I asked the following question on math.SE (https://math.stackexchange.com/questions/2420298/bvps-for-elliptic-pdos-when-do-green-functions-l2-inverses-define-pseudo-d) just over two months ago, and it only received one, rather unsatisfactory to me, answer there. I'm wondering if people here can have a look. Some related questions have been posed here and here (and I noted in particular the literature link provided in the answer to the latter question). However, neither of those discussions cover the possibility of boundary conditions being present, or possible lack of compactness.

Repost of the original question

Let me illustrate my question by starting with the simplest possible example: Let us consider $P := - \mathrm{d}^2/\mathrm{d}x^2$, an elliptic partial differential operator on $\mathbb{R}$; let us also consider the following boundary-value problem on the interval $\overline{\Omega} = [0,1]$: \begin{equation} P u = f, \qquad u(0)=u(1)=0. \end{equation} As is (I think) well-known, when seen as an operator $L^2(\Omega) \to L^2(\Omega)$, $P$ is unbounded. However, it is closed on the dense domain $D(P) := H^2(\Omega) \cap H_0^1(\Omega)$ where $H_0^1(\Omega)$ is the closure of $C_{\mathrm{c}}^\infty(\Omega)$ in the $H^1$ norm (so that any element of this space has vanishing trace on $\partial \Omega = \{0,1\}$, i.e. it satisfies the Dirichlet boundary condition above in a weak sense). Furthermore, $0$ is in the resolvent of $(P,D(P))$, i.e. there exists a bounded inverse $P^{-1} : L^2(\Omega) \to L^2(\Omega)$. In fact, in this example the inverse is easily computed: it is the integral operator defined by the (continuous, as it happens) kernel \begin{equation} G(x,y) = \begin{cases} x(1-y) & x \leq y \\ y(1-x) &x > y \end{cases}, \quad (x,y) \in \Omega \times \Omega. \end{equation} Of course, when viewed as a distribution in $\mathscr{D}'(\Omega \times \Omega)$, $G$ is the Schwartz kernel of $P^{-1}$ which we know on abstract grounds must exist since $P^{-1} : C_{\mathrm{c}}^{\infty}(\Omega) \to \mathscr{D}'(\Omega)$ is continuous.

My question is the following: in this example and in more general examples where $P$ is a second-order elliptic differential operator on, say, an open (and not necessarily compact) region $\Omega$ with smooth boundary in $\mathbb{R}^n$, and assuming that we can find a suitable dense domain $D(P)$ for $P$ as above so that $(P,D(P))$ has a bounded inverse $P^{-1} : L^2(\Omega) \to L^2(\Omega)$, does the Schwartz kernel $G$ of $P^{-1}$ always define a pseudodifferential operator on $\Omega$?

Addendum for MO

User mcd on math.SE points out that the Boutet de Monvel calculus ought to be relevant here. Aside from wishing to see exactly how this is, I wonder whether the possible lack of compactness (of $\Omega$) might cause problems in such an approach.


I have reduced my question to the following subproblem: Composition of a smoothing operator with an $L^2$-bounded operator, non-compact Riemannian manifold, as explained in a comment there.

  • 1
    $\begingroup$ The operator $P^{-1}$ is a right inverse? $\endgroup$ – Deane Yang Nov 28 '17 at 18:37
  • $\begingroup$ Yes: that is, $0$ is in the resolvent of the unbounded operator $(P, D(P))$. $\endgroup$ – JahvedM Nov 28 '17 at 18:50

The book by Melrose was almost exactly written to address issues like the ones you mentioned above, where the very first example is the one in the question. The construction fo the parametrix is the focus point of the book. However the book is supposed to be very dense and difficult to read, and it also has a fair amount of typos. Now there are more reader friendly lecture notes by his academic descendants. Depending on your exact research problem, I am not sure if you need the de Monvel calculus, which is much more difficult than the $b$-calculus and $c$-calculus constructed Melrose's book. The expert on the site is Rafe Mezzo, who you may contact directly.

  • $\begingroup$ Thanks for the references and pointers. Proposition 5.64 in Melrose's book has a statement which might be of relevance, though at this point in time I have very little idea if that's really the case. $\endgroup$ – JahvedM Nov 28 '17 at 20:35

By complementing Deane Yang's strategy with Jochen Wengenroth's observations in the related question Composition of a smoothing operator with an $L^2$-bounded operator, non-compact Riemannian manifold, I can now write a fully detailed answer to my question.

