Elliptic regularity: Does the source have to be $L^{2}$

Question

Let $D:C^{\infty}(\mathbb{R}^{d})\to C^{\infty}(\mathbb{R}^{d})$ be an elliptic differential operator (formally self-adjoint for simpliciy), say the Laplacian for example. Is the following true:

Let $u\in L^{2}(\mathbb{R}^{d})$ be such that $Du=f$ in the distributional sense for some $f\in C^{\infty}(\mathbb{R}^{d})$, i.e. $Du$ is the function defined by the relation $\langle u,Df\rangle_{L^{2}}=\langle Du,\varphi\rangle_{L^{2}}$ for all test functions $\varphi\in C^{\infty}_{c}(\mathbb{R}^{d})$. Then, $u\in C^{\infty}(\mathbb{R}^{d})$.

My question is the version above correct, or do I need to assume that $f\in C^{\infty}\cap L^{2}$?

Answer 1

Here’s the version of Elliptic regularity as stated in Folland’s PDE text (Theorem 6.33; this specific page is available on Google preview)

Let $\Omega\subset\Bbb{R}^n$ be open, and $L$ an elliptic operator of order $k$ with $C^{\infty}$ coefficients on $\Omega$. Let $u,f$ be distributions on $\Omega$ such that $Lu=f$. If $f\in H^s_{\text{loc}}(\Omega)$ for some $s\in\Bbb{R}$, then $u\in H^{s+k}_{\text{loc}(\Omega)}$.

As a corollary, if $f\in C^{\infty}(\Omega)$, then $f\in H^s_{\text{loc}}(\Omega)$ for all $s\in\Bbb{R}$, so $u\in H^t_{\text{loc}}(\Omega)$ for all $t$, so by Sobolev’s embedding theorem, $u\in C^{\infty}(\Omega)$.

You work with these local Sobolev spaces by fiddling around with smooth compactly supported cutoff functions (and absorbing whatever error terms they generate into a ‘good’ elliptic estimate).