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Claim: Let $T$ be a distribution on $\mathbb R^n$ such that $\nabla T$ belongs to $L^p(\mathbb R^n)$ for some $p\in [1,+\infty]$. Then $T$ is a temperate distribution, i.e. belongs to the topological dual of the Schwartz class $\mathscr S(\mathbb R^n)$.

I believe that the previous claim holds true, but I do not have a complete proof.

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  • $\begingroup$ Much more is true. If I remember it right, once $\nabla T$ is a temperate distribution and $T$ is a distribution, you can immediately conclude that $T$ is temperate as well. Try to look it up in Hormander. If you fail, I'll post the proof. $\endgroup$
    – fedja
    Commented Nov 30, 2017 at 2:01

1 Answer 1

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Let's do it in dimension $1$. By Holder's inequality $$\int_0^x|f'(x)|dx\leq \| f'\|_p|x|^{1-1/p}.$$ A distribution with locally integrable derivative is a continuous function, $$f(x)=f(0)+\int_0^xf'(x)dx=O(|x|^{1-1/p}),\quad x\to\infty.$$ Thus $$(f,\phi)=\int f\phi$$ is absolutely convergent and defines a bounded linear functional on $\mathscr S$, that is a tempered distribution. For higher dimension the proof is essentially the same.

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  • $\begingroup$ Thanks for the 1 D argument. You write that the same argument works in $n$ dimensions, which is not absolutely clear to me since the representation of $f$ via $\nabla f$ cannot obviously be used in the same way. $\endgroup$
    – Bazin
    Commented Nov 29, 2017 at 15:16

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