$F(t) := \int w^2 dx$ where x is a vector and $a,t>0$
Given that $(\frac{d}{dt} - 2a)F(t) \le 0$
How do I deduce that $w=0$?
My idea was to do:
$(\frac{d}{dt} + 2a)F(t) \le 0$
$\implies $$\frac{d}{dt}F(t) \le 2aF(t)$
$\implies $$\int \frac{d}{dt}F(t) \le \int 2aF(t)$
$\implies $$F(t) \le \int 2aF(t)$
Since $a, F(t) > 0 $ $\implies $$\int 2aF(t)\ge0$
$ \implies F(t) \le 0$ just realised this line is not true either
Since $F(t) \ge 0 \implies F(t)=0 \implies w=0$
The only line am i not sure about is:
$\int \frac{d}{dt}F(t) \le \int 2aF(t)$ $\implies $$F(t) \le \int 2aF(t)$
because there should be a constant of integration introduced somewhere
The soln i was given was that $F(t)$ is bdd above by $G(t)$ where $(\frac{d}{dt}-2a)G(t) = 0$ and $G(0)=0$ so $\implies G=0$ but since $F(t) \le 0 \implies F=0 \implies w=0$
I cant seem to understand this since answer since G isnt defined explicitly