I have the following problem, I dont understand why is this equality.
Let $f(x)$ be a function such that $f=\exp(g)$ with $g\geq0$ on a domain of interest. We are looking at the operator \begin{equation} P=e^{\tau f}(-\Delta)e^{-\tau f}=-e^{\tau f}\nabla e^{-\tau f}\cdot e^{\tau f}\nabla e^{-\tau f}=-(\nabla -\tau\nabla f)\cdot(\nabla -\tau\nabla f) \end{equation}
and want to show that for an appropriate function $f$ defined on a bounded domain $X$ and for $u\in C_{c}^{\infty}(X)$, we have \begin{equation} \|P u\|\geq C(\tau^{1/2}\|\nabla u\|+\tau^{3/2}\|u\|) \end{equation} for a positive constant $C$ independent of $\tau$ sufficiently large. To do that, we first observe that $P=A+B$, where $$A:=[-\Delta-\tau^{2}|\nabla f|^2]\qquad B:=[\tau(\nabla\cdot(u\nabla f)+\nabla f\cdot\nabla u)]$$
My attemp was use $\Delta(fg)=(\Delta f)g+2\nabla f\cdot \nabla g+f(\Delta g)$, so we have $$(-\Delta e^{-\tau f})e^{\tau f}=2\nabla e^{-\tau f}\cdot\nabla e^{\tau f}+e^{-\tau f}(\Delta e^{\tau f})$$ and the second equality I supposed that $$ -(\nabla -\tau\nabla f)\cdot(\nabla -\tau\nabla f)=-(\Delta-2\tau\nabla f+\Delta f) $$ But I don't see how write $P=A+B$
Any help will be appreciated, thanks a lot!!
You are only lacking on differentiability rules (in several variables of course). I will first compute it conjugating by $e^f$ inside of the operator: \begin{align*} \Delta(e^{f} u)&=\text{div}(\nabla (e^f u))\\ &=\text{div}( u\nabla e^f+e^f\nabla u)\\ &=\text{div}(e^f u\nabla f+e^f\nabla u)\\ &=\nabla (e^f u)\nabla f+e^fu\Delta f+\nabla e^f\nabla u+e^f\Delta u\\ &=u\nabla e^f\nabla f+e^f\nabla u\nabla f+e^fu\Delta f+e^f\nabla f\nabla u+e^f\Delta u\\ &=e^f (u|\nabla f|^2+2\nabla u\nabla f+u\Delta f+\Delta u). \end{align*} Observe that $\text{div}(u\Delta f)+\nabla u\nabla f=2\nabla u\nabla f+u\Delta f$ and that by changing $f\mapsto -\tau f$ (differential operators are linear) as you state, we get \begin{align*} Pu=e^{\tau f}(-\Delta)e^{-\tau f}u=-\Delta u-\tau^2 u|\nabla f|^2+2\tau\nabla u\nabla f+\tau u\Delta f \end{align*} which can be decomposed as $P=A+B$ in your notation.
Remark: I guess you are trying to get a Carleman estimate for the Laplacian. If you want to know why that decomposition, lets relabel $P_f=e^{\tau f}(-\Delta)e^{-\tau f}$ and so you can check with a little bit of algebra that $A:=(P_{f}+P_{-f})/2$ and $B=(P_{f}-P_{-f})/2$. So $A$ and $B$ corresponds to the symmetric and antisymmetric part of $P_f$, respectively. So when you expand the norm, to treat the inner product term you end up with the problem of showing some kind of coercivity on the term $(u, [A, B]u)_{L^2(\Omega)}$ to get the Carleman for your operator.