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Let $1\le k\le n$ be given integers. Define the following cone $$\Gamma_k=\{\lambda\in\mathbb{R}^n| S_j(\lambda)>0, j=1, ..., k\},$$ where $S_j(\lambda)$ is the $j$th elementary symmetric function of $\lambda$.

I have some questions about the cone, which are probably known. Any pointers to literature or sharing some ideas of proof is appreciated.

Question 1. How to show that $\Gamma_k$ is convex in $\mathbb{R}^n$?

Question 2. How to show that if $\lambda=(\lambda_1, \ldots, \lambda_n)\in \Gamma_k$, then $\frac{\partial S_k(\lambda)}{\partial \lambda_j}>0$ for any $j=1, \ldots, n$.

Question 3. In this paper by Lin and Trudinger it is stated that Eq.(7), Eq.(8) are respectively Newton and Maclaurin inequalities, quoted from Ref. [6], but I didn't find them there. Are there any other references where I can find proofs of these inequalities?

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  • $\begingroup$ For question 1, with $n=k=2$ we have $S_2(x,y) = x y$ which is positive in the first and third quadrant and negative in the second and fourth. This is not convex, and the point $(x,y)=(-1,-1)$ also creates difficulties for question 2: $\frac{\partial(xy)}{\partial x} = y < 0$. But perhaps $\Gamma_k$ consists of $2^{k-1}$ connected components, each of which might be convex? $\endgroup$
    – Sam Zackrisson
    Commented Oct 1, 2021 at 0:47
  • $\begingroup$ @SamZackrisson For $(x,y)\in\Gamma_2$ one also needs $S_1(x,y)=x+y>0$. $\endgroup$
    – Victor
    Commented Oct 1, 2021 at 4:05
  • $\begingroup$ @Victor Ah, I completely missed that it is required for all j up to k. Thank you! $\endgroup$
    – Sam Zackrisson
    Commented Oct 1, 2021 at 4:55
  • $\begingroup$ Yes, $\Gamma_k$ is determined by k inequalities. $\endgroup$
    – Steve
    Commented Oct 1, 2021 at 19:14
  • $\begingroup$ I can see that (2) implies (1). Indeed, (2) applied iteratively implies that the projection of $\Gamma_k$ on any coordinate subspace $\lambda_I$, $I\subset \{1,\ldots,k\}$, lies inside $\Gamma_{k-|I|}$. On the other hand, for $\lambda,\mu\in\Gamma_k$, $$ S_k(\lambda+\mu) = \sum_{I\sqcup J=\{1\ldots k\}} S_{|I|}(\lambda_I)S_{|J|} (\mu_J) $$ and is therefore positive being a sum of products of positive numbers. This proves (2) => (1). Maybe (3) implies (2)? $\endgroup$
    – Victor
    Commented Oct 1, 2021 at 22:41

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