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Definition : A curve $\omega : [0,1]\to X$ is defined absolutely continuous whenever there exists $g\in L^1([0,1])$ such that $d(\omega(t_0),\omega(t_1))\le\int_{t_0}^{t_1}g(s)ds$ for every $t_0<t_1$.

I would like to show that, according to the above definition, the graph of the Cantor function is not absolutely continuous. I think this is true since I read that for curves $\omega:\mathbb{R}\to\mathbb{R}^d$ we must have that the map $g$ from the definition satisfies $g(t)=\|\dot{\omega}(t)\|$. However, if $\omega$ is the curve of the cantor function, we have $d(\omega(0),\omega(1))=d((0,0),(1,0))=\sqrt{2}$ and $\int_0^1 \|\dot{\omega}(t)\|dt=\int_0^1 \sqrt{1^2+0^2}dt=1$ so $d(\omega(0),\omega(1))\not\leq\int_0^1 \|\dot{\omega}(t)\|dt$.

I was wondering if there is a way to show that the Cantor curve is not absolutely continuous (according to the above definition) in a more direct way without using the result that $g=\|\dot{\omega}\|$? I was thinking by contradiction : "Suppose there exists $g$ with this property..." but I am not sure how this is done.

Maybe it helps to know that the length of the Cantor curve is $2$ from this discussion, so in particular the length of the curve is different from the integral of the velocity.

Edit

Also I was wondering if the definition of absolutely continuous curve whas equivalent to the more natural (for me) condition that the metric derivative exists almost everywhere and for every $a,b$ such that $0\le a \le b\le 1$ we have $$\operatorname{length}_{a,b}(\omega)=\int_a^b|\dot{\omega}|(t)dt$$ where $$\operatorname{length}(\omega):=\sup\left\{\sum\limits_{k=0}^{n-1}d(\omega(t_k),\omega(t_{k+1})),n\ge 1, a=t_0<t_1<\ldots<t_n=b \right\}.$$ and $|\dot{\omega}|(t)$ is the metric derivative defined by $$|\dot{\omega}|(t):=\lim\limits_{h\to0}\frac{d(\omega(t+h,t))}{h}.$$

NB : the definition of absolute continuity has an inequality because although for absolutely continuous function $f:I\to\mathbb{R}$ we have $f(y)-f(x)=\int_x^y f'(t)dt$ this is not true for function $f:[0,2]\to\mathbb{R}^n$. Take $f(t)=(t,0)$ for $t\in[0,1]$ and $f(t)=(1,t-1)$ for $t\in[1,2]$. $d(f(0),f(2))<d(f(0),f(1))+d(f(1),f(2))=\int_0^2 |f'|(t)dt$

edamondo
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2 Answers2

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If you have a curve $\gamma:[0,1]\to (X,d)$ where $(X,d)$ is a complete and separable metric space then if $\gamma$ is absolutely continuous with your definition, there always exists the metric derivative and it is what you expect (the least function which you can choose as $g$ in the inequality of absolute continuity). For this fact if you need a proof I can write it.

Now I think that the proof of the fact that the Cantor function is not AC is immediate once you realise that an AC function sends sets of measure zero into sets of measure zero (Lusin N property): however the Cantor function sends the Cantor set to the whole interval $[0,1]$, hence it cannot be absolutely continuous.

EDIT: What we want to prove now is that given $\gamma:[0,1]\to X$ absolutely continuous (and $(X,d)$ complete and separable metric space), then the metric derivative $|\dot{\gamma}_t|:=\lim_{h\to 0}\frac{d(\gamma_{t+h},\gamma_t)}{h}$ exists a.e., it is in $L^1(0,1)$ and it is the minimal $g$ such that $$d(\gamma_t,\gamma_s)\leq\int_s^t|\dot{\gamma}_r|dr\;\;\forall t,s\in[0,1].$$

Pick a countable dense set $(x_n)_n$ in $X$ and define $h_n(t):=d(\gamma_t,x_n)$. By triangle inequality it is immediate to see that $$|h_n(t)-h_n(s)|\leq\int_s^tg(r)dr,$$ where $g$ is your $L^1$ function. By Lebesgue differentiation you also have $|h_n^\prime(t)|\leq g(t)$ for a.e. $t\in[0,1]$. Call $h:=\sup_nh_n^\prime$ and note that it is in $L^1$ thanks to the previous inequality. As an (easy) exercise try to prove that $d(\gamma_t,\gamma_s):=\sup_n(h_n(t)-h_n(s))$. We now have $$d(\gamma_t,\gamma_s)=\sup_n\int_s^th_n^\prime(r)dr\leq\int_s^th(r)dr,$$ so $h$ can substitute $g$ in the definition of absolute continuity. It remains to show that our $h$ is the metric speed: first apply Lebesgue differentiation to the previous inequality to deduce $$\limsup_{t\to s}\frac{d(\gamma_t,\gamma_s)}{|t-s|}\leq h(s)$$ for a.e. $s\in[0,1]$. Then note that by a previous exercise we get $$d(\gamma_t,\gamma_s)\geq h_n(t)-h_n(s) = \int_s^th^\prime_n(r)dr$$ so that we can infer $$\liminf_{t\to s}\frac{d(\gamma_t,\gamma_s)}{|t-s|}\geq h_n^\prime(s)$$ for a.e. $s\in[0,1]$, which allows to conclude.

