Characterization of field homomorphisms through Pexiderized functional equations
aa r X i v : . [ m a t h . A C ] O c t Characterization of field homomorphisms throughPexiderized functional equations
Eszter Gselmann, Gergely Kiss and Csaba VinczeOctober 30, 2018
Abstract
The aim of this paper is to prove characterization theorems for field homomorphisms. Moreprecisely, the main result investigates the following problem. Let n ∈ N be arbitrary, K a fieldand f , . . . , f n : K → C additive functions. Suppose further that equation n X i = f q i i (cid:0) x p i (cid:1) = x ∈ K )is also satisfied. Then the functions f , . . . , f n are linear combinations of field homomorphismsfrom K to C . Dedicated to the th birthday of Professor Mikl´os Laczkovich The main purpose of this work is to put the previous investigations into a unified framework andto prove characterization theorems for field homomorphisms. The problem to be studied reads asfollows.Let n ∈ N be arbitrary, K a field and let f , . . . , f n : K → C be additive functions. Supposefurther that we are given natural numbers p , . . . , p n , q , . . . , q n so that p i , p j for i , jq i , q j for i , j < p i · q i = N for i = , . . . , n . ( C )Suppose also that equation n X i = f q i i ( x p i ) = K has characteristic 0 (about theproblem on other fields we refer to Open problem 4 in Section 5). In what follows, we show thatequation (1) along with condition ( C ) is suitable to characterize homomorphisms acting betweenthe fields K and C . Remark . Obviously, solving functional equation (1) is meaningful without condition ( C ). At thesame time, we have to point out that without this condition we cannot expect in general that all thesolutions are linear combinations of homomorphisms, or it can happen that the general problem canbe reduced to the above formulated problem. 1ndeed, if conditions 1 < p i · q i = N for i = , . . . , n are not satisfied, then the homogeneous terms of the same degree can be collected together, providedthat K is of characteristic zero (in such a situation we have Q ⊂ K ). To show this, assume that p i q i = N i = , . . . , k p i q i = N i = k + , . . . , k ... p i q i = N j + i = k j + , . . . , n where the positive integers N , . . . , N j + are di ff erent. Let r ∈ Q and x ∈ K be arbitrary and substitute rx in place of x in equation (1) to get0 = n X i = f q i i (( rx ) p i ) = n X i = r p i q i f q i i ( x p i ) = r N k X i = f q i i ( x p i ) + r N k X i = k + f q i i ( x p i ) + · · · + r N j + n X i = k j + f q i i ( x p i ) . Observe that the right hand side of this identity is a polynomial of r for any fixed x ∈ K , that hasinfinitely many zeros. This yields however that this polynomial cannot be nonzero, providing thatall of its coe ffi cients have to be zero, i.e., k X i = f q i i ( x p i ) = k X i = k + f q i i ( x p i ) = ... n X i = k j + f q i i ( x p i ) = C ) already holds.On the other hand, if condition p i , p j for i , jq i , q j for i , j is not satisfied then in general we cannot expect that the solutions are linear combinations of fieldhomomorphisms. Namely, in such a situation arbitrary additive functions can occur as solution,even in the simplest cases.To see this, let p , q ∈ N be arbitrarily fixed and let a : K → C be an arbitrary additive function.Furthermore, assume that for the complex constants α , . . . , α n , identity α q + · · · + α qn = f i ( x ) = α i a ( x ) ( x ∈ K ) . n X i = f i ( x p ) q = x ∈ K . At the same time, in general we cannot state that any of these functions isa linear combination of field homomorphisms. In this section we collect some results concerning multiadditive functions, polynomials and expo-nential polynomials and di ff erential operators. This collection highlights the main theoretical ideasthat we follow subsequently. Definition 1.
Let G , S be commutative semigroups, n ∈ N and let A : G n → S be a function. Wesay that A is n-additive if it is a homomorphism of G into S in each variable. If n = n = A is simply termed to be additive or biadditive , respectively.The diagonalization or trace of an n -additive function A : G n → S is defined as A ∗ ( x ) = A ( x , . . . , x ) ( x ∈ G ) . As a direct consequence of the definition each n -additive function A : G n → S satisfies A ( x , . . . , x i − , kx i , x i + , . . . , x n ) = kA ( x , . . . , x i − , x i , x i + , . . . , x n ) ( x , . . . , x n ∈ G )for all i = , . . . , n , where k ∈ N is arbitrary. The same identity holds for any k ∈ Z provided that G and S are groups, and for k ∈ Q , provided that G and S are linear spaces over the rationals. For thediagonalization of A we have A ∗ ( kx ) = k n A ∗ ( x ) ( x ∈ G ) . One of the most important theoretical results concerning multiadditive functions is the so-called
Polarization formula , that briefly expresses that every n -additive symmetric function is uniquely determined by its diagonalization under some conditions on the domain as well as on the range.Suppose that G is a commutative semigroup and S is a commutative group. The action of the di ff erence operator ∆ on a function f : G → S is defined by the formula ∆ y f ( x ) = f ( x + y ) − f ( x );note that the addition in the argument of the function is the operation of the semigroup G and thesubtraction means the inverse of the operation of the group S . Theorem 1 (Polarization formula) . Suppose that G is a commutative semigroup, S is a commutativegroup, n ∈ N and n ≥ . If A : G n → S is a symmetric, n-additive function, then for all x , y , . . . , y m ∈ G we have ∆ y ,...,y m A ∗ ( x ) = ( if m > nn ! A ( y , . . . , y m ) if m = n . Corollary 1.
Suppose that G is a commutative semigroup, S is a commutative group, n ∈ N andn ≥ . If A : G n → S is a symmetric, n-additive function, then for all x , y ∈ G ∆ n y A ∗ ( x ) = n ! A ∗ ( y ) . emma 1. Let n ∈ N , n ≥ and suppose that the multiplication by n ! is surjective in the commu-tative semigroup G or injective in the commutative group S . Then for any symmetric, n-additivefunction A : G n → S , A ∗ ≡ implies that A is identically zero, as well. The polarization formula plays the central role in the investigation of functional equations char-acterizing homomorphisms.
In what follows ( G , · ) is assumed to be a commutative group. Definition 2.
Polynomials are elements of the algebra generated by additive functions over G .Namely, if n is a positive integer, P : C n → C is a (classical) complex polynomial in n variablesand a k : G → C ( k = , . . . , n ) are additive functions, then the function x P ( a ( x ) , . . . , a n ( x ))is a polynomial and, also conversely, every polynomial can be represented in such a form. Remark . We recall that elements of N n for any positive integer n are called ( n -dimensional) multi-indices . Addition, multiplication and inequalities between multi-indices of the same dimension aredefined component-wise. Further, we define x α for any n -dimensional multi-index α and for any x = ( x , . . . , x n ) in C n by x α = n Y i = x α i i where we always adopt the convention 0 =
0. We also use the notation | α | = α + · · · + α n . Withthese notations any polynomial of degree at most N on the commutative semigroup G has the form p ( x ) = X | α |≤ N c α a ( x ) α ( x ∈ G ) , where c α ∈ C and a : G → C n is an additive function. Furthermore, the homogeneous term of degreek of p is X | α | = k c α a ( x ) α . Lemma 2 (Lemma 2.7 of [13]) . Let G be a commutative group, n be a positive integer and leta = ( a , . . . , a n ) , where a , . . . , a n are linearly independent complex valued additive functions defined on G. Then themonomials { a α } for di ff erent multi-indices are linearly independent. Definition 3.
