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

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.

share|improve this question
2  
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. –  Ryan Budney Nov 4 '10 at 3:38
2  
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. –  KConrad Nov 4 '10 at 3:59
1  
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 –  Mark Meckes Nov 4 '10 at 3:59
    
I wonder what nice sets of conditions that include shift-equivariance are satisfied only by differentiation? –  Michael Hardy Sep 30 at 22:27

3 Answers 3

up vote 41 down vote accepted

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.

share|improve this answer
    
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. –  Todd Trimble Nov 4 '10 at 4:42
    
Short and sweet. And with no "nitty gritty" analysis either! Thank you so much! –  Steven Gubkin Nov 4 '10 at 6:23
    
You're welcome -- glad I could be of help! –  Todd Trimble Nov 4 '10 at 10:13
3  
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. –  KConrad Nov 5 '10 at 19:32
3  
Spilling the beans here: en.wikipedia.org/wiki/Hadamard%27s_lemma –  Todd Trimble Nov 5 '10 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 $L = {\partial \over \partial x_j}$.

share|improve this answer
    
Very nice answer. Thanks for telling me the larger story. –  Steven Gubkin Nov 4 '10 at 6:25
    
@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?" –  Ali Taghavi Feb 18 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:

$$L(f)(x)=\frac{f(x)}{x}$$

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

share|improve this answer
    
But that is not a derivation. –  Mariano Suárez-Alvarez Nov 4 '10 at 6:29

Your Answer

 
discard

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.