Bounding Schur symmetric polynomials on the unit circle

Question

Recall the Schur polynomial in $n$ variables, indexed by the partition $\lambda$, with $\ell(\lambda) \leq n$, is given by \begin{equation} s_\lambda(x_1,\ldots, x_n) = a_{\lambda + \delta}(x_1, \ldots, x_n) / a_\delta (x_1, \ldots, x_n), \end{equation} where $\delta = (n-1,n-2,\ldots, 0)$ and $a_\lambda = \det (x_i^{\lambda_j})$ is the Vandermonde determinant of the $n \times n$ matrix whose $(i,j)$ element is $x_i^{\lambda_j}$. My notation here is consistent with Ian Macdonald's Hall polynomial book chapter 1.

Also let $e_j$ be the $j$th elementary symmetric polynomial in $n$ variables, $0 \le j \le n$. These are defined by \begin{equation} \prod_{i=1}^n (1 + x_i t) = \sum_{j=0}^n e_j(x_1,\ldots, x_n) t^j. \end{equation} By the Jacobi-Trudi formula we know that \begin{equation} s_\lambda = \det(e_{\lambda^t_i -i + j}), \end{equation} where $\lambda^t$ is the transpose of the partition $\lambda$, that is, $\lambda^t_i = \mid\{j: \lambda_j \geq i\}\mid$. Thus $s_{1^j} = e_j$, for $j \le n$.

Now we specialize to $x_i \in \mathbb{T}:= \{z \in \mathbb{C}: \mid z \mid = 1\}$. Given that \begin{equation} e_1(x_1, \ldots, x_n) = 0, \end{equation} that is, $\sum x_i = 0$, I am interested in bounding $s_\lambda$.

I have a conjecture for $\lambda = 1^{n/2}$, and $n = 4m$, $m \in \mathbb{N}$, namely, \begin{equation} \mid e_{n/2}(x_1, \ldots, x_n) \mid \le \binom{n/2}{n/4}. \end{equation} This is attained when $x_i = (-1)^i$, that is, when half of them equal $1$ and the other half equal $-1$, since \begin{equation} \prod_i (1 +x_i t) = (1-t)^{n/2} (1+t)^{n/2} = (1-t^2)^{n/2}. \end{equation}

I don't know if this conjecture is true for all qualifying $x_i$'s. Full credit will be given to solve this special case. However, I am also interested in a general conjecture for arbitrary $\lambda$.

Answer 1

In the special case mentioned in the problem, I'll show the bound $$ |e_{n/2}(x_1,\ldots, x_n)| \le 2^{n/2}. $$ Let $$ F(z) = \prod_{j=1}^{n} (1+zx_j) = C \prod_{j=1}^{n} (z + \overline{x_j}), $$ where $C= \prod_{j} x_j$ has magnitude $1$. (The roots of $F$ are $-\overline{x_j}$.)

Now by Cauchy's theorem $$ e_{n/2}(x_1,\ldots,x_n) = \frac{1}{2\pi i} \int_{|z|=1} \frac{F(z)}{z^{n/2+1}} dz $$ and so in magnitude this is bounded by $$ \le \sup_{|z|=1} |F(z)|. $$

Now we use a very nice Theorem of Carneiro and Vaaler (see Theorem 8.1 there), which gives a bound for the maximum of a polynomial whose roots are on the unit circle in terms of the first few symmetric power sums of the roots. Let me quote their result fully: Suppose $G(z) = \prod_{j=1}^{N}(z-\alpha_j)$ is a polynomial with $|\alpha_j| \le 1$ for all $j$. Then for any natural number $K$ we have $$ \sup_{|z|\le 1} \log |G(z)| \le \frac{N}{K+1} \log 2 + \sum_{k=1}^{K} \frac{1}{k} \Big| \sum_{j=1}^{N} \alpha_j^k \Big|. $$ This is Theorem 8.1, display (8.6) of their paper.

Apply this to our polynomial $F$, with $K=1$. Since the sum of the $x_j$ is zero, by assumption, we deduce that $$ \max_{|z|=1} \log |F(z)| \le \frac{N}{2} \log 2, $$ which proves the claimed estimate.

Maybe there is an easier proof of this particular bound, rather than appealing to the Carneiro-Vaaler work. I didn't see one immediately, and would be very interested if someone found an alternative approach.

Edit: In the particular case $K=1$, which is what is needed for this question, zeb kindly showed me the following one line proof: assuming $|\alpha_j|\le 1$ and $|z|\le 1$ we have $$ |G(z)|^2 =\prod_{j} |z-\alpha_j|^2 \le \prod_{j} (2-2\text{Re }\overline{z}\alpha_j) \le 2^N \prod_j \exp(-\text{Re }\overline{z}\alpha_j)\le 2^N \exp\Big(\Big| \sum_j \alpha_j\Big|\Big). $$