Polynomially linked additive functions—II
Bruce Ebanks
Aequationes mathematicae ISSN 0001-9054 Aequat. Math. DOI 10.1007/s00010-017-0537-0
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Aequationes Mathematicae
Polynomially linked additive functions—II Bruce Ebanks Abstract. We continuethe study of additive functions fk : R → F (1 ≤ k ≤ n) linked by an equation of the form n k=1 pk (x)fk (qk (x)) = 0, where the pk and qk are polynomials, R is an integral domain of characteristic 0, and F is the fraction field of R. A method is presented mk f (xjk ) = 0 in for solving all such equations. We also consider the special case n k k=1 x which the pk and qk are monomials. In this case we show that if there is no duplication, i.e. if (mk , jk ) = (mp , jp ) for k = p, then each fk is the sum of a linear function and a derivation of order at most n − 1. Furthermore, if this functional equation is not homogeneous then the maximal orders of the derivations are reduced in a specified way. Mathematics Subject Classification. Primary: 39B52, 39B72; Secondary: 13N15, 16W25. Keywords. Ring derivation, Higher-order derivation, Functional equation, Integral domain, Characteristic zero, Field, Homogeneity.
1. Introduction Let S be a ring and let R be a subring of S. An additive function on R into S is a solution f : R → S of Cauchy’s functional equation f (x + y) = f (x) + f (y) for all x, y ∈ R. A derivation on R into S is an additive function f : R → S satisfying also the Leibniz rule f (xy) = xf (y) + f (x)y,
x, y ∈ R.
(1)
We say that the additive functions f1 , . . . , fn : R → S are polynomially linked if there exist polynomial functions pk : R → S and qk : R → R for 1 ≤ k ≤ n such that n pk (x)fk (qk (x)) = 0, x ∈ R. (2) k=1
This work is dedicated to the memories of my colleague and friend Prasanna K. “Ron” Sahoo, my stepson Edmund “Ted” France, and especially my mother Dorothy Griner Ebanks.
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We also study the special case of monomial linkage, that is when n xmk fk (xjk ) = 0, x ∈ R,
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(3)
k=1
for mk ∈ N ∪ {0} and jk ∈ N (1 ≤ k ≤ n). The study of such equations on the field R of real numbers began in the 1960’s with a problem of Halperin as reported in [1], followed by works of Jurkat [12], Kurepa [15,16], Nishiyama and Horinouchi [17], and Kannappan and Kurepa [13,14]. In the latter papers Kannappan and Kurepa posed the following problem and gave some partial results. Assume that Ui (i = 1, . . . , n) is a rational function on R, that pi is continuous on R except at the singular points of Ui , and n pi (x)fi (Ui (x)) = 0 i=1
holds for all x at which the functions Ui are defined. They stated that one “would expect” that the additive functions fi are combinations of linear functions and derivations. In spite of the last statement they were aware that this is not always the case, since there are trivial counterexamples such as the linkage f1 (x) − f2 (x) = 0 which only means that f1 = f2 = any additive function. Later works along the same lines, by Halter-Koch, Reich and Unger [9– 11,18,19], and by the author and Ng [7], showed that with more terms in the linking equation comes the possibility of higher order derivations (defined below) appearing in the solutions. More details of this history can be found in the author’s papers [4,5]. As an aside here we just mention briefly some similar results (see [2,3, 6]) involving pairs of additive (or quadratic) functions linked by a functional equation on a curve. For example, suppose a pair (f, g) of additive functions is linked by the equation xf (y) = yg(x),
(x, y) ∈ C
for some specified curve C. In the case of circles and many other types of curves one can conclude that f and g are identical linear functions. The current article continues research on the problem of solving equations of the form (3) and (2). It builds on the methods initiated in [4] and further developed in [5] for solving such equations. One difference between those two earlier papers and the current one is that here we allow our additive functions fi to map an integral domain R into its field of fractions F , while in the earlier ones fi was restricted to mapping R into itself. Our first main result, Theorem 10, is that if there is no “duplication” of terms (that is, if (mk , jk ) = (mp , jp ) for k = p), then each additive function fk in (3) is the sum of a linear function and a derivation of some order. The number n of terms also fixes an upper bound of n − 1 on the orders of the
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derivations appearing in the solutions. In fact this upper bound is further reduced in a specified way if the equation is not homogeneous. This will be proved in Sect. 3 after some preliminaries in Sect. 2. In Sect. 4 we make a slight improvement on a particular result from [5] to prepare the way for Sect. 5. Our paper [5] also updated the problem of Kannappan and Kurepa (from their 1970 work [13]) described above. Specifically, our updated problem asked the following question. Problem 1. Let K be a field of characteristic 0, and let n ∈ N. Assume that polynomial functions qi , pi : K → K and additive functions fi : K → K (for 1 ≤ i ≤ n) form a solution of (2). What are necessary and sufficient conditions on the set of pi and qi so that each fi is the sum of a linear function and a derivation of some order? We do not have an answer to this question. Nevertheless in Sect. 5 we present a method for solving all equations of the form (2) in the following context: R is an integral domain of characteristic 0 and F is its fraction field, qi : R → R and pi : R → F are given polynomials, and fi : R → F are additive functions. We conclude with a small observation in the final short Sect. 6.
