By $H^2(\mathbb{R})$ denote the Sobolevspace $$ W^{2,2}(\mathbb{R}):=\left\{u\in L^2(\mathbb{R}): D^{\alpha}u\in L^2(\mathbb{R})~\forall\lvert\alpha\rvert\leq 2\right\} $$ which has an inner product $$ \langle u,v\rangle_{H^2}:=\sum_{i=0}^2\langle D^iu,D^iv\rangle_{L^2} $$ and norm $$ \lVert u\rVert_{H^2}:=\left(\sum_{\lvert\alpha\rvert\leq 2}\lVert D^{\alpha}u\rVert_{L^2}^2\right)^{1/2}. $$
Let $u\in H^2(\mathbb{R})$. Do we then have that $u\to 0$ and $Du\to 0$ as $x\to\pm\infty$?
idea
Suppose, $u$ is not tending to zero as $x\to\pm\infty$. Since $\lvert u(x)\rvert^2\geq 0$ for all $x\in\mathbb{R}$, I think this would imply that $\int_{\mathbb{R}}\lvert u(x)\rvert^2\, dx = \infty$, contradicting the assumption that $u\in H^2(\mathbb{R})$ since this means that $u\in L^2(\mathbb{R})$, i.e. $\int_{\mathbb{R}}\lvert u(x)\rvert^2\, dx <\infty$.