You are given a series of vectors $u\_0,\ldots,u\_k$ of non-negative entries, each of dimension $n$, for which the sum of the entries of each vector is $1$ (i.e., for all $i\in \{0,\ldots,k\}$, it holds that $\sum\_j u\_{i,j}=1$).


Is it always true that either there exists a vector $v\in R^{n}$, such that $\sum\_j (u\_{0,j}-u\_{i,j})\cdot v\_j > 0$ for all $i=1,\ldots,k$, or there exist $\alpha\_1,\ldots,\alpha\_k\geq 0$, that sum to $1$ (i.e., $\sum\_i \alpha\_i = 1$), such that $u\_0 = \sum\_{i=1}^k \alpha\_i \cdot u\_i$.



Non-convex combination of vectors
Yes, either point $u\_0$ belongs to the convex hull $H$ of the points $u\_1,\dots,u\_k$ (in which case $\alpha\_i$ exist), or there exists a hyperplane separating $u\_0$ and $H$, when we can take $v$ as a normal vector of this hyperplane "pointing" towards $u\_0$.


