Inequality about disbtribution in Functional Analysis by Rudin I'm reading Functional Analysis by Rudin about distribution theory. I have a problem of derivation of the inequality (10) in theorem 6.25, page 166.
It first proves an inequality
$$(5)\quad\quad\quad\quad |D^{\alpha}\phi(x)|\le\eta n^{N-|\alpha|}|x|^{N-|\alpha|}\quad\quad\quad(x\in K,|x|\le N)$$ where $\phi\in\mathscr{D}(\Omega)$ ,$\eta\gt 0$,$n$ is that in $R^{n}$ and $N$ is a natural number. I have no problem with this inequality.
Then choose an auxiliary function $\psi\in\mathscr{D}(R^{n})$, which is $1$ in some neighborhood of $0$ and whose support is in the unit ball B of $R^{n}$. Define 
$$\psi_{r}(x)=\psi(\frac{x}{r})\quad\quad(r\gt 0)$$
I think the precise definition of $\psi$ has nothing to do with the following derivation. I put it here just for completeness.
By Leibniz formula, 
$$(9)\quad\quad\quad D^{\alpha}(\psi_{r}\phi)=\sum_{\beta\le\alpha}c_{\alpha\beta}(D^{\alpha-\beta}\psi)(\frac{x}{r})(D^{\beta}\phi)(x)r^{|\beta|-|\alpha|}$$
It now follows from (5) that
$$(10)\quad\quad\quad\|\psi_{r}\phi\|_{N}\le\eta C\|\psi\|_{N}$$
where C is a constant depending on $n$ and $N$. The norm $\|\cdot\|_{N}$ is defined as $$\|f\|_{N}=\sup\{D^{\alpha}f(x),x\in X,\alpha\le N\}$$
My question is how to derive (10) from  (5) and what $C$ is. I tried myself but I got a constant depending on $r$ as well.
 A: Look at $(9)$. If $\lvert x\rvert > r$, then $(D^{\alpha-\beta}\psi)(x/r) = 0$ since the support of $\psi_r$ is contained in $rB$. If $\lvert x\rvert \leqslant r$, then $(5)$ shows
$$\Bigl\lvert (D^{\beta}\phi)(x) r^{\lvert\beta\rvert -\lvert\alpha\rvert}\Bigr\rvert \leqslant \eta n^{N-\lvert\beta\rvert} \lvert x\rvert^{N-\lvert\beta\rvert} r^{\lvert\beta\rvert - \lvert\alpha\rvert} \leqslant\eta n^{N-\lvert\beta\rvert} r^{N-\lvert\alpha\rvert}.\tag{$\ast$}$$
Since we are interested in small $r > 0$, we may without loss of generality assume $r \leqslant 1$, and then from $(\ast)$ we obtain
$$\Bigl\lvert (D^{\beta}\phi)(x) r^{\lvert\beta\rvert -\lvert\alpha\rvert}\Bigr\rvert \leqslant \eta n^{N-\lvert\beta\rvert} \tag{${\ast}{\ast}$}$$
as $r^{N-\lvert\alpha\rvert} \leqslant 1$ because $\lvert\alpha\rvert \leqslant N$. Inserting $({\ast}{\ast})$ into $(9)$ yields
$$\lvert D^{\alpha}(\psi_r\phi)(x)\rvert \leqslant \sum_{\beta\leqslant \alpha} c_{\alpha\beta} \bigl\lvert(D^{\alpha-\beta}\psi)(x/r)\bigr\rvert\cdot \eta n^{N-\lvert\beta\rvert} \leqslant \eta \lVert\psi\rVert_N\cdot \underbrace{\sum_{\beta\leqslant \alpha} c_{\alpha\beta} n^{N-\lvert\beta\rvert}}_{C_{\alpha}}$$
for $\lvert x\rvert \leqslant r$. Since the inequality is trivial for $\lvert x\rvert > r$,
$$\sup \{ \lvert D^{\alpha}(\psi_r\phi)(x)\rvert : x \in \mathbb{R}^n\} \leqslant \eta \cdot C_{\alpha} \cdot \lVert \psi\rVert_N.$$
Take $C = \max \{ C_{\alpha} : \lvert \alpha \rvert \leqslant N\}$ to obtain $(10)$.