A function m : G → C is called an exponential function if it satisfies m ( x y ) = m ( x ) m ( y ) ( x , y ∈ G ) . Furthermore, on an exponential polynomial we mean a linear combination of functions of the form p · m , where p is a polynomial and m is an exponential function.It is worth to note that an exponential function is either nowhere zero or everywhere zero.The following lemma will be useful in the proof of Theorem 9. Lemma 3 (Lemma 6. of [10]) . Let G be an Abelian group, and let V be a translation invariantlinear subspace of all complex-valued functions defined on G. Suppose that P ni = p i · m i ∈ V, wherep , . . . , p n : G → C are nonzero polynomials and m , . . . , m n : G → C are distinct exponentials forevery i = , . . . , n. Then p i · m i ∈ V and m i ∈ V for every i = , . . . , n. .2.1 Algebraic independence As a remarkable ingredient of our argument, we recall a theorem of Reich and Schwaiger [11]. Theoriginal statement was formulated for functions defined on C (with respect to addition). Theorem 2.
Let k , l , N be positive integers such that k , l ≤ N. Let m , . . . , m k : C → C be distinctnonconstant exponential functions, a , . . . , a l : C → C additive functions that are linearly indepen-dent over C . Then the functions m , . . . , m k , a , . . . , a l are algebraically independent over C .In particular, Let P s : C l → C be a classical complex polynomial of l variables for all multi-index ssatisfying | s | ≤ N. Then the identity X s : | s |≤ N P s ( a , . . . , a l ) m s · · · m s k k = implies that all polynomials P s vanish identically ( | s | ≤ N ) . Now we just focus on the last part of the statement. Most of the original argument works withoutchanges for functions defined on any Abelian group. For an arbitrary field K we denote K × (resp. K + ) the multiplicative (resp. additive) group of K .(A) Let G be an Abelian group. If the additive functions a , . . . , a l : G → C are linearly independentover C , then any system of terms a s · · · a s l l are also linearly independent over C for di ff erentnonzero multi-indices ( s , . . . , s l ) ∈ N l . Note that s = · · · = s l = m , . . . , m k : G → C are algebraically independent if and only if m s · · · m s k k , s , . . . , s k ) ∈ N k . The latter is not necessarily holds in general. Indeed,for the n -ordered cyclic group Z n (with respect to addition) the statement is not true since ϕ n ≡ ϕ : Z n → C .In our case, when G = K × and the functions are additive on K + the analogue holds. Obvi-ously, exponential functions on K × that are additive on K + are the field homomorphisms of K .Therefore none of them are constant.Let ϕ , . . . , ϕ k be field homomorphisms. To show that ϕ s · · · ϕ s k k , s , . . . , s k ) ∈ N k is enough to find a witness element h , ∈ K such that ϕ s · · · ϕ s k k ( h ) ,
1. As a special case ( J ′ = ∅ ) we get it from the following statement. Lemma 4. ([8, Lemma 3.3]) Let K be a field of characteristic 0, let ϕ , . . . , ϕ k : K → C bedistinct homomorphisms for a positive integer k. Then there exists an element , h ∈ K suchthat Y j ∈ J ϕ j ( h ) , Y j ′ ∈ J ′ ϕ j ′ ( h ) , whenever J and J ′ are distinct multisets of the elements , . . . , k. (C) Combining these facts and using Lemma 2 or following the argument of [11, Theorem 6.]we get that if a , . . . , a l are linearly independent and m , . . . , m k are nonconstant exponentialfunctions, then equation (2) holds if and only if every P s ( a , . . . , a l ) · m s · · · m s k k = s = ( s , . . . , s k ), | s | ≤ N .Applying (A)-(C) we get the following statement.5 heorem 3. Let K be a field of characteristic 0 and k , l , N be positive integers such that k , l ≤ N. Letm , . . . , m k : K × → C be distinct exponential functions that are additive on K + , let a , . . . , a l : K × → C be additive functions that are linearly independent over C and let P s : C l → C be classicalcomplex polynomials of l variables for all | s | ≤ N. Then the equation X s : | s |≤ N P s ( a , . . . , a l ) m s · · · m s k k = implies that all polynomials P s vanish identically ( | s | ≤ N ) . As we will see in the next section, the so-called Levi-Civit`a functional equation will have a distin-guished role in our investigations. Thus, below the most important statements will be summarized.Here we follow the notations and the terminology of L. Sz´ekelyhidi [13], [14].
Theorem 4 (Theorem 10.1 of [13]) . Any finite dimensional translation invariant linear space ofcontinuous complex valued functions on a topological Abelian group is spanned by exponentialpolynomials.
In view of this theorem, if ( G , · ) is an Abelian group, then any function f : G → C satisfying theso-called Levi-Civit`a functional equation , that is, f ( x · y ) = n X i = g i ( x ) h i ( y ) ( x , y ∈ G ) (4)for some positive integer n and functions g i , h i : G → C ( i = , . . . , n ), is an exponential polynomialof order at most n . Indeed, equation (4) expresses the fact that all the translates of the function f belong to the same finite dimensional translation invariant linear space, namely τ y f ∈ lin ( g , . . . , g n )holds for all y ∈ G .Obviously, if the functions h , . . . , h n are linearly independent, then g , . . . , g n are linear com-binations of the translates of f , hence they are exponential polynomials of order at most n , too.Moreover, they are built up from the same additive and exponential functions as the function f .Before presenting the solutions of equation (4), we introduce some notions. Remark . Let k , n , n , . . . , n k be positive integers with n = n + · · · + n k and let for j = , . . . , k the complex polynomials P j , Q i , j of n j − n j − i = , . . . , n ; j = , . . . , k . For any j = , . . . , k and for arbitrary multi-indices I j = (cid:16) i , . . . , i n j − (cid:17) and J j = (cid:16) j , . . . , j n j − (cid:17) we define the n j × n j matrix M j ( P ; I j , J j ) and the n j × n matrix N j ( Q ; I j ) asfollows: for any choice of p , q = , , . . . , n j − n j − p , n j − q ) element of M j ( P ; I j , J j ) is givenby M j ( P ; I j , J j ) ( n j − p , n j − q ) = p ! q ! ∂ i · · · ∂ i p ∂ j · · · ∂ j q P j (0 , . . . ,
0) for p + q < n j p = , , . . . , n j , q = , , . . . , n the ( p , q ) element of N j ( Q ; I j ) is given by N j ( Q ; I j ) p , q = n j − p )! ∂ i · · · ∂ i nj − p Q q , p (0 , . . . , . n × n block matrices M ( P ; I , . . . , I k , J , . . . , J k ) and N ( Q ; I , . . . , I k ) by M ( P ; I , . . . , I k , J , . . . , J k ) = M ( P , I , J ) 0 . . . M ( P , I , J ) 0 . . .... . . . ...... ... M k ( P , I k , J k ) and N ( Q ; I , . . . , I k ) = N ( Q ; I ) ... N k ( Q ; I k ) . The idea of using Levi-Civit`a equations rely on Theorem 10.4 of [13] which is the following.
Theorem 5.
