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

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.

• This is the connection with my original MO question: there, $G$ is a Green function for an elliptic differential operator $P$. $P$ admits a pair of properly supported pseudo-differential operators $Q, R$, with $R$ smoothing, such that $PQ = I + R$. Then, when both sides are restricted to $C_\mathrm{c}^\infty(M)$, we can left-compose with $G$ to get $G = Q - GR$, and in order to establish the pseudo-differential property of $G$ it suffices to establish that $GR$ is a smoothing operator. – JahvedM Nov 28 '17 at 17:34
• Is this right: Since $GR$ is a bounded operator from $L^2$ to $L^2$, it has a kernel. Since it maps all distributions to smooth functions, the kernel has to be smooth? – Deane Yang Nov 29 '17 at 4:21
• But could you explain why you need such a general result? The operator $G$ is pseudodifferential operator of finite order, so $GR$ is a smoothing operator. – Deane Yang Nov 29 '17 at 4:23
• Don't we have a continuous inclusion $L^2\hookrightarrow \mathscr D'$? Then $G$ is a continuous map $L^2 \to \mathscr D'$ so that its restriction $C_c^\infty \to C^\infty$ has closed graph and is thus continuous (where $C_c^\infty$ is endowed with its inductive limit topology). – Jochen Wengenroth Nov 29 '17 at 13:49
• The closed graph theorem applies to linear maps from LF-spaces to Frechet spaces and continuity with respect to coarser Hausdorff topologies implies closed graph. – Jochen Wengenroth Nov 29 '17 at 15:33

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.

(Apologies in advance, this should be a comment, but I'm not able to write one.)

I'm a bit puzzled, because the note that you reference gives this equivalent criterion for being smoothing: that we have such an extension $GR: \mathcal{E}^\prime(M)\rightarrow C^\infty(M)$. But its proof also requires the hypothesis that $M$ be a closed manifold, hence both compact and complete! I don't have access to Dieudonne's text, but you should check whether he requires this condition as well.

• Since everything here is local, I doubt that the assumption of compactness is really needed. – Deane Yang Nov 29 '17 at 4:16
• The note explicitly states that the extension has to have the sequential continuity property that I mention in the text. It also claims (though without proof) that the result holds regardless of compactness assumptions. – JahvedM Nov 29 '17 at 6:46
• Yes, I see that now! The issue of compactness covered by your replacing $\mathcal{D}^\prime(M)$ with $\mathcal{E}^\prime(M)$ according to this remark 0.2 – Hadrian Quan Nov 29 '17 at 7:21
• Incidentally, similar comments go for Dieudonné's version. Since I asked the question, I found that essentially the same criterion as Dieudonné's can be found in Hörmander's The Analysis of Linear Partial Differential Operators I, Theorem 5.2.6. – JahvedM Nov 29 '17 at 9:45