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

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

2009-07-17

arxiv.org/abs/0907.3037v1

Mathematics

Functional Analysis

Scientific paper

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.

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