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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.

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I don't have time to write a proof down, but I would suggest trying to use the implicit function theorem to find a contradiction. –  Neal Feb 29 '12 at 11:43
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It can be proved that a differentiable curve's range needs be nowhere dense. Have a look here: math.stackexchange.com/a/70984/8157 This answers your question. –  Giuseppe Negro Feb 29 '12 at 11:51
    
I guess implicit function theorem would work, but that seems still quite advanced for our means. The arc length approach is interesting, maybe i'll be able to reuse this without integration, which we still haven't defined. –  Brad Moroah Feb 29 '12 at 13:06
    
@Giuseppe: a minor quibble, the proof you cited as stated works for $C^1$ curves. There is a small bit of work needed before one can say the same for merely differentiable curves. –  Willie Wong Feb 29 '12 at 13:08
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@WillieWong: Of course you're right. The thread already mentioned contains an answer by Dave Renfro which does just that: math.stackexchange.com/a/70130/8157 . This is just for curiosity since the question is now settled. @ Brad: I hope I didn't muddle you up. –  Giuseppe Negro Feb 29 '12 at 23:25

1 Answer 1

up vote 5 down vote accepted

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.

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Damn yes, what an easy counter-argument ... Thanks :) –  Brad Moroah Feb 29 '12 at 13:31

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