3
$\begingroup$

Let $I:=[a,b]$ a perfect interval and $\gamma\in C(I,\Bbb R^n)$ an injective path such that $\Gamma:=\gamma(I)$ is rectifiable. Show that $\dim_H(\Gamma)=1$.

Here $\dim_H$ is the Hausdorff dimension. My work so far:

Note that the canonical projections $\pi_k$ are Lipschitz, and because $\gamma$ is continuous and it domain is compact and connected then $\Gamma$ is also compact and connected, thus $\pi_k(\Gamma)\subset\Bbb R$ is compact and connected.

Because $I$ is perfect and $\gamma$ injective then $\Gamma$ is not a singleton, so there is some $k\in\{1,\ldots,n\}$ such that $\pi_k(\Gamma)$ is a perfect closed interval, thus setting $$ f:\Gamma\to\Bbb R^n,\, x\mapsto (\pi_k(x),0,\ldots,0)\tag1 $$ we can see that $f$ is also Lipschitz and we find that $\dim_H(\pi_k(\Gamma))=\dim_H(f(\Gamma))=1\le\dim_H(\Gamma)$ by some elementary identities of the Hausdorff outer measures.

However Im unable to find a way to show that $\dim_H(\Gamma)\le 1$. I dont have a clue about how to do it.

Some random ideas that I had: I tried to relate that $\gamma$ have a continuous inverse in $\Gamma$, or some uniform polynomial approximation to $\Gamma$, or the fact that $\Gamma$ is rectifiable and compact with the definition of Hausdorff outer measure, but I dont found something.

Some help will be appreciated, thank you.

$\endgroup$

2 Answers 2

2
$\begingroup$

Let $H^1$ be the 1-dimensional outer Hausdorff measure on $\mathbb{R}^n$. Since the curve is injective, you can show that $H^1(\Gamma) =$ "Length of $\gamma$". See for instance here.

Since $\gamma$ is rectifiable, its length $L$ is finite and by your argument not $0$. Thus $H^1(\Gamma) = L \in (0, \infty)$. However, if that is the case, we have $\dim_H(\Gamma) = 1$, by the definition of the Hausdorff dimension.

$\endgroup$
0
$\begingroup$

After some mistakes I think I found a valid answer.


First note that the function $\tilde\gamma: I\to\Gamma,\, t\mapsto\gamma(t)$ is bijective and note that for every closed set $C\subset I$ then $\gamma(C)$ is compact, and by the injectivity of $\gamma$ we find that $$ \gamma(I\setminus C)=\gamma(I\cap C^\complement)=\gamma(I)\cap\gamma(C^\complement)=\Gamma\cap[\gamma(C)]^\complement=\Gamma\setminus\gamma(C)\tag1 $$ Hence $\gamma$ is open, so $\tilde\gamma^{-1}$ is continuous and consequently $\tilde\gamma$ is an homeomorphism.

Let $I=[a,b]$. I will show that for some rectifiable path $\Gamma$ parameterized by $\gamma\in C(I,\Bbb R^n)$, there is a partition $\mathfrak Z_\delta:=\{a_0,a_1,\ldots,a_m\}$ of $I$ with arbitrarily small mesh $\Delta_{\frak Z}=\delta>0$ such that $$ 2\,|\gamma(a_k)-\gamma(a_{k+1})|\ge\operatorname{diam}\big(\gamma([a_k,a_{k+1}])\big),\quad \forall k\in\{0,\ldots,m-2\}\tag2 $$ Now choose some arbitrarily small $\epsilon>0$, then we can define recursively $$ B_k:=\sup\left\{A\subset\Gamma\cap\overline{\Bbb B}(\gamma(a_k),\epsilon): \gamma(a_k)\in A\text{ and }A\text{ is connected }\right\}\\ a_0:=a,\quad a_k:=\sup\gamma^{-1}(B_{k-1})\text{ for }k\ge 1\tag3 $$ where we used the fact that $\tilde\gamma$ is an homeomorphism, so for each $B_k$ the set $\gamma^{-1}(B_k)$ is a closed interval contained in $I$ and the set of points $\mathfrak Z:=\{a_0,a_1,\ldots,a_k,\ldots\}$ is well defined.

