Lemma 8.2.7B0643A9A1173

Proof. First take a single form $\alpha $, diagonal in an $\omega $-unitary frame, with ordered eigenvalues $\lambda \in \Gamma _ m$. For $1\leq r<m$, repeated splitting and deleted-coordinate positivity give $S_ r(\lambda )\geq \lambda _1\cdots \lambda _ r$. The sum of the last $n-r$ eigenvalues is positive, so each negative eigenvalue has modulus at most $(n-r)\lambda _{r+1}$. Thus every product of $r$ distinct eigenvalues has modulus at most

Indeed, replace its negative factors by this bound and its positive factors by the corresponding largest eigenvalues. In the diagonal frame only complementary diagonal coefficients of $\Xi $ contribute to $\alpha ^ r\wedge \Xi $. This proves $|\alpha ^ r\wedge \Xi |\leq C_{n,m,\Xi }\alpha ^ r\wedge \omega ^{n-r}$.

Apply this bound to the polynomial $(\alpha _1+t_2\alpha _2+\cdots +t_ r\alpha _ r)^ r\wedge \Xi $, with $0\leq t_ i\leq 1$. Its absolute value is bounded by the stated right-hand side, by polarized positivity and monotonicity on the cone. A fixed invertible interpolation matrix on this finite-dimensional polynomial space bounds each coefficient by its supremum on the cube. The coefficient of $t_2\cdots t_ r$ is a fixed positive multiple of $\alpha _1\wedge \cdots \wedge \alpha _ r\wedge \Xi $. This proves the mixed bound. The closed-cone case follows by adding $\varepsilon \omega $ to each form and taking the limit. $\square$