Existence of convex combination with no coefficient zero

Question

Given a set of points $ P = \{p_0,p_1,...,p_m\} $ where $ p_i \in \mathbb{R}^n $ and $ m > n+1$ denote by $ \mathcal{C}(P)$ the convex hull of $P$ and by int $\mathcal{C}(P)$ the interior of the convex hull of $P$ (that is excluding the boundary).

Let some $ x \in $ int $\mathcal{C}(P)$. Show that there exists a convex combination of the elements of $P$ such that $ x = \sum_{i=1}^{m} \alpha _i p_i$ with no $\alpha_i = 0$.

I understand that in general convex combination means $ \alpha _i \geq 0$ but I want to prove this stricter result. How do I go about doing it?

The idea is this: Suppose there are different convex representations for $x$ where some of the $\alpha _i =0 $ but no particular $\alpha_i$ is zero in all of them. Then a convex combination of the convex representations will give me a new representation where none of the $\alpha_i$ are zero.

Can anyone give me any clues on how to go about showing that there will indeed exist

1. A convex representation (I think applying Farkas' Lemma does this?)
2. In fact many convex representations (Carathéodory's theorem? Since $ m > n + 1$?)
3. A subset of these many convex representations where no particular $\alpha _i$ is zero all the time.

Answer 1

Let $\Sigma = \{ \mu \in [0,1]^m | \sum_i \mu_i = 1 \}$. Note that $\Sigma$ is convex and if $\mu,\mu' \in \Sigma$ are such that $\mu_i >0, \mu'_j >0$ then with $t \in (0,1)$ and $w = t\mu+ (1-t)\mu')$ we see that $w_i>0$ and $w_j>0$.

Let $C= \operatorname{co} \{ p_i \}_{i=1}^m$, and let $x \in C^\circ$. Then there is some $\mu \in \Sigma$ such that $x = \sum_i \mu_i p_i$.

Now choose any $p$, then there is some $\delta>0$ such that if $|t|< \delta$ then $x+t (p-x) \in C^\circ$. If we choose $0 < t < \min(\delta,1)$ then there is some $\mu' \in \Sigma$ such that $x-t(p-x) = \sum_i \mu_i' p_i$ and hence $x = { 1\over 2} (x-t(p-x) + x+t(p-x) ) = { 1\over 2} ( \sum_i (\mu_i' + (1-t) \mu_i)p_i + t p)$.

In particular, the multiplier of $p_i$ is positive if $\mu_i>0$, and the multiplier of $p$ is also positive.

Now choose any $j$ such that $\mu_j = 0$, and let $p = a_j$, and use the above process to get a new multiplier $\mu''$ such that $\mu''_j >0$, and also, if $\mu_i>0$ then $\mu''_i >0$ as well.

We can repeat the above until all $\mu_i >0$.