Functions of Sobolev space with asymptotic decay

Question

Define a subspace of the Sobolev space $H^1(\mathbb R^d)$ to be $$ X=\{u\in H^1(\mathbb R^d), |u(x)|=O(|x|^{1-d}), \text{ as } |x|\rightarrow +\infty\} $$ Is there a norm $\|\cdot\|_X$ such that $(X, \|\cdot\|_X)$ is a Banach space? I tried $$ \|u\|_X := \left (\int_{\mathbb R^d} |\nabla u(x)|^2 dx\right )^{1/2}, $$ but failed to see if $X$ is closed with this norm.

Answer 1

If $d=1$, then $X=H^1(\mathbb R)$ is complete. Below I assume $d\ge 2$.

Under the $H^1$ norm, $X$ is not complete. Indeed, let $v$ be an $H^1$ function with $v(0) =1$, vanishing outside the unit ball. Define $$u(x) = \sum_{n=1}^\infty 2^{-n}v(x-3^n\xi)$$ where $\xi$ is a fixed unit vector. The series converges in $H^1$, and every partial sum is in $X$, since it's compactly supported. But the total sum has $$\limsup_{|x|\to\infty} |x|^{d-1} |u(x)| \ge \limsup_{n\to\infty} 3^{(d-1) n} 2^{-n} =\infty$$

I would consider the norm $$\|u\|_* = \|u\|_{H^1} + \operatorname*{ess\,sup}_{x\in \mathbb R^d}(1+|x|^{d-1})|u(x)|$$ However, it requires $u$ to be essentially bounded in addition to being in $H^1$. Generally, $H^1$ functions may be unbounded. But since you insist on pointwise decay at infinity, perhaps you do not want to consider unbounded functions?