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f is twice differentiable on $(0,\infty)$ and $f''$ bounded on $(0,\infty)$, $f(x)\rightarrow 0$ as $x\rightarrow\infty$. show $f'(x)\rightarrow 0$ as $x\rightarrow\infty$.

We know $\exists M >0 $ s.t. $|f''(x)|<M\space\space\forall x\in (0,\infty)$. By MVT,

$$|f'(x+h)-f(x)|=|h||f''(z)|\le \frac{\delta}{2M}M=\delta/2$$ $\forall |x+h-x|\le\frac{\delta}{2M}$

$$\left|f'(x+h)-\lim_{c\rightarrow\infty}\frac{f(x+c)-f(x)}{c}\right|\le \frac{\delta}{2}$$

Since $f(x)\rightarrow 0$ as $x\rightarrow\infty$, then if we fix $\delta >0$, we have

$$|f'(x+h)|\le \frac{\delta}{2}$$

So $f'(x)\rightarrow 0$ as $x\rightarrow\infty$. Is this right?

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In your argument, it seems that you used $\lim_{c\to \infty} \frac{f(x+ c) - f(x)}{c} = f'(x)$, which is not correct (should be $c\to 0)$. –  John Nov 19 '13 at 6:47
    
Thanks, yes that's what I meant. would the proof be correct then? –  lightfish Nov 19 '13 at 12:32

1 Answer 1

up vote 2 down vote accepted

By Taylor Theorem, we have $$ f(x+h)-f(x)-f'(x)h=f''(z)h^2/2. $$ for $z\in(x,x+h)$. The right hand side is bounded independently of $x$, thus, the bound for $f'$ for large $x$ only depends on $h$, which is an arbitrary number and we conclude.

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Your idea is correct, but might be you can elaborate more as you need to choose $h$ a bit more carefully depending on $x$. –  John Nov 19 '13 at 6:54
    
$$ ||f(x+h)-f(x)|-|f'(x)|h|\leq \frac{\max |f''|}{2}h^2, $$ Thus $$ \lim_{x\rightarrow\infty}|f'(x)|h\leq \frac{\max |f''|}{2}h^2, $$ and we conclude. I don't see why $h$ depends on $x$. Maybe I'm doing something wrong. –  guacho Nov 19 '13 at 18:16
1  
You are absolute correct, but I will choose $M$ such that $|f(x|\leq \epsilon^2$ for all $x>M$ and get something like $$|f'(x)| \epsilon \leq C\epsilon^2\ .$$ somehow in your prove its seems that you assume limit $f'(x)$ exists already. –  John Nov 19 '13 at 20:37
    
Oooooooook, you are right. Change $\lim$ by $\limsup$ and everything works, right? –  guacho Nov 19 '13 at 21:06

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