2. Preliminaries Throughout the remainder of this paper R will denote an integral domain of characteristic 0, and F denotes the fraction field of R. Since we will be dealing with derivations of various orders, we shall call any derivation f : R → F as defined above a derivation of order 1 and write f ∈ D1 (R). A mapping B : R × R → F is called a bi-derivation of order 1 if B is a derivation of order 1 in each variable when the other variable is fixed. We extend these definitions to derivations of order 0 and orders higher than 1 as follows. (Note that the following definition is slightly more general than the one given in [4,5] in that here we allow the functions to take their values in F rather than restricting them to R.) Definition 2. Derivations of all orders greater than or equal to 0 are defined inductively as follows. (i) The zero map from R to F is the only derivation of order 0, that is D0 (R) = {0}. (ii) The zero map from R × R to F is the only bi-derivation of order 0. (iii) For each n ∈ N we say an additive map f : R → F is a derivation of order n, written f ∈ Dn (R), if there exists Bf : R × R → F such that
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Bf is a bi-derivation of order n − 1 with f (xy) − [xf (y) + f (x)y] = Bf (x, y),
x, y ∈ R.
We call this Bf the bi-derivation associated with f . (iv) Then define a mapping B : R × R → F to be a bi-derivation of order n if B is a derivation of order n in each variable when the other variable is held fixed. Note that a derivation f of any order must satisfy the condition f (1) = 0 because of the Leibniz rule (1). As remarked in [5], Theorem 4.8 of [4] has the following consequence. Any derivation of order p is also a derivation of order q for all q > p, that is, Dp (R) ⊆ Dq (R),
p < q.
We shall make free use of this nesting property of derivations. A key idea from [4] that we use here is homogeneity in the following sense. A map φ : R → F is homogeneous of degree n ∈ N ∪ {0} provided that φ(kx) = k n φ(x),
x ∈ R,
k ∈ N.
The following simple result, which was Lemma 2.2 in [4], turns out to be very useful. It was stated there for functions from R into R, but the same proof works for functions from R into F . Lemma 3. Let n ∈ N ∪ {0}. If hj : R → F is homogeneous of degree j for 0 ≤ j ≤ n and n hj (x) = 0, x ∈ R, j=0
then hj = 0 for each j = 0, 1, . . . , n. The next result is a combination of Proposition 4 and Theorem 5 in [5]. Again it was stated there for functions from R to R but is also valid with the same proof for functions from R to F . The only difference is that the constants ci are taken from F . We will derive several consequences of this. Proposition 4. Let n ∈ N and let fi : R → F be additive for 1 ≤ i ≤ n + 1. The general solution of n
xj fn+1−j (xn+1−j ) = 0,
x ∈ R,
(4)
j=0
is given by fn+1−j (x) = (−1)j
j n+1−j+k k=0
k
Dn−j+k (x) + cn+1−j x
(5)
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for j = 0, 1, . . . , n, where Di ∈ Di (R) for each i (recall that D0 = {0}) and c1 , . . . , cn+1 ∈ F satisfy n+1
ci = 0.
