# Lemma involving Sobolev spaces and compact sets

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?

• 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. Commented Apr 3, 2020 at 9:20