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Barbalat’s Lemma


Prove that if we have f: [0,\infty) \to [0,\infty) a function which is uniformly continuous on [0,\infty) with \int_0^\infty f(t)\text{d}t <\infty then \lim\limits_{t\to \infty} f(t)=0.
Proof: Suppose there exists (x_n)_n\to \infty such that f(x_n) \to \ell >0. Moreover, we can suppose (x_n) increasing and that the difference x_{n+1}-x_n is large enough for each n. Take \varepsilon=\frac{\ell}{4} >0. Then there exists n_0 such that \forall n \geq n_0 we have |f(x_n)-\ell | < \frac{\ell}{4}. From uniform continuity, there exists \delta > 0 such that | x_n-y|<\delta \Rightarrow |f(x_n)-f(y)|<\frac{\ell}{4}.
It easily follows that |f(y)-\ell|<\frac{\ell}{2} if |x_n-y|<\delta.
Therefore \displaystyle \int_{x_n-\delta}^{x_n+\delta} f(t)\text{d}t > \delta\ell,\ \forall n \geq n_0. This contradicts \int_0^\infty f(t)\text{d}t <\infty.

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Categories: Analysis Tags: ,
  1. Gabriel
    January 2, 2012 at 9:32 pm

    Hi Beni Bogosel,

    I needed a quick proof of Barbalat’s Lemma to put in my Master thesis, so I reproduced the one you exposed here in this post– of course, with the appropriated references to your blog. Hope you don’t mind 😉

    (Please mail me if this represents any problem for you)

    • January 3, 2012 at 7:04 pm

      I’m glad you found the proof useful. It is indeed a proof I discovered myself, but I cannot take credit for it, because the ideas I used are quite standard, and I think that most people who would try and prove Barbalat’s Lemma would go through the same steps I did. Thank you for reading my blog. 🙂

  2. September 1, 2017 at 12:51 am

    I think you should specify that the sequence is increasing or tends to infinity.

    • September 1, 2017 at 10:50 am

      Yes, the sequence goes to \infty and can be chosen to be increasing.

  1. April 27, 2012 at 3:10 pm

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