Characterizations of derivations on spaces of smooth functions

We provide a list of equivalent conditions under which an additive operator acting on a space of smooth functions on a compact real interval is a multiple of the derivation.


Introduction
By R we denote the set of reals, Q are rationals, Z are integers, N = {1, 2, . ..} and N 0 = N∪{0}.If I ⊆ R is an interval and k ∈ N 0 , then C k (I) is the space o realvalued functions on I that are k-times continuously differentiable on the interior of I.If k = 0, then we write simply C(I).The space C k (I) is furnished with the standard pointwise algebraic operations and hence it is a real commutative algebra.
The following theorem describes derivations over fields of characteristic zero.
Theorem 1 ([12, Theorem 14.2.1]).Let K be a field of characteristic zero, F be a subfield of K, S be an algebraic base of K over F if it exists, and let S = ∅ otherwise.If f : F → K is a derivation, then, for every function u : S → K there exists a unique derivation g : K → K such that g = f on F and g = u on S.
From this theorem it follows in particular that nonzero derivations f : R → R exist.It is well known they are discontinuous and very irregular mappings.For an exhaustive discussion of the notion of the derivation and related functional equations the reader is referred to E. Gselmann [5,6], E. Gselmann, G. Kiss, C. Vincze [7] and references therein.Recently B. Ebanks [2,3] studied derivations and derivations of higher order on rings.The "model" example of a derivation is the operator of derivative on the space C k (I) for k > 0 .Indeed, if we define T : T is additive and it satisfies the Leibniz rule: (3) Crucial results about equation (3) on the space C k (I) are due to H. König and V. Milman.We refer the reader to their recent monograph [11].They studied several operator equations and inequalities that are related to the derivatives on the spaces of smooth functions.Later on, we will utilize their elegant result [11,Theorem 3.1] regarding (3).Briefly, if I is an open set, then the general solution of (3) for all f, g ∈ C k (I) is of the form ( 4) for some continuous functions c, d ∈ C(I), if k > 0, and ( 5) if k = 0 (in formulas ( 4) and ( 5) the convention that 0 • ln 0 = 0 is adopted).Note that no additivity is assumed.
There is a natural question to characterize real-to-real derivations among additive functions with the aid of a relation which is weaker than (2).In particular, the very first article published in the first volume of Aequationes Mathematicae by A. Nishiyama and S. Horinouchi [14] addresses this question.The authors studied the following relations, each of them is a direct consequence of (2) alone and together with (1) implies ( 2): where a = 1 and n, m are integers such that am = n = 0. Further similar results, as well as some generalizations, are due to W. Jurkat [8], Pl.Kannappan and S. Kurepa [9,10], S. Kurepa [13], among others.B. Ebanks [4] generalized and extended these results to arbitrary fields.A recent paper by M. Amou [1] provides some n-dimensional generalizations of the results of [8][9][10]13].
This paper provides versions of the above-mentioned results for operators T : C k (I) → C(I).Therefore, we seek conditions which are equivalent to (3).

Main results
Throughout this section let us fix k ∈ N 0 and an interval I ⊆ R. We will study conditions upon an additive operator T : C k (I) → C(I) which yield analogues to equations ( 6), ( 7) and (8).Therefore, we will focus on the following operator relations: Our first theorem is a simple observation that some reasonings concerning derivations from the real-to-real case can be extended to arbitrary commutative rings without substantial changes.We adopted parts of proof of [12,Theorem 14.3.1].
Theorem 2. Assume that Q is a commutative ring, P is a subring of Q and T : P → Q is an additive operator.Then, the following conditions are pairwise equivalent: Proof.(i) ⇒ (ii).Fix arbitrarily f, g ∈ P .By ( 9) we get Since T is additive, then Using ( 9) again, after reductions we obtain (3).
(ii) ⇒ (iii).If n = 1, then (11) reduces to an identity.Assume that (11) holds for some n ∈ N and all f ∈ P .Then, by (3) and the induction hypothesis we have The next corollary will be utilized later on.
Corollary 1. Assume that T : C k (I) → C(I) is an additive operator.Then, the following conditions are pairwise equivalent: Our next result characterizes the Leibniz rule (3) on a domain restricted to functions separated from zero.Thus, we can consider conditions (10) and (11) for negative n, which involve the function 1/f .The situation is a bit more complicated, but Theorem 3 below has a mainly technical role.Theorem 3. Assume that T : C k (I) → C(I) is an additive operator and ε 1 ∈ (0, 1), ε 2 ∈ (0, 1) and c ∈ (1, +∞] are constants.Consider the following conditions: Proof.(i) ⇒ (ii).First, note that by applying (10) for f = 1 and using the rational homogeneity of T we get that T vanishes on each constant function equal to a rational number.Observe that for arbitrary rational δ > 0 (which will be chosen later) the identity √ ε 1 , then we will find some rational δ > 0 such that Using (i) three times together with (12) and the additivity of T we obtain (ii) ⇔ (iii).Analogously as in Theorem 2 for f > ε 2 and g > ε 2 .
If we assume additionally that interval I is compact, then the situation clarifies considerably.
Theorem 4. Assume that I is compact and T : C k (I) → C(I) is an additive operator.Then, the following conditions are pairwise equivalent: Proof.This statement is a consequence of Corollary 1 and Theorem 3. Since I is compact, then f attains its global extrema.Thus, we will find some rational r, q ∈ Q such that 1/2 < rf + q < 2.Moreover, as it was already observed in the proof of Theorem 3, each of the conditions of Theorem 4 implies that T (1) = 0 and then T vanishes on constant function equal to a rational number.Consequently, we have T (rf + q) = rT (f ) + T (q) = rT (f ) and therefore Theorem 3 applies for the conditions (ii), (iv), (v) and (vii) with appropriately chosen ε 1 and ε 2 .The remaining conditions are equivalent by Corollary 1.Therefore, we are done if we prove for example the implication (iv) ⇒ (iii).Fix f ∈ C k (I) arbitrarily and choose r, q ∈ Q such that 1/2 < rf + q < 2. By (iv) we get T ((rf + q) 2 ) = 2(rf + q)T (rf + q).Then using additivity we obtain and after reduction T (f 2 ) + 0 = 2f T (f ) i.e. condition (iii).
One can join Corollary 1 and Theorem 4 with the mentioned result of H. König and V. Milman to obtain a corollary.
Corollary 2. Under assumptions of Corollary 1 or Theorem 4, if k > 0, then each of the conditions listed there is equivalent to the following one: (x) there exists some and if k = 0, then T = 0 is the only additive operator that fulfils any of the equivalent conditions.
Proof.Consider f (x) = x on I and denote d := T (f ) ∈ C(I).Next, note that by [11,Theorem 3.1] the formulas (4) and ( 5), respectively hold on the interior of I with some c, d ∈ C(intI).The additivity of T implies that c = 0. Therefore d is a continuous extension of d to the whole interval I.

