Home » Mathematics » Fractals » Hausdorff dimension of certain sets

Hausdorff dimension of certain sets

1. Generalized Cantor set:

Fix 0<\alpha<1. Choose \eta (small) and N\geq 2\in\mathbb{N} (large) such that N\eta^{\alpha}=1. Fix 0\leq a_1<a_2<...a_N\leq 1-\eta<1.

Let K_1 be the union of N intervals [a_i,a_i+\eta] of length \eta. Let \mu_1 be the natural measure (normalized one dimensional Lebesgue measure restricted to K_1) arises from K_1.

Let K_2 be the union of N^2 intervals [a_i+a_j\eta,a_i+a_j\eta+\eta^2] of length \eta^2 and \mu_2 be its natural measure.

Similarly for each j, let K_j be the union of N^j intervals of length \eta^j and \mu_j be its natural measure.

We consider the generalized Cantor set E=\cap K_j.

(Exercise: Then E is compact perfect and nowhere dense set.)

Also note that,

  • \{\mu_j\}_j weakly converges to a measure \mu supported in E.
    • \mathbb{R}^n is locally compact and C_c(\mathbb{R}^n), the space of compactly supported continuous function on \mathbb{R}^n is a separable topological vector space.
    • By the application of Riesz representation theorem, space of continuous function vanishing at infinity is same as the space of Radon measures. (All the above \mu_j‘s are Radon measure, that is locally finite and inner regular measure)
    • By the application of Banach-Alaoglu theorem, \{\mu_j\} is a weakly convergent sequence.
  • dim_{\mathcal{H}}E=\alpha (enough to prove 0<\mathcal{H}^{\alpha}(E)<\infty).
    • By the definition of \mathcal{H}^{\alpha}, for \delta>0 there exists large j such that \eta^j<\delta. Then\mathcal{H}_{\delta}^{\alpha}(E)
      \ \ \ \leq \sum_{1}^{N^j}(\eta^j)^{\alpha}
      =N^j\eta^{j\alpha}=1

      and hence \mathcal{H}^{\alpha}(E)\leq 1 <\infty.

    • Since E is compact, if \{E_i\}_i is a Borel cover of E then we can assume that \{E_i\} is a finite Borel cover. Also since E is totally disconnected, \{E_i\} can be assumed to be finite Borel cover, with E_i‘s disjoint open intervals and their end points are in the complement of E.Let \{E_i\} be such a cover. Then for each i let p_i be the smallest integer such that E_i has at least one interval of length \eta^{p_i}. Note that there cannot be more than 2N-2 consecutive intervals of length \eta^{p_i} in E_i (otherwise, it contradicts the choice of p_i to be the smallest). Let k_i be the number of consecutive intervals of length \eta^{p_i} in E_i. Then 1\leq k_i\leq 2N-2. Hence k_i\eta^{p_i}\leq d(E_i)=diameter of E_i.

      We know that each interval of length \eta^{p_i} has N intervals of length \eta^{p_i+1} and N^2 intervals of length \eta^{p_i+2}. Similarly each interval of length \eta^{p_i} has N^{m-p_i} intervals of length \eta^{m} for m>p_i. Thus k_i intervals of length \eta^{p_i} has k_iN^{m-p_i} intervals of length \eta^{m}. There are totally N^m intervals of length \eta^m and since E_i‘s are disjoint and finite, we have \sum k_iN^{m-p_i}=N^m for large m>p_i (for all i).

      Also we have N\eta^{\alpha}=1. Hence

       (2N-2)^{\alpha-1}
      \ \ \ \leq \eta^{m\alpha}\sum_i k_i^{\alpha-1}k_iN^{m-p_i}
      \ \ \ \ =\sum_ik_i^{\alpha}\eta^{p_i\alpha}
      \ \ \ \ \leq \sum_id(E_i).

      Thus 0<\mathcal{H}^{\alpha}(E).

 

 

(to be updated)

This blog is loosely based on the notes prepared while giving a lecture series during January 2017-April 2017 at IISER - Bhopal. The definitions, the notations, proofs, theorems are from the references: 1) W. F. Donoghue; Distributions and Fourier Transforms, Acad. Press, 1969. MR3363413 2) Pertti Mattila, Geometry of sets and measures in Euclidean spaces: Fractals and rectifiability, Cambridge Studies in Advanced Mathematics, vol. 44, Cambridge University Press, Cambridge, 1995.  MR1333890 3) P. Mattila. Fourier Analysis and Hausdorff Dimension, Cambridge University Press, Cambridge, 2015. 4) Wolff T.H.: Lectures on Harmonic Analysis. University Lecture Series, 29. Amer. Math. Soc., Providence, RI (2003)

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