How is the Cauchy-Schwarz inequality used to bound this derivative?

Question

In "Hardy's Uncertainty Principle, Convexity and Schrödinger Evolutions" (link) on page 5, the authors state that they are using the Cauchy-Schwarz inequality to bound the derivative of the $L^2(\mathbb{R}^n)$ norm of a solution to a certain differential equations, but I am not sure how exactly they applied it.

Some context: Let $v$, $\phi$, $V$, $F$ be nice enough functions of $x$ and $t$ so that the following integrals are well-defined, $A>0, B\in\mathbb{R}$ be constants, and $u=e^{-\phi} v$ solve $$\partial_t u = (A+iB)\left( \Delta u + Vu+F\right).$$ Denote the $L^2$ inner product on $\mathbb{R}^n$ between some $f$ and $g$ as $(f, g) = \int f g^{\dagger} dx$, where $g^\dagger$ is the complex conjugate of $g$, and define $f^+=\mathrm{max}\{f, 0\}$.

We know the equality $$\partial_t \vert \!\vert v \vert\!\vert^2_{L^2} = 2\,\mathrm{Re}\left(Sv,v\right) + 2\,\mathrm{Re}\left((A+iB)e^\phi F, v\right),$$ where $$\mathrm{Re}\left(Sv,v\right) = \int -A|\nabla v|^2 + \left(A|\nabla \phi|^2+\partial_t \phi \right) |v|^2 + 2B \,\mathrm{Im}\, v^{\dagger} \nabla\phi\cdot\nabla v + \left( A\,\mathrm{Re}\,V - B\,\mathrm{Im}\, V\right)|v|^2dx,$$ holds true. The authors go on to conclude that the Cauchy-Schwarz inequality implies that $$\partial_t \vert \!\vert v(t) \vert\!\vert^2_{L^2} \le 2\vert \!\vert A \, \left(\mathrm{Re}\,V(t)\right)^+ - B\,\mathrm{Im}\, V(t) \vert \!\vert_{\infty}\vert \!\vert v(t) \vert \!\vert^2_{L^2} + 2 \sqrt{A^2+B^2} \vert \!\vert F e^\phi \vert \!\vert_{L^2} \vert \!\vert v(t) \vert \!\vert_{L^2}$$ when $$\left(A+\frac{B^2}{A}\right)|\nabla\phi|^2 + \partial_t \phi\le 0, \,\,\,\,\mathrm{in}\, \mathbb{R}_+^{n+1}.$$ However, I am not sure how the authors used the C.S. inequality to arrive at this conclusion, and am especially confused as to where the factor of $B^2/A$ came from, and why we only need the constraint to hold over $\mathbb{R}_+^{n+1}$ when we are integrating over all of $\mathbb{R}^{n}$, though I understand why we only care about positive time.

Does anyone have any insight here?

Answer 1

You have a typo on the $\mathrm{Re}(Sv,v)$ term, the leading $A$ should be inside the integral. The formula from the paper reads

$$\mathrm{Re}\left(Sv,v\right) = \int -A |\nabla v|^2 + \left(A|\nabla \phi|^2+\partial_t \phi \right) |v|^2 + \color{red}{ 2B \,\mathrm{Im}\, v^{\dagger} \nabla\phi\cdot\nabla v } + \left( A\,\mathrm{Re}\,V - B\,\mathrm{Im}\, V\right)|v|^2dx $$

I'll sketch the control of the term in red. The rest hopefully you know how to deal with.

Completing the square you have

$$ - A|\nabla v|^2 + 2 B\, \mathrm{Im} v^\dagger \nabla \phi \cdot \nabla v = - \left|\sqrt{A} \nabla v - \frac{B}{\sqrt{A}}~ \mathrm{Im}v^\dagger \nabla \phi \right|^2 + \frac{B^2}{A} (\mathrm{Im} v^\dagger)^2|\nabla \phi|^2 $$

Throwing away the negative terms we have

$$\mathrm{Re}\left(Sv,v\right) \leq \int \left(A|\nabla \phi|^2+\partial_t \phi \right) |v|^2 + \frac{B^2}{A} (\mathrm{Im} v^\dagger)^2 |\nabla\phi|^2 + \left( A\,\mathrm{Re}\,V - B\,\mathrm{Im}\, V\right)|v|^2dx $$

Using that $(\mathrm{Im} v^\dagger)^2 \leq |v|^2$, we further have

$$\mathrm{Re}\left(Sv,v\right) \leq \int \left(A|\nabla \phi|^2 + \frac{B^2}{A} |\nabla\phi|^2 +\partial_t \phi \right) |v|^2 + \left( A\,\mathrm{Re}\,V - B\,\mathrm{Im}\, V\right)|v|^2dx $$

from this you can conclude what is claimed.