I am reading Francois Treves book Topological Vector Spaces, Distributions and Kernels, page 532. Given open subsets $X\subset \mathbb R^m$ and $Y\subset \mathbb R^n$, the author defines a kernel $K(x,y)\in \mathcal D'(X\times Y)$ to be Semi-Regular if the map $$v\mapsto \left(u\mapsto \langle K(x,y),u(x)v(y)\rangle\right)$$ maps $C^\infty_c(Y)$ into $C^\infty(X)\subset \mathcal D'(X)$, where the inclusion is given by $f\mapsto \left(u\mapsto \int_X u f\right)$. Then, the author claims, without justification, that such a map is continuous $C_c^\infty(Y)\to C^\infty(X)$. Indeed, they claim that semi-regular kernels are elements of the space $L(C^\infty_c(Y),C^\infty(X))$ of continuous linear maps. I do not see how to deduce from the fact the image of $K$ lies in the image of $C^\infty(X)$ in $D'(X)$, that this map is continuous. Indeed, the topology on $C^\infty(X)$ is finer than the one induced on it as a subspace of $\mathcal D'(X)$. This is equivalent to the fact that the inclusion is continuous.
Is the claim that every continuous map $C^\infty_c(Y)\to \mathcal D'(X)$ whose image lies in $C^\infty(X)$ is continuous as a map $C^\infty_c(Y)\to C^\infty(X)$ true? Is there an easy justification that I am missing?
I don't know if I would call what follows an "easy justification" (or even necessarily the most concrete available in this situation) but this is justified by a general fact.
Claim: Suppose $E,F,G$ are topological vector spaces such that $F$ is continuously embedded in $G$. Suppose that $E$ is barrelled and $F$ is Frechet and that $T: E \to G$ is a continuous map such that $T(E) \subset F$. Then $T$ is continuous as a map from $E$ to $F$.
Proof of claim: By a sufficiently general version of the closed graph theorem (for maps from barelled spaces into Frechet spaces, see here), it suffices to see that the graph of $T$ is closed as a subset of $E \times F$. But from the hypothesis this is immediate. If $(x_a)_{a \in A}$ is a net converging to $x$ in $E$ and is such that $(Tx_a)_{a \in A}$ converges to $y$ in $F$ then $Tx_a \to y$ in $G$ by the continuous embedding of $F$ into $G$. Since $T$ is continuous from $E$ to $G$, $Tx_a \to Tx$ in $G$. This implies $y = Tx$ which is what we needed to check.
You are then in exactly the situation described in the claim, where $E = C_c^\infty(Y)$ is (by construction of its topology) an LF-space and in particular is barrelled. $F = C^\infty(X)$ is a Frechet space and by the Schwartz kernel theorem, the map under consideration is a continuous map as a map from $E$ to $G = \mathcal{D}'(X)$.