I am aware of a few example of continuous, nowhere differentiable functions. The most famous is perhaps the Weierstrass functions

$$W(t)=\sum_k^{\infty} a^k\cos\left(b^k t\right)$$

but there are other examples, like the van der Waerden functions, or the Faber functions. Most of these "look like" some variation of:

Weierstrass Function from Wolfram.

(Weierstrass functions from Wolfram)

Specifically, they are clearly not invertible. Since these functions are generally self-similar at many scales, this non-invertability would seem to hold essentially everywhere.

I'm wondering if it's possible to construct such a function which is invertible. Intuitively, maybe this would be "jittery" in the same way as the Weierstrass function, but if it had a slope which always increased, it would be invertible. Or perhaps there is at least an example in which the function is invertible over some segment of the range.

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    $\begingroup$ Guys, high school level amateur here, why can I not differentiate the individual terms to t and sum them to get the answer as an infinite sum over k of a^k.b^k.sin(b^k.t) ? $\endgroup$
    – HarryH
    Commented Jul 16, 2018 at 19:58
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    $\begingroup$ @HarryH: You can't always differentiate a series (a sum with infinitely many terms) term by term. $\endgroup$ Commented Jul 16, 2018 at 20:12
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    $\begingroup$ @Aganju basically, you would need to choose $c=\infty$, because the function drops down arbitrary steeply locally. $\endgroup$ Commented Jul 17, 2018 at 5:27
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    $\begingroup$ @HarryH ... good question. But in the cases where $W(t)$ is nowhere differentiable ($ab>3$ or something), the series of derivatives does not converge. $\endgroup$
    – GEdgar
    Commented Jul 17, 2018 at 21:45
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    $\begingroup$ You asked a great question - it's a great question because it has an answer which is both interesting and important. $\endgroup$
    – jwg
    Commented Jul 19, 2018 at 9:39

4 Answers 4


Interestingly, there are no such examples! For a continuous function $f : \mathbb{R} \rightarrow \mathbb{R}$ to be invertible, it must be either monotone increasing or decreasing. A famous classical result in analysis, Lebesgue's Monotone Function Theorem, states that any monotone function on an open interval is differentiable almost everywhere. Hence, there are no continuous functions that are invertible and nowhere differentiable.

  • $\begingroup$ To the reader: see this question for the proof that continuous invertible functions are monotone. $\endgroup$
    – Jam
    Commented Dec 20, 2023 at 18:04

Invertible implies bijective by set theory, and bijective together with continuity implies strictly increasing or decreasing, which imply differentiability almost everywhere! (This is known as Lebesgue's monotone function theorem.)


If $f:(a,b)\to\Bbb R$ is continuous and injective it must be monotone, hence differentiable almost everywhere.


I will assume by invertible you mean injective, i.e. bijective on its image. As others have mentioned, there are no such examples among functions from $\mathbb{R}$ to $\mathbb{R}$.

However you might find such examples in other topological spaces.

An example $\mathbb{R} \rightarrow \mathbb{R}^2$ is given by $g(t) = (W(t), t)$ with $W$ the Weierstrass function; it is obviously injective and continuous but nowhere differentiable, considering its projection to the first coordinate.

You can also build an example from $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ by considering $g(t_1, t_2) = (t_1, W(t_1) + t_2)$ or its inverse, given by $g^{-1}(x_1, x_2) = (x_1, x_2 - W(x_1))$.

An injection $\mathbb{R}^2 \rightarrow \mathbb{R}$ cannot be continuous so you won't find a counterexample there.


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