Are weak (Sobolev) solutions to a linear ODE a classical ones? Let $\Omega$ be an open subset of $\mathbb{R}$ and let $L$ be the differential operator
$$
Lf = \sum_{k=0}^{n-1} a_k f^{(k)} + f^{(n)},
$$
where $a_k$ are reals.

I would like to show that every weak (in a Sobolev space $W^{1,1}(\Omega)$) solution to
  $Lf = 0$
  is a classical one.

It is quite easy if $n = 1$:
$$
af + f' = 0.
$$
We rewrite the equation in the form
$$
\int_\Omega (af + f')\phi = 0,
$$
where $\phi \in C_c^\infty(\Omega)$ is a test function. Then, by the definition of the weak derivative this means
$$
\int_\Omega f(a\phi - \phi') = \int_\Omega f(x) e^{-ax}(e^{ax}\phi(x))' = 0.
$$
Since $e^{ax} \phi(x)$ could be any test function, we get $(f(x) e^{-ax})' = 0$, which means that $f(x) e^{-ax}$ is constant, hence $f(x) = c e^{ax}$.
But how should I proceed in case $n > 1$? 
 A: Definition: a distribution $T$ is of order $r$ if $r$ is the smallest integer such that 
$$|T(\phi)| \le C\sum_{j=0}^{r} \sup_\Omega |\phi^{(j)}|$$
holds with $C$ independent of $\phi$. 
Example: if $f\in W^{1,1}(\Omega)$, then both $f$ and $f'$ are of order $0$.
Property 1: if $T$ is of order $r$, then $T'$ is of order $\le r+1$. This follows directly from $T'(\phi) = -T(\phi')$. 
Property 2: if $T'$ is of order $r$, then $T$ is of order  $\le \max(0,r-1)$. To prove this, fix a test function $\phi_0$ with nonzero integral. Every test function $\phi$ can be written as $c\phi_0+\psi'$ where $c=\int \phi/\int\phi_0$ is a constant and $\psi$ is another test function. Indeed, the difference $\phi-c\phi_0$ has zero integral, and therefore its indefinite integral is also a test function. Since $T(\phi) = cT(\phi_0) - T'(\psi)$, the result follows (note that $c$ is bounded by $\sup|\phi|$ and $T(\phi_0)$ is a constant independent of $\phi$).  
After these preparations, proceed to the proof. Let $r$ be the order of $f^{(n)}$ (which is $\le n-1$ by assumption $f\in W^{1,1}$). By Property 2, the sum $\sum_{k=0}^{n-1} a_k f^{(k)}$ has order at most $\max(0, r-1)$. So, $r\le \max(0,r-1)$, which implies $r=0$. 
Moreover, since any derivative of $f$ satisfies the same ODE, all derivatives of $f$ have order $0$. 
By the Riesz Representation theorem, a distribution of order $0$ is a signed measure. Integrating a signed measure once, we get a  bounded  function; integrating it twice, we get a continuous function. Hence, all derivatives are continuous: $f\in C^\infty(\Omega)$.
Remark: the proof does not really use the assumption $f\in W^{1,1}$: it works for any distribution. 
