# 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, 2019 at 22:44
• @RobertWolfe On topological vector spaces you have a notion of Cauchy sequence. Mar 10, 2019 at 9:45
• @LorenzoQuarisa: For LCTVSs one usually talks of Cauchy nets rather than sequences. Mar 12, 2019 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. Mar 12, 2019 at 10:35

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 $C^{\infty}$. Mar 10, 2019 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 Mar 10, 2019 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). Mar 10, 2019 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 ?) Mar 10, 2019 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/… Mar 12, 2019 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. Mar 12, 2019 at 10:19