Let $f\in\mathscr{S}(\mathbb R^n)$ where $\mathscr{S}(\mathbb R^n)$ is the Schwartz space. Is it true that $\partial^\alpha f$ is a uniformly continuous function for all multi-index $\alpha$?

Here $\mathscr{S}(\mathbb R^n)$ is the $\mathbb C$-vector space of all the $C^\infty$ functions $f:\mathbb R^n\longrightarrow \mathbb C$ for which $$x\longmapsto x^\alpha \partial^\beta f(x),$$ is a bounded function in $\mathbb R^n$ for all multi-indices $\alpha, \beta$.

  • 4
    $\begingroup$ Any continuous function vanishing at infinity is uniformly continuous. $\endgroup$
    – Etienne
    Dec 8, 2014 at 14:15
  • $\begingroup$ Intuitively I agree but what about the proof? $\endgroup$
    – PtF
    Dec 8, 2014 at 15:40
  • $\begingroup$ Such a function is uniformly continuous on any compact set, and essentially $0$ outside a sufficiently large ball. You will certainly be able to write down a precise proof. $\endgroup$
    – Etienne
    Dec 8, 2014 at 17:10

1 Answer 1


Note that by definition, the derivative of a Schwarz function is a Schwarz function, hence we merely have to prove that each element of the Schwarz space is uniformly continuous.

As suggested in the comments, we can prove that if a function $f\colon\mathbb R^n\to\mathbb R$ is continuous and vanishes at infinity, then $f$ is uniformly continuous. To show this, fix a positive $\varepsilon$. For some $R$, we have $|f(x)| \lt\varepsilon$ provided that $|x|\geqslant R$. The closed ball of $\mathbb R^n$ with radius $2R$ and centered at the origin is compact. Therefore, $f$ is uniformly continuous on this set. There exists a $\delta$ such that if $|x|,|y|\leqslant 2R$ and $|x-y|\lt\delta$, then $|f(x)-f(y)|\lt \varepsilon$. We have to extend this to arbitrary $x$ and $y$ such that $|x-y|\lt\delta$ (that is, not only for those of the ball). We have to treat the cases:

  • $ |x|,|y|\leqslant 2R$: this is already done;
  • $|x|\leqslant 2R$ and $|y|\gt 2R$: then $|x|\geqslant |y|-|x-y|\geqslant 2R-\delta \geqslant R $, hence $|f(x)- f(y)|\lt 2\varepsilon$;
  • $|x|\gt 2R$ and $|y|\gt 2R$: in this case, $|f(x)- f(y)|\lt 2\varepsilon$.

One can actually show that if $f\in\mathcal S(\mathbb R^n)$, then $f$ is Lipschitz-continuous. To see this, fix $x,y\in\mathbb R^n$ and consider the function $g\colon\mathbb R\to\mathbb R$ defined by $g(t):=f(x+ty)$ and use the mean value theorem in order to bound $f(x)-f(y)$.


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