proof of Barbalat’s lemma


We suppose that y(t)↛0 as t. There exists a sequence (tn) in + such that tn as n and |y(tn)|ε for all n. By the uniform continuity of y, there exists a δ>0 such that, for all n and all t,

|tn-t|δ|y(tn)-y(t)|ε2.

So, for all t[tn,tn+δ], and for all n we have

|y(t)| = |y(tn)-(y(tn)-y(t))||y(tn)|-|y(tn)-y(t)|
ε-ε2=ε2.

Therefore,

|0tn+δy(t)𝑑t-0tny(t)𝑑t|=|tntn+δy(t)𝑑t|=tntn+δ|y(t)|𝑑tεδ2>0

for each n. By the hypothesisMathworldPlanetmathPlanetmath, the improprer Riemann integral 0y(t)𝑑t exists, and thus the left hand side of the inequalityMathworldPlanetmath convergesPlanetmathPlanetmath to 0 as n, yielding a contradictionMathworldPlanetmathPlanetmath.

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