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

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

where $a$ are some smooth functions. 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 $\langle\cdot,\cdot\rangle_{L^{2}}$.

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    $\begingroup$ When you speak of taking "closure of $D$ in some space, say $E$, do you really mean taking closure of the set $D_0=\{(f,Df):f\in C_c^\infty\}$ in the space $E\times E$, or possibly taking the bipolar of the set $D_0$ with respect to the duality between the spaces $E\times E$ and $E'\times E'$? Speaking about the inner product of $H^s$ suggests that you mean the latter. I do not see what would be the purpose of this. In case you mean the former, I do not see any real problem here since the closure in $H^s\times H^s$ is a subset of the one in $L^2\times L^2$. $\endgroup$
    – TaQ
    Commented Nov 1, 2023 at 20:08
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    $\begingroup$ @TaQ I mean closure in the standard functional analytic sense: If $A:D(A)\to H$ is a linear operator in a Hilbert space $H$, then $\overline{A}$ is the operator defined as follows: $x\in D(\overline{A})$ iff there exists a sequence $(x_n)_n$ in $D(A)$ converging to $x$ such that $(Ax_n)_n$ is convergent. In this case we set $\overline{A}x:=\lim_{n\to\infty} Ax_n$. In other words, $\overline{A}$ is the (unique) operator whose graph is the closure of the graph of $A$. $\endgroup$
    – G. Blaickner
    Commented Nov 1, 2023 at 21:56
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    $\begingroup$ So, in other words, you refer to the former case in my comment above, and consequently, in the case $s\ge 0$ you have already implicitly answered "yes" to your own question because of the last sentence in my comment: closure in $H^s\times H^s$ is a subset of the one in $L^2\times L^2$, and referring to the inner product in $H^s$ just obscures matters. If the case $s<0$ is important to you, then it is a more complicated matter. $\endgroup$
    – TaQ
    Commented Nov 1, 2023 at 22:45
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    $\begingroup$ @TaQ I don't think it is so easy. If you take a differential operator $D:C^{\infty}_{c}\to C^{\infty}_{c}$, then in general, there is no relation between its closures in $H^{s}$ and $L^{2}$ in the sense that neither of the domains $\mathcal{D}(\overline{D}^{L^{2}})$ and $\mathcal{D}(\overline{D}^{H^{s}})$ is contained in each other (since there is no relation between strong convergence in $L^{2}$ and $H^{s}$). Of course, $\mathcal{D}(\overline{D}^{L^{2}})\cap \mathcal{D}(\overline{D}^{H^{s}})$ is non-empty and both operators agree on their common domain, but that is all we can say in general. $\endgroup$
    – G. Blaickner
    Commented Nov 2, 2023 at 7:23
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    $\begingroup$ It holds because of the (well-known) norm (in)equalities $\|\,U\,\|_{L^2}=\|\,U\,\|_{H^0}$ and $\|\,U\,\|_{H^s}\le\|\,U\,\|_{H^t}$ for $-\infty<s<t<+\infty$. $\endgroup$
    – TaQ
    Commented Nov 2, 2023 at 22:11

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