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I am not sure if this question is proper, if not please let me know I would delete it.

I understand a function can be continuous but not differentiable. I can also understand existence of non-differentiable functions. The question: is there a 'measure' (like distance) between two non-differnetiable functions. The word 'measure' here should not be taken literally. Consider two non-differentiable function $f(x)$ and $g(x) $ the question: is there a way to compare the two functions (like a distance measure ) which tells either $f(x)$ or $g(x)$ is easier to make it differentiable? Something like a how far the function is away from differentiable?

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    $\begingroup$ I think this is a great question, and I hope it attracts lots of responses - it's unfortunate that you accepted an answer so quickly. Although you didn't want measure to be taken literally, actually, that does provide a bit of information: the sets on which functions are not differentiable can be compared by seeing which has larger measure. Functions which are differentiable almost everywhere can either have or lack more 'completely differentiable-like' properties, such as obeying the fundamental theorem of calculus - this turns out to be equivalent to absolute continuity. $\endgroup$ – Sarah Griffith Jul 19 '18 at 2:13
  • $\begingroup$ @CoryGriffith Does accepting answers reduce attention? $\endgroup$ – Creator Jul 19 '18 at 2:15
  • $\begingroup$ I've seen comments to that effect on other answers, but I don't have any way of knowing myself. It seems plausible, though. $\endgroup$ – Sarah Griffith Jul 19 '18 at 2:20
  • $\begingroup$ @CoryGriffith Ok I will wait for few days and see what happens. thanks $\endgroup$ – Creator Jul 19 '18 at 2:22
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There is Rademacher's theorem which says that a function that is Lipschitz continuous is differentiable (almost everywhere). So one way of interpreting your question is to "how close to Lipschitz continuous is a function". Lipschitz continuity is a specific case of Hölder continuity.

A function is $\alpha$-Hölder continuous for $\alpha\in (0,1]$ if

$$|f(x)-f(y)|<K|x-y|^\alpha$$

If $\alpha=1$ then we say the function is Lipschitz.

Note that if $f$ is $\alpha$-Hölder, then it is $\beta$-Hölder for $\beta<\alpha$.

So if $f$ is $.6$-Hölder and $g$ is $.7$-Hölder, then $g$ is "closer" to being Lipschitz, i.e. differentiable.

Here are some pictures of various functions that are of increasing Hölder continuity:

Hölder continuous for all $\alpha \in(0,0.15)$:

enter image description here

Hölder continuous for all $\alpha \in (0,0.55)$:

enter image description here

Hölder continuous for all $\alpha\in (0,0.75)$:

enter image description here

Hölder continuous for all $\alpha\in (0,0.95)$:

enter image description here

Note how the paths get "nicer"/closer to being able to differentiate.

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    $\begingroup$ May I ask how did you get this plots? $\endgroup$ – Creator Jul 19 '18 at 2:01
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    $\begingroup$ @Creator They are on the Wikipedia page on fractional Brownian motion $\endgroup$ – user223391 Jul 19 '18 at 2:02
  • $\begingroup$ @Creator Oh hey, it's you! I remember talking rough paths with you on here. Thanks for asking such good questions. Didn't realize you were asking the question at first, haha. Feel free to shoot me an email or whatever if you want to talk rough paths. $\endgroup$ – user223391 Jul 19 '18 at 2:04
  • $\begingroup$ This is the very reason I asked the question mathoverflow.net/questions/304654/… but it is shown to be related to Hurst parameter, but look at your plots now, so close to low and high SNR type. $\endgroup$ – Creator Jul 19 '18 at 2:04
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    $\begingroup$ . . . nice . . . $\endgroup$ – janmarqz Jul 19 '18 at 2:22

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