I'm trying to prove the following result (it's an exercise).
Let $V$ be a continuous function from $\mathbb{R}^{d}$ to $\mathbb{R}^+$ such that there exist $A$ and $m$ two strictly positive numbers such that $\forall x \in \mathbb{R}^d$, $\lvert x\rvert \geq A \implies V(x) \geq m$
Show that there exists a strictly positive constant $C$ such that $\forall u \in H^1(\mathbb{R}^d)$, $\lVert x \rVert^2_{L^2} \leq C\lVert \nabla u\rVert^2_{L^2} + \int_{\mathbb{R}^d} V(x)u^2(x)dx$
It looks a lot like some sort of weighted Poincaré Inequalities but as far as I remember they are true on bounded set and here we are working on $\mathbb{R}^d$ so how can I prove this ?
Thanks
It is enough to prove the inequality on $B:=\{x: \ |x|\le A+1\}$ instead of on $\mathbb R^d$.
Assume the inequality is not valid on $B$. Then for every $n$ there is $u_n$ such that $$ \|u_n\|_{L^2(B)} > n (\|\nabla u_n \|_{L^2(B)} + \int_B Vu_n^2). $$ Due to the strict inequality, it follows $u_n\ne0$, and we can assume $\|u_n\|_{L^2(B)}=1$ (or divide the inequality by $\|u_n\|_{L^2(B)}$). Then $(u_n)$ is bounded in $H^1(B)$, $\nabla u_n \to0$, $\int_B Vu_n^2\to0$. Now we can extract sequences such that $u_n\rightharpoonup u$ in $H^1(B)$, $u_n\to u$ in $L^2(B)$. Here, we use that the domain $B$ is bounded. It follows $\nabla u=0$, $u=const$, and $u=0$ (due to $u=const$ and $\int_B Vu^2=0$). This is a contradiction to $1 = \|u_n\|_{L^2} \to \|u\|_{L^2(B)}$.
Hence, there is $C>0$ such that $$ \|u\|_{L^2(B)} \le C (\|\nabla u \|_{L^2(B)} + \int_B Vu^2) \quad \forall u\in H^1(B). $$ Due to the assumptions on $V$, the claimed inequality on $\mathbb R^d$ follows.