Reference for a Grünwald–Letnikov-type definition of the $n$-th derivative of a function

Question

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 and the result is closely related to the Grünwald–Letnikov derivative.

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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?

Answer 1

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.

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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.

Answer 2

The problem is that the existence of the limits only sees how well the function is approximated but it does not see the derivatives. A simple example is something like $f(x)=\exp(-1/x^2)\sin(\exp(1/x^4))$: This is extremely small near zero and the limit for $x=0$ in your condition is $0$. At other points, the limit exists because the function is smooth there. However, the derivatives explode at $0$.

Answer 3

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}$Let $$f(x):=\sum_{n\in\Z}2^{-n}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big) =\sum_{n\in\Z}2^{-n}K\Big(\frac{x-2^{-n}}{2^{-n-2}}\Big)1(x\in2^{-n}[3/4,5/4]),$$ where $K$ is a smooth function such that $K(0)=1$ and $K(x)=0$ if $|x|\ge1$. Note that the intervals $2^{-n}[3/4,5/4]$ are disjoint for distinct integers $n$.

So, $f$ is smooth on $\R\setminus\{0\}$, $f(0)=0$, and $f(2x)=2f(x)$ for all real $x$. So, for $n=2$, $$\lim_{h\to 0}\frac{\sum_{k=0}^n f(x+k h)\binom nk (-1)^{n-k}}{h^n}$$ exists for all $x\in\R$, but $f$ is not even differentiable at $0$, because $f(2^{-k})=2^{-k}$ and $f(\frac34\,2^{-k})=0$ for all natural $k$.