Let the potential $V(x)$ be a bounded continuous function. I want to show that
The operator $-\Delta + V : H^{s} \rightarrow H^{s-2}$ is continuous for every $s\in[0,2]$.
where $H^{s}$ is inhomogeneous Sobolev space with norm $$\|u\|_{H^s} = \|(1+|\xi|^{2})^{s/2}\hat{u}\|_{L_\xi^2}.$$ I have proved already $\Delta:H^s \rightarrow H^{s-2}$ is continuous for all $s\in\mathbb{R}$ using the fact that $\Delta$ is differential(so it is linear) and bounded operator, i.e. showing $\|\Delta u \|_{H^{s-2}} \le \|u\|_{H^s}$. However I can't do that for multiplication $V$. I tried to show that, for any $u \in H^s$ and $s\in[0,2]$, $$ \|V(x)u(x)\|_{H^{s-2}} \le C\|u(x)\|_{H^{s}}.$$ And below is my approach:
$$\|Vu\|_{H^{s-2}}=\|(1+|\xi|^2)^{(s-2)/2}\widehat{Vu}\|_{L_\xi^2}\le \|(1+|\xi|^2)^{s/2}\widehat{Vu}\|_{L_\xi^2}=\|Vu\|_{H^s},$$ since $s-2<s\le 0$. I have stucked here.
For $u\in L^2$ we have $\Vert V u \Vert_{L^2} \le \Vert V \Vert_{L^\infty}\cdot \Vert u \Vert_{L^2}$ (Hölder-inequality), that means that $V:L^2\rightarrow L^2$ is continuous. For $s\in [0,2]$ we have $H^s\subset L^2$ and $L^2\subset H^{s-2}$, such that $V:H^s\rightarrow H^{s-2}$ is also continuous.