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Let $u: \Omega\subset \mathbb{R} \to \mathbb{R}$ be a function of bounded variation.

Question 1.

How can we prove that the Hausdorff dimension of the essential graph of $u$ equal to $1$?

Question 2.

Is Question 1 equivalent to asking the following?

How can we prove that there exists a representative $\tilde u$ of $u$ such that the Hausdorff dimension of $\tilde u$ is equal to 1.


Note. I've asked a more general question Hausdorff dimension of the graph of a BV function, which has received a very nice (partial) answer, however, I'd like to see a simpler proof in the 1-dimensional case.

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  • $\begingroup$ You can cover the graph by roughly 1/r balls of radius r. $\endgroup$ Commented Apr 10, 2019 at 18:03
  • $\begingroup$ @Riku Probably it makes sense to specify what you mean by graph, because a function can be zero a.e. and still its graph may have dimension strictly greater than 1, see my answer here mathoverflow.net/questions/327331 $\endgroup$
    – Skeeve
    Commented Apr 10, 2019 at 18:11
  • $\begingroup$ @AnthonyQuas What do you mean? $\endgroup$
    – Riku
    Commented Apr 11, 2019 at 11:16
  • $\begingroup$ @Skeeve You're right. What would the appropriate definition of graph be? $\endgroup$
    – Riku
    Commented Apr 11, 2019 at 11:22
  • $\begingroup$ @Riku there is a notion of essential range. Maybe it would be appropriate to define similarly essential graph, i.e. to say that essential graph of $u$ is the essential range of the function $x\mapsto(x,u(x))$. $\endgroup$
    – Skeeve
    Commented Apr 11, 2019 at 13:51

1 Answer 1

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I will assume $\Omega$ is an interval $[a,b]$, say. I assume this was intended as part of your question.

Let $\text{Var}f=M$. Let $r>0$. The function has at most countably many discontinuities, which are necessarily of jump type. The magnitude of the discontinuities sums to at most $M$. In particular, there are at most $M/r$ points at which the function jumps by $r$ or more. Let these be $z_1,\ldots, z_k$. Let $z_0=a$ and $z_1=b$. Then on each interval $(z_i,z_{i+1})$, there exist $z_i:=x^i_0<x^i_1<\ldots<x^i_{n_i}:=z_{i+1}$ such that $r<\text{Var}_{[x^i_j,x^i_{j+1}]}(f)<2r$ for each $i,j$, except that no lower bound is imposed for $j=n_i-1$ (that is the variation is split into chunks of size roughly $r$). Now the part of the graph lying over $[x_j^i,x_{j+1}^i]$ can be covered by approximately $\lceil (x^i_{j+1}-x^i_j)/r\rceil$ balls of radius $r$, so that approximately $N_i=(z_{i+1}-z_i)/r+n_i+1$ balls are needed to cover the section of the graph lying over $[z_i,z_{i+1}]$. Notice that $n_i\le \text{Var}_{(z_i,z_{i+1})}f/r+1$, so that summing the $N_i$'s we obtain a cover with approximate $(b-a)/r+\text{Var}_{[a,b]}f/r+k$ balls of size $r$. Since $k\le M/r$, we see the number of balls is bounded by approximately $(b-a)/r+2M/r$. Hence the Hausdorff dimension is at most 1.

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  • $\begingroup$ Thank you. 1) How do you prove that the function has at most countably many discontinuities, which are necessarily of jump type? 2) Why are there at most M/r points at which the function jumps by r or more? 3) You conclude that $\dim graph(f) \le 1$. How do you show that it is also $\ge 1$? $\endgroup$
    – Riku
    Commented Apr 11, 2019 at 21:31
  • $\begingroup$ The facts you’re asking for are standard. For 3, the image under projection onto the first coordinated has Hausdorff dimension 1, and projections don’t increase dimension (you can just project the covering to get a new covering) $\endgroup$ Commented Apr 12, 2019 at 1:28
  • $\begingroup$ I guess they are standard, but I've never seen the proofs. For (1), it should follow by considering that a BV function is sum of a decreasing and an increasing function. But what about (2)? $\endgroup$
    – Riku
    Commented Apr 12, 2019 at 9:33
  • $\begingroup$ Because if the were more than M/r points with jump r, then the total variation would be more than M/r. * r $\endgroup$ Commented Apr 12, 2019 at 20:08

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