Show that convex hull can be represented as an intersection of half-spaces Yesterday I've stuck with the following problem:
Suppose that $a_0, a_1, ..., a_n$ are affine independent points of $n$-dimensional affine space.
$H_i$ is the hyperplane that passes through all these points except for $i$ ($i=0, 1, \dots, n$).
$H_i^+$ is the half-space bounded by $H_i$ that contains $a_i$. Show that
$$
\text{conv} \{a_0, a_1, \dots, a_n\} = \bigcap_{i=0}^n H_i^+
$$
My attempt(s):
Firstly let's note that $H^+:=\bigcap_{i=0}^n H_i^+$ is the convex set contains all points $a_0, a_1, ..., a_n$ (as an intersection of such sets).
Then I see two opportunities that can bring us to solution:

*

*Let S be an arbitrary convex set that contains $a_0, a_1, ..., a_n$. We need to show that $H^+ \subseteq S$

*Show that $H^+=\{ \sum_{i=0}^n \mu_i a_i: \sum_{i=0}^n \mu_i=1 \space \text{and} \space \mu_i \ge0, \space i=0, 1, \dots, n \}$
If it helps, I'll describe my thoughts on both of these ways. But actually I didn't solve the problem.
Any ideas?
HINT from the TextBook: consider the barycentric coordinate system associated with $a_0, a_1, \dots, a_n$
 A: Let $\;g_i=0\;$ be an equation of the hyperplane $H_i\;$ $(i=0,1,\ldots,n)$, that is
$$ g_i : E \to \Bbb R \quad \text{an affine function with}\quad H_i=g_i^{-1}(0), $$
where $\;E\;$ is the affine (real) space in which we are working. The half-space $\;H_i^+\;$ is described by the inequality $\;g_i\geq0\;$ or $\;g_i\leq0$, and we can assume, for simplicity, that it is of the form $\;g_i\geq0\;$, changing the sign of $\;g_i\;$ if necessary. We have, for each $\;i=0,1,\ldots,n$:
$$ g_i(a_i)>0 \qquad \text{ and }\qquad g_i(a_j)=0 \text{$\;\;$ for$\;\;$}j\neq i. \tag{1}$$
Note that it cannot be $\;g_i(a_i)=0\;$ because otherwise it would be $\;a_i\in H_i$.
Now let $\;x\in H^+:=\displaystyle \bigcap_{i=0}^nH_i^+$. This implies that $\;g_i(x)\geq0, \forall i$. With respect to the barycentric coordinate system associated with $(a_0,a_1,\ldots,a_n)$, we can write
$$ x=\sum_{j=0}^n \lambda_ja_j\qquad\text{with}\quad \sum_{j=0}^n\lambda_j=1 $$
and therefore we only need to prove that all the $\lambda's\;$ are $\;\geq0\;$. This follows from $(1)$:
$$ g_i(x) = g_i\big(\sum_{j=0}^{n}\lambda_ja_j\big) = \sum_{j=0}^{n}\lambda_jg_i(a_j) = \lambda_ig_i(a_i) $$
and the conclusion follows from $\;g_i(x)\geq0\;$ and $\;g_i(a_i)>0$.
