Relationship between Clarke-subdifferential $\partial_{C}f(\, \cdot \,)$ and Bouligand-subdifferential $\partial_{B}f(\, \cdot \,)$ 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
 A: 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 $S=\{0,2\}\subset \partial_B f(0)$.
