$2^{-n}\varphi(x/n) $ does not converge in $\mathscr{D}(\mathbb{R}^n)$

Question

Let $\varphi \in C^\infty_c(\mathbb{R}^n)= \mathscr{D}(\mathbb{R}^n) $ be a test function. Prove that

1. $2^{-n}\varphi \to 0 $ in $\mathscr{D}(\mathbb{R}^n)$
2. $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$

2. I don't know what to do

Answer 1

1. 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 $$ \|2^{-n}D^{\alpha} \varphi\|_{\infty} \leq 2^{n}c \to 0. $$
2. 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 $\mathrm{supp}(\varphi(x/n))=n\mathrm{supp}(\varphi)$.