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$g$ is $C^2$ and I want to show $$\displaystyle \lim_{h\to 0}\dfrac{g(a+h)-2g(a)+g(a-h)}{h^2}=g''(a)$$

I'm confused in one step when trying to solve this by the mean value theorem

Since $$\displaystyle \lim_{h\to 0}\dfrac{g(a+h)-2g(a)+g(a-h)}{h^2}=\lim_{h\to 0} \displaystyle \dfrac{g(a+h)-g(a)-(g(a)-g(a-h))}{h^2}$$

Also, since $g\in C^2$ I can say $g$ is once differentiable everywhere, by mean value theorem, we have $$g(a+h)-g(a)=g'(t)h, \text{ for some } t \in (a,a+h)$$ and $$g(a)-g(a-h)=g'(s)h, \text{ for some } s\in (a-h,a)$$

Thus: $$ \displaystyle \lim_{h\to 0}\dfrac{g(a+h)-2g(a)+g(a-h)}{h^2}=\lim_{h\to 0} \dfrac{g'(t)-g'(s)}{h}$$

Since the $g\in C^2$, the derivative of $g$ should be differentiable everywhere, so $${g'(t)-g'(s)}=g''(\xi)(t-s), \text{ for some } \xi\in (s,t)$$

Thus$$\lim_{h\to 0} \dfrac{g'(t)-g'(s)}{h}=\lim_{h\to 0}\dfrac{g''(\xi)(t-s)}{h}$$

By the continuity of $g''$, I know that $\lim_{h\to 0} g''(\xi)=g''(\lim_{h\to 0} \xi)=g''(a)$

Then I'm wondering how should I deal with $\dfrac{t-s}{h}$?

I let $t=a+\theta_1h$ and $s=a-\theta_2h$ where $\theta_1, \theta_2\in (0,1)$

Thus $$\dfrac{t-s}{h}=\dfrac{(\theta_1+\theta_2)h}{h}=\theta_1+\theta_2$$

Then result becomes $g''(a)(\theta_1+\theta_2)$

I know probably somewhere I get wrong. Any help? Thanks!

Jason
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M_k
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  • Previously, it was suggested to you to use the mean value theorem so that you can control only one quantity that depends on an a parameter between $0$ and $h$: Use the MVT: given two nice functions $F, G$ on $[a,b]$ , there is $a<\xi <b$ such that $$G'(\xi)(F(b)-F(a))=F'(\xi)(G(b)-G(b))$$ For your problem, $F(t)=t^2$ and $G(t)=g(a+t)+g(a-t)$, $0\leq t\leq h$ for example. That will give you a quantity easy to control. – Mittens Oct 27 '22 at 01:35
  • @Oliver Díaz Yeah, you are right. But I delete the previous one since I realize that even if every step is right, it won't lead me to the answer. It just totally deviated from the right path. Then I tried to use another way just like I posted. Then I stuck again. I'm just wondering if there are some flaws in this new attempt or if I missed something which can be a crucial step leading to the answer... – M_k Oct 27 '22 at 01:42
  • Have you seen this link? I hope it is helpful. – Reza Rajaei Oct 27 '22 at 14:14

1 Answers1

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We can apply l'Hospital's rule. Let $u(h) = g(a+h)-2g(a)+g(a-h)$ and $v(h) = h^2$. Then $$\frac{u(h)}{v(h)} = \dfrac{g(a+h)-2g(a)+g(a-h)}{h^2} \tag{1} .$$ Both numerator and denominator go to $0$ as $h \to 0$. Thus, to check whether $\lim_{h\to 0} \frac{u(h)}{v(h)}$ exists, we can consider $$\frac{u'(h)}{v'(h)} = \dfrac{g'(a+h)-g'(a-h)}{2h} \tag{2} .$$ Again both numerator and denominator go to $0$ as $h \to 0$. Now we consider $$\frac{u''(h)}{v''(h)} = \dfrac{g''(a+h)+g''(a-h)}{2} \tag{3} .$$ But obviously $$\lim_{h\to 0} \dfrac{g''(a+h)+g''(a-h)}{2} = g''(a)$$ and the desired result follows.

Paul Frost
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