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?

  • $\begingroup$ 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. $\endgroup$ – Pantelis Sopasakis May 17 '19 at 0:30

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