After this question and this, I'd like to know an example of a function $f:\mathbb{R}\to\mathbb{R}$ which is not smooth in any open interval in its domain, but $(x,f(x))$ being a rectifiable curve in any open interval in its domain.

  • $\begingroup$ I request you to kindly provide an example for such a function rather than merely saying that it could be like this or that. $\endgroup$ – Rajesh Dachiraju Sep 17 '11 at 4:48
  • $\begingroup$ Your second link goes to an error page. $\endgroup$ – Rahul Sep 17 '11 at 5:12
  • $\begingroup$ I added what might be an example to my answer, but you may not be too thrilled with it (I'm not) since it fails to be everywhere continuous. $\endgroup$ – Mike F Sep 17 '11 at 5:17
  • $\begingroup$ @Rahul : yes I know, and have raised a request about it on meta here : meta.math.stackexchange.com/q/2984/2987 $\endgroup$ – Rajesh Dachiraju Sep 17 '11 at 5:18
  • $\begingroup$ In my second link it has been stated that : "A function with a rectifiable curve has to be continuous everywhere and has to be differentiable almost everywhere". I have re-stated this here for your reference as the second link is not opening. $\endgroup$ – Rajesh Dachiraju Sep 17 '11 at 5:29

If $f$ is increasing (ie $x < y \Rightarrow f(x) < f(y)$) then $f$ is rectifiable on any open interval of its domain. Such a function needs to be differentiable almost everywhere, but I think they can still be nasty enough for differentiality to fail on, say, the rationals.

Added. Here's an idea that might work: enumerate the rationals as $q_1,q_2,\ldots$. Define $f(q_n) = 1/2^n$ and $f(x) = 0$ for $x \notin \mathbb{Q}$. Hopefully the convergence of the geometric series will make $f$ bounded variation. However, $f$ is discontinuous (hence nondifferentiable) at exactly the rationals.

  • $\begingroup$ thank you for the answer, but i need an example. $\endgroup$ – Rajesh Dachiraju Sep 17 '11 at 4:45
  • $\begingroup$ In my second link it is mentioned that a function with rectifiable curve has to be continuous. For some reason the second link is not opening. $\endgroup$ – Rajesh Dachiraju Sep 17 '11 at 5:20
  • $\begingroup$ Indeed, usually "rectifiable curve" includes the condition that the curve be continuous. $\endgroup$ – Lasse Rempe-Gillen Oct 26 '11 at 13:37
  • 1
    $\begingroup$ Also, if we do not require continuity (e.g. if we just ask that the graph over any compact interval has finite one-dimensional Hausdorff measure), then the function doesn't have to be continuous or differentiable anywhere. For example, consider the map on the unit interval that takes the rationals to one and the irrationals to zero. $\endgroup$ – Lasse Rempe-Gillen Oct 26 '11 at 13:55

To expand on the answer by Mike, and make a function that is continuous, consider the following.

1) An increasing function that is discontinuous on a dense set.

At first it may be difficult to imagine that, but actually with the right point of view it is quite easy if you know a little bit of topology. (That's the motivation, it is not necessary for the answer below.) Take the standard "middle thirds" Cantor set C, which is a subset of the closed unit interval I=[0,1]. Let us collapse each of the complementary intervals (i.e. the intervals of $I\setminus C$), including the endpoints, to a point. Then we get a compact topological (even metric) space which, upon some thought, has to be homeomorphic and order-isomorphic to the interval I. So we have a projection map $p:I\to I$, and we can define an "inverse" $q:I\to I$ of this map, which is uniquely defined except on the countably many points that correspond to collapsed intervals. For each of these, we can e.g. choose the image to be the mid-point of the collapsed interval. Then this map is increasing (since p is order-preserving), and of course it is discontinuous at a countable dense set of points.

We can do this explicitly, without any reference to topology, as follows. Define $q:I\to I$ as follows. If $x\in I$ is a dyadic rational, i.e. $x$ is an integer multiple of $1/2^n$ for some $n$, then we set $q(x) := x$. Otherwise, $x$ has a unique dyadic expansion, that is, $$ x = \sum_{i=1}^{\infty} \frac{s_i}{2^i}, $$ where $s_i\in \{0,1\}$. In this case, we define $$q(x) := \sum_{i=1}^{\infty} \frac{2s_i}{3^i}. $$ It is easy to see that this map is increasing, and it is clearly discontinuous at dyadic rationals.

Remark. For the purpose of the proof, it might be easier to choose a convention for the expansion of dyadic rationals, and use the same definition for all $x$. But somehow it seems more elegant and natural to use the midpoint of the interval.

2) An increasing and continuous function whose graph is rectifiable but whose set of points of non-differentiability is dense:

Just integrate the function $q$ given above; i.e., define $$ Q(x) := \int_0^x q(t)dt.$$

This function is continuous by definition, but it is not differentiable at dyadic rationals (the left- and right-sided derivatives exist, but do not agree).

I hope this helps!

  • $\begingroup$ Nice work! A slightly less-involved way to get a function of type 1) might be to just enumerate a dense subset $(d_i)_{i=1}^\infty$ of $[0,1]$ and then define $q(x) = \sum_{d_i \leq x} 2^{-i}$. $\endgroup$ – Mike F Nov 11 '11 at 21:37
  • $\begingroup$ Good point - I still quite like the geometric interpretation using the Cantor set, but your suggestion makes for an easier proof. $\endgroup$ – Lasse Rempe-Gillen Nov 12 '11 at 19:35

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