Composition of a smoothing operator with an $L^2$-bounded operator, non-compact Riemannian manifold

Question

I'm trying to close in on a definitive answer to my own question BVPs for elliptic PDOs: When do Green functions ($L^2$ inverses) define pseudo-differential operators in the interior?, and think I have reduced it to the following subproblem about the interplay between smoothing operators and (bounded) $L^2$ operators preserving smoothness (see below) on a Riemannian manifold.

Setup/notation: Let $(M,g)$ be a smooth Riemannian manifold, not necessarily compact or complete, and let $\mathrm{d} \mu_g$ be the smooth measure on $M$ determined by the volume density of $g$. Let $L^2(M,\mathrm{d} \mu_g)$ be the resulting Hilbert space of functions on $M$ which are square-integrable with respect to $\mathrm{d} \mu_g$. Let $G : L^2(M,\mathrm{d} \mu_g) \to L^2(M,\mathrm{d} \mu_g)$ be a bounded linear operator with the property that $G[C_\mathrm{c}^\infty(M)] \subseteq C^\infty(M)$—if necessary for the proof, I'm happy to also assume that $G[H^{s}_\mathrm{c}(M)] \subseteq H^{s+k}_\mathrm{loc}(M)$ for all real $s \geq 0$ and some fixed $k > 0$, where $H^{s}_\mathrm{c}(M)$ and $H^{t}_\mathrm{loc}(M)$ are the standard spaces of compactly supported Sobolev distributions and of locally Sobolev distributions on $M$, respectively. In the latter case, however, I am not willing to assume from the outset that the resulting maps $H^{s}_\mathrm{c}(M) \to H^{s+k}_\mathrm{loc}(M)$ are continuous in the standard topologies of these Sobolev spaces. Let also $\mathscr{D}'(M)$ denote the standard space of distributions on $M$, and $\mathscr{E}'(M)$ the space of distributions with compact support. A smoothing operator is a continuous operator $C_\mathrm{c}^\infty(M) \to \mathscr{D}'(M)$ whose Schwartz kernel is an element of $C^\infty(M \times M)$.

Question: If $R$ is a properly supported smoothing operator [so that $R$ sends $\mathscr{E}'(M)$ to $C_\mathrm{c}^\infty(M)$], does it follow that the composition $GR$ defines a smoothing operator?

My understanding:

1. By my assumption on $G$, $GR$ is well-defined as a linear map from $\mathscr{E}'(M)$ to $C^\infty(M)$. This is a purely algebraic statement.
2. An immediate analytic prerequisite is that the related linear map $\widetilde{GR} : C_\mathrm{c}^\infty(M) \to \mathscr{D}'(M)$, given by postcomposing with the obvious embedding, should be continuous when $\mathscr{D}'(M)$ has the standard weak topology. But this is the case because properly supported operators are continuous from $C_\mathrm{c}^\infty(M)$ to $C_\mathrm{c}^\infty(M)$, while $L^2$-bounded ones such as $G$ define continuous maps from $C_\mathrm{c}^\infty(M)$ to $\mathscr{D}'(M)$.
3. As far as I know, even when taken together the statements in 1. and 2. do not automatically ensure that we are dealing with a smoothing operator proper. According to this note, we also need to know that $GR:\mathscr{E}'(M) \to C^\infty(M)$ is sequentially continuous when $\mathscr{E}'(M)$ has the weak topology and $C^\infty(M)$ has the standard Fréchet space topology. Alternatively, there is the following characterization of smoothing operators which I am adapting from Dieudonné's Treatise on Analysis, Vol. 7, (23.11.1) (and see also a similar statement in Hörmander's The Analysis of Linear Partial Differential Operators I, Theorem 5.2.6):

   Let $K$ be a linear map from $\mathscr{E}'(M)$ to the Fréchet space $C^\infty(M)$ which is continuous on all bounded subsets of $\mathscr{E}'(M)$. Then $K$ is the extension to $\mathscr{E}'(M)$ of a smoothing integral operator.

Under my minimal assumptions on $G$ I am not sure how to establish that either of the continuity conditions in 3. is satisfied.

Answer 1

Thanks to Jochen Wengenroth's comments, I can now give the full answer: the idea is that $C_\mathrm{c}^\infty(M) \hookrightarrow L^2(M, \mathrm{d} \mu_g) \xrightarrow{G} L^2(M, \mathrm{d} \mu_g) \hookrightarrow \mathscr{D}'(M)$ is continuous, and an argument using a version of the closed graph theorem in the (locally) Fréchet category then shows that simply by virtue of mapping $C_\mathrm{c}^\infty(M)$ to $C^\infty(M)$, $G$ must actually do so continuously with respect to the Fréchet topology of $C^\infty(M)$. In detail:

Lemma. Let $T:C_\mathrm{c}^\infty(M) \to \mathscr{D}'(M)$ be linear, continuous relative to the standard inductive limit topology of $C_\mathrm{c}^\infty(M)$ and the weak topology of $\mathscr{D}'(M)$, and with image contained in $C^\infty(M)$. Then $T:C_\mathrm{c}^\infty(M)\to C^\infty(M)$ is continuous relative to the standard Fréchet space topology of the codomain.

Proof. $T$ is continuous as a map $C_\mathrm{c}^\infty(M)\to C^\infty(M)$ when the codomain is equipped with the subspace topology from $\mathscr{D}'(M)$. Hence its graph is closed in the appropriate product topology, and it is also closed in the finer product topology obtained by equipping $C^\infty(M)$ with its Fréchet space topology. The result then follows from the fact that the closed graph theorem is applicable to linear maps from LF-spaces to Fréchet spaces.

Wrapping up. Since $R : \mathscr{E}'(M) \to C_\mathrm{c}^\infty(M)$ is continuous, so is the composition $GR : \mathscr{E}'(M) \to C^\infty(M)$. By the equivalence criteria I mentioned in the question, this implies the smoothing property.

Thanks to all who helped with this.