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I The Weierstrass function's' odd qualities

  1. Jul 24, 2017 #1
    I recently stumbled on the Weierstrass function, whose main claim to fame (as I understand it) is to be continuous everywhere, but non-differentiable everywhere as well. Apparently I was in good company with Gauss' and others who assumed that to be impossible!

    I guess I'm asking, is this dichotomy mostly down to the loosened definition of continuity, i.e. Hölder continuous instead of Lipschitz continuous? Not that I dispute the validity or use of the former definition, but it certainly would mean that my intuition was correct in the stricter Lipschitz sense.
  2. jcsd
  3. Jul 24, 2017 #2
    No, it is continuous in the true sense of the term. But what is d(abs(x))/dx when x=0? The problem with the Weierstrass function is that it has that problem for all x.
  4. Jul 24, 2017 #3


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    According to this discussion, it is Hölder continuous for every ##0<\alpha<1##, but not Lipschitz continuous (which would be ##\alpha=1##).
    These are stricter criteria than the regular continuity, however. The function is continuous.
  5. Jul 24, 2017 #4
    Oooh, that is an excellent explanation, thanks a lot. It's been a long time since college math, I had forgotten that a "kink" in a graph is non-differentiable. From there it's not too hard to construct an "only kinks" function.
  6. Jul 25, 2017 #5


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    Yes, indeed. Even more: By Rademacher's theorem, every Lipschitz continuous function is differentiable almost everywhere, which in a sense is the opposite of being nowhere differentiable.
    It is more difficult than it seems, since the "kinks" have to be everywhere.

    In addition to the references given in the link in the post by @mfb , there is also a pair of "Insights" written about it, but they seem to be missing a third part and I have not read them myself. Maybe @jbunniii would like to comment.
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