i=1
Since D1 (R) ⊆ D2 (R) ⊆ · · · ⊆ Dn (R), the linear combination of derivations on the right hand side of Eq. (5) can be replaced by a single derivation dn+1−j of order at most n. Therefore the gist of Proposition 4 is that if n + 1 additive functions are monomially linked by a homogeneous functional equation of degree n + 1, then each of those additive functions is the sum of a linear function and a derivation of order at most n. That is, briefly, fi (x) = di (x) + ci x,
x ∈ R,
where di ∈ Dn (R) for all 1 ≤ i ≤ n + 1. We recall one more result needed later. The following was Proposition 4.7 in [4]. Proposition 5. Let n ∈ N∪{0}, and let f : R → F be additive. Then f ∈ Dn (R) if and only if n n+1 j (−1)j (6) x f (xn+1−j ) = 0 j j=0 for all x ∈ R. Let us note some consequences of Proposition 5. Suppose d is a derivation of order at most k. Then d satisfies Eq. (6) for every n ≥ k by the nesting property. For n = k + 1 this allows us to solve for d(xk+1 ) in terms of d(x), . . . , d(xk ); by induction we can represent d(xn ) for every n ≥ k + 1 in terms of d(x), . . . , d(xk ). For example, if d is a derivation of order at most 2, then d(xn ) for any n ≥ 3 can be represented in terms of d(x) and d(x2 ). We illustrate with a detailed calculation for this case. Corollary 6. Suppose d ∈ D2 (R). Then n n−2 d(xn ) = x d(x2 ) − n(n − 2)xn−1 d(x), 2 for each n ≥ 3.
x ∈ R,
(7)
Proof. We proceed by induction. From Proposition 5 we see that for any n ≥ 2 n n+1 j d(xn+1 ) = (−1)j+1 (8) x d(xn+1−j ), x ∈ R. j j=1 In particular, for n = 2 this states that d(x3 ) = 3xd(x2 ) − 3x2 d(x),
x ∈ R,
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which is formula (7) for n = 3. Now suppose formula (7) is true for 3 ≤ n ≤ m. Then by combining formula (8) for n = m with the inductive hypothesis we compute that m+1
d(x
)=
m
j+1
(−1)
j=1
m+1 j x d(xm+1−j ) j
m+1 j m+1−j m m+1 = (−1) ) + (−1) x d(x xm−1 d(x2 ) j m − 1 j=1 m+1 m + 1 + (−1) xm d(x) m m−2 j+1 m + 1 (−1) = xj j j=1 m + 1 − j m−j−1 2 m−j d(x ) − (m + 1 − j)(m − j − 1)x d(x) x 2 m + 1 m−1 + (−1)m d(x2 ) + (−1)m+1 (m + 1)xm d(x) x m−1 m−1 m+1−j m−1 2 j+1 m + 1 d(x ) (−1) =x j 2 j=1 ⎡ m−2 m+1 − xm d(x) ⎣ (−1)j+1 (m + 1 − j)(m − j − 1) j j=1 ⎤ m−2
j+1
+ (−1)m (m + 1)⎦ .
The remainder of the proof consists in verifying the two identities m−1
(−1)j+1
j=1
m+1 j
m+1−j 2
=
m+1 2
(9)
and m−2
j+1
(−1)
j=1
m+1 (m + 1 − j)(m − j − 1) + (−1)m (m + 1) j
= (m + 1)(m − 1) for all m ≥ 3.
(10)
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To verify (9) we compute m−1 m−1 m+1−j (m + 1 − j)! (m + 1)! j+1 m + 1 (−1) (−1)j+1 = j 2 j!(m + 1 − j)! 2(m − 1 − j)! j=1 j=1 =
m−1 (m − 1)! (m + 1)m (−1)j+1 2 j!(m − 1 − j)! j=1
m−1 (−1) = j j=1 m+1 m+1 m−1 ]= = [1 − (1 − 1) 2 2
m+1 2
m−1
j+1
where we have used the binomial formula for (1 + x)m−1 at x = −1 in the next to last step. We also get (10) from the binomial formula. Starting with m+1 m + 1 (1 + x)m+1 = xm+1−j , j j=0 we take a derivative, then subtract the constant term from both sides, arriving at m−1 m + 1 (m + 1)[(1 + x)m − 1] = (m + 1 − j)xm−j . j j=0 Next divide both sides by x and take another derivative to find that xm(1 + x)m−1 − [(1 + x)m − 1] (m + 1) x2 m−2 m + 1 = (m + 1 − j)(m − 1 − j)xm−2−j . j j=0 Since m ≥ 3 we arrive at (10) by putting x = −1 here and re-arranging.
3. First main result In order to prove our main result in this section we first need to strengthen Proposition 4. We demonstrate that if there are fewer than n + 1 nonzero terms in a homogeneous equation of degree n + 1 then the maximal orders of the derivations appearing in the solutions are reduced accordingly. A special case of the following theorem appeared as Theorem 6 in [8].
