Characterizations of derivations on spaces of smooth functions

Abstract

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.

1 Introduction

By $\mathbb {R}$ we denote the set of reals, $\mathbb {Q}$ are rationals, $\mathbb {Z}$ are integers, $\mathbb {N}=\{1, 2, \ldots \}$ and $\mathbb {N}_0=\mathbb {N}\cup \{0 \}$. If $I\subseteq \mathbb {R}$ is an interval and $k\in \mathbb {N}_0$, then $C^k(I)$ is the space of real-valued 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.

Definition

(e.g. Kuczma [12, page 391]) Assume that Q is a commutative ring and P is a subring of Q. A function $f:P \rightarrow Q$ is called derivation if it is additive:

$$\begin{aligned} f(x+y) = f(x) + f(y), \quad x, y \in P \end{aligned}$$

(1)

and it satisfies the Leibniz rule:

$$\begin{aligned} f(xy) = xf(y) + yf(x), \quad x, y \in P. \end{aligned}$$

(2)

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 $S = \varnothing$ otherwise. If $f :F \rightarrow K$ is a derivation, then, for every function $u:S \rightarrow K$ there exists a unique derivation $g:K \rightarrow K$ such that $g=f$ on F and $g=u$ on S.

From this theorem it follows in particular that nonzero derivations $f:\mathbb {R}\rightarrow \mathbb {R}$ exist. It is well known they are discontinuous and very irregular mappings. For an exhaustive discussion of the notion of derivation and related functional equations the reader is referred to Gselmann [5, 6], Gselmann, Kiss, Vincze [7] and the references therein. Recently 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 $T:C^k(I)\rightarrow C(I)$ as $T(f) = f'$ for $f \in C^k(I)$, then clearly $C^k(I)$ is a subring of C(I), T is additive and it satisfies the Leibniz rule:

$$\begin{aligned} T(f\cdot g) = f\cdot T(g) + g \cdot T(f). \end{aligned}$$

(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 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 $f, g \in C^k(I)$ is of the form

$$\begin{aligned} T(f) = c \cdot f\cdot \ln |f| + d \cdot f',\quad f \in C^k(I) \end{aligned}$$

(4)

for some continuous functions $c, d\in C(I)$, if $k>0$, and

$$\begin{aligned} T(f) = c \cdot f\cdot \ln |f| ,\quad f \in C^k(I) \end{aligned}$$

(5)

if $k=0$ (in formulas (4) and (5) the convention that $0\cdot \ln 0 = 0$ is adopted). Note that no additivity is assumed.

It 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 Nishiyama and Horinouchi [14] addresses this question. The authors studied the following relations, each of which is a direct consequence of (2) alone and together with (1) implies (2):

$$\begin{aligned} f(x^2)= & {} 2xf(x), \quad x \in \mathbb {R}, \end{aligned}$$

(6)

$$\begin{aligned} f(x^{-1})= & {} -x^{-2}f(x), \quad x \in \mathbb {R},\, x \ne 0, \end{aligned}$$

(7)

and

$$\begin{aligned} f(x^n) = ax^{n-m}f(x^m), \quad x \in \mathbb {R}, \, x \ne 0, \end{aligned}$$

(8)

where $a\ne 1$ and n, m are integers such that $am = n\ne 0$. Further similar results, as well as some generalizations, are due to Jurkat [8], Kannappan and Kurepa [9, 10], Kurepa [13], among others. Ebanks [4] generalized and extended these results to arbitrary fields. A recent paper by 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 $T:C^k(I)\rightarrow C(I)$. Therefore, we seek conditions which are equivalent to (3).

2 Main results

Throughout this section let us fix $k\in \mathbb {N}_0$ and an interval $I\subseteq \mathbb {R}$. We will study conditions upon an additive operator $T:C^k(I)\rightarrow C(I)$ which yield analogues to Eqs. (6), (714) and (8). Therefore, we will focus on the following operator relations:

$$\begin{aligned} T(f^2) = 2f \cdot T(f), \end{aligned}$$

(9)

$$\begin{aligned} T(f) = -f^{2}\cdot T \left( \frac{1}{f} \right) , \end{aligned}$$

(10)

$$\begin{aligned} T(f^{n}) = nf^{n-1}\cdot T(f). \end{aligned}$$

(11)

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 the proof of [12, Theorem 14.3.1].

