4
$\begingroup$

To state the question and fix conventions I will introduce some notation from e.g. (Lin-Trudinger, Bull. Aust. Math. Soc. 1994, ``On some inequalities for elementary symmetric functions")

Given $\lambda_1, \dots, \lambda_n$ let

\begin{align*} \sigma_k(\lambda) = \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} \lambda_{i_1} \lambda_{i_2} \dots \lambda_{i_k} \end{align*} We will always consider this function as restricted to the ``positive cone" \begin{align*} \Gamma_k = \left\{ \lambda \in \mathbb R^n\ |\ \sigma_j(\lambda) > 0 \mbox{ for all } 1 \leq j \leq k \right\} \end{align*}

Furthermore, let \begin{align*} \sigma_{k;i}(\lambda) = \sigma_k(\lambda)_{| \lambda_{i} = 0}, \end{align*} i.e. the function $\sigma_k(\lambda)$ after replacing $\lambda_i = 0$. To see some familiar facts with this notation, note the Newton inequalities \begin{align*} \sigma_k(\lambda) \sigma_{k-2}(\lambda) \leq \frac{(k-1)(n-k+1)}{k(n-k+2)} [\sigma_{k-1}(\lambda)]^2, \end{align*} as well as the Maclaurin inequality for $\lambda \in \Gamma_k$, $k \geq l \geq 1$. \begin{align*} \left[ \frac{1}{n \choose k} \sigma_k(\lambda)\right]^{\frac{1}{k}} \leq \left[ \frac{1}{n \choose l} \sigma_l(\lambda)\right]^{\frac{1}{l}} \end{align*} One can observe the further elementary identities \begin{align*} \sigma_k(\lambda) =&\ \sigma_{k;i}(\lambda) + \lambda_i \sigma_{k-1;i}(\lambda),\\ \sum_{i=1}^n \sigma_{k;i}(\lambda) =&\ (n-k) \sigma_k(\lambda). \end{align*}

Now to state the actual question. Fix $n = 2k$. Given $\lambda = (\lambda_1, \dots, \lambda_n)$, let \begin{align*} A_{\lambda} =&\ \frac{1}{n-2} \left[ \lambda - \frac{\sigma_1(\lambda)}{2(n-1)} (1,\dots,1) \right] \end{align*} I now assume I have a vector $\lambda$ such that $A_{\lambda} \in \Gamma_k$ (recall $k = \frac{n}{2}$). The inequality I seek is that for all $1 \leq i \leq n$, \begin{align*} (n-1) \sigma_k(A_{\lambda}) \leq \lambda_i \sigma_{k-1;i}(A_{\lambda}). \end{align*} A few remarks. First, if helpful it can be expressed purely in terms of the vector $A_{\lambda}$, by using the formula $\lambda = (n-2) A_{\lambda} + \sigma_1(A_{\lambda}) (1,\dots,1)$. Second, it is known that the condition $A_{\lambda} \in \Gamma_k$ implies that $\lambda_i \geq 0$ for all $i$. Third, I have added the tag, ``differential geometry'' only because this question originally comes from a geometric problem, where $\lambda$ represents eigenvalues of the Ricci tensor and $A_{\lambda}$ is the Schouten tensor. Lastly, it is a fairly straightforward matter to verify that the inequality is true for $n=2,4$. Any help is greatly appreciated!

$\endgroup$
  • $\begingroup$ In other words, if, say, $i=n$ and $A_{\lambda}=(x_1,\dots,x_n)$ we have to prove $(n-1)\sigma_k(x_1,\dots,x_{n-1})\leqslant \sigma_1(x_1,\dots,x_{n-1})\sigma_{k-1}(x_1,\dots,x_{n-1})$, this is standard if $(x_1,\dots,x_{n-1})\in \Gamma_k$, but the problem is that we are not given this. $\endgroup$ – Fedor Petrov Mar 10 '16 at 8:26
  • 1
    $\begingroup$ Is there a $k$ that should be an $l$ in Maclaurin's inequality? $\endgroup$ – Brendan McKay Mar 10 '16 at 12:22
5
$\begingroup$

We use the following trick, which is standard in such questions. For numbers $x_1,\dots,x_n$ consider the polynomial $F(t)=\prod_{i=1}^n (t-x_i)$, let $F'(t)=n\prod_{i=1}^{n-1}(t-y_i)$ be its derivative ($y$'s are real provided that $x$'s are --- by Rolle theorem). Note that normalized elementary symmetric functions do not change their values if we replace $x$'s to $y$'s: if we denote $${\tilde \sigma}_m(x_1,\dots,x_n)=\frac1{\binom{n}{m}}\sigma_m(x_1,\dots,x_n),$$ we have $${\tilde \sigma}_m(x_1,\dots,x_n)={\tilde \sigma_m}(y_1,\dots,y_{n-1})$$ for all $m=1,2,\dots,n-1$. This immediately follows from taking derivative and applying Vieta's theorem. In proving inequalities for sigmas it is helpful to differentiate polynomial $f$ several times.

