# Semi-regular kernels and distributions on open subset of $\mathbb R^n$

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 $$G = \mathcal{D}'(X)$$.

• This is exactly what I needed, thank you. I see how the proof of the claim works for $E=C^\infty_c(Y)$. However, is it always (that is, for arbitrary barrelled space $E$) true that we only need to look at sequences, and not more generally, filters? Commented Nov 20, 2023 at 17:12
• Well, I suppose this could be done with any net, and then you're safe. Commented Nov 21, 2023 at 1:18
• @OrKedar Yes, you can just replace where I had a sequence with a net and everything works (originally I had stated a much more specialised version of the claim when drafting the answer and then I realised it was better to put the general one but I forgot to update that detail in the proof, sorry!) Commented Nov 21, 2023 at 7:52