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Let $E = C_c^0(\mathbb{R}^n;\mathbb{R}^m)$ be the space of compactly supported continuous functions on $\mathbb{R}^n$ with values on $\mathbb{R}^m$. There is a natural norm on this space: given $\varphi \in E$, we put $$ \Vert \varphi \Vert = \sup_{x \in \mathbb{R}^m} \Vert \varphi(x) \Vert.$$

For each compact set $K \subset \mathbb{R}^n$, let $E_K$ be the subspace of $E$ consisting of those functions that have support contained in $K$. Each $E_K$ inherits a the supremum norm from $E$.

Define a topology $\tau$ on $E$ by declaring a subset $U$ to be open if $U \cap E_K$ is open in $E_K$ for each compact set $K \subset \mathbb{R}^n$.

In the book "Sets of finite perimeter and geometric variational problems", by Francesco Maggi, the author introduces the following notion of convergence on $E$: a sequence $(\varphi_k)_{k \in \mathbb{N}}$ in $E$ converges to a function $\varphi \in E$ with uniform supports, and we write $\varphi_k \overset{us}{\to} \varphi$, if $\Vert \varphi_k - \varphi \Vert \to 0$ as $k \to \infty$ and if there is a compact set $K \subset \mathbb{R}^n$ such that

$$ supp(\varphi) \cup \bigcup_{k \in \mathbb{N}} supp(\varphi_k) \subseteq K,$$

that is, if the supports of the functions do not "escape" to infinity.

Now let $L : E \to \mathbb{R}$ be a linear functional.

Question:

Is it true that $L : (E, \tau) \to \mathbb{R}$ is continuous if and only if $L(\varphi_k) \overset{us}{\to} L(\varphi)$ whenever $\varphi_k \overset{us}{\to}{\varphi}$ in $E$?

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The space $E:=C_c^0(\mathbb{R^n})$ is the inductive limit in the TVS category of the spaces $(E_K,\|\cdot\|_{K,\infty})$. As a consequence, a sequence converges on $E$ if and only if it is included in a space $E_K$ and converges there; a linear form on $E$ is continuous if and only if it is continuous on every $E_K$, and the claim follows. Note that $E$ is not metrizable, nor first countable, but being an inductive limit of metric TVS, for the above reasons sequential continuity of a linear form is still equivalent to continuity.

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  • $\begingroup$ The first ``consequence'' is true in the particular situation but not for all inductive limits in the TVS category. $\endgroup$ – Jochen Wengenroth Jul 7 '18 at 8:10
  • $\begingroup$ yes, I know; it is a consequence for this situation $\endgroup$ – Pietro Majer Jul 7 '18 at 8:47
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Another way of seeing that the answer is affirmative is to realize that the topology on $E$ is the inductive limit of the topologies on $C(K,R^m)$ (see e.g. https://en.wikipedia.org/wiki/Final_topology). A direct consequence of the definition is that a function from $E$ to any topological space (regardless of the linear structure) is continuous if and only if so is its restriction to $C(K,R^m)$. This reduces the problem to the study of the continuity on $C(K,R^m)$ which, being a metric space, simplifies the matter.

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This is indeed true and follows from the fact that your space is what is called a strict $LF$-space and the theory of the latter. This can be found in the original article by Dieudonné and Schwartz which can easily be found using a search machine (La dualité dans les espaces $F$ et $LF$---available online at numdam.org) or in one of the more comprehensive monographs on locally convex spaces, e.g., that by Köthe.

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  • $\begingroup$ This a comment on the answers below which miss the main point, namely that this is a so-called STRICT inductive limit, otherwise their reasoning would be insufficient $\endgroup$ – Parschallen Jul 4 '18 at 9:52
  • $\begingroup$ could you please explain why? To me it seems the argument works in general: say we have a directed family $X_i$ of spaces where, say, continuity is equivalent to sequential continuity admitting a direct limit X. I claim that a function f on X is continuous if and only if for every sequence (x_n) which is entirely contained in some X_i and has a limit x there, we have: f(x_n)\to f(x). The proof is: by the properties of the direct limit, f is continuous if and only if so are the functions f\circ \iota_i, where the \iota_i:X_i\to X are the maps coming with the direct limit. The claim follows. $\endgroup$ – Nicola Gigli Jul 4 '18 at 11:04

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