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world:choquet [2026/04/10 15:14]
rdouc [Krein-Milman, Choquet and Birkhoff-Von Neuman Theorems]
world:choquet [2026/04/10 15:48] (current)
rdouc [Krein-Milman, Choquet and Birkhoff-Von Neuman Theorems]
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 The proof proceeds by induction on $n$, the dimension of $E$. We assume the property is true for every Euclidean subspace of dimension $n-1$. Without loss of generality, we assume that $K$ is not reduced to a singleton. The proof proceeds by induction on $n$, the dimension of $E$. We assume the property is true for every Euclidean subspace of dimension $n-1$. Without loss of generality, we assume that $K$ is not reduced to a singleton.
  
-Let $x \in K$. Choose $y \in K \setminus \{x\}$ arbitrarily. Consider $[a,b]$ the maximal segment contained in $K$ and containing $[x,y]$. Since $a,b \in K$ and $x$ is a convex combination of $a,b$, it suffices to show that $a,b \in \mathrm{conv}(E)$. We prove that $a \in \mathrm{conv}(E)$ (the case $b \in \mathrm{conv}(E)$ is similar). Since $a \in \partial K$, there exists a sequence $a_k \notin K$ converging to $a$ and for each $k$, we choose $u_k$ unitary such that for all $y \in K$, $\langle y-a_k,u_k \rangle \geq 0$. We can extract from $(u_k)$ a convergent subsequence on the unit sphere, and let $u$ be its limit. Passing to the limit yields $\langle y-a,u \rangle \geq 0$. Thus, defining $f(y)=\langle y-a,u \rangle$, we obtain that $a \in H_a=f^{-1}(0) \cap K=\mathrm{argmin}_{y\in K} f(y)$. The set $H_a$ is convex, compact, non-empty in a space of dimension $n-1$, and the induction hypothesis shows that $a \in \mathrm{conv}(\mathrm{Extr}(H_a))$. It remains to show that $\mathrm{Extr}(H_a) \subset \mathrm{Extr}(K)$. Indeed, since $K$ is convex and $f$ is linear, if $z \in \mathrm{Extr}(H_a)$ ​can be written ​$z=\lambda x + (1-\lambda) y$ with $x,y \in K$, then $f(z)=0=\lambda f(x) + (1-\lambda) f(y)$, hence $f(x)=f(y)=0$,​ which implies $x,y \in H_a$ and thus $\lambda \in \{0,1\}$ since $z \in \mathrm{Extr}(H_a)$.+Let $x \in K$. Choose $y \in K \setminus \{x\}$ arbitrarily. Consider $[a,b]$ the maximal segment contained in $K$ and containing $[x,y]$. Since $a,b \in K$ and $x$ is a convex combination of $a,b$, it suffices to show that $a,b \in \mathrm{conv}(\mathrm{Extr}(K))$. We prove that $a \in \mathrm{conv}(\mathrm{Extr}(K))$ (the case $b \in \mathrm{conv}(\mathrm{Extr}(K))$ is similar). Since $a \in \partial K$, there exists a sequence $a_k \notin K$ converging to $a$ and for each $k$, we choose $u_k$ unitary such that for all $y \in K$, $\langle y-a_k,u_k \rangle \geq 0$. We can extract from $(u_k)$ a convergent subsequence on the unit sphere, and let $u$ be its limit. Passing to the limit yields $\langle y-a,u \rangle \geq 0$. Thus, defining $f(y)=\langle y-a,u \rangle$, we obtain that $a \in H_a=f^{-1}(0) \cap K=\mathrm{argmin}_{y\in K} f(y)$. The set $H_a$ is convex, compact, non-empty in a space of dimension $n-1$, and the induction hypothesis shows that $a \in \mathrm{conv}(\mathrm{Extr}(H_a))$. It remains to show that $\mathrm{Extr}(H_a) \subset \mathrm{Extr}(K)$. Indeed, since  $f$ is linear, if $z \in \mathrm{Extr}(H_a)$ ​and $z=\lambda x + (1-\lambda) y$ with $x,y \in K$, then $f(z)=0=\lambda f(x) + (1-\lambda) f(y)$, hence $f(x)=f(y)=0$ ​(since $f\geq 0$ on $K$), which implies $x,y \in H_a$ and thus $\lambda \in \{0,1\}$ since $z \in \mathrm{Extr}(H_a)$. This shows that $z \in \mathrm{Extr}(K)$ and the proof is concluded
 </​hidden>​ </​hidden>​
  
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 <WRAP center round tip> <WRAP center round tip>
-**Theorem (Birkhoff–Von Neumann).** Let $B_n$ be the set of bistochastic matrices, i.e. defined by+**Theorem (Birkhoff–Von Neumann).** Let $B_n$ be the set of bistochastic matrices ​(of size $n\times n$), i.e. defined by
 $$ $$
-B_n = \left\{ A = (a_{ij}) \in \mathcal{M}_n([0,​1]) \mid \sum_j a_{ij} = 1,\ \sum_i a_{ij} = 1 \right\}.+B_n = \left\{ A = (a_{ij}) \in \mathcal{M}_n([0,​1]) \mid \sum_{j=1}^n ​a_{ij} = 1,\ \sum_{i=1}^n ​a_{ij} = 1 \right\}.
 $$ $$
 Then $\mathrm{Extr}(B_n) = P_n$, the set of permutation matrices (i.e. matrices having exactly one $1$ in each row and each column). Then $\mathrm{Extr}(B_n) = P_n$, the set of permutation matrices (i.e. matrices having exactly one $1$ in each row and each column).
world/choquet.1775826869.txt.gz · Last modified: 2026/04/10 15:14 by rdouc