If $f \in \mathcal C^k (\mathbb R^d)$ and $\alpha = (\alpha_1, \ldots, \alpha_n)$ is a multi-index of length $|\alpha | = \alpha_1 + \cdots + \alpha_n \le k$, then we write $$ D^\alpha f := D_1^{\alpha_1} D_2^{\alpha_2} \ldots D_n^{\alpha_n} f \quad \text{where} \quad D_i^{\alpha_i} := \frac{\partial^{\alpha_i}}{\partial x_i^{\alpha_i}}. $$
A sequence of mollifiers $(\rho_n)_{n \geq 1}$ is any sequence of functions on $\mathbb{R}^d$ such that $$ \rho_n \in C_c^{\infty} (\mathbb{R}^d), \quad \operatorname{supp} \rho_n \subset \overline{B(0,1 / n)}, \quad \int \rho_n=1, \quad \rho_n \geq 0 \text { on } \mathbb{R}^d, $$ where $B(x, r)$ denotes an open ball centered at $x$ with radius $r$. If $\rho \in C_c^{\infty} (\mathbb{R}^d)$ such that $\rho \ge 0,\operatorname{supp} \rho \subset \overline{B(0,1)}$ and that $\rho$ does not vanish identically, then we obtain a sequence of mollifiers be letting $\rho_n (x) := C n^d \rho (nx)$ for all $x \in \mathbb R^d$ with $C := \int \rho$.
I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Let $K$ be a compact subset of $\mathbb R^d$. Prove that there is a sequence $(u_n) \subset C_c^{\infty} (\mathbb{R}^d)$ such that
- $0 \le u_n \le 1$ on $\mathbb R^d$,
- $u_n = 1$ on $K$,
- $\operatorname{supp} u_n \subset K + B(0, 1/n)$,
- $|D^\alpha u_n (x)| \le C_\alpha n^{|\alpha|}$ for all $x \in \mathbb R^d$ and for all multi-index $\alpha$ (where $C_\alpha$ depends only on $\alpha$ and not on $n$).
[Hint: Let $X_n = K + B(0, 1/(2n))$ and take $u_n = \rho_{2n} * 1_{X_n}$].
I follow the hint of the author, i.e., let $X_n := K + B(0, 1/(2n))$ and $u_n := \rho_{2n} * 1_{X_n}$. Then $$ \begin{align} \operatorname{supp} u_n &\subset \overline{\operatorname{supp} \rho_{2n} + \operatorname{supp} 1_{X_n}} \\ &\subset \overline{B(0,1 / (2n))} + K + \overline{B(0,1 / (2n))} \\ &= K + \overline{B(0,1 / n)}. \end{align} $$
By Proposition 4.20 in the same book, $$ D^\alpha u_n = (D^\alpha \rho_{2n}) * 1_{X_n}. $$
By Young's inequality, $$ \|D^\alpha u_n\|_\infty \le \|D^\alpha \rho_{2n}\|_\infty \|1_{X_n}\|_1. $$
Clearly, $\sup_n \|1_{X_n}\|_1 < \infty$. Let's pick $(\rho_n)$ as mentioned in the beginning of this thread, i.e., $\rho_n (x) := C n^d \rho (nx)$. By chain rule, $$ D^\alpha \rho_{2n} (x) = C (2n)^{d + |\alpha|} D^\alpha \rho (2nx). $$
Hence $$ \| D^\alpha \rho_{2n} \|_\infty \le C (2n)^{d + |\alpha|} \|D^\alpha \rho \|_\infty. $$
So I get $(2n)^{d + |\alpha|}$ instead of $n^{|\alpha|}$ as required.
Could you explain how to get the desired constant?