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?
Best Answer
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 $K_k$.