After reading this question, which asked for some examples of commonly used (proper) dense subsets of $C_0^{\infty}(\mathbb{R}^n)$ with the $L^p$-norm I wonder. What are some "well-known" examples of countable subsets of $X\subseteq C_c^{0} \cap \mathscr{S}(\mathbb{R}^n)$ (the Schwartz space) for which

  • $X$ is dense in $C_c^{0}$ for the topology generated by the sup-norm $\sup_{x \in \mathbb{R}^n} \|f(x)\|$
  • $X$ is dense in $\mathscr{S}(\mathbb{R}^n)$ for the topology generated by the semi-norms $ p_{\alpha,\beta,K}(f):=\sup_{x\in K} \|x\|^\alpha \left\| (D^{\beta} f)(x) \right\| . $
  • 1
    $\begingroup$ $\sup_{x\in\mathbb{R}^n}|D^{\beta}f(x)|$ are not the seminorms you should use for $C_c^{\infty}$. $\endgroup$ Jul 30, 2020 at 22:38
  • $\begingroup$ @AbdelmalekAbdesselam thank you, I refined the question. $\endgroup$
    – ABIM
    Jul 31, 2020 at 7:21
  • 2
    $\begingroup$ The completion of your first space is the Banach algeba of continuous functions which vanish at infinity. There is a version of Stone-Weierstraß for this algebra which makes it relatively easy to construct families with the desired property. The only problem is to get a single such function to start with and there is standard way to construct a bell shaped one. One can then take its dilations and translates, and the algebra generated to get an example (taking care to keep the result countable by means of the usual ploy of only using rational coefficients and parameters). $\endgroup$
    – user131781
    Jul 31, 2020 at 8:00
  • $\begingroup$ Yes I was going to suggest this too using $\exp(-1/(1-\|x\|^2)I_{Ball_{\mathbb{R}^n}(0,1)}$. Since it also lies in the Schwartz-space (since its smooth and rapidly-decreasing) but I'm not sure if it is dense in the latter space. $\endgroup$
    – ABIM
    Jul 31, 2020 at 8:15
  • $\begingroup$ @AnnieLeKatsu Yes but how to show this? And to user131781 using Annie's example we would be taking shifts and linear combinations of this no? $\endgroup$
    – ABIM
    Jul 31, 2020 at 8:17

1 Answer 1


I don't know about "well known" or canonical answers to this question, but it is easy to construct an $X$ that works as follows.

Using the definition of Hermite polynomials given by $$ H_n(x)=(-1)^n e^{x^2}\left(\frac{d}{dx}\right)^n e^{-x^2}\ , $$ we define the one-dimensional Hermite functions by $$ h_n(x)=\pi^{-\frac{1}{4}} 2^{-\frac{n}{2}} n!^{-\frac{1}{2}} e^{-\frac{x^2}{2}} H_n(x) $$ and then the $d$-dimensional Hermite functions by $$ h_{\alpha}(x_1,\ldots,x_d)=h_{\alpha_1}(x_1)\cdots h_{\alpha_d}(x_d) $$ for every multiindex $\alpha$. These functions form an orthonormal basis for the Hilbert space $L^2(\mathbb{R}^d)$ as well as as an unconditional Schauder basis of Schwartz space $\mathscr{S}(\mathbb{R}^d)$. Clearly, finite linear combinations of the $h_{\alpha}$ with rational coefficients is a countable dense subset of $\mathscr{S}(\mathbb{R}^d)$. To simultaneously satisfy the other condition one can pick a smooth cutoff function $\rho:\mathbb{R}^d\rightarrow [0,1]$, constant equal to $1$ on the ball $B(0,1)$ and equal to zero outside the ball $B(0,2)$. Now take previous linear combinations and multiply them by $x\mapsto \rho(\frac{1}{k}x)$, for $k=1,2,\ldots$ This will give a set $X$ that fulfills the two requirements.


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