Show that a sequence is not Cauchy in $C^\infty_c(\mathbb{R})$

Question

Denote $C^\infty_c(\mathbb{R})$ the space of all compactly supported infinitely differentiable functions. We equipped $C^\infty_c(\mathbb{R})$ the topology induced by the following family of semi-norms: $$p_{n}(f)=\max_{\alpha\leq n}\sup_{x\in K_n}|f^{(\alpha)}(x)|,\quad n\in\mathbb{N}$$ where $(K_n)_{n\in\mathbb{N}}$ is an increasing sequence of compact sets such that $\bigcup_{n=1}^\infty K_n=\mathbb{R}.$ Then consider the sequence given by the following: Choose $\phi\in C^\infty_c(\mathbb{R})$ such that $\phi(x)=0$ whenever $x\neq[0,1]$. Define $$\phi_n(x)=\sum_{j=1}^n 2^{-j}\phi(x-j),\quad x\in\mathbb{R}, n\in\mathbb{N}.$$

Intuitively, since the $\phi_n$ converges pointwise to a function which is not compactly supported, it follows that $\phi_n$ cannot be a Cauchy sequence in $C^\infty_c(\mathbb{R})$ with the topology we equipped, as this topology is complete. However, I'm having trouble to prove this by using these semi-norms. Could anyone give some suggestions? Thanks.

Answer 1

You just have found a counterexample that shows that $C^\infty_c (\mathbb{R})$ isn't complete with the linear topological structure you used. That's why that space isn't usually equipped with that topology. Reference: Rudin's Functional Analysis at page 137.

Answer 2

I had the same problem, you can't easily describe the topology of $C^\infty_c$ with your semi-norms. Instead, for every $h$ continuous let $$\|f\|_{h,k} = \sup_x |h(x) f(x)|+\sup_x |h(x) f^{(k)}(x)|$$

Because $h$ can have arbitrary fast growth, if $f$ is not compactly supported then $\|f\|_{h,0} = \infty$ for some $h$.

Then $C^\infty_c$ is the intersection of all the Banach spaces with those norms, with $U_{h,k,r}=\{ f \in C^\infty_c, \|f\|_{h,k} < r\}$ as its basis of open sets, ie. the open sets are the translates, finite intersection, and arbitrary union of the $U_{h,k}$.

Then $f_n \to f$ iff for every open set $U \ni f$ there is some $N$ such that $(f_n)_{n \ge N} \subset U$.

To do : prove that if all the $f_n$ aren't compactly supported on a common interval then there is some $h$ such that $\|f_n\|_{h,0}$ is unbounded.

By the intersection and topology definition, iff $f_n$ is Cauchy in all those Banach spaces then it converges to some element of $C^\infty_c$.