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Let $u: \Omega\subset \mathbb{R}^N \to \mathbb{R}^M$ be a $BV$ function.

Is the Hausdorff dimension of the graph of $u$ equal to $N$? How can we prove it?


Update.

  1. In an answer to this post, it has been showed that there exist a representative $\tilde u$ of $u$ such that its graph has Hausdorff dimension equal to $N$.

  2. In a subsequent post If the Hausforff dimension of the graph of a function $u$ is $N$ and $\tilde u = u$ a.e. then $\dim_H \mathrm{graph} \, \tilde u = N$ too it has been showed that a function can be zero a.e. and still its graph may have dimension strictly greater than $1$. So probably this question is better formulated in terms of essential graph of $u$, which possibly is equivalent to asking for the property to hold for one representative of $u$ (see Question 2 in Hausdorff dimension of the graph of a BV function (in 1 dimensional setting))

  3. In the post Hausdorff dimension of the graph of a BV function (in 1 dimensional setting), I've asked about a simpler proof of the result in the one-dimensional setting.

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  • $\begingroup$ I suppose you meant $N$, not $1$? $\endgroup$ – Mateusz Kwaśnicki Apr 5 at 18:58
  • $\begingroup$ @MateuszKwaśnicki Yes. $\endgroup$ – Riku Apr 5 at 19:14
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A partial answer here.

Let us recall the following Lusin type result (see e.g. Theorem 5.34 in Functions of bounded variation by L. Ambrosio, N. Fusco and D. Pallara):

Theorem 1. There exists a constant $\kappa>0$ such that for every $u\in BV(\mathbb R^N)$ and $\lambda>0$ there exists a Lipschitz function $v_\lambda \in \mathrm{Lip}(\mathbb R^N)$ such that $$ |x\in \mathbb R^N : u(x)\ne v_\lambda(x)| \le \frac{\kappa}{\lambda} |Du|(\mathbb R^N). $$

Let us denote $D_\lambda := \{x\in \mathbb R^N : u(x) = v_\lambda(x)\}$ and $D := \bigcup_{\lambda \in \mathbb N} D_\lambda$. It is evident that the complement $D^c = \mathbb R^n \setminus D$ is Lebesgue negligible.

Let us show that the Hausdorff dimension of the graph $\Gamma_D:=\{(x,u(x)) : x\in D\}$ is $N$. Clearly it is sufficient to show that the Hausdorff dimension of the graph $\Gamma_{D_\lambda}:=\{(x,u(x)) : x\in D_\lambda\}$ is $N$ for each $\lambda\in \mathbb N$. And this is true because $u\colon D_\lambda \to \mathbb R$ coincides with Lipschitz function $v_\lambda \colon D_\lambda \to \mathbb R$. Thus we have proved the following claim:

Proposition 1. If $u$ is a $BV$ function then there exists $\tilde u \in BV$ such that $\tilde u = u$ a.e. and the Hausdorff dimension of the graph of $\tilde u$ is $N$.

What remains not clear to me at the moment is whether the claim is true for every representative of $u$. But note that the answer to this question does not depend on $BV$ regularity of $u$, and can essentially be formulated as follows:

Question. Suppose that $u\colon [0,1]\to [0,1]$ is a measurable function such that $u=0$ a.e. Is it true that the Hausdorff dimension of the graph of $u$ is at most $1$?

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  • $\begingroup$ Thank you. I've asked that last question as a separate post (mathoverflow.net/questions/327331/…). Why do you pose it only in $1$ dimension? That is, how can you deduce the $N$-dimensional statement from that? $\endgroup$ – Riku Apr 6 at 17:08
  • $\begingroup$ Also, why is it sufficient to show that the Hausdorff dimension of the graph $\Gamma_{D_\lambda}:=\{(x,u(x)) : x\in D_\lambda\}$ is $N$ for each $\lambda\in \mathbb N$? That is, why can we "pass to the limit" in that expression? $\endgroup$ – Riku Apr 6 at 17:09
  • $\begingroup$ @Riku because $\Gamma_{D} = \bigcup_{\lambda \in \mathbb N} \Gamma_{D_\lambda}$, and Hausdorff dimension is stable under taking countable unions. And thank you for reposting a more general version of my question in a separate thread. I do not claim that the $N-$dimensional statement can be deduced from the one-dimesional. But I think that the simpler version is easier to address, and maybe its solution could be adopted for the general case. $\endgroup$ – Skeeve Apr 6 at 22:10
  • $\begingroup$ I see. Thank you. $\endgroup$ – Riku Apr 7 at 11:36
  • $\begingroup$ Do you have a simpler proof (that does not rely on Theorem 1) in the case $N=M=1$ that the graph of a BV function has Hausdorff dimension equal to 1? $\endgroup$ – Riku Apr 9 at 13:15

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