A functional equation for monomial functions

Let 𝑛 be a given positive integer, κ ∈ K and f ∈ M n ⁢ ( F , K ) such that

holds for all x ∈ F .

1. If κ ∉ { 2 k ∣ k = 0 , 1 , … , n } , then 𝑓 is identically zero.

2. If κ = 1 , then

   f ⁢ ( x ) = f ⁢ ( 1 ) ⋅ x n ( x ∈ F ) .

3. If κ = 2 , then there exists an a ∈ D 2 ⁢ n − 1 ⁢ ( F , K ) such that

   f ⁢ ( x ) = ∑ j = 1 n λ n , j ⁢ x n − j ⁢ a ⁢ ( x j ) ( x ∈ F )

   holds with appropriate constants λ 1 , … , λ n ∈ K .

4. If κ = 2 n , then there exists a symmetric and 𝑛-additive mapping A n : F n → K such that

   ∑ σ ∈ S n + 1 { A n ⁢ ( x σ ⁢ ( 1 ) ⋅ x σ ⁢ ( 2 ) , x σ ⁢ ( 3 ) , … , x σ ⁢ ( n + 1 ) ) − x σ ⁢ ( 1 ) ⋅ A n ( x σ ⁢ ( 2 ) , … , x σ ⁢ ( n + 1 ) ) − x σ ⁢ ( 2 ) ⋅ A n ( x σ ⁢ ( 1 ) , … , x σ ⁢ ( n + 1 ) ) } = 0 ( x 1 , … , x n + 1 ∈ F )

   and

   f ⁢ ( x ) = A n ⁢ ( x , … , x ) ( x ∈ F ) .

   holds.

Proof

Since f ∈ M n ⁢ ( F , K ) , there exists a uniquely determined symmetric 𝑛-additive function A n : F n → K such that f ⁢ ( x ) = A n ⁢ ( x , … , x ) = A n ⁢ ( [ x ] n ) ( x ∈ F ). In terms of the mapping A n , equation (2.1) reads as

If we write x + 1 in place of 𝑥 in the above identity and expand the terms,

follows for all x ∈ F . Observe that the left-hand side of this equation is a generalized polynomial of degree 2 ⁢ n , which is identically zero. Thus all of its monomial terms should vanish. If α 1 , α 2 , α 3 are nonnegative and α 1 + α 2 + α 3 = n , then the degree of the mapping

is 2 ⁢ α 1 + α 2 . Similarly, if β 1 , β 2 , γ 1 , γ 2 are nonnegative integers with β 1 + β 2 = n and γ 1 + γ 2 = n , then the degree of the generalized monomial

is β 1 + γ 1 .

Computing the zero-degree terms in (2.2), ( 1 − κ ) ⁢ A n ⁢ ( [ 1 ] n ) = 0 follows. So κ = 1 or A n ⁢ ( [ 1 ] n ) = f ⁢ ( 1 ) = 0 . The first-degree terms of the generalized polynomial in equation (2.2) must vanish; thus we have

that is, ( 2 − κ ) ⁢ A n ⁢ ( [ x ] 1 , [ 1 ] n − 1 ) − κ ⁢ x ⁢ A n ⁢ ( [ 1 ] n ) = 0 for all x ∈ F . If κ = 1 , then this leads to

If κ ≠ 1 , then due to the previous step A n ⁢ ( [ 1 ] n ) = 0 , the above identity reduces to

So κ = 2 (and in this case, we do not have any information for the values A n ⁢ ( [ x ] 1 , [ 1 ] n − 1 ) ), or κ ≠ 1 , 2 and then A n ⁢ ( [ x ] 1 , [ 1 ] n − 1 ) = 0 for all x ∈ F . In general, if k = 0 , 1 , … , 2 ⁢ n , we have to distinguish two cases, depending on whether 𝑘 is even or odd.

