New concavity and convexity results for symmetric polynomials and their ratios

TL;DR

This paper extends classical concavity and convexity inequalities to symmetric polynomials, providing new power generalizations and implications for matrix analysis.

Contribution

It introduces novel power-based concavity and convexity inequalities for elementary and complete homogeneous symmetric polynomials, expanding existing mathematical frameworks.

Findings

New power generalizations of concavity inequalities for elementary symmetric polynomials

Convexity inequalities for complete homogeneous symmetric polynomials

A concavity theorem implying a known log-convexity result for positive definite matrices

Abstract

We prove some "power" generalizations of Marcus-Lopes-style (including McLeod and Bullen) concavity inequalities for elementary symmetric polynomials, and convexity inequalities (of McLeod and Baston) for complete homogeneous symmetric polynomials. Finally, we present sundry concavity results for elementary symmetric polynomials, of which the main result is a concavity theorem that among other implies a well-known log-convexity result of Muir (1972/74) for positive definite matrices.

Equations

$$[e_k((\mathbf{x}+\mathbf{y}{)}^p){]}^{\text{nicefrac}1pk}$$

$$[h_k((\mathbf{x}+\mathbf{y}{)}^{1/p}){]}^{\text{nicefrac}pk}$$

$$\displaystyle\Bigl{[}\frac{e_{k}((\bm{x}+\bm{y})^{p})}{e_{k-l}((\bm{x}+\bm{y})^{p})}\Bigr{]}^{\frac{1}{lp}}$$

$$\displaystyle\Bigl{[}\frac{h_{k}((\bm{x}+\bm{y})^{1/p})}{h_{1}((\bm{x}+\bm{y})^{1/p})}\Bigr{]}^{\frac{p}{k-1}}$$

$$\frac{e_k(\mathbf{x}+\mathbf{y})}{e_{k-1}(\mathbf{x}+\mathbf{y})}\ge \frac{e_k(\mathbf{x})}{e_{k-1}(\mathbf{x})}+\frac{e_k(\mathbf{y})}{e_{k-1}(\mathbf{y})},1\le k\le n,\mathbf{x},\mathbf{y}\in {\mathbb{R}}_{+}^n.$$

$$\biggl{[}\frac{e_{k}((\bm{x}+\bm{y})^{p})}{e_{k-1}((\bm{x}+\bm{y})^{p})}\biggr{]}^{\nicefrac{{1}}{{p}}}\geq\biggl{[}\frac{e_{k}(\bm{x}^{p})}{e_{k-1}(\bm{x}^{p})}\biggr{]}^{\nicefrac{{1}}{{p}}}+\biggl{[}\frac{e_{k}(\bm{y}^{p})}{e_{k-1}(\bm{y}^{p})}\biggr{]}^{\nicefrac{{1}}{{p}}},$$

$$x:y:=(x^{-1}+y^{-1}{)}^{-1},x,y>0,$$

$$(x_1,\dots ,x_n)\mapsto [x_1^p:x_2^p:\cdots :x_n^p{]}^{1/p}.$$

$$f_1\left(\frac{x_1+x_2}{2}\right)\ge \frac{1}{2}{f}_1(x_1)+\frac{1}{2}{f}_1(x_2),\text{and}{f}_2\left(\frac{y_1+y_2}{2}\right)\ge \frac{1}{2}{f}_2(y_1)+\frac{1}{2}{f}_2(y_2).$$

$$f_1\left(\frac{x_1+x_2}{2}\right){:}_p{f}_2\left(\frac{y_1+y_2}{2}\right)\ge \left(\frac{f_1(x_1)+f_1(x_2)}{2}\right){:}_p\left(\frac{f_2(y_1)+f_2(y_2)}{2}\right)\ge \frac{1}{2}[f_1(x_1){:}_p{f}_2(y_1)]+\frac{1}{2}[f_1(x_2){:}_p{f}_2(y_2)],$$

