It is a well known result that $H^1(\mathbb{R}^{n})$ does not have compact immersion in $L^2(\mathbb{R}^{n})$. However, if $s<t$ the inclusion map $H_{K}^{t}(\mathbb{R}^{n}) \rightarrow H^s(\mathbb{R}^{n})$ is compact, where $H_{K}^{t}(\mathbb{R}^{n})=\{u \in H^t(\mathbb{R}^{n}): \operatorname{supp} u \subset K\}$.

My question: Let $(u_n) \subset H^1(\mathbb{R}^n)$ be a sequence such that $\|u_n\|_{H^1(\mathbb{R}^{n})}\leq C$ and $u_n \rightharpoonup u$ weakly in $H^1(\mathbb{R}^n)$. In addition, suppose that $\operatorname{supp} u$ is compact. Does there exists an index $n_0 \in \mathbb{N}$ and a compact $K$ such that $\operatorname{supp} u_n \subset K$ for all $n \geq n_0$?

If not, is there any weak hypothesis that can be added for this to be true?

If that were possible, we could apply the above-mentioned result to $(v_n)=(u_{n_0+n})$ in order to obtain a strong convergence in $L^2(\mathbb{R}^{n})$. Intuitively it's look true, but I don't even how to start.


1 Answer 1


The answer is definitely no. Take a non-zero function $u_0 \in C^\infty_0(\mathbb{R}^n)$ and define $u_n(x) = u_0(x - nx_0)$ where $0 \ne x_0 \in \mathbb{R}^n$. Then $u_n \rightharpoonup 0$, but every compact $K$ contains $\text{supp} \, u_n$ for only finitely many $n$.

Where does this problem arise?

  • $\begingroup$ Thank you! I'm working with some estimates for the resolvent operator associated with the wave equation, and I was trying to analyze if the results are valid in unbounded domains. $\endgroup$
    – Math
    Aug 9, 2021 at 23:20
  • 1
    $\begingroup$ Look at recent work on the concentration-compactness principle for evolution equations by Kenig, Merle, Schlag, and others. Maybe that will give you some ideas. $\endgroup$ Aug 9, 2021 at 23:25

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