Fix $s\in\Bbb R^+$ and define the $H^s$-norm in $C_c^\infty(\Bbb R^n)$ by $$\|f\|^2_s=\int_{\Bbb R^n}|\hat{f}(y)|^2(1+|y|^2)^s\,dy$$ Then, $H^s(\Bbb R^n)$ is defined as the completion of $C_c^\infty(\Bbb R^n)$ with respect to $\|\cdot\|_s$. A priori, $H^s(\Bbb R^n)$ and $L^2(\Bbb R^n)$ are different spaces. I would like to show that $H^s(\Bbb R^n)\subset L^2(\Bbb R^n)$. Clearly, there is a map from the former to the latter. I need to show that this map is injective. Hence, I need to show the following:
If $\{f_n\}\subset C_c^\infty(\Bbb R^n)$ is a Cauchy sequence with respect to the Sobolev $H^s$-norm such that $\|f_n\|_{L^2}\to0$, then $\|f_n\|_s\to0$.
This question should be very basic and trivial. However, I couldn't come up with a proof. Thanks.
I finally managed to prove the result.
Let $L_s^2=L^2(\Bbb R^n,(1+|x|^2)^sdx)$ be the weighted $L^2$-space. Obviously, $L_s^2\subset L^2$ and $\|f\|_s=\|\hat{f}\|_{L_s^2}$. The assumption implies that $\{\hat{f_n}\}$ is Cauchy in $L_s^2$. Since $L_s^2$ is complete, $\hat{f_n}\to g$ for some $g\in L_s^2$. Therefore, $\hat{f_n}\to g$ in $L^2$. Moreover, $\|\hat{f_n}\|_{L^2}=\|f_n\|_{L_2}\to0$. Therefore, $g=0$. This simply means that $\|f_n\|_s=\|\hat{f_n}\|_{L_s^2}\to0$