Theorem. Let $M$ be a smooth manifold equipped with a smooth measure $\mathrm{d}\mu$ (arising from example from a metric on $M$). Let also $P$ be an elliptic partial differential operator on $M$, and suppose there exists a domain $D(P)$ such that $C_\mathrm{c}^\infty(M) \subseteq D(P) \subseteq L^2(M, \mathrm{d} \mu)$ and $P|_{D(P)} : D(P) \to L^2(M, \mathrm{d} \mu)$ has a bounded left inverse $P^{-1}$. Then the restriction of $P^{-1}$ to $C_\mathrm{c}^\infty(M)$ defines a pseudo-differential operator.

Proof. In two steps:

  1. $P^{-1}$ maps $C_\mathrm{c}^\infty(M)$ to $C^\infty(M)$ by elliptic regularity. Now we use a Lemma, whose proof is given below:

    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.

    Now, the composition $C_\mathrm{c}^\infty(M) \hookrightarrow L^2(M, \mathrm{d} \mu) \xrightarrow{P^{-1}} L^2(M, \mathrm{d} \mu) \hookrightarrow \mathscr{D}'(M)$ is continuous and has image contained in $C^\infty(M)$; hence, by the Lemma, $P^{-1}$ maps $C_\mathrm{c}^\infty(M)$ continuously to the Fréchet space $C^\infty(M)$.

  2. Also since $P$ is elliptic, there is a pair $(Q,R)$ of properly supported pseudo-differential operators such that $R$ is smoothing and $PQ = I + R$. This equality is valid when both sides are applied to $\mathscr{D}'(M)$, and when we restrict it to $C_\mathrm{c}^\infty(M)$ we can compose both sides with $P^{-1}$ to obtain $$(*) \qquad P^{-1} = Q - P^{-1}R \quad \text{on } C_\mathrm{c}^\infty(M),$$ and in fact the equality is clearly also valid on the larger set $Q^{-1}[D(P)]$. Now, since $R$ is properly supported and smoothing, it is continuous from $\mathscr{E}'(M)$ to $C_\mathrm{c}^\infty(M)$, and hence $\mathscr{E}'(M) \xrightarrow{R} C_\mathrm{c}^\infty(M) \xrightarrow{P^{-1}} C^\infty(M)$ is continuous by part 1 of the proof. It follows from known results enunciated in the related MO question that the Schwartz kernel of $P^{-1}R$ is smooth, i.e. that $P^{-1}R$ is a smoothing integral operator. Hence, the right-hand side of Equation (*) is a pseudo-differential operator extending $P^{-1}$. This completes the proof.

Proof of the Lemma. $T$ 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 a version of the closed graph theorem applicable to linear maps from LF-spaces to Fréchet spaces.


Assuming that $P^{-1}$ is a right inverse and $\Omega$ an open subset of $\mathbb{R}^n$ or an open manifold, then you can proceed as follows:

1) An operator $Q: L^2(\Omega) \rightarrow L^2\Omega$ is a pseudodifferential operator, if and only if for any open domain $\Omega'\subset\Omega$, $Q$ restricted to $C^\infty_0(\Omega')$ is a pseudodifferential operator. We can therefore assume that $\Omega'$ is an open subset of $\mathbb{R}^n$.

2) For ay elliptic operator with smooth coefficients $P: C^\infty_0(\Omega')\subset C^\infty_0(\Omega')$, there exists a pseudodifferential operator $Q$ such that $PQ = I + S$, where $S$ is a smoothing operator.

3) If $\Omega'$ is a sufficiently small neighborhood of $x \in \Omega$, then the operator $P$ is injective. It follows that $P^{-1}$ is also a left inverse when restricted to $C^\infty_0(\Omega')$. Therefore, $$ P^{-1} = Q - P^{-1}S$. Since any smoothing operator is also a pseudodifferential operator, it follows that $P^{-1}$ is a pseudodifferential operator on $\Omega'$.

I learned much of this from Introduction to the Theory of Linear Partial Differential Equations by Chazarain and Piriou. Other books include the one by Taylor and one by Treves (you only need the first volume).

  • $\begingroup$ Thanks for that answer. Although I don't fully appreciate the need for localizing (is it because you did not want to assume that $P^{-1}$ was a left inverse of $P$ from the outset? I was happy to assume $P^{-1}P = I_{D(P)}$ and in fact thought this was the accepted definition of resolvent), I think the approach you outline is very similar to the one linked in today's update to my question. However, as explained there I think there might be subtle issues to do with the definition of "smoothing operator". $\endgroup$ – JahvedM Nov 28 '17 at 20:29
  • $\begingroup$ I haven't worked this out carefully, but I think you need both a left and right inverse to make this argument work. I'll look at your question about the smoothing operator. $\endgroup$ – Deane Yang Nov 29 '17 at 3:51

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