Gauge_name
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  • But with respect to your previous comment, if you consider small intervals whose measure sum up to something smaller than $\varepsilon$, then how do you apply the definition of AC for each of those intervals if you don't know that the curve is absolutely continuous a priori? – edamondo Dec 05 '21 at 10:36
  • @edamondo it was a reasoning by contradiction. I deleted the comment because the question was deleted. – Gauge_name Dec 06 '21 at 11:03
  • Suppose that the Cantor curve is AC. Then take open intervals $(t_k,t_{k+1}) , k=1\ldots n$ covering the Cantor set but which length is smaller than $\varepsilon$. i.e. $\sum_{k=1}^{n-1}t_{k+1}-t_k<\varepsilon$. Absolute continuity implies $\sum_{k=1}^{n-1} d(\omega(t_k),\omega(t_{k+1}))\le \sum_{k=1}^{n-1}\int_{t_k}^{t_{k+1}}g(s)ds$ but how do you get a contradiction? – edamondo Dec 06 '21 at 16:21
  • I think you might need more than finitely many but the point is that the lhs is bigger than $d(\omega(0),\omega(1))$, while the rhs is arbitrarily small because g is in $L^1$ and you are integrating over a set of small measure. – Gauge_name Dec 06 '21 at 18:11
  • I understand for the rhs. However, how the lhs can be bigger than $d(\omega(0),\omega(1))$ if we take intervals which sum is of length $<\epsilon$? – edamondo Dec 06 '21 at 20:46
  • @edamomo Say that the curve is $t\mapsto \omega(t)=(t,f(t))$. Then $d(\omega(t_0),\omega(t_1))\geq |f(t_0)-f(t_1)|$. I chose the same notation for the component, making a mistake which is crucial at this point. But I hope you understand now what I’m saying. At this point you forget about the small pieces and you care only about the $y$ direction. You take the intervals such that you capture where $f$ grows (you can do that by outer regularity of Lebesgue measure and continuity of the function $f$) and you get a contradiction. – Gauge_name Dec 07 '21 at 07:44
  • Okay I think I might understand now. I summarized what I understood in an answer. Let me know what you think. Also I would be very much interested in the proof of the claim in the first paragraph of your answer. – edamondo Dec 08 '21 at 18:42
  • Thank you. When you say "By Lebesgue differentiation you also have $|h_n'(t)|\le g(t)$" shouldn't it be $h_n'(t)\le g(t)$. I am not sure since I am not super familiar with this theorem. – edamondo Dec 10 '21 at 09:38
  • No it is actually with absolute value, because the inequality is with the absolute value. The Lebesgue Differentiation theorem is sort of a mean value theorem, the difference being that you don’t require the integrand to be continuous but just $L^1_{loc}$. I suggest you to study this theorem because it is very important. – Gauge_name Dec 10 '21 at 12:35
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Consider some arbitrary $\epsilon>0$. Since $g\in L^1$ there exists $\delta>0$ such that $\int_Ag<\epsilon$, as soon as $\lambda(A)<\delta$ (here $\lambda(\cdot)$ is the Lebesgue measure). Now from the construction of the Cantor set, $C=\bigcap\limits_{n\in\mathbb{N}}C_n$ and $C_n$ has length $(2/3)^n$, we can consider a set $C_n$ where $n$ is such that $(2/3)^n<\delta$ (observe $C\subset C_n$). Note also that $C_n =\bigcup\limits_{i=1}^N [s_i,t_i]$. If we apply the triangular inequality and the absolute continuity on the interval $[s_i,t_i]$ we get $$f(t_i)-f(s_i)\le d(\omega(s_i),\omega(t_i))\le\int_{s_i}^{t_i}g(s)ds.$$ Summing up we get $\sum_{i=1}^N f(t_i)-f(s_i)\le\sum_{i=1}^N\int_{s_i}^{t_i}g(s)ds$. However since $\lambda(C_n)<\delta$, the right hand side is smaller than $\epsilon$ and since the Cantor function only increases on the Cantor set and $C\subset C_n$ we have that left hand side equals $1$.

edamondo
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