Let $\mathbf{X}$ be a $n\times m$ $(n>m)$ matrix with $x_{ij}\geq 0$ (not all $x_{ij}$ can be zero). Let $% \mathbf{\Lambda }=\left\{ \lambda \left\vert \lambda \in \mathbb{R}% _{+}^{n},\lambda _{i}\geq 0,\sum \lambda _{i}=1\right. \right\} .$ Consider the set $\mathbf{Y}$ of convex combinations of the rows of $\mathbf{X,}$ $% \mathbf{Y=}\left\{ y\left\vert y=\lambda ^{\top }\mathbf{X,}\lambda \mathbf{% \in \Lambda }\right. \right\} .$ This is a closed and convex set, and is equal to its closure, $\overline{\mathbf{Y}}$. Let $\overline{\mathbf{Y}}^{C} $ denote the closure of the complement of $\mathbf{Y.}$
Although I haven't done it, I guess that it should not be too difficult to prove that for every $\overline{y}\in \overline{\mathbf{Y}}\cap \overline{\mathbf{Y}}% ^{C}$ element of the boundary of $\mathbf{Y,}$ there exists $% \overline{\lambda }\in \mathbf{\Lambda }$ such that $\overline{\lambda }% ^{\top }\mathbf{X=}\overline{y}$ and $\overline{\lambda }_{i}=0$ for $n-m$ components.
What I'm really interested in is the following:
Let $\alpha \in \mathbb{R}_{++}^{n}$ with $\alpha _{i}>1,$ and, abusing notation, let $\lambda ^{\alpha }=\left( \lambda _{1}^{\alpha _{1}},\lambda _{2}^{\alpha _{2}},...,\lambda _{n}^{\alpha _{n}}\right) $ for some $\lambda \in \mathbf{\Lambda .}$ If we now think of $\mathbf{Y}^{\prime }\mathbf{=}% \left\{ y^{\prime }\left\vert y^{\prime }=\left( \lambda ^{\alpha }\right) ^{\top }\mathbf{X,}\lambda \mathbf{\in \Lambda }\right. \right\} ,$ the previous claim is still true for some elements of the boundary (in $\mathbb{R}% ^{2}$, those in the southwestern part of the boundary, so to speak). The intuition comes from some "symmetry" argument: When $n=2$ and $\alpha$ is such that $\alpha _{i}<1$ for all $i$, it is easy to prove that elements in the southwestern boundary of the set are the result of $\lambda \gg 0.$ In fact, it is easy to prove this for any $n$, but it is easiest to think of the "southwestern" boundary in $\mathbb{R}% ^{2}.$