Theorem 2

Assume that Q is a commutative ring, P is a subring of Q and $T:P\rightarrow Q$ is an additive operator. Then, the following conditions are pairwise equivalent:

• (i) T satisfies $T(f^2) = 2f \cdot T(f)$ for all $f \in P$,

• (ii) T satisfies $T(f\cdot g) = f\cdot T(g) + g \cdot T(f)$ for all $f, g \in P$,

• (iii) T satisfies $T(f^{n}) = nf^{n-1}\cdot T(f)$ for all $f \in P$ and $n \in \mathbb {N}$.

Proof

$\mathrm{(i)} \Rightarrow \mathrm{(ii)}$. Fix arbitrarily $f, g \in P$. By (9) we get

$$\begin{aligned} T((f+g)^2)= 2(f+g)\cdot T(f+g). \end{aligned}$$

Since T is additive,

$$\begin{aligned} T(f^2) + 2T(f\cdot g) + T(g^2) = 2f\cdot T(f) + 2g \cdot T(f) + 2f\cdot T(g) + 2g\cdot T(g). \end{aligned}$$

Using (9) again, after reductions we obtain (3).

$\mathrm{(ii)} \Rightarrow \mathrm{(iii)}$. If $n=1$, then (11) reduces to an identity. Assume that (11) holds for some $n \in \mathbb {N}$ and all $f \in P$. Then, by (3) and the induction hypothesis we have

$$\begin{aligned} T(f^{n+1})&= T(f^n \cdot f) = f^n \cdot T(f) + f \cdot T(f^n)\\&=f^n\cdot T(f) + n f^{n-1+1} \cdot T(f) = (n+1)f^n\cdot T(f). \end{aligned}$$

$\mathrm{(iii)} \Rightarrow \mathrm{(i)}$. Take $n=2$. $\square$

The next corollary will be utilized later on.

Corollary 1

Assume that $T:C^k(I)\rightarrow C(I)$ is an additive operator. Then, the following conditions are pairwise equivalent:

• (i) T satisfies $T(f^2) = 2f \cdot T(f)$ for all $f \in C^k(I)$,

• (ii) T satisfies $T(f\cdot g) = f\cdot T(g) + g \cdot T(f)$ for all $f, g \in C^k(I)$,

• (iii) T satisfies $T(f^{n}) = nf^{n-1}\cdot T(f)$ for all $f \in C^k(I)$ and $n \in \mathbb {N}$.

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 $T:C^k(I)\rightarrow C(I)$ is an additive operator and $\varepsilon _1\in (0,1)$, $\varepsilon _2\in (0,1)$ and $c\in (1, +\infty ]$ are constants. Consider the following conditions:

• (i) T satisfies $T(f) = -f^{2}\cdot T\left( \frac{1}{f}\right)$ for all $f \in C^k(I)$, $c>f>\varepsilon _1$,

• (ii) T satisfies $T(f^2) = 2f \cdot T(f)$ for all $f \in C^k(I)$, $f>\varepsilon _2$,

• (iii) T satisfies $T(f\cdot g) = f\cdot T(g) + g \cdot T(f)$ for all $f, g \in C^k(I)$, $f>\varepsilon _2$, $g>\varepsilon _2$,

• (iv) T satisfies $T(f^{n}) = nf^{n-1}\cdot T(f)$ for all $n \in \mathbb {Z}$ and all $f \in C^k(I)$ such that $\varepsilon _2<f<1/\varepsilon _2$, and $f^{n-1}>\varepsilon _2$ if $n>0$ and $f^{n+1}>\varepsilon _2$ if $n<0$.

Then: (i) with $c=+\infty$ implies (ii) with $\varepsilon _2> \sqrt{\varepsilon _1}$, (ii) and (iii) are equivalent, (iii) implies (iv), (iv) implies (i) with $\varepsilon _1=\varepsilon _2$ and $c= 1/\varepsilon _2$.

