Minimal norm of Fréchet subdifferential for function Lipschitz over its domain

Question

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.

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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.

Answer 1

Let $C$ be a triangle on the plane with a vertex at origin and angle $179^\circ$ at this vertex; $f(x)=-\|x\|$, it is Lipschitz with $L=1$. Then we have $f(x)\geqslant \langle v,x\rangle$, where $v$ is a very long vector whose direction is opposite to that of an angle bisector of $C$ at the origin. However, there seem to be no short vector in the subdifferential of $f$ at 0.

Answer 2

$\newcommand\R{\mathbb R}\newcommand\de{\delta}\newcommand\ep{\varepsilon}\newcommand\dom{\operatorname{dom}}$This is a detalization of Fedor Petrov's answer.

Let $n=2$. Take any real $k>0$. Let $$C:=\{x=(x_1,x_2)\in\R^2\colon|x_2|\le kx_1\}.$$ Let $f(x):=-\|x\|$ for $x\in C$ and $f(x):=\infty$ for $x\in\R^2\setminus C$. Then $\dom f=C$ and $f$ is $L$-Lipschitz with $L=1$ on $\dom f$.

Let $V$ denote the Fréchet subdifferential of $f$ at $(0,0)$. Take any $v=(v_1,v_2)\in\R^2$.

Then, letting $\cdot$ denote the dot product and letting $B_\de:=\{x\in\R^2\colon\|x\|\le\de\}$, we have
\begin{align} &v=(v_1,v_2)\in V \\ &\iff\forall\ep\in(0,1)\ \exists\de>0\ \forall x\in C\cap B_\de \\ &\qquad\qquad\qquad\qquad\qquad-\|x\|\ge v\cdot x-\ep\|x\| \\ &\iff\forall\ep\in(0,1)\ \forall x\in C\ \ -\|x\|\ge v\cdot x-\ep\|x\| \\ &\iff\forall x\in C\ \|x\|+v\cdot x\le0 \\ &\iff\text{for all points $x$ on the extreme rays} \\ &\qquad\quad\text{of the convex cone $C$ we have $\|x\|+v\cdot x\le 0$} \\ &\iff \sqrt{1+k^2}+v_1+k|v_2|\le0. \end{align}

So, $(-\sqrt{1+k^2},0)\in V$ and hence $V\ne\emptyset$. On the other hand, for any $v=(v_1,v_2)\in V$ we have $v_1\le-\sqrt{1+k^2}$ and hence $\|v\|\ge|v_1|\ge\sqrt{1+k^2}>1=L$. $\quad\Box$