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Math Help - Hausdorff Dimension

  1. #1
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    Hausdorff Dimension

    Show that the graph of a Lipschitz function in \mathbb{R}^{2} has Hausdorff dimension 1.
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  2. #2
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    Re: Hausdorff Dimension

    Quote Originally Posted by kierkegaard View Post
    Show that the graph of a Lipschitz function in \mathbb{R}^{2} has Hausdorff dimension 1.
    Clearly it suffices to check this for a Lipschitz function on a compact interval [a,b]. Denote by M the Lipschitz constant fo f.

    Subdivide [a,b] into m intervals of length (b-a)/m, we use these to cover G_f by m squares A_j with diam(A_j)= \frac{1}{m} ((b-a)^2+4M^2)^{\frac{1}{2}}, so that

    \sum_j diam(A_j) = ((b-a)^2+4M^2)^{\frac{1}{2}} < \infty

    Since this is valid indepentent of m, we see that \mathcal{H}_1(G_f) <\infty.

    Let A_n \subset G_f with diam(A_n) < \varepsilon and \cup A_n = G_f , and pick x,y \in A_n. We have,

    \left( 1+\frac{1}{M^2} \right) ^{\frac{1}{2}} |f(x)-f(y)| \leq \left( (x-y)^2 + (f(x)-f(y))^2 \right) ^{\frac{1}{2}} \leq (1+M)|x-y|

    so that

    \left( 1+\frac{1}{M^2} \right) ^{\frac{1}{2}} diam(A_n^{''}) \leq diam(A_n) \leq (1+M)diam(A_n^{'})

    where A_n^{'}=\pi_1(A_n) and A_n^{''}=\pi_2(A_n) (they're the canonical projections of A_n into the i-th coordinate). Now let m_1=\min_{x\in [a,b]} f(x) and m_2 = \max_{x\in [a,b]} f(x) (We may assume m_1\neq m_2 otherwise the result is trivial). We have [m_1,m_2]= \cup A_n^{''}. Let E_n be the smallest interval containing A_n^{''} then \lambda (E_n)=diam(E_n)=diam(A_n^{''}) and [m_1,m_2]= \cup E_n so that m_2-m_1 \leq \sum_n diam(A_n^{''}) and so we get

    \left( 1+\frac{1}{M^2} \right) ^{\frac{1}{2}} (m_2-m_1) \leq \left( 1+\frac{1}{M^2} \right) ^{\frac{1}{2}} \sum_n diam(A_n^{''}) \leq \sum_n diam(A_n)

    since the bound on the left is independent of the cover (A_n) we get that \mathcal{H}_1(G_f)>0. So it must be the case that the Hausdorff dimension of G_f is 1 (because 0< \mathcal{H}_1(G_f) <\infty).
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