Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I've seen some previous questions that show that the derivative operator on the set of smooth functions can be given by the Leibniz rule and/or chain rule and some other axioms.

Is there a similar characterization of the derivative $\mathcal{C}^1(\mathbb{R}) \to \mathcal{C}(\mathbb{R})$? Or other interesting cases? (e.g. a differentiation operator on distributions of some suitable type)

share|improve this question

2 Answers 2

Derivative on polynomials can be characterized as a linear map which satisfies Leibniz rule, zero on constants and $1$ on the identity function. This extends it uniquely to rational functions. Now in any space where rational functions are dense, such an operator, if continuous, must be the derivative.

share|improve this answer
With respect to what topology should the rational functions be dense and the operator continuous? –  Linda Brown Westrick Feb 17 '14 at 22:16
With respect to ANY topology, but they should be dense and derivative continuous with respect to the SAME topology. –  Alexandre Eremenko Feb 18 '14 at 0:28
For example, the polynomials are dense in $C^1[a,b]$ for the topology of uniform convergence (Stone-Weierstrass). But the derivative operator is not continuous with respect to this topology. This example does not contradict your answer, nor does it directly address the OP. But the thing I am confused about is under what circumstances would one be able to use your answer to identify the derivative. –  Linda Brown Westrick Feb 18 '14 at 5:34
There are natural topologies on both spaces, namely compact convergence on $\mathcal{C}(\mathbb{R})$, and compact convergence of the function and its derivatives on $\mathcal{C}^1(\mathbb{R})$. For these both properties (density and continuity) hold, so $f\mapsto f'$ is the only continuous derivation $\mathcal{C}^1(\mathbb{R})\rightarrow \mathcal{C}(\mathbb{R})$ mapping $x$ to $1$. –  abx Feb 18 '14 at 7:37

The local question may be easier to answer and, on the other hand, should be more or less equivalent to the original one. Let $\mathcal{O}$ be the ring of germs of $C^1$-functions (say, at $0$) and $\frak{m}$ the maximal ideal. Let $\Bbbk:=\mathcal{O}/\frak{m}$ be the residue field. (Here, $\Bbbk=\mathbb{R}$.) Then it's standard commutative algebra that the differentiations on $\mathcal{O}$ are in a one-to-one correspondence with the dual space $\operatorname{Hom}_\Bbbk(\frak{m}/\frak{m}^2,\Bbbk)$. Now, I'm not quite sure about $\dim(\frak{m}/\frak{m}^2)$: this must involve some analysis, and functions like $x^3\sin(1/x)$ make me worry; I cannot see right away that it's in $\frak{m}^2$.

share|improve this answer
I think you mean $\mathfrak{m}/\mathfrak{m}^2$. –  Ketil Tveiten Feb 18 '14 at 8:43
Yes, sure. Edited. –  Alex Degtyarev Feb 18 '14 at 10:33
Your worry has some basis. If $x^2\sin(1/x)=\sum f_i(x)g_i(x)$ with $f_i,g_i\in{\frak m}$, then you divide both sides by $x^2$ and take the limit $x\to0$. On the right hand side, the limit exists; contrary to that on the left hand side. I guess, it is easy to show that $\dim{\frak m}/{\frak m^2}=\infty$. –  Sasha Anan'in Feb 18 '14 at 12:38
Yes, it does look like $\frak{m}$ is infinitely generated: one can consider $x^\alpha$ with appropriate rational $1<\alpha<2$. Unfortunately, infinitely generated is a bit too much: we cannot use Nakayama to easily conclude something about $\frak{m}/\frak{m}^2$ :( The bottom line is that $C^1$ (or, actually, $C^k$ with any given $k$) seem to be much worse than $C^\infty$. –  Alex Degtyarev Feb 18 '14 at 12:46
It is not in $\frak{m}$ as it is not in $\mathcal{O}$: it is not continuously differentiable! –  Alex Degtyarev Feb 18 '14 at 13:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.