$u\in H^1(\mathbb{R}^N)\cap C^1 \Rightarrow u$ vanishes at infinity?

Question

Suppose $u\in H^1(\mathbb{R}^N)\cap C^1 $, where $H^1$ is the Hilbert-Sobolev space. Is it true that the limit of $u$ at infinity in any direction is 0?

Answer 1

For $N=1$ it is true that these functions vanish at infinity.

Assume $f \in H^1 \cap C^1$ but does not vanish at infinity. Then there exists some $\epsilon>0$ and a sequence $(x_n)$ of points going to infinity such that $f(x_n) > 2\epsilon$ for all $n \in \mathbb{N}$.

On the other hand, since $f \in L_2$ we know that $f(y_n) < \epsilon$ for another sequence $(y_n)$ satisfying $x_n < y_n < x_{n+1}$.

This implies a contradiction to $f'\in L_2$:

$$\|f\|_{H^1}^2 \ge \|f'\|_{L^2} \ge \sum_{n} | \int_{x_n}^{y_n} f'(x) dx \; | > \sum_{n} \epsilon = \infty$$

However, for $N>1$ it turns out that we can construct a sequence of radial smooth $\epsilon$-bumps that are in $H^1$ but are obviously not vanishing at infinity.