Functions like the Weierstrass function or van der Waerden's function exhibit self-similar plots. Is this characteristic of continuous everywhere, differentiable nowhere functions? Is there a counterexample to this?

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    $\begingroup$ Could you define more precisely what it means to be self-similar? Anyway, whatever you have in mind, it probably isn't characteristic to these types of functions. Here's 18 of them, check: epubl.ltu.se/1402-1617/2003/320/LTU-EX-03320-SE.pdf $\endgroup$ Commented Oct 22, 2011 at 23:20
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    $\begingroup$ The examples usually given look "self-similar" because those are the easiest to construct. But you can take any nowhere differentiable function and perturb different parts of it in arbitrarily horrible ways and it will remain nowhere differentiable, but won't be "self-similar" by any reasonable definition of the term. $\endgroup$
    – Ted
    Commented Oct 23, 2011 at 3:11
  • $\begingroup$ Since being nowhere differentiable is a "generic property" of continuous functions, I doubt that all would be self-similar. In fact, I believe (though this is in no way a proof) that most functions with these properties would not be self-similar. $\endgroup$
    – M Turgeon
    Commented Feb 24, 2012 at 14:48

1 Answer 1


The Takagi function, defined by $$T(x)=\sum_{n=0}^\infty {{\rm dist}(2^nx,\mathbb Z)\over 2^n}$$ is an example of a continuous, nowhere differentiable function whose graph is not a fractal, in the sense that it has Hausdorff and box-counting dimension 1 (see here and here).

I'm sure various things in this direction are true. For example (Corollary 11.2 in Falconer's Fractal Geometry), the graph of a continuous function $f$ on $[0,1]$ that satisfies a Hölder condition with exponent $0\le\alpha\le 1$ has Hausdorff and box-counting dimensions less than or equal to $2-\alpha$.
If instead there exists $c>0$, $\delta_0>0$, and $1\le s<2$ such that for all $0<\delta\le\delta_0$ there exists $y$ with $|x-y|\le\delta$ and $|f(x)-f(y)|\ge c\delta^{2-s}$, then the box-counting dimension is at least $s$. (Note that this doesn't say anything about Hausdorff dimension, which is never larger than box-counting dimension).


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