Let G be an Abelian group, n be a positive integer and f , g i , h i : G → C ( i = , . . . , n ) befunctions so that both the sets { g , . . . , g n } and { h , . . . , h n } are linearly independent. The functionsf , g i , h i : G → C ( i = , . . . , n ) form a non-degenerate solution of equation (4) if and only if(a) there exist positive integers k , n , . . . , n k with n + · · · + n k = n;(b) there exist di ff erent nonzero complex exponentials m , . . . , m k ;(c) for all j = , . . . , k there exists linearly independent sets of complex additive functions n a j , , . . . , a j , n j − o ; (d) there exist polynomials P j , Q i , j , R i , j : C n j − → C for all i = , . . . , n ; j = , . . . , k in n j − complex variables and of degree at most n j − ;so that we have f ( x ) = k X j = P j (cid:16) a j , ( x ) , . . . , a j , n j − ( x ) (cid:17) m j ( x ) g i ( x ) = k X j = Q i , j (cid:16) a j , ( x ) , . . . , a j , n j − ( x ) (cid:17) m j ( x ) and h i ( x ) = k X j = R i , j (cid:16) a j , ( x ) , . . . , a j , n j − ( x ) (cid:17) m j ( x ) for all i = , . . . , n. Furthermore,M ( P ; I , . . . , I k , J , . . . , J k ) = N ( Q ; I , . . . , I k ) N ( R ; J , . . . , J k ) T holds for any choice of the multi-indices I j , J j ∈ N n j − ( j = , . . . , k ) , here T denotes the transposeof a matrix. In [12] E. Shulman used some techniques and results from representation theory to investigatea multivariate extension of the Levi-Civit`a equation. In order to quote her results, we need thefollowing notions. 7 emark . The notion of exponential polynomials can be formulated not only in the framework ofthe theory of functional equations but also in that of representation theory. This point of view canbe really useful in many cases. Let G be a (not necessarily commutative) topological group and C ( G ) be the set of all continuous complex valued functions on G . A function f ∈ C ( G ) is called anexponential polynomial function (or a matrix function ) if there is a continuous representation π of G on a finite-dimensional topological space X such that f ( g ) = h π ( g ) x , y i ( g ∈ G ) , where x ∈ X and y ∈ X ∗ .The minimal dimension of such representations is called the degree or the order of the exponen-tial polynomial.Furthermore, f ∈ C ( G ) is an exponential polynomial of degree less that n if it is contained in aninvariant subspace L ⊂ C ( G ) with dim ( L ) ≤ n . Definition 4.
Let G be a group. We say that f : G → C is a local exponential polynomial if itsrestriction to any finitely generated subgroup H ⊂ G is an exponential polynomial on H .A function f ∈ C ( G ) is an almost exponential polynomial if for any finite subset E of G , there isa finite-dimensional subspace L E ⊂ C ( G ), containing f and invariant for all operators τ g as g runsthrough E , where τ g f ( h ) = f ( h g ) ( h ∈ G ) . Remark . It is an immediate consequence of the above definitions that any exponential polynomialis an almost exponential polynomial. Furthermore, if f is an almost exponential polynomial, thenit is a local exponential polynomial, too. Clearly, for finitely generated topological groups all thesethree notions coincide. At the same time, in general these notions are di ff erent, even in case ofdiscrete commutative groups, see [12]. Definition 5.
Let G be a group and n ∈ N , n ≥
2. A function F : G n → C is said to be decomposable if it can be written as a finite sum of products F · · · F k , where all F i depend on disjoint sets ofvariables. Remark . Without the loss of generality we can suppose that k = F ( x , . . . , x n ) = X E X j A Ej B Ej where E runs through all non-void proper subsets of { , . . . , n } and for each E and j the function A Ej depends only on variables x i with i ∈ E , while B Ej depends only on the variables x i with i < E . Theorem 6.
Let G be a group and f ∈ C ( G ) and n ∈ N , n ≥ be fixed. If the mappingG n ∋ ( x , . . . , x n ) f ( x · · · x n ) is decomposable then f is an almost exponential polynomial function. ff erential operators Similarly as before, K denotes a field and K × stands for the multiplicative subgroup of K .In this subsection we introduce di ff erential operators acting on fields which have important rolein our investigation. 8 efinition 6. A derivation on K is a map d : K → K such that equations d ( x + y ) = d ( x ) + d ( y ) and d ( x y ) = d ( x ) y + xd ( y ) (5)are fulfilled for every x , y ∈ K .We say that the map D : K → C is a di ff erential operator of order m if D can be represented as D = M X j = c j d j , ◦ . . . ◦ d j , k j , (6)where c j ∈ C and d i , j are derivations on K and k j ≤ m which fulfilled as equality for some j . If k = d ◦ . . . ◦ d k as the identity function id on K . Remark . Since the compositions d ◦ . . . ◦ d k span a linear space over C , without loss of generalitywe may assume that each term of (6) are linearly independent. Equivalently we may fix a basis B of compositions. We also fix that the identity map is id in B . We note that a di ff erential operator oforder n contains a composition of length n .If a function m is additive on K and exponential on K × , then m is clearly a field homomorphism.In our case this can be extended to C as an automorphism of C by [9, Theorem 14.5.1]. Now weconcentrate on the subfields of C that has finite transcendence degree over Q . Lemma 5.
Let K ⊂ C be field of finite transcendence degree and ϕ : K → C an injective homomor-phism. Then there exists an automorphism ψ of C such that ψ | K = ϕ . Further relations are presented between the exponential polynomials defined on K × and di ff eren-tial operators on K . The connection was first realized in [6] and the connection between the degreesand orders was settled in [7]. Clearly every di ff erential operator is additive on K and this additionalproperty is a substantial part of the following statement. Theorem 7.
Suppose that the transcendence degree of the field K over Q is finite. Let f : K → C beadditive, and let m be an exponential on K × . Let ϕ be an extension of m to C as an automorphismof C . Then the following are equivalent.(i) f = p · m on K × , where p is a local polynomial on K × .(ii) f = p · m on K × , where p is an almost polynomial on K × .(iii) f = p · m on K × , where p is a polynomial on K × .(iv) There exists a unique di ff erential operator D on K such that f = ϕ ◦ D on K .In this case, p is a polynomial of degree n if and only if D is a di ff erential operator of order n.Proof. The equivalence of ( i ) , ( iii ) and ( i v ) follows from [6, Theorem 4.2]. Remark 5 implies theequivalence of ( ii ) with the others. The last part of the statement follows from [7, Corollary 1.1.]. (cid:3) At first glance equation (1) itself seem not really restrictive for the functions f , . . . , f n . At thesame time, our results show that these additive functions are in fact very special, i.e., they arelinear combinations of field homomorphisms from the field K to C . This is caused by the additivityassumption on the involved functions, and this is the property that can e ff ectively be combinedwith the theory of (exponential) polynomials on semigroups. More precisely, with the aid of thefollowing lemma, we will be able to broaden the number of the variables appearing in equation (1)from one to N . 9 emma 6. Let n ∈ N be arbitrary, K a field, f , . . . , f n : K → C additive functions. Suppose furtherthat we are given natural numbers p , . . . , p n , q , . . . , q n such that they fulfill condition ( C ) . If n X i = f q i i ( x p i ) = is satisfied for any x ∈ K , then we also have n X i = N ! X σ ∈ S N f i (cid:16) x σ (1) · · · x σ ( p i ) (cid:17) · · · f i (cid:16) x σ ( N − p i + · · · x σ ( N ) (cid:17) = for any x , . . . , x N ∈ K , here S N denotes the symmetric group of order N.Proof. Suppose that n ∈ N , K is a field, f , . . . , f n : K → C are additive functions and define thefunction F : K N → C through F ( x , . . . , x N ) = n X i = N ! X σ ∈ S N f i (cid:16) x σ (1) · · · x σ ( p i ) (cid:17) · · · f i (cid:16) x σ ( N − p i + · · · x σ ( N ) (cid:17) ( x , . . . , x N ∈ K ) . It is clear that F is a symmetric function, moreover, due to the additivity of the functions f , . . . , f n ,it is N -additive. Furthermore, in view of equation (7), F ( x , . . . , x ) = n X i = f q i i ( x p i ) = x ∈ K ) . Therefore, the polarization formula immediately yields that the mapping F is identically zero on K N . (cid:3) Equation (1) with two unknown functions
At first we will investigate the case when n =
2. This case was also studied by F. Halter–Koch andL. Reich in a special situation (when n = p and m = q ) in [2, 3, 4]. Proposition 1.
