# On a technical step of J.P Rosay's paper "A Very Elementary Proof of the Malgrange-Ehrenpreis Theorem"

I'm currently studying Jean-Pierre Rosay's paper A Very Elementary Proof of the Malgrange-Ehrenpreis Theorem, in order to prepare a final presentation for an 'Introduction to PDEs' course.

In the first section, he proves the following

Theorem (Hörmander's inequality): let $$\Omega \subset \mathbb{R}^n$$ be open and bounded and $$P(D)$$ a partial differential linear operator with constant coefficients. There exists $$C > 0$$ such that $$\|\varphi\|_2 \leq C\|P(D)\varphi\|_2$$ for all $$\varphi \in \mathscr{C}_0^\infty(\Omega)$$.

From here, it is deduced that

Corollary: let $$\Omega \subset \mathbb{R}^n$$ be open and bounded and $$P(D)$$ a partial differential linear operator with constant coefficients. If $$g \in L^2(\Omega)$$, the equation $$P(D)u = g$$ has a solution in $$L^2(\Omega)$$.

A proof sketch follows: via Hörmander's inequality applied to the formal adjoint $$P(D)^\ast$$ of $$P(D)$$, we see that $$P(D)^\ast$$ is injective with continuous inverse as an operator

$$(P(D)^\ast)^{-1} : E \to \mathscr{C}_c^\infty(\Omega)$$

with $$E = \operatorname{im} P(D)^\ast(\mathscr{C}_c^\infty(\Omega))$$ and both spaces equipped with their inherited $$L^2$$ norm. Composing with $$\langle g,-\rangle_2$$ we obtain a functional $$\eta(P(D)^\ast \varphi) = \langle g,\varphi\rangle$$. By (uniform) continuity $$\eta$$ can be extended to $$\overline{E}$$, which is now a Hilbert space and so by Riesz's representation theorem, there exists $$u \in \overline{E} \subset L^2(\Omega)$$ that represents the extension of $$\eta$$. Evaluating at smooth compactly supported functions, we finally obtain

$$\langle g, \varphi\rangle = \langle u,P(D)^\ast\varphi\rangle = (P(D)u)(\varphi)$$

for all $$\mathscr{C}_c^\infty(\Omega). \square$$

Now, in section 2 of the paper, the author makes the following assertion:

Given $$0 < r < R$$ and $$g \in L^2(B_r(0))$$ such that for all $$v \in L^2(B_R(0))$$ with $$P(D)v = 0$$ we have $$\langle g,v \rangle_{B_r(0)} = 0$$, there exists a constant $$C$$ for which $$|\langle \varphi,g\rangle|_{B_r(0)} \leq C\|P(D)\varphi\|_{B_R(0)}$$ for all $$\varphi \in \mathscr{C}_c^\infty(\mathbb{R}^n)$$.

Paraphrasing, the justification for this is as follows: by the aforementioned results, given $$\varphi \in \mathscr{C}_c^\infty(\mathbb{R}^n)$$, there exists $$\psi \in L^2(B_R(0))$$ such that $$P(D)\psi = P(D)\varphi$$ and moreover $$\|\psi\|_2 \leq C_1\|P(D)\varphi\|_2$$. Then by the orthogonality hypothesis $$g$$ is orthogonal to $$\varphi - \psi$$ in $$B_r(0)$$ and so $$\langle g,\varphi\rangle = \langle g,\psi \rangle$$. By the Cauchy Schwarz inequality, we obtain $$|\langle \varphi,g\rangle|_{B_r(0)} \leq \|g\|C_1\|P(D)\varphi\|_{B_R(0)}$$.

There is a detail in this proof that I am failing to grasp: I understand one can take $$\psi$$ to be a solution to $$P(D)u = P(D)\varphi$$, but

How is the existence of $$C_1$$ (not depending on $$\psi$$ nor $$\varphi$$) guaranteed for the results above?

Any help is greatly appreciated. Thanks in advance.

• Since $P(D)\psi=P(D)\varphi$, isn’t $C_1$ the constant in the “injectivity inequality” in Hormander’s theorem for $B_R(0)$ ? Commented Dec 23, 2019 at 12:04
• The problem is that $\psi$ need not be smooth with compact support. Moreover, if we would work with compactly supported smooth functions on $B_R(0)$ instead of $\mathbb{R}^n$, I think Hörmander's inequality should work directly. But this is not the case as far as I know. The author only says that $C_1$ and $\psi$ exist "by $\S 1$" and section 1 only consists of those two results. Commented Dec 23, 2019 at 12:19
• Okay... can’t you then mimic the proof of the “corollary”, to show that the continuous extension of $P(D)^{-1}$ is continuous wrt the $L^2$ norms? Commented Dec 23, 2019 at 13:47

Such a $$C_1$$ exists provided that the existence theorem of the corollary is “quantitative”, ie we can choose $$u$$ with $$\|u\|_2 \leq C_1\|g\|_2$$ for some $$C_1$$.
To do this, let’s keep the notations of the corollary. Then $$\|u\|_{L^2}=\|u\|_{\overline{E}}$$ (by definition of $$E$$ and its completion). Now, $$\|u\|_{\overline{E}}=\|\eta\|_{\overline{E}^*}$$ (we are in a Hilbert), so $$\|u\|_{L^2}=\|\eta\|_{E^*}$$ by density of $$E$$. But $$\eta=\langle g ,\, (P(D)*)^{-1}\cdot \rangle$$, so $$\|\eta\|_{E^*} \leq \|(P(D)^*)^{-1}\|\|g\|_{L^2}$$ and we are done.