An alternative equation for polynomial functions

In this paper we prove that if a generalized polynomial function f satisfies the condition f(x) f(y) = 0 for all solutions of the equation x2 + y2 = 1, then f is identically equal to 0.

The following problem was formulated by Benz in 1989 [3]. Suppose that f : R → R is an additive function satisfying yf (x) = xf (y) for every x, y ∈ R such that x 2 + y 2 = 1 . Does it imply that f is linear? This question, together with a similar one for derivations, was answered in the affirmative by Boros and Erdei [4].
Motivated by the question of Benz, Szabó [14] posed the following problem: suppose that f : R → R is additive and f (x)f (y) = 0 for all solutions of the equation x 2 + y 2 = 1 . Does it imply that f is identically equal to zero? The solution was published in a joint paper by Kominek et al. [9], where they proved that the implication is true.
The purpose of the present paper is to extend the last result by providing an analogue of this statement for polynomial mappings. In order to formulate such a generalization, we have to introduce the related concepts.
Let n ∈ N. A function F : R n → R is called n-additive if, for every i ∈ { 1 , 2 , . . . , n } and for every x 1 , . . . , x n , y i ∈ R , i.e., F is additive in each of its variables x i ∈ R , i = 1, . . . , n . Clearly, an n-additive function is also Q-homogeneous in each variable.
Given a function F : R n → R, by the diagonalization (or trace) of F we understand the function f : R → R arising from F by putting all the variables (from R) equal: If, in particular, f is a diagonalization of an n-additive function F : R n → R , we say that f is a generalized monomial of degree n . It is convenient to assume that generalized monomials of degree zero are precisely constant mappings. If f is a sum of generalized monomials of degrees n 1 , n 2 , . . . n k , respectively, and n = max{n 1 , n 2 , . . . , n k }, then f is called a generalized polynomial of degree n. We note that the degree of a generalized polynomial is not uniquely determined in our context as we do not exclude identically zero terms. This approach is convenient when we formulate our statements and arguments.
For more information concerning these notions the reader is referred to the monograph by Kuczma [10,Chapter 15.9]. Now we can establish our main theorem. Proof. Given a generalized polynomial f of degree n, we can associate k−additive and symmetric functionals A k : R k → R for k = 0, 1, . . . n with f in such a way that for all x ∈ R. Now, let x, y ∈ R be arbitrary solutions of x 2 + y 2 = 1. If α, β are such that α 2 + β 2 = 1, then it is straightforward to check that the following identity holds true: (αx − βy) 2 + (βx + αy) 2 = 1.
Next, assume, in addition, that α and β are rationals, u is a real number such that |u| < 1 , and x = −u .
Take Clearly, for every pair (α, β) chosen as above at least one of the foregoing expressions is equal to zero.
What is more, we can find infinitely many distinct pairs (α i , β i ) such that α 2 i + β 2 i = 1 and both α i and β i are rationals, so let us take Multiplying both equations by (i 2 + 1) n and introducing the functions we have P n (i) = 0 orP n (i) = 0 for each positive integer i . Hence either P n orP n has infinitely many zeros. On the other hand, both P n andP n are polynomials of degree not greater than 2n . Therefore, one of them has to be identically equal to 0 . So either i.e., f (u) = 0 . We have thus proved that f vanishes on the open interval (−1, 1). Now let us consider an arbitrary non-zero real number x and any rational number r fulfilling |r| < 1/|x| . Then, according to the representation (1), we have . . . , x) .
The last expression in this equation is an "ordinary" polynomial with respect to r. Since it has infinitely many zeros (as the inequality condition for r is satisfied by infinitely many rational numbers), it must be identically zero, hence it vanishes at r = 1 as well, thus f (x) = 0 .
We note that our last argument can be replaced by a reference to Székelyhidi's regularity theorem [15]. Namely, since f vanishes on the open interval (−1, 1), it is, in particular, locally bounded around zero. However, Vol. 89 (2015) Equation for polynomial functions 21 every generalized polynomial which satisfies this condition is continuous. Therefore, f is an "ordinary" polynomial, and thus f = 0 on R.
Remark 1. Generalized polynomials are obtained from polynomials of the form f (x) = n k=0 c k x k by replacing the monomial terms c k x k with generalized monomials of degree k. Another way to generalize such a representation of f is obtained by replacing the finite sum with an infinite one. This idea leads to the well-known concept of analytic functions. We can, actually, establish an analogy of Theorem 1 for analytic functions: If f : R → R is analytic and f (x)f (y) = 0 for all solutions of the equation x 2 + y 2 = 1 , then f is identically equal to zero. In fact, according to our assumption, f has infinitely many zeros in the interval [−1, 1], hence it is identically equal to zero.
On the other hand, the regularity assumption that f is analytic cannot be replaced with a weaker one. Clearly, every mapping f : R → R which satisfies f (t) = 0 for every t ∈ − fulfils the condition f (x)f (y) = 0 for all solutions of the equation x 2 + y 2 = 1. Therefore, there exist infinitely many times differentiable functions which satisfy this condition and are not identically equal to zero.
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