Let n , m , p , q ∈ N be arbitrarily fixed so that n · m = p · q > and m , p. Let K bea field and suppose that for additive functions f , g : K → C the functional equationf m ( x n ) = g p ( x q ) ( x ∈ K ) (9) is fulfilled. Then, and only then there exists a homomorphism ϕ : K → C so thatf ( x ) = f (1) · ϕ ( x ) and g ( x ) = g (1) · ϕ ( x ) furthermore, we also have f (1) m − g (1) p = .Proof. Let N = n · m = p · q . According to Lemma 6 we have that the symmetric N -additive function F : K N → C defined by F ( x , . . . , x N ) = N ! X σ ∈ S N (cid:2) f (cid:0) x σ (1) · · · x σ ( n ) (cid:1) · · · f (cid:0) x σ ( N − n + · · · x σ ( N ) (cid:1) − g (cid:16) x σ (1) · · · x σ ( q ) (cid:17) · · · g (cid:16) x σ ( N − q + · · · x σ ( N ) (cid:17)i ( x , . . . , x N ∈ K )10s identically zero due to the fact that F ( x , . . . , x ) = f m ( x n ) − g p ( x q ) = x ∈ K ) . From this we get F (1 , , , . . . , = f m (1) − g p (1) = . (10)By appropriate substitution, F ( x , , , . . . , = x ∈ K , or equivalently f m − (1) f ( x ) − g p − (1) g ( x ) = x ∈ K ) . (11)If g p − (1) = f m − (1) ,
0, then f ≡ f m − (1) = g p − (1) , g p − (1) , f m − (1) , g p − (1) = f m − (1) = g p − (1) , f m − (1) , F ( x , y, , . . . , = x , y ∈ K ) , implies that there exist constants c , c , d , d ∈ Q so that c + c = d + d = c , d (since p , m ) such that c f m − (1) f ( x y ) + c f m − (1) f ( x ) f ( y ) − d g p − (1) g ( x y ) − d g p − (1) g ( x ) g ( y ) = x , y ∈ K ) . (12)Applying equations (10) and (11) we get that g p − (1) g ( x ) g ( y ) = ( g p − (1) g ( x ))( g p − (1) g ( y )) g p (1) = ( f m − (1) f ( x ))( f m − (1) f ( y )) f m (1) = f m − (1) f ( x ) f ( y ) , and we can eliminate g from equation (12) c f m − (1) f ( x y ) + c f m − (1) f ( x ) f ( y ) − d f m − (1) f ( x y ) − d f m − (1) f ( x ) f ( y ) = ( c − d ) f m − (1) f ( x y ) + ( c − d ) f m − (1) f ( x ) f ( y ) = . Since c + c = d + d = , c , d and f (1) ,
0, it follows that c − d = − ( c − d ) , f (1) f ( x y ) = f ( x ) f ( y ) . Taking ϕ ( x ) = f ( x ) / f (1) for all x ∈ K , we get that ϕ ( x y ) = ϕ ( x ) ϕ ( y ) (i.e. ϕ is multiplicative). Also ϕ is additive since f is additive. Thus ϕ is an injective homomorphism of K . A similar argumentshows that g ( x ) = g (1) ψ ( x ), where ψ is an injective homomorphism of K . Substituting this intoequation (9), we get that f m (1) ϕ N = g p (1) ψ N . Using equation (10) and a symmetrization process, φ = ψ follows and we get f ( x ) = f (1) ϕ ( x ) and g ( x ) = g (1) ϕ ( x ) ( x ∈ K )with a certain homomorphism ϕ : K → C and f (1) m − g (1) n = g (1) q − = f (1) m − =
0, then g (1) = f (1) = f ≡ g ≡ f . m < p . Then 0 = F ( x , . . . x | {z } m , , . . . , = C · f ( x ) m , for some positive constant C . Indeed each other summand stemming from f contain at least oneterm of f (1) in the product, similarly each product of g ’s contains g (1). Therefore f ( x ) = x ∈ K , contradicting our assumption. (cid:3) Main results
Firstly we show that every solution of equation (1) is an almost exponential polynomial of the group K × . Theorem 8.
Let n ∈ N be arbitrary, K a field, f , . . . , f n : K → C additive functions. Supposefurther that we are given natural numbers p , . . . , p n , q , . . . , q n so that they fulfill condition ( C ) . If n X i = f q i i ( x p i ) = holds for all x ∈ K , then the functions f , . . . , f n : K → C are almost exponential polynomials of theAbelian group K × .Proof. Suppose that the conditions are satisfied, then due to Lemma 6, we have that the mapping F : K N → C defined by F ( x , . . . , x N ) = n X i = N ! X σ ∈ S N f i (cid:16) x σ (1) · · · x σ ( p i ) (cid:17) · · · f i (cid:16) x σ ( N − p i + · · · x σ ( N ) (cid:17) ( x , . . . , x N ∈ K )is identically zero.From this we immediately conclude that for any x ∈ K F ( x , , . . . , = n X i = f i ( x ) f q i − i (1) = x ∈ K ) . (14)Again, due to the fact that F has to be identically zero, we also have F ( x , y, , . . . , = x , y ∈ K ) , i.e., n X i = (cid:2) c i f i ( x y ) + d i f i ( x ) f i ( y ) (cid:3) = x , y ∈ K ) (15)with certain constants c i , d i ∈ C .Without the loss of generality we can (and we also do) assume that the parameters q , . . . , q n arearranged in a strictly increasing order, that is, q < q < · · · < q n holds and (due to condition ( C )) we have that p > p > · · · > p n is also fulfilled.We will show by induction on n that all the mappings f , . . . , f n are almost exponential polyno-mials. Since the multiadditive mapping F is identically zero on K N , we have that X σ ∈ S N f (cid:16) x σ (1) · · · x σ ( p ) (cid:17) · · · f (cid:16) x σ ( N − p + · · · x σ ( N ) (cid:17) = − n X i = X σ ∈ S N f i (cid:16) x σ (1) · · · x σ ( p i ) (cid:17) · · · f i (cid:16) x σ ( N − p i + · · · x σ ( N ) (cid:17) ( x , . . . , x N ∈ K )12et us keep all the variables x p + , . . . , x N be fixed, while the others are arbitrary. Then the aboveidentity yields that either f is identically zero or f is decomposable. Due to Theorem 6, in anycases we have that f is an almost exponential polynomial function. Therefore, for any finitelygenerated subgroup H ⊂ K × , the function f | H is an exponential polynomial. In other words for anyfinitely generated subgroup H ⊂ K × , the mapping f | H is not only decomposable but also fulfills acertain multivariate Levi-Civit`a functional equation.Assume now that there exists a natural number k with k ≤ n − f , . . . , f k are almost exponential polynomials. Then, again due to the fact that F ≡
0, we have that X σ ∈ S N f k + (cid:16) x σ (1) · · · x σ ( p k + ) (cid:17) · · · f k + (cid:16) x σ ( N − p k + + · · · x σ ( N ) (cid:17) = − k X i = X σ ∈ S N f i (cid:16) x σ (1) · · · x σ ( p i ) (cid:17) · · · f i (cid:16) x σ ( N − p i + · · · x σ ( N ) (cid:17) − n X i = k + X σ ∈ S N f i (cid:16) x σ (1) · · · x σ ( p i ) (cid:17) · · · f i (cid:16) x σ ( N − p i + · · · x σ ( N ) (cid:17) ( x , . . . , x N ∈ K ) . Let us keep all the variables x p k + + , . . . , x N be fixed, while the others are arbitrary. Then, in view ofTheorem 6, this equation yields that either f k + is identically zero or f k + is an almost exponentialpolynomial, due to the fact that the first summand on the right-hand side is an almost exponentialpolynomial by induction, while the other summand consists only of decomposable terms. (cid:3) Remark . Note that if f i ( x ) = a i f ( x ) ( x ∈ K )holds for all i = , . . . , n with certain complex constants a , . . . , a n (assuming that at least one ofthem is nonzero), then we immediately get that there exists a homomorphism ϕ : K → C such that f i ( x ) = f i (1) ϕ ( x ) ( x ∈ K ) . Indeed, in this case equation (15) yields that n X i = h c i a i f ( x y ) + d i a i f ( x ) f ( y ) i = x , y ∈ K ) , that is, f satisfies the Pexider equation α f ( x y ) = β f ( x ) f ( y ) ( x , y ∈ K ) . This means that f is a constant multiple of a multiplicative function. Since f has to be additive too,this multiplicative function has to be in fact a homomorphism. All in all, we have that the additivefunction f : K → C fulfills equation n X i = a q i i f ( x p i ) q i = x ∈ K )with certain complex constants a , . . . , a n if and only if there exists a homomorphism such that f ( x ) = f (1) ϕ ( x ) ( x ∈ K ) , moreover we also have n X i = a q i i f (1) q i = .