$${\varphi }_{k,n}(\mathbf{x}):={\left[\frac{e_k({\mathbf{x}}^p)}{e_{k-1}({\mathbf{x}}^p)}\right]}^{\text{nicefrac}1p}$$

$${\varphi }_{1,n}(\mathbf{x})=(x_1^p+x_2^p+\cdots +x_n^p{)}^{1/p},$$

$$\text{Let}{\psi }_{k,n}(\mathbf{x}):=\frac{\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)e_k(\mathbf{x})}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)e_{k-1}(\mathbf{x})},\text{then}{\psi }_{k,n}=\sum _{j=1}^n\frac{1}{n-k+1}{x}_j:\frac{1}{k-1}{\psi }_{k-1,n}({\mathbf{x}}_{[j]}),$$

$$\begin{split}\phi_{k,n}(\bm{x})&=\Bigl{(}\sum\nolimits_{j=1}^{n}a_{k}^{p}x_{j}^{p}:b_{k}^{p}\phi_{k-1,n}^{p}(\bm{x}_{[j]})\Bigr{)}^{\nicefrac{{1}}{{p}}}=\Bigl{(}\sum\nolimits_{j=1}^{n}[a_{k}x_{j}:_{p}b_{k}\phi_{k-1,n}(\bm{x}_{[j]})]^{p}\Bigr{)}^{\nicefrac{{1}}{{p}}}.\end{split}$$

$$g_{j}(\bm{x}):=(a_{k}x_{j}):_{p}\bigl{(}b_{k}\phi_{k-1,n}(\bm{x}_{[j]})\bigr{)}$$

$$\phi_{k,n}(\bm{x})=\bigl{(}\sum\nolimits_{j=1}^{n}g_{j}(\bm{x})^{p}\bigr{)}^{\nicefrac{{1}}{{p}}},$$

$$[e_k((\mathbf{x}+\mathbf{y}{)}^p){]}^{1/pk}$$

$$={\left[{\varphi }_{k,n}(\mathbf{x}+\mathbf{y}){\varphi }_{k-1,n}(\mathbf{x}+\mathbf{y})\cdots {\varphi }_{1,n}(\mathbf{x}+\mathbf{y})\right]}^{1/k}$$

$$\ge {\left[\left({\varphi }_{k,n}(\mathbf{x})+{\varphi }_{k,n}(\mathbf{y})\right)\cdots \left({\varphi }_{1,n}(\mathbf{x})+{\varphi }_{1,n}(\mathbf{y})\right)\right]}^{1/k}$$

$$\ge \prod _{j=1}^k[{\varphi }_{j,n}(\mathbf{x}){]}^{1/k}+\prod _{j=1}^k[{\varphi }_{j,n}(\mathbf{y}){]}^{1/k}$$

$$=[e_k({\mathbf{x}}^p){]}^{1/pk}+[e_k({\mathbf{y}}^p){]}^{1/pk},$$

$${\Phi }_{k,l,n}(\mathbf{x}):={\left[\frac{e_k({\mathbf{x}}^p)}{e_{k-l}({\mathbf{x}}^p)}\right]}^{1/lp},$$

$${\Phi }_{k,l,n}(\mathbf{x})$$

$${\Phi }_{k,l,n}(\mathbf{x}+\mathbf{y})$$

$$\ge {\left[{\prod }_{j=1}^l{\varphi }_{k-j+1,n}(\mathbf{x})\right]}^{1/l}+{\left[{\prod }_{j=1}^l{\varphi }_{k-j+1,n}(\mathbf{y})\right]}^{1/l}$$

$$={\Phi }_{k,l,n}(\mathbf{x})+{\Phi }_{k,l,n}(\mathbf{y}).\text{qed}$$

$$S_n(x_1,\dots ,x_n)=[x_1^p:x_2^p:\cdots :x_n^p{]}^{\frac{1}{p}}.$$

$$\displaystyle S_{n-1}\bigl{(}\tfrac{1}{2}(x_{1}+y_{1}),\ldots,\tfrac{1}{2}(x_{n-1}+y_{n-1})\bigr{)}$$

