First derivative of cut off function

Question

I am working on proving the following: Let $\rho(x)= \frac{2}{2+x^2}$, $\theta >1$ (assumed integer here) and $B \subset H^1_{ul}$,(uniformly local Sobolev space), be any subset which is bounded in $H_{ul}^\theta$. Then $B$ is pre-compact in

$$H^1_\rho = \{ u: \mathbb R \to \mathbb R : u, \partial_x u \in L^2_{loc} (\mathbb R),\quad \| u\|_{H^1_\rho}=\| u\rho\|_{H^1} < \infty\}.$$

I need to show that $B$ admits the finite covering in $H^1_\rho$ by balls of radius less than $\epsilon $. Taking $\chi_\beta$ as a smooth cut off function and decomposing $u \in B$ into two parts $u= v+u$ where $v= \chi_\beta u$ and $w= u \chi_\beta$, and

$$\chi_\beta(X) = \begin{cases} 1,& |x| \ge \beta +1\\ 0, & |x| \le \beta \end{cases}.$$

Taking the following norm

$$\|v\|^2_{H^1_\rho}=\|u\chi_\beta\rho\|^2_{H^1} =\|u\chi_\beta\rho(x)\|^2_{L^2} + \| \partial_x (u\chi_\beta\rho)\|^2_{L^2}.$$

Here I can choose $\beta$ sufficiently large to get the first $L^2$ small as I want but in the second norm after using the product rule I obtain the derivative of the cut-off function which I do not know what it is and how to deal with it? Could enlighten me, please.

Answer 1

I constructed a function that satisfies what I want. Let $f(x) = e^{-1/x},\,\, x>0$ and vanishes everywhere. Now we define $\varphi_\beta(x) = f(\beta +1 - |x|)$ and $\psi_\beta (x)=f(|x|- \beta)$. Therefore the cut-off function is

$$\chi_\beta(x) = \frac{\psi_\beta (x)}{\psi_\beta (x) + \varphi_\beta (x)}.$$