If 𝑘 is even, then the 𝑘-th-degree term in (2.2) is

If 𝑘 is odd, then the only difference is that, in the first expression, we must not sum up to k 2 , but to k − 1 2 , i.e.,

This means that, for all k = 0 , 1 , … , 2 ⁢ n , we have

By selecting the members with indices l = 0 and m = 0 from the sums, we get that

for all x ∈ F . This identity indicates that the values of the left-hand 𝑘-variable function are determined by the right-hand side, with the aid of the at most ( k − 1 ) -variable functions, for all k = 0 , 1 , … , n .

First assume that, for all k = 0 , 1 , … , n , we have κ ≠ 2 k . Then, for all k = 0 , 1 , … , n ,

thus, especially, f ⁢ ( x ) = A n ⁢ ( [ x ] n ) = 0 ( x ∈ F ). We will show this by induction on 𝑘. Above, in the comparison of zero-degree terms, we have already seen that ( 1 − κ ) ⁢ A n ⁢ ( [ 1 ] n ) = 0 . Since κ ≠ 1 in this case, A n ⁢ ( [ 1 ] n ) = 0 . Therefore, the statement holds for k = 0 . Assume now that there exists a positive integer such that the statement holds for all indices less than 𝑘. In other words, suppose that, for all l = 0 , 1 , … , k − 1 , A n ⁢ ( [ x ] l , [ 1 ] n − l ) = 0 ( x ∈ F ). Since the left-hand side is a generalized monomial of degree 𝑙, which is identically zero, the symmetric, 𝑙-additive mapping, defined uniquely by it, must also be identically zero. This means however that A n ⁢ ( x 1 , … , x l , [ 1 ] n − l ) = 0 holds for all x 1 , … , x l ∈ F . This means that the right-hand side of equation (2.3) is identically zero, from which

can be deduced. So the statement holds also for 𝑘. Summing up, if κ ≠ 2 k for all k = 0 , 1 , … , n , then

thus, especially, f ⁢ ( x ) = A n ⁢ ( [ x ] n ) = 0 ( x ∈ F ). We now turn to discussing the case κ = 1 . In this case, we will also use equation (2.3) by choosing different indices 𝑘. We show that, for all k = 0 , 1 , … , n , we have

For k = 0 , this holds true trivially since A n ⁢ ( [ 1 ] n ) = f ⁢ ( 1 ) . Assume now that there exists a positive 𝑘 less than or equal to 𝑛 such that the statement holds for l = 0 , … , k − 1 , i.e., A n ⁢ ( [ x ] l , [ 1 ] n − l ) = f ⁢ ( 1 ) ⋅ x l ( x ∈ F ). Since A n is a symmetric and 𝑛-additive mapping, for all l = 0 , … , n , the mappings F l ∋ ( x 1 , … , x l ) ↦ A n ⁢ ( x 1 , … , x l , [ 1 ] n − l ) are symmetric and 𝑙-additive. Due to the induction hypothesis, for all fixed l = 0 , … , k − 1 , the mapping F ∋ x ↦ A n ⁢ ( [ x ] l , [ 1 ] n − l ) − f ⁢ ( 1 ) ⋅ x l is identically zero. At the same time, for all fixed l = 0 , … , k − 1 , this mapping is the trace of the symmetric and 𝑙-additive mapping F l ∋ ( x 1 , … , x l ) ↦ A n ⁢ ( x 1 , … , x l , [ 1 ] n − l ) = f ⁢ ( 1 ) ⋅ x 1 ⁢ ⋯ ⁢ x l . As the trace is identically zero, this mapping should be identically zero, too. Therefore,

holds for all l = 0 , … , k − 1 . In this case, however, equation (2.3) takes the form