$$\displaystyle S_{n}\bigl{(}\tfrac{1}{2}(x_{1}+y_{1}),\ldots,\tfrac{1}{2}(x_{n}+y_{n})\bigr{)}$$

$$\displaystyle\geq\bigl{\{}[\tfrac{1}{2}S_{n-1}(x_{1},\ldots,x_{n-1})+\tfrac{1}{2}S_{n-1}(y_{1},\ldots,y_{n-1})]^{p}:(\tfrac{1}{2}x_{n}+\tfrac{1}{2}y_{n})^{p}\bigr{\}}^{1/p}.$$

$${\left[{\left(\frac{S_{n-1}(x_1,\dots ,x_{n-1})+S_{n-1}(y_1,\dots ,y_{n-1})}{2}\right)}^p:{\left(\frac{x_n+y_n}{2}\right)}^p\right]}^{1/p}$$

$$\ge \frac{1}{2}[\{S_{n-1}(x_1,\dots ,x_{n-1}){\}}^p:x_n^p{]}^{1/p}+\frac{1}{2}[\{S_{n-1}(y_1,\dots ,y_{n-1}){\}}^p+y_n^p{]}^{1/p},$$

$$e_k{\left(\frac{x_i+y_i}{2}\right)}^p$$

$$\le H\left({\sum }_{|I|=k}{\prod }_{i\in I}{x}_i^p,{\sum }_{|I|=k}{\prod }_{i\in I}{y}_i^p\right)=H(e_k({\mathbf{x}}^p),e_k({\mathbf{y}}^p)),$$

$$\mathbf{X}\mapsto \frac{1}{e_k(\lambda ({\mathbf{X}}^p))},p\in (-1,0),$$

$$\frac{1}{k!}\mathbb{E}[({\xi }_1{x}_1+\cdots +{\xi }_n{x}_n{)}^k]=\sum _{1\le i_1\le i_2\le \cdots \le i_k\le n}{\prod }_{j=1}^k{x}_{i_j}=h_k(\mathbf{x}).$$

Full text

1 Introduction

The main purpose of this note is to prove the following inequalities:

[TABLE]

for $\bm{x},\bm{y}\in\mathbb{R}_{+}^{n}$, $p\in(0,1)$ and $1\leq l\leq k\leq n$. In these inequalities $e_{k}$ denotes the $k$-th elementary symmetric polynomial and $h_{k}$ denotes the $k$-th complete homogeneous symmetric polynomial, and $\bm{x}^{p}$ we mean the vector $(x_{1}^{p},x_{2}^{p},\ldots,x_{n}^{p})$. Inequalities (1.1a)-(1.1d) offer a “power” generalization to their corresponding counterparts with $p=1$ that have been well-known for a long time, with attributions to Marcus and Lopes [8], McLeod [10], Baston [3], Bullen and Marcus [4], among others—see also the reference book [5].

Outline: We begin by proving (1.1a) and (1.1c) in Section 2. Subsequently, in Section 3.1 we prove additional concavity results for $e_{k}$, notable among which is a new concavity theorem that implies the log-convexity result of Muir [11] as a corollary. Finally, in Section 3.2 we prove inequalities (1.1b) and (1.1d).

2 Generalized Marcus-Lopes-style inequalities

Let $\bm{x}\in\mathbb{R}_{+}^{n}$ and $e_{k}(\bm{x})=\sum_{S\subseteq[n],|S|=k}\prod_{i\in S}x_{i}$ denote the $k$-th elementary symmetric polynomial; we set $e_{0}=1$. A well-known and important concavity result for $e_{k}$ is the Marcus-Lopes inequality [8]:

[TABLE]

Inequality (2.1) is used by Marcus and Lopes [8] to prove the concavity of $e_{k}(\bm{x})^{1/k}$. We establish below the following concavity inequality:

[TABLE]

Using (2.2) as a key step, we will first establish (1.1a) and then extend it to prove inequality (1.1c).

