Most often examples given for bounded continuous functions which are not differentiable anywhere are fractals.If we include probabilistic fractals exact self-similarity is not required. Are their examples of functions which are bounded ,continuous, not differentiable anywhere and can not be modeled as fractals?

  • $\begingroup$ Do you mean non differentiable at each point? As otherwise, you can just take $|x|$ $\endgroup$
    – voldemort
    Jan 10, 2015 at 5:50
  • $\begingroup$ @voldemort- Yes non differentiable at each point, i will edit the question $\endgroup$
    – Curious
    Jan 10, 2015 at 5:53

2 Answers 2


First, you have to define what you mean by a "fractal". There is only one mathematica definition of a fractal curve that I know, it is due to Mandelbrot (I think). A curve is called fractal if its Hausdorff dimension is $>1$.

Now, back to your question. The condition of being bounded is not particularly relevant, as you can restrict any continuous function $f: R\to R$ without 1-sided derivatives to the interval $[0,1]$ and then extend the restriction to a periodic function $g$, $g(x+n)=g(x)$ for all $x\in [0,1]$, $n\in {\mathbb N}$.

Now, take the Takagi function: it has no 1-sided derivatives at any point, is continuous and its graph has Hausdorff dimension 1 (see here).

Edit: Note that Takagi's function does have periodic extension since $f(0)=f(1)$. For a general nowhere differentiable function $f$ you note that it cannot be monotonic (if it is nowhere differentiable). Then find $a<b$ such that $f(a)=f(b)$ and then extend to $R$ periodically using $[a,b]$ as the period.

  • $\begingroup$ Actually, extending the restriction periodically might break continuity if $f(0)\neq f(1)$. This problem doesn't occur for the Takagi function though. $\endgroup$
    – kremerd
    Jan 10, 2015 at 7:09
  • $\begingroup$ @kremerd: Right, I was careless. $\endgroup$ Jan 10, 2015 at 7:14

It might be worth pointing out that the typical function, in the sense of Baire category, has this property. More specifically, Let $X=C([0,1])$ denote the set of all continuous, real valued functions defined on the unit interval endowed with the sup norm. Let $S\subset X$ denote the set of all functions that nowhere differentiable and let $T\subset X$ denote the set of all functions whose graph has Hausdorff dimension $1$. Then $S$ and $T$ are both residual sets - i.e., each is the complement of a countable collection of nowhere dense sets. As a result, their intersection is second category as well and, in particular, non-empty.

The fact that $S$ is second category is a classic theorem of Stefan Banach - in fact, it's the seminal result of this type. A proof may be found in chapter 11 of the important book Measure and Category by John Oxtoby. The fact that $T$ is second category is proven by Humke and Petruska in Volume 14 of the The Real Analysis Exchange, though the title is "The packing dimension of a typical continuous function is 2" and it's proven in some other references as well.

  • $\begingroup$ Hi Mark! Two comments: (1) Second category usually means "not first category", which isn't the same as complement of first category, in the same way that having positive measure isn't the same as having full measure. See the start of my answer here for more about the nomenclature. (2) That $T$ has a first category complement was proved earlier in Mauldin/Williams, On the Hausdorff dimension of some graphs, Trans. AMS (1986) -- Th. 2 on p. 794. $\endgroup$ Jan 12, 2015 at 22:05
  • $\begingroup$ Hey Dave - thanks! Yes, I originally typed "residual" but then changed it. Thanks for the reminder. $\endgroup$ Jan 13, 2015 at 0:34

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