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Suppose $A(t)>0(t\ge 0)$, $a, b>0$, let $$ A'(t)\le aA-bA^2. $$ Prove $\lim_{n\to\infty}\sup A(t)\le\frac{a}{b}$.

Using Taylor formula $$ A(0)=A(t)-tA'(t)+o(t)\ge (1-ta)A(t) +tbA^2(t)+o(t). $$ then$$ \frac{A(0)+o(t)-A(t)}{tA(t)}\ge -a+bA(t). $$ therefore, I only need to prove $$ \lim_{t\to\infty}\sup \frac{A(0)+o(t)-A(t)}{tA(t)}=0. $$ but I have no idea about the above formula.

Could you please give me any hints? Thanks in advance!

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  • $\begingroup$ Where in your expression does $n$ appear? Or assuming you mean $\lim_{t\rightarrow \infty}$, what is the $\sup$ being taken over? $\endgroup$ – Tom Chen Apr 22 '19 at 2:59
  • $\begingroup$ @TomChen Yes. Superior limit. $\endgroup$ – Fyhswdsxjj Apr 22 '19 at 3:02
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Rewrite the inequality as $$ A'\leq bA\Big(\frac ab -A\Big). $$ In particular $A$ is decreasing whenever $A> \frac ab$.

Fix any $\epsilon>0$, then $A'\leq b\cdot \frac ab\cdot (-\epsilon)=-a\epsilon$ whenever $A\geq \frac ab+\epsilon$. So either $A\leq \frac ab+\epsilon$ for all $t$, or there is $t_0$ with $A(t_0)>\frac ab+\epsilon$, then $A$ is decreasing as long as $A>\frac ab$. Replace $\epsilon$ by $\frac{\epsilon}{2}$, we see $A'<-a\epsilon/2$, a definite negative upper bound, whenever $A>\frac ab+\frac{\epsilon}{2}$. Thus after $t_0$ $A$ will first decrease below $\frac ab+\frac{\epsilon}{2}$. It can never bounce back to $\frac ab+\epsilon$ - it has to decrease when $A\in [\frac ab+\frac{\epsilon}{2}, \frac ab+\epsilon]$. Thus $\limsup_{t\to\infty}A\leq \frac ab+\epsilon$. Since $\epsilon$ is arbitrary, we see $\limsup_{t\to\infty}A\leq \frac ab$.

I don't think one can use Taylor expansion in this situation since it involves large $t$.

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  • $\begingroup$ Many thanks!!!! $\endgroup$ – Fyhswdsxjj Apr 22 '19 at 3:23
  • $\begingroup$ what happens if at $t=0$: $A(0) \le b/a$? $\endgroup$ – Chip Apr 22 '19 at 8:33
  • $\begingroup$ @Chip: If $A\leq \frac ab+\epsilon$ for all $t$, then $\limsup A\leq \frac ab+\epsilon$. Otherwise find $t_0$ with $A(t_0)>\frac ab+\epsilon$ then follow the above argument to see $\limsup A\leq \frac ab+\epsilon$. $\endgroup$ – Yuval Apr 23 '19 at 3:24

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