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Theorem 7. Let n, m ∈ N with m ≤ n, and suppose fi : R → F is additive for i ∈ S where S ⊂ {1, . . . , n + 1} with |S| = m. If xi fi (xn+1−i ) = 0, x ∈ R, (11) i∈S
then each fi is the sum of a linear function and a derivation of order at most m − 1. In particular, fi (x) = di (x) + ci x,
x ∈ R,
i ∈ S,
where di ∈ Dm−1 (R) for each i ∈ S and ci = 0. i∈S
Proof. Let T = {1, . . . , n + 1} \ S and define ft := 0 for each t ∈ T . Then f1 , . . . , fn+1 satisfy Eq. (4). By Proposition 4 we have that each fi is the sum of a linear function and a derivation of order at most n, specifically, j n+1−j+k Dn−j+k (x) + cn+1−j x, x ∈ R, fn+1−j (x) = (−1)j k k=0
for j = 0, 1, . . . , n, where Di ∈ Di (R) for each 0 ≤ i ≤ n. Since |S| = m ≤ n we see that T is nonempty. Choosing j1 such that n + 1 − j1 ∈ T we have fn+1−j1 = 0 so j1 n + 1 − j1 + k j1 0 = (−1) Dn−j1 +k (x) + cn+1−j1 x, x ∈ R. k k=0
As the only linear derivation of any order is the zero function we have cn+1−j1 = 0 and j 1 −1 n + 1 − j1 + k n+1 Dn = − Dn−j1 +k . j1 k k=0
Since the right hand side of this equation is a derivation of order at most n − 1 we have Dn ∈ Dn−1 (R), hence each fi is the sum of a linear function and a derivation of order at most n − 1. If T contains a second element j2 then a similar argument shows that the orders of the derivations appearing in the representations of the fi are further reduced to n − 2. By induction we have fi (x) = di (x) + ci x,
x ∈ R,
where di is a derivation of order at most n − |T | for each i ∈ {1, . . . , n + 1}. Since |T | = n + 1 − |S| = n + 1 − m we have therefore di ∈ Dm−1 (R) for each i. n+1 Finally, since i=1 ci = 0 by Proposition 4 and cj = 0 for j ∈ T we have i∈S ci = 0.
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Armed with Corollary 6 we are ready to demonstrate an application of Theorem 7. Example 8. Suppose additive functions f1 , f2 , f3 : R → F satisfy x7 f1 (x2 ) + x5 f2 (x4 ) + x4 f3 (x5 ) = 0,
x ∈ R.
(12)
This equation is homogeneous of degree 9, so Proposition 4 suggests that we might see derivations up to order 8 appearing in the solutions. However Theorem 7 provides a much sharper result. Since |S| = 3 we see that each fi is the sum of a linear function and a derivation di of order at most 2. That is, fi (x) = di (x) + ci x,
x ∈ R,
where di ∈ D2 (R) for each i = 1, 2, 3. In fact, we can be more precise about the forms of the solutions. Indeed, substituting these forms into Eq. (12) we find that x7 [d1 (x2 ) + c1 x2 ] + x5 [d2 (x4 ) + c2 x4 ] + x4 [d3 (x5 ) + c3 x5 ] = 0,
x ∈ R.
The linear part of this equation gives c1 + c2 + c3 = 0, and the derivation part of the equation reduces to x3 d1 (x2 ) + xd2 (x4 ) + d3 (x5 ) = 0,
x ∈ R.
(13)
Since each di is a derivation of order at most 2, by Corollary 6 we have d2 (x4 ) = 6x2 d2 (x2 ) − 8x3 d2 (x), d3 (x5 ) = 10x3 d3 (x2 ) − 15x4 d3 (x). Substituting these into Eq. (13) and simplifying we arrive at 0 = x3 d1 (x2 ) + x[6x2 d2 (x2 ) − 8x3 d2 (x)] + [10x3 d3 (x2 ) − 15x4 d3 (x)] = x3 [d1 (x2 ) + 6d2 (x2 ) + 10d3 (x2 )] − x4 [8d2 (x) + 15d3 (x)] = (d1 + 6d2 + 10d3 )(x2 ) − x(8d2 + 15d3 )(x), where we have used the fact that R is an integral domain in the last step (plus the fact that f (0) = 0 for any additive f ). Thus we have (d1 + 6d2 + 10d3 )(x2 ) = x(8d2 + 15d3 )(x),
x ∈ R.
