# The convex hull of the vertices of an hypercube contain the hypercube

The title is clear, I want to proof that the convex hull of the vertices of an hypercube contain the hypercube itself. Geometrically, it is pretty obvious, but I can't come up with a rigorous proof.

I tried inducting on the dimension of $$\mathbb R^n$$ but this doesn't seem to work nicely. A manual approach would be better I think but for dimension higher than 3 it is really confusing for me to construct a specific linear combination.

Induction on dimension works just fine. The point $$(x_1, \ldots, x_n) \in [0,1]^n$$ is a convex combination of the points $$(x_1, \ldots, x_{n-1}, 0)$$ and $$(x_1, \ldots, x_{n-1}, 1)$$, which lie on the top and bottom faces of the hypercube and therefore by induction in the convex hull of the top and bottom vertices respectively.

The resulting convex combination is explicitly given by

$$x = \sum_{p \in \{0,1\}^n}\left(\prod_{i=1}^n(1-|x_i-p_i|)\right)p~.$$

• Thanks for answering, could you please elaborate how you got that convex combination?I think I'm missing something Aug 1 at 0:28
• The induction step is $(x_1, \ldots, x_n) = (1-x_n)(x_1, \ldots, x_{n-1}, 0) + x_n(x_1, \ldots, x_{n-1}, 1)$, iterating the induction yields more terms and more factors. Gathering up all those $x_i$ and $(1-x_i)$ factors can be written as a product like I did above. Aug 1 at 0:42

You can show that the product of convex hulls equals the convex hull of products $$\prod_{i=1}^n \operatorname{co}(A_i) = \operatorname{co}(\prod_{i=1}^n A_i)$$

Show that for product of $$2$$ sets, then use induction. Let's show $$\operatorname{co}(A)\times \operatorname{co}(B) = \operatorname{co}(A\times B)$$

Clearly, $$\operatorname{co}(A)\times \operatorname{co}(B)$$ is convex, and contains $$A\times B$$, so it contains $$\operatorname{co}(A\times B)$$. So only the inclusion $$\operatorname{co}(A)\times \operatorname{co}(B) \subset \operatorname{co}(A\times B)$$ left to prove. Now, let $$x= \sum_{a\in A} \lambda_a\, a \in \operatorname{co}(A)$$, $$y = \sum_{a\in A} \mu_b\, b \in \operatorname{co}(B)$$. Now we check that $$(x,y) = \sum_{(a,b)\in A\times B} \lambda_a \cdot \mu_b\, (a,b)$$

$$\bf{Added:}$$ The above solution provides a writing involving $$2^n$$ terms. But we can provide a writing with $$n+1$$ terms. For let $$x=(x_1, \ldots, x_n)$$ in the unit $$n$$ dimensional cube. Assume that $$x_1\le x_2 \le \ldots x_n$$ Then $$x_k = \lambda_1 + \cdot + \lambda_k$$, where $$\lambda_i \in [0,1]$$, $$i=1, n$$. Then we have $$(x_1, \ldots, x_n) = (\lambda_1, \lambda_1 + \lambda_2, \ldots, \lambda_1 + \cdots \lambda_n) = \\ = \lambda_1 (1,\ldots, 1) + \lambda_2 (0,1,\ldots, 1) + \cdots \lambda_n(0,\ldots, 0,1) + (1-\lambda_n)(0,0,\ldots,0)$$