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

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.

• What's the metric on $C_c^\infty(\mathbb{R})$ that you're using? You give an induced topology, but topologies don't give enough structure to define Cauchyness. Perhaps there's an obvious metric that I should be seeing? – user123641 Mar 9 '19 at 22:44
• @RobertWolfe On topological vector spaces you have a notion of Cauchy sequence. – Lorenzo Quarisa Mar 10 '19 at 9:45
• @LorenzoQuarisa: For LCTVSs one usually talks of Cauchy nets rather than sequences. – Abdelmalek Abdesselam Mar 12 '19 at 10:31
• @Kato yu: the reason you are getting in trouble is the seminorms you use do not define the topology of $C_{c}^{\infty}$. As Lorenzo explained below, these are appropriate to the space $C^{\infty}$ without the compact support property. – Abdelmalek Abdesselam Mar 12 '19 at 10:35

## 2 Answers

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.

• To expand a bit: the topology OP defined has the 'uniform convergence with derivatives on all compact sets'. The topology which makes it complete has the 'uniform convergence with derivatives and supports contained in a fixed compact'. So, you add the requirement that the supports of the functions in the sequence are contained in a fixed compact. With this topology, the above example is no longer a Cauchy sequence. Also, if I recall correctly the completion of $C_c^{\infty}$ with respect to OP's topology is just $C^{\infty}$. – Lorenzo Quarisa Mar 10 '19 at 9:51
• @LorenzoQuarisa Except that what you said doesn't define a topology (what are its open sets) that's the reason for OP's trouble. The next trouble is that the norms in my post won't translate directly to distributions ($T$ can be continuous even if $|T(f)|$ is unbounded on $f \in U_{h,0,1}$) which is quite different to the Schwartz space – reuns Mar 10 '19 at 9:59
• @reuns of course the convergence itself doesn't define a topology. What I'm saying is that there is a topology inducing that convergence which makes $C_c^{\infty}$ into a complete topological vector space (and it is the same topology defined in your post). – Lorenzo Quarisa Mar 10 '19 at 10:12
• @LorenzoQuarisa Well I wanted to insiste on my second point, because I don't really understand it, why $C^\infty_c$ must be more complicated than $S$ (it is about things like metrizable, locally strictly convex, or whatever abstract property ?) – reuns Mar 10 '19 at 10:49
• @reuns: You are perfectly right that $C_{c}^{\infty}$ should not be (much) more complicated than Schwartz space $\mathcal{S}$. It's just unfortunate that almost all books (except the one by Schwartz himself or the one by Horvath) give a good treatment of the topology of $\mathcal{S}$ but make a complete massacre of the explanation of the topology of $C_{c}^{\infty}$. As to the different complexity of the two spaces see this MO question: mathoverflow.net/questions/187404/… – Abdelmalek Abdesselam Mar 12 '19 at 10:27

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 $$C^\infty_c$$.

• The $||\cdot||_{h,k}$ do not define the topology of $\mathcal{D}=C_{c}^{\infty}$. The easiest way to see this is to realize that these seminorms force distributions to be of finite order globally. Take a sum of higher and higher derivatives of Dirac deltas at isolated points with no accumulation point in the domain gives an easy example of infinite order distribution. – Abdelmalek Abdesselam Mar 12 '19 at 10:19