Assuming that $f''(a)$ exists, show that: $f''(a)=\lim_{h\to 0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2}$

Question

Assuming that $f''(a)$ exists, show that: $$f''(a)=\lim_{h\to 0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2}$$

I got the first derivative by using the limit rule $(f(a+h)-f(a))/h$. But when taking the second derivative, I thought of doing the same thing on $f(a)$ of the equation, but I couldn't reach this answer.

@Bernard L'Hopital rule assumes that $f$ is differentiable in some neighbourhood of $a$, and we also need continuity of the second derivative — Jakobian, Jan 23 '19 at 00:22
The O.P. didn't mention hypotheses, but if $f$ is twice differentiable at $a$, it must be differentiable in some neighbourhood of $a$. — Bernard, Jan 23 '19 at 00:25
@MartinArgerami: Because differentiability at $a$ implies the function (here, the first derivative) is defined in a neighbourhood of $a$. If that is not satisfied, how do you define the rate of variation and its limit? — Bernard, Jan 23 '19 at 09:14
@Bernard Yeah, I meant that $f$ must have a second derivative in some neighbourhood of $a$ to use the L'Hopital rule twice, which we don't know. — Jakobian, Jan 23 '19 at 13:49
Actually, I meant applying L'Hospital's rule once, then the definition of the second derivative. — Bernard, Jan 23 '19 at 13:56
@Bernard After such correction, then your proof is correct and is the same with mine. We need to bear in mind that we cannot use L' Hospital rule twice. — Danny Pak-Keung Chan, Jan 23 '19 at 22:40
I believe that I wrote a perfect proof. However, someone gave me a down vote without any reason. I would like to call for justice and ask the moderator to ban this guy. — Danny Pak-Keung Chan, Jan 23 '19 at 22:43
@Bernard I don't mean you are the downvoter. I just mean there exists (an unknown) downvoter who downvoted my answer without any justification. — Danny Pak-Keung Chan, Jan 23 '19 at 22:58
Downvoting is completely OK, provided that this action is justified. For example, a serious mistake appears in the proof and the downvoter explicitly points it out. Every donwvoting must accompany with a reason. — Danny Pak-Keung Chan, Jan 23 '19 at 22:58

score 4 · Answer 1 · answered Jan 23 '19 at 00:45

To be precise, let me re-phrase the question: Let $I$ be an open interval, $a\in I$, and $f:I\rightarrow\mathbb{R}$. Suppose that $f''(a)$ exists, then $$ f''(a)=\lim_{h\rightarrow0}\frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}. $$

Proof: That $f''(a)$ exists implies that there exists $\delta>0$ such that $f'$ exists on $(a-\delta,a+\delta)$ and $f'$ is continuous at $a$. Define functions $\phi:(0,\delta)\rightarrow\mathbb{R}$ and $\psi:(0,\delta)\rightarrow\mathbb{R}$ by $\phi(t)=f(a+t)+f(a-t)-2f(a)$ and $\psi(t)=t^{2}$. Clearly $\phi$ and $\psi$ are differentiable. Note that $$ \lim_{t\rightarrow0+}\frac{\phi(t)}{\psi(t)} $$ is a $0/0$-indeterminate form. By L'Hospital rule, we have $$ \lim_{t\rightarrow0+}\frac{\phi(t)}{\psi(t)}=\lim_{t\rightarrow0+}\frac{\phi'(t)}{\psi'(t)}, $$ provided that the limit on the RHS exists. But \begin{eqnarray*} \lim_{t\rightarrow0+}\frac{\phi'(t)}{\psi'(t)} & = & \lim_{t\rightarrow0+}\frac{f'(a+t)-f'(a-t)}{2t}\\ & = & \lim_{t\rightarrow0+}\left[\frac{1}{2}\frac{f'(a+t)-f'(a)}{t}+\frac{1}{2}\frac{f'(a)-f'(a-t)}{t}\right]\\ & = & \frac{1}{2}f''(a)+\frac{1}{2}f''(a)\\ & = & f''(a). \end{eqnarray*} Therefore, we conclude that $$ f''(a)=\lim_{h\rightarrow0+}\frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}. $$ The case that $h\rightarrow0-$ can be proved similarly.

Assuming that $f''(a)$ exists, show that: $f''(a)=\lim_{h\to 0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2}$

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