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The proof of this fact is quite classical. I tried to find it on my own and, apparently, had "the right idea" since it seems to be the standard way to show it.

However, I am stuck with some detail I cannot rigorously justify. There are some questions about this but none, as far as I know, addresses my problem.

Define the Schwartz space $\mathcal{S}$ as follows:

$$\mathcal{S}(\mathbb{R}^{N}):=\left\{f\in C^{\infty}(\mathbb{R}^{N})\mid \forall N\in\mathbb{N}_{0},\forall \,\alpha,\beta\in\mathbb{N}^{N}: \Vert f\Vert_{\alpha\beta}:=\sup_{\mathbb{R}^{N}} \left\vert x^{\alpha}\partial^{\beta}f(x)\right\vert<\infty\right\}$$

I detail the entire proof for the convenience of the reader but my problem can be understood by only looking at the beginning and the end of this post (before the first horizontal rule and after the second one): let $f\in\mathcal{S}(\mathbb{R}^{N})$ and let $\psi\in C_{c}^{\infty}(\mathbb{R}^{N})$ (the space of compactly supported $C^{\infty}(\mathbb{R}^{N})$-functions) such that:

$$\psi(x)\equiv\begin{cases} 1& \text{if } \vert x\vert\le 1\\ 0& \text{if } \vert x\vert\geq 2\\ \in[0,1]&\text{otherwise} \end{cases}$$

Let $f_{\epsilon}(x):=f(x)\psi(\epsilon x)$ for any $x\in\mathbb{R}^{N}$ and any $\epsilon>0$. We observe that $\text{supp }\psi(\epsilon\,\cdot)\to\mathbb{R}^{N}$ when $\epsilon\to 0$ and in particular, $\psi(\epsilon x)\to\psi(0)=1$ for any $x$.


We have:

\begin{align*} \partial^{\beta}f_{\epsilon}(x) &=\psi(\epsilon x)\partial^{\beta}f(x)+\sum_{\substack{\vert\gamma\vert\neq 0\\ \vert\gamma\vert\le\vert\beta\vert}}\binom{\beta}{\gamma}\partial^{\gamma}\left(\psi(\epsilon x)\right)\partial^{\beta-\gamma}f(x)\\ &=\psi(\epsilon x)\partial^{\beta}f(x)+\sum_{\substack{\vert\gamma\vert\neq 0\\ \vert\gamma\vert\le\vert\beta\vert}}\binom{\beta}{\gamma}\epsilon^{\vert\gamma\vert}\left(\partial^{\gamma}\psi\right)(\epsilon x)\partial^{\beta-\gamma}f(x) \end{align*} Obviously, $\epsilon^{1}\geq\epsilon^{\vert\gamma\vert}$ for any $\vert\gamma\vert\neq 0$ where $\gamma$ natural multi-index. Hence, we obtain:

\begin{align*} &\left\Vert f_{\epsilon}(x)-f(x)\right\Vert_{\alpha\beta}\\ &=\sup\left\vert x^{\alpha}\partial^{\beta}f(x)-x^{\alpha}\psi(\epsilon x)\partial^{\beta}f(x)-x^{\alpha}\sum_{\substack{\vert\gamma\vert\neq 0\\ \vert\gamma\vert\le\vert\beta\vert}}\binom{\beta}{\gamma}\partial^{\gamma}\left(\psi(\epsilon x)\right)\partial^{\beta-\gamma}f(x)\right\vert\\ &\le \sup\left\vert x^{\alpha}\partial^{\beta}f(x)-x^{\alpha}\psi(\epsilon x)\partial^{\beta}f(x)\right\vert+\epsilon\sup\left\vert x^{\alpha}\sum_{\substack{\vert\gamma\vert\neq 0\\ \vert\gamma\vert\le\vert\beta\vert}}\binom{\beta}{\gamma}\partial^{\gamma}\left(\psi(\epsilon x)\right)\partial^{\beta-\gamma}f(x)\right\vert \end{align*}

The second term can be easily bounded:

$$\vert\partial^{\gamma}\psi(\epsilon x)\vert\le \sup_{\substack{\vert\delta\vert\neq 0,\,\vert\delta\vert\le\beta\\ x\in\mathbb{R}^{N}}}\vert\partial^{\delta}\psi(x)\vert=:C_{\beta}<\infty$$

and

$$\sup\left\vert x^{\alpha}\sum_{\substack{\vert\gamma\vert\neq 0\\ \vert\gamma\vert\le\vert\beta\vert}}\binom{\beta}{\gamma}\partial^{\beta-\gamma}f(x)\right\vert=:C_{\alpha\beta}<\infty$$

since $f\in\mathcal{S}(\mathbb{R}^{N})$ hence the second term is bounded by $\epsilon C_{\beta}C_{\alpha\beta}$ and thus will converge to $0$.


My problem is with the other term. How can I prove that the following quantity converges to $0$ when $\epsilon\to 0$:

$$\sup\left\vert x^{\alpha}\left(\partial^{\beta}f(x)\right)(1-\psi(\epsilon x))\right\vert$$

I cannot obtain a rigorous argument. We don't have the absolute convergence of $\psi(\epsilon\,\cdot)$ to $\psi(0)$, if I am not mistaken.

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Fix $\rho>0$ (this is the $ε$ in a '$ε-\delta$' proof). To bound $$S:=\sup_{x∈\Bbb R^n}\left\vert x^{\alpha}\left(\partial^{\beta}f(x)\right)(1-\psi(\epsilon x))\right\vert$$

We split the sup into two parts

$$S = \sup_{|x|\leq r}\left\vert x^{\alpha}\left(\partial^{\beta}f(x)\right)(1-\psi(\epsilon x))\right\vert + \sup_{|x|>r}\left\vert x^{\alpha}\left(\partial^{\beta}f(x)\right)(1-\psi(\epsilon x))\right\vert =: S_r + T_r$$

for fixed $r$, $S_r=0$ for small enough $ε \leq ε_0(r)$. For $T_r$, we rewrite

$$T_r = \sup_{|x|>r}\left\vert x^{\alpha+1}\left(\partial^{\beta}f(x)\right)\left(\frac{1-\psi(\epsilon x)}{|x|}\right)\right\vert \leq C_{\alpha+1,\beta} \sup_{|x|>r} \left|\frac{1-\psi(\epsilon x)}{|x|}\right| \leq \frac{C_{\alpha+1,\beta}}{r} $$

Now pick $r_0:=r_0(\rho):=\frac{1}{\rho C_{\alpha+1,\beta}}$ so that for $ε ≤ ε_0(r_0(\rho))$, $$S \leq S_{r_0} + T_{r_0} = 0 + \rho$$

which is what we wanted.

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  • $\begingroup$ Thanks a lot, I should remember this kind of tricks. Nice and concise answer :) $\endgroup$ – MoebiusCorzer May 24 '16 at 21:01

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