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Let $f: \mathbb R^n \to \mathbb R^m$ be continuous, and differentiable almost everywhere with $Df = 0$ almost everywhere.

Question: What is the maximal Hausdorff dimension of the graph of $f$?

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    $\begingroup$ In another question you are asking for the minimum value for dimension of singualr set. Could you please say the motivation behinds such questions $\endgroup$ Commented Nov 17 at 12:33
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    $\begingroup$ @AliTaghavi Well, for this one I want to know if there are singular functions with large dimension, despite being flat almost everywhere. In general it is natural to try to obtain bounds on the Hausdorff dimension of graphs of functions subject to regularity conditions. Eg: Lipschitz and BV functions have graph dimension equal to that of the domain, $\alpha$-Holder functions have graph dimension at most $n + (1-\alpha)m$. $\endgroup$ Commented Nov 17 at 12:38

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This problem is neatly resolved by (an extension of) a theorem by Peres and Souza. It is stated as Theorem 1.18 in the paper Dimensions of graphs of prevalent continuous maps by Balka.

The theorem states that for any compact, uncountable subset $K$ of $\mathbb R^n$, a generic continuous function from $K$ to $\mathbb R^m$ has graph Hausdorff dimension $\text{dim}_H (K) + m$.

Using this, we may show that there exist singular functions $\mathbb R^n \to \mathbb R^m$ with graph dimension $n+m$ as follows.

First we handle the $n=1$ case. We take $K \subset \mathbb R$ to be a middle interval Cantor set with measure $0$ but Hausdorff dimension $1$, and let $g: K \to \mathbb R^m$ be a continuous function with graph dimension $m+1$, as provided by the Theorem. Then in the removed intervals, we interpolate the endpoints of the Cantor set with Cantor staircases.

Finally, for the generic $n$ case, we can simply use the function

$$F(x_1, \dots, x_n) = f(x_1),$$

where $f$ is our one dimensional function. The graph is the product of an $m+1$ dimensional set with $\mathbb R^{n-1}$, which has Hausdorff dimension $n+m$, as desired.

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