This last equation is of the form governed by Proposition 4, since both d1 + 6d2 + 10d3 and 8d2 + 15d3 are additive functions. Applying that result here with n = 1 we find that (d1 + 6d2 + 10d3 )(x) = D1 (x) + bx, (8d2 + 15d3 )(x) = −2D1 (x) − bx, where D1 ∈ D1 (R) (that is, a derivation of order one) and b ∈ F . From either of these equations we get b = 0, since all other terms are derivations. Hence
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15 1 d1 = −10d2 − 35 2 d3 and d2 = − 8 d3 − 4 D1 . Now we define D2 := d3 ∈ D2 (R) and write the solutions of Eq. (12) in the form
5 5 D2 (x) + D1 (x) + c1 x, 4 2 15 1 f2 (x) = − D2 (x) − D1 (x) + c2 x, 8 4 f3 (x) = D2 (x) − (c1 + c2 )x, f1 (x) =
with Di ∈ Di (R) and ci ∈ F for i = 1, 2. The following example deals with a non-homogeneous equation in less detail. Example 9. Suppose additive functions f1 , . . . , f11 : R → F satisfy x4 f1 (x3 ) + x4 f2 (x2 ) + x5 f3 (x) + x2 f4 (x3 ) + xf5 (x6 ) + x3 f6 (x3 ) + x2 f7 (x4 ) + x3 f8 (x2 ) + x4 f9 (x) + xf10 (x4 ) + f11 (x7 ) = 0 for all x ∈ R. This equation is not homogeneous. It splits into the three homogeneous parts x4 f1 (x3 ) + xf5 (x6 ) + f11 (x7 ) = 0, x4 f2 (x2 ) + x5 f3 (x) + x3 f6 (x3 ) + x2 f7 (x4 ) = 0, x2 f4 (x3 ) + x3 f8 (x2 ) + x4 f9 (x) + xf10 (x4 ) = 0, of respective degrees 7, 6, and 5. Applying Theorem 7 to each of these homogeneous equations in turn, we find that f1 , f5 , f11 are sums of linear functions and derivations of order at most two, while the other fi are sums of linear functions and derivations of order at most three. Further analysis of the kind done in the previous example provides more detail as to the exact forms of the solutions. Now we are ready for the main theorem of this section. Theorem 10. Let n ∈ N, and let mk ∈ N ∪ {0} and jk ∈ N for 1 ≤ k ≤ n. Suppose additive functions fk : R → F satisfy (3): n
xmk fk (xjk ) = 0,
x ∈ R.
k=1
If (mk , jk ) = (mp , jp ) for k = p ∈ {1, 2, . . . , n}, then each fk is the sum of a linear function and a derivation dk of order not more than n − 1. Moreover, for each different homogeneity degree in Eq. (3) the orders of the dk are reduced further. Specifically, suppose fk appears in lk different homogeneous parts of (3) and the numbers of terms in these parts are respectively
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lk t1 , . . . , tlk . (So i=1 ti ≤ n.) Then fk is the sum of a linear function and a derivation of order at most ok − 1 where ok = min{t1 , . . . , tlk }. Proof. Let H = {mk + jk : 1 ≤ k ≤ n} be the set of homogeneous degrees in (3). Applying Lemma 3, for each h ∈ H let xs gh,s (xh−s ) = 0 (14) s∈Sh
be the part of (3) that is homogeneous of degree h. Here Sh = {mk : mk + jk = h}, and for each pair (h, s) with h ∈ H and s ∈ Sh we have defined gh,s := fk so that s = mk and h − s = jk . By our hypothesis the pair (mk , jk ) determines k uniquely, hence the pair (h, s) determines k uniquely and thus gh,s is welldefined. Now, applying Proposition 4, we see that for each h ∈ H the solutions gh,s for s ∈ Sh are sums of linear functions and derivations of order at most h − 1. Therefore each fk is the sum of a linear function and a derivation dk of some order. What remains is a further discussion of the orders of the derivations dk . Suppose fk appears in exactly one of the homogeneous parts of Eq. (3), say the homogeneous part of degree h1 . If this homogeneous equation of degree h1 contains t1 terms then by Theorem 7 fk is the sum of a linear function and a derivation of degree at most t1 − 1. Now suppose fk appears in lk different homogeneous parts with respective numbers of terms t1 , . . . , tlk . Applying the argument in the previous paragraph to each of these lk homogeneous equations, we conclude that fk is the sum of a linear function and a derivation of degree at most tj − 1 for each 1 ≤ j ≤ lk . Hence fk is the sum of a linear function and a derivation of degree at most ok − 1 where ok is the minimum of the numbers t1 , . . . , tlk . Remark 11. The non-duplication condition (mk , jk ) = (mp , jp ) for k = p is necessary for the conclusion of the theorem to hold, as explained in the discussion near the beginning of this paper. We close this section with an example. Example 12. Suppose additive functions f1 , f2 , f3 : R → F satisfy x4 f1 (x3 ) + x4 f2 (x2 ) + x5 f3 (x) + x2 f1 (x3 ) + xf2 (x6 ) + x3 f3 (x3 ) + x2 f1 (x4 ) + x3 f2 (x2 ) + x4 f3 (x) + xf1 (x4 ) + f2 (x7 ) = 0
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for all x ∈ R. Here f1 and f2 appear in the 3-term homogeneous (degree 7) equation x4 f1 (x3 ) + xf2 (x6 ) + f2 (x7 ) = 0. Thus we have x ∈ R,
fi (x) = di (x) + ci x,
i = 1, 2,
with di ∈ D2 (R) and ci ∈ F such that c1 + 2c2 = 0. On the other hand, f3 only appears in 4-term homogeneous equations, so x ∈ R,
f3 (x) = d3 (x) + c3 x,
with d3 ∈ D3 (R). Nonetheless, further detailed analysis of the homogeneous equations of degrees 6 and 5 and the relationships between the derivations and constants in the solutions leads to the conclusion that the only solution is f1 = f2 = f3 = 0.
