all x, y, z ∈ X. If we replace z by −z in (2.4), we obtain that for all x, y ∈ X (because Γ(x, −y) = Γ(x, y) for all x, y ∈ X). Now, subtract equalities (2.4) and (2.5), and use the identity (2.3) to get Replacing y by −y in (2.6) we get Since Γ(x, −y) = Γ(x, y) and Y has a characteristic different from 2 we get by subtracting (2.6) and (2.7) that which, after setting z = y, leads to Going back to (1.1), we infer that f is solution of (1.3).
We turn now to f (0) = 1 (zero or not). By using (2.2), the equation ( which means that Hence, there exists a Drygas' map D : X → Y such that The converse is straightforward. As a consequence of Theorem 2.1, we get the following result on the alienation of Drygasity and exponentiality up to a constant. (2) f (0) = 1 and there exist a Drygas' map D : Moreover, if f (0) = 0 then f ≡ 0 and g is a solution of (1.2).
For (2), in view of Theorem 2.1 there exists a Drygas' map D : X → Y such that So we get the claimed result because the map The other direction is easy to check.
The following result is devoted to study the alienation problem between exponentiality and quadraticity. g(x + y) + g(x − y) = 2g(x) + 2g(y), x, y ∈ X.
Subtracting (1.4) from (2.10) we get Setting x = 0 in (2.11) gives us which yields, by using (2.9), that for all y ∈ X. This means that g is even, i.e., g(−y) = g(y) for all y ∈ X, because Y has a characteristic different from 2. Thus, from Theorem 2.1 we get the proof of the first direction. The converse statement can be trivially shown.
A result about the alienation of quadraticity and exponentiality up to a constant will be shown in the following corollary (2) f (0) = 1 and Moreover, if f (0) = 0 then f ≡ 0 and g(x) = Q(x) for all x ∈ X.
Proof. The proof follows from Theorem 2.3 by using similar arguments as in proof of Corollary 2.2. Now, we will study the solutions f, g : X → Y of the equation where x, y ∈ X. (2) f (0) = 0 and there exists a Drygas's map D : X → Y such that Moreover, if f (0) = −1 then f ≡ −1 and g is a solution of (1.2).
Proof. Let f, g : X → Y be a solution of the equation (2.12). If we add the identity element 1 in the two sides of (2.12), we get F (x + y) + g(x + y) + g(x − y) = F (x)F (y) + 2g(x) + g(y) + g(−y), x, y ∈ X, where F (x) := f (x) + 1 for all x ∈ X. So by Theorem 2.1 we get the claimed result. The other direction is easy to check.
As another application of our first main result, we investigate the alienation phenomenon of Lobachevsky's and the exponential Cauchy functional equations, i.e., the equation on 2-divisible abelian group, where the functions f and g take their values in a field K of characteristic different from 2. We will show that (1.5) and (1.2) are strongly alien in the sense of Dhombres.
Corollary 2.6. Let (X, +) be a 2-divisible abelian group, and K be a field of characteristic different from 2. The pair of functions f, g : X → K is a solution of (2.13) if and only if it is a solution of the system of two equations Moreover, if f (0) = 0 then f ≡ 0 and g is a solution of (1.2).
Conversely, it is elementary to show the other direction.

Drygas' and the additive Cauchy functional equations
In this section we show that modulo a constant, the equation (1.6): Proof. Putting x = y = 0 in (1.6) we get for all x ∈ X. So by (3.1) we see easily that the pair (F, G) is a solution of (1.6) such that F (0) = G(0) = 0. We put G := F + G, then the equation (1.6) becomes We define for all x, y ∈ X. The map Γ, as a Cauchy difference, satisfies the cocycle equation Γ(x + y, z) + Γ(x, y) = Γ(x, y + z) + Γ(y, z), x, y, z ∈ X. Consequently, for all x, y, z ∈ X, which is equivalent to Replacing z by −z in (3.2), we get
Due to the 2-divisibility of X, we deduce that Hence g is a solution of the equation Going back to the equation (1.6), we see that f is a solution of the equation The converse statement can be trivially shown.
The following corollary states that Drygas' functional equation and Jensen's functional equation are strongly alien, on 2-divisible abelian group, in the sense of Dhombres.
As another consequence of Theorem 3.1, we get the following result due to Ger ([9]) about the alienation phenomenon of additivity and quadraticity up to a constant.
if and only if there exist an additive function A : X → Y , a quadratic function Q : X → Y , and a constant c ∈ Y such that f (x) = A(x) − 2c and g(x) = Q(x) + c for all x ∈ X.
Replacing x by 0 in (3.7) and using (3.8) we obtain that g(−y) = g(y) for all y ∈ X. So by applying Theorem 3.1 we get the desired result. Conversely, it is elementary to verify that the above formulas of f and g is a solution of (3.7).
If we subtract (4.1) from (4.2), we get Setting x = 1 in (4.3), we obtain that g is even. So by applying Theorem 4.1 we get the desired result. Conversely, it is elementary to show that the above formulas of f and g are solutions of (4.1).