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

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.

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?

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.

• The above example also works for any $n > 2$. When $n = 1$ your condition is equivalent to differentiability. Apr 30, 2021 at 15:40
• Very simple and nice! Apr 30, 2021 at 15:55
• Thank you! Yes, for $n=1$ it is just the usual difference quotient 🙂. Apr 30, 2021 at 18:29
• @MaximilianJanisch ah! you study with Camillo. Are you also working on the nonuniqueness problem, or something else with the Euler equations? Apr 30, 2021 at 18:40
• @WillieWong Yes that is correct! I sent you an e-mail to wongwwy@math.msu.edu 🙂. Apr 30, 2021 at 20:49

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 $$0$$.

$$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}$$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 $$k$$.

• @WillieWong : Oops! This should now be corrected. Thank you for your comment. Apr 30, 2021 at 15:51
• check out my simplified version of your argument. :-) Apr 30, 2021 at 15:54
• The simplified version referenced by @WillieWong. May 1, 2021 at 1:34
• @LSpice : I commented (and voted) on that answer. May 2, 2021 at 1:23
• I was just providing the link because that's my habit. May 2, 2021 at 3:06