# proof of Barbalat’s lemma

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 $|y(t_{n})|\geq\varepsilon$ 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 $t\in\mathbb{R}$,

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

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

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

Therefore,

 $\left|\int_{0}^{t_{n}+\delta}y(t)dt-\int_{0}^{t_{n}}y(t)dt\right|=\left|\int_{% t_{n}}^{t_{n}+\delta}y(t)dt\right|=\int_{t_{n}}^{t_{n}+\delta}|y(t)|dt\geq% \frac{\varepsilon\delta}{2}>0$

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.

Title proof of Barbalat’s lemma ProofOfBarbalatsLemma 2013-03-22 15:10:45 2013-03-22 15:10:45 ncrom (8997) ncrom (8997) 7 ncrom (8997) Proof msc 26A06