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 Ω of ${\mathbb{R}^d}$ the differential operator P(D) is surjective on C ^∞(Ω) if and only if Ω is P-convex. Hörmander showed that P(D) is surjective as an operator on ${\fancyscript{D}'(\Omega)}$ if and only if Ω 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.