13s a consequence of the previous statement and Theorem 3 we have the following.
Theorem 9.
Let K ⊂ C be a field of finite transcendence degree over Q . Each additive solution ofequation (1) must be of the form f i = n − X j = P i , j ϕ j , where P i , j ’s are polynomials on K × and ϕ j : K → C are field homomorphisms. Then ˜ f i ( x ) = P i , j ϕ j ( x ) ( x ∈ K , j = , . . . , n − . is also a solution of (1) and all ˜ f i are additive.Proof. By Theorem 8, the solutions f i : K → C of (1) are almost exponential polynomials of theAbelian group K × . Since K is a field of finite transcendence degree, by Remark 7 and Theorem 5 all f i ’s are exponential polynomials. Thus there are nonnegative integers k , l ≤ n − m , . . . , m k : K × → C , further additive functions a , . . . , a l : G → C that are linearly independent over C and classical complex polynomials P i , , . . . , P i , k : C l → C withdeg P i , j ≤ n − f i = k X j = P i , j ( a , . . . , a l ) m j . (16)Substituting f i to (1), we have0 = n X i = f q i i ( x p i ) = n X i = k X j = P i , j ( a , . . . , a l ) m j q i ( x p i ) . (17)Since m , . . . , m k are distinct (nonconstant) exponentials, the coe ffi cients of the terms m a · · · m a k k in the expansion must be 0. Taking all terms that contains only m j as an exponential in the product.By this reduction, we get that n X i = ( P i , j m j ) q i ( x p i ) = j = , . . . , k .The additive functions with respect to addition on K constitute a linear space that is translationinvariant with respect to multiplication on K × . By Lemma 3, we get that if P kj = ( P i , j m j ) is additive(with respect to addition on K ), then P i , j · m j and m j are additive for every j = , . . . , k . Thefirst implies that ˜ f i is additive. Since m j is additive on K that has finite transcendence degree andmultiplicative on K × , by Lemma 5 m j can be extended as an automorphism φ j of C . These implythe statement. (cid:3) Remark . It is worth to note that the role of homomorphism m lost its importance. By Theorem 9for finding a solution of (1) it is enough to find all solutions of (18) separately for every j = , . . . , k .Since N = p q = · · · = p n q n and m j ,
0, equation (18) is equivalent to n X i = P q i i , j ( x p i ) = . (19)Conversely, if (19) holds and P i , j ( x ) · x is additive, then f i = P i , j ϕ is an additive solution of (18),where ϕ is an arbitrary homomorphism. 14 emark . Our next aim is to prove Theorem 10. If we omit the condition of additivity of f i thenwe can easily find solutions that are neither homomorphisms, nor di ff erential operators as it can beseen in Example 1. Example 1.
To illustrate this, let us consider the following equation on a field K . f ( x ) + g ( x ) + h ( x ) = x ∈ K ) , (20)where f , g, h : K → K denote the unknown (not necessarily additive) functions.Let d : K → K be a nontrivial derivation and define the function a : K × → K by a ( x ) = d ( x ) x (cid:0) x ∈ K × (cid:1) . Then a is an additive function on group K × .Consider the functions f , g and h defined through f ( x ) = − (20 + a ( x ) + a ( x )) x ,g ( x ) = + a ( x )) x , h ( x ) = x , that clearly provide a solution for (20). Indeed, using a k ( x l ) = l k · a k ( x ) for all l , k ∈ N we have f ( x ) = − (20 + a ( x ) + a ( x )) x = − (20 + a ( x ) + a ( x )) x ,g ( x ) = (2(1 + a ( x ))) x = (4 + a ( x ) + a ( x )) x , h ( x ) = x . On the other hand, it does not satisfies (14). Clearly, f (1) = − , g (1) = , h (1) = f ( x ) + g (1) g ( x ) + h (1) h (1) = ( − − a ( x ) − a ( x ) + + a ( x ) + x = − a ( x ) + a ( x ) − , . This is caused by the fact that at least one of the function f , g and h is not additive. It is easy tocheck that g and h are additive on K , but f is not. Theorem 10.