An important building block of our proofs is provided by the parallel sum operation111The parallel sum operation is also defined for positive definite matrices, where we set $\bm{A}:\bm{B}:=(\bm{A}^{-1}+\bm{B}^{-})^{-1}$.

[TABLE]

Lemma 1 proves a key joint concavity property of (2.3), as well as its “power” generalization.

Lemma 1**.**

The parallel sum $:$ defined by (2.3) is jointly concave on $\mathbb{R}_{+}^{2}$. Moreover, for every $p\geq-1$, the ‘$p$-parallel sum*’ $x:_{p}y:=[x^{p}:y^{p}]^{1/p}$ is jointly concave. Also, both $:$ and $:_{p}$ are monotonic in both arguments.*

Proof.

The Hessian equals (p+1)\left(\begin{array}[]{cc}-x^{p-1}y^{p+1}\left(x^{p}+y^{p}\right)^{-2-\frac{1}{p}}&x^{p}y^{p}\left(x^{p}+y^{p}\right)^{-2-\frac{1}{p}}\\ x^{p}y^{p}\left(x^{p}+y^{p}\right)^{-2-\frac{1}{p}}&-x^{p+1}y^{p-1}\left(x^{p}+y^{p}\right)^{-2-\frac{1}{p}}\\ \end{array}\right), which is clearly negative definite for $p\geq-1$. Monotonicity is clear from first derivatives. ∎

Observe that $x:y=y:x$ and $(x:y):z=x:(y:z)$, thus we extend the above notation and simply write $x_{1}:x_{2}:\cdots:x_{n}\equiv x_{1}:[x_{2}:[\cdots:(x_{n-1}:x_{n})]]$. However, the operation $:_{p}$ is not associative if generalized the same way. Thus, a more preferable multivariate generalization is the following:

[TABLE]

Later in Section 3 we will study joint concavity of (2.4) and with it some new inequalities for $e_{k}$. The rest of this section is devoted to proving inequality (2.2).

Lemma 2**.**

Let $f_{1}:\mathbb{R}^{m_{1}}\to\mathbb{R}_{++}$ and $f_{2}:\mathbb{R}^{m_{2}}\to\mathbb{R}_{++}$ be continuous concave functions. Then, $f_{1}(x):_{p}f_{2}(y)$ is jointly concave on $\mathbb{R}^{m_{1}}\times\mathbb{R}^{m_{2}}$.

Proof.

The proof is immediate from first principles; we include details for completeness. It suffices to establish midpoint concavity. Since $f_{1}$ and $f_{2}$ are concave, we have

[TABLE]

The function $:_{p}$ is monotonically increasing in each of its arguments and is jointly concave, therefore

[TABLE]

which establishes the joint concavity of $f_{1}(x):_{p}f_{2}(y)$. ∎

We are now ready to present our main result.

Theorem 3**.**

Let $1\leq k\leq n$. Then, the function

[TABLE]

is concave for $\bm{x}\in\mathbb{R}_{+}^{n}$ and $p\in(0,1)$.

Proof.

We will prove the claim by induction on $k$. For $k=1$, we have

[TABLE]

which is clearly concave. As the induction hypothesis, assume that $\phi_{k-1,n}$ is concave for some $k>1$. The key step in our proof is the following remarkable observation of Anderson et al. [1]:

[TABLE]

where $\bm{x}_{[j]}$ denotes the vector $\bm{x}$ with $x_{j}$ omitted. Using representation (2.5) we see that for suitable (positive) scaling factors $a_{k}$ and $b_{k}$, we can write

[TABLE]

From induction hypothesis we know that $\phi_{k-1,n}(\cdot)$ is concave; thus, applying Lemma 2 we see that

[TABLE]

is jointly concave in $x_{j}$ and $\bm{x}_{[j]}$ (and thus in $\bm{x}$). Consequently, we can further rewrite (2.6) as

[TABLE]

which is clearly concave as it is the (vector) composition the coordinate-wise increasing concave function $(\sum_{j}x_{j}^{p})^{1/p}$ with the vector \bigl{(}g_{1}(\bm{x}),\ldots,g_{n}(\bm{x})\bigr{)}, where each $g_{j}(\bm{x})$ is itself concave. ∎

Following an idea of Marcus and Lopes [8] (who proved concavity of $e_{k}(\bm{x})^{1/k}$), we can now prove (1.1a), that is, the concavity of $[e_{k}(\bm{x}^{p})]^{1/p}$, by leveraging the ratio-concavity proved in Theorem 3.

