Hausdorff dimension and $W^{1,1}$ functions

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?

• To avoid problems with null sets, do you want to take the essential image? – Nate Eldredge Mar 30 '19 at 3:36
• @NateEldredge Yes. – Riku Mar 30 '19 at 10:00
• 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. – Skeeve Mar 30 '19 at 14:35
• @Skeeve That's alright. – Riku Mar 30 '19 at 17:00

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

• 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? – Riku Mar 30 '19 at 12:28
• Also, why is the dominated convergence theorem relevant? Do you know anything about the pointwise convergence of $f'_n$? – Riku Mar 30 '19 at 12:36
• $f_n'$ 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. – Skeeve Mar 30 '19 at 13:10
• 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? – Riku Mar 30 '19 at 14:12
• 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 $x\in \mathbb R$? – Skeeve Mar 30 '19 at 14:20