The unit cube is the convex combination of its vertices

Question

I want to prove this statement:

$[0,1]^d = conv(v_1,...,v_n)$, with $n = 2^d$ and $\{v_1,...,v_n\} = \{0,1\}^d$.

Geometrically this is evident, I'm looking for a pure algebraic proof.

Answer 1

All right. In two parts:

First, that all convex combinations are contained in the cube. For each $i$, the $i$th coordinate $x_i$ is a convex combination of the $i$th coordinates of the vertices $v_j$, each of which is $0$ or $1$. Such a convex combination must lie in $[0,1]$. Repeat over all $i$, and the point lies in the unit cube $\{x: x_i\in [0,1]\text{ for all }i\}$.

Second, that all points in the cube are convex combinations. For this, we just find a convex combination explicitly. Index the vertices with subsets of $\{1,2,\dots,d\}$; the vertex $v_S$ has $x_i=1$ if $i\in S$ and $x_i=0$ otherwise.
Now, to reach a point $u_i$ with coordinates $u_i$, we give it coefficients as follows: $$u = \sum_{S} \left(\prod_{i\in S}u_i\right)\left(\prod_{i\not\in S}1-u_i\right)v_S$$ Now, in order to extract a particular coordinate $u_k$, sort the $v_S$ by their $k$th coordinate. That coordinate is $1$ if $k\in S$ and $0$ otherwise, so the nonzero terms always give us $u_k$ rather than $1-u_k$ in that product. Every other term in the product freely varies between the two possibilities, covering all $2^{d-1}$ options - and the sum of those products is therefore $1\cdot 1\cdots 1\cdot u_k\cdot 1\cdots = u_k$. Done.