We define the local Sobolev space $H^s_{loc}(\mathbb{R}^n)$ as
$$
H^s_{loc}(\mathbb{R}^n)=\big\{f\in\mathcal{D}^{\prime}(\mathbb{R}^n): \forall \Omega\Subset\mathbb{R}^n \ \exists g_{\Omega}\in H^s(\mathbb{R}^n) \ \text{s.t.} \ g_{\Omega}\big|_{\Omega}=f\big\}.
$$
where $H^s(\mathbb{R}^n)$ is the usual fractional $L^2$-Sobolev space
$$
H^s(\mathbb{R}^n)=\{u\in\mathscr{S}^{\prime}(\mathbb{R}^n): \langle\cdot\rangle^s\hat{u}\in L^2(\mathbb{R}^n)\}
$$
and $\langle\xi\rangle=(1+|\xi|^2)^{\frac{1}{2}}$. I was wondering if the following equality is true
$$
\bigcap_{s\in\mathbb{R}}H^s_{loc}(\mathbb{R}^n)=C^{\infty}(\mathbb{R}^n).
$$
For example in Folland's book Lectures on Partial Differential Equations http://www.math.tifr.res.in/~publ/ln/tifr70.pdf Corollary 3.8 says that
$$
\bigcap_{s\in\mathbb{R}}H^s(\mathbb{R}^n)\subset C^{\infty}(\mathbb{R}^n)
$$
by the Sobolev embedding theorem. Also he states in the result about elliptic regularity (Corollary 4.45) that elliptic $\Psi$DO's are hypoelliptic, i.e. if $Pu$ is smooth then $u$ is smooth, where $P$ is a $\Psi$DO of order $m$. This should be based on the local regularity result in the same corollary, i.e. if $Pu\in H^s_{loc}(\mathbb{R}^n)$ then $u\in H^{s+m}_{loc}(\mathbb{R}^n)$.
I get that if $Pu$ is smooth then clearly $Pu\in H^s_{loc}(\mathbb{R}^n)$ for all $s\in\mathbb{R}$ by a cut-off argument. So from this it follows that $u\in H^{s+m}_{loc}(\mathbb{R}^n)$ for all $s\in\mathbb{R}$, i.e.
$$
u\in\bigcap_{s\in\mathbb{R}}H^s_{loc}(\mathbb{R}^n).
$$
But can we conclude now that $u$ is smooth? I didn't find any references on local Sobolev spaces or their properties and Folland does not give any proof for the hypoellipticity of elliptic $\Psi$DO's. Can anyone help me?
Best Answer
From $u \in \bigcap_{s \in \mathbb R} H_{loc}^s(\mathbb R^n)$ it follows $$u \in \bigcap_{k \in \mathbb N} H^k(\Omega)$$ for all bounded and open $\Omega \subset \mathbb R^n$.
Now, you can use the Sobolev embedding theorem to conclude $$u \in \bigcap_{k \in \mathbb N} C^k(\bar\Omega).$$ This readily implies $u \in C^\infty(\mathbb R^n)$.