Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ be locally Lipschitz sontinuous.
(1) The Clarke-subdifferential $\partial_{C} f(x) \subset \mathbb{R}^{n}$ of $f$ in $x \in \mathbb{R}^{n}$ is defined by
\begin{equation*} %\label{eq:clarke_Subdifferential} \partial_{C} f(x) := \left\{g \in \mathbb{R}^{n} \mid \langle g , d \rangle \leq f^{\circ}(x ; d) \quad \text {for all } d \in \mathbb{R}^{n}\right\} \,, \end{equation*}
where $f^{\circ}$ denotes Clarke`s generalized directional derivative.
(2) The Bouligand-subdifferential $\partial_{B} f(x) \subset \mathbb{R}^{n}$ of $f$ in $x\in \mathbb{R}^{n}$ is defined by
\begin{equation*} \partial_{B} f(x):=\left\{g \in \mathbb{R}^{n} \mid \exists (x^{k}) \subset \mathcal{D}_{f}: x^{k} \to x \, , \, \nabla f(x^{k}) \to g \quad \text{für }k \to \infty\right\} \,, \end{equation*}
where $\mathcal{D}_{f} \subset \mathbb{R}^{n}$ denotes the set of points where $f$ is differentiable.
(3) It is known that
\begin{equation*} \partial_{C} f(x) = \operatorname{conv}\left( \partial_{B} f(x) \right) \end{equation*}
holds, where $ \operatorname{conv}$ denotes the convex hull.
Question: Is there a smaller set $S \subset \partial_{B} f(x)$ such that
$$ \partial_{C} f(x) = \operatorname{conv}\left( S \right) $$
holds?
Or is the Bouligand-subdifferential $\partial_{B} f(x)$ already the smallest such set, i.e., is $\partial_{B} f(x)$ the set of extreme points of the Clarke-subdifferential $\partial_{B} f(x)$?
My intuition says that the Bouligand-subdifferential $\partial_{B} f(x)$ in indeed the set of extreme points of the Clarke-subdifferential $\partial_{B} f(x)$ but I am unable to find such a result in the literature...
Also I dont need a proof for this. I just need to know whether it holds or not...
Any hint is greatly appreciated
It's not true that the Bouligand subdifferential is the set of extremal points of the Clarke subdifferential.
Indeed, it's possible that there exists $S\subset \partial_B f$ with $\partial_C f = \mathrm{conv}(S)$. You can do with $f:\mathbb{R}\to\mathbb{R}$. Find a function such that there are 3 sequences $\{x_k\}$, $\{y_k\}$, and $\{z_k\}$ with all of them belonging to the set of differentiability points of $f$ and with all them having limits $x_k\to x$, $y_k\to x$, and $z_k\to x$ with $\nabla f(x_k) \to \tilde{\nabla f}_1$, $\nabla f(y_k) \to \tilde{\nabla f}_2$, and $\nabla f(z_k) \to \tilde{\nabla f}_3$ but such that $\tilde{\nabla f}_i \neq \tilde{\nabla f}_j$ for $i\neq j$. For instance you might find $\tilde{\nabla f}_1 = 0$, $\tilde{\nabla f}_2 = 1$, and $\tilde{\nabla f}_3=2$.
Take the function
$$f(x) = \begin{cases}3^n, &\mbox{ if } 3^n\leq x\leq 1.5(3^n), n\in\mathbb{Z}\\ x-3^{n+1} + 2.5(3^n) & \mbox{ if }1.5(3^n)\leq x\leq 2.5(3^n)\\ 2x-3^{n+1}, &\mbox{ if }2.5(3^n)\leq x \leq 3^{n+1}\\ 0, &\mbox{ if }x\leq 0. \end{cases}$$
for instance, with $x_k = .5(3^{-k})$, $y_k = 2(3^{-k})$, and $z_k = 2.75(3^{-k})$. Then $\nabla f(x_k) = 0$, $\nabla f(y_k)=1$ and $\nabla f(z_k)=2$ for all $k\in \mathbb{N}$. We can in fact see that $\partial_B f(0) = \{0,1,2\}$ but $\partial_C f(0) = [0,2]$, i.e., $\partial_C f(0) = \mathrm{conv}(S)$ where $S=\{0,2\}\subset \partial_B f(0)$.