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.