This question is about the topology on $\mathscr{D}(\Omega)$, the space of test functions. The conventional way to construct $\tau$ is to define a locally convex neighborhood base of $0$, and check that it gives rise to a topological vector space structure. But an interesting alternate claim I have seen (e.g., Exercise 10.4 of Leoni's Sobolev Spaces book) is that the topology $\tau$ can be given through a particular family of seminorms. Despite spending several days trying to prove this, I simply cannot see how the seminorm topology is the same, and I'm wondering if any knowledgeable people could chime in.
The Definitions
For an open subset $\Omega \subseteq \mathbb{R}^n$, given a compact $K \subseteq \Omega$, define $$\mathscr{D}_K(\Omega) = \{f \in C^{\infty}(\Omega) : \text{supp}(f) \subseteq K\}.$$ It is well-known that these $\mathscr{D}_K$ admit a Fréchet space structure: simply take the seminorms $p_{n, K}(f) = \sum_{|\alpha| \leq n} \sup_{x \in K} |(\partial^{\alpha} f)(x)|$. Using these $p_{n, K}$ as a countable family of seminorms, one gets a locally-convex topology $\tau_K$ on $\mathscr{D}_K$.
A neighborhood base $\mathcal{B}_0$ of the origin in $\mathscr{D}(\Omega)$ is defined as follows: an absolutely convex set $U \subseteq \mathscr{D}$ is an element of $\mathcal{B}_0$ iff $U \cap \mathscr{D}_K \in \tau_K$ for every compact $K \subseteq \Omega$. Then the topology on $\mathscr{D}(\Omega)$ is simply given by $\tau = \{\phi + B : \phi \in \mathscr{D}, B \in \mathcal{B}_0\}$.
Finally, for any open set $\Omega \subseteq \mathbb{R}^n$, there exists a compact exhaustion: a sequence of relatively compact open sets $(U_n)$, all contained in $\Omega$, such that $U_0 = \emptyset$, $\overline{U_n} \subseteq U_{n + 1}$ for all $n$, $\Omega = \bigcup_{n \in \mathbb{N}} U_n$. We write $K_n = \overline{U_n}$.
The Statement
The claim is that we can define an equivalent topology on $\mathscr{D}$ by using a family of seminorms: for a strictly increasing sequence of positive integers $\mathbf{m} = (m_n)$, and a sequence of positive reals $\mathbf{\epsilon} = (\epsilon_n)$ decreasing to $0$, we can define the ball $$V(\mathbf{m}, \mathbf{\epsilon}) = \{\phi \in \mathscr{D} : \sum_{|\alpha| \leq m_n} \sup_{x \notin U_n} |(\partial^{\alpha} \phi)(x)| < \epsilon_n, \ n = 0, 1, 2, \dots \}.$$ This can be given by the seminorm $p_{\mathbf{m}, \mathbf{\epsilon}}(f) = \sup_n (\frac{1}{\epsilon_n} \sum_{|\alpha| \leq m_n} \sup_{x \notin U_n} |(\partial^{\alpha} f)(x)|)$. The claim is that by considering the set of all $p_{\mathbf{m}, \mathbf{\epsilon}}$, where $\mathbf{m}$ and $\mathbf{\epsilon}$ range over all suitable sequences, this family generates the same topology $\tau$ on $\mathscr{D}$.
To check their equivalence, it suffices to look at neighborhood bases of $0$. One inclusion (that $\tau_{\text{semi}} \subseteq \tau$) is clear from the definition of $\tau$. The other direction is what's tricky. It is argued that
Now suppose that $V \in \mathcal{B_0}$; then for each $K_n$, $V \cap \mathscr{D}_{K_n}$ is a neighborhood of $0$ in $\mathscr{D}_{K_n}$, so contains a set of the form $\{f \in \mathscr{D}_{K_n} : p_{m_n}(f) < \epsilon_n\}$ as the semi-norms are non-decreasing. Therefore $V(\mathbf{m}, \mathbf{\epsilon}) \subseteq V$ so these two topologies coincide.
The Issue
I can see everything up to the last sentence: how on earth does the inclusion hold? Because if we expand out the definitions, we see that elements of $V(\mathbf{m}, \mathbf{\epsilon}) \cap \mathscr{D}_{K_n}$ do not necessarily have to be in $\{f \in \mathscr{D}_{K_n} : p_{m_n}(f) < \epsilon_n\}$.
