Exists a complete metric on the space of smooth, compactly supported functions?

Question

Let $\emptyset \neq U \subsetneq \mathbb{R}^n$ be an open set and denote by $C_c^\infty (U)$ the space of smooth functions with support contained in $U$. On this we have the following metric

$$ d(f,g) = \sum_{k\geq 0} \frac{\Vert f - g \Vert_{C^k(U)}}{1+ \Vert f - g \Vert_{C^k(U)}} $$

where

$$ \Vert f \Vert_{C^k(U)} = \sup_{\substack{\alpha\in \mathbb{N}^n: \ \vert \alpha\vert \leq k} \\ \quad x\in U} \vert \partial_x^\alpha f(x) \vert.$$

One can prove that $(C_c^\infty (U),d )$ is not a complete metric space (the support might accumulate at the boundary, preventing the limit to belong to $C_c^\infty (U)$). My question is the following:

Does there exist a metric $\tilde{d}$ on $C_c^\infty (U)$ such that $(C_c^\infty (U), \tilde{d})$ is a complete metric space and such that the topology coincides with the one induced by $d$?

Answer 1

No, such a metric doesn't exist.

Let $(K_n)_{n\in \mathbb{N}}$ a sequence of compact subsets of $U$ such that $K_n \subset \operatorname{int} K_{n+1}$ for all $n$ and

$$U = \bigcup_{n \in \mathbb{N}} K_n.$$

Then

$$F_n = \{ f \in C_c^{\infty}(U) : \operatorname{supp} f \subset K_n\}$$

is a proper closed (linear) subspace of $C_c^{\infty}(U)$ for every $n$, hence has empty interior. But

$$C_c^{\infty}(U) = \bigcup_{n\in \mathbb{N}} F_n,$$

so this is not a Baire space.

Answer 2

If your metric on $C^\infty_c$ is such that $\forall u \in C^\infty_c, \lim_{\epsilon \to 0} d(\epsilon u ,0)= 0$ then take $(\phi_n) \in C^\infty_c$ with disjoint supports, $\bigcup_n \text{supp}(\phi_n)$ not compact

and set $$f = \sum_n c_n \phi_n $$ where $c_n > 0$ is chosen such that $d(0,c_n \phi_n) \le 2^{-n}$.

Therefore $$\forall u \in C^\infty_c,\qquad d(u,f) \le d(u,0)+\sum_n d(0,c_n\phi_n) < \infty$$