I am reading some lectures notes about microlocal analysis and I do not understand how an inequality is actually found.
Let $D_x = -i \partial_x$, $u \in \mathcal{D}’(\mathbb{R})$ and $\chi = 1_{[-1,1]}$ (the characteristic function of the interval $[-1,1] $). Then the following holds :
$$ 2 \, \mathrm{Im} \langle D_xu, e^{-x}\chi u \rangle \ge - \Vert \chi e^{-x/2}D_xu \Vert_{L^2}^2 - \Vert e^{-x/2}u \Vert _{L^2[-1,1]}^2, $$
but I do not see why.
My first idea was using the Cauchy-Schwarz inequality after writing $2i \, \mathrm{Im}(z) = z - \bar{z}$ but it does not seem to work. Do you think that idea could do the job or is there something else ?
Any other ideas ? Thanks !
Cauchy-Schwarz works! Compute that \begin{align*}2\text{Im} \langle D_x u, e^{-x}\chi u\rangle&=-i\left(\langle D_x u, e^{-x}\chi u\rangle-\overline{\langle D_x u, e^{-x}\chi u\rangle}\right)\\ &=-i\left(\langle D_x u, e^{-x}\chi u\rangle-\langle e^{-x}\chi u, D_x u\rangle\right)\\ &=-\left(\langle \partial_x u, e^{-x}\chi u\rangle+\langle e^{-x}\chi u, \partial_x u\rangle\right).\end{align*} This is bounded below further by the absolute value of both entries. Note that $e^{-x}=e^{-x/2}e^{-x/2}$, which can be distributed evenly to both sides of the inner product. From there, one can just use Cauchy-Schwarz and the fact that $ab\leq (a^2+b^2)/2$ for non-negative $a$ and $b$. Interchanging $\partial_x$ and $D_x$ in the norms is free.