Consider an arbitrary linear partial differential operator $D:C^{\infty}(\mathbb{R}^{d})\to C^{\infty}(\mathbb{R}^{d})$, i.e.

$$D=\sum_{\alpha\in\mathbb{N}^{d}}a^{\alpha}(x)\partial^{\alpha}_{x}$$

Furthermore, lets assume that $D$ is elliptic in the sense that its principal symbol is invertible. As far as I understand, one version of the elliptic regularity theorem states the following:

Let $u\in L^{2}(\mathbb{R}^{d})$ be a (weak) solution of $Du=f$ for some source $f$. If $f\in C^{\infty}(\mathbb{R}^{d})$, then $u\in C^{\infty}(\mathbb{R}^{d})$.

Now, in some project, I stumbled over the following situation: I have an elliptic differential operator $D:C_{c}^{\infty}(\mathbb{R}^{d})\to C_{c}^{\infty}(\mathbb{R}^{d})$ and I complete it (in the functional analytic sense) in some Sobolev space $H^{s}(\mathbb{R}^{d})$ for $s\in\mathbb{R}$. Let us denote its completion by $\overline{D}$. Now, in this specific case, it turns out that the minimal and maximal closed extension in $H^{s}$ agree and hence, its domain can be written as

$$\mathcal{D}(\overline{D})=\{f\in H^{s}\mid Df\in H^{s} \} $$

where $Df\in H^{s}$ has to be understood in the "$H^{s}$-weak sense", i.e. there exists an element denoted by $Df\in H^{s}$, such that $\langle f,D^{\ast}\varphi\rangle_{H^{s}}=\langle Df,\varphi\rangle_{H^{s}}$ for all $\varphi\in C^{\infty}_{c}(\mathbb{R}^{d})$.

Now, if I know that a function $f\in\mathcal{D}(\overline{D})\subset H^{s}(\mathbb{R}^{d})$ satisfies $\overline{D} f\in C^{\infty}(\mathbb{R}^{d})$. Is it possible to argue by elliptic regularity that $f\in C^{\infty}(\mathbb{R}^{d})$?

Remark: If I would take the closure of $D$ in $L^{2}$ instead, then the question is clear, since $\overline{D}f\in C^{\infty}(\mathbb{R}^{d})$ for $f\in\mathcal{D}(\overline{D})$ (where the closures are now taken in $L^{2}$) exactly means that $Df\in C^{\infty}(\mathbb{R}^{d})$, where $Df$ for $f\in L^{2}$ has to be understood in the distributional sense. So, the difference in my question is really just the different inner product, i.e. $\langle\cdot,\cdot\rangle_{H^{s}}$ instead of $\langle\cdot,\cdot\rangle_{L^{2}}$.