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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.

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1 Answer 1

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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.

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