$2^{-n}\varphi(x/n) $ does not converge in $\mathscr{D}(\mathbb{R}^n)$ Let $\varphi \in C^\infty_c(\mathbb{R}^n)= \mathscr{D}(\mathbb{R}^n) $ be a test function. Prove that

*

*$2^{-n}\varphi \to 0 $ in $\mathscr{D}(\mathbb{R}^n)$

*$2^{-n}\varphi(x/n)  $ does not converge in $\mathscr{D}(\mathbb{R}^n)$
Definition
My definition of converge in $\mathscr{D}(\mathbb{R}^n)$ is that:
Given sequence $\varphi_n \in \mathscr{D}(\Omega)$, $\varphi \in \mathscr{D}(\Omega)$. We say that $\varphi_n \to \varphi  $ in $\mathscr{D}(\Omega)$ if
a)  $\operatorname{supp}(\varphi_n) \subseteq K$,  $K$ compact
b)$D^\alpha\varphi_n \to D^\alpha\varphi  $ uniformly over $K$, $ \forall 
 $ multi-index $ \alpha \in \mathbb{N}^n$
My try
1)
a) $\operatorname{supp}(\varphi_n)=\{x\in \mathbb{R}^n:\varphi_n(x) \neq 0 \} =\{x\in \mathbb{R}^n:2^{-n}\varphi(x) \neq 0 \}=\{x\in \mathbb{R}^n:\varphi(x) \neq 0 \}=\operatorname{supp}(\varphi)\subseteq K$
b) $D^\alpha\varphi_n(x) = 2^{-n}D^\alpha\varphi(x) \to 0D^\alpha\varphi =0  $ uniformly over $K$, $ \forall 
 $ multi-index $ \alpha \in \mathbb{N}^n$


*I don't know what to do

 A: *

*For b) maybe you should write: "since $\varphi \in \mathscr{D}(\mathbb{R}^{n})$", given $\alpha$ there exists $c$ such that $\| D^{\alpha}\varphi \|_{\infty}<c$. Then
$$  \|2^{-n}D^{\alpha} \varphi\|_{\infty} \leq 2^{n}c \to 0. $$

*Assume that there exists $\psi \in \mathscr{D}(\mathbb{R}^{n})$ such that $2^{n}\varphi(x/n) \to \psi(x)$. Since $\psi \in \mathscr{D}(\mathbb{R}^{n})$, its support is contained in some ball $B(0,R)$. Now notice that $\mathrm{supp}(2^{n}\varphi(x/n))=\mathrm{supp}(\varphi(x/n))=n\mathrm{supp}(\varphi)$. By continuity (and that $\varphi$ is not the zero function), there exists a ball $B(x_{0},\varepsilon) \subset \mathrm{supp}(\varphi)$. Then $nB(x_{0},\varepsilon) \subset n\mathrm{supp}(\varphi)$. Now take $n$ big enough such that $nB(x,\varepsilon) \supset B(0,R)$. This contradicts condition b) in your definition of convergence beacuse $\mathrm{supp}(2^{n}\varphi(x/n))=n \mathrm{supp}(\varphi) \supset B(0,R)$.

EDIT: Proving that $n\mathrm{supp}(\varphi)=\mathrm{supp}(\varphi/n)$
Let $x \in \{x \in \mathbb{R}^{n}; \varphi(x)=0\}$. Then $nx$ satisfies $\varphi(nx/n)=\varphi(x)=0$. Thus $nx \in \{y \in \mathbb{R}^{n};\varphi(y/n)\}$, that is, $n\{x \in \mathbb{R}^{n}; \varphi(x)=0\} \subset  \{y \in \mathbb{R}^{n};\varphi(y/n)\}$.
Now take $y \in \{y \in \mathbb{R}^{n};\varphi(y/n)\}$. Then $y/n$ satisfies $\varphi(y/n)=0$, that is, $y/n \in \{x \in \mathbb{R}^{n}; \varphi(x)=0\}$, which is equivalent to say that $y \in n\{x \in \mathbb{R}^{n}; \varphi(x)=0\}$. Then $\{y \in \mathbb{R}^{n};\varphi(y/n)\} \subset n\{x \in \mathbb{R}^{n}; \varphi(x)=0\}$.
So far
$$\{y \in \mathbb{R}^{n};\varphi(y/n)\} = n\{x \in \mathbb{R}^{n}; \varphi(x)=0\}.$$
Now look at the complements and take closure to obtain $\mathrm{supp}(\varphi(x/n))=n\mathrm{supp}(\varphi)$.