Denote $A_{\lambda}=(x_1,\dots,x_n)$ and suppose that $i=n$. Then expressing $\lambda_n=(n-1)x_n+\sigma_1(x_1,\dots,x_{n-1})$ and substituting 'elementary identity' $\sigma_k(A_{\lambda})=x_n\sigma_{k-1;n}(A_{\lambda})+\sigma_k(x_1,\dots,x_{n-1})$ we get equivalent form of your inequality: $$ (n-1)\sigma_k(x_1,\dots,x_{n-1})\leqslant \sigma_1(x_1,\dots,x_{n-1})\sigma_{k-1}(x_1,\dots,x_{n-1}), $$ which may be rewritten as $$ {\tilde \sigma}_k(x_1,\dots,x_{n-1})\leqslant {\tilde \sigma}_1(x_1,\dots,x_{n-1})\cdot {\tilde \sigma}_{k-1}(x_1,\dots,x_{n-1}). $$

Denote $f(t)=(t-x_1)(t-x_2)\dots (t-x_{n-1})$, $g(t)=f^{(k-1)}(t)$, so $g$ is a polynomial of degree $k$ with $k$ real roots $u_1,\dots,u_k$. We have to prove $$k^2u_1\dots u_k\leqslant (u_1+\dots+u_k)\sigma_{k-1}(u_1,\dots,u_k).\,\,\,\,(1)$$ (1) is obvious if all $u_i$ are non-negative, so suppose that $u_k<0$.

We are given that $f(t)(t-x_n)=t^{n}-c_1t^{n-1}+c_2t^{n-2}+\dots+(-1)^k c_k t^{n-k}+\dots$ for positive $c_1,\dots,c_k$, and $f(t)$ has only real roots. Taking $k$-th derivative we see by Rolle theorem that $(f(t)(t-x_n))^{(k)}$ has only real roots and they must be positive. So, $(t-x_n)g'(t)+kg(t)$ has $k$ positive roots. Thus $(g(t)\cdot (t-x_n)^k)'$ has a root $x_n$ of multiplicity $k-1$ and $k$ positive roots.

If $x_n\leqslant 0$, then by Rolle theorem polynomial $(g(t)\cdot (t-x_n)^k)'$ has a root between $x_n$ and $u_k$ (or has roots $x_n$ of multiplicity at least $k$ if $x_n=u_k$.) It contradicts to above observations. Analogously, if $x_n\geqslant 0$ but $u_i\leqslant 0$ for some $i\ne k$, we get a negative root of $(g(t)\cdot (t-x_n)^k)'$, that again contradicts to above.

So, $x_n>0$ and $u_1,\dots,u_{k-1}>0$. Since $W(t):=kg(t)+(t-x_n)g'(t)$ has $k$ positive roots and positive leading coefficient, we have $W(0)(-1)^{k}>0$. On the other hand, $g(0)(-1)^k=u_1\dots u_k<0$. Thus $(-1)^{k-1}x_ng'(0)>0$, i.e. $\sigma_{k-1}(u_1,\dots,u_k)>0$. It easily implies that $u_1+\dots+u_k>0$, thus $(u_1+\dots+u_k)\sigma_{k-1}(u_1,\dots,u_k)>0>k^2u_1\dots u_k$ as desired.

$\endgroup$
  • $\begingroup$ It is nothing but taking derivative and applying Vieta. But very useful! $\endgroup$ – Fedor Petrov Mar 10 '16 at 18:46
  • $\begingroup$ no, I do not care on how lambdas are ordered, that is why $i=n$ is generic case. $\endgroup$ – Fedor Petrov Mar 11 '16 at 4:46
  • $\begingroup$ We express everything via $A_\lambda$, as you suggest in the end of your post. $\lambda_n=(n-1)x_n+x_1+\dots+x_{n-1}$, substitute this and use your first 'elementary identity' $\endgroup$ – Fedor Petrov Mar 11 '16 at 5:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.