Indeed, if $f \in V(\mathbf{m}, \mathbf{\epsilon}) \cap \mathscr{D}_{K_n}$, then $f \in \mathscr{D}_{K_n}$ and $\sum_{|\alpha| \leq m_0} \sup_{x \in \Omega} |(\partial^{\alpha} f)(x)| < \epsilon_0$, $\sum_{|\alpha| \leq m_1} \sup_{x \notin U_1} |(\partial^{\alpha} f)(x)| < \epsilon_1$, and so on, all the way until $\sum_{|\alpha| \leq m_n} \sup_{x \notin U_n} |(\partial^{\alpha} f)(x)| < \epsilon_n$. And this does not seem to give any good control over $p_{m_n, K_n}(f)$ whatsoever.
Except for the first term, all of these bounds only give control of $f$ on subsets of $K_n$: e.g., we know $p_{m_n, J}(f) < \epsilon_n$ for a subset $J = K_n \setminus U_n \subset K_n$, and $p_{m_{n - 1}, R}(f) < \epsilon_{n - 1}$ for $R = K_n \setminus U_{n - 1}$, etc., but we never get a full bound on the quantity $p_{m_n, K_n}(f)$. We know $f$ has small seminorm on individual pieces, but that doesn't mean $p_{m_n}(f)$ stays below $\epsilon_n$ as we consider the supremum over the whole set $K_n$.
So the set $V(\mathbf{m}, \mathbf{\epsilon})$ does not seem to constrain $p_{m_n, K_n}$ itself. The definitions of $V(\mathbf{m}, \mathbf{\epsilon})$ and the seminorms $p_{\mathbf{m}, \mathbf{\epsilon}}$ as involving the supremum over $x \notin U_n$ are the serious impediments here; it means the information about the size of $f$ is only valid for a subset of $K_n$, and I'm not sure how to ensure $f$ is small globally.
If, for instance, $\epsilon_n$ is much smaller than $\epsilon_0, \epsilon_1, \dots, \epsilon_{n - 1}$, then it seems possible that $V(\mathbf{m}, \epsilon)$ might contain a test function that falls quickly as it approaches the boundary of $K_n$, but peaks in the interior, and $p_{m_n, K_n}( \cdot ) \sim \epsilon_0$ could be enormous relative to $\epsilon_n$. Then perhaps that test function might be an element of $V(\mathbf{m}, \epsilon)$, but not $V$.
So I can't see how the seminorm topology is of equal strength to $\tau$. Am I missing something obvious here? After all, if Laurent Schwartz, Giovanni Leoni, and Clément Mouhot are all claiming the same thing, it seems very likely that they're the ones in the right.
I don't know how to prove this rigorously, but I think the picture is the following:
For simplicity, assume $K_n=B_n(0)$ on the Euclidean space. Assume $N$ is an open convex balanced neighborhood of $0$ in ${\mathscr D}$, then $N\cap C_0^\infty(K_n)$ is open in $C_0^\infty(K_n)$ for every $n$. Assume $N\cap C_0^\infty(K_1)$ is the set $\{u:\sup_{x\in B_1(0), \,\, |\alpha|\leq m_1}|\partial^\alpha u|<\epsilon_1\}$. Now consider $N\cap C_0^\infty(K_2)$: the point is, this set cannot have any restrictions on derivatives of order more than $m_1$ on any interior points of $K_1$, because functions in $N\cap C_0^\infty(K_1)$ can have uncontrolled $m_1+1$ or more order derivatives at any interior point of $K_1$. Therefore, if the defining condition for $N\cap C_0^\infty(K_2)$ have any restrictions on $m_1+1$ or more order derivatives, it must be something like $$ \sup_{x\in S, \,\, |\alpha|=m_1+1}|\partial^\alpha u|<\epsilon_2, $$ where $S$ is some set in $K_2-K_1$. (This statement is the spot I don't know how to prove - The issue seems to be, in a Frechet space like $C_0^\infty(K_2)$, how to describe the set ALL continuous semi-norms.). That is, on interior of $K_1$, there is only restriction involving $\leq m_1$ order derivatives. Continue like this, given any $n$, for sufficiently large $m$ if there is any restriction on $m$-th derivative for functions in $N$, such restriction must stay away from $K_n$.
If you know a rigorous proof, please let me know.