Let $L: C^\infty(\mathbb{R}) \to C^\infty(\mathbb{R})$ be a linear operator which satisfies:

$L(1) = 0$

$L(x) = 1$

$L(f \cdot g) = f \cdot L(g) + g \cdot L(f)$

Is $L$ necessarily the derivative? Maybe if I throw in some kind of continuity assumption on $L$? If it helps you can throw the "chain rule" into the list of properties.

I can see that $L$ must send any polynomial function to it's derivative. I want to say "just approximate any function by polynomials, and pass to a limit", but I see two complications: First $\mathbb{R}$ is not compact, so such an approximation scheme is not likely to fly. Maybe convolution with smooth cutoff functions could help me here. Even if I could rig up something I am concerned that if polynomials $p_n$ converge to $f$, I may not have $p_n'$ converging to $f'$. My Analysis skills are really not too hot so I would like some help.

I am interested in this question because it is a slight variant of a characterization given here:

Why do we teach calculus students the derivative as a limit?

I am not sure whether or not those properties characterize the derivative, and they are closely related to mine.

If these properties do not characterize the derivative operator, I would like to see another operator which satisfies these properties. Can you really write one down or do you need the axiom of choice? I feel that any counterexample would have to be very weird.

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    $\begingroup$ If $L$ is a derivation on $C^\infty(\mathbb R)$ it is a Lie derivative (wrt a vector field). See Conlon's "Differentiable Manifolds" book, pg 67. $\endgroup$ Nov 4, 2010 at 3:38
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    $\begingroup$ This question is closely related to the question mathoverflow.net/questions/25054/… and take a look at the string of comments I made on my own answer there for a discussion of how the product rule (in settings that include more rings than just the smooth functions on R) can be derived from more basic conditions. $\endgroup$
    – KConrad
    Nov 4, 2010 at 3:59
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    $\begingroup$ It's kind of dense, but you might be interested in this talk (I don't think a preprint is available yet): fields.utoronto.ca/audio/10-11/analysis/koenig $\endgroup$ Nov 4, 2010 at 3:59
  • $\begingroup$ I wonder what nice sets of conditions that include shift-equivariance are satisfied only by differentiation? $\endgroup$ Sep 30, 2014 at 22:27
  • $\begingroup$ See also math.stackexchange.com/questions/1682284 (to avoid using $C^{\infty}(\Bbb R)$... which requires differentiation to be defined). $\endgroup$
    – Watson
    Jan 19, 2022 at 10:26

3 Answers 3


Yeah, these force it to be ordinary differentiation. We have to show that for each fixed $x_0 \in \mathbb{R}$, the composite

$$C^\infty(\mathbb{R}) \stackrel{L}{\to} C^\infty(\mathbb{R}) \stackrel{ev_{x_0}}{\to} \mathbb{R}$$

is just the derivative at $x_0$. For each $f \in C^\infty(\mathbb{R})$, there is a $C^\infty$ function $g$ such that

$$f(x) = f(x_0) + f'(x_0)(x - x_0) + (x - x_0)^2g(x)$$

and so $(ev_{x_0} L)(f) = ev_{x_0}(f'(x_0) + 2(x - x_0)g(x) + (x - x_0)^2 L(g)(x))$ by the properties you listed. Of course evaluation at $x_0$ kills the last two terms and one is left with $f'(x_0)$, as desired.

  • $\begingroup$ In case this helps: for every $f \in C^\infty(\mathbb{R})$ there exists a (unique) $g \in C^\infty(\mathbb{R})$ such that $f(x) = f(x_0) + (x - x_0)g(x)$. The statement I used is derived by iterating this. $\endgroup$
    – Todd Trimble
    Nov 4, 2010 at 4:42
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    $\begingroup$ Short and sweet. And with no "nitty gritty" analysis either! Thank you so much! $\endgroup$ Nov 4, 2010 at 6:23
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    $\begingroup$ You're welcome -- glad I could be of help! $\endgroup$
    – Todd Trimble
    Nov 4, 2010 at 10:13
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    $\begingroup$ There is "nitty-gritty" analysis when you prove Todd's comment on his own answer (that you can factor x - x_0 out of f(x) - f(x_0) in the land of smooth functions on R). Although if you use the fundamental theorem of calculus in a suitable way it is not too painful. $\endgroup$
    – KConrad
    Nov 5, 2010 at 19:32
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    $\begingroup$ Spilling the beans here: en.wikipedia.org/wiki/Hadamard%27s_lemma $\endgroup$
    – Todd Trimble
    Nov 5, 2010 at 20:03

Here is a slightly more general take on this question. First notice that your condition $L(1) = 0$ is redundant. That is because if you take $f = g = 1$ in your third condition, you get $L(1) = L(1) + L(1)$. It is a theorem that goes back as far as I know to Chevalley (see around page 76 in his Theory of Lie Groups) that if $M$ is a $C^\infty$ manifold, then any linear map $L$ of $C^\infty(M)$ to itself that satisfies the derivation condition (your third condition) is a smooth vector field. This means that in local coordinate $(x_1,\ldots,x_n)$ it has the form $L(f) = \sum_i h_i {\partial f\over \partial x_i}$ where the $h_i$ are smooth functions. Moreover, if we compute $L(x_j)$ using this formula we see that $h_j$ is just $L(x_j)$. So in particular if a linear map $L$ of $C^\infty (R^n)$ to itself satisfies the derivation condition and $L(x_i) = \delta_{ij}$, then $L = {\partial \over \partial x_j}$.

  • $\begingroup$ Very nice answer. Thanks for telling me the larger story. $\endgroup$ Nov 4, 2010 at 6:25
  • $\begingroup$ @DickPalais What about if we assume that $L$ is a derivation on $C^{\omega}(\mathbb{R}^{n})$, the space of real analytic functions? Is the real analytic version of Chevalley theorem, true? I ask this question because the proof of the smooth version is based on usage of smooth bump functions, which are not real analytic. More generally assume that $M$ is a $C^{\omega}$ manifold. Is it true to say "Every derivation on $C^{\omega}(M)$ corresponds to a real analytic vector field?" $\endgroup$ Feb 18, 2014 at 17:36

I remember thinking the same thing as Richard is saying and thinking about defining $L$ simply by using the third property. I believe only using this property you can prove change rule (but maybe I'm wrong, and my notes are far away).

The caveat is that if you don't use $L(x)=1$, then you can think of (with suitable restrictions on domains) the following:


which, by the way, satisfies the chain rule. :)

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    $\begingroup$ But that is not a derivation. $\endgroup$ Nov 4, 2010 at 6:29
  • $\begingroup$ It also does satisfy $L(x) = 1$, so it's not clear how you're demonstrating the consequences of not using that rule. $\endgroup$
    – LSpice
    May 10, 2019 at 12:18

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