Explicit set of seminorms on $C^\infty_c(\mathbb R)$

Question

Let $E_n = C_{[-n,n]}^\infty(\mathbb R)$ with the topology generated by the sets $$ \Gamma_n = \{q_{C^k, [-n,n]}|_{E_n}\} $$ of (restrictions of) seminorms $$ q_{C^k, L} : C^\infty_{c}(\mathbb R) \to [0,\infty[,\ f\mapsto\max\{\|f|_L\|_\infty, \|f'|_L\|_\infty,\ldots, \|f^{(k)}|_L\|_\infty\} $$ for $k\in\mathbb N_0$ and $L\subseteq \mathbb R$ compact.

We will now look at two sets of seminorms on $E = \bigcup_{n\in\mathbb N} E_n$

$$ \Gamma_1 = \{q\colon E \to [0,\infty[ \ \ \text{seminorm} : (\forall n\in\mathbb N)\ q|_{E_n} \text{is continuous}\},\\ \Gamma_2 = \{q_{C^k, L} : k\in\mathbb N_0,\ L\subseteq \mathbb R \ \ \text{compact}\}. $$

Note that $\Gamma_1$ induces the topology $\ \mathcal O_1$ on $E = \varinjlim E_n$ as a direct limit of locally convex spaces. We denote the topology on $E$ generated by $\Gamma_2$ by $\mathcal O_2$. Is is true that $\mathcal O_1 = \mathcal O_2$?

Answer 1

The topology $\mathcal{O}_1$ is the correct one. The topology $\mathcal{O}_2$ is strictly weaker and is the typical thing beginners (me included, a long time ago) in the area try in order to give explicit seminorms for $\mathscr{D}$. You can learn about a correct set of explicit seminorms in Example 3 of Doubt in understanding Space $\mathscr D(\Omega)$

To see why $\Gamma_2$ does not work. Take your favorite bump function $\varphi$ supported in $[-1,1]$ and define its translates $\varphi_n(x)=\varphi(x-n)$. It is easy to see that such a sequence does not converge in $\mathscr{D}(\mathbb{R})$ and in fact is not even a bounded set because there is no common compact containing the supports of all the $\varphi_n$. However this sequence converges to the zero function for the topology $\mathcal{O}_2$.

Answer 2

Consider $\Phi \colon (E, \mathcal O_1) \to (E, \mathcal O_2),\ f \mapsto f$. We wish to show that $\Phi$ is a homeomorphism.

It suffices to show that $\Phi$ and $\Phi^{-1}$ are continuous at $0$ as they are linear.

Step 1: $\Phi$ is continuous at 0.

As a linear map from a locally convex direct limit to a TVS, we see that $\Phi$ is continuous iff $\Phi|_{E_n}$ is continuous for every $n\in\mathbb N$. Let $U \subseteq (E,\mathcal O_2)$ be an open $0$-neighborhood. Thus there exists a compact set $L\subseteq R$, a constant $k\in\mathbb N_0$ and an $\varepsilon >0$ such that $$ B_{\varepsilon}^{q_{C^k, L}}(0) \subseteq U. $$ We know that the topology of a LC direct limit does not depend on the first few terms, i.e. $E = \varinjlim E_n = \varinjlim_{n\geq n_0} E_n$. We thus may choose some $n_0\in\mathbb N$ with $L \subseteq [-n_0, n_0]$ and obtain $$ \ldots \subseteq B^{q_{C^k, [-(n_0+1), n_0+1]}}_{\varepsilon}(0) \subseteq B^{q_{C^k, [-n_0, n_0]}}_{\varepsilon}(0) \subseteq U. $$ Thus $\Phi|_{E_n}(B_\varepsilon^{q_{C^k, [-n,n]}}(0)) \subseteq U$ and hence $\Phi|_{E_n}$ is continuous for all $n\geq n_0$. By identifying $E=\varinjlim E_n =\varinjlim_{n\geq n_0} E_n$ we may thus assume that all $\Phi|_{E_n}$ are continuous and hence $\Phi$.

Step 2: (Edit) This won't be possible. See accepted answer.

We only obtain $\mathcal O_2 \subsetneq \mathcal O_1$.