I understand that the Weierstrass function is an example of a function that is continuous everywhere but differentiable nowhere. I was thinking if it is possible to "make the Weierstrass function differentiable" somehow. One approach on this was already discussed here.

My approach is the following: The Weierstrass function (let us call it $w$) can be represented as the set $$W := \{ (x,y) \in \mathbb{R}^2 \mid w(x)=y \}$$

I am wondering if there is a differentiable parametrization $p:\mathbb{R} \to \mathbb{R}^2$ of $W$, in the sense that

  • $W = p(\mathbb{R}) := \{ p(t) \mid t \in \mathbb{R} \}$
  • $p$ is injective
  • $p$ is differentiable (almost everywhere would be fine)

Is there such a parametrization?

My guess is no, because based on a differentiable parametrization, we could probably compute the derivative of $w$ itself. However, I have not found out how to do this yet...


1 Answer 1


Of course we'd need to have $p'(t)=0$ almost everywhere.

In fact the answer is yes, at least for almost-everywhere differentiability. The "Cantor function" $x:[0,1]\to[0,1]$ is a continuous non-decreasing surjective function with the following property: There exists an open set $E\subset[0,1]$ with $m(E)=1$ such that $x$ is constant on each connected component of $E$. So if we let $p(t)=(x(t), w(x(t)))$ then $p$ is constant on each connected component of $E$, hence $p'=0$ almost everywhere.

  • $\begingroup$ Thank you for your answer! Unfortunately, I forgot to specify that my parametrization should also be injective (otherwise, it would be a strange parametrization). I think in that case, the Cantor function cannot be applied directly? $\endgroup$
    – Peter
    Commented Oct 27, 2017 at 16:25

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