In Folland's Introduction to Partial Differential Equations (Chapter 2 The Laplace Operator, p.67, Theorem 2.1)

Suppose $L$ is a partial differential operator on ${\mathbb R}^n$. Then $L$ commutes with translations and rotations if and only if $L$ is a polynomial in $\Delta$.

Here are my questions:

  • Is "partial differential operator on ${\mathbb R}^n$" short for "partial differential operator on functions on ${\mathbb R}^n$"?
  • In this theorem, is such operator of the distribution meaning or the classical one?
  • When one talks about an operator, there is supposed to be an vector space. So what's the underlying function space here?
  • Since the differential operator can also be defined on distribution via test functions, how should I understand the operator in this theorem? In other words, should I regard it as an operator in the classical meaning or one on distribution? Or maybe it does not matter in the theorem?
  • $\begingroup$ Why down vote? Is this question too stupid? $\endgroup$ – Jack Jul 5 '11 at 2:20
  • $\begingroup$ (1) Yes. (2) The theorem is probably in the typical setting of sufficiently smooth functions, but I can't really tell because I don't have the book to see the context. $\endgroup$ – anon Jul 5 '11 at 2:22
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    $\begingroup$ I didn't downvote but the question doesn't strike me as a terribly good one (since you asked whether it is too stupid). I doubt Folland commits the sin of stating theorems involving terms he doesn't define. Also, it would probably be a good idea to give the precise reference (theorem number and page). $\endgroup$ – t.b. Jul 5 '11 at 2:34
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    $\begingroup$ Jack, I think it's clear in this case that Folland is referring to operators on functions, since in the paragraph preceding the statement of the theorem, he refers to operators on functions. But I do understand the confusion because differential operators are introduced earlier in the text as distributions, and I'd expect the result to be valid in the distributive sense as well. You might be interested to know that there is a more general theorem of this nature in the setting of isotropic Riemannian manifolds. $\endgroup$ – John M Jul 5 '11 at 3:49
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    $\begingroup$ @John Perhaps write that up as an answer? $\endgroup$ – Willie Wong Jul 5 '11 at 10:03

In general, a differential operator usually refers to a operator on a space of functions that are sufficiently smooth enough in order that the differential operator makes sense. However, as you rightly point out, a differential operator could very well be meant in the distributive sense, meaning that the differntial operator acts on the space of distributions.

The term "operator" is something of a misnomer, since it's really more of a map between different function spaces! As an example, the usual derivative $\frac{d}{dx}:C^1(\mathbb R)\rightarrow C^0(\mathbb R)$ maps the space of continuously differential functions to continuous functions. So it's a linear map, but not what is usually termed as "operator".

In terms of the nice theorem that Folland proves, he proves it only as a differential operator on functions, but I'm pretty sure that a little argument can show that the result extends to differential operators in the distributive sense as well. Indeed, if a distributive derivative commutes with translations and rotations, then that means that the operator, as an operator on the space of smooth, compactly supported functions, also commutes with translations and rotations. Thus, the operator would then be a polynomial in $\triangle$.

For more general notions of a differential operator on a manifold, the go-to reference (from an analyst's point of view) is Helgason's Groups and Geometric Analysis. On pg 236, he give a somewhat convoluted definition of a differential operator on the space of smooth, compactly supported functions on $\mathbb R^n$, but then proves in a very interesting theorem that a differential operator $$D:C_c^\infty(\mathbb R^n)\rightarrow C_c^\infty(\mathbb R^n)$$ is actually characterized by the property that it decreases suppports: $$\text{supp}(D\phi)\subseteq \text{supp}(\phi)$$ for any $\phi \in C_c^\infty(\mathbb R^n)$.

This then motivates his definition of a differential operator on a manifold $M$ (pg 239):

A differential operator $D$ on $M$ is a linear mapping of $C_c^\infty(M)$ onto itself which decreases supports: $$\text{supp}(Df)\subseteq \text{supp}(f)$$ for any $f \in C_c^\infty(M)$.

He then goes on to prove a generalization of Folland's result for "two-point homogeneous spaces" (pg 288).

This is not the only way to generalize differential operators to manifolds. From a differential topologist's point of view, a first-order differential operator $X$ is a derivation (also known as a vector field), which we want to think of here as linear maps from $C^\infty(M)$ to itself satisfying the Leibniz rule: $$X(fg)=(Xf)g+f(Xg)$$

Then, to get all the higher-order differential operators, we can consider the subring of the endomorphism ring $\text{End}(C^\infty(M))$ of smooth functions $C^\infty(M)$ generated by those first-order differential operators.


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