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

TL;DR

This paper proves Trèves' conjecture that in two dimensions, P-convexity and strong P-convexity are equivalent for subsets of , clarifying a longstanding question in the theory of differential operators.

Contribution

The paper establishes that in , P-convexity and strong P-convexity coincide, confirming Tre8ves' conjecture and simplifying the understanding of these convexity notions.

Findings

P-convexity and strong P-convexity are equivalent in 

The result confirms Tre8ves' conjecture for two-dimensional cases

Clarifies the relationship between P-convexity notions in differential operator theory

Abstract

A classical result of Malgrange says that for a polynomial P and an open subset $\Omega$ of $\R^d$ the differential operator $P(D)$ is surjective on $C^\infty(\Omega)$ if and only if $\Omega$ is P-convex. H\"ormander showed that $P(D)$ is surjective as an operator on $\mathscr{D}'(\Omega)$ 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, Tr\`eves 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.