An equivalent condition of 0 is in the convex hull of a set of vectors Let $V=\{v_1,...,v_k\}$ be a set of vectors in $\mathbb{R}^d$. And for each $v_i$ we define the half-space $$H(v_i):=\{u\in \mathbb{R}^d:u\cdot v_i\le 0\}.$$
The convex hull of $V$ is defined as $conv(V):=\{u=\sum_{i=1}^k \alpha_iv_i: \alpha_i\ge 0, \sum_{i=1}^k \alpha_i=1\}$.
Edited: We can assume that $0$ is not in the convex hull of any proper subset of $V$ and $v_2-v_1,...,v_k-v_1$ spans $\mathbb{R}^d$ and are linearly independent.
One claims that $0\in conv(V)$ if and only if $\cap_{i=1}^kH(v_i)=\{0\}$. And I am trying to prove this claim.
For the $\Rightarrow$ direction, assume $u\in \cap_{i=1}^kH(v_i)$. As 0 is in the convex hull, then $0=\sum_{i=1}^k\alpha_iv_i$. And then
$$0=0\cdot u =\sum_{i=1}^k\alpha_i (v_i\cdot u),$$
which implies $u\cdot v_i=0$ that for all $i$ (as we assume $\alpha_i>0$). Therefore $u\cdot(v_i-v_1)=0$ for all $i$, which implies $u=0$.
I heard separation theorem is involved in the proof of the other direction. But I am not sure.
 A: Proof of the reverse direction.
According to Farkas' lemma (https://en.wikipedia.org/wiki/Farkas%27_lemma):
$\forall A \in \mathbb{R}^{k \times d}$ and $\forall b \in \mathbb{R}^k$, exactly one of the following is true:

*

*$\exists x \in \mathbb{R}^d, A x = b, x \ge 0$

*$\exists y \in \mathbb{R}^k, A^T y \ge 0, b^Ty \lt 0$
Let's pose $A^T = - (v_1, v_2, ..., v_k)$: the matrix of $v_i$ in columns, with a minus sign. $-A$ is the matrix of $v_i$ in lines.
Then $\forall b \in \mathbb{R}^k$, the second alternative is never possible, as either $y = 0$ and $b^Ty \lt 0$ is false, or $y \ne 0$ and $A^T y \ge 0$ is not possible because $\cap_{i=1}^k H(v_i) = \{0\}$.
So $\forall b \in \mathbb{R}^k, \exists x \in \mathbb{R}^d, A x = b, x \ge 0$. As this is also true for $-b$, we know $\exists x \in \mathbb{R}^d, (-A) x = b, x \ge 0$.
Then let's take any non-null vector $b$, and $x_1$ such as $(-A) x_1 = b, x_1 \ge 0$. We can do the same for $-b$, with $x_2$. Then for $x = x_1 + x_2$, $(-A)x = 0$ and $x \ge 0$.
We now scale $x$ so that the sum of its coordinates $= 1$. This means $0$ is in the convex hull of $V$.