Let n ∈ N be arbitrary, K a field, f , . . . , f n : K → C additive functions. Supposefurther that we are given natural numbers p , . . . , p n , q , . . . , q n such that they fulfill condition ( C ) .If n X i = f q i i ( x p i ) = holds for all x ∈ K , then there exist homomorphisms ϕ , . . . , ϕ n − : K → C and α i , j ∈ C ( i = , . . . , n ; j = , . . . , n − so that f i ( x ) = n − X j = α i , j ϕ j ( x ) ( x ∈ K ) . (22) Moreover α i , j ϕ j gives also a solution of (21) .Proof. Let us assume first that K ⊂ C be a field of finite transcendence degree over Q . By Theorem9 we can restrict our attention on the solutions f i = P i · ϕ . Namely,0 = n X i = ( P i · ϕ )( x p i ) q i = ϕ ( x N ) · n X i = P p i i ( x p i ) q i ( x ∈ K ) . ϕ has no special role in the previous equation (see Remark 9), thus ϕ ≡ id can beassumed along the proof. Therefore the solutions are f i = P i · id . By Theorem 7 we can identify f i = P i · x with a derivation D i defined with (6), where the degree of P i is the same as the order of D i . Let us denote the maximal degree of all P i by M . Note that D i can be uniquely written in termsof the elements of the basis B defined as in Remark 7.Let the elements of B be the functions x , d , . . . , d k , . . . , d i ◦ · · · ◦ d i s ( x ) that are linearly indepen-dent over C for all i , . . . , i s < n . Since every composition is an additive function on K , by Theorem2 we get that the elements of B are also algebraically independent.Now fix i such that D i has maximal order M and q i is the smallest possible. Thus it contains aterm d j ◦ · · · ◦ d j M ∈ B . Then we have that d j ◦ · · · ◦ d j M ( x p i ) = p i x p i − ( d j ◦ · · · ◦ d j M )( x ) + p i ( p i − x p i − d ( x )( d ◦ · · · ◦ d j M )( x ) + . . . Let us assume that M >
1. Since x , d j ◦ · · · ◦ d j M ( x ) ∈ B and they are distinct, the coe ffi cient of x q i ( p i − ( d j ◦ · · · ◦ d j M ( x )) q i (23)uniquely determined and it must vanish. In D i ( x p i ) q i we have only the term of (23) with nonzerocoe ffi cient. Since q i was minimal, D j ( x p j ) q j does not contain the product (23), if j , i . In such asituation however this term cannot vanish, contradicting to the algebraic independence. This leadsto the fact that M =
1, i.e. every D i ( x ) = c i · x , for some complex constant c i .This clearly implies in general that every solution can be written as f i ( x ) = n − X j = c i , j ϕ j ( x ) , for some constants c i , j ∈ C and field homomorphisms ϕ , . . . ϕ n − : K → C .Now let K be an arbitrary field of characteristic 0 and assume that the statement is not true. Thenby Theorem 8 there exist almost exponential polynomial solutions defined on K × such that f i = n − X j = P i , j ϕ j . n − X j = α i , j ϕ j . Then there exists a finite set S ⊂ K which guarantees this. The field generated by S over Q isisomorphic to field K ⊂ C of finite transcendence degree. Let us denote this isomorphism by Φ : Q ( S ) → K . The previous argument provides that f i ◦ Φ satisfy (22). Since Φ − is also anisomorphism, f i satisfies (22), as well. This contradicts our assumption and finishes the proof. (cid:3) Remark . Here we note that the proof of Theorem 10 essentially uses the fact that the field K hascharacteristic 0, that we assume throughout the whole paper.The following example illustrates a special case when not all of f i are of the form c · ϕ . Theorem11 is devoted to show that this is in some sense the exceptional case. Example 2.
Let K be a field and f , g, h : K → C be additive functions such that f ( x ) + g ( x ) + h ( x ) = x ∈ K . According to Theorem 8 define the 4-additive function F : K → C through F ( x , x , x , x ) = f ( x x x x ) + { g ( x x ) g ( x x ) + g ( x x ) g ( x x ) + g ( x x ) g ( x x ) } + h ( x ) h ( x ) h ( x ) h ( x ) ( x ∈ K ) . F is identically zero, thus F itself is identically zero, too.From this we immediately get that F ( x , , , = h (1) h ( x ) + g (1) g ( x ) + f ( x ) ( x ∈ K ) , that is, the functions f , g, h are linearly dependent. Using this, we also have that0 = F ( x , y, , = − h (1) h ( x y ) − g (1) g ( x y ) + h (1) h ( x ) h ( y ) + g ( x ) g ( y )has to be fulfilled by any x , y ∈ K .Define the functions χ, ϕ , ϕ : K → C as χ ( x ) = h (1) h ( x ) + g (1) g ( x ) ϕ ( x ) = √ h (1) h ( x ) ϕ ( x ) = √ g ( x ) ( x ∈ K )to obtain the Levi-Civit`a equation χ ( x y ) = ϕ ( x ) ϕ ( y ) + ϕ ( x ) ϕ ( y ) ( x , y ∈ K ) . Using Theorems 5 and 8, we deduce that there are homomorphisms ϕ , ϕ : K → C and complexconstants α , α , β , β , γ , γ so that g ( x ) = α ϕ ( x ) + α ϕ ( x ) h ( x ) = β ϕ ( x ) + β ϕ ( x ) f ( x ) = γ ϕ ( x ) + γ ϕ ( x ) ( x ∈ K ) , where the above complex numbers will be determined from the functional equation.Indeed, from one hand we have − f ( x ) = g ( x ) + h ( x ) = (cid:16) α ϕ ( x ) + α ϕ ( x ) (cid:17) + ( β ϕ ( x ) + β ϕ ( x )) = α ϕ ( x ) + α α ϕ ( x ) ϕ ( x ) + α ϕ ( x ) + β ϕ ( x ) + β β ϕ ( x ) ϕ ( x ) + β β ϕ ( x ) ϕ ( x ) + β β ϕ ( x ) ϕ ( x ) + β ϕ ( x ) = (cid:16) α + β (cid:17) ϕ ( x ) + (cid:16) α α + β β (cid:17) ϕ ( x ) ϕ ( x ) + (cid:16) α + β (cid:17) ϕ ( x ) + β β ϕ ( x ) ϕ ( x ) + β β ϕ ( x ) ϕ ( x ) for all x ∈ K .On the other hand − f ( x ) = − γ ϕ ( x ) − γ ϕ ( x ) = − γ ϕ ( x ) − γ ϕ ( x ) ( x ∈ K ) . Bearing in mind Theorem 2, after comparing the coe ffi cients, we have especially that equations α + β = − γ α + β = − γ α α = β β = f ( x ) = − g (1) ϕ ( x ) − h (1) ϕ ( x ) g ( x ) = g (1) ϕ ( x ) h ( x ) = h (1) ϕ ( x ) ( x ∈ K ) . q , . . . , q n arearranged in a strictly increasing order, that is, q < q < · · · < q n holds. Theorem 11.
Let n ∈ N be arbitrary and K a field. Assume that there are given natural numbersp i , q i ( i = , . . . , n ) so that condition ( C ) is satisfied. Let f , . . . , f n be additive solutions of n X i = f q i i ( x p i ) = Then f i = c i , j ϕ j if i > or q , , n − X j = c , j ϕ j if i = and q = , (25) where ϕ , . . . ϕ n − : K → C are arbitrary field homomorphisms and n − X i = c q i i , j = for all j = , . . . , n.Proof. By Theorem 10 every solution f i ( x ) = k X j = c i , j ϕ j ( x ) ( x ∈ K ) , for some c i , j ∈ C thus the statement for f if q = x = ϕ ( x ) , . . . , x k = ϕ k ( x ), equation (21) yields that n X i = (cid:16) c i , x p i + . . . + c i , k x p i k (cid:17) q i = . (26)By the polynomial theorem n X i = X J i , + ... + J i , k = q i q i ! J i , ! · . . . · J i , k ! c J i , i , · . . . · c J i , k i , k · x J i , p i · . . . · x J i , k p i k = . (27)Since we have distinct homomorphisms it follows, by Theorem 3, that the coe ffi cient of each mono-mial term of the polynomial in equation (27) must be zero. Two addends belong to the same mono-mial term if and only if J i , p i = J j , p j , . . . , J i , k p i = J j , k p j . If q i ≥ J i , = J i , = q i − , J i , = . . . = J i , k = . For each addendbelonging to the same monomial term p i = J j , p j , p i ( q i − = J j , p j , J j , = . . . = J j , k = . This means that p j divides p i or, in an equivalent way, q i divides q j . Without loss of generalitywe can suppose that q i is the maximal among the possible powers. Therefore there is no any addend18elonging to the same monomial term as x p i x p i ( q i − . Since q i ≥ c i , = c i , = x k and x l we get that except at most one c i , j =
