I am currently reading some lecture notes about microlocal analysis and the author gives a simple example of the ‘positive’ commutator argument, that I do not really understand. Let me explain it.
Let $u \in \mathcal{D}’(\mathbb{R})$ : we want to estimate $\lvert u(1) \lvert^2$ in terms of $\lvert u(-1) \lvert^2$ and $D_xu := \frac{1}{i} \partial_x u$. Let $\chi = 1_{[-1,1]}$ be the characteristic function of the interval $[-1,1] $. We have
$$ -\lvert u(1) \lvert^2 + \lvert u(-1) \lvert^2 = \int i [D_x,\chi] u \cdot \bar{u} \, dx = i^{-1} \left( \int D_x u \cdot \chi \bar{u} \, dx - \int \chi u \cdot \bar{D_x u} \, dx \right),$$ which is equal to $2 \, \mathrm{Im} \langle D_xu, \chi u \rangle$. The the author says that if $D_x u = 0$, the first equality reveals that what really provides control of $\lvert u(1) \lvert^2$ is the fact that $\chi$ has negative derivative at 1. What does it mean ?
It seems to me the derivative of $\chi$ at 1 is $-\delta$, the Dirac mass at 1. But I do not understand how it is used in the first equality, and also how $D_x \chi$ really is (in the commutator $[D_x,\chi] $ ) in general and how to find the right hand side of the second equality.
Any help will be appreciated, thanks.
Edit : and ‘positive’ means the ‘good’ sign (the one that allows us to control), which is the negative sign here..
I wouldn't read too much into this kind of informal comments. What is written there is simply the definition of $[D_x, \chi]$ and integration by parts. By definition, $$ [D_x, \chi]u=D_x(\chi u)-\chi\, D_x u.$$ Therefore, \begin{equation*} \begin{split} \int_{-\infty}^\infty [D_x, \chi]u\cdot\overline{u}\, dx &= \int_{-\infty}^\infty D_x(\chi u) \overline u\, dx - \int_{-\infty}^\infty \chi D_xu\cdot \overline u\, dx \\ &= \int_{-\infty}^\infty\chi u \cdot D_x\overline u\, dx - \int_{-\infty}^\infty D_x u \cdot \chi \overline u \, dx . \end{split} \end{equation*}
The rest does not make much sense to me, since $D_x u=0$ means that $u$ is the constant distribution.