Proof

$\mathrm{(i)}\Rightarrow \mathrm{(ii)}$. 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 an arbitrary rational $\delta >0$ (which will be chosen later) the identity

$$\begin{aligned} \frac{1}{f^2-\delta ^2} = \frac{1}{2\delta }\left( \frac{1}{f-\delta }- \frac{1}{f+\delta } \right) \end{aligned}$$

(12)

holds for $f \in C^k(I)$ such that $f>\delta$. Next, if $\varepsilon _1>0$ is given and $\varepsilon _2> \sqrt{\varepsilon _1}$, then we will find some rational $\delta >0$ such that $\varepsilon _2>\varepsilon _1+\delta$ and $\varepsilon _2^2>\varepsilon _1+\delta ^2$. Consequently, if $f \in C^k(I)$ and $f>\varepsilon _2$, then $f\pm \delta >\varepsilon _1$ and $f^2-\delta ^2 >\varepsilon _1$. Using (i) three times together with (12) and the additivity of T we obtain

$$\begin{aligned} T(f^2)&=T(f^2-\delta ^2)=-(f^2-\delta ^2)^2T\left( \frac{1}{f^2-\delta ^2}\right) \\&=-\frac{1}{2\delta }(f^2-\delta ^2)^2T\left( \frac{1}{f-\delta }-\frac{1}{f+\delta }\right) \\&=-\frac{1}{2\delta }(f+\delta )^2(f-\delta )^2\left[ T\left( \frac{1}{f-\delta }\right) - T\left( \frac{1}{f+\delta }\right) \right] \\&=\frac{1}{2\delta }\left[ (f+\delta )^2T(f-\delta ) -(f-\delta )^2T(f+\delta )\right] = 2fT(f). \end{aligned}$$

$\mathrm{(ii)}\Leftrightarrow \mathrm{(iii)}$. Analogously as in Theorem 2 for $f>\varepsilon _2$ and $g>\varepsilon _2$. $\mathrm{(iii)}\Rightarrow \mathrm{(iv)}$. If $n=1$, then (11) is trivially satisfied. Assume that f, n and $\varepsilon _2$ satisfy the assumptions of (iv). For $n>1$ we proceed like in Theorem 2. If $n=0$, then (iv) reduces to $T(1)=0$, which follows from (iii). If $n=-1$, then for $1/\varepsilon _2>f>\varepsilon _2$ we have

$$\begin{aligned} 0=T(1) = T\left( f\cdot \frac{1}{f}\right) = \frac{1}{f}\cdot T(f) + f\cdot T\left( \frac{1}{f}\right) . \end{aligned}$$

Assume that $n<-1$. By downward induction, one can check that for $f^{n+1}>\varepsilon _2$ we have from (3)

$$\begin{aligned} T(f^n)&= T\left( f^{n+1}\cdot \frac{1}{f}\right) = f^{n+1}\cdot T\left( \frac{1}{f}\right) + \frac{1}{f}\cdot T\left( f^{n+1}\right) \\&=-f^{n+1}\cdot f^{-2} T(f) + \frac{n+1}{f}\cdot f^n\cdot T\left( f\right) = nf^{n-1}T(f). \end{aligned}$$

$(iv)\Rightarrow (i)$. Take $n=-1$. $\square$

If we assume additionally that interval I is compact, then the situation clarifies considerably.

Theorem 4

Assume that I is compact and $T:C^k(I)\rightarrow C(I)$ is an additive operator. Then, the following conditions are pairwise equivalent:

• (i) T satisfies $T(f\cdot g) = f\cdot T(g) + g \cdot T(f)$ for all $f, g \in C^k(I)$,

• (ii) T satisfies $T(f\cdot g) = f\cdot T(g) + g \cdot T(f)$ for all $f, g \in C^k(I), f>0$, $g>0$,

• (iii) T satisfies $T(f^2) = 2f \cdot T(f)$ for all $f \in C^k(I)$,

• (iv) T satisfies $T(f^2) = 2f \cdot T(f)$ for all $f \in C^k(I)$, $f>0$,

• (v) T satisfies $T(f) = -f^{2}\cdot T\left( \frac{1}{f}\right)$ for all $f \in C^k(I)$, $f>0$,

• (vi) T satisfies $T(f^{n}) = nf^{n-1}\cdot T(f)$ for all $f \in C^k(I)$ and $n \in \mathbb {N}$,

• (vii) T satisfies $T(f^{n}) = nf^{n-1}\cdot T(f)$ for all $f \in C^k(I)$, $f>0$ and $n \in \mathbb {N}$.

