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 $\mathcal{C}^{\infty}$ 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?

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?

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 $\mathcal{C}^{\infty}$ 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 $$\overline{S_i} \cap S_j \neq \emptyset \implies S_j \subset \overline{S_i} \setminus S_j , \ i \neq j$$

$$\forall x \in \overline{S_i} \cap S_j, \{x_k\}_{k \in \mathbb{N}} \subset S_i : \lim_{k \to \infty}x_k = x, \lim_{k \to \infty}T_{x_k}S_i = \mathcal{T} \implies T_xS_j \subset \mathcal{T} $$ and for all $i \in I, \ u \in S_i$ there is satisfies the nonverticality condition: $$(0,...,0,1) = e_{n+1} \not\in \mathbb{R}^{n+1}$$

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?

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 $\mathcal{C}^{\infty}$ 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?