# Notion of convergence on the space of compactly supported continuous functions

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 $$\Vert \varphi \Vert = \sup_{x \in \mathbb{R}^m} \Vert \varphi(x) \Vert.$$

First question: is $E$ equipped with this norm a Banach space?

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 $\varphi \in E$ 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

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

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

Second question: does this notion of convergence somehow generates a topology on $E$? Is it metrizable?

• The answer to your first question is "no". Take a nice bell-curve, chop it off outside of $[-n, n]$ and connect short line segments in order to make it continuous. It's not difficult to show sequences created this way are Cauchy, but if they were to converge, they must converge to the original (non-compactly-supported) bell curve. Commented Jul 3, 2018 at 2:40

First question: is $E$ equipped with this norm a Banach space?

No. In fact, its completion is the space of continuous functions $\mathbb{R}^n\to\mathbb{R}^{m}$ that vanish at $\infty$ (that is, $f(x)\to 0$ as $\|x\|\to\infty$). Given any continuous function $f$ which vanishes at $\infty$, let $f_k$ be obtained by "cutting off" $f$ to be supported on a ball of radius $k$ around the origin (so, say, $f_k$ agrees with $f$ on a ball of radius $k-1$, and interpolates radially to $0$ between the sphere of radius $k-1$ and the sphere of radius $k$). Then $f_k\in E$ for each $k$ and $f_k$ converges uniformly to $f$, and it follows easily that $(f_k)$ is Cauchy in $E$ but has no limit in $E$ unless $f$ has compact support.

Second question: does this notion of convergence somehow generates a topology on $E$? Is it metrizable?

There is no metrizable topology which induces this notion of convergence. To prove this, for each $R$, choose a sequence $(\varphi_k^R)_k$ such that each $\varphi_k^R$ has a ball of radius $R$ as its support but $\varphi_k^R\to 0$ uniformly. Then each of these sequences $(\varphi_k^R)$ converges to $0$ in Maggi's sense. If this notion of convergence came from a metric, we could choose for each $R$ a $k_R$ such that $\varphi_{k_R}^R$ is within $1/R$ of $0$ with respect to the metric, and then $\varphi_{k_R}^R$ would converge to $0$ as $R\to\infty$. But this is impossible since convergent sequences are required to have some uniform compact support.

However, this convergence does come from a natural topology. For each compact $K\subset\mathbb{R}^n$, let $E_K$ be the subspace of $E$ consisting of functions with support contained in $K$. Say that a set $U\subseteq E$ is a basic neighborhood of $0$ if $U$ is balanced and convex, and $U\cap E_K$ is an open neighborhood of $0$ in $E_K$ with respect to the sup norm on $E_K$ for each $K$. Finally, say that a set $U\subseteq E$ is open if it is a union of translates of basic neighborhoods of $0$.

It can be shown that this topology makes $E$ as the colimit of the spaces $E_K$ with their sup norms in the category of locally convex topological vector spaces. That is, it is the finest topology on $E$ which makes the inclusion maps $E_K\to E$ all continuous and makes $E$ a locally convex topological vector space. It is then clear that any sequence which converges in Maggi's sense converges in this topology, since such a sequence will converge in $E_K$ with respect to the sup norm.

Proving the converse (every sequence convergent in this topology converges in Maggi's sense) is a little messier, but here's the idea. Suppose $(f_k)$ is a sequence of functions in $E$ such that the supports of the $f_k$ are not contained in any fixed compact set. Then, passing to a subsequence of $(f_k)$ if necessary, we can find a sequence $(x_k)$ going to $\infty$ in $\mathbb{R}^n$ such that $f_k(x_k)\neq 0$ for each $k$. Now let $U$ be the set of $f\in E$ such that $|f(x_k)|<|f_k(x_k)|$ for each $k$. Then $U$ is a basic neighborhood of $0$ in $E$ (this uses the fact that any $K$ contains only finitely many of the $x_k$). Since $f_k\not\in U$ for all $k$, this means $(f_k)$ cannot converge to $0$.

• Maggi says: "a linear functional $L : E \to \mathbb{R}$ is continuous with respect to this convergence if it is bounded.." and then he defines what he means by "bounded". If this "colimit topology" on $E$ is not metrizable, does it make sense to characterize continuity by convergece of sequences? Commented Jul 3, 2018 at 22:39
• I believe it turns out that a linear functional on $E$ which is sequentially continuous is automatically continuous (this is not at all obvious). I don't recall how the proof goes off the top of my head. Commented Jul 3, 2018 at 22:46
• I will formulate it as another question. Thanks! Commented Jul 3, 2018 at 22:49
• Actually, I misstated the topology on $E$. You want not the topology I described, but a translation-invariant version of it. I have to go now but will edit this answer later. Commented Jul 3, 2018 at 23:16
• Ok. Please do it! Commented Jul 3, 2018 at 23:23