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?