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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?

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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.

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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 '14 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 '14 at 14:18

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