8
$\begingroup$

For each $f\in C^\infty(\mathbb R^n)$, there exists $u\in C^\infty(\mathbb R^n)$ such that $\Delta u=f$. This, I guess, has been known well before the more general Malgrange-Ehrenpreis theorem that says the same for a general differential operator with constant coefficients.

My question is twofold:

1°) Who proved this first?

2°) Does there exist a continuous operator $S:C^\infty(\mathbb R^n)\to C^\infty(\mathbb R^n)$ such that $\Delta Sf\equiv f$ ?

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ I know how to prove the existence of a solution of $\Delta u=f$, but the solution involves partition of unity and the Wlash theorem on harmonic approximation so the solution does not depend linearly on $f$. I do not see how to prove the existence of solution to $Pu=f$ for general operators with constant coefficients. The Malgrange-Ehrenpreis theorem proves the existence of a fundamental solution, but the convolution of $f$ with the fundamental solution does not make sense if $f$ growths fast. Can you provide references to the proof of the existence of a solution to $Pu=f$ in the general case? $\endgroup$
    – Piotr Hajlasz
    Commented Jun 1, 2018 at 0:18
  • $\begingroup$ Sorry I'm late. I somehow checked some original stuff by Malgrange before I asked, and I think he claimed the general $\Delta u=f$ case. My question was itself a reference-request, remember... $\endgroup$
    – Jean Duchon
    Commented Jun 11, 2018 at 15:55
  • $\begingroup$ @PiotrHajlasz, Indeed every linear differential operator with constant coefficients is surjective on $C^{\infty}(\Omega)$ for any convex open $\Omega \subset \mathbb{R}^n$. This follows from Theorem 5.1 and Corollary 3 of Theorem 6.1 in the book "Linear Partial Differential Equations with Constant Coefficients" by F. Trèves. $\endgroup$
    – S.Z.
    Commented Feb 11, 2022 at 21:47

1 Answer 1

Reset to default
5
$\begingroup$

Assuming Question 2 asks about a continuous linear right inverse, the answer is no. In section 3, Theorem 3.3 of the article "Some results on continuous linear maps between Fréchet spaces" by Dietmar Vogt it is proved that no hypoelliptic operator on $C^{\infty}(\Omega)$ admits a continuous linear right inverse for any open $\Omega \subset \mathbb{R}^n$ with $n>1$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ I wonder if there is a nonlinear continuous right inverse. The method that I described is nonlinear, but I have not checked if it can give continuity. $\endgroup$
    – Piotr Hajlasz
    Commented Feb 11, 2022 at 22:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .