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?](https://mathoverflow.net/questions/40082/why-do-we-teach-calculus-students-the-derivative-as-a-limit/40147#40147)


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.



Do these properties characterize differentiation?
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. 


