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My effort:

$\operatorname{conv}({\{ \pm e_1,...,\pm e_d\}}) \subseteq B_1(0,1)$:

Suppose $x \in \operatorname{conv}({ \pm e_1,\dots,\pm e_d})$ then from Carathéodory's lemma we have that $x$ can be written as a convex combination of $d+1$ elements of $ {\{ \pm e_1,...,\pm e_d\}}$, from these at least one will have a negative sign. Let $k$ be the elements with postive sign and $m$ the ones with negative, such that $k+m=d+1$. We have that there exist $a_1,...,a_k,b_1,...,b_m \geq0$ such that $\sum_{i=1}^{k} a_i +\sum_{j=1}^{m} b_j=1$ and $x=\sum_{i=1}^{k} a_ie_i +\sum_{j=1}^{m} b_j(-e_j)=1$, and for every $x_n, n\in[d]$ we know that $x_n=a_n-b_n$, which means $|x_n|\leq a_n+b_n$. So, $$\|\mathbf{x}\|_1=\sum_{j=1}^{d} |x_j|\leq \sum_{j=1}^{d}a_j+b_j=1$$ So $x\in B_1(0,1)$.

Conversely, let $x=(x_i)$, with $\sum_{i=1}^{d} |x_i|\le 1$ then $$x= \sum_{i=1}^d |x_i|\operatorname{sign}(x_i) e_i + (1-\sum |x_i|)(e_j-e_j)\in \operatorname{conv}({\{ \pm e_1,...,\pm e_d\}})$$

Does this make sense? I know that there's probably something messed up with the indexes in the first part but I'm really confused about how to say my argument and I'm not even sure it holds.

Rodrigo de Azevedo
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NoetherBoy
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1 Answers1

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Yeah, your indexing has several problems. Though I'm not sure why you use Carathéodory's lemma, the same thing works otherwise: take $x$ in the convex hull, then there are $a_i$s and $b_i$s such that $$x=\sum_{i=1}^d(a_i-b_i)e_i,\ \sum_{i=1}^d(a_i+b_i)=1, a_i,b_i\ge0$$ Now, both $a_i-b_i$ and $b_i-a_i$ are smaller than $a_i+b_i$ so $$\|x\|_1=\sum_{i=1}^d|a_i-b_i|\le\sum_{i=1}^n(a_i+b_i)=1$$ Hope this helps. :>

ultralegend5385
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