Problem:
Let $f: [0,\infty) \mapsto \mathbb{R}$ be continous on the interval $[0,\infty)$, and differentiable on $(0,\infty)$. Also, $f(0) = 0, \lim_{x\rightarrow \infty} f(x) = 0$. Show there $\exists \xi > 0 : f'(\xi) = 0$
My attempt:
Take an arbitrary $x > 0$. Since $f$ is continous and differentiable on the given interval, we can use the Integral Mean Value Theorem. In other terms, there $\exists \xi \in (0,x)$ such that:
$\frac{f(x)-f(0)}{x-0} = f'(\xi)$
Since $f(0) = 0$, we have:
$\frac{f(x)}{x} = f'(\xi)$
Taking the limit of both sides as $x$ tends to infinity, we get:
$\lim_{x\rightarrow \infty}\frac{f(x)}{x} = \lim_{x\rightarrow \infty} f'(\xi)$
Notice, since $f(x)$ is continous on $[0,\infty)$, it's bounded above by some real number $N$. Hence, the quotient $f(x) / x$ as $x$ tends to infinity, is just $"N/\infty"$, which tends to $0$. And so, in order for the equality above to hold, $\lim_{x\rightarrow \infty} f'(\xi) = 0$ for some $\xi$.
Since $\xi \in (0,x)$, we have that $\xi$ is now in the interval $(0,\infty)$.
In other words, we have proven the statement.
$\square$
I hope you can give me some feedback on my solution. Whatever it might be that may be a wrong move.
Thanks!
Best Answer
If $f\equiv0$ then we have nothing to do.
Suppose, without loss of generality, that there is $x_{0}$ in $(0,+\infty)$ such that $f(x_{0})=\delta>0$. By hypothesis $f$ is a continuous function and $f(0)=0=\lim\limits_{x\rightarrow+\infty}f(x)$, then by Intermediate Value Theorem, there is $0<x_{1}<x_{0}<x_{2},$ such that, $f(x_{1})=f(x_{2})=\delta/2$, thereby, for Mean Value Theorem there is a $\xi$ in $(x_{1},x_{2})$ such that, \begin{eqnarray} f^{\prime}(\xi)=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}=0. \end{eqnarray}