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 $n > 1$?


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.

  • $\begingroup$ Great. Thank you. I would like to ask two things: (1) why every test function can be written as a constant plus a test function? (2) if we know that $f^{(n)}$ has order $r$, then by Property 2 all $f^{(k)}$'s have order at most $\max(0,r-1)$, right? $\endgroup$
    – xen
    Jun 13 '16 at 10:57
  • $\begingroup$ (1) subtract a constant from $\phi$ so that $\int \phi=0$; then let $\psi (x) = \int_{-\infty}^x \phi(t)\,dt$. (2) correct $\endgroup$
    – user147263
    Jun 13 '16 at 11:00
  • $\begingroup$ Perhaps this is easy, but how such $\psi$ is a test function? Does it have compact support? If we substract a constant from $\phi$, then it is also not a test function. $\endgroup$
    – xen
    Jun 13 '16 at 11:20
  • $\begingroup$ And what do you mean by $T(c)$? As far as I know, $T$ should acts on test functions, isn't it? $\endgroup$
    – xen
    Jun 13 '16 at 13:30
  • $\begingroup$ @xen Good point about $T(c)$; I rewrote the proof of Property 2. Concerning $\psi$ having compact support: yes, it does have it. It begins by being zero at $-\infty$, and ends by being $0$ at $+\infty$. $\endgroup$
    – user147263
    Jun 13 '16 at 13:43

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