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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.

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  • $\begingroup$ Since $P(D)\psi=P(D)\varphi$, isn’t $C_1$ the constant in the “injectivity inequality” in Hormander’s theorem for $B_R(0)$ ? $\endgroup$
    – Aphelli
    Commented Dec 23, 2019 at 12:04
  • $\begingroup$ 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. $\endgroup$
    – qualcuno
    Commented Dec 23, 2019 at 12:19
  • $\begingroup$ 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? $\endgroup$
    – Aphelli
    Commented Dec 23, 2019 at 13:47

1 Answer 1

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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.

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  • $\begingroup$ I have finally been able to give this a read, and it is crystal clear! Thanks once again for taking the time to answer. $\endgroup$
    – qualcuno
    Commented Dec 24, 2019 at 6:05

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