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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
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  • $\begingroup$ 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... $\endgroup$
    – peek-a-boo
    Aug 6, 2021 at 11:05
  • $\begingroup$ @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? $\endgroup$ Aug 6, 2021 at 11:12
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    $\begingroup$ The support of $\phi (x/n)$ becomes bigger and bigger, hence it goes outside any compact set as $n \to \infty$ $\endgroup$
    – Crostul
    Aug 6, 2021 at 11:17
  • $\begingroup$ @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? $\endgroup$ Aug 6, 2021 at 11:49
  • $\begingroup$ 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. $\endgroup$
    – Crostul
    Aug 6, 2021 at 12:11

1 Answer 1

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  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 $\mathrm{supp}(\varphi(x/n))=n\mathrm{supp}(\varphi)$.

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  • $\begingroup$ 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? $\endgroup$ Aug 6, 2021 at 12:03
  • $\begingroup$ 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 $=d(0,B)+r+2021$). $\endgroup$
    – Sebathon
    Aug 6, 2021 at 12:06
  • $\begingroup$ 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 $\|f\|_{\infty} \to 0$. $\endgroup$
    – Sebathon
    Aug 6, 2021 at 12:07
  • $\begingroup$ is f a sequence of functions in your last example? And does this property hold for converge to a non zero function? $\endgroup$ Aug 6, 2021 at 12:24
  • $\begingroup$ 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)$ $\endgroup$ Aug 6, 2021 at 12:27

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