= Solution
Use $D_j=-i\partial_{x_j}$, so the <Fourier transform> of $D^\alpha u$ is $\lambda^\alpha\widehat u$. Distinguish the full degree-$N$ <polynomial> $P(\lambda)$ from its homogeneous top-degree part $p_N(\lambda)$, the <principal symbol>. The <elliptic differential operator> condition is
$$
\boxed{p_N(\lambda)\ne0\quad\text{for every real }\lambda\ne0.}
$$
It does not require the full <polynomial> to be nonzero at small frequencies. Compactness of the unit sphere gives $c_0=\min_{|\theta|=1}|p_N(\theta)|>0$. Homogeneity and the lower-degree remainder imply
$$
|P(\lambda)|\ge c_0|\lambda|^N-C(1+|\lambda|)^{N-1}.
$$
For sufficiently large $|\lambda|$ the second term is at most half the first, so the <high-frequency lower bound for an elliptic polynomial> is
$$
\boxed{|P(\lambda)|\ge c\langle\lambda\rangle^N,}
\qquad\langle\lambda\rangle=(1+|\lambda|^2)^{1/2}.
$$
Here $\langle\lambda\rangle$ is the <Japanese bracket>. The order-zero case just means a nonzero constant and is immediate.
For real $s$, the <Sobolev space> definition with the current Fourier normalization is
$$
H^s(\mathbb R^n)=\{u\in\mathcal S':\langle\lambda\rangle^s\widehat u\in L^2\},
\qquad\|u\|_{H^s}^2=(2\pi)^{-n}\int\langle\lambda\rangle^{2s}|\widehat u(\lambda)|^2\,d\lambda.
$$
Changing the harmless constant in the norm gives the same space. A <distribution> on $X$ belongs to the <Local Sobolev space> $H^s_{\rm loc}(X)$ when $\chi u$, extended by zero outside $X$, belongs to $H^s(\mathbb R^n)$ for every <test function> $\chi\in C_c^\infty(X)$.
A <compactly supported distribution> of finite <order of a distribution> $m$ has a smooth <Fourier transform> satisfying $|\widehat u(\lambda)|\le C(1+|\lambda|)^m$, by applying the finite-order estimate to a fixed cutoff times the exponential. Thus its weighted squared transform is bounded by $C\langle\lambda\rangle^{2(s+m)}$. This is integrable precisely in the sufficient range $2(s+m)<-n$, and proves
$$
\boxed{u\in H^s(\mathbb R^n)\quad\text{for every }s<-m-n/2.}
$$
This is the <negative Sobolev regularity of a compactly supported distribution>; the strict inequality is important, and is not a claim that this sufficient threshold is optimal for each <distribution>.
To establish <elliptic regularity> without assuming the answer as an a priori smoothness hypothesis, first obtain a global constant-coefficient gain. For a compactly supported <distribution> $w$ with $P(D)w\in H^q$, the <polynomial> lower bound at high frequency gives
$$
\int_{|\lambda|\ge R}\langle\lambda\rangle^{2(q+N)}|\widehat w|^2\,d\lambda
\le C\int_{|\lambda|\ge R}\langle\lambda\rangle^{2q}|\widehat{P(D)w}|^2\,d\lambda<\infty.
$$
On $|\lambda|\le R$, the transform of $w$ is smooth and bounded. Therefore \b[$P(D)w\in H^q$ implies $w\in H^{q+N}$]. Possible low-frequency zeros of $P$ are harmless; we never divide by them.
Two elementary mapping facts supply the variable-coefficient argument. <Distributional derivatives> of order $j$ map $H^r$ into $H^{r-j}$. <Sobolev multiplication by a smooth cutoff> is bounded on $H^r$ for every real $r$, including negative ones. Indeed, for a compactly supported smooth $a$, the <Peetre weight inequality>
$$
\langle\xi\rangle^r\le C_r\langle\eta\rangle^r\langle\xi-\eta\rangle^{|r|}
$$
and $\widehat{aw}=(2\pi)^{-n}\widehat a*\widehat w$ reduce the bound to <Young's convolution inequality> with the integrable kernel $\langle\zeta\rangle^{|r|}|\widehat a(\zeta)|$. Smooth coefficients only need this property on compact subsets, where they can be multiplied by another cutoff.
Write the lower-order part as $B=\sum_{|\alpha|<N}f_\alpha D^\alpha$, with $N\ge1$, and suppose provisionally that $u\in H^r_{\rm loc}$ on a relatively compact neighborhood. For a <cutoff function> $\chi$ supported there,
$$
P(D)(\chi u)=\chi Lu-\chi Bu+[P(D),\chi]u.
$$
The bracket is the operator <commutator>. Its order is at most $N-1$: in the product rule, every surviving term has at least one derivative falling on $\chi$. Choose a second <cutoff function> equal to one near $\operatorname{supp}\chi$ when estimating the products. The derivative and smooth-multiplication bounds then give
$$
\chi Bu,\ [P(D),\chi]u\in H^{r-N+1},\qquad \chi Lu\in H^s.
$$
The global gain just proved applies to the <compactly supported distribution> $\chi u$, and yields
$$
\boxed{u\in H^{\min(s+N,r+1)}_{\rm loc}.}
$$
This is the <cutoff bootstrap for local elliptic regularity>; it works for real indices, not just nonnegative integers.
There is always a legitimate starting index. Around any fixed point choose $\chi_0=1$ on a smaller neighborhood. The <compactly supported distribution> $\chi_0u$ has the negative Sobolev regularity proved above, so $u\in H^{r_0}_{\rm loc}$ on that neighborhood for some finite $r_0$. The index need not be uniform over all of $X$. Repeating the one-step gain finitely many times reaches $s+N$, or the target is already reached if $r_0\ge s+N$. Since the point was arbitrary,
$$
\boxed{Lu\in H^s_{\rm loc}(X)\ \Longrightarrow\ u\in H^{s+N}_{\rm loc}(X).}
$$
For $N=0$ the result follows directly by dividing by the nonzero constant.
Finally, if $Lu=0$, its forcing belongs to every <Local Sobolev space>. The gain consequently gives every local Sobolev order for $u$. For each integer $j\ge0$, choose an order larger than $j+n/2$ and apply the <Sobolev embedding theorem> to a localized solution. It has a $C^j$ representative; the representatives agree because they represent the same <distribution>. Thus \b[every distributional solution of $Lu=0$ is smooth on $X$].
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