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.


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.

  • $\begingroup$ 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 '19 at 16:32

$D^k a\phi_m$ 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.

  • 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 '19 at 17:21
  • $\begingroup$ Right, I didn't see that he just requires a to be smooth $\endgroup$ Sep 5 '19 at 18:33

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