4. A slight improvement of a result from [5] For our work in Sect. 5 we need a slight generalization of Proposition 4 which we prove now. Theorem 13. Let n ∈ N, a ∈ F , and f1 , . . . , fn+1 : R → F be additive. The general solution of n
xj fn+1−j (xn+1−j ) + axn+1 = 0,
x ∈ R,
(15)
j=0
is given by fn+1−j (x) = (−1)j
j n+1−j+k dn−j+k (x) + bn+1−j x k
k=0
for j = 0, 1, . . . , n, where di ∈ Di (R) for each i (recall that D0 = {0}) and b1 , . . . , bn+1 ∈ F satisfy a+
n+1
bi = 0.
i=1
Proof. Define additive functions g1 , . . . , gn+1 by gk := fk
(2 ≤ k ≤ n + 1),
g1 (x) := f1 (x) + ax.
Then the gk satisfy n j=0
xj gn+1−j (xn+1−j ) = 0,
x ∈ R.
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By Proposition 4 there exist derivations di ∈ Di (R) for 0 ≤ i ≤ n and constants c1 , . . . , cn+1 ∈ F such that j n+1−j+k dn−j+k (x) + cn+1−j x gn+1−j (x) = (−1)j k k=0
and n+1
ci = 0.
i=1
Now fk = gk for 2 ≤ k ≤ n + 1 and n 1+k f1 (x) = (−1)n dk (x) + (c1 − a)x. k k=0
Defining bk := ck for 2 ≤ k ≤ n + 1 and b1 := c1 − a we have the claimed result.
5. A method for solving equations of the form (2) We turn now to functional Eq. (2). The question posed in Problem 1 seems to be rather difficult to answer. Certainly the condition of “non-duplication” is necessary. That is, if there are two terms pk (x)fk (qk (x)) and pi (x)fi (qi (x)) for k = i with pk = pi and qk = qi , then they can be combined into a single term pk (x)[(fk + fi )(qk (x))]. Therefore we cannot hope to discover the forms of the individual functions fk and fi , only the form of fk + fi . Yet it is also clear that some milder form of duplication is not fatal. This is demonstrated by Example 10 in [5], in which n = 4, p1 = p3 , p2 = p4 , q1 = q2 , and q3 = q4 , nevertheless all the functions fi are represented in terms of linear functions and derivations. In fact in that example f2 and f3 must be linear, f4 is a derivation of order one, and f1 is the sum of a linear function and a derivation of order one. On the other hand, the condition of “non-duplication” is not sufficient. In the following example we have both pk = pi and qk = qi for every pair (k, i). Yet (depending on R) there may exist solutions which are not sums of linear functions and derivations of any order. Example 14. Suppose additive functions f1 , f2 , f3 : R → F satisfy xf1 (x + 2) + (5x + 1)f2 (x + 1) + (x + 2)f3 (x) = 0,
x ∈ R.
The homogeneous part of degree 2 is xf1 (x) + 5xf2 (x) + xf3 (x) = 0, which yields (since fk (0) = 0 for each additive function) f1 = −5f2 − f3 .
(16)
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The homogeneous part of degree 1 is 2xf1 (1) + 5xf2 (1) + f2 (x) + 2f3 (x) = 0, which transforms by (16) into f2 (x) = −2f3 (x) + 5xf2 (1) + 2xf3 (1). Putting x = 1 here we see that f2 (1) = 0, so f2 (x) = −2[f3 (x) − xf3 (1)].
