k-th Elementary Symmetric Function : Definition

Question

My textbook provided following problem set:

Let $e_k = e_k(x_1, x_2, ... , x_n)$ denote the k-th elementary symmetric function in n variables. Show that the sequence $\{e_k(m_1,m_2,...,m_n)\}_{k\ge0}$ is log-concave when $m_i$s are positive integer.

What is the definition of k-th elementary symmetric function?

Answer 1

It's $$e_k=\sum_{1\leq i_1<i_1<...<i_k\leq n}\prod_{j=1}^kx_{i_j}$$

Answer 2

From now on, I will abbreviate $e_k(m_1,\dots,m_n)$ as $e_k$. $$ (x+m_1)(x+m_2)\cdots (x+m_n)=\sum_{k=0}^n e_{n-k}x^k $$ Since polynomial multiplication corresponds to convolution of coefficient list, this shows the sequence $[e_n,e_{n-1},\dots,e_0]$ can be written as the following convolution: $$ [m_1,1]*[m_2,1]*\dots*[m_n,1] $$ Each of these very short sequences is trivially log concave. Since convolution preserves log concavity, we are done.

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Why does convolution preserve log concavity? Note that if a sequence $(a_i)_{i\in \Bbb Z}$ is log concave, then you can show $$ a_i a_j \ge a_{i+1}a_{j-1} \qquad \text{ when $i\ge j-1$}\tag1 $$ This follows since $$ a_i/a_{j-1}=\prod_{\ell=j}^{i} a_\ell/a_{\ell-1}\ge \prod_{\ell=j}^ia_{\ell+1}/a_\ell=a_{i+1}/a_j $$ Now, let $(a_i)_{i\in \Bbb Z}$ and $(b_j)_{j\in \Bbb Z}$ be two log concave sequences. Using $(1)$, you can show that $$ \forall i,j,k\in \Bbb Z,\qquad (a_ia_j-a_{i+1}a_{j-1})(b_{k-j}b_{k-i}-b_{k-j+1}b_{k-i-1})\ge 0 $$ Now, let $(c_k)_{k\in \Bbb Z}$ be the convolution* of $(a_i)$ and $(b_j)$, that is, $c_k= \sum_{i\in \Bbb Z} a_i b_{k-i}$. Then \begin{aligned} 0&\le \sum_{i,j\in\Bbb Z} (a_ia_j-a_{i+1}a_{j-1})(b_{k-j}b_{k-i}-b_{k-j+1}b_{k-i-1}) \\ &=\sum_{i,j} a_ib_{k-i}a_j b_{k-j}+\sum_{i,j} a_{i+1}b_{k-i-1}a_{j-1}a_{k-j+1}-\sum_{i,j}a_ib_{k-i-1}a_ja_{k-j+1}-\sum_{i,j} a_{i+1}b_{k-i}a_{j-1}b_{k-j} \\ &= c_k^2 + c_k^2 - c_{k-1}c_{k+1} - c_{k+1}c_{k-1}, \end{aligned} proving the log concavity of $(c_k)$.

*The fine print is that we need to assume the convolution is well defined. This is certainly true if $(a_i)$ and $(b_j)$ have only finitely many nonzero terms, which is all we need for your question. However, my proof shows this applies more generally to semi-infinite sequences, or when $(a_i)$ and $(b_j)$ are both square-summable.