$f$ twice differentiable on $(0,\infty)$, $f''$ is bounded on $(0,\infty)$, $\lim_{x\to\infty}f(x)=0$. Show that $\lim_{x\to\infty}f'(x)=0$

Question

$f$ twice differentiable on $(0,\infty)$, $f''$ is bounded on $(0,\infty)$, $\lim_{x\to\infty}f(x)=0$. Show that $\lim_{x\to\infty}f'(x)=0$. First of all, it seems to me that the condition that $f''$ bounded is useless... Isn't it?

My attempt is to pass by mean value theorem, so i would like to know if my approach holds, please.

First, consider the interval $[x,x+a]$ with $x>0, a>0$. As $f$ is differentiable on $[x,x+a]$, by mean value theoreme there exists some $c_x \in ]x,x+a[$ such that

$f'(c_x)=\frac{f(x)-f(x+a)}{a}$.

Inserting the limit on RHS and LHS we get the following:

$\lim_{x\to\infty}f'(c_x)=\lim_{x\to\infty}\frac{f(x)-f(x+a)}{a} = 0$.

Therefore, $\lim_{x\to\infty}f'(x)=0$ as wanted.

I know there is a post already on this statement, but they use the fact that $f''$ is bounded in theirs proofs... If this hypothesis is necessary, could someone explain why or give a counter example on my proof?

Answer 1

Counterexample. Let $f(x) = \frac 1x \sin(x^2)$. Then $f(x) \to 0$ as $x \to \infty$, but $$ f'(x) = - \frac{1}{x^2} \sin(x^2) + \frac 1x \cos(x^2) \cdot 2x = - \frac{1}{x^2} \sin(x^2) + 2 \cos(x^2), $$ which is not convergent to $0$ as $x \to \infty$. The problem here lies of course in the second derivative, which is not bounded.

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Problem with your reasoning. For each $x > 0$, you find just one point $c_x \in [x,x+a]$ for which $f'(c_x)$ is small. And indeed, if you fix any sequence $x_n \to \infty$, your construction yields a corresponding sequence $c_n \in [x_n,x_n+a]$ that satisfies $c_n \to \infty$ and $f'(c_n) \to 0$. But there is no reason why every sequence $y_n \to \infty$ should be obtained as $c_n$ in this way.

In a way, this may be surprising. Your family of $c_x$ is indexed by all points $x > 0$, and the intuition may tell us that those $c_x$ should cover at least the whole halfline $[a,\infty)$. But this intuition is wrong! It would be true if $x \mapsto c_x$ was a continuous function, while it is not. The best way to see this is by studying some counterexample.

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How to solve this. As you noted yourself, this is the easy part. This question was already asked and answered, e.g. here.

Answer 2

hint

Apply second order Taylor-Lagrange formula in $ [0,x] $ for an $ x >0 $.

there exists $ c\in (0,x) $ such that

$$f(0)=f(x)-xf'(x)+\frac{x^2}{2}f''(c)=0$$