20220701 barbarat lemma proof

(Barbalat lemma )
If differentiable function $f(t)$, When $t\to \infty$ There is a finite limit , And $\dot{f}(t)$ Uniformly continuous , So when $t\to \infty$ when , $\dot{f}(t)\to 0$.

prove （ Reduction to absurdity ）：
Suppose that $t\to \infty$ when ,$\dot{f}(t)\to 0$ Don't set up , Then there is an increasing infinite sequence $\{t_n{\}}_{n\in \mathbb{N}}$ bring （1） When $n\to \infty$ Yes $t_n\to \infty$ ;（2）$|\dot{f}(t_n)|⩾ϵ>0$ For all $\{t_n\}$.

consider $\dot{f}(t)$ Consistent continuity of , according to $ϵ-\delta$ theory , There is a certain $ϵ>0$ , Such that for all $n\in \mathbb{N}$ and all $t\in \mathbb{R}$, When
$$|t_n-t|⩽\delta$$ Then there are $$\left|\dot{f}\left(t_n\right)-\dot{f}(t)\right|\le \frac{\epsilon }{2}$$

therefore , For all $t\in [t_n,t_n+\delta ]$, And all $n\in \mathbb{N}$, Yes $$\begin{array}{r}|\dot{f}(t)|=\left|\dot{f}\left(t_n\right)-\left(\dot{f}\left(t_n\right)-\dot{f}(t)\right)\right|⩾\left|\dot{f}\left(t_n\right)\right|-\left|\dot{f}\left(t_n\right)-\dot{f}(t)\right|⩾\epsilon -\frac{\epsilon }{2}=\frac{\epsilon }{2}\end{array}$$ therefore , For all $n\in \mathbb{N}$, Yes $$\left|{\int }_0^{t_n+\delta }\dot{f}(t)dt-{\int }_0^{t_n}\dot{f}(t)dt\right|=\left|{\int }_{t_n}^{t_n+\delta }\dot{f}(t)dt\right|={\int }_{t_n}^{t_n+\delta }|\dot{f}(t)|dt\ge \frac{\epsilon \delta }{2}>0$$

According to the hypothesis ,${\int }_0^{\infty }\dot{f}(t)dt<\beta$ There is , therefore , When $n\to \infty$,$$\left|{\int }_0^{t_n+\delta }\dot{f}(t)dt-{\int }_0^{t_n}\dot{f}(t)dt\right|\to 0$$, Conflict with the above formula , Therefore, the counter evidence method can prove , When $t\to \infty$ when , $\dot{f}(t)\to 0$.