Problem of convergence in Schwartz class Can someone help me to show the following: Let $\phi \in \mathcal{S}(\mathbb{R}^n)$ satisfying 
$$\partial^\alpha \phi(0) = 0,\;\;\; \forall |\alpha|<k,$$
for some integer $k>0$. Considere $f \in C^{\infty}(\mathbb{R}^n)$ such that $f(x)=1$, for $|x| \geq 2$, and $f(x)=0$, for $|x| \leq 1$, and define $f_\varepsilon(x)=f(x/\varepsilon)$, for $\varepsilon>0$. Show that $\phi f_\epsilon \rightarrow \phi$ in $\mathcal{S}$ as $\varepsilon \rightarrow 0$, i.e. 
$$\rho_{\alpha,\beta}(\phi f_\varepsilon - \phi) \rightarrow 0 \;\;\; (\varepsilon \rightarrow 0), \;\;\; \forall \alpha, \beta\; \text{multi-index},$$
in which $\rho_{\alpha,\beta}$ is the semi-norms on $\mathcal{S}$.
My attempt (copied from comments):

I don't have too much, in fact nothing. I was just trying to estimate the case $\alpha=\beta=0$ so I could have uniform convergence
  $$
\left|\phi(x) f_{\varepsilon}(x)-\phi(x)\right|=|\phi(x)||f(x / \varepsilon)-1| \leq\left(\sup _{x \in \mathbb{R}^{n}}|\phi(x)|\right)|f(x / \varepsilon)-1|
$$
  Can I get uniform convergence here $|f(x / \varepsilon)-1|$ ?? - pedrohtak 13 hours ago

 A: The correct formulation is that, assuming that $\partial^{\alpha}\phi(0)=0$ for $|\alpha|\leq k$, then
\begin{align*}
\rho_{\alpha,\beta}(\phi f_{\epsilon}-\phi)\rightarrow 0,~~~~\alpha\in\mathbb{N}_{0}^{n},~~~~|\beta|\leq k.
\end{align*}
We see that
\begin{align*}
&\rho_{\alpha,\beta}(\phi f_{\epsilon}-\phi)\\
&=\sup_{x\in\mathbb{R}^{n}}\left|x^{\alpha}\partial^{\beta}(\phi(\cdot)f(\cdot/\epsilon)-\phi(\cdot))\right|\\
&=\sup_{x\in\mathbb{R}^{n}}\left|x^{\alpha}\left(\sum_{|\gamma|\leq|\beta|}C_{\gamma}\partial^{\beta-\gamma}\phi(x)\partial^{\gamma}f(x/\epsilon)\dfrac{1}{\epsilon^{|\gamma|}}-\partial^{\beta}\phi(x)\right)\right|\\
&\leq\sup_{x\in\mathbb{R}^{n}}\left|x^{\alpha}\partial^{\beta}\phi(x)f(x/\epsilon)-\partial^{\beta}\phi(x)\right|+\sup_{x\in\mathbb{R}^{n}}\left|x^{\alpha}\left(\sum_{0<|\gamma|\leq|\beta|}C_{\gamma}\partial^{\beta-\gamma}\phi(x)\partial^{\gamma}f(x/\epsilon)\dfrac{1}{\epsilon^{|\gamma|}}\right)\right|
\end{align*}
where $C_{\gamma}$ are the binomial coefficients. Note that $f=1$ on $\{|x|\geq 2\}$, then
\begin{align*}
&\sup_{x\in\mathbb{R}^{n}}\left|x^{\alpha}\partial^{\beta}\phi(x)f(x/\epsilon)-\partial^{\beta}\phi(x)\right|\\
&=\sup_{|x|\leq 2\epsilon}\left|x^{\alpha}\partial^{\beta}\phi(x)f(x/\epsilon)-\partial^{\beta}\phi(x)\right|\\
&\leq\left(1+\sup_{|u|\leq 2}|f(u)|\right)\sup_{|x|\leq 2\epsilon}\left|x^{\alpha}\partial^{\beta}\phi(x)\right|\\
&\rightarrow 0
\end{align*}
as $\epsilon\rightarrow 0$.
On the other hand, we appeal to Taylor expansion of $\partial^{\beta-\gamma}\phi$ at $0$, we get
\begin{align*}
&\partial^{\beta-\gamma}\phi(x)\\
&=\sum_{|\omega|\leq|\gamma|}\dfrac{1}{\omega!}\partial^{\omega}\partial^{\beta-\gamma}\phi(0)\cdot x^{\omega}+\sum_{|\omega|=|\gamma|+1}\dfrac{|\gamma|+1}{\omega!}\cdot x^{\omega}\int_{0}^{1}(1-t)^{|\gamma|}\partial^{\omega}\partial^{\beta-\gamma}\phi(tx)dt\\
&=\sum_{|\omega|=|\gamma|+1}\dfrac{|\gamma|+1}{\omega!}\cdot x^{\omega}\int_{0}^{1}(1-t)^{|\gamma|}\partial^{\omega}\partial^{\beta-\gamma}\phi(tx)dt,
\end{align*}
and hence
\begin{align*}
&\sup_{x\in\mathbb{R}^{n}}\left|x^{\alpha}\left(\sum_{0<|\gamma|\leq|\beta|}C_{\gamma}\partial^{\beta-\gamma}\phi(x)\partial^{\gamma}f(x/\epsilon)\dfrac{1}{\epsilon^{|\gamma|}}\right)\right|\\
&=\sup_{|x|\leq 2\epsilon}\left|x^{\alpha}\left(\sum_{0<|\gamma|\leq|\beta|}C_{\gamma}\partial^{\beta-\gamma}\phi(x)\partial^{\gamma}f(x/\epsilon)\dfrac{1}{\epsilon^{|\gamma|}}\right)\right|\\
&\leq\sum_{0<|\gamma|\leq|\beta|}c_{\gamma}\sum_{|\omega|=|\gamma|+1}\sup_{x\in\mathbb{R}^{n}}\left|x^{\alpha}\partial^{\omega+\beta-\gamma}\phi(x)\right|\cdot\sup_{|x|\leq  2\epsilon}|x|^{|\gamma|+1}\cdot\dfrac{1}{\epsilon^{|\gamma|}}\cdot|f(x/\epsilon)|\\
&\leq\sum_{0<|\gamma|\leq|\beta|}c_{\gamma}'\sum_{|\omega|=|\gamma|+1}\sup_{x\in\mathbb{R}^{n}}\left|x^{\alpha}\partial^{\omega+\beta-\gamma}\phi(x)\right|\cdot\sup_{|u|\leq 2}|f(u)|\cdot\epsilon\\
&\rightarrow 0,
\end{align*}
so $\rho_{\alpha,\beta}(\phi f_{\epsilon}-\phi)\rightarrow 0$, as expected.
