4
$\begingroup$

Is there an easy way to calculate the Hausdorff dimension of the graph of a real "elementary" function, like $f(x)=\sin(1/x)$ ?

$\endgroup$

1 Answer 1

11
$\begingroup$

The graph of any Lipschitz function $f\colon [a,b]\to\mathbb{R}$ has Hausdorff dimension $1$ (this follows since Hausdorff dimension is invariant under bi-Lipschitz mappings). Your example of $f(x) = \sin(1/x)$ also has a graph with Hausdorff dimension $1$, since the graph can be decomposed into a countable union of curves that are graphs of Lipschitz functions, even though the function itself is not everywhere Lipschitz, and then you use the fact that Hausdorff dimension is stable under countable unions: $\dim_H \bigcup_n Z_n = \sup_n \dim_H Z_n$.

I'm not sure exactly what other sorts of "elementary" functions you're interested in. Certainly if $f$ if piecewise $C^1$ with only countably many points of non-differentiability, then its graph has Hausdorff dimension $1$, by the argument above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.