Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ be locally Lipschitz near $x \in \mathbb{R}^{n}$ with constant $L>0$, i.e. there exist $\varepsilon > 0$ and $L>0$ such that
\begin{equation}
|f(y) – f(z)| \leq L \| y-z\| \qquad \forall y,z \in B_{\varepsilon}(x) \, .
\end{equation}
The generalized Clarke subdifferential $\partial f(x)$ is defined by
\begin{equation}
\partial f(x):=\left\{g \in \mathbb{R}^{n} \mid f^{\circ}(x ; d) \geq g^{\top} d \quad \text {for all } d \in \mathbb{R}^{n}\right\}
\end{equation}
where $f^{\circ}(x ; d)$ denotes the generalized directional derivative which is defined by
\begin{equation}
f^{\circ}(x ; d):=\limsup _{y \rightarrow x, \, \, t \downarrow 0} \frac{f(y+t d)-f(y)}{t}\, .
\end{equation}
It is known that $\partial f(x)$ is a nonempty, convex, compact set such that $\partial f(x) \subset B_{L}(0)$ . According to Clarke and many other authors the Lipschitz property and the fact that $\partial f(x) \subset B_{L}(0)$ imply that for all $x_{i} \in B_{\varepsilon}(x)$
\begin{equation}
\left\|g_{i}\right\| \leq L \text { for all } g_{i} \in \partial f\left(x_{i}\right) \,.
\end{equation}
This is always used without further explanation and therefore I assume that this is beyond trivial but I just dont see it. So I am sorry in advance and very thankful if someone could help me with this.
Best Answer
Assuming $B_\epsilon(x)$ is the open ball, for all $y\in B_\epsilon(x)$, there exist $0<\delta\leq\epsilon$ such that $B_\delta(y)\subset B_\epsilon(x)$. Hence $$\lvert f(z_1)-f(z_2)\rvert \leq L\lVert z_1-z_2\rVert, \qquad \forall z_1,z_2\in B_\delta(y).$$ Hence $f$ is locally Lipschitz near $y$ and $\partial f(y)\subset B_L(0)$ by the property you mentioned. Does that seem right?