0. Thisimmediately implies equation (24).Finally, condition n X i = c q i i , j = (cid:3) Example 3.
Let K be a field. Illustrating the previous results we consider all additive solutions f , f , f , f : K → C of f ( x ) + f ( x ) + f ( x ) + f ( x ) = x ∈ K ) . (28)with f i (1) , i = , , , f i is of the form c i ϕ , then c + c + c + c = ϕ can be any homomorphism.If not, then there are two di ff erent field homomorphisms ϕ , ϕ such that f i = c i ϕ , f j = c j ϕ . for some 1 ≤ i , j ≤ i , i ∈ { , , , } are such that f i = c i ϕ , f i = c i ϕ . and for j , j ∈ { , , , } \ { i , i } we have f j = c j ϕ , f j = c j ϕ . It also clearly follows that c q i i + c q i i = c q j j + c q j j = . For instance, if i = , i = , j = , j =
4, then we get that f = c ϕ , f = c ϕ , f = c ϕ , f = c ϕ , where c + c = c + c = ϕ , ϕ : K → C are arbitrary field homomorphisms.19 .1 Summary We can assume that 0 < q < q < · · · < q n . As a consequence of Theorem 11 we get that fora given system of solutions f i of (21) the index set I = { , . . . , n } can be decomposed into somesubsets I , . . . , I k ( k < n ) such that k [ j = I j = I and k \ j = I j = ∅ if q , { } if q = . If q ,
1, then for every I j ( j = , . . . , k < n ) there exists injective homomorphisms ϕ j : K → C such that f i = c i ϕ j and P i ∈ I j c q i i =
0. If q =
1, then f = P kj = c , j ϕ j , f i = c i ϕ j if 1 , i ∈ I j and c , j + X i ∈ I j , i , c q i i = . Conversely, if there are given a partition I j ( j = , . . . , k ) of { , . . . , n } such that except maybeelement 1, the sets are disjoint, then for every field homomorphism ϕ , . . . , ϕ k : K → C we get asolution of (21) as f i = c i ϕ j if i ∈ I j and either i , q , P kj = c , j ϕ j if q = i = , where P i ∈ + I j c q i i = q ,
1, otherwise c , j + P i ∈ + I j , i , c q i i =
0. Additionally, we get that for everyset I j the system of f i = c i ϕ j ( i ∈ I j ), where c i satisfy the previous equation, is a solution of (21).This is a sub-term of (21), thus it seems reasonable that we are just looking for solutions that do notsatisfies any partial equation of (21).We say that the system of functions f , . . . , f n form an irreducible solution if it does not satisfya sub-term of (21). Corollary 2.
Under the assumptions of Theorem 11, let f , . . . , f n : K → C be additive irreducible solutions of (21) . Then for all i = , . . . , n,f i ( x ) = c i · ϕ ( x ) ( x ∈ K ) , where ϕ : K → C is an arbitrary field homomorphism and c i ∈ C satisfies n X i = c q i i = . The following statement which is only about the real-valued solutions, is an easy observation whichallows us to focus on the important cases henceforth.
Proposition 2.
Let n ∈ N be arbitrary, K a field, f , . . . , f n : K → R additive functions. Supposefurther that we are given natural numbers p , . . . , p n , q , . . . , q n so that they fulfill condition ( C ) . Ifequation (1) is satisfied for all x ∈ K by the functions f , . . . , f n and the parameters fulfillq i = k i ( i = , . . . , n ) with certain positive integers k , . . . , k n , then all the functions f , . . . , f n are identically zero. roof. If the parameters fulfill q i = k i ( i = , . . . , n )with certain positive integers k , . . . , k n , then equation (1) can be rewritten as n X i = (cid:16) f k i i ( x p i ) (cid:17) = x ∈ K ) , in other words, we received that the sum of nonnegative real numbers has to be zero, that impliesthat all the summands has to be zero for all x ∈ K . Thus the functions f , . . . , f n : K → R areidentically zero. (cid:3) As an application of the results above, first we study the case f i ( x ) = a i · f ( x ) ( x ∈ K , i = , . . . , n ) , where a , . . . , a n are given complex numbers so that at least one of them is nonzero. Theorem 12.
Let n ∈ N be arbitrary, K a field. Assume that there are given natural numbersp , . . . , p n , q , . . . , q n so that they fulfill condition ( C ) . The function f : K → C is an additive solu-tion of n X i = ( a i · f ) q i ( x p i ) = if and only if f ( x ) = c · ϕ ( x ) , (30) where ϕ : K → C is a homomorphism and for the constant c equation n X i = ( c · a i ) q i = also has to be satisfied. According to a result of Darboux [1], the only function f : R → R that is additive and multi-plicative is of the form f ( x ) = f ( x ) = x ( x ∈ R ) . From this, we get also that every homomorphism f : R → C is of the form f ( x ) = κ · x ( x ∈ R ) , where κ ∈ { , } . Corollary 3.
Let n ∈ N be arbitrary and assume that there are given natural numbers p , . . . , p n , q , . . . , q n so that they fulfill condition ( C ) . Let f , . . . , f n : R → C be additive solutions of (1) . Thenand only then, there are complex numbers c , . . . , c n with the property n X i = c q i i = so that for all i = , . . . , n f i ( x ) = c i · x ( x ∈ R ) . The above corollary shows that for real functions every solution of equation (1) is automaticallycontinuous (in fact even analytic) without any regularity assumption.21
Open problems and perspectives
In the last section of our paper we list some open problems as well as we try to open up newperspectives concerning the investigated problem.
Definition 7.
Let ( G , + ) be an Abelian group and n ∈ N , a function f : G → C is termed to be a (generalized) monomial of degree n if it fulfills the so-called monomial equation , that is, ∆ n y f ( x ) = n ! f ( y ) ( x , y ∈ G ) . Remark . Obviously generalized monomials of degree 1 are nothing else but additive functions.Furthermore, generalized monomials of degree 2 are solutions of the equation ∆ y f ( x ) = n ! f ( y ) ( x , y ∈ G ) , which is equivalent to the so-called square norm equation , i.e., f ( x + y ) + f ( x − y ) = f ( x ) + f ( y ) ( x , y ∈ G ) . In this case for the mapping f : G → C the term quadratic mapping is used as well. Proposition 3.
