A "proof" that all separately continuous maps on LF-spaces are continuous

Question

Problem

Consider the locally convex spaces $C^\infty(\mathbb{R})$ and $C^\infty_c(\mathbb{R})$, the former equipped with its standard Fréchet topology, the latter equipped with the inductive limit topology given by $C^\infty_c(\mathbb{R}) = \varinjlim C^\infty_c([-n,n])$ where each $C^\infty_c([-n,n])$ is equipped with its Fréchet topology inherited from $C^\infty(\mathbb{R})$. This makes $C^\infty_c(\mathbb{R})$ into an LF-space. Note that here we use the notation $C^\infty_c(K) = \{f \in C^\infty(\mathbb{R}) : \text{supp }f \subset K\}$ for compact sets $K$, as it is used in Treves.

Now, the map $C^\infty(\mathbb{R}) \times C^\infty_c(\mathbb{R}) \to C^\infty_c(\mathbb{R}), (f,g) \mapsto fg$ is separately continuous, but not jointly so (see Treves, Chapter 41). Hence, there must be something wrong with the very basic looking arguments in the following statement, and I'm curious if anyone can tell me which step is the problematic one.

"Theorem": Every separately continuous bilinear map $\psi: C^\infty(\mathbb{R}) \times C^\infty_c(\mathbb{R}) \to C^\infty_c(\mathbb{R})$ is jointly continuous.

"Proof": A linear map $E \to F$ from an LF-space $E = \varinjlim E_n$ to a locally convex space $F$ is continuous if and only if all restricted maps $E_n \to F$ are continuous (Treves, Proposition 13.1). Hence separate continuity of $\psi$ implies separate continuity of the restricted maps

$$ \psi_n : C^\infty(\mathbb{R}) \times C^\infty_c([-n,n]) \to C^\infty_c(\mathbb{R}).$$

But if $E,F$ are Fréchet and $G$ is locally convex, every bilinear separately continuous map $E \times F \to G$ is even jointly continuous. (Treves, Corollary to Theorem 34.1). Hence all $\psi_n$ are continuous.

But it is also true that if $E = \varinjlim E_\alpha$ and $F = \varinjlim F_\alpha$ are locally convex inductive limit spaces, and $G$ is any locally convex space, then a bilinear map $E \times F \to G$ is continuous if and only if the restrictions $E_\alpha \times F_\beta \to G$ are continuous for all $\alpha,\beta$ (Mallios, Chapter IV, Lemma 2.1).

Viewing $C^\infty(\mathbb{R}) = \varinjlim C^\infty(\mathbb{R})$ as a trivial inductive limit, this means that the continuity of the $\psi_n$ implies the continuity of $\psi$. This "proves" the statement. $\stackrel{?}{\square}$

Notes

I came across (a more general version of) this statement and its proof in the paper "Cyclic-type cohomology of strict inductive limits of Fréchet algebras" by Lykova, Lemma 4.1. I don't want to appear like I'm calling them out, though, it may very well be that only my use of their statement is incorrect. The books I cite, "Topological Vector Spaces, Distributions and Kernels" by Treves and "Topological Algebras: Selected topics" by Mallios seem to me very respected, and none of the statements seem obviously false to me, so I'm a bit confused here.

One suspicion I have is that treating $C^\infty(\mathbb{R})$ as a trivial inductive limit is the issue, but the proof of Lemma 2.1 in Mallios looks very straightforward, and doesn't seem to care whether the inductive limit is a strict one or not.

Answer 1

Already the first sentence in the ''proof'' is doubtful: The characterization of continuity of maps $f:\lim_\limits{n\to} E_n \to F$, that all restrictions to $E_n$ are continuous, holds for linear maps but there is no reason that this should be true (for colimits in the category of locally convex spaces) for bilinear maps.

This is not only doubtful but indeed wrong. For example, in an old article with the late Susanne Dierolf (Inductive limits of topological algebras. Linear Topol. Spaces Complex Anal. 3 (1997), 45–49) we constructed a strict locally convex inductive limit of Fréchet commutative algebras such that the multiplication is discontinuous (although the restrictions to the ''steps'' are continuous). I believe that Mallios claimed the contrary.