Weak version of Barbalat's lemma proof

Question

I didn't find this result in the literature and therefore I ask you to review my proof on it:

Given $f:]a,\infty[ \to \mathbb{R} \in C^1(]a,\infty[)$ and assume that $\lim_{t \to +\infty} f(t) = L \in \mathbb{R}$.

Proof that there exists a divergent sequence $t_n \to +\infty$ such that $f'(t_n) \to 0$.

Proof:

Since $\lim_{t \to +\infty} f(t) = L \in \mathbb{R}$ we have that $\forall \epsilon > 0. \exists M > 0. x > M \implies |f(x) - L| < \epsilon$.

Taking $\epsilon_n = \frac{1}{n}$ we get $M_n$ such that, applying mean value theorem: $$|f'(\xi_n)| = |f'(\xi_n)||(M_n+1)-(M_n+2)| = |f(M_n+1) - f(M_n+2)| = $$ $$= |f(M_n+1) - L + L - f(M_n+2)| \le |f(M_n+1) - L| + |L - f(M_n+2)| \le \frac{2}{n}$$

and clearly, $\{f'(\xi_n)\} \to 0$. Where $\xi_n$ is an appropriate point between $M_n+1$ and $M_n+2$.

It remains to guarantee that $\xi_n \to \infty$. To do so I think a strategy would be to redefine $M_n$ as $M_0' = M_0$ and $M_n' = min\{M_{n-1}+1,M_n\}$.

Please let me know whether this proof is right or any improvements that can be made to it.

You are technically applying MVT and $\xi_n \in (M_n+1, M_n+2)$. Something similar was done here — rtybase, Mar 05 '18 at 21:09
@rtybase thanks for the reference ill take a look at it as soon as i can — user1868607, Mar 05 '18 at 21:17

score 1 · Accepted Answer · answered Mar 06 '18 at 14:09

The solution is simpler.

One takes the sequence $n$ which obviously diverges.

For each $n$ there exists $\xi_n$ such that $f'(\xi_n) = f(n+1)-f(n)$ by MVT.

By continuity of $f$, $f'(\xi_n) = f(n+1) - f(n) \to L - L = 0$.

Note that since $\xi_n \in ]n,n+1[$ it is also a divergent sequence.

Weak version of Barbalat's lemma proof

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