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

Let $g:\mathbb{R}^n \to \mathbb{R}$ be a smooth real-valued function that decays like some positive power of $|x|^{-1}$ at infinity. Define the first order distribution $\mu $ by $$ \langle \mu , \phi \rangle = \int \limits _{\mathbb{R}^n} \big ( \phi (g(x)) - \phi (0) \big ) \,dx , \qquad \phi \in C_0^\infty (\mathbb{R}). $$ Let us for simplicity assume that $g$ is both positive as well as negative at some points so that $$ \operatorname{supp}\mu \subset [\inf g, \sup g]. $$ I'm wondering if we actually have equality rather than just $\subset $?

share|cite|improve this question
up vote 1 down vote accepted

I think so. Let $t_0\ne 0$ be in that interval and note that the interval also equals the image of $g$. Assume $t_0$ does not lie in the support of $\mu$. Then there exists a neighborhood $U$ of $t_0$ such that $\langle \mu,\phi\rangle=0$ for every $\phi$ supported in $U$. By shrinking $U$ we can assume that there exists $\alpha>0$ such that $U\cap [-\alpha,\alpha]=\emptyset$. Choose some $\epsilon>0$ such that the closed $\epsilon$-neighborhood $V$ of $t_0$ is contained in $U$. Now take a test function $\phi\ge 0$ supported in $U$ with $\phi\ge 1$ on $V$. Then $$ \langle\mu,\phi\rangle \ge \int_A 1\,dx, $$ where $A=g^{-1}(V)$. As $A$ has positive measure, we have a contradiction. As the support is closed, it also contains zero.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.