# Distribution whose singular support is open

I would like to gain intuition how "singular" a distribution can be.

Let $X \in \mathbb R^n$ be open, $\mathcal D(X)$ be the smooth test functions on $X$ with canonical LF topology, and let $\mathcal D'(X)$ be the topological dual of $D(X)$.

Let $\Psi \in D'(X)$. Can the singular support of $\Psi$, i.e. the complement in $X$ of the largest open subset of $X$, on which $\Psi$ is a locally integrable function, be open?

For example, the Dirac delta has singular support in $0$. Similarly, integrals with respect to Hausdorff measures on finite measure submanifolds of $X$ may be distributions, whose singular support is not open.

All examples of distributions I know constitute of a function "disturbed" by a "small" singularity.

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How about the distributional derivative of a nowhere differentiable function? – Nate Eldredge Dec 9 '10 at 13:47
Thanks, that is a good point to proceed. – shuhalo Dec 9 '10 at 13:59
You need to be a bit careful with the definitions. $X$ is an open domain. You define the singular support to be the complement of an open set in $X$. So unless your singular support is all of $X$ or the empty set, the set cannot be open (since it is closed in $X$). – Willie Wong Dec 9 '10 at 14:13
May be what you want to say is whether $\mathop{singsupp} \Psi$ can contain a nontrivial open set? – Willie Wong Dec 9 '10 at 14:14