Continuous positive-definite function with prescribed support - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:55:26Z http://mathoverflow.net/feeds/question/101301 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101301/continuous-positive-definite-function-with-prescribed-support Continuous positive-definite function with prescribed support Simone Virili 2012-07-04T11:15:19Z 2012-07-04T14:19:38Z <p>Consider a (locally) compact Abelian group G and a compact neighborhood $C$ of the identity ($0$) of $G$. Is it possible to find a non-trivial (i.e., $\phi(0)\neq 0$) continuous, positive (with real non-negative image), positive-definite and absolutely integrable function $\phi :G\to \mathbb C$ (in symbols $\phi\in L^1(G)^+\cap P(G)$) whose support (the closure in $G$ of the set of points with non-zero image) is contained in $C$?</p> <p>As I do not need a solution for all the compact neighborhoods but only for a cofinal subset of the family of such neighborhoods (ordered by inclusion) we can restrict to symmetric neighborhoods (i.e., $C=-C$). For such $C$, my idea was to find a compact symmetric neighborhood $C'\subseteq C$ such that $C'+C'\subseteq C$ and take as $\phi$ the convolution of $\chi_{C'}$ with itself. Since $\chi_{C'}(x)=\overline{\chi_{C'}(-x)}$, one easily obtains that $\phi$ is positive-definite. It is also easily verified that $\phi$ is absolutely integrable and positive. What about continuity? Does this work?</p>