Theorem 4**.**

The function $\bm{x}\mapsto[e_{k}(\bm{x}^{p})]^{1/pk}$ is concave for $p\in(0,1)$ and $\bm{x}\in\mathbb{R}_{+}^{n}$.

Proof.

With Theorem 3 in hand the proof is simple. Indeed, we have

[TABLE]

where the first inequality follows from Theorem 3, while the second is Minkowski’s inequality. ∎

In the same vein, we easily obtain a proof to (1.1c) which generalizes Theorem 3 (for $p=1$, this generalization reduces to an inequality of McLeod [10]; also independently of Bullen and Marcus [4]):

Theorem 5**.**

Let $1\leq l\leq k\leq n$, then

[TABLE]

is concave for $\bm{x}\in\mathbb{R}_{+}^{n}$ for $p\in(0,1)$.

Proof.

The argument is straightforward. First, we rewrite $\Phi_{k,l,n}(\bm{x})$ as

[TABLE]

Then, using Theorem 3 and Minkowski’s inequality we have

[TABLE]

3 Additional convexity / concavity results

In this section we provide some additional concavity results for $e_{k}$ before proving the remaining two convexity inequalities (1.1b) and (1.1d) for $h_{k}$.

3.1 Elementary symmetric polynomials

We begin with a simple result about joint concavity of the multivariate $p$-parallel sum.

Proposition 6**.**

The multivariate $p$-parallel sum map $\bm{x}\mapsto[x_{1}^{p}:x_{2}^{p}:\cdots:x_{n}^{p}]^{1/p}$ is concave on $\mathbb{R}_{+}^{n}$ for $p>0$.

Proof.

The proof is by induction on $n$. The case $n=1$ is trivial; the first nontrivial base case is $n=2$, which is established by Lemma 1. Assume thus that the claim holds for $\mathbb{R}_{+}^{k}$ for $k=1,2,\ldots,n-1$. Consider thus,

[TABLE]

From the induction hypothesis, we have that

[TABLE]

Thus, using the monotonicity of the $:_{p}$ operation and (3.1) we get

[TABLE]

Now concavity of $:_{p}$ (Lemma 1) immediately yields

[TABLE]

establishing concavity of $S_{n}$. Thus, by induction the said map is concave. ∎

The following lemma is immediate by observation; we state it for subsequent use.

Lemma 7**.**

Let $f:\mathbb{R}^{n}\to\mathbb{R}$ be a continuous function. Then, $f$ is reciprocally concave, i.e., $1/f$ is concave, iff $f\left(\frac{x+y}{2}\right)\leq H(f(x),f(y))=2(f(x):f(y))$.

Theorem 8**.**

The function $\bm{x}\mapsto e_{k}(\bm{x}^{p})$ is reciprocally concave for $p\in(-1,0)$ and $\bm{x}\in\mathbb{R}_{+}^{n}$.

Proof.

First observe that $(x_{1}\cdots x_{k})^{r}$ is jointly concave for $r\in(0,1)$, and thus $(x_{1}\cdots x_{k})^{p}$ is reciprocally concave for $p\in(-1,0)$. This allows us to conclude

[TABLE]

where the second inequality follows from the joint concavity of the $:$ operation. ∎

An immediate corollary of Theorem 8 is the following reciprocal concavity result for matrices.

Corollary 9**.**

Let $\bm{X}$ be a Hermitian positive definite matrix. Then, the function

[TABLE]

is concave (here $\lambda(\cdot)$ denotes the eigenvalue map, i.e., the vector of eigenvalues of a matrix).

