A function $f:S\subseteq \mathbb{R}\to \mathbb{R}$ is said to be nowhere differentiable if it is not differentiable at any point in the domain $S$ of $f$ .

It is easy to produce examples of nowhere differentiable functions. Indeed, any functions discontinuous at every point in $S$ is nowhere differentiable, as seen in the example $f(x)=x+1$ if $x\in S\cap \mathbb{Q}$ and $f(x)=x-1$ otherwise. Less trivial examples are those that are everywhere continuous, but nowhere differentiable (hence not an analytic function). These functions are simply termed continuous nowhere differentiable functions.

The prototypical example of a continuous nowhere differentiable function is the Weierstrass function. The formula for the Weierstrass function is

$$ f(x) = \sum_{n=1}^\infty b^n \cos(a^n \pi x) $$

with $a$ odd, $0 < b < 1$ , and $ab > 1 + \frac{3}{2}\pi$ .

In the following series of figures, sequential renderings of the Weierstrass function are shown, each time zoomed in on the origin by a factor of ten. This shows the fractal nature of the function and illustrates how it is continuous, yet doesn't get any ``smoother'' as one looks closer (a prerequisite for differentiability).

Notes

Alternative examples of continuous and nowhere differentiable functions are given by the sample paths of standard Brownian motion, also known as the Wiener process or white noise. It was to handle such ill behaved processes, which have important applications to both pure and applied mathematics, that the techniques of stochastic calculus were developed.

Remark. It can actually be shown that the Weierstrass function is in fact a fractal (that is, its graph in $\mathbb{R}^2$ has topological dimension $1$ but Hausdorff dimension $>1$ ). Indeed, many other examples of continuous nowhere differentiable functions can be found in functions whose graphs are fractals. An example of a continuous nowhere differentiable function whose graph is not a fractal is the van der Waerden function.

Nowhere differentiability can be generalized. For example, a function $f:S\subseteq \mathbb{R}\to \mathbb{R}^n$ is called nowhere differentiable if at each $x\in S$ , $f=(f_1,\ldots,f_n)$ is not differentiable. This means that at least one of the coordinate functions $f_i$ is not differentiable at $x$ , for every $x\in S$ . Given any function $f:S\subseteq \mathbb{R}\to \mathbb{R}^n$ , for any point $x\in S$ , we say that $x$ is of type $(i_1,\ldots,i_n)$ where $i_j=1$ or $0$ according to whether $f_j$ is differentiable or not. If $f$ is nowhere differentiable, then no points of $S$ can be of type $(1,\ldots,1)$ . Question:

is there a continuous nowhere differentiable function $f:S\subseteq \mathbb{R}\to \mathbb{R}^2$ such that for every pair of type-$(0,1)$ points $x_1<x_2$ , there is a type-$(1,0)$ point $y$ such that $x_1<y<x_2$ , and every pair of type-$(1,0)$ points $y_1<y_2$ , there is a type-$(0,1)$ point $x$ such that $y_1<x<y_2$ ?

Bibliography

1

I. Stewart, The Problems of Mathematics, Oxford University Press, 1987.

2

M. Yamaguti, M. Hata, J. Kigami, Mathematics of Fractals, AMS, 1997