# Hausdorff Dimension

• Dec 8th 2011, 07:54 AM
kierkegaard
Hausdorff Dimension
Show that the graph of a Lipschitz function in $\displaystyle \mathbb{R}^{2}$ has Hausdorff dimension 1.
• Dec 9th 2011, 01:05 PM
Jose27
Re: Hausdorff Dimension
Quote:

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

Clearly it suffices to check this for a Lipschitz function on a compact interval $\displaystyle [a,b]$. Denote by $\displaystyle M$ the Lipschitz constant fo $\displaystyle f$.

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

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

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

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

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

so that

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

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

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

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