I need to prove that the convex hull of the finite set: $$S= \{x\in\{0,1\}^d|\left\Vert x \right\Vert_0\leq k\}$$ is the set: $$T=\{x\in\left[ 0,1\right]^d|\left\Vert x \right\Vert_1\leq k\}$$
Where $\left\Vert x \right\Vert_0$ is the L0 "norm" (not really a norm), meaning the number of non-zero elements of $x$, and $k$ is an integer.
I already know how to show that $conv S\subseteq T$, but have trouble with the opposite direction. I tried to describe a vector $x\in T$ as a convex combination of those of $S$ but it seemed a little complicated to me. I would appreciate your assistance.
If $x \in \{0,1\}^d$ then $\|x\|_0 = \|x\|_1$. Hence $S \subset T$ and since $T$ is convex, $\operatorname{co} S \subset T$.
Suppose $x$ is an extreme point of $T$, then $x \in \{0,1\}^d$ (and $\|x\|_1 \le k$, of course). To see this, suppose $x_i \in (0,1)$ for some $i$.
If $\|x\|_1 = k$, then there is some $j \neq i$ such that $x_j \in (0,1)$ and hence $\|x+t (e_i-e_j)\|_1 = k$ for $t$ in some open interval containing zero and $x_i+t,x_j-t \in [0,1]$. Hence $x$ cannot be an extreme point.
If $\|x\|_1 < k$ then $x+t e_i \in T$ for $t$ in some open interval containing zero and hence $x$ cannot be extreme.
Hence if $x$ is extreme, then $x\in\{0,1\}^d$ and hence $\|x\|_0 = \|x\|_1 \le k$ and so $x \in S$.
Since $T$ is the convex hull of its extreme points, we see that $T \subset \operatorname{co} S$.