Every P-convex subset of ${\mathbb{R}}^2$ is already strongly P-convex

Abstract

A classical result of Malgrange says that for a polynomial P and an open subset $\Omega$ of ${\mathbb{R}}^d$ the differential operator $P(D)$ is surjective on $C^{\infty }(\Omega )$ if and only if $\Omega$ is P-convex. Hörmander showed that $P(D)$ is surjective as an operator on ${\mathcal{D}}^{\prime }(\Omega )$ if and only if $\Omega$ is strongly P-convex. It is well known that the natural question whether these two notions coincide has to be answered in the negative in general. However, Trèves conjectured that in the case of d=2 P-convexity and strong P-convexity are equivalent. A proof of this conjecture is given in this note.

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