What can be said about the Hausdorff dimension of the image of a set by a $W^{1,1}$ map? 


In other words, 
what is the relationship between 
$\mathrm{dim}\_H f(A)$ and $\mathrm{dim}\_H A$, where $f \in W^{1,1}$? 


Does the result also hold if $f$ is a $BV$ function?



Hausdorff dimension and $W^{1,1}$ functions
If $f$ is Lipschitz then $\dim\_H f(A) \le \dim\_H A$, since $H^s\_\delta(f(A)) \le Lip(f) \cdot H^s\_\delta(A)$ for any $\delta>0$.


However there exists an absolutely continuous function $f\in W^{1,1}$ such that $\dim\_H(f(A)) > \dim\_H A$ for some set $A$. Let us construct such a function (by a modification of the classical Cantor function).


Consider the classical one-third Cantor set $T\subset [0,1]$ and one-fifth Cantor set $F \subset [0,1]$. Let $T\_n$ and $F\_n$ denote respectively the unions of $2^n$ closed segments of lengths $3^{-n}$ and $5^{-n}$ such that $T = \bigcap\_{n\in \mathbb N} T\_n$ and $F = \bigcap\_{n\in\mathbb N} F\_n$. The standard computation shows that $\dim\_H T = \frac{\ln 2}{\ln 3} > \frac{\ln 2}{\ln 5} = \dim\_H F$, so we are going to define $f$ in such a way that $f(F) = T$.


Now let $f\_n$ denote the increasing function mapping the endpoints of $F\_n$ to the endpoints of $T\_n$, interpolated linearly for the rest of $x\in[0,1]$. Arguing as in the construction of the classical Cantor function we get that $f\_n$ converges uniformly to a continuous function $f$. 
Moreover $f\_n'$ converges pointwise on the open set $F^c$ (hence a.e.).


Let $g(x):= \sup\_{n\in\mathbb N}|f\_n'(x)|$ for a.e. $x$. It is possible to show that
$$
\|g\|\_1 \le 1 + \sum\_{n=1}^\infty 2^n \cdot 5^{-n} \cdot \frac{3^{-n}}{5^{-n}} < \infty,
$$
hence by dominated convergence $f'\in L^1[0,1]$, i.e. $f$ is absolutely continuous (and $f\in W^{1,1}$). Since $f(F\_n) = T\_n$ for all $n$ it follows that $f(F) = f(T)$.


