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Consider the space $H^s=H^s(\mathbb{R}^N)$, where $0<s<1$. Take any $u \in H^s$ and any smooth function $\varphi$ such that $\operatorname{supp}\varphi \subset B(0,R)$, for some radius $R>0$. Moreover, $\varphi(x)=1$ if $|x| < R/2$ and $|\nabla \varphi| \leq C/R$. Is it true that $$\lim_{R \to +\infty} \|u-\varphi u\|_{H^s}=0?$$

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You can assume that $\varphi=1$ on $B(0,R/2)$ and $\varphi=0$ outside $B(0,R)$, and that $|\nabla \varphi| \leq 2/R$. – Siminore Apr 22 '12 at 16:12
up vote 2 down vote accepted

Edit: Well, I think I have it mostly.

First, it is true for $s=1$, which is pretty easy to verify.

Now choose $\tilde{u} \in H^1$ (or even $\tilde{u} \in C^\infty_c$) with $\| u - \tilde{u} \|_{H^s}< \epsilon$. We have by the triangle inequality $$\| u - \varphi u \|_{H^s} \le \| u - \tilde{u} \|_{H^s} + \| \tilde{u} - \varphi \tilde{u}\|_{H^s} + \| \varphi \cdot (u-\tilde{u}) \|_{H^s}.$$ The first term is at most $\epsilon$. The second term is bounded by $\| \tilde{u} - \varphi\tilde{u}\|_{H^1}$ (since $\| \cdot \|_{H^s} \le \| \cdot \|_{H^r}$ whenever $s < r$, which is easy to verify), and this goes to 0 as $R \to \infty$.

For the third term, I would like to claim $\|\varphi v\|_{H^s} \le C \|v\|_{H^s}$ whenever $\varphi \in C^\infty_c$, as is the case here (i.e. $v \mapsto \varphi v$ is continuous on $H^s$). Since it is true for $s=0$ and $s=1$, we should be able to invoke some interpolation theorem and get it for all $s \in (0,1)$. But there should be an easy way too, I just don't see it. It's worth noting that by the closed graph theorem, it would suffice to show $\varphi v \in H^s$ whenever $v \in H^s$.

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Could you give the details for $s=1$? – Davide Giraudo Apr 22 '12 at 16:50
When $s=1$, it is a standard tool of PDE's. Just expand the gradient of the product $\varphi u$ and look at the integrals. – Siminore Apr 22 '12 at 16:56
Actually, the suggested solution is wrong: we have a function in $H^s$, but we cannot assume it lies in $H^1$. Too easy to be correct, sorry. – Siminore Apr 22 '12 at 17:35
@Siminore: Oh, good point. But I think a density argument will sort it out. I'll try and update later today. – Nate Eldredge Apr 22 '12 at 17:40
@Siminore: I edited, though I don't see quite how to do the last step. – Nate Eldredge Apr 22 '12 at 20:35

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