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We know that if $f:[a,b] \rightarrow \mathbb{R}^{n}$ is a $C^1$ path then $$\ell(f)=\int_{a}^{b}\|f'(t)\|dt.$$

Moreover, in the proof of this result we use explicitly the continuity of $f'.$

I'm trying to find an example of a differentiable function (at every point of $[a,b]$) such that $f':[a,b] \rightarrow \mathbb{R}^n$ is not continuous and $$\ell(f) > \int_{a}^{b} \|f'(t)\|dt.$$

I found this answer, but if I'm not mistaken, the Cantor function is not differentiable at every point.

Can anyone help me with this example?

Math
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  • $l(f)$ is only defined when $f$ is path $\mathcal{C}^1$ and continuous (or at least absolutely continuous), don't you mean $d(f(a),f(b))$ instead of $l(f)$ ? – Cactus Jul 18 '23 at 21:05
  • @Cactus Let $f:[a,b] \rightarrow \mathbb{R}^{n}$ a path and $P={a_0=a<a_1<\cdots <a_n=b}$ be a partition of $[a,b]$. We define $\ell(f,P)=\sum_{i=1}^{n} |f(t_i)-f(t_{i-1})|$. We say that $f$ is a rectifiable curve if ${\ell(f,P)}$ is bounded for all partitions $P$ of $[a,b]$. In this case, we define $\ell(f)=\sup_{P}\ell(f,P)$. – Math Jul 18 '23 at 23:14

1 Answers1

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Such a curve does not exist. When $f$ is differentiable, we always have the bound $$ \ell(f) \leq \int_a^b \|f'(t)\|\,dt. $$ To see this, consider a partition $a = t_0 < \dots < t_n = b$ of $[a, b]$. Using the fundamental theorem of calculus, we can write $$ f(t_{i + 1}) - f(t_i) = \int_{t_i}^{t_{i + 1}} f'(t)\,dt, $$ where each side of the equation is vector-valued. By the triangle inequality for vector-valued integrals, we get $$ \|f(t_{i + 1}) - f(t_i)\| = \left\|\int_{t_i}^{t_{i + 1}} f'(t)\,dt\right\| \leq \int_{t_i}^{t_{i + 1}}\|f'(t)\|\,dt. $$ It follows that $$ \sum_i \|f(t_{i + 1}) - f(t_i)\| \leq \int_a^b\|f'(t)\|\,dt. $$ Taking a supremum of the LHS, we obtain the inequality above.

Edit: As pointed out in the comments, we have to be careful when applying the fundamental theorem of calculus in the above proof when $f'$ is not Riemann integrable. It turns out that the (second) fundamental theorem of calculus holds iff $f$ is absolutely continuous (see Theorem 7.18 of Rudin's Real and Complex Analysis). Moreover, if $f$ is everywhere differentiable and $f'$ is $L^1$, then $f$ is absolutely continuous (see Theorem 7.21 of Rudin). Therefore, the only case where the fundamental theorem of calculus fails is when $f'$ is not $L^1$, in which case the RHS of the inequality above is $\infty$.

Frank
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  • What about $\int_{c}^{d} f'(t)dt$ if $f'$ is not continuous, how do you apply the Fundamental Theorem of Calculus? Maybe the integral isn't even defined. – Math Jul 18 '23 at 03:25
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    The FTC is still true without the continuity assumption as long as your function is Riemann integrable. – dezdichado Jul 18 '23 at 04:02
  • Differentiability alone is not sufficient for FTC. If Riemann inetgrability of $f'$ is assumed it should be made epxlicit in the answer. This is only a partial answer. @dezdichado – geetha290krm Jul 18 '23 at 04:57
  • I agree with you @geetha290krm – Math Jul 18 '23 at 12:47
  • @geetha290krm ok you're right I mixed the premises of the question and the answer – dezdichado Jul 18 '23 at 13:41
  • Thanks for pointing out the extra assumption. I've edited my answer with some clarification—the proof should work if we use some powerful results :) – Frank Jul 18 '23 at 19:09
  • The problem is that I don't want to impose any additional assumptions. I'm really looking for some counterexample, if any. – Math Jul 18 '23 at 23:18
  • @Math23 My answer (after the edit) doesn’t involve any additional assumptions, so you won’t be able to find a counterexample. – Frank Jul 19 '23 at 21:40