Give a function $u(x) \in H^s(\mathbb{R}^n)$ with $s \in \mathbb{R},$ which the norm is $$ \|u \|_{H^s}^2=\int_{\mathbb{R}^n} (1+|\xi|^2)^{s} |\hat{u}(\xi)|^2d\xi, $$ here $\hat{u}$ is the Fourier transform of $u.$ I would like to prove that $u$ can be approximated by compact support function in $H^s.$
My attempt is to choose a sequence of appropriate cut-off function $f_\epsilon(x) \in C_c^\infty(\mathbb{R}^n)$, and I hope that $f_\epsilon(x)u(x)$ tend to $u(x)$ in $H^s$ when $\epsilon$ tends to 0.
By definition, and note that $\widehat{f_\epsilon u}=\hat{f}_\epsilon*\hat{u},$ thus we have \begin{align} \| f_\epsilon u(x)-u\|_{H^s}^2=\int_{\mathbb{R}^n} (1+|\xi|^2)^s |\hat{f_\epsilon}*\hat{u}-\hat{u}|^2 d\xi. \end{align} So I would like to choose $f_\epsilon$ so that $\hat{f_\epsilon}$ is the standard mollifier $\epsilon^{-n}\rho(\frac{\xi}{\epsilon}).$ Here $\rho(\xi)$ is a $C_c^\infty(\mathbb{R}^n)$ non-negative function with support in $\{|\xi|\leq 1\}$ and $\int \rho =1.$
However, by Amrein-Berthier theorem in harmonic analysis, a non-zero function and its Fourier transformation cannot be supported on finite measure set simultaneously. Hence the function $f$ doesn't have compact support.
What should I do? Can anyone give me some advice?
