It is well known that the class of Schwartz functions $\mathcal{S}$ in dense in all $L^p$ spaces therefore for each $f \in L^2$ there exists a sequence of Schwartz functions $(f_k)$ such that $\lVert f - f_k \rVert_{L^2} \to 0$ as $k \to \infty$.

If we suppose further that $f$ has compact support can we find a sequence of Schwartz functions $(f_k)$ such that $\lVert f - f_k \rVert_{L^2} \to 0$ as $k \to \infty$ (as above) and additionally $\operatorname{supp}(f_k) \subseteq \operatorname{supp}(f)$ for all $k$?

  • 1
    $\begingroup$ Yes, we can. Meanwhile, this sort of question is best suited for MathStackExchange... This site is for fancier things. :) $\endgroup$ Dec 5, 2020 at 22:27
  • $\begingroup$ Is there any theorem I can appeal to show this easily? Also I guess I'll post on MathStackExchange first next time. $\endgroup$ Dec 5, 2020 at 22:36
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    $\begingroup$ @paulgarrett: What if $f = 1$ on a fat Cantor set and $f = 0$ otherwise? I do not think there is a non-zero Schwartz class function supported in a fat Cantor set... $\endgroup$ Dec 5, 2020 at 22:49
  • $\begingroup$ @MateuszKwaśnicki, we might be surprised... $\endgroup$ Dec 6, 2020 at 1:55

1 Answer 1


No, this is not possible, even if you define $\text{supp}(f)$ only as an equivalence class (up to a null set). For a counterexample take the characteristic function $f$ of a closed set $E$ of positive measure but without interior points (like that fat Cantor set which Mateusz mentioned).

Indeed, if $f_n$ is a continuous function with $\text{supp}(f_n)\subseteq\text{supp}(f)\cup N=E\cup N$ for some null set $N$, then $f_n=0$, because otherwise $E\cup N$ would have to contain some interval $(a,b)$ which must contain a point from the complement of $E$ and which after shrinking (since $E$ is closed) can thus be assumed to lie completely in the complement of $E$, hence $N\supseteq(a,b)$, a contradiction.


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