On an Erdős Problem about the Maximum Modulus of Littlewood Polynomials on the Unit Circle

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Tamás Erdélyi, On an Erdős Problem about the Maximum Modulus of Littlewood Polynomials on the Unit Circle, arXiv:2608.00744v1 (2026), dated May 12, 2026 on its title page.

The copy read for this card is arXiv:2608.00744v1. The folder holds its PDF and a Markdown reading copy; Theorem 2.1 was checked against the PDF's p. 3. The arXiv record (https://arxiv.org/abs/2608.00744, read 2026-10-02) names the Creative Commons Attribution 4.0 license.

Reading depth is claims checked for the definition of $L_n$ in Section 1 and Theorem 2.1 in Section 2. Its proof in Section 4 was read for the argument's structure and displayed estimates, but the cited Bernstein inequalities were not checked against their original sources and the proof is not independently verified here.

Main result

The paper uses degree normalization, not coefficient-count normalization:

$$L_n=\left\{P_n(z)=\sum_{k=0}^{n}a_kz^k:a_k\in\{-1,1\}\right\}.$$

Thus $P_n$ has exactly $n+1$ coefficients and Parseval gives $\frac{1}{2\pi}\int_0^{2\pi}|P_n(e^{it})|^2\,dt=n+1$. With this convention, Theorem 2.1 states exactly that every $P_n\in L_n$ satisfies

$$\max_{t\in\mathbb R}|P_n(e^{it})|^2 \ge n+1+\frac{1}{38}n^{1/3}.$$

The statement is Theorem 2.1 in Section 2; its proof is in Section 4 under "Proof of Theorem 2.1," equations (4.1)--(4.15), using Lemmas 3.1 and 3.2. The proof sets $T_n(t)=|P_n(e^{it})|^2-(n+1)$. Parity of the autocorrelation coefficients and Parseval give the derivative-energy lower bound (4.2). Assuming an upper excess $\delta_n$, the $L_1$ Bernstein inequality gives (4.7); the dyadic level-set decomposition (4.8)--(4.11), combined with the Bernstein--Szegő pointwise estimate (4.12)--(4.15), gives an incompatible upper bound when $\delta_n=\frac1{38}n^{1/3}$.

This is an additive $n^{1/3}/38$ gain over the mean squared modulus $n+1$. After taking square roots and comparing with E1150's $\sqrt n$ scale, it says

$$\frac{\max_{|z|=1}|P_n(z)|}{\sqrt n} \ge \left(1+\frac1n+\frac1{38}n^{-2/3}\right)^{1/2},$$

whose right-hand side tends to $1$. A fixed factor $(1+c)\sqrt n$ with $c>0$ would require a squared-modulus excess of order $n$, whereas the theorem supplies only order $n^{1/3}$. It is therefore quantitative progress beyond Parseval, but remains far short of the fixed multiplicative gap asked for in E1150.

Bears on. #1150, by giving the universal $n^{1/3}/38$ additive squared-modulus gain while leaving the requested fixed factor unresolved.