# Null sequences and uniform convergence

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.

• 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 $m\to +\infty$. – eleguitar Sep 4 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.
• $|D^c a|$ 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 $m$. – eleguitar Sep 5 at 17:21