Pome

Lemma 8.2.3F86594126C69

Lemma 8.2.3.F86594126C69 For $\lambda \in \Gamma _ m$, put

\[ e_ m=\frac{\sigma _ m(\lambda )}{m\sigma _{m-1}(\lambda )}. \]

Then $\lambda -t\mathbf1\in \Gamma _ m$ for $0\leq t<e_ m$. If $m\geq 2$, the conclusion also holds for $t=e_ m$. For $m=1$, the endpoint lies on the boundary.

Proof. The polynomial

\[ P(x)=\sigma _ m(\lambda +x\mathbf1) =\sum _{j=0}^ m\binom mj\sigma _ j(\lambda )x^{m-j} \]

is a positive multiple of the $(n-m)$th derivative of $\prod _ i(x+\lambda _ i)$. Hence all its roots are real, by Rolle's theorem. Its coefficients are positive, so all its roots are strictly negative. Write $P(x)=\prod _{i=1}^ m(x+r_ i)$ with $r_ i>0$. Logarithmic differentiation at zero gives

\[ \sum _{i=1}^ m r_ i^{-1} =\frac{m\sigma _{m-1}(\lambda )}{\sigma _ m(\lambda )} =e_ m^{-1}. \]

Thus $\min _ i r_ i\geq e_ m$, with strict inequality if $m\geq 2$. For $x> -\min _ i r_ i$, every derivative of $P$ is positive: differentiating its product expansion produces sums of products of positive factors. Since

\[ P^{(m-j)}(x)=\frac{m!}{j!}\sigma _ j(\lambda +x\mathbf1), \qquad 1\leq j\leq m, \]

all the required symmetric polynomials are positive at $x=-t$. When $m=1$, the sole root is $-e_1$. $\square$

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