0
$\begingroup$

If $F(t)$ is twice differentiable at $x$ and $G(h)=\max_{t\in(0,h)}\left[\frac{F'(x+t)-F'(x-t)}{2t}-F''(x)\right],$ where $x$ is fixed; then how can we show that $\displaystyle\lim_{h\to 0}G(h)=0$.

  • 0
    There are abstract duplicates of this question [here](http://math.stackexchange.com/q/146023/264) and [here](http://math.stackexchange.com/q/123206/264).2012-06-09

4 Answers 4

1

Hint: $\frac{F'(x+t)-F'(x-t)}{t}=\frac{F'(x+t)-F'(x)}{t}+\frac{F'(x-t)-F'(x)}{-t}.$

  • 0
    @Kns: I was being too indirect. Have changed the hint, added primes. Note that each part is related to the derivative of $F'$.2012-06-09
1

Since $h\to0$ means $t\to0$, so $\displaystyle\lim_{h\to 0}G(h) = \lim_{t\to 0}\left[\frac{F'(x+t)-F'(x-t)}{2t}-F''(x)\right]$ apply l'Hôpital's rule,we can get $ \lim_{t\to 0}\left[\frac{F'(x+t)-F'(x-t)}{2t}-F''(x)\right] = \lim_{t \to 0}\left[\frac{F''(x+t)+F''(x-t)}{2}-F''(x)\right] = 0$

ps: your tags include real analysis, I assume $x\in \mathbf{R^{n}}$, although real analysis isn't only about real numbers. I dou't know if l'Hôpital's rule can apply under other situation.

  • 1
    seems it is not appropriate here. Thanks for the reminding2012-06-09
1

Hint: By definition

$ F''(x) = \lim_{h\to 0} \frac{F'(x+h)-F'(x)}{h}$

Couple this with Andre's comment

1

Since $\,t\in(,h)\,$ , we have that $\,h\to 0\Longrightarrow t\to 0\,$ , so: $\lim_{t\to 0}\frac{F'(x+t)-F'(x-t)}{2t}=\lim_{t\to 0}\frac{1}{2}\left[\frac{F'(x+t)-F'(x)}{t}+\frac{F'(x-t)-F'(x)}{-t}\right]$ and you get what you want since we know $\,F''(x)\,$ exists, so the limit defining this second derivative exists.