Test functions are dense in Schwartz functions (with respect to $L^\infty$-norm)

Question

Given a Schwartz function $\varphi:\mathbb{R}^n\to \mathbb{R}$, I want to construct this way a sequence of test functions that converge to $\varphi$ with respect to $L^\infty$-norm: $$\varphi_n(x):=\varphi(x)\psi(x/n),$$ where $\psi$ is a test function such that $\psi(0)=1$. Now I need to prove that $||\varphi_n-\varphi||_\infty\to 0$. Intuitively since we are dilating the graph of $\psi$, the function $\psi$ will flatten around $0$ at the value $1$ and then $\varphi_n-\varphi$ will flatten to $0$ but I don't know how to make this rigorous.

Answer 1

Given a Schwartz function $\phi$ and any $\epsilon>0$, we fix an $R>0$ such that $\sup\limits_{|x|\geq R}|\phi(x)|\leq\epsilon$ (why does such an $R>0$ exist?). Next, let $\beta$ be a smooth compactly supported bump function taking values in $[0,1]$ such that $\beta(x)=1$ for $|x|\leq R$. Then, for $|x|\leq R$, we have $\phi(x)\beta(x)=\phi(x)$, but if $|x|\geq R$, then $|\phi(x)\beta(x)-\phi(x)|\leq 2|\phi(x)|\leq 2\epsilon$, by the triangle inequality. Therefore, $\|\phi\beta-\phi\|_{\infty}\leq 2\epsilon$. Since $\epsilon>0$ was arbitrary, this shows density of $\mathcal{D}(\Bbb{R}^n)$ in $\mathcal{S}(\Bbb{R}^n)$.

If you really want, you can turn this into a sequential proof by fixing a sequence $\{\epsilon_n\}_{n=1}^{\infty}$ of positive numbers converging to $0$.

Answer 2

I will preliminarily make some clarifications and give some definitions that I will use in the demonstration:

Let $\Omega \subset \mathbb{R}^n$ open, let $\mathscr{S(\Omega)}$ be the Schwartz space,

the theorem you asked for is the following:

Thm(1): $C^\infty_c(\Omega)$ is dense on $\mathscr{S(\Omega)}$

For completeness i remember that $\mathscr{S(\Omega)}$ is the Frechét built starting to the set of rapidly decreasing function, in particular the metric on these space is the following:

$$d(\phi, \psi) = \sum_{k=0}^{\infty}2^{-k}\frac{p_k(\phi-\psi)}{1+p_k(\phi-\psi)}$$

Where the familiy $\{p_k\}_{k \in \mathbb{N}}$ is the familiy of seminorms connected to the locally convexity of $\mathscr{S(\Omega)}$. In particular we can choose to define $\mathscr{S(\Omega)}$ the following seminorms:

$$p_k(\psi) = max\bigl\{\lVert x^{\alpha}\partial^{\beta}\psi \rVert_{L^\infty(\Omega)}| \alpha, \beta \in \mathbb{N}^n : |\alpha|, |\beta| \leq k\bigl\}$$ In particular is very useful the following characterization of convergence:

Thm(2): $\psi_n \to \psi$ in $\mathscr{S(\Omega)}$ iff $p_k(\psi_n - \psi) \to 0 \hspace{0.3cm} \forall k \in \mathbb{N}$

Then we can start with the proof of density:

Proof of Thm(1): Let $\phi \in \mathscr{S(\Omega)}$, the strategy of proof is smoothly cut off these function.

Let $\psi \in C^{\infty}_c(\Omega)$, such that $\psi(0)=1$, we define $\psi_n(x) = \frac{\psi(\frac{x}{n})}{\rVert \psi \lVert_{L^{\infty}}}$, then $|\psi_n(x)|\leq 1$, and consider as candidate approximating succession $\phi_n = \psi_n \phi$. In particular to show that $\phi_n \to \phi$ in $\mathscr{S(\Omega)}$, for Thm(2), is sufficient to show that $p_k(\phi_n - \phi) \to 0 \hspace{0.3cm} \forall k \in \mathbb{N}$.

