Must continuous $H^1(\mathbb{R}^2)$ function tend to zero at infinity?

Question

Here, $H^1(\mathbb{R}^2)$ is the standard Sobolev spaces for $L^2(\mathbb{R}^2)$ functions whose weak derivative belongs to $L^2(\mathbb{R}^2).$

My question in the title comes from calculus of variations. It is usually the case that a minimizer of some given energy functional defined on $H^1(\mathbb{R}^2)$ is known to be continuous (or even $C^2(\mathbb{R}^2))$. I want to know the behavior of this minimizer at infinity.

If $u \in L^2(\mathbb{R}^2),$ then is known $\liminf_{|x| \to \infty} u(x) = 0.$ But it cannot say $\limsup_{|x| \to \infty} u(x) = 0$ since counterexamples exist.

If we assume $u \in H^{1+\epsilon}(\mathbb{R}^2)$ for some $\epsilon > 0,$ then the classical Morrey's inequality can imply uniform H\"older continuity of $u.$ So we can conclude $\limsup_{|x| \to \infty} u(x) = 0$ via proof by contradiction.

So my problem is about the case $\epsilon = 0.$ That is, when $$u \in H^1(\mathbb{R}^2) \cap C(\mathbb{R}^2),$$ is it true that $$\limsup_{|x| \to \infty} u(x) = 0?$$

Using proof by contradiction, I think this should be true. Here is my non-rigorous argument.

Assume not, then there are $\epsilon > 0$ and $x_n \in \mathbb{R}^2$ such that $|x_n| \to \infty$ and $|u(x_n)| \geq 2\epsilon.$ By the continuity, there is $r_n > 0$ such that $|u(x)| \geq \epsilon$ for all $x \in B(x_n, r_n).$ Since $u \in L^2, r_n \to 0$ as $n \to \infty.$ I think non-rigorously that $$ \int_{B(x_n, r_n)} |\nabla u|^2 \gtrsim \int_{B(x_n, r_n)} (\frac{\epsilon}{r_n})^2 = \epsilon^2$$ for large $n$ and $$ \int_{\mathbb{R}^2} |\nabla u|^2 \geq \sum_{n\,\text{is large}} \int_{B(x_n, r_n)} |\nabla u|^2. $$ So they imply a contradiction $\int_{\mathbb{R}^2} |\nabla u|^2 = \infty$

I appreciate any discussion.

Edit: How about $u$ is additionally assumed to be $C^1(\mathbb{R}^2)$ or even $C^2(\mathbb{R}^2)?$ Is there any proof or counterexample?

Answer 1

It does not hold for general functions $u \in H^{1}(\mathbb{R}^{2})$, even if they are assumed to be smooth. The reason lies in the following lemma:

Let $N \ge 2$. Then for any $x_{k} \in \mathbb{R}^{N}, \; \epsilon>0$ and $\delta>0$ there exists a radial smooth function $u_{k}$ such that:

1. $u_{k}(x_{k})=1$

2. $u_{k}(x)=0$ for $|x-x_{k}|>\delta$

3. $\int_{\mathbb{R}^{n}} |\nabla{u_{k}}|^2 \le \epsilon$

Now choose a sequence of points $x_{k}$ going to infinity, choose corresponding $\delta_{n}=\frac{1}{n^{2}}$ and $\epsilon_{n}=\frac{1}{n^{2}}$.

Let $u=\sum_{n} u_{n}$.

Then $u \in H^{1}(\mathbb{R}^{2})$.

But for any $r>0$ we find some point $x_{r}$ with $|x_{r}|>r$ such that $u(x_{r})=1$.

Answer 2

This seems to be not a general property for functions in $H^1(\mathbb{R}^2) \cap C(\mathbb{R}^2)$ because I find the following remark from this book.