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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.

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    $\begingroup$ 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? $\endgroup$
    – user123641
    Mar 9, 2019 at 22:44
  • $\begingroup$ @RobertWolfe On topological vector spaces you have a notion of Cauchy sequence. $\endgroup$
    – Lorenzo Q
    Mar 10, 2019 at 9:45
  • $\begingroup$ @LorenzoQuarisa: For LCTVSs one usually talks of Cauchy nets rather than sequences. $\endgroup$ Mar 12, 2019 at 10:31
  • $\begingroup$ @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. $\endgroup$ Mar 12, 2019 at 10:35

2 Answers 2

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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.

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    $\begingroup$ 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 $C^{\infty}$. $\endgroup$
    – Lorenzo Q
    Mar 10, 2019 at 9:51
  • $\begingroup$ @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 $\endgroup$
    – reuns
    Mar 10, 2019 at 9:59
  • $\begingroup$ @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). $\endgroup$
    – Lorenzo Q
    Mar 10, 2019 at 10:12
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    $\begingroup$ @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 ?) $\endgroup$
    – reuns
    Mar 10, 2019 at 10:49
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    $\begingroup$ @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/… $\endgroup$ Mar 12, 2019 at 10:27
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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 $C^\infty_c$.

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  • $\begingroup$ 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. $\endgroup$ Mar 12, 2019 at 10:19

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