Let $f:\mathbb{R}^n\to \mathbb{R}$ be a smooth and bounded function which tends to 0 at infinity. Define, for $t>0$, the distribution $$ \nu (t) = \int \limits _{f(x)\ge t} dx, $$ and (in the distributional sense) the positive measure $$ \mu = -\frac{d\nu }{dt}. $$ By a change of variables we see that $$ \langle \mu , \phi \rangle _{\mathcal{D}', \mathcal{D}}=\int \limits _{\mathbb{R}^n} \phi (f(x))\,dx \quad \text{for }\phi \in C_0^\infty (\mathbb{R}_+). $$ In the text I'm reading it is stated (without motivation or proof) that $$ \operatorname{supp}\operatorname{sing}_A \mu = \operatorname{supp}\operatorname{sing}_A \nu , $$ where $\operatorname{supp}\operatorname{sing}_A $ denotes the analytical singular support of a distribution (i.e. the smallest closed set outside of which the distribution is real analytic). In general we have $\operatorname{supp}\operatorname{sing}_A u' \subset \operatorname{supp}\operatorname{sing}_A u$ but why do we have equality in a case like this?
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1$\begingroup$ Something isn't right here. Is $\nu$ a distribution on $\mathbb{R}$ or on $\mathbb{R}^n$? Either way, there's something fishy with the integral $\langle \mu,\phi\rangle$. $\endgroup$– Ketil TveitenAug 29, 2013 at 9:35
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$\begingroup$ Sorry, I have made an edit. $\nu $ should be a distribution on $\mathbb{R}$. I'm trying to phrase the question in slightly simpler terms than the original. $\endgroup$– flavioAug 29, 2013 at 9:50
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$\begingroup$ Ok, then I may have an idea for you: $\partial_t\nu = \int_{f(x)=t}dx$ if I'm not mistaken, and $\nu$ is singular precisely where $f$ has critical points, and the same is true for $\partial_t\nu$, and you're done. $\endgroup$– Ketil TveitenAug 29, 2013 at 9:55
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1$\begingroup$ In one dimension, I believe that you have $\mathop{supp}\mathop{sing} u' = \mathop{supp} \mathop{sing} u$ since you can recover $u$ from $u'$ by integration. $\endgroup$– Martin HairerAug 29, 2013 at 11:05
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1 Answer
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$\nu$ is a distribution on the real line and the operator $P=d/dt$ is elliptic with constant coefficients. In that case we have $$ \text{singsupp $\nu$}=\text{singsupp $P\nu$} $$ for the $C^\infty$ singular support as well as for the analytic singular support. The same equality holds for wave-front-sets (smooth and analytic) and this implies the previous equality.
