Hausdorff Dimension

**kierkegaard** · December 8th 2011, 07:54 AM

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

**Jose27** · December 9th 2011, 01:05 PM

Originally Posted by kierkegaard

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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