Alienation of two general linear functional equations

We study the alienation problem for two general linear equations i.e. we compare the solutions of the system of equations ∑i=1nαif(pix+qiy)=0∑j=1mβjg(sjx+tjy)=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \left\{ \begin{array}{l}\sum _{i=1}^n\alpha _if(p_ix+q_iy)=0\\ \sum _{j=1}^m\beta _jg(s_jx+t_jy)=0\end{array}\right. \end{aligned}$$\end{document}with the solutions of the single equation ∑i=1nαif(pix+qiy)=∑j=1mβjg(sjx+tjy).\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \sum _{i=1}^n\alpha _if(p_ix+q_iy)=\sum _{j=1}^m\beta _jg(s_jx+t_jy). \end{aligned}$$\end{document}To this end we introduce the notion of l-alienation—alienation in the class of monomial functions of order l. We use our results among others to study the alienation properties of two monomial functional equations.


Introduction
The alienation problem for functional equations was first studied by Dhombres who in [1] introduced the following definition.
where 0-additive functions are understood to be constant functions.
The main tool used in this paper will be the following result of L. Székelyhidi. In this theorem, we use the following definition Definition 3. ( [14], Definition 3.5 ) Let G, S be abelian groups, let n be a nonnegative integer. The function f : G → S is said to be of degree n if there exist functions f i : G → S and homomorphisms ϕ i .ψ i : G → G such that ϕ(g) ⊂ ψ(G), i = 1, 2, . . . , n + 1 and the equation

holds.
Vol. 94 (2020) Alienation of two general linear functional equations 289 Theorem 2. ( [14], Theorem 3.6) Let G, S be abelian groups, and suppose that G is divisible, Let n be a nonnegative integer. The function f : G → S is of degree n if and only if it is a polynomial function of order n.
For some generalizations of Theorem 2 see for example [12] or [16].

Two introductory results
Recently, several authors have dealt with alienation results with respect to two linear equations. Namely, Z. Kominek if and only if f is a polynomial function of order n − 1 and g is a polynomial function of order m − 1.
Proof. Writing (2) explicitly, we get Since n > m, all arguments of g appear on the left hand side as arguments of f. This means that we can rearrange (3) and get Now, our sum contains exactly n+1 terms and it is enough to substitute x−nh in place of x to get from Theorem 2 that f is a polynomial function of order at most n − 1 i.e. Δ n h f (x) = 0. Consequently, using (2) we get Δ m h g(x) = 0 which finishes the proof.
Similarly, it is possible to get an alienation result for the quadratic and additive functional equations which was obtained by Ger in [5].
if and only if where function a is additive, q is quadratic and c ∈ H is some constant.

The general case
In this section we will not restrict ourselves to concrete equations, we will work with two general linear equations. We will include here also the proofs of some known facts to keep the paper as self contained as possible. To use all the tools needed in our approach we need the assumption that the domain and codomain of our functions are abelian and divisible groups. Such structures are simply linear spaces over Q. However in some of the results presented below it is not essential that the constants involved are rational. Therefore we will work on linear spaces over some field which contains rationals.
Remark 1. Suppose that H, K are fields such that Q ⊂ H, K, let X, Y be linear spaces, respectively over the fields H, K, let n, m ∈ N and let the functions f i : X → Y, i = 1, . . . , n be of the form where a i ∈ Y, i = 1, . . . , n are some constants and functions F i,j : X j → Y are j-additive and symmetric. If then for all j ∈ {0, . . . , m} the system of functions f 1,j , . . . , f n,j also satisfies (7). Indeed, for every rational number r we have Vol. 94 (2020)

Alienation of two general linear functional equations 291
Now, it is enough to take any r ∈ Q and substitute rx, ry in places of x, y, respectively. Then we get This equality is satisfied for every rational number r, thus all the expressions standing at different powers of r must separately be equal to zero. It means that for each j the functions f 1,j , . . . , f n,j satisfy our equation.

Lemma 1.
Suppose that H, K are fields so that Q ⊂ H, K, let X, Y be linear spaces, respectively over the fields H, K, let α i ∈ K, p i , q i ∈ Q and let l be any positive integer. Then the following conditions are equivalent to each other: is satisfied by some nonzero monomial function of order l, ii) (7) is satisfied by all monomial functions of order l, iii) Proof. First we assume that i) is satisfied and we show iii). Let Similarly as in Remark 1, all expressions of the same degree with respect to x must sum up to zero. Thus for every k ∈ {0, . . . , l}, we have Substituting here x in place of y and using the assumption f = 0, we get as claimed.

