A dominant vector minimises its distance to a dominant weight

Since $\eta$ is dominant, the closed chamber is $\bar{\mathcal{C}}=\{x\in E:(x,{\alpha }_i)\ge 0\text{ for all }i\}$, and for $x\in W$ and a simple root ${\alpha }_i$ with $⟨x\eta ,{\alpha }_i^{\vee }⟩<0$ we set $c=-⟨x\eta ,{\alpha }_i^{\vee }⟩>0$, so that $s_{i}x\eta =x\eta +c{\alpha }_i$. Let $d(x)=\mathrm{\#}\{\beta \in {\mathrm{\Phi }}^{+}:⟨x\eta ,{\beta }^{\vee }⟩<0\}$ be the number of positive roots pairing negatively with $x\eta$.

For such $x$ and ${\alpha }_i$ one has $\mid \xi -s_{i}x\eta {\mid }^2-\mid \xi -x\eta {\mid }^2=-2c(\xi ,{\alpha }_i)\le 0$, because $s_{i}x\eta =x\eta +c{\alpha }_i$ and expansion gives $-2c(\xi -x\eta ,{\alpha }_i)+c^2({\alpha }_i,{\alpha }_i)=-2c(\xi ,{\alpha }_i)-c^2({\alpha }_i,{\alpha }_i)+c^2({\alpha }_i,{\alpha }_i)$, while $(\xi ,{\alpha }_i)=⟨\xi ,{\alpha }_i^{\vee }⟩({\alpha }_i,{\alpha }_i)\mathrm{/}2\ge 0$ by dominance of $\xi$. Thus a descent step never increases the distance from $\xi$.

If $⟨x\eta ,{\alpha }_i^{\vee }⟩<0$ then $d(s_{i}x)=d(x)-1$. Indeed, for $\beta \in {\mathrm{\Phi }}^{+}\setminus \{{\alpha }_i\}$ the orthogonality of $s_i$ gives $⟨s_{i}x\eta ,{\beta }^{\vee }⟩=⟨x\eta ,s_i{\beta }^{\vee }⟩$, and by [F2] the map $\beta \mapsto s_i\beta$ is a bijection of ${\mathrm{\Phi }}^{+}\setminus \{{\alpha }_i\}$; the root ${\alpha }_i$ itself pairs negatively with $x\eta$ but, by $s_i{\alpha }_i=-{\alpha }_i$ and $⟨x\eta ,-{\alpha }_i^{\vee }⟩>0$, not with $s_{i}x\eta$. Hence the negative positive roots at $s_{i}x\eta$ are in bijection with the negative positive roots at $x\eta$ other than ${\alpha }_i$, of which there are $d(x)-1$.

Starting from $x_0=w$, iterate: if $x_j\eta$ is not in the closed chamber, choose a simple root ${\alpha }_i$ with $⟨x_j\eta ,{\alpha }_i^{\vee }⟩<0$ and set $x_{j+1}=s_i{x}_j$. By step 2.1 the distances $\mid \xi -x_j\eta \mid$ are nonincreasing, and by step 2.2 the integer $d(x_j)$ drops by one at each step, so the iteration terminates after at most $d(w)$ steps at an element $x_m$ with $x_m\eta \in \bar{\mathcal{C}}$. By [F3] the dominant point of the orbit $W\eta$ is unique, so $x_m\eta =\eta$. Therefore $\mid \xi -\eta \mid =\mid \xi -x_m\eta \mid \le \mid \xi -w\eta \mid$ for every $w\in W$.

Suppose $\mid \xi -w\eta \mid =\mid \xi -\eta \mid$ and run any descent from $w$ as in step 3.1. The values $\mid \xi -x_j\eta \mid$ are nonincreasing and their first and last terms are equal, so every step is an equality, and step 2.1 with $c>0$ gives $(\xi ,{\alpha }_{i_j})=0$, equivalently $s_{i_j}\xi =\xi$, for every reflecting root used. Writing $x_m=s_{i_m}\cdots s_{i_1}w$ and $x_m\eta =\eta$, the element $v:=s_{i_1}\cdots s_{i_m}$ fixes $\xi$, and $w\eta =(s_{i_1}\cdots s_{i_m})\eta =v\eta$; hence $w\eta \in W_{\xi }\eta$.

Conversely, if $w\eta =v\eta$ with $v\xi =\xi$, then the $W$-invariance of the form and the orthogonality of $v$ give $\mid \xi -w\eta \mid =\mid \xi -v\eta \mid =\mid v^{-1}(\xi -v\eta )\mid =\mid v^{-1}\xi -\eta \mid =\mid \xi -\eta \mid$. Together with steps 3.1 and 4.1 this proves the inequality for every $w\in W$ with equality exactly when $w\eta \in W_{\xi }\eta$; if $\xi$ is regular then no root reflection fixes $\xi$, so $W_{\xi }=\{1\}$ and equality forces $w\eta =\eta$.