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I have the following statement:

Let $K\subset \mathbb{R}$ be a compact set and define $H^s_K(\mathbb{R})=\{u\in H^s(\mathbb{R}); supp\,\,u\subset K\}$, for $s\in \mathbb{R}$. Then $H^s_K(\mathbb{R})$ is a closed subset of $H^s(\mathbb{R})$.

Here, $supp\,\, u$ denotes the support of $u$ and $H^s$ the Sobolev space of functions in the Schwartz space.

To prove it, I have tried to take a sequence $(f_n)\subset H^s_K$ converging to $f\in H^s$, but I cannot conclude that $supp\,\,f\subset K$, although it makes sense to think that considering that $supp\,\, f_n \subset K$ for all $n$.

Any ideas on the proof?

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  • $\begingroup$ Just an idea. Choose $\psi$ to be a smooth function with compact support in inside $K^C$, the complement of $K$. Then, the multiplication operator $M_\psi: f \mapsto f \psi$ is bounded in $H^s$ and $H^s_K$ is inside the kernel of $M_\psi$. If that is true, you can cover the complement $K^c$ with a smooth partition of unity $(\psi_k)_k$ and describe $H_K^s$ as the intersection of the kernels of $M_{\psi_k}$, which is closed. $\endgroup$ Commented Apr 3, 2020 at 9:20

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