# First derivative of cut off function

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.

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