# How to prove $\lim_{n\to\infty}\sup A(t)\le\frac{a}{b}$?

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.

• Where in your expression does $n$ appear? Or assuming you mean $\lim_{t\rightarrow \infty}$, what is the $\sup$ being taken over? – Tom Chen Apr 22 '19 at 2:59
• @TomChen Yes. Superior limit. – Fyhswdsxjj Apr 22 '19 at 3:02

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 $$t$$.
• what happens if at $t=0$: $A(0) \le b/a$? – Chip Apr 22 '19 at 8:33
• @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 $\limsup A\leq \frac ab+\epsilon$. – Yuval Apr 23 '19 at 3:24