showing that the statement also holds for 𝑘. Note, however, that this means (with k = n ) that A n ⁢ ( [ x ] n ) = f ⁢ ( 1 ) ⋅ x n ( x ∈ F ), that is, f ⁢ ( x ) = f ⁢ ( 1 ) ⋅ x n ( x ∈ F ). We continue with the case κ = 2 . Recall that, in this case, we have already seen that A n ⁢ ( [ 1 ] n ) = 0 and equation (2.3) for k = 1 does not contain any information for the values of the mapping x ↦ A n ⁢ ( [ x ] 1 , [ 1 ] n − 1 ) . Consider the additive function 𝑎 defined on 𝔽 by a ⁢ ( x ) = A n ⁢ ( [ x ] 1 , [ 1 ] n − 1 ) ( x ∈ F ). We show that, for all k = 1 , … , n ,

holds with some appropriate constants λ k , 1 , … , λ k , l ∈ K . Due to the definition of the function 𝑎, this statement is true for k = 1 . Assume an index k ≥ 2 exists now, such that the above statement holds for all indices l = 1 , … , k − 1 . Note that, from the induction hypothesis, it follows that we have

for all x 1 , … , x l and for all l = 1 , … , k − 1 . Therefore, we have

for all x 1 , … , x l and for all l = 1 , … , k − 1 . Due to identity (2.3) and the induction hypothesis, we have

for all x ∈ F . From this, we obtain that the statement also holds for 𝑘. Observe that this means that

holds with appropriate constants λ 1 , … , λ n . Using this representation and equation (2.1), the identity

follows for the additive function 𝑎. In view of [15, Corollary 4], this means that a ∈ D 2 ⁢ n − 1 ⁢ ( F , K ) .

Let us consider the case κ = 2 n . We show that, in this case, A n ⁢ ( [ x ] k , [ 1 ] n − k ) = 0 ( x ∈ F ) holds for all k = 0 , … , n − 1 . Since, in case κ = 2 n , we have A n ⁢ ( [ 1 ] n ) = 0 , the statement holds for k = 0 . Assume now that there exists a k ∈ { 1 , … , n } such that the statement holds for all l = 0 , … , k − 1 , that is, we have

for all l = 0 , … , k − 1 . Since the left-hand side of this equation is the trace of a symmetric and 𝑙-additive function, we have A n ⁢ ( x 1 , … , x l , [ 1 ] n − l ) = 0 ( x 1 , … , x l ∈ F ) for all l = 0 , … , k − 1 . Using this and equation (2.3), we obtain that

for all x ∈ F . Indeed, since l + ( k − 2 ⁢ l ) ≤ k − 1 and k − m ≤ k − 1 holds for all l = 1 , … , ⌊ k 2 ⌋ and for all m = 1 , … , k , the mappings F ∋ x ↦ A n ⁢ ( [ x 2 ] l , [ 2 ⁢ x ] k − 2 ⁢ l , [ 1 ] n − k + l ) and F ∋ x ↦ A n ⁢ ( [ x ] k − m , [ 1 ] n − k − m ) are identically zero. Since the left-hand side of equation (2.3) vanishes for κ = 2 n , we have no information for the mapping x ↦ A n ⁢ ( [ x ] n ) .

Using this and computing the ( n + 1 ) -th-degree terms in (2.2), we deduce that

that is, A n ⁢ ( [ x 2 ] 1 , [ x ] n − 1 ) − 2 ⁢ x ⋅ A n ⁢ ( [ x ] n ) = 0 ( x ∈ F ). The left-hand side of this equation (as a mapping of the variable 𝑥) is the trace of a symmetric and ( n + 1 ) -additive function. However, the ( n + 1 ) -additive mapping is necessarily identically zero if the trace vanishes. Thus we have

for all x 1 , … , x n + 1 ∈ F . ∎

Notice that the above theorem does not state anything about the cases when

We conjecture that, in the case of k = 2 , the order of the higher-order derivation, appearing in the representation of the function 𝑓, can be reduced. This conjecture is supported by our results for the case n = 3 , which can be found in the next section.