If $f:\mathbb{R}^n\rightarrow\mathbb{R}\cup\{+\infty\}$ is a proper lower semi-continuous function that is Lipschitz continuous over $\text{dom}(f)$, where $\text{dom}(f):=\{x\in\mathbb{R}^n:f(x)<+\infty\}$ is a closed convex set. Is it possible that the clarke subdifferential $\partial^C f(x)$ is empty at some $x\in\text{dom}(f)$?
Here the definition of $\partial^C f(x)$ is given by $$\partial^C f(x)=\{\xi\in\mathbb{R}^n:(\xi,-1)\in N_{\text{epi}(f)}(x,f(x))\},$$ where $\text{epi}(f):=\{(x,y)\in\mathbb{R}^{n+1}:y\geq f(x)\}$ is the epigraph of $f$ and $N_C(y)$ denotes the Clarke normal cone of a set $C\subset\mathbb{R}^n$ at a point $y\in C$. The definition of Clarke normal cone is defined to be the dual cone of Clarke tangent cone, i.e., $$N_C(y):=(T_C(y))^\circ = \{\xi\in\mathbb{R}^n:\langle \xi ,v \rangle \leq 0, v\in T_C(y)\}$$ and the Clarke tangent cone is defined by $$T_C(y):=\left\{v\in\mathbb{R}^n:\lim_{C\ni z\to y,~~t\downarrow 0^+}\frac{d(z+tv,C)}{t}=0\right\}. $$ Here $d(\bullet,C)$ denotes the Euclidean distance of a point to set $C$.
The difficulty here is that $\text{dom}(f)$ can have empty interior, and in that case one cannot find an approximate sequence $\xi_n\in\partial^C f(x_n)$ where $x_n$ is in the interior of $\text{dom}(f)$ and hence $\partial^C f(x_n)$ is bounded by the Lipschitz constant of $f$. It seems that most examples I can find online to illustrate empty Clarke subdifferential involves function like $f(x)=\sqrt{x}$, which is not Lipschitz continuous on the boundary of its domain $[0,+\infty)$.