In Reed, Simon, Methods of Mathematical Physics, it defines the Sobolev space $W_a, a\in \mathbb{R}$ as the set of tempered distributions $\mu\in S'(\mathbb{R}^n)$ such that its Fourier transform $\hat{\mu}$ is measurable and
$$
|| \mu ||_a^2 = \int (1+p^2)^a |\hat{\mu}|^2 dp <\infty
$$
Let $\Omega \subseteq \mathbb{R}^n$ be open. Then the local Sobolev spaces on $W_a (\Omega)$ is the set of distributions $\mu \in D'$ such that $\phi \mu \in W_a$ for every $\phi \in D(\Omega) = C_c^\infty(\Omega)$.
It seems that if $\Omega = \mathbb{R}^n$, then the local Sobolev space $W_a (\Omega)$ should equal the Sobolev space $W_a$. However, I can't seem to prove it. Any help would be appreciated.
Best Answer
This is false. In the same way, locally integrable functions are not necessarily integrable on $\mathbb{R}^d$. You need some decay at infinity to be in global Sobolev spaces.
Take for example $μ = 1$. Then for any $\varphi\in C^\infty_c$, $\varphi\,\mu = \varphi\in W_a$. But of course $\mu\notin L^2$ so $\mu\notin W_a$.