Advanced search

Browse by subject

Browse by provider

Browse by language

Content Details
Normal View
CIP View

Every P-convex subset of $\R^2$ is alrea...

Kalmes, Thomas...

Free

Share this content

Every P-convex subset of $\R^2$ is already strongly P-convex by Kalmes, Thomas ( Author )

Content Provider	Lemonjar Open Access
Publisher	N.A
Publication Date	17-07-2009
Summary	A classical result of Malgrange says that for a polynomial P and an open subset Ω of $\R^d$ 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 D′(Ω) 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.
ISBN	-
Item Type	Article
File Type	pdf
File Size	36.88 KB
Language	English
Subjects	-
Unlimited	MYR 0.00
Total read	-
URL	https://arxiv.org/abs/0907.3037

Citation

Review the full bibliography data and make any necessary corrections before using. Always consult your library resources for the exact formatting and punctuation guidelines.

AMA 11th EditionReferences

Kalmes, Thomas. Every P-convex subset of $\R^2$ is already strongly P-convex. N.A. July 17, 2009. Accessed September 19, 2024. https://online.provdatabase.com/cite/2384

APA 7th EditionReferences

Kalmes, Thomas. (2009). Every P-convex subset of $\R^2$ is already strongly P-convex. N.A. https://online.provdatabase.com/cite/2384

Chicago 17th EditionReferences

Kalmes, Thomas. Every P-convex subset of $\R^2$ is already strongly P-convex. N.A. 2009. https://online.provdatabase.com/cite/2384

HarvardReferences

Kalmes, Thomas. (2009). Every P-convex subset of $\R^2$ is already strongly P-convex. [Online]. N.A. Available from: https://online.provdatabase.com/cite/2384. (Accessed 19 September 2024).

Harvard AustraliaReferences

Kalmes, Thomas, 2009, Every P-convex subset of $\R^2$ is already strongly P-convex, N.A, viewed 19 September 2024, <https://online.provdatabase.com/cite/2384>

MLA 9th EditionWorks Cited

Kalmes, Thomas. Every P-convex subset of $\R^2$ is already strongly P-convex. N.A, 2009. PROV Database, https://online.provdatabase.com/cite/2384.

Share this eBook