{"paper":{"title":"Every P-convex subset of $\\R^2$ is already strongly P-convex","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":["math.AP"],"primary_cat":"math.FA","authors_text":"Thomas Kalmes","submitted_at":"2009-07-17T10:29:53Z","abstract_excerpt":"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 giv"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"0907.3037","kind":"arxiv","version":1},"verdict":{"id":null,"model_set":{},"created_at":null,"strongest_claim":"","one_line_summary":"","pipeline_version":null,"weakest_assumption":"","pith_extraction_headline":""},"references":{"count":0,"sample":[],"resolved_work":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57","internal_anchors":0},"formal_canon":{"evidence_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"author_claims":{"count":0,"strong_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"builder_version":"pith-number-builder-2026-05-17-v1"}