Let $f:\mathbb{R}^n\rightarrow\mathbb{R}\cup\{+\infty\}$ be an extended real-valued function that is proper, lower semicontinuous, and * Lipschitz continuous over its domain $\newcommand{\dom}{\text{dom}}\dom(f)$* with constant $L$, i.e.,
$$|f(x)-f(y)|\leq L\|x-y\|,\quad\forall\,x,y\in\dom(f),$$
where
$$\dom(f):=\{x\in\mathbb{R}^n:f(x)<+\infty\}.$$
is a closed convex set, possibly with empty interior. Let $\hat{\partial}f(x)$ denote the

*Fréchet subdifferential*of $f$ at $x\in\dom(f)$.

**My question**: can we show $d(0,\hat{\partial}f(x))\leq L$ for $x\in\dom(f)$ satisfying $\hat{\partial}f(x)\ne\emptyset$? or in fact $d(0,\hat{\partial}f(x))$ can be arbitrarily large?

Here $d(0,\hat{\partial}f(x))$ denotes the distance from $0$ to the set $\hat{\partial}f(x)$. It is well-defined and achieved by a unique element because the set is closed and convex.

Recall the *Fréchet subdifferential* of $f$ at $\bar{x}\in\dom(f)$ is given by
$$\hat{\partial} f(\bar{x}):=\left\{v\in\mathbb{R}^n:\liminf_{x\to \bar{x}}\frac{f(x)-f(\bar{x})-\langle v,x-\bar{x}\rangle}{\|x-\bar{x}\|}\geq 0\right\}.$$
It is well known that $\hat{\partial} f(\bar{x})$ can be empty for many $x\in\dom(f)$, but all points where $\hat{\partial} f(\bar{x})\ne\emptyset$ form a dense subset of $\dom(f)$.

Furthermore, if $\bar{x}$ is an interior point of $\dom(f)$, then the desired result is obviously true because $\hat{\partial} f(\bar{x})\subset\partial^C f(\bar{x})$, where the set of Clarke subdifferential $\partial^C f(\bar{x})\ne\emptyset$ and every Clarke subgradient $v\in \partial^C f(\bar{x})$ satisfies $\|v\|\leq L$. However, if $\bar{x}$ is on the boundary, then $\partial^C f(\bar{x})$ can be unbounded, and so is $\hat{\partial} f(\bar{x})$. This means we cannot expect every element to be controlled by a constant, but I observe that I can always find one element with small enough norm so that it is controlled by certain constant. I am not sure whether it is true in general.