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is it possible for a function continuous everywhere and differentiable nowhere have a finite arc length between two points? and if so how would you find it?

it was a little weird finding out such functions exists I didn't know if it was possible or not to even define arc length.

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    $\begingroup$ I don't think it's possible because it's a fractal. I might be wrong tho. $\endgroup$
    – user808403
    Aug 3, 2020 at 20:25
  • $\begingroup$ The graph of $f:[a,b]\to\mathbb{R}$ has finite arc-length (i.e., rectifiable) if and only if $f$ is of bounded variation, and so, differentiable almost everywhere. Consequently, every nowhere-differentiable function is non-rectifiable. $\endgroup$ Aug 3, 2020 at 21:55

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Consider $\mathbb{R}^2$ with the norm $|(x, y)|_1 = |x| + |y|$.

Then $f$ has finite arc-length along $[0, 1]$ $\iff$ $f$ is bounded variation on $[0, 1]$. But functions of bounded variation are a.e. differentiable.

Since we have $|(x, y)|_2 = (x^2 + y^2)^\frac{1}{2} \ge \frac{1}{\sqrt{2}}|(x, y)|_1$, the same argument applies.

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  • $\begingroup$ I think you could make this work for the usual definition of the norm as well. For example, using the fact that all norms are equivalent on $\mathbb R^2$. $\endgroup$ Aug 3, 2020 at 21:50
  • $\begingroup$ (+1) Or simply using the inequality $$V_{a}^{b}(f)\leq\operatorname{length}(\{(x,f(x)):a\leq x\leq b\})\leq V_{a}^{b}(f)+(b-a),$$ where $V_{a}^{b}(f)$ denotes the total variation of $f$ on $[a,b]$. $\endgroup$ Aug 3, 2020 at 21:59
  • $\begingroup$ @StephenMontgomery-Smith Indeed. I'll edit it to make the answer more complete, thanks. $\endgroup$ Aug 3, 2020 at 22:02