# Minimal norm of Fréchet subdifferential for function Lipschitz over its domain

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.

• Just an ituitive idea, which you may already considered. How about consider the tangent cone $C=T_xdom(f)$ of the convex set $dom(f)$? If the subdifferential $v\in\partial f(x)$ belongs to $C$ you might probably argue similary to the case $x$ is interior point. Otherwise probably the projection of $v$ to $C$ is still subdifferential and this reduces to the case $v\in C$ (hopefully). Jan 4 at 15:42

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.
$$\newcommand\R{\mathbb R}\newcommand\de{\delta}\newcommand\ep{\varepsilon}\newcommand\dom{\operatorname{dom}}$$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 $$\|v\|\ge|v_1|\ge\sqrt{1+k^2}>1=L$$. $$\quad\Box$$