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Let $\{ U_k:k\in \mathbb{N}\}$ be an increasing sequence of open subsets of $\mathbb{R}^d$ whose union is $\mathbb{R}^d$ and such that each $K_k:=\overline{U_k}$ is compact and $K_k\subseteq U_{k+1}$.

We then have that $$ C_c^\infty (K_k)=\left\{ \phi \in C^\infty (\mathbb{R}^d):\mathrm{supp}[\phi ]\subseteq K_k\right\} $$ is a Fréchet space with the usual seminorms ($\sup$-norms of partial derivatives). We also have inclusion maps $i_k:C_c^\infty (K_k)\rightarrow C_c(\mathbb{R}^d)$.

On the other hand, we also have that $$ C^\infty (U_k)=\left\{ \phi :U_k\rightarrow \mathbb{C}:\phi \text{ is smooth.}\right\} $$ is a Fréchet space with the usual seminorms ($\sup$-norms of partial derivatives restricted to compact subsets). We also have restriction maps $r_k:C_c^\infty (\mathbb{R}^d)\rightarrow C^\infty (K_k)$.

Is $C_c^\infty (\mathbb{R}^d)$ to be equipped with the final topology with respect to the former sequence of maps (i.e. the direct limit of this sequence) or the initial topology with respect to the latter sequence of maps (i.e. the inverse limit of this sequence)? Perhaps the topology is the same?

EDIT: I believe I have answered this question (see my answer below), and this has provoked yet another question.

If I am correct, then, of the two choices presented above, we should equip $C_c^\infty (\mathbb{R}^d)$ with the direct limit topology of the sequence $C_c^\infty (K_k)$. This is indeed standard (at least, every source I have seen does it this way), but I want to know why. I first thought of the above alternative, and discounted it for reasons given in my answer. But what about this other 'obvious' alternative: for a fixed compact set $K\subseteq \mathbb{R}^d$ and non-negative integer $n$, define $$ p_{K,n}(f)=\sup \left\{ \left| \partial ^\alpha f(x)\right| :x\in K,|\alpha |\leq n\right\} , $$ and equip $C_c^\infty (\mathbb{R}^d)$ with the topology generated by this collection of seminorms. This should be the subspace topology when thought of as a subset of $C^\infty (\mathbb{R}^d)$, and so is quite natural. Why is it that this is not the topology conventionally equipped when $C_c^\infty (\mathbb{R}^d)$ is used as the space of test functions for distribution theory?

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  • $\begingroup$ I'm not sure if it has what interests you, but Manifolds of differentiable mappings by Peter W. Michor (large PDF!) contains a lot of good stuff on topologies on spaces of differentiable mappings. $\endgroup$ Oct 22, 2012 at 14:19

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I don't think the second topology is even well-defined. If your definition of a smooth function $\phi : K_k \to \mathbb{C}$ is a function which is smooth on the interior, then there's already no guarantee that the partial derivatives of such a function are bounded even if you require $\phi$ to extend continuously to the boundary of $K_k$.

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  • $\begingroup$ I agree. I think I still have a question regarding this though. Let me see if I can reformulate it in a way that makes sense. $\endgroup$ Oct 22, 2012 at 14:11
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Topology on $C_{\mathrm{compact}}^{\infty}(R)$, which is given by all the good semi-norms, is generated by the following collection of semi-norms $\| \cdot\|_{m,\epsilon}$

$m=\{m_i\}_{i=1}^{\infty}$ is an increasing positive integer sequence,

$\epsilon=\{\epsilon_i\}_{i=1}^{\infty}$ is a sequence converging to zero from above.

$K_{i}=[-i,+i],\quad K_{0}=\emptyset$ ,

$\|\psi\|_{m,\epsilon}=\sup_{n\ge0}\epsilon_{n}^{-1}\sup_{x\notin K_{n}}\sum_{m_{n}\ge j\ge0}|D^{j}\psi(x)|$

I think you can prove that the preimage of any open ball under any good semi-norm contains an open set given $\| \cdot\|_{m,\epsilon}$ for some $m=\{m_i\}_{i=1}^{\infty}$ and $\epsilon=\{\epsilon_i\}_{i=1}^{\infty}$ by partitioning any test function into annuli first then consider the convex combination of it.

