PlanetMath: proof of Barbalat's lemma

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proof of Barbalat's lemma

(Proof)

We suppose that $y(t) \not\rightarrow 0$ as $t \rightarrow \infty$ . There exists a sequence $(t_{n})$ in $\mathbb{R}_{+}$ such that $t_{n} \rightarrow \infty$ as $n \rightarrow \infty$ and $\vert y(t_{n})\vert \geq \varepsilon$ for all $n \in \mathbb{N}$ . By the uniform continuity of , there exists a $\delta > 0$ such that, for all $n \in \mathbb{N}$ and all $t \in \mathbb{R}$ ,

$\displaystyle \vert t_{n} - t\vert \leq \delta \; \Rightarrow \; \vert y(t_{n}) - y(t)\vert \leq \frac{\varepsilon}{2}.$

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

$\displaystyle \vert y(t)\vert$	$\displaystyle =$	$\displaystyle \vert y(t_{n}) - (y(t_{n})- y(t))\vert \geq \vert y(t_{n})\vert - \vert y(t_{n})- y(t)\vert \geq$
	$\displaystyle \geq$	$\displaystyle \varepsilon - \frac{\varepsilon}{2} = \frac{\varepsilon}{2}.$

Therefore,

$\displaystyle \left\vert \int_{0}^{t_{n}+\delta} y(t) dt - \int_{0}^{t_{n}} y(t... ...t_{n}}^{t_{n}+\delta} \vert y(t)\vert dt \geq \frac{\varepsilon \delta}{2} > 0$

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

"proof of Barbalat's lemma" is owned by ncrom.

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Cross-references: contradiction, converges, inequality, left hand side, Riemann integral, hypothesis, uniform continuity, sequence

This is version 4 of proof of Barbalat's lemma, born on 2005-04-06, modified 2005-04-14.
Object id is 6933, canonical name is ProofofBarbualatsLemma.
Accessed 3225 times total.

Classification:

AMS MSC:

26A06 (Real functions :: Functions of one variable :: One-variable calculus)

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