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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?

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    $\begingroup$ To avoid problems with null sets, do you want to take the essential image? $\endgroup$
    – Nate Eldredge
    Commented Mar 30, 2019 at 3:36
  • $\begingroup$ @NateEldredge Yes. $\endgroup$
    – Riku
    Commented Mar 30, 2019 at 10:00
  • $\begingroup$ Could you please clarify what are the domain and range of $f$? In my answer it is assumed that $f\colon[0,1]\to[0,1]$, but maybe you want something different. $\endgroup$
    – Skeeve
    Commented Mar 30, 2019 at 14:35
  • $\begingroup$ @Skeeve That's alright. $\endgroup$
    – Riku
    Commented Mar 30, 2019 at 17:00

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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 $f(F) = f(T)$.

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  • $\begingroup$ Thank you. That's very useful. Do you happen to know any result on the Hausdorff dimension of the graph of a $W^{1,1}$ or $BV$ function? for Lipschitz maps, we have that it is equal to $1$. Does the same hold for $W^{1,1}$ or $BV$ functions? $\endgroup$
    – Riku
    Commented Mar 30, 2019 at 12:28
  • $\begingroup$ Also, why is the dominated convergence theorem relevant? Do you know anything about the pointwise convergence of $f'_n$? $\endgroup$
    – Riku
    Commented Mar 30, 2019 at 12:36
  • $\begingroup$ $f_n'$ converges pointwise on the open set $F^c$. In fact since $F^c = \bigcup_{n\in\mathbb N} F_n^c$ and $F_n^c \subset F_{n+1}^c$ are all open, for any $x\in F^c$ there exists $n\in \mathbb N$ and an open neighbourhood of $x$ on which $f_k$ is stationary for $k\ge n$, hence also $f_k'$ is stationary. $\endgroup$
    – Skeeve
    Commented Mar 30, 2019 at 13:10
  • $\begingroup$ Thank you. What about the Hausdorff dimension of the graph of a $W^{1,1}$ or $BV$ function? Can we say that there are functions such that it is strictly greater than 1? $\endgroup$
    – Riku
    Commented Mar 30, 2019 at 14:12
  • $\begingroup$ By graph do you mean the image, i.e. $f(\mathbb R)$, or really the graph, i.e. the set of points $(x,f(x))$ where $x\in \mathbb R$? $\endgroup$
    – Skeeve
    Commented Mar 30, 2019 at 14:20

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