I want to prove the following statement:
Let $X$ be an open subset of $\mathbb{R}^n$. Suppose $u\in D'(X)$ is a distribution of infinite order. Then the singular support of $u$ is unbounded.
I am kinda stuck, and don't know how to translate properties of the fact that the distribution if of infinite order to properties of the singular support.
Best Answer
The singular support of $u\in \mathscr D'(X)$ can be bounded, e.g., for $X=(0,1)$ and $u(\phi)=\sum\limits_{n=1}^\infty \phi^{(n)}(1/n)$. However, if the singular support is relatively compact in $X$ then $u$ is of finite order: Take $\psi\in\mathscr D(X)$ which is $1$ near the singular support of $u$, then $u=\psi u + (1-\psi)u$ has finite order because the first term has compact support and the second is even a smooth function.