Segment intersections and the inversion bound

Segment criterion. If $f=g$, the open segment is empty and no hyperplane separates the two points, so the equivalence holds. Otherwise, for a defining nonzero linear functional $a$ of $H$, its value along the segment is $(1-t)a(f)+ta(g)$. This affine function has a zero at exactly one $t\in (0,1)$ precisely when its endpoint values have opposite strict signs: if it is nonconstant, the zero is $a(f)\mathrm{/}(a(f)-a(g))$, and this lies strictly between $0$ and $1$ exactly under that sign condition; if it is constant, it has either no zeros or all points as zeros. Thus the stated intersection condition is equivalent to separation. Convexity also ensures $[f,g]\subseteq \mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}$.

Count all crossings between two specified chamber translates. Suppose $f\in \bar{C}$ and $g\in w\bar{C}$. For a separating hyperplane choose a positive root $\alpha$ defining it, possible by [F1]. By [F2], $f(\alpha )\ge 0$, so the strict opposite-sign condition gives $f(\alpha )>0$ and $g(\alpha )<0$. Write $g=wx$ with $x\in \bar{C}$. By [F3], $g(\alpha )=x(w^{-1}\alpha )$. If $w^{-1}\alpha$ were positive, [F2] would make this value nonnegative. The root-sign dichotomy therefore makes $w^{-1}\alpha$ negative. Hence every separating hyperplane has a representative in the inversion set $\{\alpha \in {\mathrm{\Phi }}^{+}:w^{-1}\alpha \in -{\mathrm{\Phi }}^{+}\}$. The map from that finite root set to its hyperplanes need not be invoked as injective: the number of separating hyperplanes is at most the number of roots in the set, namely $\mathrm{\ell }(w^{-1})$.

Fixed point and arbitrary endpoints. If $h\in (f,g)$ and $H_{\alpha }\cap (f,g)=\{h\}$, then $H_{\alpha }$ cannot contain $[f,g]$, since $f\mathrm{\ne }g$ and the open segment contains more than one point. By 1.1 it separates the endpoints, so 2.1 gives the stated bound. For arbitrary endpoints choose $u,v\in W$ with $f\in u\bar{C}$ and $g\in v\bar{C}$, and put $f^{\mathrm{\prime }}=u^{-1}f$, $g^{\mathrm{\prime }}=u^{-1}g$, and $w=u^{-1}v$. Then $f^{\mathrm{\prime }}\in \bar{C}$ and $g^{\mathrm{\prime }}\in w\bar{C}$. The invertible linear map $u^{-1}$ carries segments to segments and permutes $\mathcal{H}$: by [F3], $u^{-1}{H}_{\alpha }=H_{u^{-1}\alpha }$, and roots are stable under $W$. Consequently 2.1 gives only finitely many separating hyperplanes for $f,g$. Every point in the final displayed set arises from such a hyperplane by 1.1, and each such hyperplane contributes one point; hence that set is finite. This proves all assertions.