# $2^{-n}\varphi(x/n)$ 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

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 $$\alpha \in \mathbb{N}^n$$

1. I don't know what to do
• For (b), suppose $\phi$ is non-zero on the ball $B_1$ of radius $1$. In other words, $B_1\subset \text{supp}(\phi)$. WHat can you say about the supports of the functions $f_n(x)=2^{-n}\phi(x/n)$? Are the supports getting bigger or smaller? Therefore... Aug 6, 2021 at 11:05
• @peek-a-boo They seem to be getting smaller, $2^{-n}$ is tending to $0$, and $\phi(x/n)$ seems to be shrinking to $\phi(0)$. So I would say $f_n(x)$converges to $0$. But that is the opposite of what I must prove. What am I missing? Aug 6, 2021 at 11:12
• The support of $\phi (x/n)$ becomes bigger and bigger, hence it goes outside any compact set as $n \to \infty$ Aug 6, 2021 at 11:17
• @Crostul for each fixed n, can't K change to get big enough to hold the support of $\phi(x/n)$?. And how would you write it down more explicitly? Aug 6, 2021 at 11:49
• As you wrote in your question, you say that convergence requires the condition $$\mathrm{supp} ( \phi_n ) \subseteq K$$ where $K$ is compact. This means that $K$ does not depend on $n$, otherwise it would be $$\mathrm{supp} ( \phi_n ) \subseteq K_n$$ However this latter condition is superfluous since by definition every $\phi_n$ already has compact support. Aug 6, 2021 at 12:11

1. For b) maybe you should write: "since $$\varphi \in \mathscr{D}(\mathbb{R}^{n})$$", given $$\alpha$$ there exists $$c$$ such that $$\| D^{\alpha}\varphi \|_{\infty}. Then $$\|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 $$\mathrm{supp}(\varphi(x/n))=n\mathrm{supp}(\varphi)$$.

• why are you using the infinity norm? And in 2, why should the first ball be centered at 0? This is something I've been seeing around but I can't get my head around it. I think that in principle the ball could be anywhere, couldn't it? Aug 6, 2021 at 12:03
• Yes, the ball could be anywhere, but only for simplicity I choosed a ball centered at zero. Since $K$ is compact, in particular is bounded and therefore there is contained in some ball $B=B(y,r)$. This ball is contained in some other ball centered at zero (take a ball with radious $=d(0,B)+r+2021$). Aug 6, 2021 at 12:06
• The infinity norm is the same as say uniform. Indeed, recall that for $f:X \to \mathbb{R}$ $$\|f\|_{\infty}=\sup _{x \in X} |f(x)|.$$ Thus $f \to 0$ uniformly iff $\|f\|_{\infty} \to 0$. Aug 6, 2021 at 12:07
• is f a sequence of functions in your last example? And does this property hold for converge to a non zero function? Aug 6, 2021 at 12:24
• Could you proof this part: $\mathrm{supp}(\varphi(x/n))=n\mathrm{supp}(\varphi)$ ? I was mistakingly taking the limit inside the argument and getting $\varphi(x/n) \to\varphi(0)$ Aug 6, 2021 at 12:27