Lipschitz function admits Whitney stratification

Question

I've been reading Topological Aspects of Nonsmooth Optimization by Vladimir Shikhman.

There I have found the following observation that:

Lipschitz functions $f : \mathbb{R}^n \to \mathbb{R}$ admit a $\mathcal{C}^{\infty}$ Whitney stratification.

It's not proven anywhere there, and I do not see why this is true.

We would like to partition the graph of $f$ into $\mathcal{C}^{\infty}$ submanifolds $\{\mathcal{S}_i\}_{i \in I}$ of $\mathbb{R}^{n+1}$ such that $$\overline{S_i} \cap S_j \neq \emptyset \implies S_j \subset \overline{S_i} \setminus S_i , \ i \neq j$$

$$\forall x \in \overline{S_i} \cap S_j, \{x_k\}_{k \in \mathbb{N}} \subset S_i : \lim_{k \to \infty}x_k = x, \lim_{k \to \infty}T_{x_k}S_i = \mathcal{T} \implies T_xS_j \subset \mathcal{T} $$ and for all $i \in I, \ u \in S_i$ there is satisfies the transversality condition: $(0,...,0,1) = e_{n+1} \in \mathbb{R}^{n+1}$ and $$e_{n+1} \in T_uS_i$$

Now, how can we check that any stratification of the graph of $f$ is Whitney and nonvertical?

For example, for $f(x) = \sqrt{|x|}$ the nonverticality conditoin isn't satisfied, because the vector tangent to the graph in (0,0) is $e_{n+1}$.

It seems that the nonverticality is satisfied for Lipschtz functions, because there is a double cone whose vertex can be translated along the graph, so that the graph always remains entirely outside the cone.

But I don't know how to make that argument precise.

Could you help me with that?

Answer 1

In general no such stratification exists. Lipschitz functions coincide with $C^1$ functions on sets of positive measure:

Theorem. If $f$ is Lipschitz in $\mathbb{R}^n$, then for any $\epsilon>0$ there is $g\in C^1(\mathbb{R}^n)$ such that the Lebesgue measure of the set $\{ f\neq g\}$ is less than $\epsilon$.

You can find a proof of this result in almost any book on geometric measure theory, see for example Theorem 1 in Section 6.6.1 in

L. C. Evans, R. F. Gariepy, Measure theory and fine properties of functions. Studies in Advanced Mathematics. CRC Press, Boca Raton, FL, 1992.

Therefore, you can decompose the graph of $f$ into countably many $C^1$ pieces (defined on measurable sets) and a set of measure zero. However, in general a Lipschitz function need not coincide with a $C^2$ function on a set of positive measure so there is no hope for any sort of $C^2$ stratification even for functions defined on a real line.

By the way, I checked the book by Vladimir Shikhman and there is no such result as the one formulated in the question. Instead, they consider Lipschitz functions $f : \mathbb{R}^n\to \mathbb{R}$ whose graphs admit a $C^\infty$- Whitney stratification, and that is a different story. No theorem, but assumption.