TITLE: A continuous, nowhere differentiable but invertible function?
QUESTION [88 upvotes]: 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 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.

REPLY [5 votes]: 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.