1
$\begingroup$

Def. 1. $\phi:\mathbb{R}^n\to \mathbb{R}$ is called test fuction if $\phi$ is infinitely differentiable ($\phi \in C^{\infty}(\mathbb{R}^n)$) and $\phi$ has compact support (ie the closure of the set $\{ x\in \mathbb{R}^n\,:\,\phi(x)\neq 0\}$ is a compact subset of $\mathbb{R}^n$).

Def.2. Let $\{\phi_m\}_m$ be a sequence of test functions. $\{\phi_m\}_m$ is called a null sequence if:

  1. There is a compact set $K\subset \mathbb{R}^n$ containing the supports of all $\phi_m$, $$\bigcup_{m\in \mathbb{N}}\operatorname{supp}(\phi_m)\subset K. $$
  2. For each multi-index $k=(k_1,\dots,k_n)$ $$\lim_{m\to +\infty}\max_{x\in \mathbb{R}^n}|D^k\phi_m(x)|=0$$ where $D^k\phi_m$ denotes the partial derivative of $\phi_m$, $$D^k\phi_m=\frac{\partial^{k_1+\cdots+k_n}}{{\partial x_1}^{k_1}{\partial x_2}^{k_2}\cdots {\partial x_n}^{k_n}}\phi_m.$$

Note that 2. is equivalent saying that $\{\phi_m\}$ converges uniformly to $0$ in $\mathbb{R}^n$, and so does the sequence $\{D^k\phi_m\}$ for every $k$.

What I have to show is the following

Prop. Let $\{\phi_m\}_m$ a null sequence and $a(x)\in C^{\infty}(\mathbb{R}^n)$. Then $\{a\phi_m\}_m$ is a null sequence.

proof

For every $m\in \mathbb{N}$ the product function $a\phi_m$ is smooth since $a,\phi_m$ are smooth and also $\operatorname{supp}(a\phi_m)\subset \operatorname{supp}(\phi_m)$. Then $\{a\phi_m\}_m$ is a sequence of test functions. Property 1 from Def.2. for $\{a\phi_m\}_m$ easily follow since $$\bigcup_{m\in \mathbb{N}}\operatorname{supp}(a\phi_m)\subset \bigcup_{m\in \mathbb{N}}\operatorname{supp}(\phi_m).$$ It remains to show that $$\lim_{m\to +\infty}\max_{x\in \mathbb{R}^n}|D^k\,a\phi_m(x)|=0$$ for every $k$.

I really don't know how to proceed in order to prove the last statement. Any hint would be really appreciated. Thanks in advance.

$\endgroup$
  • $\begingroup$ I think for $k=(0,\dots,0)$ it is very simple. Let $m\in\mathbb{N}$ and $x \in \mathbb{R}^n$. Since every $\phi_m=0$ outside the same $K$ and $K$ is compact, then $|a\phi_m(x)|=|a(x)\phi_m(x)|=|a(x)||\phi_m(x)|\leq \max_{x\in K}|a(x)|\cdot \max_{x\in \mathbb{R}^n}|\phi_m(x)|$. Since $m$ and $x$ where arbitrary we have $$\max_{x\in \mathbb{R}^n}|a\phi_m(x)|\leq C\cdot \max_{x\in \mathbb{R}^n}|\phi_m(x)|$$ for every $m$, where $C=\max_{x\in K}|a(x)|$.So $\max_{x\in \mathbb{R}^n}|a\phi_m(x)|\to 0$ as $m\to +\infty$ since by hypothesis $\max_{x\in \mathbb{R}^n}|\phi_m(x)|\to 0$ as $m\to +\infty$. $\endgroup$ – eleguitar Sep 4 at 16:32
1
$\begingroup$

$D^k a\phi_m$ can be written as a linear combination of terms of the form $D^c a D^d \phi_m$ where $c$ and $d$ are multi-indices. $|D^c a|$ will attain a maximum value since it vanishes outside of $K$ so you can estimate that by a constant and then just use the fact $\{\phi_m\}$ is a null sequence.

$\endgroup$
  • 1
    $\begingroup$ $|D^c a|$ will attain a maximum in $K$ because it's continuous in $K$, but we don't know if it vanishes outside $K$ ($a$ is just a smooth function). Sure $|D^c a D^d \phi_m|$ is zero outside $K$ because $\{\phi_m\}$ is a null sequence, so we can write $|D^c a(x) D^d \phi_m (x)|\leq \max_{x \in K} |D^c a(x)| \max_{x \in \mathbb{R}^n} | D^d \phi_m(x)|$ for every $x$ and $m$. $\endgroup$ – eleguitar Sep 5 at 17:21
  • $\begingroup$ Right, I didn't see that he just requires a to be smooth $\endgroup$ – Jonathan Hole Sep 5 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.