# Hausdorff dimension of the graph of a BV function

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.

• I suppose you meant $N$, not $1$? – Mateusz Kwaśnicki Apr 5 at 18:58
• @MateuszKwaśnicki Yes. – Riku Apr 5 at 19:14

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 $$1$$?

• 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? – Riku Apr 6 at 17:08
• 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? – Riku Apr 6 at 17:09
• @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. – Skeeve Apr 6 at 22:10
• I see. Thank you. – Riku Apr 7 at 11:36
• 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? – Riku Apr 9 at 13:15