T. Szostok AEM
In the next part of the proof we show that iii) implies ii). Let f (x) = F (x, . . . , x) be any monomial function of order l. Using (3) in (8) i.e. f is a solution of (7).
Finally, the implication ii) ⇒ i) is obvious.
In the next lemma we show that in many cases the solutions of equation must be polynomial functions. and for i = 1, . . . , n; i = i 0 then f must be a polynomial function of order at most k − 1, where k is the maximal number of pairwise independent pairs for j = 1, . . . , m; j = j 0 then also g must be a polynomial function of order at most k − 1.
Proof. If p io = 0 then we interchange x with y in (9) and we may use Theorem 2 directly. Now assume that p i0 = 0 and putx = x−qi 0 y pi 0 in place of x in (9). After this substitution we get Vol. 94 (2020) Alienation of two general linear functional equations 293 Using (11), we can see that the coefficient standing here at y is different from zero. Similarly and from (10) we know that again the coefficient of y is not equal to zero i.e. we may use again Theorem 2 to prove that f is a polynomial function. If there are some linearly dependent pairs among (p i , q i ), (s j , t j ) then we group the occurrences of f and g containing these coefficients into new functions so that we achieve the optimal estimation of the order. Now we will deal with function g. If (12) is satisfied then we have two possibilities.
1. If for all i = 1, . . . , n we have then we proceed similarly as in the first part of the proof and we show that g must be a polynomial function. 2. The second possibility is that for some In this case we consider the expression which in view of (14) may be written as a function of s j0 x + t j0 y. Then we show that this new function is a polynomial function. Finally from the first part of the proof we know that f is a polynomial function thus g also must be a polynomial function.
Now we give some simple examples of equations which will illustrate the possibilities described in Lemma 2.
has only polynomial solutions because of the term f (y) on the left hand side, and because of the linear independence of the coefficients of the arguments of g. Equation has solutions f, g which are not polynomial (f − g must be additive) because there is no occurrence of f or g with coefficients linearly independent from all the others-this shows that the assumptions (10) and (11) cannot be omitted.

T. Szostok AEM
The function f occurring in must be polynomial [ (10) and (11) are satisfied] but g is not necessarily a polynomial function. Namely, any pair (f, g) where f is additive and g is even satisfies this equation. This shows that the (12) is necessary to guarantee for function g to be a polynomial function.

Remark 2.
It may also happen that the assumptions of Lemma 2 are not satisfied and the solutions of a given equation are polynomial functions. Equation has only polynomial solutions. Indeed, after some rearrangement we can see that both f − 2g and f − g are polynomial functions. Then are also polynomial functions, since a linear combination of polynomial functions gives a polynomial function.
As we can see, in many cases functions satisfying (9) must be polynomial. Moreover we know from Remark 1 that the monomial summands of solutions of a given equation also satisfy this equation. Therefore we will study the alienation of linear functional equations in the class of monomial functions of a fixed order. We will use the expression l-alienation if the monomial summands of f and g of the order l alienate. A more precise statement is given by the following definition.

Definition 4.
Suppose that H, K are fields such that Q ⊂ H, K, let X, Y be linear spaces, respectively over the fields H, K, let α i , β j ∈ K \ {0}, i = 1, . . . , n, j = 1 . . . , m be some constants and let p i , q i , s j , t j ∈ H, i = 1, . . . , n, j = 1 . . . , m. If the solutions f, g : X → Y of the system of equations and m j=1 β j g(s j x + t j y) = 0 (16) are different from the solutions of Eq. (9) in the class of monomial functions of order l then we say that Eqs. (15) and (16) are not l-alien. In the opposite case we say that these two equations are l-alien.
We begin with a lemma which gives us some relation between the monomial solutions f, g of (9).  F (x, . . . , x), g(x) = G(x, . . . , x) satisfy Eq. (9). If Proof. Using the forms of f and g, we may write (9) in the form Now it is enough to take y = x in the above equality, to get (19).
In the next theorem we present the general solution of Eq. (9) in the class of monomial functions of a given order. The following assertions hold true.
1. If A l = B l = ∅ then Eq. (9) is satisfied by every pair (f, g) of monomial functions of order l. 2. If A l = ∅, B l = ∅ and the monomial functions f, g of order l, satisfy (9) then f = 0. Conversely, every pair (0, g) where g is monomial function of order l is a solution of (9). 3. If A l = ∅, B l = ∅ and the monomial functions f, g of order l, satisfy (9) then g = 0. Conversely, every pair (f, 0) where g is monomial function of order l is a solution of (9). 4. If A l , B l = ∅, A l = B l and monomial functions f, g of order l, satisfy (9) then f = g = 0. Proof. To prove the first statement observe that from Lemma 1 we know that, in the case A l = B l = ∅, all monomial functions of order l satisfy (7) and (16). This means that all pairs of monomial functions (f, g) satisfy (9). Now assume that A l = ∅ and B l = ∅. Then for some k we have a k,l = 0 and b k,l = 0. In view of Lemma 3 this yields f = 0. On the other hand B l = ∅ means that all monomial functions of order l satisfy (16). Thus the pair (0, g) where g is any monomial function of order l satisfies (9).

