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 $r\in (0,1]$.
Here im doing $N\geq 1$ case.
We have $|D_{\alpha}k_r(x)|\leq c_{\alpha}r^{-|\alpha|}$ for all $\alpha \in \mathbb{N}_0^n$. Let $$V = \left\{\psi \in C_0^{\infty}(\mathbb{R}^n)\right|\partial^{\alpha}\psi(0)=0 \ \text{for all} \ |\alpha|\leq N\}$$
I want to show for $V$ the inequalities, $x\in \mathbb{R}^n, |\alpha|\leq N, r\in [0,1)$. $$|\psi(x)|\leq c|x|^{N+1} $$ $$ |\partial^{\alpha}(k_r(x)\psi(x))|\leq c'r^{N+1-|\alpha|}$$ $$ |\left\langle u,\psi k_r \right\rangle| \leq c'r $$
Im getting better at this...but not Im fully up to speed yet. Any help is appreciated.
By Taylor's Theorem in the multivariate version you have for $|x|\leq 2$ $$ \psi(x) = \sum_{|\alpha|\leq N} \frac{(D^\alpha \psi)(0)}{\alpha!}x^\alpha + \sum_{|\beta|=N+1} R_\beta x^\beta, \text{where } |R_\beta| \leq \frac{1}{\beta!}\sup_{|x|\leq 2, |\gamma|=|\beta|} |(D^\gamma\psi)(x)|. $$ For $|\beta|=N+1$ you have $|x^\beta|\leq |x|^{N+1}$, so your first inequality.
Apply this theorem to $D^\gamma\psi$ when $|\gamma|<=N$ you see $$ |(D^\gamma\psi)(x)| <=R_\gamma|x|^{N+1-|\gamma|}. $$
Now, by the Leibniz product rule, you have $$ D^\alpha(k_r(x)\psi(x)) = \sum_{|\beta|+|\gamma|=|\alpha|}\frac{\alpha!}{\beta!\gamma!} r^{-|\beta|}(D^\beta \chi)(x/r) (D^\gamma \psi)(x). $$
Combining these two results and observing that for $(D^\beta \chi)(x/r) = 0$ for $|x|>2r$ you have
$$ |D^\alpha(k_r(x)\psi(x))| \leq C_\alpha r^{N+1-|\gamma|}r^{-|\beta|} = C_\alpha r^{N+1-|\alpha|}, $$ where $C_\alpha$ does not depend on $r$, your second inequality. Finally, as the distribution $u$ has order $N$, you have $$ |\langle u,k_r\psi\rangle| \leq C \sum_{|\alpha|\leq N} \sup |D^\alpha(k_r\psi)| \leq C' r. $$