Let G be an Abelian group and n ∈ N . A function f : G → C is a generalizedmonomial of degree n, if and only if, there exists a symmetric, n-additive function F : G n → C sothat f ( x ) = F ( x , . . . , x ) ( x ∈ G ) . Open Problem . In this paper we determined theadditive solutions of equation (1). It would be however interesting to determine the higher ordermonomial solutions of the equation in question. More precisely, the following problem would alsobe of interest. Let n , k ∈ N be arbitrary, K a field, f , . . . , f n : K → C generalized monomials ofdegree k . Suppose further that we are given natural numbers p , . . . , p n , q , . . . , q n so that p i , p j for i , jq i , q j for i , j < p i · q i = N for i = , . . . , n ( C )Suppose also that equation n X i = f q i i ( x p i ) = f , . . ., f n ?We remark that in case k ≥
2, we do not know whether such ‘nice’ representation for thefunctions f , . . . , f n as in Theorem 8 can be expected.At the same time, there are cases when the representation is ‘nice’ as well as previously. To seethis, let us consider the following problem. Assume that for the quadratic function f : K → C wehave f ( x ) = f ( x ) ( x ∈ K ) . Since f is a generalized monomial of degree 2, there exists a symmetric bi-additive function F : K → C so that F ( x , x ) = f ( x ) ( x ∈ K ) . F : K → C through F ( x , x , x , x ) = F ( x x , x x ) + F ( x x , x x ) + F ( x x , x x ) − F ( x , x ) F ( x , x ) − F ( x , x ) F ( x , x ) − F ( x , x ) F ( x , x ) ( x , x , x , x ∈ K ) . Since F ( x , x , x , x ) = (cid:16) F ( x , x ) − F ( x , x ) (cid:17) = (cid:16) f ( x ) − f ( x ) (cid:17) = x ∈ K ) , the mapping F has to be identically zero on K . Therefore, especially0 = F (1 , , , = F (1 , − F (1 , , yielding that either F (1 , = F (1 , =
1. Moreover,0 = F ( x , , , = F ( x , − F (1 , F ( x ,
1) ( x ∈ K ) , from which either F (1 , = F ( x , = x ∈ K .Using that0 = F ( x , x , , = F ( x , − F (1 , F ( x , x ) + F ( x , x ) − F ( x ,
1) ( x ∈ K ) , we obtain that ( F (1 , − F ( x , x ) = F ( x , − F ( x ,
1) ( x ∈ K ) . Now, if F (1 , =
0, then according to the above identities F ( x , = x ∈ K .Since F ( x , x , , = x ∈ K , this immediately implies that − f ( x ) = − F ( x , x ) = F ( x , − F ( x , = x ∈ K ) , i.e., f is identically zero.In case F (1 , ,
0, then necessarily F (1 , = − F ( x , x ) = F ( x , − F ( x , ( x ∈ K ) . Define the non-identically zero additive function a : K → C by a ( x ) = F ( x ,
1) ( x ∈ K )to get that f ( x ) = F ( x , x ) = − F ( x , + F ( x , = a ( x ) − a ( x ) ( x ∈ K ) . Since F ( x , x , x , x ) = a : K → C has to fulfill identity − a ( x ) + a ( x ) + a ( x ) a ( x ) − a ( x ) = x ∈ K )too.In what follows, we will show that the additive function a is of a rather special form.Indeed, 0 = F ( x , y, z ,
1) ( x , y, z ∈ K )means that a has to fulfill equation a ( x ) a ( y z ) + a ( y ) a ( xz ) + a ( z ) a ( x y ) = a ( x ) a ( y ) a ( z ) + a ( x y z ) ( x , y, z ∈ K )23et now z ∗ ∈ K be arbitrarily fixed to have a ( x ) a ( y z ∗ ) + a ( y ) a ( xz ∗ ) + a ( z ∗ ) a ( x y ) = a ( x ) a ( y ) a ( z ∗ ) + a ( x y z ∗ ) ( x , y, z ∈ K ) . Define the additive function A : K → C by A ( x ) = a ( xz ∗ ) − a ( z ∗ ) a ( x ) ( x ∈ K )to receive that A ( x y ) = a ( x ) A ( y ) + a ( y ) A ( x ) ( x , y ∈ K ) , which is a special convolution type functional equation. Due to Theorem 12.2 of [13], we get that(a) the function A is identically zero, implying that a has to be multiplicative. Note that a is additive,too. Thus, for the quadratic mapping f : K → C there exists a homomorphism ϕ : K → C suchthat f ( x ) = ϕ ( x ) ( x ∈ K ) . (b) or there exists multiplicative functions m , m : K → C and a complex constant α such that a ( x ) = m ( x ) + m ( x )2 ( x ∈ K )and A ( x ) = α ( m ( x ) − m ( x )) ( x ∈ K ) . Due to the additivity of a , in view of the definition of the mapping A , we get that A is additive,too.This however means that both the maps m + m and m − m are additive, from which theadditivity of m and m follows, yielding that they are in fact homomorphisms.Since F ( x , x ) = f ( x ) = a ( x ) − a ( x ) ( x ∈ K ) , we obtain for the quadratic function f : K → C that there exist homomorphisms ϕ , ϕ : K → C such that f ( x ) = ϕ ( x ) ϕ ( x ) ( x ∈ K ) . Summing up, we received the following: identity f ( x ) = f ( x ) ( x ∈ K )holds for the quadratic function f : K → C if and only if there exists homomorphisms ϕ , ϕ : K → C such that f ( x ) = ϕ ( x ) ϕ ( x ) ( x ∈ K ) . Open Problem . Motivated by the above open problem as wellas Remark 10, we can also pose the question below.Let n ∈ N be arbitrary, K a field, f , . . . , f n : K → C be generalized or exponential polynomials.Suppose further that we are given natural numbers p , . . . , p n , q , . . . , q n so that they fulfill condition( C ). Suppose also that equation n X i = f q i i ( x p i ) = f ( x ) = − (20 + a ( x ) + a ( x )) x ,g ( x ) = + a ( x )) x , h ( x ) = x The functions f , g and h are exponential polynomial solutions of (20). Thus it is clear that Theorem11 do not hold without additivity of f , g and h .Again, the question is, how can we characterize (exponential) polynomial solutions of (33)? Open Problem . As it already appears inthe definition of homomorphisms, the natural domain and also the natural range of the functions in(20) are rings.On the other hand, it is easy to see that there is no nontrivial field homomorphism from K to C , if the characteristic of K is finite. The careful reader can also deduce using our methods, thatalready (20) has no solution in this case. At the same time, it can be easily seen that the equationhas solutions if the functions f i : K → L are constant multiple of a field homomorphism where K and L has the same characteristic.According to this, the general question arises how the solutions of equation (1) look like in casewhen the functions f , . . . , f n are defined between (not necessarily commutative) rings? Open Problem K = C ) . To pose our last open problem, here we recallthe following. Concerning homomorphisms, instead of R , in C the situation is completely di ff erent,see Kestelman [5], since we have the following. Proposition . The only continuous endomorphisms f : C → C are f ≡ , f ≡ id orf ( x ) = x ( x ∈ C ) . These endomorphisms are referred to as trivial endomorphisms . Concerning nontrivial endomorphisms we quote here the following.
Proposition . (i) There exist nontrivial automorphisms of C .(ii) If f : C → C is a nontrivial automorphism, then f | R is discontinuous.(iii) If f : C → C is a nontrivial automorphism, then the closure of the set f ( R ) is the wholecomplex plane.(iv) If f : C → C is a nontrivial automorphism, then f ( R ) is a proper subfield of C , card( f ( R )) = c and either the planar (Lebesgue) measure of f ( R ) is zero or f ( R ) ( C is a saturated non-measurable set. As we saw above the continuous endomorphisms of C are of really pleasant form. This im-mediately implies that the continuous solutions of equation (1) in case K = C also have the samebeautiful structure.Obviously, the continuity assumption can be weakened to guarantee the same result. At the sametime, our question is whether instead of a regularity assumption, an additional algebraic suppositionfor the unknown functions would imply the same?25 cknowledgement. The authors are grateful to
Professors Mikl´os Laczkovich and
L´aszl´o Sz´ekely-hidi for their valuable remarks and also for their helpful discussions.The research of the first author has been supported by the Hungarian Scientific Research Fund(OTKA) Grant K 111651. The publication is also supported by the EFOP-3.6.1-16-2016-00022project. The project is co-financed by the European Union and the European Social Fund. Thesecond author was supported by the internal research project R-AGR-0500 of the University ofLuxembourg and by the Hungarian Scientific Research Fund (OTKA) K104178. The third authorwas supported by EFOP 3.6.2-16-2017-00015.
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