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Let $\Omega$ be a nonempty open subset of $\mathbb{R}^n$, and let $\cup_{n=1}^\infty K_n = \Omega$ be an exhaustion of $\Omega$ by compact sets. Let $\mathcal{D}(\Omega) = \mathcal{D}$ be the standard $\mathbb{C}$-vector space of smooth functions $\phi: \Omega \to \mathbb{C}$ with compact support in $\Omega$. Equip $\mathcal{D}$ with the inductive limit topology endowed by the Frechet spaces $C_K^\infty(\Omega) = \mathcal{D}_{K_n} = \{\phi \in \mathcal{D} :\text{supp } \phi \subseteq K_n \}$ (this is the usual locally convex topology on $\mathcal{D}$).

Suppose that $\mu : \mathcal{D} \to \mathcal{D}$ is a continuous linear map, and let $K \subset \Omega$ be compact. I would like to show that there exists an index $N$ such that $\text{supp }\mu \phi \subseteq K_N$ for all $\phi$ such that $\text{supp } \phi \subset K$.

This fact is stated at the end of Chapter 2 (in Definition 2.8.1) in Friedlander's book on distribution theory (in fact, Friedlander takes it as part of the definition of a continuous linear map $\mu : \mathcal{D}(\Omega) \to \mathcal{D}(\Omega)$. However, I am having trouble establishing why it is true when we begin from the usual topological definition of continuity (inverse image of open set is open).

From Rudin's functional analysis bookk, I know that continuous linear maps $\mathcal{D}(\Omega) \to \mathcal{D}(\Omega)$ are bounded (here, I am using bounded in the sense of topological vector spaces). I think this is the fact that I need to use. At the moment, I am trying to make a proof by contradiction work. Suppose that $\mu (\mathcal{D}_K)$ is not contained in some $\mathcal{D}_{K_N}$. Then we can find functions $\phi_m \in \mathcal{D}_K$ and points $x_m$ without limit point in $\Omega$ so that $|\mu \phi_m(x_m)| \neq 0$. But I'm not sure where to go from here.

Hints or solutions are greatly appreciated.

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  • $\begingroup$ I think the question would benefit from the definition of "inductive limit topology". Is it the same thing as the usual topology on the space of test functions? $\endgroup$ – TZakrevskiy May 27 '15 at 23:32
  • $\begingroup$ @TZakrevskiy - Yes, the inductive limit topology is the same as the usual locally convex topology on the space of test functions. $\endgroup$ – JZS May 27 '15 at 23:41
  • $\begingroup$ @Jose27 - This fact holds for the identity map. For each compact $K$, we need a corresponding compact $K'$ so that if the support of $\phi$ is in $K$, then the support of $\mu \phi$ is in $K'$. $\endgroup$ – JZS May 27 '15 at 23:43
  • $\begingroup$ Hint: Exhaust $\Omega$ by compact sets and argue by contradiction. $\endgroup$ – Jose27 May 28 '15 at 0:06
  • $\begingroup$ @Jose27 - I have tried to follow your hint so far: if $\Omega = \cup K_n$, and $\mu(K)$ is not contained in some $K_n$, we can get a sequence of functions $\phi_m$ and points $x_m$ without limit point in $\Omega$ so that $| \mu \phi(x_m)| \neq 0$. But I am not sure where to go from here. $\endgroup$ – JZS May 28 '15 at 12:30

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