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I got stuck in the second question of the following problem. I cannot prove anything significant other than the base case. So I think I need some help. Here is what I did:


Let $I\in \mathbb{R}$ be an open interval. Show that, if $u\in \mathcal{D}'(I)$ and $\partial u\in C^{\infty}(I)$, then $u\in C^{\infty}(I)$. \ Let $$P(x,\partial)=\sum_{i=0}^{m} a_{i}\partial^{m-i}$$ be a differential operator with $C^{\infty}$ coefficients defined on $I$. Suppose that $a_{0}\not=0$ on $I$. Show that if $u\in \mathcal{D}'(I)$ and $Pu\in C^{\infty}(I)$, then $u\in C^{\infty}(I)$.

(Assume that,$\forall x_{0}\in I,\exists \phi\in C^{\infty}(I)$ such that $P\phi=0$ and $\phi(x_{0})=1$. Then use induction. The existence of such $\phi$ follows from ODE)

For the first claim we know any two primitives differ by a constant from the text. Assume $\int \partial u=v$, we can thus claim $u=v+c$ for some $c\in \mathbb{R}$. Since $\partial u\in C^{\infty}(I)$, $v$ and $u$ must be in $C^{\infty}(I)$ as well.

For the second claim we proceed with induction. We know $a_{0}u=v$ has a $C^{\infty}(I)$ solution $\frac{v}{a_{0}}$. If $f\in \mathcal{D}'$ is also a solution then we have $$\int a_{0}(f-\frac{v}{a_{0}})\phi=0,\forall \phi\in C_{c}^{\infty}(I)$$This is impossible unless the difference is 0, since $a_{0}\not=0$. Thus the general solution has the form $\frac{v}{a_{0}}$.

Now consider the case $(a_{1}\partial+a_{0})u=v$. We have \begin{align*} \langle (a_{1}\partial+a_{0})u,\phi\rangle &=\langle u,(a_{0}+\partial a_{1})\phi+a_{1}\partial \phi\rangle\\ &=\langle u,(a_{0}+\partial a_{1})\phi\rangle+\langle u,a_{1}\partial \phi\rangle\\ &=\langle v,\phi \rangle \end{align*}

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1 Answer 1

Abstractly it follows from elliptic regularity. The operator P is an mth-order operator with principal symbol $\sigma(x,y) = a_0\cdot y^m$ Which is invertible for $y\neq 0$. Hence P is elliptic and the claim follows.

If you want to avoid using this, at the moment I have no other idea.

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This is a high brow proof. I need a low level proof. But thank you. –  Bombyx mori Jun 16 '12 at 18:13

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