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Let $f$ be a continuous, strictly increasing function from $[0,1]$ to itself with $f(0)=0, f(1)=1$. Let $\Gamma_f$ denote its graph. What can be said about the Hausdorff dimension of $\Gamma_f$? In particular, is it true that it is always 1?

If not, is there a link between $\dim_H(\Gamma_f)$ and $\dim_H(\mu)$, where $\mu$ is the measure whose distribution function is $f$? (That is, $f(x)=\mu[0,x]$.)

I would appreciate some examples if there's no general answer. Specifically, I think something has to be known when $f$ is the Minkowski question mark function, but Google wasn't much help here, unfortunately.

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    $\begingroup$ Maybe one can use the graph of a positive, upper bounded continuous function g of bounded variation periodic on $[0,\sqrt{2}] $ such that $ g(0)=g(\sqrt{2})=0 $ and perform a rotation of $ \pi/4 $ to get $ \Gamma_{f} $ . $\endgroup$ Commented Jul 8, 2018 at 20:56
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    $\begingroup$ In fact a $-\pi/4$ rotation transforms the graph of an increasing function on $[0,1]$ into a $1$-Lipschitz graph, which has Hausdorff dimension $1$ $\endgroup$ Commented Jul 8, 2018 at 21:16
  • $\begingroup$ @PietroMajer This is indeed a very nice argument. If I correctly remember Minty used it in the context of monotone operators. I can find a reference in the case you are interested. I suggest that you write it as an answer so people can see it ans learn this elegant argument. $\endgroup$ Commented Jul 9, 2018 at 1:35
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    $\begingroup$ Thank you but it is Sylvain Julien's idea! $\endgroup$ Commented Jul 9, 2018 at 5:58

1 Answer 1

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Theorem 1. The Hausdorff dimension of the graph $\Gamma_f$ of $f$ equals $1$.

Proof. Take a partition of $[0,1]$ by intervals of length $1/n$. Since the function is increasing you can cover the graph by boxes of size $1/n\times (f((k+1)/n)-f(k/n))$, $k=0,1,\ldots,n-1$ located above the intervals of partition. The diameters of each of the boxes is bounded by $1/n+(f((k+1)/n)-f(k/n))$ and the sum of diameters is bounded by $$ \sum_{n=0}^{n-1} \frac{1}{n}+\left(f\left(\frac{k+1}{n}\right)-f\left(\frac{k}{n}\right)\right)=2. $$ Since the diameters of the boxes covering the graph converge to $0$ as $n\to\infty$, it follows that the Hausdorff $\mathcal{H}^1$ measure of the graph is bounded by $2$. $\Box$

Note that the the argument used in the proof shows also that the length of the graph is bounded by $L(\Gamma_f)\leq 2$. The graph of $f$ can be parametrized by a curve $\gamma:[0,1]\to\mathbb{R}^2$, $\gamma(t)=(t,f(t))$.

In fact a stronger result is true:

Theorem 2. $\mathcal{H}^1(\Gamma_f)=L(\Gamma_f)\leq 2$.

This is a consequence of a more general results about curves in metric spaces that we will discuss now.

If $\gamma:[a,b]\to (X,d)$ is a curve in a metric space, then its length is defined by $$ L(\gamma)={\rm Var}(\gamma)= \sup\left\{\sum_{i=1}^{n-1}d(\gamma(t_i),\gamma(t_{i+1}))\right\} $$ where the supremum is over all $n$ and all partitions $a=t_1<\ldots<t_n=b$.

$\gamma$ is said to be rectifiable if $L(\gamma)<\infty$.

Theorem 3. Every rectifiable curve $\gamma:[a,b]\to (X,d)$ can be reparametrized as a $1$-Lipschitz curve.

See Theorem 4.2.1 in [1]. This is so called arc-length parametrization.

Theorem 2 is a straightforward consequence of the following more general result and the fact that $L(\Gamma_f)\leq 2$.

Theorem 4. If $\gamma:[a,b]\to (X,d)$ is an injective rectifiable curve and $\Gamma=f([a,b])$, then $$ \mathcal{H}^1(\Gamma)=L(\gamma). $$

Proof. By Theorem 3 we can assume that $\gamma$ is Lipschitz and in the Lipschitz case Theorem 4 is proved is Theorem 4.4.2 in [1]. $\Box$

[1] L. Ambrosio, P. Tilli, Topics on analysis in metric spaces. Oxford Lecture Series in Mathematics and its Applications, 25. Oxford University Press, Oxford, 2004.

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  • $\begingroup$ K.J. Falconer, Geometry of Fractal Sets, Cambridge University Press, 1985, chapter 3's results are related to this $\endgroup$ Commented Apr 6, 2019 at 12:10

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