Moreover: because $\Gamma$ is rectifiable then $|\mathfrak Z|<\infty$ and because $\tilde\gamma^{-1}$ is uniformly continuous for any chosen $\delta>0$ there is an $\epsilon>0$ such that $|\gamma(x)-\gamma(y)|<\epsilon\implies |x-y|<\delta$, so for any chosen mesh $\Delta_{\frak Z}=\delta$ we can choose an arbitrarily small $\epsilon>0$ on $(4)$ and rename the set of points as $\frak Z_\delta$, what is a partition of $I$ of mesh $\delta$.

Now note that for $\mathfrak Z_\delta=\{a_0,a_1,\ldots,a_m\}$ by construction $|\gamma(a_k)-\gamma(a_{k+1})|=\epsilon$ for $k\neq m-1$ and $\operatorname{diam}(\gamma([a_k,a_{k+1}]))\le 2\epsilon$ because $\gamma([a_k,a_{k+1}])\subset\overline{\Bbb B}(\gamma(a_k),\epsilon)$, so $(2)$ holds.

Now from the definitions $$ \begin{align*}\mathcal H_\epsilon^s(A):=&\inf\left\{\sum_{k=0}^\infty[\operatorname{diam}(A_k)]^s:A\subset\bigcup_{k=0}^\infty A_k\text{ and }\operatorname{diam}(A_k)<\epsilon,\,\forall k\in\Bbb N\right\}\\ L(\Gamma):=&\sup\left\{\sum_{k=0}^m|\gamma(a_k)-\gamma(a_{k+1})|:\{a_0,\ldots,a_{m+1}\}\text{ is a partition of }I\right\}\end{align*}\tag4 $$ we find that $$ \mathcal H_\epsilon^1(\Gamma)\le\sum_{k=0}^{m-1}\operatorname{diam}([a_k,a_{k+1}])\le2\epsilon+2\sum_{k=0}^{m-2}|\gamma(a_k)-\gamma(a_{k+1})|\le 2L(\Gamma)+2\epsilon<\infty\tag5 $$ for arbitrarily small $\epsilon,\delta>0$ where $a_k\in\mathfrak Z_\delta$ (note that $\epsilon$ depends by the chosen $\Delta_{\frak Z}=\delta$, however in any case is bounded below by zero). And because $\mathcal H_*^1(\Gamma)=\lim_{\epsilon\to 0^+}\mathcal H_\epsilon^1(\Gamma)$ we find that $\mathcal H_*^1(\Gamma)\le2 L(\Gamma)<\infty$, and consequently $\dim_H(\Gamma)\le 1$, as desired.


A simpler construction of a partition of $I$ by recursion is as follows $$ a_{k+1}:=\inf\{x\in[a_k,b]:|\gamma(a_k)-\gamma(x)|=\epsilon\}\cup\{b\}\tag6 $$ where we set $a_0:=a$. Then we have a partition of $I$ defined by $\mathfrak Z:=\{a_0,a_1,\ldots,a_m\}$ with the property that $$ \begin{align*}|\gamma(a_k)-\gamma(a_{k+1})|&=\operatorname{diam}\big(\gamma([a_k,a_{k+1}])\big),\quad\forall k\in\{0,\ldots,m-2\}\\ |\gamma(a_{m-1})-\gamma(a_m)|&\le\operatorname{diam}\big(\gamma([a_{m-1},a_m])\big)\le\epsilon\end{align*}\tag7 $$ (note that by construction $a_m=b$). Then we find that $$ \mathcal H_\epsilon^1(\Gamma)\le\sum_{k=0}^{m-2}\operatorname{diam}\big(\gamma([a_k,a_{k+1}]\big)+\operatorname{diam}\big(\gamma([a_{m-1},a_m])\big)\\ \le\sum_{k=0}^{m-2}|\gamma(a_k)-\gamma(a_{k+1})|+\epsilon\le L(\Gamma)+\epsilon\tag8 $$ Then taking limits above as $\epsilon\to 0^+$ we find that $\mathcal H_*^1(\Gamma)\le L(\Gamma)$, and consequently that $\dim_H(\Gamma)\le 1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .