# is it possible for a function continuous everywhere and differentiable nowhere have a finite arc length between two points? [closed]

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.

• I don't think it's possible because it's a fractal. I might be wrong tho.
– user808403
Aug 3, 2020 at 20:25
• 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. Aug 3, 2020 at 21:55

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.
• 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 $\mathbb R^2$. Aug 3, 2020 at 21:50
• (+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 $[a,b]$. Aug 3, 2020 at 21:59