Proof

This statement is a consequence of Corollary 1 and Theorem 3. Since I is compact, f attains its global extrema. Thus, we will find some rational $r, q \in \mathbb {Q}$ 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 functions equal to a rational number. Consequently, we have $T(rf+q)=rT(f) + T(q) = rT(f)$ and therefore Theorem 3 applies to the conditions (ii), (iv), (v) and (vii) with appropriately chosen $\varepsilon _1$ and $\varepsilon _2$. The remaining conditions are equivalent by Corollary 1. Therefore, we are done if we prove for example the implication $\mathrm{(iv)}\Rightarrow \mathrm{(iii)}$.

Fix $f\in C^k(I)$ arbitrarily and choose $r, q \in \mathbb {Q}$ such that $1/2<rf+q<2$. By (iv) we get

$$\begin{aligned} T((rf+q)^2)= 2(rf+q) T(rf+q). \end{aligned}$$

Then using additivity we obtain

$$\begin{aligned} r^2T(f^2) + 2rqT(f) + T(q^2) = 2r^2fT(f) + 2rq T(f) \end{aligned}$$

and after reduction

$$\begin{aligned} T(f^2) + 0 = 2fT(f) \end{aligned}$$

i.e. condition (iii). $\square$

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 the 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 $d\in C(I)$ such that $T(f) = d\cdot f'$ for all $f \in C^k(I)$

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 ${\tilde{d}}:= T(f)\in C(I)$. Next, note that by [11, Theorem 3.1] the formulas (4) and (5), respectively hold on the interior of I with some $c, d \in C(\textrm{int} I)$. The additivity of T implies that $c=0$. Therefore ${\tilde{d}}$ is a continuous extension of d to the whole interval I. $\square$

3 Final remarks

Remark

The inequalities between f, g and constants $\varepsilon _1$ and $\varepsilon _2$ in Theorem 3 are not optimal. This however was not our goal since the role of this result is auxiliary only. Similarly, the 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 of $f>0$, that f is bilaterally bounded by two rational numbers, like 1/2 and 2. However, since this generalization is only apparent and easy, we do not include it in the formulation of the theorem.

Example 1

Assume that $\varphi :(1, \infty ) \rightarrow \mathbb {R}$ is a smooth mapping that satisfies the equation

$$\begin{aligned} \varphi (2x) = 2 \varphi (x), \quad x \in (1, \infty ). \end{aligned}$$

(13)

Such mappings exist in abundance. In fact, every map $\varphi _0$ defined on (1, 2] can be uniquely extended to a solution of (13). Next, let $d:(e, \infty ) \rightarrow \mathbb {R}$ be defined as

$$\begin{aligned} d(x) = x \cdot \varphi (\ln x), \quad x \in (e, \infty ). \end{aligned}$$

It is easy to see that

$$\begin{aligned} d(x^2) = 2x d(x), \quad x \in (e, \infty ) \end{aligned}$$

and

$$\begin{aligned} d(xy) \ne x d(y) + yd(x) \end{aligned}$$

in general, unless $\varphi$ is additive. Define $T:C^1((e, \infty ))\rightarrow C((e, \infty ))$ as follows:

$$\begin{aligned} T(f) = d \circ f , \quad f \in C^1((e, \infty )). \end{aligned}$$

One can see that T satisfies (9) for all $f, g \in C((e, \infty ))$, but fails to satisfy the Leibniz rule (3). Thus, the assumption of additivity in all our 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

$$\begin{aligned} T(f) = f'' - \frac{(f')^2}{f}, \quad f \in C^2(I), \, f>0. \end{aligned}$$

Then T satisfies (3) for all $f, g \in C^2(I)$ 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 that 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.

