# Is the Clarke Subdifferential always defined for Lipschitz continuous functions?

According to various websites, for some function $$f:X\to R$$ we can define a map

$$D(x,v):= \lim_{y\to x} \sup_{t\searrow0} \frac{f(y+tv)-f(y)}{t}$$

and the Clarke Subdifferential (Def 1) is

$$\partial f(x) := \{v \in X^* : D(x,v) \geq v\}.$$

The Clarke Subdifferential generalizes the gradient when $$f$$ is smooth, and the subgradient when $$f$$ is convex.

Another definition (Def 2) is

$$\partial f(x) := \{v\in X^* : (v,-1) \in N_C(\mathbf{epi}~f(x))\}$$ where $$N_C$$ is the normal cone and $$\mathbf{epi}~f = \{(z,g) : f(z) \leq g\}$$ is the epigraph of $$f$$.

There is another theorem that says that whenever $$f$$ is Lipschitz continuous, then $$\partial f$$ is nonempty and convex. In particular, there is an example I have found for $$f(x) = -|x|$$ (absolute value) where the claim is that

$$\partial f(0) = \partial (-f)(0) = [-1,1].$$

(See, for example, Techniques of Variational Analysis by Jonathan Borwein, Qiji Zhu page 191; this page is available freely on Google Books.)

I am trying to understand this from the definition, but it is escaping me.

Let's first look at Def 1. Taking $$f(x) = -|x|$$ and $$y > \epsilon > 0$$ and $$t <\epsilon$$ ($$v$$ normalize) we get $$D(0,v) := -v$$. Taking $$y < -\epsilon < 0$$ we get $$D(0,v) := v$$. Therefore, by Def 1, we have

$$\partial f(x) := \{v : v \leq v \text{ and } v \leq -v\} = \{0\}.$$

The second definition is even more strange. For this choice of $$f$$, the normal cone to the epigraph at $$x = 0$$ is empty! Therefore, how can the subdifferential not be empty?

Can someone help me with my understanding here?

• You can have a look at Chapters 6 and 8 of the book "Variational Analysis" by Rockafellar and Wets where the authors explain the concept of a regular normal cone and a regular subgradient. – Pantelis Sopasakis May 17 at 0:30