Hausdorff dimension of the non-differentiability set of a locally Lipschitz function

Question

Let $f:\mathbb R^n \to \mathbb R$ and $E := \{x \in X : f \text{ not Fréchet differentiable at }x\}$. Then $E$ is Borel measurable. It is well-known that

Theorem If $f$ is convex, then the Hausdorff dimension of $E$ is at most $n-1$.

I would like to ask for a reference of the following statement, i.e.,

If $f$ is locally Lipschitz, then the Hausdorff dimension of $E$ is at most $n-1$.

My closest search is the following

Definition 1.2. A set $E \subset \mathbb{R}^n$ is porous at a point $x \in E$ if there is a $c>0$ and there is a sequence $y_n \rightarrow 0$ such that the balls $B\left(x+y_n, c\left|y_n\right|\right)$ are disjoint from $E$. The set $E$ is porous if it is porous at each of its points, and it is called $\sigma$-porous if it is a countable union of porous sets.

Theorem 1.3. Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a Lipschitz function. Then the set of those points at which $f$ is not differentiable but it is (directionally) differentiable in $n$ linearly independent directions is $\sigma$-porous.

We have the Hausdorff dimension of a $\sigma$-porous subset of $\mathbb R^n$ is at most $n-1$. However, the Lipschitz function in Theorem 1.3. has one more restriction, i.e., it must be directionally differentiable in $n$ linearly independent directions.

Could you elaborate on such a reference?

Answer 1

You can't find a reference because it's false. Rademacher's theorem (Lebesgue-almost everywhere differentiability) is the best one can do.

In fact, for every Lebesgue-null set $E \subset \mathbf{R}$, you can construct a Lipschitz function $f: \mathbf{R} \to \mathbf{R}$ that is not differentiable at any point of $E$. [ACP10]

The reference for this is Theorem 1.1 in... the same paper you linked in your question. The result is on page 1. I don't think they give a proof there, but elsewhere, Preiss gives the argument in some notes from a talk given in Helsinki. The proof basically goes as follows.

Proof. Construct some nested sequence of open sets $\mathbf{R} = G_0 \supset G_1 \supset \cdots \supset E$ that are rapidly shrinking: for every connected component $C$ of $G_k$, the next open set has \begin{equation} \lvert G_{k+1} \cap C \rvert \leq 2^{-k-1} \lvert C \rvert. \end{equation} (Let us say additionally that $\lvert G_1 \rvert \leq 1$.) You can find such a sequence because the Lebesgue measure is outer regular.

The sets $(G_k \setminus G_{k+1} \mid k \in \mathbf{N})$, together with $E$, partition the real line, and we define the function $\psi: x \mapsto (-1)^k$ if $x \in G_k \setminus G_{k+1}$. This is bounded and measurable. (We need not define $\psi$ on $E$ as it is a null set.)

Then the desired function is $f: x \in \mathbf{R} \to \int_0^x \psi$. This is Lipschitz, but not differentiable at any point of $E$.

Essentially, the reason is the following: take an arbitrary point $x \in E$, and let, for each $k \in \mathbf{N}$, $(a_k,b_k)$ be the connected component of $G_k$ containing $x$. Then the derivative of $f$ on $(a_k,b_k)$ is equal to $(-1)^k$ on a large portion of the interval. Certainly this is the case on $(a_k,b_k) \setminus G_{k+1}$, and therefore \begin{equation} \lvert f(b_k) - f(a_k) - (-1)^k (b_k - a_k) \rvert \leq 2 \lvert (a_k,b_k) \setminus G_{k+1} \rvert \leq 2^{-k} (b_k - a_k). \end{equation}

If you now repeat the same calculation for the next term, you find that \begin{equation} \lvert f(b_{k+1}) - f(a_{k+1}) - (-1)^{k+1} (b_{k+1} - a_{k+1}) \rvert \leq 2^{-k-1} (b_{k+1} - a_{k+1}). \end{equation} If you divide this equation through by $b_{k+1} - a_{k+1}$, respectively the previous equation through by $b_k - a_k$, you will see that the two combined are incompatible with differentiability of $f$ at $x$. Q.E.D.

[ACP10] G. Alberti, M. Csörnyei, and D. Preiss. Differentiability of lipschitz functions, structure of null sets, and other problems. In Proceedings of the ICM 2010, pages 1379-1394.