Can the limit $\lim_{h \to 0}\frac{f(x + h) - 2f(x) + f(x - h)}{h^2}$ exist if $f'(x)$ does not exist at $x$?

The second derivative of $f$ can be written as

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

while it can also be written as (in fact, I believe this is the definition of $f''(x)$):

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

From the second expression, it seems clear that if $f'(x)$ does not exist, then $f''(x)$ cannot exist as well.

However, the first expression does not involve $f'(x)$, so this got me wondering: If $f'(x)$ does not exist at $x$ , can the limit $\lim_{h \to 0}\frac{f(x + h) - 2f(x) + f(x - h)}{h^2}$ at $x$ exist?

How would one prove/disprove this? If this were to be true, what would be some examples?

Solution 1:

The first limit can exist even though the second does not, but this is not the definition of $f^{\prime\prime}$. By definition, $f^{\prime\prime}$ can only be defined if $f^{\prime}$ is.

However, for the limits, consider $$ f\colon x\in\mathbb{R}\mapsto \begin{cases} 0 & \text{ if } x < 0 \\ 1 & \text{ if } x = 0 \\ 2 & \text{ if } x > 0 \end{cases} $$

Solution 2:

An easy counterexample is the case of an odd function that's not differentiable at $0$: let $f$ be any odd function (defined at $0$). Then: $$\lim_{h\to0}\frac{f(0-h)-2f(0)+f(0+h)}{h^2}=\lim_{h\to0}\frac{f(-h)+f(h)}{h^2}=0.$$ So there are lots of counterexample, e.g., $f(x)=x^{1/3}$.