Assume $u\in L^2(\mathbb{R}^n)$ and let $(x_0, \xi _0) \in T^\ast \mathbb{R}^n = \mathbb{R}^n_x \times \mathbb{R}^n_\xi $. Assume I can find $a\in C^\infty (T^\ast \mathbb{R}^n)$ which is also bounded with all derivatives and $a(x,\xi ) = 1$ in a neighborhood of $(x_0, \xi _0)$. Without loss of generality we may even assume $a\in C _0 ^\infty (T^\ast \mathbb{R}^n)$ and still equal to 1 near $(x_0, \xi _0)$. Assume moreover that we know $$\ \operatorname{Op} [a]u \_{L^2(\mathbb{R}^n)} \le Ch^s$$ where $$\operatorname{Op} [a]u(x) := \frac{1}{(2\pi h)^{n/2}}\iint \limits_{T^\ast \mathbb{R}^n}e^{\tfrac{i}{h}(xy)\cdot \xi }a(\tfrac{x+y}{2},\xi )u(y)\, dy\, d\xi .$$ Is this enough to conclude that there is some neighborhood $U$ of $x_0$ such that $$\ u \_{L^2(U)} \le Ch^s \quad ? $$ If not, what kind of condition would be sufficient?
1 Answer
Of course no. Your inequality on $\Vert Op[a] u\Vert$ is only providing some information microlocally at $(x_0,\xi_0)$ and says nothing about what is happening elsewhere. For instance, with the homogeneous wavefront set setting, you may have in one dimension a (complexvalued) distribution with wavefrontset equal to {$0$}$\times\mathbb R_^*$. It is the case for instance of $\frac{1}{xi0}$.

$\begingroup$ Ok, thank you, it's what I was afraid of. Actually I also have that $u$ satisfies $Pu=g$ where $\g\ \le Ch^s$ and $P$ is elliptic in the complement of the set where $u$ is microlocally $O(h^s)$. $\endgroup$– flavioApr 26, 2012 at 8:17

1$\begingroup$ Well, in that case that's OK since your first estimate does control $u$ near the point $(x_0,\xi_0)$ and by the ellipticity of your $P$, you control as well what may happen elsewhere. You can write down the details for that stuff with a partition of unity. Best, Bazin. $\endgroup$– BazinApr 26, 2012 at 11:47