On polynomial congruences

. We deal with functions which fulﬁl the condition Δ n +1 h ϕ ( x ) ∈ Z for all x,h taken from some linear space V . We derive necessary and suﬃcient conditions for such a function to be decent in the following sense: there exist functions f : V → R , g : V → Z such that ϕ = f + g and Δ n +1 h f ( x ) = 0 for all x,h ∈ V .


Introduction
Let V be a linear space over Q, R or C and n ∈ N (we assume that 0 ∈ N). The symbol ≡ stands for a congruence modulo Z (so a ≡ b ⇐⇒ a − b ∈ Z, a, b ∈ R), the symbol [x] denotes the integer part of a real number x andx denotes the fractional part of x (so x = [x] +x,x ∈ [0, 1)). Following e.g. [10], we define the difference operator: Definition 1.1. Let f : V → R be a function. Then is called a polynomial function of degree n.
The aim of this paper is to examine functions ϕ : V → R fulfilling a less restrictive condition than (1.1), namely We call condition (1.2) polynomial congruence of degree n. This study is inspired by several works (e.g. [1][2][3][4]), in which the so called Cauchy's congruence (or Cauchy equation modulo Z) i.e.
is considered. In these works the problem of decency in the sense of Baker of solutions of (1.3) is discussed (see e.g. [1]; the solution ϕ of (1.3) is called decent iff there exist an additive function a : V → R and a function g : V → Z such that ϕ = a + g). In many cases Cauchy's congruence can be easily transformed to the con- To be more precise, if ϕ fulfills (1.3), then Almost conversely, if Δ 2 h ϕ(x) ∈ Z for x, h ∈ V , then the functionφ = ϕ−ϕ(0) fulfillsφ(x+y)−φ(x)−φ(y) ∈ Z. Indeed, observe first thatφ(0) = 0. Moreover, Obviously, if ϕ = f + g, f : V → R is a polynomial function of degree n and g : V → Z, then ϕ solves the congruence (1.2). In analogy to Baker [1], we call such functions ϕ decent solutions of (1.2).
Examples ofÁ. Száz and G. Száz from [13] and Godini from [8] prove that there exist non-decent solutions of (1.3). Thus the natural question arises: what conditions should be imposed on the solution of the congruence Δ n+1 h ϕ(x) ∈ Z to ensure its decency.
In the present paper we obtain results which correspond to those of Baron et al. from [2] and results of Baron and Volkmann from [3]. Namely, we present analogues of results from [2,3] for polynomial congruences of degree greater than 1. Below we cite one of the characterizations of decent solutions of the Cauchy's congruence from [2], because we use it in Remark 1.3: When dealing with polynomial functions the inductional approach may always come in mind. In our situation one could expect that a solution of the congruence Δ n+1 h ϕ(x) ∈ Z is a decent iff for every h ∈ V the function V is a decent solution of the polynomial congruence of degree n − 1. However, this is not the case as it is visible from the following remark: There exists a function ϕ such that Δ 3 h ϕ(x) ∈ Z for all x, h ∈ R, Δ h ϕ is a decent solution of the polynomial congruence of degree 1 for every h ∈ V , but ϕ is not a decent solution of the polynomial congruence of degree 2.
Proof. Let α : R → R be a function fulfilling α(x+y)−α(x)−α(y) = m(x, y) ∈ Z for all x, y ∈ R, which cannot be expressed as a sum of an additive function and an integer-valued function (the existence of such a function is proved in [8], [13]). Then α fulfills the congruence Δ 2 h α(x) ∈ Z, x, h ∈ R (which is proved on the previous page). Define is a decent solution of the polynomial congruence of degree 1 (for every fixed h ∈ V ). Suppose that the function ϕ is a decent solution of the polynomial congruence of degree 2. Then from Theorem 2.2, which is proved in the second part of this paper, it follows that for every v ∈ R there exist con- Then α(ξv) ≡ b v ξ for ξ ∈ Q and Theorem 1.2 implies that α is a decent solution of Cauchy's congruence, which is in contradiction to our choice of the function α.
We make use of the following, easy to check, properties of (decent) solutions of the congruence (1.2): Proof. Ad (i) The first part is a consequence of the equality Δ n+1 Observe that ϕ is of the form ϕ = f +g, with f : V → R being a polynomial function of degree n and g : Ad (ii) The first part follows from the identity Δ n+1 which proves the first part. We , which means that ϕ + m can also be split into a polynomial and an integer-valued part.
We can also notice that ϕ fulfills the congruence Δ n+1 for all x, h ∈ V , where0 means the neutral element of the quotient group (R/Z, +). We recall the well-known result (see e.g. [14], Theorem 9.1, p.70) describing solutions of the Frechét equation in a wide class of spaces. It will be useful for us in our further considerations (Theorem 2.2) and, moreover, it will clarify why we cannot use it for the group R/Z (the group R/Z is not divisible by n! for n > 1). For the simplicity of the statement we assume that a 0-additive function is an arbitrary function, whose domain is the linear space {0} (see e.g. [7]).

