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Let $f : (0, +\infty) \to \mathbb{R}$ be a bounded and differentiable function. Prove that $$ \liminf_{t\to+\infty} f'(t)\leq 0 \leq \limsup_{t\to+\infty} f'(t) $$ and deduce from that the existence of a sequence $(t_n)$ such that $t_n \to +\infty$ and $f'(t_n)\to0$.


My solution.

First of all I remember the following characterization of limsup and liminf:

Let $ \,h:\mathbb{R}\longrightarrow \mathbb{R}\, $ be a real funtion, then $$\limsup_{t\to+\infty} h(t)=L \iff \begin{cases} \text{if } x_n\to+\infty \text{ and } h(x_n)\to M,\text{ then } M\leq L,\\ \exists\, x_n\to+\infty \text{ such that } h(x_n)\to L \end{cases}$$ $$\liminf_{t\to+\infty} h(t)=l \iff \begin{cases} \text{if } x_n\to+\infty \text{ and } h(x_n)\to m,\text{ then } l\leq m,\\ \exists\, x_n\to+\infty \text{ such that } h(x_n)\to l. \end{cases}$$

We suppose initially that there exists the limit of $f$ at infinity. In this case we can apply the Lagrange theorem on each interval $[n, n + 1]$ thus obtaining the existence of a sequence of points $\zeta_n\in (n, n + 1)$ such that $$f(n+1)-f(n)=f'(\zeta_n)(n + 1-n)=f'(\zeta_n).$$ We get that $$\lim_{n\to \infty} f'(\zeta_n)=\lim_{n\to \infty} f(n+1)-f(n)=0.$$

Using the characterization of liminf and limsup we get $$ \liminf_{t\to\infty} f'(t)\leq 0 \leq \limsup_{t\to\infty} f'(t) $$ and we have also proved the existence of the $t_n$ (that is $(\zeta_n)$).

Now we remove the hypothesis of existence of $\lim_{t\to +\infty} f(t)$. My idea is to use the same proof but with some modification. We put $x_n=n$ and since $f$ is bounded we get the existence of a convergent subsequence of $f(x_{n_k})$. We apply the Lagrange theorem on each interval $[x_{n_k}, x_{n_{k+1}} ]$ thus obtaining the existence of a sequence of points $\zeta_k\in (x_{n_k}, x_{n_{k+1}})$ such that $$f(x_{n_{k+1}})-f(x_{n_k})=f'(\zeta_k)(x_{n_{k+1}}-x_{n_{k}})$$ and we get that $\lim_{k\to +\infty} f'(\zeta_k)(x_{n_{k+1}}-x_{n_{k}})=0$. Since $|x_{n_{k+1}}-x_{n_{k}}|\geq 1$ we obtain that $\lim_{k\to +\infty} f'(\zeta_k)=0$. Again using the characterization of liminf and limsup we get $$ \liminf_{t\to\infty} f'(t)\leq 0 \leq \limsup_{t\to\infty} f'(t) $$ and we have also proved the existence of the $t_n$ (that is $(\zeta_k)$).


Questions:

  • do you agree with my proof ? It is all correct ?

  • do you know any alternative proof? In particular are you able to prove first the inequality $ \liminf_{t\to\infty} f'(t)\leq 0 \leq \limsup_{t\to\infty} f'(t) $ and then the existence of the sequence $t_n$ ? In my proof I do the opposite.

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  • $\begingroup$ I don't know what Lagrange theorem is. I would use the mean value theorem with the fact that if there is an $\epsilon>0$ such that $\liminf_{t\rightarrow\infty} f’(t) \geq \epsilon$, then there is a $T$ such that $f’(t) \geq \epsilon/2$ for all $t \geq T$. $\endgroup$ – Michael Jun 29 '18 at 6:20
  • $\begingroup$ Lagrange theorem is the mean value theorem. $\endgroup$ – Ef_Ci Jun 29 '18 at 11:13
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Your argument is correct and interesting. Here is probably the answer that was expected by the teacher/author: suppose $\liminf_{t \to \infty } f'(t) >0$. Then there exists $\delta >0$ and $\Delta \in (0,\infty )$ such that $f'(t) >\delta $ whenever $t >\Delta$. This implies $f(\Delta +n+1) -f(\Delta +1) >\delta $ for all posiitve integers $n$, by MVT. By iteration this gives $f(\Delta +n) >(n-1)\delta \to \infty $ which contradicts the fact that $f$ is bounded. We have proved that $\liminf_{t \to \infty } f'(t) \leq 0$. Applying this to $-f$ we get $\limsup_{t \to \infty } f'(t) \geq 0$. For constructing $\{t_n\}$ we can use the fact that derivatives always have IVP. Let $n \geq 1$. Since $\liminf_{t \to \infty } f'(t) \leq 0< \frac 1 n$ we can find $a_n >n$ such that $f'(a_n) <\frac 1 n$. Similarly, we can find $b_n >n$ such that $f'(b_n) >-\frac 1 n$. By IVT there exists $\xi_n >n$ such that $f'(\xi_n)$ is between $f'(a_n)$ and $f'(b_n)$, But then $\liminf f'(t_n) \geq \liminf f'(b_n) \geq 0$ and $\limsup f'(t_n) \leq \limsup f'(a_n) \leq 0$ so $f'(t_n) \to 0$.

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  • $\begingroup$ Thanks for your consideration. Your solution is probably the one that was expected by the author. $\endgroup$ – Ef_Ci Jun 29 '18 at 14:22

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