Topology of ${\mathcal D}(\Omega)$ (space of test functions)

Question

I have seen two approaches to the topology of ${\mathcal D}(\Omega)$:

(i) Let $K$ be a compact subset of $\Omega$ and consider the subset ${\mathcal D}_K(\Omega)$ of test functions with support contained in $K$. This is a Fréchet space with the topology given by uniform convergence of all derivatives. Now let $K_1 \subset K_2 \subset \cdots$ be an exhaustion of $\Omega$ by compact subsets, and consider ${\mathcal D}(\Omega)$ as the inductive limit of the spaces ${\mathcal D}_{K_i}(\Omega)$.

(ii) Let ${\mathcal D}^\prime(\Omega)$ be the dual space of ${\mathcal D}(\Omega)$ in the following sense: the set of all linear functionals $f: {\mathcal D}(\Omega) \rightarrow {\mathbb R}$ such that for each compact subset $K$ of $\Omega$ there exists $C>0$ and $m \in {\mathbb N}_0$ such that $|f(\phi)|\leq C\|\phi\|_{C^{m}(K)}$ for all $\phi \in {\mathcal D}(\Omega)$ with support in $K$. Now equip ${\mathcal D}(\Omega)$ with the weak topology, i.e. the initial topology $\sigma({\mathcal D}(\Omega),{\mathcal D}^\prime(\Omega))$.

It's not completely clear to me that these topologies are the same. Can anyone help?

They aren't the same, but they induce the same notion of convergence of sequences, which suffices for many purposes. — terceira
– terceira, Commented Jan 25, 2024 at 14:14

Robert Furber · Accepted Answer · 2024-01-26 21:32:01Z

There's a general principle for proving that a topology on a vector space $E$ is not a weak topology (in the general sense, a topology of the form $\sigma(E,F)$ for some $F \subseteq E^*$).

For $\sigma(E,F)$, every neighbourhood of zero $N$ contains a linear subspace $S_N \subseteq N \subseteq E$ of finite codimension in $E$, i.e. $E/S_N$ is finite dimensional. (Easy to prove for the usual neighbourhood basis of $0$ in $\sigma(E,F)$.)

To prove topology (i) is not a weak topology, on e.g. $\mathcal{D}(\mathbb{R})$, we just find a neighbourhood of $0$ that contains no subspace of finite codimension. So take $U$ to be the set of compactly supported functions $a$ with $\sup_{x \in \mathbb{R}} |a(x)| \leq 1$ (i.e. the unit ball of the sup norm on $\mathcal{D}(\mathbb{R})$). It's easy to verify that this is a neighbourhood of 0 for the inductive limit topology (i). But its only linear subspace is $\{0\}$, which is not of finite codimension in $\mathcal{D}(\mathbb{R})$.

In other words, an infinite dimensional locally convex space which has a continuous norm does not carry its weak topology. — Jochen Wengenroth
– Jochen Wengenroth, Commented Jan 27, 2024 at 9:41

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