(17)
By (16) this means the solutions are given by (17) and f1 (x) = 9f3 (x) − 10xf3 (1) where f3 can be any additive function. If R = Q[i] for instance, we may choose f3 (a + bi) = a + b for all a, b ∈ Q. Then f3 is not the sum of a linear function and a derivation, as demonstrated in [5]. In spite of such phenomena, we do have a procedure for solving all equations of the form (2) with given polynomial functions pi : R → F and qi : R → R. (We restrict the range of qi to R so that fi ◦ qi is defined.) Since R is an integral domain of characteristic 0 it contains an isomorphic copy RZ of the integers Z. Therefore the field of fractions F contains an isomorphic copy FQ of the rationals Q. Recall that any additive function on a field containing Q has the property that f (rx) = rf (x) for r ∈ Q. This property extends, mutatis mutandem, to additive functions f : R → F in the sense that f (rx) = rf (x) for all x ∈ R, r ∈ FQ ∩ R. The method mi (i) Let pi : R → F be defined by pi (x) = j=1 aij xj for aij ∈ F , and let ti k qi : R → R be given by qi (x) = k=1 bik x for bik ∈ FQ ∩ R for all i, j, k. Then in detail (2) takes the form mi ti n
aij bik xj fi (xk ) = 0.
i=1 j=1 k=1
(ii) Rewrite the equation in the form M N
xs gN,s (xN −s ) = 0,
N =0 s=0
where gN,s :=
aij bik fi
(18)
and the sum is taken over i ∈ {1, . . . , n} such that j = s and k = N − s. (Note that when s = N we are only defining gN,N (1) and not a function
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on R.) Using the basic homogeneity lemma (Lemma 3), this equation separates into a system of homogeneous equations N
xs gN,s (xN −s ) = 0
s=0
for each N ∈ {0, . . . , M }. (iii) Using Theorem 13 (sharpened in the case of equations with “gaps” by Theorem 7), solve the homogeneous equation of each degree N for the functions gN,s as sums of linear functions and derivations of various orders. (iv) Solve for the functions fi using Eq. (18) from step (ii). These equations may be “underdetermined” in the sense that we may not be able to solve for all of the fi uniquely in terms of the gN,s . If that is the case then some of the fi will be undetermined additive functions. We have seen already in Example 14 an illustration of the phenomenon that some of the solution functions fi may not necessarily be sums of linear functions and derivations. Applying our method to the equation in Example 14 we have in step (ii) g2,1 (x) = (f1 + 5f2 + f3 )(x), g1,1 (1) = (2f1 + 5f2 )(1), g1,0 = f2 + 2f3 and g0,0 (1) = f2 (1). Then g2,1 = 0 implies f1 + 5f2 + f3 = 0; g1,0 (x) + xg1,1 (1) = 0 yields that g1,0 is linear, specifically (f2 + 2f3 )(x) = −x(2f1 + 5f2 )(1); and g0,0 (1) = 0 gives f2 (1) = 0. These equations lead precisely to the solutions we found in Example 14, in which f3 is an undetermined additive function. Now we illustrate the method with a positive example. Example 15. Suppose additive functions f1 , f2 , f3 : R → F satisfy (x2 + 1)f1 (x + 2) + (5x + 1)f2 (x2 + 3x) + (x2 + 6x)f3 (x2 + 4) = 0,
x ∈ R.
We will show that the only solution is f1 = f2 = f3 = 0. Lemma 3 dictates that the following equations hold, corresponding to homogeneous parts of degrees 4, 3, 2 respectively. x2 f3 (x2 ) = 0, x2 f1 (x) + 5xf2 (x2 ) + 6xf3 (x2 ) = 0, 2x2 f1 (1) + 15xf2 (x) + f2 (x2 ) + 4x2 f3 (1) = 0. We consider these equations one at a time. The equation homogeneous of degree 4 reduces immediately to f3 (x2 ) = 0. By Theorem 7 this means f3 is the sum of a linear function and a derivation of order 0, say f3 (x) = cx. Clearly c must vanish, so f3 = 0. Now the equation homogeneous of degree 3 reduces to xf1 (x)+5f2 (x2 ) = 0. Applying Theorem 7 again, we see that each of f1 and 5f2 is the sum of a linear
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function and a derivation of order at most 1, more precisely for some c1 ∈ F and d1 ∈ D1 (R) we have f1 (x) = −2d1 (x) + c1 x, 5f2 (x) = d1 (x) − c1 x.