Example 3

Let $\mathcal {S}$ be the space of all functions $f \in C^1((0, \infty ))$ which satisfy the functional equation

$$\begin{aligned} f(x+1) = 2f(x), \quad x \in (0, \infty ). \end{aligned}$$

(14)

Note that $\mathcal {S}$ is not closed under multiplication. Moreover, each function $f_0: (0, 1] \rightarrow \mathbb {R}$ can be uniquely extended to a solution of (714). Therefore, $\mathcal {S}$ is an infinite-dimensional subspace of $C^1((0, \infty ))$. Define $T:C^1((0, \infty )) \rightarrow C^1((0, \infty ))$ by the formula

$$\begin{aligned} T(f)(x) = f(x+1), \quad f \in C^1((0, \infty )), \, x \in (0, \infty ). \end{aligned}$$

It is easy to check that T is additive and satisfies (3) for $f, g \in \mathcal {S}$. Thus, there are more solutions of (3) if the domain of T is restricted to a particular subspace of $C^k(I)$.

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 $f\in P[x]$. Define $T:P[x] \rightarrow P[x]$ by

$$\begin{aligned} T(f) = \deg (f) \cdot f, \quad f \in P[x]. \end{aligned}$$

Then T is not additive, it satisfies (3) and there exists no extension of T to the whole space $C^k(\mathbb {R})$ which is a solution of (3).

Example 5

Let

$$\begin{aligned} \mathcal {S}:=\{f:(0, \infty )\rightarrow \mathbb {R}: f(x) = x^k \text { for some } k \in \mathbb {Z}\text { and } x \in (0, \infty ) \}. \end{aligned}$$

Note that $\mathcal {S}$ is closed under multiplication but it is not a linear space. Next, let a double sequence $\varphi$ on $\mathbb {Z}$ of natural numbers be defined as follows: $\varphi (0)=0$, $\varphi (k)$ is arbitrary but $\ne k$ if k is odd, and if $k=2^n \cdot m$ with some $n \in \mathbb {N}$ and odd $m\in \mathbb {Z}$, then

$$\begin{aligned} \varphi (k) := 2^{\frac{n^2-n}{2}}\cdot m^n \cdot \varphi (m). \end{aligned}$$

Note that we have

$$\begin{aligned} \varphi (2k)&= \varphi (2^{n+1}\cdot m) = 2^{\frac{n^2+n}{2}} \cdot m^{n+1} \cdot \varphi (m) \nonumber \\ {}&= 2^n \cdot m \cdot 2^{\frac{n^2-n}{2}} \cdot m^n \cdot \varphi (m) = k\cdot \varphi (k), \quad k \in \mathbb {Z}. \end{aligned}$$

(15)

Define $T:\mathcal {S}\rightarrow C((0, \infty ))$ by

$$\begin{aligned} T(f)(x) := k \cdot x^{\varphi (k)} , \quad x \in (0, \infty ) \end{aligned}$$

(16)

if $f(x) = x^k$ for $x \in (0, \infty )$. One can see that if f is of this form, then by (15)

$$\begin{aligned} T(f^2)(x) = 2k \cdot x^{\varphi (2k)} = 2k \cdot x^{k\cdot \varphi (k)} = 2f (x) T(f)(x) \end{aligned}$$

for all $x \in (0, \infty )$, i.e. T satisfies (9).

Moreover, one can see that (10) is equivalent to the equality

$$\begin{aligned} \varphi (k) - \varphi (-k) = 2k, \quad k \in \mathbb {Z}, \, k \ne 0. \end{aligned}$$

Therefore, we can construct a sequence $\varphi$ such that T defined by (16) satisfies (10) as well as another sequence $\varphi '$ for which T does not satisfy (10). Finally, (3) is not true on $\mathcal {S}$. Indeed, note that if (3) is satisfied by T given by (16), then:

$$\begin{aligned} \varphi (k+l) = \varphi (k) + l = \varphi (l) + k, \quad k, l \in \mathbb {Z}, k \ne 0, l \ne 0, \end{aligned}$$

which does not hold.