# Support of a distribution

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 $\subset$?

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.