proof of Barbalat’s lemma

We suppose that $y(t)\to ̸0$ as $t\to \mathrm{\infty }$. There exists a sequence $(t_n)$ in ${ℝ}_{+}$ such that $t_n\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$ and $|y(t_n)|\ge \epsilon$ for all $n\in \mathbb{N}$. By the uniform continuity of $y$, there exists a $\delta >0$ such that, for all $n\in \mathbb{N}$ and all $t\in ℝ$,

$$|t_n-t|\le \delta \Rightarrow |y(t_n)-y(t)|\le \frac{\epsilon }{2}.$$

So, for all $t\in [t_n,t_n+\delta ]$, and for all $n\in \mathbb{N}$ we have

$|y(t)|$ $=$ $|y(t_n)-(y(t_n)-y(t))|\ge |y(t_n)|-|y(t_n)-y(t)|\ge$

$\ge$ $\epsilon -{\displaystyle \frac{\epsilon }{2}}={\displaystyle \frac{\epsilon }{2}}.$

Therefore,

$$\left|{\int }_0^{t_n+\delta }y(t)𝑑t-{\int }_0^{t_n}y(t)𝑑t\right|=\left|{\int }_{t_n}^{t_n+\delta }y(t)𝑑t\right|={\int }_{t_n}^{t_n+\delta }|y(t)|𝑑t\ge \frac{\epsilon \delta }{2}>0$$

for each $n\in \mathbb{N}$. By the hypothesis, the improprer Riemann integral ${\int }_0^{\mathrm{\infty }}y(t)𝑑t$ exists, and thus the left hand side of the inequality converges to 0 as $n\to \mathrm{\infty }$, yielding a contradiction.

Title	proof of Barbalat’s lemma
Canonical name	ProofOfBarbalatsLemma
Date of creation	2013-03-22 15:10:45
Last modified on	2013-03-22 15:10:45
Owner	ncrom (8997)
Last modified by	ncrom (8997)
Numerical id	7
Author	ncrom (8997)
Entry type	Proof
Classification	msc 26A06