Let $u$ be a distribution on $\mathbb{R}^n$ with support = $\left\{0\right\}$. Then there exists $N$ such that $u$ has order $N$. Let $\chi\in C_0^{\infty}(\mathbb{R}^n)$ a smooth function with
$\chi(x) = 1$ for $0\leq |x|\leq 1 $,
$\chi(x)\in [0,1]$ for $1\leq |x|\leq 2$,
$\chi(x)=0$ for $|x|\geq 2$
Denote $ \chi(x/r) = k_r(x)$ for $r\in (0,1]$. For the case $N=0$ I want to show that there exists $c_1$ such that $$|\left\langle u,\phi \right\rangle|\leq c_1 |\phi(0)| $$
for all $\phi \in C_0^{\infty}(\mathbb{R}^n)$. The idea would be to apply $u$ on $\phi = k_r\phi+(1-k_r)\phi$ and letting $r\to 0$.
So then I guess we can write $$ |\left\langle u,\phi \right\rangle| \leq |\left\langle u,k_r\phi \right\rangle| + |\left\langle u,(1-k_r)\phi \right\rangle |$$
But not really sure how this follows…
Moreover how can i show that $\left\langle u,\phi \right\rangle = \left\langle u,\phi \chi\right\rangle = \left\langle u,\chi\right\rangle\phi(0) $?
Best Answer
The precise and quite clear proof of the statement above can be found in Hoermander "Linear PDE's" vol. 1. p. 12-13, theorems 1.5.3 and 1.5.4.
Best.