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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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1. 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