(19)
Next, the equation homogeneous of degree 2 reduces to 1 0 = 2x2 c1 + 3x[d1 (x) − c1 x] + [d1 (x2 ) − c1 x2 ] 5 6 1 2 = − c1 x + 3xd1 (x) + · 2xd1 (x) 5 5 6 17 2 = − c1 x + xd1 (x). 5 5 Therefore both c1 and d1 vanish and the only solution is fi = 0 for i = 1, 2, 3.
6. An observation The “nestedness” of derivations – that is the property that Dn (R) ⊆ Dm (R) for n < m – has an interesting interpretation in the context of derivatives. Let us take R to be the integral domain C ∞ (R) of infinitely differentiable functions on R, and let d : R → R be the derivative operator defined by d(f ) = f , f ∈ C ∞ (R). Then d ∈ D1 (R). A characteristic functional equation for first-order derivations is d(x2 ) = 2xd(x). In the present context that means (f 2 ) = 2f f . A characteristic functional equation for a second-order derivation d2 is d2 (x3 ) = 3xd2 (x2 ) − 3x2 d2 (x). Since d ◦ d ∈ D2 (R) whenever d ∈ D1 (R), we have in this context (f 3 ) = 3f · (f 2 ) − 3f 2 · f . The fact that D1 (R) ⊆ D2 (R) means that the first derivative also satisfies the same functional equation. That is, it is also true that (f 3 ) = 3f · (f 2 ) − 3f 2 · f .
References [1] Acz´ el, J.: Some unsolved problems in the theory of functional equations. Arch. Math. 15, 435–444 (1964) [2] Boros, Z., Erdei, P.: A conditional equation for additive functions. Aequ. Math. 70, 309–313 (2005) [3] Boros, Z., Fechner, W.: An alternative equation for polynomial functions. Aequ. Math. 89, 17–22 (2015) [4] Ebanks, B.: Characterizing ring derivations of all orders via functional equations: results and open problems. Aequ. Math. 89, 685–718 (2015) [5] Ebanks, B.: Polynomially linked additive functions. Aequ. Math. 91, 317–330 (2017)
Author's personal copy Polynomially linked additive functions—II [6] Ebanks, B.: Linked pairs of additive functions. Aequ. Math. (to appear). https://doi. org/10.1007/s00010-017-0514-7 [7] Ebanks, B., Ng, C.T.: Homogeneous tri-additive forms and derivations. Linear Algebra Appl. 435, 2731–2755 (2011) [8] Ebanks, B., Riedel, T., Sahoo, P.K.: On the order of a derivation. Aequ. Math. 90, 335–340 (2016) [9] Halter-Koch, F.: Characterization of field homomorphisms and derivations by functional equations. Aequ. Math. 59, 298–305 (2000) [10] Halter-Koch, F.: A characterization of derivations by functional equations. Math. Pannon. 11, 187–190 (2000) [11] Halter-Koch, F., Reich, L.: Charakterisierung von Derivationen h¨ oherer Ordnung mittels ¨ Funktionalgleichungen. Osterreich. Akad. Wiss. Math.-Nat. Kl. Sitzungsber. II 207, 123– 131 (1998) [12] Jurkat, W.B.: On Cauchy’s functional equation. Proc. Am. Math. Soc. 16, 683–686 (1965) [13] Pl, K., Kurepa, S.: Some relations between additive functions I. Aequ. Math. 4, 163–175 (1970) [14] Pl, K., Kurepa, S.: Some relations between additive functions II. Aequ. Math. 6, 46–58 (1971) [15] Kurepa, S.: The Cauchy functional equation and scalar product in vector spaces. Glasnik Mat.-Fiz. Astronom. Ser. II Druˇstvo Mat. Fiz. Hrvatske 19, 23–36 (1964) [16] Kurepa, S.: Remarks on the Cauchy functional equation. Publ. Inst. Math.(Beograd)(N.S.) 5(19), 85–88 (1965) [17] Nishiyama, A., Horinouchi, S.: On a system of functional equations. Aequ. Math. 1, 1–5 (1968) [18] Reich, L.: Derivationen zweiter Ordnung als L¨ osungen von Funktionalgleichungen. Grazer Math. Ber. 337, 45–65 (1998) [19] Unger, J., Reich, L.: Derivationen h¨ oherer Ordnung als L¨ osungen von Funktionalgleichungen. Grazer Math. Ber. 336, 1–83 (1998) Bruce Ebanks Department of Mathematics University of Louisville Louisville KY 40292 USA e-mail:
[email protected] Received: October 21, 2017
