1
$\begingroup$

Easy Garding Inequality states that if $a=a(x,\xi)$ is a symbol in $S=\{a\in C^{\infty}||\partial_{\alpha}a|<C_{\alpha} \hspace{2mm} \forall \alpha\}$ with $a\geq \gamma >0 $ on $\mathbb{R}^{2n}$ then for $h$ sufficiently small we have that: $$ \langle a^{w}(x,hD)u,u \rangle \geq (\gamma-\epsilon) \|u\|_{L^2}^2$$

My question is now precisely about what happens when the symbol $a$ does not belong to $S$ but rather $ a\in S(m)=\{a\in C^{\infty}||\partial_{\alpha}a|<C_{\alpha}m \hspace{2mm} \forall \alpha\}$ for some order function $m$ (For example $m(x) = \langle x \rangle^k$ ) Can we write the easy Garding Inequality with with the $L^2$ norms replaced by weighted Sobolev norms? Thanks

$\endgroup$

1 Answer 1

5
$\begingroup$

I understand that you are dealing with semi-classical symbols $$ b(x,\xi, h)=a(x,h\xi), \quad \vert\partial_x^\alpha\partial_\xi^\beta b\vert\le C_{\alpha\beta} h^{\vert \beta\vert} m(x),\quad 0<h\le 1. $$ According to Hörmander's terminology, we have $$ b\in S(m, g),\quad g_{x,\xi}(t,\tau)={\vert dt\vert^2} +h^2{\vert d\tau\vert^2}. $$ If $b\ge 0$, then the standard Garding inequality says that there exists $c\in S(mh, g)$ such that $$ b^w(x,D)+c^w(x,D)\ge 0. $$ The Fefferman-Phong inequality, a drastic improvement of the latter, says that there exists $c\in S(mh^2, g)$ such that $$ b^w(x,D)+c^w(x,D)\ge 0. $$ Note that tou have to verify some mild assumption on the function $m$ for the above statements to hold true, e.g. $m(x)>0$ and such that $$\exists C,N,\forall x,y,\quad \frac{m(x+y)}{m(x)}\le C(1+\vert y\vert)^{N}, $$ and this is satisfied by your example.

$\endgroup$

Your Answer

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

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