Final remarks
Remark.Inequalities between f , g and constants ε 1 and ε 2 in Theorem 3 are not optimal.This however was not our goal since the role of this result is auxiliary only.Similarly, inequality f > 0 in some of the conditions of Theorem 4 can be equivalently replaced by an estimate from above or from below by any other fixed constant.
Moreover, in the proof of Theorem 4 we showed more than is stated.Namely, it is equivalently enough to assume instead f > 0 that f is bilaterally bounded by two rational numbers, like 1/2 and 2. However, since this generalization is apparent only and easy, we do not include it in the formulation of the theorem.
Example 1. Assume that ϕ : (1, ∞) → R is a smooth mapping that satisfies equation ( 13) Such mappings exist in abundance.In fact, every map ϕ 0 defined on (1, 2] can be uniquely extended to a solution of (13).Next, let d : (e, ∞) → R be defined as We see easily that d(x 2 ) = 2xd(x), x ∈ (e, ∞) and d(xy) = xd(y) + yd(x) in general, unless ϕ is additive.Define T : C 1 ((e, ∞)) → C((e, ∞)) as follows: One can see that T satisfies (9) for all f, g ∈ C((e, ∞)), but fails to satisfy the Leibniz rule (3).Thus, the assumption of additivity in our all results is essential.Observe also that T has the property that it vanishes on constant functions equal to a rational.This fact, as a consequence of additivity, was frequently used in the proofs of our Theorems 3 and 4. Therefore, the additivity assumption cannot be relaxed to this property.
Example 2. Assume that I is an interval and T is given by the formula Then T satisfies (3) for all f, g ∈ C 2 (I) such that f > 0 and g > 0. This observation is a particular case of the second part of [11,Corollary 3.4].Clearly, T is not additive.Moreover, T cannot be extended in such a way it satisfies (3) on the whole space C 2 (I).
The following examples show that if the domain of operator T is changed, then the conditions discussed in our results are no longer equivalent and various situations are possible.
Then T is not additive, it satisfies (3) and has no extension to a solution of (3) to C k (R).
Note that S is closed under multiplication but it is not a linear space.Next, let a double sequence ϕ on Z of natural numbers be defined as follows: ϕ(0) = 0, ϕ(k) is arbitrary but = k if k is odd, and if k = 2 n • m with some n ∈ N and odd m ∈ Z, then ϕ(k) := 2 n 2 −n 2 • m n • ϕ(m).Note that we have Define T : S → C((0, ∞)) by ( 16) T (f )(x) := k • x ϕ(k) , x ∈ (0, ∞) if f (x) = x k for x ∈ (0, ∞).One can see that if f is of this form, then by ( 15) for all x ∈ (0, ∞), i.e.T satisfies (9).Moreover, one can see that (10) is equivalent to the equality Therefore, we can construct a sequence ϕ such that T defined by ( 16) satisfies (10) as well as another sequence ϕ ′ for which T does not satisfy (10).Finally, (3) is not true on S. Indeed, note that if (3) is satisfied by T given by ( 16), then: ϕ(k + l) = ϕ(k) + l = ϕ(l) + k, k, l ∈ Z, k = 0, l = 0, which does not hold.

Example 4 .
Let P [x] be the space of all real polynomials of variable x.By deg(f ) we denote the degree of a polynomial f ∈ P [x].Define T : P [x] → P [x] by