Let $U\subset\mathbb R$ be an open set. Let $n\in\mathbb N$ and suppose that $f\in\mathcal C^n(U)$, i.e. that $f$ is $n$-times continuously differentiable on $U$. The $n$-th derivative of $f$, denoted by $f^{(n)}$, then satisfies, for all $x\in U$,
$$f^{(n)}(x)=\lim\_{h\to 0}\frac{\sum\_{k=0}^n f(x+ k h) \binom nk (-1)^{n-k}}{h^n}.$$


I proved this fact [here](https://math.stackexchange.com/a/3895323/631742) and the result is closely related to the [Grünwald–Letnikov derivative](https://en.wikipedia.org/wiki/Gr%C3%BCnwald%E2%80%93Letnikov_derivative).




---


**My question.** Suppose that $f: U \to\mathbb R$ is *any* function such that $$\lim\_{h\to 0}\frac{\sum\_{k=0}^n f(x+ k h) \binom nk (-1)^{n-k}}{h^n}$$ exists for all $x\in U$. Does it follow that $f$ is $n$ times differentiable? If yes, is there an easy proof or a reference that proves this?



Reference for a Grünwald–Letnikov-type definition of the $n$-th derivative of a function
First, let's write out what your expression requires when $n = 2$:


$$ \lim\_{h \to 0} \frac{f(x) + f(x + 2h) - 2 f(x+h)}{h^2} $$


is required to exist for all $x \in U$.


Let $U = (-1,1)$. Take $f(x) = |x|$.


When $x \neq 0$, there exists a sufficiently small interval $I\_x$ around $x$ such that $f|\_{I\_x}$ is $C^2$, and clearly the expression evaluates to $0$.


When $x = 0$, you have that for any $h$:


$$ \frac{f(0) + f(2h) - 2 f(h)}{h^2} = 0 $$


and hence the limit exists and equals 0.


The absolute value function is clearly not twice differentiable at 0.




---


A little bit more analysis:


Let $D\_{h} f(x)$ be the difference quotient of the function $f$
$$ D\_h f(x) = \frac1h (f(x+h) - f(x)) $$
The expression you wrote down is
$$ \underbrace{D\_h D\_h D\_h \cdots D\_h}\_{n \text{ times}} f(x) $$


$n$-times differentiability requires the limit
$$ \lim\_{h\_n \to 0} \lim\_{h\_{n-1}\to 0} \cdots \lim\_{h\_1\to 0} D\_{h\_n} \cdots D\_{h\_1} f(x) $$
to exist.


Your condition only requires the limit to exist along the diagonal, so can be quite far from enough.