Main result
We start with the result which corresponds to Theorem 2.1 from [2]. In the proof we make use of Theorem 1.5 and the following, very obvious remark: , which takes only integer values for rational arguments, then p is constantly equal to p(0).
Our first theorem reads as follows: v ∈ V there exists a polynomial p v of degree smaller than n + 1 with real coefficients so that ϕ(ξv) ≡ p v (ξ) for all ξ ∈ Q.
Proof. Firstly, assume that ϕ is a decent solution of the polynomial congruence of degree n. Then there exist functions f : . . . , v) for an i-additive and symmetric At the beginning, let us consider the case ϕ(0v) = 0 for v ∈ V . Then we can choose polynomials p v ∈ R n [X] in such a way that From the above congruence it follows that For an arbitrary function ϕ considerφ = ϕ−ϕ(0). Using the already proved part of the theorem to the functionφ, we obtain thatφ is decent. From Remark 1.4 (ii) it follows that it is equivalent to the decency of the function ϕ.
Considering (i) of Remark 1.4 we can rewrite Theorem 2.2 in the following manner: Then ϕ is a decent solution of the polynomial congruence of degree n if and only if for any vectors v, w ∈ V there exists a polynomial p v,w of degree smaller than n + 1 with real coefficients so that ϕ(v + ξw) ≡ p v,w (ξ) for all ξ ∈ Q.
In our main theorem we apply the following result: Theorem 2.4. (Ger [6]) Let X and Y be two Q-linear spaces and let D be a nonempty Q-convex subset of X. If algint Q D = ∅ then for every function Now we present our main result, which provides necessary and sufficient conditions for a function ϕ fulfilling Δ n+1 h ϕ(x) ∈ Z for all x, h ∈ V to be a decent solution of this congruence.
Then the following conditions are equivalent: (i) ϕ is a decent solution of the polynomial congruence of degree n, (ii) For every vector v ∈ V there exists a polynomial p v of degree smaller than n + 1 with real coefficients so that ϕ(ξv) ≡ p v (ξ) for all ξ ∈ Q, (iii) For every vector v ∈ V there exist ε > 0 and a polynomial p v of degree smaller than n + 1 with real coefficients so that ϕ(ξv) ≡ p v (ξ) for all ξ ∈ Q ∩ (0, ε), (iv) For every vector v ∈ V there exist ε > 0 and a polynomial p v of degree smaller than n + 1 with real coefficients so thatφ(ξv) Proof. The equivalence (i) ⇐⇒ (ii) has already been proved. The implication (ii) =⇒ (iii) is obvious. Now we show that (iii) =⇒ (ii). For this aim, denote Ω = {ξ ∈ Q : ϕ(ξv) ≡ p v (ξ)}. From our assumption it follows that Q ∩ (0, ε) ⊆ Ω.
where E n+1 denotes the set of all natural even numbers smaller than or equal to n + 1 and O n+1 denotes the set of all natural odd numbers smaller than or equal to n + 1.
From our assumptions it follows that there exist functions m : E → Z and q : E → (−α, α) such that ϕ| E = m + q. Since X is a locally convex linear topological space and intH(E) = ∅, there exists an open and convex set U such that ∅ = U ⊆ H(E). Fix x ∈ U and choose h ∈ X such that x + kh, x − kh ∈ E for k = 1, 2, . . . , n + 1. Then 1 2 n+1 ) for x ∈ U . Thus there exist functionsm : U → Z,q : U → (− 1 2 n+1 , 1 2 n+1 ) such that ϕ| U =m +q. Now we fix x ∈ U and choose h ∈ X such that x + h, . . . , x + (n + 1)h ∈ U . Then we have hm (x) and Δ n+1 hq (x) = 0 for x ∈ U and h ∈ X such that x + h, . . . , x + (n + 1)h ∈ U . Theorem 2.4 applied for the functionq, the space X and the set U implies that there exists a polynomial function F : X → R of degree n such that F | U =q. Therefore F is bounded from both sides on U , so it is continuous (Theorem 3.7).
Obviously, G is a continuous polynomial function of degree n.
Denote Ω = {x ∈ X : ψ(x) ≡ G(x)}. We know that U − c ⊆ Ω and U − c is a convex neighbourhood of 0. We show that if W is a convex neighbourhood of 0, then W ⊆ Ω implies that (1 + 1 n )W ⊆ Ω. Choose arbitrary x ∈ W . From the convexity of W and 0 ∈ W it follows that 1 n x, . . . , n−1 n x ∈ W . Thus Theorem 3.9. Let X be a linear space and let ϕ : X → R be a solution of the polynomial congruence of degree n. Assume that one of the following two hypotheses is valid 1. X = R m , with some positive m and ϕ, is Lebesgue measurable.
2. X is a real Fréchet space and ϕ is a Baire measurable function. Then ϕ is a decent solution of the polynomial congruence of degree n. Moreover, ϕ = f + g with f being a continuous polynomial function of degree n and g being an integer-valued and Lebesgue (resp. Baire) measurable function.
If k 0 = 0, then the previous theorem and Remark 3.5 in case (1) and Remark 3.6 in case (2) implies the decency of ϕ and the continuity of its polynomial part in a decomposition of ϕ on a polynomial function and an integer-valued function.
If k 0 ∈ {1, . . . , 2 n+2 (2 n+1 − 1) − 2}, then consider the function Of course, the functionφ is a solution of the polynomial congruence of degree n and Therefore, from Remark 3.5 in case (1) and Remark 3.6 in case (2) and the previous theorem it follows thatφ is a decent solution of the polynomial congruence and a polynomial part of its decomposition is continuous, but then also ϕ is a decent solution of the polynomial congruence of degree n with continuous polynomial part in the decomposition. We proved that ϕ = f +g, where f is a continuous polynomial function and g is an integer-valued function. Since f is continuous, it is Lebesgue measurable in case (1) and Baire measurable in case (2). Therefore, g = ϕ − f is Lebesgue measurable in case (1) and Baire measurable in case (2), too.
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