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For the Sierpinski Triangle, $S$, the $d$ dimensional Hausdorff measure is given by, $H^{d}(S)$. If a linear transformation, $W$ is applied to $S$, with

$$W(x,y)=\begin{bmatrix} 1/2 & 0 \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix}$$

We get a new set $S*$. Denote the Hausdorff measure by $H^{d}(S*)$. Are there any results for the value,

$$\rho=\cfrac{H^{d}(S*)}{H^{d}(S)}$$

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    $\begingroup$ The easy estimate is that it is between $(1/2)^d$ and $1^d=1$. $\endgroup$
    – Gerald Edgar
    Commented Jun 18, 2016 at 17:33
  • $\begingroup$ Related: math.stackexchange.com/questions/107443/… $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Jun 18, 2016 at 17:43
  • $\begingroup$ @ChristianRemling ... If we are in $2$-space, and $d>1$, can't we say more? Something like ${} \ge (1/2)\cdot 1^{d-1}$ perhaps. $\endgroup$
    – Gerald Edgar
    Commented Jun 18, 2016 at 21:45
  • $\begingroup$ @GeraldEdgar: Yes, I think you're right. $\endgroup$
    – Christian Remling
    Commented Jun 18, 2016 at 21:51
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    $\begingroup$ @zachary-w-robertson: The fraction $\rho$ is not fixed and depends on the location of the set $S$. For example, both horizontal and vertical intervals of length 1 have 1-dimensional Hausdorff measure 1, but after the linear transformation the horizontal interval will have 1-dimensional measure 1/2 and vertical will still keep 1. The only thing we can do is to give upper and lower bound for $\rho$ and these bounds depend on the maximal and minimal eigenvalues the linear transformation. $\endgroup$
    – Taras Banakh
    Commented Aug 13, 2017 at 5:07

1 Answer 1

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(Just an attempt to answer my question, I'm probably breaking all the rules, but hey I'm just a naïve mathematician at present)

Insert the Ansatz,

$$d\mu(w(S))=r^d d\mu(S)$$

Assume, that since $w_*S=\cup_i w_i(S)$ holds, where $w_*=\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$, that

$$d\mu(w_*S)=r_*^dd\mu(S)=\sum_i d\mu(w_i(S))=3r^d d\mu(S)$$

$$\Rightarrow r_*^d=3r^d$$

Given that the Haussdorf measure for $r$ term shrinks the x axis, and the one for $r_*$ term stretches the y axis, which is by symmetry the same as stretching the x axis, I think its safe to assume the two constants are related by,

$$r_*=1/r$$

$$\Rightarrow 3r^d=r^{-d}$$ $$\Rightarrow 3r^{2d}=1$$ $$\Rightarrow r^{2d}=\frac{1}{3}$$ $$\Rightarrow r=3^{-\cfrac{1}{2d}}$$ $$\Rightarrow d\mu(w(S))=r^d\mu(S)=\cfrac{\sqrt{3}}{3} \cdot d\mu(S)$$ $$\Rightarrow \int_{w(S)} dH^d=\cfrac{\sqrt{3}}{3} \cdot \int_S dH^d$$ $$\Rightarrow H^d(w(S))=\cfrac{\sqrt{3}}{3}H^d(S)$$ $$\Rightarrow \cfrac{H^d(S_*)}{H^d(S)}=\cfrac{\sqrt{3}}{3}$$

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  • $\begingroup$ I think this is just wrong. $\endgroup$
    – Gerald Edgar
    Commented Jul 19, 2016 at 0:24
  • $\begingroup$ @GeraldEdgar No objection from me. I was just trying the obvious solution out...still haven't made any progress. I doubt a formula is easy to get. $\endgroup$
    – Zachary W. Robertson
    Commented Jul 19, 2016 at 3:33
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    $\begingroup$ It's good to try things like this, but it's best not to post as an answer unless you have checked that it is correct. I would suggest deleting this answer and perhaps noting it as a possible approach within the question. Having "answers" that don't answer the question just confuses people; in particular, in some cases it can make it look like the question is resolved and cause people to ignore it. $\endgroup$
    – Nate Eldredge
    Commented Oct 16, 2016 at 23:44

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