Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ a convex function. Since convex functions are locally Lipschitz, they are differentiable almost everywhere. Let $\delta f(x)$ be the set of subgradients to $f$ at $x$. Suppose that $f$ is not differentiable at $x\_0$ and we are interested in a certain subgradient at this point, $v\_0 \in \mathbb{R}^n$ where $f(x) \geq f(x\_0)+v\_0 \cdot (x-x\_0)$. 


Let $\mathcal{B}\_{x\_0}$ be some open ball around $x\_0$ and define $\mathcal{S}=\{ v \in \mathbb{R}^n : \exists x \in \mathcal{B}\_{x\_0} \mbox{ s.t. } \delta f(x)= \{v \} \}$, that is $\mathcal{S}$ is the set of all gradients in the ball. 


My question is this: Does there exist $\hat{\mathcal{S}} \subset \mathcal{S}$ such that $|\hat{\mathcal{S}}|< \infty$ and $\sum\_{v \in \hat{\mathcal{S}}}v a\_v= v\_0$, where $a\_v>0$ and $\sum\_{v \in \hat{\mathcal{S}}}a\_v = 1$. That is, can $v\_0$ always be formed as a finite convex combination of nearby gradients of $f$? 


I know that the problem is related to the concept of a Clarke subdifferential, but any other references are greatly appreciated.



For a convex function, can subgradients be formed from finite convex combinations of gradients?
The answer is "no".


Consider the function 
$$f(x,y)=\sqrt{x^2+y^2}+|y|$$
Note that $v\_0=(1,0)$ is a subgradient at $(0,0)$.
The gradient of $f$ is defined if $y\ne0$ and at all these points its first coordinate is strictly less than 1. 
Hence the statement follows.


