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I would like to know if the following statement is true.

Let $f:(a,b)\subset\mathbb{R} \rightarrow \mathbb{R}$, such $f'$ exits for all $x \in (a,b)$ and also exists $f''(x_0)$ for some $a<x_0<b$. Then

$ \lim_{h \to 0} \frac{f(x_0+h)-2f(x_0)+f(x_0-h)}{h^2} = f''(x_0)$.

Obs: I'm not assuming that $f'$ is continuos in some neighborhood of $x_0$. However, I think that requirement of the existence of $f'$ in a neighborhood of $x_0$ is a necessity for the existence of $f''(x_0)$.

CiaPan
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2 Answers2

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By Taylor's formula, $$f(x_0 + h) = f(x_0) + hf'(x_0) + \frac{1}{2}h^2f''(x_0) + o(h^2)$$ and $$f(x_0 - h) = f(x_0) - hf'(x_0) + \frac{1}{2}h^2f''(x_0) + o(h^2)$$ as $h \to 0$. This shows that the expression whose limit you are to compute is equal to $f''(x_0) + o(1)$.

Walter
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  • Could you give a reference for the Taylor's theorem with its hypothesis. In every reference I looked, they always assume that $f'$ is continuos in some neighborhood of $x_0$. –  Nov 24 '17 at 19:22
  • @Diego: That particular version of Taylor's theorem is not very popular. You may have a look at the link I gave in the comment to your question. – Paramanand Singh Nov 25 '17 at 02:52
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Let $\varphi(t)=f(x_{0}+th)-2f(x_{0})-f(x_{0}-(1-t)h)$, $0\leq t\leq 1$, then $\varphi(1)-\varphi(0)=\varphi'(c_{h})$ for some $0<c_{h}<1$, so $\varphi(1)-\varphi(0)=[f'(x_{0}+c_{h}h)-f'(x_{0}-(1-c_{h})h)]h$. For $\epsilon>0$, find some $\delta>0$ such that if $0<|u|<\delta$, \begin{align*} \left|\frac{f'(x_{0}+u)-f'(x_{0})}{u}-f''(x_{0})\right|<\epsilon/2. \end{align*} Now for $0<|h|<\delta$, \begin{align*} \left|\frac{1}{h^{2}}[\varphi(1)-\varphi(0)]-f''(x_{0})\right|&\leq\left|\frac{f'(x_{0}+c_{h}h)-f'(x_{0})}{h}-c_{h}f''(x_{0})\right|\\ &~~~~+\left|\frac{f'(x_{0}-(1-c_{h})h)-f'(x_{0})}{-h}-(1-c_{h})f''(x_{0})\right|\\ &\leq|c_{h}|\left|\frac{f'(x_{0}+c_{h}h)-f'(x_{0})}{c_{h}{h}}-f''(x_{0})\right|\\ &~~~~+|1-c_{h}|\left|\frac{f'(x_{0}-(1-c_{h})h)-f'(x_{0})}{-(1-c_{h})h}-f''(x_{0})\right|\\ &<\epsilon. \end{align*}

user284331
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  • Note that $\varphi(1)-\varphi(0)=f(x_{0}+h)-2f(x_{0})+f(x_{0}-h)$. – user284331 Nov 24 '17 at 21:19
  • This is a nice and novel proof for me. It does look a bit complicated because of $c_h$ (lot of $h$ in the equations) and you could have just used $c$ instead. But showing the dependence of $c$ on $h$ explicitly does not leave any room for confusion. +1 – Paramanand Singh Nov 25 '17 at 02:43
  • Minor nitpick. Using $\epsilon$ instead of $\epsilon/2$ in the beginning also leads to $c_h\epsilon+(1-c_h)\epsilon=\epsilon$ at the end. – Paramanand Singh Nov 25 '17 at 02:45
  • Okay, that is clearer, thanks. – user284331 Nov 25 '17 at 02:46