Let $\varphi \in \mathcal{D}(\mathbb{R}^n)$ and $h \in \mathbb{R}^n \setminus \{0\}$. For $t \in \mathbb{R}^n \setminus \{0\}$ we put $$ \varphi_t(x)= \dfrac{\varphi(x+th)-\varphi(x)}{t}. $$ The question is: prouve that $\varphi_t \in \mathcal{D}(\mathbb{R}^n)$ for all $t \neq 0$.

My purpose is: Let $t \neq 0$. We have $\varphi_t \in C^{\infty}(\mathbb{R}^n)$ beacause $\varphi \in \mathcal{D}(\mathbb{R}^n)$.

Now, it remains to prove that $Supp \varphi_t$ is compact. We have $Supp \varphi_t \subset Supp(x \to \varphi(x+th)) \cup Supp \varphi$, but the problem is that $Supp(x \to \varphi(x+th)) \cup Supp \varphi$ depends on $t$ and it seems to me that the $Supp \varphi_t$ must to be independant on $t$.

My questions are: please, how we find $Supp \varphi_t$ independent on $t$? And why the support have to be independent on $t$?

  • $\begingroup$ Why should the support be independent of $t$? I see no reason to assume this. $\endgroup$
    – Luke
    Aug 18, 2017 at 9:57
  • $\begingroup$ I think for such problem the parameter $t$ is chosen very small (less than 1) check your source again. otherwise there is no reason on why this support should be independent on $t$ $\endgroup$
    – Guy Fsone
    Aug 18, 2017 at 10:03

1 Answer 1


If I understood correctly the question, you are asked to prove that for any fixed $t\in\mathbb R^n\setminus\left\{ 0\right\} $, the function $\varphi_t$ belongs to $\mathcal D\left(\mathbb R^n\right)$. Therefore, fix once for all $t$.

You can use the fact that for any two functions $f$ and $g$, $$\operatorname{Supp}\left(f+g\right) \subset \operatorname{Supp}\left(f\right)\cup \operatorname{Supp}\left(g\right) $$ (because if $x$ does not belong to the right hand side, we can find a small ball around $x$ such that $f$ and $g$ vanish on that ball, hence so does $f+g$).

The support of the function $x\mapsto \varphi\left(x+th\right)$ is compact (recall that $t$ and $h$ are fixed hence the support is only a translate of the support of $\varphi$) hence the support of $\varphi_t$ is bounded. Since it is by definition closed, this ends the proof that $\varphi_t$ belongs to $\mathcal D\left(\mathbb R^n\right)$.

  • $\begingroup$ I think the real exercice is to prove $\lim_{t \to 0} \varphi_t$ converges in $D(\mathbb{R}^n)$ $\endgroup$
    – reuns
    Aug 18, 2017 at 20:53
  • $\begingroup$ @reuns Maybe, but it is not stated in the question. $\endgroup$ Aug 18, 2017 at 21:07
  • $\begingroup$ Well it is not hard. Let $\phi(x) =\sum_{i=1}^n a_i \frac{\partial}{\partial x_i}\varphi(x) = \frac{\partial}{\partial t} \varphi(x+ta)|_{t=0}$. Then $\varphi(x+a)-\varphi(x) = \int_0^1 \phi(x+ta)dt$ $\endgroup$
    – reuns
    Aug 18, 2017 at 21:44

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