Are there examples of continuous, non-differentiable functions whose "rational derivative" exists?

Question

Define the operator $\Delta_n$ according to the equation

$$\Delta_nf(x)=f\left(x+\frac1n\right)-f(x)$$

Observe that for differentiable $f:\Bbb{R}\to\Bbb{R}$

$$\frac{df}{dx}=\lim_{n\to\infty}n\Delta_nf$$

(Note: The limit can be evaluate from either side by changing the sign of $n$)

This matters solely because it is easier to prove that the sequence $(n\Delta_nf)_{n\in\Bbb{N}}$ converges to some limit $L$ than it is to prove that $\lim_{h\to0}(f(x+h)-f(x))/h=L$ over the reals - so much so, that it's tempting to use this as the definition of the derivative.

So why isn't this the definition of the derivative?

The most significant reason that I can think of is that while the existence of the derivative implies the above equation the converse does not hold. It is possible to have a function such that the above sequence converges when the derivative does not exist. For example, take:

$$g(x)=\begin{cases}e^x & x\in\Bbb{Q}\\0 & \text{otherwise}\end{cases}$$

The sequence $n\Delta_ng(x)$ converges to $g(x)$ for all $x$, but $g$ is not continuous - hence, not differentiable - at any point of its domain.

This problem is easily resolved by adding the qualification "if $f$ is continuous at $x$," since this is a relatively simple condition to check in many cases. So the new definition of the derivative is as follows:

For a function $f:E\subseteq\Bbb{R}\to\Bbb{R}$, continuous at a point $x\in E$, the derivative of $f$ at $x$ exists and is equal to $\lim_{n\to\infty} n\Delta_nf(x)$ iff the sequence $(n\Delta_nf(x))_{n\in\Bbb{N}}$ is convergent.

This sounds correct, but it still leaves the possibility of pathological counterexamples. Continuous nowhere-differentiable functions come to mind, but for every example I can think of, the above sequence does not converge.

Are there any examples of a continuous, non-differentiable function s.t. $\lim_{n\to\infty} n\Delta_nf$ still converges?

Answer 1

Let $f(x) = x\sin(\pi/x)$ if $x \neq 0$ and set $f(0) = 0$.

Then $f$ is continuous but not differentiable at the origin. But

$$ \Delta_n f(0) = \frac{\sin(\pi n)}{n} = 0, $$

so the rational derivative exists and is zero.

Answer 2

Let $f:(0,1) \rightarrow {\mathbb R}$ be a continuous function and let $\mathbf{h} = \{h_1,h_2,h_3,\ldots\}$ a sequence of nonzero real numbers such that $\lim\limits_{n \rightarrow \infty} h_n = 0,$ and put

$$ f'_{\mathbf{h}}(x) \; = \; \lim_{n \rightarrow \infty}\frac{f(x + h_n) - h(x)}{h_n} $$

when this limit exists finitely. The situation you're interested in is the special case where, for each positive integer $n,$ we have $h_n = 1/n.$ This is called the sequential derivative of $f(x)$ (relative to the sequence $\mathbf {h}),$ or a sequential derivative of $f(x)$ when the sequence $\mathbf {h}$ is not specified. The first nontrivial study of this notion was probably in the following paper:

Miklós Laczkovich and György Petruska, Remarks on a problem of A. M. Bruckner, Acta Mathematica Academiae Scientiarum Hungaricae [later title: Acta Mathematica Hungarica] 38 #1-4 (1981), 205-214.

Theorem 7 (p. 207) and its proof show that for any perfect nowhere dense subset $P$ of the open interval $(0,1)$ and for any sequence $\mathbf h$ as above, there exists a continuous function $f:(0,1) \rightarrow {\mathbb R}$ such that $f'_{\mathbf{h}}(x) = 0$ at each $x \in (0,1)$ and the ordinary derivative $f'(x)$ does not exist (finitely or infinitely) at each $x \in P.$

Note that the Lebesgue measure of $P$ can be arbitrarily close to $1,$ although we cannot conclude from this result that an almost everywhere nondifferentiable $f$ exists with $f'_{\mathbf{h}}(x)$ finite everywhere.

Googling the above paper's title will lead you to more recent related results.