Erdős Problem #230
DISPROVED LEANdatabase updated 2026-08-24
The statement as posed is false. A solution has also been checked by machine, in Lean.
What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).
The maximum of \(\lvert P(z)\rvert\) over the unit circle, for \(P(z)=\sum_{1\leq k\leq n}a_kz^k\).
Let \(P(z)=\sum_{1\leq k\leq n}a_kz^k\) for some \(a_k\in \mathbb{C}\) with \(\lvert a_k\rvert=1\) for \(1\leq k\leq n\). Does there exist a constant \(c>0\) such that, for \(n\geq 2\), we have \[\max_{\lvert z\rvert=1}\lvert P(z)\rvert \geq (1+c)\sqrt{n}?\] This is Problem 4.31 in [Ha74], in which it is described as a conjecture of Erdős and Newman. The lower bound of \(\sqrt{n}\) is trivial from Parseval's theorem. Körner [Ko80] constructed, for all \(n\geq 2\), polynomials \(P(z)=\sum_{k\leq n} a_kz^k\) with \(\lvert a_k\rvert=1\) for \(1\leq k\leq n\) such that, for all \(z\) with \(\lvert z\rvert=1\), \[(c_1-o(1))\sqrt{n} \leq \lvert P(z)\rvert \leq (c_2+o(1))\sqrt{n}\] for some absolute constants \(0<c_1\leq c_2\). The answer is no (contrary to Erdős' initial guess). Kahane [Ka80] constructed 'ultraflat' polynomials \(P(z)=\sum a_kz^k\) with \(\lvert a_k\rvert=1\) such that \[P(z)=(1+o(1))\sqrt{n}\] uniformly for all \(z\in\mathbb{C}\) with \(\lvert z\rvert=1\), where the \(o(1)\) term \(\to 0\) as \(n\to \infty\). For more details see the paper [BoBo09] of Bombieri and Bourgain and where Kahane's construction is improved to yield such a polynomial with \[P(z)=\sqrt{n}+O(n^{\frac{7}{18}}(\log n)^{O(1)})\] for all $z\
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research solved · 1 further variant / known partial result in the Lean file. The authoritative wording is on erdosproblems.com/230.Prizenone
Topicsanalysis, polynomials
Statement in LeanFormal Conjectures
Solution in Leanyes
Original pageerdosproblems.com/230
Attempts
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