Let $\Omega$ be an open subset of $\mathbb R^k$. we known that $L^s(\Omega)=\overline{C_{c}^{\infty}(\Omega)}^{\| \|_{s}}$ is a norm space with $\| \|_{s}$ sobolev's norm.
Too we have $$ L^s_\text{loc}(\Omega):=\{ f:\Omega\to \mathbb C \ \text{measurable} \mid f\in L^s(K) \ \ \ \forall K \subset \subset \Omega \} $$ is a fréchet space with seminorm using compact $K_{i}\subset K_{i+1}\subset \Omega$.
We know
$$L^s(\Omega) \subset L^s_\text{loc}(\Omega)$$
but my question is, what happen if $\overline{\Omega}$ is compact?, what can i say about $L^s(\Omega)$ and $L^s_\text{loc}(\Omega)$, are the same?
$L^s_\text{loc}(\Omega)$ is normed if $\overline{\Omega}$ is compact?. thanks.
In general the answer is no. Let $\Omega := B(0,1)$ be the unit open ball in $\mathbb{R}^k$, whose closure is compact. Let $F(x) := (1-|x|^2)^{-10k/s}$. Since $F$ is bounded on every compact subset of $\Omega$, $F \in L^{s}_{\text{loc}}(\Omega)$. (Here we used the fact that for every compact subset $K$ of $\Omega$, there exists a constant $r \in (0,1)$ such that $K \subset B(0,r)$.) However $F \notin L^s (\Omega)$, because $F$ grows very fast near the boundary of $\Omega$.