TITLE: Does $F(t)F'(t) \le 0$ imply that $F$ does not change signs?
QUESTION [7 upvotes]: The question is in the title, really. More precisely, suppose $F:[0,1] \to \mathbb{R}$ is continuously differentiable and satisfies 
(i) $F(t)F'(t) \le 0$ for all $t \in [0,1]$
(ii) $F(0) = 1$
Does this imply that $F(t) \geq 0$ for all $t \in [0,1]$?
My intuition is that (i) implies that $F$ and $F'$ have different signs. Since $F(0) = 1$, this means that $F$ must cross the $t$-axis at some point, but the instant it becomes negative it would have to immediately shoot up again. I think this is not possible, but I have not been able to come up with a proof or counter-example. Thanks for your help!

REPLY [12 votes]: Condition $(i)$ implies that $t \mapsto F(t)^2$ is decreasing. Since $F(0)=1$, we have two cases: $F(t)>0$ for all $t \in [0,1]$ and we are done, or there exists $t_0 \in [0,1]$ with $t_0=\inf \{ t \in [0,1] : f(t)=0\}$. But then $F^2(t_0)=0$ and since $F^2$ is decreasing, it must follow that $0\leq F(t)^2\leq F(t_0)=0,\ \forall t \in [t_0,1]$, concluding that $F(t)=0,\ \forall t \geq t_0$. Since before $t_0$ the function $F$ is positive, we are done.