Subdifferential - equivalent definitions?

Question

I've been reading an article on Clarke critical values of subanalytic Lipschitz functions. There I've come across the following definition(s) of subdifferential:

$$f: U \to \mathbb{R}^n, \ \ \ \ \emptyset \neq U \subset \mathbb{R}^n$$

$f$ is locally Lipshitz continuous

Let $$x \in U:$$ The Frechet subdifferential:

$$ \hat{\partial} f(x) = \left\{x^* \in \mathbb{R}^n \ : \ \liminf_{y \to x, y \neq x} \frac{f(y) - f(x) - \langle x^*, y-x \rangle}{||y - x||} \ge 0 \right\}$$

$$x^* \in \partial f \iff \exists x_n \in U, \exists x_n^* \in \hat{\partial} f(x_n) : x_n \to x, x_n ^* \to x^*, \ \ n \to \infty $$

and finally,

$$\partial^o f(x) = \overline{\rm conv}\partial f(x)$$

is called the Clarke subdifferential of $f$ at $x$.

My question is - is $\partial^o f(x)$ the same set as this one defined in Clarke's paper as subdifferential of $f$ at $x$?

$$S_1 = \{ s \in \mathbb{R}^n \ : \ \langle v, s \rangle \le f(x+v) - f(x) \ \forall v \in \mathbb{R}^n \}$$

I know that for convex functions, the above set is equal to:

$$ S_2 = \{s\in \mathbb{R}^n \ : \ \langle s, d \rangle \le f'(x, d) \ \forall d \in \mathbb{R}^n \} $$

Here $$ f'(x, d) = \lim_{t \to 0} \frac{f(x + td) - f(x)}{t}$$

I suppose it would also be helpful to note that $S_1,S_2$ are closed, convex.

Could you help me find a link between $S_1$ and $ \partial^of(x)?$

Answer 1

Ok I think you need to make this question more clear. So for instance, I don't understand the reason you've included the Frechet subdifferential. It doesn't seem to relate to your question. $S_1$ seems to be the classical definition of the subdifferential.

Take a look at Convex Analysis and Monotone Operator Theory in Hilbert Spaces by Combettes and Bauschke. Ch16 has the $S_1$ definition of subdifferential. Proposition 16.3 states that for a proper function $f$, the subdifferential at $x$ in the domain of $f$ is closed and convex, and therefore $\partial f(x) = \overline{\text{conv}}\partial f(x)$.

After you clarify, I'll edit my answer. But I notice the following:

1. The definition for Clark subdifferential does not coincide with the definition of the subdifferential. For instance, a differentiable function on $\mathbb{R}^n$ has Clark subdifferential that is equal to its derivative (can be proven easily with Taylor's theorem. This limit will in fact be just $0$ in all cases). However take the function $f(x)=-x^2$: It has empty subdifferential everywhere, but has a Clark subdifferential everywhere (equal to its derivative).

2. The alternate definition for subdifferential that is in the equation array just below the definition for subdifferential makes no sense. What is $U$? Why can't you just set the whole sequence $x_n=x$? As written (which I assume is wrong) this definition is simply the closure of the subdifferential.

EDIT: Ok now things are clarified, and I understand what you're asking.

Let $\partial f$ be the subdifferential, let $\hat{\partial} f$ be the Frechet subdifferential, let $\partial^\lim f$ be the limiting subdifferential, and let $\partial^o f$ be the Clark subdifferential. Just before the words "and finally" (second line of equation array) you have actually defined the limiting subdifferential. So I am using different notation to you: I'm using $\partial^\lim$ instead of $\partial f$. So my $\partial f$ coincides with $S+1$.

So I observe the following:

$$ \partial f \subsetneq \hat{\partial} f \subsetneq \partial^\lim f \subsetneq \partial^o f$$

This should be obvious (apart from the not equal to counterexamples).

1. $\partial f \subset \hat{\partial} f$ because the denominator in the fraction will always be positive for elements of the subdifferential!
2. $\hat{\partial} f \subset \partial^\lim f$ because you can simply set the whole sequence to the same element: $x_n=x$ and $x^*_n=x^*$. The limiting subdifferential is merely the closure of the graph of the subdifferential.
3. $\partial^\lim f \subsetneq \partial^o f$ because we are taking closures/ convex hulls.

These are not equal in general, in fact they can all be different in a single example. Let

$$ f(x) = \begin{cases}x, &x\leq 0 \\ 0, & x>0 \end{cases}$$

Then we have the following:

$$ \partial f(x) = \begin{cases}\emptyset, &x < 0 \\ \emptyset, & x=0 \\ \emptyset, & x>0 \end{cases}\\ \hat{\partial} f(x) = \begin{cases}1, &x < 0 \\ \emptyset, & x=0 \\ 0, & x>0 \end{cases}\\ \partial^\lim f(x) = \begin{cases}1, &x < 0 \\ \{0,1 \}, & x=0 \\ 0, & x>0 \end{cases}\\ \partial^o f(x) = \begin{cases}1, &x < 0 \\ [0,1], & x=0 \\ 0, & x>0 \end{cases}\\$$

I think this is correct. Someone tell me if I made a mistake.