2
$\begingroup$

The class $\Sigma^1$ of symbols on $\mathbb R^{2n}$ is made with $C^\infty$ functions $a$ of $X=(x,\xi)\in \mathbb R^n\times\mathbb R^n$ such that $$ \vert\partial_X^\alpha a\vert\le C_\alpha(1+\vert X\vert)^{2-\vert \alpha\vert}. $$ Assuming that $a\in\Sigma^1$ is real-valued principal type and denoting by $A$ its Weyl quantization, using the fact that $A$ is continuous on the Schwartz space $\mathscr S(\mathbb R^n)$ and on its dual (tempered distributions) $\mathscr S'(\mathbb R^n)$, we may define the maximal extension $H$ of $A$ with the domain $$ D(H)=\{u\in L^2(\mathbb R^n), Au \in L^2(\mathbb R^n) \}. $$
Then I claim that $H$ is self-adjoint. I believe that it is well-known and I am looking for a reference in the literature.

A related question is the same problem for first-order pseudo-differential operators on a compact manifold without boundary $\mathcal M$ (equipped with a smooth density): let $A$ be a first-order pseudo-differential operator on $\mathcal M$ (I do not want to assume ellipticity, but I know that $A$ is continuous on $C^\infty(\mathcal M)$ and on the distributions on $\mathcal M$) and assume that $A$ is symmetric, that is such that for $\phi, \psi\in C^\infty(\mathcal M)$ $\langle A\phi,\psi \rangle=\langle \phi,A\psi \rangle_{L^2(\mathcal M)}$. Then consider the maximal extension $H$ of $A$ with $$ D(H)=\{u\in L^2(\mathcal M), Au \in L^2(\mathcal M) \}. $$ Then $H$ is self-adjoint. Is it true and well-known?

Last but not least, dropping the compactness assumption on $\mathcal M$ in the second question above, assuming that $A$ is properly supported, can I get the same result?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

The harmonic oscillator with non-positive coupling constant, i.e., $H_0 f = -f''(x) + cx^2f(x)$, with $c \in \mathbb{C} \setminus \mathbb{R}$, is non-self-adjoint (see the book in preparation by Sjöstrand (http://sjostrand.perso.math.cnrs.fr/161014bok.pdf)). This does not answer the question since $c = -1$ is not allowed (and gives a symmetric operator).

Improve this answer
$\endgroup$
8
  • $\begingroup$ Thanks, I modified the question with a real principal type symbol $a$. $\endgroup$
    – Bazin
    Feb 8, 2019 at 15:40
  • $\begingroup$ What do you mean by principal symbol? Shubin or classical sense? $\endgroup$
    – mcd
    Feb 8, 2019 at 15:44
  • $\begingroup$ I mean that $a=a_1+a_0$ with $a_j\in \Sigma^j$ is such that $\vert \nabla a_1\vert^2$ is elliptic in $\Sigma_1$. $\endgroup$
    – Bazin
    Feb 8, 2019 at 15:50
  • $\begingroup$ In my example with $a_1(x,\xi) = \xi^2 - x^2$ and $a_0 = 0$, we have that $|\nabla p_2(x,\xi)|^2 = 4(\xi^2 + x^2)$, which is elliptic. $\endgroup$
    – mcd
    Feb 8, 2019 at 16:36
  • $\begingroup$ OK, I want ellipticity and the characteristic set to be a manifold, which is not the case of your example $\xi=\pm x$. $\endgroup$
    – Bazin
    Feb 8, 2019 at 19:57
1
$\begingroup$

The affirmative answer to your second question is indeed well-known. For example, in

Commun. Math. Phys. 308, 325–364 (2011), Lemma A.1 on page 358,

Frédéric Faure and Johannes Sjöstrand prove that every PDO of order 1 on a closed manifold has a unique closed extension in $L^2$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.