Corollary 9 in turn yields Corollary 10 (a 1974 result of Muir [11]) as well as its recent generalization Corollary 11 by Mariet and Sra [9]. Both of these results easily follow since reciprocal concavity of a positive function implies log-convexity.

Corollary 10** ([11]).**

Let $\bm{A}\in\mathbb{C}^{n\times k}$ ($k\leq n$) have full column rank, and $\bm{X}\in\mathbb{P}_{n}$ (the set of Hermitian positive definite matrices). The function $\bm{X}\mapsto e_{k}((\bm{A}^{*}\bm{XA})^{-1})$ is log-convex.

Corollary 11** ([9]).**

Let $\bm{A}\in\mathbb{C}^{n\times k}$ ($k\leq n$) have full column rank. Let $\bm{X}\in\mathbb{P}_{n}$ and $p\in(-1,0)$. Then, the function $\bm{X}\mapsto e_{k}((\bm{A}^{*}\bm{XA})^{p})$ is log-convex.

3.2 Complete homogeneous symmetric polynomials

In this section, we prove (1.1b) and (1.1d), the power generalization to existing subadditivity inequalities for complete homogeneous symmetric polynomials. We follow a proof technique very different from McLeod [10] and Baston [3], which we believe offers a much simpler proof of these inequalities.

Key to our proof is the following convenient representation for complete homogeneous symmetric polynomials (we learned of this representation from [2]):

Lemma 12**.**

Let $\xi_{1},\ldots,\xi_{n}$ be independent standard exponential random variables. Then, for any integer $k\geq 0$

[TABLE]

Proof.

Since $\mathbb{E}[\xi_{i}^{j}]=j!$, by expanding $(\xi_{1}x_{1}+\cdots+\xi_{n}x_{n})^{k}$ and taking expectations, the result follows. ∎

Representation (3.2) permits an immediate proof of the following convexity result of McLeod [10] (cf. the concavity result for $e_{k}(\bm{x})^{1/k}$, which is much harder).

[TABLE]

We prove below a “fractional power” generalization of (3.3) below; we will need the following lemma.

Lemma 13** (Mixed-Minkowski).**

Let $p\geq 1$ and $k\geq 1$. Let $\bm{x},\bm{y}\in\mathbb{R}_{+}^{n}$. Then,

[TABLE]

Proof.

Consider $a,b,c,d\geq 0$. Since $p\geq 1$, we have $a^{p}+b^{p}\leq(a+b)^{p}$, whereby $(a^{p}+b^{p})^{k}+(c^{p}+d^{p})^{k}\leq(a+b)^{pk}+(c+d)^{pk}$. Using this observation with Minkowski’s (norm) inequality finishes the proof. ∎

Theorem 14**.**

Let $\bm{x},\bm{y}\in\mathbb{R}_{+}^{n}$, $k\geq 1$, and $p\geq 1$. Then,

[TABLE]

Proof.

Given the representation (3.2), the proof is straightforward. Indeed,

[TABLE]

where the inequality follows from Lemma 13. ∎

A slightly more careful argument with representation (3.2) also yields a “power” generalization to the subadditivity result of Baston [3] for the ratio $h_{k}(\bm{x})/h_{1}(\bm{x})$.

Theorem 15**.**

Let $\bm{x},\bm{x}\in\mathbb{R}_{+}^{n}$, $p\geq 1$, and $k\geq 1$ be an integer. Then,

[TABLE]

Proof.

Once again, we rely on the representation (3.2); this yields

[TABLE]

To bound (3.6) by the right hand side of (3.5), it suffices to prove the following inequality:

[TABLE]

To prove (3.7) we use Dresher’s generalization of Minkowski’s inequality [7, 6], which implies that

[TABLE]

Now take $p$-th root on both sides of (3.8) and noting that $p\geq 1$ we obtain

[TABLE]

Since $p\geq 1$, we have $[(a+b)^{p}+(c+d)^{p}]\geq a^{p}+b^{p}+c^{p}+d^{p}$, the left hand side of (3.7) is smaller the left hand side of (3.9); this concludes the proof. ∎