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