MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $C$?

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?

share|cite|improve this question
Sounds like a homework problem. – Nik Weaver Jul 4 '12 at 13:24
no... it's not an homework... I need it just to understand a proof by Justin Peters in this paper [ Justin Peters, {\em Entropy on discrete abelian groups}. Adv. in Math. 33 (1979), no. 1, 1–13. ]. The problem is that my background is in module/ring theory... probably my question may seem stupid to experts. Anyway, any help is welcome. – Simone Virili Jul 4 '12 at 13:32
You can use the dominated convergence theorem to prove that the convolution of any two bounded functions is sequentially continuous. Probably this means it's continuous for any LCA group, but possibly there are pathological exceptions, I don't know. Your main question is more easily answered by starting with any nonzero continuous function $f$ supported on C' (which exists by Urysohn's lemma), then letting $\phi$ be $f$ convolved with $\overline{f(-x)}$. – Nik Weaver Jul 4 '12 at 15:52
If I interpret correctly your suggestion, one takes a compact neighborhood $C$ and a compact neighborhood $C'$ so that $C'+C'\subseteq C$. We can take an open neighborhood of $0$, $C''\subseteq C'$. Then ${0}$ and $G\setminus C''$ are two closed disjoint subsets of $G$ and, by Urysohn's lemma one can find a continuous function $$f: G \rightarrow [0,1]$$ such that $f(0)=1$ and $f$ is trivial on $G\setminus C''$, which implies that the support of $f$ is contained in $C'$. Now the convolution $\phi$ of $f$ with $\overline{f(-\ldots)}$ is the correct function. Many thanks, it perfectly works! – Simone Virili Jul 4 '12 at 16:13
Yes, that's right. You're welcome! – Nik Weaver Jul 4 '12 at 18:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.