In a homework assigment, we were given a certain recursive definition of a space-filling curve $f : [0,1] \mapsto [0,1]^2$ and asked to determine where it is differentiable. My intuition tells me that (as a fractal construction is involved) the curve should be nowhere differentiable.
I think I finally came up with an elementary proof for this particular curve, however it's quite messy with quite some case distinctions necessary.
So looking for a more elegant proof, my question is:
Does the mere fact that $f$ is a space-filling curve, i.e. surjective and continuous, already allow us to deduce where $f$ is differentiable?
And if so, how elementary can such a proof be? Yet, we don't know Brouwer fixed-point theorem etc.
Best Answer
As stated, the answer is no. Take any space filling curve $f:[0,1]\to [0,1]^2$. Now take any smooth curve $g:[0,1] \to [0,1]^2$ such that $g(1) = f(0)$. Define $h(x) = f(2(x - 1/2))$ if $x \geq 1/2$ and $h(x) = g(2x)$ if $x < 1/2$. Then $h(x)$ is smooth on $[0,1/2)$. By construction $h$ is continuous and surjective, so is also space-filling.
Indeed, by the above construction, for any $x\in [0,1]$ you can find a space-filling curve $h$ and a neighborhood $N$ of $x$ such that $h$ is infinitely differentiable on $N$. (In fact, for any $x\in [0,1]$ and $y\in [0,1]^2$ you can choose $h$ such that $h(x) = y$.) So there's no intrinsic argument to whether a space-filling curve must be non-differentiable at a point.
You will necessarily have to actually use the recursive definition of $f$ to conclude that it is not differentiable at certain points.