CASE[$k=0$]: Because $\psi \in \mathscr{S(\Omega)}$ then $\exists B$ closed ball such that $\lVert \phi \chi_{B^c}\rVert_{L^{\infty}}\leq \epsilon$, Moreover by construction $\psi_n$ converge locally (i.e at fixed compact) uniformly to 1,

\begin{equation}\begin{aligned} \lVert \phi_n - \phi\rVert_{L^{\infty}}&\leq\lVert (\phi_n - \phi)\chi_{B}\rVert_{L^{\infty}}+ \lVert (\phi_n - \phi)\chi_{B^c}\rVert_{L^{\infty}}\\ &=\lVert (\psi_n - 1)\phi\chi_{B}\rVert_{L^{\infty}}+ \lVert (\phi_n - \phi)\chi_{B^c}\rVert_{L^{\infty}}\\ &\leq \lVert (\psi_n - 1)\phi\chi_{B}\rVert_{L^{\infty}}+ 2\lVert \phi \chi_{B^c}\rVert_{L^{\infty}}\\ &\leq \lVert (\psi_n - 1)\phi\chi_{B}\rVert_{L^{\infty}}+ 2\epsilon \to 2\epsilon\\ \end{aligned}\end{equation}

By arbitrary of $\epsilon$ we conclude that $\lVert \psi_n - \psi \rVert_{L^{\infty}} \to 0$.

CASE[$k>0$] It is sufficient to prove that $\lVert x^{\alpha}\partial^{\beta}\phi_n - x^{\alpha}\partial^{\beta}\phi \rVert_{L^{\infty}} \to 0$, $\forall \alpha, \beta \in \mathbb{N}^n$. Preliminary we rewrite: $$\partial^{\beta}\psi_n\phi = \sum_{\beta_1 \in \mathbb{N}^n : 1 \leq |\beta_1|\leq |\beta|} \binom{|\beta|}{|\beta_1|} \partial^{\beta_1}\psi_n \partial^{\beta-\beta_1}\phi + \psi_n\partial^{\beta}\phi$$ Then \begin{equation}\begin{aligned} \lVert x^{\alpha}\partial^{\beta}\phi_n - x^{\alpha}\partial^{\beta}\phi \rVert_{L^{\infty}}&\leq \lVert (\psi_n -1)x^{\alpha}\partial^{\beta}\phi \rVert_{L^{\infty}} + \lVert x^{\alpha}\sum_{\beta_1 \in \mathbb{N}^n : 1 \leq |\beta_1|\leq |\beta|} \binom{|\beta|}{|\beta_1|} \partial^{\beta_1}\psi_n \partial^{\beta-\beta_1}\phi\rVert_{L^{\infty}} \end{aligned}\end{equation}

In particular $\lVert (\psi_n -1)x^{\alpha}\partial^{\beta}\phi \rVert_{L^{\infty}}$ tend to zero for the same reason explained in the previous case. The second term is pretty obvious that is small because $\psi_n$ is smooth with compact support and $\phi$ is rapidly decreasing, anyway in details: \begin{equation}\begin{aligned} \lVert x^{\alpha}\sum_{\beta_1 \in \mathbb{N}^n : 1 \leq |\beta_1|\leq |\beta|} \binom{|\beta|}{|\beta_1|} \partial^{\beta_1}\psi_n \partial^{\beta-\beta_1}\phi\rVert_{L^{\infty}} &=\frac{1}{n} \lVert \sum_{\beta_1 \in \mathbb{N}^n : 1 \leq |\beta_1|\leq |\beta|} \binom{|\beta|}{|\beta_1|} \biggl(\frac{1}{n}\biggr)^{\beta_1-1}\partial^{\beta_1}\psi \partial^{\beta-\beta_1}\phi\rVert_{L^{\infty}}\\ &\leq \frac{1}{n} \text{max}\bigl\{\text{sup}\{|\partial^{d}\psi(x)||x\in\Omega\}|d\in\mathbb{N}^n: 1\leq|d|\leq |\beta| \bigr\} \sum_{\beta_1 \in \mathbb{N}^n : 1 \leq |\beta_1|\leq |\beta|} \binom{|\beta|}{|\beta_1|} \biggl(\frac{1}{n}\biggr)^{\beta_1-1}\rVert x^{\alpha}\partial^{\beta-\beta_1}\phi\rVert_{L^{\infty}}\\ &\leq \frac{1}{n} M N \to 0\\ \end{aligned}\end{equation} where in the last inequality i have used that countinuous with compact support is bounded and that $x^{\alpha}\partial^{\beta-\beta_1}\phi \in \mathscr{S(\Omega)}$ and that the sum is finite. $\blacksquare$