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I believe I may have answered my question (now that it is well-posed). Please comment if you find an error in my logic.

At the end of the day, we may equip $C_c^\infty (\mathbb{R}^d)$ with any topology we please. In particular, we could equip it with the initial or final topologies mentioned in the question. However, of these two, only one of them makes $C_c^\infty (\mathbb{R}^d)$ into a limit.

We have an inclusion map $i:C_c^\infty (\mathbb{R}^d)\rightarrow C^\infty (\mathbb{R}^d)$ as well as other restriction maps $r_k':C^\infty (\mathbb{R}^d)\rightarrow C^\infty (U_k)$ such that $r_k=r_k'\circ i$. Hence, $C_c^\infty (\mathbb{R}^d)$ cannot be the inverse limit. My guess is that in fact $C^\infty (\mathbb{R}^d)$ is the inverse limit of this sequence, although I have not sat down and proved universality yet. Likewise, my guess is that $C_c^\infty (\mathbb{R}^d)$ is the direct limit of the former sequence (although, once again, I have not confirmed this).

Thus, while we can equip any topology we like, the 'correct' (in a categorical sense) is the direct limit topology of the sequence $C_c^\infty (K_k)$.

EDIT: With respect to the second part of my question, that is, why we do not equip $C_c^\infty (\mathbb{R}^d)$ with the subspace topology inherited from $C^\infty (\mathbb{R}^d)$, I asked my professor via e-mail the other day and he informed me that the reason is that because the resulting space is incomplete. A counterexample is more easily found in $C_c^\infty ((0,1))$ where you can construct a sequence of functions $f_n$ whose support is, say, $[1/n,1/2+1/n]$ that converges in $C^\infty ((0,1))$ but whose limit will not lie in $C_c^\infty ((0,1))$. For this reason, we instead equip $C_c^\infty (\mathbb{R}^d)$ with the direct limit topology, which is complete.

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    $\begingroup$ Be careful, $C_c^\infty$ with the direct limit topology is only sequentially complete, it is NOT complete as a general uniform space invoking Cauchy nets. $\endgroup$
    – Vobo
    Oct 24, 2012 at 10:56
  • $\begingroup$ @Vobo Thanks for that. I've updated the answer accordingly. $\endgroup$ Oct 30, 2012 at 0:34
  • $\begingroup$ @Vobo I was reviewing today when I found a result that superficially seems to contradict what you said. This is 6.6 in Schaefer's "Topological Vector spaces": "The strict inductive limit of a sequence of complete locally convex spaces is complete.". (He does use Cauchy filters, not sequences to prove this, so it's not a matter of him using the word "complete" to mean "sequentially complete".) Does this not imply that $C_c^{\infty}(\mathbb{R})$ is complete, regarded as the direct limit $C_c^{\infty}(\mathbb{R})=\operatorname*{colim}_{m\in \mathbb{N}}C_c^{\infty}([-n,n])$? $\endgroup$ Jan 29, 2017 at 0:18
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    $\begingroup$ $\mathcal{D}$ with the classical inductive limit topology is definitely a complete space. I do not know, what you meant with the direct limit topology here and what I had in mind in Oct. 2012 ... sry if I have messed up here some things. $\endgroup$
    – Vobo
    Feb 4, 2017 at 9:53
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Regarding your follow-up question: $C_c^\infty$ is sequentially complete with the direct limit topology. This is among other properties (like barrelled, Montel spaces) one of the important properties of the space of test functions. With the topology inherited of $C^\infty$, any continuous linear functional (i.e. every element of $\mathcal{E}'$) would be of finite order or, equivalently, of compact support, both restrictions not practicable for the theory of distributions.

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  • $\begingroup$ With metric spaces, if we have an incomplete space, we can always just complete it. Can we do the same with more general topological vector spaces? Does the naive thing (equivalence classes of Cauchy nets) do the trick? If so, can one easily describe the completion of $C_c^\infty$? $\endgroup$ Oct 30, 2012 at 0:36
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    $\begingroup$ @JonathanGleason: Yes, it can be constructed in the naive way, usually with so-called minimal Cauchy filters. See wiki's entry on uniform spaces. And no, I haven't really seen a description of the completion of $C_c^\infty$ with direct limit topology. $\endgroup$
    – Vobo
    Nov 6, 2012 at 21:41

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