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

Suppose we have a non- tempered distribution $u\in \mathcal D'(\mathbb R^d)\backslash \mathcal S'(\mathbb R^d)$. Is it possible to have $\partial_{x_1}...\partial_{x_d}u \in \mathcal S'(\mathbb R^d)$ where the derivative is taken in the sense of Schwartz distributions? I cannot find an example nor prove the converse. Any reference to suggest?

share|improve this question

1 Answer 1

A distribution is tempered if and only if it is the distributional derivative $D^nF$ of a continuous function $F$ which is $O(x^k)$ for some positive integer. Its primitive is then $D^{n-1}F$ and so also tempered. Hence the answer to your question is no. (Suitable reference: J. Sebastiao e Silva, Integrals and orders of growth of distributions, Lisbon, 1964).

Edit. I wrote up the one-dimensional case dor simplicity. If $n=0$, you have a continous function and you simply take its classical primitive which is clearly a tempered distribution.

share|improve this answer
    
I agree that there is a structure theorem for tempered distributions. In fact $u=\partial^\alpha ((1+|x|^2)^n f(x))$ for some continuous and bounded function $f$ and multi-indexe $\alpha$ and integer $n$. The problem is if $\alpha=(\alpha_1,...,\alpha_d)$ is such that $\alpha_i=0$ for some $i$: then taking the derivative of a lower order makes no sense. –  Thomas Mar 18 at 13:11
    
I have a concern: that tells me that for any tempered distribution $T$, there is a tempered distribution solution to $\partial x_1...\partial x_d u=T$. In dimension $1$ the only primitives of $0$ are constants so in fact any solution of this equation is tempered. But what about in higher dimension? The solutions to $\partial x_1...\partial x_d u=0$ have support in the union of the axes of $\mathbb R^d$ but is there more? –  Thomas Mar 18 at 14:18

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.