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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)

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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.

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    $\begingroup$ With respect to what topology should the rational functions be dense and the operator continuous? $\endgroup$
    – Linda Brown Westrick
    Commented Feb 17, 2014 at 22:16
  • $\begingroup$ With respect to ANY topology, but they should be dense and derivative continuous with respect to the SAME topology. $\endgroup$
    – Alexandre Eremenko
    Commented Feb 18, 2014 at 0:28
  • $\begingroup$ 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. $\endgroup$
    – Linda Brown Westrick
    Commented Feb 18, 2014 at 5:34
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    $\begingroup$ 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 $1$. $\endgroup$
    – abx
    Commented Feb 18, 2014 at 7:37
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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 $\frak{m}^2$.

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  • $\begingroup$ I think you mean $\mathfrak{m}/\mathfrak{m}^2$. $\endgroup$
    – Ketil Tveiten
    Commented Feb 18, 2014 at 8:43
  • $\begingroup$ Yes, sure. Edited. $\endgroup$
    – Alex Degtyarev
    Commented Feb 18, 2014 at 10:33
  • $\begingroup$ 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 $\dim{\frak m}/{\frak m^2}=\infty$. $\endgroup$
    – Sasha Anan'in
    Commented Feb 18, 2014 at 12:38
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    $\begingroup$ 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 $C^\infty$. $\endgroup$
    – Alex Degtyarev
    Commented Feb 18, 2014 at 12:46
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    $\begingroup$ It is not in $\frak{m}$ as it is not in $\mathcal{O}$: it is not continuously differentiable! $\endgroup$
    – Alex Degtyarev
    Commented Feb 18, 2014 at 13:24

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