Answer to Question #285176 in Real Analysis for shani

Question

Suppose that 𝑓 is differentiable and 𝑓′′(𝑎) exists. Prove that 𝑓′′(𝑎)=limℎ→𝑎𝑓(𝑎+ℎ)−2𝑓(𝑎)+𝑓(𝑎−ℎ)ℎ2. Give an example where the above limit exists, but 𝑓′′(𝑎) does not exist.

Answer 1

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

$=\displaystyle \lim_{h\to 0}\frac{\displaystyle \lim_{h_1\to 0}\frac{f'(x+h)-f'(x+h-h_1)}{h_1}-\displaystyle \lim_{h_2\to 0}\frac{f'(x)-f'(x-h_2)}{h_2}}{h}$

letting $h=h_1=h_2$, i.e. taking them all to zero at the same rate we have:

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

then:

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

example:

for $f'(x)=|x|$ :

by L’Hôpital’s rule we have:

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

$=\displaystyle \lim_{h\to 0}\frac{|0+h|-|0-h|}{2h}=0$

That means the limit exists at x = 0, but $f'(x)=|x|$ is not differentiable at 0, so $f''(0)$

does not exist.