3
$\begingroup$

For $K\subseteq \mathbb{R}^d$ compact, let $C_{\mathrm{c}}^{\infty}(K)$ denote the space of smooth functions on (an open neighborhood of) $K$ with compact support contained in $K$ with the usual Fréchet space subspace topology inherited from $C^{\infty}(K)$, and denote by $C_{\mathrm{c}}^{\infty}(K)^*$ the dual with the Mackey topology (or, equivalently, the strong polar topology). For $U\subseteq \mathbb{R}^d$ open, let $C_{\mathrm{c}}^{\infty}(U)$ denote the LF-space of smooth functions with compact support contained in $U$, and denote by $C_{\mathrm{c}}^{\infty}(U)^*$ the space of generalized-functions on $U$, again, with the Mackey topology (or equivalently the strong polar topology).

If $K\subseteq U$, we have extension-by-zero maps $C_{\mathrm{c}}^{\infty}(K)\rightarrow C_{\mathrm{c}}^{\infty}(U)$, and we thereby obtain restriction maps $C_{\mathrm{c}}^{\infty}(U)^*\rightarrow C_{\mathrm{c}}^{\infty}(K)^*$.

Is it true that $\operatorname{colim}_{\substack{U\supseteq K \\ U\text{ open}}}C_{\mathrm{c}}^{\infty}(U)^*=C_{\mathrm{c}}^{\infty}(K)^*$? That is, is $C_{\mathrm{c}}^{\infty}(K)^*$ the colimit in the category of locally-convex spaces of $C_{\mathrm{c}}^{\infty}(U)^*$ as $U$ ranges over all open sets which contain $K$?

For what it's worth, the motivation is that this colimit is the usual definition of the sheaf on a not-necessarily-open subset $K$ (or if you like, the global sections of the inverse image sheaf under the inclusion $K\hookrightarrow \mathbb{R}^d$), and so it would be nice if this colimit had a relatively concrete description.

$\endgroup$
1
$\begingroup$

I think that the answer is yes. For the algebraic equality one needs that $C^\infty_c(K)$ is topologically equal to the (projective) limit $\lim C^\infty_c(U)$ and this follows from the fact that $C^\infty_c(K)$ is a topological subspace of each $C^\infty_c(U)$. The toplogical equality co$\lim C^\infty_c(U)^\ast =C^\infty_c(K)^\ast$ then should follow from an open mapping theorem: The identity from the left hand side to the right is clearly continuous (functoriality of the Mackey topology) and it is open because the colimt is a webbed space (because it is enough to consider a decreasing sequence $U_n$ with intersection $K$) and $C^\infty_c(K)^\ast$ is ultrabornological (it is the strong dual of a Fréchet-Schwartz space), one can thus apply de Wilde's open mapping theorem (see, e.g., theorem 24.30 in the book Introduction to Functional Analysis of Meise and Vogt).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.