If
The third assertion is completely analogous to the second one and, therefore, the proof in this case is not needed.
Assume now that A l , B l = ∅, and A l = B l . Let for example exist k such that a k,l = 0 and b k,l = 0. Then from Lemma 3 we get f = 0. However if f = 0 then from (9) we get that g satisfies (16) and, since B l = ∅ we have b k,l = 0 for some k. In view of Lemma 1, this means that g = 0.
In the remaining part of the proof we assume that A l , B l = ∅ and A l = B l . If for some c we have a k,l b k,l = c, k ∈ A l then from Remark 3 we know that g(x) = cf (x). Using this form of g in (9), we obtain a linear equation with one unknown function f. Using Lemma 1 it is easy to see that this equation is satisfied by all monomial functions.
Vol. 94 (2020) Alienation of two general linear functional equations 297 Finally, consider the case where a k,l b k,l is not constant. Then there exist k 1 , k 2 and c 1 , c 2 such that a k1,l b k1,l = c 1 and a k2,l b k2,l = c 2 .
In view of Remark 3 and from the above equalities we get Thus we have c 1 f (x) = c 2 f (x) which means that f = 0. Now using (24) we get that g = 0 which finishes the proof.
Remark 3. Let A l , B l be defined as in the above theorem. Equations (16) and (7) are not l-alien if and only if case 4 (i) of Theorem 3 occurs.
Using this theorem, we will give some simple examples of equations that are l-alien and not l-alien, respectively. There is no need to test the l-alienation for l > 1, since f, g satisfying equation are polynomial functions of order at most 1.
The above remark was connected with the Cauchy equation, now we present a result connected with the quadratic equation.
Remark 5. Suppose that H, K are fields such that Q ⊂ H, K. Let X, Y be linear spaces, respectively over the fields H, K. Let f : X → Y be some functions, let q 1 , q 2 ∈ Q be any numbers. Equations Now, using Theorem 3, we will study the alienation problem for two monomial equations. and are not l-alien for l < n and l-alien for l ∈ {n, . . . , m}.
Proof. We will study the equation which certainly is an equation of the form (9). Moreover, it is clear that f, g must be polynomial functions of order at most m. Let l < n be a fixed positive integer and h b a monomial function of order l, then Of course, in the case of Eqs. (26), (25) considered in the above theorem it would be enough to assume that X and Y are (abelian) groups such that Y is torsion free and divisible by m!. Our assumptions follow from the fact that, in the proof we use Theorem 3.

Remark 7.
The result of R. Ger concerning the alienation of the Cauchy equation and the quadratic equation is a particular case of the above result (with n = 1, m = 2).
We conclude the paper with two final remarks.
Remark 8. In this paper we presented results where we first get the solutions of Eq. (9) (using Theorem 3) and then we deduce the alienation properties from the solutions. A similar approach was presented by Gilányi [8], however Gilányi used a computer program to solve the equations.
Remark 9. The non-alienation obtained in Theorem 4 is quite surprising, since it is much more common that two equations are alien than non-alien. The explanation is provided by Theorem 3. It is difficult to obtain non-alienation, since then all elements of A l and B l must be proportional. In the case of monomial equations this occurred because each of A l and B l consisted of one element only. If we study 0-alienation then the non-alienation effect appears much more frequently, since A 0 and B 0 can have only one element.