Show that the graph of a Lipschitz function in has Hausdorff dimension 1.

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- Dec 8th 2011, 07:54 AMkierkegaardHausdorff Dimension
Show that the graph of a Lipschitz function in has Hausdorff dimension 1.

- Dec 9th 2011, 01:05 PMJose27Re: Hausdorff Dimension
Clearly it suffices to check this for a Lipschitz function on a compact interval . Denote by the Lipschitz constant fo .

Subdivide into intervals of length , we use these to cover by squares with , so that

Since this is valid indepentent of , we see that .

Let with and , and pick . We have,

so that

where and (they're the canonical projections of into the i-th coordinate). Now let and (We may assume otherwise the result is trivial). We have . Let be the smallest interval containing then and so that and so we get

since the bound on the left is independent of the cover we get that . So it must be the case that the Hausdorff dimension of is (because ).